1.7 Knowing that the central portion of the link BD has a unifonn cross-sectional
area of 800 mm2, detennine the magnitude of the load P for which the nonnal stress
in that portion of BD is 50 MPa.
Problem 1.7
OBI> -= 50
M Po.. ::- So ~ 10'
<7
2.
0
eD ':" 000
W\W\
=- oDO)(
A
~Ct
10
-,:l-
WI
Fao = oSDABD= (.sO></O")(/?oox/6()
D",o.,.;+iI'H- bootl ,L.~V'4.~
Cj
t
c,.
Clfbod,
+, Me.=
FQD
p=-
(S'JO
=-
~D ) -
0.135 P
': 0
I.SC.'6b FaD
'P ':"(1.5b8b
p
A'B D.
0
240
(O.lfSo)
= qox/O" tv
)(4-U)( IO!)
c.;2.7)( 103. N
r;;'2.7 ldJ
-lilt
Problem 1.10
1.10 Two horizontal 5-kip forces are applied to pin B of the assemblyshown.
Knowing that a pin of D.8-in.diameter is used at each connection, determinethe
maximumvalue of the averagenormalstress (a) in linkAB, (b) in linkBC.
Use
jDi\l\t
B
C<..$-tV'ee
boJy.
10 K",s
F~e~<
fO I<.'ps
L<.1'V of
Foil'/: t +r-I'GlII\~Ie.
S iVles
FAa
&.
,'", 600
-
s,'." 'ISO
FA-G ~
AS
LI'~ k
S
-- - ,.'Iv orb
051.., ~.
F~c:. -=
7. '32 DS k"f'5
a'S Q.. +elf\~jo
~ e~
(el) stress
.;t1A8
Lank BC
,'s Q. ~()N'\pr--e-s.siC»"l
c.~"ss
(6)
s e.c+i()V\~
S1¥es.s i~ 8('
q,v-eet
p,""
,'s
Gee. .::
' a.
O.S'II\
lA\rla = (I. g - 0.8 )(O.£")
s eCJh' o
6:Aa--
kIps.
be-"
M'V'lIIJ"'"
t:tf
8.QbS8
P"'B
..
Aliff'
.- -
7;.5:(0'5""
::
0.5
Iif . f;;lf
'<5"
...
~e~beV'
A:
(f. S ) ( O. 5) =
- Fee. -
A -
o. q
"" '2.
- 3. Cf~S'8 =- - cr.q6 k.st
o. <:t
~
Problem 1.17
0.4 in.
1.17 A load P is applied to a steel rod supported as shown by an aluminum plate into
which a O.6-in.-diameter hole has been drilled. Knowing that the shearing stress
must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, detennine
the largest load P that may be applied to the rod.
--1
0.25 in.
t
Fo~
Fot"
,
~..Pv ,;nv",,-
rt2. = ~2.
LiMI'+"v\~
P
1:' ':. A.
p
:.
v~jve
A2.
=
iT" d
AI =
steel
...
"lI
t
=-
-IT
(
=
?:
P
IS
tJ,e
.IIII(0. ~)(0. '.n
L
D. ;SlfD
A, 1:'. ~ (0. 7.No¥~)
==
13.57 k"ps
,.<;) ( C>.~ 5')
1> = A2.1::2.= (I..'-Sbb )('0-)
(.)f
1T r.lt = 11
=-
I. :zSbb
i",1.
-;: \ <. S"7 k;f-S
5t'11e~jJe.l"'VCA-JlJe'"
P
'::"
':<.SI
k'I':).5
....
Problem 1.31
1.31 The 1.4 kip load P is supported by two wooden members of uniform cross
sectionthat arejoined by the simplegluedscarfspliceshown. Determinethenormal
and shearingstressesin the glued splice.
p::-
At:.:- {s.O
6=
e
14-00 110
P
)(3.0)
GOS
~
~
e
A,
::-
9()t::>
- bO.
-=
3Do
IS 1i'\1.
(/'-/DO )((..0$ 30")2.
-
Is
D -= 70.0 p.,.t' .....
T-=
P:s;",':?e
2Ao
-
(I '-tOO '\ s;V\ 60"
(2J( IS>
L
=- 4o_4 psi
.....
Problem 1.34
1.34 A steel pipe of300-mm outer diameter is fabricated nom 6-mm-thick plate by
welding along a helix that forms an angle of 25° with a plane perpendicular to the
axis of the pipe. Knowing that the maximum allowable normal and shearing stresses
in the directions respectively normal and tangential to the weld are u= 50 MPa and
.= 30 MPa, determine the magnitude P of the largest axial force that can be applied
to the pipe.
p
6mm
-Ir0.300
dlJ:
weld~
I'll
r..:= to - t::
fo:: 1'010::
O.ISo - 0.00.6 = O.It.ft1
Ao ':. 1f ('f'01..- r~~)::
Ba.seJ
\G" \
0'"
p::-
::
50
Aors
Ct>& 1
e
Bc...seJ o~ l't \ = 30
p
=
~ Ao 1::
~h
SY'f'IalJet/' vAlve
MPo.:
=
( S".~
~e
IS. +~~
~/O-g
Vt'i2..
;l.So
::
(5::
ro
CDSze
><I Q. ~ ) (Sc X' (96 ')
e.os'Z. 25'0
M~
:::
~
rr ( (). ISO7..- 0.. I"+t.i
2.)
::::- S_StJ
e
0. \5"0 M
?:. :: ;5;0
-
337 , ></03
::
4-34 X 10
'51'", 28
(Z)(s..SL/~o-~)(30xlO6)
:3
si... 500
JjJo",,~4e
vc...Jv~ of "P
I"
P = 331
kN
-'
..~
1.35 A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
Problem 1.35
p
36
AD : (,)( Eo)
G::::
(0...)
p
-
AQ CO$
t'I'1c:;..,6te", 5";
M~..
( b)
2.8
. l.
rh
-2 6'to
s-h..-e..s.
~o~t-e-~si".e.
~t
Cu
==.
- '.67
Co):''1.8
o.-t e = 90°
:;: 0
~+V"'e5.$ = <0.67 ks,.
e=
e
~
0'"
'T..."1'::' ~~o ::- (~~ ~~6)
~+
'a.S
:=
~ LfSo
3. ~3
K'=>"
~
1.46 Two wooden members of 3.5 x 5.5-in. unifonn rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 75 psi, detennine the largest axial load
p that can be safely applied.
Problem 1.46
5.5 in.
= (3.S
1\0
e
r -::::.f
'P =
'P
:51'..ec.o.sG
2A.~
S \-'1~e
'=:
'10°
-=
:<1\0 S
.
)(5.5)
- 20° -
=- I Cf.?5
j",~
70"
'J
I Y'I .-{
B
(:{)( Icr. 2 ~ )L7§)
sin 14Do
-= 44'11- jib
4. 41 /<;r.s
PROPRIETARY MATERIAL. @ 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
-4
Problem 1.51
1.51 Link A C is made of a steel with a 65-ksi ultimate normal stress and has a
t -in. uniform rectangular cross section.
t
x
It is connected to a support at A and to
f
memberBCD at C by -in.-diameterpins, while member BCD is connectedto its
support at B by a I~-in.-diameter pin; all of the pins are made of a steel with a 25-ksi
ultimate shearing stress and are in single shear. Knowing that a factor of safety of
3.25 is desired, determine the largest load P that can be applied at D. Note that link
AC is not reinforced around the pin holes.
Bin.
D
Use
+v-ee bo,ty
+J
p
MB
p
~<
B,r-4f
= 0:
(G)Ci~ FAc.) -
-=- o. Lf~
+) Mc ::, 0:
2
B.j:-
is: +
~
'B.J"). -: -/1. ';.5,1. -+ (~)~
She"-'" j", p;"s
L""
-=
r,~c.
oVlnet
F~~ =-
S~.»
eo/' vJve
Fo"o.>~
(,)
S~eo.""
in
B=
::=
-y",
F:~
.sed-I'C)"",
6' A-t
=
:II oJ7.
::
tt
ev+
Ff'u"- (:t)
AP)(.)..,j,JJ~
.
P""
P
.
P
I.e.
1.&.11'-(,.1
-
0
=
0
~
- G B:f - 4 P
A
C~s ':ill.V'3 )'
':3.~~) 4 k-s
~ '"
2.
I
.3 'P iI
p = 0.7058'-6
P
'8
(2)
=-
D.1?<t~ 5'
k;rs
<:Uo\",,1
C.
~
o. 6~.S k:rs
O.l:2.S k:ps .
p -= (O.I..f~)(0.'2.'5')
'L A
~
~F:
10 Ae.
'- Zs P
=
~~, Ah.t ~ (3~~-rtit Xi.- ;)
of' F,.<.. is
p;'" J
B-
A c..""J Co..
.
API'"
1::
Te.VI'5I'OV\
wt
p
-.3
.,.. 0
Ct~
. X
B)( =- I~ ~c
tIp
10 P
I=AC.
-+21=,,=
r 0:
By I
B
BCD.
= o. '300 k;rs
13.
~
t
1=:s.
.!LJL~
~
~S\l1[
-= (O.70S-8:8)(O.S8q~Cj)
vJve
of'
'P
;$ f~'l
S'
( 3. :<5' 1\ &f)( 1C:.)
Ly
'=' 0.4/'
OS-gqqq
.
k:
P~
k:ps
P =:
5Y>1e..U-et/' V".,YIJe,.
6~
0.300
p -= 300
k:ps
l.h
....
Problem 1.55
1.55 In the steel structure shown, a 6-mm-diameter pin is used at C and IO-mmdiameter pins are used at B and D. The ultimate shearing stress is ISO MPa at all
connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a
factor of safety of 3 is desired, determine the largest load P that can be applied at A.
Note that link BD is not reinforced around the pin holes.
18mm
+ ee
Use.
D
Front view
+'LM(:,.
6 mm
D.28b
A
'I
B
~w~'-~w
bod, A Be.': 0:
p -
O.I~O
Fe.o = C>
(I)
~ Fa.,
p""
Side view
+J~HI3
p
=0:
'oomm['oomm1
o. I(.0 P
A
B
Top view
p
C
- O.':;(
0
C-
,3.
~
,~C
=
0
(~)
JeV\'5io,,", 01'\ net- sed-;oV\ of J>;V\kB D.
~GOf
Gt
p1
~C1O-= 6" A~t
-::
~.
AMf
=- (400)((O)(6)iIO-S)(I~-IO)(lo3)
3
l:.JtO
-=
SheCti"
;VI
ry
Fs.D
$w-J
pi
==
v\s
d
8
. ::>
Api'"
1:<.1
(1 )
SheaV'
C
;V\
::
lJ.
F.S. 't
Pe". v,.}!)£? of'
~ f"O"'"
CtVlJ
('1.')
SMJJ
e'" vJVf?
N
d 2. -- ( Jso')l./O"
)(lL4 )(, 0
2,
.
l
>l
-3
0
)2
-
1>.9~~O:
,>,/03 N
Feo IS 5.'91.,,70.></0'3N..
pi \11 c.J
Wvw-.
,o?
'[).
I.G~.3 ~ /03 N
1> = (~)( Bll:f J..7ox[!cl) -:
2 rC Api...
)l
=-
C
~
~. ~c1?.'" (2)('5"0 ;(0'-
'P = (~Y1.g;('N'Ic:{d)
:-
)(~ )(b)C/O~)a.:" .Z.~~7'f .,./o'J. N
2. ,~)( lo~ tJ
~f -p i~ JJo..J ~Uf(' vc..L~"
'P =- 1.'~3
x 10' tV
'P"'" I.' 83 IetJ ~
1.65 The 2000-lb load may be moved along the beam BD to any position between
stops at E and F. Knowing that a.1I= 6 ksi for the steel used in rods AB and CD,
determine where the stops should be placed if the permitted motion of the load is to
be as large as possible.
Problem 1.65
60in.
P~t" "",dt~J
AS
~e-
be"..
(SJ...
+o.rc..e..s:
=
CD:
'-J7lo
Fc.,p
t--
10
~~
60;...
p= ~~)h
Use. Me- he'"
ib
+:> L HD
- (60
=-
)( )(i)
I. 'tD7g
k:f'.s
4.
f H.
2.DOO k:rs
..,
+ (60
Go - x:.. -=
b0
P
E?
=-
- 'X.,E)P
FAB
~
'):SF=-
=
0
(bo '>( I. 17g Joj
2. 0
0 0
35.3'f3
~G
Go ~~D-
bo...l7"
e.
0
) FAB
=
)(;i)
kps
BE F D 4.S
'P = A.OOQ
+)ZHg
)(
(F'"D ),.,"'" -= OJR Ac.p = (b
':"
t='AIJ
= (6
=- OJ.( A
-= '2'1.7
iYl.
.4
0
X-P
bO ~D
P
= 0
"=' (bo)(J.g4o7i,)
Z. 000
X-p;:"
55.:<
in.
~
Problem 1.67
1.67 Each of the two vertical links CF connecting the two horizontal members AD
and EG has a lOx 40-mm unifonn rectangular cross section and is made of a steel
with an ultimate strength in tension of 400 MPa, while each of the pins at C and F
has a 20-mm diameter and is made of a steel with an ultimate strength in shear of 150
MPa. Detennine the overall factor of safety for the links CF and the pins connecting
them to the horizontal members.
1"250 mm
1---400m~
B
I 250 ml
Use MeMbel" EiFG
0..$
~ee bcJy-
t
Fee
F
~
~F
O.Lfo ---+=-°.25"'
24kN
21l'kN
1) ~ ME = 0
O.LfC Fc'F"-(O.G5')(~'t""O!)
Fe.F '::
8a.sed
in .Pin ks
0111 te.~5ili¥l
Cb- d) t
A:::
on
= (0.0£10
- 0.02.)(0.010)
&o!} I-::;Ji? .sheC(.~;n
A
'= ~r;l
F",
~
Ar':.+,}(,v
.JJ.. t~
2 YuA
F,)
7T
= ft(o.O2o}
:=
tS
{:"Q.(;,-L;)If'
ot -se:o:fe.ty
IO~
N
~
= '20D t./o-c \"1'1" (()~ej"nk)
':"
IGO.O
"1l103
N
P,"YlS
=
314.'bx/O
-,
~
~
(2){t5D><ID')(3ILf." )C/o-') = <7't.248 )(/O!
SW1a.JJev'" \le.J~e.)
F;S.
~
0
CF
Fu '=25'vA= (2(400'tIO')('200'JC/O-~)
B~sed
3Cf)(
c
~IJ
':
'-CI="
i.e..
Ft) =- QI/.'248
'ftf.'248
:/0'3
3'" ~ 10
~/O3
':
N
N
Z.1.f2
..
Problem 1 .69
1.69 The two portions of member AB are glued together along a plane forming an
angle 0 with the horizontal. Knowing that the ultimate stress for the glued joint is
2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of 0 for which the
factor of safety of the member is maximum, (b) 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.)
Ao :: (:?D)(J.
2..5)
1::
';..~b
ia-t1-
A-t rl..e °p+,'V!f\v"",, o.YI~Je
0=:5')0- -=(FS.)t'
,,/
n
'r.::p-,
'l/,\
.
"D
f5'uA.:>
IV,O~ ClA' S V'"€.$.S' '-.J = - C<:>.$ t::7 '.
IV <S"=
L
+
( rs
-:
)
r.. . \)
Sh e.Ct"':Vltl\ sfV' e<;$:
..J
1:'.;:-
'P.s
Ao
I", e (.OS
( r-t-.
E1vJ;k:
h
'
S~J'II\ V\
Gb)
~
~:
A0
'P~
=
GuAo
'P
e
--
('" '\
;:0. It:
,:
'P
pI,)
I
1:' =
'PtJ.~
'P
--
GU-S ~
v
C oS
e
e
1vAe> --
-6IvIg (..M~e
~Ao
'PS;", e <..<)5
e
DuAo
= Tv Ao .
Pc.o.s1.<9
'Psi"lec:...>s$J
~..L
~.s e
Pc) =-fS'vt'°
Gu.s (;)
':::'
TC1V1
:::
e
1:'J
'::'
~
(J~.S}(1.Sc»
f:...o$7. ~7. 1;,'"
1.3
':'
=
1:5
,:'
c),v). 0
(~) fjo.,t--27.50 ..
7. '14 k:ps
ES. ': f'v
cry P = 7.;2.4
3.31
~