Problem Solutions
By
Dean Updike
Chapter |
Pye
Ao
ws
sss. 4.1 Two solid.cylindrical rods 4B and BC are welded together. at Band loaded as.
Problem 1.1
-~ P.~
~
120kN
50 im
e CSOY = 1963-5 inin®
Ana?
ona
shown. Determine the magnitude of the force P for which the tensile stress in rod AB
is-twice-the-magnitude-of-the-conmpressive-stress-in-t0d BE
?P
P
Aas "(963-5
iat
= $09-3x107 oP
Ag.
75 min
B
120 kN
=
Yr (75° = 4417-9 mm*
_
(220) - P
A
AB
See. -
- 240-P
44IT GF
Equating
.
750 man
pe
FOO mi
4
~ 9.0543 — 226-420 °P
Sa,
to
~
2 Og
.
£09-3 xr" Px
Problem
=
2(0-0543 - 226-4x10
1.2
~&
P)
Po = 1/259
kN
<4
1.2 In Prob. 1.1, knowing that P = 160 kN, determine the average normal stress at
the midsection of (a) rod AB, (h) rod BC.
1.4 Two solid cylindrical rods 48 and BC ate welded together at B and loaded as
shown, Determine the magnitude of the force P for which the tensile stress in rod AB
is twice the magnitude of the compressive stress in rod BC.
Roe
{a)
AB.
50 mim
P= soo kN Ctension)
Mo = Fai
6,
AB
Cb) Red
Fr
he
=_ 2.
~ BON. 1968-5 mm
3
160 x16
Tr fee
Nag | (1963-5
ot OLS MPa
LOKN
2
7 mm
A
P= 160 KN
~<a
750 mit —---~- I ere ren
<—.
— 1000 mim
a
»{
3
BC.
(60-(2)\G20)
=~ fo
_ Taee
Kec * td
One = EL
Age
.
-
mWz7s)?
onal
_=80xs07
GAIT
kw
> AGITE
e@.
Bo
EN
Compression,
min
Ope ~18-1 MPa <A
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Eoited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student asing this manual is using it without permission,
_Problem.1.3
1.3 Two solid cylindrical rods 4B and BC’ are welded together at B and loaded as
shown. Knowing that the average normal stress must not exceed 175 MPa in rod AB
and TSO MPa in tod BCY determine The Sinallest allowable valties OF J
Red
AB.
P
300 mm
=
40
S
Pe
MG
a,
= TOkKY
-_P.
Ane
._
yo
+30
=e
f4P
_~
_ Tex6
.
=
Foxlo’
N
4YP
At
wa,*
|“WOT
Geox)
S x 10"
5
amd
22.6
A,
%1loO
pe
= 22-6
wy
wy
<i
mn
=x
250 mm
Rod BC,
P=
30 kN
30 UN = 30xJo'N
.
Ce
PL.
PH
7 oe
tA?
= AP
Ta”
(4) (20 x 10)
d= fae.
°
TSO x (0%)
opt
IS--96x10
a,
Problem 1.4
=|596
1.4 Two solid cylindrical rods AB and BC are welded together at B and loaded ag
shown. Knowing that d, = 50 mm and d; = 30 num, find average normal stress at the
midsection of (a) rod AB, (b) rod BC.
(.) Rod ARB.
40 +30 = 70 kN
P=
= 70 x10°N
A = Bare E (60)= 1.9635 xlo%men = LAGS THIS
300 mam
Sas = x " eae.
= 35.710" Pa
ng = 3S.1 MPa
250 mm
(bo.
Rod
P
BC,
= ZO kN
A = Fa;
3OKN
<a
Gye =
FE.
A
30x10" N
F (a)? = 706.86 mm= 706.86x10" m*
3o xo?
706, 861s
= 42.4x10*° Fa
One
-
44.4 MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without peérroission.
<0
So
Problem
4 5
~
oe
oe ESA strain gage jocated at C on the surface of bone AB indicates that the average
normal stress in the boné is 3.80 MPa when the bone is subjected fo two 1200-N
POYEES AS
SHOWA.
ASS
1200 N
eek
a
A
Az
Geom edvy
>
=
2
a
A=
tan
2
an
1200 N
At
each
1.6
boSt
ee
poe
Pi, of the
plate
upwavel
equi
3h
dl
2
4P
TE
CAYC1209)
GET aBox 18s}
222.9 *107°
=
14,93
WwW
A, = 14.9%
lo ym
mm
a
1.6 Two steel plates are to be held together by means of 16-mm-diameter highstrength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the
average riormal stress must not exceed 200 MPa in the balts and 130 MPa in the
spacers, determine the outer dianteter of the spacers that yields the most economical
and safe design.
focatron
fowce
|.
ai
B
&,
GA
Ayo = (25x
‘Problem
a
knowing that its outer diameter is 25 mm, determine the inner diameter of the bone’s
cross section at C.
the
boft,
At
a
with
Upper
plate
is
publled
the same time
the
force
compressive
dlown
by the
spacey pushes
fe
Phat
marntain
over te
Tn
Ts.
tens;
4; lrtam
Fon
the
heft
For
the
Spacer,
Equating
GS, *
Py
Se
ana
i
As
>
athe
T
= chy? )
P,
or
P
= -
=~
S,
$ 6.
di,
|
(d,°- d,")
Pe,
EG de = P&ldeaf
be
ov
A=
de)
fi4+ 200
dg = 25.2 mm <a
-
Problem
1.7
1,7 Each of the four vertical links has an 8 x 36-mm uniform rectangular cross
ae0.4
~ section and cach of the four pins has a 16-mm diameter. Determine the maximum
value of the average norma
ting (a), points B and, D,.(b)
he links co!
|
points C and E.
Cc
0.25
body,
Sree
a
a$
ABC
bar
Use
oO ¥N
BBEN
O.ORS
i.
A
4
A
ON. —
(),
>be
B
|
c
Fee
W Fan
oO:
i
2M.
(0,00) Fi, — (0.025+ 0.040) (20x10?) = ©
Fey
=
stress
=
(Geo
Area
Feo|
in
N
m tension.
aa
ia
(0,008)(0,036.-
=
And?
BD.
BRO KIO"
in
CE
parole) Din Ks ,
2 325 IO"53
Din k
Lin’
fewsion
Sor
Drak
Any
one
fon
«1a
For twe
Mm.
160 x10”
Tensile
-=IQS
Sink
ong
of
aAKEea
Net
-
js
- (0.025 )(20 x10") = ©
~(0,040)F,
0:
=M,=
BD
Link
N
32.5 xjo
=
Feo
Com
2 \o).66 «10%
=
(0,008
Dio 2
Hales ,
pression
Com Phe 351 On.
0.016
)
B20
“jo *
,
Sep=
2
R&B
x
jo"
mi
’
For
+wo
pera
13
2
—&
(bog = RE = RS 2 wai 70 010
=
,
jOLGM
)€0.036)
A
ut
= S76
2
Le MPa a
So
=~
yw
wn
*lO
Gage -RLTM
.
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited’distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
_ Problem 1.8
1.8 Knowing that the central portion of the link BD has a uniform cross-sectional
area of 800 mni’, determine the magiiitude of the load P for which the normal stress’
P
in that portion of BD is 50 MPa.
P
a
o.7m
:
SA
=
Feo
A
.
(300 x 15%)
= (50%10*)
1,
N
lO
40
=
BD
4m
tA2m
\ |
_
Bd=./ (0.56)4 (1-92)"
a
=
DDH.
DDMe
==. Oo:
O
.
0.56
Pal
1.92
2:88
4) +4 92
GLH
Fae (HO™ (uox We ya VC.)
Ee (dow1c)AAY CH)
- P(o.74h4)
P=
Problem
0.5m
for staties,
Free Baoly AC
Use
=
m
2.00
=
1.9
2
0
s
1.9
2
.
323.) xlO'N
Ps 83.)kN
Knowing that link DEE is 25 nm
-=8
wide and 3 mm thick, determine
the normal stress in the central portion of that link when (a) 0 = 0, (b) 6 = 90°.
Use
bev
CEF
as
a
free
body
HIM. = O
Foe (0:2) (240 sin®)-(o- 4) (240 cos @) =O
~O3
Foe = —/bo sin® —3200e08 0 N
~- (0-025) (os 003)
1
>
5
mem
Spe
(a)
me
6=-
“Roe
oO:
Foe = ~320
~320
Soe *> Frye
(b)
-6
= 76 %Xto0
o
n
Pf
@= 40":
- ~f60
Coe — 7s 0e®
me
AT
4.27
N
OM
MPa
at
- {LoN
Fy
=
~2-12 MPG
—<
Problem 1.10
Ds
a
-
960 N
80 Link AC has a uniform rectangular cross section 3 mm thick and 12 mm.wide.. .
Determine the normal stress in the central portion of the link. -
»
Free
75 mm
AEREEARDA
~
BER
once
rere
SRR DRUIDS
SN
Body
Diagraw,
of
Pilate
ace nneestnnneyge -
i175 mm
wh
Note
960 N
that
Forme
the
a
+wo
coupte
oF
(960 N\(015m)
OIM,-0!
oy
Stress
: Povces
mament
|
;
= 144 Nm
44 Kin
Fre 7 6651
Aven
9bOrN
~ (Fye cog 30° (025m) = O
W
Bink!
Gre =
Age
Me.
ta
s (3 mm
=
\(l2 inm)
b6$:
—
>
36
mmZ
= 18-475 MPa
.
>
Sper 18:5 MPa
Ac
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
~-- Problem
oblem 1.14
.
. . 4,74 The rigid bar EFG is supported by. the truss system shown. Know- ing that the meraber CG is a solid circular rod of 18 mm diameter, determine.
th
HO
Yona
“bpp
dots:
HnOPMar suvess nree;
Using
portion
HOE,
ew
“Lm.
=O:
Fae =
Te
OS
Fag
-/5
&
as
a Fee
2
beaw
EFG
becly
¢
Fee
=
Cress
Fae
=
Re + U2) (4
NENENS
A
Fer
SLE
Foe
Problem 1.12
of
member
anh
USE.
254M AKIO
ce
7.42
yg
o
Z
VISEA
2
rane
CG,
in
1 oe
;
CG
= 4 (o-olb y = 25464 x19 Pin
etvess
Novw af
Fes
N\
avea
=
Keo = Fa"
¥ cer
Fi.) = ©
25 kW
sectionat
>
e
=
25 EN
9M, = 0! (2 ok
Fae
a Free. badly
ms |e~ 2 ms
Using
Fae
EVeC®
<a
The rigid bar EFG is supported by the truss system shown. Deter-
mine the cross-sectional area-of member AE for which the normal stress in the
member is 105 MPa.
Using
portion
.
a
2Or2S
WEF
Fag
Stress
EFGECB
=
Fre -
as
_
a
Free
body
=O
LEkN
in are ws Lew
AE
Gace
*
{05M fra
Ene
Cae ~ Age
Mew
FE,
ue
AE
i0 .
sam
=
25x
105 %f0
238/X/0 ~eé m2 e '
Proprietary Material, © 2609 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
7 Problem
OC
1.13
_—_—s«&LA.- A couple M of magnitude 1500 N - mis applied to the crank of an engine. For
the position shown, determine (a) the force P required to hold the engine system in
equilibriv
|
|
P
-
Use
piston, rod, and evank
together
200 vim
"
wall
ae
ME
as
peaction
reactions
ome
#)
4
G0 san
| stress.in the connecting rod BC, which has.a45Qreceou{uncsev- enews
mm’ uniform cross section.
L.
Ma
Tree
H
bady,
and
.
Aclol
.
beaning
Ay and Ay.
=
Of
(0.230m)H ~ 1SOO Nem
H= 5.3571 x10% N
Use
piston
= ©
alone
as ree
body.
Note that rod
a
two-force mem ber;
is
hence
the
oF Force
Fee
is known.
Draw
triangle
ane
sodve
ane
far P
divection
the fovee
Fre
by
proportions.
Q =f 2007 4.60" = 208.81 mm
=~
—
xo
200
4002
=
Go
(a)
TE 7
So
Rod
BC
Stress,
is &
Compression
Og.
-
~Fe
AO
member.
~
BEL
Ets
17. 86xlo*
P=
17. 36 kW
N
=a
Fag = 18.643 xO
area
—/8-643x/0*
Taso
xio®
Cb)
P=
_
is
4SOmm=
AL.4 jo
&
450 x JO” mn
Pa
See = - 41.4
MPa
<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
_ Problem 4.14
1,14 An aircraft tow bar is positioned by means of-a single hydraulic cylinder
connected by’a 25-mm-diameter steel red to two identical arm-and-wheel units DEF.
Lire tow bar is 200 kpc and its center of gravityis located at G-For
the position shown, determine the normal stress in the rod.
Dimensions in mm
0
450
250
850
500 |
FREE
Goby
825
675
- EXTRE
“tow BAR!
W = (200 kg)(4.81 m/s?) = 1162.00 N
Ay,
Ay
HFomm
—
.
“y
ty]
G
.
fF se.
2 |
om
re
850mm
Govyr
-
F WHEEL
=
BSOR - usS0(I4G2.00 N) =O
N
ZQS4,5
R=
R
BoTH
OS
A
OF
A
FREE
Ww
Atwr
UniTs 2
loo
tan
xg
=
Lye
Ts
A=
$.4270°
+5EmM, 20:
(Ecos KSS0)
~ R(S0Od) = O
S00
Feo =
$50
C05 8.4210" (2654.5
= 2439.5
S0omm
GTS mero
Fep
cp”
Aco
=
N
we
Coome.)
2439.5
N
41 (0.0125 yn)*
#
R226S4.5 kN
ooz
Hw)
- 4,9697 * 10° Pa
oz,= 74.97 MPa 4
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
~~
~
Problem
1
-
45
11S
of.
©.
Coa
| ESkN
T he wooden members 4A and B are to be joined by plywood Splice plates that
will’ be fully glued on the surfaces in contact. As part of the design of the joint, and
knowing that the clearance between the ends of the members is to be_6
determine the smallest allowable length 1 if the average shearing stress in the glue is
—_
not to exceed 700 kPa.
al
There
I
Fach
Gam
ave
of
the
foor
separate
these
15 KN
AVEAS
doade
aveas
tut
mits
ane
trans
ane
giued-
ha OT
the
Thos
F=4P
= 4(15)=
7.5 \N
=
7500N
~ 75 mm
Fo.
Po eee
aa
ret
lf
Lo
Aength
Foy
each
gqbued
Average
The
tn
shearing
3 p
or
one
.
area
Az
Y=
Shearing
4
Ls
>
ond
stress
EF
W
E
i's
>= 78 rm?0.0718m
=
Xs:
Joox)o?
1.16
15 mm
(Joox10 (0.075)
La lgap)+
mz ON4286
Steet
Shearing
.
~
286
m
=
JURA
B= 142.8546+ 14285
242mm
-—ed
1.16 When the force P reached 8 KN, the wooden specimen shown failed in shear
along the surface indicated by the dashed line. Determine the average shearing stress
along that surface at the time of failure.
\.
At
Avea
mt
»
|
90 mat
wrath.
Pa
L =
Problem
be its
E
7TS00
~
Tw
Le
\
Lw
styess’
4,
Tote? Fength
qdued
area
eDlowalhle
.
Sedv
ot
> 22
Fo.
{
Let
—}
being
A= Jommy
sheared
[Simm
1
=
1aSOumm
2
= 1350 *(/O%
m
\
Wood
stress
Force
wr - F
A
Pr
.
8~1l0°>
8x10?
(Rea
xo-6
=~
N
5.93 wl?
xfo” P2=Piz5.93 MPa
met
Proprietary Material. © 2009 ‘The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ETT wo wooden planks, each 12 mm thick and 225 mm wide, are joined by the dry
Knowing that the wood used shears offalong ifs grain when the
ress reaches’
8 MPas determine the magnitude ? of the axial load
co
‘1.17.
Problem
—
16 mm
which will cause the joint to fail.
|-~ 16 mm
:
Siy
aveas
when
the
areas
its
hag
area
must
jot
be
sheared
Pails.
Each
climeasrouns
off
&
[Gum
*
these
12 mm,
being
A= UG KIZ) = 192 met = 192 410°
At
Far
uve
F=Since
the
Force
vA
theve
Problem
F
caveied
= (@x10%)(142 xo)
ave
SIX
foe Yove
1.18
aK
+
each
oF
= 1536
Pe
AvVEaS
Aaneag
N =
GF
1m
1.536
kM
= (Y0.836)
=
F22kN
<a
1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into
which a 12-mim-diameter hole has been drilled. Knowing that the shearing stress
must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate,
determine the largest load P that can be applied to the rod.
- 40 mm 7
10m mo
by
__
For the stee? rod ,
8mm
.
o1e)
Yo.
olG
A, = TA h = (N
m= 376.99 *10°% m*
12 aml
-
—
Fe
GA,
P = (180 «10*)(376..97» 10°) = €7.86r10*N
Fox
the
a fuminum
plate,
A,= Wd,t, = (rY0.040X0, 008) = 1.005 a1nIO* m
TG
ie
w
Re
TA
= (7ox10%)(1,0053%10°°)
The
oF
Aimiting
P,
amd
vale
Por
the
Poacl
=
oe
70,372 «10° N
ts
the
smaller
Pr.
P= 67.8610" N
PeC77kN
«a
Problem 4.19
1.19 The axial force in the column supporting the timber beam shown is P = 75 KN. Determine the smallest allowable length 1 of the bearing plate if the bearing stress in
the timber
isnot to exceed’
30 MPa
SOLUTION
O,
~P
.
A
_P.
~~
tov
Solving
3
tw
RP
2
|
|<
Ow
75 ¥{O
_.
Gora? Wryoy
178.6 «107° m
[=
min
wh
1.20 The load P applied to a steel rod is distributed to a timber support by an annular
washer, The diameter of the rod is 22 mun and the ner diameter of the washer is 25
mm, which is slightly larger than the diameter of the hole. Determine the smallest
allowable outer diameter d of the washer, knowing that the axial normal stress in the
siecl rod is 33 MPa and that the average bearing stress between the washer and the
timber must not exceed 5 MPa.
A=
rod!
Stee)
Ps
6A
=
(0.092\"
=
F
BQO .1Fr1IO ow
35x10° Pa
GS =
= (25K10E
3 BB0, |e VF
Y
13,.305*/07N
Wachen?
6,
Required
=
Swipe
bearing
Pa
aves?
13.205 «1p
bh
4
’
de= d;° + te
24 +
es
HNE-S60F
“IO
=
3
3
WwW
oS
oN
2
3
8
a4
ts
fF)
A
yw
= (O.025)°
ul
1.20
Q.
Problem
173.6
1.21
Problem 1.21
A 40-KN axial load is applied to a short woaden post that is supported by a
concrete footing resting on undisturbed soil. Determine (@) the maximum bearing —
stress on the concrete footing, (4) the size of the footing for which the average
P= 40kN
bearing stress in the soil is 145 kPa.
120 mm
(a)
Bearing
on
concrete
Yo kN =
ty
P=
stress
a
A = (JOOMI2@)=
"
6 = p
Footing.
4oOx%/o* N
IQx1O we = (2/07 me
Yoxjo*
alo
_
=
3.33x10°
2.323
(b)
Footing
VO
Tae
C
Since
the
~
ey
PR
A
avea
be fA
Problem
P=
1.22
A
is
_
~
Yo
40x
-
Fe
S
«103
-
(4s*
Square,
.
= f0.27586
=
A=
N
/O>
a
fo?
-
Ss
145
kPa=
oO.
27586
mw
Pa
MPa,
45x10°
aah
Pa
Rr
b*
0.525m
b = S25
mm <i
1.22 An axial load P is supported by a short W200 * 59 column of
cross-sectional area A = 7560 mm’ and is distributed fo a concrete foundation
by a square plate as shown. Knowing that the average normal stress in the column must not excecd 200 MPa and that the bearing stress on the concrete
foundation must not exceed 20 MPa, determine the side a of the plate that
will provide the most economical and safe design.
For
the
ov
column
P=
For
the
axa
A = Since
a=
OA
the
/A
ABb
=C200%00°)( 7560%/0" )=/5/2 kN
po. te
5
eo Gi
= S25
prate
_=
GC
GS:
2 007eb/m
=
:
Aza’
is Sqoave
-40,:0756
20M;
o«n.7h
™
a275
mm
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
A 6-mm-diameter pin is used at connection C of the pedal shown. Knowing that
. P = 500 N, determine (a) the average shearing stress in the pin, (b) the nominal. .
bearing stress in the pedal at C, (c) the nominal bearing stress in each support bracket
— Problem 1.2300
é
a
——- 300 mm
|
Dvaw
P
Svee
body
Diag tom
Since
of
ACD
3- force
the
Frow
the
Cc
tow
dines
ot
oval
porn
action
of
E
the
the
intevseatian
ot hey two
forces.
Ce = {200s 125* = 325 rm,
3 ea met “y?
te
2
Tpin * “Ap:
at
divected
of
at ZF, = 0:
(2)
is a
member,
reustion
is
3 math
ACD.
Bec-P=o
- £o
AS
“Eg?
Mo
‘)
O,-oA,G-
(C)
6 ° = tC
2% AL
=
Problem
1.24
_
<
~ 22.0x10" Pa
=
!
Gin’ 43.9 MPa, a
= 24.
| «10° € Pe
(6x10?
4x lo"*)
{Zoo
.
= (2.6500)= 1300N
(2)C1300)
op ALMISOOL
~ Wlexios)?
100
£dt =
C
2c
Fan
Wa®
~
qj -
C=26P
= 2 T«lo®
Rat = (2)\Cexlo# )CSx10°? }
6
Ss
FA
= 24.) MPa wa
6,2
217M
Pa —ea
6
1.24 Knowing that a force P of magnitude 750 N is applied to the pedal shown,
determine (a) the diameter of the pin at C for which the average shearing stress in the
pin is-40 MPa, (4) the corresponding bearing stress in the pedal at C, (c) the
corresponding bearing stress in the each support bracket at C.
75 mm
300 mt
9mm
Draw
Pr
Free
Aiagvam
|
Siuce
tt 3h =0:
@ Tn=w= AE
322
“ROF
(bo) Gi=
~-
2.
©
=
*#¢
NK
GE€
OC
2dt
mem ber,
of the
of action
#Cc-Pz=o
2 BO
Ba
a
at C
ftowavel
CE =_[3007 + [25°
qeo metry >
ACD.
the
peactrou
is Aivected
5 mm
From
of
Fig
ACDis
3- Force
D
bo Ay
- f2C
om .
Lines
=
E,
the
intevsect fom
of the other two
fovces.
B25 mm
C= 26P -(26)250)=
1750 N
/BOs#se.
[Tass
ipe) 2 S57 IT tan
[ase
=. @snovGuery
I4So0
© (AMS.57> 10
peiat
* 3% 7"l0 é Fa
.
Sx10"* )
= 85.0%10° Fa
maz ds ESI may mm
G,=. 38.7 MPa <a
= 3S5.0MPa <a
problem
1.25
1.25
senomapsunsttntgsenaepseensnsus
18 mm
A
12-mm-diameter steel rod AB is fitted to a round hole near end
C of the wooden member CD. For the loading shown, determine (a) the max-~
siti AVSER RS’ HOTHNTAT SHOE IW tHE WOOA.” (BY thE TSANLE'B
Tor WHICH the’
4.5 KN
”
erage shearing stress is 620 kPa on the surfaces indicated by the dashed lines,
{c} the average bearing stress on the wood.
fy Os LN
C})
A EC
Mayinon
‘
P=
C= A
_
vy. P .
Us
Nore
ad
stress
in
wood.
RK
> FbE
P
O°
a
_
450 kN = 4¥S50¥10"N
Tiga
| 4 50
Pe
o
* 3.97«x10°
‘
Pa
4.50 x(0*
|
3.77 MPa, ai
.
REE * BilanctVeroxios) = 2O2*IO
b=
CSF *
the
Nret = (75-12 UB) = 11S 410? mom= 1.134 17
[-b
(by
CWVEVAGE
P
aE
dt
4 Sox
F
((2x107
to#
exter?x J
‘
KIO”okFP
= ROB
2
3
202mm
a 0.8
MPa
MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
«ef
Problem
1.26
1.26
-hctuatianimionm amen mmimiaodninrasancneenences
Two identical linkage-and-hydraulic-cylinder systems control the
OSTHOR-OF-the forks-of-a-forklift-irack: Phe load: supported: by ‘the one-system="
shown is 6 KN. Knowing that the thickness of member BD is 16 mm, determine (a) the average shearing stress in the 12-mm-diameter pin at B, (b) the
bearing stress at B in member BD.
t
300 mm
900mm
6kN
B,
f
*
380 nin
N
Use
one
6S
Ga, free
By
12
}.——.-____
Me
E
OR
|
~
o.
/
-lo5)l6)=
=
£
EN
0
BR. = SKN
=O:
By-
6&6
=0
Oo
—_>
=O
E+B,=
==
+t ZR
body.
=o:
:
ob E
<<»
fark
By: -E
—
By = GEN
B= (ar B = [ETT= rain
(a.\
Shearing
stvess
Apis =
=
t
(\o)
a
Dp
B
=
Ag:
Bearing
in
Tike?
oA
tae
B.
-
lide} x10
~6
mM
2.
—_
=
64
mPa
<a
B.
3
RX
o=- Bee.
At
oA
=f (oe/2)*=
tiB-fy
stress
pin
(0-012)( 0-016
40+ b ipa
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any-means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
»\
‘1.27 For the assembly and loading of Prob. 1.7, determine (a) the average shearing
Problem 1.2
:
i
stress in the pin at B, (6) the average bearing stress at B in member BD, (c) the
verage bearing stress at B.in.member ABC, knowing that this.n
mm uniform rectangular cross section.
1.7 Each of the four vertical links has an 8 x 36-mm uniform rectangular cross
section and each of the four piris has a 16-mm diameter. Determine the maximum
value of the average normal stress in the links connecting (a) points B and D, (b)
points C and £.
Use
bav ABC
as a tree
body .
Jow ho gag >-— 0.040 —
\,
.
A
°|
C
Fen
(0.040)F,
{ Fee
-(0, 025+ 0.040)(20 “/08) = oO
Feo = 32.5% 10° N
(a)
Shear
pin
at B.
where
A= HAte
~
te
(b)
(c)
Y= =
6, = Fe
. LoS
G
>
dt
=
ABC
sheow,
‘
x10
C=
dt = (0.0i6)(0.00&)-
82.5 4109) ~ 176,95 x10"
ot
830.3MPa
<8
|2R «lo
m
6,127.0
MPa<a
B,.
= (0.016)(0,010)= ]€O0x10° m
Feo.
A
Az
double
201.06%/0° m*
80.8
06 x107§} 7
(2)(201.
BD.
A=
.
32.5 10%
in
for
E- (0,016 Y* =
Bearing : Pinke
Bearing
Zh
Feo
32.5 x/0%
160 x1o-*
~
R03 x 10
6
2
6,- 203 MPa ~<a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
-
Problem1.28
.
1.28 Link AB, of width 6 = 30 mm and thickness ¢ = 6 mm, is used to support the
end of a horizontal beam. Knowing that the average normal stress in the link is -140
__ MPa, and that the av.
hearing stress_in,¢:
f the. two..pins.is .80..MP acc.
pins, (b) the average bearing stress in the link.
oO
Rot
AB
is
in
Az
bt
wheve
A?
(0,050\(0.006)5
P==-GA
=
(2)
Diametey
compvess1on,
al.
=
br 50mm
and
900x210
AT
.
m
Gum
2
-C1y0x«0%)(300%107%)
42 «10>N
=
j
aaa
1525«107°)
~
= 2.589 *}/O
a
.
Ex
42»})07
(25.85 *107 YO, 0296)
=
wy
x
o*
=
25.9
3
Ww
aa
<A
oO
Pa
ATIMIO"
S,7
271
MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=<
' Problem
1.29
1.29
The 5.6-KN load P is supported by two wooden members of uni-
__.form cross section that are joined by the simple glued scarf splice shown. De-_
termine the normal and shearing stresses in the glued spli
P
VG.
75 mm
125 mm
or
P-
Ao
6127N
=
Ce
9 = Yo” -60"°
t26)(6075)
~ (€<227
P cos"
G-
A.
«3
1.30
375
Ka? (cos
-3
xlo”
mm
30°)"
6 0049 MPa,
Kx
_ (6.227 x0"
sin 60°
(204-375 159)
aRe
.
|
T + Oe28f MPa<a
1.30 Two wooden members of uniform cross section are joined by the
simple scarf splice shown. Knowing that the maximum allowable tensile stress
in the glued splice is 525 kPa, determine (a) the largest load P that can be
safely supported, (6) the corresponding tensile stress in the splive.
75 mm
@)
P+
SA,
Cos
os
‘DO
1
iO
aes]
a
"
A, = (0-125 )L0+075) 3 9325 X10” ine
iO
‘0
Probiem
Ge
TATSRIC
EsinkO
v=
=
= 30°
3
=
~2
(28x10) (9375 x70 7)
cus*
a
3
ENG LN
wn
ot
tk
P-6162
E sin 28 - £66 62) sin Go"
we
C2) 9375 Ke)
T=0'303
EN
<a
|
MPa «lle
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed
in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problon 7 3
oo
1.31 Two wooden members of uniform rectangular cross section are joined by the
simple glued scarf splice shown. Knowing that P= LE KN, determine the normal and
_Shearing stresses in the glued splice.
O67
P
For-
Pr
Yo?
i} kN
=
=
neesnennnutttnty uns an
HE?
jlxto*N
75 mam
A
= (ISO)CIS)
oO
S
=
© ASHI
Peos'@
—
A,
.
“
—
P sin2O
IAS &10° m
#107 }cos* 45°
¥Bqwio
-
(Iheto*
ZA,
Pe
Xsin
Sz
489 kPA <8
40°)
(2X .2sxtio*)
=
Problem 1.32
©
1.25 «(or
=
T=
CS)
me
4.89xIO
Pe
Ye YQd KPA at
1.32 Two wooden members of uniform rectangular cross section are joined by the
simple glued scarf splice shown, Knowing that the maximum allowable tensile stress
in the glued splice is 560 kPa, determine (a) the largest load P that can be safely
applied, (6) the corresponding shearing stress in the splice.
A, = (ISoX75 © IL2S*
IO mm
75 mm
S=
Or
P
—
cos
ER,
= SG0~/0°
Pa
8
_ (S60 x10° Olas
~
cos * 4g?
12.60x10°
PsinB
t
mh
(b)
FP
=
& Aro
“eos? QO
Ny
@
S60kPa
= LL 2SKeIC
N
corO
A.
= S60x%]0"
»Jo”)
~ 12.60
LM) eh
_
(12. 60 “to
\GintS *Yeos 95°)
W.2S «1077
Pa
t= SCO KPa <a
Problem 1.33 0
1.33.
A centric load P is applied to the granite block shown. Knowing
“thatthe resulting maximum value of the shearing stress in the block -is'17-MPay~
determine (a) the magnitude of P, (b) the orientation of the surface on which
the maximum shearing stress occurs, (c) the normal stress exerted on the surface, (d) the maximum value of the normal stress in the block.
O-=
45°
CY
Low
plane
Cag 2 tel
of
jpl=
Coy, 7 VT MPa
mn
Ag zo-IS (015): 010225
TC mese
aA.
Tmeepe, =(2)(0+0228)(ITMO)
“
150 mii
()
Gye = EAs cos* 45 =
~
e
1.34
1,34
eo,
FP
<<
@245°
90°
sin2Oz}]
.
Problem
20%
(b)
—_
KN
= 765
3
MPa
7b6S kre
=
2. — 76S
QR. TG
SK
[0.0125)
_7bL KIO
oe
2G
4
A 960-kN load P is applied to the granite block shown. Determine
the resulting maximum value of (a) the normal stress, (b) the shearing stress. Spec-
|
ify the orientation of the plane on which each of these maximum values occurs.
Pp
Ao
(0-15) (018 )=
G
=
+
cosO
06-0225
~ 7 G0 Kto
=
souee
o
(u.)
mow
tensife
150 mm
To.
P
= BAL
°
6
©
=
stress
ot
(b)
2
a
tos O = -42671KI0
&
cos"
OL:
Mong, Compressive
{50 mm
m
3
stress
-
2 (0022S)
O=
=
@
=
4267
Yo®
MAR
=a
MPa
=a
cy”
Kio’
Ybo "
oF
ot
of
eB
45°
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~ Problem
1.35 A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by
1.35
welding along a helix that forms an angle of 20° with a plane perpendicular to the
axis
of the pipe: Knowing that the maximum_allowable
normal and shearitig stresses”
in the directions respectively normal and tangential to the weld are o = 60 MPa and t
= 36 MPa, determine the magnitude ? of the largest axial force that can be applied to
the pipe.
16 mm
dy = 0. 4OD m
Y=
fh -
A. =
eta
t=
0.200
Wrt- ne)
|
=
MPa:
0,200 ™
~0.010
=
O./90
m
10, 200*~ 0.1907)
12.25
6
69
=
=
x/O78
m*
20°
G
=
re
cos*Q
(R29.€osx10"20 U6OX IO") _ g33.x 10% N
MPa
a Ae’ Cc
P
Smaller
“
sia
Bin
QM
x}
a
07?
sia 20
)(3Ex
jo
1272
¥/o”
N
sin YD ?
26.
vafue
Problem
=
rs
is the
ebPowehble
value
ot Pe.
P=
—
—
333 KN
=i
1.36 A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by
welding along a helix that forms an angle of 20° with a plane perpendicular to the
axis of the pipe. Knowing that a 300-kN axial force P is applied to the pipe,
determine the normal and shearing stresses in directions respectively normal and
1.36
tangential to the weld,
Oar
fo = tal, = D. 200 m
do= 0.400m
0.200-0,010 = 0,196 m
Ye= VWo-t =
A. = W(We"- Wy") = 1 (0. 200% - 0.1907)
SF 12.25
B=
xpo™4
hm”
20°
= B00 xg" cog 20"
-
= Poo tag
>
30
IZ) =
§
on
—
Based
B+21.6x10°R
St
.
°
26
=-
CO =-BIGMPa
om
~300 x}o? sin Yor
(212. 25") oO)
=< 7, %7 xjo°
C=
7.37 MPa, <#
Problem
1.37
- ta
sonnei
,
pe 240 aren
—
1.37 A steel loop ABCD of length 1.2 m and of 10-mm diameter is placed as shown
around a 24-1mm-diameter aluminum rod AC. Cables BE and DF, each of 2-mm fo
aa ara ed IB ply THe Téa O KHOW!
)
atrengtiv of th
uséd for the loop and the cables is 480 MPa, determine the largest load Q that can be
240 mma
applied if an overall factor of safety of 3 is desired.
Using
anal
joint
cons
2+
B
de
Fae
Q=
‘
bods.
sgume
~Q
)
=o
& Fin
A asa Free
joist
Usina
asatvee
ran
body
and considering symmetry,
@Gamem
2-3Fin~ Fae = O
$EQ-FeeO
Based on strength
Qy
Based
=
&A
of cable
=
on streng H
Fe
“
Qe Fgh.
BFE,
& Eat =
of steed
Qu* Fav?
(ygox$F io®)
(0.012)" =
Poop,
54.24%10*
GAs $6, Fa%
= £(480«1l0%) F(0.010)" = 45,24~/0°
= £(aCoxjot)F (0.024)* =
Blow
b Le
Joan
GL =
Quy
a
is the smadbest.-.
=
1524 xl
N
a
Qeigh.
= FGA = FOFa
Road
N
|
Based on strength of tod AC.
Actua? ultimote
EZ,
3
=
.
88.22/02
OQ =
IS.ORxjJo
Qs
N
45.24~l0"
oO
N
N
15,08
kN
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of (he publisher, or used beyond the Hinvited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
1.38 Member ABC, which is supported by a pin and bracket at € and a cable BD,
was designed to support the 16-kN load P as shown. Knowing that the ultimate load
Problem 1.38
_..£0F cable BD is 100 KN, determine
the factor of safety
with respect to. cable failure... fo...
Use
member
Free
body, ancl
that
member
two-foree
0.5 m
DS=EM,
04 m|
ABC
=
as a
note
BD
P
A
:
is a
member
3%
(Pros HO? )C1. 2) + (P sin 40") (0.6) —{ Fa, cos 30° 0.6 )- (Fey sin 30° MOY) = O
l.3O4Yd3 P -— O.71962 Fen = oO
Fen
=
1.31335
Fuse
=
[00x(O*
Te
Mm.
Problem
=
Fuet
_
Fan
1.39
P
= (C.B12aS)(iG%
109) = 2.49014 x10? N
N
100
%
x*fO
S
2.4014*10%
£39
FS
=
3
Se
aS
<2
Knowing that the ultimate load for cable BD is 100 KN and that a factor of
safety of 3.2 with respect to cable failure is required, determine the magnitude of the
largest force P which. can be safely applied as shown to member ABC.
Use member
body, andl note
that
member
DEM,
0.4 n-|
as a
Free
4wo-fovce
OS sa
ABC
=
BD
P
A
is a
member.
Oo:
(Pees HOC.2) + CP sin 40°)(0.6) ~ (Fa, cos 30°100.6)~ CF a, sm 30°04) = 0
|.30493 P - 0.71962 Fy = 0
P=
ADPow adbe
0.55404
vetue
of
Fy,
,
Foo .
Py 7 (0.88 404)(3,105)
Fon 7
Fig
ne
7
SS
le
v
Pe?
a
SS
\.732 kN
Proprietary Material. © 2009 The MeGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or ased beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<8
Problem 1.40
1.40
The horizontal link BC is 6 mm thick, has a width w = 30 mm,
:
:
_ and_is.mad
is the factor of safety if the structure shown is designed to support a load
P= 40 kN?
|
|
DEM. = 0
40EN
(0-3 cos 80°) Fag — (0:45 sin 30°40) = 0
r
Fac
Ay?
300 map
KN
3$4°6
(0.006)00+035)
=
i-8
Xion#
m
G. = Fe
- Som
Nec
f
Aec. Gort ob BY 104) (450 x10") _= 2034.
Ss
Problem
1.41
FS.
~
Fee
:
3a»
& x10?
1.41
The horizontal link BC is 6 mm thick and is made of a steel with
a 450-MPa ultimate strength in tension. What should be the width w of the link
if the structure shown is to be designed to support a load P = 32 KN witha
factor of safety equal to 3?
4) 2M:
©
(O+3cos 30°) Fxg ~ (0-45 sin 30" (32)
Fas
>
gc. = Fe
_
Ars
2 Je]
=
KN
Fee
iw
. Sot
ES.
0217 xo)
wow LF+S GowFae _” (3
(pr 0eb) (450 x10)
W 2010807
=
30° 8
m ae
mm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course prepatation. A student using this manual is using it without permission.
Problem
-
1.42
-
..
1.42 Link AB is.to be made of a steel for which the ultimate normal stress is 450
‘MPa, Determine the cross-sectional area for AB for which the factor of safety will be
that the link will be adequately reinforced around the pins ata and B:
<3:50-Assume
-
i
dam
Yo0
kN4m
t O4m
044
Pe
(1.2)¢8)
tO ZM
y
Fig
~ (9.8
it
-
9g
= OF
Sin 35°)
+(0.4 (20)
Faa 7 21619
=
6
AG
tae
Aue
FE Ss.
:
Fas
ie
.
+ (0.2)(9.6)
IN =
Q
O.2m
Dy
oO
21.619«I1O'
N
~ (ES) Fa . (B.50\( 21.619 x10")
A
- Soe
kN
if
Ae
iF
SAN
AR
Gon
= 168.1 x10 Sm
HOO
“ {o*
Aggy? 168.) mm”
Proprietary Material. © 2009 The MeGraw-1Hll Companies, bre. All rights reserved. No part of this Manual may be displayed, ceprodaced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individeal course preparation. A student using this manual is using if without permission,
eat |
Problem
1.43
1.43 The two wooden members shown, which support a 16-kN load. are joined by
ae
___ plywood
splices fully glued on the surfaces in contact. The ultimate shearing stress in
the ghic is 2-5 MPa and the Clearance
between the members is 6 mm. Determine the
required length Z of cach splice if a factor of safety of 2.75 is to be achieved.
iGkN
125 wim
gre
Fack
atve
of
x
6mm
There
sheav
LOKN
SkN
Requirec
Requivad
Py
>
Ferg th
ot each,
oeTA
OF
i, A LYa
.
Length
af sptices
Problem
GEN
>
qhve
.
1.44
®P,
JGKN
of
spdiec
e
cSeaprauce.
Area
of geue,
vdimete
22
(2.75 )(8+10*) =
(2.51 O° V6, 125)
=
N
“le
10.4107" m
.
= (270.4 x10? )+ 0.006 = OY6R x10
Theve
ave
Fach
qiue
of
4
sheav
doacd.
le
ap
ta
where
L=4t(L-c)
A=
Iw
separate
avea
wk
Mm
of saletye
of
adue.
3
kM
Poad.
3kN
2
=
Jew
= 8xlo'N
yh
A\
oy
7.4
Doe and
= $@.180-0.006) =
(0.087\0.125)
Poe BAz
aveas
must transmit
=
=~
aq
0.037 m
10.375107*
m
(2.5%10°)(l0.375% 10") = 27,1875«10% N
.
Factor
Poacd.
92x10"
5
P=
th
KN
1.43 The two wooden members shown, which support a 16-kN load, are joined by
plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in
the glue is 2.5 MPa and the clearance between the members is 6 mm. Determine the
required length £ of each splice if a factor of safety of 2.75 is to be achieved.
6mm
g
8
8x«i0°N
=
.
2D 40
travemit
1.44 For the joint and loading of Prob. | 43, determine the facter of safety. knowing
that the Jength of each splice is L = 180 mm.
225 mm
en
=
(F.S.)P
=
AVER »
,
must
vt timate
j ro
thw
Le
avea
aveas oF q due.
boas,
Ps
Po
sepawvate
3
ES
=
B.
RUN oete
Eis,
=
3.40
<a
“Po
Problem 1.45
4.45 Three 18-mm-diameter steel bolts are fo be used to attach the steel plate shown
to a wooden beam, Knowing that the plate will support a 110-kN load and that the
of safety...
__Uitimate shearing stress for the steel used is 360 MPa, determine the factor
for this
design.
As
bolt,
each
For
=
7 d*
E (ig)
254-47
©
mnt
= 254.47 10% m
ce
« 10°)
1360 7
x105 .4
Poe AT, = (854
HO-kN
,
|
For
the three
Fucton
oe
boSts >
sabety
ms, =.
P
Problem
1.46
N
x lo
4.609
=
Po
(3 \(41.€09« (0°)
= 274.83 xjo? N
.
74 x {o*
FMI
110 * to?
saa so
at
- 1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden
beam. Knowing that the plate will support a 110 KN load, that the ultimate shearing
stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired,
determine the required diameter of the bolts.
For each bolt,
P= HE = 3e.667 kW
Requived
Py =
(FS.)P =(3.35 (86.667)
= 122.83 kw
wept
a A
|. BR
~ Ta
LLO KN
d=
[ARs
TT
—~
. 4
~ Wat
[(4122.93 x40*)
wl3to~x 10%)
=
-3
40.8 XID “im
A =2O.8
mm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by MeGraw-Hill for their individual course preparation. A stadent using this manuat is using it without permission.
st
_
Problem
1.47 A load P is supported as shown by a steel pin that has been inserted in a short
wooden member hanging from the ceiling. The ultimate strength of the wood used
1 AT
__.is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength ofthe steclis |
145 MPa in shear. Knowing that b= 40 mm, c= 55 mm, and d= 12 mm, determine
the load P if an overall factor of safety of 3.2 is desired.
1.48 For the support of Prob. 1.47, knowing that the diameter of the pin is d = 16
mm and that the magnitude of the load is P = 20 KN, determine (a) the factor of
safety for the pin, (6) the required values of 5 and c if the factor ofsafety for the
wooden members is the same as that found in part a for the pin.
P=: 20 kN=
2oxlo’ N
Sk
Pur
As Tat = £(0. 016)= 201.06 x10 m
P, = 2A% = GN Ror. Igxio® (145 * 10%)
a0 0m
53.336%10"
~c
F.S.
(bo) Tension
Gy
in wood
AKy
=
b=d
Py
=
wib-ol )
=
Bo
Py
W Go
+
:
O.
=
N
. Pa _ 58.336xI0°
_
=
”
P
~
2Z20x%10%
§8.336 s/o" N
where
|
Ol6
+
for
Wr ~ 4Omm
Shear
in
wood
Py
Double shear;
Py
|
C=
=
wre
_
2A
each
Po
Jw te
=
53.336*10
area
P.
2.42
A
same
F.S,
*=v
0.040nm
58.236%10°
== yo40.Bx%/0
Byjo*
—
+5 940 \(GO *1O*)
b
te
Ty = Fe
x
v>
shear
Double
v
is
N
At we
Qwe
(2V0.040)(7.
10%)5
4O.B
Lon
ewv ere Peerer
8336x103
=
=
mM
mm
same
<a
FES.
|
97.2 xlo”? m
Cz
.
97,2
mm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual
is using it without permission,
—_
|
1.48 A load P is supported as shown
ROWE by a steel pin that has been inserted, in a short
Problem 1.48
ae
So,
“48
wooden member hanging from the ceiling. The ultimate strength ofthe wood used
60 MPa
iii tension atid
7.5 MPa in shear, while the tiltimate strength of the steel is
145 MPa in shear. Knowing that
= 40 mm, ¢ = 55 mm, and d= 12 mm, determine
-the load P if an overall factor of safety of 3.2 is desired,
“fa
~ Based
aouble
shear
iv pin
Py = 2A = 2Ed*y,
uy
r mm. Ly 7
40
on
~
E (2)(o.012) (14510*) = 32.30 x10° N
Based
Po
=
on
tension
AS,
—
In
Ww dod
wilb-d)6,
= (0.040)(0.040 - 0.012)(60x 10%)
=
-_
Based
Py
t
~
Ose
.
on
double
ZAT,
=
sheav
67.2 xlo? N
in the
2wety
oa
wood.
(2)(0.040)(0.
058 )(7.5
10%)
33.0 x)0° N
smattest
P,
=
32.8
xo"
.
APowalhde
NV
3
Pr
ie
=
-
24-2
“lor
/
*
10.25" )0"
N
10.25 kN
a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without perinission.
|
_ Problem
1.49
|
1.49 Each of the two vertical links CF connecting the two horizontal members AD
..and.£G.bas.a.10.x.40-mm.unifonm
rectangular cross.section. and is. made.of.a steeh
with an ultimate strength in tension of 400 MPa, while each of the pins at C and F
has a 20-mm diameter and is made ofa steel with an ultimate strength in shear of 150
MPa. Determine the overall factor of safety for the links CF and the pins connecting
[ 250 mm
és
them to the horizontal members.
400 mm
250 nim
as free body,
t Fee
EFG
Use memher
| Fee
e._&S
LE
LL.
0.40
025)
24kN
24 kN
4) aa Me
=O
0.40 FE, ~(06S5)(a4vI0*) = ©
F, = 37x10"
Based
on tension
A= (b-d)t
in Pinks
CF
= (0.040 - 0.02)(0.010) = 200 r10% m* (one Bink )
F, #26, A= C2{400x10%)(200 *io®)
Based
on
dovide
N
=
166.0 x]o*
N
shear in pins
a
A = Fd* = 2F(o.020)* = 314.16 x10" m*
*
10°) = 74.248 10° N
0
«10° (314.16
Fy = 2TA = AY(15
fy is smaller vahve, ie
Aetuad
Fy = 94.248 10° N
;
Factor
of
satety
FS,
=
os
=
Tee
3
=
2.42
Cr
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
al
oof oocoscnm ne oon
Problem
ne
1.50
1.50 Solve Prob. 1.49, assuming that the pins at C and
with a 30: mm diameter,
have been replaced by pins
1.49 Each of the two vertical links CF connecting the two horizontal members 42)
and EG has a uniform rectangular cross section 10 mm thick and 40 mm wide, and
250
is made of a stee! with an ultimate strength in tension of 400 MPa. The pins at Cand
F each have a 20 mm diameter and are made of a steel with an ultimate strength in
mm
Me
ae
shear of 150 MPa. Determine the overall factor of safety for the links CF and the pins
460 mm
A
connecting them to the horizontal members.
~Ta50
my
Use
member
EFG
as free body.
Fee
Fee
(e
le
D2M_
pam |
=
Ot mo
24.KN
ba—o:25 mo
8
24 &A
O;
(0-65°\(24x00 )
=
O
Fi. = 39 EN
Fa dure
by
tension
in
Pinks
Net section avea for
CE,
(2
| Jinks
pavatte?
bin ks)
A=(b-d)E =(0-04-0-02\60.01) = 200 x16 "m*
Fy = 2A6, = (2) (4vox1a”)(200x
*)
ig*) = tbo kM
eet Puye
by
douhte
A = ga
Fy
Actual
=
2ZA%T,
vi timate
Factov of safety:
shear
in
pins.e
* % (0122) = 314-16 x10 m?
= (2) (31e- tence *VGS0x108) = 94.26 kM
Doad
is
the
smaller
= FS= =
ce
Vadue
e
Fy
= 94625
EN
F.S.2242
<i
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
1.51 Each of the steel links AB and CD is connected to a support and to member
Problem 1.51
a.
tsp
50 mm
ameter
soc]. pins
pins acting
iameter.stcel
acting, in.sin
12 mm
Use
——
<=
|e 450
450 pam
mum
ing.
~
ae)
member
=. 4
2
=
BCE
Of e
O35
as
Free
body,
E ep
~—aoqTs FP
=
Fep h
Pp
P
(
OFM=
&
5
:
he |
Cc
OS
.
Din
ks
have
Theve Fove,
they
by
pin
the
Same
have
the
in singe
area,
same
A=
sheav,
cdiametev
aude
3 Fre
metevial
,
foad.
Bd*=
EX G.028)'= 440-9410 me?
-d)
3x06 *m?
t= (os — owrs)o-o1zz
A=(b
tension in Sink.
by
pin
ubtimete
2
=
~~
= (2toxo®) (490-9xj6°) = 103-09 kN
Fy = tA
F, = GA
Ditimate
2
~
Ss Feo
&
;
Farbure
©
.
P=
Fark ore
i
03% g-04P =O
a
y
Both
the
ed in the pins and that the ultimate normal
stress is 490 MPa for the steel used in the links, determine the allowable load P if an
overall factor of safety of 3.0 is desired. (Note that the links are not reinforced
around the pin holes.)
.
Pp
|
= (490x10°) (3x10 *) = 147 kW
Poad
#iruk
and pin
F, 7/0309
LY
for
is the
smelber.
ADlow abte values of Fin and Fra,
Fy? fe = AGRO = 34.36 kN
Abbowahte
Aoad
For structuve
Pp: 2 34.36)
is
the
smatber
of
Ry
amel = Fy,
Ps/s-7 ky
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<li
Problem
1.52
:
1,52 An alternative design is being considered to support member BCE of Prob.
a
1.51, in which
stissmonen eas snc
link CD will be replaced by two links, each of 6«50-mm
cross
SOCtOR;. causing.-the-pins-at-C-and-D-to. be-in-double-shear: Assuming -that-all-othere-<fo.0-.-.~
specifications remain unchanged, determine the allowable load P if an overall
50 mm
factor of safety of 3.0 is desired,
12 min
L534 Each of the steel links 4B and CD is connected to a support and to member
BCE by 25-mm-diameter steel pins acting in single shear. Knowing that the
ultimate shearing stress is 210 MPa for the steel used in the pins and that the
a
ultimate normal stress is 490 MPa
for the steel used in the links, determine the
allowable load P if an overall factor of safety of 3.0 is desired. (Note that the links
are not reinforced around the pin holes.)
member
BCE
as
free.
body.
O
Use
Avea
E
492 Me=z
of afl pins
Fab pre
by
avea of
pins
by
He
2 Paks
A and B
in
tenston
in
Bink
(Fie u = (49x12 9B x16 *)
Wihimate
foad Pow Pink
Covrespen ding
Faifure
by
vIt+imate
pins
Mi
a
P=
Fre
Aner= (bd) tag = (oes ores) O.012) = 3x/9 in
CD
ic the
sing fe
shear,
(Fro), = (20x10 )(490-9x10"*) = 103-09
FacPuse
03 Fp - oar
_
F.
A iin* E f* = TE (ovasy = 490-9 x0" be
Net section avea of Prek AGE
Net sectron
Ol
Pe
P =
O
|
wip
Fas
= ei
Cand
same.
Fae). =
EN
AB,
“ay
XT, Apin
Fase =
65 Anuet
ka
and pins AB is the smaller’ (Fixg)y = 103-09 kW
Load ,
Py
=
3(Faady
in double shear.
= 68°73
CRs
EN
2 Apia
(Findy = (2 UZ10x1%) (490-9 x10" 9) = 206-18 KN
Faifuve by tension in Inks CD.
(Fro),
DOtimate
2 (440x10%) (3x10"*) = 147
Poud
Corres ponding
Acti A
Lor Pinks
ultimate
vstimate
Load
Mlowshte Load P:
kf
anef pius
doz:
is the
(Fis) 2 Sb Avet
CD
iis the smabler;
Ps = 2 Feo)y
smatter.
p- ES,tv
y 7/47
kN
> SE-B kA
9 = S88
_ 508
3.0
CR
kA
PaI0b kN aes
Problem
resent
1.53
1.53
In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-
diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all
staan
sooo CONN CCHONS: and the ultimate normal stressis 400 MPa imiink 2D: Knowing thata™
factor of safety of 3,0 is desired, determine the largest load P that can be applied at 4.
Note that link BD is not reinforced around the pin holes.
Use
Lyee
ocd.
ABC.
Front view
“18 mm
ADEM, = 0:
0.230 P - 0.120 Fep = ©
-—ia<—-— 6 mim
BB
P=
&
Fap
(1
Side view
452M,
20:
0.1460
P CFB
Top view
P=
° [
|
a
4
Tension
on
nel section
G
ef
Band
Frou
value
in
Sheav
of
Fen
is
pin
af
of Pink
BD,
ig-1oe)
2927010
= 3.9270 «10° N
N,
1.683 K 10" N|
C
20 apm = ae ay. (2) (SPO EY (exo? = 2.8274 x10" N
Cr
From
|
D.
P= (S)(3:7270xl0"} =
(1)
(2)
6. 46xjo* N
Fey = TAp.> ot Fa? = (SLAVE Vio eto?
Sma dber
O
Fas? CAnt = SE Aw
ij
im pins
=
#C
(Hognlo'(e vio
Shear
CG
—
Fae
G
O.l20
Pe Yasar}
@)
Smaller
vehue
F
P
is
aDlivus «Ade
=
212x107N
ve due »
Pre
1.692x1lo
N
P= 1.683 kN =f
Problem
M
1.54
tie
1.54
mt
Solve Prob. 1.53, assuming that the structure has been redesigned to use 12-
min-diameter pins at # and D and no other change has been made.
ho
diameter pins are used at B and 0. The ultimate shearing stress is 150 MPa at all
connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a
factor of safety of 3.0 is desired, determine the largest foad P that can be applied at 4.
Note that link BD is not reinforced around the pin holes.
Use free body ABC.
Front view
tDEM,
<— 18 nm
= Or
O.2BOP
P=
| mn
160 mre
ania
Bo
120 min _£
Side view
4 4 oa
- OL120
3 Fap
Me
fe
Ps
‘Top view
Fae
,
‘Tanston
!
on
net
Dp
P
|
doo
x
B
and
to
vabse
From (0)
«¢
Ox1O* 7 Ey
= (52
of Fen
ts
\CH Cats
Sheaw im pin at C,
Drak
BD,
Se
7
f
xJfo®?
N
-3\2
Y= S654 «(oN
2.06 xI0° N
|
C= 2UAm = Zoe Edt= = (2 Be
Smabler
PY
4.80x%lo' N,
P=(S\4.80%[o%) =
From G\,
'2)
D.
* ao a
Fen7 TApi
SmmeWevr
C
\guio® U8-i2ie
3
= 40
eA
=O
«lots FS.
= (129
th pins
3
section
~
Ve
‘Shear
- O.2200
C
f
G
Q)
=o:
9.160 P
AB
Fey = O
VE\Geiwe
= 2.8274 4107 N
P= B\(zez7teict) = 21axlN
valve
of
P
ors
the
aD Poute
be
Vodue
P-
2.6618
P= 206kN
NV
«a
Problem
1.55
1.55 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter
_ pins are used at B and D. Knowing that the ultimate shearing | stress is 100 MPa at all
“connections and that the ultimate normal stress is 250 MPa in each of the two links
joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is
.
desired.
Top view
. [- 200 mm ap
80 nam
|
ft
12 mm
Statics 2? Use
Simm
as
ABC
free
body.
7
B
ah
ozs —H
20 mm
Fa
8 mm
=
0.18 —|
Fao
8mm
HFMe,
Based
double
on
P=
12 mm
Front view
-
pis
A
Based
on
Based
on
davble
pins
tu
shean
A= Ed*= Foo} =
Eye = au
eee
Fow
compression
one
A=
Pink
Fao = “es
= Gila
P= ig Feo =
Movable “valve
in
of
O18
P
=o
Fan
3.35} ¥Jo32 N
N
3.72 x10
=
PF,
Fz-
LF,
Paes
A.
A=Bd*= $(0.008)* = 50.266 x10 * m
re ZtvA _ (A100 x10%)($0.266 ¥Jo_ 2
A
ES.
3.0
p=
0.20
OF Map= 0: 0.20 Fep-0.38P
= 2
Side view
shear
20:
a}
Band
D.
113.10 x/Q° m™
“6
es loxlons)
2 zea yig® N
Sinks
BD.
(0.020 )(O, 008)
=
xlo WigoxissS)
_
160 x/0-
mo
ae 7 #)0° N
14.04 »1o* N
P
is
smatlest.
+.
Pr
3.72
P=
*1O" N
3.72 KN
~<a
soe seseespettte te
.
Problem
4.56 In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter
1.56
is to be used at A. Assuming that all other specifications remain unchanged,
oon determine the allow:
ad.P if.an overall factor of safety. of 3.0 is desired
eenccccscrnnudeace ose ous
1.55 In the structure shown, an 8-mm-diameter pin is used at A; and 12-mm-diameter
pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all
connections and that the ultimate normal stress is 250 MPa in each of the two links
joining B and D, determine the allowable joad P if an overall factor of safety of 3.0 is
desired.
Top view
[200 mmorten 180 min
Smt <I
| 12mm
i.
;
at
as
2
ot
.
.
Statics
>
Use
A
Fp
0.18 “y
Feo
P
=
“ Mp=
Based
on
double
rc.
=
= ¥ (0.0.16)*
A=4d*
pin
1”
shear
.
RS.
Pp
Fa
fo
1y
0.20 Fep- 0.38 P= a
*¢ Be
.
Fe.5¥ x10°% m”
= ZteA | (100 xjos)( 7B.S4 xfo”)
A
3.0
P
0.20 Fy-ONRP=0
0:
P=
A. oe
body.
.
HEMg=O%
~
free
c
ee
V
12mm
as
8
H-——-}.20
8 mim
ABC
=
:
5,236
x Jo?
N
P= BPR = §.82 x10" N
Based on double shean in pins at B and D= 113.10 x0" m™
R= Bd* = Boom)
Fan = AGA
_=
Coo xI10S0 )(113.1oxlo”*) = 7.54x/0° N
=e
3.97 x107 N
P= 12 Fin =
on
one
Fov
Feo
compression
=~
in Minks
BD.
A = (0.020 )(0,008)
Pink
1Go x/o* my
285A _ (2)( 250 0‘ igo xI0"*)
FS.
|
3.0
.
‘|
Based
26.7 *10° N
P= ag Fan = 14.04 xJo3 N
AVowabde
value
of P
is
smatlest.
..
P=
3.97 *1O> N
P= 2.97KN
«a
_
Problem
1.57
*1,57 A 40-kg platform is attached to the end B of a 50-kg wooden beam AB, which
_ is supported as shown by a.pin at 4 and by a-slender steel-rod BC with a 12-kN
“ultitnate load: (ay Using the ‘Load and Resistance Factor Design method with
resistance factor ¢ = 0.90 and load factors 7 = 1.25 and
»% = 1.6, determine the
‘largest load that can be safely placed on the platform. (6) What is the corresponding
conventional factor of safety for rod BC?
DEM,
0:
(24)8P-
240-120,
For dead toading,
W,= (40)(9.81) =
W, = ($0)(97.31) = 490.5 N
Po = (3X 392.4) +E (490.8) =
For Pive
Poading,
P=
Wr=mq
mg
Design
from
~ Pr FWrEU,
392.4
N
1.0623 * 108 N
Wz = 0
which
om = 2
entenion,
+
YR
|
9 Po
=
1B
p = PP VL-%R
_
x10")
0628 ®)
~ (.25)(
rio
(0.90)LGCia
= §.920x10* N
@) Moweble load.
ms ¢ S22x10
Conventional
safety.
(b)
Pr
Rah
F.S.-
=
Pwe
Pactov
10628 “10> +.5.920%xJ0o”
_* Gaaduio®
12 x)0%
m= 362 kq ae
3
=
6:483/0°
N
ES.=el L7t8<<
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ave Porn
_.Problem.1.58...0
P
__*1,58” The Load and Resistance Factor D.
ign method is to be used to...
select the two cables that will raise and lower a platform supporting two window washers. The platform weighs 72 kg and cach of the window washers is
assumed to weigh 88 kg with equipment. Since these workers are free to move
on the platform, 75% of their total weight and the weight of their equipment
will be used as the design live load of each cable. (a) Assuming a resistance
factor @ = 0.85 and Joad factors yp) = 1.2 and y, = 1.5, determine the re-
Pp
ulumate load of one cable. (b) What is the conventional fac-
quired minimum
tor of safety for the selected cables?
YP
Py
+
=
UR
eRe
=
PR
+h
de 72) #1 5)C4 «2% 88)
Ch.
O.8S
=
Conventronal
P=
P+
c
~
.
S.
factov
P=
Po
P
283576
2&7h4
=
=2-73
-EN
<<
of safety
472+
_
ka
bap
0.75
% 2%
7
—
BB
= Ib bbs = eh 4RKN
.
e
64
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educatois permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without perthission.
-"
Problem
1.59
4.59 Link BD consists of a single bar 24 mm wide and 12 mm thick.
Knowing that each pin has a 9 mm diameter, determine the maximum: value of
“the average normal stress in link BD if (a) 0°= 0; (6) 0 = 90"
IGKN
Use
Q=
O,.
49 EM, = O02
(0-45'sin 30° )U6) - (0-3c05 307) Feg = ©
4
Fan
~ 13+PFS6
Area
for
kA
(teusiou
tension
Stress >
fo)
ABE
as free ody.
300 mat
fa)
bar
Aoad ing 5
Fao
QO =
=
A
=
A=
)
(b - d)t
=
(24.
9 (12) < 80 mm?
{3-8S6x107
Bo xto-&
O=
=
77 Mha
eh
OF 4%,
+5
Oz
2M,
Feo
Stress
Por Compression
?
GC
30°) (6)
ie.
KN
—~ 24
=
Avea
—(6-485c0s
> Feo
A
”
(+3
aos
30°)
Fa
=O
compres son,
Aoading
A=
bt
= (74)(2)
= 288mm
= mente
28910
G=- 83-3 kN
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
mg
Problem 1.60.
1.60
Two horizontal 20 KN forces are applied to pin B of the assembly
shown. Knowing ¢ that a pin of 20 mm diameter is used at each connection, de
“termine the maximum value of the average normal stress (a) in link AB, (b) in
link BC,
45 nym
Use
20 KN
(omt
Bas
Free
body.
2OKN
12mm
40 kN
40 EK.
of
Law
Force triang fe
Sines
Fas
Fe
Sin 45°
mad
Sin GO?
Sin WZ"
Fre = 284 LAY
at
Fae = 34-9 EN
Link
AB
is
a tension
Minimum
section
(2) Stvess
in AB-
%
‘
Link
BC
‘
is
Cross Sectionad
.
at
pia
a
“Vh
2.
Stress
in. BC
me net = (0:045-0101) (0-012) = 300 x10
,
Che = et
a Tomp ression
*
(b)
member
area
is
‘
Bc.
6,
“R
)
pos
6
in
747 Mpa
<
member
AF (01(010
045
/2,) 2 $4.
)0X10 em
.
-Fo
we
PE,
R=
=
=
a
34 F
.
:
Fhowoe
>
7
64
¢
2
4MPa
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced. or
distributed in any form or by any nieans, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<<?
Problem 1.61
1.61
For the assembly and loading of Prob. 1.60, determine (a) the average shearing stress in the pin at C, (b) the average bearing stress at C in member. BC,.(c).the average. bearing. stress. at.B.in member. BC
1.60 Two horizontal 20-KN forces are applied to pin B of the assembly
shown. Knowing that a pin of 20-mm diameter is used at each connection, determine the maximum
link BC.
45 mm.
value of the average normal stress (a) in link AB, (6) in
Use jomt
Bas
40
free boy.
kN
Fis.
4a
kai
Force triangde
Law
of Sines
(2)
Sheaning
Ap
z
()
=
4.
Ed
=
>
Bearing stress at
(20)
Bo
nh
= 214-1b
»
BATTO
Bearing
Stress
=2U2)@0)
Az2td
=
Bin
ee
YT
cam
480 X10"
=
member
P
care
=
T=
55-3
6, =
BC.
a,
MPa
Eas
|
.
= (4A. BIS X104%
at
a
&
member
Cin
7
be
au
= (12)(20) = 240 mn®
Ea
Ob *= Soxie®
Gy
,
j
—
a
Cc.
(2)\G14- 16x 10~%) ~ $5-338 Kio
Az
{Cy
i
—
F4770
“___
at
Pin
14)
stress
v
4°
sin
Sin 60°
Sin 45°
E
40
Fee
Fag
6, > 144-9 MPa
BC,
Sy?
—_
re
480 mmt
72-437 x10 ®
Gy
= 72-4 MPa
<a“
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
1 62
4.62 Two wooden planks, each 22 mm thick and 160 mm wide, are joined by the
glued mortise joint shown. Knowing that the joint will fail when the average shearing
Stress.in.the
glue reaches 820. kPa,.determine the.smallest allowable-length d-ofethe--2udewrn oe
cuts if the joint is to withstand an axial load of magnitude P = 7.6 KN.
Seven suvfaces
P
foul
7A
P_ 7.6 x10
=
d= f=
Ls
Fach
7% ~Gagoupery
we
=
the
toh?
P= 7.6 LN = 7.6 %I0%.
Let
we 2B
canyy
gfe
_.
RR
mm,
area
is
© S32004 WO”
Ax dt
a
|,32404 *10" mm”
(-32404«I0
22
Ss
_= gn 9
d=
60.2 me,
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course préparation. A student using this manual is using it without permission.
—=afi
Problem
1.63
1.63 The hydraulic cylinder CF, which partially controls the position of rod DE, has
been locked in the position shown. Member BD is 15 mm thick and is connected to
othe,
vertical rod. by a.
9-mm-diameter,
lt,bolt Knowing
that, P =2. KN and @ =
as ao tree body, ane
a two force member
BCD
Use member
is
AS
that
note
200 mnt
Pp
(2
Lo]
45 >
|
us
of
member
a
200
Ding
Fe 3
ABS
+
+ 4H
cos 75°) kw
(2 sm7s° JEN
Soe
a
FELT
Length
,
205
20S mm
Ht
Dz
MF
F,, )Cloo
( 402
Os
ROS
sin 20°)
84.16.96 F,4- 348.668
(loo
Faz
se
~_
20°)
Ces
-(2 eos 78° (ITS
sin
sin 75°
—
=O
20°)
=O
Ct 7S cog 207)
Fags 4.1424 kN
SZR=OF
Cy — FS-(4.1424) +1 2ces7S°=O
Ce 7 0.3717 kN
HPZR=
0:
Cc, -
Cy=
$.9732 kN
Reaction
at
Ct
5.49860
(a)
stress
Sheaning
Awe
CT - Ry
Beaving
in
(sing te
bolt
styess
6
G3. €17*1O
59860 %10> _
‘
7 Gaerxioe
at
©
=
MOTO
in member
6,
i
A. = dt = (0,007 \(0.015) =
LL
Ao *
£9RLOHIOY
iaex tore
kN
sheav).
=
#(0.009)
.
2sin75°=O
C,?
_
JE yt
4 do:
- G
EU. M24) -
Cz=C/+
Cr
—
(b)
7520.00 dooce oe nn.
determine (a) the average shearing stress in the bolt, (b) the bearing stress at C in
member BD.
wm
2
-
Pa
T= 94.| Mew
_
BD.
™ 10° m*
135
44.34 x10"é Pa
6, .= 44.3MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior writfen permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
mr
Problem
1.64
- 175 am
100 mm Soy
1.64 The hydraulic cylinder CF, which partially controls the position of rod DE, has
been locked in the position shown, Link AB has a uniform rectangular cross section
f 12x 25 mm.and is.connected_at.B.to. member, BD. by.an 8-mm-diameter.pitevudecce ce. co.
Knowing that the maximum allowable average shearing stress in the pin is 140 MPa,
determine (a) the largest force P which may be applied at E when @ = 60°, (4) the
corresponding bearing stress at B in link AB, (c) the corresponding maximum value
of the normal stress in link AB.
\
D
fe
\
200
Use member BCD as a tree body , and
note that AB is a two force members
P
Peos
i“
45 mm
.
.
°
P
Lewath
60”
sin
GO
ob meu ber AB,
Sag {200° + HS*
=
ROS
mm
HD EM= OF (BEF)
(00 ces 20°) -GE= Gi, Woo sin 20°)
=—(P eos60° (17S sin 20°) -(P sin Go" (175 cos 0") =
BY1696
fe)
A PPowarte
Pou
Fag ~
172.3414
P
=O
Pin
A
ts
Pe
at
Fagi 2.0475 P
iu
singe
shear,
Arm ba = Blo.coay = S0.26S85%10° me
Y=
14oxlo®
.
Pa
Fig
6
Apis
re
(bh) _ Bearing3
stress
adi
at B
in Mink
AB.
= (0,008 )(0.012)
Fa 7 (2.0475 (3.4375
S. =
CC)
Maximum
P
P= ody kN
a=
a Bram,
= 96«/O0-"
™10%) =
stress
Ave = (b~ ae)
- Fae
Kui
_
in
beak
AB.
7.0383~10°
[2 mw
7.0333 %/0° M
;
G,= 73.3 MPo
b= 25mm,
= (0.025 -0,008%0.0IR) =
| RO4® [O~F
t=
wm
‘
7.0383 xl0” _
Fag
7 73.3™10 Pe
A= _ “sexta
normal
Che >
2.04 75
[HOMO = 5O, 265 »(av%
P= 3.4370x10'N
= 34.52 }0° Pee
L=0.012
204/07
Se
|
mw
¢
n?
Cra> 34.5 ME
<n
Problem
1.65
_*
1.68 Two wooden members of 70 x 110-mm uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the maximum allowable
: Shearing stress-in-the glued splice-is 500 .kPa, determine the largest-axial-load-P.that----
11) nm
can be safely applied.
A,= (0.07OmVO1MOm):
70mm
8 =
v
Pp,
A,
Sin
9
Co
<
2.7%*107m
zB
4O°- 20° = Fo"
a
9
P = ggt ag GL
,73
Msoo lO) = 11.973 x10" N
“~
Sin JO?’ cos
%
7O?
Pe
H.98
kN
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individuat course preparation. A student using this manual is using it without permission.
ath
Problem 1.66
4 66
‘The 900- -kg load may be moved along the beam BD to any posi‘atE and Fo Knowing that 6. = 42 MPa for the stéel ise
in rods AB and CD, determine where the stops should be placed if the permitted motion of the load is to be as large as possible.
q_ 16mm
4” diameter
Per
wi ted
favces:
(Frelum= Ct Ave =(42800) EY(or012)"
tt
AB?
mewber
CD:
(CRaduae =
4°15 EN
Ow
|
A, = (42x00 )(E ) (0.016)?
ry 8°44kN
Fas
Foo
ht.
& jy
ie
L
aA
D
Use
membev
as
«
Free
boaly
P= 00 ky < & P24 LN
4D 2M,
P= Foo
BEFD
=
©
-C-5\ Fig t eS - Ke) P = O
a (4-75 x10)3
_ FS Fae 665)
5
Be
=
P
=
$+ P29
O+§o7
he
4) FM, =
Ks
Feo
Me
=
x493
=
Ge
673
ily
O|
=
TP
“~%P
=
O
5 Fey —Ch5)
(9:44)
=
P89
b6h45py
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individuat course preparation. A student using this manual is using it without perntission.
~t
fo
Problem
1.67
.
1.67 A steel plate 10 mm thick is embedded in a horizontal concrete slab and is used
to anchor a high-strength vertical cable as shown. The diameter of the hole in the
ength of the ste l used is 250.MPa, and the ultimate. 0) occ
24. mm, the ultimat
, bonding stress between plate and concrete is 2.1 MPa, Knowing that a factor of safety
of 3.60 is desired when P = 18 KN, determine (a) the required width a of the plate,
(b) the minimum depth b to which a plate of that width should be embedded in the
concrete slab. (Neglect the normal stresses between
the concrete and the lower end of
the plate.)
10 mm~
1%
IN
@)
Based
ou
beusion
in the plote.
Az (a-d)t
Pu =
SS
rss - Po,
fe. Slade
Soe
T
Sofving
Sov
a,
(ESP _ 6 ogy 4 (260018 10%)
as ol4 Gt
A=
(b)
Based
A
ow shedr lbetween
=
pevimeter
Re WA
od
in
Sofuing
Fo
o
*
depth
plate
=
(2a4
3
b
-
b=
" (2Z5DHIO*NO.010)
0.CUFU92m
=
and
concvete
2b)
b
2% lost)
b
A
492Gwmew
slab.
Ta
-
Q.1*1o°
Pa
Rs =
SDP
Aatbh)%
=
0.257423m
(3.62
(18
¥ (o*)
(2)(0.09992+ 0.01 (2.1 x10%)
b=257
mm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~<a
ee
Problem
o
1.68
1.68
The two portions of member AB are glued together along a plane
forming an angle @ with the horizontal. Knowing that the ultimate stress for
in tension and 9 MPair
é
of values of 6 for which the factor of safety of the members is at least3.0.
A, = (0105)(0:03) 30-6015 mM
P = lokwW
Based
Sy,
50 mi
‘
Cas
cos@ = 0-912
Based
sin
on
=
RAoTe
stress
cos’
~ SAS
=
fy
~
CU7xio
&.
)(00015)
Soe
Fa
me Oe PET
30 X10*
6222.8
WF
30 kN
@
fo
sin@cwsO
A,
_
*
= Po
AA,
sin 28
2
(2) (0.0015) C4 x10"
~
Ps
30 x07
20 = b4-16°
Hence
Ao
=
stress
@ = 22-8°
sheaning
26
on tensive
Po
26
P, =(RS)P
O= 321°
QO € 225°
228° € @ € 324°
Problem
1.69
—
1.69 The two portions of member AB are glued together along a plane
forming an angle @ with the horizontal. Knowing that the ultimate stress for
the glued joint is 17 MPa in tension and 9 MPa in sheat, determine (a) the
LOKN
value of @ for which the factor of safety of the member is maximum, (6) the
corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors of safety with respect to normal stress and shear.)
A, = 0105) (0103) 26° 0015 mn?
At the optimum angle (FS.)e =(FS.)e
Normal stress? G= F eos O x Ryo: As
(FSJe = p=
we
Shean ng
stvess$
=
Pisin Beer 6
E
toed ’ ng:
Sodving:
a8”
T old, A. (Se
=
Pein® ws e?
BENS
=
SA
Rie
=
GLA
ae too.
Pas
Paz
P
= lhc.
i
As
Psin®
ws O
5
tan ©vogue) = 77
=0°5249
(a) On = 279°me
6
|
fos
oe DES 2 Be dah
3260
=<
|
Problem 1.70
1.70 A force P is applied as shown to a steel reinforcing bar that has been embedded
in a block of conerete. Determine the smallest length L for which the full allowable
normal. stress.in-the.bar.can-be. developed..Express.the result.in.terms.of the diameter...
d of the bar, the allowable normal stress 0,1 in the steel, and the average allowable
bond stress 1, between the concrete and the cylindrical surface of the bar. (Neglect
the normal stresses between the concrete and the end of the bar.)
Tau
>
gt
Tal
A
P=
Equetina,
=|
tension,
Tah
ae
|
Ent a*)
6A =
_
For
if
shear
+
For
TuwNTaAl=
Cua Ed’
Solving for L,
Leiner
Sie AA 4 Mee
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
eee
PROBLEM 1.1
1.Ci . A solid steel rod consisting of # cylindrical elements welded together is
.. Subjected.
tothe loading .shown..The. diameter. of clement-i.is
denoted. by. di-and-theeeon.0fooue
load applied to its lower end by P,, with the magnitude P, of this load being assumed
positive if P, is directed downward as shown and negative otherwise. (a) Write a
computer program to determine the average stress in each element of the rad. (6) Use
this program to solve Probs. 1.2 and 1.4.
™ Element n
SO
SOLUTION
uN
FORCE IN ELEMENT Lz
_—
[E (5 the sum of the forces applied
to that element and all lower ones:
P,
L
Fe
AVERAGE STRESS IN ELEMENT
A;
=
i
Ta;
Ave
stress
=
o*
v
=|
Area =
6
oT]
hed
PROGRAM
QUTPUTS
Problem
1.2
Element
Stress
(MPa)
|
Problem 1.4
Element Stress
(MPa)
4
91-487
1
35-651
2
499708
2
4-2: 444
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educatars permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM
Member
A 20-KN load is applied as shown to the horizontal member ABC.
1.C2
1.C2
vertical links, each of 8 x 36-mm uniform rectangular cross section.
Each of the
four pins at A, B, C, and D-has the same diameter d and is in double shear.
(a)
Write a computer program to calculate for values of d from 10 to 30 mm, using
1-mm increments, (1) the maximum value of the average normal stress in the links
connecting pins B and D, (2) the average normal stress in the links connecting pins
C and E, (3) the average shearing stress in pin B, (4) the average shearing stress in
pin C, (5) the average bearing stress at B in member ABC, (6) the average bearing
stress at C in member ABC. (6) Check your program by comparing the values
obtained for d = 16 mm with the answers given for Probs. 1.7 and 1.27. (c) Use
this program to find the permissible values of the diameter d of the pins, knowing
that the allowable values of the normal, shearing, and bearing stresses for the steel
used are, respectively 150 MPa, 90 MPa, and 230 MPa. (d) Solve part ¢, assuming
that the thickness of member ABC has been reduced from 10 to 8 mm.
SOLUTION
IN LINKS
pean
2 Dinees A
ABE:
OF
)=0
AC
P(
.BO
p(
Fa
2
03
S
M
R
“ee
Gh) fePaps P(A, Q/2(BO)3 (TENSION)
eB
©
A
0:=2E, (BC)-P(AB)=0
26 we Om =e] tI 2M,
ey
Fop= PUAB)/2(BC) (ComP)
V2 Fe
Ed
Je. caf
(3) “oPint
+
= + Fap/Ag
ty,
B
Thickness
:
STRESS
of Member
i
Thickness =
f
miler
Ema
Aer = tw,
_
,
L
l
e =F, /(d 4)
SHEARING STRES 5 IN ARC
ATB
AC=t
S19 Bear B= Pon/(at,.)
G) BEARING
F,
Mee
Fee
ea
ag
-C
PIN
(4)
0
= lap (Td)
(5) BEARING
CE
4 Ce
App >t, (uj -d)
Cap
LINK
(2)
BD
(1) Lik
AC
» TRESS, ATC
UNDER
2 Fan
4
PIN B
fy. C (Waac /2)
e'e ~F_Zac tac
SQ Bear C= Fo /Cdte) | Spi] |
2F,=0!
25-25 lap
oegy 2 Fen© 1B
> tre he
(CONTINUED)
~~ PROGRANLOUTEUTS...
.. PROBLEM L.C2. CONTIN
INPUT
DATA
FOR
PARTS
a), (b), (c): P = 20 KN, AB = 0. 25 m, BC =
AC = 0.65 m, TL= 8 mm, WL
ad
Sigma
BD
10,00
"78.13
411,00
12.00
13.00
14.00
15.00
16.00
LTI00
18.00
19.900
20.00
21.00
22.00
23.00
24.00
25.00
26.00
27.00
28.00
29.00
30.00
81.25
84.64
88.32
92.33
96.73.
101.56
TUG ToT
112.85
119.49
126.95
135.42
145.09
= 36 mm, TAC = 10 mm, WAC
Sigma
CE
Tau
B
Tau
Cc
SigBear
-21.70
79.58
~21.70
-21.70
-21.70
~21.706
-21.70
-21.70
270
~21.70
-21.70
-21.70
~21.70
-21.70
~21.70
-21.70
-21.70
~21.70
-~21.70
~21.70
=21.70
-21.70
65.77
55.26
&F7.09
40.60
35.37
31.08
ZTTSE
24.56
22.04
19.89
18.04
16.44
15.04
13.82
12.73
11.77
10.92
16.15
9.46
8.84
IE"
:
ay ae
80.82"
TITSY
63.86
57.31
51.73
46.92
42.75
39.11
35.92
33.10
30.61
28.38
26.39
24.60
22.99
(c) ANSWER:
CHECK: For d= 22 mm, Tau AC = 65 MPa<90 MPa
0.40 m,
= 50 mm
} B | SigBear
[325
125.00
113.64
104.17
96.15
89.29
83.33
73.13
73733
69.44
65.79
62.50
59.52
56.82
54.35
52.08
50.00
48.08
46.30
44,64
43.10
41.67
3
°
Cc
CAG
2 xt
|
_
203.12
ISTTIS
180.56
171.05
162.50
154.76
147.73
141.30
135.42
130.00
125.00
120.37
116.07
112.07
108.33
16mmsd< 22mm
~
~<a
(c)
OK.
INPUT DATA FOR PART (d); P= 20 KN, AB= - 0.25 m, BC = 0.40 m,
AC= 0.65 m, TL= 8 mm, WL = 36 mm, TAC = 8 mm, WAC = 50 mm
d
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23.00
24.00
25.00
26.00
27.00
28.00
29.00
30.00
Sigma
BD Sigma
78.13
81.25
84.64
88.32
92.33
96.73
101.56
106.92
112.85
119.49
126.95
135.42
145.09
Z
CE
Tau B
Tau
C
-21.70
-21.70
~21.70
-21.70
~21.70
-21.70
-21.70
-~21.70
~21.70
~21.70
~21.70
~21.70
~-21.70
-21.70
~21.70
~21.70
~21.70
~21.70
-21.70
“OX,
80.82
71.59
63.86
57.31
51.73
46.92
42.75
39.12
35.92
33.210
30.61
28.38
26.39
79.58
65.77
55.26
47.09
40.60
35.37
31.08
27.54
24.56
22.04
19.89
18.04
16.44
15.04
13.82
12.73
11.77
10.92
10.15
~21.70
24.60
9.46
~21.70
22.99
8.84
(d) ANSWER:
CHECK: For d= 22 mm, Tau AC = 81.25 MPa<9
0MPa
SigBear
B SigBear
C
156.25
142.05
130.21
120.19
111.61
104.17
97.66
91.91
86.81
82.24
78.13
74.40
71.02
67.93
65.10
62.50
60.10
57.87
55.80
.
135.42
18mm<d<22mm
OK.
53.88
52.08
*@
(b)
(a)
Two horizontal 20 KN forces are applied to pin B of the assembly
1.03
C3
PROBLEM
. a8 eof eoroecennenn
same. diameter: and
.-$hown.-Bach-of-the-three-pins.at-A,..B,.and..G-has.the
.
in double shear. (a) Write a computer program to calculate for values of d from
12.5 mm to 37.5 mm, using 1.25-mm increments, (1) the maximum value of
the average normal stress in member AB, (2) the average normal stress in
45 mm
member BC, (3) the average shearing stress in pin A, (4) the average shearing
stress in pin C, (5) the average bearing stress atA in member AB, (6) the average
bearing stress at C in member BC, (7) the average bearing stress at B in member
BC. (b) Check your program by comparing the values obtained for d = 20 mm
26 LN
™ SEN
12.5 riot
with the answers given for Probs. 1,60 and 1.61. (c) Use this program to find
the permissible values of the diameter d of the pins, knowing that the
allowable values of the normal, shearing, and bearing stresses for the
steel used are, respectively, 150 MPa, 90 MPa, and 250 MPa. (d) Solve part c,
assuming that a new design is being investigated in which the thickness
and width of the two members are changed, respectively, from 12.5 to 8 mm
and from 45 mm to 60 mm.
SOLUTION
MEMBERS
FORCES
IN
FREE
Bopy:
BC
FRoM FORCE TRIANGLE:
~~ 2?
Fan _ Fac
sin4s? ~ sinéO sin7S°
PINB
B
60%
AND
AB
45°
Fac
FF 2P(sinhsysin 75°)
.
Fac = 2P(21h 60°/5in 75)
2F
(1) MAX AVE. STRESS INAB
fan = Width= swr
Thicknes
Inn
ee“AB> Rayan
Trae /Ange
G) PIN A
Be Teele
(4) PIN C.
C= (Pye /2)/N Td)
47/4)
STRESS
Age = ae0
Bc
Fac
,
Cy = Fag /Z)/
(5) BEARING
=L
(2) AVE. STRESS IN BC
FE
ATA
Sig Bear A= Fas /dt
(7) BEARING STRESS AT EB
(6)BEARING STRESS ATC
Sig Bear C= Fyc/dt
IN MEMBER BC
S19 Gear B= Fac /Zal
(CONTINUED)
o
PROBLEM
1.C3 CONTINUED
cece PROGRAM OUTPUTS
_ INPUT DATA FOR PARTS (a), (b), (c): P=Z0EN
D
SIGAB
SiGBC
mn
M Fe
MPa
TAUA
| M1 Po
w=4Smm
TAUuC
MPa
68 687 ps-631- , S720”
nf
t=i2mm
— SIGBRGA
SIGBRGEC
~ S/GBRGB
MM fa
Mba
M Pe;
201-890
247278
123-641
(LS
7p6si
IT-S
916972
- 68: 687
65578
Bog20
f4he dle
FPbe Gre
PP.21f
Zo
00+95
~68e087
F0:209
OI 490
nb IPS
184.544
77-272 +)
20S
S4terth
~68 687
26660
.3252%
91.7973
112-329"
Sb 201
“Maer
17-493
Chgon
PR ADB
HL DIT
37-5 |356-4G7 | 68-687
C)ANSWER:
INPUT DATA FOR PART (d); P=20kKN
2
nim
S(GAB
Mere
SI1GBE
M Po,
TAUA
MPa
w=6omm
175mm & ad & 27-6inm
t= 8mm
TAUG
MH
,
SIGBLA
M Pe;
Pa
$1GBRC
M Pe,
S/6BRB
MM fa
Ieee
FR SS2
ABS 2S 7
(28527
[S}. 420
334497
42g
2066 b4
20
{08-156
~ Os. &s7
Se. 269
b)> 440
Dto-gip
28 Rs77
(LF 7A
LS
{Ze:t&eo
-PS.§57
2i< tdi
BU RLT
168-262
yocey
103.032
22g
146-208
FHS
nord b§
28.187
(24.547
Jeg pss
F242}
37-5
18674
ESP ST
[4«2¥e
V7. 494
H21ES
(27376
6S ERS
LT
CA) ARS y Epp Bram sé ct £ Sie 20mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribtition to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM 1.C4
1.C4
A 16KN force P forming an angle a with the vertical is applied
as: shown to member ABC; whichis supported by -a-pin-and:bracket-at:C-andfor
by a cable BD forming an angle B with the horizontal. (4) Knowing that the
ultimate load of the cable is 100 KN, write a computer programm to construct a
table of the values of the factor of safety of the cable for values of a and 8
from 0 to 45°, using increments in a and 6 corresponding to 0.1 increments in
;
tan. and tan f. (b) Check that for any given value of & the maximum value
375 wim.
of the factor of safety is obtained for 8 = 38.66° and explain why. (c) Determine the smallest possible value of the factor of safety for B = 38.66°, as
well as the corresponding value of a, and explain the result obtained,
seo 450 mM
Deaw
(2)
<300 mm m+
SOLUTION
OF
FB, DiscRAM
FD
M.= OD (Ps in QQ e7s mm)+(Pc0509(260mm)
~ (F cos) (276 mm) ~(F'Sin B)(200 mm
C
|
ABC:
17
Fe p £5SinX+ 30 osx
+480 nem pC
OUTPUT
For
0
ALPHA
0.000 3.125
5.712 2.991
11.310 2.897
16.699 2.837
pe een 2.805
2365.2.795
30.964 2.803
34.992 2.826
36.660 2.859
41.987 2.899
45.000 2.946
fo
5.7L
=O
is Cosfa + fe sins
P=
/% AN
FS
11.31
VALUES OF
BETA
16.70 21.80
26.56
30.96
34.99 [38766] 41.99
45.00
3.712
3.552
3.442
3,370
3.331
3.320
3.330
3.356
3.395
3.444
3.499
3.913
3.745
3.628
3.553
3.512
3.500
3.510
3.538
3.579
3.631
3.689
3.966
3.796
3.677
3.600
3.560
3.547
3.7558
3.586
3.628
3.680
3.739
3.994
3.823
3.703
3.626
3.585
3.572
3.583
3.611
3.653
3.706
3.765
3.977
3,807
3.687
3.611
3,570,
3 558%
37568
3.596
3.638
3.690
3.750
3.358 3.555
3.214 3.402
3.1133 3.295
3.049 3.227
3.014
3,190
3.004 3.179
3.013 3.189
3.036 3.214
3.072 3.252
3.116 3.298
3.166 3.351
AND
3.830
3.666
3.551
3.477
3.438
3.426
3.436
3.463
3.503
3.554
3.611
Ful
= /o0 EA.
~~
4.002] 3.995
6.830} 3.824
B.710) 3.704
8.633] 3.627
B.59213.586
Bl5791 3.573
BVS5013. 587
8.619/3.612
B.66113.655
B.7213/3.707
B.773}3.767
|
1 (b)
(5) When (= 38,66" fanA = 0.8 and cable BD is perpendicular
lever
the
() F.$.= 3.579
le ver
arm
Bc.
arm
for
YW=266°
P
is perpendicular
fo
the
AC
NOTE:
The
value FS. =3,579
Corres ponding
Corres ponding
is the smallest of thevalves
To B= 38.66°
to x= 26.6",
/s @ saddle point’, or Minimax"
and the
The point
largest
O=26.65
of the
function
of
FS.
pf those
(3= 38. 66°
F,5 (&,f),
PROBLEM 1L.CS5
1.C5 A load P is supported as shown by two wooden members of union form-rectangular. cross-section that.are joined -by-a-simple glued-searf-splices---foor-(a) Denoting by oy and ty, respectively, the ultimate strength of the joint in
tension and in shear, write a computer program which, for given values of a,
b, P, oy and Ty, and for values of a@ from 5 to 85° at 5° intervals, can be used
to calculate (1) the normal stress in the joint, (2) the shearing stress in the joint,
(3) the factor of safety relative to failure in tension, (4) the factor of safety relative fo failure in shear, (5) the overall factor of safety for the glued joint. (>)
Apply this program, using the dimensions and loading of the meimbers of Probs.
1.29 and 1,31, knowing that «7, = 1.26 MPa and ty = 1.50 MPa for the glue
used in Prob. 1.29, and that o, = 1.03 MPa and 7, = 1.47 MPa for the glue
used in Prob. 1.31. (c) Verify in each of these two cases that the shearing stress
is maximum for a = 45°.
SOLUTION
Draw
the FB.d fag ram
& aes
Ov
F-
OG =
FSN=
)
FS.
()
OVERALL
V = Pees 0
Psink =O
F = PsinX
3
ed Sim 0
‘re. *
Shearing stress 3
FS. for
O
ab/sin Oo
=
Normal SIs
(G3)
lower mem ber:
-V+PcaosX=
47Z A = 0:
Area
of
rosea
©
xn
=
= (P/Ab) Sinx cosex
Ba
§
F/O
For shear:
FSS > CU/ EC
FS
=
FS.7
The
smaller
of
FSN
and
FSS.
(CONTINUED)
>
PROBLEM 1.C5 CONTINUED
PROGRAM
Problem
OUTPUTS
1.31
a=
150 mm
b=
75
mm
P=
11 KN
~SIGU =
1.26 MPa
TAUU =
1.50 MPa
ALPHA
SIG (MPa) ‘TAU (MPa)
5
10
15
20
25
30
35
40
45
50
55
60
65
70
15
80
85
Problem
.007
029
.065
114
.175
244
.322
404
.489
574
. 656
.733
.803
863
.912
948
.970
.085
167
1244
.314
375
423
459
481
489
481
459
423
375
314
1244
.167
.085
FSN
FSS
FS
169.644
42.736
19.237
11.016
7.215
5.155
3.917
3.119
2.577
2.196
1.920
1.718
1.569
1.459
1.381
1.329
1.298
17.669
8.971
6.136
4.773
4.005
3.543
3.265
3.116
3.068
3.116
3.265
3.543
4.005
4.773
6.136
8.971
17.669
17,669
8.971
6.136
4.773
4.005
3.543
3.265
3.116
21577
2.196
1.920
1.718
1.569
1.459
1.381
1.329
1.298
(b) (Cc)
1.29
a=
275
b=
F7Smm
P=
SiGu
=
TAUU
=
mm
Sboon
pos
MPa
{S mPG
ALPHA
St kPa)
&
4.69%
69
ag
Vb
AR
226-669
800
bc
49
PS
648-868
Tarr tk bo)
ESN
ECS
Es
©6728
i2-s7@
26408
2b Go¥
280243
e998
C98 b
L986
221-764
3. 14
4.£&b6
2. 1y
gea-GoY
2.4143
L998
£8.877
i, G19
abe 408
De
fy 2
i 619
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual niay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited disteibution to teachers
and educators permitted by MeGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM
1.C6
1.C6 Member ABC is supported by a pin and bracket at 4 and by two
~~-links; which are pin-connected to the member at'B and to a fixed support
at'D:
(ayer for
Write a computer program to calculate the allowable load ?,, for any given values
of (1) the diameter d| of the pin at A, (2) the common diameter a of the pins at B
and D, (3) the ultimate normal stress oy in each of the two links, (4) the ultimate
shearing stress 7 in each of the three pins, (5) the desired overall factor of safety
FS. Your program should also indicate which of the following three stresses is
Top view
f200 Tam «180
Simm
critical: the normal stress in the links, the shearing stress in the pin at A, or the
mein >|
[iz min shearing stress in the pins at Band D, ( and c) Check your program by using the
data of Probs. 1.55 and 1.56, respectively, and comparing the answers obtained for
'__.
Pay with those given in the text. (d) Use your program to determine the allowable
:
A
& mm
B
C
f
load Py, as well as which of the stresses is critical, when d; = d, = 15 mm, oy =
110 MP for aluminum links, ry = 100 MPa for steel pins, and FS. = 3.2.
SOLUTION
r’ (@® EB DIAGRAM
Le 8
Smm>
)
Front view
aon
B
A
mm
OF ABC:
+
ao.
Side view
> Ma=
fh
Cc
>
yp
A
200
en
rep
/B0
P= E00 lbp
my
P
p
.
2M,=0:
Zoo rc
~ {80
Fi
ive
f pin:
O)W\ For
given
d, of
pink
24,
F,=_ 2, /es)(fFd7/9),
O:
‘A
P, =. 42020 Fy
) forgives deaf pint BantD: Fin 2(F, POM: |), B= $28 fap
2) Eor given
8)
If
i
of
pi
id
for ultimate stress
|
Por
(4)
cl
:
< Fy,
Py <P,
Stress
and
c
P<
=
26
[FS
id?
- 20
tn finks BY: Fan ™ 2(G,/F5)(0,02)(0,008) , B = £00 Fan
oc
¢
:
Es shearing etrese. ip pins.
Foe desired pyerall
FE
DS
ESwt
ig
‘
Fe
critical
4
Fi,
is
the smaller
is
the
in
links
RB, stress
Py <Roand Bx 2, » Stress
PROGRAM OUTPUTS
smaller
of
on
of Fi and
E,
B,
and
Py
/s critical in pin A
ts
critical
in Pins
Band D
(4)
Frob. 1.53. DATA: c= 8mm, A,=l2mm,&,
=2 50 MPa, f= (00M Fa. F-§=3,0
May = 3.72 KN.
Stress in pr
Ais
crifPical
<a
(C)
Prob.t.54% LATA? d= 10mm, dy =]2 mm, C= Z250MPa, @ = 0OMPa.,FS.=3,0
Mayr BIZKN.,
Stress in pins Band Dis critreal
~@
a) DATA;
A, =d, >
[Sam,G,=HOMPa,
Fay 25,77 KN,
Stress
in
C,, =100 MPa
links
és
,
£S.2=
critical
3.2
<a
Chapter 2
Problem
2.1
2.1 Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter
aluminum rod. Knowing that, with an axial load of 6000 N acting on the rod, the
of the aluminum used in the rod. _
=
Ege
= 250,18 --260.00 = O18 mm
L-le
S = AL =
2 = O!8en
-
7.
A>
25Omm
ial
.
~
9. 90072
H(2Y = 13.097 mm? = [13.047 #10
To
B.
s000
© = A
m
=f
i
_
Teraapes = S3.052«10° Pa,
Ss
.052
«10°
23.052
410
= 73.682«10"
Feds
Problem
2.2
.
4
Pe
Fe72.76Pa
—<
2.2 4 polystyrene rod of length 300 mm and diameter 12 mm is subjected fo a 3-KN tensile load. Knowing that FE = 3.1 GPa, determine (a) the
elongation of the rod, (6) the normal stress in the rod.
A=
7 dd’
=
i (0-012)* = U3 1x00" mi *
PL
=
(3000 \ (03)
=
(o)
8
i){
a
6s & = 22
—
AE
\
Problem
=
(3-1x10-% )(3+1x%107)
=o2nsar
31 X1or&
2.3
8-062
0-002567m
s
Pa
=
ré
«<mt)
62265 MPa
2.3 A 60-m-long steel wire is subjected to 6 KN tensile force.
.
min
<=
Knowing that E =
200 GPa and that the length of the rod increases by 48 mm, determine (a) the
smallest diameter that may be selected for the wire, (6) the corresponding normal
stress.
P= GxlOON
.
A=
=
po
PL _ _
~
ES
Cex lo®)(60)
@)
tb)
-<¢
2 GC ainistm
2.4
d= 6.41 mn
6=160.0MPa <4
A 9-m length of 6 mm-diameter steel wire is to be used in a hanger.
P
=
@)(oo0by
S:fF
G=
37.5 «lo
~
L
g= fF
AE
It is noted that the wire stretches 1] mm when a terisile force P is applied.
Knowing that E = 200 GPa, determine (a) the magnitude of the force P,
(b) the corresponding normal stress in the wire
,
L= “7m
Ed?
~
Pa
= [GOxl0* Py
fre on
Problem.2.4
A-
E = 200xI[0%
¢
a
x1O"
SO)
de PPh =[ NG
f=
(SF
48xlo*m
x)
fo"
(200*107)(48
i
Ae Bp
cy
-
Sz
&é
=
P= ABS
=
E®
=
BY. 27x
10%
= C827 x10") (200 ¥10") (0).
(200x101) Covet
—
m 2
9
2
)
=<
244.4
mPa
6.4 EA a
—
‘Problem 2.5
2.5 A cast-iron tube is used to support a compressive load.
Knowing that F = 69
GPa and that the maximum allowable change in length is 0.025%, determine {a)
.-the.maximum.-n
the tube, (})-theormal.st
minimum: wall ress-in
thickness for
adoad-“-f<- — -
of 7.2 KN if the outside diameter of the tube is 56 mm.
E= 6COGP
&=
(a)
2
= 64107
Pa
= +H fGen >
& =
Ese
Oo
0.00025
= (€69x%107\(o. 00025)
ee
A
=
E = EI
=
17.25 «}0° Pa
& = 17.25 MPa «a
= HID BARI
mm = TL F mm”
3
~
A= Bde nds)
di8= dd SA = 508 ABER = 1963.56 me!
dz = 44.363 mm
t=4@.-de)=
4(s0-44.363)
t
Problem
2.6
=
2 .
R 2
wam
a}
2,6 A control rod made of yellow brass must not stretch more than 3 mm when the
tension in the wire is 4 KN. Knowing that F = 105 GPa and that the maximum
allowable normal stress is 180 MPa, determine (a) the smallest diameter that can be
selected for the rod, (b) the corresponding maximum length of the rod,
PRP
A
~o
6
A
a
ag.
Hf
>
- Pr
So
4xlo”
Ie
=
RAL KIO” e mw 2
1BOxKIOS
ahd
|
TO\l 92.929 elo’
- wm
= 5.82x}0
= [GAG22.222 x0
-
ES
de
YP
a
@)
A=
»
ge FE
ts
AE
x
-
OP
<i
wm
8
a
oO?
L =
Problem 2.7
mm
(22.222x10°° )(jo5 x10? (3x1 07%)
.
_- AES
5.32
2.7
Two
gage marks
are placed exactly 250 mm
L7SO
apart on a 12-mm-
diameter aluminum rod with E = 73 GPa and an ultimate strength of 140 MPa.
Knowing that the distance between the gage marks is 250.28 mm after a load
EB.
73x
to’
S =
280,28
Pa
is applied, determine (a) the stress in the rod, (b) the factor of safety.
~250.00
Lo = 250 mm
=
=
O.28
mm
=
O2BxfOo
tm
250 %10% m
5
4
(a)
O= be=
= ra eee
(by)
Ie
ops, = Se. _ Hexlo”
G ~ Bh7exlo®
4pr7y%
=
3
)
etext Pa
BLB MPa
f.T12
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, Afl rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribation to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
th
Problem
i
(a)
2.8
6, = 4ooxl0*
P,=
P,
“eu”
Pa
As Edt
= E(s)*=
GLA = (400x108 )(19, 63gK10*)
Po.
>
78S4
-
3a
7
Es,
QUS4
= 7854
|
Probiern 2.9
2.9 A block of 250-mm length and 50
compressive load P. The material to
Determine the largest load which can
must not exceed 80 MPa and that the
A = (Se\(4e)= 2000 mm — most 0.12% of its original length.
=
2x107*
m*™
adiownbte
P=
Con sidering
x 40 mm cross section is to support a centric
be used is a bronze for which E = 95 GPa,
be applied, knowing that the normal stress
decrease in length of the block should be at
Ao = (2x%10°7)(80OxJ0%)
detormatrou
= (60x10? N
z
P= AE(R) = (Qx10™)(95%109)(0.0012) = 228x107
Problem
2.10
P=160.0
N
[60%10°
P=
governs.
Jahue
smaller
The
mt
stress t
obfowebde
S = ce
kN
EF = asxio" Pa
BOxld" Pa
6 = -
245
.
Gy = BO MPa
Con sidevina
Piet
50.0 *10 mm
(4.E35%10*) (Qooxio")
AG
(9.635 mm = 19.635 #10 mi"
N
N
(24s4)\(eo)
-
§ = RE
fo)
2.8 An 80-m-long wire of S-mm diameter is made of a steel with E = 200 GPa and
. an ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desiréd,
~o. determine. {a)-the largest-allowable-tension-in-the -wire,-(b)-the: corresponding [oer
elongation of the wire.
-
KN
2.10 A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal
"stress must not exceed 40 MPa when the red is subjected to a 3-KN axial load.
Knowing that £ = 70 GPa, determine the required diameter of the rod.
{=
1.5m
Sr
[x10
6S =
mm
Deformation
>
4oxn{o
AzEkSo e
GekA
Shresst
$
=
Yo—SBU
x 10%
_
Ne
ayr
&
A
4
Pa,
Io x10
=
= quo tm =
te
ES ~ Gomd\njosy
vetve
E
Pa ,
Ae Be ~— e (Sx10*
5)
ho)
Langev
<<
~ OF
governs
-
BH,
AT¥IO
ea
Ke
UY
ae7s)
P=
75 mm
.= GE.
Um = &
75
2x/ON
27 mm
_
mm
d= = 9.77 mm
-
2.11. An aluminum control red must stretch 2 mm when a 2-KN tensile
load is applied to it. Knowing that o,, = 154 MPa and B = 70 GPa,
2.11
Problem
determine
the smallest diameter
and shortest length which may be selected for |
P=2kN
|
§ =Zmm
Cig
= (S44 MPa
C= F Ss Guy
A
=
G=
Zs
Ee
d=
=
Problem
2.12
|20x}0%
Pa
Jox]o?
Pa.
=
PLE
S* ye
@
<
~
Oza
E
Tr
ES
Lz ES
GO?
A> &
7
Zo00
=
-
iq
fe
~ 12 IBT mn
~
|
An,
= 4°07
nm
ams
Lmni
= O°9/n my
<tl
Shy
(20 x10
7
\(oe2) 9.959m
15%
x 104%
2.12 A square aluminum bar should not stretch more than 1.4 mm when it is
subjected to a tensile load. Knowing that = 70 GPa and that the allowable tensile
strength is 120 MPa, determine (a) the maximum allowable length of the pipe, (2)
the required dimensions of the cross-section if the tensile load is 28 KN.
S=
.
6L
te
4
Le
-
«j/O*
ES
m
.
Wexlo*)(.4x«io7)
reo wie
=_ ,0.817
L=
(boy
6 = fF
A =
=
4
0
A= a.
ene
|
az
Problem 2.13
A
233.389
x1o Mm =
= of 233.332
mm
~~
233.383 mm
az
lS. 2B mn
=<
2.13 Rod BD is made of steel (EF = 200 GPa) and is used to brace the
axially compressed member ABC. The maximum force that can be developed
in member BD is 0.02P. if the stress must not exceed 126 MPa and the max-
P = 520 kN
imum change in length of BD must not exceed 0.001 times the length of ABC,
determine the smallest-diameter rod that can be used for member BD.
Fon
=
0.02 P = (0,0a)( S20) = 04 kN
Considering
G
=
Feo
K
ae,
Consides mms
stress?
A
—
A=
AE
AveQ
Qoverns,
A=£qd*.-.
d = [7A
=
6°=
te
(26 Mfa
logan
“se
olePoemation?
- Foleo.
Larger
=
B17
wr
Fanluo
ES
>
9254
mmt
§ = (0.001)(3600 ) = 2-6 mm
(04-102) (350)
~ Czooxo8\(3°6)
= 19-5 inm*
= 19:5 farms
_
sue
j=
5 mm
at
Knowing
2.14 The 4:inm-diameter cable BC is made of a steel with E= 200 GPa.
2.14
Problem
and that the
190 MPa
stress in the cable must not exceed
that the maximum
,
».lonpation.ofthe.cable
must not.exceed.6.mm,
find.the maximum load P-that.can.be Joo
applied as shown.
Lge = | G*4 ee
Use
har A
OM,
Pe
Con sidering
A=
Consider
©
ae
“ "
ot Pans
3.5 P
stresst
abso
0.950%
Problem
6\=4-
Foe
Fae)?
19OxK1D®
Pa
12.566 x/07* in™
etongation
;
s
=
G
2,388 %10° N
xjo™°
m
Ra _
HI®)
ant
YOAV AS,
Fee
Qo
.
Ge
&e
Vafue
~
body.
ay
°s)
AESs a (UIRS6EHIO)(20
0 IOVS
KE
Smaller.
m
Fur 6A = G70 wo \Ua.seéxio®) =
Focbne . Fe
.
Pr
Oo
Bd" = i O.004)" =
FZ
as afvee
0.7507
abilownkde
G = Fac
c
=
=
Re
2.07)]™x
io*
Ni
= (O.ago? (2. oF not )= 1.9838 %10°N
2.15
2.0491. x10" N
P1433 KN -ae
2.15
A 3-mm-thick hollow polystyrene cylinder (2 = 3 GPa) and a
rigid circular plate (only part of which is shown) are used to support a
250-mm-long steel rod AB GE = 200 GPa) of 6 mm diameter. If an 3.2 kN
load P is applied at B, determine (a) the elongation of rod AB, (6) the deflec-
or
tion of point B, (c) the average normal stress in rod AB.
Rod AB:
Ana
P,-3200N
=
qa
=
1,,=0025m
Ce
= 28:27 X19
5)
0202
(220
Q) Sygz Ag FahEasneAce _ (2001
01)
(26:27
250 mm
cy dincber?
HoPPow
d.
d=bmm
20705in
a.
x10 *)
ggg
4.
5S
3 pq
xto
> (0r0§-(0003)(2)\=
O+04-4- ty
) me
0443
044
= 0100
A = Ed2- dz) = EB (01052-01
L = o.03m,
_
Sey:
P =3200N
PEL.
EA
— Sp= Sug t Sey =
(2)
SG 2B.
= Baa
in yoo! AB
= 010722
(4%101)(0-0004-43)
(b) DePhection of post B
Stress
an
(320e)(0-03)
370°
2827
Ke €
Lys
113
2M
ag
Ke In
or2l4-xio | a
Ps
|
a,
ee
The specimen shown is made from a 24-mm-diameter cylindrical
_2.16
Problem 2.16
steel rod with two 36-mm-outer-diameter sleeves bonded to the rod as shown.
ow. Knowing that.£.=.200.GPa, determine (a).the load P.so.that the total defor- do
mation is 0.04 mm, (b) the corresponding deformation of the central portion
BC.
|
a6 mm diameter
(a)
=
72mm
ee
E
a Ae
48 mel
AR
48
36
1017-9
0-04-16
|
Bc
2
2h | 952-4
oN SQS
CDi
48
36
004716
.
Ko-0#)(0-25347)
P = (Zeoxig?
Phe
RaeEe”
See=
Problem
=
P
lee
E
Re
7017-7
Rod
<—
sum
0-25347
3I-SbZ
«10%
it 6410
200%
\ 03
a
P= ang kN
= 3hS62- 110° N
_
.
a
© = 01025 nm
(ouster)
the total deformation of the composite rod ABC, (6) the deflection of point B.
_—
—
=I
|
2.17 Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel (E= 200 GPa) and rod BC of brass (E= 105 GPa). Determine (a)
2.17
P = 30KN
2500 ram
, mi!
L/A
Aya
.
(bE)
Ed.
AF
am
L. fon
|
=
Ze a
ES
™
“48 me
>
b3
.
P
B 24 min diameter
30 mm
Ang =
-30xI0N
Ens = 200%10" GPR
-P =
gz
O.250m
AB?
Lag =
-
~
(30) = 706.85 mm = 706.85% 10 “a
30xi0?) (0.250)
4
Sag
_
Fig
z
bao
Exe Ans
=
_f
a
7
Qoo #107\(106. 85% 107%
6
)
~ 53. 0S2«10° Pa)
on
amd
~BO mm
300 mm
Red
Fa,= 30+40
BC:
Lac 7
0,3900m
= 7Oxlo"N
= 70 kN
105x107
Fa.=
Pa
Roe = EGON = 1.4635 x10 mm = 1.IB 3S XIDwn
See
=
=
(a) Tota
(b)
of
point
(Tex to*) (0.300%
‘EgAe.
(loSxlo*)(1.463Sx10°%)
~- 01.359 x107-S mw
it = Sagt See 7 7 1SH-DIO
deformation?
Deffection
Fee Loe
B,
Sg*
See
-é
m
=~
=
Sp7
-O.15S44
0.1019
mw
mm V
<add
: Problem
2.18
2.18 For the composite rod of Prob. 2.17, determine (a) the load P for which the
total deformation of the rod is -0.2 mm, (4) the corresponding deflection of point B.
P = 30kN
2.17 Two solid cylindrical rods are joined at B and loaded as shown.
Rod AB is
made of steel (E = 200 GPa) and rod BC of brass (£ = 105 GPa). Determine (a)
the total deformation of the composite rod ABC, (5) the deflection of point B,
Rol AB:
mm
4OKN
ye
-P
—
-E,g= 200x107 (2,
F (So)= 706.
BS mm” = 706.85 *10° wm
%
eo
>
Law = 0. 250m,
4
—> {930
§
>
250 men
<—50 mm
Fag Las
>
@
}
n-
300 main”
lec ©
O04 300m
>
g
"
ECSo\*=
=
=
\ (706.85 % 10°75
P
Feehee
Fac = lOS*IO"
\.9682 S
From 1),
jo")
a)
Sag + See
P ~
1.45513 x10 PR ~ 58. 205«/0 “
N
P=44.0kN
<a
Sge2 -UlYSSIBxIO Y4a.da7™io")- 58. 205x107"
= = [QL
DePhe ction
1.9638S*10* yy?
= $8, 205xjo7*
Sy4=
- L7634)x1O"
43987107
=
¥ 107 )C1.4625%
ClOS
= 14S513x 1077 P
=
mm”
P,
(P+ do te )(0.300)
2
Tot 2 defoumeation:
-O,.2*[0
(b}
1.763841
|
EeeAge
*e
P=
(Joex(o*
Fac = — CP + 4oxl0?)
Rod BC:
ca)
P_Co.250)
Fag Ang
=<
S
se
of
point
RB,
IO" wy
Sg *
Spe
Seat
0.1222 mannY
<td
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem 2.19
2.19 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa.
ae
a
A
Force
Ane
z
(0.020)
dec = ¥(0.060)"
in member
4 on
Efongeti
AB
is
314,16 v1O
2. 3274
P
PLLae
=
§|
20-mun diameter
vam
=
H
Ang
i
(a)
iq lf
Knowing that the magnitude of P is 4 KN, determine (a) the value ofre so that the
_- deflection at.4 is. zero, (b).the corresponding deflection of B,..
m*
¥io%
tensions
=
(4x 10
3) , (0.4)
72.786
* {07>
(Foxt07)( 319.16 xio"®)
EAs
i
as
om”
mm
60-inm diameter
member
wm
Shord @ ina,
“ae
For zero
defdection at A }
2.5263x10' (Q-P)
Q
(od
=
48.3
"/o
Sig = See >
=
+ 4xto®
Sg tr
Sac =
72.756 ¢1O™
=
82.8
Q-
~ (Q-P)be_
E Ag,
Pp
Comp Mss ion «
(Q-P)(0.5)
(70x17 )(2.8274«(5")
2.$263 "10°? (Q-P}
44
|
is
Sae
Oo
%Jo°
72.756
¥ 0m
QN
P=
=
28.8 xIo®
32.8
0.0728
N
kN
mmo
~<a
A
Loan:
BC
it
in
4
Fovee
Proprietary Materiai. © 2009 The McGraw-HBl Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
2.20 The rod ABC is made of an aluminum for which E= 70 GPa. Knowing that P
6 KN and QO = 42 kN, determine the deflection of (@) point A, (6) point B:
040
hg
Zz.
Aw = Bedge*
; 20-mm diameter
Pa
Tt
= Blo. obo,"
7
= P=
bap
60-1
diameter
>
lec
i
w
g
(b)
=
Sp 7
SB
Sag
7
Sax
tC
=
°
7
* lo
90.4
m
le
Pecl ec -
(-36¥/o0° )C0.8)
Ae E
(7.8274 «15 “Jo ro")
= FO.947
4
«lo 7 wm
Q
wm
¥1O
&£
109%. 130«lO =~ 90,947
16 Sm
~
O.S
(314. 1¢+10°)(7ox10°)
a
(n2)
f P
(Gxto8 CON)
109.135
>
mi"
6xlo~ 42x iots - 36010" N
O.4 em
Aas &
©
2.8274 xio™*
6x1 N
Sig 2 Faohen
=
=
= S14. 16 ¥/0°% m
Pac = P-Q=
0.5m
x
= (0.0207
am
4
Arg =
IB.1GxloS
“4
Problem 2.20
O.018IF
~ 0.0404 mm
wm
mm
f
<l
or 0.0914
mm b Hi
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
:
Pr oblem
2.24 For the steel truss (£ = 200 GPa) and loading shown, defermine the
deformations of the members AB and AD, knowing that their cross-sectional areas
2.21
are 2400 mm’ and 1800 mm’, respectively.
oo
By
a
z
Ss
eS
of,
States?
Sas
Member
ane AQ) Py pee Z
jott
Use
40 m-—
Aas
BD
a free
25°
=
=
Os
.
at em
TF
Fro
menber,
HTT
m
=
Fae
Fy. o Cts fe)
KN
_
Member
AB
*
San
satxicC4e)
5 aD = FiE nkAres _ (2(i
00 x107)(1 800 x10
~
eh
4.717
oO
= 182.4 kN
(RISO ¥102) (4.717)
Eh.Lae ”- Gooust
atone)
Fan
Spa - 2-H mm
z-2.1/¥1077 m
)
=o
48
KN
Fae = 21510
ZF,
farce
4p
=O:
ZR
4
bodys
a
Fao
is a zero
Las = fot
Fhe
a
|
upwavel at A#C,
114 kV
are
Reactions
2,04
xf
~3
7 in
Pac
2.038 mma
Preprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers .
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
.
fo
Problem
2.22
.
-
2.22
For the steel truss (E = 200 GPa) and loading shown, determine
the deformations of the
“tonal
areds are’ t250
. I20KN
members BD
and DE, knowing
that their
im i* and 1875" nin?;’ respectively:
cross
°
_A
Free
2okN
Body:
Por tiovy
ABC
nA
of
truss
EM,
ZH
=O;
Fe. (5m)
~ (120. kK) (z+)
ie 2ehidy
~ (fa kn)(#4m) =O
NS
—wlE
Free
Body:
Fay= +192
Fortin
ABEC
kes
ot truss
{Zo kN
b2ZE
120 EK
Ido kn +120 kN T= O
eee
X2hoomm
xwin
MPa))
)(200000
(i2S0mm'
BD” = AEPL. (elt2
B
(#240 Kio) (4500 mm)
_ PL
Spe ~ NE ~ (7875mm*) (200000 MPa)
= 03
be = +240
kN
Se07mtB4%mm
G = 42.88 mm
Proprietary Material. © 2009 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
«i
‘Problem
2.23
2.23.
Members AB and CD are 30-mm-diameter steel rods, and mem-
~-bers
BC and AD are 22-mm-diameter steel rods. When the tumbuckle
is tighte"" [oo
ened, the diagonal member AC is put in tension. Knowing that F = 200 GPa,
determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed 1.0 mm.
h2m
>
'
8
|
=:
E
C
Fo
+PZF,s0:
Fre =
as
,
a Lvee
Feo
“s
TOT
%
Leap
=
(200x10')(706-Fxi0 )Co1001)
_ Edcefen
ie?
.
"
Leo
-6
4
Aeo
f
|
joint
ub
Fesler
"
Use
fam
= Edt = F003) =706-9x10 m
s
cD
he
a)
=
Che
—
(17-9 EN
bods
IeDopePae |?
~ qe
OUT-8)
O
.
wa
Ene
=
LoS
TE
Foo
HIF TZ EN,
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it sithout permission.
Problem 2.24
2.24
For the structure in Prob. 2,23, determine (a) the distance h so that
the deformations in members AB, BC, CD, and AD arere equal to 1 mm,mn yt the
-». corresponding: tension-in-member.AC.2.23 Members AB and CD are 30-mm-diameter steel rods, and members BC and AD are 22-mm-diameter steel rods. When the turnbuckle is tightened, the diagonal member AC is put in tension. Knowing that & = 200 GPa
and A = 1.2 m, determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed { mm.
(a) Statics?
Use joint Bo
.
Fro]
FanLg
Foc.
Fin.
‘h
b
as a tree body
L
hf
Ne
Fe
Fas
=
a
Fae
For. equal
Sin AB =
é
E Ans
“ee
fous
for
Fo
h = Bb
=
~~,
(b)
Set}
i
ing
Sap
*
h*
E
=
6 = OT
42 (9.4)
=
. Ase
Fee
=
E dhe.
~De
TS
dag
m
227m
“=
liam
Fe.
F Fee he.
See
b
Ane
7
See
Fas
= Re > EGR
JS
Ze
dee
b”
Geomaty
Fae
Ans
dae
L.
E Aa.
bees pA Fe
h
h_
&
Trang Le
d eformatious
Dee.
Equetin
Force
XP
.
(200x10") $ (0-022) (or00 f)
Se
E
2 Ae
Org
b
Ke
= P45 EA
1S
Fag = 2 Fa. * 77 (84-5 )
From
the tonce
Fee = Fae
yy
-
~=
USe2
kA
twiangde
Feze.
au
=z
we
rnd
”
lege F
EN
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by MoGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 2.25
2.25 Each of the four vertical links connecting the two horizontal members is made
oo
of aluminum (£ = 70 GPa) and has a uniform rectangular cross section of 10 x 40
--sMM.... For.the loading shown, determine the deflection of-(a)-point-£; (b)-point-Fy-<-Joe(c) point CG.
ae
Statics.
-
Free. body
EF Gs
250 pin
RRS
2 Fee
:
Ey
E
-
400 mem ———}—
Je
]
+ 4
300 mm
250 mm —} oy) yay
>
M
e
=
=
~ (400 \(Q Fee) ~ 504)
Aven
|
Binks
one
of
+ EM,
A = (10)(40) = YOO wan™
=
Lenath
.
Be =
L=
|
|
Defoumestre
x
300
“
mm
=
See
So
O.300
.
Deflection
of
point
EY
(b)
Deflect:
oF
point
=
OA
Geometwy change «
©
be
Smalk
im
\ (0.200)
{- 7.5
#10")
no
aoe
7(ao0
x10 x10”
Tonio
Feel_
(19.5xto” ) (0.300)
EA
(7a xto7 (400% tor* >
ee
(a)
°
—_
as
.
.
Ss
=
BO.4
Sp
-
Se
=
209
Se
ET
|
“6 m
203.93 %(0
> Spel
Ser
Lo
-
Se
bem
Wm
t
y
im
- el
<i
F
the
angde.
HOO
(c)
~m
~ RO. 357%
=
chanae
sfope
0 *
(4.5 ky = 1% S5>{07 N
wm
=
Feel
=
Ser*
let
= O :
(4o0\(2 Fiz) — (650)(24) =
e ws
YOO x1
= =72.5 x10" N
Fae z<7.5 kN
24kN
Set Se
Ler
Def Rect ian
Sa = Set
>
many ———__--—_»he-—- 250 won,
_ B0.357x107°+ 208.93x10"
a
0.400
oF
pernt
beg@
389.73
G.
=
“IO”
967
Sr
+
=
723.22x)0°° reckons
L-, 2
*)
208.92x10%+ (0.252\(723.22x10
om
a6
=
3490
Amy
<A
Problem 2.26
2.26
Each of the links AB and CD is made of steel (E = 200 GPa) and
has a uniform rectangular cross section of 6 X 24 mm. Determine the largest
--load..which.can.be.suspended.from.point.£.4f.
the deflection. of. £18. NOtLO.OX 2-04 enna.
ceed 0.25 mm.
Statias
.
Pyee
fv
c
fp
2rnaf—
¢
375 mm
=
2yommi
$OEM.
"
-
= O2
sen]
E
"
3TSiamy
4
O25,
=O
- 037EP
= hSEP
Aven of Sink:
Fag = |
OF
+DZMg=
A =(6)(24) = 144 mm*
= O
O25 FL, -ab25P
Lengths 200 mm
L
BCE,
e
8
250 min
ool
Feo = 2.5 P
=
Deformations,
Sper mae
EK
=~
&
=
EA
Deflect
peas,
Geomeiry
eS PICA
(z00x107)(44 5/5")
=
(200 x10!) (ly Kp’)
Point
B.
Se =
Sag
Powt
C.
S. = Se,
s pogi7 wio8 P
= 7-736!
Spt
(-O4tT
xia?P
x10
P
sy
S.7 1TRbI ¥lo® PL
change.
,
Se DA.
Cc
E
8
Se
6
|
Se
Se + L.@
Limiting
vatve of
275 mam
ae
BetS<_
Lae
value
|
j
-8
.
nlo P
Lem nlo Pans+ 1736s nlO OP _ gist
=
Limiting
ZlS0Omm
Se
F
of
FP?
17361 «(0° P + @37S)(o-dit lO PR) = $9027 16°P b
Set
Se = OZS x6 in
5: 9027
x1o® P
Pr 4235 N
_
=
025x167
P=4.24 kN
a
ee
Problem
22000
|
2.27 Each of the links AB and CD is made of aluminum (E= 75 GPa) and has a
: cross-sectional area of 125 mm’, prowing that
th at they s support the
th we rigid mmember BC,
oo... determine. the deflection of point Boon
opm
Fa
—
conn
0.20.
0.44 wy —o |
6
“
DEM, = of
Ree
%
Use
She
Fag Lean
and CD,
Seo
ge
=
Feoeghe Leo
EA
«
A=
125mm
= (3.9328, slo? )C 0.36 )
EA
(1s*107 KiI2s x Ilo
.
»F
(5625: xpo? )( , 0.36 )
Cis mo) Clias* (07% )
¢
n
"
Se
diaavan .
ase
TS n/c? N
Fas 7 B43
$+
15625 ¥/o" N-
Fug 2
= 12610" m*
122.06
=
.
60.00
x10ané
~jo
_
-¢
=
oe
mo
m
=
§
=~ 38
ee
“
ust
S828, Tete
O-64
Shpe O=
Qa
>
Deforma Hon
BC
5 *1lO”N
DSMg=0;.~ § (0.64) F, - C0.20Sxio*)= 0
-
her
body
Cc
- (0.64) Fa + (0.44 )(SxI0 = oO
For Pinks AB
mem
Free
NZS
«lO
pod
2,0
60.00 x10°* + (0.44 (112.5 10%)
107.5 «10° m
Se 2 0.1095 mm | aa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written perniission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
_
Problem 2.28
-
2.28 The length of the 2-mm-diameter steel wire CD has been adjusted so that
with no load applied, a gap of 1.5 mm exists between the end 2 of the rigid beam
Sees
ACR aid a contact point Eo Knowing that E = 200 GPa, determine where a 20-ke
block should be placed on the beam in order to cause contact between B and £.
Rigi id
0.25 m
vod
o
Pout
ACE
-
elaine
isto?
meves
casng Pe @
te cPase
qop-
3.75 «107° yod
0.40
Go
Hhvoug h
doumward.
3. = 0.08 © = (0.08(3.75 210%) = 800 410° m
|.
pen 0.32 nao /
:
:
0.08 m
Dep
S. = 300 X10 im
=
Ans GAs
E = 200»{0"°
geese.
Kee
Sep
Pa
ZOV B1916 mm = B.IVIC«KlO* wm
10% )C 300 w(0
E BeBe _ (00x10 4 YB. IG «KID"
—
ac
)_
753.09
N
ORS
Dse
Ec.
cof
Ay
ACR
beam
030%
Ww
as
a
Free
Ws
body.
+O ZMy =O?
O.40~%
= (20\9.81)=
mq
176.2
N
0,08 F, -(0O.40-x) Wo
=- (Q08N7S3.98)
~%2
|
0.30743 m
0.0926
m
=
92.6
ka
Few cout aed ,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Allrights reserved. No part of this Manual may be displayed, reproduced, of
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
2.29 A homogeneous cable of length Z and uniform cross section is suspended from
one-end. (a) Denoting by p the density (mass per unit volume) of the cable and by E
wots. modulus. of elasticity,.determine. the. elongation. of the cable. due. to-its-owne 0.
weight. (6) Show that the same elongation would be obtained if the cable were
horizontal and if a force equal to half of its weight were applied at each end.
Problem 2.29
(a)
For
ePement
ot
P = weight
pomt
identified
by coovélinate
y
o? portion below’ +he pd
= pg A(L-y)
as = 2% . GAC -y)yy.
EA
at}
PILY)
dy
_ (eakbee) sy . AB (Ly -4y9).
we)
(eo)
Tote!
weg ht.
.
r=
EAS
L
ss 3ft =
PZAL
GAT ee = reg at
Peaw =
Proprietary Maitcrial. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course € preparation. A student using this manual is using it without permission,
2.30 A vertical load P is applied at the center A of the upper section of a
frus
stant
the
Extend
lar ¢
eigh
Looe?
theve.
ny
“how
2°
At
~
coovd
Defonw
as
Total
P.
Ev
ot
Ay
¥
= ——t
oF
a
TE tan*d
deloumatrou,
UP
.s
S,° FE tant ol {
ay
=P
bn
TEtavh
@bo,
CF y tenet
Ty
A=
,
Hement
J
“hh
b.
y
pomt
j
1
|
P1* Fagai a
inete
ator
>
=
}
Oo
Oe
tan ol =
Fane
system
>
Me
geometry,
pn ate
.
a
sides
pF the cone qu meet
O ane
«a pout
at
face the evga im oO
hie
ee a
ye
herg ht
dy:
Pee
dy
y?*
LP
Teed yh,
© Pb)
TEeab
"~
As
a,
.
2.30
il
_ Problem
Pp
gy
* TE taala, ~b)
-—Ph
A TEab
) «atk
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
[Tt
©
if
+
.
2.31
Problem
i
2.31 Denoting by €the “engineering strain” in a tensile specimen, show that the true
strain is €,= In(/ + €).
Dn
=
(i+
®)
E,>4u
Thos
Problem
2.32
TH the vosume
Lh.
lo
= Wise)
fe * Qo
~<a
2.32 The volume of a tensile specimen is essentially constant while plastic
deformation occurs. If the initial diameter of the specimen is d,, show that when the
diameter is d, the true strain is 6 = 2 In(d, A).
Eda
is constant ,
da.
a*
(148)
=
7 J, L,
(2y*
a
=
Jn ($)
2
rr
E724
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Gam
2.33 A 250-mm
Problem 2.33
bar of 15 x 30-mm
aluminum layers, 5- mm
rectangular cross section consists of two
thick, brazed
to a center
brass
layer of the same
8
the aluminum layers, (5) in the brassdayer.
—™
250 mm
5 m0
A
Fer
5 mn
.
“
PL
[50
Poach
on
wr
=
byase
_€
ELA * EA
ISO
x107%
tay
ev-
-~ Wo
Per ee
ae R=
be
P
i
ifs
oe
8)
iin
t E
av
a
=
fonvee
PB
Pab
Oo
S
Tota?
a»
fy
alvion,
a
Ox WA
=
at PL = Peed an each alominan
Inyer
30 uur
&
Det
fuyev,
A= (30)}(5)
Smm
Aluminum —
Brass
Aluminurt —
each
3
3
Problem
GO, 2-BS.7 MPa
T-B85.7*10° Pa
Sie
<5
$5
> ~
(ol
<A
2.34 Determine the deformation of the composite bar of Prob. 2.33 if it is subjected
to centric forces of magnitude P = 45 kN.
2.34
2.33 A 250-mm bar of 15 x 30-mm rectangular cross section consists of two
aluminum layers, S-mm thick, brazed to a center brass layer of the same
thickness. If it is subjected to centric forces of magnitude P = 30 kN, and
knowing that £, = 70 GPa and £, = 105 GPa, determine the normal stress (a) in
the aluminum layers, (6) in the brass layer.
250 mm
5 mim
|
Deform
Total
ty
ation,
Pouce.
5 7-&
let
5mm
ms
each
Sayer
,
A> Gos) = SO me = [SOR
5 wn
4
Aluoiir UTA,
Brass
Alumiauwn —~
For
Br
Poad
we
ew each efumitnum Aoyere
ae
}
~~.
30 mm
Bl"
BAA
P=
PL
EWA
wt
Re
=o pe
Rh
1.2
on
=
brass
a
Pi =
2P
sexo?) (asoxton) __
7 Cloxtot)lisoxtor-* )
§=
«
Rsk
eke
3.5 P,
fay er
os
_
_ E,
Poe
2P.4+4
Road
oex
lon
”
3OG™
50.306
mn
“ta
;
2.35
=
250 man
25 min
~
Compressive centric forces of 160 KN are applied at both ends of
embly show: by means of rigid plate
Knowing that FE, = 200 GPa a
a, determine (a) the normal stresses in the stéel core and the
minum shell, (6) the deformation of the assembly.
Let
Ta
R
tk
2.35
portion
of aiak
fovee
cavvved by
it
Problem
“
povtrou
oF ania?
force
cavried
by
shelf.
Lore.
Ahiminem
shell
~ Steel core
62 won
Fab
SF EA,
Ey
SER
Talal Force
re &
— of
&
bo kA
2
A=
Ed, - A”)
As=
EAt = % (or025)*s 0100044) mw
2
2
Et (0: 062'- 0:025) =9,002522 m
G=
Ese
Gu:
EQé
S= Le
(200x109)
(- SBleSxio*)
=—
116 3MPa
= -— 4007 MPa .
'
C5 RI Sua?)
(Toxio
b
~ lboawo
~¢
= — SOLS X10
(70x 101) (0-002528)
+ (200x101) (0. 000441)
A
=
P=
t
Detar
= E,Aat EAs
tf
p*
(LY
Fa
S
P= Pat Py = (ExAat Ex Ag)=
2.
@)
= Eahkn
L.
F,
= (6025) (-SBRSwO°) = ~1¢5 K10 8m =~ OIG mm “A
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~
Problem
2.36
—
- 2,36 The length of the assembly decreases by 0.15 mm when an axial
force is applied by means of rigid end plates. Determine (a) the magnitude
of fo
the applied force, (b) the corresponding stress in the steel core.
6mm
25 mam
pane
SNS
6mm
ax
5.
25 nem
ZA
Gum
Let
NN
Steel core
}
portion
oh anial
Re
portion
oS ona d forre
g - BE
ASE.
As 7 (0:025)*=
b25x0
°
caviied by
Ay = (01039)= (ooz5)= 74¢ x00
p-
BAS
Ss
L.
m
Z
shed.
steed
core.
‘
.
(a) P= [loSxt0')(744-«10') + (200 x16) (62SX I i
EB
bvass
bn
~@
(i) 6.= Ber
by
= (BA, + BA)
R+P,
Ps
caveied
lL
AE,
300 min
foree
pe Eiko
g- ft
Eo = 206 GPa
Brass shel a
Eo-= 105 GPa
Poe
= lobG EN
= Doon’) 22eer _ 144 mp,
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
=
Problem 2.37
2.37 The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm
diameter.
Knowing that &, = 200 GPa and &, = 25 GPa, determine the normal
_Stresses
in the steel and in the concrete when a 1550 kN axial centric fforce P is
Swipe
450 mm
Let
Pe
=
porti on 6f ayia? force
=
portion
Ps
E5m
-
&
cave ec
.
EA.
the
SIX
conevete
steep Yods
p-E&AS
~
« SL
EA.
a
by
cawied
&
= Eafe’.=
Poe R = (EA + BAZ
2
PL
En A, + Es A,
Ag :
A.
¢ Fae
=
C23)"
= 3.69US
10% mm” = 3694S «10? we
4 mm”
tde ~ Ag = $0480) —2.624swjo*= 155.34* 107
=
a)
ce
a
3
x to*
- S
Ze
’
ten
—
~3
(25 «(07 JUSS.
399% 107) + (200 x107)(3.6945 «107 )
=
Ge = Ege = (200%10°YS35.31 w 10°)
= 67.1 IO Pa
S.= E.& =
= 838
Pd
=
(asx to? e35.a1x
fo7*)
«Jo
235,31!
¥IO
ff
a
‘SSO
~,
LIMP, <i
¥y
ie
5
= 153.344 */O> m*
8.33MPa <1
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
. 2.38 For the post of Prob. 2.37, determine the maximum centric force which can be
_ applied if the allowable normal stress is 160 MPa jn the steel and 18 MPa in the
2.38
.
Problem
:
oacposin
POL
in
2.37 The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm
diameter. Knowing that 2, = 200 GPa and E, = 25 GPa, determine the normal
stresses in the steel and in the concrete when a 1550 KN axial centric force P is
oe
applied to the post.
450 mm
: 5
15in
alfouahte
Stee#:ee
Ey
E>
Concrete?
SmatPer
let
PF
Re
por ti ou
=
portion
- %L
o}
Doud
canwniéel
:
=>
ve pue
B.A.
?
ee
by
six
2
+
=
”
in each
matentad.
Gow [oe
CO
x
Foon
ior = F0O%!0
Sa,
qovems
caves eof
by
Ss
—
Et
sthratn
eo
u
Determine
“&
&F
ad
TRoxlo*
2
=
720
“0°
Concrete.
steel
S* ER
Pe
S° BE
B+ BAP = BAe
pods-
ELA.E
P=Peot+ Ps = (EAL + ASE
A,
=
© 4 de
Ac = Fde*P=
.
= 3.6945 «(0° mm = 3694S R10" m
E (a8)
Ag =
F(H50)'- BEUSHISe
155.399 110 mm”
= \SEBFF MIO om
10)| (720 «10°?
107%) + (200 #107 03.6445
[ Caswto?)(1S5.349%
=
3,33 ulo*
N
3330
kN
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. Alll rights reserved. No part of this Manuat may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<0
Problem
2.39
2.39 Three steel rods (E = 200 GPa) support a 36-kN load P.
oe
and
CD
has
a 200-mm’
cross-sectional
area
and
Each of the rads AB
rod
EF
has
a 625-mm?*
_» :eross-sectional
area, ... Neglecting the deformation of rod. BED, determine. (a).the...}o 0.00000...
change in length of rod EF, (6) the stress in each rod.
500 mim
Use
member
BED
as
a
Peo py
Roo = Pre
fe
ZR +o:
Fag
+
Ro
Pr.
Spe
Fre bas
=
|
E Aye»
Since
Lag
Since
pornts
Since
menber
=
beg
P.
Sen
Ren
FE
and
Ane
and
C,
A,
Js
BED
L
Paw bas.
Las
~~
Poe vee
Faw
EAs
~~ E Aer
D
“=
>
Per
Ser -
Aco
5
.
-
Pre
=
=
~.
Pig =
Pen =
eri
lor
=
=
OQ. 256
23.810
“(O*
(200 *10*
24h
°
tt
a
“4
Tits
Aas
=
Spg = 6,
5
2er
ep
=
Sep
_
=
R00. Wo
G25
$00
) Se
= Ser
Pee
Per
1. S12 Pep
N
(0.256 023.810 108) = 6.095 «10° N
¢
if
48
«10* )
x19" )(S00
_ (6.0
QO.
Per
o
Are . Ler
Acer
Law Fer
8 * Ser * GoawioF)Ceren 10)
ov
§
-
_ (23.810 «(07 (400 ¥ O78)
-
(2)
lo)
=
+
=o
Seo
=
P= 2Pe + Per = (210.256) Pep 4 Pepe
Per
P
Ler
Dae
=
Se
=
Sg
regre,
”
Ree
E
Sag
Ss
+ Fer
2 Fre
|
are fivedt:,
&
body «
on by ZMg=©
By symmetvy,
i
)"
tree
C200
200
Be xlo
3
3
«176
=
-38.1
2
m
=
O.O0762
2
mm
~
o
-
XID)
GOUSHIO"
(6-2 x10
76.210
im
26 Sun’ P=
10° Res
30.5 MPa me
28.1 MPa
—
Problem
2.40
-
2,40
Three wires are used fo suspend the plate shown. Aluminum wires
~ are.used.at.A.and. B.with.a.diameter.of.3 mm.and.a.steel.
wire. is.used at
a diameter of 2 mm. Knowing that the allowable stress for aluminum (£= 70
GPa) is 98 MPa and that the allowable stress for steel (EF = 200 GPa) is 126
MPa, determine the maximum load P that can be applied.
By
symmetry
Adso,
Se
Strain
eI bow bbe
WiresAYBS
Wive
C3;
Eg
Pe
Da
=
>
and
Sz
=
Sa
=
Se
§
wire.
B
&. = 2
=
2E&,
stram.
On
E=
=
=
in each
Ex=
Determine
Pa
=
Ec
= oe
= hh xion®
7 x/04
> 4Bxio?
ca
2
fe
98
~
‘E,
= siptess
= 9:63
w/o
La
E,>
fin
= Ey&
wire
P=
Fe
Ce
For
RP
° ELF.
equid; beruw
$6.
of the
C
= ose
x17
governs
“*
Ge. = 126 MPa
Aaa&, * F 6.003)" (70 tO’) o-31r Ks”)
it
6,
styacn
=
[y.97
H
ADouebte
a
if. GIN
= A. &.
=
Zu
01002) (2610°)
= 395-84.N
pfate
P= Ppt Pa + BR = 707-6 N
~
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Cowith.o dcr
Problem
2.41
:
. 2.41 Two cylindrical rods, one of steel and the other of brass, are joined at C and
.
restrained by rigid supports at A and E.
For the loading shown and knowing that £,
~ 200 GPa and Es = 105 GPa, determine (a) the reactions at A and £, (6) the |
‘
en
Ate:
KE
E=
A
30-min cian.
-
2
¢o
ot
—
|
A
COkN
Ad+Bi
P= R,
Seo
Bro C:
Sec
Cte
Dt
.
er
Er
EBA=
P=
=
Ry-
_ PL
Ra (0.180)
gsi aapejos
-
=
Goxlo*
point
R, - 60x10" )(.1o0)
7 (Ra = bone
toe}
o Yo.too)
(Ra-loox1
TE Iiexlo®
Conunot
-n
= 447,47 x10”os Rom 26,848 x10%.
L=
R, - 100+]0%
E
move
100mm= O.100m
-
=
1,.34735xI0" R,
|,34735x
10
=
3.85837K/07 Ry —
pePative
3.85837%10" R,- 292, 424~lo° = O
Se
Sant Sane =
116367107
|
-
B0.841xI}0"°
L= [00 mm = O./00 m
a4
_
to
A,
Sag
2H
=
og
R, —
134,735 %]o
R4x IOS
©
=— RA AER. BBIxio°N = 62.9kN< ~e
Re = Ry-loOx]o” = 62.8x{0% fooxjo’
bY
wm
L = 120 mm= O.1120m
SeSzTTbe
_*
N
716.20 *10" Rp
Ps: R, - Goxlo*
P=
O.180
:
(Ra- 60103
\(0, 120)
ERT
los»lo?
Pa
74.220ox}]0%
I18Omm=
Ss. = Seat SectSco t+ Soe=
Since
(@)
_
RT
Ph
Doe —= EA
Ato Et
E&:
N
A = (oy = 206.86 mm = 706.86 *x/0%m™
L=
Ph.
PL
EF:
«&
251.327*10°
Ho kN
Sec = EA
Dt
Ct
—
Pa
A= Eto)= 1.25664x/03 mm = 1.256641? mt
EA=
40-11m diam,
200x]0"
= -—37,2*/0'N
t
Dp
B7.AkNe =<
Ry ~ 26.843x10"S
= (1.16369x10 "(62.831 1o*®) ~ 26.843x/0°
= 46.3x10°°in
|
S.7 YS pum > <<
2.42 Solve Prob. 2.41, assuming that rod AC is made of brass and rod CE is made of
Problem- 2.42
steel.
*
2:41: Two-cylindrical rods, one of steel-and the other of brass;-are joined -at-C and
restrained by rigid supports at 4 and £.
For the loading shown and knowing that £,
= 200 GPa and Ey = 105 GPa, determine (a) the reactions at A and £, (6) the
Dinensious in ment
deflection of point C.
| 100
r
4
Arto
t:
E
=
10S
«10°
Pa
A = Eto) = 1.25664%)0? mm = be 25E64x IO
40-im diam,
A
B
c.
<
—>
Ke
60KN
Ade
Bt
P=
D
eS
C
+0
Ho ke
Re
R,
PL
JRO mm
Re (0.180
Tan aurxi)oe
=
?(3/.947*/0%
2
E
_
=
R00
N
¥ 10
=
141,872
=
4
Fa
mm
= EGo) = 706.86
EA
b=
x/O°
.
.
= 706.86 *10m
N
OV2Own
SSB xIO - R,
Pr RR - 6ox10%
-
EB?
A
| as}
Be EA
Bi Ci
EA
30-mni diasn.
we
Le
20
mm
=
O12
OMm
Teka
Sec = cS = (Ra = 60 n10")(0.120) = FOF. 456 x10"=e RB — 54.567% - 10%
Dr
pe R,- 6oxI0°
5s
- BL
“"
Sag
+
cannot
See
+ Sen
move
+
=
Soe
red tive
3.63834x(O" R, - 167.743 x15 = Oo
Re
(oY
E
=
=
0.100
m
100
mm
=
0.
R= 42, 44) x10"
— 70,735x10°%
3.68834xj/o' BR, -— 167.748 x10
to A,
Sag
7 O
Ry = 4S 479xIST NV
HSS kN
RP, - [00x1O* = 4S. 479x/08 — Joo x 1o* = = S4.8a)* lo”
Se = Sag t+ Sg.
&
100m
707,35, 4 “10 ~t Ry
{4}. 372 * 1o*
Sse
peiat
L
.
(Ra-10
0x10*)a (O,l00)
=,
EA
Et
Since
(2)
PL.
=
bi
A te
-looxlo
mm
707.
B54 *( 7
141.372
*% lot
PPR,
Ove
100
(Ra~b0x/0%)@.100)
EA
DRE:
Le
Nf
Co
<a]
SY. SkN « <8
2.27364 %1077 Ry ~ S4. 567
x 10%
= (2.27364 «107 (45.4977 x10") ~ 54. 567» lo®
=
Y48.8x1lO~°
w
Se
YBLS Atm > a8
sae
Te
.
AX
Problem
2.43
2.43
A steel tube (£, = 200 GPa) with a 30-mm outer diameter and a
3-mm thickness is placed in a vise that is adjusted so that its jaws just touch
. -the..ends. of the tube without. exerting any. pressure. on. them..The..two.forces..don
shown are then applied to the tube. After these forces are applied, the vise is
adjusted to decrease the distance between its jaws by 0.2 mm. Determine (a) the
forces exerted by the vise on the tube at A and D, ©) the change in length of
the portion BC of the tube.
ee
Ra
mm
Rp
Z2kN
d, = 30mm
= 4,-2t
PL.
gy
ne
ootsm
R, (0.0715)
(206x104 j}( 254-5
P= Kat32kN,
u
ov)
&
ale
G
i}
OF
Ra t+
PL
_
Sy=
4-419
& K10
©
nse
Lz Qo15
(Ry + $000) (0-015)
Set
movement
lo”
Sap
=
KIO Ry + 4715 ¥I0
2
—OV
“6b
mM
*
hy
VaT3 *1O" Eat
= 4eF19KI0
See t Bo
“a
TZ
6)
110%)
45K
(250
r0
“(107)
EA
jew
Bkn,
'
L2o07Sm
C280 ¥ (ot) (254
PF
,
I-4-73 kto
X to7&)
7
EA
~4 p
bh
=
PL _ (Ryt32000) (0015)
A®D:
Rat $8+94
Good
TI xto”
Kio bh,
m
Ry + 58-94 io 7 = —Oldeg2r
Ra= ~S58S47N
R,= 58-6 KN —
Sec *( 413 x10") (-$8£47)
+ ATE YI0 fF
Ro, 50: bEN
i)
= -50597N
392410" m
u
Ry + 8000
(b)
D
.
[e.
LAI
=
EA
Bt Co
@)
7 me
= 003-2 (01003) = Qoz4dm
Fs
P=
BF
Given
re
Ba - as) = #(0c03% — 6: 024%) = 25465 X10 ° ww
=
Ch
7 7mm
FEN
For ts Whe
A to
|A
39.2x10°
<—
<i
mm
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Solve Prob. 2.43, assuming that after the forces have been applied,
2.44
2.44
Problem
the vise is adjusted to increase the distance between its jaws by 0.1 mm.
3-mm thickness is placed in a vise that is adjusted so that its jaws just touch
the ends of the tube without exerting any pressure on them. The twe forces
shown are then applied to the tube. After these forces are applied, the vise is
adjusted to decrease the distance between its jaws by 0.2 mm. Determine (a) the
forces exerted by the vise on the tube at A and D, (b) the change in length of
the portion BC of the tube.
A
R,
&
¢
Z2kEN
For
the
as
D
Re
26 LN
tube
d,
= d.-2t
=
30
fnta
= 0«03-2(0003)> F024n
2
R= Edt
- ds) = F003 -0:024*)
2 254. BRI”
At Bt
P= R,
PL
Sea © =
EA
BweC:
R
(200 x
Re
H
b47Zxto!
Rat
7165
(254-5 16%)
=
.
Ko
(73
movement
Sap
Ket 5B-44x1078
=
~0:10200
=
4
74
+ T4
1p atl
Sto> Sho t Sec + Sen © FetlAMC0 Ba + 58-94Id
jew
area"
} = o1o7gmm.
(Ra +8990) (0101S?
44(4xo!
See
=
.
RA
-
Ato D:
(hb)
Leovo7Sm
R, 4 gooon
P=
et9%)
So ° * p
(200x{
EA e
@)
ATR TA x107F
Ro
_(Ri+32000) (0.075)
b.
Given
me
(254-3 x(0")
ER © (200 ¥109 (254+5x10°)
D:
.
(0.075)
‘
104)
P= R+3rkN,
= EL =
Ss,
Cte
7
Lt Ovo
15 wm
mm
xio
~&
[
~orcoo|
R, =~ 25467 N
R,=-35°97 kN
—ant
Ro= Rat Beoo = —27967
Roo27 IT LN
ee
C1 473x60
-4\;
T)¢-
25967
-6
) + 4b ISXt0°
=7
=
-
S$ 2 KIO ; mn
Ss
63
Xe o.
Mis
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is osing it without permission,
004
Problem
2.45
2.45 Links BC and DE are both made of steel (E = 200 GPa) and
are 12 mm wide and 6 mm thick. Determine (a) the force in each link when a
'
“o2-4<KN force P is applied to the rigid member AF shown, @ the: corresponding“
deflection of point A.
Let
the wigiel
member
ACDF
rotate thyough
about pomt E-
smelf angle 6 chock wise
Pea
=
10m
~>
f
4
eh
Then
$5
100 num ~|«—-125 mm
,
-
Fer
Duker
~
=
=A
Lae
(200x10")(12x107*)l0-18)
.
o-]
bg = EA See. (ex I oN
boe
2
on
ov
= he
-¢
“0.
MONOD
Olas
«10°
cag
|
>
E A
F =
bee FOrdm
ae
©
- Spe ze oosGm
Fi
A = (010/2)(0.006) = 7210 “m™,
~_ FASe
Fee *
S
=
r
Log 0-125m
O
Ay—>
menber
ACDF
Ce Z2Me=O:
P=
-
2400
(a)
(lo)
as
a2P
a Fae7
a
free
body ,
«<<
Be
— Of Fae +005 Fog = O
Di—>
F
tee
QO
Fw
= &C m4
pad
= B64 x10° O
@ =
0-27778
Fac?
(4-4 « (08 )\C0-27778 x16)
Fec=
Fer
~(676
Foe = 1G
Deflection
D
«10° \ (0.27778 xo")
at
point
*r
7
x10°]6-d0676 x10°)O
K[o"*
P
10 6
5
Use
>
|
4 kM
<«
kN
x
A.
GS, 928 = (0:2)(027778 xlo™ )
S,¥ 01056 mm
> <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
pene
Problem 2.46
:
2.46 The rigid bar ABCD is suspended from four identical wires.
tension in each wire caused by the load P shown.
Deformations
of
be
baw
the
ABCD
O
ane
deformed
From
Since
rons
of
St
the
Potat for
$a, Se
anal
wires
Q:
A,
8,
Se
eC,
and
D.
Sake
lO
Set Spt 2LO
= 2$e- S
cn
Sy= Set BLO
=
@)
alf wives ave
ave.
Wives
the
be
Sa,
geometry,
Se =
in
Let
Determine the
3Sg-2Sm
tdenticad,
proportion af
the Povces
to
+he
deFovmations.
Use
bar
ABCD
4) SM,=0:
tPSR
+0:
Sobstitvting
-4T,
Substituting
as
a
Puree
2Te = Th.
a")
Typ >
37g
(2°)
LE
= Oo
-2T,
beady,
-2L%-LR+
Tat Te
T=
3)
+Te tH
- P=
.
4)
@!) inte (3) aud divedingby L,
+21,
=
O
Te= 27s
UV, '), anda’)
Thi Qh + 8% + 4h
(3')
inte G),
-P=0
—
«*10T%=P
Th = io P
~
Te? 2% = @G)P
B=tP
<
T=
Tt BP
~a
%+3P
=
CVEP)-GEP)
To= GIGPY- @UEPY
|
Problem
2.47
_.
2.47 The aluminum shell is fully bonded to the brass core and the assembly is
unstressed at a temperature of 15 °C.
Considering only axial deformations,
__... determine the stress
in the aluminum when the temperature
reaches 195°C.
EF=
X=
10S GPa
20.9n10°7°C
E=>70o
|.
Free
AT =
GPa
Brass
= 23,6x10°/°C
60 mim _
LX PANS ion.
(S-),
shedl:
Pp
P
Khe
compressive
Bress
core?
E,=
;
L &,. (AT)
S=L
(o-oo YAT)
auel
shef?,
in the
Fever
-
L of (AT)
in the cove
force,
tensife
the
be
let
~
.
($+) =
of shelf with vespect b the core.
Net eae pan Sion
IRo° Cc
=
IS
~
core?
A fomsnuwm
P
19S
oF the assem bdy,
Senyth
thermal
shedlf
Aluminum
be the
L
tet
cove
Brass
loSxlo*
Ps
Au= B25)" = 490.87 mm = 440,37 10 mw”
~
bE
Sek = BA
4
A\fommum
A=
shelf:
E.F
7o
{07
Fe
E(G0°-257) = 2.3366
%10 mm = 2.3366x10° m
S = (Sb + Spa
Lld,~
kK
Then
Strees
P
th
=
Pea
EA,
—— +i I
=
ExA,
EA,
=
H
where
eur):
of, MAT)
+
*
K PL
|
(losxto*\(490.37xI0°*)
25, 516K1o 7
: \
+ (70 xjo7)(2.
3366 x {o> )
No
Wly-da
iv. AT) _:
afuminum-
Pee
EA,
GO.
Qa.exlo™
= 20.2«1o
UB) = ag14104 1x10°a N
We Ele
wos
-
7
&
mo
4
seeaee
we
-
B.1S« (0°
Pe
oO
6+ 8.15 MPa =
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
|
. Problem 2.48
2.48 Sotve Prob. 2.47, assuming that the core is made of steel (E = 200 GPa, a =
11.7 x 10°/°C) instead of brass.
ostream
he alumina, shell is fully bonded to the banes
core and the assembly i |
unstressed at a temperature of 15 °C. Considering only axial deformations,
determine the stress in the aluminum when the temperature reaches 195 °C.
Aduminum
E'=
ad
=
shell
bat
70
GPa
23.
6xJo-*/°S
Lobe
Free
dength
theumaf
Ar
60 mm
the
Stee
=
of the assem bsy.
expansron,
195 -IS
cove?
(S.),
=
>
120°C
| ols (ATS
Aluminvs shell: ($,), = Lo(AT)
Net
Sep ension
P
of shebl
P
with
let
res bect
P
do the core.
|
be the
tensile
compressive
fonce
Stee? cove!
As *
Pp
e
Force
in the cove
= 490,97 mm
Pa
=
4490.87 xjo~
shebf
3
ELF
Jo
x
to"
Pa
(Sp),
PL
Gp), + Ope
L (ag- Og MAT) = Ee
4 oe
%
h
=
where
i
1
K = 5a t EA.*
= 16,2999«10"?
Then
Stress
m
EA,
=
25)*= 2.3366 «(07 mn” = 2.3366 x 1073 mt
S=
and the
in the sheff,
EL = 200x107
E¢as)*
L (a.- o WAT )
(Sp )- = Poh
EA
Pim invAa= shell
$(60*Alum inom
$=
= KPL
ta,
|
{
Caconio*\laaex1S*)* (Fo e1O NZ BREEx OD)
Qo!
~ Oa-o%sYAT)
_= (23.6«I0%
(1. x Brio
Use)
K
.
16.2999
{o-4 .
P
in
aduminum
.
S27
a
a
= Sees
,
3
=
_
”
8
\At.
4}
xo
+ §6.2x]0°
6,7 756.2
MPa
N
Pa
—sat
2.49 A 1.2-m concrete post is reinforced by four steel bars, each of
18 =mm 25 diameter.
GPa j Knowing that E, = 200 GPa, a, = 11.7 X 107°/°C and
= 25 GPa and a, = 9.9 X 10°6/°C. , determine the normal stresses
induced in the ‘stéel aud in the Concréte
by a temperature
tise of 27°C!
Problem 2.49
4 Fd? = @Ae-o18}*
2,
= O.00f0O(TF mw
,
it
12m
|
het
A-
Pz
equil;
200 mm
© 200mm
Shrains ¢
57
As
be
the
beiues
Povee
m
0*2*— 0.0010(74
=
tensi fe
with
the
aA * a(AT)
in the
Zevo
four
dota l
steef
1
oe
-
B.
Es
re
EA.
iS
the
-P.
compressive
.
= - vs t (AT)
.
+
o%.(AT
f
(Gere
foree ,
<
)
™ EAs
+
ol(Ai
~
)
(Os ~ of. YAT)
x, ) P.
+
concvede. Fos
rods
E.* ok + & (AT
Matching
= Oo ©34 m*
|
Boe
301584) * Goovicnoooar)
Fe
> (UWeT~ 94-4) ¢107%) (27)
me
Stress
in
steel?
Shess in concrete
SIRS
6.7 -
N,
&
A
~~
>
~ BIE
LRIO>
DeoraTTA
MET
“Log,
8:04
N
Mpa
6.> Fes TREC - 0.21 MPa
'
3
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission
—t,
~
Problem
2.50
2.50 The brass shell (a,= 20.9 x 10°/°C Jis fully bonded to the steel core (a,= 11.7
,
20 mm
920 my \,
YZ.
RES
egy
Sz
SSS
Fo
Steel core
Ee 200 GPa
equi libriom
compressive
250
E = 105 GPa
2
S
fm
b
Match
a
nA)
Cove.
steel
totud force
in the brass
the
shelf is
Ps.
+ ol $ (AT)
S
E. A
E, =A,
5 7
Es
zevo
the
in
)
+ OL(AT
vf
ey
+ O,(AT) = >
+ CAT)
(ol, - &, )(AT)
+ EXP
x
with
force
€.
Strain' s
:
950
2-7
Brass shell
cleveloped
force
axial
=
Fe
Let
Ban
Fi
Pan
x 10°/°F ). Determine the largest allowable increase in temperature if the stress in
..the.steel.core.is.not.to.exceed.53.MPa....
os
Ag= (0.020)(0.020)= 400 *10°° om
(0.030% 0.030) - (0.020Ya.020)
Oy - As
2x1
Po =
= (SS j0% (400%
t
E.A,*
GA,
4
EvA,
Ei
Ay.=
/°C
.
© (Zoo w107
SOO iO
|
Io)
=
*
|
22 610° N
i
{
C400 x 10°")
=
ay
2] Sour”
Clos x10" K500%I0*)
(21.65 «jo" 022% 10%) = (4.2%10°% (AT)
AT= 75.4
°C
na
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All fights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-H il for their individual course preparation. A student using this manual is using it without permission.
_
Problem
2.51
2.51 A rod consisting of two cylindrical portions AB and BC is restrained at both
. ends. Portion AB is made of steel (E, = 200 GPa, a5 = 11.7 x 10°/°C) and portion
... BC.is.
made of brass (Ey= 105. GPa, ag = 20.9. x.10°/°C). . Knowing
that the rodis..Jo00.0
.
A
initially unstressed, determine the compressive force induced in ABC when there is
_a temperature rise of 56°C.
:
;
—~ 30-mim diameter
250 mm
Fon
gal
steel
Fo4%_
Fy,
i dae =
HE (se)>
Ang? ©
-—- 50-mim diameter
Ae
=
oe
deg 7 ECBO)" = 106.86 mm = 706, 86%10" mi
=
£4635
x [O° mar = 1.935%
300 mnt
io?
m*
:
brass
Free
thermal
oy
=
©
Las
expan Stan»
(AT)
+
LacX(A
TT)
= (0.280 WU. 7 x10°%)(S0) + (0.300 20.9 x10" \(S0)
A
= 459,75 X10 m
Shortenana
doe
aa
6
8
to
induced
compressive
SOAR
Bik Ae
0.300 P
(200 ¥lO7T)\(106.86"1o*)
of
cf
= 3, 2235 xjO~"
Por
zere
net
j42.62xIlo0?
— (loS*107)VG.96 35 *]0%)
P
dPlection,
BQ2BS*IO
P=
“PL
Ent Ed
O28
~
force
:
L
Sp
Sy
P= ASG TEx 10%
N
P=i42.GkN
Proprietary Material. © 2009 The McGraw-Hill Companics, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means; without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and edacators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—a
0.
Problem 2.52
2.52
A rod consisting
of two cylindrical portions AB and BC is re-
strained
bn
—~—brass
Ae
. T
7.
Ma = Fdyg = (60)”
40.d0-mn dianeter
diamete
7777
ends.
Portion AB
is made
of brass
(E,
=
105
GPa,
a, = 23.9x 10°6/°C). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B.
77 G-n0n dinneter
Lam
at both
aluminum
EF
-
Ave
Free.
|
3
x
—
= 2.8274 H10 mm = 28274 15m
«3
io7m
C
A
mm = ov i. 1.2566
0)
y
CNet“a == E(Moyt
== 1, . 2566
x lot mm
#12 m*
thevma#
Spt Lap
|?
.\2
ex pa aston
(AT )+ Leo CATY
= (120.7 «15% X42) + 0.3 1(93.9210* X42)
= 2.4705 x/o77 m
Shortening
due
te
induced
Rom pvess ive Fovee
Plan
=
Fox
18.074% 1077
zero
net
det fection
1B.074¢%(07
Ps
See,
(6)
Se
ee
.
~
Te
e
OL
~
P
Ace e
~
"
Sie
- 3+
P = 2.2705 xio™*
[2sGa xio*® N
128.62 KIO
jo?
2.3274
125.62
Sp
LL
10°
1 2566 ¥ (O74
44 4
-
|
|
100.0
[00.0
~
«lo
¥,
Oo. Luu
R.:
ai
é
(0° Pa ==<
M
MPa.
ue
-100.0. MP,
oth,
=e
i}
Ko
BR
(72 #1071). 2566 Kt 073
P
By
ay
NPG
CfOSKLON)(ZB27KKIS*)
Ie
2L
7 VO
%)06
_ 125.
eye
0.4%16* (42)
= Goenchiaarimiesy ~ & (2
—-
«500%jO%m
=
-0.500
i.€.
mm
O.S00
mm
+
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution {o teachers
and educators perntitted by McGraw-Hill for their individual course preparation. A student using this mariual is usingit without permission.
~ae
Problem
2.53
.
2.53 Solve Prob. 2.52, assuming that portion AB of the composite rod
is made of steel and portion BC is made of brass.
BOA
“EOM Consisting
of two éylindtical portions AB
strained
-- 6O-inm diameter
teel
itm
at both
a, = 20.9X
@, = 23.9%
~ ee
ends.
Portion
AB
is made
of brass
and’BC is fe
(E,
105
GPa,
mine (a) the normal stresses induced in portions AB and BC by a tempera-
+
ture rise of 42°C, (b) the corresponding deflection of point B.
‘
~~ aluminum
Kec
T
*
cz
=
i
& dae
t
=
CAT) = (43)(23-9 «10%
dve
we
2
.
om
= 42°
x10m
x JO
= 305
C42)
.
to
iucete
_
Plee
.
compressive force
(Seec
Toted
For
zevo
nel
EAs.
Op
-4
hie
(3
P_
+
(Spec.
—_s
Os. =
(bY
Fao
ZP27akw*
Ate
*
fo"
x
jo"
Pp
P
> (1945.25
«107 ) (3-16)
Crs
a
2 ho MPa
4a
6,.°7
Mt ly
.
_..
“
Te
[-25bbX10*
WN
MP.
ao
oe
143-16
~
Pe
{6313-8
P= 3/6
U3
= - 4002 me
= -5fA
) Gg fePo.
Ate
=
1¢3b2-SE
P
Sp= Sy
(6313-8 x10 "Pr 4.946 *lo™”
Gia
~
(7axlO')zsbéxir®)
= (Se),
ded Leetion
m
.
P.
(Spl > Baye 7 Gooxio\abitaaery 7 (#53510
p
:
t Sela, = 1846 XIO* iy
Phas
.
Xto
AT
.
-b
Me .
= (NUET ¥ 10°) (42) = 0-541
Totes = SCS
,
F A2shb
Expansion,
(Sac = bec
Shortening
2
2°27%
=
(0-04)
(S2ag = ban WsCAT)
P
se
a (@ ob)" 2
~
q Daw
_.
thermal
Fvee
yy
=
Ang
—- 40-mm diameter
13m
(&)
=
t07°/°C) and portion BC is made of aluminum (£, = 72 GPa,
10°87°C), Knowing that the rod is initially unstressed, deter-
70:4 MPa
<i
= 022 «10m
Sar (Srgh t (Sp),,1 = eset uot b+ 02261074
Sgz 032 mm
ov
(Sp) ge = 4368. SBx107 (8-16) = 1626
x Io” m
Sat
(S. qf
Dg 7
+ (S,),. ¢ =
Sz
mm
4
fsorxie’ft 4
Ccoheeks
)
4626
KIS*h
+ <@
PO
SBIR DBA”
BAW
steel vailfoad
thack
GE =900 GPa ae S17
10°9C) was
laid out at a temperature of —1.0°C. Determine the normal stress in the rail
when the temperature reaches 52°C, assuming that the rails (a) are welded to
- form a continuous track, (b) are 12 m long with 6-mm gaps between them.
L=
-
(2m
S, = Lel(AT) = (2) (te Tx10) (52--1)
= 0100744.
™.
~ 2 =
T = bone boxio T
Ex = ES saan
Set _PLE_L
~f{
=
(A)
Spt Sy=0
boxio”& + 010074-4-
(L)
Sp+ Sz rook
beKie"G 4 00744
:
=O
= o006
;
|
.
6= +124 Mpa
G2 -24MPaq,
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~«
x
Problem
2.55
2.55 Two steel bars (£, = 200 GPa and a, = 11.7 x 10°/°C) are used to reinforce a
brass bar (Es = 105 GPa, ay = 20.9 x 10°/°C) which is subjected to a load P = 25
_KN. . When the steel bars were fabricated,
the distance between the centers ofthe [0 |.
holes which were to fit on the pins was made 0.5 mm smaller than the 2 m needed.
The steel bars were then placed in an oven to increase their length so that they
would just fit on the pins. Following fabrication, the temperature in the steel bars
dropped back to room temperature. Determine (a) the increase in temperature that
was required to fit the steel bars on the pins, (5) the stress in the brass bar after the
load is applied to it.
Stecl
.
af 5am
Brass
$;<
La, AT,
*1O7by*m
OWS bay
=stee#?
mml
OS ane
Sp ho F éxp
requivect
change
Tempevatove
change tor
tempevatuve
Requirecit
(@\
je
Steel
0.5 «10"* =(z00V IIo"
KAT),
AT
=
mee
O.Sxio* = C2)
7* lO WAT)
AT = 21.368 °c
(b)
Once
assen bled,
a tensife force
o tompvess've
force
P* develops
efongate
the
EPongation
stee#
Contraction
~
mae
of
brass: .
=
pL.
(S,),
"Pb
(Sp).
As =
+
Ay = (4o)UIS) +
to
9.
Stee
?:
Brass
6
~¢
600 mm
is.equal
do
the
initial
8.811xlo*®
606 x10
Sm”
BL746rlo7P
6.746 x15" P*=
=
=
P
*
amount
of
mishit
0.5 «10"*
N
+tabrreation,
~
Ss
YOO mm ® 400%/0"% wnt
P"(Q2.00}
P*
due
VCS 40%
(600 x10°F
)UlOS #107)
(Sp),
act
bvass.
PC 2.00)
(Spl
t (Sp), = O.Sxio®,
Stresses
the
“(joomo*\goouedy = 271"
ALE,
214°C.
PY devadops im the steed anol
in the brass im avden
+e
contract
of stee#:
(Sels* KE
Bot
and
amour
his
PP”
-
As
_g.8ixio*.
=
LOO
«107
*
-- =~ EBIEIO
3
‘Oo
-
AA,
O3
x10
.
Pa
.
27.03
MPa
= = 14.68% 10% Pa. = =1968 MPa
b
Ts
these
stresses
must
be
acdedaci
the
stresses due
to
tha
25 kN
Continued
foad
Problem
2.55
the
added
Fou
both
the
Aouad
stee?
and
devedupeci
3!
-
the
Miso,
dispfacement.
in the steed
RL
brass.
det
and
Py
Let
3!
Py
and
be the
additional
.
additiona+
Fovces
byvass, vespectivedy.
to xiot §'
it
L
3
Pe
Ps
sta Ceooxio"S\(los
lO”
2.00
+
Py
=
25 xjo>
S'< 349¢.65~10% nm
(31.510)
(2494. ereio® ) FF
t.4ovlo® N
xo?
=
3
P.
2
Ap
stress
"
13.986 Wl0* N
13.486
Add
3y.gy¥i0% §'
=
Ss = fo: VTi
Se
6! 2
= ASx/om
P. = (4orlo®)(349.65™10%)
=
3
N
Yo xjo® $' + B.S 10% §'
P=
Cr
AsEy
ApE,
Tota d
Same
be the
Po = AgBs 5° = (oous*)aooxe™) 5+
2.00
«
+he
~ Pl
Asks
P,
1's
cleformatron
adcditionat
the
,
|
:
-
continued
due te
_H-INOx~lo>
_
60OXx1O"*
Pa
34,97™«10"
“
18.34
Lo brecation,
xlo*®
Tetal
Pa
stresses.
6 = 34.97 x10% + 22.03 410% s S7,0x1l0° Pa
=
57.0 MPa
fH
=
368 MPa
GL =
18.36 ~/o°-
WWeaxto% =
Sean
ae
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 2.56.
2.56 Determine the maximum load P that may be applied to the brass bar of Prob.
2.55 ifthe allowable stress in the stee] bars is 30 MPa and the allowable stress in the
uadbrass bar.is 25. MPa. ooo ccece cece oppo ce get eet geeseee cape tay vet ge cettawesusunusaye fogs es
F
2.55 Two steel bars (E, = 200 GPa and a, = 11.7 x 10°/°C) are used to reinforce a
brass bar (By = 105 GPa, ap = 20.9 x 10°/°C) which is subjected to a load P = 25
KN. When the stee! bars were fabricated, the distance between the centers of the
holes which were to fit on the pins was made 0.5 mm smaller than the 2 m needed.
The steel bars were then placed in an oven to increase their length so that they
would just fit on the pins. Following fabrication, the temperature in the steel bars
dropped back to room temperature. Determine (a) the increase in temperature that
15 mm
Steel
Brass
bar after the
was required to fit the steel bars on the pins, (6) the stress in the brass
Stecl
load is applied to it.
Sotduton
te
Problem
2.65 to
Ge= 22.08 MPa
Pe
Gs
stresses?
Avaitahte
stress
&
30
6,
=
increase
397.68
MPa
ave! pabPe
_ 7.92*108
~ ES
~
=
7s
Ee
2OOxRIO?
Aveas!
As
Tota?
P=
\
P,
if
< nee
10S
* Io
yoverns
CIS}CYO)
fubweation
stwsses,
Gp
~
ate 7
RS
MPa
,-
strains.
“6
~
39.85
10
= 377.9 «jo®
“
€
=3%.85/0°°
VME&EYC40) = Hoo mm”
H
b
Ps
ThE
it
Vetue
Doad
4.638
_~ S
Shabder
from
MPa.
MP.
se, = Se ~ 39.68*I0%
6°
30
7.97
Aad+
the
- 1468 MPa
~- 22.03-
Corres p one). ing
Ee
Gs ave
H
ADPewah
obtain
=
600
mm”
ti
See
GOO «107°
i
40 nama
wa”
G00» 1O°° yn*
EF, A, &
= (200«1
C40007
wiO* )C39.85%10°) =
EyA, &
= Clos» 107 )}(G00 xJo'® )(394.85*JO°) = 2511 xlo N
ulpow« hte
P+
Por
additionad
3.18aKlo®
3.128 vic8 NV
Pores.
4 2.S11 «10%
=
5.70 ¥10* N
=
§.70
KN
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem
2.57
' 2.57 An brass link (E, = 105 GPa, a= 20.9 x 10°/°C) and a steel rod (£,= 200
o
GPa, a, = 11.7 x 10°/°C) have the dimensions shown at a temperature of 20°C.
othe. stecl.sod.is.cooled..antil.
it.fits freely. into. the.link....The.temperature.of the. .0.000.000.0 6-2.
50 an
>| me
~
37.5 mm
30-mm diameter
mm
[250
“Steel
Tritral
dimensions
Fined
dimen
T=
at
20 Poe
i
nememenen
Determine (a) the final stress in the steel
37.5 min
lpaaedianies
fm
0.12 min
whole assembly is then raised to 45°C.
rod, (4) the final length of the steel rod.
~<——
A
Bras
LL.
AT
=
45-20
=
.
“use?
Te
at
srons
as’e
Section A-A
Free
thermal
ex pau sion
each
pat
-
Brass Dink:
(SA, = bh CANL = (20. THO OAS (A250) = 130,625 x10° m
Stee?
(S4)s
vod
At the Fina?
of
oF
the
=O, CATIL = (11.7% 10% XYAS)(0.850) =
Peunp er atuve
steel
rod
and
S = Rolo
the
the
Atference
brass
between
Fak
73.125
% 16% m
the Free
beng th
is
4.73,125«10~ 180.625 x10° = 62.5% 107"im
Ade
equal
the
bracs
Brass Pink!
but opposite forces P to efonaate
Pink
and
contract
the
stee?
E = j05 x107 Pe,
Vo.
|
Az (2)\(S0 X87. 5) = 3750 mm = 8.750%107 mm”
-
PL._P@.150)
Steef
pod?
(Spe)s
Ex
200%107
Gea + (Sp), = §.
(a)
Stress
Az E@oy
-fL
~ _P
(0.250)
(200x104)( 706
EAs
26.006
Pr
in
steef
(105x107 \(3.7Sox10 *)
EA ~
(Sp)
26x}
Final
dength oF
x*jo
|. 76332
<2
“P
-1 POD
2.4033 x10" P = 62.510"
x]lo>
N
6.5
Yoo.
~
£;
~
2
- B6006x10")
= - 36.8x10"
Pa
stee#
odd |
Lp =
Ly +
|
6, = ~36.8
MPa <@
(5. \g
~ Bp).
Lp =. O.250 + 120%107° + 73. 1a5* 107° — (1.76838x 10°? (26.003 x10»
HW
(oe)
654.9210
= TOG .B6 mm = 706.36
¥ (Ow
==
0)
-
O- 250 t47 -m
Lp &
R50L147
mm
<td
Problem 2.58
2. 58
Knowing that a 0.5-mm gap exists when the temperature is 24°C,
the tempera
which the normal stress in the aluininu
ar will
de
75 MPa, (6) the corresponding exaet length of the alin
05 mm
>| | 350 min “T—* 450 mm —|
6.7
-~TS&
MPa
=~ 6A, <~(78x10*)
(p00 yi0) = 138 LN
Bronze
Aluminum
A= 1500 mm”
A = 1800 mm?
Ee=
E = 73 GPa
105 GPa
@=216X 10°C
.
Shoeten
aw = 93.2% 10°C
6.
duc
Pls
FLAn
P
~
;
P
“4
sf
fe
1
P
Pl.
. BAAR
(135000) (03S)
p (138000) (0+ 48)
(054107) (1500 x/0°) (T3x107)(1P00 x 10S)
= 7b263 X70 ae
Availabte
eLongetion
fox
thevmal
S, = (00005¢ 7Th203K10
But
—-¢
eapan
Sfan
-3
) = h2b23K/0
Mm
= bya (AT) + Loola(O7)
= (0035).
b x10) (AT) + (0 46)(23- .2K/0)CAT) = G8 X10 aT
Equetting
¢
(8 x07?
:
fo 2623
AT =
xfo">
Thet * Test + AT
(a)
* 24 470+/
Jo</
°¢
at
= 94-/ °C
Ar)
— a
S,- = Lid, AP)
= (0¢4S) (23-2 xi9-6 2€76+4 )- (138000 )Lov 4s)
(73 X109) (1866¥7/0")
herwct
2 OC FOF 0+ D7X)0°
= 0-GSoL]
d
b)
YAT
O27xI0
™~
m
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
2.59 Determine (a) the compressive force in the bars shown after a temeet “perature rise of 82°C; (b) the-corresponding. change:in length of the bronze-bare--f---6.5 mm
>| [- 350 mm
7
450 mm
|
Thernaal
e2opang ion
f
free
oF constra cnt
Se bya (AT) + Lao (AT)
Brovze
A=1500mn?
f= 105 GPa
a=216xK
‘union
A = = 1800 mm2
t = 73 GPa
18°C
= (0:35 )(2bbx10
(OES
DBI
KO *)
(82)
=PATb Yo mM
a = 93.2x loo
Cet
©) (Pz)
Constrainee
CALPA vison s
S TOS
Shortening
clue
compres 31Ve
ts
induced
am
force
P
Sp = 4 76x03 — 0: 0005 = 0°97bK1073by
Bot
» Els
Sp - Ee
EA = (eit BYP
x
(a). Equating
ore
(Gorm
ower)
+
064
f
(73%107) ( (Poe xsd BP
GS b4E7TK167 P = Or FIOXi
OF
= £ 647 x
P=/72835
f
~%.
°
Al
172-&
(b) Sor
hNl
Lyolsr) -a
s
©
3
.
_
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
~<e
Problem 2.60
—
2.60 At room temperature (20°C) a 0.5-mm gap exists between the ends of the rods
shown. Ata later time when the temperature has reached 140°C, determine (a) the
«Moral stress in the aluminum rod, (4) the changein length of the aluminum.
rod.
oj...
0.5 mm
tr
250 mm ~~
AT
= |Yo - 20
Free
therma?
Set
«= 23
A = 800 mm?
EF = 190 GPa
=
a= 17.3 *K 10°C
Sp
Se
Plea
Ple
—f
bs
ER) BA “Ca
=
B.G4Y¥7x(o | Pe
@
6 =-£. -_
(b)
Sor
A.
Io?
43%
=
O.B47
*/O* “4
Lis
BaP
ga
x lO7
R00 x 1O*)
O.847XIO*
232.3710
MAA noni anearnan sme
2000 «x {Q-¢
LyolAT) =
tonstyacnt,
meet
do
-— OG rvi@?4
0.250
(15 x 10° \(2000~«10"¢ )
Equal ing;
120) + (0,250Y}2 3x/0° Yi20)
to P
doe
{84 7*
0.308
= (
.
1.3472 KIO* wm
Shertening
_
cal pansion
= (0,300 \(2%xIO*
Stainless steel
107%
[20%
Leaul(AT)+ Lea (AT)
as
Aluminum
A= 2000 mu?
Bo
75 GPa
=
ne
—
\P
#&P
6.2 «(07 Pa
26447
N
~——- 300 mim
x10
-4
232.89x)0°
P
N
~H6.2MP.
«ab
Pha
eaten rape
En Aa
K 107 (0.300)
= (6.300)(23
x10 NC 20) ~ (232,34
0x
CIS x 107 \(200
(a°*}
363% 10%
0.363
m
Pa ny,
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
a
Problem 2.61
2.61 Jn a standard tensile test, an aluminum rod of 20-mmi diameter is subjected to
Se
a tension force of P = 30 KN. Knowing that v = 0.35 and £ = 70 GPa, determine
soar thaynatn sana wiinnrta mittens, sonn(@) the. elongation of the.rod.in.an.150-mm. gage length, (6).the change in diameter.
of the rod.
[
O-san
150 mn
P=
Zoxto*
N
A=
Bat = Eo = BI4.1S8 mm = SIYISE MIO
di ameter
~
Roxio
G; * E * Bip isacioe 7 1-498 410° Pa
-
ey
By, = ~ Vey 7
(DS,
= dey
PE AIBXLOS
1. SG4I8 x10 -$
x loe
Zo
=
205 #10. ws
BE41R*1
0° ) =
Sy = LE, = (15MI
010
7
>
p’
&
0.205 mn
(OBS). 364181?) = = 477.46 «lov
*m
= (20x1OPY-477. 46x10") > - 9.5510
Problem
2.62
at,
mm
—-~ O.009SS
,
2.62 A 20-mm-diameter rod made of an experimental plastic is subjected to a
tensile force of magnitude P = 6 KN. Knowing that an elongation of 14 mm and a
decrease in diameter of 0.85 mm are observed in a 150-mm length, determine the
modulus of elasticity, the modulus of rigidity and Poisson's ratio for the material.
Let
the yr anys
be
be
t+nansvevse.
abons
the
bength
y=
Modulus
the
roel
anol
the
= 19.0985 «10° Pa
re
= 0.093333
Si. ime
Es
of dasticity 2
=
(4079S 1O
=
204.63™10
=
- 2.82.
—_
y= O.4SS
~<#)
= = 0.0425
wer = 2 2
Modulus of Cigiclidy ‘Ge
Po
-
= 7o.31«10°
204. 63x10"
G
d
Poisson's watiot
Paw
MPa
f= 205
fy = =
x-awi's
P= Glu? N
16 IO mn”
BId.
A = Fao) = 3146 mm =
6, = k =
of
Fo.a
MPa.
Pa
<a)
|.
Problem
2.63
2.63 A 2.75 KN tensile load is applied to a test coupon made from 1,6-mm flat steel
plate (H=200 GPa, v= 0.30). Determine the resulting change (a) in the 50-mm gage
ottiaancisinaina
aia ataataenticnmamiaciianonc mona MRgth,. (B). inthe width.of portion AB ofthe test coupon, (c).in the thickness.of...J....
portion AB, (d) in the cross-sectional area of portion AB.
A= O.6\GQ) = 19.2 nm’
= 19.2 1075 w*
2.75
»10°
N
2s
62 Pe aise
/
C
/
f
J43,.229x«10%
=
6;
Ex
= =
@)
-
Pa
=
[
F
-- 50 mn -
em
2.75 kN
F2mire
;
143,229 x 10%
=
HB Aat xt
Ey =
By
2B KN
RZ
14.2 «tor*
~
A
*
.|
WGLIS «107%
PE. = ~(0.80K7I6.15¥10*)= ~ 214.8410
Is *10°) = 3. 8/%00S m
Sy= L& = (0.050.116.
Ls 0.050m
0.03858
()
We
mm
Sy>
84 « 10%) =~Z.578KIo*wy
y wé& y = (0.012\~I9Y4
0.01I2m
—-O,OO2S3
(CV
t=
0.00l6w
$+
mm
Axe
SA
©
wilt e)t.Ci+e&)
=
A~-A,
=
Wole
(Ey
a
Ee, = (0.0016\(-214.84 «10°) == 3437x10"7 wm
~ 0.0003437
dd}
at
= mt
4&4
(4 e+&
neg lig. tte
RY0.012K0.0616
VC 214.84 K1T)
fern
4 ee, )
)
x
Act
mm
<0
“paleo
Rid, Loky
=z B.ASMIOM
=~
0, 00825 tone
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ae
_..Probem 2.64 >
2.64 A 2-m length
of an aluminum pipe of 240-mm outer diameter and 10-mm wall
thickness is used
as a short column and carries a centric axial load of 640 KN. f
Knowing that E = 73 GPa and v= 0.33, determine (a) the change in length of the
#
u
2Owmwn
B(d.-di*)= F(240*- 220") = 7.2287 10% mm*™
iy
>
di = d-ab=
belOmm
mm
ZHO
o
a.
pipe, (6) the change in its outer diameter, (c) the change in its wall thickness.
1.2257 *1073
P=
4
me
(as
m*
GHoOxlo*®
N
Pit
eee
( 2.00
(640 uid’)
-
AE
3
\
2.2257 1373
w 107)
*xIO
= - 2.427%
=
-~2.43
!
“3
mm
mm
~«e
Eg
2 = ~ 2x”
= ~ VE
=
= {2133
«jo
= —(0.33)C1.2133' r/o)
400. 4x10
=( 240K
de Eat
(o)
Ad,
=
(oc)
At
= Cee
=
i.
'
~3
) =
(10) 400.4x1o% ) =
400.4 «jo *
im
—
0.00400 mm
—
0.096!
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distributed in any form or by any means, without the prior written perrission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 2.65
sce
2.65 The change in diameter of a large steel bolt is carefully measured as the nut
is tightened. Knowing that E = 200 GPa and v = 0.29, determine the internal
tet
. force in the bolt, if the diameter is observed to decrease by 13 ym.
60 mm
Oy
=
= 13 xJo°%
mm
d=
Oy.-. aio
13xlo*
Ey => 5*=
=
io =
7
~
€,
- =
76
IO"
= AGEL
Lo
=
F479
G,
=
Es
= (200™tO?
A
=
Tb
=
F=
GA
= (149.43 vio
# (60
747.13 «KIO *) =
213 «Jo7*
2.82710"
=
mm
149,42 «10°
Problem 2.66
Pe
mr
«x lo-t
2.227
=
)(2.827"IO” ) =
Yaaywio?
=
422
N
KN
et
2.66 An aluminum plate (E = 74 GPa and v = 0.33) is subjected to a centric axial
load that causes a normal stress 9. Knowing that, before loading, a line of slope
2:1 is scribed on the plate, determine the slope of the line when o= 125.MPa.
The
5
216.67 */0 -
.
we
aii+ey)
60xlorw
Ey = - vey
sdope
after
deformation
is
tan
O-
Z2Q0+8y)
I+ Ex
¥ 107
«IO ) = 0.5574
= ~ (0.33 11.6892
tan© = 261 2.005
= 1.99551
~<a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual ’s using it without permission.
Probl em
2.67
.
| 2s 67
The
aluminum rod AD is fitted with a jacket
n BC oftheused to apply
AD, (b) the change i in diameter at the middle of the rod.
360 nim
=
6
u
{
6,
=
~p
=
-42
MP,
B81;
Gy
=
o
- 26;)
500 mins
5
| -42x108 — (0°36) (0) ~ (0°36) (-42 Xo “yj
Fou
= —384x)6"
“38 mn
=
ad
- (003 6) l~ 42/0") O-loseye- 42/0) |
= 432x 1/9 °
su
Length
@)
Sys
(b)
5S,
b ject
Ley
Lo
to
steacn
= 61/296 x10
=€03) 64320159)
.
Ag,
6-3m
Lor
Eye?
_
= (0-038) C-384x fo °)
:
:
me
mM FOB Mm
-%
/
:
> O.01459 X10 M = —CI1FG
/
Wm am
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course. preparation. A student using this manual is using it without permission. - .
Problem
2.68
oO
2.68 - For the rod of Prob. 2.67, determine the forces that should be ap-
plied to the ends A and D of the rod (a) if the axial strain in portion BC of the
~™<"rod
4s 16 remain Zero as the hydrostatic préssure 18 applied, (b) if the total length
AD of the rod is to remain unchanged.
2.67 The aluminum rod AD is fitted with a jacket that is used to apply
a hydrostatic pressure of 42 MPa to the 300-mm portion BC of the rod. Knowing that E = 70 GPa and v = 0.36, determine (a) the change in the total length
AD, (b) the change in diameter at the middle of the rod.
300 min
500 nun
Over
the
press uv izect
6,= &
nen .
a
n=
38 wpe
£
=
@)
= —p
E
(2p
(fy ec =
=
+
Ru pt Sy = O
ly
pz m QV 0.36
Over
un pressurtred
(ye?
Eo
ALKIO”)
-50¢e¢24 MPa
= P(oose)*
=
is4¢ x10? in
= (1134x716?) (30-24 440") = - 34292
ie.
(b)
—- v6;)
Sy)
oO
A = ¥a*
A 6y
BC
6= 6
(-~26y + Gy
‘By =
Fe
portion
portions
&
AB
«?
3463
CO
kN
6,2
6,
Kf
Compyession
<i
=O
=
For no change in Densth
S =
Law Eylig + bucEylec + ben (Eyleo
= O
(Laet Leo) Gyo + baclGJe = O
(05-03)
+ 2? (2up ‘ 6) =0
6, - - %bYP
af
Ps
AG; tle
_
_ (0.6) (0+36)¢ 42x10")
=~ 90.72
MPa,
Oe}
34¢ 00) (-ForT2X 10% = ~ [O77
i.e
fore]
y
kal
pre $3 (0%
Lom
<<
“PO
Problem 2.69
2.69 A fabric used in air-inflated structures is subjected to a biaxial loading that
:
results in normal stresses o, = {20 MPa and o, = 160 MPa. Knowing that the
o--properties -of: the -fabric-can -be-approximated.-as..E-=87> GPa and v= 0.34
¥
determine the change in length of (a) side AB, (5) side BC, (c) diagonal AC.
mm
P00
7
6,- = oxfo* -Pa,
—
—
aa
:
<
-s
&, 7
.
a
TUE,
- YG
-—!
- vs)
eo _
.
(0.34 (igoxI10 y]
if
~ Brrig7 | 12ox
754, 02 x 10~*
=O, 6 = Koto Fa
G
€, > x (-26
1.3701
Sa.
= (8) EL
ga
0.0754
mm
(75mm) C13701x 10° Y=
0.1028
mm
20
dc = & da
100
AiFPeventials
de = 2ada+
Se
= ole
as
a,b, anda,
by ca deutus.
2b db
+ = ab
mm,
bt
da = Sag = 0.0754 man
(CY
ABC
o* = ata bt
Obtain
c
=
* 160 «IO ‘7
(100 ma )(754.02*10°) =
babel sicles of wight triangle
b
But,
- (srXize x10")
A
(b)
a
gsr]
Faer
A
Sy > (AB )e,
8
(a)
A
+ 6 )=
Jo”
q
=
~VSy
100mm,
C7 7100476*) = 125 mm
db = Sg = 0.1370
leo.
= oF
Co. o7s#) +
wm
(0. 1028)= O1220
mm —e
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
foo os
Problem 2.70
ve
2.70 The block shown is made of a magnesium alloy for which Z = 45 GPa and v
= 0.35. Knowing that ¢, = -180 MPa, determine (a) the magnitude of o, for
so» Which.
the change. in. the height. of the. block .will. be. zero, (2).the. coresponding....j...change in the area of the face ABCD, (c) the corresponding change in the volume
a
4
of the block.
ay =O
fy = 0
z(G
Ey =
- v Oy
Gy = v6,
=+63
(by)
&, >
=
(G,
- 26,
coe=
fy
ob
E
(6, ~
by
-
» 6) =
~ uv,
)
.
-2 (G4 6)
Gy - E Vb,
~
_
« 10%
_
6,
180 ¥ 1o°)
Pa
6y = ~63
/
(0.35 . usb243%10
x fOF
IS7.95
=°
)
- ve,
+ (0. 25)
“x1Oo
ee} xO"
MPa
«)
«<&
4
F-}BIxfo
re B.S]
Jo
3
A, = Lala
A=
(eel,
AAz
A-A,>
+e)
=
Kl, +e
byl, Ce,¢ 8, t&8,)
+&
%
+ eye)
Lyl, (48)
AA = (100 mmm 2S mmr) (3.512157= 1, 89x15)
c)
|
Yer
Vez
AA=~13.50mm
<f
ly lyk
Le )ly Ut gy) l,C+e,)
> Lylyl, (1+ Et8 +E, + Eby + GE, + GE
+ Eve,
ee )
AV = V-VU, = LLL, C4 & + & + small terms)
AV = (f00)(4o Was )(- 3.51 x10 4 O-1.89x10°)
AV = —-SHOomm
Proprietary Material. © 2009 The McGraw-Hil} Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A stadent using this manual is using it without permission.
<4
- Problem 2.71
oe
a
2.71 The homogeneous plate ABCD is subjected to a biaxial loading as shown. It
is known that o, = o and ‘that the change in length of the plate in the x direction
..Mmust. be.zero,.
that. is, .2.=.0...Denoting .by.£.the modulus. of elasticity.and by.
Poisson's ratio, determine (a) the required magnitude of o,, (6) the ratio ao / €.
6,=
6,
Ey?
(a)
bY
GyF
we (Ss
6,
=
6,
oO
~ vt
,
e&,
=
oO
- v6, ) = #(6, -26,)
a
a
£25 B (26 - 7h +6) E(-E-0+6)= EG
Proprietary Material, © 2069 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
v.00
:
.
4 72 For a member under axial loading, express the normal strain e ‘in a direction
Problem 2.72
forming an angle of 45° with the axis of the load in terms of the axial strain e, by
(a) comparing the hypothenuses of the triangles shown in Fig. 2.54, which
“represent respectively an element before and after deformation, (6) using the ~
values of the corresponding stresses o’ and o, shown in Fig. 1.40, and the
generalized Hooke’s law.
(a)
(a}
.
(hb)
.
(c}
Fig. 2.54
‘
Fig. 1.40 (5)
2
Ji C148')
|-2é,
}
}
(a)
Before
a
2
a(i+
tt
(47 C1 +e')]
(i+8,)°
2e'se' )t
uch
4
ag’?
4
\+
deformation
After
2 Ey
deformation
,
(\- wey
14 2a, se 41
- 20g,
=
Ex
+E
Negbect sqoaver os smell
-
2px
+
ev%e,
é,™
48 = 28-28
~aZ
.
1s? e,
Gg!
ro
e'=
=
_
~
t-¥.
E
- deere
QE
- wv
- ee
_
yy G!
=
P
ah
om
—<
Proprietary Material. © 2009 The McGraw-Hilt Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators pemuitted by McGraw-Hill for their individual course preparation. A stadent using this manual is using it without permission,
Problem
2.73
i
2.73 In many situations it is known that the normal stress in a given direction is
zero, for example ¢, = 0 in the case of the thin plate shown. For this case, which
.
wow a dS. known. as. plane stress, show. that if the. strains ¢,.and ey have been determined
experimentally, we can express
=
vik
E+
VE
Fe,
,
&
+ Vey
Ex,
=
~~ ye
Sx
a
Mobtiplying C1 by
ext 4)
@
.
pw &,)
=
y+ vee)
Tag
™~
and adleting to (2)
jw
-
=e
joy
gt FC- VE + GH)
= paar (eu+
S,
or
oy
fy tv Ex = “EW Oy
e, = ple,
&,
.
add ina to ()
aud
vy
0)
+ VE
l-
-2-*,(
Gy) =
E+ Vey
+ Ey + VE)
<a
=~ 204¥)
inv (8.48)
- = -Ey
J= we (g,a8,)
2.74 In many situations physical constraints prevent strain from occurring in a
Problem 2.74
given direction, for example ¢, = 0 in the case shown, where longitudinal
movement of the long prism is prevented at every point.
Plane sections
perpendicular to the longitudinal axis remain plane and the same distance apart.
Show that for this situation, which is known as plane sirain, we can express 02, &x
and e, as follows:
o,=
v(o,+0,)
a
5 =u
Or
i:
v*)o,- v4 va, ]
= [-vG,- V6 +
6; )
é = Zld- vo, ~ v(L+ v)o,)
ov
6,=
v (6, + 6)
~~
= ECG, - 26 - 96) = EL
6 - YG - GH G)|
-“
vy
- »(1+v) 6, |
i (- v6, +6y- 76,)= F] YG, Gy - vet6 )
H
=¢[0-»*)6,
dL
VG
-vitv)
6y |
Lenareranss
by
(2)
E
oy
>
B= EG -v6y)
MuPtip Py ing
x, a, and £, as follows:
<a
00000000000.
Problem
2.75
_..
-@.75. The plastic block shown is bonded to.a fixed base and to a horizontal rigid plate to which a force P is applied. Knowing that for the plastic
used G = 385 MPa, determine the deflection of the plate when P = 36 KN.
|
a"
*|5F
ae
F
Consider
the
plastic
-
The
shearing
Fore
is
The
Shearing
,
Sheax na
But
Y=
Problem
Stress.
.
strain,
2
~
avea
wn. 2.
-
is
Ax
biock.
e
cartied
Ps
364N
(008% )(orvaz) 1088 x10" m*
Foxe
a
A” (hopexios — S2467S MPa.
ve
B24b7S
+ 00 8433
ae
Yor =
$=
hY=
2.76
(63 \(e:008433) +
o4s mm
ae,
2.76 A vibration isolation unit consists of two blocks of hard rubber
bonded to plate AB and to rigid supports as shown. For the type and grade of
rubber used 7,4) = 1.54 MPa and G = 12.6 MPa. Knowing that a centric
vertical force of magnitude P = 13 kN must cause a 2.5-mm vertical deflec- tion of the plate AB, determine the smallest allowable dimensions a and b of
the block.
Consider the vob ber
Canvies a shearing
The
sheacin
stress
ov
required
A
But
Shearing
Bt
Hence,
force
equal
1%
77
te
$P
Te a©
-
b
bt
=
te
4bS
t
oa 05628 in
andl
t
Y = GC
= 2
a
*
=
=
25
8
nd
bain
*
$63
nin
t hS4MPa
Meee
* je
mt
|
:
mm
atram.
nal
= ee
A = Godb
Hence
Use
=
on
Gt
vig ht.
the
biock
7
OL 12QA2Z
ah
_ Problem
2.77
2.77 The plastic block shown is bonded to a rigid support and to a vertical plate to
which 2 240-kN load P is applied. Knowing @that for the© plastic used
w
G = 1050
.. MPa, determine.the deflection of the plate....
ci
tte
.
- At BOMIRO) © F.Ex/07 mm
P
2HOx%lo'
M
240x!10"
v fae
G =
1080
/d°
=
Py
231810 «10 3
,7
m
0.08
=
ho = SOmm
25x08
Pe
asxjoo
josp
{pe
- 2,
Yes
= OG 1 *
S = hY= (0.060.403.3«10?
810)
=
Problem
2.78
1.190
2.78 What load P should be applied
mm deflection?
410°
m
1 OF} vam
v
i
to the plate of Prob. 2.77 to produce a 1.5-
2.77 The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 240-KN load P is applicd. Knowing that for the plastic used G = 1050
MPa, determine the deflection of the plate.
—+
Se
i
hS
mm
=
15x10m
h = 5Omm > SOxIO™wm
Ss.
h
Pe
a
sae
G = |0S0~10°
Dimensions
in mn
7
>
CY
=
NOTE:
it
is
males
A=
(2080)
P=
A=
_
—
Soxlo?>
~
(oso
31.5
30 «lo
-3
Fa
« 1o* )(30
¥ (0°
x 1o"*
Pa
4.6% 10% man = 96X10” me
31.6 «10°NGEx(O”) = 202*1O'N
202 kN <e
Tin probdens 275 through 2.82, 2.87, 2.88, anc 2.132
assumed
the
that
retative
Y
js
the
tangeut
displacement
of +he ang te
prop ortionad
change.
+o the
This
Fovee.
7
Problem 2.79
>
—
2.79 Two blocks of rubber with a modulus of rigidity G = 12 MPa
- are bonded to rigid supports and to a plate AB. Knowing that ¢ = 100 mm
“=~ and Pos 40 KN, determine the smallest allowable dimensionsa and b of the ~°] >"
_ blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the de-
oo
WY
flection of the plate is to be at least 5 mm.
7
ae
Shearing
strain
Yo
QO.
3
=
e
GS
(12x10°)( 01005)
OT-
Tee
!
Shearing
vr
stress
_
~
4b
Problem
2
=
2068
2.80
a
ate
ra
a
>
le
A}
= &0429b6m
== 43mm.
2.80
Ao
« 10*
(20051) (1-4x108)
~0: 14-3
-
ms
43 mm
. til
Two blocks of rubber with a modulus of rigidity G = 10 MPa
are bonded to rigid supports and to a plate AB. Knowing that b = 200 mm
and c = 125 mm, determine the largest allowable load P and the smallest al-
~
LB
Pp
lowable thickness a of the blocks if the shearing stress in the rubber is not to
exceed 1.5 MPa and the deflection of the plate is to be at least 6 mm.
~
a!
i
Shearing
stress
Pe
Shearing strain
|
-
T= ae
Qbev
a
«<a
2(2)(0-2)lo-neVLS xo") =7SkN
1 = 2. -&
a. =
GS
y
=.
(10x10EB€)(01006)
xioe
=m OfOF 4 M
=
4o
yan
a
Problem 2.81
2.81 An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as
shown to provide flexibility during earthquakes. The beam must not displace
more than 10 mm when a 22-kN lateral load is applied as shown. Knowing that
the maximum allowable shearing stress is 420 kPa, determine (a) the smallest.
allowable dimension 5, (6) the smallest required thickness a.
Sh earing Sorce
Sheasina
Pr
stvess
-
420
-
Y* & =
_ §2.381xJo°?
.= £2.33) x x/O'ert?m
|
~s
-
= 466.67 10°
[Own
i
2.
OT REZ mm
3
ex
466.67 « fer4
Tv
Problem 2.82
Pa
(200 mm YC b)
A
OF
410°
52.381 <[0*% mm”
b = gee F SARS
(a)
N
a-BR.
BR,
A> a= gang
Ko
t -P.
A=
22x 1o*
21.4
mn
|
<i
2.82 For the elastomeric bearing in Prob. 2.81 with b = 220 mm and a = 30 mm,
determine the shearing modulus G and the shear stress r for a maximum Jateral
load P = 19 KN and a maximum displacement 6 = 12 mm.
Area
Force
At
P=
19xio®
{200 wer )C220
men )
N
uUxfo*
4
Sheaving
.
=
men
44xion®
re Be MeI Sys sivst Pe
*
.
7
bina:
Shearin:
22,
eas
Shearing
¥
az
2mm
-
BO
5
z “Bible”
=
is
makes
L
~
O.
a
400
1O080«(0"
=
it
um
kPa
meds dus
G = >
NOTE:
AB}
w*
1080
Pa
MP
In problems 275 through 2.92, 287,288, anc 2.182
assumed
the
that
velative
Y
ts the fangent
dis peace ment
of the angle
pro port fouad
to
emanse.
the
This
Fovee.
we
Problem 2.83...
*2.83
A 150-mm diameter solid steel sphere is lowered into the ocean
” to a point where the pressure is 50 MPa (about 5 kim below the surface). Knowing that & = 200 GPa and v = 0.30, determine (a) the decrease in diameter of
the sphere, (b) the decrease in volume of the sphere, (c) the percent increase
in the density of the sphere.
For
a sodid
7
=s &G,-
Ex
F
*
Like wise
E
=
Ex
vey
+
-Ad
(ob)
-AY=-MWe
&
=-d,€,
Let
m=
= €,
+
Paes
_
Le}
loo xre|
ye2 == U=2up= (0.4 MsoK00") |
62)
E
-6
zooyso?
z-lo0oxlo
-~
306x}o~
= ~ (#767 x07 VC -30-0 x1o~* ) =
mass
of sphere ,
PU
x 100 Te
S30xlo7!
Mm
3
~<A
= pPUZ(1+e)
ye
el
~@4e@°-e74...)
# @
as
m = constant.
Ue.
El = gogey
= (1
Eafe
=
4
=-(o/s)(
100 x10°* ) = 1/5 *lo~"Mn,.
m= pPVe=
-
- V6)
Eis y® = 767x107 m?”
3
=z
&
(a)
(©)
_
sphere
i
~
|
=z
|
~C+
-
lL.
Tee!
e*
~ e374...
= 360 xja"°
(300 10°) (00 ®) = 01032
<
Proprictary Material, © 2609 The McGraw-Hill Companies, Inc, AH rights reserved. No part of this Manual miay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their isidividual course preparation. A student using this manual is using it without permission.
| __
|
Problem 2.84
*2.84 (a) For the axial loading shown, determine the change in height and the
change in volume of the brass oytindsr shshown. (4) Solve© past a, 2assuming that the
.. loading is hydrostatic with a,= oy=
~.70 MPa.
ax
BS MPs
E=
105 GPa
“y
.
= 0.33
h.=
(85
mm
Ant
Bao
2
=
O.135wm
~
_
2
“
= E(as\= 5.6745 x10 mm = SIGTIS
HIS we
Vez Anh, ? 766.0610 me = 766.06%/0
8 w*
(a)
Gy= © , Gy -S8*IN
Pa
== ie
E
£y
(
fo
vey
+S,
&
= ~ -S3xle"
A=
eo
~
TBE (Gus Gy
=
~137.81«{o°°
inZy
Gh
Ey
ke
= Gy
-=
wey
-
OJ
-
= = $62.38 x)0°°
=
G
=
t(.
gE
(+6
+
Ah = he fy=
2 (0.34)
53 x1)
~70x*10° Pa
6
-
.=
VG)
_
=
6y + By + 62%
I-R»
E
=~ 226.67
*10
6, ) = .
met
- RIO xIO"*
Gy
-
“
(35
mm VG 226. 6?xto% ) =
- It =av (6,le + 6y+
=
ae UHI
—_
= (766. 06%(0* wed \-187.81x1O*) = ~ 148.9 pom?
Tos elo?
AY
=
+S) = NEEL = ior
(0.34 \(~Foxjo*)
e
26;
boty = (18S mm SE2.38%10°) = = 0.0746 rom
AV = Ure
(b)
_
6, a = 0
~- 0.0306 mm
2lowio4)
€.349)(lof atoF
UlQe= (166.06xt0? ma iC-CZ0n(0°
Je
=
~680x(O
ts
6
- SAI mm?
Proprietary Material. © 2009 The McGraw-Hill Companies, tne. All rights reserved. No part of this Manual may be displayed, reproduced, or
distribated in any form or by any means, without
the prior written perrnission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by MeGraw-Hiil for their individual course preparation. A student using this manual is using it without permission,
att
Problem
r
2.85
ee
Bo
_
AGEN ¢
*2.85 Determine the dilatation e and the change in volume of the 200-mm length
of the rod shown if (a) the rod is made of steel with E = 200 GPa and y = 0.30, (b)
the rod is made of aluminum with £ = 70 GPa and v = 0.35.
AGEN
4Gxlo*N
=F
=
EQQ2} = 380.13 mm’ = 280.13 xo ° mm”
121.01
2,2 BG eG -ve): &
@>
& 4 8,4
Vodame
the
Aw:
Ve
+ He,
lo® Pe
Gy = G,=
©
By 8,7 - ve, > ve
~UCp~ VC, ) F Lo dzise
AL = (380.13 mat
200 ma
Y=
cad steed: © = Aa Sele = gua v0"[8.40
(242 #1075) =
026
AV = (16.«/08
76.026 «/0" mw?
h
Ps
fe Late
yam?
(o) efantnvn® — @ = LO2IRORO) LS Siqyyort
Aw = (76.026 r(0* OS 10% Y=
BRN
mm®
hob ob
|--200 mm—+|
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reprodaced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 2.86
.
*2.86 Determine the change in volume of the 50-mm gage length segment AB in_... Prob. 2.63 (a) by. computing
the dilatation of the material,
(6) by subtracting the
original volume of portion 4B from its final vohime.
2.63 A 2.75 KN tensile load is applied to a test coupon made fiom 1.6-mm flat steel
plate (E= 200 GPa, v= 0.30), Determine the resulting change (a) in the 50-mm gage
length, (6) in the width of portion AB of the test coupon, (c) in the thickness of
portion 4B, (d) in the cross-sectional area of portion AB.
A
ave
caPeudations
in
heve
mm
| fo
2.75 kN
SO
ES
|
12mm
Age
(2M be) =
Volume
>
Ex = #6
~ v6;
- 26,) = Ee
=
€&,
AV=
From
denath
wie Eh
thickness
volume
+ &
UVle
the
Sy =
The
= -v&
f+
=
Pa
Gy
33.224 nie
=
—(0.30K716.isxlo’)
107%)
=
Prohfem
=
= 6,
TGS
=O
«10°F
~ 214.84 x1o°*
I2-
=
0. 275 mm? |
2.62
Sy 7 7 0.002578
= SOt
£,45,
WV: Lwt
te
when
W437
960 mm?
286.46
x fa*
wm
Ls lot &
Lr
F
sodutron
dimessious
We
=
143.229 x10°
= (460)0 286.46
0.08581
19.2% 107% m
LAL = (S0MIG. RY =
ee
@F
AV
Vez
5
6, = po
éy
(b)
19.2 mm™
onder the
2.75 1)
0.03581
0.00257=
mm
S,° ~ 0.000 8437 mn
IbcR
W.4974AZ
ave
mm
|
mm
1.6 ~ ©,000B437 = 1, 5496563 mm
= ($0.0888NC. 49742C. S496 568)>
V-~V, = 960,275
tt
|
§0.0388]
- Go
t
(a)
960.275
mm®
O.275 mm?
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—
[|
Problem
2.87
*2.87 A vibration isolation support consists of a rod 4 of radius Ry = 10 mm and
@ tube B of inner radius Rz = 25 mm bonded to an 80-mm-long hollow rubber
cylinder. with a modulus of rigidity G= 12 MPa, Determine the largest-allowable --|force P which can be applied to rod _A if its deflection is not to exceed 2.50 mm.
Let
80mm
T
styvess
Shearing
P
_ P
._P
(fae
ae
“ 2nvk
shearing
strain
Te~-& -* Hehe
dy
-
_P_
Tote!
3
.~ oe
¥}
@>
“~~
-”
-
0.010 wm, 5
I2x1o°
Fa,
a
anY
\2 oP
R,)
2
enn Pe - tn
din
_27h
lOmm=
_“onGh
as
_P
2nGh
_
Rt
¥
deSormatior
_
Date:
dv
Ponsth
dy
352 Yoav = ah
Leas
1s
vadiad
overt
atron
detorm
Sheaving
acdina
cova
on & cy Pindnicad
of fadsusr is
The
hotfow
Ry
VY
Ri<
cytinder
Vubber
the
Over
cosrcdinate.
a vadtad
Y be
z
Mn
a
_
ov
RyFX WSTmm = OLO8T Mm
S=
a27GhS
An
(Raf,
}
be BOmm = 0.080m
2.50%10% m
(27) C12 x 10% )(0.080)(2.$0 x/o~*)
Dn (0.025
/ 0, 010)
P=
6,460
N
16.40 kN) <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyord the limited distribution (o teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem 2.88
*2.88 A vibration isolation support consists of a rod A of radius Ry and a tube B
of inner radius R2 bonded
to a 80-mm-long
hollow rubber cylinder with a
“modiillis of tipidity °G'="10:93 MPa: Determine the required value of the ratio
R»/R, if a 10-kKN force P is to cause a 2-mm deflection of rod A.
Let
Vv
be
vubb evr
8O mm
a
vadtal
cosrdinate.
cytinder
.
Rie
r
On
&
oy
hod fow
/
styess
T
cy? fl nal nicat
fadiust
aching
suvtace
is
.P
-~_P
TH
The
the
VER,
Shearing
oo
Over
2nvrh
shearing
steam
v
is
.
v= & * Oréhr
detorm
as.
dr
rv
d5
=. 7
dedorma tie
_
S48
P
vadiak
°
r
aia
Tne
AY
+
Don sth
°
> de
“aR
r
> Un R,
- DP)
ij
dX
1
°
tl
Ry
overs
3
s
Total
ar
atron
t
Shearing
it
DiS
RZrGns
mS
J, & =
P
_.
exp (1.0988)
(AW)10.93x 10%.3.080)(0. 002)
[0 x 10?
=
= 1.0988
3.00
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
dv
“|
Problem 2.89
By
50GPa
*2.89 A composite cube with 40-mm sides and the properties shown is made with
glass polymer fibers aligned in the x direction. The cube is constrained against
deformations in the y and z directions and is subjected to a tensile load of 65 KN
“in thé x diréction. Determine’ (@) the charge in the length of the cube in the
x
direction, (6) the stresses ox, dy, and o,.
sv, = 0.254
E,=152GPa 4, = 0.254
E,= 152GPa_v,, = 0.498
Sivess- to - stran
Gy
nee
me
VypyTy
"EE,
w
6
The
constraint
Vsing
C2)
conditions
avd
(3)
EB
—
+he
ay Sy
~ 9888 6
Solving
=
- 2.488
6,
+o
Ge
CH)
ont
Ey = Zz
| l-
Ch
Pye
Pay
~
Par
aa
Px
| ©)
= (40)(40)
conc,
tiens
gives
(3)
or
0.254 Sy.
ow
S, ~ O.Y286, = 0.077216 6x
- 0.428 6
Cy
,
GL
—
G),
2
'
+ S, = 0.077216 6,
0.134993
.
4
Ey = & Sy > BS, -
6,
2
BE Se
Sy
=
1600 mm™
=
1600 x10 * mw"
3
Cy
=
PL
A
~
SS
¥XIOT
1600 x 1o-*
~
(6)
E, = 0.
(o.2s4 (0.134993) ~ (0. 254 )(0. 134993 ){ 6,
O.93142
A
and
4
E,
(7)
= o2s4 Sy,
Gin
yx
ee
nop
SY
o 2
eonstrant
2S,
simultaneous.
Using
y
BE,
@
fy =O
LG - Ye Sy = ee
“GIES
Bye
Dy
ay
we
(
ee
EE,tee
y
ave
j
E.
tee
BT
ave
with
BaP,
oe
90x) Oy
one
eq vatio AS
YO
_€25
xto*
Pa
continued
|
Problem
299
corhnved
|
12 \4e.625 10")
vib"
ey _ (0.93=o
fr
S.=
Cy
(b )
“&
—
mm
0.0303
~<}
|
MPa
4O.6
=
62asxvjo°Pa
40.
756 18 410
(HO um 0756.78*KIT°) =
Ly és =
=
:
5.48 110° Pa
Gy = Oy = (0.134993 40.625" (0%) =
=
$2.43
<a
MPa
*2.906 The composite cube of Prob, 2.89 is constrained against deformation in the
z direction and elongated in the x direction by 0.035 mm due to a tensile load in
the x direction. Determine (a) the stresses oy, oy, and @,, (6) the change in the
dimension in the y direction.
Problem 2.90
E,=50GPa — v,, = 0.254
Ey= 15.2 GPa
Pyy = 0.254
E,= 15.2GPa_
v,, = 0,428
€
oy
Load
conel:
From
equation
Sz
tron
De
e
ee
BeOy
Vey
E,
7
op
Q)
BE
a
ty
VyyDe SE vee
Vyx
}
E,
VyG,
oop
cd
VG,
Oy
Ta
ip
condition
PyxOy
Py Oe
po
ee
Constraint
we
EE,
EE,
ay
Vey
w
Pa
Ym
(6)
cece SE
OG;
te (8)
z
E,
omen
BE,
©
.
C3),
EZ
A=
§
0
s wv
E. | St
=
Siete
DA
to
(©
259018
aL
= S
2)
—
=
|
OL
O07
i
4)
We
B-E
2.
(
7
continued
Problen
From
270 contin ved
ea vahiou
&>
Eo
6y
with
E I6,-
=
tL
- wt
Ez
Ex
y
Gi)
E,
0.254 6, |
EG
- SE,
= Eli-
(0.25y
)Co. 077216)] 6%
2.98084 6,
.~a
Ke
Sy
ut,
(a)
&
Oe
eee
0. FBO34
=
Be
= O08
mm
~ (§0 «IO? )C375x 107%)
Sy
6.42039
=
=
£75 “10°
=
Pao=
44.6 MPa
iu
&y=
~ B46, +26
3.4496 «10° Pa
= 3.45 MPa
- 26,
4
—©-25Y) (44.625 10%) 4 6G . (0-423) (3.446 x 10%)
(
Sy Pls
"
iy
SO K1O®
|
IS.2¥101
- 323 - 7% x10" -e
(40 mn \(- 323.78 ¥10*)=
<m
ty
.
©
= (0,077216 \(4,6aSK1o*) =
From (@,
44 625 x10?
~ 0.0124
mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
att
*2.91 The material constants E, G, & and v are related by Eqs. (2.33) and (2.43).
Problem 2.91
Show that any one of these constants may be expressed in terms of any other two
.. constants...
For example, show that (a) & = GEA9G.- 3#) and (6) v = Gk- 2G)M6k...1.
+ 2G).
.
Ik
_-&
=
(a)
B(\-av)
aud
WE.
f+ v= - 7
k=
=
G
° yy
Pe
Shi - 28-1]
EG
EE
2(1+D
)
Zé
|
te
gO
op
~ RES
3[26-26+46
REG
Re
- CE
aescatieh,
qG-
(b)
ko.
20+¥)
Bli-
2vy
3K-
Gkv
=
&S
Sk~2G
-=
G
-
Zit
26+¢€k
Bk
-=~
3k
Assume
w
26°
.
20
‘me,
2G
2.92
2 }
26”
=
Gk +
Problem
264
*2,92 Show that for any given material, the ratio G/E of the modulus of rigidity
over the modulus of elasticity is always less than ; but more than > [Hint: Refer
to Eq. (2.43) and to Sec. 2.13.]
E
or
for
E
admost
ed
Z2Ci+wv
al?
motevnrints
and
we«<
4
ot
post hive
a
&
as
reciprocads
a
ow
+4
3 <=
mm
Ww
bounds,
[K
the
+he
min me
Taking
ing
Vv
App +y
on
bulk modudtus.
2C4+4)=3
a
3
a
z
~a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. —
Problem 2.93
2.93
Knowing that P = 40 kN, determine the maximum stress when
{fa)r = 12 num, (b) r = 15 mm.
P
=
D=/258mim
Lhe LAN
gf = 62mm.
121°
5
= “tr
=
2.616
Mein < dt =lorb62)lo-0lf)= Kilbyro
(a)
vz
[2mm
From
Fig.
+
=
=
-3
Mm
08/935
Re
2.64
b
Kez
1.44
125 mm
Gans
\ 18 mm
(b)
KP.
C44 Morte)
Aan
IS
yo ISmm
tz
=
Problem 2.94
K =
Orly
_
Sime > KP _ (686)Goxlo™)
bilbyZ
66-7 Mh,
d=
(a)
Ys:
Fig.
J = 2016,
= (oro62)(o.01P) = Ltt X10 2m
r_
/Okm
From
D
62mm
~
Aus. t= Ab
.
es
= HO MPa
D= (25mm
Le
hQé.
2.94
Knowing that, for the plate shown, the allowable stress is 110 MPa,
determine the maximum allowable value of P when (a) r = 10 mm, (b) r =
18 mm.
Cran
195. mei
. 64-5 Mpa.
hettbxro>
fe
a7
2.6494
Kr
=
2
>
2.
6. /6/
oF.
\ 18 mm
Gran,
P=
tb)
KE
A iin Change - (hilbxro io xfo4)
rr
K
p=
Ps
=
/Pam
Ann Sone
K
JF
vy.
-
Der
$8*7 EW.
= 027
i116 x19?) (110 x00)
=
~<a)
K = &76
64-76
EN
1676
Proprictary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usingit without permission.
Problem 2.9
2.95 For P = 35 KN, determine the minimum plate thickness ¢ required
if the allowable stress is 125 MPa.
5
/ P 55 muy
At the hole:
WWrl2mm
dz 55-24 23]
ry = Lamm
Ys
dat
rp = LOomm—
1m
=_
ap
og. :
0
34
eT
>
40 ram
From
Fig
zeta
K = 2-26
. KP
Swe -KP
Re .Ge KP . 8 be
= C2
200
(0031) C1ZS x08
At the Met
Tat
3
&
.
t=
=
$s
:
K
dg>
lomm
Me.
K
P
~
=
Di.
as
.
“tO
ia
=~
55
40
2
1.375
0-295
Cua ?
i. 70
(7035000)
ae. Craw
tom,
jo.
ole
2.64 b
Fig
From
D= 55m
0 2 OG tm 220+ Lem
Oeo\iG
_ Ke
Aan
ha
=
Ib
_7 AcE
KP
Gaim
fa 04) C25 x of)
The Jarger Vadveris the required wini mum plate thickness
Pr 20-4 mm
Problem
2 96
:
2.96
Knowing that the hole has a diameter of 10 mm, determine (a) the
radius r; of the fillets for which the same maximum
hole
we 2. = Sm,
FFx
d= lov -107domn
Avett
100 ram
de
=
From
10 mm
dt
Fig
SC.
(a)
Fow
Av.
=
=
From
D=lOomm
£; Det
Fig
dt*
Kew
ds
(0062) (oo)
P
2.64 b
Kite
6
(o:0004) (105 K10)
i
Sv
-Ast
°
OS
. Kinde
P
2
= 010004
=
by”
2°68
,
-
=
B3-1SEN
2.
2 frm
Amin
20'056
Avnet
= 01900662 bam
-
ee
=(0-0%)Cosoi)
2644
sae
mak
P
if the
allowable stress is 105 MPa.
For the circular
(b)
stress occurs at the hole
A and at the fillets, (6) the corresponding maximum allowable load P
LOmm
<a
Oma
x
oa
2
=
fe bf
(0100062) (168 x10")
fp% 01%d = Corih) (62) 2 N16
-
mm
fe
<a
Problem
_-
2.97
2.97 A hole is to be drilled in the plate at A.
'.
The diameters of the bits available
to drill the hole range from 12 to 24 mm in 3-mm increments. (a) Determine the
' diameter d of the largest bit that can be used if the allowable load at the hole is fo
“exceed that-at the fillets: (6) If the allowable stress in the plate is 145 MPa, what’ |
is the corresponding allowable load P?
12mm
the
Fidlets
)
Y=
Yeu
a
=
FE
FT
Aww
=
Ginax
.z
>
O12
From
(75 YR)
=
ads
=
Fey
Kr?
Joo
.
Fig
2.
wm
2.64
b
=
Pup
_-
K
900
mm
the
hoe
j
Anat
De
112.5
taken
Pron
where
K
is
Sinase =
KE
Hole ebiam|
v
>
(D-
Fig
= 6.y
+
.
FOO x{O~
ya
N
c,
= 62:1 KN
~
Yaolrus
=
of
se
civete
tzl2mm
a,
Pua ®
A\ net Cote
K
| a=D-2r[ v/d | K | Aver) | Pa, W)
}
4
12 am | 6.Onm
100.5
0.0601 2.80|
1206xJo% | €2.5xI}08
IS mm | 7Swm!
975 mm
10,077
|1oxjo® | 61.7 «Io®
Wmm
FHS mm | 0.075 | 2.71 | N34xI0°°
21
mm
| LOmn|
10,5m|
Z4 nm | 120mm)
I
2.10
7:
Y=
2.64
damm
¥ jO7*)LI4S 108)
Amin
iz Gy — F900
ar)
mam
2AO
=
= CUIx1o”
A+
ry
75 ayn
2 16
mm™
Sim
r
7S
2S
B=
N2.Smm
De
Gun
it
At
oO
412.5 man
WS
mm
12.75
[O.115 | 2:67 | 1098xlo~%
Co.7™ Jo”
54.6
» |o#
BRSmom | 0-136 | 2.62 | 1062xlo” | S8.axIo°
@)
Bit
drameten,
(b)
AlPowehte
Vod.
{=1]2mm
C21
kN
=a
«
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—
Problem
2.98 (a) For P = 58 KN and d= 12 mm, determine the maximum stress in the plate
shown. (6) Solve part a, assuming that the hole at A is not drilled.
2.98
r=
125mm
9mm
”
:
75mm
Maximum
Use
stress
Fiq.
at hole.
2.64
ve
a
Fov
ns
=
W225 - 12
Aine:
+
Cane
Maximum
Use
4
we
Fig:
-
=
Av
7 U2)0IS)
°
(a)
With
hoDe
(b)
Without
0.12,
2
4
BD. HES
=
PL.
K
:
S8xlo
1206 *10* wt
=
247%
J0¢ me
b&b
*
Sonn
(280
2.80
fi Diets
oe
~
Ke
1206 x (o7*
at
2.64
K
S-IRY = 106 me
Avert
stress
ot
0.0547|
GRU
"
a
vatues
min
ond
hole .
Geo
mm
Qo
-
PLbets
=
sx
Foo
,
= 1.50,
Ks
2.10
900 %107° mm”
3)
“
| 35.3
*
¥{o"*
nage
Gray
=
=
10
Pa
134.7
MPa
—e,
135.3
MPa
<A
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
2.99
2.99 The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Knowing that £ = 70 GPa and gay= 200
- MPa, determine the maximum allowable value of ? and the corresponding. total.
“elongation of the specimen. (5) Solve part a, assuming that the specimen has
been replaced by an aluminum bar of the same length and a uniform 60 * 15-mnr
rectangular cross section,
¢
x1
2OO
Cui?
TO:
Es
Pe
Re
xI1O”
.
Amin
(a)
=
(GO
me \OIS
mm
‘
BL BB.
a7 Go =
Sr
5
ina
P
.
-
AG.
K
¥
az
ype
6
{oO
eo GO mm ,
Y = Gis
re
dé
7 91°
Ke
From Fig, 2.6406)
d
Dimensions
7s mm y
D-
Specimevis
Lest
JOO
=
rere
Foo
v
|
6.2 KE
A
195
(400 107% (200 ¥10*)
.
-
’
on
=
42320810"
P2428
Wide
g2-
area.
xSth
Poh.
AE:
B®
-
=
Sma
Dp
Bs
Ly
Re
Sram)?
Uartorm
P=
©
q9,
aT eesais
= 92.308%10
“Jo xpO%
=
fb)
{) RE mm
74Pxjo7*
1.49S* 107? pn
[TLRS Kio
bs
4 0:300
—<®
a
4,
9,180
| Gooxlor® * TVS
Sr
kN
N
OTA
70°5
|4
mn
baw.
N64
= (F00r 107% )(200 x10%)= |30%)07N
i
%)
(0.600)
c=
BL . Go
(fox10°)
6
fm
“AE
orin
Ce
xto*s
}
.
TIVO-
P=I1BO.0kN
ma
SLT4 om
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of ihis Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers .
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
.
Problem
2.100 For the test specimen of Prob. 2.99, determine the maximum value of the
normal stress corresponding to a total elongation of 0.75 mm.
2.100
2.99 ‘The aluminum fest specimen shown
is subjected to two equal
and opposite |
centric axial forces of magnitude P.
(a) Knowing that # = 70 GPa and oa = 200
MPa, determine the maximum. allowable value of P and the corresponding total
elongation of the specimen. (5) Solve part a, assuming that the specimen has
been replaced by an aluminum bar of the same length and a uniform 60 * 15-mm
rectangular cross section.
S
ys
=
L, =
Bh
Pr
EA.
E
vor
bg = 1SOmm
Ave Ag= (iSmm
Le
.
2 A;
~=
=
A,
in om
Dimensions
0,450
{co mm K1S nw
O, 300
bs
A:
S
= O.18Om
Sma) =
)
L
= ZOO mm
2
wre
-
=
1077)
R
~
x
[Ov
lo
~3
wm
= OLS0O
1 125%
1 wm
ADox
* TLigsxjos * GOO ™
(Foxio\lo.75
O75
25 mm
900
x
>
=
_O.150
1el2Sxlo"® *Gaoxlo®
ES"
—
enh mere
m
+
io-> ww
~)
3
Ag
D.@7s
*
bo
ig.
From
~
Oynax™
Note
ELL
K Ane,
that
_,
“fe
Xe.
age
a
\*¢ 5
2.54
Uae
~—
Cinan
<
6
~
6D
Cl
x1o07*
O,
YL
Sum
,
10
AGS
87.5% 10%)
Goo
d= Comm,
D= 75mm,
concentration .
Stress
=
|BLExIO”
rs
Pa
na?
I89.6
MPa
&
Proprietary Material. © 2009 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers —
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~—a
Problem 2.101
—
2.101
Rod AB is made of a mild stee! that is assumed to be elastoplastic with E =
_, 200 GPa and g,= 345 MPa. After the rod has been attached to the rigid lever CD, |
~“$tis found that end Cis 6 mm too high. A vertical force Q is then applied at C until
this point has moved to position C “ Determine the required magnitude of Q and the
deflection 6, if the lever is to snap back to a horizontal position after Q is removed.
Ang 2 EGY s 68.617 mm? =
1.25 m
Since
vod
AB
is
fe
be
63.61? «lo * m
stvetahed!
we
permaneit hy,
(Fre mag = Ana Sy = C63.617 810° YS x10°)
=
3,
O04 m
Quran 7 Pt (21.948 xjo?) =
Fae
d.
=)
Qa
ee
¢
A
6'=
Sas.
$=
\S.97 kN
ae
Sie 2"EA
y%
Spas
\3.9¢6710>N
°
@
av
LIQ -O7F, =O
Oo:
ZU,
se pere OT ne
21.748%/07 N
=
(21.948 x10" (1.25)
(loo wOI\ERGIIIO*)
3. 08041O7>
~
2ASEIS KIO? m
wel
LILO = 3,39x10%
2.102
ath
B39 mm -ell
Solve Prob. 2.101, assuming that the yield point of the mild steel is 250 MPa.
‘2.101 Rod AB is made of a mild steel that is assumed to be elastoplastic with E==
200 GPa and 0, = 345 MPa. After the rod has been attached to the rigid lever CD,
it is found that end C is 6 mm too high. A vertical force Q is then applied at C until
this point has moved to position C ’ Determine the required magnitude of Q and the
deflection 6, if the lever is to snap back to a horizontal position after Q is romoved.
125m
“Ang = EtQY = EB.E17 men = 68.617 210% m
Since rod AB ts to be stretchel permanentdy,
6 mun
(Fig) see * Aya Sy = (63.617
=
H<——>}<— 0.7 m _!
04 m
18.9043 %J0*
49 2M,= 0
10° X250
10%)
N
LiQ-O7,
=o
Qnnog, £1
Pi (15.9043%IO) ‘= 10.12“IO* N
Eng)E lee
Ae” OAS
San: ,> VAS
®~
if
favasy
°
—
w
4
=
Q' = 2a
2S)
(1.25)y
10° teios
043
(15.940!
a Veati
Cosm
JO.12 kN
3
™m
7 PS62Sxio
x 1o> reed
2232)
2.46 x/0778 m
2.46
ram
<a
|
Problem
2.1 03
2.103 The 30-mm square bar 4B has a length L = 2.2 m; it is made of a mild steel
me
---P js-applied-to the bar-until-end.4 has moved down by an amount d,:- Determine —-the maximum value of the force P and the permanent set of the bar after the force
that is assumed to be elastoplastic with E = 200 GPa and ay = 345 MPa.
B
A force
has been removed, knowing that (a) d,.= 4.5 mm, (6) j= 8 mm.
$, 2 Le = ESF = QR CET ZION) = 3.795% 10° = 3.1957 mm
P= AG; = (400 «10° (345x1o% ) F 310.5 *10'N
IF Sm 2 Sy
7
Ts
SF
Unioatings
P.=
3,2 4.5 mn > Sy
Sp
Problem
> £.0mm
= 3.795 mm
Sy =
Ley
When
SG.
Foo
=.
= Go){3e}
=
810.5 kN
—<
2
=
BIOS kv
=
_
men
ROS
=
=
bos
exceeds
Sp,
of
the
it
Sp?
mim
om?
(b)
Sp7 6.5 mm
— Ss
yn
ON
a
YBISxIOwm = 4.3125 mm
=
a
producing
Sy, +hus
waytmum
permanent
vs
Force
(345% 10%) = ZIO.S«ION
|
Sur Sy
(a)
107%
= ASN BNE
ACG = (900 “10
Sn- SF
9o0
mes
Pp
3.5
N
2.104 The 30-mm square bar AB has a length L = 2.5 m; it is made of mild steel
that is assumed to be elastoplastic with £ = 200 GPa and ay = 345 MPa. A force
P is applied to the bar and then removed to give it a permanent set 4,.
Determine the maximum value of the force P and the maximum amount 6,, by
which the bar should be sfretched if the desired value of 6, is (a) 3.5 min, (b) 6.5
mm.
-
stretch
Sp=
3.795 mm
Bio.s oN
Pt
P=
7
OLTOS mm
=
~ 3.795 mw
2.104
A
3/0.5*10°
> Sy
Sat Ban
(b)
4S mm
=
Sperm
Sy
=
= Se
Ot Sar 8
Sp
(a)
¥j Ow
FOO
=
yam
Gop
=
= (30)(36)
A
‘
B.S mm
6. Smt
*
= Bla.5 kN
—e
Sut Spt Oy
4.3125 mm
>
FBi
ma,
weasel
4312S mm
F
10.8) mn
_
Proprictary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
2.105
2.105 Rod ABC consists of two cylindrical portions AB and BC; it is made of a
’ mild steel that is assumed to be elastoplastic with Z = 200 GPa and ey = 250 MPa.
A force P is applied to the rod and then removed to give it a permanent set dp = 2
“ni” “Determine the maximum value of the force P and the maximum amount 6,
by which the rod should be stretched to give it the desired permanent set.
40-1nm
12m
.
~ diameter
L
:
Aas
,
= *
fan)" =
706.86
ine
m7OG.9S
Ane = GY C4OY= 1.25664 110% weve”
30mm
0.8m
diameter
Pavey.
~
Drow dy
= (706. 86% lO
ae
‘
xlo~
ASCY HI
E
YASOxtOo
| ———
“
\ =
Prnone
p
gle
Piles, P'le
EAs
EA.
San Bl) ov
Bye
Problem 2.106
=
rt
VIGLTISKLO
196.7
3
N
kW
ah,
_ 76.x103
218
) (2.8)
(176.718 «(0° JOR)
§ (Loox(O4\(706.86x108) | ooxlot YL 15664 fF
= BIBS ¥(O Fwy
Se
wm
=
1848 TS men
BG HS
QH1LB4B7S
Sim= 3. BY mn dt
2.106 Rod ABC consists of two cylindrical portions AB and BC; it is made of a
mild steel that is assumed to be clastoplastic with E = 200 GPa and ay = 250 MPa,
A force P is applied to the rod until its end 4 has moved down by an amount d,, =
5 mm. Determine the maximum value of the force P and the permanent set of the
rod after the force has been removed..
40-ran
1.2m
~~ diameter
7
Nig = 4 BOY
0.8m
F 706.86 mm
2
lee
= 706.86 10% wm
wm
LR5EC4*IO®
©
Nae = E (yo
° 30-0
—
a
diameter
to
|
Powe = Avin 6o = (706.86
410 GSO 418%)
(76.T1F ¥10*}0.8)
gi= Piles , Pilee.
EAag
=
=
Si 7S!)
an”
.
.
Se
> 1.25644 x1
EAse
1h.@4 328 x)0F
=
176 TISKION
Peon 176.7
kN
ees
(MIELTISx10°1(1-2)
Qoosio?K706L8 x10) ~~ Qoox{ot )(}.a5664 «10%)
mw
SF -WB4BIT
=
18487
=
BiG
mm
mm.
Spe 3.16 mm
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc.. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
adh
|
Probiem
2.107
2.107 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm*. Portion AC is made of a mild steel with E =
200 GPa and oy = 250 MPa, and portion CB is made of a high-strength steel with
£ = 200 GPa and oy = 345 MPa. A load P is applied at C as shown. Assuming
both steels to be elastoplastic, determine (a) the maximum deflection of C if P is
graduaily increased from zero to 975 KN and then reduced back to zero, (6) the
maximum stress in each portion of the rod, (c) the permanent deflection of C.
r
190 num
Displacement
onet) - 0.23715 m
~ Q.19e\asoe
Sey = bac bys. = “EO
|
Corresponding
5* (aso rin')
*
Fre= A Oyac =(1760%10
fovce
=
—
EA Se =~ QE2Qx10 Knsoxio" NO-287S¥I5")
;
Vat?
equid beiom
Since
Ade
at
8
P=
t+
g'=
_.
97Sxl0° =
Spt
:
Sao
875 x10" N
> BIS «IC MV,
O.RA179 IO
©.
=x 250
aqa
mm
.
—_
:
MPa
—«
wt
untoad: ne
‘L
°
ER
Pl-
*
=
-3
2.
410% Pas - -307MPa
7
307.1410"
- eo
kw
'
* Re=~ Re ies
Po
(487.5 x1o* )}(0. 190)
(200 x10
C
VIS*IO'N
Cag 7 Cy as
and Forces fon
o = ER *
P=
« ys7.sxio" N
487,5uI0-975"10°N-= - $37.8 1d MV
Fe ,.
ae
= S32e%10%
lisovio*
Pel
N
yiehde.
Mayimum stresses:
Detection
a3
Re AL-Fg=
~ Falco, (537.5 xJ0" (0.190)
“ Goowto* (iso 10%)
"EA
Se =>
(c)
element
apphied Dot
Fig = Fie - P=
(b)
ot
Fin -CRy+R)SO
portion
(a)
y87.° vio
4
For
C
Somos
0.190
te
Ge he
Lae 6;
_
of At
yielding
at C te cause
_ |
190 mim
)(17SO «t0"* >
=
aa
2Py
=O,26464x%
o>
Pye= 487.5%I0° N
m
“3
_3
-3
S ' F O.2TI7I HO” — 0.26 IS” =F O.027IS*IO m
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 2.108
2.108 For the composite rod of Prob. 2.107, if P is gradually increased from zero
oo,
:
until the deflection of point C reaches a maximum value of 6, = 0.03 mm and
---then decreased back.to zero, determine, (a) the maximum value of ;-(6) the -.
maximum stress in each portion of the rod, (c) the permanent deflection of C after
. the load is removed.
2.107 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm’. Portion AC is made of a mild steel with E =
200 GPa and oy = 250 MPa, and portion C’B is made of a high-strength steel with
E= 200 GPa and oy = 345 MPa. A Joad P is applied at Cas shown. Assuming
_both steels to be elastoplastic, determine (a) the maximum deflection of C if P is
gradually increased from zero to 975 KN and then reduced back to zero, (6) the
maximum stress in each portion of the rod, (c) the permanent deflection of C._
Dispfacement
eae
atC
Qn.
=~
*
leg
Strains
Gyre
=
=
Ey,ac
1FO mm
intied
O.30mm.
_
. OBOm™
~
ot
Su F
0.80mm
2 ~ me
Ece
is
The
.
2
c7g9 Kj?
_250*10o
“Doorloi
*
T.2S x1O
Sxloe
2
|
4
oF =
Fig =
EAE,
For equi Di bviun
Pe
-3
Detection
s'=
3
Ces
=“ R="
&
= 437.610"
MV
|
-P=0
Fie ~ Fes
=e
= 990.1 X1O°N=990 KN
~<a
250 MPa
SE26 703
i
vio
=
~316
x08
Pra, =
-316
MPa
«a
ancl forces for undousl ina -
Pao Lae
- Pés les
Poe
EA
-
nae
Pie
~ Fae
te
o
Sm- 5! =
=
=”
-P
at
Pe = APL Fq90.1 ION
I0") (0,190)
49S0Sx
t. (200
K(09)C17SO x}o ~*)
Se =
.
asoxio®)
Cc,
at
Cre® Sere =
EA
Pie
)
(200x107 )(I7S0 «fo )(- 1.5789" 10" )= ~552.6 xjo*M
oF efement
CB
(Cd
(elastic)
-s
AG = (7S m1o*
=
= AC
(b) Stresses>
OD:
(yielding
Fo = 432.5 «1074 552.600?
Ker
ave
yiekel ing
~
la) Forces:
Styacne
~3
Eyeg + Gate = BUERIOE va rzsH ie
-.
covves porsling
©. 26374
&
Pee
¥{O 7
4asosio*N
oe
Mm
0.80 mm - 0.26874mm = 0.031 mm
-2
B74 man
<
“Problem 2.109
2.109 Each cable has a cross-sectional area of 100 mm? and is made of an
elastoplastic material for which oy = 345 MPa and E = 200 GPa. A force Q is
“applied
at C to the rigid bar ABC and is gradually increased from O0'to SOKN and
then reduced to zero. Knowing that the cables were initially taut, determine (a)
the maximum stress that occurs in cable BD, (6) the maximum deflection of point
C, (c) the final displacement of point C. (Hint: In part c, cable CE is not taut.)
E fon gertion
2in
let
6
constraints
=
-
°
Feo
Fee
From
(2),
Since
fa}
CE
ap
<
A6;
Monimum
=
Z4.5
x
.
[0°
Ni
31.0» 0"
foo «lore
oF del be ctien
Lap
Feo —8P
~82
-
EA
(1)
Ce
+
ARc
.
Fey
Lac
~
Fee
+
O
Lae Q=
2
Fen
(2)
AG, =(loon}8*)(245
«10% )= 24.5 «10°
}
cube
=
See =
(See maw ~ Fephes
(2),
cahte
Cable
f,,
BD
BD
fs
eburtrc
Ges
CE
F
is
mares
SY
QQ
7
=
B45
SO
N
kM.
MPa
Beanz BIO MPa
«<a
mn
3
_
x fo * m
8.1
2 Gap
=
CE?
(Seep =(S.-)- Oxtee
B.2%!0
im
C.20
2 (Serr )itene 7 Srkce
sYack
2(Q- Fee)
re
how
*
> 2IOx10° Pa
(Blo
\S1o x} % 07 fay \
*)
x 107
(loojo
(Zeex¥
So 7
UnPoaking.
Since
ARG
ee *
Len Fan + Lae
= 6.20x)o~ Biseto“@)
Pron
bar
.
of point
Hons oA ion of cabbe
(Sce),*
tc.)
bar
oce
a
ce
.
A
Manwinuw
Pevrmd ueut
=
stresses.
Feo
eo
From
o}
Ree
wigial
Linc
L Be
*
ay
caubdes.
Fey = 2(Q- Fic) = @X SOx 1o*- 34.5%/0°) = 3BlOxjo7N
Seo
(oy
is yielded ,
_
=O?
49M,
Fe
cable
See
bru
O
Assume
S20
ge
Egurt:
Ry
anyde
Las
——>|
meta]
pot fron
tayt
=
=
|
=e
2.75x1o3m
(Fiz-0
)
at
2(0-0)
ebastre,
mm
=
Sap
=
x
Qe=
O.
O
Peer
a
pte a
Cor
=
OD
tlh,
fj
: Problem
2.110
42.110 Solve Prob. 2.109, assuming that the cables are replaced by rods of the
same cross-sectional area and material.
Further assume that the rods are braced
2.109
Each
re
-
--..80-that they can.carry compressive forces.
cable has a cross-sectional
area of 100 mm2
and
is made
of an
elastoplastic material for which oy = 345 MPa and £ = 200 GPa. A force Q is
applied at C to the rigid bar ABC and is gradually increased from 0 to 50 KN and
then reduced to zero. Knowing that the cables were initially taut, determine (a)
>
the maximum stress that occurs in cable BD, (5) the maximum deflection of point
“m
C, (e) the final displacement of point C.
EPone
ation
Let
o
Yo
pee]
pp-—fa——]
8
=
=
*
z
felis
47
Q
os
riaie
baw
.
ABC.
L
See
7
Z Ver
Cy)
bay , ABC,
ao
2ZMy2
Qe
aug Je
-
= Lae
Equi hy briv»
z
:
Lae
Las
Fe
.
Sce
bag
ol
Fan
“f
vobatrou
Sec
Sep
b
const raints
(Hint: In part c, cable C£ is not taut.)
©:
Fee
Lag Feo
+ pt
Fey
+
.
Lac ee
Fie
~ Lac QO
+ 7 Fe,
(2)
Agsome cable CE is yielded.
Fig = AGy = (00x15
* (2451
0%) = 34.5
x 10% N
From (a)
Fep= 2(Q- Fe)= (a\Sox/07- 24.5%10*) = 3104107 N
Since
Fag
fay Minimum
<
AGy
Mawtmom
—
Sep *
7 .
cable
BD
is ePastre
Oce
deb fection
Feohkes
EA
ing .
Edestic
a)
= lox lo" Pa
Peint
Se. o
=
B45
t
Q'=
kN,
MPa
Seo > SIO MPa
~
—
2)x/o%
rx
S0x(0*
=
”
(Sc) = See = 2500 = G.2xlo*mM
N )
6.22 mam a
¢
See
* Se
From (1)
Sao = %
Se
= EASe
_ Qoomioty
Las
2
(ivo
io
Wee)x gv ige g!
Fan
EA Sez.
(200% (07 Yioo x15" KS) _ Jovjo*
Lee
Equating
= OY
Q=S5O
Cc.
_
_(St0xlo7)(2)
~ (200x107)Cleox1os)
E/
Fre (2)
wher
3
eee
From (1)
Dnteows
34.5x/% N
stresses,
Seo *
lb)
=
Q=
P.d+ thy
express ions
S,
4
7
=
Q
12.5108
S,
I2.5*(0°S, =
0% (0°
4 x 107% m
(2) Eine? displacement.
Se = Gq
So = SBT
LIKIs BAIT wm
2.40
ww
¥
am
—
Problem
2.111
©
2.111
Two tempered-steel bars, each 4.76 mm thick, are bonded to a
12.5-mm mild-steel bar. This composite bar is subjected as shown to a cen““tric axial load of magnitude ‘P. ‘Both steels are elastoplastic with B= 200
GPa and with yield strengths equal to 690 MPa and 345 MPa, respectively,
for the tempered and mild steel. The load P is graduaily increased from zero
4.76 mm
until the deformation of the bar reaches a maximum value 6,, = 1.016 mm
and then decreased back to zero. Determine (a) the maximum value of P, (b)
the maximum stress in the tempered-steel bars, (c) the permanent set after
the load is removed.
4.76 mm
—
-
12.5 mitt
355 min
For
the
S$,
mrkd
2
steet
LS
wlO35 5)
E
For
the
tempered
Steed
&
A, =2(2100476)(0508)=
483-6 Kio m
2od K10%
A=
A, +A,
bm.
-
= 1N86x0
= A, GS,
P= P+
Stresses
6, =
.
Ay
UnPoadi . ng
(c)
Permanent
7
KN,
= 276-8
tN.
BR 2445-7 EN
S, = fe
RB
™
(346 K10%) = BAOTS
°)x10
(63S
115") o-001
P, = FAsL Sm .(2o(A826
0v10
4)16)
iss
(o)
>
-6
The m Pd steed ytelds . Tempereed steed is elashe.
Su < Sm < Svz
YP,
OGIZXIO *h
&
| rae
0-35 )lb40K2)
avec:
) = 635 x orn
X10
*365
E
(a2) Forces
ae yso*).
206
Syy = ASre
Total
A, +(0-0125) 0.0508
-
Oy
-
= 346 MPa
= 572-4
_276M0
<
.
MPa
AF26% oe
1.
St
set
BL
EA
Sp =
_ (448400) (0355)
=
.
~3
Coeomely (1nGebxrd 4)
3m 7 S'
=.
Lot
=
"ye
OTE TKO
—oeTF
Ww
7] =9°229
.
mn,
tOr,2nm
at
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is-using it without permission,
2
Problem 2.112
—
2.112
For the composite bar of Prob. 2.111, if P is gradually increased
fom zero to 436 KN and then decreased back to zero, determine (a) the max- .
_. ,
i
deformation of the bar, (6) the maximum stress in the tempered-steel
imum
bars, (©) the permanent set after the load is removed.
2.111
Two tempered-steel bars, each 4.76 mm
thick, are bonded to a
12.5-mm mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic withE = 200
GPa and with yield strengths equal to 690 MPa and 345 MPa, respectively,
for the tempered and mild steel. The load P is gradually increased from zero
until the deformation of the bar reaches a maximium value 6, = 1.016 mm
and then decreased back to zero. Determine (a) the maximum value of P, (b)
the maximum stress in the tempered-steel bars, (c) the permanent set after
the load is removed.
ov
~b& 2
Jota#:
avce
Cy = e
P
>
Py
Let
Py
>
=
Pi
2
theroture
force
Pore
te
A
mild
carvted
cavrned
=
yiebel
Pez
A,
+A,
the
z
mid
(WP
Af
6 x10?
me
steeY
AGy, 2liiPebn10°) B4rK70 ) 23854 ha
steel
yields.
by varded
by
A, = 2.0004 16)(Ow $08) =AP 36X10 bn?
steed
Tempeved
Pan
= 63SKIo
=(0+012$)(0-.0f08)
A,
Midd steel
steel
tempeved
steel
P= AG, = (65K16%) (24610) =2IG07S N
P+R=zP
~ PAL
(OS™
wW
6:
Un Doadi
:
(c)
Sp
© (26425) (0359)
=
grag
yiatm
= oe
5
ER, “Goonred (4PEcbeiae) 7 OF MEMO mn = OFF tam
—_
42s)
|
&A, {Clb
=
449.562
463-bx10°
=
S'
"J
B= PAR = 46000-19076 216495 N
'
PL
EA
=:
MPa,
(436000 ) (6-358)
Senne
(Deoxio' ) CINB4 x10")
S.- §' , = O146 -0-692
ees:
= O6642x10 M = O 642 am
= 6-lo4mn
2O"7
wn
~
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
.
Problem
2.1 13
2,113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 x
6-mm rectangular cross section and made of a mild steel that is assumed to be
~ elastoplasti¢ With E ='200 GPa and oy = 250 MPa. Themagnitude of the force |
Q applied at B is graduafly increased from zero to 260 KN. Knowing that a =
0.640 m, determine (a) the value of the normal stress in cach link, (6) the
maximum deflection of point B.
0.640(Q-P,)-
2 0%
2M.
Statics:
A = (B285)(6)= 225 mm
EA
20
$47 2.640, Sp a= 0.640 9
Deformation?
Ps “s
2.64P,,
S. =
= 225° «10% m*
2e2 107 Maas xo. 5,226.47 vJof 5,
aD
2 ce epee Vacs 2)
—
Sas =
ibe
-
Poe =
én.
=
310.6 xio"
69.88&x/0*
6
6
(200 10TYR2E 10" Is,
=
Lae 58
1.0
¢
= 45x10"
Se
= (45»%)0° \(o.6490 8) = 28.30 x/0° 6
Qos
Fron,
Statics
,
Qe:
Pee
+
2.6%
0.646
640
7
Fe
Pao
2
>
123
Pas
+
x jo?
8
4.12
Pap
= [ago xj0%s (4.125\(64.88-108)
18 = 317.06%/0°
Oped yielding of Pak Ad
Sap > Gye 250m’ = 310.6 4/07
O, = 804.89 vio ~®
|
Qy = (317.06 10% \( 204.8910*) =
fe)
Since
Qs
Piz
From Statics,
260 xjo°
AS)
Qe,
Pak
255.2103
AD
yiedds
= (225 «10° asoxio*®)
=
N
Gap = 250
MPR
me
56.25*1O°NM
Pag? Qu 4125 Pry = 260 x10 -(4.125
(56.25 10")
Pye = 27.97 110°N
wo)
>
|
Ge = GE = ARSTO”
Les e. (2297x108 4 Uo)
Syx Peceece
© (200 110 7)(225 ¥1o*,
ER
14.3 vl
3
2
621., 53 X10
“6
124.3%
Pe
MPa
~<a
Problem 2.114.
- 2.114 Solve Prob, 2.113, knowing that a = 1.76 m and that the magnitude of the
force Q applied at is gradually increased from zero to 135 KN.
,
"2,113
The rigid bar ABC is supported
by two links,
AD and BE, of uniform 37.5 x |
-
6mm rectangular cross section and made of a mild steel that is assumed to be
elastoplastic with E = 200 GPa and oy = 250 MPa. The magnitude of the force
Q applied at B is gradually increased from zero to 260 KN. Knowing that a =
: 0.640 m, determine (a) the value of the normal stress in each link, (5) the
maximum deflection
of point B.
Q
Stohes:
FM, 20:
Co
EPastre
1.76(Q~R,)-2.64 PR, =0
Sg= 1.78 ©
S,= 2640,
Oshormetion?
ACy
Por
Po
-
Anady
$183
A = (37.5)(6) = 225 mm” = 225¥j07* m*
Va
8
ne
p
.=EA
AD
& | Se
La
SF
>
~e%
goose Vas x10")
5, = 26.47 x10" Sy
.
=(26.47x1O® }(2.646) =
G6F.88 xfo*
Gio = Gee = 3/0.6x]0" 6
Poe
EA
lee oa
=
=
Frou
10?
>
~«
= La90 wo" Maas 10")
79.2«10°
Stetics ,
6
5g
Gee
Qz
Part
2g)
45 x10" Se
*
no
os
=)
=
6a
+
= (4Sxl0c (1.76 8)
35Z%/0"
1.500
6
Pao
= [73.8 «108 +l. Soo Yo7.28x108)|O = 178.62 j0% 6
ey
at
yiededing
Oy=
od}
Dink
BE
Q*
135%
Pee = A GY
From
States
Sap
From
=
Gy
=
250
¥jO~
=
B52xJ}o'
2,
710.23 «107 &
Qy = (178.62«10%
Since
One
elastic
Serv
io®N
>
Qy.
=
dink
Pro
*
ap (Q
Pao
. S25 x10%
RBS?
amed y s's
176
of
@
AD,
=
126,86%/07
BE
= (225
x10 (aso xjo%)
;
>
\(7110.23%10°)
~
yiedds
=
Ope
=
r ate
1. 322 *10% m
.
=
> 250
MP,
=
N
52.§xlo®*
233.3xl0 ¢
8
=, 6,
56.25 x}0o*
Pa.)
|
N
N
233
.
781,
P.
MPa
29x(o
—
-§%
~ rad
Sez 1.322 mm
1=
*2.115 Solve Prob. 2.113, assuming that the magnitude of the force Q applied at
B is gradually increased from zero to 260 KN and then decreased back to zero.
-Knowing that.a =-0.640 m, determine
(qa) the residual stress in-each link, (6) the-|final deflection of point B. Assume that the links are braced so that they can
carry compressive forces without buckling.
Problem 2.115
to
ee eset no
2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 x
6-mm rectangular cross section and made of a mild steel that is assumed to be
elastoplastic with £ = 200 GPa and oy = 250 MPa. The magnitude of the force
Q applied at B is gradually increased from zero to 260 KN. Knowing that ¢ =
0.640 m, determine (a) the value of the normal stress in each link, (6) the
maximum deflection of point 2B.
2.64 m —-———
See
and
sodution to
the
Frobfem
det fection
Sa
The
7
for
pomt
Bo
norma’
Gee =
stresses
anadtysis
given
317.06/0*
Q
"
0!
Pink
124.3 xlo* Pa
in
the
sodution to
PROBLEM
2,
Lt
wep dres
|
317.06 «10°
8
_
260 x10*
~
317.06 «lo*®
'
123 «10"
ce
0.640
Res! ‘dod
-
6
820,03 */0°
= (310.6 %107\(820, 03
%/0°) = R254. 7OxIO* Pa
@ = (123x10%)(220,08«i0°) =
i}
@'
Opp, res = Gay- 6!
524.82 x1~ wm
250 “1Oe ~ 254.Jo lO”
= ~4.70«10% Pa
=
Goe,ves® Soe~ Soe
=
= Sg- Se F
i
Sep
104.9¢%10° Pa
stresses
4
YY
e
oo
m
Q
n
Ban 2 310.6 «107 B
(b)
in each
doadineg
un boudinag
Q’=
fa)
the
after
G2I.5S3 «lo °m
edastre
the
2.113
2SO x10° Pa
Sap
+0
of
124.3 10%
104.96 "0"
wm
MPa
=
19.34% 10° PR
=
19.34
E21.S3*1O'- 524.82 «10%
FO_71 ¥ (0
~4.70
=
0.0967
MPa
my
~<a
A
<a
Problem 2.116
-
-
2.116
A uniform steel cod of cross-sectional area A is attached to rigid
supports and is unstressed at a temperature of 7°C. The steel is assumed
“to be elastoplastic
with oy = 250 MPa and E = 200GPa: Knowing that:
« = 11.7 X 10°°/°C, determine the stress in the bar (a) when the temperature is raised to 160°C, (>) after the femperature has returned to 7°C.
lL
~ Te
Dehler mine
|
let
tha Compressive
to
_= _&L
‘=
LoafAT)
cause
*
250% lo!
Ea
@)
be
tenpevetore change
LPLAge+
9
(AT), = Gy
But
P
© (200x107)
Ck 7x10)
AT = f60~7
=
IS3°c
>
yiehding
(ATY
=
im
+he Vod
.
.
LataT)y, .= 6
= [06-8 °C
(AT)
6
Viel ding occuvs
Cooking z
Pore
-6y = 250mq <d
152°C.
S's S.4 si = -Fb+ Laan’ =o
6'=
E
=
— EalAT)'=
= = (200 x1oA)CIE7K15°) (162) = — 35é
(b)
Residuef
MPa.
stress
Greg © Gy - 6
= KWOK
H BSF KI0® HIP MPG
losmp,
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<_<
~
Problem 2.117
oe
2.117 The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of 25°C.
The steel is assumed elastoplastic, with E = 200 GPa and
Ax 300 mm?
A = 500 ma
°C.
Knowing that ¢ = 11.7 x 10%/°C, determine (a) the stress in both portions
of the rod, (4) the deflection of point C.
C
A,= 500%|0™ wm”
Lae F O.1SOm
A. 300n10 m*
leg 7 0-250
Constvaint
>
8
c
7
P
Pe
_
Po
FY
=
Ag OY
Py
Lae
Liew
or
O
yied#dinag
Lie
Les
Ake.
+
Ace
= (00x 1078 (2 $0 x10") =
75*10°N
\
* 1a
7 \ S0Ox1O-® = B00xJa°s | ~
x 107%)
~ (0.400)(200x107)(11.
150°C - 25°C
AT:
For AT > QT),
@)
(b)
=
2s°Q
>
Yielding
(AT),
ve
) = 90.8
0.250
4
0.150
75 x10o>
Actual
cB.
portion
it
Lae CAT)
.
E Age
(
=
eause
Plea
Lag Fol
OM = Tea (8 * Bes |
~
~
E Ase
+
to
AT
~Lise
FP
.
yielding,
De
Detevmne
ke
AT
At+
*
DCE OYS,
P= R = 75*l0 N
Sper
- te
DOO KIOME s _isoniotP “
Me 2 _ 2x10?
At
MPR
Bae F-ISOco vir
=e
Ge
Org
<a
a
For
=
_
AT
Ro.
hen
(AT )y
>
+
» eae
S
C/A
-6,
,
Yr
powtion
vemarns
AC
7-250
MPa
ebastic.
Lye A(AT)
Ac.
to
Since
-
CISxlo® \C0.150)
(aso xe
point
Scone)
A is stationary,
+
(0.180 iO. )Ciaxio*
S. 7
Seat
a6
as) ==
«4106, %I0
106.9
a6
mw
[06.9%10° m
S$.7 OL06% mm >
«ll
Problem
.
ae SAM) ant
2.118"
/-¥2,418 Solve Prob, 2.117, assuming that the temperature of the rod is raised to
150 °C and then returned to 25°C.
,
A= 300 mm?
2.2.417 The steel rod ABC is attached to5 rigid supporis and is unstressed at a
temperature of 25°C.
The steel is assumed elastoplastic, with £ = 200 GPa and
oy ~ 250 MPa. The temperature of both portions of the rod is then raised to 150
°C. Knowing that @ = 11.7 x 10°/°C, determine (a) the stress in both portions
|
of the rod, (&) the deflection of point C.
Ay. S00%{O™* wn”
Lag = O.1SO m
Acs 30010 m™
Leg7 0-250m
Constvaint
A
P
c
ie
eB
t
Deteveamme
Pe
P
Sp
+
AT
-Llxe
to
A+ yieding
P=P=
Py
_
portion
CB.
(
Ane
= (00x10
+
Ace
)
Kagoxjo*) =
75«/o*N
Le
(Se
¥ Gs |
1S X10?
[ o.tso
4 02280
_
.
(6.400)(200 x1b*VII.7 ¥Jo-*) | S0OxIO~e | Booxjor* | = 79-812 °C
joc.
1
Cooking:
pi
P
(ATY =
ty__Ebas
P
Lac
Ane
Ol -
in
2 1 a(AT)
Aetual AT: 160°C- 25°C = 125°C > (AT),
= 75% N
PPR
Fr AT > (AT)
| A
yiedding
eae 5 Lew
Las Eol
AnSy
Lae
AT) Eg
7
=
O
.
E Age
2 —P
=
eguse
EBles
E Anse.
AT
or
{o7¢
+
Zoe
Oct OVS,
256°C
°
of (ATV
+
L<o)
ee
4
ooxior\(o
. ioe)(h.78 io™*
o* i )C1as ) =
S00x
Yiedding
103,235 xlo32 y
x {ot
|
Residual
(@)
force?
Pees = Pe R=
Resiclual styesses
103,435 Mo’ 75x10% =
(tension)
3
.
Sac
fis
ce
Oz *
(bi
28.235%10°N
Permanent defection
of
point
aR
Sele
Cag?
Pres.
28.285% /O*
cm “B00 xto-*
C.,
06.5
MPa
wa
MPa
a
.
Fes bane
Sed Ste
Seg? FL)
Se?
O.0424
mm >
= =e
" #2.119 Bar AB has a cross-sectional area of 1200 mm? and is made of a steel that
Problem 2.119
is assumed to be elastoplastic with E = 200 GPa and oy = 250 MPa.
Knowing
_ . thatthe. force.F increases from 0.to 520.KN and then decreases to zero, determine. .
(a) the permanent deflection of point C, (4) the residual stress in the bar.
Ac:
portion
yield
fe
Pie=
AGY
~
300
=
R,
Pac
2
*/07N
=- R20
EA
Pe
_
—
.
(2004107 )Q200 *107*)
22a0xie8
__
Unfloadina.
-
Pre Lac
a
Ps
EA
len
EA
CF
L
Pl-
—
Peo
GS
=
bee
=
378.189~(0%~S20x(0%
37B.182K/07
_
“TRe0xlo*®
*
LIF. BIBKIO™
xo
=. ~“Tg5e
2.
Ser
o.18
¢
xJo°
an
O. 293333 X10 = 0.1 BFOM KID
Gy - Gls
.
Cec, res * Org
- Seg
Fa,
90%] xjO”” mm
-
Se!
N
Ne. 1@Zx/O°¢ Pa
aS
.
= 0.1042 x10
=O.
.
1O42mm
aso xlo* - 3IS.1S2 *1OS = -65.2 10" Fa
2
H
o
Sep =
~
= -141. 818 xI1O
BIS.\S2
(378.182 )(0.120)
~ (200 x107) (1200 x10%)
e
F
N
+
7a
=
+}
Se
©
F
"
Cee
Jhew
O.44O
Pye
=
PZ
EA
> “EA
we
bac thea
Sac
©
a2 m/0° N
- (S20 103 (0.440 -O.120) _ 359
Fle
Pao =
(2)
Lac
-
|
Pe lea
* BA)
i
Ac
9.293333 ¥18* m
2
.
©
pia
N
~ |B3.333*1O°+ 118, 182 x10%
nou
$!
©
=
300x10°- S20 x13"
x jo”
Se = 7 Few bee . (220 x10" (0.440 -0.020)
~~ Pre.
Poe
E+
Ra = Pe- Fs
=
so x 10°)
= (1200 «10°
equi hi bvun,
Foy
a.
ee
<
me
~
|
440 min
Force
Qo0oxlo
=
haa
[200
“
A
120.
a=
~6S.2
MPa
-~65,2x10*%
~65.2
MPa
3%
m
at
-_
Pe
~<a
Problem 2.120
*2.120 Solve Prob. 2.119, assuming that a= 180 mm.
. . #2149 Bar AB has a cross-sectional area of 1200 mm” and is made of a steel that. |
is assumed to be elastoplastic with E = 200 GPa and oy ~ 250 MPa.
an
Knowing
that the force F increases from 0 to 520 kN and then decreases to zero, determine
(a) the permanent deflection of point C, (4) the residual stress in the bar.
ee]
a= 120 mm
patience
A
$AG ITN
[200
—_
Force
to
tam”
yiebd
= [200 » 107°
portion
Phe
°
FF
*
Pre = AG,
AC:
= i200 x10" * Caso x1 0%)
=
200x107
For aqulibeivn
cea
N
F + Pee
—~Fan
=
OP
2
Pen = Fae - Fo = B00x(0°- S20 HI”
.
Fr = 220 “10>
N
s.> ~ Papbes
. (220
n109CO.440
- 0-180) _ 4 295333 %10"° m
EA
(200 x107)( 1200 xio"*)
-
Pea
Cee
220/07
A
1200 ¥ fo">
(83.383
410" Pa.
Dh Pood ing
gis Pretec . _ Poobes . (F-Predles .
©"
EA
EA
4 = PJ-F
7 (oo 0 "\Ciaoo
oe
+.
Sea
(A)
= 207,274 x10
:
-3
oF)
0.230485
“10 ~
= B01 273 XEo-
.
Re
A
Sep =
EA
S20 » Jo’ = ~212.727*1 N
¢ _ (307.273 «10% \(0.180}
Che
p.(be + Lee) = Flos
Fhe
(S20 xlo (0. 440-0180)
r. Fbea
Be
EA
~22.727«107
“1200 * 7ore
So - Su
256.061 ¥10° Pa
|
>
~177.273xjo
@
P
0.239333 x/0%- 0,23 04SSx/O” =
0.00738 x10m
= 0.00788
(lo)
Gres = Ser Ge = 250 x10%- 256. 06/10" =
mm
<8
- 6.06x)0*
Pa
>
-6.06
MPa
wt
Gigves = Seg- Seg = 7 19%.238 xl0" + 177.273 x10 = - 6.06 xIO Fa
= - 6.06 MPa
=a
Problem
2.121
.
"2,121
For the composite bar of Prob. 2.111, determine the residual
stresses in the tempered-steel bars if Piis gradually increased from zero to 436
ENT ahd then degreased
‘back to zero.
SA
=~
2.111
Two tempered-steel bars, each 4,76 mm thick, are bonded to a
12.5-mm mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 200
GPa and with yield strengths equal to 690 MPa and 345 MPa, respectively,
for the tempered and mild steel. The load P is gradually increased from zero
until the deformation of the bar reaches a maximum value 6,, = 1.016 mm
476 a
4,76 min
;
~
2.5 mm
and then decreased back to zero. Determine (a) the maximum value of P, (b)
the maximum stress in the tempered-steel bars, (c) the permanent set after
the load is removed.
353 7mm
Areas? Midd stee? — Ay= (010125) (01050®) = 625x108 mn
Tempered
steed
Ag = [z(oo4rey(o.
on) ==4P26K 0m
Totof>
A = A +A,= tuk bxtoe mn
Totad force te yiebel the mcf
Gy, = Pe
P> Pry
Let
e
therefore
°
Force
= force
P=
+ PF
P,
=P
AL
Un Joreting
Residuc!
ACY
= LiPo xis )C24EN0) = 285.917 EN,
mild steed yiefals
camed
b
carries!
mi tel
hy
steel
tempered
steet
AS, 2 (63810 6) (34.5K10°)= DIGo7
6,
= FH
z
Pt
steed
= P-P
B
a
2 Ade.
ga
4 G266xr0"
G's
+
+8
= 426x190"
NP &xio®
-
jo
2436
HNN
LW
xjot=
217 kA.
gata|
|
=
2 £9.28
.
Moa
stresses
mild steed
jenpered steel
Bi res =
= S = 2G 5 MPg—3h4 fifa = — 44-0F MPa
Sires =
= 4G Ro] — 284:
>
£94 MPa.
<1
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by MoGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
*2,122
Problem
2.122
For the composite bar in Prob. 2.111, determine the residual
.
Stresses in the tempered-steel bars if P is gradually increased from zero until
. .. the. deformation of the bar reaches a maximum value 6,, = 10 mm and is then...
decreased back to zero,
°2.411
Two tempered-steel bars, each 4.76 mm thick, are bonded to a
12.5-mm mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 200
476mm
4.76 min
GPa and with yield strengths equal to 690 MPa and 345 MPa, respectively,
a
for the tempered and mild steel. The load P is gradually increased from zero
until the deformation of the bar reaches a maximum value 6,, = 1.016 mm
and then decreased back to zero, Determine (a) the maximum value of P, (b)
the maximum stress in the tempered-steel bars, (c) the permanent set after
the load is removed. -
For
the
mefd
S
-
LG
E
"4
Fov
the
t
YR
Stresses
2
th
P=
steel
= Oe GIDKG?
m
2,
my
A, =2,(a1004 76) 0.0 50g) 248B-G x13 In
~ (0355) (Monat )
eo K10%
Az
-6&
(010508) = 63£Ki0
A+
A,
= 27% m
6 2,
1ias& Kio im
A Sy
= (636x10 ©)(Z4sxi0%)
EAS.
Lo
~Ceexto?) AP 6n8 Co veto16)
oss
S, = ft
2
6,
=
= Fe
2
Unboading
=
:
= (00125
The mibd steeP yields. Tempered steel ‘s elastic.
a
P,
L Gye
E
A,
(oe 35S (345x108)
‘Deo Xig?.
avea:
Sr < Su < Sy
Forces
~
tempereel
Sy
Tota!
steed
i!
355 mm
Gy
27268/xlo*
A.
GS
t.
=
48 26
Po
A
= 345
= HA07S
EN
= 27E
kA
bnfa
549.4 MPa
x10
_ Agseo
ie. oxi5”
2,
443%
MPA
.
Residual stresses
Gives
= 6, - OF = 34 ~ 443.3
= —FG-2 fs
Snes
= Op - Oe
= 1292) MPG
$7264 — 44363
~~
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
Sohn
on
2.123
anata anes
a
*2.123
os
ptatecas
A natrow bar of aluminum is bonded to the side of a thick steel
shown: Initially
at Ty =
21°C, all stresses ‘are zero. Knowing.-that the -
temperature will be slowly raised to 7, and then reduced to T,, determine (a) the
highest temperature 7, that does nof result in residual stresses, (6) the temperature T, that will result in a residual stress in the aluminum equal to 400 MPa.
Assume a, = 23 X 10°9°C for the aluminum and a, = 1.7 x 1O~9°C for
the steel. Further assume that the aluminum is elastoplastic, with £ = 75 GPa and
oy = 400 MPa, (Hint: Neglect the small stresses in the plate.)
Determine
S =
Ee
temperature
+ Loa&(AT), =
f=
change
te cause
yielding
Loa (AT),
SO = ~E(a- OAT
= -S
2 412°
jo"
(Ty)= E (=bee 06) ~ CS 109)400(23-1
1-1106)
(ay
Tay =
T,+AT)p = 214472
S
=
Sel
gic
+ Lo&(AT)
ea
vestduet
G.. =
Set
stress
Cys
.
T
A
T,
TF
T
yieka
=
Lo, (AT)
La (ATY =
& - Fe
-6, =
(b)
=
yielding
After
The
= 443¢¢
=
Lou (ATY
is
=
6y - Ela-ods) (AT)
Gy
6) - Elda Ws KAT)
26,
=
E(ol,~ Os )
=
T,
+
AT
4
>
=
C15 x107) C23D+
> 96S ’e
the
in Compress jon,
944
afuminum
Klo”
le 1) C1076 )
=
han
F6S
will
=
¢
144°C
="
°C
most
di keAy
Problem
2.124
—.- eee
» 2.124
-The aluminum
rod-ABC (E = 70 GPa), -which consists of ‘two - -
cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod
DE (E = 200 GPa) of the same overall length. Determine the minimum re-
HORN
HORN
quired diameter d of the steel rod if its vertical deformation is not to exceed
the deformation of the aluminum rod under the same load and if the allowable
Stress in the steel rod is not to exceed 165 MPa.
Deformatron
PL
eh.
.
_
An E
loxre?
70 \101
1 (0-029
Steed
|
s-
EL
EA
=
Requived
+
2
as
2
avea
A
«Ss
the
jf tA
=
vod
PL Re
PL AS
AgE
+3
vod
PL,
ES
=
atone
2
Sa
450
of
$
F (4
0+ 45
lloxie?
Pavger
= {Cave-067
Vadvue
=
,
~2
= O-EB3KIO ™.
= 0+ 68-2 KiO'm,
Citoxi8) Monde (0°78)
jefxiet
)
(0-058):
(200X107) (0-683xi57)
*
a
L
*
= 6 rode
OEb67XIO
A
—6 .
Kish
3 bk 2
= 0- 667 KIn
KIO7) = 06029ma29mm
im
~~
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any fornt or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
2.125 The brass strip AB has been attached to a fixed support at A and rests on a
/ Problem 21 25
1
ene
_ rough support at B. Knowing that the coefficient of friction is 0.60 between the strip
and the support at B; determine the decrease in temperature for which slipping will
ee"
impend.
Ww
Brass strip:
E = 105GPa_
—-
Pp
|
an
<J
100 kg
a = 20x 10°
“40 mm
Fy
nm:
tt ZFysOt
N~WeO
+, ZF,
P-ywN:O
0%
N=Ww
Pe
pw
~-foe 2b uur).
S =reg
Datat
AT— =
0,60
m=
100
AU= EAL ~ ERO
A=
ke
= yma
eh
= 4"
+ baléT) = ©
we
GOV)
= COmm
2.126
60x10 w*
Py
47°C3
(los *lOT¥60w1o*}(2Ox lo" )
2.126
=
Fo = jog«!o?
m/s"
q = 4.91
(9.6°\Cl00\(
=
_ 9.81)
Problem
20mm
a
Two solid cylindrical rods are joined at B and loaded as shown.
Rod AB is made of steel (E = 200 GPa), and rod BC of brass (E = 105 GPa).
Determine (a) the total deformation of the composite rod ABC, (b) the deflec-
7
tion of point B.
ome
Portion ABl
Pygs (TB RN
Ane = Zd*
7
Sas
=
Las= [ia
2h Gor) 0-001963 m*
Prot AS =
“se Eno Ane
~3
Sack
Lee =o78m., d= 15mm
Pa. = ~O8 EN
Pelee
Ee
@)
B=
(6)
SS,
Age.
7O0°45m Kio M
(2.00 X107) (0-019 63)
Nace Hat=H Gos) = 61004442 m?
bpm 78 kN
E,, 2200 GPa,
(itexte 3 JU)
Lom
Portion BOL
_
E,. = eS GPG
= -0-142y152 Mm
C-B8X10) (0-75)
(toSxto') (o-
a=Somm.
a6 442.)
5.4 See OFS - 0-142 = OrZOPR mm
$= 630fmm
<a
=
Sgr
Ot (42mm
—e
- Se
|
Problem
2.127
.
conti
-
be livin tae,
2.427
Link BD is made of brass (E = 105 GPa) and has a cross-sectional
«area of 250 mm’. Link CE is made of aluminum (2 = 72 GPa) and has a cross_ Sectional area. of 450 .mm?’.. Determine the maximum force. P. that can be...
applied vertically at point A if the deflection of A is not to exceed 0.35 min.
Use
as
member ABC
a&Tree body
cl
150m
225 mm -—-
»
ZM.
>
o,
»
ZMg=0
350P—22¢5
Fes
=
6
>
Fen
=
1.5586
P
~
By Aeo
So = Se = Feber _
Ese Ace
Cue SE5E P) (02225)
C10S x101) (250 Kio" 6)
(72x10) (450x109)
From
*he
Shope
8
Sa
Toc 641 &~ iol
dispfacement
p = 210035
Zweite)
Pimit
-4
Py
PT
Atag Yaw
1e- 40S 6x06"7 9
Be
P
Pog 8
:
|
-4
13-3334 K54 P + (0-125) (Josbaxto JP
=22-(1x04
Apply
* ,
Ten
+
Fig 7 0. S856.P
2687 22x10 |
Set&
.
Sa
oO,
cleformation
st
ham
"
Di a.
rs
Deformation
=
13-3334%xiw
_C0+S886)(0-15) ~
2.
s.
zzSE.
4
Loco
(2$p
s}
WY
Fa
R
5
wa
of
[25 mm
G, = 0+ 6063S
6.799
P
= 2217 Kio P
EN. = +79 EN
~
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without pertuission..
-Problem 2.128
we BB MME
~
25 mm
2.128 A 250-mm-long aluminum tube (£ = 70 GPa) of 36-mm outer diameter and
BE REN
28-mm inner diameter may be closed at both ends by means of single-threaded
SOreweon covers
of 1,5-mm pitch... With one cover.screwed on tight,
a solid brass
rod (E = 105 GPa) of 25-mm diameter is placed inside the tube and the second
cover is screwed on. Since the red is slightly longer than the tube, it is observed
that the cover must be forced against the rod by rotating it one-quarter of a turn
before it can be tightly closed. Determine (a) the average normal stress in the
tube and in the rod, (6) the deformations of the tube and of the rod.
ol
|
250 BBE erry
Dave = Ede - ad) = Flac"
= 28") = YOR 12 wwe = HORA HIT m*
Ant = Ba = EQS
.
Shube
Shoe
PL
~
ExaeAgte
a
PL
(0.250)
(Foxio?
=
P
40212
Jo" ®)
=
“jo?
3.83
Co. 280)
8.8315 x]O'P
P
or
Srod
98 +
=
1S mm =
+
0.3878 me =
4.8505x10°P
=
JO" "
=
(3.8815 + 4.8505) (io)
P
S*
375
xj0
2
0.376- «io?
=
P
B7S% 107° mm
— Sra 7
Sane
15
=—4,3505«
Eysd Ave = (105 10° (490,87 x10°*)
= (fF ten)
Spote
= 490,87 mm™ = 490.87 %10° me
27,308 «10
-S
N
3
@)
IS, > 629 +10* Pa = 6704 MPa
Stehe* Tha = ~fB0R
__P_
(bY
Croc
.
Spe
=
Ayod
_ _ 27.308xJo°
~
4490, 37 xto*
.
og
,
-SSLERIO"
(8. BBIS xO "YC 27BOBMIS) =
8
Pa
55.6
MPa.
om = O.242T mm
242.510
<a}
te
Spat =~ C4. 8808
» 15°? \(Q7, 308 KIO) 5-182. S¥ 10" m =-0.1325 mn
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
2.129
,
2.129 The uniform wire ABC, of unstretched length 2/, is attached to the supports
shown and a vertical load P is applied at the midpoint B. Denoting by 4 the
cross-sectional area of the wire and by £ the modulus of elasticity, show that, for
“'O<< E the deflection ‘at the midpoint Bis
7.
Salt
VAE
Use reprenintion,
sin
a
P
Stoties
Pas
Efongation .
Sua =
O
a
#2 F,
t+an
2
OO;
= Pu
260
PoP
.
AE
Dethection
6
‘a?
2.
LZ
2 Pas
g¢
snO-
P=0
PL
25
Ps*
ZAES
|
From
the
(P+
Sag)
3°
A
wight
=
+
trianade
fe4
28
Sap
= 228 yg (14
we
Ss
+
San’
£8)
- 4?
#20
Bg
PLe
AEBS
o
a3,
*
PR.
Ke
YP
* §* LVEF
~
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution-to teachers.
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission,
—
Problem 2.130
7
2.130
The rigid bar AD is supported by two steel wires of 1.5 mm
diameter (E = 200 GPa) and a pin and bracket at D. Knowing that the wires
"were initially taut, determine (a) the additional tension in each wire when a
900 N load P is applied at D, (b) the corresponding deflection of point D.
ket
9
be
the
vote:
Ton
of
hav
ABCD
”
Lao mm
Then
300mm — 300 mn
Se
~
+O
= 9.60
|
= (260 xto") F(o-veiS ¥' (0°36)
STE
4241;56
Se = ER
Pee
ber
.
2)
= EALeSe c - (200 x9
=
D2ZM,
Se
Lee
~
po cr
body
030
AE
EA Se
=
Using Free
+
Par Lee
p:
BE
D
Og
47/234
ate (06)
Elo00i8)
0 ae
6.
ABCD
ob fag
tO-6Re
-0:4P
= 0]
(0:3) (424 NE O) + (016) 471239
6) — (0:9) (Joe) = ©
409978
@ = Blo
@ = 1I7bKIO?
(a)
ti
Pye = (42#US)(UITEKIoW3) = 938 N.
Pee =
We)
vad
C4234)
$5 04
CLATEXKI07?)
ot
= TBIN.
@ = (0:4) (HITEKI*) = TB WG?
ota,
=.
-_—
17F
mm
Proprictary Material. © 2009 The McGraw-H&l Companics, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission..
-
Problem
2.131 The concrete post (F, = 25 GPa and a, = 9.9 x 10°/°C) is reinforced with six
2.131
steel bars, cach of 22-mm diameter (E, = 200 GPa and a, = 11.7 x 10°%°C),
Determine the normal stresses induced in the steel and in the concrete. by.a
. temperature rise of 35°C.
Ag=
6 Fd = 6 E(aa\*=
Ac =
240"- Ag = 240°- 2.2808K/0° = SS.32 x1
Sh
A
Matching
:
For
equifibrium
Se
(ZA
Force
&
deve Kypea!
BR
t+he
six stee? rods equals Pe.
in the
ae
+ utar)
~ EK. )®
= (o&% -&. )(AT
a
)
!
XSS5.3axto3 ) * ( 200.*{OT)(2 2208x110")
Fe
= 59.32 1073 m*
im Khe conevete,
tote? foree.,
zero
mm
fe 4 a(aT) = - aa mG (AT
}
| anor
with
-Feoacary 5
Es7
7
if
Straing
P. = teasife farce
compressive
bao mm
240 m7
Let
2. 2B0BxIO* mm = A. 2808 #16° me
21.61 x«x}Jo*
ne wee
a
( Re
(axon
a
as)
N
=
0.371 ¥10S Pa
=
0.391 MPa
tl
,
|
eg uayio’
Pa
=
-9.49 MPa
we
A,
5
sy
ra
ANGlKIOa
~ 2.2808 x fo"?
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may he displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
-
2.132
Problem 2.132
vo ae ii eee
P
“as
Avibration isolation unit consists of two blocks of hard rubber with
a modulus of rigidity G = 19 MPa bonded to a plate AB and to rigid supports
shown. Denotitig byP the magnitude of the force applied
to the plate and’ |
by & the corresponding deflection, determine the effective spring constant,
k = P/6, of the system.
7 om
na
stvain
Shearing
stress
Force
4 Pre
Shea
>,
150 mm
| :
30 mm .
Ps
7 =
|
2.
6
E=
=
SS
Az
aGAS
oo
Effective
spring
constant
.- BR. 2GA
2
Data:
Ge
i
/9uh-
A = (0/8 )(o01) 2 0101Smw
h = Bemm
y= BL Iurat (0018) - 19.4 /
ata
eroy
2.133 Knowing that 6. = 120 MPa, determine the maximum allowable value of the
centric axial load P.
Problem 2.133
15 nwa
At
hoJe Az
d=
20 mm
v= dre)= [Ome
D-2r= ]00-
20
= SO mme
1ROO me = 120% 107? no
Aye = dt = B0)CIS) =
100 mm
CL 1
a7
go
From
KP
P=
Commas = “Bat
A+
Ant
hole
Bt
v=
p - Cisoxlo*
K
£ls0)= 25mm
= (S0)UIS) =
“if
Aret Swe. . (220
-
750 wm
MURox1os)
‘
d=
7
ORS
Fig. 264a, K= 2.65
x1073
»!0°*)(120 x108)é
2.65
=
5S4,3x/0° NM
100-50= SO muy
= S015 om,
4.7 «102
Y= és
= 0.80,
K=
N
2.16
2.16
APPow wbDe
Value
otf P
is the
smatber,
T=
4).7x/O'N
2
UlL7
kN
<«<@
|
Problem
2.134
2.134 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 2950 mm’.
Portion AC is made of a mild steel ‘with E =
200. GPa and oy= 250 MPa, and portion CB is made of a high-strength steel with
“B= 200 GPa and oy= 345 MPa.
A load P is applied at Cas shown.
Assuming
both steels to be elastoplastic, determine (a) the maximum deflection of C if P is
gradually increased from zero to 1625 KN and then reduced back to zero, (#) the
permanent deflection of C.
DispPecement
ot C
Fe
=.
EA
Se
--
Ht
.
Sie
Qoonior — = Ooo
IQ
im
Fret
A Oye = (225 On10° QQ50
x 10°)
At.
«10% @X248.9
equidbriom
Since
portion
‘ET
=
f 00
of Ac
0. B20)(4S 0x10")
x10
.s
787.5 lot N
° «1
Yo. Hoo.
>)
~ -737.8 10°
N
0. 320
For
lf eo +P
yielding
OF se
fovce
4
Lea
°
cause
Lie
Per= bre Byte
Covrespendina
Pr
to
of
ebement
C,
1475 *108N
R~ Rye
8 Re
a7 Cg RI=O
Pz i628 «10° N > 1425/0 W,
applied Dead
Ag
at
yiefes.
Fog = Fae — P = 737,5™I0'-1625%*10°N-= - 887.5 * 10° M
5.
4
.
O.43/38ExjO
|
-320)
* 7)
(2950(0 x10)
_ Fea
o 7.5
xto10
EA leo _(o(88
Mm
,
*.
O.481
if
(a)
mm
«4
Ore7 Syre = 250 MPa
Mawimum Stresses.
887.5%103
See =- Fee
“R* _F 2 Feed
wore _7% 7_ 300.81™) O° Pa,= --30|
301 MP MPa
(b)
Delfection
ro
Poel
ee ee
aud
Forces
for
un toadking
’
P= j625xlo° =
.
‘
Bl- Py
.50
~ (200x107) (29
x107©
>
Lx
= 2P%
20)
.8
«108 (0.3
1. (8£2
é
.
=O.
4406 8x
Pye= 872.5 "10" N
/o*
hm
Sp = Sun 8! = 0.481396 (0° = O.4406BxIS* = O,0H068 10m
Permanent
det fection
ot
C
Sp
=
D.OYO7T;+mm
)
mm
|
Problem
2.135
2.135 The uniform rod BC has 4 cross-sectional area A and is made of a mild steel
which can be assumed to. be elastoplastic with a modulus of elasticity E and a
~ yield ‘strength, Using ‘the’ block-and-spring ‘system shown; it-is desired to
simulate the deflection of end C of the rod as the axial force P is gradually applied
and removed, that is, the deflection of points C and C ‘should be the same for all
values of P. Denoting by x the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m of the block,
(b) the required constant £ of the spring.
-
Fovee
-
def
Dont
Vaoes
diagran
s,
mo
pe
Cc
oft
You!
BC.
P
Pe
A Gy
=
Py
<
P
For
Pb
5
F<
peiat
4
P
ma
Far
cA
at
Pe
EA
Prone= Py
_
= AGY
N
Force- det loction
SPring
4P EFL $0:
KEE
TF
=O:
tock
He
LF
dees
Se
N=
twittal
Force
=
P-F
not
0
N=
=O
move,
diagram
ne
is
peint C' of
bhek Cann ob
ma
Pe
op
<
&N =
Fp
fm".
or
P<
wma
P= k§,!
then
P
for
em,
Eo
P = Ame,
TE +a
mg
Sys
s Pip
the
at
P=
removed ,
F,, = fama
the
occurs,
spring
returns
te rts
Pr
Fyne,
Ae
beayth.
P
+
MAT
(2)
Eqvating
od
Mes age
”~
a,
(b)
Equating
.
slopes,
m2
LOX
5
PROBLEM 2.Ci
elementn
_
"Element |
/
2.C1 A rod consisting of n elements, each of which is homogeneous and of
uniform cross section, is subjected to the loading shown. The length of element |
Ss«Ei8 denoted by Z,, its cross-sectional area by A,, modulus of elasticity by E), and
the load applied to its right end by P,, the magnitude P, of this load being
assumed to be positive if P, is directed to the right and negative otherwise. (a)
Write a computer program that can be used to determine the average normal
stress
in each
element,
the
deformation
of each
element,
arid
the total
deformation of the rod. () Use this program to solve Probs. 2.20 and 2.126.
SOLUTION
FOR
EACH
Li
J
ELEMENT,
AG,
COMPUTE
ENTER
E;
DEFORMATION
UPPATE
AXIAL LOAD
COMPUTE
FOR
P=P+P;
FACH
ELEMENT
T= PSA
5; = PLS A; Ej
TOTAL
DEFORMATION:
UPDATE
THROUGH
j=
PROGRAM
Problem
Element
1
2
Total
Problem
Element
1
2
Total
ph ELEMENTS
é + b;
OUTPUT
2.20
Stress
(MPa)
19,0986
-12.7324
Deformation
=
Deformation
(mm)
- 1091
-.0909
.0182 mm
2.126
Stress
(MPa)
90.6775
~19.9095
Deformation
Deformation
(mm)
0.4534
0.1422
=
0.3112
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All! rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Element 1
SOLUTION
WE
AND
RELEASE
COMPUTE
FoR
REACTION
THE
CONSIDER
THE
ROD
bp, WITH
FACH
UPDATE
AXIAL
p=
AT
AT
B REOUNDANT
B
Rpg =O
ELEMENT,
ENTER
LaADd
PrP;
PSA:
qT, =
6;
ELEMENT
FACH
FOR
COMPUTE
ti
-.Blement n
2.C2 Rod AB is horizontal with both ends fixed; it consists of n elements, each
of which is homogeneous and of uniform cross section, and is subjected to the
”Ioading shown. The length of elementj is denoted by L;, its cross-sectional area)
by A,, its modulus of elasticity by E), and the load applied to its right end by
P,, the magnitude P; of this load being assumed to be positive if P; is directed to
the right and negative otherwise. (Note that P; = 0.) (a) Write a computer
program which can be used to determine the reactions at A and B, the average
normal stress in each element, and the deformation of each element. (b) Use
this program to solve Probs. 2.41 and 2.42.
.
PLif
H
PROBLEM 2.C2
AYE;
DEFORMATION
TOTAL
UPDATE
5, 7 55
+ §;
COMPUTE
fp
DUE
TO UNIT
LEAD
AT
B
unit, = If Re
= L;/Ac Ec
ULMIT by
UPDATE
TOTAL
UNIT
UNIT J, + UNIT
fg =
UNIT
DEFORMATION
Se
SUPER POSITION
FOR
TOTAL
bp
+
DISPLACEMENT
Eg
ir
Ds
=O
Sp/ UN'T
dp
AT
B=
ZEPo
SOLVING:
Rae
THEN:
Ra
x
=P;
+
Ke
CONTINUED
PROBLEM 2.C2 CONTINUED
FOR
EACH.
PROGRA
i
4g
v.
Rpg unit
Gp
dj + Rg unit
a
M
+
OUTPUT
Problem
2.41
RA =
-62.809
RB =
-37.191
Element
Stress
1
2
3
4
kN
kN
(MPa)
~52.615
3.974
2.235
49.982
Problem
2.42
RA =
-45.479
RB =
-54.521
Element
Stress
1
2
3
4
cee
W.
©
ELEMENT
-77.131
-20.542
-11.555
36.191
kN
kN
(MPa)
Deformation
(mm}
~.05011
-00378
-00134
-04498
Deformation
(mm)
~.03857
-, 01027
-,01321
-06204
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM 2.C3
Element a” “Element 1)
Fig. P2.C3
2.C3 Rod AB consists of n elements, each of which is homogeneous and of
uniform cross section. End 4 is fixed, while initially there is a gap 5) between
end B and. the fixed .vertical surface on the right. The length of element.j is _ |. .
denoted by L,, its cross-sectional area by A,, its modulus of elasticity by £;, and
its coefficient of thermal expansion by aj. After the temperature of the rod has
been increased by AT, the gap at B is closed and the vertical surfaces exert
equal and opposite forces on the rod. (a) Write a computer program which can
be used to determine the magnitude of the reactions at A and B, the normal
stress in each element, and the deformation of each element. (6) Use this
program to solve Probs, 2.51, 2.59, and 2.60.
SOLUTION
WE
COMPUTE
ASSUMING
THE
THERE
ENTER
ENTER
NO
SUPPORT
AT B
AT
Li) Ap, Fi, %e
TEMPERATURE CHANGE
COMPUTE
&;
DISPLACEMENTS
tS
FoR
=
Oo
UPDATE
EFACH
L;
B:
T
PLEMENT
T
TOTAL
DEFORMATION
bp = det bj
COMPUTE
dp DUE TO UNIT
LOAD
AT
B
UNIT & = Li/AG Ez
UPDATE
TOTAL
UNIT
COMPUTE
DEFORMATION
Sp = UNIT §, + UNIT
§-
REACTIONS
FROM
SUPERP D>ITION
Re = (4, - §,
FOR
UNIT
EACH
unit Sp
FLEMENT
72
- Re/Ac
8; =
obi
T+
Ra bi/ mE;
CONTINUED
PROBLEM
2.C3 CONTINUED
PROGRAM
Problem
R=
OUTPUT
2.51
125.628
Element
Stress
1
2
Problem
R=
Element
Element
1
2
(MPa)
~44.432
~99,972
2.59
. 172.835
(MPa)
~115.223
~96.019
2.60
232.390
Deform.
(microm)
-500
-.500
KN
Stress
1
2
Problem
R =
kN
Deform.
(mm)
0.236
0.24
:
kN
Stress
(MPa)
~116,195
~290.487
Deform.
(microm)
363.220
136.7806
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manuai is using it without permission.
2.C4 Bar AB has a length L and is made of two different materials of given
cross-sectional area, modulus of elasticity, and yield. strength. The bar is
PROBLEM 2.C4
“subjected
as shown to'a load P which is gradually increased from zero until the ~-"]
deformation of the bar has reached a maximum value é,,
and then
decreased
values of
120% of
determine
back to zero. (a) Write a computer program which, for each of
5,, equally spaced over a range extending from 0 to a value equal
the deformation causing both materials to yield, can be used
the maximum value P,, of the load, the maximum normal stress
25
to
to
in
each material, the permanent deformation 4, of the bar, and the residual stress
in each material. (5) Use this program to solve Probs. 2.111, and 2.112.
\
‘A a, Ka. ya
Plate
SOLUTION
THE FOLLOWING
NOTE:
ASSUMES (9) < (t,),
INCREMENT
DISPLACEMENT
Sy = 0.05 (Ty), L/Es
m
DISPLACEMENTS
AT
YIELOING
§, = O),L/Er
FOR
EACH
IF
5, = (7), L/Fe
DISPLACEMENT
dy, < &:
g ; 2 Sm e/t
%
Fy Eft
iS
Pon
(3n/L)(AE,
+A, &,
}
IF 5, < &m< SBi
g, = (T,),
J, = bm
F,/L
PAT
IF
+ (5n/t) AE
dm > Oe?
T;
= (),
Pm=
OF
FIFST
(ELASTIC)
SEGMENT
SLOPE = (AE, + Ax Ez V/L
(Tres
Sm — (Py / SLOPE )
i\
Sp2
A, T, + AST,
DEFORMATIONS ) RESIDUAL STRESSES
PERMANENT
SLOPE
Oo =(%)2
2% - (E, Pm /(L sore))
(7 ye = Ge - Ea? /(t store ))
it
Fin
CONTINUED
PROBLEM 2.4 CONTINUED
Problems
DM
nm
PM
KN
oO
oO
2.109
and
|
-
2.116
SIGMtiH)
MPa,
= 314M Ct)
Mp
DP
nt
oO
O
oO
Si@RU)
Mfa
O
O
= StGRC2)
MPa,
o
O°368
233571
207,006
207 G00
oO
oO
09796
436.000
248000
GGRCE2
0-104
-4G092
SFP O
€ Palio
[.ofb
498-82Rs
34Si000,
S71: 398
01229
-9F.60
128838
4 P2109
had
ESb000
348.086
Joo+soo
OSG
-Woe2gp
[FF 305°
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, Alt rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it withoul permission.
-
PROBLEM 2.C5
ce
at
eves
205
vine ata te
_ The plate has a hole centered Across the width. The stress con-
centration factor for a flat bar under axial loading with
a centric hole is:
P
>
K = 3.00
K
°
— 3.13)
2r
— ] + 3.66)
(*)
Qy 2
—- fF —
3.667")
1.53
dw
—
3
1s3(~)
where r is the radius of the hole and D is the width of the bar. Write a computer program to determine the allowable load P for the given values of r, D,
the thickness ¢ of the bar, and the allowable stress o,, of the material. Knowing that r = 6 mm, D = 75 mm and o,, = 110 MPa, determine the allowable
load P for values of 7 from 3 mm to 18 mm, using 3 mm
increments.
SOLUTION
ENTER
ry Dy ty oy
COMPUTE
K
RD = 2.0
r/D
kK = 3.00 -3.13 RD + 3.66 RD -1.53 RD”
COMPUTE
Tove =
AVERAGE
STRESS
Ty14/K
LOAD
ALLOWABLE
P ov = oy. (D-2.0r)t
PROGRAM
OUTPUT
Radius
Allowable
(im)
(kN)
3
6
9
12
L5
18
16.405
16.023
15.368
14.391
13.165
11.721
Load
Proprictary Material. © 2009 The McGraw-Hill Companies, Ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted
by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
2.C6 A solid truncated cone is subjected to an axial force P as shown.
PROBLEM 2.C6
exact elongation is (PL)/(22c’E).
The
By replacing the cone by n circular cylinders
_ of equal thickness, writea computer program that can be used to calculate the
elongation of the truncated cone.
What is the percentage error in the answer
obtained from the program using (a) 7 = 6, (6) n = 12, (c) 2 = 60.
SOLUTION
IK
L
.
pa
FOR
t=I1 TO
r
I
L,
+.
=
n:
(t+a5)(L/n)
r;, = 2c-c(hj/L)
AREA:
_
.
Azar,
DISPLACEMETT:
Lil
(=
EXACT
&+
p(L/n)/ (Aa €)
DISPLACEMENT:
S exact
PERCENTAGE
2
= pL/(2.07¢ E)
ERROR:
PERCENT
=
100( 9 - Senex )/ Cerner
PROGRAM
QUTPUT
n
Approximate
Bxact
Percent
0.15852
0.15899
0.15915
0.15915
0.15915
0.15915
- .40083
-,10100
- .00405
6
12
60
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Chapter 3
Problem 3.1
3.1 For the cylindrical shaft shown, determine the maximum
ng
= is
Cnay
~
I=
Ee!
= LAS
av
Tmo
~
BY.6B2x10°
00)
wlo,o2ay
Trnag
Problem 3.2
22 mm
39.1
MPa
=e
3.2 Determine the torque T that causes a maximum shearing stress of 80 MPa in
the steel cylindrical shaft shown.
¥
Tmae™
Te
J
=
tT: Eebou=
1338.1
=
Problem 3.3
shearing stress
_ eaused by a torque of magnitude T= 1.5 kN - m.
ze
£lo.022) (80x10°)
Te
N-m
«dl
1Z%.8kN-m
3.3
Knowing that the internal diameter of the hollow shaft shown is
d = 22 mm, determine the maximum shearing stress caused by a torque of
magnitude 7’ = 900 N > m.
c,=4d, =(£Y 40) 2Zzomm
Cy = kA = (Ez)= 1
Je
mm
Ee,’- 3 \s F (z0°- wu) = 228327
-
Tra
C=010Z2m
te
(Joa\(a02z)
TT
a0329x00r2
mt
7% 6 MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=
_
Problem
3.4
oo.
,
.
3.4
Knowing that d = 30 mm, determine the torqueT that causes a max-
_ imum shearing stress of 52 MPa in the hollow shaft shown.
C= te = CE )(30)= 1 mm
J=
Et (c,"- C)
Tmaxe
Pe
40 mm.
Te
Problem
= ¥ (zo'-
CG = 0:02m
|
C22 34, =(4)\(4e) = 22 mm
s*)
= 171806 mm
~~
sl Zone
ff
171
806 x10."x18" Y¢ )(S2x
180
Zklor0°
ye
4467
Nin
=|
3.5 (a) Determine the torque that can be applied to a solid shaft of 20-mm
3.5
diameter without exceeding an allowable shearing stress of 80 MPa.
(0) Solve
part a, assuming that the solid shaft has been replaced by a hollow shaft of the
same cross-sectional area and with an inner diameter equal to half of its own outer
diameter.
(2) Sofid shaft:
T=
= Llo.02c) = 0.010m
m4
Eloio)t = (S.209e%/0°7
s
Bel
T=
czsftd
ee
-%
©
(bo) HokRow shaft:
A
=
TT (c,*-
Same
¢7
)
=
whe?
araa
c, = £e,=
as
-(4¢,) ]
FZz (0.010)
n
4,
C, ~-~= aC
‘
OS. 7oRO x10" MSeXlO*) - as cey
0,0057735
_=
sofid
=
shaft
Sic,
=
O.0HS4H7O
wa™
m
m
T= Flet-o") = Elo. 01tsa70"— 0,0067735")=
Tz
Dued
Cy
nn
.
Te 125.7 Nem
(80x10%)( 26.180 107%)
O.01S 470
z
131.33 3
26.180 «1077 in!
T=
=
“yeh
131.4 Nem
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or —
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.6
_-
3.6 A torque T = 3 KN
- m is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress, (b) the shearing stress at point D
which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c} the
60mm
percent of the torque carried by the portion of the cylinder within the iS-mm
po
30 nem
radius.
c=td=
BO
min
=
BOmIGS
mm
.
T= Fot= E(goxjes*)” = \.27235 «10% m4
m\U”
T=3kN-
“D 7
T=
200 mm
ea
Z2kKN = 3x
3x10?)
¢ _ (3x
lm, = FF
.
N
(4amto®)
(3o™
= 70.736 10° Pa
2
1.27235 x 10S
On:
(e}
x
Po
> IS wen
©)
T= 3
look
MPa.
ae
~
i=
354MPQ
=f
6.25%,
<a
~ IG
F
43
= BAG
Tos “3h
To = £ (isx 107 )°(a5.368x10°) = 187.5
Te
wex
70.7
1S e107
m= -Bx Pow .=f USnio7
eee10.736 x10)
~ 5
=
~> Bylo}
1825
(lao %
y=
Nem
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyand the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
3:7
3.7
The solid spindle AB is made of a steel with an allowable shearing
- stress of 84 MPa, and sleeve CD is made of a brass with an allowable shear-
ing stress of 50 MPa, Determine (a) the largest torqueT that can be applied
at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter d, of spindle AB.
Sheeve CO:
(oa
G2
Geekdad, = ECTS) = 37-L mw,
C,-5
=
SUS
.
&
Zid bran
of
T = E(ot- of) = T(orosist~ cose") 2 1-56 x10" Sunt
f+6am
|i
1
in
i
Powe
:
Teo
Jof
=
os
For
(b)
Te
equil,
Sobral
spindle
AB:
T=
we
TO.
2T
|
ST
wes
t
3a
Qs
Vf ae
=
3
Te
d=
2e=
Problem
3.8
,
V-B@erbo)
~ cSente
~6
.
Sexe)
becom
T=
2080
=
6
= 208 OK.
2af
KAY
ey
ety,
CO2TM
= 20mnH,
WwW CS¢
i oP)
OCC)
d,= SOM=a
3.8
The solid spindle AB has a diameter d, = 38 mm and is made of a
steel with an allowable shearing stress of 84 MPa, while sleeve CD is made of
a brass with an allowable shearing stress of 50 MPa. Determine the largest
torque T that can be applied at A.
J=
160 mm
{ete
Eman
F ts.
The
~a
CQ
ae
AB:
spindde
Sobid
td,
=
[Fmm
GoS
| Nm
4(3%)
=
E@oiayts 20407
107
te ~ (2047x101)
(894109) “-_
ooq
:
200 nan
Sleeve
C=
Je
79 pun
vasoe
oF torque
T
C=
q-te
is the
aan
SVS
ECt-S")
_= The
G,
T..
AMovehde
CD:
z
Gz
= H (or0375"
-¢,,
$(75)
Z2eE
3 yeSmm,
mm,
ootE le lS
&
_2 (tSbxid
ses )ieoxa)
te
smeBben.
=
—=
T=
x00 eat
2ofoNhw
Zt
oF
kKALMH
oy
‘Problem 3.9
.
3.9 The torques shown are exerted on pulleys A and B . .Knowing that each shaft is
O
,
. Solid, determine the maximum shearing stress (a) in shaft AB, (4) in shaft BC.
(a\Shakt AB: TgsZooNm, d= O0.030m,
Toy= ES . 2T.
(20300)
tT (o.01s)
"
ames
J
d
= 0.046m.wy
C=
=
0.023
0.023
BC:
Tac= 300 + 400=
yo TST
m
MPa
<2
36.6 MPa
=f
56.6
=
= 56.588xIlo° Pa
(b)ShaFt
67 2.01Sm
700
Nem
-
2 TO3
AT - AMO)
W (0.023)3
Caerre
= B6.626«]0°
Pa
3.10 The torques shown are exerted on pulleys4 and B which are attached to solid
, Problem 3.10
circular shafts AB
and BC.
In order to reduce the total mass of the assembly,
determine the smallest diameter of shaft BC for which the largest shearing stress in the
assembly is not increased.
T, = 300N-1
Shett
Ty = 400N-n
of
0,046m, c=
* Tot”
AT (oars y3
The Dag est
Reduce
the
stress
Tec =~FJoo
00 WeNa.
COT
2T
Ue
36.626
(56.588 «10° Pa)
dianeter
of
BC
+
oe la
Tne
*
Cr IR8ISKIO mm
e.SBB
provide
TOL
._@i7eo)
=
56.6 MPa
= 2T
.
ito)
Tes
~
(6.023)
J
>
Pa
Tee = 300 +400 = 700 Nem
Tne= L&
0.23m
, OF O.01S m
. ZT _ Geo)
= 56.588 ¥/0*
Shoft BC:
a= 0.030m
Tae? 200 New
AB:
Conny 2 Te
46 nam
d=
|
.
30 min~—
Oe) 7
~1O%
oecens
the
36.6
=
Paw
MPa
in portion AB
Same
ORT
TES
% BS*IO
~e
stress, .
wm
4
= dd = 20 = 39.79 x108m
39.8 mm
-
Problem
3.11 Under normal operating conditions, the electric motor exerts a torque of 2.8
3.11
_KN- m on shaft AB. Knowing that each shaft is solid, determine the maximumé..
shearing stress in (a) shaft AB, (b) shaft BC, (c) shaft CD.
Ty = LAkN om
56 mm
48 mia
i
plow 09 KN -an
/ ne = O.51GN
48 intr
“jf
ShaPt AB:
vw
The=
. Te
‘an ~
2.8kNem
2T
_
“owe
(ob) Shaft BC?
= 2.9/0?
(AMA B«10*)
*
(0.02%)°
(Cc) Shaft CD!
Tey= O.SkNem
2T
foo ~ Fee
See
in
the
MAK] MUI
“(Gozeye
3.12
a
1S
—
to
AB,
magnitude.
Problem
BS,
B1l-20x10° Pe
MPa.
stress
Tae= B12 "10°
ar.
eeThy,
Nem,
Yaz 81.2 MPa <a
cstel= 24mm= 0.024m
oe
“
7 C447 x10 Pe.
Tact 64.5 MPa
crtes
Utmm=
23,03 xJo* PB,
=
~
~
von
3.U
and ep.
8/,2
te
28mm= O,O2RBm
a
0.024m
Toy = 23. OMPa
<eth
3.12 In order to reduce the total mass of the assembly of Prob 3.11, a new design
is being considered in which the diameter of shaft BC will be smaller. Determine
the smallest diameter of shaft BC for which the maximum value of the shearing
stress in the assembly will not beincreased.
sheaving
~
©
=a?
= O.Sx10 Nem,
(2) (0.5% 107)
~
eodrtion
shafts
Nem,
Tag? 14 WNem = YX!
(0-4 eIO7)
AL. = “foloays
wvbee .= re®
Problem
AB mm
Lov
The
Pavgest
Adjust
is
Figure
81.2
tle
ane
for
Shearing
oliameter
maximum
occurs
ot
sheow: ne
in shatt
shatt
Be
so
stresses
AB.
Its
hat
its
MPa,
Pa,
Tae = 14e)}0°
Nem
21.
Fes
3
/(2V.4
% [0%
[Foo
(21.2% Tey
lo y
il
(0)
+ a
RAAD
“10°
d=
=
m
2¢
22.22
22
>
me
4G
U ww
1g
3.13 The torques shown are exerted on pulleys 4, B, and C. Knowing that both shafts
Problem
in (a) shaft AB, (5) shaft BC.
shearing stress
“are solid, determine the maximum
3.13
me
(a) ShePt ABE. T= 400 Nem
“erztd=
1200 Nm
400 Noa
0
nm
J
£@.080)=
¢ ct
0.015 m
Te.
a
Tr
af
.
igyt |= TST x10" Fo.
Toro00)
Tone _ (C4
30 pres
ct
12
ar
=“Wes
Toe
i
>
0,020m
=
if
Nem
300
Tr
BC:
- (hb) Shaft
800)
(AX
W (0.0%0F*
63.7x¥0*
=
Pe
Crags = 63.7
Probl em 3.14
.
-
40 mm
v=
6
FE
L8m
=
W Tone
=
(6.39
1¢
™ te.
Tan
Bey
30mm
27
Tat
=
*{/Ri4oo
Vr leoxlo*)
xlo*
m=
dne™
[|
—~=a#
(a) Shaft ABZ
T= 400 Nem
Ln = CO MPa: = GOXIO* Pa.
:
AOU N ern
MPo
3.14 The shafts of the pulley assembly shown are to be redesigned. Knowing that the
allowable shearing stress in each shaft is 60 MPa, determine the smaltest allowable
diameter of (a) shaft AB, (5) shaft BC.
mon
;
1200 N -1m
MPa, -e
75.5
Crests
16.19
R07
mm
S240,
T= 800 Nem
(ob) Shaft BC?
Tue = GOMPR = Goro’ Pa
)
Geox10%)
- [BIWT (GO
cw~ Vtf 2h
wes
=
20,40 xO
m
=
20.40 mm
Oe
2c.=
4HO.Smm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this mariual is using it without permission.
~te
Problem
3.15
3s The allowable shearing stress is 100 MPa in the steel rod AB and 60 MPa in the
brass rod BC. Knowing that a torque of magnitude 7'= 900 N - mis applied at A and
neglecting the effect of stress concentrations, determine the required diameter of (a) rod
AB, (b) rod. BC.
.
Te
Cnap®
Shot ARB:
T
_ wf
’
J=
2
c
)
Cio = 100 MPa se
8/7214 60)
Teodor
a>,
_=
)
a=
His
IT. 374210
GOMPa=
So wm =
=
2c.
+3
ALA
W (G0 «10% )
1789
=
25,8
me
win
QZL2AG men
me
KIO
da. = 205 424m
Problem 3.16
ae
Coxlot &
_ 3/400)
cs
242T
TT Cane
loowlot Pe
dhe
Shaft BE!
~
<e
3.16 The allowable sheating stress is 100 MPa in the 36-mm-diameter steel rod AB
and 60 MPa in the 40-mm-diameter rod BC. Neglecting the effect of stress
concentrations, determine the largest torque that can be applied at A.
te TS,
Shaft AB?
Se Ret
T= Etc?
Taye 100 MPa = 100%/0*% Pa
a=
the =
HBC)
= IR mm
The™ % (looxio* O.01a
Shaft BCL
Tae
6OMPQ
=
=
=
AOE
9/6 N-m
GOx/0* Po
C=Zhe EGO) = 20mm = 0.020m
~
Ter Z(COmio* o.og0) 3 = 1.754 Nem
The
abboural,be torque
rs the
smobber
of "Tap tonwl Tre _-
T=
4
Nem
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distributed in any fornt or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=e
3.17
_ 3.17 The solid shaft shown is formed of a brass for which the allowable shearing
Stress is 55 MPa. Neglecting the effect of stress concentrations, determine the
_
a
"smallest diameters dag and dgc for which the allowable ashearing stress is not
exceeded
i
dan
750 mm
.
Age:
ic
.
Lo
ns
Boh
CoE
3 CIC
ZooN
c= Cer)
TSS ¥1OF
=
Mivir mom
Shedt BC!
c
- |
=
Ty.=
One
°
-400
800
4
2L.09¥10
=
Re.
:
Nem
oe
“m = 21.0yn
42.0
mm
400 Nem
C400)
Gee)
WSS
EER
\200
=
Tan
°
AB
Shed
“600 om_|
Pe
10°
tC
J
.
B
SS
5
MPa
= SS
TC ses
Ty = 1200N-m
|
_ Problem
o,
wIO
)
—_
16.667
«flO
-s
wy
-
Ie.
eT
-
YH bet
Yim Ue
dae = Re
=
B3.34mm
<t
3.18 Solve Prob. 3.17 assuming that the direction of Tc is reversed.
Problem 3.18
3.17 The solid shaft shown is formed of a brass for which the allowable shearing
stress is 55 MPa. Neglecting the effect of stress concentrations, determine the
smallest diameters dig and dgc for which the allowable shearing stress is not
exceeded.
dyn
a
7505 om]
ro
Og¢ ge:
had
sreft AB:
as
Shaft
a
J
BC?
tm the Figure to the Left
vevevsed
.
Tag
600 mn
=
| :
~
it
J200+400
ieee)
=
TSS eI)
Tact
Tne?
=
26.46
SS MPa
x
ESxI0°
=
- ie.
2
aiaed
J
Te?
1600
Nem
¥% io
m
=
26.46
Nem
_ Sok
04)
T (SE
=
16.667%10
am
=
c
TV Tae
mm
=
ee
26
=
52.9
mm
<a
|
16.67 mn
minimum
Pa
~2/ar.
Min rw uw
YOO
been
the divectton ob Te has
Note thet
oom
=
2o. = SBa.h mm
A
‘Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or -
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission
Problem
3.19
.
3.19 The
allowable stress is 50 MPa
in the brass rod AB and 25 MPa
m the
aluminum rod BC. Knowing that a torque of magnitude 7= 125 N - m is applied
_ at A, determine the required diameter of (a) rod AB, (b) rod BC.
Abeminani
To,
; Brass
:
c
Trae
T
31.69 Mm
=
316A
Problem 3.20
ise
~
+
8 (2002S
Aye=
mm
ag =
2
, Brass
Lan= 25 *10% Po.
Ten
C,
=
oh
:
C2
4
te.
C=td=
we
Ty
= 5O*10% Pa
$ (0.025)
=
tet
=
m
wt
<4
w*
=
63.4 mm
<_<.
£(C,
QT
reTe Co
G= 7.597 x10
m
(bo)
ALowabde torque .
m
et) Le
)
Nem
Tee = 132.936 Nem
c,=
?
2
¥ (2V132.536
=
g=Ee"
132.536
oO
0.0185
~ — (50 nI*)
me
(Co
O.ORS
C
T= te
0.015m
= F (o.as fPasxio® ) =
fate
tor®
.
Solid rod BO:
C,=td,
lew
mm
T
Hollow vod AB?
TI Tay
3.20 The solid rod BC has a diameter of 30 mm and is made of an aluminum for
which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an
outer diameter of 25 mm; it is made of a brass for which the allowable shearing
stress is 50 MPa. Determine (a) the largest inner diameter of rod AB for which
the factor of safety is the same for each rod, (5) the largest torque that can be
applied at A.
Alumina
Toe = Dette
aT
7 15. 15° ¥ Jo
50.3
=
Ze
31,831 x"Io*
=
c
oc “T(SOMI)
mm
>
ae
Js
=F
ABl
Cc? = GXise)
(bk) Rod BE:
C=
25.15
=
m
2s. ISxid
2
ca Kea
\
CF
:
ee
ZS%mm
d=
0.025)
2ce, =
=
3.3203 wig”
IS.1B
mm
ayom
«tl
Tay= 132.5 Mem oa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.21
3.21
A torque of magnitude T = 900 N « m is applied at D as shown,
Knowing that the allowable shearing stress is 50 MPa in each shaft, determine
the required diameter of (a) shaft AB, (b) shaft CD.
Gp
Too Nm.
loo=
ob (400)
160
C109)
=% Tre
~~
Tas 7
Ve
40 min
= OOON + tit
aa sD NM:
40
Tue = SOMPA
es
afar
(a)
c
“TV Trge
Shaft
AB:
3
CF
w (Sexo) -
;
(b) ShaFt
CD:
.
c
_ 510306
AGEN
-
2016s
wz
dhe = 20= bh2mm
.
{C407
V Fen0e
Lo,
= 9.0225 m
ys
Tr
v1):
100 mm
L2S
Aw
dep? RO
Problem
3.22
<<
45 mn,
xt
3.22
A torque of magnitude T = 900 N° m is applied at D as shown.
Knowing that the diameter of shaft AB is 60 mm and that the diameter of
shaft CD is 45 mm, determine the maximum shearing stress in (a) shaft AB,
(6) shaft CD.
Teo = TOON
r
The =
ra
=
(a) Shaft AB:
Crue?
wats
Y
re (Joo) =225aNm-
c= day =
=
¥F = 900 N >in
ZOomm
100 mm
~
"D
To
(A) Come = 5305 M%G
¢
¢
Fi (6.0
Emme
&
2 tt
= $3.65 Wha
Lean * (292250)
3)3
(b) Shaft CD:
- 40mm
3i a,
~aT
.
= we*
~
<=
220
QCD
Wh (osor2gy®
—08
mn
_eE.
sue
= So.3
ery
Mim
(b)
Tray
= BO°3 MIA 0
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation.
A student using this manual is using it without permission.
Problem
3.23
3.23 Two solid steel shafts are connected by the gears shown. A torque of
magnitude T = 900 N - m is applied to shaft AB. Knowing that the allowable
.
' shearing stress is 50 MPa and considering only stresses due to twisting, determine
the required diameter of (a) shaft AB, (4) shaft CD.
240 tes
The
=
Tv
>
895
Nem
Teo = Go The > “55 (200 = 2700 Nem
n
(a Shaft
UN.
_
3
AT
(6) Sheft CD
=
2
es
Tet?
p AAC Fee }
“TUSDx Oe
)
~_
SOxsO° Pa
2f8T
€
Wine
-3
=
“
22.5 5«1S "yy = AASS mm
A ge,? RE
.
|
\.
Tog? SO MPa=
"
Ct
Problem 3.24
AB:
<2 It.
Tr
.
Rio
=
USL)
me
mm
Tome = SO MPa
3 {2)C2700 §
TT (S0%10°)
=
32.52
“3
x]o
aan
nw,
7
FR_52R
=
Apt ROT
CSO mm
af
3.24 Shaft CD is made from a 66-mm-diameter rod and is connected to the 48mm-diameter shaft AB as shown. Considering only stresses due to twisting and
knowing that the allowable shearing stress is 60 MPa for each shaft, determine the
largest torque T that can be applied.
Cnn = G6OMPa.
240 mes
= COxI0* Pa
Shaft AB:
TT
Vnnse 7
T=
Sheff CD?
= dale = 24mm = 0.024 m
ar
=
=
Fo?
a
F (Co x} Of (0.024)
ce {eko=
= 1303
23 mm=
z (6oxf0*)(o,033) = 3337
TS The
B+
ple
Woah
,* 346 (3387)
Ie
coteus ateoi
torque
is the
=
1124
smablev
Nem
0,033m
New
New
of the
two
values.
Ts
Hat Nem = LVKN-w
ee
Problem
3.25 The two solid shafts are connected by gears
of a steel for which the allowable shearing stress is
diameters of the two shafts are, respectively, dace = 40
determine the largest torque T; that can be applied at
3.25
as shown and are made
50 MPa, Knowing the
mm and dgp = 30 mm
C.
Kime =< SOMPa
100 ram
T=
60 pin
crid=
dea = Fomm
BC:
Shaft
Loew
26 mm.
T nee? =% (SK10") (0102)
= 628 Am,
Jt.
Te =
ae
statics
Te F
4
E (oxo )rois)= 265N.m,
i
By
= BT &
15mm,
crtds
pp = 30mm
Shaft EF:
oT,
=
~~
26S)
=
44/-7N.M.
2
APPow able
vadue
of
Te
is
the
smatler,
Te = 444-7 Nin ae
|
Problem 3.26
3.26
The two solid shafts are connected by gears as shown and are made
of a steel for which the allowable shearing stress is 60 MPa. Knowing that.a
torque of magnitude T, = 600 N + m is applied at C and that the assembly is
in equilibrium, determine the required diameter of (a) shaft BC, (b) shaft EF.
100 mnt
Une,
tT, ()
60 min -
Cc
bo
Te
Lianne
.
-
a
gt Te
ex
(-
T. = 6O60Nm
BC:
Shaft
~
tha
_ 2T
=
J
—ew
a>
tic?
Z) lbse)
Y Wl boK10® )
penn
4: LT
=
on
i? The
O10;"S™
che = Ze=2 7mm
.
Tee 2XS Tz = fe (be0) = B60N
m,
Sheft EF
A
é
oe
3 J2
i
seromartatmernene,
$e
*IDBbe)
~
“WF Coo
KO"
>
€
O°OI1S&
hw
dee =
ae
7
Bie2inm
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course prepatation. A student using this manual is using it without permission.
<i
3.27 A torque of magnitude T= 120 N - mis applied to shaft AB of the gear train
shown.. Knowing that the allowable shearing stress is 75 MPa in each of the three
solid shafts, determine thet required diameter of @) shaft AB, (b shaft CD, (e)
|
- shaifF.
3.27
re
Problem
75 mm ~
STATICS
30 rn
Oe
he = h oT. 2.7.
Shaft Ag,
,
an
_
Go
od
=
V2
le
~
25
|
”
CD,
— Tee
°
2.4
TT
~
Ve
-
25mm
.
Gears
D
ane
Force
a
Y=
39
mm )
on gear cirches,
Tzroe= a7MewTe 27 EF
9 (24T)
Sheft EF,
REQUIRED
DIA
_
Frmay =
Ca) Shaft AB,
Thg =
ShaPt CD,
dey. = Z
C=
thee =
37
T
aT
j=
F120
QC)
WH (7S*1O% )
9-2
A= ~ 2e72
.
ZOVIO Em
=
Tee = (4)UI20)
é
==
we
6T
6T
af (2\ (720)
TC7S*1OF 7)
(4
Fe
Pa
Mem
Te, = @HMi2e) =
3[(Z)(aasy
C2N CRBS)
(c) Shaft EF.
Fag= ie =
ERS
te.
yy
> 2v
We
Pra ® a
Vie = (Samm
Tee 7 Te = Te =
Corus= ISK1O%
(bY
~
ee
Te
° me
N
Shaft
60 3am
gear ewedés,
{}
Ie
oo
mm
760
. Te
Fs. . Te
on
oa
vere
Ke
= 25mm,
oi
Fovce
Vg
Cc,
uj
and
BS
Geags
i}
eZ
26.4
=
298 Nem
.
x10”
he = 20.1min a
|
|
m
Ceo
=
26.4
mm
72OMem
:
.
=
36.6x10%wm
ep= 36.6mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Tne. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
th,
Problem
3.28
©
3.28 A torque of magnitude 7 = 100 N - m is applied to shaft AB of the gear train
shown, Knowing that the diameters of the three solid shafts are, respectively, dig
= 21 mm, dcp ~ 30 mm, and der = 40 mm, determine the maximum shearing
,
stress in (a) shaft AB, (6) shaft CD, (c) shaft EF.
75 maw,
Do
STATICS
30 mur~.
Aw!
Shoft A@,
a = Ge Te eT
Gears Band C,
a= 2mm, V2 Comp
‘Force
on
Te
fee
, Oe TO
Ts 2Fas
Shaft
wm
“6
,
qear
ewelés
Fae * ie
= wt
|
2.4 7
=
Te
Y=
30
Tey
CD,
.
#4
=
Tp
=
T
25 mu
Gears
D
ana
:
Force
on
,.
EY
Gear
D
MAYIMUM
SHEARING
~Pyne 2=
Foe
Vi.
.
=
= 7S mm
Ty
Y
_
=
Te
Kt
EF,
Tee = Te = Te =
GT
STRESSH&S
(b) Shaft cd.
__
7
EF,
Pa
SF5.0x10°
=
fo?
T=
YF
-
45.3
)Cf00\
()C 600)
H(goxlo-tys
Pa
x{O
MPa
<a
;
_
@vaue.
Cetd
.
Tinev= 55.0
T= (2.4\igs)= 240 Nem
es tds [Simm = IS*IO%m
“TUE
Tomo
Cetd = 10.5mm= 1O.S*(Om
T=100 Nem
Toone 2 SENT
ot
L mage
©
)
es , 20
TOs
(a) Shabt AB,
(c) Shatt
Foo.
circtes
wm
5
= GT
RAT)
Vis om
Te = Tt
Shaft
D
Cire =
45,3
M Po ~te
47.7
M Fa
Nem
= Goo
= 20mm = 20x10°w
=
47.7
G
x10
Pa,
~~
~
bmnax =
-«
Problem 3.29|
3.29 Whilé the exact distribution of the shearing stresses in a hollow-cylindricat
shaft. is as shown in Fig. P3.29a, an approximate value can be obtained for tinax by
assuming that the stresses are uniformly distributed over the area A of the cross
section, as shown in Fig. P 3.296, and then further assuming that all of the
elementary shearing forces act at a distance from O equal to the mean radius Ac,
+ cz) of the cross section. This approximate value ™ = 7/Ar,, where 7' is the
applied torque. Determine the ratio tax /% of the true value of the maximum
shearing stress and its approximate value m for values of eile respectively equal
to 1 00, 0.95, 0.75, 0.50 and Q.
ta)
(bh)
~
=
Pow a bellow Shek
By
ho Dow
def.
artion
'
-
T&
_
Ke
)
7
2T
Toa = OTE = TEEN)
f+:
”
JT
=
ANY
j
Problem
| [028
3.30
w=
Ww
w=
=
2TH
* betel
|
(6)
T7w),
-
| 0.0
|
ltz0
| 1.200 | ho
|
|
Pength
weight
| + Cy
Lg
>
T
W
™
speci Fic
echt
,
> PELA= pga
Brace
-C
Cz
_
| 0.5
weight pper unit
.
T - (c,* +02") Veg
(a)
Ww
C)=0 for sete shaPhe
(-r /w)r
| 0.75
3.30 (a) For a given allowable shearing siress, determine the ratio T/w of the
maximum allowable torque T and the weight per unit length w for the hollow
shaft shown, (4) Denoting by (TAv)» the value of this ratio for a solid shaft of the
same radius cz , express the ratio 7/v for the hollow shaft in terms of (T/Av)o and
e/e2.
T
1+ (A/a)
| 1.0 | 0.95
Tian Te | 1-0
«
1+ (G/ca)
C2 #0
e,/¢,
.
2
:
A (C24;
Dividing, Soe = G20C24G)
te
2
2T¢,
AGP?)
aT
Te
TE ONG
=
Ze
QZ
w(¢?
-_
_
Pye
4, *Vez -¢,")
A
a\he
hog te
(a),>
(7)
(e,* - 62)
an
Chola
Csobeed
Fy (i+ &.)
&
TAL
:
shaft)
shaft
~
)
—<
oO
3.31 (a) For the aluminum pipe shown (G= 27 GPa), determine the torque To
causing an angle of twist of 2°. (b) Determine the angle’ of twist if the same
‘torque Ty is applied to a solid cylindrical shaft of the same length and crosssectional area. —
Problem 3.31
49 nm
50 vm
ih.
@) Ce
SO mm *% O,08Om 2
J =i¢ccf-e")=
= 5.7962
2° =
Gt
27x10"
(b)
iH
of
Radius
Js
Pe
piper
2.13851 x]0"
tTles-c:):
of
sedial
of
Same
2
are
Pa
al
oo
= 2.8274 wr
7lo.0sa'- 0.045")
Cyty
°
<8
Tri= 214 bem
New
™
8
107 )(2.5)
.1
5t*
(2
(a7xto4 JOL. 272 35%10°° )
o
= 25
&
2c
A
=
0.030m
ee
|
oo
1.27238%10 °° mw
=
£0,030)"
Zz
—= Teh
T
A
34.907%
10" raed
2.5
2
Area
£(o.0s0"- 0.040")
- (27«1ON S746 2«Io* )(34.907*I10* >
Ie
To=
OLOYOm
107° wt
pt
4}
G* EF
6. = 4Omms
2
a
0.15902 ved
—
P>
m
4.4
o
3.32 (a) For the solid steel shafi shown (G= 77 GPa), determine the angle of twist
at A. (b) Solve part a, assuming that the steel shaft is ‘hollow with a 30-mm-outer
. diameter and a 20-mm-inner diameter.
J
30 mum
(T=
250N-m
0.01ISm,
c=td=
(QQ)
x10
79%. $22
=
=
_
280
m*
“|
S50) (1-89
Je Bet = Floors)"
ore-
180 _
xJ0°% 4
ee C134
¢, = 0.015 w,
- E(o.015"-
+
C,=4d,= 0.010 i
0.010" )e
(2500.8)
(77 X{07)(63.
314
=
3.
T
B10"
i
oF
P=
?
Pe.
ss
TS NTMO
i
wad
a
J=E(c,"-¢,")
wm?
GSB KIO wae
G = 77x/07
Lb
xig"?
(77 «107 )(74.S22
g
(b)
bm,
L=
QD
=
= ee
B21. saxto* )= 525°
ol,
|
Problem 3.33
. 3.33
The ship at A has just started to drill for oi! on the ocean floor at
‘a depth of 1500 m. Knowing that the top of the 200-mim-diameter steel drill
pipe (G = 77.2 GPa) rotates through two complete revolutions before the drifl
bit at B starts to operate, determine the maximum shearing stress caused in the
pipe by torsion,
TL
~ . GIP
PGS
v= If
S
~
=
:
GIPC
St
Q2yev
zs4ds=
Gr
yz
L
=
.
2RAT)rad
ioo
mm
=
SPS
12.566
0.100
ved
m
7722GPa = T7210"
(PE 2x10 02. 566 0.100)
Cy
yr jos
. [S00
Cs
Problem 3.34
3.34
Pe
64.7MPa
=a
Determine the largest allowable diameter of a 3 m-long steel rod
(G = 77 GPa) if the rod is to be twisted through 30; without exceeding a shear-
ing stress of 84 MPa.
iL
P=
Bm
vu
9*
84.M
~
re
or
gor
30° 7 * AS
§ = 0.52360 pao
Pq.
T
ca el BHP)
-
oF
ot
C3
C17X1o%) (0:$236)
,
J
JL
2 O00b25mm = b20 nw
Lo
GQ
d= 2Zast2=Smu,
—e
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and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3. 35
3600 Nom
3.35 The electric motor exerts a S00 N - m torque on.n the aluminum shaft ABCD
when it is rotating at a constant speed. Knowing that G = 27 GPa and that the
torques exerted on pulleys B and C are as shown, determine the angle of twist
200 N-m
and C, (6) B andD.
(a) B en
betwe
(a) Ange twist
between
Teg= 200 Nem,
Lat bh
std = 002m,
(b)
Th
ve” CF
Ang fe
2m
44 mm
G*27x/0"
3
2
of twist
lm
between
Baw
C2
_
(87 p07 \CSAN.1S3&10")
=
~
G = 27%/0" Pa
E=tels 0.024m,
Lep 0.9m;
(S00 \ (0.9)
Peret Lem
Psion
Pae® 1.384"
D,
= Fe% = Flood)’= s2is3xto’ m
*
Z
@oeyi.a)
.
¥ 1078 wad
(27 #109) (367.97%107) ~ ZY 1ST
Ter = SOO Nem,
Pn
Lam
40 mm
367.97 "107 m
Je." Fete
@
Banol Ce
%
BL WOxlO
24- ISTx/o + Bl. 980 x1"
=
rd
56.137v10 rad
“=
Pero
B.22°
<h,
800N-m
Probiem 3.36
3.36 The torques shown are exerted on pulleys A, B, and C. Knowing that both
shafts are solid and made of brass (G ~ 39 GPa), determine the angle of twist
between (a) 4 and B, (6) A and C.
{a
Angle
Tra
of twist
=
Yoo
Nem,
Ter BOT
.
Lag?
A and
12
B a
400N<m
\,
Angfe
18m
= BFx/a? fa
WAS VIO” a
Th.
Ga
(4o0)tt.2)
(39 1073079. S2%107)
-
7 O.1S4#T72 ved D
(b)
30mm
yw
G
= 0.01m
= tol
Pasa *
belween
40 man
1200 N.m
of twist
between
Tac = BOO Nem
|
Pam =
A alt C
=e
’
lat Bm,
Sz Ze" = %(0.020)" =
8.87" 9D
er gd =0,020m,
251.327«107
)(1. 8B)
Pore =_ (Baw (Boe
iotV(as7.
327x107)=
G = 39x/0'fR
m’
ICVIZwed
Pre = Pare + Pac 0.1S4I772.-0. G2 = 0.0078
GO ved 5
Dey O4S0°D ti
Problem
3.37
3.37 The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB (G= 39
GPa). Knowing that each rod is solid and has a diameter of 12 mm, , determine the
angle of twist (a) at B, (5) at C.
Both
povtrous
Te
200 mnt
Brass
__
c=
Bete
Pp =
|
—
=
mat
Clee
~
Rod
Bc:
Lag 7 0.200 m
(34 %107)C2.03575
x}077 )
Ge.*
26
Paz i4.43°
x107
(102 \(0, 3001
0.25141 + O.5667%
=
wm
= 2.03575 x1077 mm"
veel
=
=
Pac
Pat
= Gxio®
Geo )lO.200)
=
Gel ~* (26*1D*) (2.03575 *107")
=
De =
(bo)
Pe
TOL
= 0.25191
Vege
mm
Gia= 3410" Pa,
Aluntinuin
100. N «im
G
[00 Nem
T=
(2)
>
Bexwt)
Rod AB:
300 mm
gad
Pa. ,
OS
L
Bo”
oO.
200
<a
re
aol
ob 14°
= O.8137O rd
G.=
469°
—m
Problem 3.38
BO mun
3.38 The aluminum rod AB (G = 27 GPa) is bonded to the brass rod BD (G = 39
\
GPa). Knowing that portion CD of the brass rod is hollow and has an inner
Tz = 1800 N.m
diameter of 40 mm, determine the angle of twist at A.
Red AB:
G= 27¥10" Pa,
‘T =
800
J: Fe'=
BCL
Tr
Ge
34% 10" Pa
_
oF
E
Bae Oy | Ue BIS HI
Ls
0.375
m
eda:
J To
_ (4400)(0.375)
(Faxo*V02788 deo)
6, = tal, = 0.020 »
Feet CD*
re
Go
Argde of +wrst
.
(329x107
at A »
\(f.O2102
Pa
=
« fore)
Pant
a“
om
x
nim
735 vim
wn
ning
O.C3 Om
= Eo c30)=
127234 JO
wn!
FT IBIS7 IO * peek
Elce"-a")= Fo. 030°- o.onc") eo, OR1OR x}
TL
(A#00)(0.250)
_
-3
*
a 400
Sond
C, 2 £4, = 0.030m
t
Hf
Pein
AIOE
|
é
Ay
-
Bee t/600 7 RYOON-m,
Pare =
T=
164.296 «1077 v,
($02 )(o. Yoo)
\
T, A = 800N-m
O.0Bm
=
Crd
Fools)
TL,
Pave =
Pet
Nem
“
36mm
m
0,400
L=
=~
Pet
1S ,OGZx/Oo
5
L=
m”
0.250m
.
rad
Ric
[05.080 x10% yael
p,
*
6.02"
|
Problem 3.39
3.39
Three solid shafts, each of 18-mm diameter, are connected by the
gears shown. Knowing that G = 77 GPa, determine (a) the angle through
which end A of shaft AB rotates, (b) the angle through which end £ of shaft
‘EF rotates,
—
Geometyy:
Lage hZm
, leptoam
.
_
Stetites:
Ty
70
Geav B.
c
se
5
Nm
Te = 20
Nm
Co
~YeF,
Th
|
“
~
2
+ Ta
=
4/0
Ty, =20N"
ox
©
20
F,= 2778S
P= 2778N
= Ot
49 EM,
Gear FL
%
T,=10N-
45 2M,
= 0:
r
.
ee
Lep 2 h2m
D
~ 01036 [-
A
pes
Ve 70-048
1", 70-+44m,
Vg= 0036m,
0.9 ta
e=0
A+
Geav C,
+ > M te =o
.
. ~
at z= O
pane
.
Feoo
N
4167
4167
=
2
“~“nRF-
oe
+T=0
wR,
F, 2 FIGIN
Deformation
For
= 0-144) (2778) - (0-144) (10 ) + Te
$c
t ti = /o-306Kto-?
J-
~- Taabae
Pere =
Qs
ere
~
= ERE
=
Kinematics!
Get
(@)
(b)
One, =
Py +
EQ.
= Ved.
m+
=
pols.
(771101) (10-306 x18)
~
‘2
aie
=
002
ate
Pes, =
VeP3= VR
Min
= 414
Te
m
(7 U+2
2
GL
Para
T0009
Cztd
shafts
off
=
O0469s
9
aD
vad
= ones
D
Dy? = @, = 140-0465) = 01878 ved 5
Drie
>
Pp= Dit Pry =
01878
+ 0.01512
= 0.20292
Py = 16°
red >
PereegQ.*= aoo.
0-104 0469s)= 0-Moes rad
o:04e
O-teoes+ 0.03024
= 0-17109
wud
Sat
D
q-p°*)
sat
Ty,
Problem
Cobeodation
3.40 ‘Two shafts, each of 22-mm diameter are connected by the gears
shown. Knowing that G = 77 GPa and that the shaft at F is fixed, determine
the angle through which end A rotates when a 130 N - 1 torque is applied at A.
3.40
of tovques,
Civeumterentiad
between
Ir
cowtact force
gears
B
ote
. Tevy
= YY,
The = (3
oud
E,
+. Yo.%F laa
Nm.
Tey? FE (3¢) = 17763 Nin
Tw
ist
in
shat
FE,
L+Beomm,
T=
C=tol
Ect
~
Jb.
GI
Rotation
at E,
Tangent f
=flmin,
a
G=T76Pa
C1]2)(0+3)
= ov
CTIXI0AV
C23 x071)
Oe
=
displace ment
at
150mm
= E (owt) t= 23x10 Imt
Pere
Rotation
,
B
Pe
Pe ip
at
=
gear
=
@,
(05
O03
Vad
cirche
~
rad
S =
(e103)
=
Ye Pe
Ya Pa
0164-09
vad
a
Twist
tn shaft
BA,
l= Osim
. Tho
Pave
Rotation
Li3o)(o-35)
GJ
at
T= 23anig
CTIX (6%)
A.
Pp
=
-
(23X07)
Pa
+
Dui
=
~
Ora2ed
0
ons]
6106046
rad
tat
vw
ved
Pp = HF"
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the liniited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usitig it without permission.
atte,
240 mm
.
.
3.41
Problem
3.41 Two solid shafis are connected by gears as shown. Knowing that G=77.2 GPa
for each shaft, determine the angle through which end A rotates when T, = 1200 Nem,
Cotcudod jon
of tovaues.
Civcumfeveticl
qears
Band
contact
Cl.
pr delle
Te
_
4
leo =
a“
wm shaft
Js
ts
A
anadte
Civcumteventias
Ve
0.030m,
=
C
\
L
>
Von
meout af
Lam,
Ge 77.2*10" Pa,
=
HS.981x10
IO")
Rio
=
contact
48, 931 «10°
poids
oF
3 vael
yod
years
B
om
C
Pa
ar EG. = SE les.agixio*)= 13).442*10*
vad
~
g
Twist in shat AB:
cthd = O.02’im, b= lom,
T= Bet = E(ooar)’
Rotation angde
Lt
§,27234 xloe wm?
(3600)(1.2)
Rotation angle f B.
=
1.6m
(77.2% JO? Cj. 27234
Aispface
p.
42 mm
Nem
c=ted=
eA
<
.
= 3600
CD:
GI
=
las
XN
_
(i200)
~ Th.
$
4.4%
Bot = Elo.ogey?
Pew
Rotation
between
1200 Nem
The =
Twist
v
fovee
TL.
2
oA A,
geo
CG
772xl0" fe
= 305.49 «107? mt
j (1.6)
Pr=
Vao+ Para
=
*
a
SESIax«io
3
AZ.
354 x10 rad
Pa = 12.22°
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior writien permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
<
_.
Problem 3.42
3.42 Solve Prob. 3.41, assuming that the diameter ofeach shaft is 54 mm,
240
mum
3.41. Two solid shafts are connected by gears as shown. Knowing that G = 77.2
for each shaft, determine the angle through which endA rotates when T, =
1200
7m,
a
GPa
Cabeu Petron
ot
torques
Cincom? eventiad
Gears
¢ ontact
Bad
Tas
re
Ke
F
‘Vag ¥
Force
belween
™ 80 min
C
Tep
tee
v.
1200
K
Tor
cn
54 mani
Ys
Lam
Tae
Neva
Ter = S22 (i200) = 3600 Nem
Twist
in
sheft
CD:
¢ z=td
= 0.027 m,
T= Het = £(0.027)" =
Bory
* Th
GS
.
Le
L2
m ,
GF
324.744 1077 mn!
@600)(,2)
|
(77.2%107)(834.78
1077)
77.2» j07 Pa
= 67.033 x10* veel
Rotation amate at C.
fe = Posy = 67.083 %107* pad
Circumtevarrad displacement at contact points of geavs Band C
S= VAR = Ved
|
Rotation angle at B,
Jj =
Pra?
Rotation
AB:
cr gol=
234.79 %]07"
2. Th.
L
(77.2% 107 (834.749 « 1077)
at
A
,
3
6=77.2xI107 Pa,
m
001.6)
EF
GI
anghe
0.027 m,
oS
3
Pa ® ie Pe : £40 (69, 033nlo7)= 201, 104/53 pad
4
Twist in shat
a
Dy
ad
Pz
+
Pre
=
=
230.29
29. 7¢2«1lo?
R100
x10
rad
wd
P, = 13.23"
. Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
=
Problem 3.43
-
3.43 A coder F, used to record in digital form the rotation of shaft
4, is connected
‘
to the shaft by means of the gear train shown, which consists of four gears and three
solid steel shafts each of diameter d. Two of the gears have a radius r and the other
two a radius nr. If the rotation of the coder Fis prevented, determine in terms of 7,
4G, J, and nthe angle through which end A rotates.
Teo
-
eo Tae
=
Tha
Dod
ary
n
i
pe
=
— Teeter
Pe
Der
Gol
> &
4
=
vo VE
Pas
= Ten Pes
“
G
=
Ve.
=
¥,
N26.
Th
=
~
n° GJ
Tal
YiGQT
Py + Per
-
Pe
Th.
“
_
n
ot
Pe
.
~
Ve
~
-
|
Te?
Thad
WIGS
©,
He.
Tt
=
i
* NGF
a
Ty
|
fat
(as+#)
.
Bl#té)
Vag = Lon fan, - Taf
Qs=
Pat Paw
=
BL (+
Problem 3.44
3.44
ksi)
|
~é
For the gear train described in Prob. 3.43, determine the angle
through which end A rotates when T = 0.6 N+ m,/ = 60 mm, d = 2 mm,
G = 77 GPa, and n = 2,
3.43
Acoder F, used to record in digital form the rotation of shaft A, is
PROBLEM
connected to the shaft by means of the gear train shown, which consists of four
gears and three solid steel shafts each of diameter d. Two of the gears have a
clev edop ment oF
3.4 2 Fe
equet ' on
Fo * Pa
radius r and the other two a radius nr. If the rotation of the coder F is prevented,
determine in terms of 7, /, G, J, and a the angle through which end A rotates.
to
sofution
See
\e
o'
:
.
:
aS
fh = Et ( [+ hi + tr)
Datat. Tre orb Nm
f= bomm,
Cztd=
Imm
4
6 = TI x10 Fa,
~“f2
Y=
2
Wo?
2
.
do
=
Eo
=
E
(001)!
=
bS
7x10
Lay.
a fe
|
d
ety (ee Trem) (It Ht ie) = O24 reH+
(a:6)(0.0 &)
22-4"
af
Problem 3.45
3.45 The design specifications of a 1.2-m-long solid transmission shaft require that
thé angle of twist of the shaft not exceed 4° when a torque of 750 N - m is applied.
Determine the required diameter of the shaft, knowing that the shaft is made of a steel
with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77.2 GPa.
T= “Iso New,
Tr
4° = 69.813x)]O ved
AOMP
Basect
FOwlO®
ou angle
.
C
=
Based
VRTE
_+
shearing
35°"
2T
Tz
ce
=
stress |
“i =
Pe
a
(CQNTSOV2)
ie
NT yee
(2\(750%
=
IO?
=
|$.06xI0°
18.06
x10
m
AG
lon?
Vass
Use darger vervee
Problem
ad
TCT7.2*
IO VGF. 813%
3h
T= bet
G¥77.2 GPa= 77.2107
of twist.
| #35
on
Pa
Le lagm,
17 SIO"
m
C= 1B.06¥IOim = 18.06 mm
d= Rar 36.) mm <f
3.46 A hole is punched at A in a plastic sheet by applying a 600-N force P to end D
of lever CD, which is rigidly attached to the solid cylindrical shaft BC. Design
specifications require that the displacement of D should not exceed 15 mm from the
time the punch first touches the plastic sheet to the time it actually penetrates it.
Determine the required diameter of shaft BC if the shafi is made of a steel with G =
3.46
77 GPa and ti = 80 MPa.
Tez
VP
= (0.300m\(Goon)=
180
Nem
‘SheFt diameter based on ais pdacement
p= 2 =“ {omm = OOS yard
ae
2Tk
©
QWB0 VlO.S00)
WEQ
CF
C7
3B
[L,0dS «107% m
CG
.
=
=
the
2T
fn
Tr
2 Te
cue
(77109) (O08)
diameter
Shaft
Use
f
.
oe
basecl
on
(2) C30
Value
Mm =
te
~
styvess,
—EB2B4I
IN.
882
both
x10"
Az
P=
80x10"
KIO
VL 273 mm
meet
206
w~Ge"
2 jy p89 x
11.04Sm
-
G\tiBe
WUL273 IO
Aanger
=
.
Piva te
Pa
r=
= ate
we
ot
Pimcds.
20.5.28.28 mm -
Ze Ft 22.5 rm
A=
22.5 mm
aa)
Problem 3.47
°
wey
3.47 The design specifications for the gear-and-shaft system shown
require that the same diameter be used for both shafts and that the angle through
which pulley A will rotate when subjected to a 200 N « m torque T, while
pulley D is held fixed will not exceed 7.5°. Determine the required
diameter of the shafts if both shafts are made of a steel with G = 77 GPa
and Ty; <= 84 MPa.
,
Statics.
+)
Gear
Z2Me
B.
Te
—
Yi
—
—~
om
«
*
Pp
=
=
Pe+ Pare
on
Diameter nased
é
Cm =
—F
tt
*
Cc
Diameter
p=
c*
a=
WT,
eaLb.
=> baba
Gt
2
=
Tab
oF
2
O
P.=
Dor
By
=
Ye
= Fp 7 2-5
Te
-
wt,
OF “BE
Mee
GT
-
2
based
on
votatvon
Dimit,
Gre ite
|
Lh)
27) Tab
Te? C
=
"6p
Choose the fanger
Tene xlO§O.1309)
diameter,
Ts
TT,= 200 Nin
De
o:01Fb
P=
2
Thy = Ten =
*
torque
| AX25\CQx107)
ae oe
TT (84 poK1D>)
= WTp
2
Th = Tou = B4MPa ,
ant
T
TAL
(NIN
EF
27
Lovgest
Sivess,
~~
ay va
wenn
Cm
Tal
GT
nip
TE27L 2
Pet
Ye
RT HNR
Pe os
Pa ~=~
~
Pe =~lePe
Py
5
Teo
SS L
=>
Peo
>.
=
“Tie
Te
Sd
Tk
=
Tas
shafts,
Pare
gs
n=
a
.
Deformations
.
as
&
_
= nty
= weSh
iv
Kinematics
,
_
= VF
Tergqves
Tr,
Tp
F
= ©
Tp
as
200 man
F
ty
4SEM,=0O:
a
400 navi
“
T
~ta
F
F = Tp/ta
Var- T 20
>
F
=O:
Gear C
oe
150 min
75°
ad = Ze
>on
010312
0.130%
vad
mM
=
m
Orr obm
La+t
TOGISS Tm
= f=2e= a03lm
d= 3himmn
«1
Problem 3.48
3.48
Solve Prob. 3.47, assuming that both shafts are made of a brass
150 md
with G = 39 GPa and t4, = 56 MPa.
va
3.47 The design specifications for the gear-and-shaft system shown
require that the same diameter be used for both shafts and that the angle through
which pulley A will rotate when subjected to a 200 N - m torque T, while
pulley D is held fixed will not exceed .7.5°. Determine the required
diameter of the shafts if both shafts are made of a-steel with G = 77 GPa
a
400 mn
a
we
200 mint
and t,y = 84 MPa.
States
Geary
Ss
Ts
Ci
ADEMg =O:
AN
=
=
D
_
Dey
Deformations +
Pasa
ae7
.
Kinemefics
2
Q
et
2
=x
Jal
ae
Tra
f
“
er
+ TaL
TF
“EE
Pet
—.
TTm
base
on
ails
\
tL
votatyon
.
&b MPa
Se nO
$ imit
farger
P=
Sa Tike
~ al C3400 *10 100. 1309)
dtameten,
4
;
7.5°
1h
=
T=
_
1, = 200
Now
inane
Ai2e =a-0IS6m
~ aO171Fm
2SVTaAL _ 4f G7. 2512x104)6)
USP
nehe
“ato
-
~ \V WCEheoxlor),
2
TAL P)
(WHIV Ts
(ost)
Lovgest. tovque
.
.
Tint
"WTAL
O+
~“N@
*
> Wa
2
fan Ve _ FF ARSIQ~ 107) _
= fe
the
3
=
nts
=
Te
25
5
Pot Perm =
~ .
=~ K Fe
ante
anne
af
a
Choose
ies
Pao =
Pa
Fe
=
Diamete
=
based on stress,
Tat . 2
Diameter
©
\
ot
are =- na
Py == Pat
Tm
‘er
=
Teo =
Te
ee
-
%
n=
=
Th
=
Gf, = O
VePe,
=~
Pe
-F
te
Tap
shafts.
in
Teragves
Tp
F
hn
©
-
A
fF
E = Ty/te
Vari= Th, 20
Geax C
EM.=0:
ey
Yet
Ty
=
O.1304
pm ere OG
= OOiBlS
vad
ob6m
d= 20 = 00563m
a=
363m
<e
Problem
Ten = Ty
=
Y;
wet}
Tag* %
For
3.49 The design of the gear-and-shaft system shown requires that steel shafts of the
same diameter be used for both 4B and CD. It is further required that ¢,,,, < 60 MPa
and that the angle ¢, through which end D of shaft CD rotates not exceed 1.5°.
Knowing that G = 77 GPa, determine the required diameter of the shafts.
3.49
1000
100
len *
design
40min
Nem
G
(iooe)=
——
based
Sarger torque.
2500
=z
on stress,
Trg=
vd
T = 1000 N-m
Nem
vse
A500
Nem,
“~a400 man
ye
Te , at.
J
-
Ta?
Cot et
ns
CF
VBR
xO
Design
based
n,
. TL
CS
=
29.82
rotation
1900
6
_
2500
100, toca
Desi gn
1OCO
L5°
—
26.18 vio“ vad
+
~
Pep
_=
42502 ,
oT
L= 0.6m
+
600
602
.
GI
-
_
TG Dy
— W77 veo )Cacaxto?)
B46 HI mm
use
Danger
= BL4G mm
Vvodue Pow ol
3109
.
3/00
GEc*
GCS
»
~
4
“4
2
&)(siee)
(a\€3loc)
most
=
mm
~ GF
Nem,
{ (ooe)(o.6)
=
GT
-
CF
=
57.6
l0eo
Teo =
_ Th
Pep = GIT
C
im
L= Ont m
“
te * Yor
Pe
an
od
cp
~
Py
_
oy
-
Ve
«
Pe =
y.
ana de
on
3
d=2Ra=
-
Pas
Po
mm ,
_ @so2)(0.4)
.
Pe
Shaft
> 2-586 v/O
Tig= 2500 Nem |
Dar
Gears
=
_
™ gan
A
.
_
Theo ony
on
Shaft AB:
L
:
(22500
2T
a.
ae
777,06 ¥ 10
wm
a FAC = ELI mm
d=
62.4 mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using 1 without permission.
Problem
3.50
3.50 The electric motor exerts a torque of 800 N-m on the steel shaft ABCD when
it is rotating at constant speed. Design specifications require that the diameter of the
shaft be uniform from 4 to D and that the angle of twist between4 and D not exceed
15°. Knowing that 7, s 60 MPa and G = 77 GPa, determine the minimum diameter
shaft that may be used.
300 Non
500 N-m
Torg
ves
Tag= 300 + S00 = B00 Nem
Tee.
=
500
My
™)
Tex
=
-~lt
ts F
. 2T
s_ 27
* es
based
Paice
AMBoo)
om
defor meton
Oy
7
Veelse
= Sng ee
_
=
Goo
Day =< Teskse
- (400)(0-4)
_
gro
00
~
QS
Ot
)(0.6)
Gr
~
© *
C=
Design
must
'
_
~
= 8438x108
15°
=
26.18 *1073 youd
.
Que = “ert
4.
O30 |
d= 2e = 4HO,8 mm
20.40 mm,
=
=
-
.
CN TE * WESREET
C= 20.40 x{07 m
Desiqn
oT
TU = 60%*/0* Pa
based on stress
Destan
©
(2)(E20)
4G Poia
vse
Paraer
GAO
TT? KIS VRE. 1B x1077 )
vabse
21.04 mm |
oF d,
_
(2X20
-a
|
(A620)
m =
204%
620
oy
|
A= Re = YR.1 mm
d = 42.1
mm
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
a
The solid cylinders AB and BC are bonded together at B and are
3.51
Problem 3.51
attached to fixed supports at A and C. Knowing that the modulus of rigidity is
-
26 GPa for aluminum and 39 GPa for brass, determine the maximum shearing
“stress (a) in cylinder AB, (b) im cylinder BC.
Abnninnin-
T=
Bet
a
=
T @025)'=
= Tee b
-
Gy
Mat ching
_
Gr
foun
Pe
Lto3m
= Oralf fae
Tag (23)
= Sf
=
7e
56-37 ¥10
Talos)
tov
of connection
Tas + Tae
Substituting
—
48.
*
18-8 * 10°
Pa «
Sb-3T
jo°
Vee
» 10" ° Treg
=
(he
%{0° Te.
Tae
fa)
= 300
Maximum
(o)
Cs <7
=
T=1z00
Nm
(2
inte (21,
C —
stress
Tee €
TF
=}-2%10°
39984 Tre
Tee.
stress
J
Maximum
Tag t Ige-T =O
B:
at
Nin
lag
SoS
The
Q)
= h2é/0%
CQ)
in
=
cy Piypder
(300 (0-019)
ZoH7 nod
In
&
613-6x16
4 m4
Tac v 47984 The
Equibiberum
1,
L = 045m
(39 108) (613-6 X17)
express [ons
ad
~ (26x (01 \Za4-7x10~7)
c= gal = 0-025m
J > ¥ ct
stateudly
ane
BC
votation
= 704-7 x10? mt
- too
BC%
3
er
AB
Pe
Cyfinden
and
cytinder.
ch
Cy finder
SO ram
the
, Mateh
in determinate.
ea
AB
in cy Rivclens
tevques
The
“en
cy Pinder
(900 ) (0-025)
B15 6 xi07
_
=Joo
Nm
A&,
.
2785 MPa
“~
227
Baye 2
4
MPa
<td,
Bc.
=
3 2667
MPa
T, 236-7 MPa
6c
Proprietary Material, © 2609 The McGraw-Hill Companies, Ine. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=i
Problem
3.52
3.52
ne
Solve Prob. 3.51, assuming that cylinder AB is made of steel, for
which G = 77 GPa.
a .
3.81
The solid cylinders AB and BC are bonded together at B and are
Alansinam
damm
attached to fixed supports at A and C. Knowing that the modulus of rigidity is
26 GPa for aluminum and 39 GPa for brass, determine the maximum shearing
stress (a) in cylinder AB, (b) in cylinder BC.
Pe SEN of
Fe
The
Brass.
Mea
cy
cae
450 rani
.
Cy binder
T=
Po
Cy Linder
BC
T=
=
Beosyt
GL
express /Ons
inte
Ci)
Substituting
Get
= 204.726
Im *
= Tonk
_ _ Tha @8)
GI
t)
Maximum
«
=
=1%03 x10” Ty
~ C97 *107\(04-7x10°7)
°
pee
~~
Int
Pe ;
19.03 x10
To.
wt
~
Tag = 18610
oS
ae
Te.
(i)
,
That Tae T=
Bz
at
©
Ts lz00Nm
(2)
ZO1ZZ5
(2)
in cyfinder
Tas® — /596)(010/9)
-=
I
stress
Teac C
oS
~
-
“
Tas
= {200
Tae 7503 Nm
stress
Maximum
t.
L= agm
L = 045m
Tag = 576 Nm
Tag=
d019m
= [200
Tae + Ta.
@)
Pow
:
Pa
= gisexidt m4
tor
of connection
Equilibrium
3 d=
Tae
160/223
=
Tee
c=
~ (349 «tot (613-6 x10°7) ~ 8B RIO”
‘
Matching
AB
Tac@-45)
_
iecb
-
fa =
BC ave stared y
votation
the
Matel,
es
c= yal = 025m
Bet
AB and
in ey Binelevs
torques
in detevminate.
n
4 ‘ ef
i
L047 K10-7F
in
z
AB.
.
$5.32
cyfinder
( 603) (0:025°)
.
p.
MNS
Tas
2
S5-3 MPa «aA
5973
Mla
BC.
“
24-57
MPa
Vac
=
zheb
MPa
bi3exto?
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and edicators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using.it without permission.
)
Problem 3.53
TO
-
3.53 The composite shaft shown consists ofa S«mm-thick brass jacket (Grass = 39
.
|
GPa) bonded to a 40-mm-diameter steel core (Gg = 77.2 GPa), Knowing that
the shaft is subjected to a 600 N -m torque, determine (a) the maximum shearing
stress in the brass jacket, (b) the maximum shearing stress in the steel core, (c) the
angle of twist of B relative to A.
2m
T
~ Brass jacket
a
° meX
’
Steel core
I= ¥lc'-
Total
carried
bovave
P
_
by
=
wo,
T
=
Shearing
M enet an un
won= GO Yor
=
Marimunm
(2)
Avale
shearing
=
w
GY.
CI
Cm
= O.O0R5Sm
20S = 25 mm
+72
-
|
Goo
19.4025 x{O3 + [F1283xl03
stress
=
iy
(6J,
brass
+
z.
6.7,)#
=
12.894 «fo
8
|
reds
™
joekedt,
= (349x104) Co. O2syCi7. B44 x10 3\
Pe
stress
GG, ef:
27.6 «lo®
Te = G4,
7,
G, oF
= I74Sx10%
(hb)
GA, +
14.1283 %/0% N-m*
brass jacket
T
6S, + Gh
lL
(a)
,
Nem"
ye T (g,0a5"- 0.020") = 362.265 %/0°7 m*
268 HOT)
Gad, = (39 *107)(62.
Tergve
Cc)?
br
2 O- 020
== 20, mm
hd
jacket?
“T=
by steel cre
careied
- Torque
—
om?
19,4025x10°
=
677, 2%10 7) C25 1.3310")
6,5:
Brass
,
RO mm = O.020m
C= 4a =
= E (o. O20)" =
25}. 32» Ion?
Steed core?
J, = Eo
L brass
in
stech
= 12.45 MPa. <8
core,
(77. axjo7)(o. on
C17. B74 ¥ 10 3)
Let
Pa
27.6 MP. =
of tuist,
p= LE = (RUI7, 89410) = BS, 78B¥IC red
Proprietary Material. © 2009
The McGraw-Hill Companies,
Inc. All rights
9: 2.05"
reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
: - Problem 3.5 4
o
3.54 For the composite shaft of Prob. 3.53 the allowable shearing: stress in the brass
_ , jacket is 20 MPa and:45.MPa
in the steel core. Determine (a) the largest torque
UA
which can be applied to the shaft, (6):the corresponding angle of twist of B relative
to Ay
_ 3.53-The composite shaft shown consists of a 5-mm-thick brass jacket (Ghrass * 39
GPa) bonded to a 40-mm-diameter steel core (Gye = 77.2 GPa): Knowing that
the shaft is subjected to a 600 N - m torque, determine (a) the maximum shearing
stress in the brass jacket, (6) the maximum shearing stress in the steel core, (c) the
. angle of twist of B relative to A.
G Yoice
Ton ane =
ress
jacket? Ty = 20-109 Pa,
Duy
20x 10°
.
Pon
each
Pon
GC; J, 2
TOF
7
Carey +e
&
-
Tess,
Co Coma.
Pe
L
7
material
each,
meter iA
C,= Wmm= 0,020 m, Cys 2045 25mm= 2, 09Fm
weal /m
RO. SIB* IO
(34 x1o7 (0.045)
L
T
:
.
Jn= G(Gl- oc) = Z (0.025% 0,020") ’
262.265»
{0 ~~ im
Stee?
FE
caeteacttteetreerenecireamarrnaasin mam arene
cove:
Ty=
2Fel
Simnctb er
vadue
Terque
of
T,=
oe
by
by
GS,
| y=
tag fm
ate
sieeve
.
RO, 5134
wip
peut / ny
.
265% 107 (20.513 x10) © 289.8 Nem
(39% 10°)(362.
steet Core,
ad © (77.a%107 (25! 3ax 107 Q0,513*S* ) = 898.9
(@) AdPouabhe torque,
DMPA? e
sy
251,33 «)07 m*
governs
brass
&& Bu
Lavypied
,
R29 VIS IO
= $lo.020)" =
cartied
Ts
~
¥
€F 0,029 m
—
(77. Rx lONlo.0405
J, = det
(b}
~%
4S
™ 0% Pay
usw joe
L
Torque
=
twist
Ps
L
T= 7 4T,
7 687.8 Nem
Nem
eS
amste,
QVM RO.S
IO) = 4), 026«15 re
~3BN
Py = 235°
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem 3.55
o
“
_
3.55 At a time when rotation is prevented at the lower end of each shaft, a 80-Nmtorqueis applied to endA of shaft AB. Knowing
that G = 77.2 GPa for both shafts,
determine (a) the maximum shearing stress in shaft CD, (6) the angle of rotation at
.
r= 50mm
Let
240 mm
Th?
Tae
t torque
Ten
=
Statics
Th
at A = 50
appdied
tovque
torque
im shatt
AB
in shaft
CD
Nem,
:
~ Taw
-
Fy
=O
Teo Fr = 0
a ~The) = § Ta-Tas)
Teo?
¥\
Kinematics °
Yn Pa
=
of twists
Anges
VaR
Dr = Taal
Pa = Ya
Po =
Task. 2.2 (TanTan)b
Gee
2%
2%
Gla”
==
(b+ sen )Te =-(($4(18)') 4 Te
= (0.47960 Y80)
Ta
tO.
Tag
FT,47
= 38.368 Nem
Ten = §(80- 38,368) = 27.755 Mem
(DY
Mayimunr
.
Teo *
(b)
Angle
shearing
Teo
ck
ot
C
.
Z Teo
wet
rotation
Pp, = Tek
GLa
stress_in
.
ad
~~
shalt
CD,
2)(27.1SS5
)
10,0075)?
=
YT,
Fv 10’ Pa
Teo?
ALG
MPa
ate}
A,
Ate
mGet l.
,368.)(0.24°)
(2(38
0(77,.2x107)(0,009)1
|
yy St vo? red
Q,- 0.663
Proprietary Material. © 2009 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manuat may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or tised beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
—=t
-
' Problem 3.56
Cees
3,56 Solve Prob. 3.55, assuming that the 80-N - m torque is applied to end C of
shaft CD,
Se
3.55 Ata time when rotation is prevented at the lower end of each shaft, a 50-N -m
r= 50mm
torque is applied to end A of shaft 4B. Knowing that G = 77.2 GPa for both shafts,
determine (a) the maximum shearing stress in shaft CD, (b) the angle of rotation at
Let Te= tongue cipplied at C=
Ten
=
torque
in
shaPt
CD.
ShaP+
AB,
SO Nem
240 ram
Tag
= -porqve
Statics
Te
Tag =
%
(Te - Tey)
Kinematics?
=
- Teo
+
-
~ mF
3 (Te - Teo)
ee > eh
=
a
|
= he
iL
oe
Ga.L
Te
QF ~
¢
ay
Angdes —ot twist
/
tw
Pa = lis
Tot
_3.2
E-kh&
J
(2
2
+ +) Too
2.
9\7T
-{(12)4,
} +4) leo =
.
~ 22 “Sts
Gi,
>(E
= (0.52040 (82) = 41,632 Nem.
Te
Tes = 0.4796
ae
27
Tag = EC FO - 41.632) = 57.552 Nom
Maximum
v2
“e
{b)
Ana ie
3
~ A =
shear'ng
Hef 2Jee
of) Yotation
Tae L
Stig
z.
.
stress
2 Tee, -
62 guint
“,
or (o.0075)}3
62.8
M Po. ma
A.
2 Tag
TG
CD.
(2VG4632)
TCS
ad
in
J
uy
@)
rong
>
(2\ 57.582
)(0. 240)
TF (77.2* 107) (0.009) 4
i7.36¥lo
~3
rad
Q,= 0.995"
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem
3.57 and 3.58 Two solid steel shafts are fitted with flanges that are
then connected by bolts as shown. The bolts are slightly undersized and
permit a 1.5° rotation of one flange with respect. to the other before the flanges
‘begin to rotate as a single unit. Knowing that G = 77 GPa, determine
the maximum shearing stress in each shaft when a torque of T of magnitude
570 N + m is applied to the flange indicated.
3.57 The torque T is applied to flange B.
3.57
Shaft AB
, bag*
Tra
T=
Saat
had
.
c=
,
6m
-
015m
=
36 mm
Bet = Eloors\" = 79-5 K107 Tmt
“Taha Las
fe
GT,
“AG
T=
Gna
AB
9
- (Wo
Lae
Gasn0
-49
30 man
op
o'b
= 10202-5 Pg
Shatt ED
T
x
Jeo
~~
.
leo,
Lea >
> Ee."
>
O:4din
)
C=tol
2
rool
(0.018 \" = 1649 x10! mt
~ Teo bes
Pe
°
Mer dred Terave
‘GSeo
rotation
Tovqve te vemove
7
(164- a
Teo = Sosa ds = 71 vi
Clearance
0.6m
-
for Mange
T
=
57 oNm
Lp. = motte
Bt
gt
i-S°=
26,18*107° nad
clearance 7 Tha = (1oz02-5)(26,\8xIO°)=
267-1. Nm
TT" = 570 - 267+} < J02+4 Nh
Torque Tbe cause additional votation gy":
T" = Tag + Teo.
\P"s (4108+
302-4 = (t0202-5
2
)¢"
Qa
410°? pad
1thb
Tho= (Vez02-5 \12-46 xlow*) = 127-1 Nm
Tes = (/4108.1\(12-46 « o™) = 175-8 Nm
Max imum
a
lag
Maximum
ae
shearing
=
AS
—
c
sheanng
dest
Nee
stress
(
+
stress
in
;
7°.
AB,
}
in
(175-8 \Coro18)
164-9x1o te
a
q
The
38
M
Pa
—
Tt, AR *
Jk-4 MP,
q
ctl
ED,
= 19-19 MPa
|
Rot
°
19-2MPa
we
Problem
3.58
3.57
.
3.58
Shat AB
TF
Tas
) bag?
and
3.58
Two
solid steel shafts are fitted with flanges that are
then connected by bolts as shown. The bolts are slightly undersized and
permit a 1.5° rotation of one flange with respect to the other before the flanges
begin to rotate as a single unit. Knowing that G = 77 GPa, determine
the maximum shearing stress in each shaft when a torque of T of magnitude
570 N - m is applied to the flange indicated.
Obm
>
o>
ade
The torque T is applied to flange C.
Odigm
36 mm
Jag = Zot = Eloais)' = 79-5x107 mt
9
“Tao Las
30 me
Glan
8
mo
The
ae
mm 10202-5
Fe
Pe
0.6 pr
ED
Shaft
T
Q
CTI x10! )79-5 018%),
6Tas Ps i
yo=
=
Teo ,
i
I
ben = o-Gm y
Eo"
_
= ¥-(0-018)"
Cotas
>
Tckes
acl8m
Apphied tovave
[64 DK 107 Tn?
T=
fe * EK.
The
=
Galen Te
~
CTT
q
rio
77
LOSE Tx18
Clearance totetion for Phange C2
Torque te remove clearance °
<4
Le.
|4log-1
Dy
Qe = 15° = 26.18%10° yacd
Tog = (4108-1
Torque T" te cause additional
=
57°Nm
rotation
(26,18 yo*) = 364-4
g's
T's
570
- 3674
Nm
=
200-6
Nin
T"= The + Teo
200-6 = (/0202-5
)p' + (H4108-1 ) p”
The =
(702025 \(#25/5 xjo*)
SQM = 8.2515 «107 pad
942
Nim
Teo = (4108+ \C8-2515 KIO) = 11b-4 Nm
Maximum
shearing
Cig ® Tage
y=Po
shearing
Feet.
ke
in
- (44.2) (0-015 )
Sha
Maximum
stress
AB,
=
19:5 Kio
stress
in
(5-89 Mfa
Tag * IS 7 MF
«4
CD.
boas164.9x10
b4Vlo-018)
_ 53.93 mea
4
V,<53 MPa
et
The steel jacket CD has been attached to the 40-mm-diameter steel
3.59
Problem 3.59
shaft AE by means of rigid flanges welded to the jacket and to the rod. The outer
diameter of the jacket is 80 mm and its wall thickness is 4 mm. If 500 N+ m
torques are applied as shown, determine the maximum shearing stress in the
,
jacket.
SoPid
shalt:
czas
0.020 »
Je = Bet = Blo.ow)* = 251.33 vio! mn"
Jacket:
C,- £ = 0.040-0.004 = 0,036 m
C,=
carvredd
Torque
carnied
by
by
Shalt
Ts
jacket,
Ty
Is
Js+Jr
Meazinum
TS
vs
Shean ua
_
GIy P/E
Problem
.
(38290) (S00 5
3%10*
1.39
F252
*8
2
1.*1IO
Stress_in
Vaaayxio-*
3.60
¥
423.1
Mem
jacket.
- Tee.. 423.1004)
“RF
«. sg -
= Wei GG
Tote) torque, T= E+
=
1.3829 */6% m*
7 GIs O/L
bt]
Torque
E (0.040'- 0.036") =
4
Ty = Efe," 4")
¢,2$d = 0.040 m
22
4yjo* Pp.
12.24 MP2, <0
*
"
“
3.60 The mass moment of inertia of a gear is to be determined experimentally by using a torsional pendulum consisting of a 1.8-m steel wire, Knowing that G = 77 GPa, determine the diameter of the wire for which the torsional spring constant will be 6 N + m/frad.
Tovgion
Spring constant
Ket:
Pp”
a
cts
ts
h/6S
2LK
TO
C= 0:00307m
Kr
&
BL
hn
.= “wl77
@ycuaé)
107)
Ne
/racl
mG
ZL.
.= 89.292 "1
d=ac26
tz"m
mm
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No pact of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Hmited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<=
Problem
3 61.
.
3.61 An annular plate of thickness / and modulus of rigidity G is used to connect
shaft AB of radius r, to tube CD of inner radius r,, Knowing that a torque T is applied
to end A of shaft ABand that endD of tube CD is fixed, (a) determine the magnitude
- and location of the maximum shearing stress in the annular plate, (6) show that the
angle through which end B of the shaft rotates with respect to end C of the tube is
Use o Free bod
conststing
of sheoFt AB add an tnne
ortion
outer
| of
the
radius
ot
ns
portion
ye
The
force
A.
SMe=
per
“ss
v
a
+
\
\
.
v
7
Xe
x
'
|
t
0
TE Campie
_~_
onit
of circumference
Pength
UA
Sate. BC, the
being
Tv
Ome = a{t-4)
AnGt\ 7?
~ T=090
ft.
autp~
shearing
fa.) Maximum
‘
.
Sheaning
strain
~
~
at
ocaurs
stress
t.qe “*
The
rh
aa
ae
Tineae =
p= i,
,
GES
relAive
radiat
aS =
circumterertiad displac ement
fenath
ap
Y dp
=p
is
do
ap = r#
>t
ap
wo)
anGtp~
@=
Foie
'
\
dep
a
a
(Loe
sree
e?
alt
ft
~ 2auGt { are
_
TF
” 2uGE
|
2
fay
ans |
p3
Tt
anGt
LT
nO
j
(to
or
aap*
r
* ANGE
fb
~4anGe Lis
My
fy
lt
at
=
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.62
_
.3,62. An annular brass plate (G = 39 GPa), of thickness ¢ = 6 mm, is used to
_ connect the brass shaft 4B, of length L, = 50 mm and radius r, = 30 mm, to the brass
- tube CD, of length
L, = 125 mm, inner radius r, = 75 mm, and thickness ¢ = 3 mm. *
Knowing that a 2.8 KN + m torque T is applied to end A of shaft AB and that end D
_ of tube CD is fixed, determine (a) the maximum shearing stress in the shaft-plate-tube
system, (4) the angle through which end A rotates. Hint: Use the formula derived
in Prob. 4, 1 to solve part 8.)
Shaft
AB,
a
ar.
at
Tes
Ty 3
_2)
ee)
T(0.030)8
=
Plate BC
BC,
€6,0IO
=
(See PROBLEM
Cr S|obEve*:
‘ Pa
3.6)
fo Jeccvedkon)
eed
9 .030)*
6Y.0_0
2(0.00£8
82.5% 1° RH
= 92.5 MPa
Ret = O. ons + 0.003= 0,078 m
Caz
C2 Mt O.07Tm,
Shakt CD,
Z60,4- 6,4) = £(0,07¢'-0,075 ‘ys B.4421 x10°S m®
T
=
Tan
Jt
~
~
(2800)(0.078)
_=
K44RZL «10%
(ao) Larsest stress,
Shaft
Png
AB,
ee
25.
q
¥/(?*
Fo,
|
~
= Thy
GS
se
Tran = 82.5 MPa oe
ATL
TEC
_ (2)(2800)(0,050)
~ WS4xiat)}(O.030)4
= 2.921 <107% ved
Plate
EBC, A See
Pee
= ah}
Wet
= 0.889
Pro
hem
-
"* “ie
10
3.61
MPa
25.9
=
,
Fan derivation
2800
of equation
}
ee0.075=f*
47 (89x 107 (0.006) ( 0.030"
ved
bePow,)
/
.
Shaft
(b)
Tote
CD, >
potud
Len.
jou eng
wm.
-
fe.
==
The
J
P,
=
De
(7800) (0-125)
--
(39x10" x B4uah 10" yO
+ Pre
+ Dey
=
4.773
=
:
[L063
x10" *
3
*i0
rae
ad
P= 9.213"
<4
Problem
3.63
.
.
_.,
EN
Co)
£2
C)
”
.
3,63 A solid shaft and a hollow shaft are made of the same material and are of the
Same weight and length. Denoting by 2 the ratio c)/c2 , show that the ratio T/7;, of
the torque T, in the solid shaft to the torque T;, in the hollow shaft is (a) J (i — n*)
/ g + n°) if the maximum shearing stress is the same in each shaft, (8) (i- ny +
un’) if the angle of twist is the same for each shaft.
by)
|
cy = Ny
For equal
wecght
and Fenath , the ane us ave equal,
Tee = Wleat=e7) = TeNM-n*) 2 C= Gq UI-nt)”*
Jg= Feo = Bel (-nt)*
y= Bl0")
at
= Eot
~i(a)
For equal
vt
TsSo.
~
IC
J,
7
~
Ts.
Ge.
Th
(b)
stresses,
_ Fa"(i-n’)*e,
Wyte
x
Fov equad
p=
ages
Gh
op
_i-n*
Taye ny
(1-n27
pene
trict .
Lf Tk
Gos
Sie
Ts 2 dL
tT
NG”
_
Fel"
| d-n*)® _
Vy7 eFimy
7 Tent
Problem 3.64
3.64
ae n?
_
Fane
Determine the maximum shearing stress in a solidshaft of 38-mm
diameter as it transmits 55 kW at a speed of (a) 750 rpm, (6) 1500 rpm.
e@
fe
i20
7
12.5
He
wm.
PF
2.
_§S«jo*
f*
anf
_
*
grizsy
7
y
.= Ts
20%
ont , @(Toe)
F(oveidy
(b) £ = (S00
T
;
TooN
.7
m
C5 MPa|
LEP
~- me. bE mPa,
<a,
He
- es xloFd
an@asy
ee ~ZU BSP)
H (610 14)*
350 Nim
ga
Mie
T= 22WSMPH <a
‘Problem
3.65
' -
3,65 Determine the maximum shearing stress in a solid shaft of 12-mm diameter as
it transmits 2.5 kW at a frequency of (a) 25 Hz, (5) 50 Hz.
u
@ $2 25We,
9 Te she = BBS
|
T
(6) F = 50 Hz
2(7.95 77)
T > “Floooeyt
Problem 3.66
= 15, 4155 Nem
_
2US.9185)
le .7 AT
Tet” , Floor
“FO
-~-
v
2500W
P= 2SkwWs
= 0.006m
cr bd= Gmm
7 16410
soo
>
Qaleoy * 7.9577
*
23.5 4108 Pa
‘
.
v= 46.4 MPa =e
Pa,
Nem
23.5 MR aes
” 3.66 Using an allowable shearing stressof 50 MPa, design a solid steel shaft to
—
transmit 15 kW at a frequency of (a) 30 Hz, (6) 60 Hz.
Lys 50 MPa = SO»10% Pa
(oa)
Fz
30
Hz
-
Te = stp
gnF *= Bele
Qa Cees
Te
Lan
Ft
tf
Ts
.
2/37
THe
3
P= 15 kW = ISx/o™ W
= 79.577 Mem
2&2
Fea
~*fQ004s77)
“PC S6 x1)
_=
10.044
oe
«(Oo
Mm
*
10.044 inn
a=
(h) f=
T=
Go
We
2 Pe.
2nF
~
ISxl0
wzn(ée0o)
=
2e*
20.lmm
aAz= 202
18.44 wn
=att
.
39,789
Nem
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
Cy
3.67 Using an allowable shearing stress of 30 MPa, design a solid steel
shaft to transmit 9 kW at speed of (a) 1200 rpm, (6) 2400 rpm.
3.67
3 WRe
P= FLW
Tea
k
=
ie
=
the
m
eo
Be
(a)
Fe
(by f=
eee
=
Je
-
SEP
T=5h
20
= /@NTnEZY
=
= yo up
=
c=
He
O.ollS
m
d=
eo = 236M
meats
|
ye
=
38-81
35-81Nm
=
Problem 3.68
@r604126
|
m
A=
2e=
(9-25
mn
Ah
3.68 As. the hollow steel shaft shown rotateat
s 180-rpm, a stroboscopic
-
measurement indicates that the angle of twist of the shaft is 3°. Knowing 2 that G =
‘77.2 GPa, determine (a) the power being transmitted, (6) the maximum shearing
stress inj the shaft.
|
a“
1
2 d, = BO vaya”
Cy
>t dy
=
I. 2BHK IDE mara” =f
234 <¢/0°% pw"
Te Hegde i= E Lesoy" Cs Y]
Qe
T
25 mn
=
Anuguban. Speed
(a)
Pow er
b eng
~
ere
Poe
tyan gm
3° es
[BOupe
tte,
0.08236
C77. 2210%
=
a
vad
Mi, 234sie”
3 vev/see
r
P=
e.05%6)
oF
—
997.61
Nem
ZH2
QnkT = 20 (3V( 947.61) = 18, R0K/0" w
"
60 mm.
125 wise
P=8.20kw
(ob)
Maximum
sh CAVING,
stress.
Ton =
Ter . (497.6
JT
7
Goxlo-®
1.224 «107°
= 24.3 xo" Pa
2 24AMPR
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~—-
Problem
3.69
3.69 The hollow steel shaft shown (G = 77.2 GPa, t= 50 MPa) rotates at 240 rpm.
Determine (a) the maximum power that can be transmitted, (6) the corresponding
angle of twist of the shaft.
6, 2 EAL F BO mm
y=
C, = GAL = 12. Svim
Bet - oc): coy -aa.9°]
"
"
= 1.934 X10% pm
T., = SQxloe
60 nm
eee
te
vw .ztTt
= -
ct
Bug vbax
speed
Fe
240
;
6
a:
079“6) _-=
TF _= GOrioy.234
te
Zo mo
=
vow
=
Ynev/see
Peantt:
Power being transmvited,
(a)
Pa
ne
~
Tv
C
25 my
= 16234 210 om!
=
4
bes
y
Angle ftwist,
Qs
ar@)\l2aose.7)= $1.7 ¥lo* W
)
(4°52. sao)
Te .= (amo
. i
SL?
kW
_70-1078
pad
id
- Problem
3.70
;
Nem
Ha
P=
Ly
(L)
R056.7
Glui7?
eo
a
3.70 One of two hollow drive shafts of a cruise ship is 40 m long, and its outer
and inner diameters are 400 mm and 200 mm, respectively. The shaft is made of a
te
YO
steel for which 7, = 60 MPa and G = 77.2 GPa. Knowing that the maximum speed
of rotation of the shaft is 160 rpm, determine (a) the maximum power that can be
transmitted by one shaft to its propeller, (5) the corresponding angle of twist of the
shaft.
wm
C,= Gd,=
200mm
T= Eel)
e,., W
CO
F =
TO
(a)
M cag tevin
P=
(o}
, de
Ang
f
C2 tay = 100 mm = O.J00m
+ £(0.200'- 0.100") =
2.3562 «10° an"
%
~
LS
me
- 0.200 m
‘Ts
2.66667
2
JH
(2.3962™1o*)(Gox100)
0.200
Come
) Hz
P=
pewer
~
anf
|
=
:
FOG .8E%f/O”
T
10°
aw (2.6667 (706. 86 xIO*) = 1,84% 4
Ww
P=
= (55.44 x10"° red
(77. 2 #1072, 3862 4 10°°
?
bey
ot twyst
Alem
L384 MW
«a
Th
p= . ae
= (Fee, 86x10")C 40)
=~ gatoI
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.71
3.71 Asteel drive shaft is 1.8 m long and its outer and inner diameters
are respectively equal to 36 mm and 42 mm. Knowing that the shaft transmits
180 kW while rotating at 1800 rpm, determine (a) the maximum shearing stress,
-
(b) the angle of twist of the shaft (G = 77 GPa).
Lz
ben
P-
1okw
£
C, 244, *
~ 1800
GO
.
°
BO
28 min
He
T
oe
c= dd, =
oP
Almm
2= “2HBOK
(Be)
2aFE
=
9550
Nin
I= Fle" ~ 6,1)= E@v028"~ oor“) =bboyro mm 4
EAC)
we
eo
te
Pr
.
dwist
_ C455)
G8)
Aeenanertinmnenemnaninanmntsnuitntirtiainininene
CT? xXtat iC bbe x10 4
Problem 3.72
c= 405 Wry
“St
WwW
oo
of
Tee
do — Mla
A55)(0s028
Le
Ang
vs
S
(oe)
stress.
Ww
pe
sheaving
°
Masimun
}
{1
f
x
&
bag
&
<cea
3.72 A steel pipe of 72-mm outer diameter is to be used to transmit a torque of 2500
N - m without exceeding an allowable shearing stress of 55 MPa. A series of
,
72-mm-outer-diameter pipes is available for use., Knowing that the wall thickness
of the available pipes varies from 4 mm to 10 mm in 2-mm increments, choose the
lightest pipe that can be used.
Cet bade= 86mm= 0.086m
J
y
vo
RT
C2
Te
10%
=
C) =
2BL26%
m
Be
Cy-6, 7 BC mm
29,26
W(c,3-C,7)
(AAS 0e \(0, 035
0,036°=
/
T (ES
x 10%)
=
G37. 875 «10%
mn
28,26 mm
=
7. 74am
Use
C= 3mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced,
or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~e
Problem 3.73
3.73 The design of a machine element calls for a 40-mm-outer-diameter shaft to
transmit 45 kW.. (a) If the speed of rotation is 720 rpm, determine the maximum
shearing stress in shaft a. (5) If the speed of rotation can be increased 50% to
1086 rpm, determine the largest inner diameter of shaft b for which the maximum
shearing stress will be the same in each shaft.
(a)
>
= 12H2
P= 4skW = 45x/0° W
4s fo%
TT _ a “Saliay 76-88 Nem
(b)
(a)
=
-
v
_ Te> _* Wee
27. .= TWe.o20)=
(2526.93)
Cb)
1o80
“EG
=
~
FS
Ie,
Hz
cts
.
E
TY
Nem
om
-4
C= 1S.20 «10m
b=
requirement
{
GetD
L
|
Pr
i2kW=
mt
Uxfotw
t= 30/0" Pe
G = 77.2>10" Pe
. tea
.
3
20
=
62.722 Nem
ue
P=
,
|
7 Wes
aT bh
.
WT C17, 2107 Yo, ov
TG?
el.ofe eae)
(ays)
albPourahte
torque
Minimum
frequency
4
}
te transmit
is the
smebfer
a fpower
value
of
=
72,305
Nem
T= 62.742 Nem
{2 kW.
anf Tt
|
¥
“ue. . <8
Bren=_ 47.5 MPa.
d= 2c, = 30.4 mm
Mawimumn
Pe
|
We
= $3,333 x/o
67
T= Evct=s E(Bonto
oon
anate
ae
»m
ote AVC
= 0.0lw
requirement :
Twist
0.020
3.74 A 1.5-m-long solid steel shaft of 22-mm diameter is to transmit 12 kW.
Determine the minimum frequency at which the shaft can rotate, knowing that G =
77.2 GPa, that the allowable shearing stress is 30 MPa, and that the angle of twist
must not exceed 3.5°,
a=hd= tlm
Stress
=
397.89
a
|
B.5° © GILO87xIO” vad
Pz
me
= 15.20 mm
3.74
L=1.5m
AO
+
'
CERT icom
C, 7 IS.20n10"° m
Problem
|
tT (¢,¢-¢,4)
87)(0.020)
M347.
0,020"~ ee
ct =
=
47.494 x10"¢ Pa
Ts She
.~ &2£Te2.
J
_
td
_P.
24
_
i ~}o?
ZW (62.722)
=
B0.4 Hz
f= 30,4 Hz ~<a
“Problem 3.75 _
3.7§ A 2.5-m-long solid steel shaft is to transmit 10 kW at a frequency of 25 Hz.
Determine the required diameter of the shaft, knowing that G = 77.2 GPa, that the
‘allowable shearing stress is 30 MPa, and that the angie of twist must not exceed 4°
P=lOkW = loxlo Ww
V
lo ¥1o>
= $5F
Stress
“an (as)
Fe 25 ve
7
63.C62
requirement,
Y=
.
N-
i
=
at
ee
Sf aMes.ce2)
cr JA
Sar 2= YEO.
~— wist
engde
a
Th
Ga
z
~
QE. CLIT L SY
TGQ
Use the
|
Ios mm
I]. O5S#10otw my == NL OSS
rguivecseed
o- aT
P= Hor C4BIBKIC vad
ATL
A
= 11.709 wm
W 77.2% 107 KGF, 813 *10°*)
havger
vedue,
G2
ih 7OF
3.76
Problem 3.76
wm
Az
20>
23.4 um <i
The two solid shafts and gears shown are used to transmit 12 kW
from the motor at A operating at a specd of 1260 rpm, to a machine tool at D,
Knowing that the maximum allowable shearing stress is 55 MPa,
determine
the requixed diameter (a) of shaft AB, (6) of shalt CD
(Q) Sheff ABs P= IZEW
f=
Ee = al Hz
7y,
=
ae
Ps
ar
».
Te.
ts
_
eso"
=
analy
:
2£T
“-
T+
Wes
7 @AD
|
c*
©
=
Ze
=
201%
$SMPpa,
Tim
aT
ET
VF Bexeey 7 Oo lor
dag
*
|
wry
(b) Shalt co
Ang = ZO
t,
4
ets
= IS(7 Alia
Tre
€
NIST)
Ja
dep
My
- £(
= Re
TC EEK Oe)
=O.624
™
6\Ol206 uy
deo® 24. pm
et
Problem 3.77
3.77
The two solid shafts and gears shown are used to transmit 12 kW
from the motor at A operating at a speed of 1260 rpm, to a machine tool at D.
Knowing that each shaft has a diameter of 25 mm, determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
(a) Shaft AB:
Pr iz kw
fz Rea. x
21
Hz
12
x {o>
The * 20F
= 2WCAY)
cetd = (25
ev:
le,
T*
mm
£7
MW
Fa.
CD
(b} Shaft
co
FIinm
J
Wo?
YT)
_
TF (@.oizs)> 2466
-
Teo?
=
Vs
a
Tra * 34166 mA,
the =2CA1)= 1SET
aes
ag”
-
°
3
=
(2
yeste7
of
)
wines?
<<
AN
Nm.
= 49°44 fa,
Lop? FU44 MR, mt
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educatars permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.78
3.78
The shaft-disk-belt arrangement shown is used to transmit 2 kW
from point A to point D. (a) Using an allowable shearing stress of 66 MPa,
determine the required speed of shaft AB. (6) Solve part a, assuming that the
diameters of shafts AB and CD are, respectively, 18 mm and 15 mm.
SOLUTION
V2
v=
b6Me,
=
le
=
-.” aFES
&T
Abbouabfe
lomm
C2
P= ZkWds
Ts
« Ee-3
ZOU
torg ves
diameter
fmm
)
shatt
Todt * E (ood)? (Lexie)
=
S3Nm.
[Simm diameter shaft
Cz9mm,
Tae = FE (01009)* Go vi0") = 786 Nm,
Statics:
"s
Te= TeCR-R)
Tb
mF
Te
=
Ve
=
a |e
30
io”
Ye
|e
°
e
02727 Te
.
(2) Adbfowabte torques
Assume
Tee CR-FY
Tae = SNM
To = 756m
Then Vy
Toa = TS 6AM.
= (02727 )CK-6) = 20-6
2
<
S3Nm%
Pe aakT = te "aie 7 Litedy
~
(o)
>
Abbowehfe
Assome
~
torques
2060
Ta, ade = 756 Nm,
T, = S3Nm
P= 2nefT
~
Ta
AIT
.
Okay )
hese
45 Ha =e
Toe
= S3NM.
Ty = (0¢2927 )(<3) = 14% Nim
Then
-~-R.
Nw
|
7
__peco
TCH)
< 756 Nm
oa,
| Fe = 2695 Hz ee
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
Do
3.79
3.79
A 1.5-m-long solid steel shaft of 22-mm diameter is to transmit
_ 13.5 kW. Determine the minimum speed at which the shaft can rotate, knowing that G = 77 GPa, that the allowable shearing stress is 30 MPa, and that
the angle of twist must not exceed 3.5°.
Twist angle requirement
it
£
.
62-7
Nin
= 35°= 610871 red
eg:
WC 77K107 Core) (c1.087 x10)
—
U2
= Jaof
Pe
=
TGV
Te GSP
LO
Mewimun
=
= E (Bowe V( owoll)
Eye
T+
qe fe. 2u
20Mfha
Ce
requirement
Stress
Pe (3-0 EW
Hmm
cefa=
Le bom
-
(AW
Alm
cblowable
dorque
is the
smePher
velse
T=
62°7Nm
2nfT
P
_
“ant
13eS° x
fo*
~ 2n(€ 62-7
)
=
34.27)
Ue
Po
34e3 Hz
=
Problem
3.80
.
-
|
2, 05%
rpm
3.80A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz.
Determine the maximum power that the shaft can transmit, knowing that G = 77.2
GPa, that the allowable shearing stress is 50 MPa, and that the angle of twist must
not exceed 7.5°.
artde
ISmm = O.01Sm
=
’
:
Stress
yequiverce”,
.
7
Le 2.5 m
=
Soxlio®
¢
Pa
@:
Te
we
o
2. Ere* + Foro Xo.ois) = 265.07
efCc
T=
:
Twist amade
requirement,
p=
BTL
te.
6
P=7.S°*
130,90 ¥/O” rad
N-m
G=77.2x10" Pe
TGct
tT: £6c'g = E (77,210 )(0.015) 130. Fo¥/6") * $03.62 Nem
Smaller
value
ol Tis
the
mapimum ablinie hte torque.
T= 265,07 Nem
Power dvansmiHed at f= 30Me
Pe Anh T=
an (s0V165.07) = 49,96r10"W
Pe 50.0
kw
«=
Problem
3.81 —
°
;
.
3.81 A steel shaft must transmit 150 kW at speed of 360 rpm. Knowing
that G =
77.2 GPa, design a solid shaft so that the maximum shearing stress will not exceed
SO MPa and the angle of twist in a 2.5-m length must not exceed 3°.
:
Pr ISOkW > ISOxl0* W
a
|Tae
Qn Ft
|
Stress
iSQ* 10%
ot
aa
3
Ane De
of
2
=
Vf
twist
_
» %X10"
3.9787
3
che
New
TF S0x10'P
.
p=
-
=
= 87.00 ¥/0%m
3°=
$2.36
[0
2 PONE STIS
SZ S)
y VWE¢
te
.
GXSATs lO"
T (50x {0%)
requirencF
oe PATO
Use
=
(6)
reguivement,
oxy SE
Te
= Boe
~3
= a7
= 37.00 mm
Vad
e235. 39010? m=
35.38 mm
WT7.
10°) ($2.
86% 10°)
2*
Rarer
vatue, —
C+
BOO
mm
d=
2e*740mm
—xa
3,82 A 1.3-m-long tubular steel shaft of 38-mm outer diameter d; and 30-mm inner
Probiem 3.82
diameter d, is to transmit 100 kW between a turbine and a generator.
Determine
the minimum frequency at which the shaft can rotate, knowing that G = 77.2 GPa,
that the allowable shearing stress is 60 MPa, and that the angle of twist must not
aE
exceed 3°,
Leism
P23? = 52360107? pad
Crt Gd, ANI mms O04 m O24 bzt 1Smm= 0.015 m
d, = 38 mm
Je
require
Stress
.
“4
i
“ae
60
T=
ment,
Eee,f-¢,4)> F(o.0141-0,
018") = 125. 186 107
wor
T=
Pu
of
C
3.96.32 Nem
re SE. Mas lero Meer)
2
ye
Menimum
| Power
*
requirement,
an 4 De
Twrst
GIP
FrequencyN
-
Pp
nT
1S
is the smabler vale,
T=
Pe {00 kW = fooxlo*W
mitted
£
te
pe
. O7ZxI0% X125.186"101)(52.360x10)
L
ebfovable toryve
trans
Ts
T
loox {o?
2h 33). Be)
—
.
2.2
=
3232.35
Nem
337.35 Nem
Pe ant T
He
=47
42472
u
Hz —<—m
_ 3.83 A 1.5-m-long tubular steel shaft of 38-mm outer diameter dj is to be made ofa
Problem 3.83
_ steel for which ta = 65 MPa and G = 77,2 GPa. Knowing that the angle of twist
must not exceed 4° when the shaft is subjected to a torque of 600 N - m, determine
the largest inner diameter d, that can be specified in the design.
Leh
dy = 38 min
Sm
c,2
dh dat
J * F(c,’-¢,")
+
vos
a1 Ve SF
war
WE GUY E yr er
4
fe
Cc,
4
~
q
21c
-
a
~
-
2.014
rust engle requirement.
f°
—_
2.948
Cy
Use
4 ATL
C, “162
sr able
a)
-
.
es
- “(ase
.
.
1.68410,
“So m=
ms
1EG84-mm
rhe
27TL
rae
a
f
~t
Problem 3.84
.
(77. 2x 109 G64, BIB x L0"*
12.4493
CG.
’
(Z)(Go00}(0.019)
ibe)
°
*10° pm
Jabye
Wet-,4)
_
=
4
|
D>
GE
*
FOCERERT
0194 . 2\C620 YCIS)
.
_4
Cc, *
a
OL.O14m
2Tez,
=~ 1S
J
’
=
D4? = GF.BIB «10 rad
2 6S¥l0% Pa
_.
l¢mm
*
11.689
a: = 26, = 23,4 me, ae
ma
3.84 The stepped shaft shown rotates at 450 rpm. Knowing that
r= 12 mm, determine the maximum power that can be transmitted without
exceeding an allowable shearing stress of 50 MPa.
t
125 mm
150 mm
+
ad =DS me
—D -fJOmm
ye
(2mm,
t
pr
R ~ {£2 = 1.20
a
iz§
For
the
smabler
ween
A
_ 12
“iz
»
ectgd-
side
if (@:0b28)*(gox 0%) _
CATER TS
“
YOO
Power
vpn
1S
From Fig 3.82
01096
62-5
[Yor
mig
“x
Ke 1.35
KTe.
J
2KT
“te
ENaa
Hz
P= anfT =. 2an(zsKH2
10) =
667 bW/s
p= 669kuy
Proprietary Material, © 2009 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ey
Problem
3.85
3.85
The stepped shaft shown rotates at 450 rpm. Knowing that
r= 5 mm, determine the maximum power that can be transmitted without exceeding an allowable shearing stress of 50 MPa.
d=
Da _/sSv
.=z
“3S
Te
_
fox
Power
pf.
A TS
|.20
Teer
assy
pin
>
_
0,04
WT @vbesPsoxoe)
RK
450
=
PD = MOR
Hypo m
Ya
Fig.
7.85
_
Te
Chem,
De
K
=
3.32
ee
1.55
Ze
-
: Kis
= 63S wm
crtd
side
smaller
For
(2046
ENE.
Wz
Pe QntT = an(nsX i294 xto®) = IY
kW /s
P-
Problem 3.86
the radius of the fillet is r = 8 mm and the allowable shearing stress is 45 MPa,
determine the maximum power that can be transmitted.
60 mm
:
x=
_KTe
Tt
30 mire
_ 2KT
—
-
TO?
T
se
oO
A Powel Le
Fig.
3.32
~ 353
tovque.
power,
Tr
Pe
*
tsi)
anfT
*
P=
B mn
Sf.
From
-
o
CPgaleiSmm
D=GOwm
a
aK
.
BO vam
Be
oe}
=
.
a=
Maxi mum
ey
3.86 The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing that
7
;
SPG EW
=O. 26667
Ke
Joi
.
(Cas
10°)
.
=
= (aw)\(50)(202.17) =
202.17
Nem
63,5x10>W
P-
63.5
kW
<a
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers _
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3.87 Knowing that the stepped shaft shown must transmit 45 kW at a speed of 2100
3.87
Problem
‘
rpm, determine the minimum radius r of the fillet ifan allowable shearing stress of
50 MPa is not to be exceeded.
:
,
60 mm
p=
30 ram
P=
smedtler
side
. ment
Ks
ue
2Yv
*
Vis
o
2
OWT
Problem
=* 4Sam (35)
1o* 2=z
=
Sam
204,63
© 0,015 m
Nem
ts
From,
= (0.171680)
3.88
.
Fre.
=
|
Kye.
ee
|
QVMLOY, ERY
2
&
Ww
wlSox~iot(o.ois)3
~
£2.
A
aap
crtd
—
a5 ue
45x10?
2
vv
row
EEO.
=
3.32
1.295
ra
=
0.17.
SL) am
vc S.1 my ~<tll
3.88 The stepped shaft shown must transmit 45 kW. Knowing that the allowable
shearing stress in the shaft is 40 MPa and that the radius of the fillet is r= 6 mm,
determine the smallest permissible speed of the shaft.
60 aim
30 mm
/
Teer
Tes
,
Dig
da ° 3
& 20.2
¥.
d~ 30 70%
Co.
From
For
Fra.
3.32
smadler
We
rele
1 2E
cztol<
vy. KDe
. 2KT
wet
0)
\CHo
¥ 108
P=
ak
VE)
45 kW = 45%]0%
Pe
ph.
ant
-~ lO
anliee.sox108 )
bes
oh
=
[68-30% 10°
Pe 2a0ntT
2g
Simm = O.91S m
7
(0.018)
WlO-O1S
"#
ie
3
.
Nom
£2 42.6H2 <a
Proprietary Material, © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.89
.
3.89 In the stepped shaft shown, which has a full quarter-circular fillet, the
allowable shearing stress is 80 MPa. Knowing that D = 30 mm, determine the
largest allowable torque that can be applied to the shaft if (a) d= 26 mm, (6) d = 24
From
Fig. 3.32,
Smahber
ne .
5
Fall quarter-circular
men
From
2
|Bmmt
Wes
TEE
2K
203
es
Ke
R (0.013'°(80
(2)U.38)
x 108}
N-wm
-LDed)
°3(D-d)
== 8mm
Fig. 3.32,
Ter
Tage
9.013 re
\
T=
32.
FU 7 18s
Problem
Ce $a
ae
fille
to edie of barger shat
QD,
pr
1.36.
2KT
S
=
(eo
side,
Ke.
<
K+
“Te 208
a
13h.
<a
Far = OLN c
ergd= IQmm= O.0!1Us,
Tf
80x }t
4 (e.g) aon)
> 16S. 8 Nem
3.90
New
- 3.90
T= 1653 Nm <a
In the stepped shaft shown, which has a full quarter-circular fillet,
D = 30 mm and @ = 25 mm. Knowing that the speed of the shaft is 2400 rpm
and that the allowable shearing stress is 50 MPa, determine the maximum power
that can be transmitted by the shaft.
Dp. 22 |
a
r=4(D=d)
~
a
v=
at
-
3(D-
a)=
Ke
1.34-
2.8
mn
at ge tel
For
Fig.
3.32
smaller
v=
ns
Full quarter-circular fillet
extent to wdite of larger shaft
2
us
From
ve
-
side
KTe
c=
Te
J
Tr
Se
tad
”
Ke
WT (e-0126)*(50 Xo")
=
+ (265 mb
Ter
2¥*
Ge
NM
(AGL 34)
+
Pe
vr
2400
pm
*
uO
He
antT= an(4olive-s) = 2% ew
P= 28h kW
mee |
bien.
Problem
3.91
3.91 A torque of magnitude T = 22 N « m is applied to the stepped
shaft shown, which has a full quarter-circular fillet. Knowing that D =
_
25 mm,
determine the maximum
shearing stress in the shaft when (a) d =
20 mm, (b) d = 23 mm.
(ay
Bee
Se
a
a
we
Tt
h2F
26S”
keen
ot
Oe
pe kDa er 2S mm,
IZ
20
Feon
Fig.
3.32
Fer smebler
KIe
vy:
yt
Fall quarter-circular fillet
a
>
K
ver =
j-3]
= lGmm
c=td
side
AKT.
-
(20312)
= 838 Me
“Sperone
PTNG-3M Py
extends to edge of larger shaft
p-=2.
(b)
i.
3
Ge
-
O043
From
=$(o-d)=
r {104
.
Fig.
kK
3.32
=
Gg
saint
pe BEE, AMEE HD ost a
For
smedfew
side
c =tel
h4
Problem
a.
3.92
o 1S
mam
Ze
:
3.92 The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with ty= 145 MPa. Determine the magnitude Fot the applied torque
\
when the plastic zone is (a) 16 mm deep, (b) 24 mm deep.©
oe
es BZ mrn = » 0.82m
yeh
-
+
(a) tps 16 mm = 0.016 m
Fe 14S * 108 Pa
Fexy-F E(0,032)*(I4sm1o*)
4,6402x10*
a
7.4634x]/0% Nem
Py = €~ bp = 0.032-0.016 = 0.016%
Te By (1-$EeV = Ses
(b) te
:
oe
= 32 mm
=
{min
N-m
Ztmm= O.O24 wy
BBR
:
Ts aed KN-m
Pr? c~ t, = 0,032- 0.024
<e
= 0.008m
47-4 4 fe \= £07 46341- 4,4 2928)
=
4.4123 «10? Nemo
.
T=
FF
Nem
=
3.93 A 30-mm-diameter solid rod is made of an elastoplastic material
. with ty = 35 MPa. Knowing that the elastic core of the rod is of diameter
Problem 3.93
25 mm, determine the magnitude of the torque applied to the rod.
o-olfm
Ty* Le.
Ty2
ECT
t
Bz dd=
fry = kay = 010125
¥ BSK0%) = IEG
E (wis
No
#086) - 4% OBE
ovoic?
2167 Nim
3
T= $5 (1-4 &)
38 MPs
af
..
&
Tr2le7J
mh
3.94
Problem 3.94
Nm.
<@
A 50-mm-diameter solid shaft is made of a mild steel that is as-
sumed to be elastoplastic with ry ~ 140 MPa. Determine the maximum shear-
ing stress and the radius of the elastic core caused by the application of a torque
of magnitude (a) 3 kN « m, (b) 4kN +m.
,
Ce
+l
cs
2Shmm
Compote Ty
(a)
(b)
T=
doo-o
T=
ts
Te
#000
The
maximum
t
Wye Str
<
|
3
Cfastic
> Ty
Sheaving
e
~
4
Sy
Pr=
oF
~
tT
plastic
stress
Pp
= 122-2 Mf,
=
eee
= an
f=
ike mG
.
Ty
Th
(1~
#2)
-4. 3
_.
.
z
Bette = Elowss foro’) = 3436 N
at
region
with
is
y_ (itor)
ease
= (ak 26)
rQs
25mm
Comp (2202 M Py, et
ebastie
cove
Toone ® vy
|
= OS 076
a)
atleo fa fia «oth
Pt card
<<?
Pez emma
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. .
Problem 3.95
o
.
3.95 The solid shaft shown is made of a mild steel that is assumed to be
shearing stress and the radius of the elastic core caused by the application of a
torque of magnitude (a) T = 600 N « m, (b) T= 1000 N- im.
2
tae
IS win
=
IS¥1O*
m
B=
at ousef of yielding
3
(2) T=
600 Nem.
(ea) T= 1000
Meei mum
Radius
ot efastic
3
=
pr =
0.46004
c
= (0, 46003)
=jds
fe
Pe
30 m7
Sa
theve
plastic
none.
Tm 2 148510 MPa <s
Ue |
Ty
Li
- (3 ]
2
0.42003
G.4O mw.
pr>
Sum)
<i
3.96 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic
Determine the radius of the elastic core caused by the
with ty = 145 MPa.
application of a torque equal to 1.1 7y, where Ty is the magnitude of the torque at the
onset of yield.
3.96
,
f
Problem
Glee
168,71
= f-
f&=/y-2t y
T
3.3MPq<e
Tis
Pe = 15.00 mm
T=
core.
3
New
768.71
2
|
Tr,
ve
stress.
Shearing
)
pre
Since
|
Nem
Pus sto
NB.3x1o* Py
cove
of elastic
Cg
wm,
= +
KECD)
Ln 2 Ee + HHO
=
Since T< Ty, the shaft is efastic.
e
stvess-
Meximum shearing
j
Tso”
=
s—oy
Radius
~3
:
2T
.
.~ 1
|
- a4.
30 mm
Determine the maximum
elastoplastic with G = 77.2 GPa and ty = 145 MPa,
O
.°
=
fu
15 wm
re
3%
O.83790
=
«
Wo
47\
\
(3 )\(t.1)
~ (EF
=
|
0.38770
7 (0. 83790 \UIS mm )
Pr
=
13.42
wim
at;
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.97
.
3.97 It is observed that a straightened paper clip can be twisted through several
revolutions by the application of a torque of approximately 60 mN - m. Knowing
that the diameter of the wire in the paper clip is 0.9 mm, determine the approximate
value of the yield stress of the steel.
= ta
“Wmm
=
Tidy $i
wet
me
So
Te
Bt,
=
Problem
O.45%107%
Tp
$ben- Fee
uy
= GOwmWem
+ Soria”
them
T
(2) (Goxto7?)
=
mw
3.98
‘
=
3.98
-
TT
2
,
3l4
<a}
The solid circular shaft shown is made of a steel that is assumed
to be elastoplastic with ty= 145 MPa and G = 77 GPa. Determine the angle
of twist caused by the application of a torque of magnitude (a) T = 8 kN- m,
(8) T = 13 KN +m.
(ered
= dlo07)
= 0:03Gm
L=sam
76 mm
‘
Tt
Torque
J=He'=
p:
at
onset
Ty —= Ted
= _=
yt
2)= 14S MPa
oF y
oo
Cesx 10°)
Hoos)" = 3275x107 4
. l¢
- 2S
"4
me
(a275 «18 7)
“swe we
* ae
T
c
= (5-929 km
@) T= 8xio*nm
Siwce
T#
(alo )a)
~~
C77 xlol \Ga75xio?
-
p
Ty, the shaft is Folly efastre.
=
39-069
X10
u
it
i &
P
(e829lotx10”
U2)
\ea7sx0r
Cire
7)
P =
tad
-
,
(6) DES %lo Mm
Tek
.=
GT
“s
9g = =
[> Ty
T=
ag,
“3
75-324 x10
29/8
ee
—_—
$m ]1-&)]
vad
3
3
H- 3x ly = 2/4
@laHOD
_,,11538
+
15-829 alo?
iv
vs
Proprietary Material. © 2009 The McGraw-Hill Companies, Tac. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers
and educators permitted by MeGraw-LHl! for their individual course preparation. A student using this manual is using it without permission.
~~
Problem 3.99 _
3.99 For the solid circular shaft of Prob. 3.95, determine the angle of twist caused by
the application of a torque of magnitude (a) 7 = 600 N - m, (6) T= 1000 N- m.
3.95 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic
with ry= 145 MPa.
Determine the maximum shearing stress and the radius of the
elastic core caused by the application of a torque of magnitude (a) T= 600. N- m, (6)
T=1000N-m.
sid
at
Tongue
‘
()
T = G00
ow
te.
P*
(b)
or?
T=
[ooo
T "3%
.
Lei
Nem.
Pp
uO
-
=l1Sxlot
~-
w-
of yiebeling.
onset
1S
“ al
Ie
_ Since
RTL
Te
Ty
the
\leoe)(l-2)
shaft
=
is
ree
Nem
768.71
efastic
OCLIV2S
wa;
-
WEG ~ TUS Ko) H(77.21OY) | PANIED aed
Vem
,
Tv
\ |
—
¥ “GT
Ly
¢
mm
—
-3\3
sseict)x
E+ 2mfr Tsao)
Te
0
tS
(G= 77.2 GPa).
(Sy ]
2Tb
> Te
zone
has
developed.
Bs
jt-38
~ CA\CIEB.TINC ALL)
_
“nIe
y
177. 2x107~ O-15026 ved
Wet SM CISKIO*
aa
16B. TE
alee
A. phestre
PE GeIe
9
.
~
O.|o2G
0.46002,
O-46 00%
—
—
pa.
O- 22668 rad
~1R.717
pee
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
mn
Problem 3.100
3.100 The shaft AB is made of a material that is elastoplastic with ty= 90 MPa and
G =30GPa. For the loading shown, determine (a) the radius of the elastic core of
the shaft, (4) the angle of twist at end 2.
12 mm
@\
a=
IRmm = 0.0IRm
es E (0,012 (40x10) = 244.29 Nem
on
Te
300
Eee Fowl oe &,
Nem
>
Ty
,
plastic
req
tow
with
elastic
Cevz
reg t(iy ie) Gee to Shs (GUE =o sises
SS = O.GBIOR — Py = (0.68102) (0.012) = BI7x15°m
P= B.D am
(by)
Le
2m
|
~
pr
stds
|
T.012)1(S0x
oe
eh
|
5.500
0.8000
~~
107)
2.5090
*
pile
Problem 3.
oblem
3.101
L=
(2)\(ge0)
Pa.
D
-
yp
3T
(AV 294.242)
~
nwoiG
-
IG
MW.Pp
G = BOxjo?
. AWE
~ oh
d
.
ow
4
a
0.63120
=
0.73
¥y
*
ral
ad
pe
a
3.101
A 30-mm-diameter solid circular shaft is made of a material that
is assumed to be elastoplastic with ty = 125 MPa and G = 77 GPa. For an
2.4-m length of the shaft, determine the maximum shearing stress and the angle of twist caused by a 850 N + m torque.
,
(fmm,
Ge 77 6fa,
B4y,
T=
— Te 12s Mm,
@50Nu
Ty = GE. Betty = Flos Pasnet) = 662-7 Hm
T
>
Ty,
me
r
S
Ty
che
plastic reqron
.
* Dy =
Ny
2
“EE
=
with efastie core
bh
-
Emax = Tp tI25MR,
LGA U2sK0F)
CH
_
~ (ots UA7TK07)
~ Or26
ved
ae
a.” Ty (1-48)
att
Py
=
3f Uy
t
-
=
3
z
~
4
-
r
P=
16873 Py
= C1873) C026)
}
&bt-
=
.
(2 (BED
soem
_
{
‘
§73
TO
O42 7
red
.
Pe 214?
=
Problem
3.102
3.102 A
is assumed to
1.2-m length
angle of twist
Ty = [YomMPa,
eng. -
18-mm-diameter solid circular shaft is made of material that
be elastoplastic with ry = 140 MPa and G = 77 GPa. For a
of the shaft, determine the maximum shearing stress and the
caused by a 200 N « m torque.
Jmm,
Le f-2m,
T=
2e0Nm
Tye Utes B27, = EF (00009 (ironies) = (eo Nin
T
> Ty
Plastic regvon
Te 2b.
PW *
63
~ Te'GS
with
,
;
iy
=_
Py
od
or 242.
0-63
cd
He. 0462
~
Oe ZF
+
Pa
)
gz
léo
Tron? Ty 2140
= 0-242, -ved
Tloeod yt 77x04
4 3T_. y- B20).
f *= o79087T1
core
Av bev rz)
Te $7 0-43.)
()
ebastic
|
Fr ==
rea
#2
22°
)
es
3.103 A solid circular rod is made of a material that is assumed to be elastoplastic.
Problem 3.103
Denoting by 7y and gy, respectively, the torque and the angle of twist at the onset of
yield, determine the angle of twist if the torque is increased to (a) T= L.1 Ty, (6) T=
1.25 Ty, (C) T= 1.3 Ty
,
4
T= $1-48)
gk
@
.
(a)
=
b)
ce
(
Vy
¥
2
Vt
aT
Pp.
So
f
Gare Ty
= I.lo
Ot
ee
has
=:
@p
w
t
Te -(3)as)
@
ge
Py
%
=
V4 (BCBS
T
(CVoayrle
Ty
Phe
|
hie)
«1G
= 1887
=
245
Dz llaePy
mt
= 1.537
<6 |
?
?
p
= 2
%
IS Py
<a.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.4 04
.
(3.104 A 0.9-m-long solid shaft has a diameter of 62 mm and is made
of a mild steel that is assumed to be elastoplastic with 7, = 147 MPa and
G = 77 GPa. Determine the torque required to twist the shaft through an angle of (a) 2.5°, (b) 5°.
L=O9m ,
T=
Ty
C=yd=oo8im
z ct
£ (e-031)" = -4cx
Tye
it
=
~~
T.
.
FL
P=
7
( 6-876
25° =
S°
rR
(1-46 %16 Y4T» 10%)
nneceermeneninraimemmnamannns
Oil
Ip? \(oF)
_
7 SE4AT
P<
43.633x10° vd
=
9 =
=
IO
DL
SS
4
8-876
Bye
vad
io"
3
Nm
_
= 3-176
.
the shaft remaus else,
CTP Vote 1a404V(43,633x10)
pe GIP,
L.
p= EE
GI
(bo)
Tt
mt
FOr ONG se)
Pe = ST
(2)
-
= % =I MPa
=
S3Es3
87. REEXIO
x1O°
raed
Tors
Num
|
7] >
Py
A plastic
Zone
kNmM
ae,
occurs.
r= ET 1- 80" ] = § eore
x wf - SREY
>
3
= 866
x{O>
Num
S5-427«/0'°
T=@6
3
ENM
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
«ts
Problem 3.105
3.105 For the solid shaft of Prob. 3.95, determine (a) the magnitude of the torque
T required to twist the shaft through an angle of 15°, (4) the radius of the
corresponding elastic core.
BAS The solid shaft shown is made of a mild steel that is assumed to be elastoplastic
with ty = 145 MPa. Determine the maximum shearing stress and the radius of the
elastic core caused by the application of a torque of magnitude (a) T= 600 N- m, (2)
T= 1000 N-m.
(G 277.2
c= bd 15im
Bm
P=
Py
| Y
me
a
~
=
>
Lz
be
CG
Py
2
rest
Te
=
there
is
a
plastic
Ce)
476.6
LY,
Pr
=
_
p
_ ~~
om
a ©:
Oo.
IS
\
26
Vad
*
Zone,
[1 - HORRY |
Nem
PrP
¢CPr
)
. Tsnion
Mss!
Jye Cug x0 “) = 763.7) Nem
T= $y[1- 4G) |= See.7
=
15 «1O7*pw,
0.2613 vad,
(1. 2CIHS
*[O
(Sx1or5
(7.2K
~
,
15°
—
=
Ga),
=
_
Ts:
4a77
Nem
wa)
Pr
= B.6\
«»,
«i
Cy
(1s xjo7*)(0.
180 26el }
Oo
2613
=>
B.62«I0 > m
Proprietary Material, © 2069 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. .
_ Pp
am
3.106
roblem
3.106 A hollow shaft is 0.9 m long and has the cross section shown. . The steel is
3
">
assumed to be elastoplastic with ty = 180 MPa and G
= 77.2 GPa. Determine the
. applied torque and the cotresponding angle of twist (a) at the onset ofnfyield, (6) when
the plastic zone is 10 mm deep.
tay. At
the
elastic
onset of yledd,
di iste\ bution
the stress
with
Conos
distribution
i's the
> ow
my Cty = 0.015.
Fee 035*-0.018")= 2.2777 los m
= TH. (2297x107
V180 #08)
44 197 Wom
~G
= NLT) kNem
~
Fe,
Wt
(ob)
Ge
Q
0.010 mn
=
hk
.
GPr
Tor que
)
-
Pr * C,-t
UBoxlot)(0.9)
~ (972.2% |NLO.025)
T,
Carn eof
ot
p=
T=
PY
by
ebestic
-
ne
Sy
.
~ SHAG MIC” Yad
7 2xjoC2777 x10"*)
= GI
Pr
xto® Y0.9
=
06,0385- 0.0/0
— 23.94 x10
rae
portion
<P
q
Ute
where
Fe
=Zuyyeo
°
amet
= 0. O25wy
a
|
Pe
ez
pre-e")
J, = £(0.025*- 0.015") = 634.07« 10" m*
Tt:
'
Tow
ST
Pr
v
qve
Ty
_ (534.07¥1077)
(180 xlo*)
0.025
c
a
ried
b. 7
Pp
Dastic
P
.
3.845 «lot
Nem
or tien
c
3
= a (180 x1o* )(0.035% - 0. 02s?)
Tote?
tor
fi
di @ (cf - pe)
10.273 lo” New
a
T= anf" TP dp = 20% EY]
gue
T= T.+T, = 3.84S¥lo 4 10.273 lo
=
14.12 «107 Nem
4.12
KNem
ae
3.107 A hollow shaft is 0.9 m long and has the cross section shown. The steel is
—
Problem 3.107
assumed to be elastoplastic with ry = 180 MPa and G = 77.2 GPa. Determine the (a)
angle of twist at which the section first becomes fully plastic, (6) the corresponding
oe
magnitude of the applied torque.
:
Crake = 0.035m
G HFA 0.0m
(a
For
onset
v= %
Q =
$e
-
O.VY
-
[80 x10°)
(0.015 {77.2 4107>
cS)
r © 27 v Tp dp =
of
plastic
Folly
T= &
:
Ef
= 134, 90x10" vad
yielding ,
Pr * =
€,
= o.
|
ant & |= Bae
OB.
02°
tt
-c3)
= 2 (190 w/o" )(o.035? - 0.015°s = 14.84 «102 Nem
= 14.89 KNem
=
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.108 A steel rod is machined to the shape shown to form a tapered
solid shaft to which torques of magnitude T = 8500 N +m are applied. Assuming the steel to be elastoplastic with Ty = 145 MPa and G = 77 GPa,
- determine (a) the radius of the elastic core in portion AB of the shaft, (5) the
length ofportion CD that remains fully elastic,
3.108
(2)
In portion AB
Tye taste
zt - 3/inm.
4g,
= Fguesitoevet
=
b7OX Hem.
= $7, U- & C)
be -ft-Far
[4
QE)
eT
Pr 0123
(b) For yiekdling at point
T=
-e%Cy . H,?
q Cub
|
{
‘
|
’
b—~
6osf
|
8
Using
Tt
12 Sim
¢ 2(01623KB1)= 193 mm
Ve Ty, A= Gy,
Cy
= 0.623
x¥ foO*
O°
T= S000Nm
- lez
20 .= {@QEeEsx0O)
“Wh ligexsce) —=
proportions
from the
27-3304
~ — X
37-6- 3]
12
g.
660334
pm,
sketoh
|
_
——
= TERT mm
= 78°79 mm.
|
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.109
3.109
If the torques applied to the tapered shaft of Prob. 3.108 are
slowly increased, determine (a) the magnitude T of the largest torques that
can be applied to the shaft, (b) the length of the portion CD that remains fully
elastic,
3.108 A steel rod is machined to the shape shown to for
a tapered
solid shaft to which torques of magnitude T = 8500 N +m are applied, Assuming the steel to be elastoplastic with Ty = 145 MPa and G.= 77 GPa,
‘determine (a) the radius of the elastic core in portion AB of the shaft, (b) the
length of portion CD that remains fully elastic.
ty
The
davgest
makes
For»
Polly
plastre
sheFt
torque
portion
In
portion
Ty
we
AB
AB
: Tot
prr?@
that
my be
ap pA red
Tubby
P astie.
aztal
zs 3lmm
= Bros
‘Ts32 Ty
(ase)
point
Fer yielding at
- lm,
at
20
Iie.
Ty
]
J,
to?F
4
ede KN
Te Ty
C,
= 6ETE5 Nw
(ima Zz fe):
T= £66785 vJo*) = 1046 Nm.
QD
js that woh vel
Cr
Cy,
T=
may
904607 Nay
aso 41m,
)
= “(ar - s (ANTe 9046.7‘P=
0103
Cx
TL,
TIE
x76")
010%
Using
proportions Fpom
erst
3S]. $— 2}
2 a
the
sketuh
3L=
6S.
4 tay
25°
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ti
Problem
3.110
~
3.110 Using the stress-strain diagram shown, determine (a) the torque
that causes a maximum shearing stress of 105 MPa in a 20-mm-diameter solid
rod, (6) the corresponding angle of twist in a 0.6-m length of the rod.
+ (MPa)
112 | -~--~-—~~n
wee n-ne
84 | -------~
56
,
a1 -ri
0
i
i
0.002 0.604 0,006 0.008 6.010
TS
Let
where
the
Evaluate
iS
tule
iwhearal
tr
sed
usin
, The
T=
where
the
Wiisa
O-
I
=
te
ane? | Z¢d2
is
given
method
of
integration
weiahtin
by
l=
numencad
Porwuda
given
Facshoy.
m Phe
is
VY | TM)
Dsina
table
=
2nrerer
c
zy
ht:
met
bo“,
“4
oO
0.000
0.25 | 6.002
56
O.S | 0:00+
Az
wl
|
0.00
a5
4
4
84.
Zi
2
42
0.75 | 0.006
49
$slZs
Ly.
10°28
1.0
fos
ol
'
os
10.008]
~
=
272s <t—
te)
T=
(0.285 )(27/-2 8)
3
=
Sim
Son's
“Pp
0.25,
we,
get
Sw7er
ance D= gn (oorP(s26x0*) = a2 Wm
gp = LYm
-
(@6)(0.008)
Cc
*
IF
22-6 MP
Yawe = &B
Note
-
wore, Mfa
}
a)
dz.
bebow
277 MP)
0,000
Te
0.008
owz7
valves
Z 1
y
Vornye 2
diag nam
wing
L
a
ZF
:
an (" pt de =
- stypain
stress
the
From
ooim
cetds
Trnnge = 105 MPa
Q)
Answers
|
_ yp
ara]
may
in reading
adrfher
the
“
shight By
stress-strain
ero pad = 275
dve
tes
conve.
AitPerences
°
=e
oF
opinion
Problem
3.111 A hollow shaft of outer and inner diameters respectively equal to
15 mm and 5 mm is fabricated from an aluminum alloy for which the stressstrain diagram is given in the diagram shown. Determine the torque required
3.111
+ (MPa)
to twist a 0.225-m length of the shaft through 10°.
0
0,002 0.004 0.006 0.008 0.010
¥
ad, = 25mm,
"fl
OO:
P = 10° = 174.5310" rad
L
CQ.
Nein
Let
2
Tz
2T
Tow
g
the
where
&Q
C;
Pp
247
t dp
=
\
is Given
-
integral
os
21,
'
225
( 250743107)
-
L.
£.
=
x
=
=
MITI.S3x1o")
.C7S
= @
oe
C,= £4, = 75mm
=
2285
&
+
=
&
Zrdz
=
Ss
=
I
by
ang
I
z2°*7 dz
Eveafuste I using a method of numenicad integration, LF Simpson's
rule
is
used,
the
Ie
w
where
in
given
is
the
integvation
formuta
is
2 > wz ve
hhng
wer
a
=¢
Az
Using
t,mm|
2¢,mPa|
w
S6
$22
}
bu
2 | 0.0027]
70
Vs
uy
70
2/3 | 0.00383 |
Bes
35°79
2
Y.Sb
S% | 0.00485 |
41
b3-44
4 | 252-76
43
98
1
0.00582
!
Note:
QnceST =
Answer
in
valves
q8
IP SH. pe
T=
get the
|wz*tMea
0.00 194
43
we
befow.
table”
z|vr
Factor.
may
= 134 Nm
rn( “ors YP Gr70*)
ditfen
sfightdy
Wedding the stress-stvatn
FO
dve
te
al Perences
cove.
—
of opinion
Problem
- 3.112 A solid aluminum rod of 40-mm diameter is subjected to a torque that
produces in the rod a maximum shearing strain of 0.008. Using the t-y diagram
3.112
+ (MPa}
150
shown for the aluminum alloy used, determine (a) the magnitude of the torque
applied to the rod, (5) the angle of twist in a 750-mm length of the rod.
125
Vong = 0,008
100
5
L
50
@)
25
0.008
0.006
0.004
0.002
0
0.010
cegd=s
O.OROm
=
hed
22%
Te
20
(
a
= £
ne
0
7 do
=
S
2” da
q
3
§ 24 de
+
Simpson
= 2nre*tl
where
the
Evabvete
woke
integ rat
L
is
Ir
Using
used,
aa.
the
is
given
method
by
ol
integration
Ie
yum evita!
furmuta
inteqvation.
5
ts
I= A235 w2'v
where
w
gives
wm
is
the
a
wer ght ing
tube
z | Vo | t, MPal
Using
Az
= 0.25
2°¢, MP)
0
43
0
3.0
|
uy
Oo
12.0
O.S | 0.004
38
22,0
2.
44,0
WS
\33
G4.7
123.0
+ |
\
[0,006 |
0,008 |
p=
Te
0:28
447.8)
>
327.8
2n(0,020)(37.2¥
10%) *
To
we
get
the vadues
w | w27, MPa |
0
oO
0.25 | 0,002
0.75
b)
factov.
bofaw,
MPa
=
1.376 10%
7258.8
133.0
Hy7 g
37.3
I0°
Nem
<— Zw7z?r
Pa
Ts 1.974 kKNem we
Sf
gq: LE -
Oe eeee
\
.
ved
= 300%10°
a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course. preparation. A student using this manual is using it without permission.
TMP
MPa)
- Problem 3.113
: 3.413 The curve shown in Fig. P3.1 12 can be approximated by the relation
.
mo
-
r= 27.8 “Oe
x 10°y ~ 1.390 x 10
ne
150
125
Using this relation and Eqs. (3.2) and (3.26), solve Prob. 3.112.
3.112 A solid aluminum rod of 40-mm diameter is subjected to a torque which
“100
produces in the rod a maximum shearing strain of 0.008. Using the t~y diagram
shown for the aluminum alloy used, determine (a) the magnitude of the torque
applied to the rod, (6) the angle of twist in a 750-mm length of the rod.
75
50
-
25
0.002
0
0.004
0.006
0.008
yY
=
L
0.010
or Cee
+
z=
- £
dp = ane? ( zt dz
t=
where
~ 1.3% «Jo?
Br
2ngriot
I
4
y
i
2ne° (2 (B%02
Rh of
.
tn
O. 7 SO
The gives stress-strain curve is
T=
(be)
mm
.
(a) Let
Te
750
|
Bue (2'de
ane 4+ BY.
and
Ce
& Oe
2 de
+ CN
t£0%we
BY C=
BYU
4 CK 2”
|
|
| |
Catde &
|
a
2K (o.020)°) $ (27.8 jo? )(0.008) + £(-1, 370x/0*)(0.008)° ¢
1.900 «fo?
1
New
1.900 kNew
—<o
iF _¢
Se
p=
LY,
0.750)\(0.008
btw
= | (2.750
Y0.008))
300% (68~ pad
Pririq?
te
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educatars permitted by McGraw-Hill for their individual course preparation. A student using this niapual is using it without permission.
Problem 3.114
3.114
The solid circular drill rod AB is made of a steel that is assumed
to be elastoplastic with ry = 154 MPa and G = 77 GPa. Knowing that a torque
T = 8475 N - m is applied to the rod and then removed, determine the maximum residual shearing stress in the rod.
SOLUTION
Cz
Zeimm
T=
Bet
Tj = Ste
=
ke
P@oH*
fom
b
/
= 1-272 «16 mt
~ C272
e272 x13
x0) LB KIO)yo.6£3,0 Nm
Bros
Loading?
Ts SY ay
Ts #0 (- 4&)
bre
4. StTY ©
Cc
. BWSIED _ 4165
‘t
65-30
Bt 2 0.47349
Unloading
.
be“
+f
*
At
prc
t=
At
p* fr
of
Resicuaf:
Tes
Pr = 0°4739
wheve
= 0-10
Gt
= M42
mm
T= P4¢76 Nm
(878)
.
Fitiyi(0:03)
ee 7200 Maha.
@ CPGIS) (010142)
haem &
*
Veet
*
_
¢
A
7
S11 Pa
=
LF
x’
At p=c
AY p= Pr
Yr = NUR
ILE
M fa.
wmentimums
Tres©
59m fro
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any forn: or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
wei
Problem 3.115
3.115
In Prob. 3.114, determine the permanent angle of twist of the rod.
3.114 The solid circular drill rod AB is made of a steel that is assumed
to be elastoplastic with ry + 154 MPa and G = 77 GPa. Knowing that a torque
T = 8475 N + m is applied to the rod and then removed, determine the maximum residual shearing stress in the rod.
From
the
sefution
to
don.
ter
:
Yr
dosding
(296
Pod
Doring
eg
i
os
&
27097
-=
iit
=
p>
4%
>
j=
C17 x104
Permanent
twist
a
>
Point
JC- 272 yt" &)
2
BG
q
(efustic)
Te S475 0.
=
=
0-B6S3
rad
=
16
oma
$o
.
Yoo
(S415
xl
U0)
p=
LE&
=
pd
UyoFs
pr =i4,2 mm
;
20-4734
> _ Pr
5
x10
on ford ina
=
?
isex10%)
tot
=
3.114
oq2K18 & in 4
Js
CF Samm
AF
PROBLEM
A4GRL°
tb
ang be
~
Dr
bkbo8er-
F653
rosysr
|
gr
Bie]?
ae
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No past of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
sFyi
Problem
.
3.116 The hollow shaft shown is made ofa steel that is assumed to be elastoplastic
with ty = 145 MPa and G = 77.2 GPa. The magnitude T of the torques is slowly
. increased until the plastic zone first reaches the inner surface of the shaft; the
torques are then removed. Determine the magnitude and location of the maximum
3.116
:
‘+> _ residual shearing stress in the rod.
tegad
60 amn |
When
the
212,95
plastic
mm
Zone
C,2
4a,
=
80
mm
reaches the inner sowkace,
soon
the stress !s equal to Tr. The covvesponeine
:
tongue
= g
dp
Q2r%y iw
T=
,
Th
"
in fequation.
T, (ef -6,3)
=
7606410"
76064 10? Nem
Gasxid*))=
= 2 (juswi0t)[(a0«107*
Dntoad ing
by
p%lanp dp) = ant, prop
ptdd=
AT=
caobcvdeded
rs
Nem
J = Ee, - c, )= Fiaoy-casy] = 12234x10%
mm = 1,234 x10 m*
pF
a
Ter
tS
Residval
stress
O22. S10?)
r-034 «107
_
/
‘
= TP.O5¢ 410" Pa
(7.606410) (a0x1o73) _
“1.234 * 1O7F
"
it
nS
. (7.6064
THOS
192. 63x10" Pa = 192.63 Ma
.
Innen surface?
Tee= Te- U,'= (4S-77.05
= 67.95 MPa
Outer sudace?
Tres= Ty
=e47.63 MPa
Maximum
MPa
a
1. Tle,
residual stress:
T=
[4S 192.6%
68.0
Mle
at
inner surface,
act
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
| «3.117 In Prob. 3.116, determine the permanent angle of twist of the rod.
oe
3.117
Problem
3.116 The hollow shaft shown is made of a steel that is assumed to be elastoplastic
with ty= 145 MPa and G = 77.2 GPa.
The magnitude 7 of the torques is slowly
increased until the plastic zone first reaches the inner surface of the shaft; the
‘y"
torques are then removed. Determine the magnitude and location of the maximum
residual shearing stress in the rod.
t,=
When
60 mm
35 fn
the
the
AT=
ptdd=
= 42, S mm
plastic
Stress
toque
gd,
rs
cna;
Zone
is. equal
reaches
to
catewbatedd
tr.
the
= 30
mnenr
mm
surtwce,
The covvesponding
by intequation.
pT (arp do) = 20%, prep
= 1 4, (eb-¢3)
Ce
= ant, | p
of ‘Cinsw108)[(goxio")- Ciaseig*Y] = 76064 x10% Nem
Rotation
ang Pe.
at
maximum
Ci Prp
2 Fs
ies
f=
TL.
mote
G
Un dowding
Cc,
_
-
JUUSs1oO"®
7, 6064
xO”
t
ToL
Pecmanent
Prog
.
a
(7.6064
xO?
10
@'=
Vad
New
1.934 x{O%mm = 1.234K/O" oe
=
}C5)
~ C12 «107)(1.284x[oF)
angle
oO,
IS130
)
Fle, - c,) = FiG@eoy"-U2.s yf
GI
Prerm =
(ysxio® CS)
(7, Qn
Tor
Ni
Q's
7
tongue.
~
0.
34922
rac
of twist.
O.7S13B6
- O<.39922
>
O.B5ROB
vael
Pperm = 222°
Proprictary Material. © 2009 The McGraw-Hill! Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced,
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
a
or
,
Problem
3.118
_. 3.118 The solid shaft shown is made of a steel that is assumed to be elastoplastic
with ry= 145 MPaand G = 77.2 GPa, The torque is increased in magnitude until
the shaft has been twisted through 6°; the torque is then removed. Determine (a)
the magnitude and location of the maximum residual shearing stress, (4) the
permanent angle of twist.
06
AD EE
~~
-
Cz
O51
m
=
16 mm
|
Tray
=
a?
=
(o.
:
am
=
0.00187382
O0ofm
fe'-¢ CO.016)" = 108.444 -Jo"7 m”
= gC oy, = Bo, 016) (1¥sx)0° ) > 932.93
ET
At end of Iouch'sog. Tua
_
At p=
At P= Px
ge
°
wrk
“Sd
Residual :
At p-c
Tle
elas tre,
©
t's:
wot.
1
Tle
Nem
ss xo
(14a
_ (1.14855 l0" )(o. O16|
BG
07. 944 x O°7)
lies 7 Unt
-— T'
l4Swlos~
178.52xlo*
TE
TEs
FI
x lO" Tres 7 19S
Marti num
120,381
RIO
-
Previn”
=
=
°
sCreed
©
"
—_
Dred
47?
4.97
i
P
~33.5axlo” P.
=
At p> Pe
6g
S
[02.444 xo"?
(‘178.52 x]0" Fo
ES fee 7s. 52x 10° (0.67433)= 120.38xI0° Pa
(772.207)
Tres
1-4 .67433)°]
4 (432.98
(1-4 f=
_¢ Se
19355 10*)(0.6)
v
Unboading?
Nem
hIgss x/oF Nem
=
(BY
0,0021925
YL, * “o.coa7egg * 0 67260
Ty : vt v
@
=
:
Ea
_
Me
fr.
J:
*
= 10472 “lo pad
IE Mio. -72%*/0° *)
HS
Vy
6°
T
~ 33.5
RH. ED?
MPa
x 10° Fe
24.6
MPa
residual stress? 335 MPa atpzibmm
QDpary = \O4STAXIO= BELT & [0 = TTB KIO red
Penn? 082°
—
Problem 3.119
3,119 A torque T applied to a solid rod made of an elastoplastic material is increased
until the rod is fully plastic and then removed. (a) Show that the distribution of
residual shearing stresses is as represented in the figure. (b) Determine the
magnitude of the torque due to the stresses acting on the portion of the rod located
within a circle of radius zo.
,
bsp
By
‘qe?
R esreduad
(a)
Atter
oan, $Rey >
Poudina
Dn Poaabing
Residual
Te fnd ce.
-
te
38%€
Cows *
G-
Set
Te
O=
aud
z7O
2
a a (* p"
cal
=
-£%
at pre
- x(1-#)
hE
I: #2
T
if
(ob)
- 2¢ Tet)
= ZT
ts
a OL,
ple
Cos eC
%4(1- 3
) dp
= ganze)
4 (y+ Key
i
4 - A yc = 0.2209 Be
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.120
3.120 After the solid shaft of Prob. 3.118 has been loaded and unloaded as described
in that problem, a torque T; of sense
the shaft. Assuming no change in the
which yield is initiated in this second
which the shaft started to yield in the
“0.6m _ ~
opposite
value of
loading
original
to the original torque T is applied to
gy, determine the angle of twist g; for
and compare it with the angle gy for
foading.
3,118 The solid shaft shown is made ofa steel that is assumed to be elastoplastic
with ty= 145 MPa and G = 77.2 GPa. The torque is increased in magnitude until
the shaft has been twisted through 6°; the torque is then removed. Determine (a)
the magnitude and location of the maximum residual shearing stress, (5) the
permanent angle of twist.
~~
I
16mm
the sofution
fo
PROBLEM
TH = IS «lo Pa,
The
residual
For
stress
Poading
at
in the
= Te
+
Tees
a
of twist
at
- Hb
:
A
ONS
is
Tus
L=0.6
yielding
(7
=
the
33.5
MPa
change in stress to
19S x)0% - 33.5« 10° = tht.
S xO
LS xfo%)
2 to
Tt. G02.944
of
bun'st
JI7
Nem
unden
reversed
x10? ) (0.6)
foygus .
a
* (T1107 YX} 02.4440" 1)
for
&% ~ &
Toe
yiekeling
in
ovrginad
hy
, . CG
LT. .(0.01
(0.6.E
x 10%) _
IT 1*let)*
Py
Pa
0. O§€
=
a
red
S4.1¢x/lo
= Zio
~
Angde
m
is
Cc
-
Ang be
pre
yielding
9
Tv,
C2 0.0 IG m ,
opposite sense,
reversed
my
produce
3.113
ST = 1020944 10? am"
#t
From
Joacking
.
.
7O.4Z4x10
3
yaed
y+
;
“oy
,
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
«a?
~ Problem 3.124
So
3.121 Using ray.= 70 MPa and G = 27 GPa, determine for each of the aluminum
”
-
bars shown the largest torque T that can be applied and the corresponding angle
of twist at end B.
|
45 mm
Lys TOxlO Pa
G=27«10" Pa = LL = 0.400 m
F (a)
5
to.)
25 nim
45 mm
P
P*
Th
3
te
7
Ta bfe
wy
4
OQ, 208
Gy;
ot
=
all
t*lor dD
yen
=
7
°
ey
Ts 700 Mem Ls 0.900m
os
T
NY
Frown
\
v
4?
mr
_
|
Table
-
i
2.
‘
“Tv
ec,
B= 27«10% Pa
=
Oy
.
Q2é7
C3
bT 25 mm,
TT.
Toman = Zabt
G,abbG
263
C, tl* (0.267) (0.045) (0.015 )*
= 74,0108 Pa
Tag TH0 Mi
(200)(O.9o00)
ab*G ~
(0,263) (0.045 )G.0lS (27
= 166.9410 rad
25mm,
- 0.
200
Th.
PG
TL
OA4OE
45 mm
900 om
'
<-
3.122 Knowing that the magnitude of the torque T is 200 N -m and that G = 27
GPa, determine for each of the aluminum bars shown the maximum shearing
stress and the angle of twist at end B.
jen
De:
7
> Cat
a
rayujory | SF
(6.106 VO. 025 (0.02
-
_
se
9.05
3.122
aD
f(b) =
?
=
Ts 22B Nem
Nem
_
(227-5 (0.700)
pa
Gabe”
Problem
O
= (0.2080. nas\(o.028)" C7ox1O%)
= 227.5
=
7
$7.94 10 rae
Fron
rho,
¢
cab’aiw
Ts
Caan e ay
:
;
..
(134. 2 1(0.900)
kos IE mms y
ma
Aras
| (2)
a 0.263
alt
(0,263Xo. ors lors FQTsIOn
—
Crab?S
BO
T = (187.2N-m =
”
-
=
(0.015 )(Fo uO) = 1849. 2N- mm
T= 0.78 £0,048
t
pm
Th
=
$
4,7 0.267,
:
C-
25mm
Cp
15 mm
TT
‘
500
90
b=
From Table 8-1,
T
LS mm
or
107 )
Q= 456°
E10. From Tahte 3.1,
7 0.208, c= 0.1406
200
“
_
(0.208 (0.025 (0.0a5)* ~ GISx10"
(Q00\(.920 )
~ (0.1406 \(0.0a8 WO, 025) (27
* 107%)
<i
ep
Fa
Conan” GIS MPa oa
= [21.4ulO real
P=
G45"
Problem
3.123 Using +, = 50 MPa and knowing that G = 39 GPa, determine
for each of the cold-rolled yellow brass bars shown the largest torque T that
can be applied and the corresponding angle of twist at end B.
3.123
‘I
|
,
Cap?
35 non
50 mm-~
/
(a)
50 Mfa
:
«40mm
b = SOmm
a
>
From
ToAle
.
Le
orgy
a
a
Cc,
=O.
~~
Coma.
7
-
ho
208,
C,7
_
C,
a
L*
T
O.V406
2
-
&
ak
Conant
400 min
T=(O. 208) O05) O0S) (40 x10°) = (300NA
Tx eZ LNm =
tb)
_ThL
.
P= C,Ab>&
=
(b)
a: 7okm
br 3am
Conse ape
~ (0.1406) O0SYouS)
{6/73
ce
( fo3«to” (0-4)
|
(29x07)
x to’* rad
20
From
pr
Table
Bl
C,7 O.246 , ©, 50.2249
|
Tk.
(hosrxto® )(O4)
CrObL®G ~ (0.229 Xow) O02)? (ZI)
|
Problem
? me
TE AbQ Caan = (0.24 GONGLED fro) = LoL New
~~
p>
oof]
T = tosy kN.mM mq
=
~ ISTP M10
Ss
“
rad
©
PF O402"
8.124 Knowing that 7 = 800 N- m and that G = 39 GPa, determine
3.124
for each of the cold-rolled yellow brass bars shown the maximum shearing stress
and the angle of twist of end B.
T= Goarlm
fe)
Le orgm.
GgeS0mm
b= Somm
From Tebbe
Tonal
!.
=Ca
3.1,
=
=
1.0
€,2 0.208,
250.1406
MGeOSYE. cyt =
©. 2OB)fo
bb*
20,77 M Poy
Tonos 30°17 Mf
re
b)
($00 )(o-¥)
P= C,08 & ~ (0.1466)
G0 S\Q0lFBI000= 10")
TWZ4 410 wed
(b)
ae
7ohm
be 25mm
*
C,ak*
~The.
?* Bave
2 2.0
From Tebte 21
~
(0+246 ) (0:07) Coro
x)”
-
m,
27.0
_
Sov
an
Tomas
2
P= oS35°
$7
$ao)( 0-4)
(02240167 (0:03) (39K?)
9
mta
=
ats
0.246, C,= 0.224
a
Lirasf
olf
= 27M
PC
.
Tied
P= 0 boy
coe
|
Problem 3.125 —
3.125 The torque T causes a rotation of 2° at end B of the stainless steel bar
shown. Knowing that 6 = 20 mm and G = 75 GPa, determine the maximum
shearing stress in the bar:
Qt
BOmm
De
B22
= 0.030 m,
SHIFT”
Tee
P *eabse
vores
caLF
sor
iS,
.
vad |
Ke
S0.6°S0
O, =
3,1 }
C2 b GP
=
(0. 221) (75@ «K1074)
Problem 3.126
.
O23!
07 x10)
_ (0 19SB) CLO» 107?) O75 21073 (34,9
mee
0,020m
pe 2.2?
*
=
—~Table
Frem
bz 20 mm?
@,5 O.19ES
59.2 %*10°
Pe
Toa? S42.MPo, <0
3.126 The torque T causes a rotation of 0.6° at end B of the aluminum bar shown.
Knowing that 6 = 15 mm and G = 26 GPa, determine the maximum shearing
stress in the bar.
umm
4
20
=
Ton
*
a
wb
ie
_.
= Z6,
O,030
m
Cc
From
Table
3. ly
T
>
bt
15mm=
0.018m
10,472"107" vad
GaseG
750: min ~
6
4
S
;
6
Oo”
~
®
ar
2
*
=
£2
ek
abe?
-
&
q,L
Cab’
Cab
_ 0.246
c, =
|
=
Cy, =
0. 224
Lone = S07MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permisston.
<4
Problem 3.127
3.127 Determine the largest allowable square cross section of a steel shaft of
length 6 m ifthe maximum shearing stress is not to exceed 120 MPa when the shaft is
twisted through one complete revolution. Use G = 77.2 GPa.
L=
Gm,
We
irowlo®
Pa
r
ce
<
C\0.\™
p
a
ekeemir ate.
Divide
ay
COLT
Bee
b=
Fron
Ty
me
;
then
solve
2
& Square
“
x107)
3.1 28
Sovve
tow
on
vad
77%
Gy
Lb.
‘8
(Aan)
b=9mm
by
2
Q2
section
= 2m rudd
¢Z
=
a
2
SRG
ab L
os
abe
rk
1.0
s
RR2OxIO
7
m
b=
|
2.20 vam,
|
3.128 Determine the largest allowable length of a stainless steel shaft
of 9 X 18 mm cross section if the shearing stress is not to exceed 105 MPa
when the shaft is twisted through 15°. Use G = 77 GPa.
Cran >-—f
cabs
( 2)
for
b,
@
CKOSS
~~ 2
iv de
Divi
Pelrev
C,= O.208 , Cr, = O.1MO0G
(0.14906 C77.2
18° =
P,
G
~
Q=
=
2h
Gab?
3.1
a= 1@ mm,
b
-
~ (2.208Y(6)URZ0“105)
Problem
Cl
G = 77,2107
~
For
Table
,
=
Emme
= 105 MPa
0, 26180 vael
tb
P= Davie
C1)
+ °
ef tuned te
Le
= R
,
T.
Yn
2.
=
C2)
c.abi
G
c abil
= abe
Eb
{bEG
C, Goran
Vable B.1 gives
.
(Q.22gXor0ed
77 xo" 4 lo, 2180)
(0.246 \Ues« io *)
C= 0.246.
_ 1609 m
Cy= 0.2249
La pgm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, Alf rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<i
Problem 3.129
3.129 Each of the three steel bars shown is subjected to a torque of magnitude T=
275 N«m. Knowing that the allowable shearing stress is 50 MPa, determine the
required dimension b for each bar.
TT?
IS
me
(2) Square? ack
Torna?
From Toble 3.1
Cong “Gab
ae
£0
Erbe
Pe
¢,2 0.208
yo tT b*
3
b
«to*
215
fe,
Ton
exe
= 24.%x107*>m
(b)
Civeke
2-
Ge?
c= f2E
e® =
(2) Rectangle’
az
1A2b
From Tahte
3.1
4b
Corns ™ fs
= 25
= {BQISy
= 15,18 . x10 wm
“(SO * 10")
ISBwH
q7O.21%
3
b= TE e.7
b= 24.8 mm <8
b= Re = 30.4 mm meta
Jt.
Tne” Eabe
* Tet
tit
*
C2 VO-217(SOxloe) = AZE x10” n
be 27.6 mm <8
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be dispfayed, reproduced, or
distributed in any form or by any means, without the prior-written permission of the publisher, or used beyond the fimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
3.130
3.130 Each of the three aluminum bars shown is to be twisted through an angle of
2°.
Knowing that 6 = 30 mm, ta = 50 MPa, and G = 27 GPa, determine the
shortest allowable length of each bar.
P= 27s BY G7
HI wd,
G
>
For
27107
Square
Pe
ons!
From
(c)
Se
73
[= (O14 260, 030K R27! 0% (34.907 «1077 ) =
() Civelet e=4b>o0ism
Dive de
to
ebrwivate
pie CEP
.
,
(2) RectangJet
t= Te
i3
then
a=12b
Sox
m
l= 382mm <a
g-Te
Solve fon
F128
los
L
‘
From Table 31
9
too
1. = (01661 K0.03027 x 10%asf(34.907
#10
beg
Lo
£
= aaF
_ G.01S)(27%109)(B4.I07%}0%)_
~
Le
_
._
38210
CO, 208)C50 xlo*e)
yin ©
OF 0.208, G27 2.1406
3.1
Tohbe
then sodue fur
oc
to ePrice
Severe:
m
- th
a
Q._cabl
o)
0.030
retavade
ye TT
Diviele
= SO0x10°
Pa
b= 30
~
=
a
REBy¥IO
om
wm
-
Le
2383S
mm
<8
6,5 0.914, €,7 0.166)
.
bavi
Im
«LY
mm ae
(0. 219)(50 x [0°>
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
3.131
Each of the three steel bars is subjected to a torque as shown.
Knowing that the allowable shearing stress is 56 MPa and that 6 = 38 mm,
determine the maximum torque T that can be applied to each bar.
Problem 3.131
b+ 36mm
True. * 56 MEQ,
(a)
Sqave:
a= b=0038m
Fvom Table
Bl
Tonge =
CO, = 0.208
T=
eo
F=l.o
Concur
C,lo*
Ts (0.208) (01038) (21038)* (6 xto® )
T= 639 Nin
(ob) Circles c= 4b = 4mm
- le . 2v
eR
Ton = EF
< F (01019 )3 (56x10 6)
(c) Rectang fet
From Tabfe
T=
=U 2) 3BB)= 45:6 mm
C,= 0.219
.
3.1
Ob Tou
3.
-
<a
Te?
TE ZO" Tone
T= 603 Nm <a
= 1.2
= (0.219 )Co.04:96)(o. 038) (610%)
Tr
P07Nm
Proprietary Materiat. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Mariual may be displayed, reproduced, or
to teachers
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
permission.
and educators permitted by McGraw-Hill for their individuat course preparation. A student using this manual is using it without
<0
Problem
3.132
‘3.132 Each of the three aluminum bars shown is to be twisted through an angle
of 2°. Knowing that 6 = 30 mm , t, = 50 MPa, and G = 27 GPa, determine the
shortest allowable length of each bar.
P= R= 34.97
xId ved
G
= &47xlo?
For
= SOx10* Pe
Pe.
b= 20 mm=
Square owned reotangde
ws oa
Divide te
rake
ePiabicts
T3 then
Qi
1ebrl
"G2 ab?GS
(2)
o
Sevaret
(C .%24)( 50 x10%)
Dive de
=
G=tb
Civefe
E*
p=
T=
O.01F m
te ebiminadte
T:
then
=
bee
~ Oe
282010" rm
Qe
fon
L.
£.
L= B82 mm a
ie
= a
>
a
|
Lew £EP . C01) (Q72107)(34.907IO?)
7
te)
.
Rectangle:
SO « 104 ‘
°
Tr
a=t2ab
& =4.2
From Table
4
re oe -
ee
x
“=
:
3.1
ty
Meh dente
w
233x160
€,*
O.214,
.
=
L
= 0.208, ¢,>0.1106
Js
solve
sofue for
lo
From Table 3.1
0 OF (34.707 109)
YX 27d
[= LQ.1426)0.03
(b)
0.030m
424W(5 m
L=
mm
283
<8
Cyr 0.1661
;
Lo= 424 mm
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<
Problem 3.133
3.133 Shafts A and B are made of the same material and have the same cross-
.
sectional area, but A has a circular cross section and B has a square cross section.
Determine the ratio of the maximum torques 7, and Tg that can be safely applied
to A and B, respectively.
\e+
cz
of
square
Fou
foalius
of
civcubar
section
equah
A
and
B.
aveas
c=
section
be
side
oe
TO
4
=h
he
Tr
Ciecles
T
Retio: es 20%
For the
% b*Ze
same
"
Problem
Cab*
&
Tar™%,
ere
JR.
=
3.1,
Table
:
Tg
ba
b
C
8
Tr
ta
20,4%
Cb To
= 7G
Y
|
2 F-mata.
stresses
e
Ie,
~ trier
*: Th
fron
C, ib?
3
I,
2
.“
te
v=
Le
.
are
0.208
C,=
Sqvare.-
;
27:
NH = Es =
-
——
ayoaoayt
7 Se
3.134 Shafis A and B are made of the same material and have the same length and
3.134
cross-sectional area, but _4 has a circular cross section and B has a square cross
section. Determine the ratio of the maximum values of the angles g4 and gg
through which shafts A and &, respectively, can be twisted.
Let
of
yadius
a=
square
For equaf
af
section
evees dos
section
aveas
B.
TC’=
Circle? You> TB-= SM
Ti
~ ile
2
Te 3
Te
Gabe ~ 4.203 53
Op, =
8”
Redo
Tel.
Gabe
®
Pa
o.nzon
_
ofechte
~
> Lt
CG&
be
side
2 Qe ET
C2 = 0.106
TeB= = 0.2088
% Ce
O-ROBL
_
a,b
+. bG
For equal stresses
and
b =4T&
b=
¢,20.208,
TabSe 31,
Square?
A
LATTE%,
%= Te
=
=
1.4794
LG
0.616
Lb Pe
b T%
& a
-
=70.676 Tt
Tn
Te
Go. o.ciedt = 1198
=
Problem
3.135
3.135
A340 N « m torque is applied to 4 1.8-m-long steel angle with
an L102 X 102 X 9.5 cross section. From Appendix C, we find that the thick- hess of the section is 9.5 mm and that its area is 1850 mm. Knowing that
G = 77 GPa, determine (a) the maximum
the angle of twist.
14x4x2
:
=
(OH)
(6)
=
A = (380mm, b= 45mm,
ed
=
=
Trex
T
C,Aab*
p=
I
Cc,
wo
=
a=}. Ee
C,=4(]
- 0.630
340
b)\=
oe
— (0, 2230 )(0° 1947 )Ca1e09)*
—Trhk
?
Note?
HO
’
~ S79
(342) (WP)
-
shearing stress along line a-a, (2)
OO, 3230
,
:
Mbhe
—«
= (47-4 via" pad
GAbl’G ~ (0-323) (0-1947)(000F5
17 )>
x107
P= Sins?
Le hom,
Problem 3.136
= /DheT hm
at
3.136 A 3-m-long steel angle has an L203 x 152 x 12.7 cross section, From
Appendix C we find that the thickness of the section is 12.7 mm and that its area
is 4350 mm’. Knowing that ry, = 50 MPa and that G = 77.2 GPa, and ignoring the
effect of stress concentration, determine (a) the largest torque T that can be
applied, (b) the corresponding angle of twist.
A=
4350 tom
Equivatent
&
1,203x 152x
12.7
be
rectana
fe
12, 7mm
,
Os
az?
Be ase
= 26.47
Oo
= B42
5S Imm
.
\
.
L
Gt 0,2 $(1- 0.6302) = 0.32555
a
-
-
T= Gab’, = (0.32555)(26.47%16"
27 1s*) ($0 x/0*)
=
&)
?*
70.807
Nem
o- lees
Gabe
=
0.15299
|
(70,207) (3)
T=
70.8 Nem
=e
_
Crass WAG. F7XTO™ W12.7% 1079 )\(77, 2x 107)
rad
Co
P=
3.77°
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distributed in any form or by any means, withont the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
i
An 2.4-m steel member with a W200 X 46.1 cross section is sub-
3.187
Problem 3.137
jected to a S60 N « m torque. The properties of the rolled-stee! section are given
in Appendix C. Knowing that G = 77 GPa, determine (a) the maximum shearing stress along line a-a, (b) the maximum sheating stress along line b-b, (c)
the angle of twist. (Hint: consider the web and flanges separately and obtain a
relation between the torques exerted on the web and a flange, respectively, by
expressing that the resulting angles of twist are equal.)
SOLUTION
.
W200 x 46.1
.
A=
Frange!
20s
b
>
frm
bY
S= 0-322
e.27C= - a et(;~
230 ¢0.680 2)
c,ab*
*
Tre
Kp, = (0-322) 0203 )CorotlPP
Ww
C=C, = $C1Tw= Gab’ SR
aKp+
=
~9F
Dr
7,
“=
Ky
Te
Gab®
9.29757 (56OS0) gt 61°
SiMe
Pa, _= 04
O
Teac
SP.
=
* 66. 3249 Yo: +}8/)\( 0.0072)
lL
~
?
=
VGT 1181) +2.
.
3/-44-MPa
.
=
tear
ol
(480 )O-A
P * (77x10 TYRE 7xjo VN +20 ase
EF]
i)
wait
:
2007
MBla
where
[. =
ad
=
62
96 x/5
CFM,
_
62*7
C,ab*
te
“
bm,
24866
y™
~Gagayorosvaou
lw
)
(b
=
Kw
AKG+
“
”
Kw7
2+
Tr,
dKe+ky
Tp
>
* QP Ix 16°) + 21901061
~
P
=
Pu
=
|
w
3
m
24e9S Xfo
C8 TKE MM Ge)
nee = 2 LHPs
(a)
Ch,
ao f+ (y-
AL
a.
tm
BO
Te
b*
a
b = 72mm,
~ Ket
_T
L
-
Ke SE
T= 21p + Ty = (2Kea
Tetal torque
GO.
(Ps 4X
ebRS
QS
=
=
97 x10 on
ang des
tui'st
motehing
Oe ~=
-
&
Ky = (0.3249 VloelFI)(o.0072 )*
For
os
7
b
o6sed)= 0.2249
Du = litle
Shere Ky = cab
= KEP
}
~
~
=
Ke
(Simm,
203-@CM)=
as
eh?
.
where
Kp Ge
=
She
2
mm,
=//
10" oad
—
24
=
a
my
°
™
Problem
_
¢ | > I
3.138 A 3-m-long steel member has a W250 x 58 cross section. Knowing that G
3.138
_= 77.2. GPa and that the allowable shearing stress is 35 MPa, determine (a) the
largest torque T that can be applied, (4) the corresponding angle of twist. Refer to
Appendix C for the dimensions of the cross section and neglect the effect of stress
concentrations, (See hint of Prob. 3.137.)
W250 X58
Flange
:
QA:
203
mm
a
br
13.
S mm
C,= C, = $(1-0.6302)-
-teh
+
Pe GARG
2
7
Q = 252 -Q IBS
=
Ket
Tote? torque?
im.
be
2
m*
Bmm
C=C, =4(1- 0.682)b = 0.2259
at
oT,
matching
C,ab* Ge
tuist
=
Ky SE
37.54 «1077 m"
angles
Perr
Qu =
PQ
T = 21 +Tw =(2Kp +k) SP
GO... T
tage
>
APPow oh be
159.53x/o"
WS
Kw = (0.3269\(0.225)(0,008)°=
For
©
Toe cal Se se Se
7 Te = C,ab
2228.18,
Put
IS.O4
0.3194
Ke. = (0.3194 XO. 203(0,0185) =
Web
fs
a
Vabve
for
a
T
eT
Te Tag
4 TTA
Tw
= Ve Se
=
based
.
oR
Kw
on
_ 2ket
Kw
~
~
abPours be
2K,
Te
re
Ty
vadue for Tr:
Te = cab’? = (0.3199)(0.203(0.0135) (aslo) = 413.6 Nem
(ug)
B75 3)+
T = @US4.5
=
924.6
Nem
iS4.53
ADow able
valve
Sor
T
based on
athowoh te
vatue
a~
Choose
p=
. (2)089.53)4 37.54
=
Sey
smadler
TL
(2 Ke+ Kw )G
64.25)
valve
_.- (TRS) 3.00)
(356.6 x10°8) (77.2 ¥f0%}
u
Tw = cab = (0.32589)(0.
(0.008) (aso)
225
IS6O0
For
Twi
= 164.25 Nem
.
N-m
JT =
2 N0O7vIo md
F25
Nem
est
= Oe S77? met
Problem 3.139
3,139 A torque T= 5 KN - m is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points aand b.
.
10 mim|
|
|
Te
5
IO?
Avea
New
bounded ley
center Dine
OL= bh = GAs): 7 F354 ros
- 1.938 x«1O8 mt
AF
IL.
— 7
Te
mm
|
.
-
peind
A
qe
L
=
b
ml
.
a
te
Guam
Z
S10"
=
T
= 0.006 m
24tQ ~ (ZYo.006)(7.435x1O*)
.
62.5 MPa <a
= 52.5 «10° Fa
b
~
Zt
:
{OQ
O.0!0
oy
mn
we
:
_
_&xlo*
(2Y(O.010Y 7, 935%
14)
_
Bl_-5
Pe BLS MPa
Pe
*/o
<i
3.140 A torque T = 750 KN - m is applied to the hollow shaft shown that has a
Problem 3.140
——
point
uniform 8-mm wall thickness. Neglecting the effect of stress concentrations,
‘determine the shearing stress at points a and b.
--- 90 ym ————>+
Ke
>|
Detai?
ES
30°
:
#
in L
2
Ly
*
of corner,
= @ tan an 30°
_
c=
_t
2Q tan 30°
= Phan gow = GAB mm
8
_—
b = 90-26= 76./44 mm
Avea
bounded
Q-1b#h
by
- Bye
= 2510.6 mm
b
ws
rT
z tQ
=
=
certer
=
Pine
- Berea
2870.6 x1o7>
m
%
0,008m
TSO
()l0.cedsj\Qsioxio™®)
= 18.67«I0°
Pa
=
13.67MPaet
Problem
3.141.
3.141 A 750-N - m torque is applied to a hollow shaft having the cross section
shown and a uniform 6-mm wall thickness. Neglecting the effect of stress
concentrations, determine the shearing stress at points a and 6.
Gum
,
Dine,
ceater
by
hounded
Avea
Q = 22 (a3y +6016) = 7381 tom”
60 ann
= 7188) x )0
EL
Tt
77?
=
ZEA
°
Problem
©
.
750
potats
(20.096 }(738Ix}O)
3.142
re
a. and
.
oo
at bath
m
= 0.006
Then. at
7
8.4
7xfJOo
.
B
a and b.
lo,
7
A,
, 8.47 MPa
3.142 A 90-N - m torque is applied to a hollow shaft having the cross section
‘shown, Neglecting the effect of stress concentrations, determine the shearing
stress at points a and 6
L
{
J
“4
f
Sporty mend
N\
”
4mm
{
om
ji
4 min--
\
2am
40 mm
at
7
55 mm
SLmm
b
a
{
nn
|
—e) \
2mm
—s6nm———-
j
?
I
So
RT
|
LZ men
SS mary
i
|
SS pun
AREA
BOUMYED
sz
st
BY
CENTERLINE
39.
-|
3
Le
fe ( }’s
39
37
SLXSR ~ BIK3G+ 2069)" = 2378 mat s 2298
Q=
3
x10 van
T= GO Nm
7a
=
TT
oT
LEG
t
9O N-m
3
X18 “m?)
29410 m)E2,378
T
7" 2aQ~ 2(2x0
=
PO Nom
MN 2.398 X10 *9m?)
Ria
a FPS MPs
52 24M
.
<
Problem
3.143
3.143
A hollow brass shaft has the cross section shown. Knowing that
the shearing stress must not exceed 84 MPa and neglecting the effect of stress
concentrations, determine the largest torque that can be applied to the shaft.
Cahevdate
12mm
150
' i,
~ 38 min
of
the
The
Bey
aven
with two
38 mm
the
wal
avea
bounded
cevess
section.
by the center Pine
ik ae vectangde
semi-eincutaa
cutouts.
12mm
7
[eS
=
<— 125 um ———»
3mm
bras
FF
=
ttomem
"
hz iSO - 12 = 138 mm
VF
GQ=
38 - 2-3
a(zT v")2 = U20\Ct28)— 1 BS)*
bh-
Tv
|
= 35-5 mm.
= 12600 mm*
®
|
b
= 16 XI0-3) Mm
2
,
Tinin = 21805 hy
Tones= 84 MPa
Vmen 3OE.,
T= 2Q ten Coe = (Qimbx16 0005 B4x10%) = 10584 NM
Tt
(058
kAm
Proprietary Material. © 2009 The McGraw-Hill Compantes, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<<
Problem 3.144
bk
3.144 A hollow member having the cross section shown is formed from sheet
metal of 2-mm thickness. Knowing that the shearing stress must not exceed 3
MPa, determine the largest torque that can be applied to the member.
50 rom —
| ho mm
48
Avea bounded
50 mm
centerdine
by
ie
O. = (48)(8)+4oX3)
2
Te
ij
704 mm = 70%%/0% m*
0.002
Zt ar
(2)(0.002 \704™10* }( 2x Jo*)
8.45
Nem
“T= 8.45 Vimeen
3.145 and 3.146
A hollow member having the cross section shown is to
be formed from sheet metal of 1.5 mm thickness. Knowing that a 140 N«m
torque will be applied to the member, determine the smallest dimension d that
can be used if the shearing stress is not to exceed 5 MPa.
Problem 3.145
Avea
bounded
OL
by
center Pine
= (0°/48S) (010734) on (e:o8fS) A
t-
- man
to
m
i
vente
"3
|
]
t
3
-
bS iam,
vs $Mfa ,
200109]
Ts
-010S/La
(40m,
we mle
Ue
BS
Abe
nA.
Q
Sf
=
3FF
9° O109;.a¢08'ISd
GES
Spe:
a
72
pmo
A = 22HSE
x
ito
(2. Co oof SYCKj0%)
es 000 2.66 a
=
60
732
d= Soe bin,na
Problem
3.146
3.145 and 3.146
A hollow member having the cross section shown is to
be formed from sheet metal of 1.5 mm thickness. Knowing that a 140 N+ m
torque will be applied to the member, determine the smallest dimension ¢ that
can be used if the shearing stress is not to exceed 5 MPa.
Avea
bounded
by
center
CL = (0-148
) (00786 -d) + 0 1048S h = 0101091 — old.
be kS mu
>
x
tt
ats —
SMPa,
T=
(oN.
Q
Tv
j
|
tc
OL
ABS
”
QE:
liye-o
é.
©+6f041 ~ ofa
1GheS
we)
= (ZYoroctS (Sx 0°) = 0.00953.
Os
dz
Wwe
~_
ool Sf
Oe
=
j
0F0/LPF
.
wm
as/KeP mr aut
P
Problem
3.147 A hollow cylindrical shaft was designed with the cross section shown in
Fig. (1) to withstand a maximum torque 7) . Defective fabrication, however,
resulted in a slight eccentricity ¢ between the inner and outer cylindrical surfaces
of the shaft as shown in Fig. (2). (a) Express the maximum torque T that can be
safely applied to the defective shaft in terms of 7p , ¢, and 7, (@) Calculate the
percent decrease in the allowable torque for values of the ratio e/f equal to 0.1,
3.147
0.5, and 0.9.
let
Te
Con FEC
Forming the ratio
(b)
Reductt ou
Expressee!
pt,
OS
tH,
75
-T
the
“
(StS
Hiickness
ean mu. Thickness
-T.
=
be
C4~Cy
= b =a
torque,
th.
te
Qe
conPiquration s
Fee @)
ot
For (1)
The
both
ay
’Q)
_T
Gu Z(6-2)
's
ot
t-
\"
Ct).
2.
260
. Th
ST
EET Se)
T=Ta-?)
=<
é
-
Zs veductionw
=
£*
100%
bad
(a) Fov
- Problem 3.148
3.148 A cooling tube having the cross section shown is formed from a sheet of
Se
stainless steel of 3-mm thickness.
‘The radii c; = 150 mm and c, = 100 mm are
measured to the center line of the sheet metal.
Knowing that a torque of
magnitude T = 3 KN - m is applied to the tube, determine (a) the maximum
shearing stress in the tube, (4) the magnitude of the torque carried by the outer
circular shell. Neglect the dimension of the small opening where the outer and
inner shells are connected.
Area
bounded
by center fine,
Q= T(cf-¢,) = 1S0"= loo) = 39,27%10% mm
=
t
lil.
=
0.003
B9.27
4
if
T,
(Qregt¢co)
m~
.
»
3x10?
(OT FEe * Doves yarimer)
(b)
xio*
~ -73xl0*
.
.
T= 12.76 MPaow
+= are tr
27 (0.180)°(0.003) (12.73 105)
=
5.46 xlot Nem
T= S40 KN mae
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3.449 Equal torques are applied to thin-walled tubes of the same length Z, same
thickness ¢, and same radius c. One of the tubes has been slit lengthwise as
shown. Determine (a) the ratio 1; “4, of the maximum shearing stresses in the
Problem 3.149
tubes, (b) the ratio 9» /g, of the angles of twist of the shafts.
Without
Area
Pa?
3h
Q: anc,
With set?
ES
=|
s
2
Racge
Gh . CES
3TL_.
Qn,
_atct
=
aneteG
Ty
_—
-
36=
3c*
re
i
te _ .3T
etl?
3T
.
A
-
ratio:
en
Pr
;
patio?
Te”
a)
.
(b) Twist
TT
=
Q=
iS
(ad Stress
=
GC,
:
OEE
$=
b=2t,
ne
“|
A?
=
=
centewt
4
Cc,
BF
tie
by
we
aT ¢
aw
Loundecl
SS
J
shit
Proprietary Material. © 2609 The McGraw-Hiil Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3.150 A hollow cylindrical shaft of length Z, mean radius c,, and uniform
Problem 3.150
thickness f is subjected to a torque of magnitude 7. Consider, on the one hand,
‘the values of the average shearing stress tye and the angle of twist g obtained
from the elastic torsion formulas developed in Secs. 3.4 and 3.5 and, on the other
hand, the corresponding values obtained from the formulas developed in Sec. 3.13
for thin-walled shafts. (a) Show that the relative error introduced by using the
thin-walled-shaft formulas rather than the elastic torsion formulas is the same for
Tave and g and that the relative error is positive and proportional to the ratio ¢ /¢,,.
(6) Compare the percent error corresponding to¢ values of the ratio ¢ /c», of 0.1, 0.2,
and 0.4,
C,
aD
Let
oe
=
ovter
Pastivs
7
Cu4+
+2
and
C,
=
ranes
radius
=
Ome
at
J=Fla- cf) = Betsotc e,)(e,-6,)
= | Cmt Ct t GEE Cen Cab fh
2i
(o-
vos
Te.
™
J
O
=
-
Ares
4+
eb*)out
“
=
IG
bounded
9,
centerdine |
S
Q:
Tn
te
=
§ 4
.
tllanc./t)
TL
ZINC tG
”
®%
@ceetGS
QF Ot
TL
Lz
tet
Tame
Te
a,
__
4(ncat )b S
Tie.
tee
Ratios?
C.
ARE ICL
2b6Q © anest
z
(Q)
Th
ATG
by
ae
we"
TL
an(Ca+ GL YE
Tk
we
Rey) E
TT
~
|
+
Cn*
gale TL ets 2 yy 48 To.
qe
_%
0.0025}
| 0.25R|
<=
~
0.01 | 0,04
.
1% | 4%
—
3.151
Problem
3.151
Knowing that a 10 mm-diameter hole has been drilled through cach
of the shafts AB, BC, and CD, determine (a) the shaft in which the maximum
shearing stress occurs, (2) the magnitude of that stress.
Hole: Gadd, =4(io)= Simm
WON-m
270 N - 1a
9ON
«are
depcD © 38mmm
Shaft
AB?
C, =
+d,
T
eC
B
|
fOmmnm
=. ent
JT
COrsi) = bei MPG Varyst)
= (Ro
Folloret)
14-273 xto)
Cy
Shaft
Tom
Tr
BC 2
-90+270
006%)
= 4-13 Kd 4 Tht &
Lae
ds, = 20 mm
=
A6Nm.,
. Ba! -c,") = FE (o.oit
‘dye = 25 mm
A
Te
*
tal, =
izeS ty
= [80Nm
T= Elql- 6%) = E@orst-aoos*)= 37-37 x1a tnt
o Te, . CiBe)(o-0125)_ fn. 2
TR
Sgro t= OOF
MPA
ShaPt CD! To = -Aor2tottto = 240 Nm.
T= Ble 6%) = Elooist-oioes") = 18S x1 tnt
cio
vo
Tee
US
- Mert!)
19-S4X1T4
Ansvers?
2
erg
Cae ade = IS mo
MPa
(q) ShePt AB
(b)6G1MURg
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
mt
Problem 3.152
3.152
4
i
A steel pipe of 300-mm outer diameter is fabricated from 6-mm-
thick plate by welding along a helix which forms an angle of 45° with a plane
perpendicular to the axis of the pipe. Knowing that the maximum allowable tensile stress in the weld is 84 MPa, determine the largest torque that can be applied
~
to the pipe.
300 am
From
equation
H
hence,
6 mit
oy
textbook,
v, ™ 90
i
as
(3.14 } ot the
.
Conn
=4
a,
=
c
~
84.0%
+ Geo)
Cy =
Cy
£2
re
Ele,- c*)=
riso0mm
om& =
E
144mm
ous)
ay]
= 1198 x10 nt
_ (BAO $\(119.bavi!
Problem 3.153
3.153
For the aluminum shaft shown (G = 27 GPa), determine (a) the torque
T that causes an angle of twist of 4°, (b) the angie of twist caused by the same
torque T in a solid cylindrical shaft of the same length and cross-sectional area.
Pp
._ TL
= ET?
OP
8
(a)
= 47 =
~
}
€4,813%10"
= 27 GPa
GIP
L
>
pee
,
Le
= 27 «107 Pa
I=Ules-¢')= $@.0s*- oi}
- 18mm
T
=
avneas
T=
Bets
gt
TL
P= GE
ty
= feet
A=
=
199.5397
me?
N-m
Ts
2
=
OL OLBHIG
= 50.894 x10? m
(1.25)
(27107 \GO. 894 x 1077)
199.5 Nem —<m
= m7 (eQ- 67)
f0.018*- 0.012"
Flo.ogiel
. 98539)
32,524 410 7 !
(27 «107 ) (132.424
“oC. 813 “o)
igs
=
lbs Matching
hasm
m
4
11.514 «10° pad
p= 10.40"
=
Problem
3.154
3.154 The solid cylindrical rod BC is attached fo the rigid lever 4B and to the
fixed support at C. The vertical force P applied at A causes a small
displacement A at point A. Show that the corresponding maximum shearing stress in the rod is
|
Gd
‘
Toon A
- 2ba
where d is the diameter of the rod and G its modulus of rigidity,
Lever AB
angle @ te
im the
shown
Fiquee.
aunt Piary
Vertical dispfacemen? is
from
toms Hiro ogh
sition A'B as
which
A= a sing
@
=
avesin
A
The moytimum Shearing stress in rod BL
GV nnn
Coe =
For
mal
are sia
4,
small
7
=
d
GSE
Ce
se
A
a
is
_ &
~ Gd
_ A
ave Sin Gr
A
OTK
aL a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.155
.
3455 Two solid steel shafts (G = 77.2 GPa) are connected to.a. coupling disk
B and to fixed supports at A and C. For the loading shown, determine (a) the
reaction at each support, () the maximum shearing stress in shaft AB, (c) the
maximum shearing stress in shaft BC.
250 mint
200 wis
r
a
Shaft AB
‘
_
Sr
© thd =2Smm=0.025m
= O.200m,
T = The 5 Me
|
38 aM
wets lO. 02s)=
613.54 0107 mn"
L4kN-m
_
Pa
Toss Soe gy = OMS
“Tee
3
og
Pa4 = 286,817%10"
.
BC
y=
Lar
ES,
— (77.2 «10613. 59*IO7)
GA,
Shatt
“Tas
os
bee=
,
tal
. C
O.250m,
=
=
Pum
.
O.01ITm
y= ee
Jee = Eels E@.oi\! = 204.71 «16? m*
Tee
=
QO,
ae
=
G
T=
disk.
230 B47 «IO
LY xlo™*
Tig + Tee
x10” Pp
Po + 68.2
4 6657107 ° pool.
if
Pe
Tae
(236.847¥/07)(4.6657™l0")
-
= (63,214 * [0° (4. 6657» IO?) =
(a)
PD,
to®
2\4»
«
of coupd ing
Equi A ibrivm
63.
*)
Fivio
“Nae
a
(77.
a
at
Reactions
supports
1050610"
Nem
294,94
Ta
.
Nem
Tis Tee =
shearing
(bo) Massimum
Tro?
Tee *
- 4: 1o526 HV
ies
(0) Manin um
Shearing
Jet
Tee
stress
in
Num
=
245 Nem
=
JOS
= Tag?
AB,
ons)
45.6 wio® Fa
Ther 45.0 MPa —a
stress _ m BC.
O- O18) _ 27.4%10"
= (294, 74 VC
x1O""
204.71
Cae? 27.4MPQ
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
we
Problem
,
3.156
,
3.156 The composite shaft shown is twisted by applying a torque TF at end A.
Knowing that the maximum shearing stress in the steel shell is 150 MPa,
determine the corresponding maximum shearing stress in the aluminum core,
Use G = 77.2 GPa for steel and G = 27 GPa for aluminum.
:
SOLUTION
let
and
40 mm
G,) Jy and T, vefer te the adum ‘num core.
Gz) Ja; and, refer to the steeh shelf-
At the outer surPace on the steef
Alomintm)
%
=
oa
L
At the outer
x
r
’
_
tL
¢
t,t
C,G,
“|
Solving
Fov T,,
T,
_ & 6
=
aq
a
%,
£
L
surface
Cc.
Matching
~
Pon
a
Te
>
of he
&
L
=
both
nN
G,
=
shelf,
te
Ci Gx
afuminum
x
core
2
ct
CG
Components,
C,G,
_
*
29.030
27xf0?
S04o” grrle.-
1ISO*1O* = 39.3%/0°
Pa
t= 37%3MPa 0 (ed
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution te teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 3.157
3.157
In the
gear-and-shaft
system
shown,
the shaft diameters
are
dy = 30 mm and dep = 38 mm. Knowing that G = 77 GPa, determine the
angle through which end D of shaft CD rotates.
CascodedA
1o%
oy
tore
Circumferential
between
gears
p> Ts
%
contact
B
= Je
amel
fone
Cc
Tha
Y
% lee
Teo = 60 Nin
The? Zee) = (SOON
C= ddhy=
Le 645m
Twist tn sheaf} AB:
SJ= Bet= F@orsy* = 613.6110 wi.
_ Th
_ (600) 0:48)
Ex ~ C17 8109) C613. 6 xi)
=
010143
Pn = Pan 7 C0143
rad
Pen ~
Rotution at B.
Tangent if displacement at
Roteahiou at
Twist
iw shalt
I
>
Je
Poe
Pe?
c.
ot
bh
ET L*
Rotation AD.
7
cirese
&
>
ET
(oaiayt
S=
(e043)
r
Vad
OO3S5S7
=
UO
rad
C= tal, siting
Los O6m
CD:
-
gear
pad
=
rofl
OOO
FMAbaar
tat
.
Lg,
= O102LY val
fe= Bt Poe * C105BS vack
Qt 25°
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=e
Problem 3.158
3.158 A torque T is applied as shown to a solid tapered shaft AB.
integration that the angle of twist at 4 is
Show by
O= 12xGc*
Show -ve
ag
y
coordinate
Introduce
n= &&
Twist
yo
oO
\ coo
\
: \
yet
=
pn
in feng th
Tey
_ Ta
. 2£tTldy
- Tech y
” ei
2. yori! ay — 2TL
* dy
AP=
P = ,
\
Y
x
Ye
1Gc
*
~ T6ct
- 2te ( t
~ 7Ger(
=
»
* wees
a
Wee?
dy"
oath
3y*§,
atl".
fay
ee
ay
“BES * 38
iT.
(2413
<h
muGec
ae
Problem
3.159
3.159
A38-mm-diameter steel shaft of length 1.2 m will be used to trans-
mit 45 kW between a motor and a pump. Knowing that G = 77 GPa, determine the lowest speed of rotation at which the stress does not exceed 60 MPa
and the angle of twist does not exceed 2°,
o etd =l9mm
meat
require
Stress
Te
Lehem
Ys
bo
bre = Ebon
Twist angde
p=
T=
Monimum
Pe
_
EGet@.
2b.
Bowebbe
=
cp = R=
ae
64¢6Nm
B4.FOTHIO”®
AT be
TGet
alae) (oot)
(AYI-2)
tovque
=
Tr
YO
MPa
on?
requireme.T
Gt
P= 45LW /s
ved
(24.9078)
is tho suntler
yodtre.
2 GSP
Te
Nm
458 Nm
antT
Po.
Pant
YC lo®
qalacpy
—
* N
Ie
He
Fe a3bepm
am
3.160 Two solid brass rods 4B and CD are brazed to a brass sleeve EF.
Determine the ratio do/d; for which the same maximum shearing stress occurs
in the rods and in the sleeve. .
Problem 3.160
Let
Cc,
=
Z
Shft AS.
Sheev
4
C,-¢,
let
Solve
AT
nce
stresses |
4
x=
by
=
Cz
T -
ta
Zo
=
TC
a
GB
.Z7)
7G
=
eT)
ate
M
FG
oo
4
cc,
c
x'-f
= x
=
b= V2 = bisa,
x
td.
.
2Te,
|
= We,t2c,7)
ov
Successive cpp roximatians
y= 2.20
=
=
|-221,
1.221
%, = 42184
x2
iad
stop ting
with
#
equad
aud
|.216,
2
ft
Fov
EF.
2
a,
12}
Xo2
lo.
72.216 = 1.220
}
X= {2-221
OS
= 1.224
|
(conversed) .
Ag
'
Proprietary Material. © 2009 The McGraw-Hill Companies, Tuc, All rights reserved, No part of this Manval may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, ar used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course prepatation. A student using this manual is using it without permission.
Problem
3.161 One of the two hollow steel drive shafts of an ocean liner is 75m long
and has the cross section shown. Knowing that G = 77.2 GPa and that the
shaft transmits 44 MW fo its propeller when rotating at 144 rpm, determine
(a) the maximum shearing stress in the shaft, (6) the angle of twist of the
3.161
—
r
shaft.
580
580
mm
mr
L.
=
75
P=
wm,
f
44 MW
=
Lo:
Prarhrs
de
= S80,
240
mm
=
O.240
m
(bo)
9
08° CS)
(2917871
,_ (774
x10
¥ [0.08
\0
Tk
~ Ol
Problem
29,
A. Square?
~]
~
2-1)
lt
“Gab*
.
eo
TORO
=
5 °
Ba d pad = 16.1
6.48
=a
The cross section of shaftA
|
ied
Te
.
JT
=
~
€,20.208 (Table S14)
7
~ go 708 b3
ew -ie. 20
Ee
4x10”
aa
83.9 MPa.
is a square of side 6 and that of shaft B is a circle of diameter 6. Knowing that
the shafts are subjected to the same torque, determine the ratio t/ ty of the
maximum shearing stresses occurring in the shafts.
T
Retio
$0.08 «io? m
3.162 Two shafts ate made of the same material.
3.162
p
)
2.9178 «10% Nem
|
83.9 «10° Pa =
10.0% xJo"8
Nn
T=
(2)
_ (2.9178«10‘)(0.
2490)
“= ie
. 4
4a x 10% -W
é
Vs
iF =~ THO”
2ir(2.4)
T= Eich-c) = F (0.210"- 0.160") =
Tez
rpm
~ wes
9300Sar
[6T
Tw b3
= 0.744—a@
I
C,
}44
gYUS
320 mm
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or .
distributed
in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
‘PROBLEM 3.C1
iement x
3.C1 . Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Its end A is fixed, while its end B is free, and it is subjected to the loading shown. The length of element i is denoted by L,, its outer
diameter by OD,, its inner diameter by /D,, its modulus of rigidity by G,, and
the torque applied to its right end by T,, the magnitude 7, of this torque beirig
assumed to be positive if T; is observed as counterclockwise from end B and
negative otherwise. (Note that 2D, = 0 if the element is solid.) (a) Write a com-
puter program that can be used to determine the maximum shearing stress in
each element, the angle of twist of each element, and the angle of twist of the
entire shaft.
(6) Use this program to solve Probs. 3.35 and 3.38...
|
Element 1
SOLUTION.
FER Frew CU INDeIcAL
Aes
AND
ORs
9
1%,
PLEMENT.
Gy
ENTER a
G
COMPUTE
J; = (7/22}(087- 107)
CBOVOT LINE
OF
UP DAFE
OND
FROGETIAM
TeR& LE
Te Th I
COALS
av.
= 7 (02,/2)/J,
pi
= Th (Eg 4:
ANGLE OF TWIST OF ENTIRE SAAT,
WITd
Gz0,
UPPATE
C=
FROLMAM
Problem
3.35
211.9575
23.0259
twist
Problem
Element
1.00006
2.0000
3.0000
Angle of twist
PH;
Maximum Stress
{MPa}
1,0000
2.0000
of
BVA ELEMENT
OVI PUFF
Element
Angle
@t
THROU GL
STARTINE
for
entire
Angle of Twist
(degrees)
1.3841
1.8323
shaft
=
3.2164
°
_ 3.38
Maximum Stress
(MPa)
87.3278
56.5884
70.5179
for entire
Angle of Twist
(degrees)
4.41181
1.0392
0.8633
shaft =
6.0206
°
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
3.C2 The assembly shown consists of n cylindrical shafts, which can be
solid or hollow, connected by gears and supported by brackets (not shown).
End A, of the first shaft is free and is subjected to a torque Ty, while end B,
of the last shaft is fixed. The length of shaft A,B, is denoted by L, its outer diameter by OD,, its inner diameter by /D,, and its modulus of rigidity by G,.
(Note that ID, = 0 if the element is solid.) The radius of gear A, is denoted by
a,, and the radius of gear B, by b,. (a) Write a computer program that can be
used to determine the maximum shearing stress in each shaft, the angle of twist
of each shaft, and the angle through which end A, rotates. (b) Use this program
to solve Prob, 3.40 and 3.44,
PROBLEM 3.C2
SOLUTION
FORGUE (4 =
ENTER
Teh
Trap= 7% Piss 1B)
FOR EACH SHA _
Compre:
ENTER
Sp = (732) oD, oa 1D,’)
Tau; = 7 (OP; /2)) Jj
7%: /G; de
PAL, =
FANELE
OF
PROFATION
COMPLI,
ATF
END
A,
ROTATION AT THE "AO END GE FRCH SHAFT
SIAR
wit
FROM
77
ANGLE > FHL,
AND
UPDATE
TO é, AD
ADD Feu,
Anete = pneee (AJB, , + PHL,
FYCOGLAM
CUIPLT
Shaft
No.
a
2
Angle
through
Max
Stress
Angle of Twist
(MPa)
(degrees)
62.17
1.472
84.80
1.7193
which Al rotates = 3.816°
FROG CAM QTPT
Shaft
No.
1
2
3
Angle
Max
Stress
(MPa)
382.17
Angle of Twist
{degrees}
22.403
191.09
95.55
through
which
Ai
11.202
§.601
rotates
=
22.403°
.
PROBLEM 3.C3
3.C3 Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Both of its ends are fixed, and it is subjected to the
loading shown. The length of element i is denoted by L,, its outer diameter by
OD, its inner diameter by ID,, its modulus of rigidity by G,, and the torque ap- plied to its right end by T;, the magnitude 7; of this torque being assumed to
be positive if T; is observed as counterclockwise from end B and negative otherwise, Note that 1D, = 0 if the clement is solid and also that 7, = 0. Write a
computer program that can be used to determine the reactions at A and B, the
maximum shearing stress in each element, and the angle of twist of each ele-
:
Element ni
ment. Use this program (a) to solve Prob. 3.155,(b) to determine the maximum
shearing stress in the shaft of Example 3.05.
SOLUTION
WE ConsperR
THE REACTION
REDUNDANT
COMPUTE
FOR
Ly
PACH
9
OG,
ELEMENT
O8%
AND RELEASE
wire Te
AT BAS
THE DuET AT LB,
tO!
» FHTER
(2;
Gy
;
7
(son 7, =
%O)
CLOMP OTE
Jy =(/22X oot ~ nf)
UPORTE
Foreaa@uk
72 7t%
AND
COMPUTE
Fale FaAcK’ FLENENT
7au,
= 7 (an/2)/d;
PHL
= Tht [Gide
LOM Pl FE
2,3
UPDARNL
THROUEH
BIA
PME
29
WITH
aro
OQ; = Gy + PH
iy
TAU,
pp
OO YNIT
Gro,
AT &
T2RAVE
= One /2Jf
unix PHL =
For
£
Caz DUE
COMPUTE
pYD
ELEPAENTS
Li leek
En ENT;
UNTO, (t) = UNIT Ey /L) + UNIT Fats
-SULER PON FPO
FOR
TAL
ANZZE AT ETTORE
BFR,
Opgt
& lun
Br
JHE
Element
~“TE7AL
1B =2
TA
tau
max
Problem
TA
TR
= Fille
FH,
12105 kN*m
kN*m
Angle
of
(degrees)
~45,024
27.375
3.05
=, lumT en)
“0,295
(MPa)
1
2
~0.267
-.267
=
=
&
TAU, = TAL: + 7g (um7 TL, )
Tome
ANELE: OF TWIST)
Problem 3.155
_ SROReem
sake
=
Jp z BT) + 7B
FOR Eker ELEMENT 2 pAOx S7pexs!
PRospan Ovrpor
Gl»)
-69.767
+-50.232
Nm
Nm
Twist
+7@ (ot'7 PH1;)
PROBLEM 3.C4
oie
e
a
.
.
.3.04 The homogeneous, solid cylindrical shaft AB has a length L, a diameter d, a modulus of rigidity
G, and a yield strength Ty. It is subjected
to a
_ torque T that.is gradually increased from zero until the angle of twist-of the
:
shaft has reached a maximum value ¢,, and then decreased back ‘to zero. (a)
Write a computer program that, for each of 16 values of ¢,, equally spaced
overa range extending from 0 to a value3 times as large as the angle of twist
at the onset of yield, can be-used to determine the maximum value T,, of the
torque, the radius of the elastic core, the maximum shearing stress, the per_
manent twist, and the residua! shearing stress both at the surface of the shaft
_ and at the interface of the elastic core and the plastic region. (b) Use this program to obtain approximate answers to Probs. 3:114, 3.115, and 3.118.
SOLUTION
AT
OF
ae
LOADING ! 47h,
a
de Bp = Cele -
:
7
(FY
I
$7
Tue
y
_—
.
JM
Yy
RL
cG
£@.-C1)
|
.
F.(2)
fo
Ae
:
YIELD”
Y«
2%BY. y< L
yen
AL
7
/
ONSET
|
UNLOADING (ELASTIC)
TL
R
#y= 27
Py=
PNELE
OFF
PWIST
For
YINLoAdin¢
1¥
hy
Te 7
7
%
Bae
r=
KR
I=
‘
PAU AT poe
TRU
AT
Phy
T
SUPER POSE
fae
_
|
WHEN
LOADINE
GeO
#
<
1
ft :
wien b> gi
FESIDURL +
ANDO
UNLGADING
d= Bdy
mG
7-7
7
vent
-
O2Fy INCREMENTS
}
P
=z.
a
(Y
4
use EGC)
Po= Pu-Py
ze
(% UE
7eq* are
Dy
Pye
:
EG(2)
Pag = By
fy
Y Fey
ee
.
CONTINUED
PROBLEM 3.C4 - CONTINUED
interpolate between values at the values of T,,, Or Bux indicated
Problem 3.114
PHIM
and
TM
3.115
TAUM
O
Ry
mm.
3O64P0
24+ b3S
46044
Zar 4PO
Firoly
LS 147
e177
21676
[Xf b90
kNm
deg
0
PHIP
M Po
deq -
TAURI
TAURZ
MPa
M Pa
oO
Oo
oO
Oo
Q
2G
202399
~ 326139
$6. bbe
¥e7Rb
136843
fh 6%
74-372
61+ Ft7
~ 4b]
HB.
929914
{o: lee
(SAbI6
6 & 138
B4898
49693
tT 5
Problem
Tinos
TEATS
kai im,
3.118
PHIM
deg
™
kN*m
0.000
0.807
1.614
2.421
3.228
4.9036
4.843
5.650
6.457
7.264
8.072
8.878
9.685
10.492
11.300
12.107
0.000
0.187
0.373
0.560
0.746
0.933
1.064
1.132
1.168
1,191
1.205
1.215
1.221
1.226
1.230
1.232
RY
aan
16.000
16.000
16.000
16,000
16.000
16.000
13.333
11.429
10.000
8.889
8.000
7.273
6.667
6.154
5.714
5.333
TAUM
MPa
0.000
29.000
58.000
87.000
116.000
145.000
145.000
145.000
145.000
145.000
145.000
145.000
145.000
145.000
145,000
145.000
PHIP
deg
0.000
0.000
0.000
0.000
0.000
0.000
0.240
0.759
1.405
2.114
2.859
3.624
4.402
5.188
5.980
6.776
TAURI
MPa
TAUR2
MPa
0,000
0.000
0.000
0.000
0,000
0.000
7.198
19.486
31.542
42.197
521.354
59.184
65.901
71.699
76.739
81.152
0.000
0.000
0.000
9.000
0.000
0.000
-20.363
~30.719
~36.533
~40.046
~42.292
~43.794
~44 837
-45 .583
~46.132
-46.543
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3.C5
PROBLEM 3.C5
The exact expression is given in Prob. 3.158 for the angle of twist
of the solid tapered shaft AB when a torque T is applied as shown. Derive an
approximate expression for the angle of twist by replacing the tapered shaft by
n cylindrical shafts of equal length and of radius r, = (n + i — 3)(c/n), where
i=
1,2,,...,n. Using for T, L, G, and ¢ values of your choice, determine the
percentage error in the approximate expression when (a)n = 4, (b)n = 8,
(ec) n = 20, (d)n = 100.
SOLUTION
Fike FRog.2158
EXACT
£r PR ETE
po Le
OR,
ea eee abet
Nes
fz
eee
LONSIPER
Fy
(S
(27rG eet
Teh
@er
=
O/E56E8
i aennad
enn
JYP ICAL . Cb
Y=
Fe
TL.
FEF
Eo nereenaes
SHArFT
(7 ri -£)(V>)
Le fn
Up = £&)
7 (*%)
Gk
Az
LN7En
veuz
(NOTE!
LEW FETE
yvaAewes
PP FOL
VALUE
7
CAN
tf,
EN tere)
PEO
OF
LOre
op
SAGE IS
p
UFO TE
0
Arp
CE
CYLIMPIICAL
= NUIEEL.
é =)
For
TL GG
VALLES
1
FANTEle
OF
Seb
papt sp
OvFfus
oF
PLCS
AA
COEFFICIENT
Exact
coefficient
Number
n
4
8
20
100
of
approximate
0.17959
0.18410
0.28542
0.18567
from
of
TL/Gc*4
Prob.
elemental
3.158
disks
exact
0.18568
0.18568
0.18568
0.18568
=
is
0.18568
n
percent
error
~3.28185
~0.85311
~0.13810
~0.00554
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ION!
3.C6_ A torque T is applied as shown to the long, hollow, tapered shaft
AB of uniform thickness ¢, The exact expression for the angle of twist of the
shaft can be obtained from the expression given in Prob. 3.153. Derive an approximate expression for the angle of twist by replacing the tapered shaft by
# cylindrical rings of equal length and of radius r; = (n + i — 3)(c/n), where
i= 1,2,...,”. Using for T, L, G, c and ¢ values of your choice, determine
the percentage error in the approximate expression when (a) 2 = 4, (b)n = 8,
PROBLEM 3,C6
(c) = 20, (d)n = 100,
SOLUTION
vce
VRE
SW8ET IS LONE
CC <<h 4
JS
B40Rt
aN WE CAN USE
OF AS
COL
THE
YN
rr"
\
YEO
RIAL
p
L|
ANbLE SY
Titrck ESS
RINGS, .
For,
\
Tae
RHE
e
O<k
Ez
ze~
Lan @= aoe
=
=
be
\
yl
L
fn
tle
n
LAE
yalate-Z)(F)
4 ae (agen) 73” = Cr né)r = wre ry
. 2%)
GJ;
Ay=
ENTER
UMT
(Nott!
VALLES
ENTER
(NWT
LIV IER
M2
fon
FoR
<PRerIC
C=
TL,
VALUES
VALLE OF
NUMGER
1 ve
OF
2,
G
2, AMPK
CAN G& ENT&R2EO
BHEO
Fok
fh
CYybriORICAL
UPonre
IF DESsy2e0)
RIMES
cp
cpr prh¢
QUIET
oF fleagleany
COEFFICIENT of TL/Gtc*3
Exact
n
coefficient
Number of
approximate
from Prob. 3.153 is
elemental disks = n
exact
0.05968
percent
error
4
8
0.058559
0.059394
0.059683
0.059683
~1.883078
~0.483688
20
100
0.059637
0.059681
0.059683
0.059683
~0.078022
~0.003127
3
Chapter 4
Problem
20
40
4.1 and 4.2 Knowing that the couple shown acts in a vertical plane, determine the
stress at (a) point A, (5) point 2.
4.1
20
For rectangle
M=I5KN-m
Be
I
=
Quteite vrectanafe:
x
bb?
I= eBoy (120)*
T= 1.62% {0% mm? = 1S2*1T% m
Cutovt:
I, = % (4e)(30)*
¢
Ty = 1.70667%)0% tom”= 1.70667 x10 * mt
Dimensions in vim
Section:
(2) yr HOmn + 0.040m
I
=
I,
-
+
=
4. 81333KIO~
= — Mya
- ~ S10"
1.040).
_ ey Exjo*
Lr
481333
107°
:
Oc)
yee-COmmt-0060m
Js
Problem
Gar
Gl.6
MPa.
ol
6,2 —MYa 2 _ (I5*10®C 0.060) 2 ay zxlotB
eo
TFT
F,31333%(0%
“
6g2 F.1 MR
en
4.2
4.1 and 4.2 Knowing that the couple shown acts in a vertical plane, determine the
Stress at (a) point A, (2) point 2.
Cor
30 mm
30 mm
—
m’
the
L is
solvel
zs
bh°
rectangle
=
|
zt (0-12) (o-ab)*
= 2-16 X10 “Gay
120 mm __.|
For
civcufar
one
cutout,
Tz
=
ora
er"
2
¥
F (0.019)
102-85 %16 Un*
For the
(2)
Tn
=
Ds
section,
B08ym
Sa
|
I - 21,2 216x0
t- oye
=
_,
aooy(ar03)
~6 ~ (a\Clez3sx16 ?), sh4ss x15 “pe”
oc
1985x107 ©
(b)
Yea* ~ 71019m
.~.
MY,
Gg =-—*
=-
(2000\-6:0/9)
«= 30697
MPa
6.
we
MPa
zo-7
in
BOF
A
= 19.44% MPa
1.9ssxie ®
Gg
= (24
Mh
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individuat course preparation, A student using this manual is using it without permission.
|
Problem
4.3
43 A beam of the cross section shown is extruded from an aluminum alloy for
which oy = 250 MPa and oy = 450 MPa. Using a factor of safety of 3.00, determine
the largest couple that can be applied to the beam when it is bent about the z axis.
stress = oe
of
Problem
~
aheut
Z
aasts,
mm*
T= KOO =
43.64 «107
mm"
T, = 682.67 *10* mm"
I, 4+ D4 Ty = 1.40902 410% mm = L4o4go2eta~ int
with c= £(80)= 4Omn = 0.040 m
“
é Ase
(Luogoanic”
4.4
inertia
632,67 10%
oe ve
M = is.
150 MPa
Fs bcelso¥=
Ta*
T=
=
Sox {0% Pa
5
Moment
> “28.
4
APourckbe
¥
xio*)‘ = 5 23x10" Nem
M = 5.23 kN
«al
4.4 Solve Prob. 4.3, assuming that the beam is bent about the y axis.
4.3
A beam of the cross section shown is extruded from an aluminum alloy for
which oy = 250 MPa and oy = 450 MPa. Using a factor of safety of 3.00, determine
the largest couple that can be applied to the beam when it is bent about the z axis.
APlowahte
Moment
stress
&
=-
Sv
=e
.7 Zoo
459
+
{50
¥}o*
rT
Lz
with
Me £&
= ,z
2°
i500 MPa
Pe.
inertia aboot yronis ,
354.987 «10° mm’
I, 2% yey + (Ro\b igs
T=
.7
E32
= 10.923 ¥ 10° mm"
a= 1, = 354.987
% 10° mm"
Do 42,44
c= ¢l48)
= 720.9
4/0 mm
= 24mm
0" Usa «10*)
(7Z0.2x1
.
B.on4
=
= 720.9410" m’
= 0.024 m
4 S1xlo
%
N
-
m
|
.
|
M
=
4.51
kN
.
™
wna
:
Problem
4.5
4.5 The steel beam shown is made of'a grade of steel for which oy = 250 MPa and ay
= 400 MPa. Using a factor of safety of 2.50, determine the largest couple that can be
applied to the beam when it is bent about the x axis.
16 wm
—F
The
momert of inevtra
that
A]
of
a
rectangle
|-—200 min —|
with
to
a cutevt
b,
=
y
n,
16 mim
Libr
Le
T,
rectangse
“
b, =
I, =
Section $
=
=
Com = Se
6.
Ms
olf
ae
[BO wm
=
0.130
Sen
=
[CO
=
=
190mm
= is
T,
3
by
by?
260-
(20016)+ RQQBmm
187.662 "10% umm*
105.271 *10°% mm” =
=
T-I,
mm
292,933%10%
mm
200-10
= 121L(140)(228))
260
>
eb
>
ot
2 (200(260)° =
Srag then
262
h
oe
Yj
7
G Ly
lb,
mm
ROO
b, T
reetangle:
FE
V77A
;
4;
le— p,~3I
Cr
equivePent
260 ram
Z
Laver
Ty cis
105.271 x10°° wt
w
MP,
=
(60%/0°
Pa
)( 160 10%)
Mm = SSw 2 (108.271*10°*
eC
x
=
]29%564 «10°
0.130
Nem
M,
*
1Z%G
kNs
wr
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. Ail rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
att,
4.6 Solve Prob. 4.5, assuming that the steel beam is bent about the y axis by a couple
Problem 4.6
of moment M,,
4.5 The steel beam shown is made of a grade of steel for which ey== 250 MPa and oy
yp
My
= 400 MPa. Using a factor of safety of 2.50, determine the largest couple that can be
applied to the beam when it is bent about the x axis.
16 mm
For
mm
one
be
Wange
=
16
Lp
mm
™
hy =
(6 )(Q00)8 =
Tre
For
,
the
web
I, = 7s
5
b,he
200 mm
10,6667* 10% mm’
bbw
b, = 260- Q)(1e)
= 223 mm
Iw = ik (aaa 0
T=
Section?
=
Siu
Supe
oo
age
=
=
Pe
Mes
ty
QI, + Ly
100mm
=
|
=
= fe
me
J
=
=
[oO
mM PA
= 19210" mm"
21352 x10 mm! = 2Q1.352% 10° m*
.
0100m
[60 MPa
DS.
hw
~
=
Pa
160%/0°
Se. 3sr rio" C60 x108)
~
mG
“
O.}0°
Sw
34. 163é¢ 10>
New
My = 34.2 New
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed, reproduced, er
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=a
Problem
4.7
4.7 through 4.9 ‘Two vertical forces are applied to a beam of the cross
section shown. Determine (he maximum tensile and compressive stresses in
portion BC of the beam.
ma
payee
75 mm 1
i
Ye
i
A | yo | AY
O
©
oe Ky
40
EL
i
I, = Kb
3
+ Ad?
[1/280 | 12S [40 6280
®
juz]
fo
=e!
28] 280
25
F726
= [r2{vu
=
Y,
y
> po
®
Nevutval
above
aE mu
EI
the
a
DS ben.
hase.
4d
fay? (0-18) 3 t Cole) - (OOS)
= shtp(0007S)
2
Dies
ARIS
.
;
a
.
= 44-22
Kio~&
a
~6
4
4
28) (008) o be- 47 Ko Me
I= aebhds A,dp = ty (0-225) (ors) i+ (oro
Ts Tad = 74Tx0” u*
Yep 7 12Sme
_
t
yug FTL em
M
")
M-
Paso
Me Pa = (60)
(1) = bo EN.
ep kaa
&
= - My bp 2
P
1
Gy = ~
‘yet -
bev-oo) (0-129) 2-441
19°70 F
~Coseee) (<0
MPA (compressvon)
=the l MPa
8
(heaston)
tf
719-7 Xo
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.8
4.7 through 4.9
[wo vertical forces are applied to a beam of the cross
section shown. Determine the maximum fensile and compressive stresses in
pe
|
portion BC of the beam.
4725
7
wo 9 ane
A
my
She2s
[P4326
“jeg
Yo *
© ©
B20 4E4UL Kio
td
the
Pres
ie
o
base.
a &
Mm
&
ry
$442Z2x¥1t0
Gb
“™™
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‘
te bahs + Ajely = i (loo) (28)? +2. Se) Liob4)*= 2. 617 Kta hs
I+ 2, + I, - bo-X9 x10”
Fie = F0r6° two
|
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=
-
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-
ma *
Yue t= 4 & min
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M=
tea a
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Ll,
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[T=
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[= bbs + A.A, = az ee)(25)3-+ (Soo) (631) *
of Zo
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(compre
ssvon)
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a
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é
(tension)
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student usitig this manual is using it without permission.
=e
4.7 through 4.9 Two vertical forces are applied to a beam of the cross section
Problem 4.9
shown. Determine the maximum tensile and compressive stresses in portion BC of
the beam.
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Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.40 Two equal and opposite couples of magnitude Mf = 25 kN - mare applied to the
channel-shaped beam 4B. Observing that the couples cause the beam to bend in a
horizontal plane, determine the stress at (a) point C. (4) point D, (ce) point £.
4.10
Problem
120 mim
=o
__.
180 mm
:
adil
Y,
c
- 36 mm
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mn
D
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aa
Sa
MPa
Pa
aah,
4.11
Problem 4.11
Knowing that a beam of the cross section shown is bent about a
horizontal axis and that the bending moment is 900 N +m, determine the total
force acting on the shaded portion of the beam.
The
stress
cross
distribution over the entive
sechion
is given
by the
hending
stress
Lon muta
~
where
yo
is
a
the neotral
ot the
con the
L
coord inate
axis
entive
shaded
distribution,
and
with
tts
I is the
cress sectional
ow'gin
moment of inertia
avea.
The force
ts cabeudated from this stress
Over an aren ePement dA
dF= &dA = - Waa
The total force
on
on
the force is
|
the shaded
area, is they
Fe (de = ~(ivdA = -B(yda- ~M5*pe
where
shaded
y * is the
portion
centrotdad
and
A*
{Ss
its
T=
7
|
axis
_—_
|
|
@
BS
axis
ay
CVO A»
ik
bh? ~iebh,
= &(46)(62)"
|
|
of +he
i,-t,
=
®
~-
O
.
cooveinate
~~ (ge) (46)?
= Or 67025
Ktg” tain
yA = ZA+ As
= (27)€23) C8) + (eS) (23) 9) = 708& for
F = My*At
TA
(Fen)
ee tiie(084,10
O+ 6702S Xo
3,
|
a) EW
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc, Ail rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachors
and educators permitted by McGraw-Hill fer their individual course preparation. A student using this manual is usingit without permission.
~
Problem 4.12
4.12
Solve Prob. 411,
assuming
that the beatn is bent about a vertical
axis and that the bending moment is 900 N «im.
|
_.
|
4.11
Knowing that a beam of the cross section shown is bent about a
horizontal axis and that the bending mornent is 900 N -m, delermine the total
force acting on the shaded portion of the beam.
Snr
46 mart
vg. Samin
ole
30 iM
- § min
distribution over the entine
section
is
given
by
the
hending
stress
Loan muta
whene
the
|
is
a
neutral
coordinate
axis
and
I
with
is
the
its
oveg in
moment
an
of inertia
of the entive cress sectional avea. The fovce
on the shaded ts cabecudated From this stress
distribution,
Over an avea efement dA
the force is
dF= Q dA = - Yaa
The
total
force
on
the
shaded
area
is they
Felder - ~ (My dA = - Bi yda
shaded
y’
ts the
portion
centroidal
and
r
A*
a
where
Oo
if
a
tS
~EYR
coovelinate of the
its
OVER 6
I,- 1,
&
bh?
”
bh.
= 1 (62) (46)? - 4 4b) (30)?
i
= 0.371402 10"
*
H
y* A*
H
7
stress
cross
ase eS
Samm
The
c=
YaAa
iam t
4 Jers
Cite) (8) (23) + C9) (22908)
§424
My*A*_
+
minr?*
(400)(S426x00|)
+ B4°}4.03 X10
= (2-2 EN,
=
Problem
4.13
«216
4.13 Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 6 KN - m, determine the total force acting on the top
mm
|
flange..
evoss-section
:
Gx =
72 mm
nevtva?
on
the
of
inevrtta of the
oO
is
LAK
2S
|
Fy
= S4-
1&
ad, = 54 - 36-54
Moment
T=
L=
Re
[i
is the
shaded
stvess
an the
avea
moment
sectional
the
total
fovee
shacled
portion
element
dA
avea
is theu
Fe lar =-SHeaA= -Blyda--By
=BEmm
the
ynis +he
centrotdal
coovdinete
ot
por tion and A* is its avea.
shaded
of entire evoss section.
= S(aleyss)*+ (gic sec
=
10, 41 25 10% mn!
hl +AAat= Ara Vioa)’+ (72Wi08 36)" =
T+2,
2k
I
its onrgin
this
on
Fron
The
where
bbb +A
Ths qb
and
entire evoss
Over an
mn
of mertira
bending
ag = Gada = -AYQa
4,
= 36
with
distributvon.
the force is
=
=
28.5535xd0% mmt
For
the
shade
N=
@ie
yr
26
ary’
—|
17. 6360 10% mm"
28. 55352#/0° mn!
avea
a6) = 7776
wn?
rm
mm6 =
= 219,X10
93
p -.|MATY"
VM
d,
cakeuPakd
——_-
ORS
oxers
Force
The
avea.
y
entire.
M
is a coordinate
y
where
|. »|
by the
(s given
Poumu da
stress
the
avew
distribution
stress
The
10
279.936%
- (6x10? (274. 936%10-)
i.
58.Rx1O°
2% eSE AS x fore
N
F=58.3
kN
—<
Problem
4.14
4.14 Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 6 KN - m, determine the total force acting on the
shaded portion of the web.
216 mm—»~
.
—
:
{
a
36
54 mm
poe
|
The
styess
drstrcbution
- section
'
mm
CVSS
FT
stress
Ox
whawe
Y
:
aver
1s
on
the
trertia
a.
coovel
The
7——_Ny
Or
=
the
fovee
"A,
The
toted
dad.
a7
1B
=54
> 8G
+ 36-54
inn
where
=BEmm
I
the
avess
on the
Fron
is
Oe
moment
sectronaf.
shacled
+hes
in
portion
stress
Overan
avea element
dA
i4
fovee
the
yrs
an
the
shacled
the
shacleel
aved
tS they
centpotedel
per too
and
coovlinate
A*® is ts
of
avea.
be bho tA
cvoss sectton.
= A(aie)36 )4 (216 Maeacy =
(72\(joB 036)" =
Tas wbhy+ Ads BOAR)
i+,
7
78.5636 x08
For
Ld
We,
4gry
don
|
mot
=
shacted
oe
Y=
4S mm
he
F
=
=
2G),6«%]07
[mer t
10% mam’
17, 636.0%
yw"
avea
(72.\€40)
ANY
10, 487.5% 10% mam!
23. 55a5/0°
A*
t]
Loeb, 77
*
fo
the
|
2
and
its
t
Moment olijueetita of entine
L=
mn 4
Fa Sar = ~StgA= -Elyda--FyYA*
.
1, =
bend
des G&A = -o
|
a, =z SH-
wth
entire
force
cakcudated
5
the
inate.
axis
of the
disterbuhion.
< LE.
by 4
entive
Lt
nevtyal
avea.
the
My.
=
oF
1S
ta
..
108 mim
to
;is
Poumu
avey
G48O0
ham
_
ww = 291.6 x10 °m
(Gx 107 (24,6
x10 )
6.3107 N
HESS 35 xf OF
F=oi,3
kN
ae
Problem
-—
4.15
Knowing that for the casting shown the allowable stress is 42 MPa
in tension and 105 MPa in compression, determine the largest couple M that
can be applied.
4.15
125 mm —|
=
[
Ize mm
7
mr
dL
ree
#29 WIM
J2 am
5d mn
tT Gk
A yam
y, mm
©6
1576
30-45 nm
iofmem F
SESS mm
x’
139500
AGS
4ASSO
é
3600
3000
187650
=
7
a
« bnenS
187650
-
ZAY
mm
Bo00
° A
Ki
LGkbh + Ad?)
He 282) + (5 \12 045
n
KH
x
Ay pom
93
600
Mi1@
]
Foo
|»
+ bU2 Vs Y + (12 VIS\Gs-08y"
soy 12)? + (80 liz Ysbess)*
+
aA
rye
ASSUMES
Oy =
+ 42MPa
=
=+or..
*
OTs
~™
>
ASSUME
&.
3 4iXid
~&
= ABKO )Te3baS
~Hy
A
mM
= 2-T1XI0
=3999Goommt
= 4505 hm
= 76
BB KO
M
CHOOSE
=
THE
x
oe ——
,
= (105%10°)
SmaLuEe
Mm
+
ae rere
tA
So igee
—
F
Mau™
a
6563-5
4-5
Nt
km
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, Afl rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<
Problem
—
4.16
4.16 The beam shown is made of a nylon for which the allowable stress is 24 MPa in
tension and 30 MPa in compression. Determine the largest couple M that can be
applied to the beam.
40 mm -—-~ “|
u}
602 |
@®|
°
|
AS
20-
ev
ma
FAS
nevkya?
OO.
ovis
ny
Owes
|
we
2 (0°
=e
Bet
Y=
The
8
¥ 107
15.2
5.5
UG
=
13.sxfo%
22.5 |
2 | 4o
|
V tonTF
> mm
AYo
@
“30 mim
w
3
-
Yo 3 im
mm
A
eT
rae
= 30 mm
Pies
VUSmm
= 11S mm
above
SAAS
V bot
the
Toe
me
O,
L,2 te bh?s Ad? = &4oisY +( 600) 5)* = 26.25 €10% mom
I, = re bh + Ardy = i (2005+ (300 V0)”
Loz L,4 Dy 2 Gh 81seio wom!
lol= |e
FOS
1
4
mm"
= 61875 «10 en?
side
©
_ (24x 0* C61. 378% 10)
=
GI DTS
Ie
r
Bottom> compression,
Choose
ry
Tv
ced
OK
OWS
mM = [So
:
_
‘Top:
2S 61x12
pottom.
smalbler
M
of.
~ Oh
= LZOXIOE) "G1 BASH ITT)
va due.
0, 0175
>
[13,8
N-m
'06.1 Nem
Ms:
106.1
Nem
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jinsited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this matiual is using it without permission.
<a
Problem 4.17
—
40 nan
ae
a
4,17 Solve Prob. 4.16, assuming that d= 40 mm.
4,16 The beam shown is made ofa nylon for which the allowable stress is 24 MPa in
ol
tension and 30 MPa in compression. Determine the largest couple M that can be
applied to the beam.
15 mim
i
d= 40mm
A
a
t.
|
17
|
|
@
) wm
RSIS x10"
Bottom
Choose
wm
mm
above the bottom.
m
60.827 x 10% wm
C
0. 91)"
+( 500)
b(2orl25y'
=
85.556 * [0% mm!
tf
Aydp =
N
oo
ho +
= | My)
M
I
tension
23.4)
=
¥
Be
Aa
x
i
i
at
Pe
23.4)
=
Ph 4 (600 09.09)
eAd = G4ONIS
=
side.
M
¢ Compress1an,
smalber
M
vabue,
=
q
= 196,383%10 |
|S)
é.
‘
Top
\\oo
Dies
B.25°K10"
25,75 [o*
2.8
m
De I, +L, = 146.983 x/O% mm?
is)
mn mao
soo
30
$
bh
T=
~ 0.0234]
=
mm
23.4]
7
=
Yoot
O.016S59
;
@j\|
>i
neutvas avis
=
mm
16.54
=
23.4)
AY
Coe | 225 | 12s x10%
o”
=< 4O~
mm
®|
=
The
Yeo,
(aabero*
Verne
&
= (S007) 08
2
38
‘
1, “4
2
ah
SES
4 '
=
ZZ
~
189.6
N-
ye
Nem
M= 187.6 Nem <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill fot their individual course preparation. A student using this manual is using it without permission.
Problem 4.18
4.18 and 4.19 Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa in compression, determine the largest couple M that
can be applied.
A,
@
©
nm
M1
arseo
@{
x |
~3O
ay
Yet
S4- BO = 24 mar
mm
7
=
Dies
Xr
aebh, +A
T=
T.4T,
z
1080 | 36
33330.
3
3240
497200
47200
BO
“Bie
mm
'
5 25816210"
mm
Choose
side
Compression
the
~
26
the
bottom.
=
SHH 32x10°
nm!
=
218, 4x to? mn”
75816 x10 om"
\mis [S|
M
» Wore
M
=
.
bottom:
above
7
OL024 wm
= gz (yo) Cse? + bao ls)?
M
tension
me
- O.030m
ist = [S|
top:
a.
3
T= abhi
+ Ad” = 7g (Honsd)? + yoXsa\(a)*
Tit
mm
y
Yip =
axis
Aye,
53320
Y"
nevtval
mm
27
Ys
The
Yo
smaller
as
6De1gs 1G
6
¥lO%}
¥to
(150 210*
)(
758.16 Alo *)
sag
0.
Mase.
Mat?
=
3.7908
=
32.7908
*jo"
x io
Alem
N.
m
3.7908 210° Ne,
‘Miyp
23.79
kien
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ao
4.18 and 4.19 Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa in compression, determine the largest couple M that
can be apphed.
Problem 4.19
@
veckanole
50 mm
@
circedar
7.85398 x/0% wm”
Aa=-T(S0)* =
| 15mm
OO mm
=
y
2964602 wio® mm
=
A, —Ag
A=
f Vi
Yo= 7 50 mm
As
=
(3.2463
Le)
ky’
ye
VY Ae
QTLL)
(37.§x10* )(o )4 (-7.85393 x1 (50)
29.¢4602 ¥]O?
Y=
In
-
Y= ZAY
mG
|
37.5%/07 mm
A, = Uso) (Qag0) =
150 mm
cotout
mm
= Z(T+Ad*)
=
I,
-
FT,
= [ ob CisoX2soy 4 (325 lo?
3. 2463)
~ I z (So)* + (7.85398 x07 (so+ 13.2463)" J
=
Top?
201.892%/O%= 36.3254*]0°
(Tension
side)
Cr
M
I
Bottom?
L
13.2463
165.567%1O%
mm = 165.567
%)/0 wm
=
Hil
6s, 56?x 10
S783) mm
Te
_
~
Gc
*
O.ul?sS
=
177.79 #10 ™ Nem
M ~=
Cr
siele )
Le
C
DIT wy
1254
13.2463
ao
OLL%B2ZS
= 138.2463 mm
Mm
.~ (68. 567*'o* \US0 «1o®)
O.\322S
=
Choose the smaller. |
=
20 «10*%)
=
(Compre s$i'on
Me
@ .= tS
125~
=
179.64 x 10" Nem
Me 177.8*lo” Nem
M1778
kaw
<i
Problem 4.20
kK een
4.20 Knowing that for the beam shown the allowable stressis 84 MPa
in tension and {10 MPa in compression, determine the largest couple M that
40 pana mo
can be applied.
{2mm
T(12\* = 226.2 bs
.
Yr
oa.
a
i
Mi
>
>
T
~
.
y.-2ZAY
— 226.2 = £7966 mn
“ar
abave
HeGium
inertia,
about
[te
the
pam
bottom
the
vomecd
of
base
|
= 22819
peo
He
Bolom: compression
(£72
P IED)
mmt
=
Pel
inm
Ist _= |)|).
tension side
4
inertia
Ty- AY” = fofo7e
Top?
ohm
tbh?-Ert = £4eee- Eliz) = jofo7s mm
seonbeosdel
=
2.
ua
(Pov (10) — (226 2)L56))
Dies
oF
>
Ytep
A
$73.2
axcs
event
bod
some
up _ (4)U2z)
ZA
Nevived
2
cutoul
2
mus
Goo
=
(40) (20)
v3
>
= semy- civeudar
@
=peckanabe
a
3
~
M
Yost
=~
FINS
mn
3
ot |
*)
(24x 10° (oapian
GtOagy
= 236-4 Nm
= (ons )l22814 wo) = 20s Nin.
Geohf
Choose
the
smaller
valve
M=
210-9
Nw
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.21
4.21
A steel band blade, that was originally straight, passes over
200 mm diameter pulleys when mounted on a band saw. Determine the maximum stress in the blade, knowing that it is 0.5 mm thick and 16 mm wide.
Use £ = 200 GPa.
0.5 mm
Reelrus
>| |
cuvve tone
of
CHEE
PE
QO
M eux
.
tm
4,
Mayinmium
££
oy
2
2
—-
ene
®
styarn
Stvess
=
Ge
rz
putter =
of
Radius
¢
+hickness?
bia ce
Band
oy
= OSanm
+a
center
=
[Gomm
Bone
oF bhcte?
= l0025mm
OzZE iam
&
Orns
pe
7
EE,=
D gtCoZ4TA
(2eoxio
2) (0002444)
6,5 490-9 MPa
Ore 418-8 MPa
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manuat is using it without permission.
Problem 4.22
4.22 Knowing that o, = 165 MPa for the steel strip AB, determine (a)
the largest couple M that can be applied, (4) the corresponding radius of curvature, Use £ = 200 GPa.
T= cbh* =(4)(25)(6)'2 deo mut
,
6 = ae
~~,
G1
fo.)
Ma
a= (Fb)
.
.
>
oF
= Bam
|
~H2
(168x108)
7 ESV x10 |)
a
=
&00%
M= 24.74
Nin =
¢.
&
3:
==
eC
E
()
Ee
(220y 10") C0003)
7 oe
ese
ow
(6S kr o®
p
=
“=
"
4.23 Straight reds of 6-mm diameter and 30-m length are stored by coiling the rods
inside a drum of 1.25-m inside diameter. Assuming that the yield strength is not
exceeded, determine (a) the maximum-stress in a coiled rod, (6) the corresponding
bending moment in the rod. Use & = 200 GPa.
D
=
Q
Ht
Let
Pp
=
insite Dismeter of the
diameter of pod >
fadivs
ot
Pp
Lz
(b)
i
z Glo.c0o3)t =
4
Ge =
Ec
M
=
ET
Pp
when
of
center
dine
bent,
63.617 xJO.
_ Qoox]o*
oon
—
covvetore
s
0.622 m
i
7
aeD-~ gh = gUas)-gEro*)=
Trde’:
ot
Vou
chum |
Cc: 4,
(0.003)
= 965% 10°é Pa
(200x%107)(63.617%/5") |
3 CLL
=
20.5 Nem
S965 MPa
<a
M=: 20,5 Nem «4
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using if without permission.
Problem 4.24
.
4.24 A24KN - m couple is applied to the W200 x 46.1 rolled-steel beam shown. (2)
Assuming that the couple is applied about the z axis
as shown, determine the maximum
stress and the radius of curvature of the beam. (6) Solve part a, assuming that the
couple is applied about the y axis. Use Z = 200 GPa.
For
W20046.1
steed section
=
USS
44g xfot mm”
=
443 10%
[5.3% 107% mn"
1ST 10% me?
Ty = S.3xlot mm" =
Sy = IST ¥ 10% wm =
fe)
24
M,=
2410" Nem
=
kWems
‘
o =
PB"
a.
2
3
=
° aoe
ET
.
MM
SB.6x10° Pa
=
My
=
24
kKNem
=
=
a
”
M
379
f
EL
~S
MPa
mm
a
af
ak
m
= 24x(07 Nem
24
.
=
: Gee
>
.
=
.
lo?
x
s
o
$3.6
=
“Qoonoiasentow) 7 FOS7*IO™
24
P=
(o)
x 1o% wy?
«10% mm*
Ty = 455
Sy=
volled
158.9 «10° Po
-
¥fo*
~
§ (ZooxlOt}Uls.3xjo*%)
aa
158.9 MPa
=
=
784 IO
127.5
—_
ai
7m
om
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ae
Problem 4.25
4.25 A 60 N - m couple is applied to the steel bar shown. (a) Assuming that the
couple is applied about the z axis as shown, determine the maximum stress and the
radius of curvature of the bar. (6) Solve part a, assuming that the couple is applied
about the y axis. Use £ = 200 GPa
12 mim
(a)
Beuding
60 N-m
T=
20 nun
bh? =
Zr
Le
M
fF
EL
=
2
ont
Pp
_
m
26.7
yrom's.
* se
ZF = Gmm =
(RGA)
3
=
2, BB*/D mm"
. 2.BBx 10
me
0,006m
6 = a (ooo. O=
- a
= 750 MPa
a%
37.5% 10m
(jooulo? VBn1o=*) 7
bh?
2.
_
_
0,010m
JoOmm=
75,0 «10° Fa
Go.
ahout
Bending
3x 107mm
3xXIo 7 mwm"
=
>
Se
ops
=
ATS.
tU2)20) =
:
ce
(b)
about
125.0 «10° Pe,
(Roon1OR B&x 107%) Ral OMIT
1
= 125.0 MPa
wi"
P*
9.60
m
<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill-for their individual course preparation. A student using this manual is using it without permission.
Problem 4.26
4.26 A couple of magnitude Mis applied to a square bar of side a. For each of the
orientations shown, determine the maximum stress and the curvature of the bar.
I=
c= 2
b=
fod
= &
a
|
EL ER
p78
For one titanate the
momet
inertia about its base
is
J
I, = bh*= Wie (R= &
I,
co 2
6
T=
T=
T+,
Me = Mel BBM
bi: eB
x
4
=
o
og
au
Be
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hilt for their individual course preparation, A student using this manual is using it without permission.
Problem 4.27
'
4,277 A portion of a square bar is removed by milling, so that its cross section is as
shown. The bar is then bent about its horizontal axis by a couple M . Considering the
case where # = 0.9h, , express the maximum stress in the bar in the form o,, = ko,
where 0, is the maximum stiess that would have occurred if the original square bar had
if
L
uv
been bent by the same couple M, and determine the value of &.
“4,4
(Veh he + @\VQl2h.2h )Ch*?
i}
his thh?- $h h? = Shh?- h
Me
PST
kor
the
Co
&
Co
-
hr
=
(4h,-3h}h?
G=
0.982
©,
20,
original
he
Mh
“Shhh
h
_
~ (Fh-3hyhe
square,
2
(4ho-3h.) he
(Hho~(BYO.Mh\O.TKE )
“
3M
h=he ,
az
h.
3M
ne
ra)
q 5 Oo
Kk = 0.959
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
|
Problem 4.28
4.28 In Prob. 4.27, determine (a) the value of # for which the maximum. stress a,
is as small as possible, (5) the corresponding value of k.
4,27 A portion of a square bar is removed by milling, so that its cross section is as
shown. The bar is then bent about its horizontal axis by a couple M . Considering the
case where 4 = 0.9/, , express the maximum stress in the bar in the form @,, = ka,
where a is the maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of &.
I=
Oe
is
=
t
al,
=(4GE) hb? «(2 )(8)(2he~Zh hE
= Ght- Shh gh? = 3Fh
<=
h
Cz
lla
©
41,
maximum
at
Al
Es ghighy -(ghy = fh
For
af
the
G6,
SIVA
,
h=
=
- h J
Shh?
h
a
orginal
thoh -
O
on tee
£hh-Sh*=0
2
bh
-
—a
2h,
._ M
he
ce
he,
a
= 2
h
>
»2 +332
Me .= See
3M
eT
o
he
Ze GS. Wt,
ms
7
Rt.
0.944
Kt . 0.949
=a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student asing this manual is using it without permission,
~ Problem
4.29
4.29 A W200 x 31.3 rolled-steel beam is subjected to a couple M of moment 44 kN m. Knowing that £ = 200 GPa and v = 6.29, determine (a) the radius of curvature p,
(b) the radius
of curvature p' ofa transverse cross section.
For
W
200%3!.3
T=
3)
4 .tot mm
a.
@)
yvobled steel section
M.
P
= 311o4 mm"
4s x [0*
EE
© (200410°)(31.4x Jor*)
=
c
71T¥IO*
2134.6 m
aL
(b)
=
fr’
ve
Problem
= (0.29X7.17
«1S *) =
2.07 «10m!
f=
Bl
"
mm!
=e
mn
=<
4.30
For the bar and loading of Example 4.01, determine (a) the radius
of curvature p, (b) the radius of curvature p’ of a transverse cross section,
4.30
(c) the angle between the sides of the bar that were originally vertical. Use
E = 200 GPa and pv = 0.29.
From Exampfe 4.0/
(
fo
0)
P
Ob)b
M
os
EL
=
=
v
e!
=
oF
= Ye
~2
e)
Problem
M=3ZkENm,
(3.xi0%)
~
(zea
Gere?)
j
E&
2B
= Be
.=
I= 362000 mu®
mv!"
Geeused
Cc
2
=
gh
ye
= (0.29)
(6+ 04167) ty" = a: 0/208m
OZ
. MOF.
yyy,
bL
4.31
th
4
|
Ue,
ave 2
ofhave
Senath
fength
= 2.
p
4.31
od
.
pe Fa 7L mm.
-€
¥1O
vad
= O.OISSO
—e
°
—lt
For the aluminum bar and loading of Sample Problem 4.1, deter-
mine (a) the radius of curvature p’ of a transverse cross section, (b) the angle
between the sides of the bar that were originally vertical, Use E = 73 GPa
and vy = 0.33.
From
Sampde
p
Probtem
44
To
toe ML forsee
fa)
(732x107)
But
p
=
ro
}
Vis
™
3b
(4
(0.33
x1076)
Ore
432 xem?
O34
6
=
359
=
)
3
nl
Me
fof kw
hn
Oolong.
nt
<t
Pp’
tb)
QO
=
Jength
of ave,
vad Sus
*
=
zi
fy
fe
GX
OOP
* a7
|
= P242x10 “vad =
Oc oG- Tp?
—_
Problem
4.32
4.32 It was assumed in Sec. 4.3 that the normal stresses 9, in a member in pure
bending are negligible. For an initially straight elastic member of rectangular cross
section, (a) derive an approximate expression for g, as a function of y, (6) show that
(Bmax
* ~(C/2PMOJyax and, thus, that o, can be neglected in all practical situations.
(Hint. Consider the ffree-body diagram ‘of the portion of beam located below the
surface of ordinate y and assume that the distribution of the stress a, is still linear.)
Denote
Senath
~
OB
Diag ram
bed,
the free
Using
b and
beam by
the
kb
os
cos 2
with
above,
|
= 0
2 Si Gibdy seQ
Gb +t
ZR=0
of the
the width
by L.
¥
Bu
(a)
S,
The
)
(Gy
sing
=e
Gy
(7
a0
dy
Cx
6x
dy
>
-44
8x
dy
Gy = (ae SE
(Sx
= arr
~
may
mactimux
nae
*
=—~
S Y yay
vetlue
(Gr) mee
*
ral
“Se
(So) sae
eceurs
8
=
Sk
>
an
+\
aft
y
~
=
hoe
Gro
~
(y*-e*
wt
O
} range
= Celeees
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution ta teachers
and educafors permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
it
Problem
4.33
10mm
mm
O
4.33 and 4.34 A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below, determine the largest
permissible bending moment when the composite bar is bent about a horizontal axis.
I]
|
Aluminum | Brass
Aluminum
{
UmUA
Modulus of elasticity | 70 GPa
| Allowable stress
105 GPa
100 MPa
| 160 MPa
Brass ants
Use
ad dor in dire
Fou
‘a
For
OS
oP serine
brass ,
Valves
the
yebev2v
,
ne
Le
matkevtiat.
:
a)
E/E,
of n ave
2
shown
108/72
onthe
>
hs
Plaure,
~“
For
the
teanstor mee
Te eibhes
t
n
L=
T,=
15S
=
I,
283.333%]0°
=
7S
;
bh?
Afumonume
¢
(Zoe y*
=
(G0 x10"
{\>
LO,
neies
Msthe
-
a
wm"
mm”
lyt>
30 von
=
%/0°")
iy!
=
ROmm
7
(160
x 10°
1to” )
2 )( 666.67
(
U.5 C0. 020)
smahller
value,
0.0
=
(1.0)C6, 030%
2
7
St
m= | [T=
(foo x(0*) (666.67
Mos
Choose
Ee do\Goy+(LooyickasyY = 252,33310" nent!
1
\ol= _inMyi
l=
Brass
section,
mAd* =
3
30m
Cz
2222%«Io*
O,1OPD
=
,
wm
5
C=
B.5556%10°
Mz 2.2210"
Mem
toe xic”
Pe
Nem
Igox10°*
We,
Nem
M = 2.22 kev ~<
Proprietary Material. © 2009 The McGraw-Hill Companies, ine. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.34
4.33 and 4.34
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below, determine the largest
permissible bending moment when the composite bar is bent about a horizontal axis.
8 mm
38mm
ae oeecvenene 32, TT men
32 ann mmr
oe
-
ena
Modulus of elasticity | 70 GPa
Allowable stress
100 MPa
32 mm
adurain
For
~ Brass
brass,
Vasues
57
)(gaPr
Se2 bh, = £E(a2
I, =
on the
shewn
5
LS
figure.
wit
~[O*
Al. B45
=
G2)
.
&h,
~ ix
Am
af
10S /70
z
-
oe
%
bet
3
n
pad
|.o
E,/E,*
ave
metevial.
section
trangtormect
+He
ne
yi
oy
fe
om,
105 GPa
160 MPa
reteremece
as the
afawinum
Dse
For
Fov
:
Aluminum | Brass
_—_
Aluminuns
/
-
= 131.072%10° mm"
ce
5
DT, = 21.8453 %10° mm!
T= T+ ED, 4¢ Lg =
174.7626 210° mm
_|@r
Me ay
_ 1M
lol = [ee]
A fami now.
*
M =
= 174.7626 %10" mi
ne
oO
>
ly!
=
{G
mm
=
~4
(100 10° UIP. 7EREX1O™)
O.01G
m
Ls
,
gg
G
==
[00%{0"
xJ0®
Nem
(1.9 \lo.oléy
Brass?
M
ne i5,
x
Choose the
.
lyl= l6mm= O.0l6m,
(Go ¥ - 10% (174.7626
% (O07)
:
.
(5 )(e.016))
smaller
vadue,
Pa
Me
=
Ge
*
[Heo
1.0942 «jo
65
New
160x]0% Pa
1x
[o
a
*
Nem
M= 1.092 kN-m =
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distcibuted in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hilf for their individual course preparation. A stadent using this manual is using it without permission.
Problem
4.35
4.35 and 4.36 For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
,
of Prob. 4.33.
4.35 Bar
4.33 and 4.34 A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below, determine the largest
permissible bending moment when the composite bar is bent about a horizontal axis.
10mm“4;
10mm
>|
>]
{
10mm
Aluminum
Modulus of elasticity | 70 GPa
Allowable stress .
100 MPa
2|
For
the
Lm
a
transtormed
IT, = a bh = 2
brass,
ov
fF
~
NF
Je
na
shown
Ls
10§$/70
=:
on the
|.
=
Figure.
section,
Roy4ot)
=
106.6667* 10* mm"
Los Mi Lhe + A = $2 (Ho\(Ioy'+ G.SY4oWoYisy
T=
moatevrat .
Vzho
los /
bf
of n ave
Values
105 GPa
160 MPa
vevevence
the
akuminum,
Fou
c
bk
con
ee
10 mm
~T
as
inven
aku
Use
40 mm
|.40
-
Aluminum | Brass_-
= 140, 10% wa"
IT, = 140
x {0% mmt
T=2«I,+t,
+1.
=
nM
ist= . |(>>
Abuminum?
Mm =
Brass?
M
Cheose
the
10° mm
=
293.667?xlo"
yw?
~ | ol
m= |S
n=
LO,
ly]
(199 410° 336.667%10")
(1.0 JL, 020%
n=t.S)
-=
396.667
ly]
=
=
RO
=
20 mm
157) _=
10%), 396.662
(60x(.s\C0.020
5
smaller
value.
M=
wim
= 9,020)
1933
G
=
loo x/O° Pe
Nem
= O,020m,
§ =
1GO
xjo*
Pa
.
2062 Nem
1933
Nem
Me |.433 kN-m <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.36
4.35 and 4.36 For the composite bar indicated, determine the largest permissible
'
8min
bending moment when the bar is bent about 4 vertical axis.
4,36 Bar of Prob. 4.34.
8 min
32 mm
or
.
MMominum
Modudus
Adowable
32 mm
Use
For the
I=
Brass
transformed
T= Gh be = BE
T=
t =
T=
I,+ 2,4 TE,
edowrinum
Fos
brass,
Vedues
of
,
=
hE
n ave
materiad
1.6
JE.
shown
MPa
loS /7o
on the
=
1S
Figure.
Le (aa\(3)"+ (10 (22 i(s}a0y"
=
103.7653x{0% mm
zy = 131.072
* Jo" mm
=
33% 88 «lo mm!
= 328.58 x/0 7 m!’
oy: (S|
ne lO,
lyl= 24mm= O.0244m ,
(leox. lof< 338-S8x)0%) = 2 yy
ne hS,
_
[60
as the neterence
(1-0 (0.024)
M
lOOME,
j03.7653%10° mm”
jot = OMy
Brass?
105 GPa
sectiou
wh, b+ WA, a =
Mdominum
m =
FO GPa
stress
alumnum
For
Aluminen)
of ehastic ty
Brass
(160 ¥t0%). 333.58 e1O)
Choose the smatler valses
100%10* Pa
ay.
* 100
lye iG mm = O.0lgm,
(.5 (0.016)
6
=
= 2.257 10"
MF LAlEx
iO? Nem
Now
I16Ox/0° Pa.
a)
Now
Me hull
kNem =<
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
4.37
- 4,37
Three wooden beams and two steel plates are securely bolted
together to form the composite member shown. Using the data given below,
determine the largest permissible bending moment when the member is bent
about a horizontal axis,
:
.
Wood
AR$
:
Use
man} tram | mm
6mm
Properties
ok
the
I=
Steel:
-
Wood?
Tran
Iwr
sfovmeed
=
Danas
sect
Wi, I,
+
wood
as
For
woo a
for
steel
(6
i4 GPa
200 GPa
Allowable stress
14 MPa
150 MPa
vetevence
Vig:
=
We
.
ESSE = 200/74 = head
mm
|
/
ye ortosso)@sv®)
= /9&3/29 Xo © pte
rm
tow
9) (/9 5:2 28 S10)
We Dy = (16 IA) (18+ bas x1) ¢
>
Y
iba
t= [SE
Wood +
nal
M
=
:
lyl=
(28 iam
Gwe \Gibek nig
y= /4e24
lyl r 12
in mm
smother
vadve
1¥mfa.
&=
Aim
(SO
= 25210"
C4229) Corr2©)
the
OF
= Abe G x10"
(ise - xpo°) (48 bx 10”)
Cheese
{
(& bat xr0
+6)Qso)?=
.
Stee@
mateviad
section.
= G/§$- & Ko"
jet =
Steel
Modulus of elasticity
the
qeo metric
ae
Ms
24.2
Xie
MPa
Nay
Mf ha
M
= 3&2
LAL.<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All nights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.38
4.38
,
For the composite member of Prob. 4.37, determine the largest per-
missible bending momnient when the member is bent about a vertical axis.
e
4,37 Three wooden beams and two steel plates are securely bolted
together to form the composite member shown. Using the data given below,
i
determine the largest permissible bending moment when the member is bent
about a horizontal axis.
ea
250 mm.
i
Wood
i
le
[
I.
|
Modulus of elasticity
Allowable stress
_—
Stel
14 GPa
14 MPa
200 GPa
150 MPa
50 [2]
50.
mm)
mm
mm)
Use
wood
as
Fo v
wood
n=
]
steel
n=
E/E,
For
Properties
of the geometwe
the
refevence
motevial
[It
= Do
seetion
Tote 5
Ty, =
tb hb?= £ (2s0)(162)= PPL73N00 rm
Steeh?
Tee
i h(B-b*)-
Wood:
L,= L,-T,
Trans Formed
Lims?™
Ms
lg+ mT 1
(iv2nyestiove) PLL OSD»)
lyl = Fl ain
Ciyxio® \C120 418°)
(i) Core $f)
Ms
Cheese
8)
o’
= Uen20K
Zt)
Ci4.29) Core
the
smafher
= [20X10
fam
=
G = 14m Pr |
2a
a“
/
x]jo”>
vatve.
Nor
JO"
Net
6 = [> Me
n=l249 — lyl > Simm
Steeh?
|
hrm,
= Pé212¥00
= | SEI
n=l
M-
= 23 Elove pacim
£EsV(E25- 56)
section
ist= Joey]
Woe dt
|
LG oF
ae
6mm
4org ~lO”
Mz
s
20°74
Nem
LN
Ms 20° 74iNm
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, réproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<é@
Problem 4.39
4.39 and 4.40 A steel bar (£, = 210 GPa) and an aluminum bar (£, = 70 GPa) are
bonded together to form the composite bar shown. Determine the maximum stress in
(a) the aluminum, (b) the steel, when the bar is bent about a hortzontal axis, with Af=
200 Nm.
12 mnt
Use
12mm
cdumrnuin
ml
ne
otumsrnum
For
stee?
\
A
mae
n=
mead
section
mom"
nA
mm
metevrad.
|
210/70
Yee
nAy
= 3
;
nad.
@ | H32
b2%e
13
23328
@
4¥32
©
REA
i432
25420
1728
The
|Summ
=
veterence
ne E/E,=
Z.
Y5 r
the
Foy
Transhor
®
aS
neutral! axis
lies
[= Bb he4 OA, ob = 2 (edli2ysoe i29eLt
Sy =
Imm
above the
bottom.
27.216 « 10% mmm"
T= Bebhet nye > ay (26 (1274 432A = 40.176 «10% mn"
DT = L.4+2, = 67.392 xto° mm!
(0) Aduminum?
ne
ly
yt
|
0.015)
= — GYQCV™
)o
ET BMI
(b) Steel
we
Ca"
n=
3,
6. = _ £2)(200)C0.009)
G7.34a%lo?
y*
Imm
=
67.372«107 int
= -O.0ISm
STEx IO" Pa
Gum
6.= . 44.8 MPa
<<
6,7-30.1MPa
~<a
0,007m
=~80,|22%x10° Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.39 and 4.40 A steel bar (£, = 210 GPa) and an aluminum bar (£, = 70 GPa) are
bonded together to form the composite bar shown. Determitie the maximum stress in
(a) the aluminum, (b) the steel, when the bar is bent about a horizontal axis, with M=
200 N- m.
Problem 4.40
12mm
/
Steel
12 mint
Use
aduvainum
12mm
For
[a
®
7
A
ELS
X
TT Siam
7 eS
n=l
=
mat
yveterence
)
y=
Zlo/to = 3
section.
yum
8
NA,wmm*
Yo, mm
nAYe , mm |
©}
432 | 432
&
2542
@\|
138
2%
18
Igy
@)|
(44
432
1B
TI1G
1152
=
13.5 wm
The
Tt Pa bh
nA
neutve?
IS 552
auis Jies 13:5mm above the
Y=
ay & (288 VS
= Bea
bottom.
+ (482) (75) = 2d4BAKIO® mm”
(Belay
T= Beb hes navel=
materia,
|
n= E/E,
|
Y,
the
rao
Transtormect
ne}
7
asum
Fov stee?,
36 mim _.|
2)
as
% mn"
4288610
Tar Bebhs nAgde= 2 (i2via+ (432045)= 13.932 ¥ 102 me
T=
M=
D42,+1, =
GO
62.704
xlormm*
=
52.70% /0" m?*
New
oe =. omy
I
(a) Afowinomt
Sa
y=
1B.5mm
___CHCoo0X- 6.0135) _
Foy riot
(b) Stee?
2.
OF
n=l,
N= 3,
y*
- 0.0185m
‘
.
Gz> Si-2MPo,
SI 2H IO Pa
\O.5 mm
_ B)\Qe2)(0.0105)_
704 «lo
=-
=
(19.5%
=
O. DIOS
[0°
6
Pe
wat
mw
62
;
HIT. S MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem 4,41
a
4.41 and 4.42 The 150 X 250-mm timber beam has been strengthened
by bolting to it the steel reinforcement shown. The madulus of elasticity for
wood is 12 GPa and for steel 200 GPa. Knowing that the beam is bent about
a horizontal axis by a couple of moment Mf = 50 KN > m, determine the maximum stress in (a) the wood, (b) the steel.
250mm
Use
woe
aS
For
eeseel
50* 12 aan
4
the
vel
Oo,
erence
n=
|
3
j
“50 x 12mm
125 X 12 nan
Far
steed
Yn = Es.
Sw
4
®
.,
|
Transformed
See ~ an,
Aynm'|
n Aymm
|37500|
37500
6027517 B
Bee g”
@) | 1800
25000 s
= 73 inin
nevtral amis
ax's.
ae b, h?
_ Map,
pe
In>
yh,
+n Ala,
jp oyh,
+ ny A; a,
(a)
50
kEN”M
Wood
=
6,2
#7
oy =
137
da:
| é
*
2
=
(6-667
am
| $13°71500
37
hoo [4-9
@
150003
6027517-9
the
bottom,
bhmm
73
=
~
73
|x 3b mm
-
7%
|=
&7 mm
wi-(150)(250)"+ (37500)(64)° = 348-4 x10 mont
16667 (24) (50)? + (20000+4)(.36)* = 30-09.Ki0° mm”
ra
fb
cel
G2s)\(z )* + (25000-5)( 67 Y= 112-$3x10 mm"
i
|
-6
Ig = 4491-52510" mmt = 490-63x10 in *
De=I,+tI,+
M=
nA ay”
+
abeve
=
a
4
nevivad
Tis
732 mm
Frum
Distantes
v=
Aes
=
137
| 82500+4
uy
The
” 20a
4
¥,mm | HAY
@ | 1200 | 200004
y=
>
materiale
nH
=
|
.
- OY
y
a
= 262-73
=
_
60) Sox10°) (0-194
(89 mm
==~
OfPFm
~ (9.225 MPa
Gy #-!92 MPq
dl
6. = 122.8 MPa
«hy
ADLER xron
(b) Stee??
Gs
=
=
bMy
n= 16-667,
=
—_
(/6:667
yF
-
) Gono? \(-
z
~ 8073
2073 )
m
me
423.797 MPa
4AUSAXIG
Problem
4.42
C20
®
:
4.41 and 4.42 The 150 X 250 mm timber beam has been strengihened
by bolting to it the steel reinforcement shown, The modulus of elasticity for
wood is {2 GPa and for steel 200 GPa. Knowing that the beam is bent about
X 17.1
a horizontal axis by a couple of moment M = S0KN - m, determine the maximum stress in (a) the wood, (/) the steel.
aK
Use
wood
as
the
petevence
250 min
4
@
For
150 mm
Fou
woot
For
stee#
|
For the
,
above
Transformed
3
°
seetton
the
Pies 25d
+ SsG-
(Eebs
base.
Yo
the
Section
A
grin”
= £39400
centyor
24/
= 242mm
min”
Jos
mm
oF the
for
channe?.
nAyYe yin?
T26Ny
®
1Pd-Jimm
adhove the
les
73674 |
bein
mm
DD | a7 [Bb | 242 | B957169
|37Se0| 27500
in
2Al7o
13412669
The neutral omnis Lies.
.
46P 7500
13412.669
bettem.
|
o
Ly=
|
=
ny Ty + MAdy = 16007) (290000) +36174) (2412 — 18961)”2 S132 X10 bor
I,
ut bh, + NAc
y
¥
2
#
L
(a)
2
=
_.
(50 2(280)3 +(27 S00) (182+ /-/L 5) es BIT SF 16
&
me
¥
TAT, © 45269 ¥700° mm?
Lo kNm
Wood:
|
n=l
ww
Stee?
:
an
GO, =
yy?
eo = - EE.
~ $.433 in.
3
~2
RIDE FINED gs
6,2 2e°/M Pa
apy
et
48909 Ke *
-
(bv)
nA
2
A
Jy
mom
He
LE
camposite
channel
section ,
channel
= lbe67
= dee jiz
E/E,
n=
Ew = £06 pin
Le
,
nel
ll. s
CBE
matevicd
y=
167
y?
Div ¢ Sob
~ /f2e]
= 735
AP
~~
(Ibe LIL £0 Kb CTRL KIO
Tent iek
/34.2mM MPa
=~-~ 13462
66 =~
=-(38:3Mfa
13403 MPa <t
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.43 and 4.44 For the composite bar indicated, determine the radius of curvature
Problem 4.43
caused by the couple of moment 200 N - m,
- 443 Bar of Prob. 4.39
12 mn
I = 67,392
12 mm
le
36 mm
EL
~
«ONG
(JO
“so
_
200
iL. Mo
P
E. = 70 x10" Pa
Nene
200
Me
|
Io"
ot ZL.
codcudatton
4.34 fon
Problem
to
sojution
See
7. 892
“
Y
«TOT
H2,.39G“1D
ft 23.6
4.43 and 4.44 For the composite bar indicated, determine the radius of curvature
caused by the couple of moment 200 N - m,
Problem 4.44
4.44
12mm
See
Bar of Prob. 4.46
solution
te Prubfew
4HO
I = S2.704*10" m4
Steel
MM
=
R00
for cadeutution
of I,
E..= 7oxio* Pa
Ne pn
M
36 mm —|
cal
200
=
P~ Er ~ Goxlot\(er7oux io)
SH 21]
w
fr
a}
BAS,
<a
4.45 and 4.46 For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 50 kN + m.
4.45 Beam of Prob. 4.41
Problem 4.45
See
sefution
Te
to
A415 3x10 ©
Probdten.
U4!
4
Ew
Foy
= fh
Catevdatt ov: of I,
jo?
PA
250 ram
M
+
= Soxjo3
ie =
Mi.
Seno
%
=z
125ty X 12 mm
Problem
p*
4.46
sane
4.46
i
XY
150 men
EF
0
a!
Le
tly
See
sedution
to
To = 45290
Bar of Prob. 4.42.
Problem
£42
mis
M
=
cakeu Pation
ob
tT.
S0kVm
5
bret
To
fay
E,,= 12 GPa,
250 pam
i
UTG gm
-3
4.45 and 4.46 For the composite beam indicated, determine the radius
of curvature cansed by the couple of moment 50 KN +m.
4.45 Bar of Prob. 441.
7 €200 X VTA
Mf}
= 24767
2) O
C2107) (qars3 x10)
EL
soxiamm ©
6
Nm
SO¥iID
Cex X4cr49y
09
oe
=
00092 Mm
p> (o8:7m
=i
Problem
4.47
4.47
Aconcrete beam is reinforced by three steel rods placed as shown.
The modulus of elasticity is 20 GPa for the concrete and 200 GPa for the steel
Using an allowable stress of 9.45 MPa for the concrete and 140 MPa for the
steel, determine the largest allowable positive bending moment in the beam
400 mim
22 mm diameter
|
p—_
200 mm
Bo
E.
50 mm
a
Ze0 Kio?
=
Es
n=
|
\o
X10
114004 man
A, = 3 Edt = 3(B\e2y=
|
2.
:
(t4o4
=
nA,
e-2o0-of
Locete
x
nevivad
a
avis.
v
axTs
Aaecomomemenean
—4o¢
ol ve fon
xX
Xn
©
- A&GIbaD =O
levy+ 404
NAs
4 2.
4+ 4047 ¢ (4/00) (Axl bp )|M16
/
Yim
Z Cleo,
400
I=3 +
~ (leoy)4oo-xY=
200% %
4o0
-x%
=
236mm
200 x* + nA, (400-«)"=
= 246.15K10° mm?
isl = [QE]
x (2o/6PY + (40
sme ae
p= Ca ENE Cy
Conevete *
n=
lo,
Steel :
n=
10 ,
23L)
le] =
l\yl= (64mm,
ly! = Z3binm
;
—&
M = (148 000x107
V2Ib: IS 07) 8
=
44S MEA
YompPea
(eb LN
Qo )Ca+ 232)
Gheose
the
Proprietary Material,
smatfer
valve
M=
i7-]
ENn
© 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced,
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
or
Problem 4.48
540 mm
4.48 The reinforced concrete beam shown is subjected to a positive bending moment
of 175 kN - m. Knowing that the modulus of elasticity is 25 GPa for the concrete and
260 GPa for the steel, determine (a) the stress in the steel, (5) the maximum stress in
the concrete.
25-mm
diameter
~
Y}
300 mm —>|
As
x
axis
.
200
Ee.
~
25
=
4.
=
nA,
L- 200 >|
Es
‘Locate
ta?
GPa
GPa
= (EZ
x10"
16.708
the
8.9
nevt red
1Gsy
=
L4685
x10"
hm
am"
AxlS,
_t
H36
ns
a
~ (18.703
xJO )C4B0-x)
=
9
150 x*4+ 18, 708xJo* x ~ 15393 «10%
M
fon
xi
6
=
~ 15,708 10? + f (IS. 70810°)* + 4)C50)(7.5399 x 10°
x.
AVIS)
HBo- x =
302,13
mm
at
a
q
tH
Solve
300
x )*
3 300 x° 4 (45.708 x0? )(ugo4 (g00)( 177.87) + (1S. 708 «10° 02.13 }*
1.9966 *10° met
=
19966%1O>
m”
ge - -aMy
I
(a)
Steed:
YF
7 362.45
mm
=
G = - 8.0K
17S x10*)(-0, 30245)
1.9966x Jos
(b)
Conevete
G=-
:
YF
177.87
U.09CI7S x10?
1. 9966 & jo
wm
=
(0.17787)
0.30245m
_
212xlo° Pa
O.17787
= 212 MPa
mm
m
=~|5.5Gxlo° Pa = - 15.59 MPa<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
'
Problem 4.49
4.49 Soive Prob. 4.48, assuming that the 300-mm width is increased to 350 mm.
4,48 The reinforced concrete beam shown is subjected to a positive bending moment
of 175 KN - m. Knowing that the modulus of elasticity is 25 GPa for the concrete and
540 min
200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in
the concrete.
25-mm
diameter
4
60 mm
_&
-350 mm —|
n=
A, =
}
4
.
Ee
200 GR
-
25
4 Ed®
NA, =
_
Gra
7
8.0
1.9635 «10% mm
= (NF (as) =
\3.709 xjo?
mm™~
HZO
_|
nA,
Locate
bai
the
%
250 xX
neutw?
axis
;
- (15.708 x%jO° )(480-x)
175 ¢* + )5.708%*Jo"x
Solve
Pov x.
ge
_
= O
- 7.539ax%}0°
2 lS.703xI0 + of 15708418)
=o
+40 75)C7. 5378 x10% )
Q)AT7S)
X=
I=
#67. 48mm
,
4BO-xX
=
BIZ.52
mm
~- x)
$1350)? 4 (15. 703 x/0? C480
4 (350)(167.48)* 4(45.708 xfo? )\(aia.52 )*
1.0823
x(o?
mm?
= 2.0823
x/O”
m*
G = -nMy
ia
= y= ~ 312-52
(a) Stee:
G
(b)
ro
Concrete:
S -
mm = -O. 31252 m
(8.0 )C17g5
x1o*® cae}- cae©. 31252)
ae
—
2.0323 xo
y
>
1G74B
mm
=
O.16743
21075*
103 io?
CO. 16748)
4 oene
=<
2
loxjo°
oO
c
;
=
Pa = 210
Ki
Pa
eon}
m
Ls
gg
yjo*® Py
=
~I408x10°
Pa == —]4.08
—|4.08MPQ _
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.50 A concrete slab is reinforced by 16-mm-diameter rods placed on
140-mm centers as shown. The modulus of elasticity is 20 GPa for the concrete
4.50
and 200 GPa for the steel. Using an allowable stress of 9 MPa for the concrete
and 140 MPa for the steel, determine the largest bending moment per foot of
,
16mm...
width that can be safely applied to the slab.
Es , Zookeo! _
ne
140 mim
Lao mat
a
z
Fay
As
i 140 niin
NA,
=
10
208x191
Constdev
.
140 mm
“ee
E.
Section
2
w
(6)
= $
Dolo
[£6 fm
20]
wide.
Mot
man”
Locate the nevtw? ois.
-——
5,5 —>
Oat
nevtwed
ONS
~
N°
Sefuing
|
H-x
2% =
fame
-
(106 ~%) (2810)
Jo x*+ 2010
pe oo
LLL
CE.Me a ZL’
nAs
x
ons
for
=
%~ 201000
x
.
fe fmm
= oO
:
foo ~x
= L9 Imi
I= d0/40)x® + (2010)(/00-x)”
= (Ho)4r iY +(2010)( 599)"
,
.
(10st
M =
K15 "G4
x10")
G.0\Coro4iry
Stee? s
n= 16,
_
= 2236
y= LR Gmim ,
m= Lleravo Coxe)
Clo) (0106 $4)
Cheese
with
M
the
smaPbev
vabuve
as
Nm
Se fo Mes |
2428
.
=
6 = IML.
yr 4fhelmm,
nel,
Concrete
zs
—
lel = ney |
® Mat
Kio42
= 10-
the.
vn
DPrur ah
te.
moment
|
2236
Nm
= 2:24 bi
for
a
[Yo br.
Problem
[<—
4.51
4.51 _ Knowing that the bending moment in the reinforced concrete beam
: 125 min
750 mm
"
stress in the conercte.
mum
:
600 tem
25-mm
diameter
is +200 KN > m and that the modulus of elasticity is 25 GPa for the concrete
and 200 GPa for the steel, determine (a) the stress in the steel, (6) the maxi-
y=
.
Es=>
+:
E.
Zoox’
aS
.=
Kt of
2
“S
A, = 4 4d* <4 (B)Os)\= 1963-5 am
NAg = /S70F mm”
Lecete the nevtvel
(780 )UZ8) (x
axis
12 ol
®
Yn
929750 x + CP LIL7S+ISOX
|
Fur
[sok
x
x= ~ 1094S
Aly -
=
og
O
— ES/OP20
& IST OFX ZO
.
As,sonic
Solve
6-5) + 360% */2
— SFP)C GIL -~ x)
®
L.
avis
x
=
+
1097¢S£%
-b39¢¢6:=
tL liogsey” t (um)
2 C/¥D)
God
O
(65944
=6
Pry
mm
T, = oho + Avd* = EC7Yja¥ +¢7s0K/25)(69. Y= SIP
CKO nat
Ins dbx? = £(i2K6Y? = 86H mn®
Ty= nAdd = (Us708 409)” = 2627 6S HO nd
Te D+ I,+Dy
=
OY
(a) Stee
G=(b)
Concrete
O=-
= 3618¢§ x00)
ment
hee | Me 250 Em
|
n=8.0
ye
odin
T7079)
(G0) (200 x10) (4.0
Be SUL 10°?
= Ooh
yF
{2] mm.
(1.0) (Zeke? (orfB)
gg
nz
oO,
3: 1SY4Ex 6°?
=~
Mw
M
MPa.
~<a
<_
Pr 2mpa .
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.52
4.52 The design of a reinforced concrete bean is said to be balanced if the maximum
stresses in the steel and concrete are equal, respectively, to the allowable stresses 0;
and o,. Show that to achieve a balanced design the distance x from the top of the
beam to the neutral axis must be
x=
1+
d
oF,
oF,
where E, and £, are the moduli of elasticity of concrete and steel, respectively, and d
is the distance from the top of the beam to the reinforcing steel.
6, = oe
SH
Os
d
gr
_
_
n M(d-~x
(d
)
n Cd-x
)
i Ss
l+n@
|
+
:
6, =x
EMG
=
a
I+
EF.§
BE
Ex Gs
Es Ge
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4,53 - For the concrete beam shown, the modulus of elasticity is 25 GPa for the
Problem 4.53
concrete and 200 GPa forthe steel. Knowing that 6 = 200 mm and d= 450 mm, and
using an allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel,
determine (a) the required area A, of the steel reinforcement if the beani js to be
balanced, (5) the largest allowable bending moment. (See Prob. 4.52 for definition of
a balanced beam.)
nt—
Ex
_ 200»e ior
|0"
rae
~—
Qs
aM
=
a-x)
CG.
Ir
Gon
4.
(d-»)
be
)
——b#
.
le we
= | * ge
=
S
“
x4 nA. (d-xY=
M
Steehe
=
=
187.5
1674 wm®
=
mm
(a) Ag 1674 ane eat
Is
ny
y<2
lO
J
=
IRS
(1.3623 xfo"*)(12. § ¥10% )
(j.0
0.1875 )
mm
=
=
=
both
volves
are
the
Same
G= 12.5 «}o* Pa
2.1875m
FO.3
We B.0
y= U.S mm = 0.2685mn
= 13623
AIT * (HOw 108 )
that
=
1.3623% 10° Mm
(2.0)(0O. 2625)
Note
= 4-46
4 (0087.5)+ @.0V1674) (262.5)
r
:
z
(28.0) 262.5)
nMy
Concrete
gsvioe
= (O.41667)CUSO)
(00 \C187-S)
we
1.8023 x% 107 mm)
=
ldo Ion
(d-x«)
an (d-«)
T= $b
Lr
Otis.
x o- nA,
A=
Mx
a)
© = O.N16ETd
nevtvel
=
<
1&
x7
Locate
* 8.0
5
Nom
&= 14ox% 10
90.8 x10”
f.
-
sion
3
Pa
N-m
Ladamcerd
clestan,
(6b) M= 90.3 kN+mn —«8i
Problem 4.54
4.84 For the concrete beam shown, the modulus of elasticityis 25 GPa for the concrete
and 200 GPa for the steel. Knowing that 6 = 200 mm and d = 450 mm and using an
allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine (a)
the required area A, of the steel reinforcement if the beam is to be balanced, (5) the
largest allowable bending moment. (See Prob. 4.52 for definition of a balanced beam.)
Es _
ni
200
E.
“*{o%
aswjot
~
8.06
6 = MMG@ex)
e~
Me
Ir
Ge
. on
e°
a ..
I
(dex)
7,
A &,
x
Locate
nevtval
nen
_
lt we
= O47
d
=
4
A,
= 1+ go
d
fe xin
gsvoe
~ 4-40
= (O.4IEE7\CUSO) =
:
187.5 mm
anis
bw ¥~ wA, (a-x)
wor
by
=
As
(Z00 Aas
=
an (d-»)
=
(A\8.0) 262.5)
{G74
”
wm
2
(od
Aw=
_
*
1674
mm
2
<a
T= $b x4 nAg(d-x¥= 4 (go0ia7.
sy + 01674) (962.5 )*
ee - omy
2
Concrete
2
M <=
=
M
Note
M= = if3S
n-l.o
y*
187.5 mam =
(1.3623 xjo"* (12. § ¥1O% )
(1.2 0.1875 )
N=
Steef=
1.3623 % 10° w*
i
a
1 8623 % 107 mm»
that
BLO
2E4 Sam
ye
(1.3623 x1
(140%
Vvohues
are
6=
= O,R6LG5yw
|
10°)
dhe
G= 12.5 ejo* Pa
40, Bx fo”3 New ape
=
(2.0 )(0. 26255
both
O.18275 m
Same
1Ho~¥ lo" Pa
i
F0.B x1O° Nom
f-
bakamceed
design.
(b) Me 90.3 kN-m <a
Problem 4.55
4.55 and 4.56
Five metal strips, each of 15 x 45-mim cross section, are bonded
together to form the composite beam shown. The modulus of elasticity is 210 GPa
the steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that
beam is bent about a horizontal axis by couple of moment 1400 N - m, determine
the maximum stress in each of the three metals, (4) the radius of curvature of
composite beam.
15 mm
Use
aluminum
Es
Y=
=
n=
as
the
veference
for
the
(a)
the
material.
210
=
Fo
1-0
=
3.0
in steed.
tn afowinuur.
AXIS
For
the
transform eel
ho
I=
section,
,
eb
+
nA,a,
= HP (ys \(is)¥+baX4s)(s\(30)” =
T,>
px bh, + MAA
Ne
%
sj
T,= - Ghshy
FL =
5
Te
21;
=
@) Adorinum?>
Brass *
o:
w)
4-HET
P-
LS
w\?
Bl
GECYSeNrey?
tS)
==
7
Se
37.97 x10"3 mm
mm
FT
=
y=
n=
oe Pee
F RE7HIPa
2S mm = 0.0075m
h = 3.0
(Joxto4 )CL.77189%x«loO7e
.
pw”
= 29.6%10
Fa
n=ius
108
, we
lho
RBS mm = 0. 22S mn
TY = (ao KwiooXa.0076?
=~
.
4
LT TIBI «IO
O.037 5m
nMy_ (0) 400%(0, 03757)
F
| 371B¢«lo-e
YF
245. 80%! me”
\ f
Te= I, = 622.16 «10% mm”
4.771
BG 10S mm
37.5
Ce
Steeht
= FRCHSNISY + USVISVUS VIS) =
2
246.80 #10% mm”
T=
y=
620.16 x10? wm"
= 17.7310Pa
=
1.28710 mw
6,5 24.6MPaes
Ge 26.7 MPa
GF 17.738 Moy
Problem 4.56
4.583 and 4.56
Steel
15 min
Aluminium
15mm
Brass
Simm
3
Alumina
Five metal strips, each of 15 x 45-mm
-beam is bent about a horizontal axis by couple of moment 1400 N - m, determine (a)
the maximum stress m cach of the three metals, (4) the radius of curvature of the
composite beam,
{5Rasy
mm
Steel
Os 2
[Saam
[5 sm
|
abou
y=
Es
N=
=
A=
:§
Fon
Lo
3.0
the
ud
"i
¥
hh,
T,=
er,
T,=
DD, 7
bs
(Q)
hes
=
3
3
TT
LS
=
_
L
4
15°
in
od
u
[
xJO"”
mm
a
.
=
I
6
e
12,43 «10° wm
4
“y
lL. BGO4GF 107
=
i o69oOx
lo *
0.03875mw
n=
3.0
O,O22F
C.eCi4eeVo.0225)
4.0640 x [O-¢
=
wm
wo
M
Po”
EI
1Yoo
(Fo x}o4 \(4. 06%0 xfoOrTe)
4
G, =7 SBIMP,
2)
= 2THx}0% Pa
> 3.370 10° Py
“40690 ¥io°e
snny
wa!
38,7%*/6% Pe
~ rMy . (5 \IH00)(o.0075)
o£
IE 4.53 x}0 2mm”
ed
~
mm
G.0)(1¥00) (0.0375)
2S mm
O56
section ,
nels
YF
b
deer
=O, 0075 m
Brass *
:
steed.
2:2. (4s Ys)" 4 (3.0 uss \Goy” =
22.25 mm
_
nMy
eda
nA,4,
4Y.0690 «{o7*
J
C
ra
7
37.5
nMy
b)
=
moteviad,
LO 7,
= ge
(svUs) 3 +6 Lows yaisyasy
4H. OGFO
y
ov=
©
oh
=P sits)
7
=F
Steef:
210
™
164.536€ 10% mn”
2];
G=
2
+n Ad,
oc
.
fa
traustovmedt
n,
I
T=
2.0
10S
Le
yebevence
=
X,
La
the
fo)
a
Ee “
3.0
“6
as
Yiduy
Ee
L
cross section, are banded
together to form the composite beam shown. The modulus of elasticity is 210 GPa for
the steel, 105 GPa for the brass, and 70 GPa for the aluminum, Knowing that the
G.= 214M ia
<<
Gy> 3.87 MPa, <i
4UIS2ZXIO
P=
wm
LOB
Mm
Prob! em 4.57
4,57 A steel pipe and an aluminum pipe are securely bonded together to form the
,
Aluminum
composite beam shown, The modulus of elasticity is 210 GPa for the steel and 70 GPa
for the aluminum, Knowing that the composite beam is bent by couple of moment 500 N
+m, determine the maximum stress (a) in the aluminum, (4) in the steel.
/
3am
Steel
—6 mim
a
aluminum
N=
Wn
Stee:
LO
=
Es
En
yetevence
T=
7 181.73 xto* mem!
T+,
Fd
2.0 .
:
iw
de 16")
U.0MG00)(0.010)
.
Cz lGmm = 0.016»
aMe _ (3,0\(500 (0.016)
=
2 bMe.
o-"t
“f=
‘V81.73 xIO™
1BLT3x IO
steeD
180.8510% mam"
=
§0.38 «10° mm"
= 18/..73%1077 m*
c= 1mm
= O<.01F m
Afominum:
wmatevrad
tad gm ind we
Ale
76
1, ¥(n- ts" )=(2)5
D,7
(b) SheeP:
~
the
n, E(vd-r*) = B.o)#(16"-10") =
I, =
Aduminum?
(a)
bv
OS
$2.3
rraxm
|
P, =te
x10" Fa
132. 1x1o* Pa
i
Use
52.3
MPa,
132.1 MPa
Proprietary Material, © 2009 The McGraw-Hill Companiss, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
mi
—e
Problem
4.58
.
Steel
458 Solve Prob. 4.57, assuming that the 6-mm-thick inner pipe is made of aluminum
and that the 3-mm-thick outer pipe is made of steel.
4.57 A steel pipe and an alurninum pipe are securely bonded together to form the;
3mm
composite beam shown. The modulus of elasticity is 210 GPa for the steel and 70 GPa
Ahiminan
for the aluminum. Knowing that the composite beam is bent by couple of moment 500 N
“6mm
sem, determine the maximum stress (a) in the aluminum, (5) in the steel.
~ LO min
se
38 mm ~--—»
adomrnom
m=1-O
nz.
&s
. 36oO
E.. - AIO
40
n, $C"
I,+ I,
Afuminum =
(b) SteeR =
o*
©
oC =
nMe
I
6") =
- 1") = (Lo ECG
*- io”)
; nv
steed,
152.65 10% mm"
=
43.62/07
mnt
196.27«10*% mm’ = 196.27%1O* mi"
16
_
ram
=
©O.0OI16
mw
(1.0)(Se0Vo.016)
182710
i}
(2)
Li
pefevence martenrtad
abuminvrr
J, = a B(n"-n*) = @o)FA*-
Aduminum:
I=
the
Un. 8xio’ Pa = 4o.g MPa
=e
WW
Steef:
ja
as
145.2xlo
Pe = 145.2 MPa
“<
C= IF mm = 0.01%m
Gr avs
~ (3.0)(s00 )(0.019)
196,.27«(0-%
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.59
4.59 The rectangular beam shown is made of a plastic for which the
‘value of the modulus of elasticity in tension is one-half of its value in compression. Fora bending moment M = 600 N + m, determine the maximum
+o
1.
(a) tensile stress, (b) compressive stress.
oxis_|
;.
—
F
©
‘
n=l
Locate
-
H
stress:
-
v= -7¢ Ch-x)
|,
$8.579
=
=
n=4
M
ny
=
a
,
Yt
1.1844 x10% mm"
= STS] KIO mn”
(690
)
0. OS8S5749
= ~0.O0S38579
(b) compressive stress°
M
nity
=
n= 1
7
Pe
6.15 x10"
4.85 FS xJo°e
6.15
2.8545 10
sy
-
|
om
B.64 x10”
-«
Se = 7 8.69
nth
MPa
y= AL42) mm = OORT
C0 Goo) Co. 041421)
m
6‘
= ¢
)
G.=
==
wy,
mm
7 S8.S7I mm
(0.5 \(ea0 \(- 0. 0585
t
oO
.
im
HI4AZT
= 2.8595 w10% mm = 28595 x10* m*
* D4 Dy
CG
=
058.874
Tp = Ni& Kh-w8 = VAs
tensize
=
aYP
= (NGO)
T, = a tbx?
sidle
=O
4442)
O.41421
Th
ool
h-x
(a)
nb h-x) te
xr = 3 (h-90"
m=
mertye ¥ Onis.
axic.,
boxt
tb (h-x)*
- = 0
@|_4
of
side
the pension
on the compression
nevtred
n), bw
ved | hex
L
on
4
y=
Dune
&
P
Fa
MPa.
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
el
*4.60 A rectangular beam is made of material for which the modulus of elasticity is
Problem 4.60
:
.
_
:
.
é,in tension and £, in compression. Show that the curvature of the beam in pure
bending is
1M
Ps,
where
4EE,
Use
Seen
+
|
Ey
The,
as the
En
=
Locate
bd
lI,
reference
nEs
neutral
onis,
nbxX% = bh-x)
ol
nx
-(h-x)
~
p34
3
bx
"i
,
£
mtn
3 (w+)
i
=
pe
M
=
Ep Des
Ey
bh
4
x)
(
h-x
[4
ri
2.1 nci+tn
3 fn +1 )3
M
Dns
E./ Ee
° (VE7e+1)*
_
&
=
fn
th
La
2.
bh°
=
n
+i
a(qn+t)*
Y
|
2. 7n
3 (fier\?
=
3
bh”
3
bh
?
t
12
= (h-x)
\+ (2)
woh een
chs’
x
=/BoEN
E, =
T
YE,
dn
.
AbLtheyy
bth
= 0
so
eT
Dyrans
modudbus.
a»
bL
>
4 Ee be
* Cestey
Proprietary Material. © 2009 The McGraw-Hill Companies, lac. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.61
4.61 Knowing that the allowable stress for the beam shown is 90 MPa, determine (a)
the allowable bending moment A/ when the radius r of the fillets is (@) 8 mm, (6) 12 mm. —
8 mm
yp
= abhP- fe S)Ue)= 42667 x10%mm! = 42.667 x10" m!
Di.
BOwm
a
(a)
5
= eo
=
OZ
5
=
Vem
= O08
Frq.
&. 00
4.31
K=
1.50
(L503( 0.020)
128
From
Ms
m
107)
. (40x10%V(u2.067«
Seek
Ke
~
Mx
=
(b)
.
domme
From
KMS.)
>
6
were
0.020
=
20mm
C=
2
N-m
Fig
ab
(Foxlo
~
4.3]
Ke
1.36
4
42.667 x10")
=
142
Nem
a
O.35 )( 6.020)
4.62 Knowing that M= 250 N +m, determine the maximum stress in the beam shown
Problem 4.62
when the radius r of the fillets is (a) 4 mum, (6) 8 mm.
/ 8 awn
My
T=
&bb
Xd
c=
20
Bb aHoY = 42,667 xlo%mm"= 42.667 x0 I!
=
mm
=
0. ORO va
BO mr
5
(a)
$= iis oto
Gra = K -
&)
B ow,
$= Yr
_
0.20
wMe
Shay? Ke
.
QO mm
A.00
K=
From Fig. #3!
* Ce
=
219 «10° Fa = 219 MPa
From Fig. 43!
_ (1-5@150)(9.020)
= SS CeTmio™
1.87
K=
_
et
1.80
in’ O, =
176 «10° Pa = 176 MPa
<<
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem
4.63
4,63 Semicircular grooves of radius r must be milled as shown in the sides of a steel
.
member. Brewing tnt M= 450N- “tm, determine the maximum stress in the member
To3
>
Go
Fri
I=
bh
C=
td
6
ty
@3
KMe
Pyhow
pe
G
_
=
4q. BA,
= Os!
Ke 2.07
= 38.3 x10" Pa
«FO -*
Kis
=
KMe
mey
Problem
aS
Ge
m
= 108 -C@VMB)= 12am
rig.
o=
0,045
[1.0 FBS
£82
5
Ls
a
Fig. 4.32,
(2.07\(450 \o.045)
I
d= D2
t2oO
= dlieqoy = 1.0935 vlo% amt = 1.098SxIO
mr”
= 4S mm
“e
[0B-YWG) = 0 mm
A= D-2v.=
Qa)
B-esi5s
a
FO
1. Gl
é=
SS9.87x%107% mm
=
£- B- ozs
a”
TZ
ma
=
OLO36
S59. 87 6K107*
m
m?
<a
Samy = 46.6MPa.
Pa
= 46.6%10°
elOn4
AT G
RE
4.64
za
=
0.61 )\C4YS0 (0.036)
TF
Gray = 38.3
M Pa <a
4.64 Semicircular grooves of radius r must be milled as shown in the sides of a steel
member. Using an allowable stress of 60 MPa, determine the largest bending moment
that can be applied to the member when (a) r = 9 mm, (5) r = 18 mm.
18 min
,
fa)
f
Dd.
M
yy
a”
From
Gom:-
[0B -A2V4A)=
d= D-2eer
Oo
[of
qo
Fig.
4.32 |
Yew
A
K
=
FL
Gp
=O!
2.07
L= £ liso)? = 1.0935%10% mm
= 1,0935%10” wm
czatd=
Yom
§
_ CIT_ ]oxio%Vl. 03S)
K Me
"Ke
°F .
OF
OL,OUDTWH
=
5
72mm
(bY d== 10%108-@Xi8)=
c= del = 36mm
te (is )(7ay
i=
_&il
Me
_
>
0.0236m
(2.°7)\(0, OY )
Ps _
From Fige4 32,
= $59.87%(8 mn
(Coxto® )(S59.87™ (O77)
(1.61 )CO,036)
108ois ee
7
_
M=70o4
= 7O4
Nem
18.
f= er. Bsoas
Ko
1.61
7 mt
S59.87%/0
=
pe
= 5 BO
-
a
|
'
M = 586 Nom <a
“4.65 A couple of moment 44 = 2 KN - m is to be applied to the end of a steel bar.
Problem
4.65
_
Determine the maximum stress in the bar (a) if the bar is designed with grooves
having semicircular portions of radius ry = 10 mm, as shown in Fig, 4.65a, (6) if the
bar is redesigned by removing the material above the grooves as shown in Fig. 4.650.
M
For
beth
Db:
150
£2
1O
Pp.
i¢
Fon
100
mm
me.
ve
*
SO
o
*
For
ads
fee
ft
4
me
lew
3
M
confi gurations
18 im
i
=
o,.1/0
con Figur ation
4.32
gives Rae
contig
T=
@),
.
uration
bbh?
Ch,
.
na.
2216
Fra.
:
(a)
:
4.3)
(hb)
\
gives
Ke
= & (is)oo)* = LSxlOo mm
1.79,
=
1.810% my"
cz tad= $0 mm = O.28m
(a) 6
ves
62.21 [Be eae
(b) 6 = SES = CL Ale O28
147 «0° Pa = [YT MP
=e
WY MPQ
we
= 1G doo Paz
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiled distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.66
4.66 The allowable stress used in the design of a steel bar is 80 MPa. Determine the
largest couple M that can be applied to the bar (a) if the bar is designed with grooves
having semicircular portions of radius r= 15 mm, as shown in Fig. 4.65a, (6) if the
bar is redesigned by removing the material above the grooves as shown in Fig. 4.658.
,
bodt-h
DPD.
aT
¥.
For
~
ie
loo
.7
1S
s
joo
M
0.
gives
IS
(a) , Fig
K=
cortiauve
-Waam
tion
1.92.
(o>
{a)
Fie
4.3)
gives
Ky =
L=
bh?
= $Us)loof
cz
yds
lay
(ib)
M
1.50
contrguration
4.32
Fov
que tions
[Smm,
yr
Aa
cont
Ge
.
SO
KMe
Ape
M - GE
ee
1.57.
mm
.-
_
(b)
-
Me
= 1.5 x10% mm"
i
For
|.5x Jo ~ m
¥
O.908Om
GI
~ OL
|
. (80x/0% (15x10)
(1-92 (0.08 )
(8oxlo*)U1.sxlo%)
(E7000)
~_
pee
=
em
125 kNem
|.53x]o Saban
Nom = =
.
1LS3 kKNem
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers _
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission, ©
at
=
4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic
with oy = 300 MPa and is subjected to a couple M parallel to the x axis. Determine
the moment M of the couple for which (a) yield first occurs, (4) the elastic core of the
bar is 4 mm thick.
M
“
@
S
‘
= SiI2x1O
M.
a
a
Y
bhe
LE ram
*
O£,004
Mm
w
iexioo')
<= Gch . Boorse
Oc 4
0.
C
38.4
q
ue
.
I= gbh- pat 2B) 9 - 512 rm
Mem
My
it
4.67
33.4. Num
4
Problem
oO.5
tl
SN
it
ey
(b)
tJ
Oe
on
i
af
rc
“<
x
8 mim
iw
4
=
12mm
Yy an
4 C4)
3s.4)[ 1 -
=
2rm
“is
=
&
Les) | = 52.8 New
M=
Problem
4.68
4.68
52.3N-m
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic
with ay= 300 MPa and is subjected to a couple M parallel to the x axis, Determine
the moment Af of the couple for which (a) yield first occurs, (8) the elastic core of the
bar is 4 mm thick.
T-Abh?= & GR)
J.1§$2 ~10°
"
@
ce
the
M,
=
mm
=
per”
jisaxio?
ww?
O,O0Gm™
G&L _ B00 10°. ISAx1 4)
:
12 men
|
8mm
O. 00G
£ 57.6 New
eter
~<a
83.2N-m
~<a
L
3
yr RU)= 2mm
Hs
7]27 = 88.2 Nem
1- $$6)
M-$m[1-4@)] = 2 (57.
Ars
(b)
Mees
M=
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hilf for their individual course preparation. A student using this manual is using it without petmission.
Problem
4.69
For the steel bar of Prob. 4.70, determine the thickness of
the plastic zones at the top and bottom of the bar when (4) M = 30 N.m,
4.69
{b) M = 35 Num.
S intis
4.70 A bar having the cross section shown is made of a steel that is
assumed to be elastoplastic with £ ~~ 200 GPa and oy * 330 MPa. Determine
the bending moment M at which (a) yield first occurs, (P) the plastic zones at
the top and bottom of the bar are 2 mm
thick.
=~ i42, bh? = E(MAY = alncoeal
c=
-
My
!
Ged
= Seb
My
(a) M= 30Nm
oor
=
_ & Boxset (B41 5X00 oi = 22 16N im
é
28
,
_
Ms: 36 tm
-
&.
4.70
In) *
A bar having
mm
“
67
bp r- yy
|
C4) 228 me
4.70
tp 4-2673 = ork T mm
BEY.
2 -f3-2
Yr los
Problem
M
dkef = [3 - 2(45) = o-72y
Yr = 0+9394) (4) = 2073 mm
(by
1g
Sa
-_
7 LIZ OH
the cross section shown
is made of a steel that is
assumed to be elastoplastic with B= 200 GPa and oy = 330 MPa. Determine
the bending moiment M al which (a) yield first occurs, (b) the plastic zones at
8 pam
the top and bottom of the bar are 2 mim thick.
ia
Sim
I= a L bh? =
c*
My =
Zh
6,
xt
=
(PIP = 3412 ment
tint.
(330K 10%) (Hee syx7o")
te
f=
+ 0.
OBIE,
yt a8 bi. meg
Cb)
yy
*
c~
tp
Me = aul
=
Gd
z= amm
| - s(2Y¥]
= £ Ge)
1 - 3(Z7]
Mp= 39-72
Nm aaa
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or _
distributed in any form or by any means, wilhout the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.71
4.71
The prismatic bar shown, made of a steel that is assumed to be
elastoplastic with E = 200 GPa and oy = 300 MPa, is subjected to a couple
of 162 N + m paraliel to the z axis. Determine (a) the thickness of the elastic
core, (b) the radius of curvature of the bar.
y
2mm
(ay
T=
¢ (ori2)(ov1s) = 5:875 «1
M, = Sx
15 mm
0075 Mm
=
4 (e018 )
CF
. (300% J 0° \( 3375 «bor 7)
Cc
~
POO7TS
z
=
M
a
Y
M
743 = aye
Yy
F
= (0'779-6 C0007)
Threkness
(6)
{3 -
yy
= &P
fe
- YrE
_
& p
_
*
3
~
BMF
{ayes
~
135 Nm
£8
7
|
= O-1746
= 581x107 Fm
of
=
-~
mt
easdic
3
cove g>
= /h6é nm
<
7
£
(88Ix10
\Qooxlo )
300» 10%
2 yy
= 3873
P7hIM
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
at,
.
4.72
Problem 472
Solve Prob. 4.71, assuming that the 162 N - m couple is parallel to
the y axis.
,
4.71
The prismatic bar shown, made of a steel that is assumed to be
elastoplastic with £ = 200 GPa and ay = 300 MPa, is subjected to a couple
of 162 N + m parallel to the z axis. Determine (a) the thickness of the elastic
¥
core, (6) the radius of curvature of the bar.
12mm
>”
fay
JT
i (owts\(ovo1z)* = 208 x ion tm"
CaF
My
2
YO
GG \G@ror2
=
&
I
a
= 01066
&
Geomio®) (218
_
.
x10?)
2066
= 109 Nm
= 3[1- 3Q)]
mM
=
[3-28
Ye
Bn:
= 016549
(01689 \Co.006 ) = 4, 9¢aKI6
Threkness
(b)
bz
a
=
Ey
Pf
=
Pp
of etashe
cove
2p
= zmm
p>
ob mn
*
YE
&
4 4$4 x10 *\ 0200» 107)
300 DOF
- >,
= ob626m
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual i is using itwithout permission.
—/
4.73 and 4.74
Problem 4.73
A beam of the cross section shown ig made of a steel which is
assumed to be elastoplastic
E= 200 GPa and a, = 240 MPa. For bending about the
z axis, determine the bending moment at which (a) yield first occurs, (5) the plastic
4
zones at the top and bottom of the bar are 30-mm thick.
comm
a
4}
:
BLA
axts
as
=
YS
~ Gr
My
60 mm —>|
zth
=
0.945
iS m
So
m
seid
0,045
”
¢
2 R: |
t
mn
~ (24ox0*)(n
mae
80mm
3.645¥/0 ° m"
10% mm =5%
= 2.64
2 coq
T= bbs
(@)
;
IqHawlo
-
My
N
™
=I244 kN
met
,
é*
930 )
R= - GA, =_f(Z40x10°Y(0.060)(0.
=
RQ,
y,
432
«/0O°
)Smm34
>
N
=
1lSmm
a“
0.030
wm
Re = SCA, =(4\(240%10* )(0.060)(0. 215}
Le coa_sl
= {o¥ xfo* N
Yo = ZS mm} = JOmm
2 (Ry, : Rey.) = 2L(3210* (0.080) + (148 x Jo (0.010 9]
if
(b) Mz
= 2.0lOm
28.08% 10° New
|
Mea
28.1
kKNem
Proprietary Material. © 2009 The McGraw-Hill Compautes, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
<a
Problem
4.74
4.73 and 4.74 A beam of the cross section shown is made of a steel that is assumed
to be clastoplastic with E = 200 GPa and oy = 240 MPa. For bending about the z
axis, determine the bending moment at which (a) yield first occurs, (6) the plastic
y
zones at the top and bottom of the bar are 30 mim thick.
30 mm
—+
& Int ebb,
.
:
4; (30) (40)?
if
Wy
30 mm
%
=
30 mm
15 mm
30 mm
To
Le
-
Te
15 mim
°
1 BAS
ig by hy
lO mn
%
4
= @ (B03)
f-
le
M,=— of
Set
=
LBGxlOsmn
ue
NE
jo¥fo®M8\C1,
is
29
f, BY «io*
~
Ok
C=
19 _0B x1” New
14
>
.
ad
=a
SYA,=
=F
ey”
Yu
(240x108) (0,030)(0.030) =
IS +lS
= BO
mm
= OL080
=
lOmm
=
=
1ORxIveN
0.010m
u
Y2 7 ECS)
216 xlo®N
m
Ri 7 F6,A, = $ (a4ox 10*\(0.068)( 0,015)
x
O.O4T m
Ro,
NS
|
2
4S mt
My = 10.08 kN-m
Fd
KNEE
Yo
wlD® wm”
Tt
7
AKAS
oS
ww
—
()
=
= 2[ (2icx1o”
wn
=
ELIZ
3
x [O°
Wo.030)+ (168x10*) (0.018 )]
%
New
M
= 16.12
KNew
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
«
Problem
4.75
4.75 and
that is assumed
ing about the z
which (a) yield
4.76 A beam of the cross section
to be elastoplastic E = 200 GPa and
axis, determine the bending moment
first occurs, (b) the plastic zones at
shown is made of a steel
ay = 300.MPa. For bendand radius of curvature at
the top and bottom of the
bar are 25 mm thick.
25 pan
50mm
(or
Smm
| —-»|
50 mm
|... 12.5 mm
=
B
bh,”
To= bh 43
I,
=
'
eH
12.3 om—|
I,
c=
I,
=
+ A,
a,” = L
eysy(as} g(ac)tes iia1.s)'2 224376 ma”
,
= & (40) (50)2 = 520723
mm
2724270
I,+1,+I,
print
= 5997603
mmt
Somm
GT
“eo
i
i
~b
ot
| (genxso ) (5989893 x10)
=
050
= 33°97 kin
we
R, = 6A, = Goo x10 (0.0 75)(0.025) 2 SOS LY
NALS
oo
avis
H
Ce
Ys LZERIBd
Re LOA,
3 BTS
mn,
4 (300 10°) (010 S) (0-028)
Ye
(b)
M=
2(Ry,
i
[E705 kAL
ZOE)=
+ Raye) = 2 [l6be-s) (or037s) + (EMS
(667 tm
001667) ] 2 48S Lahm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—
4.75 and 4.76
Problem 4.76
A beam of the cross section shown is made of a steel
ing about the z axis, detetmmine the bending moment and radius of curvature at
which (a) yield first occurs, (6) the plastic zones at the top and bottom of the
bar are 25 mm thick.
T= bh? + Aa = A079 09% (78) (2s) (37-8)
y
= 272437 6mm
25 mm
~~ »
Leaf
le. 25 am
T= dbhy
= ACS) C50)3 = 2604/7 mat
25 mir
I,=
1, = 2734375 mm
T=
I+ 14
|
|
4
= £73967
me
.
mm.
£0
c=
&
hn
RE
J
|
GLA, =
,
,
:
.
axis
R,
ow
RL
[3o0Xi0 (0078)
(0.028)
.-
LSP]
2S RII My
ih
|
R
ft
CL _(300xX1o") (672.167 ¥10)
KOA, = L£esee xo") (20 410°)"
=
93 <7" kN
Ye = FUSV = 167mm,
)
Ms
a(Ry,
*Ry) = Q[(sbo-oe3 733 + (13-79) (0° 0/667)| = 45-3 Lin.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual niay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem 4.77
4.77 through 4.86 For the beam indicated, determine (a) the filly plastic moment
M, , (b) the shape factor of the cross section.
4.77 Beam of Prob. 4.73.
From
PROBLEM
4.73
E = 200 GR
and Gy 7 AY40 MPa,
A,
)
Ro
Mp =
(by
Ts
Cr
beh® = bE oo
mm
My = Set
y=
A700
=
=
2700
¥jO”
m
nana
1
OA,
£48 «lO°
N
d = 45 mm = 0.045 m
= (648x/07) (0.045) = 24.16 KIO" Nem
Rd
4S
=
i
—_
(a)
(g0)( 49
= (240 xia" )(9709 «10"*)
F
LILA
axis
42
ae)
tf
"Tr
90 mm
Ma= 29.2 komen
= 3.645% 10% mm! = ZC45x/0 Fm"
O.O4Swe
(240s Oak ts 15)
Me
- 29.16
My
19. 44
=
19.44 x 1D® Nem
kel
“
SOO
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reser'ved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
«mR
Problem
4.78
.
4,77 through 4,80 For the beam indicated, determine (a) the fully plastic moment
M,, (B) the¢ shape factor of the cross section.
4.78 Beam
From
of Prob. 4.74,
PROBLEM
4.74
Es 2006Pa
R,
=
YP
KNOB
AKNS
“
R
.
R 2f
Ky
‘ate
SA,
if
(a40x10% )(0.030)(0,030)
216x107 N
§4
at
if
©
30
*
Ww
~
=
i
Ya
=
0.030
mw
Ax
Ov
=f CRYO
oo
and Gs 240 MPa
¥]
6 {O.
.0
(0, 0600
0° io,
iS
2G x{lorN
7.5 nm
=
0, 0075 nw,
we
1
MpF
2 (RRS, + R,
=
ye ala 16x 1O*}(0.020)4
16,90 v[0* New
tb)
oO
Mo=
Tet bhp
’
CPUS mm
=
Sel
M
M,ae
KNew
—=ed
eZee)
=
67,5*10% mmm"
t
Doe Ty te
My =
16.20
Ty = pbb = Gellqoy = 1.8225 %J0% mn"
oe
KK
(216 xj03)(o. 00785) |
=
.
roxio
16.20% [0%
LR
inode 1S
‘6
= 1.89 10% mnt =
=
Lexie
1.84 %10% mn"
O.O4Sm
-t
~=
10.08*|0O° Nem
k=
~
1,607
Preprictary Material, © 2009 The MeGraw-Hill Companies, Ine. AH! rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it withaut permission.
aes
Problem 4.79
4.77 through 4.80
For the beam indicated, determine (a) the fully plas-
lic moment M,, (b) the shape factor of the cross section,
4.79
ne
12.5 mm —
j___|
to
NX
WI
L
Na
5
7
6
R= 6,A
|.-195 mn
50 mm
¥
axis
25 mm
onl
= 300 mPa,
E= 2910% and Gy
Fon PROBLEM 4.76
smn
—
Beam of Prob. 4.75.
.
= (300%10) (0-678) (0:025)= £b6I0kN
'
ev
LSA
f deg
2
DP
An
R= Op, = (doe wo®)or05)lo-d8)
= 3 7EN
Z
,
Yn?
MAN
&
£G@S)=
+
28
Je
o
jaeL mm
j
Mp
=
2¢Ry +
+2 ) :
| (625) 000276) + (376)(0-0128) J = Sle6 EN
(ob) J, 7 dbhe a Ade = F(IDOSY#79 28) BS)
I= tbh,
9734374
L-1,+T,+
My
S0pMm
~
oot
heed
K
273 4375 mnt
= 6605) =490033 met
T, > JT,=
Cc =
~<a
M
wien
My
I,
|
nm
= 4909983
by
= (eonse)
Lh b
renee
(34
pa.
(SOER"
=
—12)
Moe
f
“7
.
2
35.4
LN
fy.
:
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written perthission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=a
Problem
4.80
4.77 through 4.80
,
For the beam indicated, determine (a) the fully plas-
_ tie moment M,, (b) the shape factor of the cross section.
4.77
Beam of Prob. 4.73.
4.78 Beam of Prob. 4.74.
.
25 nm
r
4.79
Beam of Prob. 4.75.
4.80
Beam of Prob. 4.76.
From PROBLEM 4.75 E =200xK10 6fk amc 6) = Seah,
50 ment
25 mm
|.—25 mm
|.
.|
25 mim
.
.
R,
é
g
25 min _|
;
6, A, = Goonte )( 4076 }lo-025
= £h25 by
Ye AL FIDL SPS moy
ZAC
2.2 6A, = (360 xe ) (0026)(o025')
~~
KS
=IkFe5
Ny
NS
LN
NN
¥2?
£025
\=
12°
ates,
Mo = 2 CRY, + Raya) = ROS (OusIE) HZ) (O0I2S)) = 4] LN im
I,
Bb he + Ad,* = ZV bs)}?+(78)
Oe) BES)= 2734378 pol
tt
(6)
c
=
= abot mn!
DT, 7 3724370 punt
i
ds
T
t
%
1, = thheh? = FEN)?
I,+I,
:
+ I,
=
72
C
Fle7
fur
aft
Warne
LO
_ GT
Me
mt
|
_ Goayist) (onU67 No) |
"
“
Ook
AAT.Yy
k= teMy . 8a
34
4-
LN
mM
1 303
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Probiem 4.81
4.81 and 4,82 Determine theplastic moment M, ofa steel beam of the cross séction
——-
shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa.
50 tam
Tota?
NA
Co
mm
aftA
=
30 mm
4,
(50)(90) ~ (30)(30) >
1800
=
10 mm
lw.
| |__|
A=
.
:
ZA
ae
“am
1Oainm
aveat
wm
u
36
-
1200
"SO
*
=
8600
pam
R,
_**
Ls?
Q
*t.
SS
Koy
As
Ry——>
~N
kb +l
(Ro)( 30 )
=
600
nm ,
Ya
fE0iCle;
=
foo
hoe,
Ya =
Avy,
>
1B
AY, =
Aly:
700 mY,
=
7mm
=
29
mm?
324 «10%
v
2
(SO)C4)=
tt
YF
4.9 x 107 inn ®
Ay
+A,y,
Avy,
Mp
=
Oy
4
Ass
ZA:
+
mm
A, Ys
UG aw,
Aaya
79,.2x10°
(240x)0%)(79.2 107")
Me=
=
& lo*
24.5™
FIR
=
mm
V7 U
it
A,
4
»
>
A, = (60)(36) = 1800 me
lo-
jo” *
a
me
m
19,008x10°
19.01
yr wr
kNem
3
Nem
ity
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
ard educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.82
4.81 and 4.82 Determine the plastic moment M ofa steel beam of the cross section
"shown, assuming the steel to be elastoplastic. with a yield strength of 240 MPa.
oy
Ry
36 mm
Perera
ne
yt
Ry
hoy
y
£92
NI oA
ne
Y¥3
<—-30 mm _
Tote!
areat
tA=
A=
270
wm
=
os
A=
A
ee
A,
= 4aby
B27)
=
y
=
SHO
4
2=
2 y?
3y,
=
AS.
YSS3
=
kh
my
2]0 mm
= (QIRIBA)\CBS= ASLYSSB) =
RR.
A
5
W
ree
H
=
Jo?
6,
-_
A,
A;
_
=
A,
240
=
46.324
“10°
G4HBxio* NV,
ay
=
G4SS3B
BA,
IZ,
==
y*
& Cat. 4iBav2s.4SSB)
=
eam
A,
,
A, =
Since
2
z (30\(3¢)
=
22BLTG
=
46,224
=
21,2132
Z70 xjo~*
wm”
«jo
& (36 - 2AS4SSB) =
nm”
= 2AIB.6
TG 10% >
*
wa"
Az
Ry» S3,6822x/0N,
mm
mm
>
Ry M178 x10
BoB
SS x(OT®
SLQ27Rl am
F
S272
X [Omng
Ja = 3 (G6-25.4558) = 7024S mm = 1.0295» 1
Mp
=
Ry,
+
» Ya + Raya
=
Fi Nery
Mo
= GLP
Nem <a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers:
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
4.83
4.83 A thick-walled pipe of the cross section shown is made of a steel that is
assumed to be elastoplastic with a yield strength ay. Derive an expression for the
plastic moment MM, of the pipe in terms of c), c2, and oy.
D>
i
Ek
oo
(b)
Problem 4.84
4.84 Determine the plastic moment M, of a thick-walled pipe of the cross section
shown, knowing that c; = 60 mm, cz = 40 mm, and ay = 240 MPa.
See
the
FodPowing
Mp =
PROBLEM
expression
4
3S
cy
240 MPa
Cc,
=
GO
mm
Cy
=
4O
mm
=
3
4.93
Par
derivation
oF
the
for Mp?
3
- oS)
.
*
240x]0°
a
Gyr
to
O.260
m
a
Datai
sodution
0.040
vi
Pa
Mp = £ (240 #10% \( 0.060" - 0,040) = 48.64
* fo" New
Me =
48.6
kNem
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
8
Problem
4.85
-
4.85 Determine the plastic moment M, of the cross section shown,
assunting the steel to be elastoplastic with a yield strength of 330 MPa.
LAR
50 min
==
A
= (78) (iz)
A
avec=
Vata?
2mm
r
:>
Yo
fe7$ bow
M25)
($0
+
=
aA
Bu
ola
2 Joomm , y,t i225 mm,
A, 29902)
min
jy.
=
~
.
2
ny
32,
Ajy, = HozS mem
ALY," 492 od mim
A, = 43-18) 09) = 1043. Siam? Jat PEIN
ALS, 224258
_
ene
.
3
i
B
*
Ze
ABIS
a4
at
x“
Me
*
Sy (AY,
+
1 Ye
+ As Ya)
“J
;
= 3ox10"(Hoaet 489.3 + 22925.) (ine ) = UIT EN
Problem 4.86
sutmine4.86 the
Tota? avea:
Determine
plastic moment M, of the
shock to be the wlactoptastie
with a vield strenoh
mm
of 250
MP.
2
:
2,
Ay,
A, = (50mm,
Y= 6025,
Aaya > 1705 pon?
A, = 750mm
Ya = 3 12&
Any, 52343 7S wnt
=
Je
bESS
Aayy
:
2,
Boo imi
Mp = Sy (Avy, + A2y2 + Agy,
+ 22200
oe
400 .
Mm
3
J. TAPS5
t
= izeo ees
>
A,
b
cross section shown
Ax [tvvx 12 )+ (7S) C12) + (56 )Ciz) = 2Jo0 men
$A = 1350
:
~~
assuming the sfeel to be elastoplastic with a yield strength of 250 MPa,
&
A
(ore
SE
min
Ay 2 (6:25) (28) = S626 mm, yy, = 2-125 mm
-
ANE
2180
=
men
4 Aa ¥« )
re
.
= (280x00 ) £22200 +937 42343754 Goo) ( X10") = 24-4 km
a)
Proprietary Material. © 2609 The McGraw-Hill Companies, nc. Al rights reserved. No part of this Manual may be displayed, teproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3
Problem 4.87
4.87 and 4.88 For the beam indicated, a couple of moment equal to the full plastic
moment M, is applied and then removed. Using a yield strength’ of 240 MPa,
determine the residual stress at y = 45 mm.
'
4
4.87 Beam of Prob. 4.73.
ye
I
60
un
>|
AFG
=
Mp
90 wn
=
om
¥1O°
(See
Nem
a.gaswio’®
w?
=
MoeaY
240 MP
=
2
Hee
sofutions
cr
of
O.O4f9
y
360 MPa
>
4.73
Prob fems
fo
GC
cred
4,77)
m
=
AS
wm,
120 MPa
i
+
LOADING
G
1
-
UNLOADING
(91G*I0") (0.045)
Gres =
3 bdFxfoe
G'- OY
=
*
360¥10
RESIDUAL
STPCESS ES
Fa.
360xl0%-240r10%= 120%10° Pa
20,0 MPa a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.88
4.87 and 4.88 For the beam indicated, a couple of moment equal to the full plastic
moment M, is applied and then removed. Using a yicld strength of 240 MPa,
,
determine the residual stress at y= 45 mm.
4.88
Beam of Prob. 4.74.
30 ram
30 mm
Me
=
i
1G.20x}o°
New
PROGLEM
4,78
from the
sovPutiou
to
‘|
|.
30 mm
O.O4S
5mm
m
15 mm ->
lL Boxfo.*
¥
tH
30 mm
1
(1G.20x10° Yo.o4s
BRS.
x10°
Pao
=
B85.7
335. 1MPa
LOADING
At
Res
UNLOADING
GS =
y 248
eluat
stvess
Ches
=
MPa
(45.7 MPa
RESIDUAL STRESSES
6! = 385.7
MPa
240
)
|O~*
LB
+
r
=
=
Mee
=
Mra
6!
rea’
C'-
6,
MP.
Gres
=
(4S.
TMP
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
a
Problem
4.89
4.89 and 4.90 For the beam indicated, a couple of moment equal. to
the full plastic moment M, is applied and then removed. Using a yield strength.
of 290 MPa, determine the residual stress at (@) y = 25 mm, () y = 50 mm.
y
25 min
125 mm-—
|__|
50 mm
50 mm
Mp
. | 25 snm
I
L195
mm
=
4.89
Beam of Prob. 4.75.
4.90
Beam of Prob. 4.76,
bb
tw
= 9k 9892
oR
re
,
}
Mase
Tp
(See
nat
Y
-
sobshous te Pehleu's 4.16 owed 4. fo)
GQ:
Reyes
M,c
wet, CEhB
07} (0°0
K5)
S2imm.
.
a, 7 Mm
6! = “Saparpaciote 7 A307
43007 MPa
LOADING
Mys2Smm
(a)
|
@)
UNLOADING
Greg F300
=ge
7 dec7
Oe
RESIDUAL STRESSES
2/5036 MAY,
= 7)
Sle 4 4300
FINGER
At yr Somme
:
MPa,
— BH 6
MPa,
6 = Gio]
aa
MA
Gig, = Boo t Y3007 = 730-7 MPa,
y
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usingit without permission.
Problem
4.90
4.89 and 4.90 For the beam indicated, a couple of moment equal to
the full plastic moment 47, is applied and then removed. Using a yield strength
¥
of 290 MPa, determine the residual stress at (2) y = 25 mm, (6) y = 50 mm.
4.89
4.80
25 mm
“re
Mp = 46-7 ENM
[25mm
25 mim
|
25 aa
|. ee 3mm
T-
£9724167
|
Beam of Prob. 4.75.
Beam of Prob. 4.76.
(See sofutions to Peolfems 4.75 ana 4.79)
mnt
cr
,
6'
=
1
Moy
Samm,
7
=
Me
at
(469x103) (00H)
y
(a)
.
-
UNLOADING
At ye ZE ity = te
Ghug
6's
c.
oy
Aoy MPa
LOADING
=
Ear tot hn
_
RESIDUAL STRESSES
+ (409)¢ 204-6 mPa
— 3B00 + 20465 © ~9S € moa .
Ab ys Sommee
Gun * = 300 + 404
ls Goa hha
= 109
MPa.
.
RGN
:
a
a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without perinission,
|
Problem 4.91
4,91 A bending coupleis applied to the beam of Prob. 4.73, causing plastic zones 30
mm thick to develop at the top and bottom of the beam. After the couple has been
removed, determine (a) the residual stress at y = 45 mm, (4) the points where the
residual stress is zero, (c) the radius of curvature Corresponding to the permanent
deformation of the beam.
y
90 rim
See
SOLUTION
stress
~— 60 mam-->
to PROBLEM
distribution
M =
4,73
duving
28.0b%I0 Nem
E = 200 GPa
10% (0.0
28,08
(a)
6!
=
o wu.
=
¢
My
_
(28,08 1o*)(0.015)
lone
T
At yF yy
yy =
[Sime = 0. 91S m
6, =
240
B46
7
=
og
.
Pas
|
116.6 MPa
1OB.T MPR
346.7 = 240
«Cig?
Ges = O'- 6, =
US.6 —2Ho
Grae 124. 4.HPA
346.7 MPa
pn oe
—»
106.7 MPs
~ 124.4 MPo,
124.7
=
Oo
BDNLOADING
“
fee
_ TG
Je
MPa
~ 106.7 MPa,
LOADING
Gres
ae
ee
240 MPa
(b)
MPa,
346.7
Gres = G'-G.=
240 MP
anc
MPa
Fa
x}o*
IIS.6 x10"
4 Caee
yc,
At
=
Mo.o3s)
fae
—
bending noupde
Cc = 0.045
3.645% 10m
T=
tov
Boasting.
a
SG
=
RESIDDAL
0
_ &easxis*)(ayorlo*)
M
-
STRESSES
23.03 X102
Ons.
= Z3bIS
lO 'm
= BlLIS mm
Yo 2 7 3ie1S mm, O, BINS men
et
Shes 7 — 124-4 * 10% Pa
Ge
Ey
2.
P
+ Ey
=
(Zooxlo*
~ 124.4
(0.015)
xJ{o®
-
prttim
a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alt rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.92
4.92
A bending couple is applied to the beam of Prob. 4.76, causing
plastic zones 50 mm. thick to develop at the top and bottom of the beam.
After the couple has been removed, determine (a) the residual stress at
¥
-
curvature corresponding to the permanent deformation of the beam.
23 mm
50 mm
|
See
95 mm
25 ran
|
iL
SoLuTIoN
st ress
| 25 mm
M=
E
(a)
6’=
oe":
Me.
(4S
i
M3
S32
My
-
To
Za) hy.
Yr
= rendPs
x70")
3 N10")
Le-
Pa
4803
te PROBLEM 4.75 for benching couple and
distys bot i an
oF
gpa e7io
G, = 300 MPa.
a
=
X 10%)
(o- o1S)
SJz7
=atf mm.
345-3 MPa
a
FID
TT ee
TUas
At
yre
Grea= 65-6,
= 3983-300
=
At
oy? yy
Ors = O"- Gy
= /9T7- 200
= — fo203 MPa
Soom
fe
2524953 MPR
+.
93 nla,
~
p GB MPa
— (O23 Mfe
=
{02.3 MD,
TLS
LOADING
(bd)
Gy =zO
Yo .7
UNLOADING
RESIDUAL
+ Mie - 6 = 0
1S
Gt
Yo
ans.
(c)
At
y=
yp,
oe--f
Pp
STRESSES
.
(epqernis”)
WSte e0K0)
:
BMLY
|
==
te
U°03279M.,
= ~37dmam
Oo,
,
279mm
eh
Gres 5 ~ (02-3
Mfg
:
p> ~ Ex,
Oro net) (0025) «49.9m
(02*3 Xlo
:
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ab,
Problem
4.93 A rectangular bar that is straight and unstressed is bent into an are of circle of
4.93
radius p by two couples of moment M, After the couples are retnoved, it is observed
that the radius of curvature of the bar is gg . Denoting by py the radius of curvature of
the bar at the onset of yield, show that the radii of curvature satisfy the following
relation:
Let mm demte 7
M: 2M, (1-3#:)
Problem
4.94
(a) + =
ms
4,94 A solid bar of rectangular cross section is made of a material that is assumed to
bé elastoplastic. Denoting by My and py, respectively, the bending moment and radius
of curvature at the onset of yield, determine (a) the radius of curvature when a couple
of moment M= 1.25 My is applied to the bar, (+) the radius of curvature after the
couple is removed. Check the results obtained by using the relation derived in Prob.
4,93.
m= 2mM,(i- $4.)
£ =13-2m
-ZU-3f)
P=
bh
(b)
fetpet-M:.
P
EL
=
0.16421
aon
biome
P
ET
het ms Ho= Las
2.
O.TO7L
ail-m.
Pfr
Pa = 6.07 py
0.70711
=
Pr
wt
0.70711
<a
Py
. be
fe
|
~
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.95
AS
4.95 The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which E = 200 GPa. Knowing that the radius of curvature of the bar is 2.4 m
when a couple of moment 4f= 350 N - mis applied as shown, determine (a) the yield
strength of the steel, (4) the thickness of the elastic core of the bar.
an
M- 3am, (1-4#:)
3 EG )
-$ G8)
%
~o
os
20 mm
(
Defeat
[| - f- Sr)
E = 200x10"
k=
20mm
|
| -
Selving
bo)
yr
~
&p
.
E
750
by
Pa,
slo”
theckness
420
YWoxjo”
6y |
=
Gy
_ (292 *10°)(2.4)
.
Cube
MF
dural
200
x{O4
of
etastic
=
equation
Nem
P=
sgh
= 8mm
6, |
= aso
273.44
292
~JO°
_
=
R Yr
10
=
for
24
Cy
m
© 0.003m
x lo °
Pa
Gy =
BL
3.504
~
cove
o™
M
o
i.)
=
= 0.020 m
(1,.2%%10°°) 6, | 1 ~
Gy
2
be’ 6y
(a)
i
N
L6 man
fy
oO
oo
C
io
uv
= $92 (1-4 Se)
3.
“*™
7. Ol
SOY
QZa2r2
MPa
Wy et
wm
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
it
Problem
4.96
4.96 The prismatic bar AB is made of an aluminum alloy for which the tensile stress-
strain diagram is as shown. Assuming that the o-e diagram is the same in
40 mm
compression as in tension, determine (a) the radius of curvature of the bar when the
B
maximum stress is 250 MPa, (6) the corresponding value of the bending moment.
(Hint: For part 6, plot ¢ versus y and use an approximate method of integration.)
M
“ey,
.
6
o (MPa)
60 mm
200
(ay
Gy
250
E.. 7
MPa
=
O,O064
250x107
Frou,
Pa,
eovve
Cc
b
100
;
0
0.005
0.010
sf
200
th
az
= 20mm=
>
YO
mm
Lo.
a
€
€
=
=
=
0.050wm
OLOYO
Q,06e7
wm
=
6.21333
mw
P=
distyrbution.
Bending
where
the
s
J
using
psec
w
is
the valves
| uv
O
O.28
let
I
the
is
given
by
whee
in the table
Oo
flo
OS
10.0082
1380
O.7S
1.00
[O,OO48R
(0.0064
225
250
§
=
et) (i218)
BIS)
o.a5
=
J
= 2o-
du
vi6l
u I6l dv
nteqva tion.
xf
Simpson's
is
Deing
Av
= 0.28
ulol,mPal
oO
AVS
qo
168.75
Aso
01.25
MPa
M= (20.04)(0, 080) (101.
25 10°
we
get
beton :
w]
wolelpry
|
ye
O
1,0
2
130
4
|
675
aso
RIS
J= =
ve
wuiel
| IG1,(™MPa)|
©
|O.001G
,
formuta
a weigd ting factor,
Given
2ber
oF numeriaf#
in tegration
F
®
yleidy
a method
J: 42
where
zb\
=
inteqnaf
Evakuate
yude
~S.U
=e
couple,
=-\ ySbdy
M
eB: - &.
"
Strain
ok.
(b)
4.69 m~
=
10)/.25*10%
= 124210 Nem
aye
&
?
wo
jo}
Pa
M= 7.24 kN ema
Problem
4.97
4.97
The prismatic bar AB is made of a bronze alloy for which the ten-
sile stress-strain diagram is as shown. Assuming that the o-e diagram is the
saine in compression
as in tension, determine
(a) ibe maximum
stress in the
bar when the radius of curvature of the bar is 2.5 m, (#) the corresponding
value of the bending moment. (See hint given in Prob. 4.96.)
350
.
280
ao
é
a
|
210
140
™
Cc
Zoe
ie—
ILVO
Te
f
!
the
From
=
curve
[FS pm,
CQ
be Do rm,
mm,
20D
f>
. {a)
o (MPa)
-
0,006
G6,=
J00MmPa.
i
/
/
10
0.004
(hb)
0.008
Strain
€
oistycbudion
Bending
Bs
~ emt
+
-& U
where
Ur
Z
foup le
M =-S yobdy = ab | yleldy + 2be ) vi6ldv = 2be
where
+the
integuad
Evaduate
tude
ww
J
using
Useol
I
the
wig
the vadues
u
fe]
oO
O
@
qiven
a method
form
Fehr,
tn the tlle
oO
Wo
1.00 |
300
0.006
|
f7&
262
0.75 | O.OO4S|
82)
>
ule}
do
ts
Using
|
Av
w
|
oO
4
2
(75
2SB
i
SYo
g
=
[3ac8
M
= (2) (0.02)( 08 05)” (log
we.
get
lwo lel,
O
2a
aS
|
4303
iL’
300.
=
below:
jl
.
2.26)
uta,
lel, m@[v Il, wl
0.25 | O.001S |
OS | 0.003
S
wolel
wergd hing
given
by
of numevad integ vation . LF Simpson's
integration
J = aS
wheve
is
Zoo
567
<+—
wu
F|e)
w1 Pa
K10*)
SLITT
KAM
|
Problem 4.98
4.98 A prismatic bar of rectangular cross section is made of an alloy for which the
stress-strain diagram can be represented by the relation ¢ = ko” for 0 > 0 and e =—{ka"|
for a <0. Ifa couple M is applied to the bar, show that the maximum stress is
_ 142” Mc
a
Stroin
distributions
=f,u
where
-{ yobdy = 2b) ylsldy = 2be*{
in
2
Zbe*( uli du
th
~
For
=
=
Then
vv
i
Io aly
n
K6
=z
Oleg
m=
=
Oo
couple,
if
Bending
6 2-82
3h
5
WU
En, 7
ft
(2)
Kou
os
IS) =
—
Cut
Mz 2be* | uG,U*dv = 2b¢6,) B'* du
> abero,, 24h
ME} - ant]
20 pete
6, = antl M
Gm
.
*
antl
3n
Hin
{
|
b@cy¥
2
fF| bGcl
. 240°
i
Nv
I
E=
ooO
Recadd
Goji
ba*
Me
I
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manuat ts using it without permission.
Problem 4.99
150 mat
4.99 Determine the stress at points A and B, (a) for the loading shown, (5) ifthe
60-KN loads are applied at points 1 and 2 only.
a
150 numa
ta)
Loaweing
is
P=
tendyra.
180
kN
= 180x10®.N
Bad & 60-£ =lOHDY -9.s0 0° Re
= (FO\240)
=
2.6 oO? mm
= ALE xO
=-~8.33
Lf
“
i
(b)
Eccenfivc
MPa
wt
AD
foading
120 mm
120 nm
N90 snan
P
aw
120
kN
-
120
ai
(Goxio®
At
_
Gs “FE ~ se “~
MB
6p
Problem
o
%
es ~ Goon
Wve)
Ne = ON eee
%
2.0
*iO° Nem
to)
GO KN
150 mm _
:
xfo?
:
boa
3
A
=
ro
num
At
:
ing
Cb)
.
i
lee
120mm"
~-P
S90 mm
Cc.
Me
2 _ boxio®
SP mR 7 FF
ce
A and
Eccentric
M
i8‘
Cb9@24)
Es
120 mm
~ 15.91.40" Pa =~9715
MPa,
7 P8CK/O"Pa = 486 MPa me
oy
4.100
Determine the stress at points A and B, (a) for the loading shown,
(b) if the 60-KN loads applied at points 2 and 3 are removed.
4.100
MGkN
BON
ay
AR
01% «107 )»
fg bh® = & CoN 240) = 103.68 x70% mm! = 103.68 x10 ms"
.
A
«fo*
tie.
w
=
P
~
9162/6
B
- oe
=
18s
k
Af
a
=
-
3
G=-83
Poad ing -
P=
Mea
60 kN
= 60\@IB) = 9 kNm
& bh?
=
dor
9 ¥o.z4P = 163-68 Xto
-6
in
= alzm
(9x01)
Weenie” jodaxee7__ 3)OITM Pa
Me on«-£,Rs MeBE = — foxto,
Siig fdaenorla) = Het A
Oya=— JBIh 2h2Mha
Gs= 76 Mfa 76h
Problem 4.101
4.101 Two forces P can be applied separately or at the same time to a plate that is
welded to a solid circular bar of radius r. Determine the largest compressive stress in
the circular bar, (a) when both forces are applied, (b) when only one of the forces is
applied.
Por
a
eotid
erreutar
Compressive
section
etress
S
>
Aste
x
- -F
we
(2)
(b)
forces
P ,
apphied.
force
One
Both
Gr-
alll
.
wr
apphied .
ie
ey
arr
Ms.
. 4M
orn
FrAazP,
M=0
Py
Me
4Pr
C=-~SP/ny?
ryt
<a,
4.102 Knowing that the magnitude of the vertical force P is 2 kN, determine the
Problem 4.102
siress at (a) point A, (4) point B.
A=
Q4)G8)"
432 mm?
= Y32x¢I0° wm”
= BAW = 1, 664%107 mm
4
c
* (ig)
i}
=
3018)+ 40 = 4G mm
{8 inm
40 mm
we
:
12mm
e
(b)
Oe
j
Fi-o
+
4
Gy
= O.009
mM"
m
= 0.049m
12 min
Mt
(a)
= Imm
11,664 x10
Mc
—
oe
TL
*
oe
2%1Q3
A
*
4 (13
Y0.004
432mlo-* * Weed xjorw?
--P_ Me ~_2x!0*
Eo
Pe = (2}0*)(0.049)
432x107%
_ (@8\o.00%)
1é64xlo-
= 99 Nem
-
= N.oxlo’
_
Gy
9
Pa
= 70
MPa. <<
- * BO, 2% 10% [%
6ea7
-%0.2MPR,
<a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
~
Problem 4.103
4,103 ‘The vertical portion of the press shown consists ofa rectangular tube of wall
thickness t= 10 mm. Knowing that the press has been tightened on wooden planks
being glued together until P=20 kN, determine the stress at (a) point A, (B) point2,
Rect emaud ew
cutout rs
GOwm* 4O Wate
LD
&Go)(a0? - Feo eoh = 4,84%/0% wom!
=
“ofa
200 min 801m
Ms
(a)
(b)
ge By
Gp
A
Se
1
Me . Roxlo*,
T
-
mm
OL.O4UOm
C= 200440
ygaxlo®
=
m
gp 4
MPa =
eg,
FA
{o
G6.0
~
lore
L.34<
Pa,
CGa=tl2.8
20x” _ Y.8riot Moot _
¥(o®
O.240
to.
12.710
.
i
2.4
24mm?
Nem
G.elo* oot) _
24vlor-
TT
A
$
YO my
4.Bzlo*
La, 240) =
s- Bi. Me.
C.
[g4xlo*
Section @-a
(20 x10
=
Pe
m
24x10"
mms
Ar (Bo)(60) - (GoY4e) = 2410
some?
-
Gg =- 96.0. MPa <08
Problem 4.104
4.104 Solve Prob, 4.103, assuming that f= 8 mm.
4.103 The vertical portion of the press shown consists of a rectangular tube. of wall
thickness ¢= 10 mm. Knowing that the press has been tightened on wooden planks
being glued together until P= 20 KN, determine the stress at (a) point 4, (6) point B.
Rect
congu
cutout
few
HY
*
64min
1s
him,
.
A = (80\(60)~ (C444) = 1984/07 rm
1.984%/07°%m
=
Renee
200 uu
seationsa
Ls
Sf
sR 1A
gy l6ol(eoyGG)
SO min.
B= ROOX4O™
Me: Pe = Goxlo* Vo. 240) = 4.8% 107 Nem
|
)
>)
*
P
See
rere
,
qT
=
Me
ter:
7
7
Fe
a
20¥10°
oenrnimrmemnimner
ta
L484x(o%
SM
ayy
f. Sq 281
x
lo’
mw’
.
ot
(4.3% cementing
107 (0,040)
pean yer
z
Z4O mm = O. 240m
.
xo rat Looe
254
aah
_-.
.
= 1.59881 }0% an
=
C= HO mm = 0.004 ty
iw
3
S488) % (O"
1B 0.2410" Pa
f
110.
oO
.
‘
x10" Pa
627-100
MPa—
4,105 Portions of a.12 X 12 mm square bar have been bent to form
the {wo machine components shown. Knowing that the allowable stress is 105
MPa, determine the maximum toad that can be applied to each component.
The macimom
Ss
ws
stress occurs
at point
RB
=~ 105 Mpa
~-~k~/§
Met ._2_ Pee where
Ke
7+
= (0-012) (0-012) = 44810
A
Lt
EC
-b
,
pn,
& Goa) (oor2)* = +728 x! ot mt
i}
Problem 4.105
=0:024
{QA)
mm
CC FO006
+
= enna
K
Tibxiot
902778
P=
o+olZ.
a
for
SFUALS
vbh
T=
a
rs
the
centroidal
ame
anes,
£729 xlo™ 1
»
+ 222#\(0 00848) = (24722:
K *ighxire*
P
Lr
U6aN
om
TL O10 REF
a
C=
For
2
* 902778 m
Se.
ee = . Cresxi®)
~
(ob)
.
.
(0024) (+006 )
>
~
. C&
OR
728 «lot
12262
“= (-po5x10) i
(24 722-2
P=
842N
<<
Problem
4.106
4.106 The four forces shown are applied to a rigid plate supported by a solid steel
post of radius a, Knowing that P = 100 KN and a = 40 mm, determine the maximum
stress in the post when (a) the force at D is removed, (4) the forces at C and D are
removed.
For
a sottel
civcodar
A= Wa®
(cv)
3P
My,
“
Fovees
f=
=
=
of
=
D
Se
My,
Z2P
D
At
Answers?
(Qi)
t
. _ 2P
are
=
0
4a*
2
M,
TQ?
moved.
-
=
Pa
rs
Moe f
aouf te
P=
UP
GPa)-a)
M,
Me _ ~&P
fiPanw
G6: -~.Ff - g -._ 5S:
7S _ Ta
Numericead?
=
removed,
Po
bending
Resoftand
FE
‘fs
Ta?
are
100xJO°
N
(HY¥iooxio®)
M,~
= 42
_ _ Sa
244k
OQ
=
—
Oa
=~
‘7
Be
P
Pa
- 2.487 Pla z
OLoyon
= 79.6% 10° Pa
'
:
|
(by @=- DELI! = 139 axto8 Pa
©
&
7
at
Po,
.
C
Ly
er
Force
. SP
. Mez
e=-~-E
(c)
=
a
g--£ 2. ae
Co)
of pedrus
l= fat
Centric
—
F=
section
~T9.6
MPa
~<a
137.3 PR
3
Ee
= = 152.3
¥l0” Pa
~I52.8MPa
<a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.107
4.107 A milling operation was used to remove a portion of a solid bar of square
cross section, Knowing that a= 30 mm, d= 20 mm, and oy = 60 MPa, determine the
magnitude P of the largest forces that can be safely applied at the centers of the ends
of the bar.
A=
ad,
=tad®
c=gd
a,
ex 3-4
i
S
£-
p>
6oxto*
3la-d)
ad=
j t
.= oy
wm
¥
=
a?
wt
P= 14.40 kV =<
[4.40 x10" N
~
“Bice7v[os
K
K
dis 20mm= 0.020
(B)CO.010
f
=
G+ Se
where.
KP
0.0%0m
Z0mm2
az
Ontas
_=
ana)
~4)
3P
+ Sb
o- =
6Ped
P .
.
Me
ec - fhe:
4.108 A milling operation was used to 1emove a portion of a solid bar of square
cross section, Forces of magnitude P = 18 KN are applied at the centers of the ends of
Problem 4.108
the bar. Knowing
that a = 30 mm
and oy, = 135 MPa, determine the smallest
allowable depth d of the milled portion of the bar.
PQ
ad El
ov
Solutug
Datat
fon
d,
a= O0.080m,
d=
3
P=
IBxio*N,
= 16,04 %fo-*
|
4* Gas wir) Ail
Bono
)*
~3P
+ 02 PO
& =
3
3P(ad
. By
ad*
aa
rads
| BEY
NUR 104)
asad
qod*,
r ©, Pe@-dtd
se
'
-
Q
F&F. . Pes
it
MH
ear,
Res
o- $-4
u
&.
ks
os
’
20
= ae
¢
135x]o%
ey
| t (RMB vIO (13s 1108)
Pa
| 6A fo?
0.036
d= 16.0% mm <a
Problem 4.109
4.109 An offset A must be introduced into a solid circular rod of diameter ¢.
Knowing that the maximum stress after the offset is introduced must not exceed 5
times the stress in the rod when itis straight, determine the largest offset that can be
used.
|
| For
centric,
Poadi , nq
For
etcentrre
|
Gi
P
A +
+
Phe
.=
°F
Phe
Sh
=
S
et
S
+
PL
A
Phe
7
6;
-P
bh
..
A
Problem
Se
fou ding
wethine
1= yk
aa
warn
On FeP
-
tf
CA
. 4d"
ac) \e 7
>
La
ed
2
>
|
O.500
few
4.110
An offset h must be introduced into a metal tube of 18-mm outer
diameter and 2-mm wail thickness. Knowing that the maximum stress after the
offset is introduced must not exceed 4 times the stress in the rod when it is
straight, determine the largest offset that can be used.
4.110
d
c= td = Tamm.
q=
C-t=9-2=>
a
A~ W(c?-c)= n(4*—_
=
leo. cme
= E(c- co. j= E(t 75)
=
Fow
4
Been
.A 3
he
rT
on
Ey
A
h
=
Fox eccentric
f
6.7
fouding
ceutrre
Ir
The
ot. -
Ae
-
ye
city
6
eX sn61- 3x0
)( 0.00%)
CfooeS X15"
a
3267.3 mm
doaching
Bee.= £
_
h = loe$
+ Phe
x”
|
fun,all
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or.
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers.
and educators permitted by MeGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
4.111 Knowing that the allowable stress in section ABD is 70 MPa, determine the
largest force P that can be applied to the bracket shown.
Problem 4.111
:
eer
H
i
A
Hy
~
a
30 wni-
=
a
24 mun i
i,
~
isnml
CS
1
where
P=
KK = x * a * ser
a
K
.
Jo xiv"
8680-6
(103\(o1024)= 12x16mn?
2, (0108) (01028)? = 34-56 x 107m"
pt @024)
= orol2 m
= 01045 — U02c
2 o-o2zim
a ee
, ezti\@olZ)
+ Sst
word
~
Lact
8680-6
in
P=
G06 kN
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
aati,
Problem 4.112
4.112
A short column is made by nailing four 25 X 100 mm planks to
a 100 X 100 mm timber. Determine the largest compressive stress created in
the column by a 64-KN load applied as shown in the center of the top section
of the timber if (a) the column is as described, (6) plank | is removed,
64kN
(c) planks
| and 2 are removed, (d) planks
1, 2, and 3 are removed, (e) all
planks are removed.
(0.)
| Centrica
Joacling .
Ap
LA
(b)
=
.
17800
:
=17500 mm
e
_ 8:93 min
|
“
(3)
(50\(fe0)
= 1S@oo mm
62-5
x,
Joo mm
M=Pe
UV) (oo)(25)
= [ae
= $0
6= tf
tl
_ Yes.
a
= 12 e8Kr0* mm?
|
mm
NA
Somm
ii
v
WZ
Sz -43MPq
Poad ings
A = (f00\(100)+
Cf
ay
G2-2
~ Seto"
Exccentere
.
O = ~ 526 MPa
(2) Centric Poading, Mzo
l2einm
zy
bh Geodesy’ + (odes) (53-15)
3x10"*)
DPOF x1p°$
A=
<a
,
64ul0F — b4yio*) (a 93x10 (58.
6
ROA
_ ES \edlers)
A
-32M~a
P _ Pec
-K 7
C=
Pe
M=:
Ose \lroo\? 4 (150\00\(0-22) +
oe
Z
G+
-
17500 xie
rl
pe
6=- sexier :
= (te0\(00) + (AY 25\(¢00)
2b-0S x70® mm*
GO =
om
= 22104 mm™
A
ZAY
= ~*
A = Yood + VG
doading.
Eecentre
yo
IT = S(L+Ad*) =
M=0O
pk (100) (128\° = 16-276 x10 Smt
_
_ b#% (0?
(64%1 04 ) 00125)
125x183
16«276K107@
Joe
Ss
cn
+0625")
GO 2z~&Z
(e)
foo mty
>}
o0minl
fo?
1, ~~
LLL
Centric
Poadins.
Mz
O
Mfa
atl
< — 64MPa
=e
=
S
_ £
A
A= (/00\ (oo) = 110% mm?
64
l
x{O
C8 Ra
2
-
Problem
4.113
4.113 Knowing that the clamp shown has been tightened on wooden planks being
glued together until P = 400 N, determine in section a-a (a} the stress at point A, (6)
the stress at point D, (c) the location of the neutral axis.
10 mm
f° ponerse *|
4mm
|
Locate
centyo ia.
S12
3
6
@ |
oa
or
Y, - 123
72D
13
Ho
, ©
—
= 1846 my
1232
lo
The centroid dies 11.846 mm
Eécentyreityt
Bevwling
A =
lodmm
=
[04x10
couple ;
M=
Pes (oo) (~s8.1s4xlo*>
=
=~ 23.262
Nem
m
1.5632 «10% mmt
(a)
at
Stress
point:
My_
_P_
CaF RE
()
Stress
Yoo
Jowsioe™
at point D.
~P,
On
My.
A 7
OF
11.3846
2O-
ye
A.
mam
B.1IS4
=
| (23. 262¥aiso®)
y=
400
mm*
23120 x10"
= 3.8802% ID” mt
3.3802x 10% mm!
©
DT +In
above point Dy
@ = (50+20- 1. B44) =-SBIS4 mm
T= kbs Ad? = ROY + Gol(E.isdy =
Toe oe bh t Ady = RCs? 4 (4X8. 346) =
D=
322
Ayo, ram?
Yes mm
A, mm”
Pert
Section a4
3802 10-7
_
vn
%, 1S4xjo°
=
SRTHION
:
Pa
MPa,
Ca = 52.7
=I
B46 mm
=
~ .B46%iO%m
_ (- 23.2621.
B46 e107)
lo4xtor®
3, BRO2«
x 10
Jor1
7.2%
10"
Fa
6, =-67.2MPa.
c)
C=
RP
7
My
_
KR - >
J
rc
Tr
Ae
oe
=
P
A
~
Pey
© ©
L
Z.8802% 17
Cod xfoOes8. 154 vjo%
Pies
[1.846 -0.642
=
ae
.
=e
=
Neutral axrs
<a
G=0
axis.
nevtvel
of
beeetion
~<a
11.204 mn
0
642
-O,
642
x10
|
m
rn
He 20 mm
adove
) <4
4.114 Three steel plates, each of 25 x 150-mm cross section, are welded together to
Problem 4.114
form a short H-shaped column. Later, for architectural reasons, a 25-mm strip is
removed from each side of one of the flanges. Knowing that the load remains centric
with respect fo the original cross section, and that the allowable stress is 100 MPa,
F
determine the Jargest force P (a) that could be applied to the original column, (5) that
can be applied to the modified column.
5
~ 6A
Eecentyic
i”
L. 125% 10°N
webs
2.75
@)
.
\
:
t
tentrorel
12 eb
i,
=”
a
3
ky
b,
Pres
be Ad,
hes
lOL.937S
mm
16.5625
O
10, 4375
|-37, 5
|-213.75
43,4375
109,375%107
~
jo.coxto®
Fyrom
t+he
mtd
_
~
10.1375) mm
polut
of
+he
web,
HUSO\UAS)+ (8.75 «107 (16.5625) = 29.177 10% mm"
=
de (as iso)’+ (3.15% 1 )l0.9 375" =
‘
7.480% 10% mau!
z
Ts? ix bhs 4 Aya =
A(loo (ary 4 @. 50 %tF V98,4375) = 24.355% lot mm”
i=
>
I,+1,4
1,
C=
(0.9375
M=
Pe
wheve
6 = -=&
Ss
a
pa
L,
=
54 01210
+ 1S + BS
SS
@6-
=
)
=
an
Ae,
Mh ba
ay
—s
\09. 275
SAF _
FA
(10° mn )
section
O
Zr | jo.co
~
The
Ay
evess
87.5 | 328 ies
2,50
Y=
Ie 25*IO om
PeHQa5 kN
C ceduceet
Yo win?
@ | 278
mid pont
ot
foacding
10° mona
OQ |
+
aw
= =(-]00x
10°) (1. 25 x Jor?)
mine
_
=
VL 2S«1Ot mms
(2)US0)(25) =
A=
(b)
S
Junding
Caatric
)
|O.4375
p ~ fee
Tp.ooxiost
122.465
= S54. 012 «10° w
~¢S
= 1OL4BIS
wom
@r
-Gigenie)
wm!
=
mew
=
=
O.1/09387S5wm
[0.4375 x/o"°
ro o- KP
yr
A = [0,.00%10
wi?
(lowa37s% 10? (0.09375) _
RI7MIO>
S4.01a *10O-*
No.
~
Pe
122.465
BITkn.
m
-2
ml
Problem
4.115
4.115
In order to provide access to the interior of a hollow square
tube of 6-mm wall thickness, the portion CD of one side of the tube has
been removed. Knowing that the loading of the tube is equivalent to two
100 mm
equal and opposite 60-kN forces acting at the geometric centers A and E of
the erids of the tube, determine (a) the maximum sttess in section a-a, (b) the
stress at point F. Given: the centroid of the cross section is at C and
i= 2% 10° mm’.
Aven
Ax (2)(foe)(6) + (84) (6)
= 1728 mmr
M=
Pe = (60)(d0S-0:636) = 08% EN,
Moget mum
where
F
|... 100 min-»|
Cina
:
s
stress
Y
P
KR
=
jog
occurs
aE
=
of top of section
6Yy-
M
boro?
ui
i728xe?
+ LY
rue.
(tt) (0106)
-
Zxlo “e
Cora
(b)
At
point
E
Po.
Ge
zoe
A
yr
+
<<
a=Zbnm.
My _ oc _bewo? ,+ - (P¥o)(-01034)
Tt
= Gle 64Pa
28x
ha
£107©
-
=
ar, j
Op = 1% 6 MEa :
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ae,
Problem 4.116
4.116 Knowing that the allowable stress in section a-a ofthe hydraulic press shown
is 40 MPa in tension and 80 MPa in compression, determine thelargest force P that
can be exerted by the press.
Like
25 mm
|
[=-250 mm
Locate
250 iim
25 nm
centroid
E |12So0
LY ZIS7S «107
centrocd
2.4 2187S % 108
193.78 mm
Yo 2 SSeS
The
262-5 |1 e4o628 ~io*
QL S250]
oe
~
Iies at
peint
T= Hob +A a’ = LASIa5)*
=
vy
62.093 “jo
T=
==
T=
A
ing
mom.
ent
o---tY
Stress?
Then
Sz
:
24.367
12500 mm”
eee
pressive
K
about
=
£4
4o
x /0°
—_
P=
Choo se the
mm"
.
7 125 x407*3
Cr 300 425
the
centroid
4}. 60x10 «ww
m*
4250 -193,75
Mo:
is
= 331.95 mm
~ Pe.
K=x+
where
Kp
-
Fey
;
Pa.
ys
91.960 «/O~¢
12.5xto°>
P=
Com
¥ {0°
(338b2sxio Mabwxtio*)
-!
Ke
y
brea
St
P= f
stress
Tensile
+ (orso
oa 18."
6 bh: +Ads-= A AsoNas Y+(6zs0\ee.75y"
Becentoicsty
bene
C.
Ti+ I, = 82.194 «10% mm!
=
vam?
| anes xio*
Ol ezso | BS
clea rsletigres|
Ay. »
m
Jo,
Aj rom”
Ye
Section a-a
Dimensions in min
section,
.
c
The
of evoss
|
i
12.5 «105%
=
x io? vn
HIE BS
m
.
96.8x%10" N
¢
stress
$1.95
a
7
= =80x108 PA
y= -193.75 x10? m
(381.2510 E-193.75xl0*)
a
2
om
728,25
=
=
1.90% 10"
=SexlO = 110.6% 10" N
strabler
value
P=
%_o xto*
N
Pe
96.0 kN
Problem 4.117
4.117
The four bars shown have the same cross-sectional area. For the given
loadings, show that (2) the maximum compressive stresses are in the ratio 4:5:7:9, (6)
the maximum tensile stresses are in the ratio 2:3:5:3. (Note: the cross section of the
triangular bar is an equilateral triangle.)
a
Stresses
6, -- 5 ~ Bea
ATA
BES)
SRC
See
aheats)
-
)
-@ =a
Cr Cet $a,
Lewa',
Aza,
ee
ACs
= ral
-
-44
.
Pp
=
2 Fe
Armes a se: # ,
. , (a(meee)e) : - -5F
6°74
Rm
con B(OH
5,2 -£
Sa
Ay
G
°e
=
&
I
Ba
cz
Qe
A,t
Me
a z(s\(Es) r 3B s*
=
ees
&
(1+ Gane)
BI
(8
as@s}
G+
Ct
Eh)
ae 4
)
)
<a
=
_
~
at
& £-
_
He
Qt
2 e
)
Ye).
a
=
ere
aa
-7§
(1 le ltallde)y
ti
|
28
-\) =
Rae
\
Pecs
4
-E
3
Oe,
AB
A ®Ca
-£
=>
*
Bs"
ae
-9F
<
3£
~—_?
Problem 4,118
4.118 Knowing that the allowable stress is 150 MPa in section a-a of the hanger
shown, determine (a) the largest vertical force P that can be applied at point A, (6) the
corresponding location of the neutral axis of section a-a. . -
Locate
t
°
¥
40 mm
80 mim£
7
¥
@
20 tom
Aw” | Jomm |AYormn|
© |
@®
1200
1200
oP
Z|
2400
60 man
The
t
”
40 mor
So
Io
5 BAY,
36x10%
84 « fo?
o
— H2Ox 703
centres}
Pres
SO
rw
te
=
zh
exe
=
SO mm
the
tra ht
of
3
+he
section.
edge ol the
AetHt
|
- L
ceatvord.
z
— §0) pmb
Bending
eB
Section a-a
I, =
Qo Yeo
coupdat
M
Pe
WO
=
40450
>=
=
me
=
O.0OFO
w
(zoo (20% = 840% 10% mm’
Tae ok (6020 +(1200}20}r = 520 «10% man"
to)
Basecl
=
1,
T+
Te
on
tensrfe
o=7 -EY
Pe
(b)
+
Location
t.260x«Joeum
=
stress
KP
at
The
=
40.3x10°
of netrad ox ist
. Bilas vis \00-90) =
avis
Pies
DPef4 ed ge:
A= 2400 x10 * m™
Y
= ~SO mm = -0.050
|
ee
© Peete
neutvad
1.360107 8m?
|
-
y* ie
=
6.30 mm
C=
NO
=
40.3 kN
~d
o
;
6.80 110"
to the right
Op 56.30 mm From thle deft Face.
9
|
m= 6,30 mm
of the centverd
|
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.119
4.119 Solve Prob 4.118, assuming that the vertical force p is applied at point B.
4,118
Knowing that the allowable stress is 150 MPa in section a-a of the hanger
shown, determine (a) the largest vertical force P that can be applied at point A, (4) the
corresponding location of the neutral axis of section a-a.
Locate
[.
|
40 mm
80 mw
.
|
20 yan
SAY.
y
Ayo mm
A, mm | Yo,mm]
|
|
"40mm
®
{200
30
@|
1200
70
84x 10%
Zl
2400
—
centyore
the
_
'oxlo®
“=
SO
120 ¥ Jo?
aito0
60 mm
The
TA
&
26103
Dd
2
centro'd.
Ires
2&bed
SO
edge
of
mm
to the
the
am
waht
oY
section.
60 mim >|.
20 mm
on
s}ress
at
DeFt
o= EF ~ Rey.
X
Ko:
P=
Based
nn
-
+-
ey -
&
.
KL
on
Raoonee
-~!80orfo e
ad wight extae
a secHon
Cheose
(b)
A
Location
Ss
epsnassaapanssqanamapayenansamayenr
~
a4u00x
ISD * 10 ¢
3
* geese
the
smadicr
of
nevtre /
Pey
6: f-5&
.
|
=
L_€y¥
1
.
Pies
?
_¢o,070\o.980}
1
-
y
SO mm
=
0.5350 wr
ey
x {os
SO-
.
+
4608
{oO
4 2
m
.
M
N=
64.6~1O
P=
w
~
fo-*
67.6 kN
ory
G=oO
Sl=
=o
*
76.5 «loS
value
.
le
~
{.360
1o°@
*
axrs
___1-36o0
=
VY" BRE * GiooxtoF}Co.o7e)
Nevtred axis
N
KP
~
Pt
.
~ 2.15697 X107 wr*
1.360% (o-*
64.6 *Io°
nN
y = 790 man = = FO m
0.070)- 0,080)_
~
+
-
al section:
_ ©
—!
7” =2.1869x1o3
stress
=
eolge
KP
6 R- EY:
Ke
|
a
”
4
hy
¥E
Baseci
th
m™
1.360 %/0% wat = 1.360%/0% m™ > A = 2400710"
=
T,4 I,
7 O.070m
=
mm
-~7O
JRej}* = 520% Jp? mm
Ge Qo} + (1200
Tar
(2)
= BYOx 107 mm
A(zoi(éo¥ + (i200 20>
=
50) RO
er
Pe
Mes:
coup det
Ben cling
Section a-a
8.10 >
.
.
~
~ BIO xlomm
AILF
OO
*10°
= = B.}O mm
wm From Deft face,
~tt,
Problem 4.120
4.120 The C-shaped steel bar is used as a dynamometer to determine the
‘magnilude P of the forces shown. Knowing that the cross section of the bar is
a square of side 40 mm and the strain on the inner edge was measured and found
to be 450 yz, determine the magnitude P of the forces. Use B = 200 GPa.
AL
the
stray
492
Vocasdticon
.
G > Ee =(260x%0U( FSOKi0
=
40mm
.
vo
A=
(¢0)( 40)
:
I
@2=
CF
&
~> P,Mc.
A+
7
LL
K>
yt
-
P
ec
Ry,OK
7
_
LC”
ftd}0
fe
f
£
FoodexlOF®
_.
K
Lo0to
=
|
=
a
(oo
9/3333°3 inet
nm,
hun.
(oN
fée0
x; 0°
*
=(b6c0 mut
i (40 W4oy
Pe
TF .
+
90 MPa
~ by
KP
|
Corgz)
213333-3xio'*
GoooN
.
~ foooo
22
mM
P=7k&N
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the liniited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
«dt
4.121
Problem
4.121
Ashort length of a rolled-steel column supports a rigid plate on
which two loads P andQ
are applied as shown. The strains at two points A
and B on the center line of the outer faces of the flanges have been measured
and found to be
€, = 400 X 10°¢ mm/mm
€p = ~300 X 10°Smm/mm
Knowing that £ = 200 GPa, determine the magnitude of each load.
W
Ga = E&
*}
gi
E &,
gages
B From strain
and
at A
Styesses
(Zooxro!)(-400 x10 * )= —Fo MPa,
(200 410° X-300 xJO*)=
—
bo MPa,
i 250 nem >|
Centre
force
Beneling
A = 6450 um?
Cs
lat
Ff
4 Me
‘
-
Fe
P+
Q
M:-6P-6@2
cou pte
mm.
= 114 x 108mm?
=
6,
an
.__P+@
+ (oP
~0.15@Q Yoizs)
UY kro”&
b4a0 x10"
Vy
Porro = WE PH 3/9L GE,
= "e.
Ge
A
°
~b6oXie
and
(2)
tr
Lys
~ 219 P
_ @/YP-o/TQ)C?S)
Nex sor &
Ko”
£9
a
QR
simud taneous by
P=
(os-y¢ kn
b
Cl)
=
._ P+Q
Qs:
o5bed kA
A
Solving
&
Me
Proprietary Material. © 2009 The MeGraw-Hilt Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
4.122
4.122
An eccentric force P is applied as shown to a steel bar of 25 x 90-mm cross
section, The strains at 4 and B have been measured and found to be
25 may
&4 = +350 x
\
Ep = -70x
Knowing that £& = 200 GPa, determine (a) the distance d, (6) the magnitude of the
force P.
30 mim
ht 1S + 4S 4-30
7 90 wm
b= 25 mm
90 mt
es
=
Ya 7 €O- 4S = 1S mm = O.0ITm 5
from
stvain
E@,:
G,
=
of
K
and
ra
»1o7*
7ox/o*. Pa
Pa
ue
ON)
G,-
6,
=
|
e)
7 HS
.
ECG -6e) _. _ tis reasxio)(84«105) _
°y
(QQ)
- Yeon
~ 2885 Now
nog
Mens
wes
m4
B:
(200r10' )G70 jo * )=- 14x10
Subt racting,
~~
gages
1.51875
yg % IS- 4S = - 30mm = - 0.080m
(2007107 )(as0xj0"°)=
+ Ee,>
Gp7
= 2.045 m
A= bh -(25\(90)> 2251S
mm = 2.250"
bh + & @svaoy = 1.51875 «105 am?
15mm
Stresses
th = AS om
=
Va
(4
aud
a)
by
Je
an
ol
subtracting,
-
- ye) =
pe. ACSa-yG) . (z 25 x10°)\ (o.018)(- “i410 - 0. ose
=
(co
lb
70-105
0.04
Ya- Ve
94.5 *1O°N
M--Poa
2
de -2
- Po
mm al
d= 30.0
= 0,030m
P=
F4.5 kN
<1
4.123 Solve prob. 4.122, assuming that the measured strains are
Problem 4.123
é4 = +600 x
én = +420 pt
4.122 An eccentric force P is applied as shown to a steel bar of 25 x 90-mm cross
25 mn
section. The strains at 4 and 2 have been measured and found to be
&4 = +350 pe
ep = -~70 ut
30mm
Knowing that E= 200 GPa, determine (a) the distance d, (6) the magnitude of the
force P.
1448
We
90 y
Yo: vam
=
+30
=the “4s mo?
a
b = 245mm
Ya?
60-48
Stresses
=
IS
6, = E&
Qo
Se
-
7
P
=
at
x to
= S182
15 +45
= = BOmm=
Sm
-0.03Fm
Aavnel B?
= (2ooxj0')(600 «jo } =
120% 10° Pa
)=
Fe
B4xI0%
7 >
m
M
BEB.
M
_
—we
A
- nla
Se> . _Us1a2sn10" V(ge ¥ 108 ye
Ja- Ye
Mo Dt: phy i me
(2)
Se - yeS
YA
by
ane
Ci) 2
= (Ya - Ye ‘
_ (2. F
and
Ye
)
subtracting
oflone \(e4x1o*)~ (~0,030)(120 ¥108) |
an
=
RIS Nem
6.0457
6a - eG)
p- A
(2)
me
Moye?
- Sa
6,
Subtracting,
ws
ye =
0.015 m ,
gages
2.25% 19" vane = LAS KIS™ mn”
= 1.51975K10 mm!
0x1a
10" \(aow
7 (g0
E&s
6,7
mm
stray
From
GY
EQS
T= kh’:
bh = 45900)
"
A
1 mn
O. 4S m
O.0#sS
ag ere"
N
M =-Pdl
|
(as
(b)
-
So
d=
y..M_
PP”
-12Is
” Fae ¥ To
7=
SrJo
“8
om
d=
Ps
.
5,00
2H2kN
mm
ts
=
Problem 4.124
4.124 The eccentric axial force P acts at point D, which must be located. 25 mm
below the top surface of the steel bar shown. For P = 60 KN, determine (a) the depth
d of the bar for which the tensile stress at point A is maximum, (4) the corresponding
stress at point 4.
b= 400m
A = bd
czutd
g
\ 20 nam
oeia
(a)
Dept
h
(
ry ew gt as Pet
Ss
a
4
-
foxjo-> ( 1S*1o%
Problem 4.125
bla
;
-
a}
eveHade
A=
= o
es
hs
a
hee
LS 4 _6af
rr
Gox1o? §
Cn
‘
inked--a Gd)
2(gd
Pow
£5
be
ts
dd
b)
a
=
T=¢bd?
ezdtd-a
(f)(25x103)
with
yes pect
+o
of.
—e
= 25 mm
3@
,
= 4WOxu.lO é Pa = 4OMPa }
C25 ¥ 1°53 }*
MP
—e
4.125 For the bar and loading of Prob. 4.124, determine (a) the depth d of the bar for
which the compressive stress at point B is maximum, (4) the corresponding stress at
point B.
4.124 The eccentric axial force P acts at point D, which must be located 25 mm
below the top surface of the steel bar shown. For P = 60 KN, determine (a) the depth
dof the bar for which the tensile stress at point A is maximum, (4) the corresponding
a=
25mm
stress at point A.
P
=
ta
Cc
20 min
-«\bd) pe
- PS
42 _ (aadta\bd se\ed)
Soh
Ef R-Ge]-0
AS
(b)
do
maximum
for
2B.
6 6 = YO62x10"
x ton > t {505
*fo"S
td
-
Po bec
Se
@ Depth
eat
=
A=bd
Le
Gg:
r
P}-22
+ &2ca
ae certo
DifPocactinte
-
with vespect fo
~
a,
ws
a ; = -IOMPa. a =e
(ASO
a
(iso «jos)yt |. iovio'P
Problem 4.126
4.126 through 4.128 The couple M is applied to a beam of the cross section shown
.
ina plane forming an angle # with the vertical. Determine the stress at (a) point A, (4)
y
- point B, {c) point D.
T= GMI)
x {0% mm = 4.2667 e105" mt?
I, = Blloo)(go}4 = 4 2667
|
ya
OBO NNo aysm
MMoc:
= 250
401mm
D
_.
40 mm
*7Ye
;
Bye
My = - 250 sin 30% = - 12S Nom
a.
fe)
A
Ss;
-
Maya,
i,
+
MyZe
Gh
_
a
2
_.
Mave
TT
Ga - Meye y My
>
~ Yo
t
~ 20
(4116.61) (0,050)
6. 6667 xia*
2 Bovio®
SO mn
=
.
YOmm
ob
|
(-1as CO. OF0)
4 2667 « 1O7*
Px
- 2,80
MPa
wnat
(1as) (0.046)
267 1B
0.452107 Py
zo. (216.81
L
=
M, = 250 cus Zo" = 2G. ST Nem
eee rete
O.YS2
MPa
=x
2.30 MPa
<a
C0.050) | Gas C0, o+0)
66667 ¥ {0%
i
eT.
.
Za
My Ze. _ (216.51(-0.050)
y
Oe
(c)
= 6.6667 210% wm” = 6.66C MIOms
2.30%/0°
Pa.
A 2667
fons
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights teserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Probiem
4.127
4.126 through 4.128
The couple M is applied to a beam of the cross
section shown in a plane forming an angle 8 with the vertical. Determine the
stress al (a) point A, () point B, (c) point D.
M=45N.m
4 lo) (30)? = 22500
I, =
Z2,a
=
4.8 tas 60° >
LOS
Mey,
(a)
Oy
2 = Wlgy%
i,
°
7
~Ze
tr,
Gye
~ Heya
rT,
a
7
Tf
= _
+ My Ze
Ty
-Z2y
MF
=
(b)
=
Am y
MyzZ, An
~ Yo
=
Yo
>
ya
My
7, Ge) Clo) >
z
Ly
2 ZSvo
FC
fatm
ma’
WM
= (bU2)
> Bes
(= B9)COCI8)
ZESOO
YX It
M Pa
FAY (01S)
BBL OOK OTe
.
%
=
fmm,
gra 60°
=-
29M:
Cz. 2) (ervey).
pSCORIO,
2dO KF O~!2
,
(22S) o.oo)
"a kbD KT Ate
2-I9Mpaq
|
CM CorelS)
6, =— tes Myke
4%
t,
v
©
22S%0 Xia"?
-U/MP,
—a
~
(22-)(- 0.006)
vows
iff
ZL Ki gre
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, Alll rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, wilhout the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.128
4.126 through 4.128 The couple M is applied to a beam of the cross section shown
ina plane forming an angle f with the vertical. Determine the stress at (a) point
A, (b)
point B, (c) point D,
M=SkN-m
OM, = = 5
My
tte
D
62 mim
62 mm
Za
T,7
ia
I=
os G50)(250)* -~ 2[ Faas
(250)(150)° ~ al Fat]
a) Gs TR
+ Mle
@ 2a
i}
_ Cae
uf
4
bn 7x10"
3930) (0-125
/27.2K10©
Gr
Ln FRIAYO OTE)
ay
hE
oo7e)
= re72 X10" mm"
6-016 MPa
bP TKraF
o-7 Mla
__Ca2t#)(-01078)
}
& = hve
» Me
7 ~ Zp F IZ5mm
qesyGzY ]
Zz
ce)
Ente
= 697x100" mm"
Ui
Myze
3nf3
Ya = Ye = “yo = 78mm
25 mum
125 mm
Gye Mele
-
HO*
ees
£
=
sin WO? = = 52/4 kit
ERT X07 8 |
—~ both
MPa
5
602 MPa
<A
(3830) (-arl25)
tarzxi0-6
6a7 079 MPa
<e
( 3830 \(- 01125)
epanion’é
Op= - 6:02 MPa <M
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, of
distributed in any form or by arly means, without the prior written permission
of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.129
4.129 through 4.131
The couple M is applied to a beam of the cross
section shown in a plane forming an angle 6 with the vertical. Determine the
stress at (a) point A, (b) point B, (c) point D.
Locate
centroid
.
@
oe
50 men
(
©
160 mm
QD
5,
The
centyoid
& 00
250
|
COD
O
ay, MO 8 mm
”
Zp, = O
I
So
a
2,7
Ze
F LOO mow
(
cog ROP
>
OFF
I
osin 20°
=
66342 EN,
~~ Me ye
i
Gg2- Mave,
6F-
3.
£ (SB) (loo? = 28x10 mt.
(20)? +0)
£20
i,
lc)
[AZ mm
-2£0 odo
Yp > TOP Mm,
ag
tb)
pefat
dL
|Z,mm.
lpoooo | -218
Ya = — Yer 2mm,
x
a
Ty
S
at
ee
|/Sooe
ye
Lore
(Sd)? = 34-37
72 (los) ?
72 (& )(2+ 60)
Ia
(2)
Pres
Ajmm
M:Yo
T,
—
, Mize.
Mize.
(0:028)
+
at yio
Poe TE x of
2.
Cive)eort}
Ses
Te 4607
.
B42\(-0-1)
_ itoyl-ow2s) , Cynon
Jy
dy
(440)
Boe 70 400"
4;
, MyZ
ENM
3.05
Py
~
&
_ 0
OR btn
me
=
mw
20x10&
4
(G42)?
20
2-73Mp,
hIO
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, wilhout the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.130
4.129 through 4.131
B= 30°
erm
Fenge Ty 4 (200) Cit)’+ (200) (12) C19)”
}
=
2mm
Web:
The couple M is applied to a beam of the cross
section shown in a plane forming an angle 8 with the vertical. Determine the
stress at (a) point A, (b) point B, (c) point D.
T=
RCS) (226)2 79S
+f
|
$40/5260
min *
Cir) (200)? = Exeo® mm ¢.
ma?
T= (226)(8)) = 1643 hm
TteP?
I, = C2) 04018200)
+ 169SUST = TSTSOT mm”
=
a
cos
30°
sin
Bo?
j
Ss
ce,
=
nN
\
x=
~ Yo
=
Ye
y
Ya =
(20(&x10°) + 2643
op
Ty=
= [b00964¢3
=r
=
MEd
= f/4
- 2,
Zt
[IS mms
mm’
(COMM,
- 2% =
7
LANE.
kN mL
cay © =~ Mate, Mee - Geren
d (ors), ltl)(0
tT,
(hy
G2 -
M
ty
ISTIEBSI YO
Mee + Mize » Pbpsoslorbs)
«
2
3
W872 Rx
vio
-
Cy
Gp*
~ Maye
at.
+ Mies
“y
7
(HDS
1660964} x/9
(Meow)
~
fm fe
es)
IST2ZERSIN IG
=
:
Yn.
.
inty
16596
EL xg
je f
-
Oe
2127.0
M Pa
f
feve
an
ge
ou
16609643 «7g
{2-70
Ma
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. -
=
Problem 4.131.
. 4.129 through 4.131 The couple M is applied to a beam of the cross section shown
in a plane forming an angle f with the vertical. Determine the stress at (a) point 4, (8)
point 8, (c) point D.
M = ISkN. aa
T=
7mm
ECHO
DY MCAVOY = 2.7817 x10% roma”
2.7817
50mm
107% w®
= t(70\io)* -~ b(so\U2e)" = 3.g067x% 10% mm
L120
min —»|
|
yy
= 8.3067 ¥ 10" “mt
~——140) min —>
M , = Main
My?
oy
Ya
=
Je
=
Za
7
~2a
=
15° = (15% 107) sim 15°
Mees Is® = (i5x 10°} cos IS”
Ge
On ©
Neda
TL,
~ Yo
F
72
25
7
q
3.3323 ¥ 10%
0.035
=
0.970 wm
Nem
| (14.49
¥ 100.970)
89
8, 2067 x10~*
Fe
Sy
O
6
a
6:
°
2: - Maye
IL
Myo
L
»
4
-
My’e
T
My
3
Jo? Vo,035)
— (3.332
2.75 17 x 1O°%
i
bb)
~ (64.5 x]o* Pa
2
_ (3.3823*10°)(- 0,035)
dy
2.7817 ¥10%
7 = 65,.8™10°
—
m
(4, 48B9x10> Nem
2.7517 * 10°
5.8*x{O°
wm
7Omm
4 MeZe 2 _ G.3823% 1070035)
ff
A)
4
Pa
65.3MPa
<<
, C14,4834x109)C 0,070)
Z.BO67™
[O-S
Gy= -164.5MPa xt
C14,.438¢%/0°)C 0,070)
2.8067 «(07%
6,2 -65.2MPa
Proprietary Material. © 2609 The McGraw-Hill Companies, Ine. Alf tights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any forin or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<M
Problem 4.132
4.132 The couple M acts in a vertical plane and is applied to a beam oriented as
. shown. Determine (a) the angle that the neutral axis forms with the horizontal, (6) the
maximum tensile stress in the beam.
W310 X 38.7 se
For
W 810% 38.7
‘I, =
Ty=
165 mm
Ya
85.})x1]0° mm
=
85
ca
Bb
bY
x
w
C1610")
eos is?
he
Ss
(\6xto*) sin IS
Ca)
ton
a
Oz
et
J
vert
eG
FPR
Ya
=
-” Yo
=
~ Ye
= Z_ = ~Zy 2-2,
=
ISL SS«PO™
Nem
=
4 idttxto™
Nem
« BEIT
hn Os
1xio~°
.
mt
7.27 *10" mm? = 7.27%10° ont
Lice
sore
=
1's
QO
>
=
i#XS10)
2
[$5
pam
= AMS) = 825 mm
1
\
robbed stee/ shape,
ges
8.1888
72.3°
a ~ 72.3-15 = $23"
/.
~~
72.3°
Maximum
.
=~ Mave
i,
stress
, Mrz. 2
I
H
oe
tenste
+t
(b)
occurs
ot
port
(15.485 «10° )(- 155 x15")
88.1 x/0-
75.4 x10"
Pa
E,
, (heat vjo® )(82.5x10"7 )
7 A7x 107
6-5 15.1 MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers .
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.133
4.133 and 4.134 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (2) the maximum tensile stress in the beam.
J Lae
C200 X 17.4
For
eo x
Moe 2SKN-m
:
+
.
C200
17.|
vobled
steef
1.
T,
I,
14.4 ma
= O.S38
JY
=
410% mm)
4a
(3.4xvJo%wm?
Ze =<
2m t
Yo*
Ye
~ 2,
=
(ay
+:
(2.810% )eos
JO"
My =
(2.8410? Vin
107 =
tan
op
v
JZ.
tan®
y
=
2.181810"
486,21
+ BeBEE
w
ten
@
10
°
Tt
c=
=
0.539
=
13.4
7 -2y
— 14.4
=
ee
x10
x for’
£Q0R)
sn!
mt
=
[OS
mm
mm
Ye = Ya 2 SP-144
M,
sh=pe,
7 min
= 42.6 mm
Nem
Nem
12°
=
0.007074
O.4YOS56°
107.0,4056°
os 9.845
e
NA
ol
(b)
Mawimuw
tensrfe
stress
oceuvs
at
point
dD.
<3
Sh
owe Mz Yo
ih
+
Ey ze
F
aio
ne
SRS
O.538K10
73.807 y¥10%, 3.682
xto*
C486, 2L )C 9.1018)
,
13.4 *«fo
=
717.5x%/0% Pa
6p
=
77.5
MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. -
=
Problem
4.134
,
4.133 and 4,134
The couple M acts in a vertical plane and is applied
to a bearn oriented as shown, Determine (a) the angle that the neutral axis forms
with the horizontal, (b)} the maximum tensile stress in the beam.
For
Ty =
Gell
I,
o°782 Xo pant
Ze
Ya
17 sin go’ =
M,=
M,
=
@)
tan
le] cos
20°
=
Pp =
2
6
=~ 2p
>
Ye
Meni
mow
6,°-
°
shape
x10 © mint
~
=
Zea
Yo
=
2p
“Ye
&
4 BEY = 42-5 mm,
+ (S2)
=
76 boa,
0-S3] kNm
166
7 % tanQ@
LN
m
— Alb
* 6-782 tan (90°- 20°)
=
we
Oy
steed
a
2
volted
4
7 wD
1? 6b
ul
52 mn
SISOX
tensive
stress
eceurs
at
point
32.0\
p=
6B-2.°
os
$e.2°-7o°
-=
Jy
Guxig ©
<
D.
__> G8Id)Go07e) | Clhoo)(9:0425)_=
Maye ,+ Myzo
f
tL
($-2°
ort ge Kio”
fe
6M
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course. preparation, A student using this manual is using it without pertnission
ae
Problem 4.1 35
: 4.135 and 4.136
The couple M acts in a vertical plane and is
applied to a beam oriented as shown. Determine (a) the angie that the neutral
axis forms
¥=254N
,
om
emg
100 rom
Iy = 28 X 108 mm!
Sen
He °F XlO "han!
ae
190 mm
Ne
Zn
26
Sin
4S?
*
=~ 2eP
tensile stress in the beam.
>
22
Iy.= 2° Px 10° min’
PLin
—loo%12
=
.
~T7F
“loo my Yet lOom,
YA
=
(6) the maximum
Ye
1, = 8.9 X 10° mm!
My’
the horizontal,
22 mm
=
Adg
K
with
Yo
=
-(2mm
— (098k kilo.
Mi = a2 cos 45° = 1 9P Lim.
tan
7
p
= ay
dy
tan
6
4
{a>
oct
tan
(~ 452 )
(bY. May imum tensrfe
=~
i,
, My%
i
stress
-
“
tf
°
Mz Yo
— 3°17Eb.
P= - 72.5°
Oo
6G
=
o
occuvs
C19&0)
at
72.8°=45"
=
275°
ee,
point D.
(-9«012)
$F Kio
L7EF
=
ef
i9fo) (-0+ 078)
aePxjo~*
MPa
~<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by avy means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.1 36
4,135 and 4.136 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (6) the maximunt tensile stress in the beam.
\
A
M=
Ty = S3.6x10% mms
120N- a
we
53.6%/07
my”
- Gino
ly
*
|
we
1
7710
mina”
=
14.
72x
Oona
6mm
\
\
7 oy ob aeant
Psa x 10 nn!
\ im
WV
jo"
ya”
.
M3«
=
{ro
Sn
7D
>
$12.
764
Nemw
My = 120 ees Te" = 41,042 Nem
o
I mn
y
fa)
B=
20°,
y’
4
Tan
i,
=
M of 2
z
2!
5
@
a
536 x/O7
4.0
OD
132084
Vet —
Y
NAA,
>
§2.871°
+ 52.371- 20°
(bh)
»
2 a
etn
id, 79" [oO tan
and
The
maximum
tensife
7
~
9.915 ym
Ye?
~fGmm
ge
- Me e , Myze
F
t.
stress
Ly.
occurs
|
at
2,7
point
10
mm
_ WU2.763\G0,016)
53.6 = for?
= 61.448 x1o° Pa
> 824°
al
E,
=
0O.01Owm
(41,042 \0.010)
IYW.77x (or?
G, = 61.4 MPa «el
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual! is using it without permission. -
Problem 4.137
B
*4,137 through *4.139 The couple M acts in a vertical plane and is applied to a
beam oriented as shown, Determine the stress at point A.
‘|
A
a
2 afb90 \. 00)" = 25-02 x14! mnt
—F
rt
4
—t
GOO mm
|—}
ke
6
3
T=
afix (46)( 180) + (bo (180 y(30)4 277.16 xio
Fo
24(be)( 180)(30)(30) } = 19-44 x10 mn
mm
60 mo
60 pin
_t
fo
Using
60 mim 60 aan
Mohr's circle determine the
princtpgk
Yu
6
(42x00 mie,
Z2
(71: 76x10% mn, 19 BLE mn +)
Ec
(sh 84 x 108 mm*, oO)
oes
moments
&
IP 44X10 nm *)
.
oF
.
aid
princi pol
jnertia.
tan 2
Oo
bY.
1944
DE
2ZE:Fz
eT
m
26,,,= 36.89°
O, 2/8 43s?
+4
x
=
6
By” = 22uhyso° mm
K-j{DF*4+
&
Lyx
(S4:94-32- hie
2 libh wre nm
Ly = (St84 + 32-4) 10%
= 84. 24a KE
Mu =
Mar
MM
mm 4
14 sin (8.43S° = 443 EN
/& Cos (8,435?= 183 END
\y
Up, 2 O-/Z cos 18.4357 4 0:0bsin 1B.4BS"
F
2 0+1329.M
U,* ~ 0-12 sin [B.43S° 4 006 cu5 1B. 435 = 061897 my
¥a7lZomm
M
=H
Gr
a
|
4+
Ma
=
(3300) (0-132)
GA-24X10-*
-
~(bibh
|
Z,2fomm
Us
MPa
Wh
(4430)( 0151897)
19-44 xia
:
Oy =~ 168 MPa
<a:
Problem 4.138
*4,137 through *4.139 The couple M acts in a vertical plane and is applied to a
beam oriented as shown. Determine the stress at point 4.
A
ng
;
40 fim
Dsin
—_ 10mm
axes
me
m
OL,
40aamm
nan.
LO am
<7
mme|
Mohr's
anct
the
my rode
;
,
deterwine
prone “pad
the
moments
of
prin
ipa
mertra,
/
|.-10 tam
I,» 1.894 x 10° amt
1, = 0.614 « 10° mmt
I, = +0.800 X 10° mm
Y (1.894, 0.800) «10° wr”
Z (0.614, 0.805) 10° wm"
~E C.as4, 0 )* 10% mn”
F2t =f
0.640"+ 0.800% JO * = 1.0245 «10° mm”
= 0.2295 «10° mm = 0.2295 10° m*
In = Cb2s4 4 Lo2ts) x1o® mm
9,
ct
Ou
Zs
Sin 6,
|. 0816 x}0% Nem
0.5193» 10° Nem
US cus 25.67~ HS sin @Sl72° =
ZILO7 mm
4S cos 25.6794 WS sta 25.679
GO,0S
=
Am cos OF,
x
_
My
Up
t,
.
Movs
I,
=
=
Ya sin Ow
+
-
{1,081
%/0?)(2).07*Jo"")
0, URIS IOS
Pa
W3.0x10*%
4
(-0.5198
4
y
cos
2 OA 2e10" ) sin 25.692
¥
28.67
Bm *
i
=~ Mesia
Ma
2.2785* JO” mt = 2.2785 *(9- m
Ch 2x10* eos 26.67% =
a8
9,
i
Mees
‘
My
=
ft
My
>
= “Ohonio®
TE
KO, *
tan
t
* 10% mm”
Ly = (1.254- 1024S)
uy
ers
Wi
R=
x10® (60.08
mm
x1o7*
/
£.2 788 x (O°F
6,213.0 MPa, —f
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior writteit permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
*4.137 through *4.139
Problem 4.139
The couple M acts iti a vertical plane and is applied
to a beam oriented as shown. Determine the stress at point A.
¥
18mm,
|
27 min
Using
52 nine
a
f
aves
150 mim
1S mi
Ty
|
Mohv's
and
py ‘ne! pad
moments
oF
nerpad
ft mer
arden.
fio®) trim f,
“
¥
>|
~e--- 100 am
cwede | detevon ne the
|
1, = 3.65. 10° mnt
1, = 10.1 X 10° mm!
= 3.45 X 108 mm
Y
(366
Z
(loi, "2.4£) (0 © aoa
E
3e “S io" mat
(G85, 6) 10% pot
dle
EE = 2eZ2€¥10% nt
FR Yar Er Kotmme
BD
_
tea
at
ue
tan 26.. = Fa
* EF
Del
x6” ned,
=
267/
if
A
aS
&
Te = (b+ Pas - seme
if
Oy. = 22x”
z= #72810
(6:3) sin 23.5%
4
R=
G2) cos 28.5% = ba
= 20
é 22s
Lobe
T, =@ PI t 4s72)10° = Nob ¥ 10% mn
kNm
EN tn
U,=
yf cosO, + % sin 6, = ~9P os 25° = 210. fo ~- foorb pein
Vat
ZycosOm — Yp Sta Ow = 27 (ys 236° $ IF Sih 236° = Mo bm
©,
=
~
Mvp
dy
+.
Mua
L,
_
(x71 0) (010 lee)
_(6240)(~ 0</00 6)
He xtoe
T2-1 Mba
ders ye”
|
-
|
~t
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, Ali righis reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.140
4.140
For the loading shown, determine (a) the stress af points A and
B, (by the point where the neutral axis intersects line ABD.
Add
y and
Z
aeses
as
showw
A = Uo0NYX) = $800 nm?’
600 N
T,:
Ty
Pe
2000 N
Lo00
+ Leo )C¥s) 3 = ILIBDL fan ©
-
i (402) (160)? = 2079S
10° mm?
bending
coupdes
N
y
Pe
A
Ge0+2000
Hood
“1
M wl (600 )(0e22.8)-+(2000 ores +(loodloras)]l Into
.
=SEN mm
M, =C6e2) (0-08)+ (2090) (0:08) Cloo0){orar)
e
v
fa) Gz
(b)
qT,
A
Ly
-~P_
Maye
Intensection
of
Gs
=
La. Mie, yz
FP
= 36ce0 N
__B
~
z
and
Resultant force
.
=
one
(S¢)(or0225)| (8x) (0-68)
3be0
Moensé co Is9anqge
cis
with Pine
y 7O1022Sm,
_P. May
My
BeIsx10~&
~2
bb} By xe
=2 66-7b Fa,
fe
TY
i "= Espeyiee
neutral
*
_ (S¢)(o-0228) 4 @ooe 2)_
Myze _ 36e0
lL
fat/m
AB
or its
&7m fa
<a
enttensron.
z=
24?
(loro)
Foz
+
°
A
*
4,
ASvoxie ©
JO937SKI5" — RIE KIG
Zzeoroz7Tm
0 8x0 + H+3%32z=o0
Intevseets AB
Intersect ion
G =
of
nevtrad
0
2=
oe BAY,
A
I
2
ais
~—Lomm.
MeL
I,
with
Aine
BD
ot
ov
ids
Jax-L mM
fnew,
A
<é
D
~cte
erteusion,
yr?
_Be00
&sew Xie?
= 62667 x6" Telit! y =O
Sy, Pedd-ves)
TS92ISKO
JY
70 002A
Trdersects
BD
3.70 M07
mm
at
PETS pom,
fron
Problem
4.141
4.141
Solve Prob. 4.140, assuming that the magnitude of the force
applied at G is increased from LO KN to 1.6 KN,
4.140
For the loading shown, determine (@ the stress at points A and
B, (b) the point where
Add
600 N
i45 mm
2000 N
y and 2 ofeS
A= =
(LooXYs)
<
aS
shown.
¥SO0
mm
T, = x ((60) (4e)32 169318 mnt
oe
1000 N
the neutral axis intersects Jine ABD.
T, = C#s)
Clow 32 27x06 mint
A
J
8
Result ant
force
and
Pe Gee +2ec0
2
bending
couples
+ lboa= ¥200 N.
M, = —Ulee(o0228)+ (200030228) ¢Ikee) (010228)
D
= 6TeSNm
ly # (bee) (0-08) + (2000) (or 0) —~({be0 Cor os)
= Lo Alm.
of,(2)
PoM
—~A
r
S.2
A
Yo,
Mye.
~. -“ PA . “X,
Se
(b)
Intevsection
GS
—
=
o
oF
?
t
Mz,
ysa
*
neutre?
J
~=
Obey ys "a7
poorer
axis
.
78 x 6
with
in,
bine
2
.
AB
?.
=
Oe
4 ookfa
BPE?
ov
.
its
=
/
ae
73M
Je
(So) i-v10%) _
MATING -fL
SS
O.F
(50) (0:08)
ENE
cone
“2.60
M2s
iy
+
4200 _ Gu) (0029)
Ty
73
Pq
<4
enteusion.
.
o RM
y Mae
. pooh
Hee _corsoons),
Soa
A
OF,
q
TSB 1S!
Be TSK ee
2 = O08 ma bomm
~ O66 TL + 133333220
Noes
Telersection
S*
of nevbal
Oo
KR
OF
.
May
006667
.
with
Pine
Ze-Zin,
iL
x
axis
My
Ty
~ PbkPED
yj?
OOo4S”
intersect
|
AR
ov its exteusion.
?
4 (20) loro)
_bRsy
4200
=O
BD
not
ze7y xe &
Teag7eKiee
ys
eood
“Intersects
BD
m
= 3am,
«at [¥eImmfnon
Bo
—at
Problem
4.142
4442 The tube shown has a uniform wall thickness of 12 mm. For the loading given,
determine (a) the stress at points A and B, (6) the point where the neutral axis
intersects line ABD.
~
Ada
y~
1S
9s ky
ELBE
L
vA
Y
y
Z
wm
be’
Z~
[Ci
GHes
as
125 am
mw
showy.
Cress
rectanate
with
rectangular
sectian
ou
cutout.
J
= 22781 X10
A
= (as)ias5)-(81)U01) = 4,294 x10" nam
~<
4
wm
4.224 xJo73 m*
=
So
ya
anal
Wuom
10% mm
T= ECs asy = §(SINGeN? = 7.9283
= 7.9282 x10 mt
x 108 mm
yl Koi SIP = 3.2781
T= kdasias
y
ns
z
Sl
Fs
"a8 kN
aes
B
mm
@
Y
M
y
:
Z
i
3
Resultant
LY
70
=
14428423
Pz
LAKE
A
anel
Povce
bending
Jox}O*N
=
kN
couples:
Sm 23 kN)
mm 2B EN) + C2.
mom VOY kN) + (62.5
M, = ~ (62,5
= 2625 Nem
My = = (37, Smm C14 EN) + (32S mm LAB KW)+ (39. Sem (2B kN?)
= = 525 Nem
Meyn
~ P
TE
(Q) QFE
4aa4xlorS ~
Ty”
Ft
Miye , My28
Ty
T.
6. = P_
a” A
Joxlo*
, My2e
=
3). S2uxlo%
_
~
7oxlo®
WY OF
(525 (0.0375)
(2625 )(- 0.0625)
3.a781x10~°
¥ JO
7.8283
G5 31.5MPa
Pa
(2625) (0.0625)
7.B283%107°
+
(525 \(0.0375)
ZaQ7RIXIO-=
= = 10.39% 10° Fe
(b)
Let peint
Hl bethe pointwheve
Zy =
o>
R
I,
G2 -10.34MPa <a
the
nevives
P ,
6,
amis
=
ZI.St
*
Y
O< O31Stm
4
62.85
AB.
=O
i (f , Mz \ 2 7.9283x/0%[ Jox1o®
f H.2agulo™?
26a5
Ty
MSA
YH
mtevsects
Ms 2
Mew
_ Bu
Ya =
0.0375,
«dl
=
=
31S
FW.Owm
carsilo.o27) |
x JO
3.278)
mm
Answer?
94,0 mn, above poiut A, <i
Problem 4.143
4.143 Solve Prob. 4.142, assuming that the 28-KN force at point £ is removed.
4,142 The tube shown has a uniform wall thickness of 12 mm. For the loading given,
determine (a) the stress at points A and B, (b) the point where the neutral axis
intersects line ABD,
Add
yr
:
Stam
SS
}
~~
28 kN
5
>
=
?)
A
Z
YY
asy=
angvtar
a
cutout.
FGNie)\®
28282xi0~%
= 72283 Xlof mm"
wm”
= 3.2781 10% mm*
3.2781 Ilo mt
P= =
bending
and
force
Resodt ant
CC IIIS
5
vect
section
with
rectang Le
{25mm
(Olam
Cross
F L224 x1? mm
y
Z
Shown, -
A = Crs\ins) -(51Yor) = 4 224107 wet
4
|
a3
TL, = re Cias\Gs Ye-eConysry?
y
B
—
[eed Cay)
M SY
*
L = ws
TB aim
2- e725
TSmm%*
a
iS
125 mm
ane
[44+ 23
=
kN kN
42
=
couples
te
x
+
*10* lo” N
My = = (62. Sm MIM NY + (62, Senna (2B kN)
=
875
New
¥
My = ~ (37. Smee IY KN)— (32,
5 mm) 28 kv)
=
S25
_ 4a«to™
y
~
=
6.
SA
KR
Let point
zy =
poral
K be the
fy
Yu
4.229xlo-3
= O
1.3283 xfo-*
22.93S5xJo*
W224¥1o"S
=
Pa
the
.
6, = 22.4 MPa
3, WBIK/o°e
Py
Gg7
neutval axis
025m,
O41 =
<a
(525 (0.0375)
7, B283 x10"
8.qa3lxJo*
where
3.27BAx Jo.
75 VC.O62S)
_
| (Sas \(0.0375)
intewseets
3-46MPa
=<
BD,
O
BL Maye
yp MyFy
Tr
OR
2H
9
tb)
— (875)6-0.0625)
42103
1 MyZea
= PMY.
Nem
4y (Mew.
¥} _ 3.278 Ixla® {i975 (0.0645)
F ~ OL01B4ESm
BAS + IB.465
=
= ~ 18.465 mm
S6.0mm
—_
42x Jo
|
|
Answer? S6,0m to the right of poind B. ~<A
Problem 4.144
4.144 Anaxial load P of magnitude 50 KN is applied as shown to a short section ofa
W150 x 24 roiled-steel member. Determine the largest distance @ for which the
maximum compressive stress does not exceed 90 MPa.
al
75 mad
Add
y-
For
Ca
6,
=
a.
13.4
10°
mm’
Ty =
1.83 *1O° mm =
7
=
~
160
> Mele
dy
M, Ya
=
2060%10
7
men
wm”
13.4 %107* nw,”
|. BSxlOye
be
=
80mm,
102
Zy5
N
?
M75 *107*)
M,7 = (Sd«lo
My
yvolled- steef section
wm
Soxlo*°
P=
24
3960
$2
a ke
A
=
d=
z————}
pil
axes.
W ISO
A
y
and 2.
mm
= 5) mm.
|
* ~ 3.75 «l0* Nem
~ Pao
G, 2-90 «10° Pa
+ See
Pp
My = al tL tK * OA
1.3 ulo~® | 63.75«10° \C8orI0"} p
13.4 «107%
Si x1or
* ie
=
= 1.8398
+ (-90 x10
. )
5
“6
= Lo
S220”
3060 x jor *
$ 22.388
lO?
16.340
+
- 40 §
x10‘
Nem
3
a=
-
oe
°
- ope
=
36.8 xlo*
m
mw ~~
Q= 36.8
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
_and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.145 A horizontal load P of magnitude 100 KN is applied to the beam shown.
Determine the largest distance a for which the maximum tensile stress in the beam
does not exceed 75 MPa.
Problem 4.145
Le
eat
a
the
CAA 4 rorel °
XY
©
=
y
Vs
B)
700
Arm
2
@!
2090
10
gox!lo!
@®l
i209)
-10
|- 12x10?
|
CaO
re Vina
Cooydinades
Bend rng
te
ove
ot
Pord
gin
to
se
Ay
3209
the
Xp
My =
v2 2
%
J ram
gx {o>
$20”
=
25 wm
Bio?
ce tyoid
point:
coupdes
A
~
A
mS
Dimensions in pym
Meve
P
=
Oa
,
3
Ye
YeP
e
~AS
Miter
My = ~aP
ler 1 f ‘loo Yas} +-(Zo00)(7.5)" +. #(Go)Qo) 4 (1200)(12.5)°
0.40667 ¥[0% ram!
ry
= 0.40667%/0°
Ty = 7% Go)(loo?
G =
yui ao
" + ey
Moa
Me,
Pr
’
B25
:
wb LIMO
{00 ¥!0
Z£OR67 K/O* m4
0.40667 «(O-%
+
3°
3.537
.
-
75 bx io®
~3
{7.1 Kloe mn
P= looxjo® N
-
-
(-2.5 (00 #108 )(-2.5*10* )
3200 ¥ [07>
£0267 x10
x10sf
_
$0 ¥lO7*
>
-
‘
uf
50 xf 07
-
2. ORLTRO Swat
6,
18 10% Pa
2: O26] «10° | 4990"
J
oz
:“
ye
FP.
~-
as)
ir ,
=
~
7S x) «|
°
~ILTNLx
.
GQ. >
:
{
lo"
Nw
\hilw,
Proprietary Material. © 2609 The McGraw-Hill Campanies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
4.146
A beain having the cross section shown is subjected to a couple
M, that acts in a vertical plane. Determine the largest permissible value of the
moment Mp of the couple if the Thaxinelin sstress in the beam. is not. fo exceed
4.146
84 MPa. Given: I, = I, = 47 X 10° mm*, A = 3064 mm’, kay = 25 mm.
(Hint: By reason of symmetry, the principal axes form an angle of 48° with
the coordinate axes. Use the relations Ji,7 : ALE ag and Linn “& finan Fy + 2)
; 6
inM
I, + I,
#
iy
M, sin 48"
0.707
Me
0,707!
Me
My
M, cos 48°
i
Alen = = (3064) (2592 121K 0" pm ©
a
ATT 7
I-91) t0® =
qe 4ES X10" mun¥
6,
Ly
=
0.70711
Ss
7 Shey,
, Move
n
Zs sin 45°
Zg cos 48° -Y~_ sin 45°
= - Mute
. 8
M,
Ye £o8 45° 4
t
Vi
1
Ss
-£
j
4
pont
My
BY
325 main
coeds?
i
sinds? = —Ebmm
M,)- Ye.
ve,Me
cooth
oro 8
hOUI6”
PaBrtic”
Sex to”
* Steg
2
O4- gag HS? = (~£ Visi ds® = Fomm.
. 0.7071
Ly
M, |-
~8F
2.07
FRE LN
To
i
mn :
=
, Lat
2454-2
.
Me
|
:
~~
Proprietary Materiat. © 2609 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.147
Problem 4.147
Solve Prob. 4.146, assuming that the couple My acts-in a hori-
zontal plane.
4,146
Mo
A beam having the cross section shown js subjected to a couple
Mg that acts in a vertical plane. Determine the largest permissible value of the
12 mem
moment My, of the couple if the maximum stress sip the beam is not to exceed
84 MPa. Given: 1, = I, = 47 & 10° mm, A = 3064 mmr’, Ky, © 25 mm.
(Hint. By reason of symmetry, the principal axes form an angle of 45° with
the coordinate axes. Use the relations fyi, © Akiy, and Tyagf Dna Gt ED
iE
ie
My
saw
=
Me
cos
45°
if
My = =M, sin 48°
Toni
Loan
=
'
z
A
Kain
(0 by) (ax)*
=
Ty + LL
DF
~ Tn
- O. 7071!
M,
= (978X108 mm
= Joh (eT # EoD
b
1415)* 12 GEE X10 pom?
Yo Cob 4ST Zp sin 4SS = 24 L058 45° 4C 94 gin 45° )= —E bleh
By con 4S? ~ Yo sin 45°2
Bem net (F wn ig? = Pomm
fo 7
6
O.7TO7TILM,
195 man ——
%
36 mm
a
‘£25 ene
=
= My Us
+
My Vo
Iv
=
O.7071)
~
Me *
O. 7071)
Me |- Coe ob) .
[« qISXI0"
_
PAELYES
2
my
12
” 24842
+
Me |-2a2
4
Lunia
Vo
|
Tana
20k
| = 4.543 Me
ee oF
WARS X08
geltnz
KNM.
.
ath,
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
4.148 The Z section shown is subjected to a couple My acting in a vertical plane.
Determine the largest permissible value of the moment Mp of the couple if the
Problem 4.148
maximum stress is not to exceed 80 MPa. Given: Imax = 2.28 x 10° mm’, fnin = 0.23
x 10° mm’, principal axes 25.7° “Wand 64.3°.z.
fF
AQ mm
“* 10 ram
40 mm
Ty=
’
Tow,
2.28
«10% mnt
.
0.23 %lov mm
1O mm--~|
=
2.28
¥/0
a~&
“mM
4%
= 0.23 %I0° Mm”
My = M, cos 643°
v
we SN
Moz
hd
&
Me
sin 643°
G43°
=
han gr
=
tov2
= SAREE fe ene 20.50
Q = 37-22°
Poivt s
Ug
Vg
A
and
=
Yp ces
=F
~~ FIL OS
=
Z,
+
Lacthest
+
Prana
neuvtra/
axys.
7g sin 64.3° = (- 4S )cos 64.37 +635) sin 64.3°
min,
tos 64.3° ~ Ya Sin 64.3°
25,37
the
=
635) cog 64.37 ~ C45)
54 $4.3 °
mm
_Myde
, MoVs
v
dy
_ (Me ces 64.3°)(- 51.05 xo")
M
3Ox%lo®
Mo*
ave
643°
”
orn
B
2.22 x107*
93.81
Ro
x lo*
109.1 xlo#
x lo*
+
(M, sia 64.3° (25.37 «io:
6.23
*lo~*
;
M,
M,= 133 N+ rn
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, witout the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.149
Mo
4.149 Solve Prob. 4.148 assuming that the couple Mo acts in a horizontal plane.
oe
woe
4)
4.148 The Z section shown is subjected to a couple Mo acting in a vertical plane.
Determine the largest permissible value of the moment My of the couple if the
mun
ao
maximum stress is not fo exceed 80 MPa. Given: Imax = 2.28 x 10° mm, Imin = 0.23
{mim
x 10% mm‘, principal axes 25.7° SQ
and 64.3°.<7..
40 pan
ut
10 mim!
<- FO nim-n| |.
1Omm
Donn = 628 xfO* mm” = 0.23 «10% m4
T yo
dy
|
U
>
M, cos 64.3°
My:
My
=
M,
@=
Bin
64,3°
64.3°
tan P
=
LZ
iy
tan an ®
0.23x10°*
pee
=
?
Points
D
¥
Up =
=
Vp =
x
G.
ep
and
E
ave
From
~
the
tan 64.3"
°
= 0.2096!
H.BY?
neutra?
axis,
a5.7° ~ 4Ssin 25.7°
-F
Yp C08 25:72 — 2 Sin 25.7% #(eos
~
Zp cos 25.79
32.3%
.
pm
24.92)
mem
= — Mute, Mue
To
Zo xto*
Favthest
2.232% 10° mi
= 2.23% 10° mut =
Day
=
+ Yn SIMASL2P =
HE 006 25.7% + (-5) sin 25.72
.
.
€0.43x10oF
(M, cos. 64.3°X-24.02x167) 4 Mesin 64.3°)(48..32x10°)
”~*«C‘RS FOS
2.28 x lo-*
Ms,
M, = 1-823 «107 New
Me 1323 kN
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is usingit without permission,
Problem
4.150
|
4.150 A beam having the cross section shown is subjected to a couple Mo acting ina
vertical plane. Determine the largest permissible value of the moment My of the
couple if the maximum stress is not to exceed 100 MPa. Given: f,= = 6°/36 and I,.
= BY72.
MA.
20 mm
My
:
wt
“b=p= 60160 mn
g
et;
~
y
.
L
bt ie
:
Tn?
20 mm
b= 60mm
4
7
-
be
Principad
$0"
6
4
gg 7 Be * 0-940 "10 wm
2
.
7
=
wrt
“-
.
q
£0.
AX2#5
BO
=
ar2
OMS
410
¥
,
rane
o*
symmetry
O¥ES
y
«
of tnertia,
moments
prvepal
the
detevmrre
circte
Mohv'’s
Using
bY
~
.
a
Tyz
> I, | ©
Tye
VW
Fyide
¢
we?
O.1BOx*iO
4%
m
%
0.70711 Me
71.56°
MM
2
Io
7
oe
7
.
Upt
8:
it
"
~
Go
,
“Rw
Mle
0, 180 x lo7*
,
Ty
M uve ..
Vv,
= 42. Pans
. ©. Jo7'! MB
uv
Mo* Titvio® ~ THT eT4
value.
N-m
Yoo
.
Hit xto* Mo
Gs
_1ooxlo®
the amaller
.
-
WLTe®
V1}. Hada®
«1*)~4 | ~titeto M,
6 , (0.707 Me C2
= foo x}o%
Ss
wb
Me
Choose
4
5g.
Mv,
8B
ONBOXx1O
VF ~ 20% mm
Ue O.
Point At
S
R
SRO>IG*
pr
Point
bth
jong © Be tan 0 20028 hag 6% =
O= 48
Cn
+R
My = Me cos US? =
Mo >
©7071
sint#S® =
Mo
=
My
we
0.540 *16 mat = OSYOXIO mm”
*f
T=
0.180%10°
4 ©: 70711te
»I6”
O.SY4o xo" *
Fe x0 -F)
©. foro”
:
|
JOO Nem
M,=
Foo Nem
=e
Problem
4.151
4.151 A couple My acting in a vertical plane is applied to a
W310 X 23.8 rolled-steel beam, whose web forms an angle @ with the vertictl. Denoting by oy the maximuin stress in the beam when @ ~ 0, determine
the angle of inclination @ of the beain for which the maximum stress is 20).
Fox
W
B/OK 236.
vobbed
steet
section
T, = 42¢7xtoemm& Ly = b/6 x10" mnt
d = 20mm
Am
tan
bp = fof mam
~g
2
D
=
z
tan
6 =
Point
A
is
farthest
F
4
te
= Mea
S
26-8f
nevrel
tan®
ayers.
M,= Mo¢cos8
2 Med cas
4
Mobt
sin
Tbe
tan @)
Se
4
; (1
4
0
f,.
(1+
=
y
S
For
i
ai
tan 8
faamthe
My = Mesmné
Gy =~ Ale + Ma
be
Eg tn 8) = 265,
Ld Ad
tan 0 = - FT,
_ Cht6) (30S)
” (42°7 )iat)
> 068204
§=
4.70°
~~
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution 1o teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.1 52
4.152 A beam of unsymmetric cross section is subjected to a couple Mo acting in the
vertical plane xy. Show that the stress at point 4, of coordinates y and z, is
yl al
o,5-
et
Lt, -0
z
where i, /,, and J, denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and AZ, the moment of the coupte.
The
styess
G,
goog jaates
Force
ts
vanies
Y
ZEVO
ceutvoi da?
zZ. Since
the
7
with
the
anal
the
awraf
Z-Axes
AZ
axes.
where C, and Cz ave eovstonts..
Cy +@,z
SzG4A = AS yzdA + CaF z%ei
4
CaF
eS
and
,
Dineardy
Dye)
=
Cae
iH
it
M,=-CyG,ez
LG
|
_ Lye
WG
-C,GytaA
~
©
=
Ly ¢,
+
+ C, Cyz4A
Tye
~
ty
Ty
C,
.
.
=-(lyt,- Ty.)
TM,
C,
=.
Ly
Ty
6,
=
A
I,
M,
7”
-
Cy?
Dy."
J,
-
DM,
ly
~ Jy: 2
Jy
4:
Ly.”
M
I,
~ Jy2*
—
:
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.153
4.153 A beam of unsymmetric cross section is subjected to a couple Mo acting in the
:
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
_ at, ~ Wy
oe
‘ “7,1, -72
~ VE
,
where J,, £, and J,, denote the moments and product of inertia of the cross section
with respect to the ccoordinate axes, and M4, the moment ofthe couple,
The stvess
€,
Coove inates
force
ig
Centyor
G7
Gy
+z
zie,
cia?
Pinearde with the
Sinee Fhe axtad
t+he
yo
ovd
2Z-ayes
ave
BREE,
where
M2 =~ SyG%jdA
wanes
and z.
C, ancl
= -CUyteA
Cy
ave constants,
- C, (yz dA
c+ Eee,
0
=
~ Ty. C,
TRC,
- . ly:
My
0, Syz dA
=
\ 26, dA
=
Dy2
=
L, My
C,
6
+a
C,
~Lve BEC,
(Wi-) ¢
ELM:
dy
=
XK
—
t22
IyI.
Tyzre
-tyy.
-
Tye"
C, (zt
+
.
My.
+ Ic.
C,
-_—
.
-
Ly2 My
Ty
=LR
ee Jy"
ati
L
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual j is using it without permission.
Problem 4.154
4.154 (a) Show that the stress at corner A of the prismatic member shown in Fig.
P4.154a will be zero if the vertical force P is applied at a point located on the line
x
Zz
+
bi6
hié
=
(6) Further show that, if no tensile stress is to occur in the member, the force P must
be applied at a point located within the area bounded by the Ime found in part a and
three similar lines corresponding to the condition of zero stress at B, C, and D,
respectively. This area, shown in Fig. 4.1540, is known as the kern of the cross
section.
I,7
Z7
Let
P
hb
IL.e bh
b
-H
be
A=bh
Mer mg
the
doad
M, =-~P2,
peiwT.
My =
P2p
2 - Bo Me%m _ Mum
~~ L(2P
y
x68) _ Pz -$)
bh
&
hb?
te bhe
-~ w~f | ,~ Xe.
~
bh
bY6
OO
~%
LEDS
/2
12. =
Es
Z=0
A+
point
F:?
x>OQ'
the
will
x
x
point
Pine
within
/
=-¢
AY
TF
A
~4%
Ze
hile
the
of
portion
occur
By
ton Sidlering
the
other
ave
action
at
portions
(Xp,
marke
Cornet
Gg
Xe
=,
4
= bée
Behe
2p )
Ta
bres
,
a
densrfe
A *
6,79,
producing
avd
tenside
SO,
stresses
ident iFieel .
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.155 .
4.155 (a) Show that, ifa vertical force P is applied at point A of the section shown
the equation of the neutral axis BD is
where k, and k, denote the radius of gyration of the cross section with respect to the z
axis and the x axis, respectively, (6) Further show that, if a vertical force Q is applied
at any point located on line BD, the stress at point A will be zero.
Defindvous:
(a)
My
P2,
Gus
-"
°
.
M2
Mz Xe
Maze
I+
My
.
_2
T
“Eli ++ Ge) Me
(b)
F ~ Pxa
2)
=
+ (44) z
FO,
M,
Pz
Mam
Gar nK se
by
v(Zs,) 24
=
~ Mem
|
- he
equatiou
ale
Khe
iF
a
.
- Pg
PR
A
Frou
Py ze
=O
(Be)
.
Piyre _
_
E
Dies
2
.
PxeMe
AK?
on neil axis,
od
So
— Pet
Ak
Part Ca}.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form ot by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teactiers_
and educators
permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
~—
- 4.156 For the curved bar and loading shown, determine the stress point
A when (a) r, = 30 mm, (b) 7, = 50 mm.
Problem 4.156
A= (30) Go)= boo mu
h= 2emm
Ya = SUIS
~ 20 = BIS
mon
(68> (a0 0915)
- Mya
MPa
_ 4.07
(600 &6°)( 0100 BF )(0102)
Aes
hI Me
a
We
if
by
=
Y
et
LOmm.
Wt
TOmm
po
Ine
= BPE
Ya * S%EG
A
Sy
roms
7
men)
¥Yi-
‘Mya
vy
Ae
Wha,
34.15
=
“OO”
=
Gs
¥,= Semin
Vie Zemin
@)
2.
R=
OS
— Sd
o-
&
Ge
8s
bon
men
wm.
= FoF
mm
(68) (01009 44)
(G8x15* )(0+ 66 088 \(a0S)
Wwe
_.
.
= 50am
|
ZR. 2. MPa
~ 38-2 MPa <8
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.157
4.157
For the curved bar and loading shown, determine the stress points
A and B when r, = 40 min
h=zomm
We:
iN GC)
G rim
Re
= 600
20
5
Y=
Liner)
e@=<
P-R =
49-3261 ~ ‘to40 = 9 4326 min
Ya = ADI3LE|
—
‘Mya :
(OSX 009 42 61)
~
Ae@Vs
Jo = €432b/-Go = 1006739
6g =- Mye
Aeve
mm
= 49-3261
C6) C-t0- 6739/03
—
bai
= 50mm,
067389
mes
40 mm,
Wy 2
wwe
“C600X15°) (0:0096734 )(o ou)
|
bomn.,
>
TEs
bn Go
We
ZN
MP,
a2
mes
392M Pr <0
ot = bo mm
“(boo xo") (010006789 )lo- 06)
a
.
460mm
=
+904
Mo
O69 WG <i
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem 4.158
-
4,188 For the curved bar and loading shown, determine the stress at points A and B
when A = 55 mm.
it,
.
24 mn
600 N-m
~LoLe
wee
0
no
55
A=
(24X55)
GOON
h
Rae
SO
74.13025~
Ya?
G=-
Mya.
Y=
EM AR) F775
@=
~-R
24.13025
=
8
mm
3,36975 mm
Vy = SO
mm
(Geo (14. 13025 x1”)
(1. B2O m1
AeVv,
= 4, 13025 mm
7 FL BE
=
mem
_
~ 65.1 «10° Pe
\B3BCF7SKIO) Goro)
Ont
Yo
ese
e
*
TH.1BOVS
Mee
Aets
+
los
=
~ 30.
jos mr |
me
1.320 x1?
SS
yt
)
50 wm
Ye
mm
h=
86975
man
,
Va
7
_
CANG 30.9697 *10"°)
(320-1? KASEI IO? WOT *1O*)
LOS
~ 65.)
MPa
=e
MPa
=f
man
34,7 vlo
=
G,=
347
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.159
4.159 For the curved bar and loading shown, determine the stress at point A when (a)
A= 50 mm, (6) A= 60 mm.
24 mm
Tel
(ay
Ae
SO yw
Y=
ome
(1)
Gr
- Mya.
‘
Aen
“
?
60 Wane
he
R=
v
5
-SO
yA
mm
=
,
HO
nae
2
A:
~My
.
Aes
mm
= 77.3 «l0® Pa
G,2 777.3 MPa
(24) (60%
2
et
x1” maw
[440
TEL OF TIS
Po
$0
a
|
~
Cz F-R
C= A(KN4h) = FO mm
Ya = 76,09796 - SO = 26,097796 mw
6
SO
(1. 200%1O* )C2.8652%107)\ (50 ~ IQ")
50
oo
= Simm
(GOO MAZISITS x18)
=
mr
= 28652 mm
mm
BQBNIT
100
72 IBY IT mm
In BE
In
e@= ¥-R
72,1249
V2
2
5
Fe EMH)
2
mea.
A = (24h )€s00 © “ 1.20010 pane
.
Ya
SO
;
>
50mm
ee
SAL
BOON
GOON m=
hz
(
a
2k
mm
= 3.90204
MF FO vam
(E00 \(26,.09796 *1O}
=
£7 ul0* Pe
|
U-4 9910" (3.70204 (O° [Ox 1s®)
G,=
~S5S.7
MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<&
Problem 4.160
4.160 The curved portion of the bar shown has an inner radius of 20 mm: Knowing
that the allowable stress in the bar is 150 MPa, determine the largest permissible
distance a from the line of action of the 3-kN force to the vertical plane containing
the center of curvature of the bar.
25 mn
‘)
Reduce
t
the
internal forces
Wensmitteel
3 am
+0
a
acvess
force
centroid
bending
of the
AB
SY stew at
section.
couple
M=
section
- couple
the
‘The
is
PCa+F)
For the rectangutar sechion, the neutral
axis
Por
bending
Rope
The
at
aK
Data:
h
= AS
Ka
R= J
BE
mm ,
T
=
E
o
@=
I+
——
Vi, = 4S
at
Aev,
Ya 7 ROM
ny
mn
0
82-6 ~30,828k =
= 32.5
mm
1.6712 mn
mm
Nem
6, = - 1S0x10*% Pa
- 6,a
A _ _ (So «10*)(625x 107%)
=~ _
oceurs
rr teh) )
j +
=
Bxtos
at
stress
is given by
A
with
20 mm,
@=¥-R
A= bh=QsW2s)= 625 mm = B2Sx/O~ m™
[0.8283
P = 3xlo®
Koo
=
7 20.3238 mm)
na
b= 25mm,
R-
VY,
Also
fies
. . P _ PCatP)y,
Aer,
=K
Thos,
ob;
maximum Compressive
My
A
P
-
point Ae
6, = ~2
coupde
(K-1)e,
R-v,
_
.
93.37 - 32.5
(30.2590. 671220).
10, 32388
BI2S
=
73.37
mm
=
60.4 mm
<a
Problem 4.161
4.161 The curved portion of the bar shown has an inner radius of 20 mm. Knowing
that the line of action of the 3-kN force is located at a distance a = 60 mm from the
vertical plane containing the center of curvature of the bar, determine the largest
compressive stress in the bar.
re 20 nun
P=3kN
25 mm
¥
Reduce the interna? forces _
Jransmittecl acvoss section AB
+o a Force- couple. System at the
centroid of the section. ‘The
25 mint
bending
coup fe
M=
is
PCat)
or the rectanquiar section, the nevtvet
we
Por
bending
coupde ondy
Roig.
The
at
wmaxtnom
I+
Aer,
_
Thos,
rl =
f=
aS
Le
mm
= 20, 3238 mm
bh=
ee
Az
A=
COmm,
OAK = FF mm
P=
Sa
J.6712)C 20)
3xio?
=~
Proprietary Material
(72.5 )(10.8288)
{+i
|
KE
A
+f
G25
ICR-")
mun
RV =
_=
R-¥,
GY,
37.5
~ 39.8288
As \(as )=
as, ma,
K
*
= 20 mm, Vee 4S mm,
,
b=
_
=
,
Aev,
Ya
ii
Data:
K =
occurs
by
~ pPtorPby
A
with
= SR
P
stress
is given
.P
-BoMy
at
= F-R
Compressive
point A.
A
Also
Ries
0 = 32.5 mm
=
=
1.6712
B2SxIO~
mm
m
10.8288 mm
»
30.968
WN
. _ (80.968 )(2xto®) _ -14B.6
EAS
x lom*
«[o% Pa
GO, = 7148-6 MPa
9
© 2609 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission,
Problem 4.162
4.162 For the split ring shown, determine the stress at (@) point A, (6) point B.
Lye EGO) = OS one
Nr £40 © 20mm)
2500 N
to
808m
A=
h*
RB
R
=
e
yr -R
mm
as
;
RE 7 808285
r
4
BSO
SIKH)
At
Sto
14mm
K
zim
(us) =
>
=
-
O,8288
|
me
32.5 mm
LETI2
mm
Reduce the internal fovees
dimmnoamited acvoss
section
AB to a Force - coupte
system at
the
centvord
bending
couple
nN.
M=
Pa
of +he
cross section.
= Pr
(2500\(a2z.Svlo*)=
(a)
Ast
Point
Gi,
é
- PP.
A
~
M Va
AeR
=
A’
=~ 87.4 “lo” Pa
Beth
__ -
N-m
mm
[0.8288 mm
62-824 MP
=e
Yor BO.BRBB-4S = 14.1712 vam
x 1)
.25
\C- 14.1712
(81
2500
~
3Se0Klo®
Ui
6, a?--F.~Mye
20
81.25
. (81.25% 10.8288 » 15 )
Ae «to? )
(50 «10 )0.6712 KIO
£509
350 yjorS
Kaz 4S mm
(bY Point 82
Ys =
BO.9283 - 20 =
a =
The
rs
20.6
~
xlo®
(880K 10°V.6712
xt? 4S *10 )
Pa
6g= 36.6 MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any nieans, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem
4.163
4.163 Steel links having the cross section shown are available with different central angles 8. Knowing that. the allowable stress is 100 MPa, determine the largest force P that cau be applied to a link for which 8 = 90°.
Reduce
couple
-
.
;
rage
Se Pog
of
j ~g 2 ms’ Lye°
ren oe we estat tae oe ap ws tee wt ae
'i
eo
mM
bending
6
Pr
a
P
For the
N
ee
FCL-
coup te
section
Ap
cos £)
is
Mz
rectangy far
~ Pa
Poe
bending
Section , the
couple
R= Te:
point
A
the
tenarJe esteess
£ - My
Sa
=
KT
Hew
=‘
+~ Zee
ey
where
K
Data :
1 =Z2 mm
5
t 2226
a. =
22
PB
oe
pe
Also
Pays
~ RS
= 2B
Ren
ona
VWrlSmm
Ya
Pies
er-R
aya
R14
b= (Smear,
R= jae
= ote bob mm
Ya
3
6-44
a)
= R-t
, Y,=3omm
> 0-36 mun
(}- cos 45°) =
ands,
nevdvad
is
= (15)(6) = omen*,
:
cwvess
Force to a, force =
at G, tHe centroid
.
3f
bets
At
the
a=
The
section
Sy stem
ae 64-15
bis
=
6mm.
6+ Ye mtn,
man
OGLE OY)
(0-36) (5)
P= (40x16") (tooxso) |
Fe 9Q
/609
|
Af =
LEN
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed tn any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
Problem 4.164
| 4,464
Solve Prob, 4.163, assuming that B = 60°.
4.163
Steel links having the cross section shown are available with different central angles 6. Knowing that the allowable stress is 100 MPa, determine the largest force P that can be applied to a link for which 6 = 90°,
6 mm
15 ween
Reduce
.
section
couple
of
aA
ie
:
"22 me
nn
ns
eo
ome
The
P
PS
el
“
anis
vr (fj -
couphe
Fox
cos £)
is Me
to.
G,
A. Pavece. -
the
centvord.
bending couple
~
~ "Pa.
nevtvad
, +he
sechon
far
vectamay
Fon the
}
=
at
at
the cross sectitna AR.
OT
bending
Ponce.
sy istem
ony,
fies
of
~X
At
the
A
ee
=
where
K
Pp
Po.y
“AS
Ou
[+ Soe
_ AG
P
>
Aeyv,
=
:
Rev
ond
Ye
on
f=
aya
x
oe)
K
LF
FT R-"
22mm,
We
hmm,
(57 20mm ,
¢
(6)
22—
2 =
2u
=
tt
Joomm”
=
CIN)
Qe
_
Bp
p(t+
K
Data:
K
e@=zf- fe
densi fe stress
Aya
f+
Aso
.
RTs
N
H°b64
(f-
=
cos 30°)
Orban
=
~
(0-3)
P = Grows’ C10)
4. bo]
&
2
ln 7e
= 2-6 4mm
Ya * 6G
29S
(29S) COC)
-
fe-=
|
= IC pwr g
I=
bs 6mm
|
& 64 me
|
mm.
4627
faux
)
No 44S ex.
-
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thé limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual! is using it without permission
Problem 4.165
4.165 Three plates are welded together (0 form the curved beam shown.
For the given loading, determine the distance e belween the neutral axis and
fhe centroid of the cross section,
__ 12mm
f
50 mm
i,
78 mm
12mm
75 mink
e7
Cen
*
Ee =
oe
BIOO
O | 7x
| iz
@® | So
ly2
'
R mm
= &L
~-
~—e
~
_
ZA
2b, Bate
a
.
| &
|iei3/S | BE | 72400
[Feo
[hoo | 41983
[20.7787
WOH
c.
M1.
2
>
So
oe
e
Cc
Zboihz
2 boyz Dan Tats
\\;
2C4dA
_
Ne
|
ZA
R
él
-
_ 226900
_
[143 | P6Fo0
22X97 00
a,
GS fam
-<l
Problem
4.166 ‘Three plates are welded together to form the curved beam shown.
For M = 900 N + m, determine the stress at (@) point A, (b) point B, (c} the
4.166
centroid of the cross section.
50 mm.
B
ZCfdA
-~
r-
2
=
=.
* ZA
R
=
bzh:
Zh, tia
ZA
Eb, Bate
.
AF;
FA
[te/3/5 | Bt
[900
12
| 4429
Lo | Geo
[goo | 41993
j2
112 | 67200
[143 | &<F00
228900
20-7789
WU O0
e = HAEIe.
[72400
10757 mam
M = - 200 Wn. |
(a)
ye
R=
=
loheT ~ 78
Ye
*
4
+
R-
vi
_ Myc
Aer
=
=
2247
MPa
[O77 = 149 = = 4793
2
e
ae MY
Aet,
Gt
“<
A)
*
y
nor
or
tc)
2667 mom.
a Bier o oI OTE
Sa = a
Oe
we
4793)
200 (00
(ow 06s) (014-9)
(2e0rie®)
=
ig,Rp
~
_
Z MPa.
-@ +
zo
Me
AeF¥
-—<
=
mM
.
Af
—
eo
(2400 x10") (040757)
Proprietary Material. © 2009 The MeGraw-Hilt Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.167
4.167 and 4.168 Knowing that f= 20 KN - m, determine the stress at (a) point A,
(b) point B.
150 min
Tissue = @ = eA
5
bch
ES 4AA
2 FA
Sb
dn
Sb 0,
ZAG:
FA,
P= nl Fyn Ayn? EBA | Fn] A wn?
1® 108 | 45° | 4R60
75353
172,5
|
@
|5o
|
VG
|e
asoOL 86
ls
.
R
71729
|
Tapp
@=
K-R
Y=
)
7
=
7
Re Y,
~My
Aev,
—
:
mm
(207 lo® NSS. 606 x 107° )
+64.) wio%
POS
2
ITS.
M = 20/07 Nem
£9720 x10"* Xt. 894 xto®
if
BT
=~Roe Zl4.
telo® _
apg
mm.
--
=
ZY. F 10°
205.606 = ISO = $E.605 me
Ro, F
>
Yo
18nseH | 262.5 | 1275.75 ¥10”
9720| 47.2747
205.606 mm
Aer,
(b)
[195 | 4869)
|
TfL894
- Myn
=
6,
| 838.35 %10%
ean
.
606
~ 230
SO » to)
6 =-C41.IMe,
Pa
Te
[24
BIY
=e
many
4
107%
{20 «10% )(- 124.39
to
(9720
FC 894
65.2 «10% Pa.
}( 230 x10 )
& lo™*
Gg= GS.2MPa —
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by MecGraw-# Lil for their individual course preparation. A student using this manual is using it without permission.
4.167 and 4.168 Knowing that f= 20 KN - m, determine the stress at (a) point A,
(2) point B.
Problem 4.168
150 mim
135 mm
R
_
“435 mm
*
36 mm
v
PT
iso
am
TA
raw)
¥
A
=
22
(@)
-R
=
p.
te
bz Pn oe
“SD ye
2b: dns
te
| losly
EH sean!
7Y,mm
mPa
mm
-
Ay,
mm
4s | 4860]
15,2332
| 307.5 | 1.49 445% (08
9720|
38.9374
={= _ 2.S515x1o*
aIRo
raw
1288S mm
|
M
=
20x10%
2.5515 x0*
|
-
RERLS
mm
Nem
yy = R-W, = 244.615 - ISO = 99.615 we
6
(20«107\C 99.65 «lo? )
A
Aer,
(4740 xjo~* \UZ. BBS KIO? VIS * (O-3}
s
()
SA;
217.5 | 1.9705 xto®
og£49,615
=
mn
>
.
23.1067
>
- 17h0
38.4399
2A
~
Sbhehe
36 | 135 | 4Réo]
240 @® | ios}
R
2A:
bmmlVivam
veel Amir
aas ©
_
yer
6.
e”
R-Y
~ MYe
=
a
Aey,
=
=
{06.1% (0%
S
- B30
2Y9.EI
(20103
Pa
=
6,=
-106.| MPa ~se
Sgt
38-9MPa-—md
~ 80.385 mer
)(-B0.385% 157 )
ad mI)
(4720 #10" VR BBS xO?
38.9 «10° Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted’ by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
4.469
4,169 The split ring shown has an inner radius r; = 20 mm and a eircular cross
section of diameter d = 32 mm. For the loading shown, determine the stress at (a)
point 4, (5) point B.
2.5 kN
a= td
Vt
=
KtGe.
R=4
(a) Point At
oe
Sa
(b) Point 8
».
Yi
,
¥2 = nae
yy = R-0,
= S52 mm.
BS mr
f{P-e
NZHS
20 mn
|;
436
+ 1367-16
-R=
|
,
wn.
5.8755 wm
yt = Fae) = 804.25. mm = BOH.ZS« jw
M= PR =(2.5R105)36x)03)=
N
2.5107
=F
i
Yr
P=
ta
BE
=
@
léwm
= SHINS
= 20
FO Nem
14 IRIS mm
(oC. 1atsxta™?)
PB My Lo aSeo®
~K “Ker, “gon asnion® ~ (8044S *}0°* V1. 87SSeIO™ \AOXIO™)
= 48.2
0°%Pa
yy= R- Me= B4I24S-S2
PR _ Mya . 25x
Ge=-K
G,=~ 45.2 MPa
F 17,8755 mm
410° )
~17(40)(8755
JO™
BOHATaIT® — (BOY. 2x1 YX. B7STR
VS2*
lo”? )
~ Rev,
=
740
nto” Pa
-
Gz = 17Z.HOMPqa
=a
Proprietary Material. © 2009 The McGraw-Hif! Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
4.170 The split ring shown has an inner radius r;= 16 mm and a circular cross
Problem 4.170
section of diameter d= 32 mm. For the loading shown, determine the stress at (a)
point 4, (6) point B.
2.5KkN
C=
td
IG mm,
PLC
FB
t=
ié
ren
Y,
=
Vito
= 4g. Han
|
= BR mm
A
24.85
FS
R= tle. sre “|= al ons fare]
~ “Ro
-
Bam
"i
a”
P=28~lo”
M=
N
G.--f£- Mya
A
A
AeYx,
=
21486 mm
= EGt) > 304.25 mm = BOL AS*IO” wm"
Pr = C2SxJ0° M3210") = 80 New
--
(90 13. 856F x10")
2S 10%
$OU.AS xo =
(RO4_ ZT KOE (2. 43S x1O
|
= =~43,3 10% Pa
ge. 2 -P-Mye . _ _25xlo*
6
A
—- 804.2SxIoe
Ket.
=
|
Qyz7-43.3MPq =
(30)(-13.1436x10™)
| CRAZSx1O™ (2 936¥lO
4 HS xJ0° Pa
IG 10° )
=~ 18,1936 mm
= 29.8564 - YB
yy = R-W,
(b) Point BS
13.8564 mm
=
= 24 8564-16
yy = R=,
(a) Point A>
CYt mut
4B
IO)
Cpr I4N3MPa
ae
Proprietary Material. © 2009 The McGraw-Hiti Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual j is using itwithout permission,
Problem 4.171
4.171
The split ring shown has an inner radius 7, = 20 mm and a cir-
cular cross section of diameter d = [5 mm. Knowing that each of the 500-N
forces is applied at the centroid of the cross section, determine the stress at
Va=
(a) point A, (5) point B.
500 N
eae
S00N
Vato = 20
=
a
+
V3 )
=
r
> Of
+ IS
+4
t2hS- 27
= SOON
fa)
C=
2
estar
176-7 mm*
=
OS
A2Ze,
= OF
fon
75mm
sofial
ctreutar
section
= 27 nm
min
to:
RrQ4+M= 0
P+-Q =O
P= fooN
M =-24Q = -QYn0275)( 500) = -27.5 Nin
% = 2Omm
A”
= Py
A
ul
6
nm
¢*- a | = Alars4 lore -Ga\* |
-R
OEM,
= 325mm
= 27-5
WLS)
-
MM%,-R)_ — s00
4. 275)
( 0102 - 61027)
Aev
trgre
TFyto-®
aso0
Noe ova2)
ACM
L167 S14
(176-7410-6 VW\(a-0005
11-77 x 0* Pa
Gy te
bh)
Pot
y=
6. - 2,
sy
SA
et
2F mm
Miter R) _ _ 400
Ae
m
Ma
ts
1her kro ®
(-27-57)(a-035— 0°027)
C1767 16 &)(0:0005) (01038)
b69°32K jo
Gar — 48:3 Mra
lll
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
4.172
5SQ0.N
4.172
Solve Prob. 4.171, assuming that the ring has an inner radius
r, = 15 mm and a cross-sectional diameter d = 20 mm.
4.171
500.N
The split ring shown has an inner radius r,; = 20 mm and a cir-
cular cross section of diameter d = 15 mm. Knowing that each of the 500-N
forces is applied at the centroid of the cross section, determine the stress at
(a) point A, (b) point B.
»
gy
& mm
r=
VY,
Ve =
= 1 + 20
(ra
c
= 35mm
v=
oOK+ fa)
A=
We’ = Tllo)* = Bik 16 mm
Q=tfR
|e |
= 2mm
|
free]
i
8
sid
= ohm
for sofid civevdar section,
s4 [24 [easy -o* | = 2 mon
erV-R
= 25- 2 = fmm
Q
= S00
N
ZF,
HD EM, 501
@
K=WF
S
=
fb)
Poa
BWQ+M=0
RY
3-sof e108
Pa
Ta
P-Q=0
P = S00N
M=-20Q= -(2)or00))(S00) = - 1 Nm
mm
M4:
Ly
= A
= 0
ACV,
ae
506
3th fbie®)
(~ 1) @<018 = 21024)
th Tbxiso\(oreas
Jeo 1F)
6, = 3-5 MPa
FY
= 25mm
oe. = Ba M(fe-R)
Ret,
RB” A”
= ~ 059 x108
520
3416x6314
(- 1) eS
- 01024)
Sex1e*\ (0100 1Yoros5)
0,7 -06MR
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—<m
Problem 4.173
4.173 Vor the crane hook shown, determine the largest tensile stress in section a-a.
35 mem
Locate
25 mm
4
-
f
centroid.
A
@
60 mm |
@
Section a-a
235—4
Force - coupfe
systew
at
centvoid=
mm
4 h*( b+
loso}
go
68 xi0%
150
Z0
Go
=|
1300
yor
Maximum
Yat
=
R-W
7
=
103 ¥ Jo?
P=
iS xlo%
1900
=
68.33
a
ram,
No
= -1,025%/0% Nem
b. )
4452
tensite
x10
103x107 |
Y
gg ae
((45Ktoo) ~ (z5X40)| P, BF - (eo)(as+asy
fF +-R
mm”
@
(by - bn) ae - hie be)
_
O60 (35+25)
&=z ~“
Ay,
@M|
M= -PR = -(1S 103 )68.33
18°)
3
Rs
Yo Zz. mm
Oe
wm
stress
oceurs
at
poiat A.
23.878 mm.
i.
.
6 A - B_My
A
Kev
'Sxlo%
I8o0x1or&
84.7 xJo*%
¥ 10% )(23.878 «(0)
— (1.028
(JRO
¥ LOS (4.4952
* 17? Wo x fom?)
Pa
G, = 84.7 MPa =a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, ot
distributed in any form or by arly means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4,174 For the curved beam and loading shown, determine the stress at (a) point A,
.
Problem 4.174
(4) point
B.
i
8 [" a
20 mm
ho tat ec entvoid,
=
30
\
ae
a |
LL
aH
Sona
©
|
be)
zh io
Cv, - bt, Dn
~
@
-h(b,-
R
y=
Ss
yn =
=
- (Ho
pit.
6, + - Aew,
1.8608
300 | SS -|
51}
G00
fe
43. 5x Jo?
deo
me
= 48.333 mm.
=
- Zo)
46.8608 mm
M 2-250 Nem
|
ee
mm
(-250)( 11.8608 x10" )
(900
°
O, =
63.49
MPa,
7
t = 18,1892 mm
(257 ne
x0 WA TAS
62.4 ¥I0" Pa
=
:
)( 35 *10%)
(Foo v107*)\(1.47aSxlo™®
= -
16.5>103
|
=&
LL.
AeY,
RV
.
1472S mm
=
F-R
a
@!
b)
Teortes) ~ (20)(35)]
~My
(b)
:
©.5 \cor" (HoDn +20)
=
e =
mm
©} ceo | 4s | 27 x10"
.
R= —
AY,
hmm
A, mer
“re
ss!
&
se
wet
Nl”
Mau”
250 N-m
G:
HIT
-3"
(es wor
ae
% 10° Pa
$2.6
)
Gn
z- 52.6 MPo. <a
Proprietary Material. © 2069 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem
4.175
4.175 Knowing that the machine component shown has a trapezoidal
cross section with @ = 90 mm and & = 62 mm, determine the stress at (a)
point A, (b) point B.
SHE NS. gr
Locate
centroid
,
ME
ee;
A mm
150 mm
100 tam
~
fmm
M|
61g0 |
@!|
4650 | 200
150 | hoje xo”
0-F% x 10°
2 | i/4eoe
<s
~
R=
th (b, + b2)
AY, mu
19425 60
_
~
(942500
Yoo
Fond
1693
2l7o-4iann
|
(bt, - BY) Dn R > bCb,-
(a+5) cisoyd Got 62)
[6790(260) ~ (62) (000)
In EX” —~ lize) (Fo~ 62)
F-R = fel
|
M: 9kW
@=
(2)
Ys 2 RP,
S;.
‘ot
*
(b)
Ye =
(SG 3- loos F403 bm
Mya
le
Aen,
R- Vy
6, > - Mie
Aevt.
Mw,
—
(2009 (0+
= 893-280
-
0$93)
.
Gtecoxte¢) {40199 (0-1)
wm
422
MF,
oS
2» GoeT mn.
212907)
(o)(~oo
)(a°2$)
C1408 X10°) (OLIot
on
ge 8 iP
|
Proprietary Material. © 2009 The McGraw-Hitl Companies, Inc. All rights reserved. No part of this Manual inay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using it-without permission.
Problem
4.176
4.176
Knowing
that the machine
component
shown
hag
a trape~
(a) point A, (Bb) point B.
) 800 N+ im
Locate
centvyerd
697406
(s0
Abso |
Mi
® @|
~
Ay,
iv+3
>t
="
Aji .
‘67s0 | peo | h3SxKr0°%
“Goo
~
2 OE4E-IEDO
204 7SVO
Nt eee
|
1749.A © by inary
th? Cb, +b.)
_
© (Bfe> beMh) In & = bh Cby~ by)
=
[ G2) (280) - (94)Li00)) fa ZS2
(18%) (62-4)
too
Q@r
(a)
O
. (oS) C10) ( 62+ Fo)
VP -R
= dm
R-Y,
yo
Je =
—_
/6§:
¥ ity .
Mz
24K
_.
1 Mh
48-2 Mia
G8e% min
=
ae Mayen . _ __ (hoe0) (0:06 FY)
On
Rev,
leon x1d® (avery
£01)
(vo)
a
7
R-v,
=
MYyse
.
—-Ple&
.
am
mm
(4000) {~ 0.0816)
_
et
Proprietary Material, © 2069 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual miay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem 4.177
Ron
os
Use
<s
ae |:
50 min t~
75 mm
ul
~
iy
|
a
62
4.1777 and
and 4.178
4.178 Knowini
Knowing that
that Af
M = 560 N + m, determine the stress
‘at (a) poiat A, (6) point B.
Sori
BD mat
b,
Foeumula for
trapezoid
e
Boh = $ (be)(15)=
|
2 GZ rum
=
2325 man
7E mm
) V, = BO rare |
with
b, = o,
b, *
— &h* Cb. 4 ba)
(b, 4
7
be!)
dn Me
™ hb,
- ba )
(o-&) (7s (bite)
ae
Tpel
2/161
4 4 am
,
= 3-86 mm
R-n
+-
—
“Ye
Aevy,
y=
6
~
= hIG
Mye
BR].
~
_ _
M=
S60 Nm
mm.
(£80) (o.eduU4)
(2326 1G)
0.003
|
hey.
2b
)(O«08)
at
MPa
— §2-FE mm
(Kbo)(-0+ 05386)
(23254168) (61003 86)(04128)
v
Wl
yy= R#1
bb)
~R
#]
“sy
4
@
L( 629¢ 12) ~ (0) (52)]
tn 2 — 75( 62-0)
2)
f, [28mm
2b Vth
~
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.177 and 4.178 Knowing that M = 560 N+ m, determine the stress
at (a) point A, (b) point B.
Problem 4.178
2
A= 4 (62)(7s) 22225 mm
62 mm
a
Use
Uy
ao
75 ram
2 foo wer
SOTSD
=
ro) >
¥,
riD hun , by
Pormu da
Son
(oes) C789 (ot 62)
~~
er
_
with
(20 pana,
b, = On
7
fe
-
rn.
Aer,
M-<=
646H)o-oS)
)
(0100403
(86
e
yy,
MAe
~
n
{To
Nm.
Lb Y- fin
(2325%)5%)(6+0
~
~ 28
__Mys _ ___
Se
be
tar
3-4mm
NF
R-
Yaz
Ye
=
[ corcree)— (629
$0) ] Ln BE - 75 lo = b2)
~-R
Ga
(b)
trapezoid
Y=
4 HC b, + be)
(byte- bet)
fa — h(b,~ by)
R=
2)
62mm
.
6
oa
=
;
62+] MPa
a
mm
)6)
(s(-h6+e028
125)
(2328110) (0. 0636) o-
- ea
5*3
MPa
.
cat,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No pact of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using if without permission.
Problem 4.179
4.179
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
A
b,
inf}
2)
ry
| 23}
%
by
|
3
where A is the total area of the cross section.
rp
+.
|
#3 “ora
A
TA
‘}
Q
=O RAA
~~
Note
that
Vy.
each
A
Zone
pectang/e
Z bz dn Se
=
Nyaa
A
dal
:
v
,
. Le
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
—<
4.180 through 4.182 Using Eq. (4.66), derive the expression forR given in Fig, 4.79
for
—
*4.180
A circular cross section.
Dse
-
podav
coota'nete
width -
ow
v
[
¥
| Folge
uy
:
Zasm
W
Problem 4.180
~~
Aa
=
c
vr
- C cos f
C cos fp
Qe
v CC)
cate)
CU -- cast)
5,
shown.
f
wares
v
jaa.
4s
6 sink dg
dA=
Sa
f
2e*s8n"B
sin*
-~ C
Bh
Ap
cos f&
(@*- e) ¢2)
Pi -¢* cos
A — (FO
de
=
(i
v-6
cosB
- 26 (Fee ena) de - at ‘-e) Va oy
Ri
~
vr
2.
= t
tf
~ 20-7)
ain
av (1
b)
any
A
eg:
=
7
Zr B| + Ze snB]
4
=
20 (o-
tan
Jer? -c*
-
a
o)=-
tand
ae
Hf Ft~
ct
(Z-0)
- 2nJ/e?~—c?
we
AL.
To*
4
AWK — Qn fer
—Ce
-1
oe
aye
vit
2
6 +t4et ee
ar
£
zee
y*
V4
~
¥
‘me.
47t-ar)
ye
dé
~ 07)
\
+
-
{pe
—ce
tt
a2 (Fh
A
~
a Jpt-ec
ct
)
ae
Problem
4.181
.
4,180 through 4.182 Using Eq. (4.66), derive the expression forR given in Fig. 4.79
for
4.181 A trapezoidal cross section.
w=
C,
+C,%
web,
ot
FHV,
a
,
Cot
GY,
Cot
OY,
o
N
varies
wo
wielth
S@ition
The
v.
THK,
at
web,
ava
with
Aineards
b-b, = G(,-%,) = -¢,h
qe
feb,
7
-
b, - b
ope
Vb,
=
Vi-,)
Co
=
h Ce
h
aA
(se
=
ow
ete
~ dv .- J, fa ata
\
-
$(b,+ b
ge: 64hA -
¥2
dn
NT,
“
Co
\.
=
Co on
-
fb. ~ fi be 2
In
a
v: ” ” Y, bz
9p
&
oly
+
Cc
r[e
+ C, me -¥, )
Bias
YX,
_ (b
h
i)
yh
+ ba)
(by”
4h
(rib,
- Wh)
dn
~
h(b,-b,)
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers .
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.182
e180 through 4,182 Using Eq. (4.66), derive the expression for R given in Fig. 4.79
‘or
4,182
A triangular cross section.
The
7
b
;
a
vavies
Ww = b of P2V,
y
COE
w
Sineaad,
with
Vv,
+40
4
ZY
W=
fl
Fe
width
Cor
Oo
#
T
section
Co+
b
=
CY,
Cf,
Cin - Vv.
e,=
avd
)
wo
7
-£
af Ph
Ch
and
Cer
7c
fh,
=
bY
4
¥
I
rn
ga.
|
S¥
\
Ctw
rar
Cot¢¥
.
= der sol
Ow
an
n
= Cooly¥ 4 0,(%-%)
Ye
&
“
}
&
»
pert
a
oN
o-
HP
oO
t
S\e>
x»
ie
hy
1
bh
om”
q
-
bh
pk
y
i
r
%Q
ae W
R=
= \
A
SH
-
tbh
bE)
~
h
BR
4ae
.
—a
Proprietary Material. © 2009 The MeGraw-Hal Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—
. *4,183 For a curved bar of rectangular cross section subjected to a bending couple
M, show that the radial stress at the neutral surface is
Problem 4.1 83
At
y
ea}
a
dance
a (% 4
Is
iM ( Y- R
Sas
few
= M.
|
Aer
portion
he
axts ,
HS
and compute the value of o, for the curved bar of Examples 4.10 and 4.11. (Hint;
consider the free-body diagram of the portion of the beam located above the neutral
surface.) |
¢
. MR
Ke
Foy
x
above
the
neutral
is
force
resultant
ate S, Be b dh
. Mb ¢*y
we
~ “Re dy
= mb
eee
=
% rash
gn)
of
JS
Gy
* de
|
MBLR
(i-¢@
aS
=
ta
¥,
R
=F
~ dn
Gh!
ces
B
aA
“Sf Gy cosh b RAB
a
£
lé
=
For
ay
bR
equi fit orium%
gin
.
he
=
yesuttant oF
2
r
Fy
~
2H
26, bR sin Fo
sin &
=
oO
ee ( - & - Ing) sinZ = 0
w(1-
- 98)
oe
-
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.184
4.184 and 4.185 Two W100 x 19.3 rolled sections are welded together
as shown. Knowing that for the steel alloy used oy = 250 MPa and 0, = 400
MPa and using a factor of safety of 3.0, determine the largest couple that can
be applied when the assembly is bent about the z axis.
Properties
See
of Wioox (9-3 rolled
Appencix B
Aves.= 2480 tam™
=
Ly
For
one
rodfecdt
section
depth
6.40 =.
Maw
&
Ee
=
6
+
Depth z fobmm,
he IT 810° prim
inertra about
|
OHAG
A-&
is
b
#77 800° + (2480) (52 )* = hee HO front,
= 22-95 x10" pant
21,
= 106 mm.
_7 we
30
Sut E Tr
Problem
of
pag
1,
both sections
c=
moment
Ug + Ad® =
Loe
For
,
section
|. 7
133.33 ayMh
(133 3. sMo
Yo
Hee 7.
4.185
9LKIOare )
.
28°49
:
—
Lm
a
4.184 and 4.185 Two W100 x 19.3 rolled sections are welded together
as shown. Knowing that for the steel alloy used oy = 250 MPa and oy = 400
MPa and using a factor of safety of 3.0, determine the largest couple that can
be applied when the assembly is bent about the z axis.
2.030
| Properties
See
of
W (0041923
AppendixB
voded section
p50
Aven = 2480 min”
Width = loZ mm
Ty = fe6rk10° tant
For
one
‘I,
rolled
section
= Tt
width
n
of
inevtia
about
axrs
a Kio) mm ”
I, = 2I, = [6376
= 103
Cu = Es * ro
My = Set
moment
Adl®* = bbrxco®s (248) (Shs\* =
For both sections
c=
,
mum.
oO?
8187SF0 mm”
6 OE
£3333 Ne Dem eewet 2 U-QkN im.
01
is
oo
93M
7!
b-&
:
~
Problem
4.186
4.188
It is observed that a thin stecl strip of 1.5 mm width can be bent
into a circle of 10-mm diameter without any césulting permanent deformation,
strip, (2) the magnitude of the couples required to bend the strip.
wot
\
P*
C
(0)
Gaon = eS
(b)
M=SEPp
=
2 D
¥
G
Legbh? = gles Waizas? = 244 kro
4
> Or062ZT
fo my
q()=
wn
x1" F)"4
Y¥ a:s GELS Ale
= Geonst4e-
4xle) _=
. (Zeon 24
.
*
eat
|
= 1260MP,
Os OG4EP
Ahm. .
“s
antl
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 4.187
8 min
4,187 A bar having the cross section shown has been formed by securely bonding
brass and aluminum stock. Using the data given below, determine the latgest
permissible bending moment when the composite bar is bent about a horizontal axis.
8m
an
ee 32 mm —-—- of
Aluminum | Brass
Modulus of elasticity | 70 GPa
105 GPa
Allowable stress
100MPa
| 160 MPa
+
fs mm
-
-
aris
32 mm
Vs2
8 ini
Aluminum
\. Brass
aduwinum
For
adam inuw
Fo
brass ,
Vesues
For the trantormecd section,
T=
Bohs
SEG
of
-
GayP =
as
Vi
Y=
the
}
oe
materiel«
Y=}.0
105/70
-
E/E.
shown
ave
vefevence
on
the
LS
sketch,
.
|
.
32.768 x10% mm"
y+ Pe by (HE he) = HCN Caat— 1) = T6891
Iy=
De
DT,
T+
=
32.768 »10° mm"
Te Ty FH
ist = ("Z|
995 10% mm?
* 10" mt
= 141.995
me]
S= 100«10° Pa
Afuminums N= .0, lyl= [6 mm= 0,016m,
M = Coste Ca tse) = $87.47 Nem
Brass?
Choose
= 160%/0°Fa
one leS, lyl= lGmm=> O.0l6m,
M = SAY LIS 1 “4 ys gug.63 Nem
the
smaller valoe.
M=
837
Mem
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
A
Problem
4.188
:
4.188 _ For the composite bar of Prob. 4.187, determine the largest permissible
bending moment when the bar is bent about a vertical axis,
8mm
4.187 A bar having the cross section shown has been formed by securely bonding
brass and aluminum stock. Using the data given below, determine the largest
-- 32 mut moe]
permissible bending moment when the composjte bar is bent about a horizontal axis,
Simm
feof
ee
Aluminum | Brass"
Modulus of elasticity | 70 GPa . | 105 GPa
ts mm
Allowable stress
32m
. | 100 MPa’. | 160 MPa
*
| &mm
Brass
,
Aluminum
Use
ehaunin ven
For
For
Vatues
aS
the
relerence
afominum,
brass,
of nand
N=
n=
E,/E,=
shown
matenad,
|.0
|OS/7o-
on the
1.5
sketch.
For the transformeel section,
42 (e2)(48*- 32%) =
sie x5
w
MJ
f+
T= ah (@%-b?)=
ha b,”
Te = Ty =
T=
()(a2)*
I,
|
Afominum*
Me=
Brass?
T
_
|
M
n=l.0,
ly} =
IGmm
Csiro
-4
= 0.016nm
ds
valve,
ele
-4
ds
Gr
100 «/0° Pa
2. 2187
¥ 10% Nem
yl = 24 mm = 0.024 m
M = eo stot Hest
smabler
Ss
ag
Hoowt
n= 1S
the
It
”
6\/
Cheese
mm*
+ 354.99 410% wm = 354.99 "1077 wm"
nMy
™
= 2).8453%10°
2). 8483 "10% vom”
T4U,4
er.
IS |
= EO
311-296 ¥/0% mm"
€=
160 ~Jo° Pa
.
1.57773 «10? Nem
M = |.S7773%10°
Nem
Me 1.578 keN-m
<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
As many as three axial loads each ofmagnitude P = 40 KN can
‘4.189
7
Problem
be applied to thé end of a W200 X 31.3 rolled-stcel shape. Determine the stress
at point A, (a) for the loading shown, (#) if loads are applied at points J and
4.189
2 only.
Zooomm
A=
At
poiwl
A
@)
|
(bo) Eecentrre
doading 2
ge = Foxe?
4x/o7 >
Centric
_
(20 0v0
4000 %50
GR
bx10")(o105)
y=
gd
Cc
aives
_ &.
Ty = 3-4 x10
= fof mm.
L
foadings
fF = 2P = Fo kM
Shaye €
A=Momm,
‘
A
90 mm
om 0
A ppencdix
Woeoxrshrs
For
FaizokN
> Me ©
G=30MPa,
M=
ae
~(40)(v:09) = 26 kNA
G= Zanfa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or -
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=
P
lem 4.190
4.190 Three 120 x 10-mm steel plates have been welded together to form the beam
robie
|
shown. Assuming that the steel is elastoplastic with E = 200 GPa and oy= 300 MPa,
determine (a) the bending moment for which the plastic zones at the top and bottom
of the beam are 40 mm thick, (b) the corresponding radius of curvature of the beam.
120 mm —
“10
min
120 pin
— 10am
R,<
Or
D-~
R, +
Rs <
Q-
~
fs
O-+
> 2,
ay
A=
(120 (10) = (200 mm?
A,= (80)(1e)=
Boome
A,=
300mm
Go)Cle}?
R, 2 6,A, = (300 210% (1200 10° )= 360+10 N
R= 6A, = (300x(0% )(B00 xl“ ) = Joxfo" N
R= *6A,
=
+ (Ze0x10* )Go0x10"*)
= HfxJo
6S
Yat
4S mms
Ya =
20 mm * 20% 107* m
M=
mm
[Ry
=
6S
x1o*
485
xlorfm
+ Ryet Rays) = 2 {(3600(65)
+ (90 (4S) + (HSIAO)
p?
Eye
S67
Nem
(200 x107 )(30 x10" 2)
=
$6.7 bem
20 m
l
2
nie
T
(b)
WW
m
A
@Y
YF
Y
300
xto*
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. .
4c
Problem 4.191
4.191
'
A vertical force P of magnitide 80 KN is applied at a
eee
ee
te ABPAee Ae Porn point
Cn
C Jo-
cated on the axis of symmetry of the cross section of a short column. Know-
ing that y = 125 min, determine (a) the stress af poiut A, () the stress at point
B, (c) the location of the neutral axis.
Lotate
cen tyoia
Part | A,nine| Y, mm | AS mans
y
: 73 mm
® |
7s0e
we
@)i
Sore
| So
>
|f[zsvo
ZA.
= L[§7500
a tee * qo mm,
jrsvow |
1/8 7S00|
a
(s0ggq* = P:3I2CK16F ml
= Aleaee}4
1, ae
(52) (100) + (sees)
Y
es
|
T
:
I,
+
I,
“
22
25 mum
-
PL
°
A
Pec,
fam
fo
~ 2
-
A
~
TE
=
i200 45°
._f0000
.
Pea
3
een,
Mh
&4
D266
26x”
°
han
Dies
.
.
befow
aaheve
.
paint
|
o
©
1D
D4
atl;
Fa
f=0
eC
_
ott
sme
zt
_ to,
boo3
mA ee
_
(B0000)LO+ 6} )CO10S99
(12800) (20) ~ bo 3 min
axis
2407
=
IT
Ae
_
axis:
etree
Ly
-
dL
=
5
B26
*
“i2doo x5"
of nevive?
A
Neotrad
”
s
+
Pe
at
0980
Petr
eere
FS mm.
GMm.
2 ST
Cg 2 (SO -9
“
Location
Cyt
(80000 )(0:03) (01095)
.
of B
.
os
I
*
Sivess
°
XK te”
A
of
Stress
(a)
Sp
= 409917
410 pn:
50 pram
p
50 mm
@ = nw- 9s =Jomm
Eecentrrerty of Poad
min
(C)
| 37 se0
75 mm
40 mm
()
y _ ZA
aden
A
a
centretd
av
9S ~ bo-
3
A.
: Answer
3&0]
mm
Prom
point
A
<<
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manna! may be displayed, reproduced, or
distributed in any form or by any means, without the prior written perrission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using # without permission.
4.192 The shape shown was formed by bending a thin steel plate. Assuming that the
thickness ¢ is small compared to the length a of a side of the shape, determine the
Problem 4.192
stress (a) at A, (b) at B, (c) at C.
|
(b)
(Cc)
"
(a)
fh
Avea
Pec
A=
(22 EY F )
. PL
tr
Pec
r
le
>=
6 .-P-
.
ig tae
~
P_
+ PEGE)
knee
SI
ot
Laasa
SS
0.
oR
C*
~ PGi \GE)
+ta®
2
>= gat
zat
rat
ple
Sh
ae
A
Sv
centveid
BI
about
tt
inertia
bey
of
"
Moment
A”
oO
6
Ce
Qab
> aP
nat
ak
~-
rat
~
Proprietary Material. © 2009 The McGraw-Hill Companies, ine. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.193 The couple M is appliedto a beam of the cross section shown in a plane
forming an angle # with the vertical. Determine the stress at (a) point A, (6) point B,
Problem 4.193
(c) point D.
T= 2v'(Ee VEYwy? cf= (GFTe.- e
=(0.109757)(20)4= 17.5611 410” names 17 SEI ¥ 10"mn"
.
D
a
Ty
-
ir
r= 20mm
_
_ . 4w
Yo = Yo =~ 3H
Ye?
M,=
My
(a)
GR)
=
HLS
2Omm
ces30*
[00
sin
G = - Maye
30°
=
°
tr,
.
~ 62.8321
oe
mm =
62.332%/0"
z
-
L
ww Y
So
TBARS mm
mm
86.603
=
$0
y My Ze 2
=
6 2 - Maeve
T (Qe)
ZO
q,
“)
_
5 7 Be
100
“
?
QO- 8.4882 5
Z, 2 -2y=
y"
Nem
N-m
(86.603) (-8,4883x10°) | (50 )(20 ¥lo*)
17, Seu 1074
$7.8 «10% Pa
5 My ze 2 L (86.603 USI 7¥10>)
V7SEN «ton!
GZ
=
56.8 n10°
Pa
62,332x/074
6, = 57.83 MPa
<<
(50 )(o)
62.832%lo*
6, =
~56. 3 MPa
<=
(e) 6 = - Maye y MyZo 2 _ (36.603 )(-s.4saseid") , (So-2ox10*)
°
I,
17. Stl lon
Ly
25.9
x lo*®
Pa
62. B32xlo
6 =
25.9 MPa <a
Proprietary Material. © 2009 The McGraw-Hi} Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem.4.1 94
4.194 A yigid circular plate of 125-mm radius is attached to a solid 150 x 200-mm
rectangular post, with the center of the plate directly above the center of the post. Ifa
4-KN force P is applied at E with @= 30°, determine (a) the stress at point A, (4) the
stress at point B, (c) the point where the neutral axis intersects line ABD.
P=
dxfo*
:
= (4x 10% (aS *107*) sin $0
Mz
=
250
~
PR
y
Ni
}
sin 30°
-PR
i}
a
M,=
N (compression
+
Nem
cos
30°
- Cx JO? IQS Kio’ J cos 82"
~ 4382 Nem
‘
(200 WSO)? = 56.25 418 mm! = SEAS x10" yn"
D
Ty =
|
T,
A
G
+
B
=
Xg
*
Za
=
2g
=
75
A
= (200)(1S0\*
.
6,=-5
4,
4,
A
6
2
Maze
30x[0°
_ (2507SIe
3
x
56.25
633xjo° Pa
G
het
be the
peta
AB
on
100xlo-*
mm”
8
=
ZOx/O° wn
“fe
) , 433 %i(- 100x185 )
100 x10
[07°
—Joo x lo"
56.45 elo"
= ~233 kPa
nevtraf am's
urheve
intersects,
Xt F
a
= ef
Ke "M
CA
46.2
=
, Porat
G
Dies
xts??||
sey(rs
(-aSE
drt,
t’) SOx
. won
4 Maze]
BS
19°
433
ti
=
«IO Mm
WG.
a
mn
ya?
_ Casod(aswid*), (433 loo xlo)
Yxto®
Zox1o™
Ze 2 1S mm
G9
=
= 682 kPa
= -233xlo° Pa
(2)
mn
mm
3axto?
=
10010"
-
—Xp
a;
Ty
=
mm
JODO
Mem 2 HO”
Mey
Gye ee
A
b)
(180 i(Z00)°
.
Z
(a)
=
46.2 mn
Pron
porat
A
*
a
Problem
4.1 95
4.195 The curved bar shown has a cross section of 40 x 60 mm and an inner radius
ry = 15 mm, For the loading shown, determine the largest tensile and compressive
stresses in the bar.
40 mm
HF
A= GO) 40)
=
2400 mm™=
SS mn
2400 «107° wa”
sy
eo
os ~
™~
RGB
i
120 Neon
r=
ISmm
—
C=
=
e
“
y
FO T8E mm
Ry ee
wy
4o
r, )
£4
4.214 mm
R=
f=
@t
At
VY)
[Smm ,
}
¥
and
60 man
YO wn
.
= 90.786- 15
=
7
-e
e+
Aer
15.786 mm
(120 (15,786 «10°?)
(2400xKO
‘4.214% fo75
mm
BS
int
_.
JUS
is
Heat
* 1s*)
=
Pa
~/2.49
MPa
we
(compress fon *
At = SS mm
|
6=-
YF
JO.786 - SSF
(120 \( 2421x107?)
z
(4400% 0
= QA ZIY mm
ICH 24> lo-* )}(S8% lot )
~=
§.22*10° & P
=
&. an
MPa
at,
(tension\
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
4.01
Two aluminum strips and a steel strip are to be bonded together to
form a composite member of width 6 = 60mm and depth h = 40 mm. The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the aluminum.
Knowing that M@ = 1500 N + m, write a computer program to calculate the max‘imum stress in the aluminum and in the steel for values of a from 0 to 20 mm
using 2-mm increments. Using appropriate smaller increments, determine
{a) the largest stress that can occur in the steel, (6) the corresponding value
of a.
PROBLEM 4.C1
ot
Aluminum
.
a OF
Steel
h=40i0m
“—“_
|
<b = 60mm
‘SOLUTION
Tieansronnes See rion
fe
I ~yflnb-b
+
|
é
za)
Alu
2
Toe
2:
AT POINT
-
SSNS
been
NO
Fows7 fe
ERR
AT
h-2a
\
FoR =O
Tegrh ey [2 (s)ora- qe
"
n°
é
ni
qe
A (nb--b)
NELT
een RA
.
oe
h- | 2
|
n= am
AuuM
4
TT
a
|
h
h-$ HO
(are mo sree)
—
7° LO mm
USINE Bam [NTERVALS COMPUTE t “, LG,Sunn: Veveee.-
be
60 um
Moduli
of
h= 40mm
0.000
2.600
4.000
6.000
8.000
10.000
12.000
14.000
416.000
18.000
20.000
Find
= 200
0.8533
0.7088
0.5931
0.5029
0.4352
0.3867
0.3541
0.3344
0.3243
0.3205
(0.3200
35.156
42.325
50.585
59.650
68.934
77 586
84.714
89.713
92.516
93.594
93.750
Max
0.4804
6.4800
0.4797
Steel
=
7
stress
112.572083
121.572159
112 .572113
111.6
Aluminum
= 75
93.750
102.580
107.914
111.347
110.294
103.448
90.361
71.770
49.342
24.958
0.000
62.447
62.494
62.540
Stress
Corresponding
Aluminum
steel
MPa
‘a’ for max steel stress
and the corresponding aluminum
6.606 .
6.610
6.620
GPa
sigma
aluminum
MPa
%
m*4/10°6
mm
Steel
sigma
aot
PROGRAMrn OUTPOT
tA
M = 1500 N.m
elasticity:
MPa
stress
occurs
= 62.5
when
MPa
a
=
6.61
mm
GPa
4.C2
PROBLEM
4.C2
A beam of the cross section shown, made of a steel that is assumed
to be elastoplastic with a yield strength oy and a modulus of elasticity Z, is
bent about the x axis. (2) Denoting by yy the half thickness of the elastic core,
write a computer program to calculate the bending moment M and the radius
of curvature p for values of y, from4d to { d using decrements equal to 3ht.
Neglect the effect of fillets. (6) Use this program to solve Prob. “4.190
SOLUTION
COMPUTE
Sete
SY [ez
||
bert
ant
(CON SIOER
WWER AND
OF
STIREES AT JUNCTION
2.2 s
Maximum
YIELDING (tH THE FUANbES:
Poe
MOMENT
@2F
IN &271A
I.
bd fm 1 (tg-2Xd- -2t,)
ae
£L.AsTIC
jomenr
MOMENT:
se My=
Mom teWy
ct
tapay
UPPER HALF OF </2ass seerion)
£* d&
FL ANEE
a~_ (44)Ue
Yy
Y
RESULTANT
LETHIL
Tt
OF
STIESS
,
DIRE SAM
—
ty
(FORCES
Wy
“4
%
2°
98
i
Tp
|
%
|
> + (e+4y)
g* Jy”
~
|
S1ul-(@-le.
BL Sy
R=fF 5 [ &ele-F)]
a
Ro ET be fy, Cer
|
R=
Pr
& Geb L4y “(2 t)]
a
£T,
fe.
Sy
(«
~L,)
1
BENDING MOMENT
.
a
=] yyy le)
Onez Un
dy
/
O,7
et ——
M=2 2
K
Ay
.
(ADIUS
OF CURVATURE
: 2 (e-z,)
gt
47
sz &
YP"
=
YT
HP>
Sy&
s
=
vy
CONTINUED
PROBLEM 4.C2 - CONTINUED ©
;
(CoveipeR UppER HALE
Be Ty bp Ug
iy
Fc
see7 ron}
Skekocs
|
<= Vy
|
[~~ by a
-_
FOR NIELOING IN THE WEB
Ree Gt
R=
BENDING
Fare
W=
(ete)
PROSLIA
P
dy *
ar
le- a ) Je
er
Lar Yy
a, and R, forms) To 2.
A, FOR nal Tot
«2
ar
-zp/z
AVE 2 Ble a, fay
(uN
AD
\ALUES
NUM EET CAL
_ We
-ee/2° OECREM ENTS
MALE Ry
COMPUTE
WEOT
€, P
KEY 1 2xPRESsENS
FoR
ps <7
COMPuTE
Frere
ZY
fAOOMENT
FREMWES. OF CURVATURE
Facet
e-te 4)
We
UP p= on | THEN PRD
d,
CLERENIEN TS
ne SF 7OD, AD D>
) CHB
a
Plt N7
Y
APP HA
fag
OUR US”
For
Depth
a
d
beam
Thickness
of
Prob
= 140,00
of
am
flange
I =.0.000011600
m to
Yield
Yield
strength of
Moment MY =
For
yY (nm)
yielding still
For
70.000
65.000
60.090
yielding
60.000
in
4.190
tf =
the
Steel
49.71
the
in
10.00
mm
4th
sigmaY =
kip.in.
M (KN)
the flange.
web
Width
300
of
Thickness
MPa
rho (m)
49.71
§2.59
54.00
46.67
43.33
40.00
54.00
40.00
55.900
54.58
36.67
50.000
55.10
33.33
45.000
40.000
35.000
30.600
25.000
55.58
56.00
56.38
56.70
36.97
30.00
26.67
23.33
20.00
16.67
flange
of web
bf
= 120.00
tw =
10.00
mm
mm
PROBLEM
4.C3
4.03
A900 N- m couple M is applied to a beam of the cross section
shown in a plane forming an angle PB with the vertical. Noting that the centroid
of the cross section is located at C and that the y and z axes are principal axes,
write a computer program to calculate the stress at A, B,C, and D for values
I,= 0.62 x 10° mmt
SOLUTION
X1013 )
)
WARE
Ci
PLL
« 2Q)=-
p(ejn~50
Ds
26°
|
E
yl
Bl
baal
al.
iam’
Bye
ap”
~
20 mm
Se
71 2}2-3o
-
“s°
LOA EANEINTE ole
30
3
Yerrla} jx,
35
Je*
2
39)
FO /BO° VEING YO” INCREPAENTS,
=O
Fehe pe / TOY USINE WHT [CLEMENTS ,
FoR
“VAL URE
ER
ASS
PRINT
AND
STRESSES
RETIEN
lee Tyr
FRBERPIY
OUFPUF
Moment of couple M = 700 Am
Moments
of
inertia:
Iy
= 2-£9 XroO° in ©
eb
Iz
= O662K/0
‘Coordinates of points A, 8, D, and BE (X/o”)
Point
A:
2{1)
= 5v:
y(i)
=
35
Point
B:
2(2)
=~:
y(2)
=
3°
Point
D:
2(3)
==-30:
y (3)
==3S7
Point E:
~
= 30:
OB
at
¥(4)
Points
= 35°
-~-
"=
befo
= A
dey
MPa
MPa
~O-1g}
-Lated
= S266)
-Sh30
-2R672
4S CSL
= 34.26F
BSR
BABS
40
|
2(4)
« + Stress
40-11-1136 13
Hn
Mzg=Mcos @
TOs= Ma gn), My 2»)
La HSE pags £731
~3s"
_
AM,
ee
PREbRAIO2
bars
p= = y=
3/6) =
(4) =
OFF, Be,D
Db
E
MPs
MPa,
ba 16}
[30
(% 46]
47,093
-26-146
40.308
180
Sao tel
Qo-f6t
= -$2+ 1b]
~£2+ (bf
Pr
.
RL
PROBLEM
4.C4
Couples of moment M = 2kN+m are applied as shown to a
curved bar having a rectangular cross section with A = 100mm and
b = 25mm. Write a computer program and use it to calculate the stresses at
points A and B for values of the ratio r,/h from 10 to | using decrements of
i, and from | to 0.1 using decrements of 0.1, Using appropriate smaller increments, determine the ratio 7,/h for which the maximum stress in the curved
bar is 50 percent larger than the maximum stress in a strainght bar of the same
cross section.
,
4.C4
SOLUTION
AyfuT:
hz 00mm,
b= 2S onan,
fee s7rarur ene! Ty 2 22
Do
Neration
pe her
eer
Sikessec;
Since
3
Sye - Wy
MOG
hh=]0O 7m,
FUOERAM
OF Se,
AIS,
KEY IN
R= h/faly-r)
R)fner)
a
fh #40,
ti = 1002
70400
AT
ALSO £inunve:
FETURH AND PEPEAT
=
Lt.
Se =
6M.
4g
MPa
FOlLLewInss
2
@rr-k
a 54-2800 (T)
pe y=: M (n-Rine .re) \
VN)" (200m,
USiné L@unvon: OF Limes
PYOERAM
JHE
Feith
ns
FoR
Fore
3
6
Nae
Fotiowmé
s
> 2AM
ALSO
G/h2/0,
9, 2/00
= (eo CkewemENTS
I AyTL £VALvaTE
ratio = &, / TSoraunrFOR
Gx (00
Fe SO wy
ty RyF, 6 T, wt,
_
DENERENE
~k
OUTPUT
M
=
Bending
rl
nm
rbar
mm
1000
920
800
700
600
500
400
-300
200
100
30
80
70
60
50
40
30
20
10
ri/h
{ Note:
for
GSMS
48.00
=
5.730
eR MSS
SS
RSS
Ses
5.730
6.170
6.685
7.299
8.045
8.976
10.176
11.803
14.189
18.297
desired
having
8.703
8.693
8.683
for r1
ratio
=
MPa
100.000
=
ri/h
a rectangular
A
=
2500.00
rifh
-
mm*2
ratio
-
«49.57
-49.74
~49.95
~50.22
-50.59
~51.08
-521.82
-53.03
~55.35
46.51
46.36
46.18
45.95
45.64
45.24
44.66
43.77
42.24
10.000
9.000
8.000
7.000
6.000
5.000
4.000
3.000
2.000
-1.033
~1.036
-~1.041
~1.046
“1.054
-1.064
-1.080
~1.105
~1.153
-61.80
38.90
1.000
~1.288
1.000
6.900
0.800
Q.700
0.600
0.500
0,400
0.306
0.200
0.100
-1.288
-1.316
~1.350
-1.393
~1.449
-1,523
-1.631
~1.798
-2.103
-2.888
6.527
0.528
0.529
~1,501
~1.500
~1.499
Rte
38.90
38.33
37.69
36.94
36.07
35.04
33.79
32.22
30.16
27.15
straight)
-72.036
-71.998
~71.959
52.8 mm
is
in.
sigma2
MPa
~61.80
~63.15
-64.80
~66.86
~69,53
~73.13
~78.27
~86.30
~100.95
~138.62
(sigma max)/(sigma
94
94
94
is 1.5
h
sigmal
MPa
0.794
0.878.
0.981
1.212
1.284
“1.518
1.858
2,394
3.370
144
134
123
113
102
91
80
68
56
42
103
103
103
stresses
The
kN.m
e
mim
144
MARRS
150
140
130
120
110
100
90
80
70
60
52.70
52.80
52.90
Ratio of
2.
1049
949
849
749
649
548
448
348
247
150
SRIRAM
=
beam
R
mm
1050
950
850
750
650
556
450
350
250
100
Find
Moment
straight
in
Stress
valid
=
or
1.5
35.34
35.35
35.36
ri/h =
for
cross. section.
-
any
]
0.529
beam
PROBLEM
4.C5 The couple M is applied to a beam of the cross section shown.
(a) Write a computer program that, for loads expressed in either SI or U.S. customary units, can be used to calculate the maximum tensile and compressive
stresses in the beam. (b) Use this program to solve Probs. 4,7, 4.8, and 4.9.
4.C5
SOLUTION
INPUT:
/ ven!
Yu
foe
AAREA=
AF
BLENDINE
MIoMENT
ENTER
4, ha,
Ay.) +(hy,
fA
Fm,
é,,
AND
2
+ hy/2
(PRINT
[MoniNT OF RECTAN SLE ABOVT
Am =(AAREA) Ay.
[Foe wHat® C2OSS SECTION|
MmemtAm
LOCETION
3
Akba= WER + DARE
CENITROD
OF
ABOVE
BASE
4 = %/ppea
MOMENT
ME (NEl2 TIPE
For msl
ABOUT
7am!
HOI BONTAL
(PRINT)
CENZF(2OINGL
AXIS
y= Ay.4¢+Ay 2 + ty fe
AI= bn hifa + bby to) G-&m)
L= Tral
o
LCOMPLTATION
.
7FORRL
HEIGNT?
SIRES AT
.
(Fb
wel
AT
MG
ff~-
FOR
@
/
(Per)
BOTTOM
g
PALE
Jan
TOP
Mryg =~
STCESS
NEXT
(FEIN7)
.
OF SHIRESSES
SEE
BASE |
.
FRIVT BUTS
CONTINUED
PROBLEM 4.CS5 - CONTINUED
Problem
Summary of Cross
Width (m)
4.7
Section Dimensions
Height
tn)
0.225
0.05
0.075
0.15
Bending Moment = 60,000 Nm
Centroid is 0.075 m above lower
at
at
top of
bottom
of
beam
of beam
Problem
Summary
of
Cross
Width
(m)
Inertia
OR
Stress
Stress
Moment
#
Centroidal
Centroidal
Stress
Stress
is
at
at
Section
~94.103
(56.462
MPa
MPa
Dimensions
Height
« 60000
| 0.1194
Moment
(m)
top of
bottom
Cross
(nm)
Nm
m
of
above
Inertia
beam
of beam
PROBLEM
Summary of
Width
x 10-6m4
0.025
0.15
0.025
Moment
Centroid
79.
4.8
0.1
0.025
0.2
Bending
edge
is
4.9
=
=
lower
is
edge
60.59
-79,.815
118.237
.
x 107%mé
MPa
MPa
|
Section Dimensions ©
Height (am)
50
20
10
50
Bending Moment =
1500.0000 N.m
Centroid is
25.000 mm above lower
Centroidal Moment of Inertia is
edge
512500
Stress
at
top
Stress
at
bottom
of
beam
=
-102.439
MPa
of
=
73.1471
MPa
beam
mm74
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
PROBLEM
4.C6
4.C6
A solid rod of radius ¢ = 30 min is made of a steel that is assured
to be elastoplastic with & == 200 GPa and oy~ = 290 MPa. The rod is subjected
to a couple of moment M that increases from zero fo the maximum
elastic
moment My and then to the plastic moment M,. Denoting by y, the half thickness of the clastic core, write a computer program. and use it to calculate the
bending moment M and the radius of curvature p for values of vy from 30 min
ven een
to 0 using S-inm decrements. (Hint: Divide the cross section into 80 hortzon-
ial elements of -mim me
My=%
SOLUTION
Pg
CONSIDER
LET
7o
€-1
(0+0 3)3-
gre 42 (2490K18) # (0-03)
y 3
FOP HALF OF FOO
€ = NUMBER OF ELEMENTS
i
n=O
3 .tonondd
rr
= Tv
Bh
FO
)
HE 7H OF
61s
Mfa
104
=
44Nh
I TOR HALE
Fact FLEINENTS
Aha
+
STRESS OAM SLE KENT
s7er4
Yar (Ay
e=[ <7 = {inrosjahy |
(F 4
dy
wpe
wy
GOP
106
200
GO 7TH
(nea
106
dh
a
FAT
MDBEINT
OF
&LEIMENFT
Fete
Yy? 3a
wt SVIRESS IN ELASTIC CORE
fr
a
RO
i
T= Ty
SEES
i
FLASTIC
A MOMENT
DECRE MENTS
(AarEA)
se ALOKRCE
(x He SIE
fy
Me: M+ AM OMENT
P= ye),
PRINT
dy, A,
AAD 0,
NEXT
FILES
yo
By= Oo
AT Biam
BONE”
AnREA = 2 2/b4)
AFoRCE= Tp
AEPERT
OOF PUT
Radius of rod = 0.03 m
Yield point of steel = 290 MPa
Yield moment = 615 MPa
Number of elements in half of the
Plastic moment
rod =
40
P (mm x 103}
26.690
17.241
43.793
10.345
6.897
3.448
0
=
104.4
KNm
PROBLEM
4.C7 The machine clement of Prob. 4.178 is to be redesigned by removing part of the triangular cross section. It is believed that the removal of a
4.C7
sinal] triangular area of widih « will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to calculate
the maximum stress in the element for values of a from 0 to 25 mm using
2.5-mm increments. Using appropriate smaller increments, determine the
distance a for which the maximum stress is as small as possible and the
SO nt» -<-——— 75 nm. mot
corresponding value of the maximum siress.
SOLUTION
=
SEE
M=5lonm
FOR
FIG 4.79
rg 2125 min
AxDB
ne
Ze
3A
AO
FPCE 287
b,= 62 nm
AT
Dil WTEIIALS
Atm
6= b, (a/(n+a)
Agen
4 tb)(h/2)
5:x zo-t/s bh(a)+
$ b,4 CK) | faven
rr
¥y~(h-
x)
sh'lb, + bo)
Re
(5r.- 45) ty B = hlb-
8)
e=7rR
Min
me
£)f{ ance cer) |
ctf
PRINT
ANB
(rz)
kETUEN
PAOERAM our FUT
mr
pe
0
5
WF-FIT
98273
pg
A
tha
a
+ S8ebeb
-So- 127]
[4 4849
O
folkb00
14-955
Ae31P
tol SSB,
3-693
SSE
12°
YT
AAE DED
{66990
10:668
(02870
3.023
Zo
ioR-0S7 = - LASTS
dor £83
ITol®
104.394
9389
(08 UF
[4],
24
.
103.86]
Tm
~4P022
ee
Determination
a
R
min
mm
(8°
of
:
236792
eae
a
the maximum
sigma D
AboFe
an
compressive
SiG ha B
mf
a
a
a
ee
stress that is as
by
ame
a
MPa
bm
pln
ain,
fed+ 669
44/9400]
1§:2200
L32f
fo- 33$
i 61S too b60
G-277
-44-8992
[$a7bb
[32
foe33%
92.77
4468994
183346
1324
932%
0-977.
1SSTS
Answers
oor BG
When
as/Sb28
Iam
tho
compressive
cess
is
ee
small as possible
44-9 Mf,
Chapter 5
| Problem
5.1 For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (4) determine the equations of the shear and bending-moment curves.
5.1
;
Reactions ,
|
wh
oe
DeEM,g=
-AL+wlhkso
DEM-o:
BlL-web-0
be
oo
f
Free body
for
B
ali agram
ret *)
LY
Afr
x ——
weL
4
clatermini ng
ZN
ae
wh
M
uv
x.
at
:
.
dist
buted
concentrated
Load.
Replace
Road
by equivatent
|
V
&
2.
aS
wl/2
M
7 .)
AT
sectfon
v
wx
reactions,
V
¥
BF
wel
Place
M.
| -
fr
-
J
Of
beam,
whole
Over
Az
wh
a
<t
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.2
,
5.2 For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (3) determine the equations of the shear and bending-moment curves.
Reactions,
LA- bP
DIM:o:
LO-aP-0
From
A| to B:
vo ip
A
HEF,
c+ 2a
O«usé
:
|
DIM,
|
xX
From
B&B
to
co
M
— Pab
L
0
a.
L
~
.
0
“SEL
=o:
DEMwyso:
Pbx
a<%<1L
j
(k
|
M- Ebxyso
.
Ci
My
”
=o.
|
s~
-V=
b
L
3
a
PkTc
se
‘
A
Bb
A=
a
QO
GZM.=O:
°
LY
e
1°
be- 1 -x% —>{ Pa
vi fo:
O
L.
ve-$
-M+ F8(L-x)
At section B:
M=
= 0
ie
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill
for their individual course preparation. A student using this manual is using it without permission.
_ Problem
5.3
$4 through 3.6 For the beam and loading shown, (a) draw the shear and bendingimoment diagrams, (5) determine the equations of the shear and bending-moment
curves.
Fyom
PL
A
to
.
M
—)
[a
Be
On
2H
atehso:
ots <D
.
-P-V=0
a
.
tDEM,=O2
Put
«<.
Y
Krom
a
M=zoO
M=-Px2
Bote
Ct
QO Kw
~<a
€ RA
=P
.
Yh
Po
HDR,= OF
or
VV
4D EM,
FO:
~P-P-V=zO
Ve
Px + Plx-adb+t
~2P
Mes
-"
O
M=-2P%+Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers,
and educators permitted by McGraw-Hill for their individual coutse preparation. A student using this manual is using it without permission.
=<
Problem 5.4
5.4 For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment curves.
1WexX 5 |
ZFS- O 0 gx
V
Vv
_
.
.
x?
§
~
=O
we
W, L
Moen
= “5
wit
&
x3
6L
hb
=a
©
_ w, L®
| Manan
CG
A
.
w,L
Vem
=
M
M
x=L,
L.
” |e
Domo | Mad
At
»
ob We
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
§.1 through 5.6 For the beam and loading shown, (a) draw the shear and bending-
Problem 5.5,
_
moment diagrams, (b) deterniine the equations of the shear and bending-moment
curves.
Using
entire
$9M,=2
beam
O01 -AL
ABCD
+
asa
PiL-a)+
Free
body ,
= Oo:
Pa
A=P
Me
Vv
OF
DL
- Pawr
PCL-aly=
|
e
RK
DFP.
Cheed:
a
Vie
Fiream
H#2F, =
Ate
B:
P-P-P+P
Qn
P-Vsto
malt
M
=o
Lea
tT SE =O:
I
-P
pe
©
Ve
Po
=
4
Pa
Vv
+5 2M,
70
> -Pe4
M
From
Bete
P
|
C?
tha tS
ra"
ik
Pe
acexn<
M =O
=
Px
a
L-a
=
O
P-P-V=z
a-
V
V
Oo
=
oO
<2,
a
49 FMy
-~Px
=
O
+ Plrw-aliM=o
Mo=
From
C
to
"cla
D:
Lema <n
Pa
‘4A
eb
Vv
ley
+TZF tO!
aE
Ve P=
My = OF -M4
O
PCL-x)=FO
ve-P
hPa)
«a
5.6
Problem 5.6
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (5) determine the equations of the shear and bending-moment curves
Caleodate
feac tions
distwboteci Poael by
Goncentratec
oad,
Reactions
A=
From
after
an
veplacing
equivadent
are
D
= 4w(L-2a)
A to B:
O<x<
a
$w(L-2a)-Veo
PEF,= 0:
>i
WV
=
ava)
DM
=O:
dw(L-20)
—
~Lw(L-2a)+Me 0
Me dw(L-?a)x —«
M
From B te Co:
w
Ba
fas
3 W(L~2a)
Place
section
cut
X&.
Replace
:
VY mo
-tw(L-2a)x
D M,= 0:
M
roe
?
hoad
by
Vv
equiv.
:
cove.
Ava.
a
Vzwle-x)
=
w(x- aX 833)
+
Tb yl
?
X= 2
tw(l-Ro)
diste: butec!
bw(l-2a)-w(-a)-V
s 0+
+f ZF
at
4
a<xu<tL-a
.
+M-eoO
M = dw lL-2a)x —&-a)*]
From
< bo
x <
Lea
C to D?
‘Me
V
=O:
WEE,
.
V+
lex Tyw(e-2a),
Vv
DEM:
o:
=-M+
ae
4
CAR K= BF
Mm?
¢w(L-2a)
WCG- 3)
:
a
z
<0
=o
=
(L-2a)
dw(L-2a)(L-x)
=
Me dw(t-2a)(L-x)
=<
Problem 5.7
§.7 and 5.8 Draw the shear and bending-moment diagrams forthe beam and loading
shown, and determine the maximum absolute value (a) of the shear, (6) of the
bending moment.
20 KN
920 kN
Shear
0.225m 03M
Diagvrame
A tec.
CtoD.
G225m 03m
s
VV => Z0kN
= O
#2Fy=0
-20m@Y
APZR roe
~204+-43—-VzO0
=a? kN
Dto
V. (EA)
-20149m20-V =O
+7 Fy 7 OF
E,
2g
VFSEN
|
E+
B,
0
- 20~V
— 20,492
+ fZR +0:
8
Db
Bending
E
-
~
12
M (kif)
§-7
At
As
At
Cc.
+5 2M.
20
O
(20 \lou26\ + Ms
M z—45 kNin
oD
2f
-
'
*
:
‘
A
ME,
48;
452 My
At D.
20
a4
“lt
_oM
Cc
Db ‘
= OF
(20\o-s2s\-(48\(0:3\4M
= O
2 3d
END
mM
—
Vaaseos 4
(2010-15) - GE\G-828) 4 (Zo\O225)+M=
HOZM, = OF
“|
M
E »)
¥
|
[P7786 Ja.0-3
e775)
225
O
= 5:7 ENin
tIEZM, = O+
(20) (1-08) ~ (49) @+825) + (20) 0-815) + F0\(o-3\4M
“E: J - =
ke Zags
0
M=
c
Db
26
Diaqvam,.
M
20
2
ce
Moment
q
~ 20
A
Vir -iZz EN
By
\
Cc
A
O
=
M
V
Frow
+he
= 2-1
=
O
kNm
diag rams ,
)
(b)
[Vinay 7 29 EN
1M
| nee
=
4-7
ete,
kNin
<*
- 5.7 and 5.8 Draw the shear and bending- moment diagrams for the beam and
Problem 5.8
loading shown, and determine the maximum absolute value (a) of the shear, (6) of the
bending moment.
24kN
24KN
24 KN
24KN
Reactions
DEM_e
FE.
= oO:
(09s
|
0.75 m
Wz)
=
O
Rat
Deh,
0
A ond
+(225) (24) +C1.S 024) +(0.75 4)
BR,
-— 4@0.75m=3m
et
BO
xn
t
=O!
(O79) 24) ~ UL SOXR4) = (2. asNes) + 3Rp
-(3.75)(24)=0
24
Re= 66
Shear
Ate
C.
Ve
30
Cte
D.
Vz
30-24=
Dt
Ed
E,
F
Vs 6-24= -19kN
VrurlB-24
=> -H42 kN
Fr
B.
ft
KN
-424 66
AtA ood B,
At
GkN
f
Vet
fq
-13
WN
diag rom.,
Cc.
Maz
OS
c
= 24 kv
M2
Mg
O:
~(0.72S\(20)4
ce
=|°
M, = O
Me = 22.5 kom
1D Z Mo
~(l. sol(20) + (0 asya4)
+My
M,
At E, —b
4g
}
°
9
y
F,
27
kN-m
-(2.a5)Ge) +0. 50)Q4) + (C79) Gx) + Me = 0
From
=
18.5
kN ems
|
the
diagnaws
|
-ot
~ Mp -(0.75)(a4) = ©
Me
=
ye
‘CE
DIM,
Oo
DEM, =o:
Me
At
=
ad
{1g
kN
-m
CO) [Vlog = 42.0 kN
(b)
IM hme >
27.0
kNew
<2
ae,
Problem 5.9
5.9 and 5.10
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (6) of the
bending moment.
30 KN/m
roncentrated
Ffvee
ee}
otter
dvawing
A= 72 kt
SR
0%
2ZM,=
+
>
on
UMU6)-
-
B60)
B= 4ekyt
="
VCkWN)
+he
OF \e0)+ @\Go)=0
-S5SA4
Or
eB
®
©
*
dond
equ! vaheut
au
ACOB.
4D EMg2
|
al
body
by
Poad
buted
digtns
72-60-Co+HR
Check AZ R=Ot
=
O
O
72
wut YY
V =(72 - 80 % ) kN
45 EM, = OF
- ‘2
* ) kNewm
72
= 1S
M =(
q6
34
Cto
Ba
D.
“
kNim
TAN
4tZR zo:
w%
(x =2
m
eq
43 ky
-xK ym
4 DIM,
72 -(0\@\452M;
V+ 4a o
Mz
+4a(s-y%)
= o:-M
.
240
- 48x
”~ Oo
+P ZF,
—
Vzo
= on
Tam + Boyarte-l)+M
_
,
zo
M=(12% 460) KN-m
= -423 kw
A
© Bm
T)
ar
&
rm ©
&§
x
14
ke s
Mero
~2%x%+@0xrx)\E+
M (kNem\
M (e
0
72- 30%-V=z
be"
%
[bp
Ke
€ 2m
O<%
¢
Ate
From the
diagrams
2)
\Viu.= 720 ky
<8
(b\
[Minow = 96.0 KNem
<0
Problem
5.10
5.9 and
5.10
Draw the shear and bending-moment diagrams for the
beam and loading shown, and. determing the maximuni absolute value (a) of
the shear, (b) of the bending moment.
45 kN/m
120 kN
Reactions
DEM.= 0
=1GA +(0-4)(81)~ 0.420)
Az
SLEN
{|
|
je (VS
|
FC ~ (0-9)08I) ~ (2.2) (20)= O
|
C, 2 220°5EN.
DZMy=O
n0kW
-/45EN
©
4
Cc
A te
vy
ful
.
[hk
= 195)
8
2X
O<
FRY
7 O
DIM,
bom
BEX.
1
4%,
e
“(5 —-4Sx-~-V=2
|
OD
>~-(95~45x
~ (45x \($)
=f/eSK
C to B
foBme
|
EN,
yx 227m.
M
kAlin
kN
~M=zo6
22-567
M sassy
8 x
“lof
+m
HAY
eB
V
M (kNm)
ct
<
|
oe
= 0
x
;
ype,
.
tPSR
“fogs
le
C
ABEN|m
i20 kal
~
Cer
VE R27 =X
PER
=o
V-i20o
|
DZM,=0
the diagramed
Vi= 120 EN
-M — @2xX\i20) = O
M=/20X
From
=O
- 3224 kNm.
(o.\
Vln?
Zo kAl
(bo)
IM)
= 10h kN a
<a
=e
Problem
5.11
5.11 and 5.42 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absohite value (a) of
the shear, (6) of the bending moment.
&ry
LOKN
G4kN
. 20 nin. BY)
8
20 mm
=
M,
=O
~hko-
50
{ZA
.
BG EA
leb
VAN)
] x
EF
A
~hb
-/GEN.
AhE
Ve
Er F
FRs
Va -PEN.
Veo kal
and
At A
AD
M (him)Mm
PN:
@:
A
Nees
|‘
4p
MOY’
E
=9
4@-tb
64
=O
Gyz7Z EN
~C4+Gr= 0
ASE =O
ok
300 nt tn! 300 oun 300 mun 300 mma
~ (36a) (tb)
(300 ) (b-#)
+
(900) Uh 6)
+
~40€
:
4%
M=0
B
DAM, = 0
== O
Ms zo ~ Oe04h ENM
(“3)(ko)+M”
*)M
OB
kos,
Wy
OEMs
i"
O
(0-3)Lleb)~ (00272) + M 20
| ore
M=
0°96 ENm
TZ
@\ Maximon IVI
=O
(066) C06) - (rortqzJ +M=
M = 064g Em
2
iml
=fe92 kein
DIM,
M
=
= BEN
(b) Maximum
wet
ALE
—
~~
Gb
|
AP FA ["
MCQjes
4
~M ~(0.02) (12) ~ @3)(fe6) = 0
MOF?
EM, =0
(6
M
y pe37)
Moz
= [242 kA.
-~M ~(@3)(-6) = O
M=-0-4G2 kN im.
0
Problem 5.12
3kN
"5.0 and 5.12
3kN
450N-m
i300
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (2) of the
bending moment.
DSM_g = Or
on
(709)(3) ~ USO 410 (>= 900A= ©
Az
"200° 200°" 300
Ft
B=
Dimensions in mm
At AL
Vv (hy)
2.85
2
Vs
E
At C,
8
?
M=0
Vez SKN
CC.
At
3.45 kn
2.55 kN
9)
~ (700 (314 1009B sO
- 480
-(B00)(3)
=O:
DEM,
2.55
kN 4
s)
DSM,=
j
~O.45as
‘leona se) +M
+ S00
20
%
2.55
Vv
Msc
765
Nem
—~ 3,45
M Wm)
C to E,
At
Yr
-dO.45
.
D,
|
3
New
S)
DSM,
~{500) (2,55 }+ (2008 )
J
300 > 200
2.55
°
iy
~
+M
=
Mz
At Dt
oO
675
Nem
DSM,=
~ (Sa0\(2. ss) +200 )(8)
¥
4.55
Et
At
v
&,
Yor
E.
3-450
AV
Mr
= S45
DEM,
C
*M
= 0
1125 Nem
kN
20:
-M + @e0\(3.45) =
Lsoo of
M=
3.45
At B,
Ve 345
kN,
1035
Nem.
M=O
@)
Maximum
|=
3.45
kN
~
(bY
Mavimum
|M\ >=
1125
Nem
~<a
_
Problem
5.13
§,13 and 5.14 Assuming that the reaction of the ground to be uniformly distributed,
draw the shear and bending-moment diagrams for the beam AB and determine the
maximum absolute value (a) of the shear, (+) of the bending moment.
Over
the
whole
beau,
0310
‘a
8
TAKEASERIAS|
=O:
2n,-
Vi =
©
V =(A%) kN
.
+9M, 20: -Ox\E)4 M=0
atm
|
|
Ms (27) ken
_
[“b
OB
w>
Ve
|
aes
<
£%
O.am
D.
+e
C
0.6 KN,
Sky
si
kNem
> IO Nem
12m
+f ZR,
——
=O:
2,-4.5
per tre
lew
M = 0.040
20
-V
V2 (Rn -w5)
Hy
KN
—(AxV%) Hi S\(%G-0,3) 4 M=O
s9EM,z0!
Me ei ~ 1.5% 40.45 ) kNow
At
the
ceuten
of
the
Y= oO
beau:
O75m™
Miz = 0.1125 ken
=
m2 0.8m,
%F
-—TH2.S
At
Ch
(2)
MagimimlV1=
0.9kM
(by
Magimun IMbs
112.5 Nem
New
Vz=- 0.9 kv
=
GooN
A
Oe
}
Cc
At
M (Nema)
+f2k
(aM! 7]
OG
~O, 5
Um
Of % SZ O38 mm
. Ar—;
im
LS peo
fy
2
A
v CRN)
we
A te c.
ts lav
1.5 bo
=o
LSw-LS-1LS
+f ZF, zeOr
0.3 mn
a
5.13 and 3.14
~ Problem 5.14
36 kN
36 kN
10 kNAn
determine the maximum absolute value of (a) of the shear, (4) of the bending moment.
lokNim
: 0.9 rm
V (KN)
, 0.9m
‘ 0.9 xm
13
BE w -(0.9\Ir)-36 -(0.9\l0) = ©
We
E
Cc
,
Oo<x <o9
m
45"
: Tr?
8
D
IS kW/m
A th
» ALS
;
tA
Ww
|
=
afZ
beam
whole
Over
“09 m
Assuming that the reaction of the ground to be uniformly
distributed, draw the shear and bending-moment diagrams for the beam AB and
*
mY
|
Vz
|
x=
Vz
C+D
Om
- {ORN/m
ivi)
“offi
MIM
ih
4SKN
2.025
€
4M
x
<
kN-m
L Bm
_ pee
Maximuin
ZR = 0
S125
iml
KN-m
X%- ONS
See
BA,
ISx-%-V=0
Vz
(b)
4+ Meo
M= 25x°
Mz
(a) Maximum |V1
= 43 kW
Sx
allox)k
- (Sx) %
Dz Mz 20
Af
O
Sx -lOx-V=
MEkpz0
-18
1Sx-9
DEM,=O — ~0SxYE) + 9 (x-0.45)
~l
+M
Me
AD
Vz
M
Dte
B
2.5x2- %x
=
©
+ 4.05 =0
18 kN
=
[2.15
Dse
S
the
ethene
KN-m
mnetyy to
cabcubate
al ‘Fendling
moment,
5.18 and 5.16 For the beam and loading shown, deterinine the maximum normal
stress due to bending on a transverse section at C.
5.15
Problem
JOKN
1
3kNAn
Dsing
20
ange
ESm
15m
DEM.
—~M
2.2m
as
+
2.2
§S
©
for rectanade
modudus
bh”
= b(too)(z00)* = 666.7% /0% mm
~
Normal
ae
&
666.7 K10°* m3
stress
M
G
Problem 5.16
=
7.260108 Nem
=
S
body
(22)(3ue* Mit)
Section
CE
a Fre
>
Ms
m | nny
Vi
CB
_ 7.26
«10°
CecTxioe
=
10.89
>
7
«10-8710
iy?
Fa
MPa
ah
_ 5.15 and 5.16 For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
3RN
3kN
18 kNAn
Reaction of A,
+9Mg2 OF
oe
~4.5 A +(3.0)(3) + (.S)(3) +0. Bd. s(27sv=0
A> 705 kn f
Use
AC
1.8 W/m
TTT
as fyee
«Me
body
49ZML= 0:
‘
—
At. iste
&
10s kW
ML = (70s )CLS) + (8X5
0-75) = O
I=
ib
lh
> 7g (30)(s00)
News
"
8.55 x10
mF
ML = BSS KNe
= 180 x io* mun =
IB0O
10°
wt
= $(00) 2 1SO mm = O./5O%
Me
Se TD
_ (3.58%!IO*)(0.15e)
7
186610
_
7.1a5*1O" Pe,
=
713
MPa
<<
Problem
5.17
5.17
For the beam and loading shown, determine the miaximuin normal
stress due to bending on 4 transverse section at C.
1O6kN
LOOEN
WAl0 X 114
Ry + (376 )(100) + (4X feo)
4S
~
A
a
oi
~
.
M=
144616 EN m.
3.
S= 2200 ¢10? mm
vodbed stec# section
W4iox 4
For
fe 0-7
=O
+ M
~ (75 )L194-2)
a
DHIML=
C,
ak
moment
LA
(92:2
Ra=
= Oo
(75) (228125)
+
Bending
-o
IDEMe
A.
at
Reaction
75 KNam
v
(Fa-2kN
V
Normed stress
at
- A
c
~
3
[FEIS¥10
Ss
Problem 5.18
Ca
22
Kio”
6
3
MPH
-
<=
5.18 For the beam and loading shown, determine the maximum normal stress due to
bending on section a-a.
30KN
50 KN
50 kN
30 kN
W310 X 52
a
te
2m
Reactions :
ft
+
_.|
+~——- § @ 0.8 m = 4 m———>|
36 50
A
ZR
Using
DEM,
M
)
By
=O!
Symmedyy
A=B
Bett
held
of
A:
=
bean
8
g0 kN
as
free
body,
= 0:
:
~(go)(2) + (80)01.2)+(S0)lo.4) + M = 0
Fakedal |
M=
%°
For
104 Kem
W310 ¥ 5z)
= \04xIo* Nem
S = 748 x10? mm’
= FUR v jo* ww?
Novwe/
——
stress,
G6
.
=
M“
&
.
7
Jou
¥1O%
Sagoo
7
139,.O-}0°
é
Fe
O=1370MPa
<t
. 519 and 5,20 For the beam and Joading shown, determine the maximum normal
Problem 5.19
stress due to bending on a transverse section at C.
SkN
3kN/m
chet
Use
tad
AR
B
= = 24.86
<
nition
Mo
Free
body.
For
4,2 ————->
W 360% 57.8,
-
SF
OF
S>
24, BG «to
tee
~saaxjow
BAGKIO
7 27-7 RTT IO” IO
27.7
6=
Problem 5.20
an”
© BIG* [O* wn?
Coz et
stress,
Nowmal?
kNem
~ 2H.B6 */O° Nem
uy
a ky
kV /n
Vv
OS
—~M— C.1)(3(2.2) — (2&2) (a) =
meme 2D
q
cB
a =M. 20:
I a
W360 & 57.8
———
portion
Fe Fa
MPa
«et
5.19 and 5.20 For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
25
25
310
kN
KN
KN
10
kN
410
KN
Use
entive
»
2M,
beam
ag
—
~ (0.78 )(lo)
A:
-
Norma?
portion
AC
—(0.375 47.534
For S 200 x 27.4,
stress,
.
2.25 A ~ (87s 2s) = 0.5 Vas) = (1.125)00)
~«-6 @ 0.375 m = 2.25 m—>
ft
Body.
F ©
$200 X 27.4
Use
free
as
M
-f0.378
Io)
=
O
47.5 kN
free body
= 0
Sz
(235x103 mm
~~MM _
Cre
17.8125
Staxio®
Mr
=
_=
17.8125
kN-m
285 10% m?eye Pat
w.8x0°
6=
75.3
MPa
«df
Problem
5.21
5.21 Draw the shear and bending-moment diagrams for the beam and
loading
shown
and
determine
the
maximum
normal
stress
due
bending
) =M, = 0
Ct
5310 X 59
I
0.3m 0.6m
0.6 m
>)
>
Me.
+ Ac4){100)+ (0-6)(los)=
26
(2-3)100)—
|
,
4
2fo kA
=O
(loo) (008
= (feo) )
(0-6) ~ (24)(eo) +3
Be Fol
V(LN)
O
co
B=
O
ite
.
Arc
p
Lt
foo
M
o
aoe
Shear
8
E
Ato GC
Ch
~Fo
E*t
£6
Ve
= leotal
D
Ve
E7
Ve
fo kK!
Ve
-~Fokk}
D* te
(LN m)
|
B
Bendi ng
HotN
moments
é
~ 30.
¢ <)'
DSM.
(03)(joo) + M
ee
= 6
Moz
- 20bNm.
(oo
APD
= 0.
pe
v)
|
94)
DIM, +0
(6-9 )i(o0) = (026) (210)
(Oy
Wy
+M
M
At E
MCE
yf
@
4
Fou
YSMe_e
=0
~M
,
+ (0°6)(%0)= O
= 0
= 26bAIm.
Ms: Se Em
Ga:
max
IM}
=
Fon
S3fox 6.
,
Normal stress
SY
EWM.
%
vobled steed section
_ AMI
6 =
S$
~ _ S4KfeF
* Gey”
S=
8
86.4
640K
3.
mm
Mf
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—“
to
Problem
5.22
§.22 and 5,23 Draw the shear and bending-moment diagrams for the beam and
Jeading shown and determine the maximum normal stress due to bending.
.
40 kN/n
Reaction
at
A.
: DIM,
=
O%
~3.6R, + 25 +(24)24)Yo) - 1S
giey
K&N
6¢.778
=
Ry
x 38.7
Reack ion at
= O
=: O:
+) Ma
B.
+O
as -(-2)(249)G4e)-15 +3.6R,
Ra = 29.222
kN
HO
On
nE
'
:
HI2E
2.4 wm
.
Cee
.
= Ot.
2
.
te
XT
yy
CL.772
66.7738
=~ 24,222
Vi=
~
4Ox
-V
=O
G6.77B~ ton
Vz
O
ch
+2 ZMy2
et
0;
ZS +(Hox)¥
M=
1.6744am
- 66.78%
tox
+
66.778
+ Meo
“« -2s
M =—(20)U/.6944)° + (66.778 101.6444) - 25
20.723
=
24S
KH <
M
M=
W
Zio
x
38.7)
ro Med
=
Wt
Ve
Mme
.7
“Ee
30.728 ¥10°
“Gayiore
_
=
w
~ 29.22%
section,
kN«em
af
472.4
ke
m
kNem = 30.728 4/0" New
S-
S44
x(0°
= S49«/o*
Gaz
16944
29.222(3.6- *)+ Is
20,067
steel
«of
3.6
Mange= 30.728
For
Unv-wm
tam
wm}
ey Pa.
$5.97 ¥l0
G. =
56.0 MPa
-—
Problem 5.23.
5.22 and 5.23 Draw the shear and bending-moment diagrams for the beam and
loading shown atid determine the maximum normal stress due to bending,
YkNAn
H)EMeg =O:
— -GA + (\4XS5)
A = 20 FN
W200 & 22.5
+ 30
= 0
6B
=o
EM,
=O:
~(2)layiiy+ 36+
Bzr-2kn
A
to
C,
F ku/mn
,
_
UN-m)}
Zkny
O<
x < 2m
THT
.
xX
ie
5
LM
643
i:
2EL
=O
,
R0-Fx-V
ZO
:
V= 20-94x
-»
Mi (l-m
2b
v
a2
V
DEMpso:
zo
-20x
.
+ (9x)
Mer
AY
C
Vz
2 kN
+M
= O
2Ox~
Mr
22
48x?
kiem
ASF, = or
M
20 +18 -Vzo
IV
QSEM,=
Of:
—(4\204
M
“
=
awieooaf2ZF
Ce—
>|
DEM,
=
Fov
rolled
Normal
steef
stress,
section
:
G=
IML
S
—
W
V-2=:0
OF
- @(220
)
Mz
26x/0*
KN-m
sor
~M
26 WNem =
26
M=o6
V=
2 kv
yie Z
max IM\ =
& Xia) +
-~ BEN.m
Nem
200%
a2.5,
Sr
194 xlo°
=
= Bex 10"
194 ¥ 10
3
=
194 *]O°°
134.0 x%/0% Pa
~
men
a6
=
m
3
184.0 MPa
-
—
§.24 and 5.25 Draw the shear and bending-moment diagrams for the beam and
Problem 5.24
loading shown and determine the maximum normal stress due to bending.
300 .N
300N
20 mm
300 N
4ON
Free body
EFGH,
Note that
Mg =O
dua
to hinge,
—@7 @
Vv (ny)
om dn
200 vam = 1400 nan
=O%
DEMe
0.6 H- (0.21,C4e)
_
2
Ve
=
126-67
-2\3
Bending
M. (Nem)
61.6
¥
cai
ot
New
Nem
N-m
F.
DIM
=O:
Me ~(0.2)C126.67)
=
5
Me
by
{
.
126.67
86.67
Viz
Vo = -213.33
moment
E
|
Vi =
G.
H-
te
to
F
G
~43
= B42
N
E te F..
Sheav:
x
8
40-3004 213.3370
Ve
ATE = of
ase
=O
N
213.33
=
H
0.4)(400)
Nem
25.33
\26.67
Bend ing
Me> -
moment atl G.
OTM,
=
0
~ Me + (0.21¢€219.33) =
42.67 Nem
Me 7
PR
213.32
Free body
(26.67
ABCDE.
tOF Mg = Ot ~ ~0.6 A +(0.4)(B00) +(6.2)(300)
— (0.2\0I26.61) = 0
OF M,= O02
Az
K
|
Bending momert at B,
zs of
DIM,
¢
i
VO
0;
Vv
k
% «108
Normal
~(0. 4)(952. 78) +(0. 2)(00)
+ Moz 0
43.11 Nem
mn>
=
3x10
* mw?
stress,
6 = 2156
3x lo
-G
= 17,19 x10" Pa
f=
213.33
Ces
Nem
Bending moment al C,
4D 3 M,2
M.=
Mp
5ILSE
N
2b h* 2 £(20)(30)"
=
A_8y
468.89
IM
max
Mg = SI.56 Nem
*e1-78V|
1m
D=
~(0-2C257-78) +M,=0
y
M
$30
06D
+
4 p12
257.78
300) ~ (0.2 )(126.67)
—(0.2i-(g0
(0.4)(0)
Beneing reoment
1D 2M, =
aD,
- My -(0. (413.28) = °
Mp2 - 25.33 Nem
17.149 MPa.
as
5.25
Problem 5.25
20 kN
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to bending.
401
Reaction at ©
OM,
(S44) (20) ~— 249
°
W360 X 32.9
yk
Reackion
4
8 )l4o)=
+
Me
2-4) (40) 4
VEN.
O
C= 43</kAL
a
ct B
ChS)( 30) ~
=O
=O
4B
BF
sr
O
1669 EAT
23-f
Shean
~ibe4
.
Vit moe EN.
Ate Cc
x
oD
alaqvam.
Cite DU
VF ~2c4 G8)
Dit
Vir
A+
B
A and
2bte4o
B
f
2
JA
cM,
Sx
~ 33
bay
Moe
MM
283
ym Phy
M=zo
Cc
M, kaim.
= gs-peny
Me
=O
(2005)
4
M2
+
(6)
©
Vv
Mp
D
g
—~ Mo
(16-4)
-
Mo = 26-35 ENMm.
v
IM ue
For
occurs
W2bexX 32-4
Normal
stress
oA
C
robbed
16-4
(Mins 30 kvm
3
stee!
Section
. M_
G =
Ss
Bekm
* Wyaxige
S$
ATM XI0
|
,
= 63.32
mts
MPa.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond ihe limited distribution to teachers
and educators permitted by McGraw-Hill for their individual cours¢ preparation. A student using this manual is using it without permission.
<<<“
5.26 Knowing that =
Problem 5.26
12 KN, draw the shear and bending-moment diagrams for
beam AB and determine the maximum normal stress due to bending.
+4 Zee
or
A~82+12
WSLO
;
rey
poe
m
©
2
Vr
A te+e C7
D
=
Ve
Vz
Dit EF
E* 4 B,
Q2kN
CG kN
=)
wet
GkN
-2kW
a
—_
-
yr
Ct
im
—
V (KN)
=
X 23.8
Shear
din
-3+8
R= Be 2kN
SAN
ous C
B®
A=
}
symmetrs
By
2
A
tDEML
= OF
~ Ya)
=o
«a
2 kNem
Mo=
M (kNem)
= Of
+ 2M,
4
M- @)@)+ @0)
z
M=
4
©
kN
em
wa
2
By
mow
For
|M\
WSlox
“
Normal
stress
=
4
:
OCAUWS
kNew
Sy =
a3.8B,
.
=
Ae
Qkvim
Mr
symmetry,
MI.
=
=
at
2BOXI*
4x 10>
SSonlee
M>st
2kNew
<8
E
am?
=
=
wm
RBOWIO™
€
oy AX 1).
lo”
14,.294X
Cvnae
Pa
=
[4.29
M
Pa
aati
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All tights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
5.27
.
.
5.27 Determine (a) the magnitude of the counterweight
for which the maximum
absolute value of the bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (Hint: Draw the bending-
moment diagram and equate the absolute values of the largest positive and negative
bending moments obtained.)
By
W310 X 23.8
A
Im
im
im
Im
@
Symmetry
j
+f BR
20%
AS
Bt
Ben
ding
|
A
B
A-R4w-34+B20
8-0sW
mome-f
Me
=
oot
C,
495
-(8-05W)LV4
tiny
M.
Me
Mz
= (@-O05W)
2
Oo?
O
kNem
B-0.5W
Berding
moment
4
fe tds
act
D.
“"
-DS2ZM,2
0%
-~-(e-08Ww)(2)+6U)4mM,
.
tm
--W)
= (3
My
= 0
kNe=w
B-O.5W
Equate,
- Mo
=
M.
|
We
(a)
"We
M, = - 2.6667 kN-m
My
B=
8B-aAsw
10.6667
kN
W=(0.67kN
A
= 2.6667 kNem = 2.6667%/0% Nem
[Mug 7 266667 kNem
Fer
W3lOx
23.23
volhed
S, = 2BOx10 mm
=
VM Venere
Yr
steel
2.6667
« lo®
-
.
=
shape,
2BOx/O.
=
~&
m°
‘
.
°
eh,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part
of this Manual tay be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or
used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student
using this manual is using it without permission.
Problem
5.28
5,28
Knowing that P = Q = 480 N, determine (@) the distance a for. which the
absolute value of the bending moment in the beam is as small as possible, (5) the
corrésponding maximum normal stress due to bending. (See hint of Prob. 5.27.)
.
?
ae BOO in
50
Q
yy
-fi-— SOO nam
i:
:
P=480
N
Reaction
ot
~Aa+
Q=
480
A.
430
+9
N..
2M,
(a.-8)-
=
Oo?
420(]
- ce)
Az(qco - B2) xn
moment
Bending
A
c
|
A
moment af
Bending
Vv
y
Moet
|
Q=4AON
Equate,
-M,
V
section,
Crea
=
S
64.3!
Faaxfont
7 643x{o~
Me
=
©
OSA
_
-— 480(l-a\=o
New
- 3°
480
— My =-64.31
64.31 Nem
Nem
ae
S=
S= L(ia\is) = GHB mm?
Sn
+
Oe
%
= oO
480(1-a)=
= M,
M.=
rectangular
IM
vane
A
=
=u,
= Gao
- 8) ne,
|
Fow
O.5
|
49
My= -480(I-a)
|
A = 123.62 N
(b )
_
M.=
Z2Ms,
-M,
Lee
\-a-t
(a)
45
D.
—
Cc.
M.
4
eZ O.Sm
at
2 O
bh
z
= 64R%(O mM
|
7 F420” Pa
|
-
|
|
a le
Cig? WEUMP
‘ Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rightsreserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.29
5.29 Solve Prob. 5.28, assuming that P = 480 N and Q = 320 N.
5.28 Knowing that P = Q = 480 N, determine (a) the distance a for which the
absolute value of the bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)
Ojonow
P=
|
¥
r
~<—- 500 mm —fe«--- 500 mm--~
Q=B20N
RON
&
+
Aa
M>
45S
A.
at
Reaction
=z
O°
20
320(}-a)
4B0(a-0.5)~-
A= (300- S22) w
-~O.5AK + Me
C Me
A
= 0%
+2 EM.
Cl
momenfat
Bending
05A
M.=
. —?
#82 ) em
- 00
= (4
v
‘
= 0
Of
moment at
Vv
D,
+5
My
Q*320N
&
M, = (~3204 3200)
* Ly aal
Equote.
-M,
210a
=
+Z0a
-— R80
GQ. =
|
=O
(b)
For
116,014
N
reetangudar
Mos 58.0287
section,
S
MI en
:
3, 0065
= 38.0088.
+
- afe
foot
m , - 1.06873
,
a2
M,= -5B.006
Nem
m
8!Imm
=
Nem
Lh
= 648x107"n,
S = b(21GBY= 648 mm
6.74
=
0.81873
negative
Nem
HOO
=
-B20&
B20
Me
Reject
Az
=O
a20(\-~a)
_
~M,
A
My o> |P
(0.
Z
W
Bending
=
89.5%10°
Pa
= 87.5 MPa. xl
Snag,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. .
Problem
5.30
§.30
Determine (@) the distance a for which the absolute value of the
bendig mornent in the beam is ax sinall as possible, (6) the corresponding
20 kN
:
maximum normal stress due. to bending. (See hint of Prob. 5.27.)
(Hint: Draw the bending-moment diagram and equate the absolute values of
ces
5
.
tet
the largest positive and negalive bending moments obtained}
AOKN
Kk
J,
Reaction at
W360 X 32.9
200.
<a
+9 ZM.
~ (440) +
3-4 Re
zo
=
O
Re = sq(4b - 200)
Bending
M,ENm
(°
26-7]
2
x
moment
ef
Mo
42ZMy
O
M
M, = e
45+ HEReR a= ES, To(4b
moment
af
Doth
AY
=
*
ism
Bending
D.
C
4M.
Cote
Equete
20 O.
=O
Zoa+M,
~ Me
=
ee
N
ow
m al
stress
Ss
(a5
- Moz
IML
= 2667 kal.
Kt 0?
2.87
TSc’
SM * dimy
O
(96 = 200.)
Then
toe
=
Ms
a > fo33m
Fox W360X32°9 wihhed steel section
~ 200.)
Q/33m
My 2 (20) (133) = 2667 afm
$= 44x10 mm
Oe
eb,
2 mp
_
( b)
i
th. 3 fa. Pa.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
at,
Problem
5.31
5.31 Determine (a) the distance a for which the absolute value of the bending
moment in the beam is as small as possible, (6) the corresponding maximum normal
. Stress due to bending. (Sce hint of Prob. 5.27.)
20 kNAn
W360 101
Hinge
4 a=]
SEGMENT
GC2
8)
yt
ape
:
Vek
30 .| KN hm
-]
b&b —Ale
~ 20
+
Mz = O1-Va x +(20x KX) 4 M
SZ
"Vv
= Vaex% — /0X>
M
hp
4b
2
-rox”
= tobe
M
Mae Sh
ZOkN fm
MTT
"wy CELE
—
20a
Aj
Vv
-
FI TM,
-8b
=O?
2
=O
kN
Row,= Ab
O
=
a fob-20Xm
eit :
=
Veg
= fob
®
B ZAIN
20h
Va B + 0. -
:
5
Te x
VI
ER
RNa
Mma
(TT
ASE,“y =*
+
=
Ve
Syme.
By
in
a)
(5-4~
=
b
Let
|
5.4m
|
16 96X10 mon
5,
Jot)
W 30x
For
= 4b
.
Vy la-x)-M = 0
~20(0-2\%X)—
arm
N
M = ~7o(a-x) + ob(a-x)
Ve
yo
.
Vv;
Mmay™
~ oa
[Manne
|
=
~j00.
%2
at
ect ows
|
|Man
3
at
-
toab
=s100?~10a, (4-a)
(a} Equate
the
two
valves
StQ.>
Zai
—-—
2b
a+
a>( 4
@) IML
= 5%
—
WI
1 Mine | «
RGarey
==
=O
76-37) [5
=F4AQ
,
| ar
=
86926
mm
~
27a
Mei
+e
al
az
o9zm
@ © SokNM
UML arase
OS
z
-
of
OO,
~
So x10?
=
Tetaxson® * 7958 MPa
6,7 226 MPa
=e
Problem 5.32
6.32
A solid steel rod of diameter d is supported as shown, Knowing
used if the normal stress due to bending is not to exceed 28 MPa.
Let
W=
Uy
Reaction
A= 4W
Bending
We
= ALY
at
tote? weg bt
= FALY
A
moment
at
center
of
beam
DEM, = 0
wv.
Wy L.
ems - o
Gye)
-HE)+
&
Faq
yt)?
a
= FaAlr
-3
Ms . Who
Vv
Ww2
= ¢a)
Ce
Tw
Ad
ye
l=
Normal
2
stress
“
oe
M_
#4?"
nt
SS
Sodving
for
Data:
L=
£a?
a
~
S
&.
a
Fe
3
. Ty?
“ax
ror
d
_
d=
vr
SC
3m
Y=
(7860x981) Nfy,?
6 =
oo M fa |
60 x 28B/)
a _ €3)38*(1
x19 &
660248 mz
1248
mm.
th,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
5.33 A solid steel bar has a square cross section of side b and is supported as shown.
Problem 5.33
Knowing that for steel p = 7860 kg/m’, determine the dimension 6 for which the
maximum normal stress due to bending is (a) 10 MPa, (6) 50 MPa.
b
weralt
°
Let
L=
total
deusity
Aength
at
Bending
:
M
|
|
C
moment
Yo
3
= b'Lpq
and
Dz.
at
C,
C=
Dem,
Pas
M
¥
DBR) - RE) +H 0
Vv
ae
For
a
fe
square
Me
L\yw.
=
:
WL
oF
= bees
section,
stress,
Novmed
=
DEMero?
wo
"
| =
¥
-~Ke
L
\M
D=
O2
ne) :
~ We
may
|
ot beam.
we UY= ALPS
Reactions
ne = PY
6
S=
=
IM |
s
¢ b?
=
b' L*eq
£3
/18
2
1
At
Zs
Solve
tor
Data:
@
0.
“k=
3.6 m
@)
Gs
ber
sfc
p=
loxlo® Pa
bt ($61 a6o(9.81) =
~ BLY 7860.9),
(b)
b>
(3\(sex tot)
786°
kg Zin
g*
4.8! m/s*
(b) © = SOxjO% Pa
33.3x10%m
~8
6.6610
m
b> 33.3mm—
b=
cémm wl
el €.66
Problem 5.34
5.34 Using the method of Sec. 5.3, solve Prob. 5.1a.
5A through 5.6 Draw the shear and ‘bending-moment diagrams for the beam and
- loading shown.
;
Ae Mt
DIMyr0: -ALewi$-0
BL- wl-bz0
DIM,:0O:
i> -w
Va
~ Swede
=
Ve
t
Vz
x
=
wx
=
- wx
A-
wx
= %-wx
rf
V-
7
:
.
=
.
Bs
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or.
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.35
5.35 Using the method of Sec. 5.3, solve Prob. 5.2a.
=
Ms
>
Tbe. Pab
IM
=>
Ey
iy
TL
mip
~~
rT
&
CW
|g
Fab
{
-
3
oA
<=
i
Me.
RB
CG
j
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
O
= Fab
Proprietary Material, © 2069 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission:
=
Problem 5.36 -
§.36 Using the method of Sec. 5.3, solve Prob. 5.32.
51 through 5.6 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (5) determine the equations of the shear and bending-moment
curves.
At
-P
Ys
AL
=o
Yaw
AB,
Over
=
_
+P
=
Ve
M=~Px+C
M=oO
Co
HzO
at
- Pre.
Me
At point
B.
At port Bt
Over
Wee
BO.
At
- Pa. =
MoC,
a0
sew
x= 2a
tO
M=- Pe
~2Pa
=-2Py
= -2P
Ca
use
B.
M
Vs -P-P
-2PH4
Ms
M=-Pa
mee
ae
~—-
+ C,
G, = Pa
+ Pe
<*
M=-3Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. . — .
Problem 5.37
. ,
5.37
solve Prob.
Pro 5.4a.a
5 Using
i the method of Sec,. 5.3,5.3, solve
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
- loading shown.
iM
M
w=
w,%
Vaz
Oo
av
*
oye
Wok.2
Ve
Ma,
-
= ©
WX
ela
W
t
~ Wer
7
+
.
aL
—
ab
IM 2 Ve
A Mex
M- My = J Vda
= - 9 Me ay
An
Ee
x
a
*b
MO
ol
x
2,
Se |
- Proprietary Material, © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.38
5.38 Using the method of Sec. 5.3, solve Prob. 5.5a.
5.1 through 5.6 For the beatn and loading shown, (a) draw the shear and bendingmoment diagrams, (6) determine the equations of the shear and bending-moment
curves.
A=8
symmetoys
By
Reactions.
A-P-P+ B= O
HATA =O?
A=B=?P
‘Over
each povtion
AB, BC, and CD
V = constants
qv =o
Shear
diag vam,
Ate Bt
Blt:
Ve P
ve P-P
Bending
MzO
Riagraw
moment
aM
on
to B: .
M=
O-P
Vz
C to D?
A
we0.
VP de
.
=o
~
-«
= -P.
<
«
we
Vz
P
= Put,
Thess
«#20,
«at
|
M=
C,=0
i
Px
ae
ced:
ce,
M= Pe
teve-P
M=-(Pae
=
-Pei
Gy
Problem 5.39
5.39 Using the method of Sec. 5.3, solve Prob. 5.6a.
5.1 through 5.6
loading shown.
Draw the shear and bending-moment diagrams for the beam and
At
A,
2
Vaz Az dw(L- 2a), My = 0
Ve-
Vp
Ve=
.
O
zs
Ww aby
so
¢.
Va
= &RWCL-2a)
a
a
Me-Ma= Sve = S$) dw(L- 2a) de
~¥ (Lae)
4WCL-2a)o
Maz
—B
w(k
i
WzOo
O<x< a
B.
Ato
8(L-20)
Y
4w(L- 2a)
A=D-»°
Reactions.
Ww
w=
L-a
a<x<
te C.
7 &
N=
=
Voz
ap
=
—
Ve
'
wat
=
~wlx-a)
dw (bL-2a)-wl-a)=
-
Vz
.
s+
-{
dw (L- 2)
dwlb-2x)
x
~M, > (Vv de =
4 wbx =x),
x
M
ff
= kw(bx -x*- lata’)
dw (L-2a) ce + # w (Le ~ x?
|
= hw (bx - x*- a2)
Vi
C
+o
D,
Ve
Ve
=f
~ kw (L- 2a)
- $w(L-
Mos
La +97)
4 (b-2a)a
2a)
At x= b.
Mz
(ee = 9) a
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc. AlJ rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course prevaration. A student using this manual is using it without permission.
Problem
5.40
20 kN
Ak
|
5.40 Using the method of Sec. 5.3, solve Prob, 5.7.
§.7 and 5.8 Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (5) of the
bending moment.
F20 kN
Ly
0.225 m 83M Yag5m 03m”
Over
portions
Y=
V. (44)
Shear
22
$
C
\
8
DE
-
-12
|
AC, CO, DE,
-w
=.0O
At
Cz
Vr
= 20 ky
Cto
D-
Vr
204
Dto
E:
Vr 2g-20
E +t
Be
Vi
& zo => -IZEN
Ovraw
5°]
|
D
te
Aveas
3-4
he
Vo teerstant
dragvam,
~20
M CEA)
ad EB
shear
of shear
49
aliagram
= 29 EN
= 8ENX
as
Shown,
Aiaqraw,
Ate C:
Ane. = (-20)(0:228) = ~ 4S Nm
_
2+]
C +e De
Aey = G88)
=
e~
Dt E:
Ave= (8) (01228) = 18 kKm
Ede
Acg, = (-12)(0e3) 2 = 36 EN
B:
= 4
bm
an Hoek
©
Since
Me
7
O-
My,
2
~ES + 84
Me
=
37
+
68
=
&7
r
Ms
m oments °
SJ
_~ 26
2
s.p kKNm
Me
4S
V
is
coustant
= -#&E
kim
over
= 3
ENm
portions AC, CD, DE
wel EB, elraw “the
ENh
@ach
of
the
bending” moment dt
with
ovey
(a)
Vinee
cb)
IMI
=
Constant
slopes
portions.
these
28
£7
KN
kam
A
Bending
<=
Problem
5.41 Using the method of Sec. 5.3, solve Prob. 5.8... -
5.41
QAKN 24 EN 24kN
_ 5.7 and’5.8. Draw the shear and bending-moment diagrams for the beam and
_ loading shown, and determineibe maximum absolute value (@) of the shear, (6) of the
bending moment.
24 kN
React
TOMS
ot
ADEM,
Fy
=O:
e
=O
~(O.75Y%24)
75m
R= ; 30 kN f
+ 2M,
a4
30
and
~BR, + (2.2594) + (1. ICQ) + Co.75)(24)
<4 @0TK m= nnn
V,a kN
A
“env.
(seven)
(i.
~ @ asat) + 3Re
~(2.75)at) =O
Shear
ie
maa
Mae
|
.
22.5
13.5
*
Me
My
BOkN
AwC.
Ve
Ch D.
Ve B0-24=6kN
Dt
E,
Ve
G-24
E-
F,
Ve
-!18 -24
* -/RkN
= -9Qkn
:
Vs -42 466 = 24 kN
diagram.
“shear
of
Areas
_%
My
diagram.
FRB
°
\4
Bending
Re= 66 kN ft
|
|
.
A+ C.
Ane = (0.78 \20) = 22.
€
Aco
to
D,
=
(0.75)(@)
-
kNom
4S
bem
DteE
Roe = (0.75 618) = - 13.5 kNem
EnF,
ep = (O.75)@42) = = 31S km
Ft
Are = (0.75 )(24) = 18 kN-m
&.
moments.
=O
—)
a
= OF AVS = 21S knew
= 2284S = 27 bow
Me
Me
7? 27-13.8 2
= 19.57 3h5
M,
= 7!34+18
13.5 kom
= 18 kNem
= O
(by
[Vinay
4A
IMI
= 22
KN
att
OkUem
<8
Problem
§.42 Using the method of Sec. 5.3, solve Prob. 5.9...
5.42
-
5,9 and 5.10. Draw the shear and bending-moment diagrams for the beam and
. loading shown, and determine the maximum absolute value (a) of the shear, (5) of the
bending moment.
- Reactions
A
pte
De yyy ee
GOW
“The
2a0-—
GOK
B
$9
Mp
=
A
amd
By
=.
~SA+ (2\(aeyC4\ + (2\(2)
C2
.
Az 72 knf
49M, = 02
SB
~(3)60)- (1X02)
B=
Check:
=
tf ZF,
48 kw
=
f
> 72-()G0)-0 412
=0
Shear diagram.
Va = 72 kN
Ale C.,
O<%
€ 2m
we 20 kN /w
—~Go kv
Ve~ Vp =~ S$) 80de =
Ve = 7R- Go
M Cite).
=
I2kw
Cto
D.
V=z\
Dt
6,
Vo=zV60 = ~4Q kw
; Aveas
of
Ate
C.
sheanw
Cle DB.
B.
Dt
kw
diaaqvaur.
Avg 7 (-48)2)
= 84 kN
12
=
Me
9
=
(e)
= - 96 kNem
M,ze°
Mo = B4+
@)
= BARN
= 12 kven
Azy = HAW)
+ 84
Mo =O
4WG@-
12
Ane = A(TR 412)
Bending moments!
=
=
Fk
©
as
expectest.
Wham? 70 KN
1M
leno
©
46.0
kNem
—at
—
Problem
5.43
©
_—__—Ss_: 5.43 Using the method of Sec. 5.3, solve Prob. 5.10,
5.9 and 5.10 Draw the sheat and bending-moment diagrams for the beam and
45 kNé/m
loading shown, and determine the maximum absolute value (a) of the shear, (5) of the
bending moment.
120 KNY
ZMe=
Os
A=
-15
4DEM, =O:
VEX)
C
{20
|
Va
x
(~108)
EN
ie.
Itsky
14C-@78)-@NCi20) = ©
rz2205kN
fF
Sheav diaqvam.
(08)
(75
~18A 40-7K 8) -0:9\120) = O
~ 16S
=
Ate Cw.
EN
Of‘ X<18m
he.
= -45 kA[m<
Ve- Va = ~So wax = [45d
Vg=
M CEN ®)
IPS
7195—91 =
Cte
B.
Aveas
Vr
of
Ato C.
—fo0-5 kN
-/00-5+ 220-5 = Jzo LN
sheav
diaqvam.
SVde = ($)(-195-100:68)
>
~ fo8 kim
Ct
RB.
Bending
=~ BF EN
{08
kNin
Cv el, = (o9)(120)
= 109 kwh
moments,
a
Ma
=z
Me=
Me
=
O
M+
S$Vdg
=
O
- 108
er
M.+
SVas
refosSris8=e
= 108 BNIn
O
Maximum
[Vl = f20 kW
Mayimum
IM\=
to9
kNIM
~~
~<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.44
240mm
240mm
loading shown, and determine the maximum absolute value(aoof the shear, (6) ofthe
a
240mm
Gn
soos
* $44 andi 5.45 Draw the shear and bending-moment diagrams for the beam and
—
bending moment.
a
a
-
DEM,
- 0.72 A +0. #2120) Ho. audio)
7 OF
:
-
60 mm >|
1.4 Nem
-
hom
joo N
D
1
M (N-m)
=
oO
64.3)
—
~ Yo
-Z2A+
072 B+0
l¥ont
Sheav dagen .
|.
Ato
C,
Vz
loo
N
Ct
DD,
Ve loo-l20
5 -foN
= -14ON ~
Vr -#0-120
Dt B.
Aveas of sheamoiagram.
.
, 100.
x
|
Be
0.24
0.24a}
20N
Qon
_
(24)
.
2
02 — -lo.au\izo) - 4B ia0)-7.2
1.2 Nem
Vw)
“20
Pu
L20.N
DEM,=
,
2- -
A=toont
120.N
A
7.
_
1633.0}
A to Ce
pe XK
QV ak = (u.2¢y luo) © 24 Nem
Cte D.
Dt Be.
Svan
SVde
Beudiug
moments.
Ma
= Coau\hac)= -48 Mm
=@.24Li40)= - 33.6 Nem
=O.
Me = 0424 = 24 Nem
Met=
244722
My = 842-4.8
My
Me
=
=
31.2 Nem
= 26.4 Nem
26.494 +7.2 = 33.6
33.6 - 33.6 = 06
ta)
Maximum
|vl =
- db)
Moximum
IM|=
[40
Nem
N
33.6 Nem
2
~~
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. -
‘Problem 5.45
5.44 and 5,45 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (6) of the
bending moment.
3.5 kN/m
Reaction
+)
Ms,
—l-o9 mf
¢
AA
Beam
.
(1.5VG.0)6.5) +(L5)(3)
6.75
at
KN
B=
ond
6.75
doadina,
ky
See
6.75 kv
oF
Joued
Ate C,
C +
sketeh.
diagram.
(2.4)(&.5) ~ 8.4 kw
B,
(0.6 VES VT
2.1 uN
(0.4% 3) = 2.7 kN-m
.
Shean
V, kN
diag Tam,
Va
=
6.75
Ve
=
-h6S-%B
Ve
6-75 kN
k Pe ALO
=
kw
6.78 - 8.4 = ~1 6S kN
Vg = -465-2)
SL
c
8B
(—0.8387)
4
.
Areas
3kN
“Cs
= O
‘
By.
ACE
6
_
.
Ot
.
did
lA
chs kN
7
A=
0.6 m
Reaction
3.5 kN
A.
~ 3.0 A+
3kN
eter 5
2
at
Over
At
Ak
G
C,
Vz
3+
~HES
kN
= = 6.75 kN
6.78~-3.52%
VtG67IS-B5xX,
=
Oo
Kg. = 1 I28E m
Aveas od sheae cliaavaw.
A te G. & (1.9286 (6.75) = 6.5089 kNew
& 3.42]
G te C. 4 (0.47 J) és) © - ©3284 Kem
Cte B 4 (0.6)- 465-675) = - 3.492 LNem
M,
KN. m
Beud ine
6.50849
kN-m
6.12
B42
Ma-
motets.
= .0O
Me = O 4 6.5089 = 6.5089 leN-m
M- = 6.5089 ~0.3389 = G.12 kN~w
M+ = 6.12- 2.7 =
3.42
Me = 3.42 - 3.42
=o
kN-m
|
) IVhame= GIS KN
<a
Cb} [Muy = GuSTRN=tm <a
Problem
5.46
5.46 Using the method of Sec. 5.3, solve Prob. 5.15.
5.15 and 5.16 For the beam and loading shown, determine the maximum normal
JORN
stress due to bending on a transverse section at C.
EO
3kNAan
100 mm
200 mm
cad
es
om
V (en)
SEM= 0
“0.5 io) + 3C-@172)(3)2
.
ee?
Shear
2)
Ate
D
Vir
24ge
Dite
Co
Vo=
2S8~-lo
undev
—Aveas
Chew)
Ate
D
Fon
=
7.26
kWem
=
veclangular
cvass
stress
~ 7.42
KN
YAS8) > 3.87 Kem
Dr
Cc
SVds
Cte
Bo
SVdu = (F)2.2K6.6o} > 726 kN-m
7 (hs) 292 )= -11013 Kem
moments
Ma
zx
«O
. :
M,
7°
0 4 3.87
Me
Mg
=
=
3.87 -1LIBt
726 ~-7.26
=
5
3.87 Kiem
- 726
O
KNem
7% 26x10" Nem
section
|
Normad
-
diagvam
Shear
QVde = US
Bending
IM}
kw
Vo oF +2.42 414027 6.60 Ky
Viz 660-(22¥3) = O
ct
BD
“Fay
M
0
kw
Jor
C=
2%)
(3.81)
2.58 kN
A =
6.60
2.53
zO
~ BA +(1.SMI0) ~ (1.1)(2.2)(3) 2 ©
.
ao
15
ism
OSM,
S=
tbh*
= (4) (100) (200)
oe 666.67 xo? me? = 66C.67x/0 Sm
C
IM}
-“e
_=
7.2610" *
Ceee7xo
‘
10.89xJo Pa
=
10.89
MPa.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribation to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~—
Problem
5.47
5.47 Using the method of Sec. 5.3, solve Prob. 5.16.
5.15 and 5.16 For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
80
mn:
ih
.
aah
+P ZF
:
=Ot
A+tB-
00 mm
A
15 ee t1.5 m—>
4.35
_¥y
Nee
At ct,
Vz 435-3
Ct tod,
WE
78S
At Dt
4, OSG25
Draw
'
,
Aveas
Bending
moments$
Cr
E+
E
D
Dt
@
Mer O+8.55 = 8.55 kNem
Mg = B.SS+ 0.50625* 1, 0562S kom
M,= 9.08625 - 0.506255
3-55- 8.55 =
Maz
©
af
Aiagram.
&.55kN-m
sheav
3.58
x
lo”
Vem
oF CD.
&(7.05+4.35)LS)= BSS kN-m
£1.35 )(0.75) = 0.50685 knew
£1.35 )C0.75 ) > - 0. 50625 ken
F(-435-7.05 U5) = - B.55kNem
a vectangudav
section
S= 4 bh* =(L\80)}(Soe)
a
mmm = LQxlO rm
= 1,2%10%
normal stress at C,
Mesimun
B.55%102
Me.
oe
”
Laxio->
= 7.128 x 10° Pa
1
=
|
diagram.
S
Me.
|
E, the mea point
point
’ Fon
Ma =o
|
Vez -435 -G.5)CL 82° = 1OSkN
of +ho
KtoC
SE
ae
= - 4,35 kv
w= 1.8 kN
shear
VezO
‘
135 kM
LB KN /m
VF -38-3
0
B,
He
=
Ve £3S- (LS \B) 2 - 1.35 kW
Di + B.
At
BASS
KN/m
Viz Z0S- U-8)0.5) = 4.35 kv
A+ D7
M (kh-m)
kN
ZOS
1B
we
kn
At Cc,
“ESS
~2L0s
705
Va =
A to Cl
7.05
Br
diagnann ?
Sheav
Vile)
3-32-45)li.3a)20
if
LSkN/M
S= 2I3MPa
Problem 5.48
TOGEN
§&A8
Using the method of Sec. 5.3, solve Prob. 5.17.
8.47
Por the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section af C.
Took
75 EN
40S
one]
Sete
~4°5A
je—0,75 mi
A
uneler
Area
My
covve
sheav
=
= 0
x @75)(loe) 4 (3)Ueo) + (18) f2s2& UNS) @
i9F2 we kb Al,
Shear
A
bC
V = 192-2£4)
Ato
SV ely 2
(0°75) (/92«2)
=
Maz
=
Em,
©
O04
Me =
For
14Gi&e
a
W 40x
Novmad
stress
(4G1S
© [elf LN 3.
3
voMed
steef
& > f
=
section
ee IX Xe
S=
=
2260x10
£t-
hmm
Mpa.
ea,
Z26-0 Kié
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem 5.49
5.49 Using the method of Sec. 5.3, solve Prob. 5.18.
5.18 For the beam and loading shown, determine the maximum normal stress due
30 KN 5SQkN
5SOKN
30 kN
to bending on section a-a.
th
W3lo x < 52
Reactions:
By
af ZH
Vv Ck)
Shea,
B0
Ga lo)
C,
C
to
B
.
60
dD.
Ve =
80-30
Dt
E.
VY = S0-50=0
Aveas
of
.
to
A=
A= B= R80 kW
V=
So
\
=o:
diagvaw
A
symmetry,
kW
)
= 50
ky
.
“%
|
shean
diagvaw.
Ate
C.
QVdx = (800.8)
64 kl-m
Ch
DD
SVaw = (o)(0-8) =
4o kN:
DRE
Bending
Ma
SVdx= 0
moments,
=
©
He:
Me = 0 +64 = 64 kUem
Mo
64 +40 = \04G kKN-m
Me
M;
For
=
1o4¥
W
kNe
m
=
Blo“ S22,
,
Normal
stvess,
S
104 «10>
S=
THR
.
IMI,
-
Ss
=
Jot
=
1o4 KNem
Nem
10° mm
J04~103
-
+O
743 <lo7*
=
749x/07o
m®
=
139% 0%/O° Pa
S& +
13% 0MPq
«it
" Proprietary Material, © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
- §,50 and 5.51 Determine (a) the equations of the shear and bending-moment curves
Probiem 5.50
for the beam and loading shown, (6) the maximum absolute value of the bending
moment in the beam. .
a
=
-w
ie
~ Wo %
FAME eC
=
M= -4u2 4x4
M=0
of “=0
G20
M=0 ot w-l
te)
=~ £l™+ CL
Ve -tw, Se bw,l*
=
~bw, %
+ kw
i
\M
=
Va
ange
gw, (2-32 )Zb
=
=
dw (le
- x8/L)
“
6
Ww
Ms
M:+O
L2 \ |
,
af
sx
7
at
x
ot
xF lL
=
QO
C,
=
-Ww
>
WwW.
<a
WL
~
Mee
.
Cos
Fx
oz, OX
sin
TD
+
zx
Mut
L
2
&
web
oh
=
.
Cy
+Cx4
C,= 0
O=O0+CL+O0
at
©
a
tb)
Wy
642
da
.
C,
Wolk,
FT
M = wel
V
a)
UX
sin
-W,
V=a
=
oO
= ©
(a)
af
tc
Murex
bending moment in the beam.
V
O
—
3
E
av
7
(=
“oO
curves for the beam and loading shown, (5) the maximum absolute value of the
a
M
?- ax220
5,50 and 5.51 Determine (a) the equations of the shear and bending-moment
Problem 5.51
te = 1, sin
”
CG = El
=
bb) Mog ocows when 94 = Vv = O
Rom
C,
2
t
ax
sin US
.
|
at,
L
=
io
.
Sm
xr
CoS
Minne = Mee
i
5.52
3.52 For the beam and loading shown, determine the equations of the shear and
_ bending-moment curves and the maximum absolute value of the bending moment in
the beam, knowing that (a) k= 1, (b) k= 0.5,
w=
WeX
_
kwe(k-x)
tf
Problem
.
kw, ~ Ci+ke) Mex |
L
L
av
mW
x
Vz
kwyx-
Cia k\ ee
(14k) eX
ee
~ kw,
+C,
C0
adMeo
on
Vz
M-
Kwa x
(a)
%
=
- Ci+k) ex
— (i+k
W,
kwix*
2
- (+k
2
+ Cy
C,= 0
|
we x?
6b
=
Ww, X~
M =
%
~a
w, xX?
net
Weds
Wex=
VF
kel,
3
oe
x= O
ot
M=O
M
kwx
a
3L
%
Meximum
(e)
M
_
Vz
0
oat
x=
at
Mex
Viz
4,
k=
occuvs
L,
Beer
_
WeX* _
ar
ne
mht
wet
az
onan,
—
-
WeX®
23 |.
At xe$l
ie WABO’ wel
Atye
M:z
lb
w=
a
3
a
MeL’.
603704 wht
0
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usingit without permission.
Problem 5.53
5.53
oe
we
we
Determine (@) the equations of the shear and bending-moment curves for the
beam and loading shown, (4) the maximum absolute value of the bending moment in
the beam.
:
Iy (
v2
4 5)
Welw
one
2
Ve >
V
s em lie Be)
KEYS.
=-w,lx ,+a)
+ C,
=
oO
at
>
L,
Oo
ahs
Ms
M=O
-
we ($l - x- 4%)
wol$ix-d¢¢at
v- Ll,
Cas Sul"
IM).
occurs at mFO,
“a
L#)+c,
wff$U-40-§L)+
&,
=O
Me u,(Ele-de- 42-30)
|
Mb
FR
we
a
Proprietary Materiat. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, teproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this mariual is using it without permission.
5.54 and 5.55 Draw the shear and bending- moment diagrams for the beam and
Problem
5.54
loading shown and determine the maximum normal stress due to bending.
IG kKNAx
42
2
Me
=
2S
O3
.
AG CLISVULS MIE)
|
=
0
S150
* I8.6
Ar
| Shear
a
o
|
kw.
DEM, = O:
-(.75 SVE)
Br
22kN
Vv Cen)
L 16.8
A
\6.8
IN
le~ gd —sls-d<
|
~7.2
diagram.
V, =
16.8
=
16.8
Ve
Vee
Locate.
M (kKN-m)
=O
42.5B
ky
-(5)%le)
D
where
’
Lo~
8
a=
kM
-7.2 kw
point
3
-722
=
> Ta
1.05 m
A reas of the
Ato D.
Vz
a4d =
Lg-d
shear
diaghaw
Ow
25.2
= 0.45 m
e
GVele (AVC
05) (16.8) =
8.82 KVem |
Dito C.
QVax = (£Yo.4s 67.2) = - 1-62 kum
Cc
S$ Vay=
bh B.
OIC722)
=
8.82
kKNim
- 72
kNem
Bending moments.
Ma
=
O.|
Mp
7
O
Me =
“Mes
Maximum
IM)
For S 150*18.6
Normal
stress,
=
8.82 kN-m
+
8.82
882-
LOR
=
= 7.2
Kem
742-7220
= 8.82/03 New
vodfed steed section, S= 120x108 mm = 120 «10°m
OF
oO
“SmM) Seats
* z.
>
3
73.5x10% Pe
6=73.5
MPa
<i
Draw the shear and bending-moment diagrams for the
5.54 and 5.55
Problem 5.55
and loading shown and determine the maximum normal stress due
beam
to bending.
.
45 kNAm
IGkN -m
.
>) 2 Mg
By
)
a
7
hea
=O
=O
~ 36h + [49U24)(0-4) - 16
{250 mmm
he
a ‘iin
DM,
67-6 4iN
=
A
=0
40°4EN
B=
6726EN
=
Va
Sheav:
= O
- 16
(HEZ4D (2) + 3668
Vi, = 616 - (45) (204) = ~ 40° EN
Ve~ 40-4 EN.
Ct. B
Locete
ad _
naaeepreeaperent
Cibo
M
pein}
IM\=
Se-7
diagram
shear
under
z (d\irs)(676) = 50-7 EN mM.
GV dx
SVae = (B\0-9)(- go-aye - 12 1e Em,
Vde = lO 4ed) = — FP 4F BN
BS
Ct
= 507 Nm
M, =
0 + 507
Mes
Sef me 188
=
=o
M,
moments?
Bend ina
Me
Fat
Bh md = 06d
ds Sm
Dt
Maximum
V=o6
ek
Gore
At+toD
149 N
where
24-0
tn
Areas
(kN im)
D
2
Fafa kN
32°52~ 4o.gp2 — 19°96 kN
ENm.
mm
77/250
=
(e\7s)0s)
=
gbh*
So
section
evess
For reclanguilen
TK
{Hi
:
Normaf
stress
@r
srr
=x
=
64-9
3
MPa
Proprietary Material, © 2009 The McGraw-Hitl Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
a
Problem
;
- - §.56 and 5.57
5.56
120. kN
GOKN
Draw the shear and bending-moment diagrams for-the beam and
loading shown and determine the maximum normal stress due to bending.
4°)
=Me
-
oO
~3.0 A +(26)(6e)
+ (2.2) (60) 0.8Xize) = O
Waso x 49.
pepe
Ar
i228
kN
Ld ma
0.4m
8m
+ 3s
Mn
=O;
~(0.4)(Ge) ~COBXCO) -(2.2Yi20) + BOR
V, kN
7
123
“4
(51.2
AL __ 68
a
kW-ha)
Shear
2
M
kN}.
diag Fame
| GY
|
Fe
VA,YZ
.
V=
128-60
=
E+ bh B,
Ve
9-120
< ~H2 kM
Aveas
Vz,\
of
Vz
sheav
Ate
GC.
Dt
BE,
CD.
—
mn
:
Bending
qM
34,6
Araqvane
0.4)022)
=
UYV(BY
=
stress:
I7nZkNem
NL QkNem
~ 84.6 kN-m
Oo
.
Mes 0451.2 = SLD kik
Mo = SI2422%2 = TRY New
—%
Novmel
ST. RkN-m
(0.8)C-H2)=
7 78.4411.2
M, = B4.6-846
W250%49.1
Bkw
mowevits,.
=
Me
For
EG&RN
G8-60 >
(0.4)(68) =
E +o B,
2
72.4
V = 123 kM
Gite Dy.
Dib EY
Lal eresen
Linn)
2 kv
A to C’
| 27.2 kN-m)
io
B=
=o
Mb = 89.0
volbed
C=
steed
M
section
8%.
ial Sie
S
kNem
=
89.6 Kew
0
= B46xIOF
Sr
*
=
=
SILLS
N-m
rave > S72 %16* vn
IS6.47%1O° Pe
SG
=
156.6MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<d
Problem
5.57
5.87 — Draw the shear and bending-moment diagrams for the beam and
loading shown and determine thé maximum normal stress dué to bénding.
:
HDEM,= ©
— -OSA
4 Y40IGN(Ous)
=O
AE 204 EW.
Ny = 2UGEN
29-4 ~ Ced(o-s)= -Yorb
B™
Vg
Bt
Vyas = -#01G 4 (684)= LTP EN
Co
NM, 527-8 - (¥2)(0-2) GO
Locate
point
a.
=*
aac4
D
where
Gok
j ¥
703
asa
one
7
do
|
Or29m,
|
Svan (Geo 8)(o-29)=-8:88-7 EN
Dt Bo
SVden (42a (orz) = 2078 be -
OC
Maz
¢
moments
Bending
107
= (h)(2t4)(o-r2s) = 2-087 Ene
Svea
Ato D
=
clhaqvam
sheav
under
kW.
Vzo
d = Or2sm
Bb
O
B= bhoGkN
Sheav:
Areas
=
~- (1662) (007) (0°38)
Orel Be
=
Ma
Ds
8
My= 0 4 3:087 2 3-087 km
Me7 3-087~ 5¢887
Met
Bo.
Maximum
1
I
For
®
[
|
iM]
Ye
+
:
:
4
29P
HO
kim
+ 3-087
centroid
Locate
|
I
w 28
= -2°F kim
Ras
each trang be
ot
cvoss
jm
2 S 2100
S ine
26
HAE bo
of
inertia
T- ZI+ZAad
+ —
Amit
bottom
& bh®
T=
Mom ent
:
from
33-72 mm
7H = HLSGE2
9-4fntat
194
= 45-lo0
Ad*,
fart
section
AS,mm
dinm
sn
Tn!
$6250 1623 29816 361863
$2500 BOVE:
ISFLBE
108750
HTB
656250
i
W067 13
Novmaf
g=
stvess
Me.
(32087910) Bhe2Bxi0)
ax
= 26.9
468662 X1o-(®
wha
.
Problem 5.58
5.58 and 5.59
Draw the shear and bending-moment diagrams for
the beam and loading shown and determine the maximum normal stress due
to bending.
Reaction
at Az
4DIM, = 2
|
2 OA 4 (90) (24-2) -GlYlot)= a
Reaction
+5
at
M,
D,
=
oO
~(40(24)(1-9) + 200-3036
V, EN
Aveas
792
sb
.
ot
A]
c
&G
of Poul
C+
|
Sheaw
8
diag ras
D
to
VF
72
2 O
D=/728EN1
~
(90)(2-4) = 276 EN.
Av ag vou
A
|
A= THIENY
.
C
Vie 72
- tO 5
LN
~1386P EN,
Viz -/3heP+ 17268 = 36 EN
V= 3¢EN.
D+ B
Over
CD
Ve
~ F602-0:6)
Vie 742-Fol(te-0:b6)20 He- obs oF m
G@
At
Areas.
of
shean
Cte
GC
= 34 840 bio
A (08F)(192)
Gio
D
L (P80) (168) 2 10369 8 bln.
Dt
B
(36)(0+6) = Al &
mom
= O + 4762 © 4792
Mg = 41-52
=O
Lm.
+ 34.848 = Ba-36h Lain
(03-96G
Mo?
82-368
~
Maz
-a66
+ 346-0
=
=die b& bALm
Mg = 826368 ENMm
W 310x387 roBed
Normal
LAI m.
Ma
euts
y
M.
Fou
.
EN Mm.
. f066)(74%2) 247-52
Bending
=
diagvan
CC
A to
IML
792
stress
steef
Ghe 7
section
S
= 549
Mme
£2-368x103
S
F549 X (0
a
Ko
3
pr
6.
™
=/50
mPa
§.58 and 5.59 Draw the shear and bending-moment diagrams for the beam and
Problem 5.59
loading shown and determine the maximum normal stress due.to bending.
140 mm
HIM,
= 0:
(ayr) - (BdCAAY
B=
A}
Cc
|
Ct
S
2
e- J
a
M (cNem)*)
4-d
a
GS
~5.5
U-of = 3.3333m
m
SV lx = 4(2.16667)6.5) = 7.0417 kNem
D.
Cte
26
SVder C2000)= -2.0 kom
hte CC,
“Y.
oO.
sheav aiaavave z
of the
Areas
S.O41F
2.1667
kN
= - SS EN
Id =
S25
d=
6.5
-”
V=
where
D
point
Go.
3S
6.5 ~ (@4)
V
Locete
~2,0
V~2kN
zt eQ+
D:
By
3.5 kN
~
AteCr
Shear?
6.5
kN
4DEMe = 0: (62) +(8)H)Q) -40 = ©
eal nee Sa
160 mm
V kN)
5.5
+HB=zO
S Va
Dte B
= $(2,83333.\CS.5)+ - SLOUTT kom
~ 2.02
=
Me
0
- 2,0
= SLO4l7 kom
= -2.0 + 7.0417
M,
=O
Mg = S.O417 - S047
Meximom IMI] = 5.0417 kom
For
pipe *
tf
C.7
Novmafl
3 (160)
=
=
BO mm
(cf-.") = B[e@oy'-(70)") =
«- +.
S*
td.
S, 0417 x10”
Co
stress?
IB. SI2SKIO®
=
Bo
S
M
S
IGG,
4og™
5.0417
féc,
406
.
ki}-m
- 2,0
>
0°
=
M,
momewts*
Bending
Nem
od
Cy
F A
iF
3 (to)=
7O
um
13. 3725 10% mon"
LO°mm>
lo?
toe
~~
166.406 KIO wm
30.3 x]0° Pa
G=
30.3
MPa
<a,
Problem
5.60
5.60 and 5.61 Knowing that beam AB is in equilibrium under the loading shown,
draw the shear and bending-moment diagrams and determine the maximum normal
stress due to bending.
400 KN/in
-
tt Lt
lp
t
|.
0.3 mo[e—0-4
aN ZF) = os
B
CNd=w,)
(0.4400)
me» fe-03 m»|
Va = ©
Va = ~ 48 +@0.3)Ci6e) = oO
d
Litti titty,
" te
point
Locate
[eff
Vy ket
4B - (0,3)(400) +(0,3)UG0)= - 48 kv
it
2
a
J] J>
Novmal
22.5
stress?
72
=
kNem
Cher
700.2048)=
E 4D:
De Br
¢(0.2)(-48) = -4.9 bem
£(0,.3)C48) = - 72.2 kvem
M,
7
moments
zo
kw
Me=
22448
= IZo kn
I2.0-42
= ZQkw
1
O47.2
x
= 72
4.8 Nem
M. =
0
W200%
diaavaw.
4 (0.3)C48)
7.%4-7.27
e-<
Fov
of CD.
A to C2
Bending
kMem
midperat
is the
E
shear
of
Aveas
V=O,
where
E
syrametry
By
"oe
IM le (2.0
48 lw
=
0400360)
Vi=
400 kN/m
cl]
0
Wy* 160° kN /m
fey W200X 22.5
Shear diagram.
a
=
0
= 12,0 x10 Now
ro Pad
stee#
oO =
Shaye,
\M|
“Ss
>
or
Sy
=
Oxo”
en
144
=
¥[O me?
61.4x
=
{0%
G = 61.4
ig4xX1O-
a
Pe
MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, inc. All tights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers’
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—<
Problem
5.61.
6.61 Knowing that beam AB is in equilibrium tinder the loading shown,
draw the shear and bending-moment diagrams and deterinine the maximem
normal stress due to bending.
rey oe TERN8
f
+.
w =7S0 (19%) = 750 - [816x
A
me
ow
av
Ond mo
snes OA trom
Orgin .
OK< %<
C
Ate
:
NE
= O4
*
x ~7S50)dx
Cars
Va +
Ve
|
V (kn)
~JS0
%
ME
=
K™ ~]50%*
G3
= dM
»
= 1
M=
My, + SV ax
=O
+
.
GC 9316S x” ]SOx)dx
= 3/206 x* ~ 379 x”
Cte
8B
Maximum
Cross
c= £=
section
M==— 40 Nm
Use
IM)
~/S0 N-
Vo
At xso4¢m,
symmetvy conditions,
=
4o
kh,
(£)(20) = lo maw
T= Beh = €)0)
= 7084 mut
Normal
Mc
stress
S
=
Zr
4oV(o-o!)
=
eee
/
=
S09 MPa
.
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the linited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
|
Problem
*3.62
Beam AB supports a uniformly distributed load of 2 kN/m and two
concentrated loads
P and Q. It has been experimentally determined that the normal
stress due to bending in the bottom edge of the beam is ~56.9 MPa at 4 and ~29.9
MPaat C. Draw the shear and bending-moment diagrams for the beam and determine
the magnitudes of the loads P and Q.
5.62
|
Qo
P
I "
Q2kNAn
“4
4 (\2)(80)"
T=
“on
R
CF, 984 *10" mm
-
Ct td= 1g um
pe
O.lm
O1m
0.425m
S
me
YOON
:
¢
4b
¢
Lanas
goow
9
D
B
=
=
3,883 x10? me
At
A,
Maz
At
Cy
Me
=
3.888%10°
wm"
S&q
Nem
M,= (3, 888 «10° ) CSE. % )=-221.25
2
S
6.
Nein
— 116. 25
Me =. B88x{07° )(-29.4)=
Q
\°
2
—_
q
FO
nga
DO
ED
DZMy =
O.1 P = 0.226Q =
aso
Solving
G1)
DR
4OO-
R,~
Vz =
M,
zo
P=
Soo
N
at
Qs
ASo
N
~<a
- 250
=
S00
R,
2 SO
HWSO
ON
221.25
Mex - (6.26
Mp?
-
A
ak
Ponce
Reaction
3125)
(@)
simuttaneous Ay
Q)
and
106,25
=
0.225 Q
O.1 P+
pre
221.25 Nem
.
116.25 ~(0,05)(QQ00)-
oN
(1)
= 181.45
Q
=O:
+ OM,
Miso
Ae
+ 0.325
O.AP
VW)
z0
0.2 P - 0.325Q
221.23 - (0.1 (400) -
New
oO
|
Vp = 250
Nem
Nem
Nw
IV
= SON
aed
IM leew= 221 New <a
|
*5,63 The
to bending
Draw the
maximum
Problem 5.63
0.2 m
|
0.5 mote
0.5 m—->
beam AB supports two concentrated loads P and Q. The normal stress due
on the bottom edge of the beam is +55 MPa at D and +37.5 MPa at J. (a)
shear and bending-moment diagrams for the beam. (6) Determine the
normal stress due to bending that occurs in the beam.
T=
24 ime
Gh (2il6oy® =
432 ¥107 mm"
CF
30
mm
S*
s = 14.4x/o°mme =
[HAx/O° wn?
M= S&
At De
My =. 4xlO°)GSx 10%) = 792 Nom
Me = CH.4x 10° 1325«lo*) = SHO Nem
F.
At
Using
Free
booty
Fe,
:
V,
S4p
Br. ~ or:
0.3>
*
Using
_
,
beam
face
eutirve
45
ciaqgvam.
and
Ve 3600
E*+
Vz
Ct te EF
B,
endin
M,
=
Vr
mom
its
N
A ~ 3240-21604
M.= O+720 * 720 Nom
Maz
Feo~Foo=
O
=
3600
N
N
Aen (0.5)
Aca
r (OS
360)
=
{Roo
Je
[80 New
-—FWO
New
“IMI,.. = 400 Nem
Se
©
7eo+!BO
1IJ0O =
©
Age? (0.2 1(8600) = 720 Nem
360~ 2160 =-|800 N
Met
32240
N
=
P
aveas.
Boon —~22%0 = 860
=
+U.2\igeo))
A
Ate Cy.”
Ot
M2
~0.2P — (0.72160)
{800
= 0%
Ot
2s
RIGON
Qs
he os 4
i 0.2 le OS
2M,
45
~7942 - 0.3. +(0.8)(1800) =O
6
|e
P
Shear
Igoo N
DEFB,
body
Pree
y Loa 0.5 —T ig00
+2)
O¢
8
me
Using
>
-S40 + 03B=0
.
Ce
DIM,
o>
mM
} =
ieee
=
= FooNem
oO
6ZSXIG Pe
°
(by
Guy. = 625 MPe
at
6.64 The beam AB supports a uniformly distribated load of 7 kN/n and
two concentrated loads P and Q. The norinal stress due to bending on
the bottom edge of the lower flange is +100 MPa at D and +73 MPa at E.
Problem 5.64
{a} Draw the shear and bending-moment diagrams for the beam. (b) Determine
the maximum normul stress duc to bending that occurs in the heam.
For W200 X46] wolled steel section
Ms
S36
AED
At
.
Moe (Aabxio Visor’) = deed EN
EF
Me = (44ex 1o°©)73 xi’) = 2267
My= 44:8 Lim
Use
fyee
body
OF
Me
DE
wDEMz © 0
~ 448 t
ENm
TEN jm
pete: ELS
=©
370?
BleT
bm:
Pog mal
.
.
Vv
EAS
aM
>
Ve
=o
AEG
4327
~ CD09) logs) ~09
Ve
Use
= 32°67
.
VAN wim f9e29
45 M,
kNm
Aad ENmITTTi TTT]
.
327 (DOA) (045)
Ve,
S= 449 yo" men
free
body
ACD
4DEM,
=
=
~f66G9
7) l0°75) (0378)
4 (0°78) (70:29)
¥
Pe
joe29
WZ
EN.
©
0°45 Pt
, AGS
YzO
+ 44
F O
MRA eM
Y
0)
A ~~ (7)(0°75) + $0«29 ~j/2°34 20
A = fo7-29 EN fF
Use
Sree
22-1.
¥
CE
y¥
jusg
body
a
EFB
DS
Ma, =
0
OASQ + C7)(O7TSJ (0-278)
TT
4
(0°78) C1689)
- 327
=
Oo
q
,
GL
®
=
4-07 bl
bea.
B= le SF~ (0615)(1) - LobY = O
Coutinved
Probdem
Areas
S64
ot
Shear
continued
aqvam
Poad
(e46)(7)
=
Cte Fo
FB
fes\(7)
(ant)
=
61S
=
EN
188 EN.
ISK
107-24 EN.
My =
Avagnam
Adect
=
107-24
Vs
=
81S
jy Lal.
Lode
Ve = (04/4= 112633 =. 819 EN,
Ve = ~PlG ~ to = -1F EDEN
Vp = ~ 18:69 ~ 4064 = -£9:33 kN
—
Ve = 7 543R
Aveas of shear diagran
Ato
Cho
4(0-45)(I01244 1044) = 42512 kam
Fo
Fide B
Bending
moments
Maza
AMS) (-8:19- 1869) = — 2016-ENm
4 (0-45) (-5933-62-48)= - 27-407 kin
=e0.
Mos
= — 62-49 EN.
#IS
:
04 47572
= ADST2 kAlm .
Me= 42872-2016 = 27-412 Am
Mg = 2412 = 27407 % O
Mavimum
IM|
oecurs
} m
Mazimu
stress
ot
C=
C
Mla, = 47572
|M]
S
mee
00
ILKIO!
ATSRET
7 .G4Rx (oP
OL.
EN
06:2 Mia,
—_
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission..
Problem 5.65
L&KN
5.65 and 5.66 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 12 MPa.
3.6KN
.
Mom
FD EMp = O35
|
H24A
+ (16874 CO3)G.C) 50
Nt.
:
an
Hn
0.8 m
:
+O
|
(1.42) |
og
|
=-
D
Construct
a
(ons
24
kN
M A
=
of
-@.3)U.8)
_
08m
WN
V, Cun)
M (kM-m)
Ar
One
3%
3.6)
. 24
D-o
KN
shear
ond
bend ny
moment
diaevame
(~ 2.4)
1M
xe
| wax
Sy
.=
Sain,
.
>
I2
2.4
MPa
S:
200*/0°
2 bh?
-
-
=>
mm
=
2.4
12 *10°
-
lO”
Ne
¢ Pa
;
~&
$
2
d$(4eyh*
(6 (260 »f07)
io
>
2.4x1o?
LM dma
=
kW-m
+
Z3oxrlo
2oox!lo"
mm
h =173.2 min tt
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.66
5.65 and §.66 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 12 MPa,
ISkN/m
120 mm
&
Az
+
.
2m
|
CP
Ve
i!
Aveas
Ines
mow
C
°
.
ents 2
7.5
=
25.05 kNem
Mo =
25,058-7.59
=
17.55 ken,
a
h-/
- 17.55=
=
5S kN m
New
“12.55
anti-sy mmetyy
By
MIO
S = ac + 2808
lax 4O
=
3(I\Us)
©
M
6S
t BC.
EF is the mrelpeinof
= 25,08 x107 Nem
25-O5 kNew
rectangudar
Oo,
Vz
[7.59 kNom
17.954
For
|
©.90S)4+ 20.9) Q1IS-IS) = 17.55 km
to dD,
Me =
s
= - ISO kN
M,70
Mag = O417.5S =
17.58
ISO kN
BRC. 40NGis)e -25Ky
tN,
E
kN
of shear Aiaqvar.
Bt Es.
i \
where
symmetry,
Ate B.
{
E
point
By
< 25-05
go =
21.75
-18.0~%(0.9XiIS) = -275 kN
Locate
=21.7
1.05
I5.0 - (Mis)
Vo =
|
IM ve 2
DF
Va = RES
- Hodis) =
+t
|
=
+
Vy > 21-75 kv
Shear diagram:
Is
Mo
o
=
p
4 (0.9)
O.9 m1
A>
Bending
0D.
EE ZE,s Of As do.ayisds @US)
)
0.9 En
sy mmmetyy |
—
sect (ow
(6X 2.0375 ¥ [0%)
atl
12°
S
=
3
2.08785 KIO
”
wm? = 2.0875 «(0% mun
.
= t bh”
B23
mm
he
323mm
<i
_ §.67 and 5.68
Problem 5.67
20 RN
Fo
QO kN
equadibrivm
B
06m 0. on m
diag
raw
Ate BTS
061 06m
V= = 20kN.
Brte C~
Ve
CD
V, EA.
D+
Ve-F+f%
£7
Aveas
of
sheaw
Bt
dia
Qt
Bend ing
D
(e4){o)
F
lib) (20)
= e
6MP)
=
= Bokal
balm.
Beh tN.
bf Mm
= iz LAL
i |
moments
a
~~ ZO
(2
ROMA
(Orb) (8) =
EG
Eto
2 4
(ab)f-2e)a-/2
C
Dt
O
g bAN,
Vz
Ate Bo
=
Ve
E* + F
eh
kal
/2
Foe
=
Sheav
06m
For the beam and loading shown, design the cross sec-
tion of the beam, knowing that the grade of timber used has an allowable normal
stress of 12 MPa.
oO
.
Meg = O- (2% e-ILENm.
M, = ~12 =~ GY = — fbe§ Lm
Mp = -4P
+O
= fb
bafm.
M, kA in.
Im),
Me
=
Me
7 - 1%
=
j6-8
Requived
S*
For
a
-bS
for
Equating
000%
the
O
S
16 e107
wm,
OO auco® * (Goo x8 fra
rectamaatan
oe wk. 7
Sto
= -{2.
=
ian
vabue
a lMI ne
éy
+ 4h
+12
bh?
EO
turn
section
>
T=
ik
Cb)z e24>_ =
erepve ss ious
b= Moons&
2O1%6 mm
fee
bho
Cn
060096
a=th
|
S
b = ($b Ab,
Proprictary Material. © 2009 The McGraw-Hill Companies, inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any nicans, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usitig it without permission.
~
Problem
5.68
5.67 and 5.68
For the beam and loading shown, design the cross sec-
tion of the beam, knowing that the grade of timber used has an allowable normal
stress of 12 MPa.
M
20N/m
120
min
27 Oo:
ttZF,
*
Reactions
Axl!
\M20)20
Az 20N
lh
MU ESIEO
~My Cool
ADEM HOF
A
My, =
— 3ekNm
V (kN)
26
Shear
Va
=
(4 )(20)
AV. to B=
ie
=20kNm™
.
Myr
Bending moments:
3
»
ECI X20)~ 1okNm
Cc.
Br
MICAN fr)
= ZOLN
O
Alaqnaus
shear
of
Aveas
C
©
~
Ve
V. = 20- (/¥%20)=
t
A
dvnqram:
=
Ma
30
-
M,*
36
—
=
20
4
-/o+
=
- 30kNm
loki
©
IM wee = 3OKNIn
Sy.
S
~ \Mhnee
”
=
(Mi >
Sane
30x} id ;
+am
bE
Fde
a
rectauqulav
,
=
[$2
-
12 x108
section,
G
\@-£x10°)
120
~—
«
aS¥Yio
Sc
-7 3
+
= 353:6mm
in
3
bh”
h=
354mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<a
Problem 5.69 —
5.69 and 5,70 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 12 MPa.
Bec
B+C
sPZRso02
0.6 m
B=C
_
:
- 0.6m
—
=
a
Ve
685 kN
=
25+
6.8
=
Areas
oF
Ve -946.5 2 -25 KV
the
Bee,
SVav=(d)5)) = 6.75 ke m
Coe =
WMPa
Sra
For
Mi.
[Mune
Guy
~StC«i
I2xlo*
‘l
Me
=
LS
=
M,=
LS
=
1S
4
kNew
46.75
=
8.25 kNew
8-28 -6.75
=
hS
RSS
=
|
4
O
8.25 kNem
=
moments.
=
My
= 1S Rue
Svdy = - 6.75 kNem
SVde = 7S kom
Me
Mo =
=
diagram.
SVdx = (6S)
Bending
iM]
shear
Ate @.
Edo C.
CD,
Mayimum
4 KN
= 7K
Ve = 9 - GD
CHD.
25 kN
Vez
Ate B.
Shear:
V (kV)
+ 2.5425
+6 (ce)
=
8.25x10*
kom
:
©
Nem
Pa
825x100 3
eNOS
=
a rectanqu fan section,
C87.5xto = (t)Cleo) h*
687.5x10 ; m=
>
S=
687,510"
+ bh?
mm
3
|
hos GENER? Seton)
h=
3
203
YAS x10" mina
a
—_
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.70
:
3kN/m
bh
.§.69 and 5.70 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 12 MPa.
:
49: Mg = 02
150 mm
Re
:
4(0.6X3.6)(3)
~24A
,
= O°
49M, = Of
.
~U8 3.63) +248 =O
Vy = 2.7 uN
Ver 2.7 -(2.4)3)=
Vte-4.64 8.1
7
Vee 8.6-(2)1(3) =
Locate point
(A
|
Aveas of the
ss
oe
21s
09m
Ato De
~— Dt
Bending
Sin
For
7
216 kNem
=
12 MPa
7 AMI
«
Buy
.
-
(
s
Iz¥ Joe
Peh*t
bb =
24-d=
Sm
diaqvam.
4
1. 21S kNem
GOS )E4S) = - 3.375 kNem
Ma
+ LQIS
=
1-218-3.375° =
=
Oo
L218
-2.16
=
2.16
kKMem
.
bem
kN-in
= 0
= 216 %107 Nem
oe
|2Oxlo
.
om
So
>
180
x lo"
.
mm?
section,
=
(
zs
|
|
Svdv = G)(0.9)02.7) =
O
I2xio® Pa
216-10
pect angqu tar
Sz
Ge
M. = -21¢+216
=
vw
ads
moments:
Mg=
Gy =
Vi = 04
Svdk
= E126)
:
Mi, =
P2146
shear
BGs
Bre.
Pd
.
Mavimum IMI
sn
~&5 ku
3.6 kw
©
i
M Civ.)
wheve
Rt
de
ta de Z4-dol
D
:
BBN
ay
Sheavi
A =2.7 KN
P*bliso)*
x
(o?
E1180
x10")
Isgot
=
itox/o*
b - 438.0 mm
—~m
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or .
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers _
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
5.71 and 3.72 Knowing that the allowable stress for the steel used is 150 MPa,
Problem 5.71
select the most economical wide-flange beam to support the loading shown.
100 kN
J00 kN
DIM =O
= 3.6 A +(8 000) 502.4 )400) + (1.8 Kioe) = O
A = 200
+ OM,=O
Shear:
£90
|
P
kz
a
Aveas
foo
120
|
IRo
Bee
V = 200
CrD
V2 ]002-{80
DE
V =O-Joo
= -100 kd
RB
Gins
Swain
\60
wg
IMI
Gy
_Shepe
tga
MPa
[80
x fb?e
7 | dow
1340
.W 360%*79
LAZO
W250
# Jol
=
kN
©
cliag Naw
§ Vek =(.6)(ioe)=
§Vd&ke= Oo |
60
DRE
SvVdx = (L8)Cloo)= -162 kNem
moments
=
120 Kiem
RVD4+6O
*
186
=
Mo
=
19a
Me
=
'80.-180:
1[8@« {0°
«10%
=O
120
M.
=
7
M,
+0
kim
kNewm
* BO kWew
0
Nem
Pa
NNAR iO -3 Um 3
=
3
125
«10% mm
3
§ C0? m2}
W.530x 66
Wr 460* 74
- [po = Joo
BC
CHD
Mp = O4
160
ky
[20 kNem
kKN-m
+
shear
200
CVdx = @.6E@o0) =
MX
IMl=
jpokv
V=
Bending
Maximum
=
AdoB
Under
Ate
M Ckv- im)
;
3.6 E = C.8)G00)-0.2 Gon) (0,6) od) = ©
E
y}
kn
1460
Lightest
wiole Ffange
W 530 *GE
@ C6
beam
kg/m
Z4o
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ty
Problem
,
Knowing that the allowable stress for the steel used is 165
| 5.71 and 5.72
5.72
MPa, select the most economical wide-flange beam to support the loading shown.
TOOKN
VFMEZO
40 NA
-7T2A + Fo
Ar
SDIMg2O
T2)(56) 4 (oe
pobs
72 © = oN Paleb) ~(102)F72) = oO
Coe
[BS Lay
Shear:
Vy= 20605
Ver = 2065 (4027)
Vets
Ve
Gy
foo
thE
© 98-8 LN.
es =f
SEA.
m (40) (45)
.
8 -18er
under shear diaguam
A fe BR
Bto C
~{fie™
Feo =m
7
Areas
4ib) = Oo
LEA.
Bending
Mp zt
Mo:
SVdx =(Q\G7) (e065 +-98+5) © 411075 EN
§ Vax = (A )4es -S- BFE)= = 46 75 EN
moments >
Mazo
O + 475
= 411-75 ENK4675 - 4007S FO
Maximum IM)= 41le7s bam.
Sue
|
fe
IME
"Bat
gS
has
ee
.
| Aife7 Ko?
168 x/0"
nen
|
2 244K XO
| SGmmate)
Shape.
Wwb9o
le]
wetok Xin
25/0 &
3830
pus sox ISO
S37z¢6
13
24 0a
a1 6
xe
W 36
BB.
w¢box
=
7?
20
bean
gam
fef keg [mm .
mm
Phawee
wrele Plaige
higah test weele
vy GloxXfef
@
‘
ly
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written perntission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem 5.73
-
§.73 and 5.74 Knowing that the allowable stress for the steel used is 160 MPa, select
the most economical wide-flange beam to support the loading shown.
50 kN/m
Hz
M,
=O?
- ZB2B
OO
I
|
0.8
2.4m
V Ck)
|<
08m
DEM,
Or
32D
5
~4YO
Ke
us
o
Vp =
poiut
Locate
7
e . £4-€
#0
~
er
| Bending
a
Ma
192
Inoe
2-4-2= 0.8 m,
SVele = (4YO.8Y(-#0) = - 16 kN-m
S Vee = (089-40) =
:
moments
O-
iY
Guy
Sin
2
160
MPa
Sa
°
=
Shape
W 3iox 327 [
> W250" 28.4 |
AZ
kN-m
=
48x%fo*
IG
~
=
Mp = 32-327
mM]
M,=O
=
16
Me = ~164+64
Me
4YQ- 16
Maximum
0.
knew
48 kV.
32 kN-m
O
Nem
= 160x/(0% Pa
ee ioe
=
Cio
41s
308
+
mm* )
=
GY kNem
§ Vax = (KY0.8\G4) = - 16 kvm
Coto D.
-I6
-
~YORN
S Vdy = (AY1.6 C80 )
E.
ERC.
.
=
V=
where
E
A to B,
B to
~4O0+0
40
1.6 m
Areas:
4g
fo
8O kW
=
Ve = G0 - (24)50)= -4O kM
m)
am
~ (0.9V(2.20(50) =o
D = Ho kv
Vet = ~40 4+ 120
|~4oS——
et
kN
Ve" = 0 ~(0.8)(50) = -4o kw
Dp.
oc
BE
120
O
7
Va
=
Shear
°
M (lle
B=
z= Oo
|
a
A
+ (2.4)G.250)
300 Kfo°w*
=
300xl0\mm
Lightest wide Phange beam
- 32 kv-m
Problem 5.74
§.73 and 5.74 Knowing that the allowable stress for the steel used is 160 MPa, select
the most economical wide-flange beam to support the loading shown.
ISkNAn
MzeBy)
= (E4 2%) kW /ny
dM _
Fan
Cu)
A
8
.
M
=
M=
~72
M (leds)
Me
_
le
~
+
Cy
x=0
C, 20
-3e°-4x°
af
=
Gm
IM\=| -@\ey-
ey] = 180 kNem
oe
:
Shape
,
i340
w46ox74
460
WHO
So
x BS
[Sto
Lig htest
[280
oT,
IS9o
W 250 * fo]
I2Yo
MPa
=
I€oxlo®
a
Oy
acceptable
.
Nem
Pa
= 1125 %10%
Tear 10
=
<—
W 360 « 749
W ZIP
x [07
aoe
Shon yal = 20M"
|S ibm)
W S30% 66
|
Sin= 160
a
180xto*
=
|
~180
0
x
43
,.t
y
oS
z
ad
®»
1
_
CHe-
-—-3%
0
©
u
8
_
V
[Ml ocers
A
t
y
=
S
G
V
j
if
xz
we(G+
wide
Llesxio®
Prange
wm
bean!
.
W530 ¥ 66.
Lge
an
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
:
a
Probiem 5.75
5.75 and 5.76 Knowing that the allowable stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
80kN
100 kN/n
9s
0.8 A - (0.4¥2.4 (loo) -(16 Y8e)}= Oo
M., =o
Kz
+D2M,=0
280
Vis
>
z=
-— 280
Goo kN
tt
— Z60
4+ 600
under
shear
~ 360 kw
=
QYyYoO
”°
240 - (1.6 )Cfoo)
Aveas
?
kW
~ 280 ~ (0.3)\Cloo)
4
R,
<
ao
4
0.2 B~(.2)0.4 Mleoh- (2.4 (20) = 0
B=
Shear *
kN
kN
> 80
KN
At agnan
AteB
(4Y0.8)6280-860) = - 256 kNem
Bre
(z (6 VQ40 + 80 ) z=
Bending
=
© ~ 286
>
Ma = ©
*
~256
kN-m
v
Meg
moments
RSG KN-w
Maximum IM} = 256 kWem = 256*10° Nem
Giy
Sun
Shape
= 160 MPa
=
IM)
Guu
-
1Goxlo*
:
256™Ito
\Gox
3
joe
>
| éxio?
pS
S Clot me)
S10
S
60x
*« FB.3
jog
=
{600
a4
x[O my
|
Lightest
S
Po
S-
seetiou
14so
1685
S SID *« 73.3
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
a,
Problem 5.76
5.75 and 5.76
Knowing that the allowable stress for the steel used is
165 MPa, select the most economical S-shape bean to support the loading
©
shown.
Krom
stodies
the
Az z24in
3
0.6m 06m
VEN
Area
D(-
of
Pood
Sheav ding
A te
0.6m 06m
a1
reactions
at
b
B= 224cn
diag tan
CD
F
ave
f
UP)U60) = 288
kN
awn.
eee
Bib
C
At
Dk
D
ET
Viz
224 EN.
V = 224g—-Box 144 LN.
28 PR 2 TEEN
V = yqg.—
Vee fag EN
PO = -224EN
Vz 1g
F
Eb
over
A and
(ce
RR
Af
€
EE
F
V = [44 -—fée(%-h2)
CD
Over
At¢G
O
=
Vso
G
Aveas
or
shear
beac
forb)ietug)
Gt Db
DebeE
fob)-2)
E to F
2E6-4
Bending
Me
[2GebENO
hoot) (da) F
Ch G
M, EAM
BG END
(or EEE) ©
BteC
“224
(063224) = (94. EN,
B
Ab
mm heey
Avaaram.
moments .
7 (2962A,
* ~ Ph ¢kAlm,
= - 134.4 KN,
M,=
oe
= O +1244 = 13464 LN.
M. © le4:q4 BOG = 22006 LAm
Me = 2208 41296 =
250-4-kNm
My = 350-4 -24-67 22068 EN™,
Mp = t208 ~Bb4= (346& EM,
= O
Me = 144-1344
[Mls
.
Shape
BS
DS, (ate min )
Sain * Wb
"Gun
SSlOxX2§ |
2550
250-4
ENm
2 BBO
(65x
Use
,
Ke?
geyzixio
~3
3
2421 K1OF mm?
S Slo x 128
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
att
Problem 5.77
5.77 and 5.78 Knowing that the allowable stress for the steel used is 160 MPa, select
the most economical S-shape beam to support the loading shown,
FOKN
TOKN
45 kN/An
By
Reactions 7
Sheavt
Va
Q02.5
ne
\
C
D
-—7o
-7O4 272.5 =
V0 =
202.5 -G4S)
Vit =
-202.5
.
Vo =
70
kN
+70
+
+0
f
kM
=
Vg"
7°
+
=
=Jo
>
Ve
vi (kw)
=O
kN
272.5
Beas
a
BoE
Bzc
-To+B ~ (\W8) +0 -70
+T2F, FO:
A
Sy romety
1OALS kN
= - 202.5 kN
4272.5 7 7o kw
kN
.
~Jo
— Reaw
—2O2R5N
shea
where
Avag ran,
Vz
E
Aveas
Locate
pont
E
oO,
of the
is the
micipoint
of BC,
shean atiagvam.
( Vdx = (3)-70) = -210 kNem
§Vdy = £ldls¥202.8)s 485.625 kom
Ate B.
Be EF.
E to 0. S$ Vans $04.5)6202.5)= - 455. 62k om
Co
Bending moments?
Mg = O- 210
My=
D,
SVde = (3)(70)
©
Shape
= -210 kNem
Mer - 210+455. 625 = 245.625 kv
RYS_GAS - 455.625 =
My
=
~210
4210
~RlO kN
_ (Ml
6
ee
=
MPa
|
New
S Clo* vam
S610 *\NG
S Ste» 98,3.
2830
lqsSO
S 460r
1638S”
lod
<<
Oo
=
IML. = 245,625 kMem = 245.625 x07 Nem
iGo
= 2lo
=
248.625 %10%
leoxlo*
[GO
=
IS35.2.« to? mm?
x fo*
$510
Fe
1.5352 X10
Lightest S - shape
on
x» 938.3
a
Problem
5.78
5.77 and 5.78 Knowing that the allowable stress for the steel used is 160 MPa, select
_ the most economical S-shape beam to support the loading shown.
,
80 kN
4) 2M,2 02
7 (kW
|
BteC.,
BF
D
A 65
Vy
Aveas
=
F the
Gur 7
>
Shape
S 3a0%c4 |
S 310% 473]
Sors2 |
diagram
= -
6S
kM
KN
:
*
B25) = 66-15
kNem
SVdx=(0.9\- 6.5 )* -5.85
kyem
(09
Mat
OF 66.15
Met
66.15
Mo?
66.90 ~ 66.30 = ©
kom
=
ees
KN
13.5
Nie e.57-6as) =~ 60.30 bem
\GO MPa = 160xI10°% Pa
G66.1& x{o4
1M |
Shain 7 “En
f
:
moments:
Magimum IM] = 66.15
kw
+80
Bending
D
€-
60.5
= 6,5 ~(.8)Go)= - GOS
sheaw
CD.
8
7
Vi
SVdx
Bre.
A
=
Vt7B5
Ate.
~ €0.5
D
AtoB.
Shear?
= 0
3.6 D ~(0.9)80)~2.71.8XG0} = ©
|
Nae
.
,
~3.6A +(2.7Y¥80) 4(0.9)C1-8¥90)
At 73.5 kN
4DIM,= 0:
YS
co
,
30 kNAn
=
Mazo
= 66LIS kNem
- S85
C615
= 60.80 KN-m
«10° New
.
BLY «10S wm? =
.
413.4 * 10" mm
S Co? met )
971
543
482
—
lightest S~ see tion
§ Blox 473 C473 Qi
me
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
' distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. -
Problem 5.79
5.79
A steel pipe of 10-mm diameter is fo support the loading shown.
Knowing that the stock of pipes available has thicknesses varying from 6 mm
to 24 mat in 3-mm iicrements, and that the allowable normal stress for the steel
used is 165 MPa, determine the minimum wall thickness 7 that can be used.
|
AtoB
Bee
(hz) (2) 2-264 Lm.
(feZ)¢-
Bending
4)
=
moments :
Maz
if
-24- 42
0
i
Gott
1b”
Ml
Mfr
Sain = Ml.
Oust
Ele,-¢,")
Fog
4
-6
lb
w
Kio”
peep ma?
svt-o*
,
3
C;
= Ca-
C,
Zimm
$0 = 4dt7
me
2 wr 2:eats
IM = 4368
3
2
fF, coe™
GF
destyn
=
420
3
7
frin,
Avha.
£
=
G
in
im
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitied by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
3
XO Inge
C,2 dd = Stamm
og ee
¢t-c¢"
4.
LIMO
2 AR ”
bY Ko
c=
S= G5 FSS
= -2R2ENm.
72 balm.
\m \ 2
Maximun
Demag
balm
Me = 0-24 2 -24 EN.
Moe
Lain
~~ Lee F
tt
+
>
“ZEN
Bt GC Vs -2-22
-4 EA
Aveas:
T=
Vi:
B
A +o
:
Shee
100 mim
5.80 ‘Three steel plates are welded together to form the beam shown. Knowing that
Problem 5.80
the allowable normal stress for the steel used is 154 MPa, determine the minimum
flange width 6 that can be used.
sokM
B
a
03
3D
Bt
C.
i203 kN
[30-3 ~ 32 2 99-3 EN
[Bin
Ct
D.
92:3 = i295 — 29-7 EN
V, kN130-3
Dye
EG.
Sedin [24m
Ae2in|
A
3
~247
- 1877
Ms ENm
- 29-7 - 1285 - 157-7 EN
Ato
Aveas?
>
3
PB
A Ww HB.
Shear:
}
A
Ce
(2\(-247) 2
—
(24-1577)
= ~ 48733 kN
DD.
moments
et
€
D
(h3 (1303) © 169-34 kNM
(4.2)( 99-3 ) = 412-86 kim
Bending
c
Be
BC.
DtoE,
sa2.26
¢
A 8
O+
Ma
=
Me
= 16939
>
+ 412-86
124.74
kNM
& 58225
a
[-
=
=
©
Sy = 15% MPa
-
3
£6225
54x10¥10
=
2875+
Ou.
7
t
28
b
x10 m3 = 31908 KIO
31808
& Ua) Gs? 4 2[ ab yesY + (b\Gs\(zs0)" |
Cr
kNm
457-51 kNM
Maximum [M| = 582-25 bem
IML
6uy
km
Ma=
167-39 = 169-34
My > $82-25 — 124-74 7
Mp 7 45751 ~ 45733
mtn m =
4
kN
13003
(392)
)(32) CLS C 28)- (9-7) (9-71) C128 C128) 5 = O
126 ab Ex Bae (103
E = istt enf
128 EN
J28EN
=
Map
+5
1mm
S Fe
4.2 mh 4.2m > Son
ae
=
A
“re
19 pum
f
|
|
+ O71 128 AGIVI2D =O
ib A + Ub YR2)
i Pm
»
+t) 2Me=O8
Reactions,
[et
PQS kN
PIQSEN
32KN
mm
= 189688802 + 3127604b
2 262-5 mm
= 646433-54 1194-7b = 2.7908 x10"
b = 263 mm
Proprietary Material, © 2009 The McGraw-Hill Compat nies, Inc. Alfrights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
ah
Problem
5. 81
oo
5.81 Two metric rolled-steel channels are to be welded along their edges and used to
support the loading shown. Knowing that the allowable normal stress for the steel
used is 150 MPa, determine the most economical channels that can be used.
QOkN
20kN
20kN
By
symmetry
|
sf ZR, 202
Ar
"Bt
Jo
A
cop
8
sg
—Avecs:
“te
-
At
E
.
.
Be
~- 20- 20- 20
=O
30KN
BB.
V=8okw
GC.
Ve30-20 = IOWW
CHD.
Vs
D
V
te
E,
At
Bending
=
A+E
&
Sheav:
‘tn
A
BL
lo-20
#-lo-
oe
©-lokw
20
(0.675280)
zs
= 30
.
=
70.25
kW
|
ktm
BwC.
Cte DB,
(0.695 Mle)=
6.75 kNem
(0.678 )Cle) = - 6.75 kvm
Dt
(0.6769 }G3e 3 = - 20.25 Wem
EF,
moments ¢
M,2
©
Mg= © 4 20.25 = 20.25 kW-m
Me & 20.25
+ 6.25 = 27 kim.
Mo = 27- 6.75
Me *
Maximum
6
For
a
section
consis
. 1Mt
Swin *
For
each
_
“Ei=
iSoxio®
channed,
Shape
C
80146
C
lSonlh. 3
ting
27 «jo?
3.2
|
>
of
tw
_
>
= 20.45 bem
20.25-~- 20.25
IM) = 27 Wem
150
MPa
+
=
oO
= 27xJ0*
1802
10%
Nem
Pa
channe és,
-
180x107" m*
(& (80
«107 )
=
102
180x {0° mm
9O
x10°
3
ram ©
S C08 mn )
FER
93.6
<
Lightest
chawnnet
C
[80*«
section
14.6
—,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manuali is using it without permission, :
P.
roblem 9.82
,
OkN
4.5 kN¢n
§.82 Two L102 x 76 rolled-steel angles are bolted together and used to support the
loading shown.. Knowing that the allowable normal stress for the steel used is 140
MPa, determine the minimum angie thickness that can be used.
:
rT
,
‘on
Reactions
i
102
Im
.
By
tt ekyeo:
mm
sy minetoy |
A
= Cc
A-@us-q+e
-
= 0
.
4
1m
=
|
kn
G9~U)C4s)
=
45-9
~45
yy
Vai
<
Vp
i"
—
Sheer?
Aveas
of
4&5 kN
=-4S EN
- C)C4.S) = ~
kN
dragvaw,
§Vede= E0VG44.5) = 6.78 ee
BRC.
Bend ing
2
shear
A toB,
=
SVde= RNC 4S)= 675 kN im
moments t
My £0
Mg = 0 + 6.95 * 6.75 kNem
Me
=
Menimun
Sy =
For
\YO
the
MPa
section
=
675
- 6.75
Im |
=
6.75
=
©
kNem
=
6.1Sx1O°
Nem
14oxlo* Pa
of +wo
angfes, ,
IML
Sma ® “Guy
=
6.75%0%
lo
wor
| =
4 Qi»
lo’
|
wm
= 43,21* loo mm
Fo
each
ang be ;
Shape
Sin
=
S (103 ma
—
L JO2* 76
* 12.7
L 1orx 76% 9,5
L lo2x 76 * 6.4
Sil
24,0
16.6
£ (42. 21)
=
| 249,
oo
105
x io”
—
Lightest anate
<<
rom
is
L 102% 76™ 12.7
Tam = IR
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
a
<4
Problem
5.83
5.83 Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 170 MPa, select the
most economical wide-flange beam to support the loading shown.
Total load = 2 MN
Downward distributed food
Shean:
Va
=
0
Ate BR.
0.6
MN
(4)(0.75)(0.6)= 0.225 MN-wm
(4050.6) = 0.150 MN-m
GYos\-o6) + -0.ls0 MN-m
(4)(0.75)-0.6) = - 0.225 MNem
moments?
Ms, 20
Ma = O + 0.225
=
0.3 MN/m
7
0.6 -(-O)14.8) = ~ 0.6 MN
~0.64 (.75)(0.8) = ©
Me
2MN/m
MN/m
1.2
V.7
Vo
nels
ie
BC
O4(0.75)(08)>
Bt EE
Ete C.
Che D.
=
over
Vaz
Areas:
A
Joa
Net distvi buted
V (MN)
a3 2S=?
distributed reaction
Vewavel
OTS m
075m
w>—2—=
bo
=
0.225+ 0.150
0.225
MN-m
= 0.375 MNem
Me = 0.375~ 0.180 = 0.225 Mem
M, = 0.225 -0.225 = oO
Mavimum IM) =
Gas=
i7o
6.375 MN-m =
MPa
Sunmin 7 ML.
Cant
=
I7oxlo*
Pa
Sisxlo"
\7ox lo®
Shape
W 6490%125
375 */0° Nem
2.206 x10 > me 3 &
a3
SlCr)
3510
W 610% Jo}
2530
W 530150
3720
W 4¥60* N13
2ioo
2206 *J0" mm”
shtest wicle Hange section
<
W
GIOx]o!
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc.All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using if without permission.
—_
Problem
5.84
SOOkN
800 KN
5.84 Assiming the upward reaction of the ground to be uniformly dis‘tributed
‘
ane and knowing 8 that the allowable normal stress for the steel used is 165
MPa, select the most economical wide-flange beam to support the loading shown,
Distvi lu eek
ance
12m
“12m
Roekn
A
By
Shear:
fe
Ver
12m
Reokn
>
=
A
O+F Uaf~Fgee)
5
7 OZG)
(B62)
D
moments
Wie XG
Wasp .f67
3)
>
EN,
Met
Bto
- fo
Me?
*40
4
Pe
326
-
336
=
© — 320k,
Ma?
©
= Z20kAsm,
4320
Sinia 7
|S
° jn)
WS30x4q2 | 207oa—
W4bex 12
12400
Walex fq
Webex]22
2 O
+02 (b4p4)
- 2. CoN m |
-
290EN
=
Re,
O
F 320bNm
160 fa
Shape
= = 5230kN
Ce
Mea = 0
Gate?
Zbb-7 EN,
(k)O4) (--24b1) = - RON.
G)leb) (2667) = Selva.
Bending
IM|
=
EAL
Bre E
FRC
Mop
LN Jn .
A te B (4) (fh2)/6333) = 320M
Aveas*
Maximum
bf 2444)
~ $2338
=
Vy
= EE.
= H Zhe]
266-7 — of
st
O44 4 EH Ig
rece
+ S32 2BbN
S333 Oh
Vit =
THOTT
ee
Vaz 0
Vet =
cy
Veachen
|22060
[2010
6ia4
a
Lightest
2208 F fe 390d
hn?
21929103 brie
W- shaped
W520 x42
section
|
=
Jaiku
1 2ove
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without perntission.
Problem
5.85 Determine the largest permissible distributed loadw for the beam shown,
knowing that the allowable normal stress is +80 MPa in tension and ~130 MPa in
5.85
compression,
<a
60 mm
}
Reactions.
By
tTZR2O0+>
symmetry,
Bee
@r
Shear:
Cc
>
=O.
O.4S us
,
Ur o
Vez
O-
O.2w
FT -ORw
Va 7 -O.2w+ 0.45 w
= 0,25w
Ver F
5H
0.25
Veis
w
-O.5
w
OL.2Tw
~ O,2Sw + 0.4$5w = O.2w
zr
O.A2w
- O.2w
—Aveas: Ato By
Bt
- O%w
BzC
EF,
>
©
+(0.2\-o2w) = -0.02w
¥(0.25)(0.25w)= 0.03125 w
Bending
moments 2
My = ©
Ma = O- O.O2Rw
=
~O.02w
Me = -0.02w
Ceutrorad
be — GOnm——>
:
i
O
32 ram
®
+
TO mm
ZO
lo
Dum
®
mm
juevtia,
\2c0|
To
B+
aQ
4BZO0
30
36
20
Kee
2
y
{2460
‘
ob
,
120
460
|}
.
WO
B60
:
4
OOD
oO. “Siee7
fROKlo"
ys
+_ 50 mm
L=
1360x108? me
FAd*4+ ET =
if
LAy = (1360 *10) /30 = 4S.333%107 wm = HS.333%10° me
Bottom? TA = (1360
#10 )H-50) = ~ 27.210
Bending
moment
Tension
at
Dinits
Band
CMe=-
Ct
~ 6,02 we
Compress tan at Band C
Compression at Ft
smallest
-
)
anol
ford
bicnits
WwW,
we 181.8% LOSN/m
—0,02~ = — (130%
10° )(27.2%10°)
wie 176.8." 10? N/m
wt = (Box so8)(27,.2% 107%)
0,015 we ~(-130x10)(45.333%10"*)
allowable
|
ol’
mm 2 ~-27.2K107o m
(80/0 \(45,33210°)
BOS
Tension at E?
The
ot
@\1200]
al Zo mme
Top?
mone ut
Pavit] Aint] Sumo AS (omit ll cl, vam| Acd (iO mm 1 (10° wae)
i
4
and
+0.03I25w= O.ONLSw
doad coutvols.
We
wee 193.41
Um
w 2523.8x/0°
N/m
176.810" N/m
W= 176.8 kN/ma
Problem
5.86
.
5.86 Solve Prob. 5.85, assuming that the cross section of the beam is reversed, with
the flange of the beam resting on the supports at B and C.
5.85 Determine the largest permissible distributed load w for the beam shown,
knowing that the allowable normal stress is +80 MPa in tension and -130 MPa in
compression.
EME
.
.
+f ZF
60 mm
|
hI
By
=O:
Symmetyy,
.
BtC
O,2ui
8 [™
<P
Aveast
E
Ate
Bw
Bending
-0,O0w
~O,0Lw
ToT
5Omm
of}
SO tam
a
Part
10 mu
= -O.02w
A mr
|i200}
5S
|2400
Ma
GLOZw
7
O
=
~0.02W
= 0.01 2S w
of inertia.
ds mm
A di (10% mm")
20
Go
to
-O.02w
{2
20
72
360
HBO
“4D
G60
—4oo
.
ZAd*s
ST
=
1360K/O° mn
3
Bending
Diwit
5
2 - 45.333 % 10m
Tension et
Baud
(M
C2
Compression
The
smallest
Oe
-614)
- 0,02w
Compression af Rand Ct
Tension at E:
=
aud
Pood
We
~0.02w= ~ (-430x10* )(45.333%/0°)
O.O1AS wT
eMow able
s
= ~ (80 ¥ 10% )(27.2%16°)
0.0 12s w* e(30%J0*)(~45,333 xfo7*)
et E+
toad
~(-130 x1
contyols.
mm)
480
x10) A 030) = ~45.333 «10% mm?
Bottom: I/y= (136
Prats
Tb
30mm
H*1O mm
TA 2CS60K/07)/(s0. = 27.2 x07 mm = BAZ
moment
t
2O
+ 0.03185w
moment
a=
I=
Top?
6, How
io?
Yo
|
-~O.2w
O,2u
-
Y,rom Ay, (Pwr)
-
Gonn —af
=
=
#(0.2\G0.2w)
2
Mg
®
| 4
[L—
O~
}i200 | 50}
D
Bomm
J *
moments
=
ane
,
C
Cc.
- 2
= ©
£(0.28)(0.25
Ww) = 0.03125w
Me
Centvorcl
f
©
TH.
EL
~
I
A
O,2w
.
=
~O.2w + O45 w = 0.25w
O.2Sw "O.5w
7 -O,95w
“ORE
W*+ OWS w = O.2w
Wf
Hoyo
oy
og
O-
B
-O%4w
4
oO
~
Reactions
lt
To mn
tt
GOram
\(27. 2*
lo7*)
w=lO8
wt
08. Bxlo* Nea
w= 294.7 x10 N/m
W 322.4%(0° N/m
wW
2314. 3K(0°
N/m
Rx10> N/m
W=108,3kN/ny
«lt
6.87
Problem 5.87
Determine the allowable value of P for the loading shown, know-
ing that the aHowable normal stress is +55 MPa in tension and ~ 125 MPa
in compression,
BzD
_
Reactions.
Shear
[25.mm
te Co
C*
D
we oh 25 min
175 mr ~
Areas,
Bend
ina
a
Cr -
Tep , Ten sion
Top,
Bot.
Comp,
Smetlest
to
—jixto’> =
Teasion
Bot. Cow p
$8 Me
Cryio¥=
~f 28 Ko”>
yohve
of P
V2 O.5P-P
on
(026-Pi=
-a-25P
GFVoesP)
-
Cw
D
Che X%osP)
Dr
EB
f2e)(P)
O.5P
—
=
Vr-o.SP4l5P
BkC
=-0.5P
P
_
—
vEP
= -orSP
= ozsP
moments.
Me
-
O-
Me
= -02FP
4 erieP
oS P - 675P
* ~6:25P
Lavaest
positive
Lawqest
negotive
and
=
© -de P
+ 25°
beuding
oO
oyby
moment
bending
FP
=
Moment
osP
2
me dc 25 P
moment of inertia,
Yo, PAM
AY.,
Yeo mm
td, 2 mm
Aa? mai’
LL. 3 mut
O | 225
BIS
PTBABT
4207S
598% x10"
4: OL4 Kio
{2° 5
547
3125
G
FEO0
Zhe
2
= ABTS
+
21
Cas P ore b 2-5)
144-4 xi¢0 ©
S
oN
P
70” may,
be
~~
_
P4ed LAN
P+ 246LN
l4eS4y x16 ©
CORPY-9eG 7)
Pe
14SYL KIO
-Bhoonkde
flix 554K
Pe 20ef EA.
leesUgxiae
is the
min,
a
Oe 228 Kt”
1eIG4MO 1 43x06
BO8 124
32
Py 24.
“Seoe
¥
os
PactiA, mm
i=
MI.
eS P=
Ate Bo
Me
y= 106025 hn
Top?
Bottom t y= 4.3575 mm
Ve- PO
Vr~ P+
kD
bE
Mp >
Centvotd
L.5 PT
At 8
diaqnaw.
B’
—=
uve
fbb
LN
Pe
Sef
Af
wt
§.88
Problem 5.88
B25
;
AR
Solve Prob. 5.87, assuming that the T-shaped beam is inverted.
§87 Determine the allowable value of P for the loading shown, knowing that the aHowable normal stress is +55 MPa in tension and 125 MPa
.
:
in compression.
min
-
Reactions
Poy
1.5 mm
Lor
Bt
|
se
I
9.5?
Ly.
Lavgest
Lawaes
.
Tm
Me
~OvL.5sP>
@
VE O.SP-P = -0.5P
P
©
=
M,
|
¥.
Centrored
P= o1sP
os
me 7
x
|
o.sP
~625P
Me =O ar Pe
zt .605P4 078P = aosP
ehFO
fon eel?
2?
T-O.SP4L5SP=
Bend ina, mom eats.
M
Vr+
os
~O.5P
J P
ft
Gas}(-P)= -025P
US VesPi > os P
UWSGOSP) = -e 76P
= 025P
(e2as\(PY
Ate B
Be o
Ce D
b+ EB
Areas.
p
P
Vr~-P4hns P=
&
+
=hLS
A te 3
ct
Cob DT
D*
Vv
BrD
diaanm.
Shear
175 mm 22
|
.
Q
=
OveP
+
P
= - os
osP
bending moment
positive
bending
neqadive
ana
moment
moneet
of
&= ww fle zeP
inertia.
st
Bavt |, tora | Yoo mm|AFe, me [dy mm [Ad? mot E tom’:
Topi
:
= #3078 iam
.
Bo tom
IPE
+
y
| BPS | 625 | tgsse2
lasts | ys7-s|
boi 662 | 3128 |4273 x10
=
|7&ee
74 E87 F.
Ye es
=
T9GEF
carxvot
Top , Comp.
fA EXO
~ eee ees) )
Bot. Comp
-i2eqiee= ~ oS
of P
Pe Teh EN
~~ fereeee)
tyxre® = ~ Co
yolve
é
is the
14e4yKi0° mot
FAs+ SI =
Bot. Tension
Smethest
Oe 228-x10%
Hen Kto%,
fob
.
T=
oe - &
Top Tension
102841
oe
®
ES MAI
te Sb
| 43678 (5-48) x10 Avob4xio"|
San
Q
Pe
PNGaxtopes 1
Bowahte
$3)
EN.
Pp e186] LA
Pr 68-4 LN
value
Pe
{fol kN
ttt
5.89 Beams AB, BC, and CD have the cross section shown and are pin-connected at
Problem 5.89
,
Band. Knowing that the allowable normal stress is +1 10 MPa in tension and —150
~
MPain compression, determine (a) the largest permissible value of w if beam BC is
not to be overstressed, (b) the corresponding maximum distance a for which the
' cantilever beams AB and CD are not overstressed.
12.5 mn
te
i 200 mm->|
|
{ay
150 pun
|g ole
7 a ye
ms a
Me
Oo
cn || fe
Vez Ve
==
Centroid
oe
anel
x
moment
—.
~@
41878 | 7s
+ 2
Lass.
v= 531250
FAd*
Ie
6G43w=
.
so
186.25
| |
~—¥
|iwoces | 4c. 43 | 4,042 «jo ] 3.51
x 10°
531250
FINS mim
+ ZI
=
Bending
moment
allso (\o"* wi?)
dims:
M-
-6I/y
lo) (- 97.4710) = 9, 622-/0" Neve.
= (HOw
= ( 156 mio“) (AS8.6/0°*) = = 38.84/0 N-m
Tens
act at
Et BE:
Compion
A4D:
-(Hoxjo*)
(258.6% 1o* da
G.48w = %622*10°
~29. 45% jo"
Ne
x10)= AZ. 121% Jo* Nem
Wee 1 48S «1° NZ
w=
Ate
10.621 xio% wm"
Comp. at ASD: - Cisox(o (87.47
co
©) AMowable doad w.
Avea
47,073 «101 3.548 ee
top
| 4L07]
258.6
bottoml- 121.43 | ~ 87.47
Tensiouat
.
L.
at A.
6.48w
of inertia
Location | ¥ Gum) | D/y llotmm)
[200 ———
w
W.OT
Shear
3.6w
1156.25] 340625 | 34.82 13.03) wi0°]0.0326 *1o°
Ve Saag
"
2.2) w =
Ragett | Aton | ¥ Gomd| AF Crom | eb Gund | Al Grit] E (mmm)
—® | aso0
17.5 al
©
of shean diagvam,
Bto E
Mg = O+
&
od
=
)G.6(3.6w) = 6.48 w
a
ar)
M,.
Aveo,
12.5 9m
PD
<«
L485
KN/m
<a
Vz, = (Q23.6)w
oF shear
Bending moment
diagram:
of A Gadso D):
Za(V+Vs) > galar72)w
M2
-¢ala+22a)w
~talat2Z2)I4gs
io?) = - 13.1z1«)o*
(b) Distance ©,
KA 43.60
- 8.838720
a=
1.935_m
~—
Problem
5.90
5.90 Beams AB, BC, and CD have the cross section shown and are pin-connected at
Band C. Knowing that the allowable normal stress is +] 10 MPa in tension and —150
MPa in compression, determine (a) the largest permissible value of P if beam BC is
not to be overstressed, (8) the corresponding maximum
cantilever beams AB and CD are not overstressed.
distance a for which the
12.5 on
mm—|
[eee
Me
=
Ve
=
150 mm
le
7
de
24m
Vv
24m
at >|
|
2.4
Aven
om] be
12.5 mm
A
Me
8
#&
2
ee,
F
M
“P
©
[2500
@
\.
=
3.
Ys
XI
#200
oy
531250 |
AMowable
foacl
P,
24 Ps
jo.c2pxjo® m4
Clow)
< alse Clow!)
Piwits
Mere co
,
|
9,c24x10" Nem
~GisoxioX2sa.exto*)= 38.8%1o* Nem
Tension af At D:
at{AaD:
mm
- (ioxio*)(~ 97.4716"):
Tension of EAP:
Comp.at EME!
Comp.
|Z078x/o*|3.S4B>fo°
— 37.47
moment
Bending
4
@
= Me
258.6
fF 141.493]
156.28
IbO
se
T/y
41.07
boHom
#F
7S
5
= 24P
[9043 [4042 10°43. SIC «soe
S124
+ BIT
.
top
ar
wy
(21.43
|
ckiagvam,
84.82 |3,081 x/0° |0.0326>10%
jidoezs
§31250
SRS
£Ad*
>P
an
al.
shear
of mertia.
370625
Jas
Location | Ydew)|
—
-
of
-P
anel moment
14375 |
Ie
I
~
= Or 249P
(sere
jia7s
id
€
e
/
-
| Act? Gret)] TE Come)
¥ Cam AG ean?) | ch Crm)
Part | A leant)!
,
24P
/
A__8
Ceutvoid
=O
Vv.
BhE
2.4
Pp
M.
-
~ (Haxto (258.6 «10 = ~ 28.418
-Clsoxio* )- 87.47 xf"
9.624 10%
P=
= 13,12, «fo
4.orxio
N
P=4otk\
Shear
af
Avena
AtoB
Bending
A
Yr
of
moment
(b) Distance a,
-j[§|]}§}
P
sheer
af
Nem
em
diaqvam
A,
3
My.
avs
=-aP
=
aP
=-4oirlo
-40lw lo a= ~ 13.1ab«107
a
= B.27m
=a
Problem
5.91
5.91 A 240-KN load is to be supported at the center of the 5-m span shown. Knowing
that the allowable normal stress for the steel used is 165 MPa, determine (a) the
smallest allowable length / of beam CD if the W310 x 74 beam AB is not to be
240 kN
overstressed, (b) the most economical
W
shape that can be used for beam CD.
Neglect the weight of both beams.
Fou
rodlecl stee!
section
S = 06a xlo' mm
Guy 7 IES MPa =
Allowable
bending
W 31O%74
of
©
[060 x10
[os R10. Pa.
moment
beam
AB
mm?
,
Be
Mas = Sig = (lo6oxo°\iesxio® ) = 179.9
x Jo" Nem
= ITH. KNem
Aye
oT]
@) Beam ABL
Aven
a ae ee
,
Ato C
Bencling
Yoo
A
o
Geometvy:
6
ee
(4)
E
Beam
k
7
20 | yorrs | Lows
Ji20
Berding
Me
aS.
|
E
diagram
=
[20
@
oD
of
[25.1 kom
W460 % S2
WHlo
W360
* 46.)
*528
(lo
&
De 52a. = 2085Sm —e
E)
shear
at
cdlaaqvaw
©
125.)
.
kNem
E .
= 125.1%JO® Nem
Sain? Gehr 7 BS LntSe
S
RO
= 1457S m
(10425 C20)
|
C
midpoint
E
moment
‘Shape
c
C
Cte
=
Re kv
M (kim)
at
Rat Pes
CD
Area
Vi un
moment
QO ae ITH4
C
€
oF sheav
it
A -si
= 758.2*lo*
768.2 102 mw
vam
wm
G42
THe
B49
W 3lo x 60
851
W250x
67
Zot
Wloor
B6
853
—
Ans wev?
W4lo«4é.]
«—_
.
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
5.92
Problem 5.92
Beam ABC is bolted to beams DBE and FCG. Knowing thal the
allowable formal stress is 165 MPa, select the most economical wide-flange
shape ihat can be used (4) for beam ABC, (b) for bearh DBE, (©) for beari
PCG.
C= OSkN.
to)
Beaw
6S tn
ABC
8
t
+
be
2c
A
mabe —2.40—al
Be p30 pa
M, Ei.
eh
kA
|
Mg = —(b5 (za) = 186 bm,
ee
2
_-
ig
= [6S tx
Shape
&
WESOK EL
WHbOX7He |
ober bal
= (9x ENm.
IMI
|
Sain =
(C)
MY ere
aa
Beam
,
[4-£xr0*
- fake,
L$
3
,
¥to”
rie
:
{240
‘fe
y
S_
Bi
a
is40 < (b) W53ex 66. <<
W AEG RTE
W4ro x88
1% be
iSte
\Al 3b x74
128o
Iso
1245.
Shape | $,aee mm)
3
y
WHEOx 52)
Wr FIORE
;
MEN
692
W3iakyyesS|
634
Ww 200K 7/
A 4IES
GIL EAI.
aeae
Senin = Mls
Ca
_ 27€00
GS Keak
3.
= £qo1
je am
9F4F2
|
8
GBT 4 CC) VGC XESS
wstoxay |
W25exF@
1M Vaso ?
%
3
(ko mu)
Ws 30x bb
W25oxsei]
3 ne
.
(A) Whjox bo =a
—
W2SeXie}
W3Exe]]
FCG
To PLN
7340
Me bo
[860
fuze
fobeo
|
o~
) (Ke ian)
S
W 410k be
WECK EF,
W3IeXTe]
Shape
ne
.
iL Rie?
j
,
.
kiln
86
a
DBE
Bean
(b)
693.
(a7
att
5.93 A uniformly distributed load of 66 KN/m is to be supported over the 6-m span
shown. Knowing that the allowable normal stress for the steel used is 140 MPa,
Problem 5.93
66 kNAn
determine (a) the smallest allowable length / of beam CD if the W460 x 74 beam AB
is not to be overstressed, (b) the most economical W shape that can be used for beam
66 kN/n
CD. Neglect the weight of both beams.
W460 X 74
Ae
5S = JY60X1O%mm= HGOMIO’*
Cut = 40 MPa=
w
140% 10% Pe,
Mug = S6iy= (460 x10 )GYO x108)
V Ck)
= 204.4 x}O? Nem
Beam
YW
A
Reactions?
B
JN
¢
ead
M.CkW-m)
204
By
=
=O:
+2
Sheav
and
At C,
.
a
—p!
[10,S¢ kv
berding
=O
Oo
in beaw
AB,
19-66%
x=0
By
kN
bem
22g
—~17Bat2o4.4
Mu
GE
774
643
W3lox52
W2S50x53 |
W 200 * 7!
743
693
Joa
shear
beam
CD,
=
S (10% mm)
W 360% 44
+
anel
Vez
427
6-R2as
32.35
C7 D=
110.56
kN
bendl ins
O
<
Y2.502
=o
1.32437m
ot
point
moment
BE’
Area Peon Ato FE. . (Vde=
»D
W4lox 46.1)
198 a-33a>= 204.4
geomelry
Draw
Mice
KN +n
19Ba~- 33a"
From GC)
96, 602
ME
M=
O= 1G
|
66/
moment
Vz
Set M=
“A
Shape
O
Mt 19Bx-33%°
LTT
M
CFD
=(33 2) kv
Ge aR / ws
HOSS kM
»
198 kN = I19B8*/O7.N
CaDdD-
O<x<a,
3, 35m
A=6B
A+tB-Giee)=
= ot
A=8
+E Sy
4
Symmetyy
C= D
ft
= 204.4 Kew
AB
kN-m,
diagrams
the mial point
w
~a
Fon
of cD,
£Mlo.s¢o\(
42)
.
Me =
F2,6ORKMw = 92.602 "1O” Nem
_ Me _ 92,602 10"
Sin = 6 = Ygou[oe = 661.44x10°° wm
=
CEI.
44 XIo nam
Use
W3G0xr 44
<a
|
Problem 5.94
*5,94 A roof structure consists of plywood and roofing material supported by several
,
timber beams of length Z = 16 m. The dead load carried by each beam, including the
estimated weight of the beam, can be represented by a uniformly distributed load wp
= 350 N/m. The live load consists of a snow load, represented by a uniformly
distributed load w, = 600 N/m, and a 6-KN concentrated load P applied at the
midpoint C of each beam. Knowing that the ultimate strength for the timber used is
Wp + wy,
| | | | | | |
ou= 50 MPa and that the width of the beamis 5 = 75 mm, determine the minimum
allowable depth 4 of the beams, using LRFD with the load factors yp = 1.2, y, = 1.6
and the resistance factor ¢ = 0.9.
.
L=
16m,
Wes
a;
kNim,
2.8 kw
(Diie)(0.as)-
(2.8)
= H.R kom
vot
2
on
at
HL Qxfo"
=
ALM
Nem
Atel
J
a
‘
-
2.8K
/
B ” —
of
sheav
| okv
didl
Bencling
|
Sm
Bom
——s
7.8UN
aw
|
For
h=
~2akY
43.2 ww.
|
.
a4 rectangutan
GS
{ory
.
48.2
»
kN-m
43.2
YM, =
= 8347 x1O m=
3
m~
+
S$: To Mo + TeMe _
9 Ox ,
VALTBkY
(43.2)
Yo Mp
x
Desig
A
Cl
af
moment
Magram
=
$)(8)(7.843)
piside
728 ke
=
R= tlacyoedsc]
Live Poad:
G
Avea
M
kw
B
A
TB kV
€
,
mom
Bending
{1.2
0.6kW/m
Pz
shear Magnan e
2.8 kV
Vv Pewee
le
0.35 kv f/m
of
Cl
(A)
29 kv
=
OG
>
Ado
Avea
Ey
SSO NM
=
nn
CITA
29 kN
N/m
Dead Poad:
00 EF KN]
Wenn ——
Wyr
600
kNem = 43.2%] New
PM,
DP 6.9 S
=
(2 U2 w1o? 4 (1.0)(43.2
#10")
:
(0.9)(.50 * 10%)
1.8847 x/D% mm”
section ,
S=
tbh
/ (6.8347 «10%
sas
he
BES
mm <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem
5.95
*3.95 Solve Prob. 5.94, assuming that the 6-kN concentrated load P applied to each
beam is replaced by 3-kN concentrated loads P; and P; applied at a distance of 4 m
from each end of the beams.
*§.94 A roof structure consists of plywood and roofing material supported by several
timber beams of length L = 16 m. The dead Joad carried by each beam, including the
estimated weight of the beam, can be represented by a uniformly distributed load wp
= 330 N/m. The live load consists of a snow load, represented by a uniformly
distributed load w, = 600 N/m, and a 6-KN concentrated load P applied at the
midpoint C of each beam. Knowing that the ultimate strength for the timber used is
oy = 50 MPa and that the width of the beam is 6 = 75 mm, determine the minimum
b
.
allowable depth 4 of the beams, using LRFD with the load factors yp = 1.2, y= 1.6
and the resistance factor ¢ = 0.9.
0.35 kN fin
L=
Liytititily
Lae
Cc
A to
Avea,
Live foad:
of
D
73
-(4)(0.6) = S.4 kN
Shear
of
DP
SHY
-
ASF
Aven Dto@
3
=
AY
KM
SKN y79k
= (4)(4)(7.245.9) = 26.4 kNem
’
(4)(4)(2.4)= 4.8 Nem
Takw
Vi os uw
> 31.2xj07?
rato,
Mp =
YoMo+%
Desiqn:
5.4 ke
ag
c= foMo+Ye Me.
Ge
=
For
a rectanqudar
6s
n-4&
_
-{
|
Nem
OM= PSuS
(1. 2)(U. 210%) (1.6 ¥81.2% 10%)
(6.4 (50 x0*)
14OSx low
_
26.44+4.8 = BLQk Q2kN-m
moment tat at ¢ C =
Bend
ending
3
3
ILQxJo? N-m
GC: 1L2kNem=
8
COT
M
M2 knew
Shear
Avena AtoD
A
=
Ry= d]Ucd(0.6)4 343] = 728 kN
0.6 UN /m
on
“tn 3kv
28 kN
Avaqnam
shear
of
moment af
Bending
c
P= 3KN
.
(E\(8 28)
A
Wp=3SO0N/m= 0.38 kiv/m
& = G)U6)\lo45)=
poad:
Dead
28 kN
2.2 KN
VY
az4m,
W,= 600 N/m = O.6kN/m
sl
m
2.3 kv
lem,
=
h4OR*IO% tum?
se ction
S = tbh*
/@)U.4e8xto*)
1S
h=B2Gmm
—ald
Problem 5.96
*5.96 A bridge of length Z = 15 mis to be built on a secondary road
whiose access to trucks is limited to two-axle vehicles of medium weight. It
will consist of a conerete slab and of simply supported steel beams with an
ultimate strength oy ~ 420 MPa, The combined weight of the slab and beams
can be approximated by a uniformly distributed load a = 11 kN/m on each
beam. For the purpose of the design, itis assumed that a ftick with axles
located at a distance a <= 4 7 from each other will be driven across the
bridge and that the resulting concentrated loads. P, and P, exerted on each
beam could be as large as 9S KN and 25 KN, respectively. Determine the
Most economical wide-flange shape for the beams, using LRED with the
ifm
Bor
96 4N
P, =
as
We
when that load is located to the left of the center of the beam at a distance
equal to aP{P, + Ps.)
gm -
Ry = Ra = (ZS) CI) = B28 kas
Dead Poad:
BA
MEM Jan
Avea
AtoE
of sheav diag tom
(4NCt-5) (82-5) = 309-4
tt kM
fi -
TVA
TTy
oe
Mm
RL
Eo
Xt.
Tt
=
Peg
5
+ 4
15 ~ 1-083
Ry, 2 S6°7
= 7.093
m
1h O83 tr.
3°92
Dk
kay.
Vr F6-7 Lv
Ve $67 ~-95 + 283
Akw¢
Ct D
Sheaw =
= am
B
G3+3
(1093060607) = Hore
Ate C
Bewling moment:
Design:
aD
-15 Ry+ (7-90)(98) + (F922) = O
EMg tO
Aven
OCT
71088
L-X-A
Faq| Baim
E
ot point
kite
309-4
Monee 7
uM
Qe
fy
Lt
M+
M.=
KM
4ohG
Em -
= PMY
= PO
Lun
Suen
a
2.
Se
< Yo Mo
nD
Em
180.475
Db
Saat
:
~ (1.25
(309-4475
(0.4 420x108)
2kP2.G& KIO
F 2ER2-4 K10° m=>
~6303B
4e0h&
+ Vi M.
8
Shape
S
M4 0L6K00 J
z mm 3,
ra Mm)
W &Foxire}
2870
W GfOXikN|
4270
W 538x160]
3720
3340
WAboxiSh]
WHbOK U6 | 2800
ah
WEFoK/2 5”
att
Problem
5.97
*5.97
Assuming that the front and rear axle loads remain in the same
ratio as for the track of Prob. 5,96, determine how much heavier a truck could
safely cross the bridge designed § in that problem.
*5.96 A bridge of Jength L ~ 15 mis to be built on a secondary road
whose access to trucks is limited to two-axle vehicles of medium weight. It
will consist. of 4 concrete slab and of simply stpported steel bearns with an
ultimate strength oy « 420 MPa. The combined weight of the slab and beams
can be approximated by a uniformly distributed load w =~ 11 KN/ém on each
beam. For the. purpose of the design, it is assumed that a truck with axles
located at a distance a * 4 m from each other will be driven across the
L=
19m
a=
Pp.
bridge and that the resulting
4m
zt
AILEN
Pi
load factors yp ~ 1.25, y, ~ 1.75 and the resistance factor @ = 0.9. [Hint
254A)
It can be shown that the maximum value of 1A4;! occurs under the larger load
LIEN / 10
See
Sobvtion
My =
For
AMowable
concentrated loads P, and P; exerted on-each
beam could be as large as 95 KN and 25 KN, respectively. Determine the
most economical wide- flange shape for the beams, using LRED with the
when thatPAP 1S PL
equal to aP3(P; +
Pa).
to
‘Pro
309-4
yvolled
Live
blem
5.44
Lm
steed
foad
for
Mit
sectiurn
moment
éadcu
to the left of the center of the beam at a distance
tation
o}
the
4olkG kaAlm
S = 310
W6Iexies
FoMowine
*
,
hm
Kio
ue
Yo Mo + VME = PM = PCS
|
Mit P&aS- YoMe _ (0.9 )eHront)(2s:0%/8)~ (1.25 \(30I4 xe)
Yt.
~
.
= E3702
Ratio
e
Tnerease
ths
*
Bee
:
an
73228
2.
140-338
y
7S
NM,
.
33-8°7
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
anid educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~t,
Problem
5.98
.
5.98 through 5.160 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown. (6) Use the equation
obtained for A/ to determine the bending moment at point C and check your answer
by drawing the free-body diagram of the entire beam.
(le
WwW
(p)
=
We
~WoX%
M-
~tw,¢
pois?
C
x=
8
Hw
wpa]
a.
2°
ey
Me
-~
M.
.
.
&
—_
_.
= 0
$w,a*
(B)
__
Me =
WoKkx- a7
+
ja.
point
functions, write the equations defining the
and loading shown. (6) Use the equation
moment at point C and check your answer
entire beam.
~ WOK ap = We dx-ap = - 4
- . WX
Vt*
At
3
+ He <x-ay
Cc
x
was
Ga.
*
>
Wi
'
4 ae
2
<™
.
ast
=
ap
dv
Te
3
WA
hh"
+ Ean
aa
war,
2°
wae
ba
Me
et
(2)(dma) + Moz 0
We
Me
c
B
g
- gma
5
:
‘we
M . = ~ ex
Check: 4DZM.= O
zt
a
wee
(a)
4u
2a
§.98 through 5.100 (@) Using singularity
shear and bending moment for the beam
obtained for M to determine the bending
by drawing the free-body diagram of the
5 99
Problem
=
G#)\ma)+
soe
.
.
-~7a
Wadk- apo
2
=)
=
rtwidx-a>
SO
y
A
+
Check: DEM. : ot
T
dx-apz”
Met ~ dw, (2a) +e
(by
A
Wy,
Ve
At
W.
~
1
=
7 Ewa
o 5.0"
woo
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
:
5.100
—s-:
5.98 through 5.100 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown. (6) Use the equation
obtained for M to determine the bending moment at point C and check your answer
. by drawing the free-body diagram of the entire beam.
wet
We
_
Ve
~
Ae
We
Ae kx ~~ay!
WX”
=Wwx
w,
+ 2a
~ $e
Moe - wx",
~“pR 4 wx®
Set
At
Me
by
Cheek :
tDZMe>
Of
potat
Cc
pol
~oQa),
=
=
du
_
Lx-ay!
@M
=
aN
_ pee Xx- ay
x2 2a
Meo
Mea), wad
+ we(za)?_
wo"
M.-F Bay
|
at
=e
(Salgwa) + Me =
M. = -2 wc”
c
:
A
%
.
oo
2
j
D
WoO.
20
Problem
5.101
-
5.101 through 5.703 (a) Using singularity functions, write the equations defining
the shear and bending moment for the beam and loading shown. (5) Use the equation
obtained for (4 to determine the bending moment at point EF and check your answer
by drawing the free-body diagram of the portion of the beam to the right of £.
D
FD EM,
= 0:
-da+
Sapte
@)
Ve
Pkx-ay
Lae P-
Péx-ay
M = 125 Px~
(b)
At
M,=
tPZB
Meer —
= Or
49D SM,
ka
peint
E
1.25 P@a)-P Ga)
—~Me
P
- PKx- 247"
=
- P<x-20>
<<
-~Pla)=
D=
+ O.750Pa
:
0,750 P
O
KH=2 Ba
A~P-P+D=0.
=O
=
1,25
-
A
il,
Reaction
2aP
O.150Pa
o.7s0Pf
= O
M,_=OQ.780
©
FP
Pa
<A
Problem
5.102
:
- .
_
tty
» .- §,101 through 5.103 (a) Using singularity functions, write the equations defining
. the shear and bending moment for the beam and loading shown. (5) Use the equation
obtained for M to determine the bending moment at point £ and check your answer
by drawing the free-body diagram of the portion of the beam to the right of E.
DZM.=0
~4aA4 (Bai) (Qaco, ? —
A: 3wa
WE We = We lantey =A
M
Ve
? a] c
(). Vv
aa
.
nae
“ME
Wode-2ay
'
4 3Fw,
yo
~M
ZW oO
(Weap Xt
2
=-dwxtiw, dx- 2a
—
pied EF
DIM
DZ
=
5.103
hwon
Hac -(a2aw) =O
©3
—
Cztuwa
-Me + (a\bue ) sO!
Me
Problem
+0
ZwaGa)
Me=
= 02%
B-
xe 8a
.
Mg
=
4 Bwiax
by Mes -dw, (3a + dual +
.
aM
-
=
+
Wi;
an
ea,
. $101 through 5.103 (a) Using singularity functions, write the equations defining
the shear and bending moment for the beam and loading shown. (5) Use the equation
obtained for 44 to determine the bending moment at point Z and check your answer
‘by drawing the free-body diagram of the portion of the beam to the right of £.
DEM =O -mA-(F(3aw,)=O Az-Fmo
ni
mC
—
AWDEM,=O
.
tf
Me
_
t
200 4 (S8VGaw,)\=0
oO
woe
/
o
.
Ws dx- as
=
C= Ema
dv
~
@) Viz -wdx-07 ~ gwas Hwa
dx ays gt
= ~dw,(x-a> - Funax +48 Wacx-2a7'+ Oo
AL
point
E
x=
3a
(oy Me = -$w, (2a)*- Gwalsal+ fwa tad
.
.
.
4
.
Check:
+ 2M,
= Ot
-M,
Me
+ a(ma)=
-4
Ww, Oo”
O
a
8
Mg=-3 Wp 0.
ae
Problem
i.
5.104
—
3.104 (a) Using singularity functions, write the equations for the shear and bending
- moment for beam ABC under the loading shown. (6) Use the equation obtained for A¢
to determine the bending moment just to the right of point D.
wa Vz -P - Pie Be
=
a
Ms ~Px + 8b - péw-Shy - Ebina”
:
|
€
i.
a
.
DD |
TEA
LBS
P
Oe
au
Problem
. Just
do
the
waht
|
of
D
v=
4
F-Pe)- ~ & &
-PG)+ BE
Mis +. p(y,
+
Eh,
»!
3
EB
Pe
BAPL
De
A|
;
m= -Hek =
B
5.105
§.105 (a) Using singularity functions, write the equations for the shear and bending
moment for beam ABC under the loading shown. (8) Use the equation obtained for
to determine the bending moment just to the right of point 8.
(bo
dM.
Pdx-ay?
Me
- Pdéx-a>'
Mt
=
Pkx-ay?
Vv >
Just
SS
a
Lend
ey
} 0
@
to
the
wig ht
-O~-
Pa
|
~ Pa<x-a>”
ft
B
—t
xrat
M=-Pa—«
Proprietary Material, © 2009 The McGraw-Hilt Companies, Inc. All rights reserved. No part of this Manuat may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for theic individual course preparation. A student using this manual is using it without permission.
Problem
5.106.
5.106 through 5.109 (a) Using singularity functions, write the equations for the
shear and bending moment for the beam and loading shown. (4) Determine the
maximum value of the bending moment in the beam.
HDTMe= Os
-45R, + (Z.0X48)+ U.5)(60)-(0.9)
(60) = o
R,
oe 1
me
(a) V=
>
Yo
kN
fe 15 me
0.6m 0.9m
Ho ~ 4B dx-1.8>° ~ Co<x- 305° + GO<K-3.6>
kW
ad
M=
Yoy
kN- m
al
PL
A
xx Cm)
o
8
~
48
<x~- Ls>
~ 60K%~3.0>
M (kNv-m)
Oo
C
3.0
1s
Go\.s) =
D
E
3.6
4.5
(Ho (3.6) - (ajar) -(60Ko.6) = 7.2 kKNem
(4o)C4.5)~ 49 (3.0) -Ge(|.5) + 60)(0.9) =
M
(b)
40 kN
kKN-m
(42)(3,.07 =(48XLS) =
HE KN m
ell,
§.106 through 5.109 (a) Using singularity functions, write the equations for the
shear and bending moment for the beam and loading shown, (6) Determine the
maximum value of the bending moment in the beam.
5.107
,
80.KN
HOKN
+92 Mg = OF
A
13m
12m
0oV4-5)~ 3:66 +(B0XK2-4)+ GGT =
B
|_|
‘09m
O
kN.
60
ann
Problem
Go
+ 60<¥-3.2>°
12m _
C=
130 kN
@) V = -404 130 (x- 0d — BoKe~ 21Y
80K %~33>
Me=
-40%~%4
[30 kn~ aoF>
kN
a (m)
co)
MCEN
Oo
4
D
3
~(40Q-) +Ci0)(h2)
E
33
~G9G-3)
45
ls
m)
Cc
Bo
nat
_ 80K %-2:1>°
— 80¢ %-3.3>! knm
Porut
A
OC
~(40\04) = — 36
+(130)(2-4)
= 72
— (o\l-2)
= 84
—
~GadGs)4(120) 36) ~ BARA) - GAD = O
(bo)
Maximum
IMI ..2 84
kvm
<=
5.106 through 5.109
Problem
5.108
(a) Using singularity functions, write the equations for the
shear and bending moment for the beam and loading shown. (6) Determine the
maximum value of the bending moment in the beam.
1500 NAn
wes
(
1.5
kN
By stds,
C-D= 3kNt.
Ve
+ 3Xe-0.85+
= ht
M = =~ 0.752" 4+ Bln
Locate
point
st
were
V=O,
Ass ume
~(0.75(0.8)
+ O04
O
<< Be
he
>
<8
KN m
<8
HK,
2.0
m
uf
=
0.600
kN-wm
= - 0.480 kNem
IM Terme = 0.600 kom
Pr. oblem 5.109
KN
= —~ 0.480 kN-w
~(0.78 (a0) 4+(B)0.2) + O
Me
My = ~ (0.75K3.24 Bad) 40
(b)
+ BS%-B.2>)
he
~1S te + 3,-0.3) 40
OF
M.-
E
0.8)
3¢0-3.27
GOO Nem mes
5.106 through 5.109
(a) Using singularity functions, write the equa-
tions for.the sheat and bending moment for the beam arid loading shown.
(6) Determine thé maximam value of the bending moment in the beam.
.
4S kNAn
09m
From
S2kN
12m
Symmetry, A oy [C32) + 2(48) 60-9] = EL CEN
45 kNAn
12m
| 09m
WwW =
AS
— 4S Cp
199" +
VE £68484 Sk =o I~ 32K
ey
=
BSL ,-22°
DI
- 4
= AK P EN
M = £657= 22-90 4 are C098 — a2 dobead) — 2heKee=n. LNM
at
Ve 7 £60~ 45)0-9)= 16 EN
V5 2 £69 — (48) (ar) + (4eo(pad = 16k AF
mor
Chamyers
Sigh
il
HE
Va F £6L ~ HEE + EZ) — G2 TM LbEN
We = £65
— (9 GY t4yeg— Je = mE LAS
Vg = £65 — (FENG)
+ GE) G2) ~— 3% — 8) (0-9) ©
GQ)
IMAL
=
M, = 66 9G
Cee Vee
G2y- 0-0
$6"
=
EA
ObP20 Lm
=e
Problem 5.110
$.110 and 5.777 (a) Using singularity functions, write the equations for the shear
and bending moment for the beam and loading shown. (6) Determine the maximum
normal stress due to bending.
24 kN
24 kN
C |
|
24 kN
Jt
= Os
4DEmM,
}L_ we50 x 28.4
: |
4 © 0.75 m= Bn-rlen
DOM,
—
24 kN
-BR,
-
=o:
+(225)24)
~ (0.7 MAY) ~ O.59Q4) ~ (2.259904) + BR.
Re =
(ce) Vt
BO ~ 24€x- 0.789 — Wie-bL sy -
30%
amagow?
M=
M (kN mm)
0.78
(30)(0,7S)
[5
228
(Go Vfi.s) -(29\ (0.75) =
27 UNem
(30 0(2.259-(24)0.5) ~ Galois) =
3.0
(Ze )(3.0) ~ (29) (2.48)
~
378
Mentmum
Fou
whe-2esd
x Cm)
(30)(3.75
IM)
=
22.5
vobfed steel section
=
mm
Normaf
,
stress,
=O
.
.
4 cedx-3>°
KN
+ 66%x-3>'
C2915)
13.5
- C24) (0.78)
2.45) > (ACLS)
<a
KNem
met
kNem
=
~ 18 kNem
466 Mors)
= Oo
27x10? Nem
W250 % 22. 4
S=
308¥10° mm®
>
(b)
= 6
KN-w
(24 X3.0)- G4
27. kN-m
- GISKa4)
66 KN
- 24dx-0.78) - 24u€x= 1.5) - 24€x-2.257
=
~ Cr 5M24d)
~ (©.75X14) +lO.Is HAA}
3okW
R=
IML se Sh
27103
© rege
S * seriIo%
z
308xjo"*
$7.7?»1o° Pa
m3
= 87.7MPa =
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ali rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educaters permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
-
Problem
an
5OKN
§,116 and 5,211
5.1 11
125 kN
(a) Using singularity functions, write the equations for the shear
and bending moment for the beam and loading shown. (5) Determine the maximum
normal stress due to bending.
5O KN
4D FMy
=
07
Ge Aso) ~ 0.48 +(0. sas) (0.250)
B=
=
Ma
DS
0.2m
“
04m
0.3
125
kn
=O
ft
O:
(0.3 ¥So) ~O.4)025)+ 0.9 D -C 1.1 50) =
fa)V = - SO + 1A5<%-0.3>
c
Ms
125K%- 0.77
dm)
0.3
0.7
2
B
C
D
Masimow
S
Fr = 1S kN-m
~-0%0.2)40-040
=Sex.7HATO4)-o +0 = IS kNem
=10 kNem
= (Bo.a)s Gasyo.%)- Cas Kes) + OF
1M
|
-
is
kN-m
velled stke/
ISO» 18.0
Novmad
stress,
_
° &
0.2) =
+ (las) ~(128)(0.79 + loo
>
IS» jo°
=a
(Rox/o"
Sz
section ,
ah
>
ISx 10
20%
1oe
0
(checks)
Now
a
{)
KNem
+ 1OOKK=-1.25
MM (kN-m)
= oN)
bd
EB
<4
kN
100 L%- 12>
+
-SOm + 1256%-0.3) - [25K<%-0.77
Point
For
~
kN ft
[00
D=
mm =
120410"
‘
=
I2Sx{oO
Pa
= 125.0
MPa.
<a
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manuat may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.112
5.412 and.5,.113 (a) Using singularity functions, find the magnitude and location of
the maximum bending moment for the beam and loading shown. (5) Determine the
maximum normal stress due to bending.
40 KN/aax
IS kN - ny
F ssto
x s2
CoS
é
4M,
= 0%
IB - 3.6R,
L 1.2 m->}«.-——— 24m >
Vr
24.5
Point
R, =
- 40 <x-12>'
D,
Vz
oO.
(a)
- 18 + 29%.5x
Ma
st
M,
= -18
Me
= — 13 +(29.50(3.6)-@oMa4y
Manimum
Fer
Normal
=
= ROMK- 1.25"
1.9375
m
kN-m
kNem
+ Q9.5\0.9375)
IM] =
S 310«52
28,478
stress:
- (20)(0.737§ )* =
kis im
=
-27
at
volled steel section,
co
Cb)
kN
29.5 - 40 (x%,- 1.2) = 0
M=
19
245
- 27 = ©
kw
XxX,
~
4+ (2244)
OF
.
_
Imi
=
7
28.278 v/0!
Citxiove
xe
28.272
kKN-m
Nem
9375
S=
wm
ath,
G25 x 10% mm?
=
625
j07°
m*
456 2yjo*%
Pea
GO=4S52MP
<i
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed
in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
—
Problem
5.113
5.112 and 5.113 (a) Using singularity functions, find the magnitude and location of
the maximum bending moment for the beam and loading shown. (6) Determine the
maximum normal stress due to bending.
BOkN
HOkN
40 kN/u
aay
Me
=O:
:
~4.5 A +1245 4S
1
Lt
5
A=
wss0x6s
138
Br
462kN
Vr
~4o xm 4 138 - 6ole-1.8> -Codu-3.6>
M=
-20n' - 138%
~ Godx-/.8>'
=
~(40)().8)+138 -Co
=
Vie =
—~CHo)(3.6) +138 -€o
=
Locate
pomt
Ve
E
where
V
v -40%e +138 -~60
For
W
530
x66
(bd) Normad stress!
voltedt
GF
= a
6 kn
-66
O°.
40
kw
Tt dies
20
184x107 Nem
=
beturcen
|
at
section
steel
im)\
oo
4.5 B =
¢F
~ Gode- 3.6)
Me = ~(20
95) +(138)(1.95)
K.
-ColCnas-8)
(OD \MVeee= IBY kNew =
40) + (a4 (60) +(0.9 eo)=
kN *
tO ZM = Of
— = (B25 V4. SX 40)— (1.8)60) - (3.6604
0.9m
Ve
:
1B4
x Io”
Tau0n10*
C amd
her
= 184
©
S=
D.
1.95 m
kNew
<
1.4950 m
1340 «lO mae = 134010 - we
* 1223x108 Py
(2137.3
MPa «eo
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hilt for their individual course preparation. A student using this manual is using it without permission.
Problem 5.114
5,444 and §.415 A beam is being designed to be supported and loaded as shown. (a)
JOO KN
SON
Using singularity functions, find the magnitude and location of the maximum
bending moment in the beam. (6) Knowing that the allowable normal stress for the
steel to be used is 165 MPa, find the most economical wide-flange shape that can
be used.
BOKN
~(4-8\Go)- 26B 4 (2-4) Jo0)— (2) 50) =O
B= ~ 1667 KN
B= le67ENy
HDZM_=Or
‘L2m
24m
2m
ADT Mg = OF
=
—Walso)-U-2Vied) + 24D - GAO)
O
D = 6-67 kn
Check?
+8
2 Fy =
S0-1b67~
foot ebT-5o=
Vo = S0-—IbbR x=)
= look en 2AP 4
500 —1667C eb 2>— loo er?
M=
O
~ AY
661K
1661 e489"
At
AL
%=0,
M=o
At
At
Be.
CL
wsh2m ;
%e24m ,
M = (se)(t-2) = bo ENim
M = (SOGA = (6ENGZ
At
DL.
weer,
M = (50 Vag) ~ (66 TY3-6) ~Coe\(2-4)
At
FE.
t= bom
Mz (50) (6)- (be Nas)~CooBb) + GeNG2)
(et)
(bo) IMI,
_
6 =
n
Swan
IM yng
= - 60 ENM
120
= O
EN mM
Cehecks)
ot Cael
= 165 MPa
Sue
7 feo kam
(Ml
F foo kN&
_ [MI
Ee
Frem
_
=
be0000
= 60606Ki0
teenie
Appendix
Ly,
“G3
mm
Aightest
)
3.
= 60606 X10 Mm
wide
Flange.
3
she pets
W410x 38-3
at
or
Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced,
Proprietary
to teachers
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
without permission.
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
Problem
5.115
ne
oO
5.114 and 5.115 A beam is being designed to be supported and loaded
as shown. (a) Using singularity functions, find the magnitude and location of
the maximum bending moment in the beam. (b) Knowing that the allowable
-
normal stress for the steel to be used is 165 MPa, find the most economical
wide-flange shape that can be used.
DZMe =O 4S Ry + (45 MENRIS
+ (F0)(46
)) = O
.
Ra =
Ws
Vo=
11328
= 45 x
Location
O =
Mz
ot point
{73-26
4S kN] im
- Jode-~a9>”
D.wheve
~ 2205x*
Ab pond D
~
(xe BS m)
=
A
EW.
Veo,
Assume
= 48° %, - Fe
(73-2SK
(7225 BAL
O9e¢
Ky
«FS
Mp ez 1hSam
Gokx-o-9>
kNm.
2
Mz
(178.26 E685) (21-5 M185) “od lo 75)
= ISP EN&
(@
Maximum
ym
|
=
MgB
Simin =
Shape
S Oxo
WS30X66|
(BO
Ww Y60xX7¢)
[60
w 4lioxébol
feGo
W3G0X6G)
foZBe
ot
ie? |
©
iiteat
Cum
28
(b)
(S8 Ein
)
fem
Ke
6
FST
kta
LES m,
_
a
ttl
GST-6 XO
3
Prin
3
oe
Answer
Wiaxgeo
de
W 310K 7g) 60 bo
W2Sexlot)
(240
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual tay be displayed, reproduced, or
distributed in any form or by any mncans, without the prior written permission of the publisher, or used beyond the limited distribution to teachers |
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
§.116 and 5.117 A timber beam is being designed to be supported and loaded as
Problem 5.116
shown, (a) Using singularity functions, find the magnitude and location of the
a
maximum bending moment in the beam. (6) Knowing that the available stock consists
of beams with an allowable stress of 12 MPa and a rectangular ctoss section of 30mm width and depth / varying from 80 mm to 160 mm in 10-mm increments,
480 N/a
determine the most economical cross section that can be used.
|
HEM,
= OF
HR:
Rp
we
B48
~
7
Ws %
Vez
0.645
M=
“
0.645
~ 0.16 Xx.
0.645
- 0.16%,"
D,
kn
4
kv/m = - 2
~ 0.32<Kx- 1.5)
0.82%
=
0
V
kN
LSm
Assume
<
Xp < 4m,
+ O16 OG-1.5)"
40.16 &p ~ 0.48 Xp +
0.36
My (0.645 (2.09375 )- (6.05333
)(2. 04375) (0.05333)(0.59375)
=
Mp
Swin >
a
Ey
-
At
Next
*
O.B8721h
0.372}] «10°
7
Tg wioe
rectangular
Pein
(b)
OO. 64s
=
0.645 x - 0.05333 x? +4 0.05333 <x-1.55 9 Nem
(a) At pont
For
t
Vzo,
whee
D
if
0
+ (B1E)U-S(OM8) + (1.25 (2.50.48)
_
kx- 65>)
ago N/m = 0.48 KN Jn
= O16 x7 + O.164K-15>"
peint
Locate
o.48
|
é
ChOSS
2.6753
{E72
darger
—
OT 78-6 1SB IO
section
S$
=
Log
+b
*{oO3
C759 +10")
[O-wm
kN-m
mm = 72.6758 10 s wm 8
h”
h=
&S
.
=
120.56
increment,
mn
h = 120 ww,
~weh,
Proprietary Materia}. © 2009 The McGraw-Hill Companies, Inc, All rights reserved.
No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ae
Problem 5.117
5.116 and 5.117 A timber beam is being designed to be supported and loaded as
shown. (a) Using singularity functions, find the magnitude and location of the
‘maximum bending moment in the beam. (6) Knowing that the available stock consists
of beams with an allowable stress of 12 MPa and a rectangular cross section of 30mm width and depth A varying from 80 mm te 160 mm in 10-mm_
determine the most economical cross section that can be used.
30 mam
a
500 N/m
HEM,
increments,
a
=
a
O15 kN /ws
=08
~4R, + (B.2XKLCYMO.S) + (-6)(4X24)(0.5) =
R, =
Vz
0,980 —- O.5x% 4 O.104167 Cx- 16>
My
Va
=
=
0.880
0.980
Locete
post
D
Chetween
5
kW/m> -_dv
kN
{ Sian
0,080 KN
Bad)
KN NS
- O.S20
=
+(O.10d1E7 ZI
0.980 ~(O.5VA)
=
Ve
kw
-(O.5\(1.6) =
kN 4
Le?
833
0.20<x-
OS-
=
0.5 - S8&-16)'
0.880
©
where
Vz
change
Oo.
O = 0.880 - O.6 Xp + 0.104167 (X%- 1.6)
1.14667 = O
+
— 0..933383%,
O.NO4167 x
Ki. tce7)
0783383 #7. 83333)" ~ Woodie?
x, =
—
QYo.1049167)
2.2842
HOt
=
—
m
1.7658
,
CATR
=
kW
0.880 x -— O.25x* + 0.347222<x= LED”
My = (0.880¥1.7¢658) -(0.259.7658)* + (0. 34742\(0.1658Y = 0.776 kv-m
Mz
@
Sin
For
-
-
Mvene
Gust
kN-m
0-776
=
Mime
>
0.776
VQ
a reed angudar
«(o*
joe
At
_
64.66
=
cress section,
| ey
Dynin =f EGE enter?
(b)
X*
at
nex
hig hee
Jo- rem
=
VW3l7
ineremeut
* 1/0
SF
“6
~<a
m
1.7658
yn
3
©
>
G466*1IO
3
wm
3
h= &s
t bh*
mm
hr
120
mm
—
Problem
12kN
51 18
—
—
a
5.118 through 5.121 Using a computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increment AL,
starting at point 4 and ending at the right-hand support.
AL=04m
I6kNAn
ADEM
=07
(.ayi2y~
A
Br
1.2m
/ a
+ DEM, *
.
w=
x
4#B
476
4 @Yeyie)
kw
16<%~12>
~
{.
°
a
.
©
=
©
of
(L2rtizgy - AOC)
C=
=
ft
28.4
kN
+ 4G
f
x
av
Viz
w ledu-1.2>'
- 12 4 47.6<~-12>°
Me
~ BXe-1ay
~ IR
+ YPC <H-12>
x
rn
KN
M
kKkN-m
0.0
0.4
0.8
-12.0
~12.0
-12.0
0.00
-4.80
-9.60
1.2
35.6
~14.40
2.6
2.0
29.2
22.8
-1.44
8.96
2.4.
16.4
16.80
2.8
3.2
3.6
4.0
10.0
3.6
-2.8
~9.2
22.08
24.80
24.96
22.56
4.4
4.8
~15.6
-22.0
17.690
10.08
5.2
~28.4
-0.00
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.119
.
.
° 5,418 through 5.121 Usinga computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increment AL,
starting at point A and ending at the right-hand support.
AL =0.25m
36 kN/m
DEM,= Of
~ 6 B+ O20)4
=
89
_
kw
>
we 38 <x-3>'
12<x-3>
Vv
mn
QUOI
UU
Pe
kN
KN«m
.
.
.
.
.
89.0
89.0
89.0
89.0
89.0
89.0
89.0
89.0
~31.0
~31.0
~31.0
-31.0
~31.0
-31.4
-32.5
-34.4
-37.0
-40.4
-44.5
-49.4
-55.0
-61.4
-68.5
~76.4
~85.0
0.0
22.3
44.5
66.8
89.0
111.3
133.5
155.8
178.0
170.3
162.5
154.8
147.0
139.2
131.3
122.9
114.0
104.3
93.8
82.0
69.0
54.5
38.3
20.2
-0.0
6
So PWWWWODNN
EEE ROO OO
M= B9x- 12OSK2 >) — 26K- 38>" KN-m ~
M
m
>
kyo
‘
G<x-3>
»
-
@
~ [20 ix-27
8
89
e
Vz
:
.
:
.
‘
©
Ra
U)4)]BME) = O
BUWOGUIW
120 kN
/
SOOMUMWOMUMYWODUNWODLUWS
7
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem
5.120
6118 through 5.121
Using a computer and step finctions, caleulate the
shear and bending moment for the beam and loading shown. Use the specified
increment AL, starting at point A and ending at.the right-hand support.
50 kN
DEM,
~ 3 b R, 4+ (BoB)
R=
wt
so
=
So
69
EN
-~
Zo
7s
- [til
x
7
4 C20) (81 4)0) =
+
ZO
Fe
/y
K-18)
.
X + MM k~ pa>'
V = 69 - $0K 4 S5Ox*~ Sesode-rg>” EN =e
|
s
Me 69«- 25x + 1-852 x5 —18-524x-1.9) kNm met
x
mm
0
Vv
EN
69
o-&
4{
}2
17
M
ENM.
O
32-§
VB
- 3B
<4
2°4¢
~-20
468
2-0
-39
ad§
3-6
-47
OO
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
.
Problem 5.124
BIB
AL = Io
ie,
'
da kNvin
LOREM”
co
w
IGEN
through $121
Using a computer and step functions, calculate the
shear and bending moment for the beam and loading shown. Use the specified
increment AL, starting at pointA and ending at the right-hand support.
=
:
"
Ft
‘
Se
:
»
ae Mem
FS {0-137
—~fE km- (-387
-d
= 33-33% ~4rlu-}. 27> - 33-33h%- 13S) =
an “vlc leas
0,45 th
0.45
Ie:
Vie~-16 67m + Aha
138 4 6 67Kol-|- 355 ~ “oto
mPPY
M = —g.5b0> + 22-8h- 3sS+ osbln- easy -/6<%- 68>
x
V
in
kN
oO
kim,
O
0° b.
“5.32
3£
27
al)
27
M
“12
= 13-74
42-04
~ 4,
- 714.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hii] for their individual course preparation. A student using this manual is using it without permission.
°
= Problem
5.122
—
ee
__ $122 and 5.123 For the beam and loading shown, and usinga computer and step
_..
,
SkNéw
3kNAn
TTT
|
|
3KN
HIM,
W200 x 92.5
bt 5m
ismbism
,
furnetions, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x = 0 to x = L, using the increments AL indicated, (6)
using smaller increments if necessary, determine awith a oe2 percentow accuracy the
maximum normal stress in the beam. Place the origin of the x axis at end A of the
”
10.2
Me
IO2%-
x
=
AL=025m
—
Ra
LEX
- dx-ay
_M
sigma
MPa
0.00
0.25
0.50
0.75
1.00
10.20
9.45
8.70
72.95
7.20
0.00
2.46
4,72
6.81
8.70
0.0
12.7
24.4
35.1
44.8
1.75
4.95
13.26
68.3
2.00
2.25
2.50
2.75
6.45
5.70
10.41
11.92
4.20
14.40
2.95
1.70
0.45
15.29
15.88
16.14
53.6
61.5
74.2
78.8
81.8
83.2
(3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
-0.80
-2.05
-6.30
-7.55
-8.8Q
-10.05
-11.30
-12.55
16.10
15.74
15.07
13.34
11.30
8.94
6.27
3.29
83.0
81.2
177.7
68.8
58.2
46.1
32.3
17.0
5.00
-13.80
-0.00
-0.0
2.83
2.84
0.05
0.00
16.164
16.164
83.3
83.3~<—
2.85
-0.05
[6,2
16.164
= 0
+ (LSY3)
kN
3B + 2€x-29°
~ 3¢x-3.5)
KNem
1.25
1.50
= V
.
=
:
.
4 (1.5)
:
Bx — 2dx-2y! - B<x- 355°
| kN
m
+ Mopzoy(3)
~SRyp
We
Vr
= 02
KN/m
ON
~e
Kem
ae
For vodled
W
200
- - 4
stee! section
* 22, Ss,
S = 194 «(02 me»?
<4
= 1G4<107°
6
““
=
MM swre
S
= 16164 MIO”
194
¥ to
= 83.3%)0"
-
33.
33.3
‘
s
Pa
MPa
aa
83.3
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
§.122 and 5.123 For the beam and loading shown, and using a computer and step
Problem 5.123
functions, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x = 0 to x = L, using the increments AL indicated, (0)
using smaller increments if necessary, determine with a 2 percent accuracy the
maximum normal stress in the beam. Place the origin of the x axis at end A of the
beam.
20 kNAn
50 mm
om)
J) [00m
Mz
Re
L=6m
—3 me
~ 54
Co
HEM, = OF
~4Re +G6XSd+ @5SKal(20) = O
AL=05m
WSLH- ZF
We
ky
4S
>
20 kx-27° — 20LK- S59
— 20LK-29' 4 QOQK-5>)
kW
~Sx + 45éx-2> = 1o<x-29° 4 1OKK- SY
x
m
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
6.00
Vv
M
kN
KN-m
~5
~5
-5
“5
40
30
20
10
0
-10
-20
-~20
-20
0.00
~2.50
-5.00
-7.50
-10.00
7.50
20.00
27.50
30.00
27,50
20.00
10.00
0.00
sigma
et
For
10.0
sera
26.7
13.3
0.0
30
xX
KW oy
4.0
wm
cvess
section
S = Ebh*=(2}(52)(g00)"
=
750 ¥ |/0* mm?
=~
40.0 =~
=
re ctanay das
"13.3
36.7
et
im)
7
0.90
= - 4
Nem
Maximum
-3.3
KN/y
75oO
«jo
°
m*
c,,- Mee. Sexto”
.
Whee,
S
=
750 ™(0
4oxjo* Pa
Gent
4OOMPO
<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.124
,
§,124 and 5.125
For the beam and loading shown, and using a com-
puter and step functions, (a) tabulate the shear, bending moment, and maximum
norinal stress in sections of thé beam from x = 0 to x = ZL, using the increments AL indicated, (6) using smaller increments if necessary, determine with
a 2 percent accuracy the maximum normal stress in the beam. Place the origin
of the x axis at end A of the beam.
30 kKNAn
18 kN/m
493M,
= 0
© PE Ry 4 Co) O4SIA-27D 418)(0-6)(0°75)+U-2 09S) = O
Le
45m
1im
,
AL-= 0.07E m
Ra 2 I72BL EN,
We 30 - 12x ~ BGO = {PLK~ OS)? LN] m,
Voz
M
79235-Zox
+
Iz Kx - 04>
4
18 L~ fof)
P2de- pol > EN
= pne3ex -ISX+ 6d x- 0.0554 PX x- 0$> - 1-2 £x- ford
Vo
by
|
M-
km
sigma
MPa
0
b
a
1636
0+375
$36
4.39
0-66
O>
SAE
6-66
O75
[44
S09
6-6
-7-4%
34S
Le.o7
«7-44
GO
[126
HWE
|
166
a
.
Meximom
at
|
IM)s
A=
-
kNm,
=
|
3/5 £m
OG66Mm
sectiou
Rectangqu fae
£0 K Beamm
S
a
x.
m
~—«
z bh* = (P\40) (300)
2 7S0080
|
e =
M
2
mm >
LAS Kro*
FVFodaeo Kiet
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
5.124 and §.425 Yor the beam and loading shown, and using, a computer and step functions, (a) tabulate the shear, bending moment, and maximum
Problem 5.125
ments AZ indicated, (b) using smaller increments if necessary, determine with
a 2 petcent accuricy the thaximum normal stress in the beam. Place the origin
of the x axis dt end A of the beam.
70 kNAn
SO kNém
P WRK 445
SN SMe
areoatm
tT
:
3m
.
|
= ©
8°75 Ry 4 (TOMISIG-7O +(FOMUBMIS)= 0
L=45m
O75 m 0.751
Raz
To
- 20 <x-159
EN..
EN] Mm.
~38 xX
x
m
/ Yv
kal
4.
166 <x-~0.7S>
_M
balm
O
sigma
(0 &x--E)
-
ENA
Maxi mura
at
oO
O
6S
h2
44.39
£968
226
27
(902 73.2%
OTB ME
7-5
(23
B00
-12-F
4S
276%
O
TS
9G (26-27
0
+
Pov
A
10 KX ¥ (68 4-0-7089 + 206K-1-6> EN
=
i
+t
w
165
M
a,
=
78 +f)
EN mm
xX * 2eTm.
robbed steed section
S?
624 x10°
WRIOX 4S
iam?
— Maximem normal stress
_ M78
x10e
oO
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
-
Problem 5.126
5.126 and 5.427
The beam AB, consisting of a cast-iron plate of
uniform thickness b and length L, is to support the load shown. (a) Knowing that
the beam is to be of constant strength, express # in terms of x, L, arid Ap. (b) Determine the maximum allowable load if L = 0.9 m, hg = 300 mm, b = 30 mm,
and oy, == 165 MPa.
HAY
=
Rat
{WI
A
whi
iM
booed
Clu
Wyoy
*
wh
Wee
Me
(b)
Sofurng
3
oh
hor 7SRE
¢
we
w
for
so
Raz
Ra
= we
px
WKS
4
+M-re
Oo
Wx (L-x)
2G.
$= £bh’*
he Pee
Equeting
AR x= &
wl
wx(L-™)
L-*)
For a reeta navdav cross section
;
x (L
ebh* oon wxCeow)
@)
7
am = =SM,= 0
,
—
Re
57
BZ
nat
)]
he [X¥(-#
Hle
4Gurbhe
- ay"
aC
“wntiséentyo sia
(3)(0-9)*
|
L
~
*
=: 733-3 RAY,
Problem 5.127
5.126 and 5.127
The beam AB, consisting of a cast-iron plate of
uniform thickness b and length 1, is to support the load shown. (@) Knowing that
the beam is to be of constant strength, express # in terms of a, L, and fp. (b) Determine the maximum allowable load ifL = 0.9 m, 4a = 300 mm,
& = 30 mm,
and oy = 165 MPa.
P
yP
“M
Ve
S
For
a
vectangqulan
.
Equeting
CVOSS
section
bppr.
t bh’ = Px
Sy
h-
At xe
S
"
Cutt
Solving
for
P
Pe
Gurbhe
a)
@)
h,=3SPet
©) Divide! Eq.(0 by Eq. (2) amet sodve for
()
=tbh*
_({GPx\t
“(a
)
h
--Px
[Mz Px
WM@, 7° a%
LP.
he he (2/4)
hh
= LUE K709) (9-23) C0- 3)
C6)(0+°9)
= P2-0
~<a
Ve
EAS
ae
Problem
. 2
5.128
Oo
—
-
5.128 and §.129 The beam AB, consisting of an aluminum plate of uniform
. thickness
b and length
L, is to support the load shown. (a) Knowing that the beam is
to be of constant strength, express A in terms of x, L, and Ay for portion AC of the
beam. (6) Determine the maximum allowable load if L = 800 mm, Ag= 200 mm, b =
25 mm, and my = 72 MPa.
A
x
bs
|
|
Oe
> jenn Pf hmmm
AY
-
- Bx +M- 0
M
he
|
(b)
SoPving
>
&
for P,
2
replace ~
P=
_
x
lo<x<
ox
he
#)
atx
=
ocwch
by Lex,
26iabh,*
3L
=
ns be VE
(@) ARx Fhe ho= YSPE.
For
OF
S= z bh
cross section ,
tbh”= -
Equating,
DIM y*
hy
<x
f
Fov a rectangutar
FID
v
_—= (a)(72«10°)(0.025
Yo.200)* -=
\(0.8)
P=
Goxto®
N
GO uN
—e
Proprietary Material. © 2009 The MeGraw-Hitt Companies, Inc. All rights reserved. No part of this Manuat may be displayed, reproduced, or
‘distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators pertnitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
Problem 5.129
-
. $428 and 5.129
_
The beam AB, consisting of an aluminum plate of uniform
thickness 5 and length L, is to support the load shown. (a) Knowing that the beam is
to be of constant strength, express A in terms of x, Z, and Ap for portion AC of the
beam. (6) Determine the maximum allowable load if L = 800 mm, fy= 200 mm, b =
te
.
25 mm, and a,y = 72 MPa.
4wl+BpB=o
A-
=O:
tT DE,
B
Az
Symmetyy ,
By
= wl f
A>=B
ya
A
Fou
“~
M
AV
5
ee
Hy.
=
X
# Web
4
Vor
nw
a
= Vo
At
WX
C=
n=O
we
Viz
w= 8
For
con stant
C,ztwl
C4 prole ~5 er
M=O
gy
S
C, =
©
(3t'x- 422)
eS
Me Mos
& ML
strength,
gwol
pwol ~ Har”
=
w,
AY
2
AEs
x =0
M:
eee
=oF eee
“Tr
Ww
At
Mz
“a
~
<= 5 y
~
aC)-4EGP]
Se~
6,
.
=. dwt
. Me
e,
.=
Me
Ey
SF tes te (te ¥x8)
For
a
rectangudar
section,
S
=
¢bh*
=
(b)
Data?
Se
Me
w=
=
h
(7)
b= 800 mm
tbh,
=
=
Con
he= 200mm
$(as\(200)*=
.
S.
=
(72%10°
he
ot
5%
b= 25mm
166,
667™
667
lo)
«10°
=
<A
ax
Clys 72 MPa
mn
{2
OG tabeye 7 RRS MIE Why
WRMe , U2 \l2xo*)
ee
166.
>
d bho
5
aT
z
@)
$.7
«
%1G*
[66,667
xjo*
Dew
w= 22S RN my
wm
Problem 5.130
$.130 and 5.131 The beam AB, consisting of a cast-iron plate of uniform thickness 5
,
and length 1, is to support the distributed load w(x) shown. (a) Knowing that the
beam is to be of constant strength, express A in terms of x, Z, and Ap. (6) Determine
we = wysin gSF
the smallest value of Ao if L = 750 mm, b = 30 mm, wo = 300 KN/m, and og = 200
av
Bk
epee
toe
X
|
Wit
;
-
THX
We
Siar
Vr
ZMel
cos
K+
Vz
OO et
x=
O
aMxy = 2Wb (| 257)
aL
©,
-—>
C= "=
ax
M
S$
Por
-_
4Wol f
=
=
AWE
iM)
Gu
a
(x
==-
.
-
QWol
TT
Gu
rectanqutar
h = few
Cy.
WO. b
At x= 1
heh
Data:
"
~
oh
2L
~
(x
[M/=
26.5,
na
sim
cvoss
0%
2L.
2Wol
7
sin
Wx
TX
*
)
|
)
Section,
=
=
T
t bh*
ngab
*yaeee
12 Web”
We L*
2
2
Me
= 4, 178.) ok
-
wa
i*
se ]
he LE - HsinFt)/C- 2))" = 1.659 [HS
L= 780 mm = O.76m,
b= 30 mm= 0.030m
We = 300 kW/m = 300%10* N/m,
Gig = 200
_
(b)
AL.
(~~
gbh* = 2reb
(y - 2 sin Po
Equating,
(a)
TX
26 sin
he= Ld 3 (Goceteve.Is
= 197.610m
MPa = 200x/0* R,
h2 \97, 6mm a
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5:131.
5.130 and 5.131 The beam AB, consisting of a cast-iron plate of uniform thickness 5
and length Z, is to support the distributed load w(x) shown. (a) Knowing that the
beam is to be of constant strength, express / in terms of x, L, and fp. (6) Determine
the smallest value of Ap if L = 750 mm, b = 30 mm, wo = 300 KN/m, and aay= 260
MPa.
av.
ay
VF
Me
For
a vectanqulav
=. y
4;
At
Data:
ebh
L=-
75Q
We= 300
Wx
- 3
~ . WeX*
wx
+
O.75
KN/m
=
0.75
h, sy) Boone Nee
.
m
WoX
IM\=
z
We x?
Ss
é bh’
heer
he
@a
|
b=
30
S00x/0? N/m
)*
dM
ay
= fee
te,
h, = Vee
mm
Wx
Ca > Cee
-
imi
Section ,
=
_
"er
.
h>
x=1L,
=
SF
Cross
Lope
Equating,
(b)
SS
WF
mm
>
ho (#)
-
= 167.7% 107m
,
~
0.080m
Gin = 200
2
|
3ft
.
MPa= 200x/0° Pe,
167.77 vm
Vig on
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form of by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using tis manual is using it without permission. -
Problem
5.132
_ §,132 and 5.133 A preliminary design on the use of a cantilever prismatic timber bearn indicated that a beam with a rectangular cross section
50 mm wide and 250 mm deep would be required to safely support the load
shown in part a of the figure. It was then decided to replace that beam with a
built-up beam obtained by gluing together, as shown in. part
of the figure,
five pieces of the same timber as the original beam and of 50 X 250 mm-cross
section. Determine the respective lengths. J, and J, of the two inner and outer
pieces of timber that will yield the factor of safety as the original design.
.
P
ia
HEM;
54
b- x
At
B
At C
At D
L
_
=O
Pk+AM
as
IM)
IMla-
IMl=
=
lg (2 7 Se 7 as
.
Se.
~-Px
Px
Mine Xe Z2
IME Mave Xp
/2
Sas Ebh= ¢- bby =
Be?
S.=
$ bGY
= LE
S,
$ b (aby
= db?
4,
Mp.
Ms
Minue
Xe = AUG).
IMI]
=O
Oem
mz
=
~
m
08
e192
V1
wea
Het EYE
= 0-72
D |m
-
2).
4,
>
oR
-g.J2
=
1.28m
rou
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.133
5.132 and 5.133
,
matic
&
timber beam
A preliminary design on the use of a cantilever pris-
indicated
that
a beam
with
a rectangular
cross
section
S0 min wide and 250 mm deep would be required to safely support the load
7
shown in part a of the figure. It was then decided to replace that beara with a
built-up beam obtained by gluing together, as shown in part b of the figure,
five pieces of the same timber as the original beam and of 50 X 250 mun cross
section. Determine the respective lengths 7, and 2, of the to inner and outer
pieces of timber that will yield the factor of safety as the original design.
LTA <
KO
ey
wee
O
(2)
So
* B22
GY
2
AY
At
B
C
IML
ALF
A+
D
IMi,> IMI. O%
7 2.)°
(Miu
Maat, 72)"
S.=+tbh*-
£40b-
ob?
&
ChDbD
S,2tbh*>tbG@b=
2b?
it
=
4
At C
Li
S,=ybht = Lb(sby = 286°
Om
foe
2B - OG
me
Wxe
At Bo
Xe? ger
Mle 2 / Xe\*2
©
3M:
.
\
v
o)
we CMe
a
fen
me
“PZMy7
Xp 2 Git
=
fbAr
a
Ofm
<0
= Id
JB, 2 Zn fe
2
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.134
5.134 and 5.135 A preliminary design on the use of a simply supported prismatic
timber beam indicated that a beam with a rectangular cross section 50 mm wide and
200 mm deep would be required to safely support the load shown in part a of the
figure. It was then decided to replace that beam with a built-up beam obtained by
gluing together, as shown in part 5 of the figure, four pieces of the samie timber as the
original beam and of 50 x 50-mm cross section. Determine the length / of the two
outer pieces of timber that will yield the same factor of safety as the original design.
4
re
x
ay
DIM;
Bending
M
PL/2
o<u<§
moment
= of
diag ran
2
AAC
S= Soho
AAD
Soe Ebhg —
2
-Fxams=
is two
oO.
straight
dines
~
Mee Mra
Mp= Mou%
.
, neg
Xt
2
Sp.
=
No
(p=)
.
rel
Dhes-
Se
LQ-Xe
100
mm
OF
=
4
4
=
Mp
Me
=
Xp
Xp
2
O.3
m
i.
Az \,800 ww
ma
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individuat course preparation. A student using this manual is using it without permission.
5,134 and 5.135 A preliminary design on the use of a simply supported prismatic
_ timber beam indicated that 4 beam with a rectangular cross section $0 mm wide and
200 mm deep would be required to safely support the load shown in part a of the
figure. It was then decided to replace that beam with a built-up beam obtained by
gluing together, as shown in part } of the figure, four pieces of the same timber as the
Problem 5.135
original beam and of 50 x 50-mm cross section. Determine the length / of the two
outer pieces of timber that will yield the same factor of safety as the original design.
(a)
{b)
.
pbk
OB8WNW
= “3
Shear:
Ate C
V=O4w
Areas:
Ate C
(0.2)({0.4) w
Dr
Ct
V
Bending
\ O.4Ww
(0.42w)
A
|
B
.
“on \
*
A
(4) (0.4 }(0.4 ))w
Me
=
M
=
te
Se
Xe
=
0.32
O,OB8
w
W
he
_
.
/ 100 me,
oon.)
0.25 m
£ =
wx
Me= Mran = 0.40 w
S.-tbh,
ne
w
0.40
2 ¢bhe
F
Se
O.4O
.
At+c
At
Vr-O.4w
E
&
C-
—
‘moments,
Cc
At
.” "
Xe
0.4 w
=
rf
=
Ry = Re
a
~O.8 mee 0.8 mep<- 0.87
Mp
*
~
4
= 0.40
a. Me.
4
~%e
M
=
O.9S
WX
O40w%e
0.40 W
va
Lz1.900m
~<a
Proprietary Material. © 2009 The McGraw-Hill Companies, tne. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written périmission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manuat jis using it without permission, —
Problem
5.136
5.136 and 5.137 A machine element of cast aluminum and in the shape ofa solid of
revolution of variable diameter d is being designed to support the load shown.
Knowing that the machine element i is to be of constant strength, express d in terms of
x, £, and dp.
Draw
Oo<
& <Z y
M=
|
= PL/R
M
PLA
constant -Hvength
|
ce M
%
an
Cenk ey
For FeudS
Cx
aa
s- 2.
‘I= Fe! = ga"
For
Pre
section,
For a solid civevdav
‘piv
|
deargn,
© = constant.
|
|
aes
Ee
Gad
ge dt= egw
L,
At pmtc,
fat = &
"Dividing EqeClad by Ey. (2),
O< veg,
E(lb)
Exel,
Gb)
(a)
aa in
de ds(lax/.y
Dividing
-
M = LO)
/
BeXeL,
diagrams.
moment
bending
and
sheav
by
=a
Eq.(2)
~ 20-2)
L.
d=ds [2tt-~/.] > =e
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission...
5.436 and 5,137 A machine element of cast aluminum and in the shape ofa solid of
Problem
5.1 37
revolution of variable diameter d is being designed to support the load shown.
Knowing that the machine element is to be of constant strength, express din terms of
we
x, £, and do.
|
- wh
OR,= Ra
=3
(yyw
‘
Ss
-
\M\.
a
|
we
WX (Lm)
2
Rbu
7
Ouu
For
_
sobre
Easing
RZ
L.
DEM, =o
2
~My
4 wKE 4 M =O
Ms 2 x(L-x)
:
.
Citevtar
WAT,
Peete
2
x),
n
Gross
section
WxG-xd
2 Guu
~a =
(whe
~.4
To
Cy
T=%c
S*
s- fh.
me2.
a”
d= fis wr
wx there
(L-%a
.
T Gu
ys
X
(j-%
a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problern
5.138
-
'
§4138 A cantilever beam AB consisting of a steel plate of uniform depth 4 and
variable width d is to support the concentrated load P at point A. (a) Knowing that the
beam is to be of constant strength, express b in terms of x, Z, and do. (6) Determine
the smallest allowable value of # if Z = 300 mm, be = 375 mm, P= 14.4 kN, and ony
= 160 MPa.
ODZM,
Ca
IM\=
Vv
mf
= Or
Ce
{?
S*
al
For
.
Solving
for.
Data:
h,
L=
P=
h=
300
ma,
14.9
7
kN
>
-.
iM!
JO.
P(i-x)
&,
a
Gass
rectangular
@
,
bh’
™
section ,
cross
S=¢bh
bz
al
=
Puen)
~<a
b= b (l- X)
PL
ete
0,300
=
my,
b,
14 xlO?
=
375
ram
Nem
(160x10°)(O,375)
=
OL.375
Guz
(EV(14.4* 1O* (0.300)
\y
M=s~PCL-x)
os
aa
b= k=
PUL-x~)2 0
Pex)
Equating
At x20,
-M-
mM
160 MPa = 160 «10° Py
3
~
ZOLA
IO
nH]
y=
20,8
mm
—/
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
i
Problem
439
5
. 5,139 A transverse force P is applied as shown at end A of the conical taper AB.
a
Denoting by dp the diameter of the taper at the A, show that the maximum normal
. stress occurs at point H, which is contained in a transverse section of diameter d= 1.5
do.
ve-p
Fou
|
a.
= 4H
solvad
c=IsG _
a =
Stress
a
St
M=-Px
edveusen
section,
s-F- Loav*L
Bs 2+
= am
TeEeh-i#
key
ar
ag (to + kee)
O'=
_
ae dul
ht
ae oA
kos
a7
Fe.
6.
oe
= Bd
JE
K
-
mn Hak
k%y= gat = Shot kvty) — kxyr Fale
d=
dt4d. = 24,
d=1.5d, —#
Proprietary Material. © 2009 The McGraw-Hill Companies, luc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.140
5.140 Assuming that the length and width of the cover plates used with the beam
of Sample Prob. 5.12 are, respectively, /= 4 m and b= 285 mm, and recalling that the
thickness of each plate.is 16 mm,
500 KN
determine the maximum
normal stress on 3
transverse section (a) through the center of the beam, (6) just to the left of D
—p->il6mm
fee?
Raz
Re
7 250
1
a
M
eb pp eden
]
secret
a
ae
center
MAD
beam )
x2
center
oF
beam
2
I=
ry
(3-
Team
4
+
)
>
2m
+
M
M= 250 x
v
M. = (ase\l4) =
Suny
x= £(8-2)F
~250%
>
My
1000
os
=
©
kNem
kKNem
500 kN vn
ZT
NFO xso® + ZL (aes Wie Vl C78 , Jc)"4 deve
QRZ KIOS mm
c= S28 4 16 = 355mm
s-2 = GHYSHIO
i
uw
At
at
“DZM;
*)
Ke
250 kN
A}
kw f
|
(®
Novmad stvess;
At D,
(b)
S =
Normad
.
M
=
{ooo x(o#
Ge & * Gqyewrore
83510 {8 mm
=
stress:
300% wpo
3Mo, * Zao
G= =
~
7
mm
644 x j0o°S wm?
P.
(158-2 10° Pa= 185.2 MPa
3510x100
mm
00
x1o*
7 RAOPa = 24 MP
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written: permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual i is using it without permission.
Problem 5.141
_
9.141
‘Pwo. cover pilates, each 12 mm thick, are welded foa W690
x 125 beam as shown. Knowing that / =. 3 m and b = 260 mm, determine
the maximum noftmal stress ona transverse section (a) through the center of
the beam, (2) Just. fo the left of D.
om
i2mm
i.
=
R,
=
Re
°
3S¢e
EN
M
j
KE
fIEM, = oO
-gsona Mz
f
=
BSBEN
M
[OSPEK-m
M =
At
oc
Xe
M
DD
xP eT am
3m
Me
M7
At
center
of
beam
Te
Deco
+
Normal
mm
(b}
peint
Normad
EAm
«
fosSoa kNew
$25 kA
:
ae
2
4 de (2beleiyy
¢
stress
GS
=
Me
1
At
x
(SP)4 12 = BS) Am.
ot
(
o
223
36°
2 Tptte
I= (UG0X10) +24 C12) 260) (b+
= /F82-F x10
O
D
stress
BB
=
MM
S
Co. jite
=
/Joe.- TMP
~<a,
942+ Pio?
2870 %10 mm
Se
GS
~ (esoxte
= SES x0"
Bre Kie”
eB,
= BSfo K fo M&
=
14-9.6
MPa
mem,
Proprictary Materlal. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the linvited distribution to teachers
and educators permitted by McGraw-Hill ‘for their individual course preparation. A student using this manual is using it without permission.
Problem 5.142
5.142
Two cover plates, each 12 mm thick, are welded to a W690
X 125 beam as shown. Knowing that og, = 165 MPa for both the beam and
the plates, determine the required value of (a) the length of the plates, (6) the
width of the plates.
TOO EN
Ry =
W600
Ra
ww &
Aifow
ahde
a
Pe
At
center af beam
S*
-
C
M
Sua
=
,
~
benelng
Z-
Xyor
Rx,
fos? xot
“fet yr oe
(B+).
4,
7 6 364
BL]
moment
Kr
[466m
= FEFMm.
M = (383)(3) = [OLB
3
Mm
= Ga
S = Cree
x 108) (B10 x/a‘)
= S79SS bNa,
TESTI
BEOXK_
Mis
ENm& |
M = 360%
S = 3slo¥ig
Man
=
Des = ©
VY
ALD
My
EN,
—~220x%
+ M +
wD
260kN
Set
3D
bes
185"
1 x
>
LAlim
é
3° In
= £-36¢ re mat
Se
=
°
jam.
&
Requivect
Bot
moment
I
+
Lyeau
of
inertia
+
2
Tr
(b)
2233-8
kro
pum!
+
Toate
2232-Pxio’ = CilToxte D4
.
=
=U 4oxto
of:
Laren
2-82 70°
b=
ay +
1
ceca} }
(4)
345+3 mM.
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
:
5.143
Problem
5.1 43
Knowing that a,y = 150 MPa, determine the largest concentrated load P that
can be applied at end Z of the beam shown,
Plis x 220 mm
4D EMO?
-4BA = ZAP
HO
Az- 0.45832 P=
nd)
=
a
2.25 m
4+25M,207
43D-~
A= 0.45833 P§
7.0P
=
|
{1.2
O
|
D= {453833 Pf
4.8 m——>-
Sheav
* Ato
4,
Vz
— 0.45833P
CH
E,
Yu
P
P)
=-2.2P
NV
—e_
Bending
moments !
M2=
oO
«
|
~ 0.45333
|
©
(2:25)45 \(20p)
beset
of P
vetoe
on
strenath
[Sowlot> i6BISE
WHIOXBS 2 —
seotion
Cr
Adt =
t
T=
ROBS
7
keh= 20B.Smm
.
Ty = SIS*1O% mm”
ob = 208.5 +(E)UB)= 217. 5mm
106.9axlo° mm"
Ad?= 187,333 *10% mm”
187, 440% 10% mmt
Sig x]0&+
+ 1B
=
= G84. 88 x 10%
R265
(2 \IBwWYFOX[O®)
150% 10° * gous BWI
= §$7,83
«10° mnt
22ES mn
=
30WS.2
¥10*% mw?
“4m
Mbowable toad based on stremgth at C.
The cmalter
S
{Mel
C=
at Be
BCD:
dA =(18)(220)= 3960 mm*
2 T+
Fer
over powtroy
A= YT mm
= (220 Vays
S= 1510x[0'
mal = ISIOx[O mm.
P= 193.8%[07N
al
pve pewties
-O.95P
|Ma!
Fon WHIOX BS,
Poe
22P
2 O
Maz (ES54** \-2ap) =-1iea7sP
IM,] <
Section
,
,
Mer
ADowable
= ©
+(4.3)-0.45833
Me = -22P4
P
M,
*
Bows 8 x10
7?
el
9 & “ee
P= 207-* 710% .
chlowrte Doad controls.
P= 193.8*/0°N = 193.2 KN —t
; Problem S. 1 44
5.144 Two cover plates, each 7.5 mm thick, are welded to a W460 * 74 beam as
Shown. Knowing that /= $5 m and 6 = 200 mm, determine the maximum normal stress
_.
oe
on a transverse section (a) through the center of the beam, (4) just to the left of D.
40 kN/aa
:
,
:
.
.
75mm
t bo ey
leo un f
+0ZM;
=0%
~|60x +4ox)¥+M=o
M=
At
center
At D
of bean
x= dm
Mo=
x= X(B-P)=
Sm
of beaw
—6T
At center
320
Tel
KNem
195 kMem
Mys
=
160 x - 20x"
kN-m
+ ZT pan
= 333x106 + af (2007.5 WET EE)’ + doo Kas y
>
494.9 % (0%
qt
>
Normal
At
(e)
D
stress
S =
C=
18747.5
M
Se
3S
1460 ¥/O* mm
Normal stress
= B=
*
=
mm*
tT
236
3
B2Ox{0
TT
209%«10°¢
mm
5S?
CUT
152.6% JO
Zz.
=
&
Pa
.
2097 x 107 mm*
2097" 107% w*
.
= 152.6 MPa
S
~t
1é60~/0% mw”
SM
:
133.6 10° Pa
=
-
&=
133.6
MPa
~h
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
5.145 Two cover plates, each 7.5 mm thick, are welded to a W460 x 74 beam as
Problem 5.145,
shown. Knowing that o,,= 150 MPa for both the beam and the plates, determine the
required value of (a) the length of the plates, (5) the width of the plates.
40 kNAn
x. p>} 7.5 mM
popper
~ 160% + (40x YE) + Meso
Ms: GOx- 20 x* kiem
F¥ werner
penne
™
<i
{GO kv
20 kW
For
W
S =
460 x74 vodted
1460x107
Adkowabte
steed
beam
1460 «jo m*
mm
bendin
=o}
ADEM,
iI Wn “f
Td
W460x 747
Pd
4
momeal,
0x)
Mins GieS = (150™]0°X14610%
To
Pocate
points
\GOx
D
ond
E,
‘set
M
=
219102
=
Mie»
x=
[Go +f
=
2D
\co* -(4#\20¥214
=
(aire)
@)
Xp =
At
1.753 mM
center of beam,
Me= 6247
Mr
320
Mm
=
1.153
m
6.299 m
B= Me-Xy =
bN-m
kN-m
+ 219=0
ROx*-160%
~ 20 y*= 219
Nem
320x(0
NT
im
Mem
x 10% mm!
io me = 2133
S* Ea = BRO*ICT 5 213gx
c=
4874725
Requirecl
moment
= 236 mm"
of inertia.
T=
Ser
503.4
Sua.qulo% = 933 xjots 2f(bY7sy¥ 4244)
.
= 333x fot +
(wb)
a
xjo*
mm”
+ Lb Cs¥ 3 f
809.2 /0* b
b=
Zi] mm,
=o
- Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
5.146
Problern 5.146
Two cover plates, cach
ETT ET om
ithe beam, (5) just to the Jefe of D.
760x147
Rae Re =
“2s bal
ASD EN LO
Dz
ELS
Me
M
~
Moz
At
center
of beam
At
potwt Dd
Xt
da
(1660
T=
Normal
a
X10)
(b)
point
stress
Normal
S
.
Stress
&
2&0 x
+Me
~ 22g X*
oO
ENm,
xe atm
M.= 1406-20 Ebm,
=
go
o
kale
mm
.
wy
~=
&
D
gro
3
¢
2 JUIOM S00) P+ Ey 4 Llsooyeey’} = $079 26 Kol mnt
mm
;
pI (03925)
(O89
Me _. (yos2cne)
2£
B
—
At
+
Tyean + 2Ipat
cs (HA)+ fe = 292-0
(a
=o
7 ls- 3) = fa
M,
.
Ms
{PEER
tiak bal
[tha 6-28 Léim
7
thick, are welded to a W760
the maximum normal stress on a transverse section (a) through the center of
150 kNAn
At center Pheam
16 mm
=
20792
6% 6"
¢Glexie
CG
—
=
We
Ss"
_ (79.2 MPa,
<a
mm:
geexre”
B4foxie
©
= 2B
«f
4 Fa.
a
Proprictary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved.
No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 5.147
§.147 ‘Iwo cover plates, each 16 mm thick, are welded to a W760
X 147 beam as shown. Knowing that oy,
150 MPa for both the beam and.
the plates, determine the required value of (a) the length of the plates, (b) the
width of the plates.
Ra
= Reg
=
1128
EW.
4z0EN/m
Hs
sititttiey
aire
| aa
M=
HIG Lal
For
My
«+
40%
esx
+ M = ©
= 228m"
EN
¥
W7hex
St
147
vodbled
GUI0X/0"
ino
stee#
section
moment
bending
Deo
AMowah
*
~%b
Mui = Sle S = Cisaxc08) (4410 x00 mi
66'°§ EN
Te
Jocate
pomts
HS
x - 2eeu*s
D
avd
E,
set
664
Ms
Mie
225%" — 19%
+ bE
=
O
?,
nothALE
iZsy-
Bote
C4)(228)
(661
Sf, ) fe
=
4t2m,
.
(2) (278)
Me-
Q=
(a)
#t
CF
Required
Bt
M_
Sa
Nin
. feeb
28n0" = F- 270 KIO”
ie
ny
t
S
fe4 Ee 2 3q0.5 mm
moment
of inertia
mw
T=
Sa
I= Tyem + 2 Ipnk
= Ibbo X10%+ Ae Taf ox
bh =
FeBIS
3
KID mor
.
2679.7 wee = 660 8)$ 2{ C8) 10) (84
(be)
aly
F B-b4om
AreBp~ O68
ft
M = 46-26
beam
of
Center
Xp
arsdm,
= 3879-7
*
6
moot
TY o£ easyer
b
4.3.6+9 ty fr,
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—
Problem 5.148
5.148 For the tapered beam shown, determine (a) the twansverse section
in which the maximuin normal stress occurs, (2) the largest concentrated load
P
iy
M=
Ex.
CO<
“<
3)
4
Foy a. tapered
For
a
.
Bending
stress
To
Jocation
Fined
=
~
Data:
heast
rectanquiay
SFlo
dé.
ax
(@
beam
fb
.
G =
at
C
=z (00mm,
hat
Q+¢
Suz
EbW
bending
b
kx
oe
_
Ke
=
lis
2M
¢ bh*
stvess
=
& bas
kx \*
set
ds
oO
kx,
z
(at ky y4
Co
.
Mm
Beye’
FSO
+
&
fe prim fines
mm,
= Zoo
a
Mmm..
= (E)B\(200.© [20K/0 mm =f29K/2
2
Mi. = Ga S. = (bfx!)
lipo
,
=
x- 2 las kok
~ BP (ag koP~
(a+ kx
Xm 7
S
_3BPx
Blatlwy=
may morn
(atk )*
a. -
section
M.
'S *
tee
dx
3aP
|
cress
ky
(N99)
4
3
~b
m
3.
= 1908 im
ng,
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
|
§,149 For the tapered beam shown, determine (a) the transverse section in which the
-.
Problem 5.149
_ maximum normal stress occurs, (4) the largest distributed load w that can be applied,
‘cnowing that a4; = 140 MPa.
we
20 min
R, > Re
L
M
wi
= 4wl
= # x(L-x)
For
cress
Bending
stress!
To
focation
Finel
the
section
S >° =M.
of
al+ktix
3w
(a+ kx}?
al
FayKE
hm =
_
=
QtkX,
Sn = Eb
APPowsable
tbh*
bending
valve
tbhlas
stress
mm /m
kx)*
set
gs
Oe.
(at ew )* (b= 2x )~ (Lat 9V2 (oben
au
b
f
Cat kx 8
:
BO4
+ Deg
kx }?
|
U2ze O12)
(AK 120) + (360 YJ-2)
=
=
300
3w bx-x*
%% tare)
*
_ Bw 5 ol - (Zatkb)x
~"b-
kx
300-]
-2ax - 2hex* ~ aklLy
(a+
b
= +
kx}?
(at
b
X=
>
h
k= Soerke.
(at kxVL-2% )~ 2k (Le - OY
~ Swf
@)
beam ,
wm
S
)
(a+ kx)y*
=
taperec
{20
maximum
- ex*
Lxl:
dac
es
AS . Su gh
ax
:
| DAZM;= oO:
Vueud by 5
a2:
rectanqular
LE 2am
“paws mae =o
M= Z(Lx- xe)
N
dul
x
For
= dwelt
O:24
(800)O.24) =
aa
192% mm
= J(z0)14zY = 122.88x10* mm = 122.88x10"°
of
Mv:
MF
$.Sia
=
CL) Allowable value of wt
|
we
=
(122.88xto"*
) (140 « ot )
17. 2032
x}o*
Nem
Mm _ _ G7. 2082 to*)
Xm (L-¥m')
=
(0.24 \C0.96)
144, 3*Jo°
N/Amn
wz
199.3 kN/m
<0
5.150 For the tapered beam shown, knowing that w = 160 kN/m, determine (a) the
transverse section in which the maximum normal stress occurs, (5) the corresponding
value of the normal stress.
Problem 5.150
1
20 mm
wheve
For
the
OL=
For
aw
rectangular
,
Bending
stress
To find
e>
cross
~=
6
Docation
ot
We
M.
5
Sw.
b
Maximum
tapeved
120
section
160 KN/m
beam,
mm
S
auel
(at kxfs 7
a+
kK = S205 8°.
=
4
b h*
=
z
kx
300
b Cat
iixx?
(a+ lex)?
bending
12m.
h =
mu/m
kx \*
|
sTvess ,
set
a
=
2°
ee
Lx - x f= 3w
dS | Se gfe
a
Le
(a+ kw )4
(as kx) CL- 2x) - 2K (Ly = x?) t
w
* ns
bb
la+rkyy3
_ aw { al+ KLy ~ 2ax —~2kx* - 2uLx + alee |
~
eB
(at ky y?
- 3w
jol=-
b
-
h.,
Sh
ah.
Ma:
(b)
Maoci mum
2
5
Kx)3
C120)C.2)
ty, = OL2HO m ait
* @\Gaei~ (Boo\(1.2)
a+kx,,
z
Zax-klx!]
(a4
2at KL
rf
Xm
#
(J
.
bh.
=
(§o +(300)}o.24)
= $(20)192 * =
[22. BY
=
x jo?
#u(L- “Xm ) = isorie’ (0.24)(0.96)
benching
stress
2
s.
fond
192
mm
mm
=
=
122.88 *10* py*
1@.432x/o°
N>- m
5.” Taasaeioe ~ S010 Pa
Mem
{g. 422
lo
3
¢
6,2 1S0.0MPa
«lt
|
5.151
Problem
5.151
.
_2000am_
250 mm
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (5) of the
bending moment. —
250 mm_
Reactron
at
A,
- 0.750 Ry,
BSN
.
\
O
Mrz
B,
omol
A
MON
|
°
DZ Me
BS} e-0, 25-51 Vv
M=o
-(0.25\(@s)+
oy
M = 21.25 Nem
.
*
boy
‘
=
4
5!
ES
Just to the DePt fC,
-~65
“m
Just
1D EM
Nt
Vz ION
V=-6SN
}O
At
ES
R=
Adse
Ve
D.
B
C+
Db
.
+00. SS0\C8) +0. 300 \C7S) = Oo
Abo,
ge
VIN
= Oe
BSN 4
=
Ry
;
.
sm,
.
to the right
oF
C,
k
ai
|
y
Ry
Vv
(Oo. volas) +O. os0)0s) eM =
M = 17-50
Just to
the Pedt
ADEM,
=
ot
Nem
of Dd,
A
A
:
V
~(0.50 (35) + (0.30075)
+
Mo
M
=
op gM
—_
=
WS
O
ZO
Nem
Va
M
to
the
H=ZMs
=
Just
Os
Le0.28 —»
~M +.2sves) =
Ms
I6.25
B
C
of D,
right
.
BD
6
s
kN
(a)
Vio.
(b)
IMI
= 3600N
= 225 Nem
=
=
Problem
5.152 Determine (a) the magnitude of the upward force P for which the maximum
‘absolute value of the bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (Hint: See hint of Prob. 5.27.)
5.152
OkN
kN
A
B
W310 x 29.8
.
By
Im
y symmetry
tymoretey, the
tm
im
+125
>
O:
at
| A~-@94P
-4
A
)
A
~s
“1A
portion
A
\7
+4
ACD
noe
7
M.
M.
=
Mp
Normol
stress!
.
&
|
_M
>
F
|
_
=
Me.
M. =tA
bocky ,
= Of
= - M,
|
9-2A
A= 3kni
P= 12.00 KN f —e
@)
kN
(AXYS)-9
=
steet
5s”
18-2A
ZkN-w
[Mlnee
robled
oO
M,= 2A -4
12@- @\(3) = JR
Ww 23.4
4 equel,
~RA FONG) + Mp = ©
IA =
W310
ave
©
a tree
DEM,
Wy
Equote
Fov
-
body |
Mo =0
as
A
Then P=
B
-
M.
moments
:
.
VV
Using
+B
4DEM, = Of
ALERM,
v
A and
P=
a Free
as
AC
portion
Using
ing
bend!
Symmetry
by
Ase,
3
3
.
Fo-RP
As Br
M. k&N-m
°
reactions
A=8
Jo
im
/
BkNew
seetjon
3xto*®
2B0Klo*
|
)
|
-
=
=
S=-
\O.71
-~3S iNew
3xlo N-w
28010"
{0
(b)
C
mm?
= 2B0¥10
mm
3
Pa
6=\07iMPa
Proprietary Material. © 2009 The McGraw-Hill Companies, fnc. Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
=
Problem
5.153
5.153 Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine.the maximum absolute value (a) of the shear, (b) of the
bending moment.
fer
753mm
B
DEM= 0?
|
Ay 300 N
|
200mm
0.075. Fe
300 N
!
Fer
pc
600
3.210?
2 Fao:
Z
4 § HERO:
240
j
= O
N
200 mm
+
i
- (0.2800) -(0.¢)(300)
=
:
.
;
Aun
300,
Ay =
N
G
,
i
My = (0.07573.2~10")
-
Oo
Goo
Ay = 3.2x10°
.
Couple at D,
VO
2 O
hy 980-8000
ae
7
300
Fer
=
R40
Nem
:
Sheay Aids row .
AtwcCc.
1 0)
Aveas
|
20
of the
Rte C0.
ise
Mt te
V
B.
Ct
(60)
Vz
Ma
x
sheav
SVelk = (6.2)(600)= 120 Nem
Nem
B. |
S Vole
= (0.2 YX3e0) = Go
Nem
moments.
O
O+# RO
=
Mg = (22 +60
=
diagram.
60
Me
=
Ms
N
=C@.2)Geoo)*
z
o 7
= 800
= 600 -300
SVdx
Bending
~60
N
D.
Cte
Dte
60o
IZ0
- 240
-G@0460
120
Nm
= 180 Nem
=
>
-6GO
Nem
©
(a)
Maximum
IVI=
G00 N
(b)
Maximum
IM|=
180
Nem
h |
“900mm
ee
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.154
9.154
beam
and
Draw the shear and bending-mosnent diagtams for the
loading
shown
and
determine
the maximum
normal
stress due
to bending.
:
Ba kN
30 kNAB
yIM, = 0
“396A
W200 & 46.1
+ 9M,
4 (3e)ih8)Q-7) ~ PH)lo6)
A
=
=z
6
SO
~ Go trPMO4D)
4:
D
*
iefel
Shean:
e
|
=
Vat
Bite
+ 26D
Xk
|
.
BOL
~ C4) 42) so
EN.
Ve. = BGS — (Bo) (eR) = JPL
.
Cte
Die
“IES
D
B
Locate
Vir
Mitr
point
ee.
E
Areas
where
. f&-e@
under
EN
wlio EK
lP bp Geez
Y=
les
Thee * TRE
Bw
= &
LN.
$e
fe AD
2YEEN
oO
ve.
= 6&7
fPue = OP mm,
sheav
diag
a wy
SVelx =
Eto@
SVady = )(osp tins) = - $2075 Eb
Che
D
SO Vae = CRE) RR) e ~ Ble
Dt
Bo
SVdx = (24)(00b) = 140 Y LAI A,
Ch )(ez2) (265) = 22-27 balm,
Bending
Mi
+f
y
Ate EF
moments =
Ma?
km.
2
Mp
Makimum
iM|
=
f
Mp
A
©
a
Mg = O 4 22272 22-27 EN &.
BLY = LOTS = 17 (AL EN,
1419S = Bhi 14030 EAL mm.
~A/Ge34 thn
=
O
22°27 EN m
3.
For
WZo0x4é-!
Normal
stress
volled
G =
steef
IM)
€
section
S27
2227x104
Habre ©
ZA
3.
44¢&xra
mm
]e7
MPa :
we
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
5.155
5.155
Beam AB, of length Z and square cross section of side a, is supported by a
pivot at C and loaded as shown. (a) Check that the beam is in equilibrium. (6) Show
that the maximum stress due to bending occurs at C and is equal to wol’/(1.5a)’.
Replace
ton
L
IK 3
|”
we
A
Te,
Dood
at
by ‘equivalent
the
centvyorel
:
se
(a)
Vv
+4
2 Fy
=
©}
0%
O
QO
«x
- .
Ry =
Wo
Ry-
»
equid,’b
.
-
Sust
to
“pt
the
Ve
Sust
wz x=
2
Ce
deft
Ff
EL
C,
ZS wolk
2b
fo
Note
Wer
We (24/3° _
the
nig hf
of
Ve -§wbl+ Ros
sign
change.
4
C
wel
Maximun|M)|
aA C,
Me = ~ we(ae/sy’ = ~ gt web”
For
i
sqvave
zt we L*
cross
6,; Mees.
section,
gure
T=
vivm
We x”
ex"
Ma -
=
Wel
re
oe
[MI
he
kwh
DZM.=
Maximum
oF
«
Aiagraw
For the trcangqudar distr: butron
tentvotd
Pies at
xe 45.
te
2b
3
ted
oad
of the Poad
anea
the
|
distesbo
centvroted
ko"
C2
{a
. £ me. ey a
occurs
5
‘
Problem
5.156 and 5.157
5.156
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable normal stress of 12
MPa.
10 kNAu
/
120 mm
j<-
fh
Reactions:
A= B
WAZR7O:
Ar
V kw)
From
&
benel ing
4
Sinim
:
712
-(SXi0) = ©
+ 25
kN
mae
Guu
r
M (yW-m)
- SEIZE Mm
lh
a.
(6
=
-
12%/0°
BSi.2s*
"12
«106
Pa
1o*
>
2.604 ¥IO > m
2
= ($)lia0yh*? = 2.604 u/o*
(2.064 #105)
20
diagram,
kNem = BL
25 x/0" Nem
2.604 ¥10° min
S= tbh*
.
moment
MPQ
TM
symmetry
A+B
IM\y 2 SRE
Oe
by
_
og
130.21 «JO° mm
h=36l
x
mw
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to. teachers
and educators permitted by McGraw-Hill for their individual course preparation. A stadent using this manual is using it without permission.
Problem
5.156 and 5.157 For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable normal stress of 12
5.157
MPa,
steibuted ade we 2
ierewes
M
ae
Ve-sesc,
HeMa o
25 N/m
util
Dist:
bute
1a
.
s
tot? kaw
We
=
1o.417
M wae *
C,
—~ 26.833
Nj
S$
-
3
Sut
.
wm
B3B5224 Te
_
1O.023xI0"
ax toe
Mee
i
4
-4d
a
a.
3
ret angudar
section
-r_.
Ss
erpvess
(Ous.
945.2% x 10°
7
we
z
io. DAP
the
N-m
10,023xlo°
=
KN-m
10.023
= B3S.29 21D
For
0.407 kw
= 14484
TO
.
Required
>
~F(.4434y + (J0.417)G.4434)
=
%
kV Aw,
C,=0
HSRVIONIT
OO
&
4+ ¢,
~2¢
Ve
lo
=
;
) == O
x (LSY 3 + C,QS
— Sx 4+ 10.417
_
20.833 kN
VK
kN am
we
9
.
“=0.
M=0O of
Voz
M,
aS
Ce
-04
M=
y
‘oadke
,
L:
'
¥
x(taVa
2
for
3,
d= 216 mm
«<
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manuat is using it without permission.
5,158
Problem 5.158
Knowing that the allowable stress forthe steel used is 165
MPa, select the most economical wide-flange beam to support the loading shown.
PAM,
=
(260) (St
3
)— 2668
4 fr )l250e)
=
Br
O
;
i
SEN
452M,
=
4+(260)(51)
Ce-4S8-3EM
Shear
Zo§-3.
A
te”
87
B*
te
Cc”
Aveas
Bending
R
~ 378
Me
5|
%
M,
Ata.
456-3
(ee
© GER3 = ~ZEOEN
-286) =
D
tee)i-2G0)=
— 375 KKm.
Geb) (2083)
O-
.
74. 90S Nm
- BIS kN Mm,
Ma
375
= 26963 KAS.
Fawn,
B
moment
Me =
~2564
Vr zoh3
C
Cte
Me
YL
Vir - 250 EN.
of shear
Bt
te
M > kN in.
O
ov Ce GSh3R EN
Vou
D
Ate
~ 250
2
dragqra Wr e
Ct
~25 0
4
O
CBO KE) 4+ B6C
VEN
45%3LN
=
©
© =375iNM.
~3275 + 74¢98P 2 2740F8F LM
374§ -2795 #O
VM Neue > 376 EN
Required
Sain
Mio.
Suis
wbgo xin
w6to xk ia]
S Cin
2727410
oC
«d)
2576
2274
of 3
Xio
mW?
me
2
2836
WSsex
ISO
S726
w4bo x
HB
2Y4on
2800-
W360 x 316
ee
of
Ls
|
Shag @...
Ca
3
Lightest
wizle
Wbfextol
ange
©
beaw
fos Ep jm.
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the lintited distritrution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
5,159 Determine the largest permissible value of P for the beam and loading shown,
knowing that the allowable normal stress is +80 MPa in tension and -140 MPa in
Problem 5.159
compression.
|
6 mn
vp LQ TUT
Jk
<=} ——
0.25 m
0.5
PZM,=0%
48 mm
-O15A
+OT
Mz
“
O.75C
= Og
V
Sheay:
°
A
A te
Be
8
Ip
4
~0.5383P
Cte
~
1.53333
P
O.46667
P
0.46667 P-~P
-
0.9
~ 0.53833
V=~0.53833
P+ 1. 53383P
oO
P
D,
(ois)
P =
=
ONS P
P
JN
A
Ps
0.25 P
P
a
Avena:
Ate Be
(.25%646667P)- 0.11667 P
Bote C. (0.5 9.53333'P 0.26667 P
M
O.1667
e
Cte D.
Cc
Vz
C,
-
:
Cr
O.4CTP~
=O
A= 0.46667 P
12 mm
0.15 m
+ O0.6P-O1SP
ending moments:
Ma, =O
Matz O £ 0.1/667P = 0.1667P
Me = OLINME7T
P - O.26CL7P = - 0.15 P
P
68
My
7 -0-ISP + O.ISP
=
O
|
~ Ot
i
;
%
PH
i
Dic
Centvorel
LL.
°
FT
°
HA wae
@
® | usa]
sy
@ | 576 | 244
11728
=
[62208 | 19
113324 | 20
176032
| S200 | 13824
| 230400 | 110592
B4S600 | \2441G
|
T723
FI
= y100lb me
¥ = TORI 2 Qa.87E uo’ mm = 29.876 x10 < we
4 = 120018 © \0.682%10° met =~ 10.682 115 om
Top? y= 1G me
= 44 me
BoHom? y=
moment
inertia.
Y= 2082. umm
[T= SAd*+
Bencling
ot
ve
J
al izke
momeuat
_PantlA con® )] 5 mmm YAS Coom® | hlrmm | Act *Conma” ) | I Cone }
54
uM
and
Xi mits
’
Mic
ie
Tension of By =(-10,682«10)(s0xJo*) = 854.56 Nem
Comp. of BL = (29.376 w jo" V-Hoxjo®) = 4.1126 xto*® Nem
Tensiow atc,
Comp. of
AdPowable
ee
~ (29, 376» to°© V(80 «o®
Cy
Ss
- 2.38 élo*®
<i 8
Nem
<ic¢
to VE No xfot) = = 1.4955 x10? Nem
- 10.682»
Prat:
O.N6bTP = 854.56.
-O.ISP = -149gs5*IO*
The
smatter
vadoe
P=
P=
is
Pr
7,3%™1o* N
4,.97%/0°'N
7.32
kN
ety,
Problem 5.160
5.160
(a) Using singularity functions, write the equa-
tions for the shedr and bending moment for thé beam and loading shown.
(b) Detertnitie the tnaxinwm value of the bending moment in the beam.
+ZOMy =O
@V
=
M=
tbo~ koly-o67°[box
Valves
of
A= [60 kN.
go Xx-0-"
8oky-0.6>
~ Bo <x 1-2)
V:
(60 LY
BRC
Vi:
léo-~ For
Ve
(60 —-f0~
8 = Oo
DE
Bending
C
D
~
~
Bodn-hB>>
802K-16 BD
oe
= ©
kA
at,
kA m,
exe,
V
A+B
Ct
hy AP
—
-3-6R 4(3 (90) 4 (249014 U0)
Bo kA
Viz (60-§0-fp-8o
moment
Kel2m
is constant
= ~&o AN
and
meximom
over
C te D.
Ms (16\2)-(P2\@:0)- 0-0 = 44 ENm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
saat
Problem 5.7161
+
te
Wy sin?
. 9.161
x
qv
zw
dis
M
M
=
The beam AB, consisting of an aluminum. plate of uniform
thickness 6 and length Z, is to support the load shown. (a) Knowing that the beam is
to be of constant strength, express 4 in terms of x, L, and A, for portion AC of the
beam, (6) Determine the maximum allowable load if Z = 800 mm., A, = 200 mm, b
= 25 min., and Oy = 72 MPa.
a
=
vy
=
~We
-
Wel
sin 9%
cos
Wels
sin Fe
=O
O=
O+r
=O
O =
wat
=
“ee,
TM!
+
4 6,
C,x%+C,
O+C,
sina
C,20
cl
C,70
sin
constaud
For
2 vecdonsvfar
“ey
St
strength
For
an
= 7g be,
section
;
$
S=
=
2
Ebh*
a
Equating
At
&\
Egt)
—
2
Sefving
DAat
Eq,
GQ)
Gy = 72x10% Pa,
br
25> mm
lo =
W,
ww
fie
=
1
Q)
sin ri
(2)
h= ho (sin &
so s sin
=
Me
ey
az
2
(b)
ae
+ bh? = ee
Eq. 2
by
< bh* =
S
h= h,
w=
Dividing
far
the ture CMY YESS POMS
T’ Cab he
6 |?
b= 800n6
= ©.800m4, h,= ROO ma = 0.200 m,
O.025w
ee
)(o,025\(6.200)*
ee
YO.2001"
I8S_1 x1O7 Nin
{85.1
kN/m
Proprictary Material. © 2009 The MeGraw-Hilt Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, wihout the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~<a
Problem
5.162
5.162 Acantilever beam AB consisting of a steel plate of uniform depth
#and variable width } is to support the distributed load w along its center line
AB. (a) Knowing that the beam is to be of constant strength, express b
in terms of x, £, and 4). (6) Determine the maximum allowable value of 7 if
DEM,= Oo
Vv
M
C
At
I
S$
w
tithe
let al
For
Selvin
a
Cur”
ASM
rectangudtav
cross
wlexy
Aghis
A
obs
Ww
b,
=
W
ae,
2
= Sab.
3L*
-
2
Ss
J bh*
3wlt-x)
- ea
e
b
h*
WE**
section
be
ey
:
b,
(i-
#)
<0,
(68X10) (0+2)(0-02)*
3 (a-G}>
~
\
3
For
b=
O
¥=
~WtL-xY
ype
2
IML
w(t-»}*
a
mM
-M~ wCh-x) EE = ©
276o
AN/
ro,
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM 5.C1
5.C1
,
Several concentrated loads P;(i = 1, 2,...,#) can be applied to a
béam as shown, Write a computer program that can be used to calculate the
shear, bending moment, and normal stress at any point of the beam for a given
loading of the beam and a given value of its section modulus. Use this pro-
gram to solve Probs, 5.18, 5.21, and 5.25. (Hint: Maximuin values will occur
at a support or under a load.)
SOLUTION
FEACTIONS
ad"
nae
A
AND
B
MEM, 20% Ryl- 2 F;(@,-a)
R pi (Vt) 2 RX, 4)
AL,
R
AT
ES
"
Rp
WE
USE
STEP
WE
DEFINE:
R,== =P -
b
FUNCTIONS
[Fx
>a
(See
THEN
Ry
bottom of page 348 of text.)
STPA=!
ELSE
STPA=O
[FX2 a4¢L THEN STPB=! ELSE
IF2x >, TAEN STP(L)=/ ELSE
STPB=O
STP(L)_=b
V = R, STPA+ Ry STPB ~ 2 Fi STP (2)
Me iz, (x- ajSTPA+ hi (2- a-L)STPB~ 2. Fy (%-%) STP (AT)
=
m/s,
Where
§ #5
obtained
from * Appendix
C,
PROGRAM OUTPUTS,
Problem 5.18
RA=80.0
KN
x
Vv
M
kN
kN.m
2.00
0.00
104.00
pi = 4808
"
kN
m
Problem 5.25
a
Problem
RB=80.0
MN
R1
Sigma
bn
139.0
o
oe
a= Me 92EN =
‘
fe
22«to
3%
1694
2533
Loh
16-94
O
-30
= 210 kA
MPa
|
a
O
~63+29
S356
O
|
5,21
ws}
53
r2
. LW
=
Go LAj.
- Ne
- for
o
110-231
tH
,
>
| Won
Oo
48
ca
a
PROBLEM
5.C2
5.C2__A timber beam is to be designed
to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of its uniform
rectangular cross section has been specified and the other is to be determined
so that the maximum normal stress in the beam will not exceed a given allowable value a), Write a computer program that can be used to calculate at
given intervals AL the shear, the bending moment, and the smallest acceptable
value of the unknown dimension. Apply this program to solve the following
problems, using the intervals AL indicated: (a) Prob. 5.65 (AL = 0.1 m), (6) Prob.
5.69 (AL = 0.3 m), (c) Prob. 5.70 (AL = 0,2 m).
SOLUTION
oy,
ip
REAcTIONS
+
472 My= 0.
TT
aul
ges
b
USE
SET
~a)~P 2 (4, - 4)
) (Zz 43 +H lx.O* -0
ie
4+
| FA (t,-a + 5 (%,- B+ FZ Lo (x.-
Re
W/E
Ke L- Fx,
~~ 4 rtx
Los
B
FND
A
AT
a. “PEP Paw (ty ty) Re 2B
FUNCTIONS. ( See bottom of Page 348
STEP
> (ath+k)/Al
fOR
=O
WE
DEFINE:
TD
nm:
X%-28
of text)
.
a=
Abye
IF x >a@
THEN
CTPA=!
ELSE
STPA=O
IFAS a+
THEN STP B=! ELSE, STB=O
IF A 2H,
THEN STP] =/ ELSE
STPI -O
[PX 2,
THe
/Fx2
THEM
VA 2 xs
THEN
nx,
STP2=]
ELSE FTPL= 0
STFU st
ElSe
STP3 =! ELSE
STPZ =O
STPY=0
V= Ky STPA + Rg STPB~ PSTPIPR STP2
— Ww (A - 23) ETPZ+ W(L-Ly) STPY
M= Ry (X-a) STPA + Re (2-a@-L)STPB
-Pi
Onin
' [MI
2) ST PZ — 5 W(RK)" STP3 +4 40 (X—Ly)*ST
PY
/ C4
[F_UNKMOWM DIMENSTLN
From
Soe
[F_UNKNOWN
Fron
S-Z
— P, (L-X))STPI
th
We
,
Js Ae
DINEWS)/ON
bh*
We
A
have
heise
15
=
VES/t
E:
t=
6S/ ph?
(CONTINUED)
~ 2.50
2.50
9.00
7.20
5.40
3.60
1.80
~0.00
-1.80
-3.60
-5.40
-7.20
~2.50
~2.50
0.00
QO. 00
61. 24
86 .60
Problem
ao
OOGOKHFNN
.00
20
40
60
. 80
.Q0
20
-40
-60
.80
-00
@
m
2.70 KN
Vv
kN
¢
=
x
ioio¢
RA
kN
COOK
W
USAT
A
OAS
WE
oO
ram
140. 18
170. 59
189 34
199 75
203 10
199. 75
189 34
170. 59
140. 18
86. 60
61. 24
<@
06
5.70
WWNDNYEH
.00
.30
-60
- 90
-20
50
-80
10
40
70
.00
.30
-60
.90
20
6. 50
H
-490
60
WW
40
60
.80
. 00
FD
0. 00
54. 77
77. 46
94 87
109 54
122. 47
134. 16
144, 91.
154 92
157. 32
159. 69
162.
164 32
166 58
168 .82
171. 03
173 21
162. 02
150. 00
136. 93
122. 47
106 -07
86 60
61. 24
05
6.50 kN
Vv
kN
,
0.000
0.240
- 480
- 720
- 960
.200
-440
-680
-920
-980
040
-100
-160
-220
. 280
340
.400
.100
.800
-500
.200
- 900
-600
300
.000
mm
kN
OUTPUTS
5.69
OoOore
kN.m
3.00
H
ab
RB
M
WWNNN
EEE OOOO
40
.40
40
-40
-40
-40
.40
.40
-69
.60
60
60
60
60
-60
.60
00
-60
00
-00
00
.00
00
00
00°
KN
Pw
2.40
Vv
kN
' Problem
WWWWNHYNYNNN
ME HOODOO
HE
OOo
mOOoOOO
POO
EE
RRP
RE
RE
5.65
COCCOHFBHENNNNNNNNEE
BEER OOO
.00
-10
20
30
40
oO
MRP
.00
-10
20
30
-40
.50
-60
70
.80
- 90
-00
.10
20
30
-490
50
-60
.70
.80
NNNNNHP
an
Problem
PROGRAM
5.C2 CONTINUED
QODOOOCSTOONNNNNNNND
PROBLEM
70
-10
50
-90
-30
.30
-90
-50
-10
70
.30
90
-60
-00
40
. 80
20
60
.00
8. 10
kN
mm
.00
10. 67
18. 67
24 .00
26. 67
26. 67
24 .00
18. 67
10. 67
.00
13 33
29.
48 .00
33 .33
al. 33
12 -00
.33
.33
00
<i]
PROBLEM
5.C3
w
|
| | | |
PEPE
8.03
‘Two cover plates, each of thickness 1, are to be welded to a wideflange beam of length L, which is to support a uniformly distributed load iw.
Denoting by oy the allowable normal stress in the beam and in. the plates, by
d the depth of the beam, and by J, and S,, respectively, the moment of inertia
ay
and the section modulus of the cross section of the unreinforced beam about a
horizorital centroidal axis, write a comptiter program that can be used to cal-
culate the required value of (¢) the length a of the plates, (b) the width b of
the plates. Use this program to solve Prob, 5.145.
at
a
(a) Re
£
seen
ene
SOLUTION
|
vired Length of Plates
. {wrx
ND
D
Ra
But: Kyz5 oh
Divide
and
Mi = SO,
A-LA+K=0
Compute
(4) Required
At
midpent
SRP
WE
y
A
C
z
~
L
M+
whe
~2Xx
<i
fF beam’
FB=ACAHY2M <0:
compte
M~
gars
*
Compu C=t+ id
Cay = “ee
Be
P42)
lh fdtk
PROB.
|
and
Prom
PROGRAM
£5 Val
Width of Flates
t 5
Sy
k=
£-VL?- 4k
w=
the que dratic:
by Zur:
Set
Lt + (256, feb) =0
2
Aw
Solving
=O
FR=ADIDZ M, =o! My t Oz (#)-
But
,;
I = ree
y
bt (Ith)2
bean Fae Lt 2 [fb
I=fF.
Sing for b:, 6 =
]
compute
=
€ (T-Te)
ttt
3+ th]
<a
OUTPUTS
5.1452 W 460x74, Gy =/50MHx| PROB, 5.) 43! W 760x147, Cay = 150M
t= GORN/M, = Bm, b> 75 mn
» > 4S0EN/M , [Adm b= 6mm,
A> UST en , T= 333 xj0 in, 55/46
0x (Oem) = 37¢0mm, I, = 1660x709 ot
S=9315 K1 0% mm?
Problem 5,145
Problem 5.143
a= 4.49 m
b = 211 mm
a = 3b6Em
b =G2beFmm.
PROBLEM 5.C4
100 KN J 1.8m J 100 kN
§.C4
Two 100-KN loads ave maintained 1.8 m apart as they are moved.
slowly atross the 6-m.beam AB. Write a computer program and use it to calculate the bending moment under each Joad and at the midpoint C of the beam
for values of x from 0 to 7 mat intervals Ay = 0.45 m,
SOLUTION
L = Sn
of beam=
NOTATION: LengB=th P=
lootw
loads:
Px
Iheteveen
Distance.
loads
138m
~ d=
A<L/2
We note that
G() FRoM x20. TO x= da:
| pda
L~ x)~ ink 0.
+ 3 2 Mg = O: P(L~
M = Ky 2
Under Fo
ALC: Me = 2, (4)- P(£-x)
Q@)FROM Z=d To x= £4!
4) SM x0: PL-x)
+ P(L-X+d)—-R,L=0.
Ka= P(2L-2x+d)/L
Under
Pi:
M,= R,%~ Po
Under
RB:
M, - ®,
(2A) rom
(x~ al
X= d TO 22 L/as
Mo = Ry ($)- PE -x)- PCF ~x14d)
= Ra(L/2) ~PUL ~2X +a)
(2B) FROM X=L/2
To X-bfe +d?
Mo = Ry (t/2)~ P(& -x4+a)
(20€) FRom a= L/etd To X=L:
Mes
RL/Z
(3) FROM A=
+5 2M,
R=
Under
TO x=L4+d:
0? PUl-X4d) - Rb
=O
P (h-kA
dD /L
B 2 M,
= Ry (x-d)
(CONTINUED)
PROBLEM 5.C4 CONTINUED
PROGRAM
ou-TPUT
x
MC
ML
M2
m
kNm
kNm
kNm
0
0-9
0
—
B08
O
©
BY 75
0
o
Qo
[0fe7
1B4+6
267
Zo3- 4
2-036
[256
26
03-4
203-&
DOS
4S
20564
138° b
5-0 3.4
beh
lols]
0
3.56
Proprietary Material. © 2609 The McGraw-Hill Companies, ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM
5.C5
a
w
|
§.C5
Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply this program with a plotting interval AL = 0.9 mto the beam and loading of (a) Prob.
|
5,72, (b) Prob. 5.115.
SOLUTION
atc
he
TI
4
& 4
y
REACTIONS
HEM, Ot Rgh~ Pb- wa.(a/2) = 0
ApoB
|
Rp
eg
wa’)
R,=(I/e\(Pb+i
pe.
vntF
-
R
le,
P
b
B
FE DIAGRAM OF BLAM!
USING
|
AND
ATA
,
A”
el
b
‘B
of text).
NS (See bottom ofL=page(A348
WE USE STEP FUNCHOFor
i
L=OTOn:
SET
WE
m=L/Al,
DEFINE >
IF X2Q
(F 200>6
THEN STPA=1 ELSE
THEN STPBz=/ ELSE
sTPA=U
STPB=O-
Ve Ry -awtne + W(X-a) STPA~ PSTPB
Mz Ryx - pan? +f (Za STPA ~ P(2-b) STPB
LOCATE AND
SEE
NEXT
PRINT (X,V) AND (x, M)
PAGES
FOR PROGRAM
OUTPUTS
(CONTINUED)
PROBLEM 8.C5 CONTINUED
PROGRAM
RA
=
5.72
206.5
kN
RB
181.5
It
Problem
OUTPUT
kN
Shear Diagram
259
172 +
V (KN)
86
~86
~1724
~259
x (m)
Moment Diagram
455
Z
260
195
4
[{om—~
|
‘N
f-
\
M_
s 130 |
|
65 if
\
to
0
~65 0
= 410.75 kNm
. Westen
o>
€
/
4
©
390
305
x (m)
(CONTINUED)
5.115
RA
173.25
=
KN
RB
=
119.25
}
Problem
FOR
OUTPUT
PRogram
—
PROBLEM 5.C5 CONTINUED
kN
Shear Diagram
215
172
129
x (m)
Moment Diagram
182°
156
130 |
E 40.
104
£ ig
= olf
|—7
}
—
°
|
56 It
0
0
}
15
3
T
N
\
\
\
48
6
PROBLEM 5.C6
§.C6
_
3
Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply this program with a plotting interval AL = 0,025 m to the beam and loading of Prob.
5.112
SOLUTION
REACTIONS
AT
A_AND
B
+3 @Mp= 0:
Earn rN (0
‘
Ap lh +Mg ~My > 0 (b-d)z (A4b)=0
BS
ony
R= C/) Mam Mp 4 7 (B= OD]
Ry = “0 (b-a)~ Bp
”
TT
M
=
Qs
ti B
gL.
Al
WE
USE
STEP FUNCTIONS
SET N=L/AL,
WE
DEFINE,
|
~~
ly
(See Botton of Page
FoR Loo TON!
[F 3% >Q
THEN
IF 2h
THEN
2
42
348 of text)
xX=(ALE
STPAS]
STPB=/
ELSE
ELSE
STFPAR
STPB =O
V= Ra ~ 40 (K~a) STPA tas (X~b) STPB
M> Mp + RyX ~£ uF (4-a)* STPA + pus (x ~b) STP3
LOCATE
:
AND
PROGRAM
PRINT (2,V)AND (X,m)
OUT
POT
ON
NEXT
oS
(exe
PAGE
(CONTINUED)
PROBLEM 5.C6 CONTINUED
FROGRAM
Problem
5.112
RA
29.50
=
kKN
RB
QUTPUT
66.50
kN
Shear Diagram
‘N
-80
x (m)
Moment Diagram
40
30 20
aaa
Jf
ASS
28.3 kNm
at 1.938m fromA
10
x (m)
Chapter 6
Problem 6.1
6.1 A square box beam is made of two 20 x 80-mm planks and two 20 x 120-mm
planks nailed together as shown. Knowing that the spacing between the nails is s = 50
mm and that the allowable shearing force in each nail is 300 N, determine (@) the
largest allowable vertical shear in the beam, (b) the corresponding maximum
5
shearing stress in the beam.
20 ram Li
%
4.
>
s0imm]
hy
12 b,
I=
20 pin
i
4
{
~ it b, h,
i!
3
= aq (120\(120)° ~ & (Bo(80)¥ = 13.3667%/0% mm
13.2667 x10. m"
Wf
120 nmi
Pal
(2)
LLL LLL LA
,
A,= (120)(20)
~
$0 yom
2400 mw"
-
Q,= AV¥ = 120210%
mm =
a =
Fue
$
2 Fas
-
.hs “Sxiot
(2300)
r
BE.
>
A
Dr
|
(b)
7
1a
Q,
3
NN
[zo xto7*
4
(20rt0%
=
xio”
|
1.89667xl0°
Q=
Z HO prin
rol
_
liario*)(i3.9667x10%)
Q
-
1204/0"
VQ.
y=
GLEE
:
:
:
_
a
“y
=
50 mm
N
=
1.387 kN
(2\(20e)}4o) (20)
+
3B2xI107
=
ISR *10* wn
= 152 107° wn?
vy
ma
-
¥e
TE
=
_ (1.38667 ¥10" isa to-*)
(13,8667 jo°*)(2 x 20x10")
Z3R0xIlo*
Pa
380 kh,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ats
Problem
6.2
6.2 A square box beam is made of two 20 x 80-mm planks and two 20 x 120-mm
planks nailed together as shown. Knowing that the spacing between the nails is s = 30
mm and that the vertical shear in the beam is V = 1200 N, determine (a) the shearing
force in each nail, (6) the maximum shearing stress in the beam.
,
20 mm”
80 mm,
t
=
x
bh,”
-~*
bh,”
(aeKieoY= h(a = 13. 2667«18 mnt
== A13.8667
«(0° m*
io
120 vin ~
(a)
LLL
1
A, =
"
20 inn
(120)(20)
J = SO me
Q, = NS
= 2400 wn”
= [ROI mm? = 120 10% wm
- Va. Ciz00 Roxie ) - 0. 385%/0" N/w
$5 *
(LLL
A
T
Ae
2 Freie
Fret
$
(Jo. SBSuio) “VWSeorno*)_ seeg yy ae
=
RQ, +
BRei4e)(20)
° | . = Troi
0 Bhete? = 152 */0' mm
= ISR «(07S wn
vw.
ue
VO
Tt
, (Roo)lis2x los)
BELT IO“ Kaw Roxjo"?)
= 324x107 Pa
oo
BAW kPa aa
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual i is using it without permission.
~~
6.3 Three boards, each 50 mm thick, are nailed together to form a beam
that is subjected to a vertical shear. Knowing that the allowable shearing force
in each nail is 600 N, determine the allowable shear if the spacing s between
‘the nails is 75 mm.
Problem 6.3
Ts
kbh?+Ad*
50 mm}
~ %US9EOS
4 Usnysayl7sy
r ABIEX IO. nat
= 407 gto” m wt
= (oo)?
&bh?
L,=
100mm
|
Ts = I, = 4378x108 mot
rom
= 497-67 Ko nk
T= J,+4,+13
MN
150 ml
OLT7—+
©
e-~l
e~E_]
;
+t
GFF
Fei
Dividing
ye
+ S62500 pom
Ay, = (e)aDN7S)
Q=
cH)
a
Eq (2)
by
Eq
*
9
@
Fug Ed 2 (booyate7eic®) 2 954 p/
Qs
(462560) (75)
Problem 6.4
5
Baie
L
“
6.4 Three boards, each 50 mm thick, are nailed together to form a beam
that is subjected to a vertical shear of 1.2 KN. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s of
the nails that can be used.
+
A
d
2
4
ak bh?
50 mm
I;
q
100 mm
“K (52)(s0)3 + GsoXS0)(75)~
BI
50 mm
ie bh?
o
JI
2
=
4375 x 10% mm
4
= rt (s0\(t00)? = 47x10 °mm*
43:75
x0"
mm *
T= T+ Te4 Ty = 967 «00° mm"
150 mm
o
Len
aT»
—_
Q=
Zs?
A, x
Freie
=(iF0)(S0)(75) -
C1)
3g
Fb2-Sk10 mm
g2 2
=
3
@)
evsmeseanenearmar
S
:
FuakEF
VQ
( 400\(41- 67x10")
2-5 )
©” (t200\(s6x10
S=
5E3
om
«ell
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Al rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.5
Problem 6.5
The
composite
beam
shown
is fabricated
by
connecting
two
W150 X 37.1 rolled-steel members, using bolts of 16-mmni diameter spaced longitudinally every 150 mm. Knowing that the average allowable shearing stress
in the bolts is 73 MPa, determine the largest allowable vertical shear in the
beam.
WiS0x 3712
Sl
yF
N= 4370 mm’,
sa
£2:
ye
MAJ
Bolts:
.
Shear?
dalbmm
t/t
_ VQ.
YF
>"
v=
= A Fiat
. 1g.
a (18)
=
=
Ss
ayene7)
353970
iSomm,.
2606 mm?
(20hobxie®)
x
198-49
. (fol-743xio’\(i9s-69)
a ~
x 10% pam?
Thy t 72 MPa.
Fecer = Toe Anor = (7x0
%
104743
= (4370) 81) = 353970 mm3
Ay
Ave
_
4
2 [zzaxo®s (4310\cety |
=
- Blinm 7
Y=
1
Imm
T= 222K10 mm
= Blinn
Composite
Qs
6
d= 1bZ mm
= 14677 N
Nj mtn
Vz sb2kh
«te
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution fo teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Probiem 6.6
6.6
A column is fabricated by connecting the rolledstee] members shown by bolts of 18-mm diameter spaced longitudinally every
125 mm. Determine the average shearing stress in the bolts caused by a shearing force of 120 KN parallel to the y axis.
Cabcofate
moment
Part
Amin”) | a Com
op plate |
*
Ff inertia.
3870
[7/32
1.6
Cob
4750
Le
C2gex37|
4752
oO
= [32mm
Bot. plate
2500
|*i32
:
d=
ty
ZAd*
Ate
-
+4 ZL
L
NVQ
Aus
F
$b62000
Fer
Avett
7£ BER x10 &
dg
Inm 3
ee
bal
Dib e
Bet EN
= E)(aPooge)loses)= (7-575 EW
2
3704 x10"
(21968 xj0" + FRESE RIO” = 9 TE2 Kio prank
_ 20) (462000K%;0V _
tn
>
=
29167
2764 x10"
Gor PREKO | 24167.
‘ (21-9688
= (3800) (732)
y,
$°
TF
Fun = £GS
Tost
[eo 9Ge x06]
Zz
wo
I Gon)
* Go|
[Ad
= His)
2/7 srs
2SY-b7
fs
s
= 204-67 rans
x10”
=
9 §&
MPa
=
Proprictary Material. © 2069 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribation to teachers
and educators permitted by MeGraw-Hill for their individual course preparation. A student using this manual is using it without perniission
Problem
6.7 The American
attaching to it two
longitudinally every
the bolts is 90 MPa,
6.7
~16 nun & 200 min
Cafeudate
Part
d=
=
Standard rolled-steel beam shown has been reinforced by
16 x 200-mm plates, using 18-mm-diameter bolts spaced
120 mm. Knowing that the average allowable shearing stress in
determine the largest permissible vertical shearing force.
moment
ob
A Gent) | ACrmom | Acl* Comm)! I (10% mn?)
plate | 3200
a +>
= 160.5 mm
inestra.
S3I0*52|
6650
Bot. plate
3200
|*ico.5|
32.43
1GO.S|
22.43
0.07
164. 86
@as.44
| oO
2
Te
ZAd*
+ <I
=
260.3
«/o"
= (Z200OAS)=
A belt
> | (18eJ0*)”
Fett
~
=
F dust
Ton Avot
!
Q= Apete dus
260.3%
2849.47 x1o*
gs Ree
.
m
45.3
10°
$13.6 x1 mm*
(9ox10% )\(254.47«)0%) =
-
gS 5 RF
gv
mus
0.07
mw!
= 513.6%10° m3
2
22.90"10°
N
Cre 7 981-7
< 108 W/m
pe Cee GETS 2 198.5108 w
“6
x
V=
193.5
kN
=
Proprietary Material. © 2009 The McGraw-Hill Companies, Enc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without-the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. ~~
Problem 6.8
6.8 Solve Prob. 6,7, assuming that the reinforcing plates are only 12 mm thick.
12 X 200 mm
6.7. The American
Standard rolled-steel beam shown has been reinforced by
attaching to it two 16 x 200-mm plates, using 18-mm-diameter bolts spaced
longitudinally every 120 mm. Knowing that the average allowable shearing stress in
the bolts is 90 MPa, determine the largest permissible vertical shearing force.
~ $310 X 52
:
Cabesdate
momeut
Pert
B05.
d= 4
.
> 15B.5
A Crave™ )
- Top phate |
e
s12
of
2400
a (wm)
*ises . |
S Blox S21 6650
0
Bot : plate | 24oo
mm
inertia
C=
=
2
=
218.94%10° mmt
.
(10% me"
Pf I
60,29
“ss
=
t (igxro*)
=
(10° ment )
0,03
.
0
45.3
0,03
:
4S. 36
[20.58
Aprte Date = QOOMIAVUSE.S)
Ago > bday
Ad
|" tsa.s: | 60.29
.
2
L= FAs + SE
.
= 215.94 %/0% mi"
380.4 x 107 mam? = B8O.4«/0°° mm”
a
254.47 «07%
we
Fre * Tey Ane = (40x10 V254.47 «10%) = 22. 902x107 N|
3
7
381.7. x/ oOo” a
ao : nee?
= - . (2Y(22.903x10%)
¥ tT QF
Gs =+ 2 Fun
_
_ Va
%
T
VV:
N fi
4
_ (218.94 x10 °)(38!.7 to)
Tq
4
-
230.4
im
RV7x
xK{o°*
Vi
=
217
io
N
kN
Proprietary Material, © 2009 The McGraw-Hill Companies, Fhe. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—_
. 69 through
Problem 6.9
.
a
6.12
For the beam
and loading shown, consider section 4-1 and
determine
(a) the largest shearing stress in that section, (6) the shearing
stress at point
a
:
|”
O5m
Bees
etn |
:
ALS
:
C
Tae
8
-2.3 A 41.5 C72)
A =
At
Dime ATISIOTIS 31 UY
Section
Cobeodatemoment inertia.
TF
n= ny
Vet
aLkcsruo'']
46.957
Ae
=
kN
46, 457
7
VN
+ ala Cs \Ga''] + i (selene)
= 5.76 ¥ 10° wm
“a8
.
+) EM, = 0:
“aye
15,15 3}. Bo
_,
how
gm |
AO
A
wi
kn
=
5.76 x1o*
Lis
P
20
a
FT
S
A
At
a
REARS
tea
Qa.=
za,
:IZ tALidiiiid
2e
>
|
de
,
=
=
Ow
_— VQ.
= 30x%[0* mm
.
.
.
_ (46.457 «10°)(30 ¥10°)
It.
~
(8 76%/07*
gis eiot
yo
20xID* + 36» lo”
%")
x10
VQe _ (46.957xes10
~
I ty
he
it
t
At NA?
(GQ)
.
:
Po
SO
‘wr
m
o=
o.0307
BLIS MPa
t,5 Comm = 0.060m
Q, > Q,4 (co x20)(40 ) =
Qn.
030
ox)
wyhe
30x10
>
bi
min
(30% 20\(S0)
=
Te =
At
30
Tne
f, Dh =
(5.16 ¥ (07 )(0.060)
90 mm
Q.+ (aox20Xie) =
VQu
Thy
occurs
|
= 0.040
- (46,957
* [o* \(B4x 1S) _
(5.76 xto* 0.090)
at
am.
tp
= 66x10
e
—
m'
8.97 MPa
8.97% IO" Fe
m
66% 10 4- 1Bxlo?=
*
GE¥X1O° mm
“fo
24 IO men
1.61
10° Pe,
b,
(b)
= 84x10
Mm
= 7.61 MPa
Cnet ™ 8.47 MPa
<a
Tr
a
B.ISMPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, ‘reproduced, | or
distributed in any form or by any means, without the prior written permission of the pub! isher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.9 through 6.12 For the beam and loading shown, consider section n-n and
determine (a) the largest shearing stress in that section, (b) the shearing stress at point
a.
Problem 6.10
U3 in
4,
eo
12 in
E50 mim
12 mm
At section
Vir
I, + 4,
Ny
39.
Q:
©
mm
+
A,d,)
39.58 ¥10~ wm"
v
2A):
3C4.05 x1I9%
100
if
t
mm
rT
(1ox103 \(364.05 «10"*)
(34. 58107
Q
D
®
+
(loo )(75 (37.5 ) + (2)(S0)C12)(67)
=
t. - ¥@.
; 7
58x/0°
Ay,
i
oO
az bah,”
2a.12sxjo" +4] O.00mrJo° + 2.8566 «10% ]
i
(b)
4 (
(100) (isor + H)GENGONIAN + Goya yen)" ]
st
Cnn _ FFVQ
b,h,* +
"
a
#
3
hr
©
ES Se
kW
|. nat 200 mar ~--
n
|
L5m
lO
n-n
)(O.100 )
100 wm
%
0.100
m
= 920 «10° Pa.
Loong = TAO kPa, et
Ay + 2A,
(loo)(4o)(ss) + @I(SoI2)(C4)
3062.8
=
364.05 x10* m°
=
mm?
* 10°
+
4
Cio x1o* (302, 8x10)
(34.58 x10°* }( 0. 100)
mm
0.100
=
302.8% 10°°
m*
m
165% 10" Pa
Y= 165 kPa, —e
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.9 through 6.12
For the beam and loading shown, consider section
a-f and determine (a) the largest shearing stress in iat section, (b) the shear-
Problem 6.11
ing stress at point a.
250 mm —>
Mey
25 meni
O16
|
Jo=
Zoo
kA
E
15 mm
[
250 mm.
10 man ~~
200 EN J
15 mm
Part — |AGnoS| ddnnd|Aad® (mnt) | E Gam
oO
Web
ZLOO
Fhange
S780 | Up
5
(a)
i
= {03
=
$8-7B233
Q
LOA She os 7%
6
10254687 901398
TO32 &
[IS | 5177049TS|
Flange | 37?
T= FAs + ZT
.
mm 7
2°58 KO
111944578 | TOR eS
103 Sep 67S | FOZ GSE
LILO
re
LIZ
:
a
3
:
Aaye
& (ie) U8)(St) = Foif7e
&*
t=
fom:
4
Tome
~ VQ _pooyo2® JCFOMES
CUi-Lbxe®)Covoty
It
KIO)
— ~ 87 MP,
a
Poe; Mia
<a
—
a ‘®
+
AY;
= CBPIMUFS)
R®
(b)
-
Q
ea
7
|
As Ys
Ary, +
Qe
a (37.0) (UFZ) + Coj)ijo) C/O}
=
Lor
;
ASA
:
fem
3
fémam,
aaa;
O° ° Tt © * Cizefense5) (eet)
=
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to.teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without. pemtission.
§.9 through 6.12
Problem 6.12
40kN
mm
40kN
7
0.2m
|
12 mem
Seaton
100 mm
04m
For the beam and loading shown, consider section
a-n and determine (a) the largest. shearing stress in that section, (6) the shearing stress at point a.
By
symmetry
af ZY¥y =O:
Rp
“ Re
Ryt+Re-4e-W
= Oo
. ~100. ram-+|
03m OAM
Ra
Re
= 40
[A
AO
EN
From
40
“40
A,
NZAZLL
re
NY
NE SD Peps
MN
wy
ms
3
V
at
n-n.
Ary + Andy = (76 (12) (44) + 2(s0)U2)(28)
Qe
70128
=
oe
a: 4
)
tt
rs
~
eezarr
Qe
oe
()
eliag nem ,
Js Hoo\(on’- ab (70) (768 = 553152 mot
Y=
fear f(r
:
VQ
$
(a)
shear
T= &bh2- Fob,
t
*
the
”"
Vv CEN)
=
Tt
24 mm,
(40\(Te 128 X;07 7)
me
(SES BISEX 16") 2 B10")
AY = (leo) (/2)(44) = 52800
TH 2 OS Mh
~—eh
*
3
= {24/2 =2+ mm
VQ. 2 -
Lt
Bo
S28 osxre Fy
(SE8R1E2 x10!) (1024)
2 [Ge8S Mra,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribation to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
te
Problem 6.13
6.13 Two steel plates of 12 x 220-mm rectangular cross section are welded to the
W250 x 58 beam as shown. Determine the largest allowable vertical shear if the
shearing stress in the beam is not to exceed 90 MPa.
—
220 mim: —>}
3
W250 & 58 -
2 min
St
52.
Cabeslete
= 12mm
moment
Part
of
| AGEL od Gram) Ad? (10%!) | F Clo% msc’)
Top pla | 2640
;
252
ed = Seg
2
W 250x588)
j* 132 | Hs.989
F420
0
DAd* 4 SL =
Tine
OCCURS
at
ayis.
wee
t=
Pat
NG
® Top
45.44%
41. 444
—
B.0 mn
=
Amr)
phate | 2640
® Halkweb |
B.O% lO
¥en)
@ Top Flange | 2740.5 |
cf
in
mm
©
0,032
87.363
laxaao;
wm
Ay (10% me)
iB2
34g. 42
119.25 |
326.805
Ao | 5e.2s|
50.625
s
Dimensious
87.3
179.362%10b
mm! = 179.362% 10° om"
neviv?
L®
0.032
o
«Bot. phate | edo |* 142
=
|
|
L=
SST
inertia.
725.491
©
13.8 x ROB
@)
8.0%
N25
Q= ZAY = 725.41% lo” men = 725.91" 10% w
ye FEY _ 79, 36210) (8.0107 (70 x 10%)
y2>V¥Q2
725.41 » tO-*
Q
It
=
(V7.4% 107 N
Vzi22R.G
kN
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<8
Problem 6.14
6.14 Solve Prob. 6.13, assuming that the two steel plates are (a) replaced by 8 x 220mm plates, (6) removed.
6.13 Two steel plates of 12 x 220-mm rectangular cross section are welded to the
W250 x 58 beam as shown. Determine the largest allowable vertical shear if the
shearing stress in the beam is not to exceed 90 MPa.
(a
Cadcudate
moment
Part
;
7420
0
|* 130
et Adt 4 SE
Donat
Occurs
at
ayers.
|
t=
_ Peon
SD
rL®
BO
mm
F
8.0
Qe LAY
vy:
(Zo
®
SE25
Top Mange
HalFwe
var!
Oo
Bx
T=
400
326. 805
S0.645
.
220
J
@
13.5%
£OD
5
@
606,23
B.0
x25
|
ye LEZ _ (ie, 807x107) (8,0x(6* Vtoxlo*)
It
Q
G06. 23«[O7*
Vzi74.4
kN
=
removed.
87,.3xK/0"
mit
=
B7.3K/0. yn
Q = (326,805 + 50,625)*lo*
yx
Z2B.B-
mm = COG.23 X10" my
= 60623410
YB
phates
x10
2740.5 | 119.25 |
= 174.4 x1o"N
tb) With
87.312
Aro) |G Comm) | AY (lo® me)
>
mn
0,004}
54. UBS
® Top plate | 1760
@
|
Dimensrens
87.3
146.907 x10* mm = 146. BO7*1LO
=
neviv
io
O-
29.744
|
|
0,00%4
“24,744
130
Bot. plate | 1760
=
Le
*
1760
W 250x581
*d= 4s4 +3:
inevtia.
| A Guu | db Gam) | Ad? 08m | FE (0% mm’)
plate
Te
of
=
397,43x]0"
t=
me 3 =
b.0%(0°"m
377.43 K10
mm
3
Ete. (97.2x10°)(8.0
x10" )(40 «(0*)
Q
377.43 x {07%
=
(66.5% 10° N
Vtl66.Skv
<i
6.15
Problem 6.15
For the wide-flange. beam
with the loading shown, determine the largest
load.P that can be applied, knowing that the maximum normal stress is 165 MPa
and the kirgest shearing stress using the approximation 7,, = V/A,@,is 100 MPa.
O.6P
Rs
Draw
shear
ancl
benebing
[Vlwe= 0.6P
Law
o4
Bending .
ic§$
Pp =
W6é/ex (65
.
0.6 oePlas
Gun S
0.6
*
(o~ ode
.
Avot
{foe )
.
Pe
The
smafber
vahoe
a
[V
evonre
~
ENx
6S
LN
o. &
ow
P
Auer
Awee
Ae
6497
Cy 12°77)
FIRGC7
T
=
4220 KIO" Inher
dtu
La
Sheav
S=
_ (6S) (4220 xo?)
bag
ciianvawns
at}
For
5 = Mb
Cee
|
OFF
=
moment
Ml. 2 0.6 Plas
W
OEP
oO
-
2 IP
‘
~ 45k,
On
=
2 Mo
49
eb
f Pes the cobbou hte
‘tye
VIE 4
are
value.
FED
~
;
(292
inl
P= bys Li, =
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual js using it without permission.
Problem
6.16
6.16 For the wide-flange beam with the loading shows, determine the largest load P
that can be applied, knowing that the maximum normal stress is 160 MPa and the
largest shearing stress using the approximation ty = V/Awep is 100 MPa.
P
W360 x 122
Ra
j
K
0.6m
t
t
|
\
0.6 mi
0.61
wm
,
45 ZMg2
i 2.
Draw
Mp
V
aP
=
Or -3.6R, 4+ 3.0P4249P +18P
2PF
shear
=
= ©
andl
AP hig
bemePing
5
M.
=
lViw = 2P
momeut
Me
=
diagrams.
7
BPlag
[Mle = 3Plac
P
Mw
Bending.
For
W
360
«122
Ss
2010
x/O°
inva?
= 20lox¥lo* w*
+ Lai
“PE
3PL
es
M
* “Cn
7S
_ Gat _ (6oxlo*\(Zotexto)
:
3P lie
oa
A
B
¢ Bb
P=
Blig
Shear
Aves
E
=
=
BNt.€ 5
(
7
aly
=
=
AND
178.7»}0
N
(363)3.0)
xO me =
4TGAxIOMm
y= Whe
2°
p= Auk . doowo Wants )- cagwid W
Aust
The
smalber
valve
oS Pis the
aPforehbe
Aes
one.
3
P=
178.7
kN
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ail rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this mariual is using it without permission.
Problem 6.17
oo
,
Z4kN
‘6.17 For the beam and loading shown, determine the minimum required width 6,
:
knowing that for the grade of timber used, o) = 12 MPa and t,, = 825 kPa.
48kN
Z24kA
4
A!
mn
[150
j
Jy
A
4ABSka
ye
72kKN
Je
D ye
|
4
dp
tw
tw
lw
o.3m
/
_
2
|
|
AZM, = O+
. TER.
2kaNT
A=
.
e-w
Bending
-_=
GC
MSs
a
_%
.
lb =
Shear i
6s
i
_
6 \GBooxto®)
a ve rangudav
As .~ 4bh,
~
Vv.
PF
TE
otcurs
|
at
section.
‘Ll
-~ oot
gdh,
@= _ Ay=- gbh
Vibbh)
GEbLAL)
©
(37.2
on
|
2
io
bL=b
.
©
3M.
Abb
«10° )
(2\150 x10 (82S * 10% )
Zhe
f
A OY
Fon
of bendling
I= gbh?
S
¥Q.
b h*
t
mm
B80
=
stress
avis
newtval
vy-
_
(ise)
Masimun shearing
the
St
300%1O mm
%
ZBOOXIO* wn
section,
ve A angular
a
*
ator
Smin * es
For
E
-3.6
Nem
3.6%10°
=
kN-m
3.6
a ee
o
¢
B
A
= Z2*lo>N
2k
Vie
WA
M, kN
diagrams.
moment
bencling
amel
sgheav
the
ne
=
ap
©
~ 3A +QV24) + YV4EB) -O.S\V72) =O
Deaw
7.2
V, kN
_
05m
=
3
37.3 x10" wm
b = 87.3 mm
The
required
vatue
of
b
rs the
Parqer one.
bh?
87.4%
wien
ome
Problem 6.18
6.18
;
120 mm.
“eh
lee
10 kv
For the beam
and loading shown, determine thts nina
depth i, knowing that for the grade of timber used, ay, =
= 0.9 MPa.
Fatt
Ra = 20 LAV
Reaction at A
=
Vina
SOLA
(&j=
(io
Pood
Toted
EM,
2 £
Moe 2 RCEE)CIS)o BOT
Bending.
Se
Se
Mees
h =
les
[estos
Led bh*
al
Y
4
Ce
©
FR
)
GSOecq)
, B)lzajf
- 3¥ee
Rb Tay
davger
Va hue
of hh
is
the
4,4,
~ 3472
section.
266 & am
for rechangodan section.
& bh
Le pal
nh. =
fox rerbangudar
= Bae xo" = 2 bo 416) mm*
Y~
Shear.
& bh?
BNE
Pepe
bh
Ay = $bh°
- VQ.
BV bw
Te”
2bh
n
Cet
The
required
12 MPa and
fam
Mii MUM required
depth.
he
360+F
Pllet
Proprietary Material, © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced. or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.19
Problem 6.19
and the width b of the beam, knowing that L = 2 m, P = 40 KN, t, = 960 kPa, and o;,
¥
jonni ne
LQFJD
4
me
Be
A timber beam 4B of length Z and rectangular cross. section carries a single
concentrated load P at its midpoint C. (a) Show that the ratio t,/¢, of the maximum
- values of the shearing and normal stresses in the beam is equal to 24/7, where A and
L até, respectively, the depth and the length of the beam. (b) Determine the depth h
Usff
mm
|
= 12 MPa.
[bl—
ee I
Reactions=
R,
= Re
r
P/2t
2
~P/2
PL/4
Vinx
=
>
(2)
A
(3)
v.., = 2 Vir
OM)
Mo
Solving
forh,
bh
Ft
(5)
S
=
)
6. =~
Lin
(b)
OR,
§
(Q)
2
fon
rectang udar
*
3F
section.
for rectangudar
cs
+bh*
Mer
—S
for
.
vectangodav
equetion
b- 3P.
tht
(3)
=
section,
BPL
2bh*
h
he
Zhe Tm _ AVAMIeowle™) _ 225.15?
wm
on
12 xlo®
fow
b,
h=
Seluing
section.
@B)(4oxJo*)
— (F\(
vie 296
B2
0 x03
0 )
=
97.7xio*
B20
mm
it
m
bt
97.7
mm
=a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
.
7
Problem 6.20
6.20 A timber beam AB of lengthZ and rectangular cross section carries a uniformly
distributed load w and is supported as shown. (a) Show that the ratio t,/a,, of the
maximum values of the shearing and normal stresses in the beam is equal to 24/L,
where / and L are, respectively, the depth and the length of the beam. (6) Determine
the depth h and the width } of the beam, knowing that L = 5 m, w= 8 KNAm, ty =
1.08 MPa, and o,,= 12 MPa.
wh
Ren Roe
v
From
vue
wh
|
7,23
For
Ve
Ve
32,
32
Solving tov
Soluing
eg, (3)
he
hie
f,y
bh?
L@m
Lb:
2
L
.
Sh T.
=
| (4)
eves
section,
sy
= awh,
eq. (6),
(§)1.08x108)
.
diagram ,
|
eq. (3) by
Dividing
(ar
(3)
|
S= ¢bh*
Ahn
@)
oe
moment
a sactonqe tes
G6. F
Gy
We
A=bh
Swh
- whe
IMI,
a
|
ve»
neni
F vom
~ xk
Mm
(b)
[Vive
shear chiag ram,
For veclangutar section,
KO
~ oe
.
|
oe
&
()
= zh
|
~a
225 ¥107 > m
@d@xto45)
@ (225K j0°2 )U.08 * 10% )
61.7 ¥lo*pn
be Glo?ww
—
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved: No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
6.21 and 6.22 For the beam and loading shown, consider section n-m and determine
Problem 6.21
the shearing stress at (a) point a, (6) point d.
- se -160 mm
ISO EN
|
ce
He— S00 mam. me
300 maa —
20mm
|
100 mam
ld
Re,
30 win
30
202m
Draw
Fok
the
shear
diagram.
WV pra
2
Go
kN.
Part | AGuwt)] GGmm IAS Clctar?)] A Cre) Act? Clotmm')| I (le me)
-@
12200 | 40
233 | 25 |
-@ | 1600 | 4a.
40.
|}Goo]
®
= | GHoo
|
GY
|-45 |
64
|-25 |
2,000
|10.J067
I.ve0 * | 4.9533
t.000 | 6.9533
41g
4,000
1. 2133
7 ZA = HISxto®
T=
(@)
Ad? + ST = (40004 1.8133)¥/0% ww"
= £.81383«{0% mm? = 5.9133«1lo% mi
[600 mm”
Az (80)(QQ0)F
25 em
Y=
a= AZ = Yoxlo* mm? = 40x10"we
y=
¥ Qe. (40x10*VYorio™)
a,
Th
8?)
oxio
\Q8
9*1
~ (6.10"
V2
tb)
Az Gola) =
yr
68
alro le
le
Lgl
Ce
Qy =
AN
Ry
4
-!§
AY =
Yr, = V Gh.
b* TH
=
= Bloxlo’
MPa
31.0
é
Pa
<a
B00 mm*
=
SO
mm
ne
BO x1O% mat = BO x0" mn?
(Foxto \Goxts*)
Ce81g8x1O IROxio?)
TT
=
= 23.2«lo Pa
23.2
MPa
—dt
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc, All rights reserved.
No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
e
LOO EN
6.22
.
6.22
For the beam and loading shown, consiiet section a-a and determine the shearing stress at (a) point a, (b) the shearing stress at point b.
$100 KN
~~
KR,
180 mm
a
it
a"
, 500mm.
2
a
|_|.
:
;
—
500 ram
20 mn
Lae
|
50 mm
*
Re
=
joe
KAS
—
ut
i
38 mn
.
At section
;
nn
Vz /ookN
0 ru
/
<---200 inin-—~
Locate
cent rorel
ancl
moment
G
124208 | 170 | S44¢Q0e|
4~Gy~
@
7JLOo
Fo
OS
Tne)
Ad’ (m im’)
ACn)
$$-4 | 821812 | 10 b6eT
[AIST G21 9440000
L48 006 | ZaeG
Leer TPAEG LIS EEGET
HALGOC
|
(G4E6
Y= Re
2 UN 4eb mm
4
7 = 330725 XO
+ 45-666
T= ZAd*+ ZT = [4/186
la)
Ay
Qt
Uoeb
= (co) BE) (146-19)
= 72666
trav
l tf
ve ED
*
t
=
w
(b)
*
re
Ay(nm’)
ro
}
L
of inertia,
¥ (iam)
Peat | ACra)
?
9
caheu bate
—
>
“
aks
*
3G
>
vty
VQ
.
mS
TE
Q,7
Ay:
to
20
‘be
°
TE
MPa
~ pared
(33-728 x10¢) (20)
(20) (76) (litte b= 38) = 1h bY 32 ot
mm
.
)
(foaso0) (1/6432
VQ
ne
ss,
MP
(feeoeeK( 72666)
_
=
(3-728 x10) (20)
-
e
“Fa
IT 2 MEG
att,
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyand the limited distribution to teachers’
and educators permitted by MeGraw-Hill for their individual course preparation. A student using this manual is using it without permission. ©
Problem 6.23
6.23 and 6.24 For the beam and loading shown, determine the largest shearing stress
in section n-n,
[100
ISO kN
|
nun
|
_ 20am
100 mm
]
—
500 nim ——
| _}s0
mim
500 mm
30mm
\
30 mm
20 band
Dryw
the
shear
diagram,
Vax
qo
ky
Part | A Cun?) Flom Ag Cio’ nird)| d (in) Ad? (1o%m!)
x
©
©
|3200|40 |
|1600) Yo
£
bfoo
Fo Ww
&) | 1600 | 40
Teak _—
|
gy
ap tap
|
|
EAR
CF wm
=
$=
4
T (ont)
|
29¢ | 25 | 2.000 | 0,J067
649 |-25 | 1,000 | @,8533
f,000 | 0.8533
|-2as |
bY
Ale
Yoco
|b. 9133
Yb eto?
BH ete = SF mm
Da SAd*4 EL= (4,00041.8133)x 10° innf
8133 x10% rm =
Port | A Com?)
O
2200
®
©
300
300
133% (0° mt
¥ lm)
Ay (10* man?)
425
go
7.8
7.5
=
24S
2.25
f
834.5
Qe LAY = BESO meek = BES KIO” mn"
L
=
AX 20)
=
4¥O
mm
4Oo
=
x fo’
m
rae BR = eM MEI connate me 227 Hh
3
6
‘
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.24
6.24
Vor the beam and loading shown in Prob. 6.22, determine the largest
sheating stress in seclion a-n,
6.22 For the beam and loading shown, consider section #-2 and determine the shearing stress al (a) point a, (#) the shearing stress at point &.
TOO KN
GON
rr
=
Re
=
(60
kA
__y8S nim
mom
Lacate
centvoted
ane
f-
calcudate
moment
n-rn
Vez
foo LEAS
inertia.
Almm)
€
Z20d | 170 | S44gve|
EG
Ylmro)
of
At section
Payt
en <b
t
Hoe
alpen
Ay (nm)
[7r00}
Fa | 649000)
(6460
HG26Ce
al (inn)
Ad’ (nin)
Tm)
She4 | W21biz | le be eT
2466 | 49ST 621944 C000
pT BEY
j3s46o6']|
y . ZAY . 1197008. 2 [1Gsb vm
= SAR
oped
T=
Langest
shearing
evtire
cross
stress
SAA
occurs
7
r
+
Ql ih bm
feo
A
=
on
= [¢/ T8464 + 198
section
through
bC6 7 2 336725 ¥ roe
centreted
of
section.
|
omIn
v=
[fe
=
Det
prinl
avdy
VQ.
It
a
= [xa)(tee O)( TET) = IZ 13 3KG
Ay
Qs
va
+ SL
i
llecese) ROD
st
3
2
‘
(33-728 Kip
©) C22)
gg pa
-
me
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.25.
..
+625 through 6.28 A beam having the cross section shown is subjected to a vertical
oo
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (6) the constant & in the following expression for the maximum shearing
stress
yt
€ sax
7”
A
where A is the cross-sectional area of the beam.
A= &bh
For
a,
I=
x bh?
cot at
Pocation
Ye
AG)= 2(B y= b¥
YQ)?
Zh - Sy
Qty)= A= BY (h-y)
tTly)=
by4
Va.
V *h-y)
find
ar
®/
.
Tn
-
. AA
Maximum
of
deo
cation
>
ig
(h- - 2yn)
12V
bh?
(hy,
- yu
of
Tr
5
YuF
=
)
=
WVV
bhS
Vy Ch-y)
bh?
.
(3% bh4) Sy
Tt
@) To
an
[ah
set
+ h
x
hy]
-
eV
(hy- y*)
bh?
—
=
a
ie
= cm
Oo,
mrd- heeght
at
_
“
3
3
x
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ai
Problem 6.26
-
-
6.25 through 6.28 A beam having the cross section shown is subjected to a vertical
' shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (5) the constant £ in the following expression for the maximum shearing
stress
V
=k
T nox
A
where A is the cross-sectional area of the beam.
ol
A= 2(hbh)= bh = Teadble)- £bh?
For
a
cut
et
Jocetion
Yo
w heve
ys
h
Ag)= £0y > Be
i
E
po
te
4
yQ)*
VY
. h-§Fay
a
_
Qy>
by?
Ay:
L
-
ay
3
tty) = BY
VQ
Te
Tiy)=
(a)
To
Find
the
At
—
d
Y
wm
lyds
\66...
"CR
yv
a
Poet on
V
.
Bhe
r
ae
by | by’
By . V_ | 3(¢)
by 27 3h”
of
4g
.
het!
"0 |
=
a
mum
ot
AL
set
h
.
TL3@)- 2G]
©)
Ye
0
°
a
Bo
4
.
-
FE
- ay |
Obh
= urask¥
r.O,
wo
+4 a h
From
e
nevived On
S 6
eens
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
to teachers”
and educators permitted by McGraw-Hill for their individual course. preparation. A student using this manual is using it without permission. ~
—_
=
Problem
6.27
6.25 through 6.28 A beam having the cross section shown is subjected to a vertical
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (4) the constant & in the following expression for the maximum shearing
stress
V
Tmax = & a
|
semi civede,
For
+J
'
Qe 2
-
(a)
Tam,
Tin
)
~~
VQ
-
Ay -
-
Zc
He
O° + RG
-
center
ay
Bre
en
Hort
Problem 6.28
£
at
z
t
-
ocevrs
v-$c3
sl
Ag
-
35
3c
£3
where
SA
=
—
Y*
2
oO.
Te*
=
A
and
¥c*
+
i
be
where A is the cross-sectional area of the beam.
2M
f=
V
2c.
—
|
k=grl333 0 —
ste
6.25 through 6.28 A beam having the cross section shown is subjected to a vertical
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (5) the constant 4 in the following expression for the maximum shearing
stress
=k—V
Tax
A
where 4 is the cross-sectional area of the beam.
For
a
thin walled
Jz
a.
Avo =
—,
20k,
y
>)
(b)
a
For
VQ.
Dornre "TE
°
Tea
sewerrcotayv
Ag
7K ty
B=
2t.
N(antte)
Crean)
aTrvyty
A=
section,
cinco far
+= weet,
Q= Ay
at
nevtred ams
5
¥
ave,
Vo.
~ THER
=
= Afr
Yr, t,. ak
- 2vit,
where momimum oecurs,a
&v
OA
k= 2.00
th
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual! course preparation. A student using this manual is using it without permission.
Problem
6.29
6,29
,
The built-up tinther beam is subjected to a vertical shear of 5 KN.
Knowing that the allowable shearing force in the nails is 300 N, determine the
largest permissible spacing s of the nails.
50 rum,
I,
250i
= tbh? + A.A.”
= of (50)50)+ (50)(SD) Llow)” = ZEE20833 bork
.
FR
fr
167187449 vk
H
H
=
> £8104167
= iH (60) (250)?
wie
Q, = Ay, = ($8)(S0 )t00) = 2Sdowe mH
VQ, .. (S000) (260 ows)
GB
e~
°
Feat
a
rz b, hy
tH
J,
q
Lena
= F-4p
167187499
tL
nm
SF
Fraie
3
a
°
Beg
EE
~
F
,
Lash
.
M inom.
anti
nin
C
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution 10 teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
-
.
6.30 Two 20 x 100-mm and two 20 x 180-mm boards are glued together
as shown to
_ form a 120 x 200-mm box beam. Knowing that the beam is subjected to a vertical
shear of 3.5 KN, determine the average shearing stress in the glued joint (a) at A, (6).
Problem 6.30
20 mm >
[ 100 mm >
at 8.
none
2
T=
i (120)(200)" —~ £(go\(ico = 52.693 ¥/0% mm”
= 52.698 x/O* m*
180 min
LLL
(a)
Q:
40 man
20 mm
_
|
tn = (R)(20)* YOum = 0,040 m
|
CLL
(go)(zo)(Fo)= 14Yx 10 ae.
= ING do me
gy 10"*)
VGa _ (Suto
T=
LID,
A
T
La
~
(52.
6I73 xO
DO,
= 239 x}o0 Pa.
(b)
Qe= (120
2040)
==
O40)
= 239kPa, <a
ZIG «10% mm? = 216 KIO
mS
tg= (2)(20) = 4Omm = 0.040 wm
VQa. (3.5%10")(21g% 1S)
T°
Th"
_
(SeFaxlo*)lo.o90) © 997"!
apd
Pa
T=
Problem
6.31
T
oC
00 mm
1, “2 ~
O!a,
15050 mm] m | L6100
0 mmm
s
fe
100 um
72
3
yy
k .
Ad:
a
ss,
~
=
L (So Yoo y*
+ (geNo0 Vas)"
,
50 min
50 mr
I,
50 mm
22,+
20,=
Q=
Q =
AY,
:
go OP
qs.
4
=
ip bh®
=
(50) iso)
= 14,0625 « 10% mm”
Te
z
~ll
§.31 The built-up timber.beam is subjected to a 6-KN vertical shear. Knowing that
the longitudinal spacing of the nails is s = 60 mm and that each nail is 90 mm long,
determine the shearing force in each nail.
somafe
50.7
354 Pe,
92.71 %10° wm
= (GOM100
aN"
4
‘
Frit =
op
=
92.71~10% m*
75) = B75 10% tam
aS
24.2710" N/m
= 375 10% m*
S* Gomm= GOxt6?
YSFOS * FYROM
DAT
APIO” 6Ox1o*) = 723N
—<s
6.32 The built-up wooden beam shown is subjected to a vertical shear of 8 KN,
Knowing that the nails are spaced longitudinally every 60 mm at 4 and every 25 mm
at B, determine the shearing force in the nails (a) at A, (b) at B. (Given: £, = 1.504 x
:
60 mm
RS ram
Q.2 Q,=
(2)
0.060
m
O.O25
m
Ay, = Go KI0e
750
{07°
se) = 750 *J0* mm?
mm?
4
a
= [SO4riO~ m?
it
1 S04xJo' mm’
¥
x
i
400
wv
10° mm*,)
if
300 ~~» 50
On
[50 [+
6.32
a.
Problem
Fy
=
$a
Sa
VQiSa _ (8/07) (750 «10° (0. 060)
|
|
|
>
(b)
>
RQ,
+ Q,
H
Qe
_ VQSe
I
me
IISoY= 264S% IF me
Q,= A,y,= (B00)S
i
[-
1504 «xlon*
7
To.
2349 N-
_ (8%10*)(4125%
107% (0.025)
1504 x 1O-*
HI2S «107 mm?
4125 x loo m*
ere
Rr S4tN
=e
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.33 The built-up beam was made by gluing together several wooden planks.
Knowing that the beam is subjected to a 5-KN shear, determine the average shearing
_ Stress in the glued joint (@) at A, (b) at B.
Problem 6.33
b, = 16 vam
hh, SLA C4416
F
a
.
= Imm
mm™
46
S x 10%
Tis tbh?a 1.17
i
b,= 30480430 = 14Omm,
a,
AA,= = b, h, == 2240 mm 2
Dimensions in millimeters
ab
the
21
I:
(a)
TIT:
I
2.63179
421g = FEUIHIOwm" = 4, CALI Ow"
*
Ae
Q,>
Ty t
z®
1G.2%12 mun = IF.2 m0"
m
16 m
0.0
=w
IG
CF. 6224 x107°}(0.016)
It
= C24
= 624» 10* Pea
(80)0IG)=
¢
-
40
(A\CIE) = B2mm
-
Ite
_ Go
(4.6229
kPa
mun
1
Ye
+f
attr PY
9
RaYa = SI-2*10° mm? = S12 *10° wm
Vg
te
1280 wom?
m?
V& - (Sx10* V4. 2 #107)
ee
a
(b) Ag=
K1ocmm?"
A,= GoXie)= 480mm > Sy = FO mn
1
1
=
A,d,
h,? 4
|
Se
2
i
~
lz = 16 mm
= YOu
64,1 oy
Ste
= OL.OBIW
VS12x1O)
od
_
10% )(0. 082) —
BS]
lo
Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<8
Problem 6.34
-
:
6.34 The composite beam shown is made by welding C200
17.1 rolled-steel
"channels to the flanges of a W250 x 80 wide-flange rolled-stee! shape. Knowing that —
the beam is subjected to a vertical shear of 200 KN, determine (a) the horizontal
shearing force per meter at each weld, (8) the shearing stress at point a of thene flange
of the wide-flange shape.
Az 21TO mon,
For C200K1212
channed
the
2
b=
0.533 #10" mm!)
ly =
For
man”
10% »
tpt IS.Gmm, Tye 126
ds 256m,
80:
Fen W250
oA
mm
beam,
composite
the
4
FT. %mmt,
170.6 mm
Je¥ BES 4.57 - 144 =
Few
=
tps
beaw.,
composite
the
in
57mm,
Tz 126xjo%4 2[0.538%10%4 QI70\LI70, é) *]
253.
=
(2)
Fou the twe
Qy=
=
34 ¥ 10° mot
253.39
mt
x10"
wees ,
Ay? (2179)1170.6)
= 370,20 103mm = 370.20 x10 * m?
_ VQ _ (200I10*) (370.20 x10")
%°
292 «10°
H
.
153.39 «(O°
.
I
N/m
|
aa
teal
HZ
[12 may
For
LT
ew
Fou
-
weld ,
d
=
IH6.1%107
Ni /m
Ye
mm
Shearing force per metev of wet:
|
tb)
one
at
cots
a
and
a!
togethev,
Aye Q(t AICIS.6) = 34944 mm”
BE = 120.2 mm
Gye BE
.
Qn 2 370. 20¥10'+ (2464.4 )(120.2) = 790,23x1O° mm” = 192,28%10
Since there
ve
a
ave
V&
THE
evts
<i
[46] KN/n
ata and a,
E22 tp 2(2\IS.6)*
109)
23 90.
(Qo0x10?\(71
~ @53.39%10°)(0.0312)
19.99
-¢
m
31.2 mm = 0.0312 m,
10° Pa
U%,= 14.99 MPa
~<
Problem 6.35
6.35 Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa in the hat-shaped extrusion shown, determine the corresponding shearing stress
o
at (2) point a, (b) point b.
[~
40 nim --»
Neutvad
6mm>
60mm
ox
pies
30
4 mm
a|
a ——
fee —--
20mm
| ag
28mm
mim
a,
T.
?
Th
2~
a
Q.
E°
te
Qh
rom
bt
3
t
7?
°
Us.
bottom.
|
2, = VO
Tt,
>
Q
NNo
©
te. =
A
a4
Q.=
Ce
(6)(30}05)4
6
Lt
t
tat
ee
ic
Ub
04)(4)(28)=
mm
4260
:
mm
.
UNCC) =
=
Q,
-
.
oR
co
¥Qe
Tk.
above
$$
20mm
Ab
A.
Ls
Ce! =
muy
1568 mm"
mim
aN
min
La
Q©.” . beth
wo
nN
et
2
ft
xS
~~!
a
4]
t
42to
-
[868
We
4
R
7
3) s&
ot
ROR
I"
oan
%tt
=
Oo
Q, = (0464) (28) = 1568 me
.&
Ss
=
x=
414
MPa
4hY
MPa
~ath
Proprietary Material. © 2609 The McGraw-Hill Companies, Ine, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.36 Au extruded aluminum beam has the cross section shown. Knowing that the
vertical shear in the beam is 40 KN, determine the shearing stress at (a) point a, (b)
point b.
Problem 6.36
Gram
i
60 mn
~~~
mm
|
|
+100 mm
:
—F2 min
|
30 mm
.
|
T
=
~ y2)*
8 (t00\(60\* — iz ( teo-25\( 66
7
= {109800
(@)
mm?
Qar A,F, = Yo0~251(60(30- zaby
= 1215 frm?
a
Q
ZS
|
Ta
A ©
><o
Le
te
a
=
(Q)CE)
_
VQ.
=
TE.
= {Zinn
:
Gore \G21 50x16?)
Ceadooxid*) (eaiz)
a
Ta.* 3b5 MPa
ere
I,
(bo)
Q, = 121504 (arizy(2o22V)
= 22950 mm’
t,
4,
=
@)UZ)
VQ,
Tt,
= 24mm
(4ox10°) (22956 Xo 4
” (¢108800K16'7 lo. 02.4)
tT, = S4ASMPa
«ea
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manua! is using it without permission.
Problem
6.37 An extruded beam has the cross section° shown' and a uniform wall
thickness of 5 mn. - Knowing that & given vertical shear V causes a maximum
6.37
shearing stress 7 ~ 60 MPa, determine the shearing stress at the four points
i
t
1
1S may |
15 rans:
indicated.
15 amar
Q,* CE VES VGIE =~ betS)
S
[O83 512.0 me
15 mj
1S man
Satan” 38 maa
constant,
Qa=
Qo +.
to
Us.
+ ber zh
Sr
2)
24
* ESP
Aa ie?
.
+ BE SEIGHS) & BF Ibe S 65 prot
:
3.
Q,
Ue
GSR iD
ta
nq 448
607 M fas
Problem
VMAS
x
is proportional
TF
CEOS
Qy 7 20+ AO, + (TEMS CES) = 18318. O16 mm
Qu = Cys CSW ZEW le) = 19 BIS 626
ha
ve
Since Vi iTyand t
Care.
Qy=
Bape. 8b3
The
GeimPa 3
6.38
Lin
1pai5- bob ~
Tr
be
a
or
ohbMfas
Tye
Sbe4 MPa
act
6.38 Solve Prob. 6.37 assuming that the beam is subjected to a horiyoutal shear ¥.
637
7 nmap
15 mm
sand
An extruded beam has the cross section shown and a wiiform wall
thickness of 5 rom. Knowing that a given vertical shear V causes a maximum
shearing stress 7 -« 60 MPa, determine the shearing stress at the four points
“
indicated.
15 moun
a
1 mut
Q,
i
eae
Ti
=
Cf2: BYES
5
Sy
224
9G
pro?
Qy = (ASM EVGSHYE 2219-15 ten?
Qe? Qa + Q, +O MGs-s\C20)
Qa
ve
Since
ee
22B Ik
Te
a
Vv,
I,
=
anel
he
Eh
PURIST TS
tbe 7M Pa 3
The
0
t
Qe
AVS
«
iG TMPe
constant
.
a
O
,
T=
,
bm.
Q.,
=
& THIS inna?
GL
Y
Ss
.
ba MPa,
a
19874
mbm
proportioned
;
tes
O..
be
TFRTE
Tr
2 oO
«iit
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribation to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
-
6.39
,
,
6.39 Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa an extruded beam having the cross section shown, determine the shearing stress
at the three points indicated.
——a 120 enw
——mane
[
L
.
vt
It
>
kT
QQ?
(20)
tet
a
[
7
Ly
/
R2SO
:
..
(Q/t),.
te
Qn/ba
we"
Q.+ QOVGOMSE) = 17.5 «10% mm”
RO
mus
.
4
=
3375
mm
1RO mn,
EAE
a
Qlt.
Git 2Q, + (ael(aollis)= 20910 mane
.
;
+o
10mm
Q,/t,
Point at
propertionad
af
ot
_
oo.
e
"f
Eel
ft
Je
Be
is
-
(7s? © 22.5% 107 mame
Q./t,=
4
;
.
.
VQ.
i
50
:
Q,/7t.
eceurs
7
1741.67
at
by
Tr.
mn
= Tr = 7S MPa
te,
-
QA, /t,
ITHLGT mm” ~
7
Oe/te
BB7S mm*
~ — AASO mm?
Te
33.7
MPa
t: 75,0 MPa
T7
<«—_#
<a
4B SMPR
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.40
.
6.40 Solve Prob. 6.39assuming that the beam is subjected to a horizontal shear V.
6.38 Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa an extruded beam having the cross section shown, determine the shearing stress
at the three points indicated.
(a
¥e
TZ is
proportion ad to
Q/t
Point 6? Q.=(loWao\as) = 7.5 %10% mm
Ce
=
Qe
yf
o,f
Porat
bt
[O
/ be
mm
=
Q. = Q.+
Dinensions in mm
ty
<
20
750
wun
GolGol(so)=
57.5 x10" mun
miner
:
.
@,/t, = 2875 mm
=
COR
GQy= 2Q,% (goX6oXso) =
Pointa+
GOmm
=
te
N
223K 10% mm
(Oat = 3716.7 mn”
Tif)"
WY
occuvs
(Q/t)\,,
~
Ca
Olt.
~
spa
S116.7
mat
at
in
Q/t,
Te= 1S MPa
T=
a.
.
2 Me
BIS mm
Te
Q/t
VSO
ma*
7 75.0
WRa
—
%, 2 58.0
MPa
a
T, = 18.13 MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. .
<=
Problem
6.41 .
6.41 The vertical shear is 25 KN in a beam having the cross section shown. Knowing
that d= 50 mm, determine the shearing stress at (a) point a, (5) point b.
5 ayn
fT
£20 rani
I, = kOe
(Soe) se)" = |.25653 410% min”
I,
=
E72 \eo)*
x {0°
men”
21, = 15.3933 x/o' on?
41+
Dz
——
5,184
=
1S.3433
x /o"*
mn
¢
Q,= A.y, = (S0.e\se)= 22. 4 ¥ {07 we
=
\
©,@BSXS
co
=
7}
)
Q.:
Q,
|
b=
pe 2 Vn.
“
Tt.
Qype
2Q,4+Q,
Ly
72
mm
It.
%
a
72 xlo”*
:
127.6107"
mw
Bxl/Ooum
4 $5 «10% Pa,
wm
Koz
4, ss MPa,
it,
3
My
:
174. 4107 )
(15, 3933 xlo*
|
=
129.6 «10% nm.mm
— G8.3438+}0°)(Bx
Jo-® )
THY IO
(45 «10*
VO.
-
=
Bim
mF
BSxI0*) (22.4 x10)
=
(ob)
= (72)(eov (20) =
Qt Ay,
—
22.4%/07*
72*to~3
+
oe
=
3.93x¥{0
.
Pa.
T, = 3.43 MPa
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or”
distributed in any form or by any means, without the prior written permission ofthe publisher, or used beyond the limited distribution to teachers.
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.-
|
6.42
Problem 6.42
8mm
+
Q, = (d\s(96)
Q.7
: 1|
ann» fe~ he
|
|
8mm
=
2Q,
=
72
+ Qy
B46
J4
129, 6 */oO*
min
mem
.
CF
443
laa) + GYaXsey =
I, =
t= B mm
Tt.
Since
@)
124.6 * 10" mmm?
=
_ Vx
Ve. *
mm
Wed
QA=
Or,
Le
(bo)
448 ad
Qa= (AVGOMBOY=
Wl
120 in
alee
The vertical shear is'25 KN in a beam having the cross section shown.
Determine (a) the distance d for which 1, = ts, (6) the corresponding shearing stress at
points a and b,
Tr
d .
os
_
Vs
TT, =
Ite
*
oh
346 ol + 127. Hto
41.3
mm <a
» wm®
1.03696/0%
=
25.18) d
A=
T= $a\coY = 5.1844 10% mm?
T=
42,4 22, =
Qa
44B
t=
d
¥@e .
E.
(4.5158 «105 mm?
= GBI. 263)
©
= 14,5158%10
% m”
IB.48S7 x10 mm
(28 Jor Y18,4357«10%)
(4y.SIS8*to® )CSxlo*)
_
=
1B4RS7TRIOS m
.
3.98
lo” Pa
Vo.
=
3.48
MPa
Proprietary Material, © 2009 The McGraw-Hill Companies, Tae. All tights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the.publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.43
6.43 Three planks are connected as shown by bolts of 14-mm diameter spaced every
150. mm along the longitudinal axis of the beam. For a vertical shear of 10 KN,
determine the average shearing stress in the bolts.
100 mm
125 mm
|—
==
125 mm
fant Fp
gn 2
|
Letate
100mm
&
‘peutpal
Moment
of
ars
cand
compute
inertia.
250 mm
12500 | 200 | 2.9%10° | ars
Nd eons!) | Tamm")
yisietxio® | 0.446710"
zsooo |
125 Ia.vze «10% | 37.5
fas.ise xi0"] 130.208 x10
42800 |
200
LAC vom’)
©O
|@;
Y (own)
AY bon
| 2.5x10¢
50 000
iF
—
¥
LAI
=
“Sk
=
)
A (eves)
|
37.5 | 17.576) x10%] 10.4167 x10"
8.125 «10°
B.12S*10°
“so
So viet
ee
=
162.5
70.312 10% [VBE BAT KO"
.
mm
T=
YAd*
+ ZTE
aU
_¥
221.35 % 10° wm”
Spi
e
Roo
162,95
J
b
$
= (12500)(37.5)
- V@ _
463,75 «107 wm
—46B.2SKIOe Mm”
Cloxto*Y¥4ea.75%107°)
Ir
*»
5
fF
Ay,
*% 107° w*
y
221.35
221.45 x%x 107°
_
~
[177 "10" W
Fu = gs 2 Cal. 177x107 )(180 10°) = 3.1765xlo"N
Aso
7
3,
Bayer
.
Tas
Aust
Is3, 928 mms
=
ey
.
176S
1s3.
.
lo?
1s3. qeBxIoe =
439 «lo
20.6
v lo”
A
t
Pa
Tin = 20.6
MPa
ait
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Al rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.44
6.44
A beam consists of three planks connected as shown by 10-mm-
diameter bolts spaced every 300 mm. along the longitudinal axis of the beam.
Knowing that the beam is subjected to a 10-KN vertical shear, determine the
average shearing stress in the bolts.
50 nm
ous
a
|
50 rap
“1350
mra-
Lagoa
t
j
/
.
.
T5BO | 25] Seasv0|
{667 | 2091676 | 196 b2 86-0
50 mim
7566
(25)
133-3 19376675 | 162560
4
Tec
| 75| 562500,
isOmm
7 F
187506
22500 |
Locate
[7
131280 @|
neutret axrs
[2671678 | [FoG?2suvea
[12S x10l
and
compote
19969
X 106
moment of inertra.
Part [A Gino) ¥ Crm WAY pel dt Com) Ad (nH IT Cine’)
0
©
fo
®
Y
'
Oy
12
3
36 | 0467)
&
\2
'
1a
2 |
36—
J.
2883
16
:
me
L= SAd* +51 = 42:19 x10% met
©
Q=
A,
~
y;
=
Gatos)
Font
.
Att
weg
=
Fntt
Abit
4
2
gS
(vc)
2
£%- 3)
12£ 256
©
42 1Fcrob
24. TN Emm
(29°F) (200)
=
be
2
£ Fre
1d LY
BAG)
KAY
2
.
Bu
Via
2
F
"3a
16
VQ _ Go) (s28256)} |
~
Avot
Ba
aykoo
SR
Y=
0.667) 6.333 | 3¢
84
(2/2560
ZA.
4
21.333
1.333;
| 3¢ |
i2 | 3
5.333 | 36
FF: SY
fur
fi 34
Ma fe
ty
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their iridividual course preparation, A student using this manual is using it without permission.
Problem 6.45
6.45 Four L102 x 102 x 9.5 steel angle shapes and a 12 x 400-mm steel plate are
bolted together to form a beam with the cross section shown. The bolts are of 22-mm
diameter and are spaced longitudinally every 120 mm. Knowing that the beam is
subjected to a vertical shear of 240 KN, determine the average shearing stress in each
bolt.
mm
Angde + A= 1850 mm,
d=
200-29
I= 1.93x10% mm", y= 24mm
= [Timm
Ty = T+ Ad*
Qr
¢°
Fa
41,+
|
1,
U8SOMITI) =
V@
= 58.7926 %/05 mm"
&Galwooy = 64/05 mm!
|
4
T=
Ie
tt
Phdet
|
287.7 %1O%
mm = 287.7x10° m”
4
12
316 .35%])0° mm * + B16.38Sx]0. wi
a
ce
= (Ao nloP Jie. BS ue")
%
400 nm
|
=
268.9 10% N/m
FT gs ° (263.9x10°)(1a0¥to*) * 31.668x10"N
Aun = Bed?
= Bea) $80.13 mm = 380./3«/0% me
4
LC.
Fist
bot ~ Ky
Aun
— 81.068 %/07
Pr
380g ripe * 89-3 x10
é
Toe =
83.3
M Pa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manuaf may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
—_
Problem 6.46
6.46 Two 20 x 450-mm steel plates are bolted to four L 152 x 152 x 19.0 angles to
form a beam with the cross section shown, The bolts have a 22-mm diameter and are
spaced longitudinally every 125 mm. Knowing that the allowable average shearing
stress in the bolts is 90 MPa, determine the largest permissible vertical shear in the
beam. (Given: I, = 1896 « 10° mm‘).
-
THrErE
Flange:
*
qe
23,431,341,
SS Sy
: SSS
ees
S2EFa-,
Viens,
9",Ax
M6 ¥1O%
= 44]
mm
oe
= due
mm
I +Ad*
|
.
HG (0% + EYROV(BONY = 187.4 x10 mm"
(4$0aoVlass
2
= Elza)" =
380.1
mm
=
Fut = 2%A, 129
10%) (330.1
0»
x 10°) =
= 1896 «10° m*
=
2NS*)O°
mm”
s
_
68.42/08
ona
0018S
547,36
xfo*
380:1 «107% ,,?
-
68.42% )0" N
N/m
va I tn eats pd
I
1896«
107* ($47, 36 x108)
u
Fett
Ss
qe 8 Et
pM
A= SH20 mun®
{BO b
2AS- UNF
(S420)(180.1) =
476 */0* mm*
Qe
Q= Qp + 2Q. = HOCTxIO*
mm = 4HO67x(0 om
x
.
151.9 10% mm"
>
1896 xlo* mm
Qe
Li
Aw
=
a
Tr
ib God(480)® =
OT
Ce pul
T=
Me
I>
Ps3
£0)
"Web:
Ip = + (iso Kacy + (asoY(20\lass)* = 497, 3xj0 mm!
=
255
kN
atl
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
8.47 A plate of 6-mm thickness is corrugated as shown and then used
_as a beam. For a vertical shear of 5 KN, determine (a) the maximum shearing
6.47
stress in the section, (6) the shearing stress at point B. Also sketch the shear
flow in the cross section.
dD
,
40 tar
,
Leo
:
Leo
ft
i
im Tithe
ne
ate a nbeso
30 aim 30 rim
Locete
peutval
= fl2ect
+(+0)*
7
mm->|
Aan 7 66)
0 50) = Boo mak -
axis
ana
Compute
moment
of
ft Eon
Cs Ay= (mn?men? d (med. |A a af, *Cnn*}
WY (ava)
Part Aim NS
AB
gob
oO
BD
SoG
ae
DE
50
2o.
EF
See
-
fo
Zaova
neglect
SEG2
ic
Zoocda
¥ AGQ000
Baca
fo
ge oue
* 40006
oO
{Zoa0ea |
[2600
jZoo
|
+
Qhe
>
(56)(6) to) =
Qac
.
iasy (6) C8)
3
27
S
in
we
Soro
Aun
3275 m™ we
.
Pe
V m
Qu _ leo) (2278) epee
i="TMQ
Fa
(Bx0e0)(b)
t
(bY — Qg=
Qyg
* Bove nm?
Ts
T,
Ve,
-~ Ft
7
O
(802002)
6b)
(200050 )
=
sy ieS
foe
+
Meu ged
Que
=
AGOGO
.
2Ad
j=
Baave
Qha
£ (3e04aey
iq
=
>
ten.Ae
Se
Qe
= Jo me
Ay - CALE
Fe
neghed.
2 eo0a
Lo
oO
oO
ietin.
p
MFq
ar
thm
me®
6.48 An extruded beam has the cross section shown and a uniform wall thickness of
3mm. Fora vertical shear of 10 KN, determine (a) the shearing stress at point 4, (b)
Problem 6.48
the maximum shearing stress in the beam. Also sketch the shear flow in the cross
section.
,
‘
30 nim
tan
&
a
os
Side:
toma eam
ol=
At
(3
Ee
sec
2%.077
&)(3e)
=
102
mon”
BB seek) (B0)* = 7.6493 2/02 mmit
Fact | Alot) | Gylrm) {AG (lot mat) | Gm | Aa
ict amt) | T (lo me!)
Top
180 |
30
&.4
11. 432
Side}
102
is
1.53)
3,077
Side}
102
is
1.53
3,077
0.966
Bot | = 384
°
o
1.077
27.449
= | 468
Y,
(2)
8.46
SAY
neglect
0.966
7.64948.
7.6498
neglect
5S.008
3.46 «107
> SR:
48.627
“yes
7
«18.077
mm
|
15,2496
.
T=
SAd*
+ ST = 70.3tu lot we = 70.8) «Jot!
Quy
(8070
t=
(2)axio*) = 6xlo*m
VQ _ (loxlo? )(2.1477¢~I5*)
T *
TE
982)
=
2.19776 HOF mm
~ Go.simlot \(Culoey
Cb)
Qn=
an
Qe
=
11.982
L
=
toe
7 SOTKIO’
2.14776 «J/0> + 424.06
(7o.3)xto* (Exlo-)
T.= SO. MPa, sa
.982)
2.631%
‘
* 2.410"
Qe
To
(Moa)
|
Pa
x1o* mm
6x ]o7%w
OOx10* (2.6318 « 107%)
Stivess
‘
=
C4
2.6312 xK]O° m
t
Sheartng
214977C IOS
(2B sec ob) (1.932 XE
V Qu.
Tn
=
Pa.
(23)(3)08.077) =
.
&
_—
=
36.0
MPa-
To8iG2.4 MPa ae
1.51847 «107 mm
LSIPUPxfo®
Qe tem Gin77e~fos ($7)
Qn
Problem
_ 6.49 Three plates, each 12 mm thick, are welded together to form the Section shown.
For a vertical shear of 100 kN, determine the shear flow through the welded surfaces
and sketch the shear flow in the cross section.
6.49
—_
—
Locate
mm
50 mon
)
ZA
nevtvel
axis,
= U2)(Z00)+ (2942)
=
_
~
180 am
100 nub
2AY .
Y¥ ."ga
7 8
Ta
Q =
The
VQ
F
manimum
dow
3S 2.8
-
10.984 «lot mm
(94)00)0 25.77) =
_ (100% 10% 24.02%10)
(6.934 «tone
shear
x le
3
.
iy
the
cvoss
section
10.934 x10° m
=
24.07-10° mm = 29,07%10 "wm
2 gee no? N/in = 266 KN/m 0
|
|
ot
eceurs
the
amets.
nevtpsl
min? = I2TIB HIS
Que = WAV 2423 CREE) = 42.74Ol82
go
3
mim
lo
+ @NCRV2.77)"|
+ aL & GGA
=
-.
mm”
gelizi(2o0)* + (iz (200 (24.23)
|
5°
4656
48)
92.7105
_ VOQn _ (eoxto?V
10)
D0.
484
od
wet
ith
= 843 x10% N/m
B49 kN/m
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reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<0
_
Problem
Se
6.50
&.80
A plate of thickness / is bent as shown and then used_as a beam.
For a vertical shear of 2.4 KN, determine (a) the thickness ¢ for which the maxi-
.
mum shearing stress is 2 MPa, (#) the corresponding shearing stress at point
[* pomnns 150 fame
&. Also sketch the shear flow im the cross section.
dD
E
”
Leap
120 tint
2
Lee
°
of 207 +
Neotved axis,
Cabcudoate
dies
Se
at
*
[Bain,
Goimm
above
AB,
ld.
Tyg = (1st )lbey
2 2Jeoc0t
Ten = if (30t)lite)*s ISbevod
EF
Tee
at
f
Tog 7 (Bot) (60)? = S46000T
Ipn
7
(Fhoowt
=
Tng
7
2 %ov0et
T= 22
ta
At
pein}
c
5
Q.t
Te
(b)
T=
(/242000
Qae
+.
*~
b)
MS
=.
etress
=
_
(75
te)
ZL
+
(2G00\(64sot)
47189 S20
(6s
Ee)
_
_
=
6450
t
mapa
Gee t Qe
O + (1s Yssb)lbe)
iy
wo
(292000t
BT * Tayeanooby? SSemm
V Qe
Te
a
Qe
VQ
©"
Qe =
Shean
*+
=
t
.
(2406)(zs020)
(7789520)
|
26620 ma”
>
SEL)
=
jo
_.
Mog
CHR)
Proprietary Material, © 2009 The McGraw-Hitt Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the fimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.51
Lo
«6 51 and 6.52
. An extruded beam has a uniform wall thickness ¢’ Denoting by V the
vertical shear and by A the cross-sectional area of the beam, express the maximum
shearing stress a8 tmx = A(V/A) and determine the constant k for each of the two
orientations shown.
(a)
tb)
(a)
—
: (hat VS)
T..2
-"
Tee.
SBA
Tat) OTE .
(Fathi
5
k= 62
Gat~
-
2.08
3
Fat
vA
ok
we
|
voy
a:
ype.
“" T@ty
ae
“
eto,
rTM,
20 at
-kx
=
jt
Vigat
(ge
@t)
(ZenGr
wyv.aV
20 Gat
k=He
10 fk
210
‘wel
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
Problem 6.52
6.51 and 6.52 An extruded beam has a uniform wall thickness ¢. Denoting by V the
vertical shear and by 4 the cross-sectional area of the beam, express the maximum
shearing stress a8 ta, = k(V/A) and determine the constant & for each of the two
orientations shown.
&
h=+t/za
(at \($)’
4
‘Le $t@y > het
(dat) (Bay
Df GE
OT
u
ian)
"
(b)
y
= et
-VQ@
_ V@att)
_
Tmar = Tat ” (ZaEYRE) ©
«= 4V¥
4M
jor
.
=kX
=
—
Mt
o.
1
hb
4
—
yi
. ¥Q . Vien att)
me * Tet)” Eabyan
.
32
-
SBR
=
2.24
* 4M
FA
ks 9.danas
»-
tart
@,= (f4atY2)=
Q= Q+2Q
O&
Q, = lat YZ) = gat
Sof
&
$a°t
tf
23,+41,-
DP
I=
VV
3h
V
“gar A wat
Vo
+
yy
K- 28-27
kk
-—<
Proprietary Material. © 2609 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.53
6.53 (a) Determine the shearing stress at point P ofa thin-walled pipe of the cross
section shown caused by a vertical shear V. (6) Show that the maximum shearing
stress occurs for @= 90° and is equal to 2/4, where A is the cross-sectional area of
the pipe.
are
a attevdar
for
“>
Ieis-mt
amit
JT: Avi:
amv£
at
2ret
Ay
=
Zvtsnd
.
| (V avtsne)
(ME )CAt)
VQe
Laat)
Van
ME
(Te = Bare = OR
~
RN
Problem 6.54
6.54
60 um
a
it
h
4
L
=
Te
+
Tar
1220 Wo
=
64912
7
+
+ vo Kli2 yh
+ 1729 b*
aw
T= $0220) = 1728000 be
IT#IBébO
Qe (teXA)h
=
pave
(320 h mm”
V/320h
CE
meulmize
ISSO
VevticaD phate?
one hor'zontel pled
Bo
& plete:
!
a
47
_ve.
al
BUC W2y
=
12 min
For
Tye
__
section!
|
2V
The design of a heam, requires welding, four horizontal plates toa
Horizont
12 min
:
he D101 om L_sroom
Whobe
2
yertical 12 X 120-mim plate as shown. For a vertical shear V, detetmine the dimension # for which the shear flow through the welded surface is maxiinurn.
12mm
60am
sin
<
Cie BF 17973 60
3
set
de
2 ©
(69 l2h* +17 4/360) = 12894
eo
-
66912
A 4 179)2 60)”
a
O
elfey
mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the timited distribution to.teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
3
6.55 For a beam made of two or more materials with different moduli of elasticity,
show that Eq. (6.6)
vO
Problem 6.55
Fave
=
It
remains valid provided that both @ and / are computed by using the transformed
section of the beam (see Sect 4.6) and provided further that ¢ is the actual width of —
the beam where 7,,, is computed.
E,
of
lenbo|
I
GB
ehastietty.
Wb
fo
Widths b of actuaf section are
Actual
beeen YO ym
Transformed
n's to obtain the
by
muttipfied
The bencling
section.
transformed
Section
Section
*
- E,
n, ry Ever
Ct
*
I
where
is
the
measured
is
horizontal
the
shearing
Gny 4A
Shear Mow:
=
fest
et.
in the
bution
if
seation
section »
over Heng th
Ax
and
is
(AM) ny aa =_ BOM
Qc)
AG
n (Am)
ya —=
(REMY
momeut
cress
_nMy.
cross
fovee
ae
given by
of the tyans formed
over
Cr
y
of -the transformed
of dvanstovmect section.
9 Re AT
g is distributed
Es
Ew
inertia
.
(as dA =
AW =f
Qe
's
2°
centvore
moment or
from
Yi.
3
dist
stvess
. section
The
moelutus
reyerence
~
gE
y
«
be
Eves
Let
actual width
FOE
t,
thus
x= $.
- VQ
+e
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
:
.
Problem
:
6.56
6.56
A steel bar and an aluminum bar are bonded together as shown to form a
composite beam. Knowing that the vertical shear in the beam is 20 KN and that the
modulus of elasticity is 210 GPa for the steel and 70 GPa for the aluminum,
determine (a) the average stress at the bonded surface, (6) the maximum shearing
stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)
* Aluminnre
.
..
re
net in alumina
210
ao
Steel ne
_
FE won
,
J
Wh
stee?
in
2.0
ye
30mm
a
Locate
centrord
tompote
inevtra
20 mn
cn of
|
.
4
moment of
of trans formed
“sSeetion,
.
mm
A
Pant
.
Afum. | \200
Steet. | GOO
7 oe
nA,
ot
mm
1200
a
(a)
_
;
Ho | ¥QZooo]
| [800 | 10 | 18000]
t
Det
dem,
At
the
navtyeal
wa
nL,
Ae wom
kt
vm
iGox 10”
1a | 25a2"10"|
Ox [07
~ | c4axto® | 22010"
Qe (Bo)(UO)\U8\s 216K 10> wm
7 26 xIO” wm 3
(868 xlo-7)(0.0B0)
Axis
A
4
388.3%107
18
_
|
x
Io.54¥10%
C=
(b)
.
> BEBx/O*mm = BEBXIO™ yn"
_ VQ _ (2oxlo*Varexio)
s
r
| e@gooo]
oof + & Ad
At the bonded surtace,
=
.
mm
nAY,
yy
| 3000
T=
*
Qs
(30)(38\(8 )
=
o*
Pa
16.54
Mea
<ee
21.66 » 107% mun
= 21.66% loom
Tz
VQ
Tt
_
(ROxlo\(21.66x1lo°)
“(8EBx1O7V(G_ 030)
i6.64x 10° Pe
T=
16.64
MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by. McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<_<
Problem
6.57
_-
6.57
to form
A steel bar and an aluminum. bar are bonded together ‘#8 shown
a composite
beam.
Knowing
that
the
verlical
shear
in the beam
is
I6KN and that. the modulus of elasticity is 200 GPa for ihe steel and 73 GPa
for the aluminum, determine (a) the average stress at the bonded surface, (b)
the maximum shearing stress in the beam. (Hint: Use the method indicated in
Prob. 6.55.)
Steel-
yor
Alaminunt ~
|
in
cum
num
n=
“oe
Fact had G15 ConnAZ Cn
7
R74
steef
A GoD Ad tnd?
ITS1Z0
f206 | FO | 260300 | Bra
Stee}
m
aE Com)
tio F4S#3
Alum| GSO
[le | BIS | ah| 984646 [49479
Zz 4 Gps.
272071€
2976 124062
Sef
5
SnAF _
272078.
7 irae
44.2
Ts
ZnAd°
+ Prnl
> Gl oR bats
2263938
>
moat
b
seen
@)
At
T*
VQ
_—
man
=
| = 30s
CISC) BIT)
Ubosc)l3oys) _
Tt * Gageeyzes S 6 ute
(b) At the nevtvsd axis
t
Q.
surface?
honeled
the
VQ _=
It
Qe
C7
(f6000)(49396)
:
an
ota a
(22638328) (2)
CLM
-
r+
2
7-2
.
fren
5
|
}
“
8) >3eef ) = 49326 te
We
MPA
Mf
2
ve RR,
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.58
150n
r
-
6.58 A composite beam is made by attaching the timber and steel portions shown
with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of
elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4
vn
“|
SaeaeE
KN, determine (a) the average shearing stress in the bolts, (6) the shearing stress at
the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)
250 mm
Luan
Let
Fug
Ng
2 —
= Et
oF
Widths
.
+raus
lo GPa
GPa
forme}
section:
u
Q,=
(@)
*{O%mm*
=
~
gS
=
.18
v¥1O* ) =
(200 7
«lo?
= (23
> (B)Gay"
=
;
Qe
t=
ro
ve =
Q,+
[S50
8 * wm?
238.io
Ss
ISL 1
mm”
=
13.1
«Jo%
150 xtor?
(7,
ms
23 axaxl"jot Pa > 23.3 MP
28.3 MPa
=
—s
244.4 x10% mon”
= RIELY IO wn?
m
VQ. __ (4x0? \(274.4xJo*)
Lt
2,635 x]Jo" N
(7.5)(1as\(62.5) = 238.8%10" + 58.544%10%
mms
me
ae
Tay
bot > Eett
Awl 2 Z68S¥IO
3.1% Jone
(ob)
#10°*% m*
1819S 10? Nf us
assign ic’? De
Unio0 AEE
i
7.584
;
=t
3°
a
Awa
mm
9.766 «10°
235.8 %10%
) ©
US0Giz)(iast+e
=
Fut
= 7.5
by =G5)U1so)
mm
71-589
=
20
2 [0.0216 «10° 4 30, 89010}+
—
eo
=
& (7S )asoy
to
Z°
aL
Zoo
2[&Use (Ry + UseXiaz\Gas+6)*|
Fi
I=
allers
- E.
be ,
150
=
by
PUTZET AEE HEN
ny
.
|
=
200 GPa
84x10 Vts0 to?)
-
:
loqaxle®
Py
109.7 kPa
Proprietary Material. © 2009 The McGraw-Hitl Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
te
6.59 A coinposite beat is made by attaching. the timber and steel porfions shown. with bolts of 16-mm diameter spaced longitudinally every 200
Problem 6.59
min. The modulus of elasticity is 13 GPa for the wood and 200 GPa for the
steel. Vor a vertical shear of 16 KN, determine (a) the average shearing stress
in the bolts, (b} the shearing stress al. the center of the cross section. (Hint: Use
the method indicated in Prob. 6.55.)
tel
om
oo
ko L. |
Eve
N.
=
Ey,
=
13 GPa.
Bs.
=
289é
‘3
timber,
lw
=
pea RE
75 mm 7Sma
Fov
one
Larios
‘f
:
+ (75) (see Ui ee)
&
¥y
FA2k X16 inn”
For ste? plete,
I. = ik (12)(300)% = oT x00 Mm *
6G
(a)
trans Forme al
Q,=
Te
n, t,
+
VQ,
¥
-
(Ue) (786 eee)
I
Aunt
*
Vane
7
lb<d
Toe
Fut = $s =
Te
t,,
=
AL fim
be.
74a.
4X
to
ne
E
([62)000) 23240 N.
E (tv)
=
ad
Fett
Acs,
BT
OF
7
De]
32-46
Sap
fr
CLG
QyF ZQ, + Cis -38e)
2) (ISB) IE)
t
4
(1860) (lens 2 10000 ,.,3-
(bh)
section,
y
For
=
>
42
MPa
38 TEU
Ite
frm
VQ. _ (iheco) (BE : TEPID)
Tt * (yaguxoe)iey)
Te
= O°
~
~&
MPa
Pp,
|
-
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the fimited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.60
P
6.60 Consider the cantilever beam AB discussed in Sec, 6.8 and the portion ACK
of the beam that is located to the left of the transverse section CC’ and above the
horizontal plane JK, where K is a point at a distance y < yy above the neutral axis
(Fig. 6.60). (a) Recalling that o, = 0, between C and E and a, = (o;/y,)y between E
and K, show that the magnitude of the horizontal shearing force H exerted on the
lower face of the portion of beam ACK./ is
Plastic
4
H=~bo
2
~ a
"
vr
Vy
©) Observing that the shearing stress at Kis
cant fee
evra
~)
(2c.
Das.
Ty
fam a = tym LO. LOH
=
AA
A
=
pm
hb
Ax
b
x
and recalling that yy is a function of x defined by Eq. (6.14), derive Eq. (6.15).
P
6
c
_a dy
———
“;
Le
ae
e
kL Ho
Gy
y
Pot
Oo
K is deceted a clistance
above
the
nevtved
ayis.
ee
4
The
6 = od
For
For
equi dibriom
of
OS
¥ < Yr
reente’
and
Povces
stvess
distvibution
G6 = 6, for
<9
on
iS
given by
yr<ey <c.
ACKS,
H= |e da ==" Sey Baye J Gp bdy = Seb. (yeYGL)4
e
Gb (e- ye)
L406: ( 26 -
-%)
|
,
—
ry ER = 46 (-Be ER» - 46 (1-H
Note
that
yr
iS
oa
function
- 1 OH H _
OY
pieedining,
But
ne
ayy
Ax
Then,
Pes Bm
:
Po
Ye My
al
x,
+d
_.
Pe EM SREY
ga)
cz
=
Yeorbe® ~
_ Pet
*\ 32
ty? - £6 (1-44)
oh -
=
~
#7 6ybyr
2 aa
x
\a
—=# @)
* 6.61 through 6.64 Determine the location of the shear centerO of a thin-walled beam
of uniform thickness having the cross section shown.
(arbyt(B)+
D2M,
x dx
(
= Vat
ot
>
4
x
x
=
o
NV
ad
‘
<
x
tr
x.
»*
ce
ar
ss
Lt
o-
>
Ry
Qu.
herons
t
F, =
.
=
Ts
Le xsl
ws"
'f
BD:
*t
Pir}
Vhx
TRE
7
(MERE
>
F,=) vol
.
FE
~
dthx
x
vy: V&
be
,
=
"eae
lex al
bt fable
AtleasGb+h) ht
Q= tx:
ck
Evi
e
Pot AD?
lat bt?
21
I=
% ths
So
Nb
=
<
q
Ter
‘
im
7we
Problem 6.61
-D2 Maz
Ve
=
F,n-F,
he
__Vhit (ta)
4
e-
t (Gatebth) h*
3Cb* 2’)
Vhtlbta?)
iz
- 83V brea?)
Ga+Ghs
h
~
G(atbd+h
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimited distribution to teachers
,
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.61 through 6.64 Determine the location of the shear center O of a thin-walled beam
6.62
of uniform thickness having the cross section shown.
Lae?
dee
=
ate
3
Tog
Tee = 2ata®s datt? x 2ta®
=
-
J=ztI = ¥ta?
Leet bt @avs ta’
A=
AB:
Part
ve va
Tt”
= Ve gty’ . .
rot
Qzyty*
yr
Ly
B
“y
2vy
Aad
~ 28t
aV
a
t = ov
> Mys
r= (rsa = [vty
Past
BD:
D
2a
3V (at ax) dx
-. ( SX,
_ 3Ve
a\\“*_
BV
<u
Ne=
Ds Box (20 Hols eV
Gg (ane)
DIM
rd
DIMy:
Ve
~
w
|
=
(Ra KAE,) + (2a VF, \
$Va+
¢ Va
=
# Va
e= $a
={.250 0
tl
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.61 through 6.64 Determine the location of the shear center O ofa thin-walled beam
Problem 6.63
of uniform thickness having the cross section shown.
Te =I
= at (22)"+ hat
Tees Tre =
gto?
nat? s Fla!
at(ys
Ty t BEGa = Fta3
ant
A
.
Az
* poma®
tx
v-
ay ke
T-2t¥2
VQ.
Tt
3a
Q:
Zatx
V-Batx
evx
Brat 29088
(da = (SSM tay = Beev Vxdk =- EV
"rome
s itVQ. Vi
tate
2 2Vx
29. a*t
eae
axe
§&dA
Ve
t dx
oy 5
ee
tke
: 23 V
=VEMy:
=
F, Galt
er
BO
2a
7
iy
DIM,
Z
= . Ave
=
‘f
FL=
Q= atx
j-Z
Arte
Part DE:
Vat
aq Vaz
0.345 a
eva
ee
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
6.61 through 6.64 Determine the location of the shear center O ofa thin-walled beam
of uniform thickness having the cross section shown.
Problem 6.64
ted s tay
The = Tee=
Toa = Lep
os
(Rat)a’+
2 (a) to
lee = A tGay = 2ta
|
2act
T- ZiI-
7
38 tas
Zary
8
73
jr
+ (Za- y)
A
Pat AB:
zta®
=
Qe AY = Rta WGasy)
fat
= $t
ue -y")
VQ
“ae (4a
-y')
o
= = (4ety -¥) = via
52 /uy@)-@ eee]
>
2 VE
sit
fart
DG:
on
7
Q: tar?
+ txa
TMB e Ye (82,,)
|
Fe QUan = (OM (Sax)tale = WoO "(86 a yeydn
+}
-
- £=V
wea (See FYI = Ge [Go By
&
~-
Vto*
. 35
Sta
OEM,
Ve
=
AV
=DSMy:
-
Fe (2a) - 2F Qa) =
§ Va ~
va =
e=> fax O7IMa
2Va
<_<
Problem 6.65
6.68 and 6.66 An éxtriided beam has the cross séction as shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing
stresses caused by the 12 kN vertical shearing force applied at O.
50 mm
4 ~
~“
bee
350 tum
|
¥= 12kN
¢=
BD
ABEG|
Soo | 7S |
75% | oO
50 min
EF
206 4 FE
mi
2
2,25
Fae
%
O
2298606 (392 628U
_
Fao
Part BD
|
mal
=a]
Fer
Qa@=
GOO
Qo
Fe = Mee
mt
= 758ut0
SI
DAdts
Fee
Fee
8 78e0] go
0
246628t
[f6$ 70Gb,
3imin
Le
a
‘
Prot [A Gan? dor Ad Gnol | T Corre?)
C
See
up
% olits = SHE
0 tx
.
(oro)
By
Moments
Ve
<*
(b)
At
A,
Y
of
method
same
Fae ,
—= 2F ; TROVE
£
4
about
,+ 2S
2
.
sbaS wht _ GbadSweyay
r
BD,
Fr
ASSO
and
Gt
Qy=
the
weg ht
ce
befow Be
Q
at
~
on
Qs?
V@i
Tt
{3
H
Poa
be
O
BZsoe0 VE
bn
AEE VE
:
TE
Qut+
Ta = V@Qs
Tt
(Mer)
Qg.
txice
4
Lr
&
as
Sbaleeyt
=
—.,,
ABR mim
Th
.
a
Tp
=
Vr.
ote
=
“=
©
ey
loo) = cooet
Qe
“=
oy
- 2% 28 Mo VE
BZEVQ«
VE
_threec) (seee04)
729712. 565 EB
VOz
2
-
ae
AV
E
=O
Qyg = (Sot
-~
Just
Fee
mer
~=
My
ew
ft
Fee, and
T=
te
>
wy,
Just above Br
Tost
ot
Fer = RBISKIO OVE
eM
2aEe80
>
Tits momest abot Ht (Mandy = 25 Fog7
Fart EF:
;
Ts
=
Qeo
>
OOVt
_ U2ece) (78004)
.
caapeor
= [2660
_
9 .
§.2
Pe.
M fa
lfioo) 2
=
IL
oD
«(Nema
£
AMA
CO t
.
.
ae,
€
.Cipecor(PSoot) _ 266M
© G2g2sot
|
at
.
Contimved
ov"
Pe
next
xt
Paqe-
Prohdem
Hw
At
6.65
Crevtred
continued.
=
Cy
By
Sym etry
>
Te
=
Tt
20.6
Mh
_
MPa
a
2 12.4 MPG
<a
T=
U,
=
—_
82 MPG
6,03 and 6.66 An extruded bearn has the cross section shown, Determine (a) the location of the shear center O, (b) the distribution of the shearing
stresses caused by the 12 KN verlical shearing force applied al Q.
—
~
bsany
4
(3)
Ban
07)
-
mt
BLED
m
Tew = 7(/oo)(2)? + Ved (2) (TSF & 6877S mek
50 rom
Tye = fk (210)? © 24378d
L
i= Brom
Ler = "Les =
(bP 77AS par
=
41P 70 mov
=
$060950 mot
Tee = Ing
Le
+
= 252
UT,
Tag
V=12kN
_
€
C1212 £0) t
=
CEN 37G ) = (S3DE
T=
Problem 6.66
-
ot 1LS00t +
~ (2900) CISBI2E
VQu
Ts
Qgy
+
Qs,
>
Que
)S
exis
@) Part ABs
SI
ty+
QYy)=
“-
Fas *
Pe ayidy
= 0.8 ty?
y”
2S VE
=
gf)
iri
= ove
i“y “Ay
© Foal 2: e VET
Its moment about His foo Ke* 7031250 YE
Qe = ©.5 EIEN =
Past BD:
Qk)=
Qat
4)
2oE
XLV:
(281204 75x%) t
g(x = Yeh) . VE (282-0475)
fee
Fep = y ge) dx = eS
Its
moment
aboA
tH:
78
Lo.
“(282 S410) dy > gstzso Vb “-
Feo
Qp = [2826+
IO UeojJt
Continued
¥ 49r19 7180 VE
B 3
on
next
5
&
page.
Problem
6.66
continved-
Port EFt By symmetry with part BD,
Tts
Pant
moment
FG:
Ths
Moment
By
H
about
about
of
H
7S Vee
with
His
Fore
=
pat
100 Fee
49218750 VE
Fig = 708i YE
AB,
= 7o2ize0 VE
por
in pact
DE
is
&
// ‘OSeXf,‘io f
wy LU2E NOONE)
z
fobnGey
=
Qp
=
“5
zeve,
My =
- V@Qo
It
M
~ VQu
It
SOb2Ib0E
ay
o-7 MPa,
fo3i2z-st
1}
a
me
u
at
=
It
H (nested axis):
Ty
xz
2812657
Qg
rm
Go
u
At
xo be VE
tM, =O
VQs_ _ (/2000) (2P12 4) _ op 4
Qpy=
4
= SG
7 mm
Qu= Qe =O
Qe
5S
49219780 +703 ;250]) = Ua
= \ a (21AG
Zm,
er=
is
symmetey
momest
Ver
(b)
about
Fue= orsoVk
= (12000)(10312°S¢)
Foblabok
Qy=
Qo + ba)
_ (7800) U3I266)
LObLaLOL
ou
te
yf
= (2/206
_ gp
fp)
«| wire
3
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyand the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.67
6.67 and 6.68 For an extruded beam having the cross section shown, determine (a)
the location of the shear center O, (6) the distribution of the shearing stresses caused
by the vertical shearing force V shown applied at O.
12mm
B
YJ + 421"
G2 )C1arC#
(72 02+
r= 21G)
6 rom—>f<—
O
=
er
Ly
V=_ Lk
.
oO
192 mm
x jo°
PetAB:
+
e
17. 4844
a
=
1%
A> iax
hex ol
.
tet
5
=O
x 1o~*
wm?
Q=Ay= cant) = sax
7;
s
dads
VS.
=
t
usa
Vx
at point A,
= bye = 72mm at point B,
gd
. C USAV% y
sa V 2)"
Lr
a
tae
IT4349
x jo®
= 49M,
Point Bin
= (0.153246 V)C142)
Part AB.
oF 72 wan
= (isa \0ra) = 32. G4YxJorwm + BZ. GUY x10
=
|2mn
7
Ve.
a”
Te
=
Part BDt —
t
]
~
ES
Point
.
(ioxto® )(82-744x10~)
Tg=
in AG
QF BL. IWY%1O* mins BRIA KIO “an
Gem
=
VQ
_
0,006m
(flo
*x{0°)(€82. 944%
107%)
Ye,
te
6 mm
0.006
it
o*
(ite. 512x107)
VQ . Uloxt
9
TE
(9.484
x10* )Co.006)
me
mw
* (0° mm =
4 = 110.592
Klo + (EVE)
Qz 82.94
>.
“"
397.0 MPa
fa= TE = (19-4349% (0°°)(0.006)
.
‘
= 78,0x10° Py
te 78.0 MPa inBD
Ce
3
OLOI2m
349.0 «10° Pa
I
‘
7
Cc
Jd
NA Wt
NA
=
m
~ (14. 4844 x 10° Co. Ory
Point BZ y= 76mm
f° (IEE
ll
Bt 24 Yn
mm
AYL4ABR
C=
(a
0.153246 Vy
Ve
wt
IM,
Vio
ae.
‘
Lot blo, Pe
0.S98K 10m
7
Testor MPa =
Problem 6.68
6,67 and 6.68 For an extruded beam having the cross section shown, determine (a)
the location of the shear center O, (b) the distribution of the shearing stresses caused
by the vertical shearing force V shown applied at O.
\5-043]
Hy
12 mm—»>
a LUM ~scarGX
i
Ht
61mm
Br] + Ada ard?
K10% mm?
=
IS. 0431x107 ‘wt
oO
Cc
192 mim
6mm
art
6
AB
oven
A
:
=
|
6x
.
Qe
YQ
.
$*
FO
=
point
A.
kx al
_t
BD
ae
S
J
AY“
S76
=(6x) \B)
=
SIGH
Vx.
Ir
.
%
= O
‘at
we
Dew
=
F2 mm
at
poimt
8
72mm
F = Gegde
CMa
Vio = 0.094247 V
” (a8 gee
E
F2
@\)
@ = {7.0555 mm
(b)
Pomt At -%=0
B
Point
FE
Part
in
@=
t=
Gum
AB?
=
BD:
Point B:
4° pezzzza"
Te
Co
ye
te \2mm
= VQ
%
Point
72mm
Te
Wh
Gt
TE
=
yro
L.=
~
Om
4
nm
2 50.5 MPa, oat
Pa
Q=HLATAxIO mm = 4), 472 *1O “wm
= OL0I2m
_ (iloxwtot (41.472 x10")
~6
=
ClS.043i1o* Yo.)
26.271e 10°
Pe
Tet
25.3 MPa
=e
O, 012m
447A x10 + U2 GE ) = 96.768 <I m=
VQ _= Uloxto*Y96,768nlo*)
CS TE TUs.onsrdto loan) ~° $8-VF7x10
T=
~
O, 006m
= 50.8 x]o%
t—
%-O
~(1S,043)~(07*)(0.006)
Ta * Tt
y.
=
«dl
_ (iloxto’)(41. 47210")
- VQ
1G
g=O
QO,
19.06 mm
Cy= (760072) = 41.472 0107 mm? = ALY72 «1
'
Fact
=(0,099247)V (192)
Ve
49M.%
49M. =
2 pe
oo
96,768 X10 mr
Tee STON Ba
Ps
.
rr)
Problem 6.69
6.69
Determine the location of the shear center O of a thin-walled beam
KOR? = 10974 nm
P
tf
Fi
>»
of uniform thickness having the cross section shown,
uf
L = 2949S reo!
+ Agoh* = 4C4
6) 3B)"
”
Ee
a
o
Le
Ago = (14016) = ASE mee®
Tomi
2Miet7eG4) + AN 2194 FE) = CSB FEE
Pact ABS
Az Gy
Jrty
rn!
Qz=A¥=3y*
4
-
VQ.
to
v*
Vey
|
469
(6 5846G-)(6)7
x76 4
mre”
e
7
39
B= Sr da = f D 7e9xso Wy ody)
3
28
= pS1axie®VY | o
= 0:0833Y
DM,
>
DM
>:
Vez
2
F,
(76 sin Go")
ae
Ve = (2 (0.08333) V (76 sin G0")
ae
‘y
@ = (2)(0, 08333 )( 76 singer)= 10°77 iy
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission,
6,69 through 6.74 Determine the location of the shear center O of a thin-walled
beam of uniform thickness having the cross section shown.
oO
Le
35 mm
£GMaS) =
8S.75 KJ" mm"
8
35 mm
"
Ht
Problem 6.70
{ov
Line = TO mm
Aggy? (70XE Y= YRO mm
Tag = 4 Anah® = (2)¥20) (35)? = 171.5x/0% mam *
Te (2Xa6.75x10° )4(aY\7h.e«lo®) = O45
Port AB:
A= ts =
Ry
Q:= AZ = $5?
3
B=
«Jo? samt
_—
(vdA= ae
BYS" 44.
= SY
oO
Jo
s? As
~ (BIT
WLM,
= OIMp?
Ve = 2(F cos 60°) (70 sin go?)
=
Dividing
by Vv,
C=
20.2
290.2
V
mm
~<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manualiis using it without permission.
Problem
6.71
6,69 through 6.74
Determine the location of the shear center O of a
thin-walled beam of uniform thickness having the cross section shown.
ee = [38
=
Igo
+ 50%
3 Ag h* .
I~ > Rng +4 2Tso ©
Pant
38 sin
-
136766 for pt.
t%
Zin.
gor)
-
Symmetry,
= 4D2ZM,:
iP.
gach:
EF:
Ve
= feo Fo
Dividing
=
by
V,
F,
st
42M,
by
YEO)
Fe250
if
wise y
= 2S 1%
;
J
= BA oven,
aye"
.
Nt
Like
= 130833 inm
say
(157
QWs ACY = Sm mud
"mmm
__
Kew = (618A) 2 [FTO
= E208 min
AQ:
ABs
.
(381205) = 2 27SHT Mord
FE CMGEBE
Tug =
¥
Forsooo %
@
=
pretV
+ (2425 me
Proprietary Material. © 2069 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual mtay be displayed, reproduced, or
distributed in any form or by any mearis, without the prior written permission of the publisher, or used beyond the lintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
"
Problem
6.69 through 6.74 Determine the location of the shear center O of a thin-walled
beam of uniform thickness having the cross section shown. .
6.72
“AE
60 mm
en 2
Zig
= (Hot) (Coy=
me
hon
=
.—
Tow
=
/30* + 60°
=
F
v
~<— §0 mm
|
4
Ave
jadwio®
he
=
3
b
[00 mm
Asaz
(100 t) (co)
>
loo
I20 «10°
£
t
40 mm
L= 2]g+ ZIne = Seexiot
T
Part
AB?
Qs
120 mm
ve
A=
a]
DIM,=
DIM, :
Y=
mm
Ay
:
60
Ex
Ve
;
Vv
(co ty)
It
F,
tx
Tt
,
CR
60 mn
_
CoV
:
lex
|
x
Tr
San
= ("SY 8 te = SMEaa
Govt x|" «
i
af.
Ve = (0. 061136 V Yi20)
60.30)" VF _ 5 osiize V
(2\(528 x 1o*)¥
°
@= (0.051136 \(I20) = 6.14 men
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using
it without permission,
.-
Problem
6.69 through 6.74 Determine the location of the shear center O of a thin-walled
bearn of uniform thickness having the cross section shown.
,
6.73
For
whole
cross
section,
A=
I: Act = anabt
¥
Use
polar
st
%=
a se
Yor
sink
Q:
Ay:
> at
rv: ¥eMe:
=
Gavda
>
=
aot
a& = +0
a, Se
a Sint
2sn°2
~
At
2Zsina
= aft (1- ces)
|
: ( Ma (1- coso) ta do
Ve
partial cvess section.
S = are dength
where
» ave T s Zav
M.
for
aft
YE (1-05 0)
¥
But
Ieaq- matt
coordinate @
A=
2rat
=
Va"t (6 - sinf)
"
.
>
hence
@=
2a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved: No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. —
a,
Problem
6.74
6.69 through 6.74 Determine the location of the shear center O of a thin-walled
beam of uniform thickness having the cross section shown.
o
For
a
thin- wadled
T=OA
~~
For
we
the
Uee
=
hodPow
2no't
halP-pipe
podav
=
Cc
Cross
section
Ie ¢J=
section
coorelinate
A=st
eircu far
I
@
fo
at
=
#
.
A=
ata
natt
st
partia#
CKOsSs
.
S*
section,
an dength
-
Y
>
¥
:
Sing
where
copa
=
y
QO.
of
=
Sin
&
ese
= Ayr abta Setesd = oft (2 sind cosol)
> at singh = abtsin @
Y=
M, =
VQ
z
dt
Va’
sin
Pt
(atdA=
as
So MA
sn®@
tad
a
Ps~
But
My
=
Ve >
2
=
Va't
+
- cosO
:
Vatt
F :
=
hence
#
.
o
Va
e=
ta.
=
4.273
@
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers,
and educators permitted by McGraw-Hill for their individual course preparation. A student using this. manual is using it without permission.
Problem
18 mrp»
6.75
:
.
6.78 and 6.78
A thin-walled beam has the cross Section
Determine the location of the shear center O of the cross section.
shown.
be
T= bth?+tth
12 min
200 on
Right
Flange:
A= (zh, -y)t.
=i
het
Y)t,
= £(th,-yGhtyit
= 4( 4h,
Y=
re
aE
-y')t,
he -y7)t
Fr Stan =f ECG yay = YB(thty-¥)]
= MAS phe he a(Rya ths 4a = eae = Vth
DM, =sDMy:
-Ve =-Eb = = Ve
thi
O8)C180)°(260)
#)(2 isa) *
00)
Ehis th3 ” CiP\(2£1
G3 nm
Proprietary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators perinitted by McGraw-Hill for their individual course preparation, A student using this marual is using it without permission.
mm
Problem 6.76
675
and
6.76
A thin-walled
beam has the cross
section
shown.
Determine the location of the shear center O of the cross section.
A
het he AB) hl= DE, and hy = Fe.
T= £t (h3+ h? +h)
100 atin
.
.
:
Pact ABs
qacenecened a
"t 80 mit |
ai
rh
LY
a
2
Az(th-y)é
yr 4l¢hey)
,
at (dh-y th+y)
Likewise,
for
Part
DE*
J
a)
io”
ais
ol—,
=
t
hn
a
i
ie
HSed
a]
|
Ni
th?-y*)
.
and Tor Part FG*
¥REM,
Ay h, + & bay
= 47M, 2
Dividing by V,
ho+ ho +h?
e
Pal
=r
On ha 3+ As hg?
h#+ht+ he
3
(18) (120)+ U2
iL loo)
©
= S Je]
am
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used. beyond the Hmited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
6.77
6.77 and 6.78 A thin-walled beam of uniform thickness has the cross section shown.
Determine the dimension 6 for which the shear center O of the cross section is
located at the point indicated.
60 min
Fart AB
FA
60 1mm
s
Def
AG):
£5
y@)>
Ya-
a
Q{s)
4 Us
=
ts
A(s\ y(s)
Fag =
7.
- dts*
ys -~ts*)
t * ls! ds
.
AY Br
|Fis
Fep
By
neo
a8
Symmetry,
3
~ 4s)
“
v
Qet typhea — REM
Fee.
=
Fre
Pet BD:
Atxd=
tx
|
QO) =
Qa t+ yA) = Qg + Lye
qo
Yeo)x)
|
Fre
B
2
(
~ ME
A
6!
7
%
|p
F
Ly
q(s) = vee) = YE
Ye
a!
=
ex of
=
.
¥
(Qq + Lyx)
|
b
Fao * J gO du = FVv (Qb
|
4 4at ye b 2 )—
by
By symmetny,
Fag = Feo
Faz fs wot required) since ths moment aboot 0 is reno.
DS 2M,
= Ot
b (Fie + Fee) — Ya Fen
+ Ye Fer
z 0
2b Fra ~ 2s Fao = ©
ab. VE ( wale Des) yg ¥ (Qgb +4 byob")
BYE fleeME TE - AEEIng - 2k vob eyebY = ©
Dividing by
SUE and subslitoting numereal data,
fs (so) (Goy- EGO
I2exjo2 bh
b
|
~ [egoycce) ~ Heer"? (sob tl(zcyib
= tlogxjo’bh
IS xioi bh - 4se b= Oo
+ 450 b*=
oO
b= O
= 0
|
od
b=
4O.0 mm
mid
6.77 and 6.78 A thin-walled beam of uniform thickness has the cross section shown,
Problem 6.78
Determine the dimension 6 for which the shear center O of the cross section is
located at the point indicated.
20 aM
Y
1
Part
|
AB:
A
160 nim
0
A= te,
&
mim
s¥* WOmm,
= VQ
lex at
b
.
Q>
lod Vtx
Tr
T
he
Fe
20 mm
lL
eo |
Pheeweeseey
—
“~
4
Fe
Part
DE:
A
lex a
%
=
Vet
=
_
RO
mn
% ox
vt
tr
[BOxlo
(ROoY1R0xK(07)
il
4
$F
vertical
=
=
5
oF
Tt
¢
.6.
deg’
a
7
dz
a Eid
Prob.
(a3 - 2a
+9?)
feg:
Ttdy
(a?
=
=_
t
Q*
oFVE §
80
te
a
x dx
L
a8
= ©
a
2° 78.0 mm am
6.79 For the angle shape and loading of Sample Prob. 6.6, check that /g dz = 0 along
the horizontal leg of the angle and /g dy = P along its vertical leg..
Sampde
C
7 SCRE mm
6.79
horizenta?
Sqa
ia
F, -= \ gd
_
ar
Problem
to
J
(200) (i80x1o®)& ~ (120) (yoda ) ¥B
a= Gee caey
2
tx,
= (40 a*)
492M,= 0%
Along
(100)(60)*
“|
vr
x
bs
F
F
{ q dz
= loo XE
=Es
F, [He
Alfons
Go
q lx
606 mm
| Nene
Reter
Se
_
=
loo tx
=
+ 2a
3P(a-z2)a-32)_
= Be,
4tta®
49az+32z")
dz = ge
>
- £28)=
4az +32")
(az -4a2 + 32)|
6
—
3Pla-~y)(a+$y)
ios
al
3P (ats
(a*-
tay SP.
4a?
Sy )dy
a
_ aEga
3P 6% 7.2 + Yay
et
Sy’)
= an ; (aysdad -
>
sf)
~aet,
Problem6.80
=
6.80 For the angle shape and loading of Sample Prob, 6.6, (a) determine the points
.
-.
where the shearing stressis maximum and the corresponding values of the stress, (b)
verify that.the points obtained-are located on the neutral axis corresponding to the
given loading.
Refer. to “Sample
(a)
Along
vertical
ie
=
a
.
3P(ai ta’ y(arsy)
+
“3p
Ores
,
T=
3P(a-2Mla-azy
=
(a-32
ones
+ Hay
“Sy*)
|
zy. 70grb
ta
+{ ard z a)- ~(5)(% 0)" | = ao a
Ateq:
(a
ga
=
.
== O
loy)
Cio
herszontad
g
a
Pegs
Tn‘ = ce 3 fo
Alon
Prob. 64, :
3
(a‘- Haz
or
a
+32"
ae z=2a
+
Pp
Ce dele-onge)iogal] “86k b=
gi
At
eos (- Ha
ta
the
cornen?
tan@
= #
+ Gz).=
tan
= F
@ =
14.036"
O- P=
4S- 14.036
~. SAy_
5?
—a
25
wv?
oO
» 2F
oo yse
“>
=
mame
EAE
_ _at(a’)
a:
EA
> ae
Pak
y?
avis
y+
intersects
7. 92
ta.
verticed
Z tan 30.964"
intersects
at
0,400 a
=
axis
beg
kk
= (44 btand0de4?)az
Neutral
30.964?
at@a) _
5
Nevtval
—_
BO
hovizonta? teg at
z> 245 tenl4s?s gp)
-
ue + $ tan $9.036" Jas O.6667T Q
$a
—a
Problem 6.81
*6.81
‘The cantilever beam AB, consisting of half of a thin-walled pipe of
30-inm mean radius and 10-inm wail thickness, is subjected to a 2-KN vertical
load. Knowing that the Hine of action of the load passes through the centroid C
of the cross section of the bearn, determine (a) the equivalent force-couple system at the shear center of the cross section, (6) the maximum shearing stress in
ihe beam. (Hiat: Vhe shear center O of this cross section was shown in Prob.
6.73 to be located twice ax far from its vertical diameter as its centroid (.)
Prove
the
a
ne OF - Pipe
of the
2
vis
=
section
6
c
Ls
(a)
THs
Pine
Prom,
about
ts
shear! ns
Konce V
Datat
the
fPovee
passes
cattou
The
momeart
sheav
ceater
° FA-Fa2
system
at
Pe 2 kA!
(b)
>= 3O hem
Long
rect
due
bn,
Date!
=
f=
c
te
___.
au
guile
torque
MeOTE
Was
(2)
section
Me
2s
)
Me
QEN
=
3K
where
wl3e)
Cc, >
>
of
&(i-
Beg. 2S mm
2 AV)
el
_
vESSES
Due to Vz
Doe
foo
Few
h
T,
2P
Wat
Mo,
4
and
=
V Quoare
Tt
_
2600)
~ Tfeee3)(o-ai}
erei) 7
€
the
wrath
£
the
shearing
0.630%)
ie
Y=
T+
T=
Aee Lily
f8in in
+
(2,03
4en LEME
dovegus.
C,>
o«3
fi
= [3-62 MPG
vo
Seper position
st
fs
2F200
By
)(26e0
tT, < (Piatt)
Yt)
" Cast
Xt)
Fat
A
£
a
F
Sheaving
(1)
~
6
to toe
stress
O.
>)
J
\
Far
is
</
<
o
~e
Le
=
O
M.< Va = # Po
Ve
M.
avn
4
= 2a
“ coupde
V=P
M,
the
ceateold
the
of
C.
centvatd
EquiveXent
B
dis dane e,
fo the
beam
moment
d=: = e@-%a
+H
the
a the
Po
te
the
its
ab sind
O.
equal
through
|
seml-civete
At eack section
A
G.73,
Qroa = at
center
of
Dew
7a
For
1s
Prob
Q:
xt
a=
to
Bait
if
T=
so futiou
2
ffe3
Kl
Pa
Problem
6.82
*§.82
.
Solve Prob. 6.81, assuming that the thickness of the beam is
reduced fo 6 mini.
*6.81
The cantilever bear AB, consisting of half of a thin-walled pipe of
30-mm. mean radias and 106-mm wail thickness, is subjected to a 2-KN vertical
load. Knowing that the line of action of the load.passes through the centroid C
of the cross section of the beam, determine (@) the equivalent force-couple sysiem at thé shear center of the cross section, (b) the maximum shearing stress in
the beam. (Hitt: The shear center O of this cross section was shown in Prob.
6.74 to be located twice as far from its vertical diarneter as ifs centroid C)
From the
sodution te Prohtemn 6.73
T=
tat
Qe at sino
e-
ta
Qreoge
For a haDF- pipe section,
center
ot
the
AE
the distance from the
semr-circte fo the
cestrod
is
X= Ha
oe
At each section atthe
}s equal to P
Tts
Hiyough
of its
the
(@)
centvoid
moment
-
beam , the shee: ns Garee V
Pise of aettou posses
abavt
C.
The mament
the
shean
2a
2H
@.%
«© &
Equivatent fovee ~ coupe
system
O
is
at Ov.
M.= Va = % Pa
v=P
Date?
arin
center
my
Be
Az
2kAI
P=
V2
2. EAI
~<a
M, = 38-2.im xt
Cb)
Shearing
A
(i)
Dee
Sivesses
to
V.
. (Pet)
2Po
te
Mo,
% * Grastye) * Tab”
Doe
(2)
Fo
a
stress
by
Detat
Pong
olve
2
fe
torque
le
yess
oF
section
Me
where
C,=
BF206
,
©3 &)
(0.31996 6941257)
Super
position
Tz
Aenath
n(B°)( 6) = 767M
wedth
TL
the
shearing
& (\- 0.6305 )
=
rr
Ls & nm
CP On BIVGE
3 £&. 1GGq AthyMea
a
Je
oF
+
3
af
the tevque.
and
A
yee) ow
ts
f= tae wOZ0)> Geratnm
Tn " =
By
neck aun quien
== V ttClase
YH
gel
=
4G2<3g
MPa
Probl
6.83
consists of a.Z shape of 6-mm thickoe te*6.83 The, cantilever
ae beam shown
;
we
eat
ae
ness. Vor the given loading, determine the distribution of the shearing stresses
along line A’B' in the upper horizontal leg of the Z shape. The x‘ and y’ axes
roblem 6.83
are the principal centroidal axes of the cross section and the corresponding
moments of inertia are 7. ~ 69.2 X 10° mnyt and are Jy °$.7 X 10° mm
~
V=
IZEN
B= 22.8°
300 tom
Vist
J
|a\
upper
A
Ny
Var t
T,
$LS
Wy
Ce)
xi
to
=
td
=
V ces A
Jeg
use
(~/Sdumn€ %
£0?)
ma.
t
gr
(504%) na,
pe Or.
X cosB
J!
Doe
%
6 (564%)
rt
%
Vy?
hovizontef
coordinate
tt
V
In
V smh
y
+
J
cos Bi —
X
sinB
smP
Meee
ve (Vin Mb) Us04 x [4-150 + x)c03 8 + 156 sin 6]
i"
($7 £106) (6)
= S.asbxsot (1504
Due to Vee
J
pp = WAY.
.
Yt
MeF + O.4CITHX)
WV cos BCG )(iso 4x Vd os R- Z(- 5d +H) Su A]
(692. x18 ©) (6)
= bor x10 (ser
[ers
~ 619134
X |
x|
Totalt U4 Te = (S04) [000/72 + 9-00034/5
x Cin)
cimPay|
| - 1s
o
|
“12S |
~/66
|
-7
|
= SE
|
-a.e |
oO
| ~0-b42| -o.gag |- 0-63) | ovci2 6] boe3 | 2-58 |
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without perniission.
Problem
6.84
*6.84 For the cantilever beam and loading of Prob. 6.83, determine the
distribution of the shearing siress along Hine B'D’ in the vertical web of the
Z shape.
*6.83 The cantilever beam shown consists of a Z shape of 6-mm thickness. For the given loading, determine the distribution of the shearing stresses
along line A’B’ in the upper borizontal leg of the Z shape. The x‘ and y’ axes
are the principal centroidal axes of the cross section and the corresponding
moments of inertia are J» = 69.2 X 10° mm? and are J, ~5.7 X 10° min’,
Vs
=
Vy!
For
Vi
V
Be 22.5°
J2EN
Vic
ys
fh
Vees
=
Xt
x
For
6B
A
X=
vig b(t
til,
ib)
_
c
z,
Vio?
to
(VsinB)
Ve
900)
78 cas Bt
[Sd sinh)
(877%
Zs
Vet
sk
'
+
;
Vy! Aas ge
+
oes Bo
y
+y)
sh
Jen
~~y)
4(igns+y)sin
|
10°) 66)
© 5.6347 © g-000/54) Y
+ Aw y!)
A|
(VJ cos BY (Fe OID cosR + 7S Sin) +O CSS -— Y)ZCISVFY) cos
+ 6246!
(Veos B)LM708G
(67: 2X10") (G)
~
y" 7
2772
(49:2 x10") @)
Toteht
y
+ OCD
= (Vsma [ose 760BHe2+ Kio
25931 ~ 1I48YN
Doe
cas fB
y=
£096
Xa
+
iy
A
wey
By
y=
O
x=
oo
Dee
fra®
2180
y
7oam,
eb
A= &(/sb-y)
T
ey
pert
=
(6)(/St)
Ax
AB’
part
&
M4
¥ can |
= S697
Uy
°
| #50
-
2 22KIeF
= @ 66
-
Behorxie® y*
y
| e1e0 | ti£o
|
tiMPay| 8-692 | &121| be Ht] 3-56|
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
|
Problem
*6.85
6.85
Determine the distribution of the shearing stresses along line D’B’ in the
horizontal leg of the angle shape for the loading shown. The x‘ and y' axes are the
principal centroidal axes of the cross section.
B= 1S.8°
Vy = Pes
Aly) = Ca-y)t
B
Coord inehe
y=
dn
¥'
transte rmetio “.
-(i- ta) 5B
(t-ta)\eos B 4 ty- $a )sm B
eo
(J-#a)eos
= (ayt
=
us
At)
¥y +
(R-~tareosR
=
(£2) cosR
if
T,‘t
A
O.1361Y
:
-
(x-ta)sinh
- (-La)
£a)cosf
O48
oy
Iy't
x= 0
pavticuday,
1, = 1.428ta?
1 O1B8tta?
Vv
yr g(aty),
> Fa )eosR
w=
uf
Vi =-P sin
0.366
>
2a
(¥-Falsing
+ (ay+ 4a)
yo
1 nf
sing
6.069614
:
(Pees2 }(2a-y)(t)(0.136INy ~.0, 06961)
(0.1557 ta? Vet).
+ (-Psing)Qa-y)(o.4gily + 0.86612 a)
(1.428 a? EAA
.
P (2a~y)C 0.150 y.~ 0.500 a)
ta?
mofofe{[
ef of aefe] ey
7
(£)
~[,oao
“0.47
|
oO
|
0,250
0.333
O.250
oO
|
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.86
..
©: *6.86 For the angle shape and loading of Prob. 6.85, determine the distribution of
ee
the shearing Stresses along line D’A' in the vertical leg.
*6.85 Determine the distribution of the shearing stresses along line D’B’ in the
horizontal leg of the angle shape for the loading shown: Thea x' and y’ axes are the
principal centroidal axes of the cross section. —
"I
B=
is8° .
AG)
y=
Pees
ve
a-xVt
Vy
w= dar),
| Coord inode ctyans formation
)
os
yr=a
«
FQ Fa) P -(%~% ta)sing
a
a
0.342a\
-
2
a7
ee,
x
Ta
I= L42sia*
==
e-
if
Yo
(~ $2
eos B
~ 0.136149
- Gee
=
$a)
2
sing
0. 73.2Rta
Q
(wsda)eosA
O.4 BIL
+
Ba)
6 + 0.18922
a
| WAGLY
AG)
Te t
ly t
(Pees Bann )E) (O48 (Ilu+ O. 197224)
Co. ISS ted ME)
+ (Paina la-x)(t\-0. 13614 X= oO. 7204
.
=
-
a)
(1.428 ta® CE)
Plan) 2. Oak +h. (ooo)
tas
o
veo | o | & |
us (§)
a
= (%- ta) es@ + + (F~~Zalsmp
"
y
Le
3
<e
_—
s(y$a)enh
partievdan,
*
v -
Laden
ye (J-#a)essp ~ (%-_4a) sin
Ly = 0.155710
+h
- Pas
| 1-000
| 1,250
es]
ei]
I ises |
ase]
ee]
ei
L000 | 0.888
| OF
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No. part of this Manual may be displayed,
réproduced, or
distributed in any form or by any means, without the prior written permission of the ‘publisher, or used beyond the limifed distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
‘
' .*6.87 A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel
shown. Knowing that the vertical loadP acts ata point in the midplane of the web of
the channel, determine (a) the torque T that would cause the channel to twist in the
same way that it does under the load P, (6) the maximum shearing stress in the
channel caused by the load P.
Problem 6.87
Use
reautts
mm,
foo
h=
of
Eyampde
3b
30mm
It
kth
(Gbah)=
——+
Vz
Ca)
°
.
>
o7
|
Yyv
=
7
doe
sx
4.
Ts
4th (hs 4b)
$(8\le0)} loo +4) (30) |
lo?
SLE,
)(2
=
By lO
of
2b+h
=
1GQOmm
=
O.160
T
c,a
‘
(b)
By
super
t*
pesition,
=
-
(0.3223
|
Cray
=
M
Ty
_
Ty
g
,
)(o.1Go 81075
+
m
= 0.3228.
TH4, 64
=
mm
wm
c= £(i- 0.680 £)- $]1- ese) Be]
vy
5
= 22x10" mm = ZAI
t
Stress
> 144.64 Now
due bh V:
brbs tayh
=
|
VQ
at nevtral ants
Q=
Vv
>e |<
N
T= Ve =US#10° V4. 642415
T=Ve
p
= 1.86667 x lo mt
15*10*
Stress
3
66424 *loy mm
iz (3)(Jooy [6 \ae) + joo |
= 186667 ¥/0% mm?
——-
min
.
=
FBR mm
(BY35)
30
|
_
J22
Q+
"ae
be
_
“Bo
b
~-
with
8mm.
be
ane
G.O6
~
es
)
=
43.76
x {oO
= 48.76
fe
C5.4
Pa
MR
MPa
Proprietary Material. © 2005 The McGraw-Hill Companies, Inc. All rights reserved. No part ofthis Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. —
a
Problem
6.88
—
' *§,88 Solve Prob. 6.87, assuming that a 6-mmz-thick plate is bent to form the channel
‘shown.
*6, 87 A ‘steel plate, 160 n mm wide and 8 mm thick, is bent to form the channel
shown. Knowing that the vertical load P acts at a point in the midplane of the web of
the channel, determine (a) the torque T that would« cause the channel to twist in the
same way that it does under the load P, (6) the’ maximum Shearing stress in the
channel caused by the load P.
100 mm
h = 100
0
eee
2 +
I
Ps I5KN
_
-
.
30 mm
mr,
and
tse
mm :
a
= CH mm = EIS
Lo
= 8
loo
2+
ah
rn
(3X30)
I= ie th
(¢ b + h)=
a (e)(j00)* | (620)
=
1.400% 10% mm?
=
G
be 20. arn
with,
6.06
Example
of
‘resulds
Use
Vi=
+ 1oe |
1 400x10%
mt
1Sxlo* N
T= Ve= (Sno? )(4.6424+10" = HY L4Nm ae
@)
Stress at nevtval axis due to V
o
~
oO
Q=bt4 + 2(by)
|
~
Vy
T= Ve
v
|
|
= £(6)le0)f
100 +4)(30]
|
=
t
Ty
Sivess
=
Uist U6S*IO)
Va.
* Ti Goovlg Venlo)
Tr
clue
¢,2 $C
To
ty = Cat®
(by
>
T
to
By superpesitien
.
= dth(h+ 4b)
107 mn”
Exlo°
29,46 x10
=
w
eo.
Pa
of
~
=
abth
=
16.5
160
mm
=
=
=
IES
ng ae
29.46
~
144.64
(0.32542 Vo. ieoNeslo*)*
Vomoe =
a,
+ t
-
MPa
0.160wm
0.63808): Sf1-@stoV Soi]= 0.82546
-
O° mm
|
= 77. xl Pa = 77K MPa
~ Come,*
[06,6
MPa <a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is.using it without permission.
Problem
6.89
6.89
Three boards, each of 38 X 90-unm rectangular crass section, are
najled together
60 aI
a
38
6 mm
°
eet
to a vertical shear of 1 KN.
ay
.
6
mw
1 = & bho = ECGOWIEY = He tsa $0"
ran?
ak
388mm?
fe form a beam. thaf is subjected
‘Knowingg that the spacing8 between each PBpair of nails is 60 mm, delermine the
shearifig force in each nail.
{
ot
ho
8mm
7
7,
te
és,
A=
LA ¥,
:
:
2 MOGI
$°E
gS
= 2 Fas
Face 7
LE
3400
BY IQ
eH
hin
X, = .3P mm.
|
3
Qe Ay, = [24966 nm
$
ke 90 min=a
(9038)ye =
=
|
- (Me Eee Ed
KO“Gg)
Genaxee
=
77 Nim
8 iM
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc, Alf rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
i
A column is fabricated by connecting the rolled-
6.90
Problem 6.90
stee!] members shown by bolts of 18-mm diameter spaced longitudinally every
125 mm. Determine the average shearing stress in the bolts caused by a shearing Terce of 120 KN parallel to the y axis.
Geom ely
y
f= (2) +(te),
.
Sot 5 eT = [346Tmm-
RF
1BeF
may
Jot
Pa
©
(B47 AIF = foaebnm.
+--+
OA
:
Determine
moment
inertia.
A GrAala@melAd (mef)|
Part
4
ot
C2ovy2e | 2666 | izor$ | 28
é xb
S 2g0x37-F | 7.46
C2exy2oS |
o
| o
zbée
(2ec§ | 39+ R16 Flo
71°64 xp or
‘2
-
MG
TS
Frost
at
b*
Aus
=
.
fee) Grae
t
:
wht
(42.9
=
date
Fett
—
/B0Ki5 ©
448
“
aexte
6 -b28%0g
F235
x10
221328 prim?
A y, + @bbe) (i908) e
Q:
O° b25 Kin
|
&
|
(77.634 225)10%2 Mo x10
>
ZAd 4+ SL =
T=
T Gam?)
40
= 296 6 EN]n-
6b Yoos2s)
E (is)
[8X0
Senet
ISU FF KIO &
TP
ef
GT
SY
72 °
é
7
Petey.
kAf
2
far"
.
Mpg
.
~<a
Proprictary Material. © 2069 The McGraw-Hill Companies, Inc, Ail rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.91
[pe
gl
6.91 For the beain and loading shown, consider section n-n and determine (a) the
largest shearing stress in that section, (b) the shearing stress at point a.
7
16
n
|!
| !eo lew
,
]
a
j
Bo kw
L - ~0.9 m—H<~-0.9 _t
AY
Zo kv
section
n-n
V
=
26
Dimensions in atm
.
@
kW
,
-
re
.
Con sider
LYhoSs
section
as
Compose
cf types DO, @, ond @.
Th®
T= am Ua)" + (1a\(so)(a0)* =
L,= & (0) (a + GRO
Ty
i (eles? =
T=
4I,+
22,
ot re ctang des
8.28 +105 mam”
e)Ha*
= 5.14176 # Lo mm!
479. 24 x 107 mam!
4 22, =
Y4.274¢ 10% mm?
= 44.274%10°o m*
at
netvel
Qe
ce
Co)
2Q,
+
Q,
2a;
86.4 «108 eam"
=
RO. %
=
C80 ¥Cle)C4a)
GEXB4VIT) =
312,
256
re?
(44, 27410
x10?
“ -
VQ
It
:
QQ,
Vaxleriory
x10%
mn
F-24B* 107 mn”
mm
=
=
~ YQ _ __ (80x107)(312,256%10%°)
At pet a,
“"
+
CaSs
(2)(a0)(40) =
4
Q
1E
Cakeod ute
& ~ &
(a)
~
312.286xi0% m*
‘
2
17-68 x10 Pe
"7-63
MPa
“a
BEY R/O™ mm) = 86.4
x 10S mm"
(Box Jo" \( 86.4» /o*)
(uy_a74uto® Kian to)
=
-Olx]
¢
13.01% Io” Pa
3.01MPa
3.01
MPa
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Afl rights reserved.
No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
~lt
6.92
Problem 6.92
For the beam and loading shown, consider section 7-1
and determine the shearing stress at (a) point a, (b) point
S mm
Ra
°
Rs
=
SbIN
100 mim
Dyaw
Kor
Aina tan.
Vz 50 EN
Determine
Part
Section
[2sv0 | feo
@ | Seer!
Zz
voperties.
pp
A {m wy Yo a Ay bn m? 3 (™ m) Ad
(Gy
$d
shear
jaxevcd|
1
Yr
2Ay.
=
sh
I Gin 2
[6280660]
-25 | Biz Leve| esses7
378060
GFEqSCS a 12 Fedo.
32 Sbea
Fie
=
|
2083938
| igsovt|
| 7Sde
Y*
Sd
(nit
F
SD£K iin,
T= SAMs SE = Jaesxro’mmt
x
be fay
¥
fo)
A
tL
a
Ce
te
()
A
t=
(be)
28mm”
=>
y
z
EPL
Avin
Q,.*
Ay
= SHEETS
wit
2S fan
~
b
(a\
s
VQs _
It
= (Sb
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¥
inn
ie
27S
Qit
.
ge
~<A
— Se 7k Mia,
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= 42748
pe
26mm
F3TS
EBeOe
E)
7, = Vb 2 CERN
It
~
Gins ke) CS)
=
.
oars
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
et
Problem 6.93
.
8.93
For the beam and loading shown, determiite the largest
sheating slress in section #-n.
GO EN
25 rin
495 mm
Pe
ono 1), LA nye}
,
L
‘ sea
ioe
1.4
R,
=
Re
=
BEN
yy
Draw
50 mim
re
shear
Araataw-e
u
Y.
0.25 m
: esa
Vie
4 -
mia
.
Li
fa
V CEN)
Eb
Determine
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~ &y
S¢erion
Part AGnn? Yeon Ag lor
.
fog
@®D | sove |
as [lz Sevel-2¢
5.
se | brs0000|
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Zof3323)
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4g7hove!l BizSeae
ETSéOO
SN
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:
I>
[esvoeel
[275006
ZAY
¥"
fay
ACmm yan)
@) | 2500]
2 | 7Svo
&
properties.
TAI™ +
ee
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é
125
le
imme
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6
Q-
A, yi
=
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i
’
Lb pon
er
|
_ VQ _esreeeperasee)
Loe
Tt
*
Greweiczsy
|
* 2° Moa.
Proprietary Material. © 2009 The MeGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<<
Problem 6.94
6.94 Several planks are glued together to form the box beam shown. Knowing that
the beam is subjected to a vertical shear of 3 KN, determine the average shearing
(90 |e
60
|
20 [
stress in the plued joint (a) at A,
(6) at B.
T,= pbh’+ Ad*= bloc? *Ceekire\soy"
32. od xio*
I. = BW
Diraensions in mm
A
|
Te
ae
bh? = EGA
4
i
=
21,+
= 0,04
x 10% mm
= £G)M20) = 233 #106 mm"
Tg4 220, =
(1.88x10% mm?
—F
=
_t
Qat Ag: (60\(as)(s0) =
8
Qe
ul
(a)
Ts
”
Lt
(b) %e- ¥S
.
xlo*)\Gexjos)
|.
(it. 88%1o* K4oxfo?}
=
= 0
m
GO Jo! mm? = GOx/O7* m!
4 = 20mn4 20mm > 4YOmm = dow tom
. VQ.
-
88%10°°
|
3
37710
tm
=
379
Pe
kPa
moo
<=
Proprietary Material. © 2009 The McGraw-Hill Companies, inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachersand educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
.
«A
6.95 Knowing thata W360 x 122 rolled-steel beam is subjected to a 250-KN vertical
' shear, determine the shearing stress (a) at point A, (6) at the centroid C of the section.
Problem 6.95
105 yom
For
W360 x22 |
H
t
Z65%10°
to)
A= 363.mm ; bp = 257mm,
mm
= (B6Sxio° mm”
_
-
at
7
Ta TF
Ya F
4.&
tyre
10°(3%.8¥I08)
(250%
(BES 1S VALTHIOF
*
Y= f- B= BB 287
Mz)
A,=
ty (2-
2RTBLS mm
.
170.65 mn
388.8%pommt = 388.8 x/0* we
Q).7O
=
=
> 362 22
mm
21.7xjo°%
=
m
ae
io aay jo¢ Pa = 12.27MPa
A, > beLp = (257)(21.70=
:
VSan
(b)
OS (270
tos te
j
3
A=
|
Gat Aly
¢
{
Up 21. JOwen , Lo= 13.0 me
577
mm®
5 170,65 mm
t, )= U3.00(159.8) =
2077 mm*
ys
VQ.
=
Ite
, (250
(86S
2 (§£77X170.65)
+ (2077)07.4 )
by 7 13.0 mm
103)(1117.7« 10)
Io *)\(i3% io?)
oe
= 13%/0%wn,
2
CT
Z AY
HAR
10? mm
tp
te
gd
Q.
wo
Yo = 3(2-te)= 724mm
1127 «10°% m9
=
= 589x10°Pa = $8.9 MPa
~
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or. .
distributed it any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution
to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
6.96 A beam consists of five planks of 38 X 150-mm cross section connected by steel, bolts with a longitudinal spacing of 220 mm. Knowing that the
shear in the beam is vertical arid equal to § KN and that the allowable average
Problem 6.96
shearing sfress in each bolt is 50 MPa, determine the sniallest permissible bolt
diameter that may be used.
Peet (A Cn FEMI AS Lae YF (on mf AY? Conte)| Tem mat
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S7&e
IDs
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Between
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anel
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Re
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RF TEKIG ©
7
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ve
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6
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7 GER = ovomsh [mm
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.
=
=
bdid iment
= 9mm
at
Proprictary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual ntay be displayed, reproduced, or _
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jimtited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Lo
‘Problem 6.97
ce
- 6.97 A plate of 4-rrim thickness is bent as shown and then used as a beam. For a
-. vertical shear of 12 KN, determine (a) the shearing stress at point
A, (6) the maximum
shearing stress in the beam. Also sketch the shear flow in the cross section.
Ee
‘
|
ten
F
Lagrlggle satya
=
ae
Top:
Te
QSWEVQ4) =
Masimum
shear
ny
eLeUVS
at peint
Th
Qg
-
.. Uzxto®)(s.cusxfoe)
It
~
Grosxiotextors)
Ts,
Shearing
styess
=
Ta
(mPa
a4
mm
ehove.
=
23.2 Meo 2 eet
the
bottom,
= B.S48x 10
Vu
Q4
|
= ZHWIO™ + 1.248 HIO™ = 3.648 x/0% mm”
Qn = Qt CH seek ANA)
-_
3
7 23.210" Pa
M,
|
310.9 %10%mm! = 310.8%10" mt
24x 10% m
a eet
ca
an
(b)
21,4 Tp+ Te =
\(2.4« 107)
103
7
a4} = 11S, 46 ¥/0o% mm”
Ig = LT, = 1S.4GxI0% mm
2.410% me =
12%
:
I, = eGo + CoM
toe 4mm = 4x(o%m
VQ
-
Ter AC secol)l48)? = 39.936 x/0% mm!
Bottom:
Age
22.62"
SPanted side? Ag (4 seco! HB) = 208 mnt
Dimensions iu mim
)
m=
=
23.2
.
~
MPa
SS.2xio
co.
Fa
835.2
MPa.
mM
—_
Problem 6.98
6.98 and 6.99 For an extruded beam having the cross section shown, determine (a)
the location of the shear center O, (b) the distribution of the shearing stresses caused
by the vertical shearing force V shown applied at O.
anne
6 rn -
7
Tha = lay = ft GGY
30 mm
ii mm
PE GoYays BoX
¢
Loe v Te
s Boas)
30 mm
4mm
Fov
(a)
uo
/
y
Mange ,
~—T
f
1.449 % 10° mot
I
vara *
x 10°
Kt
Ats)
=
is
Qs)
=
yts
as)
=Yao)
¥ Vp
ES
b
Fe
feats
A gl? B
0.02716 ¥10 mm?
|. 14882 x08 rang?
a typice?
.
30 min L.
0,345 x10* min
Jus} s
0, 2645
yr
& G KA0)"
[rarer
30 mm
;
'
6mm
|
=
Diy
=
g q(s) as
_Vyt bt
AL
.k
Flange DE:-.
&
7
— &
4
4
4S Fae
x
(b)
Ceébevbition
Vis
Fianse
FE?
Feo
F
©.023502 V
Flange
HI:
Fas
=
©, joS76
t 15 Foe
OO238SO2RV
<
—
VvV-
B,&,6,
At
A ad
IS Frye
+
=
4S Fig
V
fo.223
@ = 10.22 mm
of shearing,
3sxlo*
At
t
V,
by
Dividing
Vase
idesawy ioe) _7
= Frye = UN
42ZMy?
=
42M,
Ve
WaRniae) = 0-10576 Vi =“GG C@F)
Fig™ Vs
cp
AB?.
Flange
ee :
ob
N
and J:
Hs
a
stresses .
T= (14882 *lo% m*
—
tao
Qz Go\lel(as)= B1kto* we
Bp KIS
=
a
be 6x07m
<
_ (28x10?) (8.1% (0%)
_ VQ
> Waster) 7 HT le’ Pas
CTE
Continuvedl
on
next
.
HIT MPa =e
paye.
above
tontinved —
D
ane
Jest
he bow
Fe
13.5 lot mm = 13.5%107% m?®
Q= BIxfo + G(solGo)=
6 Kio
m=
VQ
TE
rv
Just
D
just
H
1.8 ¥ 10"
D
and
jest
above
13.8 *]07 + 1.8 %10%
t
6 x1o?
8
t
=
.
F
Lt
Q=
=« V@.
TE
1.8% jo*
fam
te Hf 17
my
6
Qe
K:
_
= 68.5 MPa
He= PUI, « ince ee = oor Hr
befow
z
Pa
weg ht
the
fe
%
VQ
At
and
Gol4)\(is)=
"
&
m
é
2;
. Gsrio® Vissi)
Pa
«10°
68.5
wis)
Ye
*
© Ch B82xto
oF
rrykt
to
f
Just
te
F-
=
15.3% 10% mm?
15.3 «105%
=
m
(45 108) (18.3 «16
* (1. IHBBARIOT YE
1o"5)
Pa
77.7 *1O
©
1s. 3x10” * “(e)usi7.s) = IS.- G95 HAF mam?
(asxio*)(IS. 975x16°)
(C4884
mm?
.
_
_
IO“) (exto3)* | Ble]
;
77.7 MPa.
ut
Jost
6.93
M
Pro h Fem
_
)o “Pa
os)
1S. FS «15° we
= B11
MPa
<<8
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
- distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the liniifed distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 6.99
6.98 and 6.99 For an extruded beam having the cross section shown, determine (a)
the location of the shear center O, (5) the distribution of the shearing stresses caused
by the vertical shearing forceV shown applied at O.
14 mem
Lag
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je
mnt
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yu
b
53
g
c
—_—
497M
Qts)
Ag
:
(s)
%
>
FD
A
yts
=
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as e2.
-
z
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Vy
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*
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we
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=
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Flange
J
eng
=. Ls
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.
Ftange
Hb ——m
AG)
veuted
OMS
.
Dh
.
Cee - “Tt.
le S
|
.
% 10° mem?
Flange, .
&
i, == 0.933
0.933 x< 10° 10% mim
A
0.3645 *10% mm”
0.9324
ty picad
0.04104 {0% mm*
ZMys
Ve = 4S Fae + 1S Foe 4 (8 Fee + 45 Pus = 9.0167 V
Dividing
5)
by
V;,
Cabcud ation
a
@=
shearing
stvesses
V = 3Sxlo®? N
At
At
SJ:
A end He
4° “e
7
the
Ye
<<“
.
;
Veo
i
Q= (Bolas) = S.4x%107 mm = S.4x10% mw
.
te
mm
LT = 0.9329 «lo m"
B,E&,G,ad
Just
FIZ
.
(sxe
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2
oe
is. tule
A
ane
=
7
262.57
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tle
,
Y
d= 202.54 %10> N/m
Le
4x/o%
50.6 x{o*
Continued
wn
=
Pa.
ON
$0.6
MPa
ney+ page.
|
atl
continued
6.94
Prohtem
Just
bedow
A
v=
Jost above
Q=
D>
to
$
anol
jest
=
jest
above
H:
ts
222594
19" 2
6x¥ion*
be Pow
ana
D
oy
eight
™
~t
t=6xio7*m
10.8 «/o% mm?
=
_= VQ
suito*)Cie.gxlo%)
+P? . (6.
9atnlo \enlo®)
the
6xlo"
aa 9 xto* Pao = 32.2MPa
Fes
+ Gileo\(Ze)
5.410%
Y
Just
and
|*
te
Es
= 10. 8%/0% mm
67.8 x10"‘Dp Fe
L+ea
62S MPR
=.
GuIO tw
= 2.7 x10 me
Q = Gelle)lis) = 2.7™10% mm*
_= 1 B8xl0lol Pe,Pa == 16.88 MPa - wm
)
(sua?(ened)
ye
FE * Gaasqusy
Us VQ.
Jost beDow D ond just above Fe 9 t= 6x/c%
Qs
YH
Ts
At
Ke
Q:
4
x{o°
10.3
G2
(
- a
13.5-(o°
2.7xto*
3
(
eae
2.5 «10°
2
=
4 € 1ISX7.S
mm
IZR.Sx jo* wn?
=
* 84.4 10° Pa = 844 MPa tt
=
o* Mr7sxis)
VQ . (3Sxl
a =~ TE
© (0.4329xlo ¥extot)
THIS x10" mee
~ 88.6 io
cp
* foo me
[FAIS
.
fe
=
88.6 MPa
Proprictary Material. © 2009 The McGraw-Hill Companies, Ine. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-HiH for their individual course preparation, A student using this manual is using it without permission.
ma
6.100
Problem 6.100
Determine the location of the shear center O of a
thin-walled
|
beam of uniform thickness having the cross section shown,
Lan 7 [foot 4 IS* = 11.5 m m
Ane “BS t.
Tau = Ang h + Angel » U2se ered sas yo oy*
i
1308594L mm
LC VUGO% = 193093 4 mm”
i
Len
T=
20
In
+ Ten =
part
&
28 K/0%E mem
BD:
Q-
Qa
+
Qe
erty)
Q = Uastyoc)s (65-y) tld
= 2S00t
+
(¢bnes.
VQ
Feo
=
6
|
\
+
v olA
6s”
22st
hy
Vt
tee
tae SAY
|
- tty*
ay.
yz .t dy
H
he.
6s
,
vE
o|:
aw
(4bILS
~ ay day
7
=
Vt+
15
if
Or
DIM,
= Dz My:
binge
-
Y¥uc)- SEY
Ve
VE
= 1| =
7 a2 |(relesXos)
[ger
ous
sy
- styl,
VElsIq
Eee E22 )
O67? Vi
~V
(#67 -e)
=~ 86-7 (0.678 VD
es
867]1- 0678) = 27-9 mm
Note
that
ane
Forces
Foe
have
the
pass. through
zero
of Fre
Dines of action
moment
point
kK,
Thos,
about pomt
these
kK.
PROBLEM
6,C1
6.C1
A timber beam is to be designed to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of its uniform
rectangular cross section has been specified and the other is to be determined
so that the maximum normal stress and the maximum shearing stress in the
beam will not exceed given allowable values oy, and 7). Measuring x from
end A, write a computer program to calculate for successive cross sections,
from x = 0 to x = L and using given increments Ax, the shear, the bending
moment, and the smallest value of the unknown dimension that satisfies in that
section (1) the allowable normal stress requirement, (2) the allowable shearing
'
stress requirement. Use this program to design the beams of uniform cross sec-
tion of the following problems, assuming o(,,= 12 MPa and 7, = 825 kPa,
and using the increments indicated: (a) Prob. 5.65 (Ax= 0.1 m), (6) Prob. 5.157
(Ax = 0.2 m).
SOLUTION
See solution of P5.C2 for the determination of Ra Pp, V (2) ard MQ).
oe
We recall that
V(x) = Ry STPA4 Rg STPB- PSTPI- & STF2
Ly) S TPH
4 0a (X-)
STP3
263
— AY (2|
M(x} = Ra (A-@ STPA + Rg (x-a-L) STP B -P (x-%) STP
where
STPA, STPB, STP, ST PZ, STP3,and STPH
functions
detined
(1) To gATISFY
{[f Unknown
i
Peon
THE
ALLowner£
dimension
onknown
Jim
(2)
We
TO
Aimernsron
= =|[M|/Gy
.
SATISFY
THE
vse ky.
ffunknown
unknown
(6.10),
is
ste,
TP
STRESs REQUIREMENT?
t:
S=z
th®
ALLOWABLE
%
a7k:
dimension ish:
dimensron
are
Se it h wehae Ky=h=VESft
From
Pave
KieMAL
J
is h:
Sain = [MY Cgy « Front
J ¢
(4-5) STP3 43 (A-2uy STP4
- fa
~ A, (4-42) 5 TPL
iS
b!
we
have
SHTARING
t>
==f-
STRESS
6 Sf/h*
REOUIRENEVT!
- =¥gM _ 3M
fp>h=
+t
—
t
3M
3M
(CONTINUED)
PROBLEM
Problem
“RA =
6.C1
PROGRAM
CONTINUED
5.65
2.40
Problem
KN
x
Vv
m
kN
RB =
3.00
kN
_M
HSIG
kN om
mim
RA
RB = 25,00 kN
|
Vv
kN
0.00
0.20
0.40
0.60
- 0.80
1.00
1.20
1,40
1.60
1.80
2.00
2.20
2.40
25.00
23.00
21.00
19.00
17.00
15.00
13.00
11.00
9.00
7.00
5.00
3.00
1.00
0.000
4,800
9.200
13.200
16.800
20.000
22.800
25.200
27.200
28.800
30.000
30.800
31.200
2.60
-1.00
31.200
2.80
3.00
-3.00
.-5.00
30.800
30.000
0,000
0.00
109.09
2.40
2.40
2.40
2.40
2.40
2.40
2.40
0.60
0.60
0.60
0.60
0.60
0.60
0.69
0.60
~3.00
~3.00
~3.00
~3.00
~3.00
-~3.00
-3.00
-3.00
0.00
0.240
0.480
0.720
0.960
1.200.
1.440
1.680
1.920
1.980
2.040
2.100
2.160
2.220
2.280
2,340
2.400
2.100
1.800
1.500
1.200
0,900
0.600
0,300
0.000
54.77
77.46
94.87
109.54
122.47
134.16
144.91
154,92
157.32
159.69
162.02
164,32
166.58
168,82
171.03
173.21
162.02
150.00
136.93
122.47
106.07
86.60
61.24
0.05
169.09
The
sutatlest
<j
allewakle valve
3.20
of
M
kN.m
-7.00
28.800
3.40
-9.00
3.60
-11.00
3.80
~-13.00
4.00
-15.00
4,20
-17.00
4.40 - +~19.00
4.60. -21.00
4.80
~23.00
5.00
0.00
27.200
.25.200
22.800
20.000
16.800
13,200
9.200
4.800
0.000
A
18
the
largest
HBIG
mm
0.00
141.42
195.79
234.52
264.58
288.68
308.22
ATAU
min
378.79
348.48
318.18
287.88
257.58
227.27
196.97
166.67
-136.36
106.06
75.76
45.45
15.15
15.15
45.45
75.76
106.06
136.36
166.67
of the values
lagt two columns.
A =he
= 173.2
A
~
Nh
Prob. 565
For Prob, SI59
kN
™
2.40
For
= 25.00
x
o,10
9.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.1310
1.20
1.30
1,40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
iv the
5.157
HTAU
mm
0.00
Shiown
oUTPUTS
ig
379°
wim.
mm
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution te teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using thig manual is using it without permission.
PROBLEM 6.C2
6.C2 A caittilever timber bearn AB of length Z and of uniform rectangular section shown supports a concentrated load P at its free end and a nniformly distributed load i along ils entire length. Write 4 computer program to
deiérmine the length L and the width b of the beam for which both the maxiyaum normal stress and the maximum Shearing stress in the beam reach their
largest allowable values, Assuming oy “* 12.4 MPa
and 1) * 827 kPa, use
ms * progr am to determine the ditnensions £ and b when (a) P = 4. 448 KN and
~ 0, (b) P= 0 and w = 2.2 N/m, (¢) P © 2.224 KN and w ~~ 2.2 KN/m.
SOLUTION
occur af
A,
We have
Viz Pe ah
My = Plat
TO
SATISFY
Oy
=
Lhe
eo
ALLOWABLE
+ a,
T0;
SATISF
Y .
SFY
T HE
ALLOWAGLE
THE
m= fe
We € use
Cy
Et
=
moment
dendivie
MaXtA
the
ancl
Shear
maxtmven
the
Both
(E.10),
gv
8
Z
3
~
face
STRESS
REQU) REMEWT:
bao [2 M4
Jka
a
aa
SHEA RING
s ESS
STR
‘aa
/
REQUIREMENT:
STP:
Va
2
NoRMAL
4AY Le
3 Va
b(Bh)
b,
ols
=rer
16 b*
b*
Call
PROGRAM
For
L=0,
2 tating
Lou
Vy=
vit
L=0
be >0,
and
My =0
and
ba
equal.
We then
privt
For
OLA,
WE
L. and
OUTPUTS
Pp = A 44Pill,
Increment
while
Using Merements AL=- 0,001 M.,
9 wd de become
PROERAM
For
P and
=
= 952mm,
w=
0.0 EN]in
Zim
P=
Increment
b = BRITS min
increment
L=
=
w= 2-2kN fim
2mM
(S579 mM,
b=
25°FEmmM
2e2kyf{in
= Zmm
Lo= (7PGMM,
For P = 22244,
w=
b=
24T]7 Mm
fNCre ase
6.
PROBLEM
6.C3
6.C3
A beam having the cross section shown is subjected to a vertical
shear V. Write a computer program that can be used to calculate the shearing
stress along the line between any two adjacent rectangular areas forming
the cross section,
Use
this program
to solve (a) Prob. 6.10,
(b) Prob. 6.12,
__. (c) Prob. 6.21,
|
SOLUTION
V and the
A Enter
|
et
3. Determine
the area
A; =bshye
4, Determine the elevatfon
of each rectangle.
of the centreld of cach rectang le:
at By
(and
nl of
by and he
For iz ton, enter the dimensions
2,
nuinber
- 05h;
the chntion Z me the
a
centrocd. of
the
tative
Section:
gr (EA BSCEA)
5 Determine
Section:
the centrond al moment
T=
Lie
he +A.
6, For each surtace
Cleterm ine
Q;
Separating
of the
area
of inert la of the entire
-4y
two rectangles
belwthat
¢ and é+l,
surface
:
é
ay =
<7 " "
7, Select
The
band
for
be
he #)
the
smaller
shea ring y ress
i+!
6f b,
on fhe surface
and
brgye
between
the
rectanghks
is
“=
—
VQ:
It
<i
b
(CONTINUED)
PROBLEM 6.C3 CONTINUED
PROGRAM
{
(5)
"| 4
@,
| 20m
c
:
, 35 mm
°
+
@
i
_|
F
am
200mm —-—_——__s|
“a
j2mm
63 ma
a}
be
i2 mm
of
@)
~}
fe
WB)
Problem
38mm
(Z) -
>
38am
~~ T# j2mm
base
elements
418.39
Ll and
kPA
2:
elements
919.78
2
and
3:
Between
elements
3
and
4:
4
and
5:
TAU
=
Problem
KPA
KPA
765.03
=
Between
elements
418.39
<4 (0)
<@ (b)
kPA
6.12
Shearing force
YBAR = S0¢0
= 40
LN
56583 X10
fo mart
Between elements
1
TAU =
(5°85 MPa
Between elements 2
TAU
@
10 kN
above
mm4
Between
TAU =
‘Between
TAU =
l=
hc
6.10
Shearing force =
YBAR =
75.000 mm
I.=
39.580*10°-6
TAU
[2mm
OUTPUTS
=
and
2:
.
and
3:
and
4:
2/-65 wha
Between elements
TAU = (8-3
MPa
3
@
(0)
@
(b)
“Problem 6.21
Shearing force =
90 kN
YBAR = 65.000 mm
I = 58.133*10°-6 mm*4
Between elements 1 and 2:
TAU =
23.222 MPA
Between elements 2 and 3:
TAU
=
30.963
MPA
¢ (b)
@ (a)
Proprictary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the Jintited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using if without permission.
PROBLEM
6.C4
6.C4
A plate of uniform thickness f is bent as shown into a shape with
a vertical plane of symmetry and is then used as a beam. Write a computer
program that can be used to determine the distribution of shearing stresses
¥
caused by a vertical shear V. Use this program (a) to solve Prob. 6.47, (b) to
x find the shearing stress at a point & for the shape and load of Prob. 6.50,
assuming a thickness tf = 6 mm,
SOLUTION
For each element on the rioht hand srde,
we compute
Leng {h
of
_
e fentent
=
Li
en.
Ln
-
alec;
Xtes)
Cfor
wh
i=!
+ (4;
to m)~
z
~Y,,)
Acea of element = A; = EL,
— where t = &mm
°
Dislaue
a-h —1f2
clement
=
Gs = 3 (4:+ ¥i,,)
of Section)
Distace
Note
That
Monje
:
Laxig
from
# aris to
Fron
%, 20
to
and
ef tiretla
108
centroid
centroid of
that
ef
Hy.
Sect for
Ying, =O
alovt
centrovdal
nxrs?
LeZEA Le (ty) t (Fa) J
-
.
(
Z
Cempeoiation of GQ at point P where
Q= af, (4; - q )
between
ane
ened
where
of
section
Sherine stress at
P?
Cr
MOTE:
éum
Cray OCCUIS
and
ooo
>
2
stress is desired
extends
to
point
BP
the ateas: located
- VO
Tt
_
on neutral axis,
6,6, for
Gp = 4.
OUTPUTS
CROGRAM
Part
I
=
(a):
0.2
Part
X10*°
mm’
Taumax
= 14.06
TauB
12.5
=
MPa
I
MPa <@
Gj
=
Tauk
(b):
8.352
«10°
= 1.293
MPa
mm’
PROBLEM 6.C5
6.C5 The cross section of an extruded beam is symmetric with respect
to the x axis and consists of several straight segments as shown. Write a
computer program that can be used to determine (a) the location of the shear
center O, (b) the distribution of shearing stresses caused by a vertical force
applied at O. Use this program to solve Probs. 6.66 and 6.70.
SOLUTION
SINCE
CPC TION
1S SYMMETRIC
COMPUTATIONS
WIL
BE
WITH % xs,
DONE
Fore
FP
HALF.
Fok
€=/
Fo mtl
Zs
- ENTER
Compu %
For
LENCTH
ca?
Le
a.
.
Aye
|,
ed
CAL cotarve
+AyF
MOMENT
~
de 40 (See
5)
SEB
ENT
yA
OF
LONSIDER
5
OF
boy
FRCH
Bien 92
( Ax
j7 Ons \
d
OF
ts TAE c3l2fetn)
70-7
ay =
jv
(Nove: ne
Keo Ye
For
oO
ERCH
10a
La
/
NEA TIA + T.
.
SEBMENT
EQUAL
»
AS HAQDE
PARTS
Fon
Aaren = £; t. foo
{=/
CMe
Ye
7m
/G0
Se +Ay,.(4-0.D)/r06
AL=(aAren) y*
T=e
ne I,+Al
ee!
SINCE
LALCULATE
SEEMENTS
fer
d= 21,
SHE PRIRINL
AND
S7I2ESS
SHEAR
le!
USED
WAS
HALF
FOP
ONLY
AT
Forces
ENOQS
Ton
A AREA=
Ly 2 Joo
Fold 4 =/
4
TO
’
100.
toe
Tn eye
4, + Ay, (4-0. s/o
AQ= (AaneA)ry
Tan Trew 9 OF OF4R
OO) eae
nen = VOL,2;
Vave
T=
= on S{ Bld
Tt
OF
iM SESM ENTS
+ Zou)
Vane
CONTINUED
PROBLEM
6.C5 - CONTINUED
Lt
Fe
Force ; = 7A aren
(%
7 = VG SL, Z,
(Joncenr )y ~ vas
@, =9
AOIAC ENT
ore,
Cnet l Trasacerr
Lh)
LOCATION
OF SAEMe
LAL CLLATE
Forces
Fon
CENTER.
A OENT OF SWERR
AROUT ales giKn
=}
Tan
(Ax);
(F.), = Foret
= forcé ; Ay¢)/2,
(Ky
pamenrye Ey 3p HANH
t+ MOFMENT,,
= MoMEy?
MOMENT
MOM
WHOLE SEcTI0M
For
AFAR CENTER
IS APY
PRDLRAM
ez
OUTPUT
Prob.
6.66
X(K)
¥(K)
L(K)
nm
min
mm
mm
feo
00
15
{0-0
70
log
-00
7s
16
.00
.00
VY
yw
3
4
Moment
of
inertia:
Janction
of segments
Mom er jy
T (KR)
1
2
Ix
=
$1063
Tau
Before
MPA
Qo
MPAX*
Shear
X16% mm?
Tau
After
PA
1 and
2
9435-7
666
& 66
O47
3
309395
24AAG
24 hy
4640
3
4
39875
3kff
Moment of shear forces
+ counterclockwise
Distance
from
origin
to
about
shear
Bel
origin:
center:
=
2EN
Force in
segment
LN
2 and
and
LNT = 2 (somen7)
6:4 bf
M
=
foo
LNa
e = 666667
mm
CONTINUED
PROBLEM
6.C5
- PROGRAM
Prob.
PRINTOUTS
CONTINUED
6.70
T(K)
mm
1
2
X (K)
ram
6.00
6.00
3
of
inertia:
Junetion
of segments
1 and
2
2
3
and
60.62
00”
Ix
=
oO
70.00
35.00
06
mm**4
Shear
Tau.
After
Force in
segment
MPa
kN
mm* * 3
MPa
7350.00
2.38
2.38
11025.00
335.01
3.57
3.57
666.27
Moment.
of shear forces
+ counterclockwise
Distance
514487.
Tau
Before
LL (K}
Tm
.00
35.00
00°
6.00
Moment
¥(K)
ram
from
origin
to
about
shear
origin:
center:
M=
e=
=
1000.000
20.309
Nm
20.309
mm
N
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part of this Manuat may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
PROBLEM
6.C6
6.C6 A thin-walled beam has the cross section shown. Write a computer
program that can be used to determine the location of the shear center O of the
ctass section. Use the program. to solve Prob. 6.75.
SOLUTION
Distribvtion
Let
of shearing stresses_jo element L
e: > Shear
in
cross
Section
= Centreidad moment of inertia of section
t.
The
Fy
en
Fa
LEE
fy
ih ' 4
oS
to V af sheer center.
: ? =v
2Yhata
oa) ) 0 Deez
ZMa=ZM: neve tasib,=eV
. en
Ct; a? ba
e-
™
OUOTPUT:
Problem
For
t
=
6.75
lSmr.
For
element
t=
fmm,
Answer:
.
Zt; ay
element
1:
a
=
{og
Mm,
b
=
0
2:
a
= 7mm,
e = bY
mm
~~
ee”
t
ae
>
i
oh
Ques}
i}
of the forces F, must be.Spolvolent
Divide (2) by (1) i
PROGRAM
=>
>
SK
—|
cv
We have for shaded area
b= 260mm
A
3
01)
(B
«
Problem Solutions
By
Dean
Updike
Chapter 7
7.1 through 7.4 For the given state of stress, determine the normal and shearing
stresses exerted on the oblique face of the shaded triangular element shown. Usea
7.1
.
Problem
method of analysis based on the equilibrium of that element, as was done in the
derivations of Sec. 7.2.
45 MPa
us He
27 MPa
ug Acos |S"
IBMPa
inroreeerspernnnomns
169
18 MP2
2
r
|
Po
A co
[
GS
13 AcosiS”
[5°
eon
ennnnielh
aera
Is :
A
ISMPa.
>)
TA
Asint$?
Stresses
2TA sinlS?
SA
Areas.
1B AsintS”
Forces
|
+f2FFO%
OA + ISA cos IS? sin IS"+ US A doslS% 0s IS* = QTA Sint 1S? gin IS + 1B A sin [Sos IS? =O
G = = 13 cdeiS*sin IS? - 45 eog"]5° + 27 sin |S? ~ 13 sin [§°ees (57
C= 74% 2MPo
+ “NTE = =O Ot
[BAces!S*cos IS°= 4S A cos |S? sin 15° — 27
A+
Asin IS tog IS® —
rs -13 Ceost 1S? = sin ls?) + (45427) coslS?sin 15?
Problem
7.2
7. through 7.4
.
so
[PAsin (S?sniS’
=e
24)MPQ
T=
For the given state of siress, determine the nornial and
shearing stresses exerted on the oblique face of the shaded triangular element
shown. Usé a method of analysis based on the equilibrium
of that element, as
was done in the detivatioris of Sec. 7.2.
GO MPa
3
6A
%
<e
TA
| ao Asinse
SG MPa
Go
GO
Stresses
+K 2F+o
GA — 90 A
6
=
180
Aveas
+
T=
Gok
Yo
Aces 30°
Forces
0
sin 30° cos 30° - TOA ces 30" sin 30° + GOA cas 30°cus 807 =
sin 30° cos
30°
=~
60
cas? 80°
=
32.9
te SF =O
TA
Aces3e°
MPa
|
|
Sin 40° win
40°
90
(cas * 30° ~ Sin’ 30")
A
cos 30 cos 30°
+ GO
_ GO
cos 30°sin
30%
Aces
Bo°sin 30%
=
71.0 MPS
=
Oo
~<a
Probl
7.3
roplem
7.1 through 7.4
/.
For the given state of stress, determine the normal and shearing
"stresses exerted on the oblique face of the shaded triangular element shown. Use a
method of analysis based on the equilibrium of that element, as was done in the.
derivations of Sec. 7.2.
60 MPa
45 MPa
I0MPa
(QO MPa
2
T
O®
.
GS MPa
—
,
AsinZo
ant
60 MPa.
A con’
L
Areas
Oo
ae
Forces
45A sin 20”
GOA
Sin
20°
=O:
ty SF
20 "cos 20° - YA cos 20°siv 20° ~ ASA si'n 20 cos 20° -G0A sin 20° sin 20” =O
~120A cos
SA
B=
I20.A coo"
$
<
-
Stresses
A
a
(20cos* 20° + Y4Se0s20°%sin 20° + YEsin 20° cos Ad” +608in* 20"
G=
14,14 MPa, <8
ASF =0:
TA 20Acos20"sin 320” - Aco 20'cosl0°+ 45A sin 20'sin, 20° ~60A sin Ja"cos 20° = ©
VT = 100820 sin 20°
YE(cus* 20°- Sin*.20") + 605i 20'cos 20"
Problem 7.4
etn?
Geen
Certd On teen
T= 15.14 MPa
ait OF sires, determine the notmall fn
7.1 through 7.4 For the given state of stress, determine the normal and
shearing stresses exerted on the oblique face of the shaded triangular element
shown. Use a method of analysis based on the equilibrium of that clement, as
was done in the derivations of Sec. 7.2.
,
54 MPa
SY
d3 MEPa
a
°
<<
Stresses
SS?
42. K sinSS°
Areas
GA
Forces
2F=0
—
84 A cos 5§°cos 5° + 42 A sin SS? sin 5S?
G
Ay SFr
TA
cos
Eerste
¥
A
So
CA
A
$4
|e,
2Miby
pn
+/
i
-o
MPFe,
=
£4 eos* SS?
~ #2 sin? SS?
=
=O
~ OS46
MPa
o
— £4 Acos $S°Sin 58" ~ 42A sin 55% cog SE?
T=
126
cas S8°
sin S§°
-
F922
MPa
—_
Problem 7.5
-
7.8 through 7.8. For the given state of stress, determine (a) the principal planes, (4)
the principal stresses.
40 MPa
|
MPa
Gx = -60
“F935 MPa
+"
(\
tan
2Gp
= Gi-Sy
@YX85)
_
Ghani =
by 7 3S MPa.
gen
Go +40
3.50
+
a
\* 4/35)"
i
=fe-1e
Vy.
+
+ |(Se-8)"
ons
| 53,0°
= 376°
Op*
74.08"
20, =~
(6)
MPa
_
2%
_
60 MPa
Gy = ~4o
2
~SO
+
36,4
MPa
Gina
vain
mt
~—B86.
<A
So
MPa
7.8 For
io 7.5 through
;
= the given state of stress, determine (a) the principal planes, (4) the principal stresses
Problem 7.6
140 MPa
aon
6, = 23 MPa
25 MPa
weap
=
11. 6OMPo
$2 MPa
(@}
tan 26,*
295
-
Gy + /4oMfa
e2 be oa
.= (2)-A2)
53 Tho _ O.78§0
103.93"
Op > 18.43%,
3c.87°
Ey, +S.
6) Srna = AG
BY (SGS)
if-a
+ (222, <!f2)
=
+ To
+(-42)*
2
"f
2h
Thar
Tay ~42mMta
§42 7o
=
ig MPa
\
Goatn
1
p54 MPG
Grow
Proprietary Matertal. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
nd educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission
si
is
j
is
i
i
i
i
iss
ic
le
7.5 through 7.8
Problem 7.7
For the given state of stress, determine (a) the. prin-
cipal planes, (b) the. principal stresses,
42 MPa
=
6,
Ty
6) =~ 42 mea
.
£3MFa
* 2PM PS
28 MPa
63 MPa
2)
yan
26p
x
k bxy
(2X28)
eae tee eerie
6-6, * Fe
tarde
AO, = 28.07"
.
2
0.533
_
Op = 14.04°,
104.04°
(bY Grnaymin = 22%
4 [Ga Ory* 4 4
-
G,+
tf
.
:
. -6y
a
2
—~
a
4 992
63-422 [{Zt42)*
Ommese
TO
Me
- 49
Gonin
Problem 7.8
b
lor St EGE
A
th
Zz
0G
7,5 through 7.8 For the given slate of stress, determine (a) the principal planes, (6)
the principal stresses.
30 MPa
6)
= BO MPa
(2X-4)_
= SSg57
Toy = -F9MPa
F_ 7 (0.750
Oo= 18.49 1OB8.4°
tf
ft
DY rresaig® SEES af
e42e
St
+
15
(R82) a os
(S*Y
+ (45
Gim
7 23.0
e MPa
Snin
=
$.00
MPa
=
<a
Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 7.9
40 MPa
35 MPa
19 through 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (b) the corresponding normal stress,
60 MPa
.
tan
(a)
~ &
15.95"
cn [EH
yf (FREY
@)
©. 28 2387
98.07
Q.= 9.0,
4s)
7.9 through
7.12
=
Come 830-4 MPa
6'=6,,,=Gt& = a
Problem 7.10
mee
an
~§04+40
@VGS)
OS
s
rt,
26, t- Th
20,=
Try 2 3S MPa
6, = - 4O MPa
Gx = 60 MPa.
671 = = S00 Mie
For the given state of stress, determine (a) the
érientation of the planes of maxinium in-plave. sheating stress, (b) the corresponding normal stress.
140 MPa
Gy = 4oMfa
Gy= AFINPA
entree
ana
28 MPa
@) tan 20,2 - #5
A%y
A2 MPa
28, >~53. 13°
Ty = —42 Mp,
. ~ 28-I49 «1 3333
G42)”
QO, -26.$7°,
63.43"
=a
le) Tine > f( 5%)" + Tes
- { 23-14 oy" (- 42)" = Joma
(c)
S’
>
Bur.
* ox
-~ > astute
=
~<
FGM fy
Proprietary Material, © 2009 The McGraw-Hill Companies, Inc, All rights reserved. No part of this Manual may be displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for ibeir individual course preparation. A student using this manual is using it without permission.
Problem 7.1 1
7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maxinium in-plane shearing stress, (b) the corresponding normal stress.
42 MPa
23 MPa
3 MPa
:
«Gy
a
(2)
= 63M
|
tan 26,
RO, =
Oy >~ 42MmMPy
6.-§
x
Ie
z,
~~
=
oe
~ 61.93?
63442
@ aH)
Dey = 25 MPG
Stowe
1875
Ze
QO, =~ $0.96°, 59.097
me
) Tran= 25S)" + 7
2
-
(CE™)"
.
(c)
Problem
7.12
——
—Q
2
=
Seta
G7 EMPa
6 MPa
ae
6a
= 59-5 Mfy
5 4B
_
J0-5
Mh
me
79 through 7.12 For the given state ofstress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (5) the corresponding normal stress.
30 MPa
wd
G6'=
09"
C= SOMPa
Hy= ~ 4 MPa,
Se=8y *- 7 ~ £530
@) Tan 105 == ~ 9h,
Gea)
2G, = - $3.13
9 MPO
@,* ~ 26.6", 63.4°
3
+
Cave
Tome? 1S.00 MPa, <a
oN
(S32) 4-2)"
a
-
if
G's
=a
+ 2%
a
rae -| (S587)
(b)
37- >_ 1 633332
6’ =13.00
MPa «ad
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
7.13 throagh 7.16 For the given state of stress, determine the normal and shearing
Problem 7.13
stresses after the clement shown has been rotated through (a) 25° clockwise, (8) 10°
counterclockwise.
.
§0 MPa
|
GyutSy
|
“3
50 MPa
4b
.
Or
~
Sy:
6, = - 80 MPa
9
Ox =
. 4O MPa
Be
-
Tay’ =
6p
(a)
Q=
- @%
Ge = - 40
Sy:
(b)
zs
@=
Gy:
=
-4o
10°
=~ 4o
By >
Sy
=
-~Y4o
28
+
=
SEM
-
Oa
&
Sie ~ Oy
SG
cos 20
+
Sin
4 Tey cos 20
os
26
20
-
Ty sin 20
By
sin
26
= ~-50°
40 cas (-S0°)
~ SO sin (-S0")
-_ 40
Sta (-S0° }—5O
~ 40
cos C50°)
20
+
=
cos (~S0°)
+ SO gin
Cs0*)
6,2=
28:0 MPa
hysx
- L.S5MPq
Gy
=
~o
-/04%0mMPa
<
20°
+ 40 eos (20°)
- SO sin (20°)
~ 40 sin (20°)
- 50 cos (20°)
~ Yo cog (20° } 4.50
sin (207)
6°
Tey
6:
>
“145 4Pq
—
-
a
60.4
MR
= 60,5 MPa —<m
- Problem 7.14
7.43 through 7.16
For the given state of stress, detertnine the normal
and shearing stresses after the element shown has been rotated (a) 25° clockwise, (£) 10° counterclockwise.
54 MPa
ips n
On
Gy
=
Tey
Oy
=
~ 2$°
-/¢
Toy! >
Ba
~/#
(bo)
6 =
wet
10°
714
+ 10
26=
cos (-S0°)
eA
Troy zo
MPa
8s
Cos 26
4 Cry
.
~ On
By
Sin 28
4 Try Cos 20
= Gxt by
2
— Sas
tos 2O
=
By
-
42
=
63.) MFG
mm
-Jo sin (-50°) ~ 42 cos 50°)
=
26-6 MPR
me
~Yfel MPa
met
20
sin
26
sin XO
~50°
gin(-So*)
~To cos (-§0")
42 MFe
+ Be
4 42
sin (-§0")
= 20°
4710 cos (20°)
- 42
sin (20°)
=
STA
MPR
ue
YE MPG
es
~ 6504 MPa
at
bry =
—J0 sin (20°) ~ 42 cos (20°) = ~
6y
~ 10 cos (20° ) + 42
= ~/4
FF
Seo Sy = To Mf;
7}
@ =
sz
= ~/EMPR
&y
(a)
Gy
6, = SGM
sin (20°)
=
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
atid educators permitted by McGraw-Hill for tlicir individual course preparation. A student using this manual is using it without permission,
Problem 7.15
_
7,43 through 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (5) 10°
counterclockwise.
90 MPa
{30
HH
MPa
6, = - GO MPa
60 MPa
;
Cry
ty *
6y
©
Gy
= ~ 26°
=
If
Bey =
Gy
=
()
1S
26
oy
=
+4 Su
+ 6y
_
cog £0
4 Try
Sin 20
4 Ty cos 26
Sin 26
Gy - Gy
6,12 +SC.2MPQ
+ 75 sin(-$0°) + 30 cos (-50")
Tey! = ~88.2MPo
+75
cos GSO") ~ 30 sin(-S0")
6)’ =
oe
86,.Q2MPa —<
26 = 20°
+ 30 sin(Zo°)
+ 18 sin (20°) + 80 cvs@o")
4S
Gx
= - 50°
IS - 78 cos Qo?)
Tey *
6,
MPa,
SoS = -75 MPa
~ Sx; Sy
~
Ty = 32
= TS cos(-$0") + 30 sin(-So*)
7 10°
Gy =
Go MPa
OstSy = 1S MPa
Gy >
ta)
SF
18
cos (20°)
-—
Bo
siul2Zo’)
G.1= -4S.2 MPa
Tay’ = S33 MPa ~<a
G!
=
75.2MPa «6
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i without permission. _
Problem
7.16
7.42 through 7.16 Yor the given state of stress, determine the normal
and shearing stresses after the element shown has been rotated (a) 25° clockwise, (b) 10° counterclockwise.
AG MPa
alee
NY
|
6, = 0
One Sy
6y = Sbmh,
= 2& MP,
ey = 35M PG
S= Oy
>
+ 28M Pe
Gy = Ft8e 4, S1=SY cos 20 4 Ly sin 20
Tey! =
Gy
@ = -25°
(0)
- Oe Ox sin 2
- Ou +6y - Que Sx Cos 20 - Tey sin 26
26> - 50°
Sy 7
2% = 28 cos GS0°) + 35 sin SO")
Tey
28 sin G50")
=
Gy =
4.35 cos CS0°)
=
= = feF mPa
[0S MPa
M PQ
2% 4 2% ees GS0°) ~35 sin (-So*) = 72°F
6 = j0°
(bo)
+ Ty cos 20
Ge
= 22
Try
7
Gy
=
2%
a
me
2@ = 20°
- BP ews (20°) 4 3S sinQQ0°)
5
/3°7 MPa
—
29 sin (Q0°) 4 3S cos (Q0°)
=
425 mPa
—s
= 43-4 MPa
—
+ 2F cos (20°) — 3S cos (20°)
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 7.17
7.47 and 7.18
The gtain of a wooden member forins an angle of 15°
with the vertical. For the state of stress shown, determine (a) the in-plane shear-
ing stress parallel to the grain, (6) the normal stress perpendicular to the grain.
weeeeemmacecrn
ot FMP A
@
GEILE
Tony = 42 MPG
6, = 0
©
6, =
|
|
= - 30°
2Q
= -1S°
wm
Problem 7.18
Gy
it
~O
Gi
+ 42
3B: 64 MPG
= Set Sy
5
CO
=
= Qf
+
20
20
Cos
Try
cos (-30")
~<a
Sx- 8)
4+
+0
H
(ob)
Sin
~ S493
=
xfyt
(2)
cos 20
42
+
Ly sin 20
sin(-30")
MPa
~<a,
7.17 and 7.18 The grain ofa wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (6) the normal stress perpendicular to the grain.
L5 MPa
6
=
— 25
MPa
GQ, =
-|.5
MPa.
Try
* 0
28 + -30
Q + -15%
@) Try = — Se = Sy sin 20 + by cos 20
See
—_
-<
1.5)
om Ces)
RSS
sin. (-30
ve
)
-
oO
Tey! = - 0,250 MPa alt
(b) Gyr = Oxt&, + Ox=6y, cos AO + Ty sin 20
2.54 (le). . 228215) eos (-30°)
+ O
Gut = - 2.43 MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
ay
Problem
7.19 Two members of uniform cross section 50 x 80 mm are glued together along plane
a-a that forms an angle of 25° with the horizontal. Knowing that the allowable stresses
7.19
. for the glued joint are ¢ = 800. kPa and 1 = 600 kPa, determine the largest centric load P
that can be applied.
«es
yi
{PA
For
plane
an.
Q=
65°,
50 im -—
G
=
6,
cosO
4 Sy
AS
Pe
T
Sin’ O
O
>
O +
-
3.490x10" N
+ Try (cos @- sin’ O@)
=
= (Sdxl0"*)(20 x107* 600 xl").
sin 6S" cos 65°
Aowabte
cos
vatue
Pon?
6S°4
A
O
syn * 65°
~ Sy) sinQecosO
At
-~
sinQ@
(S0%J07* (30 x1o7* j(800x10")
sin® €s°
= ~(&
+2b ry
f
sinGS coslS+ O
g a7 u Jo? N
BIN 6S” cos 6S"
of P is the smadler one.
P= 3.70 KN
Proprietary. Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be-displayed, reproduced, or
distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation, A student using this manual is using it without permission.
~<a
7.20
Problem 7.20
A steel pipe of 300-mm outer diameter is fabricated from 6-mm-
thick plate by welding along a helix which forms an angle of 22.5° with a plane
perpendicular to the axis of the pipe. Knowing that a 160-kN axial force P and
an 800 N + m torque T, each directed as shown, are applied to the pipe, determine o and 7 in directions, respectively, normal and tangential to the weld.
6mm
:
Weld“ U
de
=
O&3in
CQ,
=
Ca-t
A
=
Wc,
Te
hoc!
P
zat
6
Cy
=
=i,
°
a-l5in,
=
psoobhin
-¢,)=
Te“
-
ol4e? Y= SS4EP xs07* wr
Flot et)= Fost = oma” ) = 9-eni0-6 m!
oe HD
kS4i xo ~*
L
o-144
StressesSs
60 Kto
me
A
;
‘29-99
MPa
e002
( Boo \( 04415)
t=
6,
=
oO
Oy
J
=
J
Tse
>
—
29:PF
MPa
}
Ts
XG
at
ao
Tey
v
Gx=8F
ihe)
tt
q
- Sa
Se
sin
~E-Capee
—~ fe
cos
~ EG
= 25:36 MPa
if
poo2
MPa
Mfa
the x’ ane y
¥
|
pespe elive 4
aves
aud novmeh
te
wehel.
the
and Toe Tey’
6 = 22.5°
Sy = StS)
s
=
Then, Guz Gyr
4
72.5°
ty’
;
tangential
'
——
G
MPa
Choose
\~
\
MPa
|
= food
DE x10"
=
MPa
20
~
Day Sin
29
Weg 45° 002 sin 45°
O=-
20
+ Try
25-4Mfa
eos 20
of SI 4S” +
f002 cos 45”
wi
DS
MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
<=
Probl em
B
7.94
7.21 Two wooden members of 80 x 120-mm uniform rectangular eross section are
.
joined by the simple glued scarf splice shown. Knowing that 6 = 25° and that
centric loads of magnitude P = 10 KN are applied to the members as shown,
determine (4) the in-plane shearing stress parallel] to the splice, (6) the normal stress
120mm
perpendicular to the splice.
SO
Pavces
A = (Bo.i2ac)
(by
+f Fy
= 0:
as
(a)
= 4.6xl07
462K.
0%
Tz
RA
Problem
7.22
= Chas
*
N=
>
m
(Or ga
3)
Psin= (loxto'ls
2S? = 4.226x10°
in N
Age
S-PeesR2o
:
-
mat = 9.Ex/o° we
N= PsinBrO
ane
Areas
=
186.0%10 Pa
S=
si
G.0631(0") sin2S
=
(36.0
kPa
wet
PeosB =(VOwlo
jas 29%
% = F.063%19
=
399
10° Pa
=
39% kPa
N
=e
7.22 Two wooden members of 80 x 120-mm uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that 2 = 22° and that the
maximum allowable stresses in the joint are, respectively, 400 kPa in tension
(perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine
p
the largest centric load P that can be applied.
120 mm
. ry
P|
N
80 mm
;
3
A
A«,
aS
ae
™,
x
Forces
A=
Nae
(g0\Ua0)
= &eoxfot ee
fw.
Gy AAR
=
¢
"
Aveas
= 26 HO
focusLO ICC46.6 nie ety =
10.251 «10°N
»
+f
Zz Fy
70
$
N-
Psin
Bro
P
334.
=
to
-3
The
Fe
~=
O:
smabler
-
mo
Ss
vadue
Pees B=
Tv
Oo
P governs.
P-
.
ea
=
27.4
xl"
N
-
Sue= Ta A/singg = MIEMOT?)
C22 SmOP RL?
+2
=
Le ange
ot N
S
asf
_7S
IS 3% vio
P=
=
“4
16.59*ip
16.58
kN
N
<a
p
7.23 A 19:5-KN force is applied at point D of the cast-iron post shown. Knowing
Problem 7.23
that the post has a diameter of 60 mm, determine the principal stresses and the
maximum shearing stress at point 1.
asin
oct
F
D
components,
2
and
¥>
%;
imto
force
kN
19.5
the
Resofve
325 mm
=
300)" 2 2sy
hoe =
300 nnn
F,=
~
O
125 min
S)=-7.5 kN
B24.
A=
~
sH+IQ kM
3q¢ (19-5)
Fy
VE
a
a
.
.
390
.
—I9xiON
=
r= 7SKIO
325
sy stew
the force ~couphe
Determine
the
poimt
on they ents
whene - it
containing
plane
the
itensects
H and K,
ebement’s
1X
Fyt
O,
Fee
Fyre +75 >10°N
- IBx1ION,
Matz (1S KN )( 300mm = [00 mm) = = LS XID Nem
(75 KN )CUSOmm)
My
12S %*10* Nem
=
Mote ~ (1B kN )CISOmm) = -2 7107
Nem
ST
a
Properties
of
A=
Te
Tet =
Bet =
S=
Ret
=
(Semrcivele)
=
ZGo)
at H
=
H
o= Kk
|
.
t *
i}
|
ener
|
_ Fy_
GOwms
mM
6Oxlo72
Me . ~—/8xto®
777
eae
CBG. 17 eIo™
7 27410
= 64,870x10°
a
[29%Jo* mm = 18% 107% net
Q=%e? = #@Gc)*=
tad
Stresses
| o
c= Fal = BOmm= 30 x10 > m
(Civcde)
x10 tn"
2.3274 x1Io> mm* = 2.8274
636.17 x10 m"
636.17 *Jormm! =
o
235%]7m*
i. 27235 %1O wm! = 1.27
section.
=
Toy
#(30)" =
Pa
OO
+ OES
26.5@6*%10°
=
Pa
G4.37
edt
MPa
(1. 1RS%107)\ (30 “io? )
ATIBS KIO
= 26.526 MPa
Continued
My
N
Probfem
7.23
continued
S
as
6, sstjmin 7 Sue Oy & (Sas Oy Yur*
Sonex jwnivn
_ OF64.37
=
7
a
= 64,37
+P
ents
y+
~32.18S5 4 41.707
(26.526)*
MPa
Gwen?
Brarn
Mavimum
sheaving
siwess
Problem 7.24
Ton
[ESV
Ty
73.697
=
Dane?
~4.52
MPa
MPa
4 TMP
7.24 A 19.5-KN force is applied at point D of the cast-iron post shown.
—“
matt
Knowing
that the post has a diameter of 60 mm, determine the principal stresses and the
maximum shearing stress at point K.
fog = e007 US)
y
Resolve
tuto
19.5 kN
the
¥,
>
= 325 mm
14.5 kN
and
Zz
force
—
_
308
Fy = Spe (17-5)
Fs
ae
Na
poi nt D
‘
= 1g kw = -axiotp
~ ae C4. S) =. 75k
Determine
the force ~coupfe
= - 7.5/0? N
sy stew af
the
point ou the yoaars
whene
intersects
the plane Contains
H
Fee
at
Fes 0
300 nani
BB ng
F
Components,
and
K,
it
ehements
0,
Rye --t8x10°N > Fy -75 *lo?N
Mite (1.5 KN (300mm - 100 mm) = = 15 «107 Nem
My?
(25 KN )CSOmm) = 1125 *%107 Nem
M,= ~(18 kN) 150 mm) 2-2. 7-10? Nem
continued
Problem
7.24
Properties
of
re" =
TGoy
Bet=
F(30)t =
|
(CCinede)
c=
Sel
2 20mm
2.3274 K1D> mm”
yi
i
= 30 xlo
= 2.3274
36.17% Jo% mm =
a
T=
section.
i)
A
continued
wm
«107 Fant
636,17 «107 m"
u
"
if
1. 2723S *1O% amt = 1.27235 10° m*
J = Ret = Fo)
(Semtcivefe)
Q = ¥e? = BGoy* = 13x/0% mm” = 18x10
mm
tos
Stresses af K.
GO
mm
oo ey
[s
“A
RaQ.
vs ~ Bee
.
It
|
Principal
o~
22.84
= ~66,.845
GO
x{o"™*
DT
Mt]
ARE
-
(2.5 *107Y(3x10)
<4
=o4
+ 73.29%
Maximum sheaving stress
30.06 MPa
Pag =
+
1.27235 Kilo
26,826 «10°
Cnt 8s
oty
(1.125% 10° (30 « to")
eee
(636.1 xjo"4}(Coxio*)
30.06x]/0°
Smotymien =
636.17™ |o-4
- 133.6% MPa
=
Pa
=
Ox
6 2.7*107 Veoxio™)
BR 827¢* lo
3.368xlo’+
+
m
Mad. IBKIO”
=
stresses
=
st
{33 6%%/0°
=~
OT
Crnaey erin
d=
(30.
+
{(PSy
o6)*
G.4S MPq
<e
Tong 7 73.3 PQ
<
Sma
Mea
Cig | CSS
+ Toy
Lop
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem
7.25
7.25
,
fhe axle of an. automobile is acted upon by the forces and couple
shown. Knowing that the diameter of the solid axle is 30 min, determine (a) the
principal planes and principal stresses at point H Jocated on top of the axle,
() the maximhtim shearing stress at the same point.
200 mir
150 ol
c=td=4(3e)
xk
Tevgton
Noe
280oso Nini
2.4 kN
a
vy: ~
*
T= =~
J
.
AT
To?
LPa
T= bet = 3976) mm!
Gz
ata
|
8
6
2.
ee
Wy = -
Hf
‘Tep View
teTt
|
ve a
= (160)(t4o00)23bhoo0e Nam
Samm
(2280)
po}
Tis
7 52:8
—-
Bending?
=
= 13S FS MPO
"1,
(0.625)
52-f MPy
aati
13.5°§ MAG
—
350mm
Gy
=
— 135°? Mpa
Gy
@
=z
O
Tey
= 5o°8 MPy
= 96 Py
R= [(HE Ys ay = feng
e(aaay
GC, =
Gut
oF ef
6. =
Cae~R = -672-9 ~ 36 = ~ (5364 Ma
2b
R
= -67-9
y
fan 28 p= Gy-~Sy
Op
=
resses
- 67-9 Mhe
=
Bie = 76 +05)
St
—»
~
~ 28° 6° and
4+ 86
(2 YU S2-F)
= 6749
—_
~a
b6ES?
E1Y?
<<a
(8! | hfe
29-6"
1f369 MPa
(bh)
Une =
R=
¥6
MPG
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
aah
Problem
Rep Pace
7.26
forces
on
7.26 Several forces are applied to the pipe. assembly shown, Knowing
that the inner and outer diameters of the pipe are equal to 38 mm. and 42 mm,
respectively, determine (a) the principal planes and the principal stresses at
point H located.at the top of the outside surface of the pipe, (b) the maximum
pipe
DCE by an equivalent
smearing
:
foree ~ coupde
sy stem
int,
* Same point.
shearing stress at the sa
at
Cc.
T=
(260)(200 )= 40000 Ninm
Me=
(120)(4e0) 48600
300 min
a
/
HN mm
F, = 2ooN
(Me),
=<
de
a
£(,4~¢,")
Le
4S = 039)
¥
=
C,7 de
i/9mm
la0732
At
mot
200 N*
"BSG N
mnt
c,-¢,
=
2mm
the section covtaining element +t}
Ts 40000 Nam,
M, = 4000 Nom,
Stresses.
Aobyo)
(24
Torsront 4 Ty2-FS =~ eed)
oe
Bending?
t
G,=-
Trensverse Sheart
“
Tat A :
6,70,
nb
Emme
:
My:
7%,
wx
=
VS.
MPa
h2
~
.;
Op
=
MPA
cooley
og,
4
7 Oe MPa
Ug
2 HP
2 To
Be
AV
Set)
Oye
Se
tan 2p
wa
ww ZO
~feosantay
Oy =~ Zomrme y
terre
a
~
(3.5°
and
Tos?
= ASI MPG
an
Pew
Ors
Zr~axss
.
=
Give = £(6,+6, \ = 4(O~ 20) = # lem pa
MPG
Tue = R
.
TOE)”
V, = 200K
gs mapa
. . (4beee)
(2)
E0397
20MFa
Eel
(b)
«2imm
Qy = E(c2-¢8)= £28 - 14°)= bol mn?
t=
Z
Pe 120 8
T
C,F
section,
Cross
|
.
z
F,
—
-
Bt
Get R = beMra
O°
Gave ~
Ror
af
oA
-2)/:2MPy
.
OO
135°2 From zemis,
Tes?
3D Prom
Zones,
Tony 7 ite? MPa
<
~<a
~~
Problem 7.27
—
7.27 For the state of plane stress shown, determine the largest value of o, for which
~
the maximum in-plane shearing stress is equal to or less than 75 MPa.
6,7
GO
lel
w= Sus Sy,
- 20 MPa
MPa. ,
=
6,
©
=
4
}
By
Then
~
=
20
Gy=
MPa.
6, -2u
60 MPa
Jut+
Ty
G,~2u
Largest
Problem
vatue
of
7.28
vs
MPa
or
205 MPa
=
G6, = 208 Me,
MPa
Gy
= 76 MPa,
=
(ay
Gare= £ (6246,
C=
Sant R=
=
6a.-
Sy
=~ SEMPa,
Ty
7
?
Ta = Roo (EGEN Gp = /[=EHP + Gp
70 Mita
6.
-B4.6
= 12.284 MPa
7.28 For the state of platic stress shown, determine (a) the largest value
of 7, for which the maximum in-plane shearing stress is equal to or less than
84 MPa, (b) the corresponding principal stresses.
ey
R
fe
+ Ty
Tey = f 8H -
63°
= §&¥ hy
= 66.46
Ma
74+ 8H = Fi Mia
=
—
= 7MPE
tt
(bY
=
+./ 757-20"
vequived,
J-
$4
th
-
60 ¥ (2\72.284)
=
uf
Us + 4/ R*- by
Gy *
75 MPa
4
R=
~77
-
MPa
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distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
Problem 7.29
7.29 For the state of plane stress shown, determine (a) the value of t,, for which the
in-plane shearing stress parallel to the weld is zero, (6) the corresponding principal
stresses.
IMPa
6,=
12 MPa
IZ MPa,
yy
Ty =.=
G)=2 MPa,
yoy
Since
Tey
=
°,
is @ peincipod
Op
ise *
F
X-cdinectrou
direction,
= -15°
°
mo
A bey
= &8,
tan 20
Tey
=
4
(6x-
Raf Ssh
6 ) ton ZO, = £(12-2)tanC3) = ty - 2.89 MPQ
+ ty = [ste aad = 5.7735 MPa
Gave = EGA)
=
7 MPA
GS, =
Sue - R
=
6, =
5.7735
7-
7.30
Problem 7.30
6.2 12.77MPa =
|
lb) Got Suet R = 745.7735
1.226MPa
105 MPa
SoMa
Gye?
» 6 = 105 mh
let
ue SxS
Tg = BEING
6, >
Torre
p= 3fR6, = Sy
+20
AP owabte
7
(OS +
ranae
(AN42)
2iMPa
=
Determive the range of values of o, for which the maxiruin in-
plane shearing stress is equal fo or Jess than 70 MPa.
= FO MPR
tp = 4 fot
= (ST MPa
ov
2)
MFG
= 6B, <= (84 MPa
Gy + Zu
se
u
(a)
|
_
|
|
+42
MPa
Problem 7.31
7.31 Solve Probs, 7.5 and 7.9, using Mohr’s circle.
7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (6)
‘the principal stresses.
40 MPa
35 MPa
. 7,9 through 7.12 For the given state of stress, determine (a) the orientation of the
| 60 MPa
_ planes of maximum in-plane shearing stress, (6) the corresponding normal stress.
“|
.O,
2 - Go MPa,
Gave
Moby's
Pon
Plotted points
= Sxt
Gy
Gy =~ 40
=
~
5d
MPa,
Loy = 35 MPa
MPa
cincle.
K 3 (6x,- Gy) = (- GOMPa , -35 MPa.)
(- 4o MPa ,
)=
Cy
CS,
Y:
35
C+ Cie, 0) = (-SOMPa,Oo)
@)
tan
B=
B=
ae
z
=
= 3.500
74,08"
Og = ~ 48 =
a=
180° B
- 37.03"
=a
R= /66*+ 6x* = fior+ 35*
Sein = Gap -R > ~S0-36.4
=
Suny
(a’)
8,
Gay t R
Og, + 48°
=
O,=
(')
—
> 105.95"
O, = Lg = §2.97°
)
MPa)
OQ, + 45°
Tran =
KR
=
Cae
6
F
=
36.4 MPa
Sag = = B64 MR,
Shae =~ 13.6 MR, —e
~SO4 36.4
=
=
|
7.97°
= 97.972
36.4
|
<<
6,7 %3.0°
~—
= 36.4 MPG et
Cue
MPa
* - 50 MPa
Oy = g.0°
|
S =-50.0MPa =
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distributed in any form or by any means, without the prior written perrnission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
7,32 Solve Probs. 7.6 and 7.10, using Mohr’s circle,
Problem 7.32
AO
75 through 7.8 For the given state of stress, determine (a) the principal planes, (b)
the principal stresses.
MPs
HO MPa
7.9 through 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (6) the corresponding normal stress.
28MPa
Gy >
28 MPa
6.
=
oy
ave
3
+ Sy
2.
%
=> — tomy
=
- 54 MPa
Tay =~ Az MPa,
;
Ty (MPA)
. Photted
points
fon
Mohs
eivede ,
X * (62,~ Ty) = Caaf, 42mPa)
V2 (Gy, ty) (0 mra,~ 42 MPa)
Cc: (Sn, 0 )= © s6ura, ©)
@\
~FX
. #%
=
tanh = ee * FR
ol t 26572
Ox *
~$o
=
~
B=
1B0°-a
Op
$f= 16-72°
13.297
* 153-43°
R=fCCrV +x)”
(b}
Cn
Same
=
Os
= flea) Caz>
Say
*!
tR
= - $64 934
Gan = Sic, = Sve ~ RF
(a)
93.9 Mla
Gree = 379 Mfe
Gos,
2 74D
elt
Mfg
Oy,=
O,4+ 48°
=
3r7?
O, = 31-7°
<
O.=
O, + ¥S°
=
12h72°
O- = 12h-7°
=<
Tune = R
(b)
5b - 2349
=
G’=
Cie
> 939 MPa
<A
6! 2254 MPa
—_
Problem 7.33
.
7,33
Solve Prob. 7.11, using Mohr’s circle.
79 through
a
42 MPa
7.12
For the given state of stress, deterrnine (a) the
otientation of the platies of maximum in-plane shearing stress, (b) the corresponding normal stress.
,
26 MPa
63 MPa
G,
-
>
634Mfa
Gy = - 42 MPa
Bys 2F Mfg
Bae 2 StS - 105 ahs
tM
fa)
D
co
Points
Y
= (64mfa, 28MPa)
X: (6 -%)
g
4! (6y, Uy) =(-42 mPa, 22 MPa)
Ct Bae, 0 ) = Uoremm, O)
>
R =
Tmax
45°
+,
GO
@,-
|) CF’
™
= 6.
6'
6
(4 Pa)
0-9333
= Ih04"
8, = g&
Oy=
la
.
28.07°
A=
6,
2 =
*. Brg
EX
Ce
tan a -=
__
R
45°
=
$4.04°
= -30,96°
4 Fx* =f 525% + 29%
-
545
MPa
Ss MPa
2 lo
si
<a
= 59:5 MPa
—_
~<a
Proprietary Material. © 2009 The McGraw-Hilt