Solutions to Problems in
Strength of Materials
SI UNITS
Solutions to Problems in
Strength of Materials
SI UNITS
Y. BRONDIAL
R. GALLARDO
A. SY
HI
Copyright 1984 by Y. Brondial, R. Gallardo and A. Sy
A ll rights reserved.
PREFACE
This volum e contains solutions to problem s in Strength o f Materials taken from the te x tb o o k “ Strength o f
Materials” b y Singers and Pytel, 3rd Edition, as w ell as solu­
tions to selected Board Problem s given over the past years.
This review er is designed especially to facilitate self-study
fo r students and graduates o f engineering w h o w ill take the
board exam ination. T h e unique feature o f this review er is
that all problem s and solutions are w ritten co m p letely in SI
units. T h e authors believe that once the fundamentals have
been thorou ghly mastered in one system o f units, extensions
to oth er systems should n ot be difficu lt.
The authors wish to express their indebtedness to
Mr.
Ram on
Po
of
the
Popular B ookstore and to their
colleagues at the De La Salle U niversity. F in ally the authors
Wisjr to acknow ledge the patience and forbearance o f their
fam ilies during the many hours required to prepare this re­
viewer. A lth ou gh great care was taken to elim inate errors, it
is inevitable that som e w ill still be found. Th e authors appreciate being in form ed about these and w elcom e any suggest­
ions fo r the im provem ent o f this w ork.
Y ola n d a Brondial
R on ald o Gallardo
A rsen io Sy
Manila, Philippines
A p ril 1984
V
CONTENTS
Chapter
J*age
1
Sim ple S t r e s s ..................................................................■• •
1
2
Sim ple S t r a in ..................................................................• • •
23
3
Torsion
...................................... ................................... • • •
77
4
Shear and M om ent in Beams
.......................................• • •
103
5
Stresses in B e a m s ........................................................... • • •
131
6
Beams D e f le c t io n s .........................................................• • •
183
7
Restrained Beams
.........................................................• • •
219
8
Continuous B e a m s ........................................................ •••
235
9
C om bined S tre s s e s ........................................................ •••
257
10
R ein forced Beams
.................................................. .. .• • •
281
11
Columns
........................................................................•••
295
12
R iveted and W elded C onnections
13
14
i
........... •.......... .. .• • •
307
Special T opics
............................................................... •••
327
Inelastic A c tio n
.............................................. .............. •••
341
A p p en d ix A — Board Problem s
................................357
A p p en d ix B — SI Units, Conversion Factors
V II
...........•••
379
1
Simple Stress
PROBLEMS
103. Determine the largest weight W which can be supported by
the two wires shown in Fig. P-103. The stresses in wires A B and A C are
not to exceed 100 M Pa and 150 MPa, respectively. The cross-sectional
areas o f the two wires are 400 mm2 for wire A B and 200 mm2 for wire
AC.
Arts. W = 33.5 kN
W
Flgww P-103.
Solution:
Pa
w W
Applying +he law o f sines !o ihe force triangle-.
W
J in TS*
.
Pac
J in
66“
-
Pab
Sir, 4 5 °
'
2
pAC = o . s j s e ,
PM
*
tv
O. 7 3 2 W
= _ fk .
^ ab
to o
X
/O c
0 .7 3 2 W
Pa b
s
-p o o *
/O-
/o -6
</oo x
} AC
/S O
X /o‘
O. 89*>& W
=
200
X / O '*
200
X
/O'
3 3 , V 6 0 /V
'• IV = 3 3 , V G O N
= 3 3 . -Pfe A/V
105. For the truss shown in Fig. P-105, determine the cross-sec­
tional areas o f bars B E , BF, and C F so that the stresses will not exceed
100 M N / m 2 in tension or 80 M N / m 2 in compression. A reduced stress
in compression is specified to avoid the danger of buckling.
Ans. A be = 625 mm2; A BF = 427 mm2; A CF = 656 mm2
Use
t h e m e t h o d o f s e c t i o n s in ci'iviclincj the.
tr u s s . T a ke t h e r ic jh t s e c t io n o f t h e ir u s s
fro m
a
- <x .
o
•,
•
PCF (8) - 4 0 ( 3 ) - 5 0 ( 6 )
Pcf ^ S Z .S k N ( c )
(* f £ M 6
=
=O
o
4 0 ( 3 ) - PtF ( - * = ) ( 3 ) = O
PaF
= 4 2 .7 2
It? ± M F
o
pBe
«)
kN(T)
~ 5 0 (3 ) = o
PB£ = (, 2 . 5 h N M
S =-£
A
SO X /O 6 = 5 2 ,5 0 0
* 0 = X / O '*
/00
X 106 -
4 Z .7 T .O
Pi Be
/OO X /O 6-
;
= 6 5 6 .iS m m J
Agp - ^ 21. 2 mm 2
X /£>‘
(, 2 , 5 0 0
AB£ X /0 -< '
= 625 mm*
4
109.
Part of the landing gear for a light plane is shown in Fig.
P-109. Determine the compressive stress in the strut A B caused by a
landing reaction R “ 20 kN. Strut A B is inclined at 53.1° with BC.
Neglect weights of the members.
Ana. o *■ 65.7 M N /m J
Hollow strut
OD = 40 mm
ID = 30
200 mm
ngura P-10*.
Solution.
Q
±M C
0 .6 5 (20 ) = R a0 S in 5 3 .1 ° ( 0 45)
Rad
=
3 6 .1 1
klM ( C )
*AO
A*a -
* A0
A aq
J [o o 1 '
I D Z]
= ^ [
s n f t 2 ~ a o 3 x]
5
A ab -
S .S
X IO ~ * m *
~ 3 6 . 11 X I Q *
S .S
=
X /£>-*
7
*
110.
A steel tube is rigidly attached between an aluminum rod
•ml a bronze rod as shown in Fig. P-110. Axial loads are applied at the
|M>»itions indicated. Find the maximum value of P that will not exceed a
iiiru m aluminum of 80 MPa, in steel of ISO MPa, or in bronze of 100
MPa.
Solution:
2m
------ *|
-------3P
Aluminum
/)- 2O0tn2
&ee\
/MdOrtl1
2.5 w
2P
Bronze.
To find the net force acting on each section.,
isolate each section and determine the forces act­
ing on each to be in equilibrium.
P
6
Pa, = P ;
Ps =
For M
so X IO6 =
2P;
P0 = 4 P
---- f ------
P = /6
OQ N
2 0 0 X / O '*
For S+ee.L :
ISO
- /O6 s
/u
2P
------------- 7
4 0 0 X «> -*
*
P = 3 0 ,0 0 0 N
For B r o n z e :
t o o K /0 6= ---- ±P_-----
p_
SOO X ZO -‘
P=
I2 ,S O O N
-
i2 g o o N
•
IZ .S
kN
112.
Determine the weight of the heaviest cylinder which can be
placed in the position shown in Fig. P - l 12 without exceeding a stress of
50 M N / m 2 in the cable BC. Neglect the weight of bar AB. The
cross-sectional area of cable B C is 100 mm1.
n#u.-» p - t i t
Solution:
Taking the FBO of AG :
^AC —
»
PBC -
Sgc
too
X /O '* ( s o X / o 6)
5 ,0 0 0
/V
£M a = O
A/(<t) ~ 5 0 0 0 ( 8 )
N = 1 0 ,0 0 0 N
=
O
7
F o r th e c y lin d e r :
8
i F y -- O
to, o o o ( o . ( , ) - W = O
w
-
6, OOO N
113.
A 1000-kg homogeneous bar A B is suspended from two
cables A C and BD, each with cross-sectional area 400 mm2, as shown in
Fig. P —113. Determine the magnitude P and location x o f the largest
additional force which can be applied to the bar. The stresses in the
cables A C and B D are limited to 100 M Pa and 50 MPa, respectively.
Ans. P - 50.2 kN ; x - 0.602 m
1.8 m
1.8 m
Flgur* P H 3.
Solution:
Pac
P
ba
w
kV =
PAC
=
-V
mg
= 9.8 k N
-
1,000 (9 .8 )
=
400 x /o-‘ (IO O X IO•>) - 4 0 k N
8
*eo S so =
Pbo -
400 X I O * (S O X IO‘ ) = 2 0 k N
40+20
- 9.8 - P = O
P * SO. 2 KN
iM „ = O
10(3.) - 9 .6 ( l ) ~ P ( x ) ~ O
X = 0. 6016 m from ft
114.
As in Fig. l-10c, a hole is to be punched out of a plate
having an ultimate shearing stress of 300 MPa. (a) If the compressive
stress in the punch is limited to 400 MPa, determine the maximum
thickness of plate from which a hole 100 mm in diameter can be
punched, (b) If the plate is 10 mm thick, compute the smallest diameter
hole which can be punched.
Ans. (a) I — 33.3 mm; (b ) d * 30.0 mm
Solution:
P
£ 400 MPa.
St < 300 MPa.
punched hole ha* tw o area*
at shown., -the area. X to P U ~q-dx
while ihe a rea parallel At P is
•trot.
a.)
P,I
=
3
-Jj
^ d 2Sc - 'iTJiSs
Ac
0.1 (400 1 I O* ) = t (30 0 X 1 0 *)* 4
t - 0. 0333 nrt * 35.3 m m
b)
^£d*Sc = TTdiS,
d(<toa x to *) - o o /(300 x io * )x 4
d -
O. O S m
as 3 0 m m
9
119. The end chord of a timber trass is framed into the bottom
ohurd as shown in Fig. P—115. Neglecting friction, (a ) compute dimen•fcM b if the allowable shearing stress is 900 kPa; and (b ) determine
,lim«nsioa c so that the bearing stress does not exceed 7 MPa.
Ana. (a) b “ 321 mm; (b ) c — 41.2 mm
Solution:
krt
The.
a.)
-fo r c e ,
a c tin g
o n
t h e
e tu i
ch o rd , s h o w n ,
S , < 9oo kP*.
A ,
=
O .1 5 b
s' - i r
'
90 0 .0 0 0 -
Peas
3 0 '
O.IS b
= 0 .5 1 0 7 5 m
I
b )
S L
<
7
= 3 2 0 . 75 m m
M Pa .
= 0.15 C
^ P„
A*
7 x
10 4 -
S O , OOO
',''
C
=
o ./ s
CCS
30 °
c
0 .0 *1 2 4 m
=
4 1 -2 .4 m m
is
P u u 3 o " .
10
117.
A 750-mm pulley, loaded as shown in Fig. P-117, is keyed
to a shaft of 50-mm diameter. Detennine the width b of the 75-mxn-long
key if the allowable shearing stress is 70 MPa.
Arts. b ~ 11.4 mm
10 kN
PUptem P-117.
Solution:
tO
fetMo = o
1 0 , 0 0 0 ( J L Z S . ) - b f i o o (JL2 S . ) — P ( & £ 1 )
P = 60, o o o N
Ss < 70 MPc*.
s
-
p
AJ
7O X IOk =
y
-
As = O. 075 k
0.07Sb
O . O n 4 3 m
-
II. * 3
m m
11
118.
The bell crank shown in Fig. P-118 is in equilibrium, (a)
Determine the required diameter of the connecting rod A B if its axial
ttress is limited to 100 M N /m 2. (b ) Determine the shearing stress in the
pm at D if its diameter is 20 mm.
,200 mm
30 kN
P-11S.
Solution:
cl)
£M 0 = O
s i n t > 0 ° ( Z ‘t 0 ) -
3 0 ,0 0 0
P
= 31, 1 7 6 . 9 N
5 117 6-9
ir
d
b)
*F f
P (2 0 0 ) = O
-
0. D Z m
-
=
too
X IOL
2 O mm
=
D k -31,176.9Dx
~
3 0 , D O C cos 60° = O
1 7 6 .9
N
ify = O
Dtj - 3 0 , 0 0 0 sin 6 0 ° = O
Dy
D =
=
2S, 9 8 0 . i N
i/ fl,1 +
0/
= J ( f 6 , 176. i ) 1 + (2 - S , 3 9 0 - 8 ) 2
-
S 2 ,9 8 + / J
S in c e the. r i v e t a t D is d o u b le s h e a r e d , t h e r e
a r e tw o s h e a r a re a s .
«— ~
Ss = * ? . 3
=
s z ' 9* 4
z x 3 £ (o .o z )* -
12
121. A rectangular piece o f wood, 50 mm by 100 mm in cross
section, is used as a compression block as shown in Fig. P-121.
Determine the maximum axial force P which can be safely applied to
the block if the compressive stress in the wood is limited to 20 M N / m 2
and the shearing stress parallel to the grain is limited to 5 M N / m 2. The
grain makes an angle o f 20° with the horizontal, as shown. ( H int: Use
the results of Problem 120.)
Ans. P — 77.8 kN
P
100 mm
Figure P-121.
Solution:
C o n sid e r a. yrcvLn a n d d eterm in e- the components
o f P perpendicular an d p a r a lle l to the. q r a t n .
P^ =
Psin 2 0 *
Sc
2 0
^
U P ol
13
sc = _ *
20 X IO* =
0 .1
P =
( O .o s )
/kA/
Ss = 5 M P ,
F ro m ~tHc -fig u r e :
A s = _ f i _________Q ! ( Q
OS)
20*
20 *
A,
S X IO‘ =
Psin 2 0 “
0.1 (o .o s )
COS 20*
P = 77.79 k N
P = 7 7 . 7 9 A A/
125.
For the lap joint in Problem 124, determine the maximum
■afe load /* which may be applied if the shearing stress in the rivets is
limited to 60 MPa, the bearing stress in the plates to 110 MPa, and the
average tensile stress in the plate to 140 MPa.
Arts. 56.5 kN
Solution:
s* - - V -
3 x ^ J *
O
6 0 x /©‘
=
3 X ~ (0
O i ) 2-
56. SS k N
<L f c -----------S
5 tcL
HO x /0‘ *
pb =
3 (O .O Z S )(O .O i)
/CS k N
(p - n.d)"t
14
m o X !O i =
Pt
=
(0.13 - O.OZ)(O.OZS)
3 3 5 kN
. \ P -= 5 6 . S'S k N
127.
A 22.2-mm-diameter bolt having a diameter at the root of
the threads o f 18.6 mm is used to fasten two timbers is shewn in Fig.
P-127. The nut is tightened to cause a tensile load in the bolt o f 34 kN.
Determine (a) the shearing stress in the head of the bolt, (b) thr shearing
strew in the threads, and (c) the outside diameter of the washers if their
•nside diameter is 28 mm and the bearing stress ia limited to 6 MPa.
Figure P-127.
Solution:
P = 3 4,000 N
d - IS .(? mm
0 = 2 2.2 mm
34.000
=
VO W
MPa.
7 r(o .o z it)(o .o /z )
7Tdt
31,000
7T(0. 0!86)(0. OlZ)
c)
= 3 6 .3 7 MPa.
The area to be considered is the area of
the washers JL to P as shown in the
shaded figure.
14
Pb = P = 34,000 A/
^
=
- Dt * J
a.
t> X /£* =
3 4 ,0 0 0
-----------1
~ [ D ox (0 .01 8)1]
D0 ~ o. 0834 m = # 9 . ^ m m
128.
Figure P-128 shows a roof truss and the detail o f the
riveted connection at joint B. Using allowable stresses o f - «« 70 M Pa
and ob - 140 MPa, how many 19-mm-diameter rivets are required to
fasten member BC to the gusset plate? Member BE1 What is the largest
average tensile or compressive stress in BC and BE1
Ans. For BC. 7 rivets: for BE. 5 rivets
I
/
6m
c
^
sS E
Jw. 4 m '
4m
r
96 kN
4m
'
200 kN
F
G
'
N. H
4 m /_
96 kN
Figures P-12S
Solution:
S y m m e try
Ra -
R H - - i| * - = 0 6 k N
Using the Left section, of
the cut truss :
= 0
B £ ( } ) ( « ) + BE ( ± ) ( 5 )
BE = 80 k N ( C )
16
Using joint C :
BC
ZFy = O
= 96 k N ( T )
BC
Ar
fcN
For B C :
P = AS
(S h e a rin g )
96.000 = n. x 3 L (o . O n y (l O X I0L)
= V. #3 7 rivets
n.
(Bearing)
96.000 = f t *
■
5 rivets
0.00(> ( O . 0 l 9 ')(l ‘/0 x io L)
n
= fe .o * =5 = 7 rivets
use
7 r ive ts for B C
For B E :
(S h e a rin g)
60,000 =
n
(B e a rin g )
n x JL (0 .P I9 )Z(70 x 10*)
— 4.03
8 0 ,0 0 0 =
n.
•
=
rivets =£= S rivets
HX
O. 013 (o . 0 l l )(l 4 0
2.3/
rivets ={J= j
X IOL)
rivets
u se 5 rivets f o r B E
To get the Largest eu/erage ietvsile or compressive
stress In BC and 0 E /
P
The area- o f the angle section is taken from
table B - S , p. 6 4 *
For BC.
^aurtq\* - g6V x 10'* ntl
" 7
_5l£dt
96 X 10*
'(* 6 4 X /O-6)-(o .O I9 )(0 OO6)
ST =
For
BE :
A.
IZ 8 M P a
u = 1,750 x IO~*mx
90
5C =
=
X IO *
i,iso x io -* -(t > .o i* )(t> o n )
52. Z MPm.
17
I \1. A cylindrical pressure vessel is fabricated from steel plates
»lin It have a thickness o f 20 mm. The diameter of the pressure vessel is
Mm nun mid its length is 3 m. Determine the maximum internal pressure
Wln< li i nn he applied if the stress in the steel is limited to 140 MPa. If
»h* Internal pressure were increased until the vessel burst, sketch the
•vi,r " f fracture which would occur.
Arts. 11.2 M P A
S tilu llo n :
Ui'mtj th e g i r t h j o i n t :
\
-
F°
‘t t
m o x to* p
p ^ °'s^
if ( O.Ol)
2 2 .1 - M P a.
=
t A in j the lo n g itu d in a l j o i n t :
l'/ o x /Ob =
p ^ ° 'f ^
1 (0.02.)
p
m ax
U
=
11.2 MPa.
p -
I I .2
MPa.
ihie in te rn a l pressure u/ere. in crea sed u n til
I hr vessel b u r s t, t h e longitudinal, j o i n t will j i v e
uitij tin d
/
th erefore
th e fr a c tu r e is a l o n j th e
luidincbl j o i n t .
135.
The strength per meter o f the longitudinal joint in Fig. 1-16
u 4HO kN, whereas for the girth joint it is 200 kN. Determine the
ni» hi mu m diameter o f the cylindrical tank if the internal pressure.is 1.5
M N / m 2.
18
Solution:
L o n g itu d in a l
——
=
jo in t
480
:
kN
G irth, j o i n t :
*
-f o
200 k N
F or the Longitudinal joint: :
2
P
pD
=
L
: - z ( { ) - p D
Z x 4 8 0 x 10* =
1.5 X lO^D
-
D
O . b l m
F or the girth j o i n t :
P =
p
x JL p *-
200 x 10i =
/)
Z/j£
=
<7.5-33 m
0 = O .S 3 3 m
-
533 m m
136.
A pipe carrying steam at 3.5 M Pa has an outside diameter
o f 450 mm and a wall thickness of 10 mm. A gasket is inserted between
the flange at one end of the pipe and a flat plate used to cap the end.
How many 40-mm diameter bolts must be used to hold the cap on if the
allowable stress in the bolts is 80 MPa, of which 55 M Pa is the initial
stress? What circumferential stress is developed in the pipe? W hy is it
necessary to tighten the bolts initially, and what will happen if the steam
pressure should cause the stress in the bolts to be twice the value o f the
initial stress?
Ans. 17 bolts; 75.3 M Pa
Solution:
D = D„ - I t
=
D -
O.VS - 1 X 0.01
0 .1 3
19
n * X — tig^Sg =
p X -Q D 1
PD’
nB =
da 1
3 . S' jt IQ* (p . t i l )1(D . O V )* [(e O - 5S) K IP 1-]
nB -
ib .i
17 bolts
2t
ST =
3.5 X \Pk ( 0 - H 3)
= 75. 3
2 (0 . P I O )
77>e bolts are -tiyhte-ned initially to create a.
6ea.liny pressure on the gasket. I f the stress in the
bolts will be twice ihe initial stress, ihe bolts will
hat/e lengthened by the amount originally compreainq
the gasket, so that the ie&lrng becomes ineffective.
137.
A spiral-riveted penstock 1.5 m in diameter is made o f steel
plate 10 mm thick. The pitch o f the spiral or helix is 3 m. The spiral
seam is a single-riveted lap joint consisting o f 20-mm-diameter rivets.
Using r = 70 M Pa and ab = 140 MPa, determine the spacing o f the
rivetsalong the seam for a water pressure of 1.25 MPa. Neglect end
thrust. What is the circumferential stress?
Ans. 43.7 mm; 93.8 M Pa
Solution:
To -find th e r iv e t capacity :
Ps = A s Ss = - ^ ( o 0 i ) l (70 X /0‘ ) =
Pb = te tS k ■■■ P -
2 2 kN
O .oi { P o z ) ( l ‘fP x I0 6) = 28 k N
22 k N
20
U&intj a.s a ba.su i h e le n g t h L a io n a t h e
lo n y itu d in a .1 j o i n t :
, ------ >P
/
ff—
F= p PL
l P =
Z (i2
— yp
pDL
X lO 3 ) = I t S X IO t ( l . S ) L
L -
0- 02.IS m -= 2 3 .S rnm
For every o n e complete, rota tio n a p o in t on the
spiral - r i v e t e d j o i n t m oves a. d ista n ce e q u iv a le n t
to th e p itc h ..
T h e re fo re :
TrD = V ( I S ) = V . 7 I Z m
By th e
Pythagorean Theorem , the hypotenuse
is J (V- 7/Z)1 t ( 3 h
=
S .S 8 m
U s m tj s i m i l a r t r i a n g l e s
_b_
S.Sg
L
r. s g ( o . o z n s )
St
= f D
it
=.
O .O V 3 7 m
= (llU L X /o6) ( , s) - 3 5 . 8 MPa.
2. ( 0 . 0 ! )
140. The tank shown in Fig. P-440 is fabricated from steel plate.
Determine the minimum thickness of plate which may be used if the
stress is limited to 40 M N / m 2 and the internal pressure is 1.5 M N / m 2.
21
Solution:
Girth
Join t:
(A $ ) m
~
( p A ) flu id
[t . (o . b ) + i f ( o . t ) ] t SL = p E o . b f o . ^ t j - Z o . v ) 1]
[ l (o .b ) t
l f ( o . ^ ] t ( H O x . I 0 b)
=
t
S .S S m m
0- O D S S S m
=
2 P =
F
Z(A S )m
2 + LSr
~
=
I S X IO * l0 .k [D .< i) + T t (o .H ) * ]
=
( p M fluid
p [0 .{,
+ O.f] L
Z t ( t o x 10* ; = I.S x IOk ( l l
i
• t
= 0.0lt>15 m = !».7 S n ,n *.
HJSm m
2
Sim ple Strain
203.
During a stress-strain test, the unit deformation at a stress
■I 1*1 M N /m 2 was observed to be 167 X 10 6 m/m and at a stress of
140 M N /m 2 it was 667 X 1 0 '6 m/in. If the proportional limit was 200
M N /m 2, what is the modulus of elasticity? What is the strain correiponding to a stress o f 80 M N / m 2? Would these results be valid if the
pioportional limit were 150 M N / m 2. Explain.
Arts. E = 210 X lO’ N / m 2; e = 381 X 10~6 m/m
Solution.
S, -
35" MR*.
Sjl=
140
€,
= y = /67x IO-* m/m
E
= slope o f S - V curve -
MPo.
e 2=
fcfc7 Xlo ‘ n'/n
AS
= ZIO x t o * Pm.
6 6 7 X IO -*
E -
210
S = E€ ;
167 X IO -*
-
Ans.
6 - s- E
80 * IO‘
38 IX
210 x I0 »
Ans.
The results would siilL be valid if the propor tiona.I
lim it w ere 1 5 0 G P a . since all stresses given, are
belou/ this lim it .
204.
A uniform bar of length L , cross-sectional area A , and unit
mass p is suspended vertically from one end. Show that its total
elongation is 8 =* pgL 2/2E. If the total mass o f the bar is M , show also
that 6 - M gL / 2 A E .
I
24
Solution:
dS =
Pci#
. _
where. P= axial load acting
over the length. */x
p
= n v
P
- p j (Ax)
_522
J _t cJy
^ !
*
dS
ptj A x d x
AE
1
Aw
&
- - f
°:L
i * -
P 9 * * iL
Z„ Bn L
Ans.
= Z iz
2£
M
M
=f>V = p A L
AL
t - iftZ J L *
2B
m 3l
<i*v -
2AE
Ans-
205.
A steel rod having a cross-sectional area o f 300 mm2 and a
length of 150 m is suspended vertically from one end. It supports a
tensile load of 20 kN at the lower end. If the unit mass of steel is 7850
kg/m3 and E = 200 X 103 M N / m 2, find the total elongation of the rod.
(H in t: U*e the results o f Problem 204.)
Ans. 8 = 54.3 mm
Solution:
L= Isain
T
P’- -2P kN
= JUi
2E
25
_
7,9 5 0 ( l . S ) ( l 5 0 ) z
2 0 , 0 0 0 (I S O )
2 X 20 0 X 10s X IOA
0. 05*13 m
St
=
3 0 0 X !0~b x 2 0 0 X I 0 t X I O4
JV.3 m m
/lW4.
206.
A steel wire 10 m long, hanging vertically supports a tensile
load of 2000 N . Neglecting the weight o f the wire, determine the
required diameter if the stress is not to exceed 140 M Pa and the total
elongation is not to exceed 5 mm. Assume E = 200 GPa.
Solution:
ZOOO
P
A
w/mr
< /VO X 106
HA
d
/0ft!
> O.OOHZb m
d > *t.26 mm
___
Sr
A£
< a oos
- y d x X ZOO X 109
d
> O OOSOS nt
d
— 5. OS m m
I n o rd er n o t -to overstress -the. ma.teria.L oro v e r de. f o r m -the.
th e large-r d ia .tn e.ier
m u s t b e u sed .
___________________
el = S' OS m m
Ans.
210. A n aluminum tube is fastened between a steel rod and a
bronze rod as shown in Fig. P- 210. Axial loads are applied at the
positions indicated. Find the value o f F that will not exceed a maximum
overall deformation o f 2 mm or a stress in the steel o f 140 M N / m 2, in
the aluminum o f 80 M N / m * or In -the bronze o f 120 M N / m 2. Assume
that the assembly is suitably braced to prevent buckling and that
E , - 200 X 103 M N / m 2, Ea = 70 X 103 M N / m 2. and E„ - 83 X 103
M N / m 2.
Bronze
A -450 mm2
Aluminum
A = 600 mm2
^P
3P '
■*—0.6 m-
Steel
A - 300 mm2
AP~+—
-----1.0 m ------- Figure P-210.
2P
'
'
_J
----- 0.8 m —
26
Solution:
Aluminum
Bronze
A = ^Cm m 1
A = Hsomm'"
A =*
H~ - ..:
Taking
p,rn "*■
-the -free - b o d y dieyrmm. o f each, r o d -
Zf>
< — F=SE=
At-
:* = =
The d a sh ed a rro w s rep resen t -the- intern*.!, -forces
exposed f o r ee^uilibriu-m o f tfac/i rod.. The a.xia.1
-forc.es on each. r o d A r e :
Pk = 3 P ( C )
Pc
=
i p
( c)
Pi
=
2 />W
^ 5 S K m « j com p ression , -to he n e ja tU /e a n d
e-longa.tion. p o s itiv e , -the n e t deformaA-iorx.
w ould be :
PL
: 3P(0.t>)
2 P (l-o )
4SO x IO-* x S3 X 10*
(,00 X I0-*X70XI09
300x lO-^XZOOx io9
< 0.002 rn
P £ 2 8 .S 2 4 .b b N
s=4 3P
Sfc
=
h 450 x /O'*
s* =
S, =
2P
< 120 x 10* ; P < 19.000 N
—
< 80 X ID * ;
P i
—
300 X 10
< 140 X 10* ,
P < 21,000 N
600 x /0
ZP
2 4 .0 0 0 N
I n o rd e r not 'to overstress Any ma.teria.1 n o r exceed,
the n e t d eform a -tion ., i h e ma.xtnn.un*. oJlou/aJoLe
1sa.lu.e o f P nvill be -the tm a.Uest f r o m -the a bove
ca.leu. lotions :
__________
• ■ \P= 16.POO N | fins.
27
211. The rigid bars shown in Fig. P—211 are separated by a roller
at C and pinned at A and D . A steel rod at B helps support the load o f
50 kN. Compute the vertical displacement o f the roller at C.
Arts. 2.82 mm
E = 200 X 109N/m2
A = 300 mm2
L = 3m
| 50 kN
2m
B
2m
{
f t
1.5 m
3m
Figure P-211.
Solution:
-5"0 kfsj
F B D
I so
kN
tIWe.
RTc
FBD
B ) S jm m + try ■
o f CD:
tI\RRv
8c s
*
Rc =
R„ =
o f AB :
Pa
■3m
25. OOO N
,t ) E M , = 0
Kc =
jy
z s ,o o o (q .s )-P B ( 3 ) = o
PB - 3 7 . 5 0 0 *
j
I
5.0 00
*
PL
AE
3 7 . S O O ( 3 ) ________________
_
300 x 10- * x 200 x i o *
£ g = 0 .0 0 /8 7 5 /rt - 1.875 m m
The exaggerated, d e fo rm a tio n d ia g ra m if sh ow n below :
3m
\-S'm
B y
r a t io a n d p r o p o r tio n
_
3
~
Sc
4.S
Sc
=
ISSs
:
Sc = 1.5 (O . 0 0 1 6 7 5 ) = 0 .0 0 2 0 1 3 m
Sc ~
2 .8 1 3 m
A h -s .
2H
ill
A uniform concrete slab o f mass M is to be attached, as
'hown in f ig. P-212, to two rods whose lower ends arc initially at the
mime level. Determine the ratio o f the areas o f the rods so that the slab
will remain level after it is attached to the rods.
Arts. A J A , — 8.57
Figure P-212.
Solution:
///////
So th a t the slab u/iU reynainL e u e l a .fie * r U
U a tta c h e d ,
ALurrr
//////
L_ -
to th e r o d s ,
noGFb
St - Sa
B = ZC>o
L = 3m
(J L L ) = (J L L )
_______S' tv._
I A t. Is
-Sl‘ r
-r
I A t L
P i (3 )
= _A M
Aa.X 7OXI0*
A ix 2 0 0 x 10*
-
W^Mg
+ )Z M
&
A t X 2 0 0 X 10*
X Ps X 6
A » X 7 O a IO*
= O
c.-3-
P ,C 5 )-
=»0
p
r* =
- -2l 6
* + )£ M = O
Ps (3 ) P « -(Z ) = 0
Am.
bo
Ai
7
A*
A*
0.5 7
A n *.
29
214.
The rigid bars AB and C D shown in Fig. P-214 are supiwrted by pins at A and C and the two rods. Determine the maximum
force P which can be applied as shown if its vertical movement is
limited to 5 mm. Neglect the weights o f all members.
A/is. P = 76.3 kN
//^//Aluminum
L - 2m
A - 500 mm2
E = 70 GPa
Steel
L - 2m
A = 300 mm2
E = 200 GPa
c
3m
3m
D
IP
Flgur* P-214.
Solution:
////(/////
ALum
-0
Ps
- & fj) = O
P* = 2PS
Sk
V
£\>'+
^ )r M c = 0
P ( 3 )~ P t ( 6 ) - O
Ps = i P
By rrttio A n d proportion-,
P*. * P
i a - Sb
3
6
N o te ■■ Due to Ri, ihe aluminun,
rod n/ill d eform by Sti .
Correspondingly , B wttl
move down by Sy due
However, due to /2 , D will
not m ove down by SyonUjt
b u t by St T Ss.
Sy =
St „
Sj
3
Sb
6
St s 2-Se.
+ Ss =
Ss ~ 1 (0 . OOS)
z fIL )
Z lAEU
+
= 2 (^ P )
l£ L )
IA E/s
= O.O!
30
PX Z
500 X10-*x 70 K JO9
3 0 0 X /0 -*K 2 0 0 X /O *
P=
7 b .3 b 3 .b V m
P=
7 6 .3 6 k N
Ams.
215.
A round bar o f length L tapers uniformly from a diamet
D at one end to a smaller diameter d at the other. Determine the
elongation caused by an axial tensile load P .
Ans. 6 - A P L / irE D d
Solution:
— H k— x —*
and proportion:
+
-4 k
«
w
- JA
u
X
ijL - d L
%
=
Dx-^x
s -
w
L i - i-i-1
TTE(D-d) L D
d J
4PL
TTE(D-d)
^PL
P t^ J
/l«s.
TTB.DJ
217. As shown in Fig. P-217, two aluminum rods A B and BC,
hinged to rigid supports, are pinned together at B to carry a vertical
load P — 20 kN. I f each rod has a cross-sectional area o f 400 mm2 and
E -> 70 X 103 M N / m 2, compute the elongation o f each rod and the
horizontal and vertical displacements o f point B. Assume a -* 30° and
9 “ 30°.
Arts. 8h - 0.412 mm; 6„ - 3.57 mm
F ro m the 'force 'triangle. ;
P ab
S 'm 6 0 “
_
Pec
2 0 ,0 0 0
S in 6 0 *
S in
(> O c
PAa = 1 0 ,0 0 0 * ( t )
Poe
= 2 0 ,0 0 0 * ( C )
The d efo rm a tio n s are
r?- P L J1
Is
AB
2. 0 ,0 00 ( 3)
400 x iO~h %70x ID9
2.143 x 10 * *
AhS.
( le n g th e n in g )
t
-
20.000 ( z )
* B C ~ t o o X to -*X 7 0x1 0 *
= [1.421 X I 0 '3n>
A h S.
( Shortening )
To a n alyse the. effects o f t h e d e formant io n s o n B's
m o v e m e n t , im agine. A B a.nd B C to b e d isc o n n e c te d ast
Q to ajlou/ th e d efo rm a tio n s ('g r e a t l y e x a g g e ra te d ) shoHun
below. To re fasten, t h e b a r s , rota te t h e m cdoout A a n d C ,
respectively , t o m e e t a t B “- S in c e th e arcs g e n e r a te d ,
are $0 S m o -ll, th e y m a-tj be. re-p Laced b y stra ja k t lin e s
th a t a r e clr*.tun, pe.rpenA 1 0 u.Lajr to A & a n d B C , i n t e r s e c tim j a t 8 ' ; SB =. B B
S b p ’ =*
. CaC>
75#l
\?' W
Is*?
.W/
/
*+2 .9 men
33
The de formations Sab And <Sbc a-r * dra-Ujn to *l.
la-rjer- s c a le on. t h e r i g h t .
p r o je c t i n g Un- th& cLirec-tion, o f
y
- ccs 3 0 °
-
S BC Si**. 3 0 ° - S a b
COS 3 0 ° - 3./</3 Sin 3 0 ° -
2.14 3
tj ~ 3. 3 m m
P r o j e c t i n g ***■ t h e d ir e c tio n . o f <fBh ;
Sgh ~ y Sim 3 0° - SBe Sim 60“
«f*fc = 3 3 s i" 3 ° ° - 2-1*3 Si~ 6 0 ° = 0. m i
P r o je c t in g i n t h * d i r e c t i o n o f
Ant.
'•
SB„ = tj cos 3 0 ° 1- Sac ° ° s (>0°
- 3 .3 cos 3 0 ° y I. * 1 1 cos b O ° ~
A L T E R N A T I V E S O L U T IO N
F ro m r i g h t -trtA n jL e -
8 D8 ' \
c o i (6 0 - J i) =
-£ l»
From r ig h t triebngLe- B E B ‘;
COS
(6 0 + f i )
a
&B
Ant.
34
5b "
&A0
Oac
cos ( b O - f i )
COS ( t > O f f i )
2. 143 X 10' 3
_
Iff.2.9 X /O'3
cos ( G O S o L t /’i n y
-f o r
a > s (G O + f i )
w h ere,
COS ( G O - f i ) ~
c o s GO cos f i
+ iin
GO s in
astd c o s (G O + /3 ) — c*>i 60 COSfi — Siw (j O si** fi
ta + tfi = £•
=
£
(,.5 S 3 °
__
£ab
.2./V3 x /O'5
cos (6 0 ~ f i ) cos (6 0 ~ 6. 5&3)
SB
■•
-
3. 5 9 6 X /O '* t n
= Js
^
JSJi
= J. 596 X /0-J f*V* 6.
f H#i
= O . V / T X 10-3m.
j g(i = o. mz mm
*
A h .s .
~ &B "=*/3
J8„ = 5- 596 * / £ '’ «> « 6 .5 B 3 °
i Bv
-
3.572- k tO * n
S s,
=
5 5 7 2 H im
4/»S-
219. A round bar o f length L , tapering uniformly from a diame­
ter Z> at one end to a smaller diameter </ at the other, is suspended
vertically from the large end. I f p is the unit mass, find the elongation
caused by its own weight. Use this result to determine the elongation of
a cone suspended from its base.
Solution:
////?//////_
¥ r~ »\
p zZ Z EZZ p ZZ Z ZE^ ~
dx
35
Frona
P-2/5,
dx
- (o - d )
Loa d tsLve to u /ei^h t on -the dLfforeniia.L e le *n e n t
dW ~
j>£jdv — p j Adx
d W * p y ^ f f 2dx
J d W = Jo P9 ^ y 2dx
W
-
J
p i j l f
t j2
( p - d
)d tf
a
P 9 ir . _ f i _
D-d
w
=
w
= _£L
II
. -± _
D -d
p t , jh b . - J ’
i<f
■o
I 9
Jd
r - il
L 3
r y s -js l
L
J
STJ
P9l ±
r ° / y 3 - <j3|
33( D
( D- d
- d) *) zEE ^
1L *
I
V*
'
,0
3E(D-d)x X
■
'J
’d 7' » - £
y* M
n LX
[
3 £ (D-d) 2 L 2
n~L£
1
3 E( p-d) 1[.
+ JJ] P
y Jj
Dl + J* - J1-
i
D
2
J
rsL*
rp1- j 2i
r
' 3 E(D-d)*l 2 J 3 E(D-J)*L“
n L'
PI1-* \
~ 6 E(D-d)* i“ °nuTaj - 3E(D_d)2i
Pj Lz(d+d)
K L1J1 [ D- d]
6 £ (D-d)
3 E(D-d)*l D J
f j L z (D * d )
6 £fp-j;
3ED (D-J)
A»s.
36
223.
A rectangular aluminum block is 100 mm long in the X
direction, 75 mm wide in the Y direction, and 50 mm thick in the
Z direction. It is subjected to a triaxial loading consisting of a uniformly
distributed tensile force o f 200 kN in the X direction and uniformly
distributed compressive forces o f 160 kN in the Y direction and 220 kN
in the Z direction. If v » j and E = 70 GPa, determine a single
distributed loading in the X direction that would produce the same Z
deformation as the original loading.
Ans. 410 kN tension
Solution:
Fjj = 160 kN
2 O O .0 O O
- S 3 .33 M P * .
0 . 0 7 5 X O .O S
t
= - 3 2 MPa.
- 2 ZO , OOO
= zze> kN
0 .0 7 5 X O .IO
- -21.33 MPa.
7Ox 109 [-•2 9 .J 3 X / 0 * - - $ - (5 3 . 3 3 A 1 0 * - 3 Z x IO G)
£ z = -S .2 0 b x !0 -4 ">/«,
V
- ( - 5 - 2 0 6 X JO * )
y3
€ x'= I.Sb2 X to-3 m/tn
Sx = € X'E = I.5b2 x ID'3(7 0 x to9) = 1 0 9 .3 3 MPa.
px
~
= (0 . O 7 S X O. 0 s ) ( / 0 9 . 3 3 X IO k)
Px = V IO , OOO N
Px = 4 I O k W
Arts.
I
37
224.
A welded steel cylindrical drum made of 10-mm plate has
an internal diameter o f 1.20 m. By how much will the diameter be
changed by an internal pressure o f 1.5 MPa? Assume that Poisson’s
ratio is 0.30 and E — 200 GPa.
Solution:
21
I.Sx/0*(l.Z)
St = — „
Z ( o .o i) ----- = 90 MR*.
SL = p ° - T' S t
Vt
2
St = VS M Pa.
U s in q Efy ( 2 - 9 ) i ta n g e n t ia l Sira.in. is ■
e* ‘ T
- ■ ' T '
A P
P
W D ' - 1Td
€ *= ■
L
AD
D
=
le c iin g t h e effect o f end. -thrust i S L ~ 0
A D =
d [ j (S t - ^ S t > ]
A D - 5 .V X 10 * m
A D ~ 0.50- m m
A n*.
C o n s id e rin g -the e f f e c t o f end. ik r u s t,
A D = p [ j & t -v -S t ) ]
A D = <-2 [ *00X10* ( 9 0 x / 0 * - Q . 3 x K x J 0 * ) ]
A P
= <1.59 X I O ~ * m .
AD -
O.M-S9 m m
/4m(.
38
225.
A 50-mm-diameter steel tube with a wall thickness o f 2 mm
just fits in a rigid hole. Find the tangential stress if an axial compressive
load o f 10 kN is applied. Assume v - 0.30 and E - 200
X 109 N / m 2. Neglect the possibility o f buckling.
Solution:
U itn g £ ^ - ( 2 - 9 )
= - j (£ t ~ * s L )
\.
_
.
€ ± = O because o f th e r e s tr a in t
of the ru^ld. hole..
From.
e.xjtrt*eL by th e u/all
(z -l i .)
= - j [ St - r ( S L + S j ] = 0
NegLec-tvrrg the, rajsUaJ. strops
(u fua l to contact
pre.s su.re m Pu.tsid.e~ o f tu b e b u t z e n » a t in s id e
o f tuba<)i
St - v $ L
c
P L ~ A
&L =
,0<000
1r o t
10• 000______
=
ir(o .o s )(o .o o i)
31. 03/ MPa.
■ s t = 0.3 SL = 0 .3 (3 1 6 3 1 ) = \S.SSMPa\
A hs .
* contact p ressu re p can be solved fro m
5*
Zt
p (o .o s )
‘t .s s x /o‘ =
z (0 001 )
p = 0. 7b M Pa which if ima.ll co m p a red
to th e va lu e o f
= 31.33/ M P a .
226.
A 150-mm-long bronze tube, closed at its ends, is 80 mm in
diameter and has a wall thickness o f 3 mm. It fits without clearance in
an 80-mm hole in a rigid block. The tube is then subjected to an
internal pressure o f 4.00 M N / m 2. Assuming v
and £ = 83
X 103 M N / m 2, determine the tangential stress in the tube.
Ans. 8.89 M N / m 2
jy
Solution:
et
- jfe - v S * )
Cj
=
O
becau se- o f C o m tr & in i
PD
4*t
~
_ U P °\ _ ±
/ ? X / 0 4 x 0.09 )
S* - v S L ' 7 l * i ) - T z ( ™ o . o o 3 ----- /
St = B .8 6 1 MPa.\
A n t.
Contact pr& uure « U » A» solved by iuparpetition.
s
- l ,|
£ J.
" it
-
V*
X ID*- -
«.
V*
o. Ob
H X. O. 0 0 3
( *
*. I D * - f i . )
fa = Z . 6*7 MPa.
233. A reinforced concrete column 250 mm in diameter is de­
signed to carry an axial compressive load o f 400 kN. Using allowable
stresses o f ae — 6 M Pa and a, — 120 M P & determine the required area
o f reinforcing steel. Assume that Ec — 14 GPa and E, — 200 GPa.
Arts. A , — 1320 mm2
Solution:
Sf = Sc
(P L
\
_
[ a e )s
(P L _ \
.
Lf -
Lc
l A£l c ' JL= S
3c
Et
•Sj
E,
Sc
X /O *
7 0 0 X /<>»
Sc = 0 .0 7 S s
I f St = /^£>
I f Sc =
6
i
= d- V WAk. (ov*rfUrtned)
i Ss = 05. 7/ Mfi* (Crovernt)
P ^P c-rP t
P = A c S c + A t S%
Ac * A t >"%(*■**)*
Ac = 0 - 0 * 1 ! - A t
* 0 0 , OOO - ( o . O V 1 ! - A t ) ( * * t o * ) + A i (»* ■ 7 / X / £ * )
40
Af =
1.323 X t o
*m z
A i — 1323 > » / » *
A ft.
239. A rigid block of mass M is supported by three symmetri­
cally spaced rods as shown in Fig. P-23S. Each copper r6d has an area
of 900 mm1; E ” 120 G Pa; and the allowable stress is 70 MPa. The
steel rod has an area of 1200 mm2; E — 200 G Pa; and the allowable
stress is 140 MPa. Determine the largest mass M which can be sup­
ported.
Arts. M - 22.3 X lO3kg
M
Copper
160 mm
Steel
240 mm
Copper
160 mm
'//?/////
V777777777/
FlgurM P - t M and P -2 M .
Solution;
W=* Mg
•s
>c
St (0 .2 t)
Pc.
Sc
If
_
Sc (O H )
200 < /O 9 /20 X fO *
9
=
= It-O MPa. , Sc = /26 MPa.
I f S c = 7 0 MPa. ; S $
T .y = O l
=
77. 78 MPa. ( G o v e r n s 7
2 Pc + Pf s A f j
M
^ AsPf - M y
zfroo'x. /0 -*)(7 0 X IO *) + 1200 x / O '* (77.76 X /<=>V = M g
M = 12360. I f
A/ =
2 2 .3 V X
K3
/O * K g
A us.
41
237. The lower ends o f the three bars in Fig. P-237 are at the
»*me level before the rigid homogeneous 18 M g block is attached. Each
•teel bar has an area o f 600 mm2 and E — 200 G N / m 2. For the bronze
bar, the area is 900 mm2 and E = 83 G N / m 2. Find the stress developed
in each bar.
Ans. a, = 124 M N / m 2; ab ■■ 32.0 M N / m 2
Solution:
i s = Sb
m
- m
t
st (I.o) _ st (/.(,)
200 x /O9
81 x /O*
SA = O. 2513 7S S*
Sy =0]
=
& (A S) s + (A S )h = /.76V x /<?*
2 (b O O x I O '* )S t ■* lOO X IO b ( O. 251375S t )
= /. 76V x / »*
Ans.
239.
Three steel eye-bars, each 100 mm by 25 mm in section, are
to be assembled by driving 20-mm-diameter drift pins through holes
drilled in the ends of the bars. The center-line spacing between the holes
is 10 m in the outer two bars but is 1.25 mm shorter in the middle bar
Find the shearing stress developed in the drift pins. Neglect local
deformation at the holes and use E, — 200 GPa.
Ans. t = 66.2 MPa
42
Solution:
£0 + S i *
/•
_________ P. (.to)_________
* /*>■* *r>
S ^ n il
g f t f/ g - / .* g x /o~J>
( o . / x o . o i s ) ( i o o x /*#•)
=
£
x o - o t s )(2 o o x //>*;
/ .2 ff X / * - *
= 2 0 ,8 3 S' H
(>70 /y
Pi = 2 P. c
*«"
Po
>1
66. 3/ MP A
-
% (o .o z )*
finS.
(finale, fh
Ss
Pj
2A
_
H-lf,7Q
2x 2j.(O.OX)*
6 6 .3 / MP*.
Attf.
(MouJtle thmajr")
240.
As shown in Fig. P-240, three steel wires, each 30 mm2 in
area, are used to lift a mass M . Their unstretched lengths are 19.994 m,
19.997 m, and 20.000 m. (a ) I f Af — 600 kg, what stress exists in the
longest wire? (b ) I f M *■ 200 kg, determine the stress in the shortest
wire. Use E — 200 G N / m 2.
M
Flgur* P-240.
43
Solution:
//////.
L* = U . I l f ^ t
Lh =
{*)
Lc = ZO~*
Sb)
Cc)
(a .)
IV =
- S 8 S O A/
7^ » load nemdad- to carry aU mfirm* A>
M m a Lenyth o f 20 nt If.
« ♦ *
_
Q 0 (,)(3 0 * . fO ~ * )(t O O X t o 9)
1 9 .9 9 V ->
_
0 0 3 ~ )(3 0 X //> -*)(i O O X IQ 9)
H117
2 <?
«
=. /,**>/> * */>* =
>V
W ke+i a t ih & M m c le n g th n / n M ^ jf, a a cft n/irm- a / ill
th e *i c a r r y ‘/ i o f th e r e n i u n u lo a d .
Pc
F in a l lo a d in U/irm- ( a ) = H O O + IOkO s 1.&LD At
Fd-nal loa d . in . n/Lre- ( i ? — I P O * - IDI> D - H t O Af
F in a l lo a d M t m /ira ( c l ~
«Sc =
Pc
/MO
A:
3 0 x /O'*
.
30 x to
= —
Aa
(6 ?
I f
a
_ £ £ £ £ --- s
35 M P a .
30 X IO-*
IV = 2 0 0 Ky = H t,D A/
i n it ia l Ipeuk a rC ’
7 )W
/w v|
UkO
=
S
/0t>O Af
^
2,7 0 0 Af
- 2700
_ _ 2tf.L t>7 H
44
Cc) w i l l bu.tJtle. o t t i o f A c t i o n **id .
lvlIL
W ith .
reifn uM s Laok..
I*. = ‘f OO N t o re*uM . le n g th - o f ( b ? ;
2- w ire -S ,
900
530 N
‘IP O * S3D = /V 3 0 N
fl =
f>
1430
A*
3 0 x /£>'*
530
5i = ~At.T
Sc =
= 30 x /<9-‘
V7. 67 A//V
-
/ 7 .6 7 A / ^
O
241.
The assembly in Fig. P.-241 consists o f a rigid bar
(having neghgible mass) pinned at 0 and attached to the aluminum rod
and the steel rod. In the position shown, the bar A B is horizontal and
there is a gap A = 4 mm between the lower end o f the aluminum rod
and its pin support at D. Find the stress in the steel rod when the lower
end o f the aluminum rod is pinned to the support at D.
Arts. a. = 174 M Pa
r
1.2 m-
0.6 m -
AC
r\
///)/
O
Aluminum
Steel
A = 300 mm2
A = 400 mm2
E = 70 GPa
E = 200 GPa
L = 1.5 m
D
777}/>7?77777777/7///////V/V/’W / V
F lgu f« P-241.
Solution:
Exojjgeradod, deformation.
A = 4 mm
f
45
<Ss
0.6
_
o• o o i 1.2
2$s
= «■ 004 - S*.
So. +
0 .0 0 4
=
(-rT f l * 2 (t f ), =
2 Pi ( I.S )
Pa. ( I S)
V O O x
IO -»x
7 0 x 10*
And,
j^ZMo =0
'
3 0 0 x
= 0. 00* - 0
I0 ~ * X 2 0 0 x 1 0 °
- £ ( O.b) + f k ( l . l ) = 0
Pi = 2 Pa.
-®
Solvung iutni/llnM ou sly;
Pa. = 2 b 0 4 b. S N
Ps = 52093 N
c
1
- JL ,
S1093
As
300 X /0-
/7J. 6¥ fk
Ans.
244. The bar shown in Fig. P-244 is firmly attached to unyield­
ing supports. Find the stress caused in each material by applying an
axial load P “ 200 kN.
A n s.
-200 mm-1
o. -
62.8 M N /m 2; a. -
120 M N /m *
-300 mm
Aluminum
Stee,
E « 70 GPa
£ = 200 GPa
A = 900 mm1 A _ 1200 mm1
n gu rM P-244 and P-245.
Solution'.
Si!
■ZOO
Rem ove ’th e w n tf in. iA e d ir e c tio n o f H t*. mfpLitd. lo a d
A n d a.llow fre e ojcuiL d t f o r m a t io n .
46
|£Sr )
—....> P
k
Rb
Briny back 4k*. uW / Ay Hfplyvruj the- W*lI I rMif/i'ow.
'2S*b
M
’
Front, "the-
deforma-tLon
;
£5p 2 TISR j — R | i l ^
ZOO. OOP Co t )
=
900 x /0-‘ x 70x /0*
o.z____________
R . [ Wflx /o**x 10x to9
0.3
tZOOx. to-*x. l o o * to9 J
Kk
Z* = o]
=
IV 3 V 3 7 .7 S N
200,000 - R0 - Ka = o
Ra = S 6 .S 0 2 .2 S N
■
To g e t the- m u a I lo * d eu-tiruj on ea-tM, nuxA^riaJ., we. d m w
ihe- fro e b o d y d*A^nun.
2.00 , OOO N
<--------------1
__
AL
"■■>
RA= S6SOZ -ZS-N
/V
pg =.
----------- ?
S Z .S -C 2 . 2 « - A/ / r }
---------
3t
r *3 ,¥ *7 . 7S N
/g = /-«3
[ « -
F?b = W 3^37.7S- /V
AI ( c )
PA ]
fTO,2.2S
s*= 5~6,
900 x /0-‘
«, ■
('r ts a ^ tio n
b ch u a e.tr
AL <*nd S t b od s )
1,200 x /O'*
6 2 .7 6
-
Am.
r^ -5 6 N n.
L—
_______
AnS.
246.
A rod is composed o f three segments shown in Fig. P-246
and carries the axial loads P, — 120 kN and P 2 = 50 kN. Determine the
stress in each material if the walls are rigid.
Arts, a, — 122 M N / m J
47
Solution:
1
Br
Al
P, ~
at B
30 kN
ITO kN
R e m o v e - th e - Lva .lL in - t h e c U r e c -tio t* - o f th e - a p p lie d
e u td a llo w
fr e e a. « < i
/ W
d e f o r m a .t u »v :
------
—
Bring ba&k th e w all by a fp k ji*ij the- w all ree&tion, :
------*|£S$k—
0.3
0-4
120,, m [ j 100 ( t o - * ) (70 X to 9)
(t>00x!0-*)(200 x 1 0 •)]
0. 6
50.000(0.3)
(600 x IO-*)(ZOOx I0 »)
It
0.3
600(100 x 10 * )
+
1,100(70 X 10*)
9.964Z& X 10-* *
Ka r
400 (S3 x 10*)
]
1.02731 X I 0 'm RA
Ka
= 96906-4 N
R0
= 110,000 N
K.
= 73013 .i> N
Drawing the- F B O i o f the, rods io g e t the axixl loads,
1 30, OOO N
Br-
XSo/M •6N
48
50
----------
AL
■
-
2 3 *3
7 3 <=>/3 ■t> /Y
■*■------- 4
2 S O/S . o M
si
73013 . b H
i-
7 3 013. (SV ------------- -----------1
. . Pst
~
7 3 <3/3 . <S-V
[* = % ]
_ 1 bib b . 4
8 "
_
ho.m
M P»
Am -
1*1.16 MPa.
Am.
\ m . 69 MPa. |
Ans.
zw odo-*)
130/3. b
A =
1200(10-*)
S* =
73 0 /3 . k
600 (/ O -* )
= [
247. .Solve Problem 246 if the left wall yields 0.60 mm.
f
—300 mm—r
Pi~<—
P2M r-
/
/
f
Bronze
Aluminum
Steel
A = 2400 mm2 A = 1200 mm2 A = 600 mm
E - 83 GPa
£ = 70 GPa
E = 200 GPa
Figure* P-246 and P-247.
Solution:
R&fer to P- t47
Allowing free a.xia.1 deformation. : ,
B rin g in g bxck -the wo,11 a-n d A llo w in g
fo r the- tjccld. ( s//ringing D w t)
4‘>
SSp m
* y
<). 9b^ie> * 1 0 ' * = /. OZ731 x 10'* Rfi + b K/0- *
Ra
=
39565. 93 M
R$
—
131*14. 07 N
IZO ,CCO
3&SBS.
■ Ffc
hs =
•93
38
3 3 AI
8/ u w . oV
s o , o.-x>
At
i3±4’± ° n— j r
r^} »
07 N
/^/.•V/V . o'r
ISIHIH
= ITS! 4tH- . ‘ST' A/
ov
[s = P/Aj
S« =
3BSK5. 9 3
2 4 0 0 ( / O '* )
81*1*.07
s* =
°s
/ZOO (/O'*)
_ J 3 >*1+ pl_
6 0 0 (1 0 -*)
16. 06 MPa.
Am.
6 7 . 0 5 M Pa .
A ns.
_ j Z H .O Z MPa.
I------ ----------
Ans.
248.
A steel tube 2.5 mm thick just fits over an aluminum tube
2.5 mm thick. I f the contact diameter is 100 mm, determine the contact
pressure and tangential stresses when the outward radial pressure on the
aluminum tube is p = 4 M N /m *. Here, Es = 200 X 109 N / m 2, and
E , - 70 X 109 N / m 2.
Ans. pe - 2.% M N / m 2; a, - 59.2 M N / m 2: <ra - 20.8 M N / m 2
Solution:
Sromm
FB0 o f SUeL.
ZPs = P*
2 /? = P^OL
*
-± £ L
50
F B D o f a.Unm.l-njA.'m :
1 Pa. -
P -P l
Z P-. - C p - p J P L
/k = (P ' f^ DL
K
m
-
s*
i -
m
t i l
- a i
=
(p -p c l j f x
x *
70 X 1 0 *
iJBQs
X * 2 0 0 X ip *
20 p - 20fc
20p
=
7f*
= 2 7p0
s= * - f T o . ™ i < ,
SmSl * :
*
‘
n
i /I
Z ( o■00257
m
I "
^
i * -
11,11
= f2 0 . » M K
1— ------------
Ans.
249.
In Problem 248, assume that there is a radial clearance o f
0.01 mm between the tubes before the internal pressure o f 4 M N / m 2 is
applied to the aluminum tube. Solve for the contact pressure and
tangential stresses.
Solution:
I f there. is a. cheurcLnc^e-, pa-rt o f -the pressure
t v M be used, to ma.ke. contact.
S =
TfAD — TfCZx 0.0! x /0~3) = - ~ ~
y
p'O
2 X I0 -*/ e —
P *
r
2
(jf D)
—
~
x ID ’ * M C Z . S X I O - * X 7 0 * I O * )
(o .O *
51
/>' =
0 .7 nPcL-
p =
y - o .i
Fr*>»r. P/-06
= .J.J Af/3»^
c
27 '
= _Z<L ( 3 y
Z7
/2. = 4 .4 V MPa-
[* - £ ]
Ans.
_ 2 .W ( o . Q k IQ6 _
sf ' T J i s l T i o ^ l
_
i-W fo -O x io * _
* * = 2 f2.S x /o-*;
MP a
An£.
A/ii
250.
In the assembly o f the bronze tube and steel bolt shown in
Fig. P-250, the pitch o f the bolt thread is 0.80 mm and the cross-sectional area o f the bronze tube is 900 mm2 and o f the steel bolt is 450
mm2. The nut is turned until there is a compressive stress o f 30 M N / m2
■n the bronze tube. Find the stress in the bronze tube if the nut is then
given one additional turn. H ow many turns o f the nut will reduce this
•tress to zero? Take £ as in Problem 246.
Ans. ab — 15A M N / m 2; 1.66 turns
_
f'e/////////////////////,
r~ p
j- H
e
r( 2Z
-L = 800 mm Ftgur* P-250.
Solution:
1—
Lo a d s in d u c e d
SB +
S,
b y one- -tu rn :
= ? *
Pi - py
* *
______p ( o e > )
-----------fa00 X I O -* )(01 X 10*)
(450X IO '*)(iO O x.lO *J
P=
VO, ei 1. 6&N
52
_ 100/9. fee
ASk =
900 X to
Fina.1 S fc =
-
=
V5\ 56
/ -turn
30 * VS. 31> =
75". 36
To re-Heve. sb-e-ti , n- =
M P a.
Ans.
o f im^n r
7^. 5A
VS. St.
/. 46 "turns
A nt.
251.
As shown in Fig. P-251, a rigid beam with negligible mass
is pinned at 0 and supported by two rods, identical except for length.
Determine the load in each rod if P = 30 kN.
Ans. PA = 9.10 kN ; PB - 11.94kN
*•---- 2 m---- ►-•---- 2 m-----►■*-1.5 m—*-j
/v
S- - - - - - - - - - - - - - - - - - - - L = 1.5 m
L =2m
A
/T, T 7
B
Figure P-251.
Solution:
U
=
m s -Sa
Pa ( z 1
_
“V
--------------P„
[£ M < ,
=
l.7S(PA * l . 5 l
I-3IZS PA
= 0]
P(Zl = PA ( Z l + Pv (3.S)
3 0,0 00 ( z ) - z &
+ I.3ZS
Pa
=
?•/ k N |
Pe
=
| II.
A/vl
PA ( j . S j
Ans.
Am.
\
53
252. As shown in Fig. P-252, a rigid beam with negligible mas*
11 pinned at one end and supported by two rods. The beam was initially
horizontal before the load P was applied. Find the vertical movement of
/' if P = 120 kN.
Ans. 2.92 mm
WS|H
Steel
Aluminum
i4 = 900 mm
E = 70 GPa
i-3m
/I = 600 ram£ = 200 GPa
L - 4 m
P
------ *i-»— 2 m —
U --------3 m
— -*■
jlm
P
Figure P-252.
Solution:
/muni
AL
2m
3xrt
lm
i-z o .a o o n
A
3
ft. (5?
—
% Sf
_
l ( P & K 4?
100x70
Pa.
bOO x 2.00
= 7/5 Ps
[ZMo = 0 ] % (3) + P*(bl = P(S)
3PS +
b l-^ P s )
=
Ps = 5 Z C .3 l.se - N
= 73b»^.Z! N
= 3.57 /urn-
Sj
=
/•75” m-m
n o , 000 ( s ')
54
= Jl
h.
s
Sp — Z.1Z mm
A hS.
253.
A rigid bar o f negligible mass, pinned at one end, is sup­
ported by a steel rod and a bronze rod as shown in Fig. P-253. What
maximum load P can be applied without exceeding a stress in the steel
o f 120 M N / m 2 or in the bronze o f 70 M N / m 2?
Steel
yy/yy
A = 900 m m 2
E = 200GPa
i
Bronze
A = 300 mm2
E = 83 GPa
L = 2m
= 3m
1m
k 2 m-
-3 mFigure P-253.
Solution:
/////////////
St
UUUlll/l
Br
2m
im
Ss
IXO M P a.
S«,
70 M P a .
Jk.
z
S
f
h
=
f
l -
St £5)
_
0*(Sbx2)
Z 00(1o y
83 (ID?*
sk = /.sstzs Ss
If
>S4
= IZO M P a .
Sb »
Tf
Sb
75" MPa. (overtfrviltdl
= 70 M P a .
s. =♦«.»»
,
I
55
[ZM0- 0]
P ( b l - P s C H - Pb ( s ) = O
GP =
6P -
2/>s5s + 5-Afc/i
Z (<iDO %
* 10 * ) +
* ( i 00 x /D's?(70 x 10*1
P =
3 0 .H H
N
Anc.
254.
Shown in Fig. P-254 is a section through a balcony. The
total uniform load of 600 kN is supported by three rods o f the same
area and material. Compute the load in each rod. Assume the floor to
be rigid, but note that it does not necessarily remain horizontal.
Ans. P B = 183 kN
F ig u r e
Solution:
P -2 5 4 .
56
i t __
*2_
6
3Se - 3 S+ =
XSc - I t * ,
JSs =
lie
+ S-.
ie>PB = izz + sP*. ©
[ x y =o] PA + pg * n = boo.ooo
®
[ I M = 0] 600, 0O0(?7 = Pg (41 + Pc ( t ?
100,000 =
+ J/£
Pg = *50,000 - \.S Pc
S u b liitu -tln j ®
^
©
~ * (§)
+ ±50,000 - 1-Sfl + Pc = 1,00,000
Pa -
0 .5 Pc.
=
IS O ,000
ie>(*50,000 - I S Pc) = AZ
* 5 fli
Substituting (s )
(j)
5PA + W Pc - t>. 100, 000
@
X.5
(£) * /
®
®
5 P ^ - Z . S P C = 750,000
SPq. + M P c
-
3,100,000
m .S Pc = 7,350.000
Pc =
Pk =
I / 77./ /
/tjy
Anf.
Z3B.5S kN\
Ant.
m . i t aa/
Ant.
255. Three rods, each with an area o f 300 mm2, jointly support
the load of 10 kN, as shown in Fig. P-255. Assuming there was no slack
or stress in the rods before the load was applied, find the stress in each
rod. Here, Es = 200 X 109 N / m 1 *nd E b = 83 X 109 N / m 2.
Solution:
\ i/
\i/
be/~
\6b
//** S s\\
X
'■ 'N
Coi * 0 °
y
-
s<
S„ = Ss c * 3 V
r - ft i* ^
P„ CJ / CCS 30-7
_
<^ 00 x I D '* J («i x I0a)
pe
At th e -J "*t :
—
(500 x IO-*)(ZOOx 10*7
=
0.31115 Ps
ZY - O
I Pg cot 10° * PS = /0A/V/
X[0-3UZS Pjon 3 0 °+ Pf =
/<?**>£
£ = 6*17.3 N
PB = 2 0 11 .3 N
[S
= P /a]
c _
*
5, =
* m .3
J00x /P'*
201Z.J
300 a 10-*
2 / . MPa.
/»„f
t>. n* m p *.
ft™
58
256.
Three bars, AB , A C , and A D , are pinned together to sup­
port a load P = 20 kN as shown in Fig. P-256. Horizontal movement is
prevented at joint A by the short horizontal strut A E . Determine the
stress in each bar and the force in the strut A E . For the steel bar,
A — 200 mm2 and E <= 200 GPa. For each aluminum bar, A «- 400
mm2 and E - 70 GPa.
Ans. P AE =* 180 N
Solution:
45 =
cot 3 0
if
SP
Oo
=
- Sc COf V-S
Po C 5 /u>( 45?
( t o o k i o - * ) ( rro*. t o » j
P0 = D.35 Po
Sg -
Sc cot 30
(M ‘ (U - »
_
E, ( 3 ) OH v s
( io o x io - lJCzo o *. /o’ ?
I
59
p^(3?cci 30
{ZOO k 10*7(200x 10*)
P„ (3 /u > s 30?
(#00 X IC>~b) ( 10X 10*)
pg =
0 .5 1 5 Pc
[ s y = O]
Pg cos 30 + £ + PQ cos HS = 1)0,000
O.SZS
cos 30 + Pc * 0.3S P0 cos HS Pc =
//7V?.0V N
P„ =
H U Z .m N
P„ =
U f bf N
20.000
[2 X = 0 ]
fp sin H5 ' Pg $ui 30 + Pa b * 0
Ans-
fys ~
[S = P / A l
15.V i M P a .
Ant.
u m q .& i
zo o a to -*
- [ 50.15 M P a
A*f.
u n z . mt
= 400 A to-*
M Pa
Ans.
^5 =
t o o * to -*
Sc =
S°
261.
A steel rod with a cross-sectional area o f 150 mm2 is
stretched between two fixed points. The tensile load at 20°C is 5000 N .
What will be the stress at —20°C? A t what temperature will the stress
be zero? Assume a = 11.7 / im / (m °C ) and E = 200 X 109 N / m 2.
Ans. a = 127 M N / m 2; T - 34.2°C
Solution:
= JL00O—
ISOa I O *
mt 2 0 ° c. } S = - £
’
A
U
=
5 5 .5 3 M P a .
(in itia l stress)
=
cK L A t —
PL
_ SL
A£
'
£
A S = (200 A / 0 9 )(H . l A /0 *)(2 .0 + Z O )
AS
=
S3. t. M P a .
60
>S fin a l 3 3 .3 3
33. 33
T =
-
|IZ6..93 MPa.
Ans.
io*ycn. t X / » < ? fr - 2 0 7
C 200 x
=
+ 93. t,
3 4 . 2 S a c, |
Ans.
262.
A steel rod is stretched between two rigid walls and carries a
tensile load of 5000 N at 20°C. If the allowable stress is not to exceed
130 M N / m 2 at —20°C, what is the minimum diameter o f the rod?
Assume a = 11.7 /um/(m ■°C ) and E = 200 GPa.
Solution:
St *
A-t = too x 10* 111.9 x io-*)(uoj
St -
S = St +
13-b MPa.
P/A = 11. b x !()•> +
d
-
0 . 0 I3 Z Z
d
-
1 3 .2 2
= 130(10?*’
w ,
m m
A nS.
263. Steel railroad rails 10 m long are laid with a clearance of
3 mm at a temperature o f 15°C. A t what temperature will the rails just
touch? What stress would be induced in the rails at that temperature if
there were no initial clearance? Assume a = 11.7 /im/(m • °C ) and
£ = 200 GPa.
Solution:
St
0 .0 0 3
T
= «* LA-fc
=
1 1 .7 X
l0 - * (l0 l(T - I S )
H-O.
=
Am .
s t = e<*At = 200 x io*(it.n*to-*i(w .b* -is)
St
=
bO MPa.
Ant.
264.
A t a temperature o f 90°C, a steel tire 10 mm thick and 75
mm wide that is to be shrunk onto a locomotive driving wheel 1.8 m in
diameter just fits over the wheel, which is at a temperature of 20°C.
Determine the contact pressure between the tire and wheel after the
assembly cools to 20°C. Neglect the deformation o f the wheel caused by
the pressure o f the tire. Assume a = 11.7 /tm/(m °C ) and E =*
200 X 109 N / m 2.
61
Solution:
■St = £ < *A t = ZOO x. I 0 9 ( l l . 1 x l O -L X lO -Z O )
St =
MPa
lb 3 .B
pCl.rt
ib i.tfC io )•>
Z ( O . O ')
f s t . t l MPa
An(.
26S.
A t 130°C, a bronze hoop 20 mm thick whose inside diame­
ter is 600 mm just fits snugly over a steel hoop 15 mm thick. Both hoops
nre 100 mm wide. Compute the contact pressure between the hoops
when the temperature drops to 20°C. Neglect the possibility that the
inner ring may buckle. For steel, E = 200 G Pa and a — 11.7 (im /
(m- °C). For bronze, E — 83 G Pa and a = 19 /tm/(m • °G).
Ans. p = 2.86 M N / m 2
Solution;
Pb
Ps
Stj, ~
-
$
&*(
&S
py =
ft
/* x ID-6 (110) - (0 I x 0.0i ) ( t >3 X 10* )
i o -b (u o )
// 7 x
+
~
(0 0l5 x o . i m o o k i o * )
t.03 x /O'* =■ 1.3S7VZ X IO-* P
P
=
e r a w .ib u
IP
*
p PL
z(»S B m .tb )=
p
-
f(o .b H o .i)
1. &b M P a .
Am.
62
266.
A t 20°C, a rigid slab having a mass o f 55 M g is placed upon
two bronze rods and one steel rod as shown in Fig.' P-266. A t what
temperature will the stress in the steel rod be zero? For the steel rod,
A “ 6000 mm2, E ” 200 X 109 N / m 2, and a = 11.7 /wn/(m-°C). For
each bronze rod, A — 6000 mm2, E = 83 X 109 N / m 2, and a =
19.0 /im/(m- °C).
Ans. T = 129°C
55 Mg
250 mm
777777Z
I
7 *7777
50 mm
T
Figure P-266.
Solution:
cM fc I
?Y>
Pb
2M500M
213500
Sty =
^ &
L A T )e = f«<LAT), +
i<ix/o-t(o.zs?(aT)= 11.7X IO -*( 0. J ) ( AT? +
,
(.00Dx/0 *(Ai %/Q*)
A t = 101.llaC- ( rite.?
T
= I01.ll+t0» | tM .H ’ C
Ani.
267.
A t 20°C, there is a gap A «■ 0.2 mm between the lower end
o f the bronze bar and the rigid slab supported by two steel bars, as
shown in Fig. P-267. Neglecting the mass o f the slab, determine the
stress in each rod when the temperature o f the assembly is increased to
100°C. F or the bronze rod, A — 600 mm2, E = 83 x 109 N / m 2, and
a = 18.9 /im/(m- °C ). For each steel rod, A = 400 mm2, E = 200 X 109
N / m 2, and a =*~11.7 jim/(m • °C).
63
Solution:
Stb - o .zO o -’ ) - S r t '
8 L 4 * > , - 2 x .»-< - ( % \
M 'l o - M K t o )
-
l l .l I t o -'H l . t l l t o ) *
^
Sts + S,s
-
‘J ° c el e , ; ^ , -
* (% L
400 f. 1 0 * (ZOO* I09)
Ps = b.MO.SQN
Pb -
lt,3 6 t.lb N
Ss = P*/As =
IS-W MPa .
Atu.
S> = P>/Ay
10.W MfW
Ant.
-
64
268.
An aluminum cylinder and a bronze cylinder are centered
and secured between two rigid slabs by tightening two steel bolts, as
shown in Fig. P-268. A t 10°C no axial load exists in the assembly. Find
the stress in each material at 90°C. For the aluminum cylinder, A —
1200 mm3, E = 70 x 109 N / m 2, and a = 23 jum/(m °C).. For the
bronze cylinder, A = 1800 mm2, E = 83 X 109 N / m 2, and a * 19.0
/im/(m-°C). For each stee! bolt, A = 500 mm2, E = 200 X 109 N / m 2,
and a = 11.7 /un/(m • °C).
Ans. a, - 33.7 M N / m 2
20 mm
20 mm
Alum.
Bronze
Figures P 288 and P-26S.
Solution:
S
0*L A T ? a -t- ( « L A T ) g - («L A T ? S =
[2 5 x /o-« (o .0 7 s ? (e o )] + [/ 9 x I O - * f 0 . l ) ( B O ) ] +
2 £ (O. I)
M O D x 10-* C0S X ! 0 9)
[J = P / A ]
lf>62t.73
fOOX /o -*
+
Ps ( O Z I S )
SOO(ID-*?(ZOQ X /O*)
65
m oo x io-f>
3 3 6 S ^ .* b
s * = I B OO *
I D 1-
««. OS M P a
Ant.
1 9 .1 0
Ans,
M Pa
273. The composite bar shown in Fig. P-273 is firmly attached
to unyielding supports. An axial load P = 200 kN is applied at 20°C.
Find the stress in each material at 60°C. Assume a = 11.7 (im/(m- °C )
for steel and 23.0 (im /(m • °C ) for aluminum.
Ans. oa — 18.7 M N / m 2; a, — 181 M N / m 2
-- 200 mm-*
-300 mm-
Aluminum
70 X 109N/m2
A = 900 mm2
E -
Steel
£ = 200 X 109N/m2
A = 1200 mm2
Figures P-273 and P-274.
Solution:
Al
— ^
St
1
200 Kn
1 *
|............. ........
R,
?
—
Sts
tip .
'4.
F *.
Sfcx + Sts +
2 3 x / 0 -* ( 0 . Z J ( 4 0 ) + //.7 x I 0 - * ( 0 . 3 ) ( ¥ 0 ) +
I TO O * to-* (1 0 0 * . ! » * } .
R, = 4 l b , 9 l 5 . U N
Getting the. F 0 D o f e+cM. ma.iej'ia.L to
l i d # * r n t s r i#
t A #
Lo A d . -
<fOOxlO-l> (1 0 x 1 0 * )
0.3
0.Z
.1 0 0 x to-6no mo*)
200000 (0.*?
66
21(>815. 21
2 I6 8 I7 .2 J
— b
I&6I5.2I
____________ _ 200000
Aluminum L ____
;/Ll
L«s =
p/ a
P3 r u&aiF.2iR
•• fit* I68!?,Z J m
2lfr,8l£2l
]
'Ss =
■SA =
zie>eis. X!
tz o o x t o - *
i l &i s z i
_
•
160. 7
MPa.
Ans
I0 .b g
M Pa
A n !.
9 0 0 X / £ -*
275.
A rod is composed o f the three segments shown in
Fig. P-275. I f the axial loads /*, and P 2 are each zero, compute the
stress induced in each material by a temperature drop of 30°C if (a) the
walls are rigid and (b) the walls spring together by 0.300 mm. Assume
a ■» 18.9 fim/m- ° C ) for bronze, 23.0 /im/(m °C ) for aluminum, and
11.7 /im/(m • °C ) for steel.
Ans. (a) a, = 118 MPa; (b) oa = 40.0 M Pa
-800 mm -
-500 mm-
- 400 mm -
Aluminum
Steel
Pf
Bronze
A = 2400 mm2
A = 1200 mm2
E
83 X 109N/m2 E = 70 X 109N/m2
E
A = 600 mm2
~ 200 X 10''N/m2
Figures P-27S and P-276.
Solution:
AL
BEL... i —
(CovtH.rac.tion d o e to t/te therm al d s o f)
(cx te vn o n c(#e-& coatf reaction)
!
(e^j je—S-SRi
67
5<St =
^
/0.9 x t0-*>(0.e)(30> + 23 * ID * (P.S )(3 D ) + H.7 X /O '* f0.l)(3D)
t f f ______b l ___________
=
'i M O D d D - i - W f U O V
+
________ * £ ________
l20D(IO-t)(7 O )(IO ‘>)
0.1__________ 1
+ too(io-*)(2ooXioV
9 .39 x / P ' *
(iu ?
if
J $
1,33010 x 10- * R, t ^
=
W O .IU CkAC r i g i d . j
tj -
O
fi, - l o s e i . e s u
Dra.u/Lruj -the F R D o f Mch. mA.te.riaJ to
dit&rnxune.
lo cu is '•
7O t tl-V S
2H00 x 10
=
fyrome
H------>
?O50 1.95m
fcTC & l& H
2 * ? . 1 M Pa.
An.s.
70511.15
1200
4-—— Alur«iniv»n(------- *
10531.35M
10531.95m
X
/ 0 «
50.03 M/k
Sa -
/>*f.
7 0 5 4 1 -1 5 '
Sj
=
tp O O X / O ' *
----- l^teel 1-------- >
70501 .*35
Ri*
//7 - fr*
/)n£.
t f walls spring togctk.e.rj tj - P. DO03
q.S*) x ID-* = 1.330!S x IO-» K, + 0. 0003
R, = 1B03&. hi N
•S i m i l a r ~to
(a . ) :
f$ 0 3 S .b l
3b ~
XWD x I D *
~
HB03&. I
!9nn «■ 10/0-c
/200x
4 B 0 5 8 .i > !
$s =
t o o x /*-*
M P a .
AnS.
HO. 03 MPa.
Ans.
\0O .O b MPa.
Ans.
2 0 .0 2
_
276.
Solve Problem 275 if /*, and P 2 each equal 50 kN and the
walls yield 0.300 mm when the temperature drops 50°C.
Solution:
( c o n ir a d io * d o * to ie*n f>en d u re,
dscap)
iI
( extc*<s/ffx due- io loctdx P,a*c/ fz
a nd arsum/ng u*x//f apr/iyi 'ny t* )
y * x s Ra + I S P =
3
10
+
0.0__________
_______ 0\s________
* L Z ‘fDO(lO-t‘) ( 0 * ) ( I O ) 9
IZ O O ( IO * X 7 0 X IO V
f
+
b O O a O )- L (t O O )(I O r > \
*
_____ ° • 4__________ _ ■] +
r 6oo(it>-*)(2ooX,c>9) J
l.l* 0 0 (l0 -* )(7 0 )(/ 0 »)
5*0000 ( o .
(c o o )(lt > -* )(z o o )(/o 9 )
= [/ e .9 x to-L (o .g)ls o )\ + [ z 3 x !0-*> (O.S)(SD)]
i- [ 11.7 x ID-1- ( d.D (S O ) I
Ra =
47t,bb.S6 N
Oraui/Ln^ th e F B P <?f eax,h maJ&rtA-l it>
gLeA&rm in.es
tfA e
LaJ.
Loads:
______
69
<-------
50000
8r#w26
-------->
■'• P** 4 M A .3 B *
. goooo
!
* --------- ALuwiinu*!1
- - - - - - - - ■? l 4 ? , C , G , W f l
••■ R,*- q?M>G>.3«N
'S a ^ J h u M .
- L
IO W
■t - W
[ 3 = P/A ]
47bbb.5&
11.0 b MPa.
A mS.
‘Wbbb.JQ
n o o x //>-<
01-31 MPa.
A ns.
l V M M . 50
to o x lO -l
m b . II MPa.
Ans.
=
Z4 0 D * / £ -*
S* =
5S =
277.
The rigid bar A B is pinned at O and connected to two rods
as shown in Fig. P-277. If the bar A B is horizontal at a given
temperature, determine the ratio o f the areas o f the two rods so that the
bar A B will be horizontal at any temperature. Neglect the mass o f bar
AB.
Ans. A ,/ A a - 0.516
Aluminum
E = 70 GPa
a = 23.0 *im/(m* C)
L= 8 m
P I«U f» P-277.
Steel
E - 200 GPa
a = 11.7 pm/(m*°C)
L • 8m
70
Solution:
1 Moa0
3 ? a --^ s
F or tkui bar to be- horizon-taJ. act <*-U 'te-m.pajroLtu^es, the*
ilongcttlorv o f the- rocLi due* to tL clui*%ge~ ***. -t&mD^rvLtU*-*sMu^st be- x&roed, b y ~tl*& con'tm .otlon- cLu& ~to the* Loa.tLs
0*v the, rod s , t- &• Sj- — Sp 'f&r- both. r W
bt*
V*.
A*. (.'JO x/ o9)
Pu =
I b I DODO A a - A T
S ts =
&?s
P itx j
HI X !0-bQS)[ca) =
,
^
As (zoo X 103}
Ps
= Z3VOOOO ASAT
[j/ k = V Ps ]
3 (iblOOOO A c / r f j =
q d m 0000 A s p rf)
= I
fth I
Am .
279.
For the assembly shown in Fig. P-279, determine the stress
in each o f the two vertical rods if the temperature rises 40°C after the
load P — 50 kN is applied. Neglect the deformation and mass o f the
horizontal bar A B .
Ans. as — 134 M Pa (tennon); oa — 11.3 M Pa (compression)
/////
Aluminum
A - 900 mm2
E - 70 X 109 N/m2 3 m
a • 23.0 nm/(m ’ "C)
Steel
4 m A = 600 mm2
E = 200 X 109 N/m2
a =11.7 #jm/(m • °C)
to ............
.1 fl
A
— 3 m---- ► ----3 m-----*!•— 3 m — A
Figure P-279.
P - 50 kN
71
Solution:
= O
Pa. ( 3 ) * Ps(i>) - 50000(9) = 0
P a .-r
S t«. +
SpA
_
i P s - 150,000
&ts
+
&Ps
3
6
ZSta. +
Sftk.- S ts + &ps
2(2.3 x /O'4* 3 f. HO) + 2
[ IS POOD - 2 P t ) ( 3 )
100*10-1(70x10*)
PsW
[S =
/I =
6012*1.36 N
pA =
- I0 Z 5 8 -7 2 N
P/A]
^S, =
5a =
80119.36
bOO x /£>-«
IO Z 5 8 .1 Z
900 X /0-<>
I33.5T5- M P * .(T )
Ant.
II. HO M P a ( C )
A af.
280.
The lower ends o f the three steel rods shown in Fig. P-280
are at the same level before the force P — 600 kN is applied to the
horizontal rigid slab. For each rod, A *■ 2000 nun2, a — 11.7 jam/
(m • ° Q , and E — 200 X 109 N / m 2. Determine the relationship between
the force in rod C and the change in temperature A T, measured in
degrees Celsius. Neglect the mass of the rigid slab.
72
Plgu r* P-280.
Solution:
U y = o]
Pa.
+
Pb *
Pc
=
bODODD
Q
[X M = 0 ]
= b 00000 (3 )
z Pb + 3 P C =
900000
PB = *50000-1-5 Pc ©
■Su.bsictu.tLng ®
&
—? O
:
= bOODOO ~ PB - Pc
= 600000 - 1/50000 + 15 Pc ~ Pc
P* = /50<W? + 0.5T /£
+ StB^-CStA ■+ <5iv) , _
A
tftc + £pc ) ~(&iA * &Pc?
b
73
5<£ffe + 3
u J S »
= &toL + &Pa
+-+
2-&Pc
“ * » , » *> * , ( " 7‘ ' ° - , ™ < * r > *
»». 7 a r ‘ M O T ,
+
‘O w a -W X M *
u jZ Z S ju ,,,,.,
•Substituting © And. (3 ) t» tk U z^uktLon.:
4 S (IS O
- 1 -5 P c-1 +
S'. B S x ! 0 ~ s A T
/.
2 1 . OS X I O - * A T
+
=
/. 25" X 10~ * ( 1^0000 + tf.J-
X to -* A T
+
■* 3 X I O - * Pc.
O. 0 2 0 Z S - b . 7 S K t o * Pc. + 21. O b K I O S a T
-
S . e s x t o - S A T +■ I .& 7 S X t o - * + t . 2 S X IO 3 P c +
l-* 0 4 X IO -* A ±
-t 3 x tO
L 0 3 7 S X / O ' 7 Pc, = 11 7 X / O -S A T
Pc. =
H Z . 77 A T
* Pc.
+ O. O IB 3 7&
+■ 1 7 7 1 0 6 .9 3
\
fin s
281.
Four steel bars jointly support» mass of 15 M g as shown in
Fig. P-281. Each bar has a cross-sectional area of 600 mm2. Find the
load carried by each b a r after a temperature rise of 50° C. Assume
a - 11.7 |tm /(m -°Q and E - 200 x 109N /m 2.
Ans. PA ■■ P D “ 21.5 lcN; PB ■» P c “ 67.3 kN
Figures P-2S1 and P-282.
S5
(ft
U
r.
n
M
C*
(M '■fc
*4
*>
♦
3> s.'*
II it
©
<*
Ki
K»
45
>3
■fe
£
S>
frs
s s
*
$
+ +
<a
«>
9- «?5
S'
S» 5
£3
N»
<fc
a. a»
*o
$
*A
%
t
0*
S>
■m
*5
(M
•*
*
*
fr-
2
©
I
■<
X
o
=1 5
n
■<1
X
■*-
*
5
+
»
0*
3
X
$
A
*
X
«
X
>3
•fe
fe
I
a
s
<2>
§
II
s
*>
-ft
«."»
O
<S
§
II
X
<Ji
§
»\
x
5
i
A
*
fr.
X
§
©
I
»
f
\
s
*
§
X
5
v5
un
§ -~>
ri \
*
S
§§
X
<0
V-/
-J
<-n
Torsion
305. A solid steel shaft S m long is stressed to 60 M Pa when
twisted through 4°. Using G — 83 GPa, compute the shaft diameter.
What power can be transmitted by the shaft at 20 r/s?
Ans. d = 104 mm; 9 - 1.67 M W
Solution;
Jr _
100 ~
i>Px 10* ( ? )
0 .(0 3 * 10°)
P
1f 0 3 . f
-
Z
m m
1
A r% 5 .
so
________
■fTT
100
5T ( ° l03srs) * ( t 3* 109)
r = is o e i.w n - m
or
s
=
5
T (° U O M L )
kO x I0b
T
p= i V T f
■—
z
-211
32.
=
I5 0 «/
N -w *
= n r(i3 o & 0 (z o y
p
=
i,b * n w
fyns.
78
308.
Show that a hollow circular shaft whose inner diameter is
half the outer diameter has a torsional strength equal to -jf of that of a
solid shaft of the same outside diameter.
Solution:
J = i
h o llo w
0
solid
D
SJ
T
p
si
II
rs -
Ss J s
P«
_ U ’/3Z ( D + - P V i l )
J h _
Ts
TT/^SZ ( p 4 )
Js
Th
Ts
>S
Ans.
It
309.
A steel shaft with a constant diameter o f 60 mm is loaded
by torques applied to gears attached to it as shown in Fig. P-309. Using
G «■ 83 G N / m 2, determine the relative angle o f twist o f gear D relative
to gear A .
Ans. 9D/A- 2 M °
800 N-m
1200 N-m
1000 N-m
Figure P-309.
79
Solution:
c l200N-rn B
IOOON-i*r
EE3I
Assuming
aA civcku/uie ( * > ) ' ■
800
800
— H
1200
<r+
K800M-IH
1000 N-M
looo
0C
000 N -* n
Tab •BC
800
1000
1000 N - m
Too -
CO
--- *
1000
1000
J _
JZT [O .D b J *i -
t/^5 =
I 05b X 10s
32
1 .1 7 2 3 4 X / O '*
0o o ( z ) + 100 ( 3 ) - i o o o ( z )
/. 0*6 X //>*■
>c
0 .3 7 0 77 r a d
~^/o =
- 2-/7'
A mc-
310.
Determine the maximum torque that can be applied to a
hollow circular steel shaft o f 100-mm outside diameter and 70-mm
inside diameter without exceeding a shearing stress o f 60 X 106 N / m 2
or a twist o f 0.5 deg/m. Use G -* 83 X 109 N / m 2.
Solution;
J -
(O. I * - O. O T * )
=
7.4LD 3 x r o - * bn 4
[ * ■ - J? ]
bO X ID* =
r =
T (o . o s )
7.
x //>-*
0. 9ST.Z K/V-m
[*■ £ ]
O.S-’x —
/0<?
XL
=
T
-
T/_
7 .4 b 0 3 x ! O - * (0 S x I O * )
SM04
K ti-n ,
Chcofe ihe- Smaller ■value-:
T =
A nt.
311.
A stepped steel shaft consists o f a hollow shaft 2 m long,
with an outside diameter o f 100 mm and an inside diameter of 70 mm,
rigidly attached to a solid shaft 1.5 m long, and 70 mm in diameter.
Determine the maximum torque that can be applied without exceeding
a shearing stress o f 70 M N / m 2 or a twist o f 2.5 deg in the 3.5-m length.
Use G - 83 G N / m 2.
Ans. T = 4.01 k N •m
Solution:
1m
1.5m /'-)- T
*
O.lrtl
;
33.
( 0 . 1 0 . 0 7 + )
V
Js zz - j £ ( 0 . 0 7 ) 4
Tu =
Ss d
-
=
'J . f b 0 3 X /O - 1
2 .3 S 7 I7 X I O 'h
7 O a 10* ( 7 . 41,0 3 X IO ~ b)
O. 0 5
10.44 k » -#
r„ -
T< =
Ts 5
70 x t o 1’ (Z . Z S 7 1 /
X to -* )
0.0 i s
4.7/•/■ U N - m
7X1
Z.S *T f
J6 J
100
VCZ)
7. « ( , 0 3 x ! o - b (0 1 X t o 9 )
rct.s )
2 . . 3 f 7 f . r x t c , - * i s r x / O '* )
T =
4 .0 K N - n t
Cltoofc -the te a .it i/ulus. :
T =
4-0 K/V-m
312.
A flexible shaft consists o f a 5-mm-diameter steel wire
encased in a stationary tube that fits closely enough to impose a
frictional torque o f 2 N m/m. Determine the maximum length o f the
shaft if the shearing stress is not to exceed 140 MPa. What will be the
angular rotation o f one end relative to the other end? Use G = 83 GPa.
Ans. L « 1.72 m; 9 - 33.3°
I
81
Ssd
<; n d *
JS 32
S s tfd 3
^/jL
lb
r
a io 6 Crf)(o.oos)3
2L =
lb
L =
A ns.
2Ld L
- f
#J G
L2
-I J G
■/-
J6
(1.716) 1
(0 .0 0 5 )4
- 0. SB ra d
(6 3 x IO 9)
Ans.
315.
A 5-m steel shaft rotating at 2 r/s has 70 k W applied at a
gear that is 2 m from the left end where 20 k W are removed. A t the
right end, 30 k W are rem oved and another 20 k W leaves the shaft at 1.5
m from the right end. (a ) Find the uniform shaft diameter so that the
shearing stress will not exceed 60 V !N / m /. (b ) If a uniform shaft
diameter o f 100 mm is specified, determine the angle by which one end
o f the shaft lags behind the other end. Use G = 83 G N / r a 2.
Ans. d — 69.6 mm; 0 = 0.448°
Solution:
4-4—
20 Kw
« -
70Kw
30 Kw
20Kw
ZOOOO
rAH= z trc z )
ABCD
20 KW
T ab =
IS 9 I S S N - m
20 Kw
T*, =
SOCOO
zTT(z)
Tar = 5970.07 A/-m
30KW
— hH
cpO
30 Kw
e f
7L, =
30000
2
= 2 .3 0 7 .3 2 M - m
I n order "toarrive. at the Safe. D Such that
s-trvss ii not exceeded Choose the
la-n^e-s t T ■
,
82
••• T = 3 1 7 0 . 6 7
S i, W
7r
-t C M
=
N -m
t Px/ Ds
32.
b9. 6V mm
~®T ~
~ ~fyc-
fins.
'&i-0
I S - V / . r & U ) - 3 e i B . B 7 ( i c ) - Z 3 3 7 - 7 Z C/.S^)
-&-T =
in. (o .o + c& i x t o * )
32
-&T
=
-
O PO70H 7 X
-O. » ( f 8 °
-&-r =
rr
knf.
316. A round steel shaft 3 m long tapers uniformly from a
60-mm diameter at one end to a 30-mm diameter at the other end.
Assuming that no significant discontinuity results from applying Eq.
(3 - 1 ) over each infinitesimal length, compute the angular twist for the
entire length when the shaft is transmitting a torque o f 170 N m. Use
Ans.
d /z
(■Olx)*
D. Dis­
J
O.OIX
J = 3.125' X /0~‘°TfX 4
32
3Z
X
ci Te/x
J6
no
. r 6-d il
x4
3 . n s x t o - ,o Tr(0 3x i o y>J
j
x
'
- i . P B b m i z i _x - 3 i t
fjl
U *
- J.
3 ‘s3J
9 = 1.29°
83
= O.O22S36000 *
100
rr
-&■ =.
Anc.
318. A solid compound shaft is made of three different materials
nnd is subjected to two applied torques as shown in Fig. P-318. (a)
I >ctermine the maximum shearing stress developed in each material, (b)
l ind the angle o f rotation o f the free end of the shaft. Use Ga = 28
(IN / m 2, G, = 83 G N / m 2, and Gh « 35 G N / m 2.
75 mm dia.
100 mm dia.
4.0 kN ■
m
1.5 kN-m
Aluminum
Bronze
2 m—«•*—-----
-3 m-
1.5 m
Figure P-318.
Solution:
Al
I,? ftl-Hlr
f
|.7KN-m
Tb r 1.5 K N to
1.5Rn-m
—
st
X = 1.5 KN
2.5 K*-m
j.fKN-m
T^ =
AL
I S (IQ) 3 ( 0 . 0 7 5 / 2 . ' )
TT ,
-
Z.SKN-n,
10.11 M P *
A hs.
10.11 MPa-
An S.
12.73 MPa.
Arte.
(0.07S)*
Sc =
l.S(to)3 ( 0 . 0 1 5 / 2 )
(0.075)*
Z.S(IO)3 (0-QS)
f z (o .n +
84
TL
JG
-e-b =
\
(t o )3 ( I. 5 )
---------- * o .o io b is
■Jl (O- 015)4(35% / o V
, i - f t i M U ------------ ,
3X (0 .0 75)* (0 3 x lO* )
-1 .5 (IO )3 CS)
■e^ -------------------- -------
- -O.OZ7Z017
( o - l ) * (ZB x IO V
&T ~ -&B -t &s + -&A
I0O
-Oy = O.OOSOV7 X -^p0 .2 6 9 °
A *S .
319. The compound shaft shown in Fig. P-319 is attached to
rigid supports. For the bronze segment A B , the diameter is 75 mm,
t < 60 M N / m 2, and G = 35 G N / m 2. For the steel segment BC, the
diameter is 50 mm, t < 80 M N / m 2, and G = 83 G N / m 2. I f a - 2 m
and b = 1.5 m, compute the maximum torque T that can be applied.
Flgur** P-319 and P-320.
Solution:
Sr
r = ts
“St:
+ tb
c " > i +
= -©s
85
_________________
(O .O t S " ) C-3S X t o * )
S8
I f SB =
5S
-
C o .o s ) (8 3 X 10*1
=
O. 4P7V-# Sf
bO MPa
126. q -76 M p a. ( o i 'e r s f r e d t d )
I f Ss = gO M Pa
-j
S B = J7. ‘iSZ MPa J 6overns
BO X /0* ( ]£ )(O . O S )4 +
T _
31.e)S2.x.lDl>(Tfz )(o.07S')4O. OIS / i
O. 05/ z
T =
5/07. * 5 A/- »»i
An*.
324.
A shaft composed o f segments A C , CD, and D b is fastened
to rigid supports and loaded as shown in Fig. P-324. For steel, G — 83
G N /m 2; for aluminum, G — 28 G N / m 2; and for bronze, G = 35
G N /m 2. Determine the maximum shearing stress developed in each
segment.
Ans. TB = 472 N -m ; t a1 = 9.3 M N / m 2
Tc = 300 N-m
Td =700 N-m
/i ? C-----------------------f
Aluminum
DI t
Steel
25 mm dia.
V/
50 mm dia.
-1.5m -
---- 2 m—
B
Bronze
> 25 mm dia.
' *
b
/
/
/
/
— 1 m----- «/
Flgum P-324.
Solution;
300 N-m
SUtl
TOO r+-m
b
Alum.
Bronze
f r ,
T*+ T> *
___________
300
Steel
Ta
T« f p
M - T*
TOO
------H
300-Ta
Alum.
Iooo-Ta
Ta
1000-T,
T p * Iooo-Ta
86
I>*/« =°]
( 3 0 0 -T A) ( l . f )
rAw
7r
sz ( o . o s ) + ( i e x io<>)
(0.025)4(03 X 10*)
(lo o p - Ta ) ( i )
=
o
- £ ( P . O Z S ) * ( 3 S x I P 9)
32.
T . = 5Zfi N Tb - 47Z N - i
[ - t ]
S
-
S
S '^ ^ (c>ozeA ' !
n Z . ! MFk.
Ans.
^ C o o l s ') *
(300 - 5~4g) ( ° - os/x )
Ans-
3 £ -(o .o s )+
3a
J
( in z ) ( oozsM
=
_
JS-J.BMP a
Ans.
3Z
325.
The two steel shafts shown in Fig. P-325, each with one end
built into a rigid support, have flanges rigidly attached to their free
ends. The shafts are to be bolted together at their flanges. However,
initially there is a 6° mismatch in the location o f the bolt holes, as
shown in the figure. Determine the maximum shearing stress in each
shaft after the shafts are bolted together. Use G = 83 G N / m 2 and
neglect deformations o f the bolts and flanges.
Figure P-325.
Solution:
+
-0fc = b ° -
T (Z )
~Si (O O5-)*(03x / P9)
^
rad
T ( i)
■Ji(O .0 *)4,(0 i X l o V
Tf
30
87
T = 12 0 0 7 9 N - m
1200. 7*? ( ° - os/x )
1200 .
40.52. MPa.
Ans.
S (, A7/=V
A/ic.
/z )
(0 .0 ^ 4
326.
A flanged bolt coupling consists o f eight steel 20-mm-diamrter bolts spaced evenly around a bolt circle 300 mm in diameter.
Determine the torque capacity o f the coupling if the allowable shearing
Ans. T ” 15.1 k N • m
»tress in the bolts is 40 M N / m 2.
Solution:
T = - f < J * s s Rn
= ^ r ( O . O H ' ) 7- (4 0 X I O * ) ( d ■ is )(8 ')
H
T
=
I S , 07*?. 64
A/-*«
Ah s .
327.
A flanged bolt coupling is used to connect a solid shaft 90
mm in diameter to a hollow shaft 100 mm in outside diameter and 90
mm in inside diameter. I f the allowable shearing stress in the shafts and
the bolts is 60 M N / m 2, how many 10 mm-diameter steel bolts must be
used on a 200-mm-diameter bolt circle so that the coupling will be as
strong as the weaker shaft?
Solution:
Ss TTci3 _
bO(lO')‘’ (V U O -Q ‘3'>3
/6
T soi id
0S'68. ?5 N - n t
Ss Jh
~
_
b o O o )* C r r / s z X o -i4 - 0 03+ )
CVO/z
4051 -40
r
s
JT c/^SsRh
40SI.48* -^ (O .0 l)z ((>0X /0h)
or
n z
balls
Ans.
88
328.
A flanged bolt coupling consists of six 10-mm-diameter steel
bolts on a bolt circle 300 mm in diameter, and four 10-mm-diameter
steel bolts on a concentric bolt circle 200 mm in diameter, as shown in
Fig. 3-7. What torque can be applied without exceeding a shearing
stress o f 60 M Pa in the bolts?
Ans. T = 5.50 k N • m
Solution:
Pz _ P,
R,
fiz
P,
Pi
o.z
z
2.
p
Pz = —
3
T -
P, R , n , +
- P ,(R ,n ,
+
P^ I?2 m z
|
R i * iz
= ^ r o .o O z ( ^ x l o O [ ^ T
* *+
nr
U * T X* )]
Ans329.
Determine the number o f 10-mm-diameter steel bolts that
must be used on the 300-mm bolt circle o f the coupling described in
Problem 328 to increase the torque capacity to 8 kN ■m. Ans. 10 bolts
Solution:
From froh . 3 20
T - P, (R , n , + -%-Rz M* ?
0OOO -
^ ~ ( t > . O l ) * ( b t > x /<?*?[
h, -
^
I\0 bolts
x4) ]
A n t.
331.
In a rivet group subjected to a twisting couple T, show that
the torsion formula r = Tp/J can be used to find the shearing stress r
at the center of any rivet. Let J = X A p 2, where A is the area o f a rivet
at the radial distance p from the centroid o f the rivet group.
Solution:
Torcju/! capacity af a single, bolt,
T - PR = ASs/o
For ayitt nuyrt b*-r o f boH%, it is
T - E P R = Z flS s/o
89
Stresses ajid strain* tune propo'iiona.1 to t h e i r
ra d ia l distasic.es; th e ra tio Ss /p it e»nsia.nt.
T- * * ( f e ) r -
t
ZAp*
S ,= 2 >
J
Am.
332.
A plate is fastened to a fixed member by four 20-mm
ilmmeter rivets arranged as shown in Fig. P-332. Compute the maxi­
mum and minimum shearing stress developed. (H in t: Use the results o f
Problem 331.)
|16kN
x 100 x 100 fa 100 a
^ mm ® mm 'P mm t
16 kN
t
t
Figure P-332.
Solution:
T -
I 33A\
!t>x lO3 X 0-3
JM n
—
—
7" '
r
t
f£
o.o2 m
|<— 300mm —^i^Kh
J =
= — (p .p i)t-\o.os’z n Z + 0 .IS * X 2.1
4
J = S X IO '* T f
ss= Z L
( i b x i o * H 0 .l 0 ) ( 0 . /sr)
S x io-*TT
A/ax Ss =
4 r.
90
M in Sg -
(lt>X ID 3)(0 .3 0?(0 OS)
5% i o ~ i i f
IS. ZB MPa
h in S j -
At%f.
333.
Six 20-mm-diameter rivets fasten the plate in Fig. P-333 to
the fixed member. Using the results of Problem 331, determine the
average shearing stress caused in each rivet by the 40-kN loads. What
additional loads P can be applied before the average shearing stress in
any rivet exceeds 60 M N / m 2?
Ans. Max. r * 45.9 M N / m 2; P *■ 55.4 kN
40-kN
p
Y mm T mm T
i
4
<
- 4
Si 1
- - X
1
— 250 mm—
£ 75 V
tI/ 75 ly
X
viT
"
f
kwA»A\VAV
!
/
/
p
40 k N
Flgur* P-333.
Solution:
J,4 o Kn
_
i 7fmm ■
l-cp-
-9 -
d: 0 .0 2 m
J = A t-p 1 = A [£ x z + Z ij* ]
= t\[0.07Sz ( ‘f ) +
0 .0 5 z ( 6 ) }
- 0.0375 A * 0 .0 3 7 S (~ )(0 . O Z ) *
J = 3.7S * I O - * l f
91
For 4 corner- rivet* ••
40000 (o.ts)(o- onoW )
3.75 x to~bV
Tp
=
M*-x Ss = M C.II H P *
Am .
For -two center rivets:
Tp _
Sr - —
h0000( 0.t5)(o .o s )
3.7$
- IT
i f . <tb Mf W
Am .
b) T 0.15 f - 40000 x O'IS =
pz
6£X tOk 13.15m o - ’’ ’i f )
ff,3 *7 .»/ V
O. 01014
Ans.
334.
The plate shown in Fig. P-334 is fastened to the fixed
member by three 10-mm-diameter rivets. Compute the value o f the
loads P so that the average shearing stress in any rivet does not exceed
70 MPa. (H in t: Use the results o f Problem 331.)
Ans. P - 7.12 kN
/ r / w / r jv r r r w fh tr
Figure P-334.
Solution:
////////////////tffUU.
77777777777777777777777
92
J
=
A (z x *
=
A (Z x
+ Z i ]* )
0 0 S Z + 0 -1 0 * -+ 0 .0 7 5 x x 2 )
= O -O Z b Z S A = 0 . 0 Z 6 Z s ( ^ ) ( 0 .0 1 )*
J = b.SLZS X ID-ffT
T=
g* J
r
7o x io * u ,.s f > z s x to -T r r ')
o .io
o.zz\r p
P=
6 * 1 4 .0 1 N
A*tf.
335.
A flanged bolt coupling consists o f six 10-mm-diameter steel
bolts evenly spaced around a bolt circle 300 mm in diameter, and four
20-mm-diameter aluminum bolts on a concentric bolt circle 200 mm in
diameter. What torque can be applied without exceeding a shearing
stress o f 60 M N / m 2 in the steel or 40 M N / m 2 in the aluminum? Use
G, - 83 G N / m 2 and Ga = 28 G N / m 2.
Ans. T - 5.94 k N • m
Solution:
p i /a e
L =
_p
AE
L
s
E.
J>S- «
EjRs
_
Ss
05 A 109(0. 15}
Ss
If
If
2 8 (1 0 )9 (0 .»
=
4 . 44t,4-Z8S7 S «
Sa =
40
M fi»
Ss =
/77. 86 M fii (ovm rslren td )
Ss = 6 0 M P a
5a
=
/*•
} G o v e rn s
93
T -
P*. R a no. + H R s Ms
A * 5a R*.
+
As 5 j Rs ms
= & ( 0. 01) * ( O . O W ( 13-4 4 x t o * ) +
¥ ( 0 . 0 I ) * ( 0 . IS )(t> )(6 0 x !Ob)
4
r =
I
a »«j.
337.
A torque o f 600 N • m is applied to the rectangular section
ihown in Fig. P-337. Determine the wall thickness I so as not to exceed
■ (hear stress o f 60 MPa. What is the shear stress in the short sides?
Neglect stress concentration at the corners.
30 mm
Solution:
[ s s = r / 2 A t\
60 x !Ob
(,0 0
2 (0 -0 3 )(0 - O i.)t
i, = 2 . 76 m m
Ans.
338.
A tube 3 mm thick has the elliptical shape shown in Fig.
P-338. What torque will cause a shearing stress o f 60 M N / m 2?
Ans. T = 3 . l8 k N ‘ m
94
Solution:
■150mm--- *1
[ * ■ iifel
7"_______________
bO x /tf6 =
2 x 3T ( 0 ! S ) ( 0 . 075X0.003?
4
T
=
3.1B K N -™
Atts.
340. Find dimension a in Problem 339 if a torque o f 600 N •m
causes a shearing stress o f 70 M N / m 2.
Ans. a = 55.7 mm
Solution:
70 k IOb =
.2 [JTT(O- O Q 2 + *7. 02
y y . 7^ rYtm
343. Determine the maximum shearing stress and elongation in a
helical sieel spring composed o f 20 turns o f 20-mm-diameter wire on a
mean radius of 80 mm when the spring is supporting a load o f 2 kN.
Use Eq. (3-10) and G = 83 G N / m 2.
Ans. Max. r = 121 M N / m 2; 8 ■“ 98.7 mm
Solution:
c-m = %
m -
«
'* » / * * .
&
P R *n
64 C Z oooU o. OB')3U p )
03 X IO *((? .O t)4
95
$
=
0. 0787 m =
\lB.7rv,,
+
/ (> (2 C O O ) ( 0 . 0 8 1 [ 3 2 - 1
Ss ~
7T(P- 02 )3
5S =
120. bO MPa.
Ans.
0.6.IS
L 32-4
Ans.
345. A helical spring is made by wrapping steel wire 20 mm in
ilmmeter around a forming cylinder 150 mm in diameter. Compute the
number o f turns required to permit an elongation o f 100 mm without
meceding a shearing stress o f 140 M Pa. Use Eq. (3 -9 ) and G — 83 GPa.
Ans. n «■ 17.9 turns
Solution:
6f
*
=
0
P R 3*
&d<
6 4 (P )(D . Q 7 S )3n
8 3 X 1 0 ^ (0 .0 1 )4
Sc
=
WO x
lb PR
7TJ3
-
I b P (0 -0 7 5 )
jr c o .o z y s
=
2.748. 87 N
1 1 . 8 1 turns
/+
o .o z
4 (0 .075)
]
Am .
348.
T w o steel springs arranged in series as shown in Fig. P-348
support a load P . The upper spring has 12 turns o f 25-mm-diameter
wire on a mean radius o f 100 mm. The lower spring consists o f 10 turns
of 20-mm-diameter wire on a mean radius o f 75 m m I f the maximum
shearing stress in either spring must not exceed 200 M N / m J, compute
the maximum value o f P and the total elongation o f the assembly. Use
Eq. (3 -1 0) and G — 83 G N / m 2. Compute the equivalent spring con­
stant by dividing the load by the total elongation.
96
Solution:
200 x /O* =
lh P ,(Q .l'f
r32- /
D. t>!S
[
TT(o. £>2 S )i L 3 Z - 4
]
8
/? = Slg2.2<1 N
.200 x /0* =
H> PX(Q.Q7S) |~30- /
L[ 30- 4<
7T C o.oz)^
+
0 .6 / f
7. S'
J
& = 3 4 1 8 .2 8 N
\ A7a*
**
2 # /V
=
^
Z
A us.
t>* P R *n
SJ4
6 ‘K 3 ‘/‘j f . 2 g ) ( o . l ) 3 ( l Z )
+
&3 X t O * (D .O Z S )+
(>V ( 3 M i f . 2 & )(Q . 0 7 S r ) * ( / 0 )
83 X 1 0 9 (0 .0 2 ) 4
St
=
0-
3118.28 N
_
0. ISU —
\ x 2 T lb .l
A H i.
349.
A load P is supported by two concentric steel springs
arranged as shown in Fig. P-349. The inner spring consists of 30 turns
o f 20-mm-diameter wire on a mean diameter o f 150 mm; the outer
spring has 20 turns o f 30-mm wire on a mean diameter o f 200 mm.
Compute the maximum load that will not exceed a shearing stress of
140 M Pa in either spring. Use Eq. (3 -9 ) and G = 83 GPa.
Ans. P = 9.05 kN
£
(2
- (1 )
W*
77777.
Figures P-349 Mid P-350.
97
II
v/T
'•oluiion:
I
/,
&<t«
A
PzRz^nz
P> R, 3» ,
G .d ,*
fj (0 .0 7 5 )3 (3 0 )
_
S3 X I 0 * ( 0 0 2 ) +
P,
H ( o . i ) 3c z o )
8 1 X //?9 CO. 0 3 ) 4
0 .3 1 X 1 * 7 5 3 1 Pz_
=
P = P, + P i
Ib P R
TTd* ( ' * & ) ■
5 -
p,
l * o x 106
Pz
px
P,
=
2
/V
/£ Pz (o . 1) f . . o. 0 3
'I
— :- 11 + --------- J
TT(0.03?3 I
4x^ 7./ /
—
= M O V -.ZO Al
.2.0*1
= 6 < t0 t.2
ON 1
y<
r &ove-rns
= 2.ISS-. 13 N
j
Ans.
351. A rigid plate o f negligible mass rests on the central spring in
Fig. P - 3 5 1 which is 20 mm higher than the symmetrically located outer
springs. Each o f the outer springs consists o f 18 turns o f 10-mm wire on
a mean diameter o f 100 mm. T h e central spring has 24 turns o f 20-mm
wire on a mean diameter o f 150 mm. I f a load P = 5 k N is now applied
to the plate, determine the maximum shearing stress in each spring. Use
Eq. (3 - 9 ) and G = 83 G N / m 2.
A ns.
Central spring: M ax. r -= 170 M N / m 2
Figures P-351 and P-352.
98
Solution;
IP^OOOH
] 55mm
A m
T
A 5
^
C
whni/nimiirnm
&A + 0 .0 Z - £e
Ph -
Pc
2-Pa -* Pb = 5000
\
6* P R *n '
bA * A
bd*
(
bn Pb 10. i s P l i f ')
Pa. 6 4 (0 -0 5 ) 3( i f )
0 5 X IO *(0 .0 1 )*
1 .7 3 5 x
1 .7 3 5
l O 't p a ,
+
G J* /B
03 X t o *(0.01 ) *
'
0 .0 1
-
4 .9 7 4 *
X I O - * Pa + 0 . 0 1 =
1.735 x to -* Pa + O.Ol -
X IO -* P B
x I 0 S ( 1 5 0 0 - IP*.)
* .6 7 4 5
0. m m 51- I - 75103b x IO~SP*.
P4 = 0 *6 .2 Af
Pg
*
3 3 * 7 . 51 A I
lt> (g U .l)(0 .0 5 ? f , . 0.0/
SA - Sc =
7T(ooi?} ( 1+ ¥x0.os
5a - Sc 5. =
110. 11 MP a I
An*.
O.Ol
£b =
I70.
J
H% 0.075
T (0 - 02P 3
MPa
Ans.
353. A rigid bar hinged at one end, is supported by two identical
springs as shown in Fig. P-353. Each spring consists of 20 turns o f
10-mm wire having a mean diameter of 150 mm. Compute the maxi­
mum shearing stress in the springs, using Eq. (3-9). Neglect the mass of
the rigid bar.
j ns
Max. r = 46.5 M N / m 2
99
Solution:
Sb
z
4
iso
= IS
lo
ZK
d
Sa
-
2-6 b
( k H M lt i ) - o ( k l l E l i L )
I 6d1 4
I
A
/>* =
zn = o
fk (z ) +
+ 4f2&) =
IPPx
-
1 0 (6 )
Pa = ^ .0 A »
=
117-6 N
100
l6 0 l7 .6 k o .Q 7 S ) r
7r(o.oi>3
L
Max Ss = Hb. 42 MPa.
j 0 .0 1
1
4( o . o i * ) ]
Ans.
355. As shown in Fig. P-355, a homogeneous 50-kg rigid block is
suspended by three springs whose lower ends were originally at the
same level. Each steel spring has 24 turns o f 10-mm-diameter wire on a
mean diameter o f 100 mm, and G = 83 G N / m 2. The bronze spring has
48 turns o f 20-mm-diameter wire on a mean diameter o f 150 mm, and
G ■* 42 G N / m 2. Compute the maximum shearing stress in each spring
using Eq. (3-9).
Ans.
For bronze, Max. r *■ 9.93 M N / m *
Solution.
Z £ b —3 So. ~
3 iy
-
S c * ZS*.
_ (6 4 />***. }
3 * b * P e Re *t,
6J4
I
Gd*
+ „ ( f t PR** }
(
/C
GtJ4 Ja.
i x J ^ K Py (0 .0 5 )3 (2 4 ) _ ^ < P c. (O .Q 7 S )* (W >
03 x IP9 (0.01)442 * IO» (0 . 0 2 )*
+
2 XJ>*X P *(O .O S )*(Z 4 >
03 x i o * (o .o iy *
/• 08*3173 x 1 6 s Ph - 3.01339*0 x IO~l f l +
7.220 915(, K lO -^ fk
l* y = Oj
f k + Py + *2. =
C ?M = 0 ]
P b (» + (id ) Pi + 3 d
4 1 0 (1 .*?)
= 13S
Py = 735 - 3Pa
PA = 410 - 73S + 3 d - Pc.
P^=
dPc ~ Z 4 5
7.1 (.10711 x IO * - 3.2530/19x 1 0 '* Pc. =
3.0133920 X IO-*> Pc. + 1-1457011 x IO~s Pc.
n
/. 7Y/0843 X /O - 3
=
H 1 .9 I n
PK =
/SVJ.SS A/
/i =
144. (.2 hi
_ H>d44-f>2 )CO.Q5) (
5fc =
+
V
I T (0 .0 1 )3
0 .0 !
\
1 * 0 .0 5 /
s« = 3 * . b l M P + .
A ns.
It- (150- 56. X o . OS?
(
1T(O .O O *
1x0.05 J
S* =
40. 2f> M ftT]
s
lt> (M 4 .* > lX / > 0 7 f)
V (o. O t-)3
Sc. -
V
« A »fJi
p.O!
I
+
p .Q Z
\
(
4 x 0 .0 7 5 J
4
Shear and Moment in Beams
404.
Beam loaded as shown in Fig. P-404.
Arts. M cd = ( —4x + 28) k N -m
10 kN
-3m-
,c
Solution:
-2 m-
#
40 kN m
(0 Km
D
104
£ F tj -
o
= 1 0 -6
= 4 Ac A/
Va b •
-1 0
H ab=
-!D %
Vgc -
- ID + 6 = - 4
-I0 t
+ I, I t - l ' ) -
-4 X
-1 2
V c = - IP + t =* - q
M<,0 = -I O x + b / x - Z ) + 4 D =
405.
-v x + ze
Beam loaded as shown in Fig. P-40S.
^ n i. W jc * ( —5 x2 + 44x + 60) kN .m
30 kN
H
10 kN/m
-10 m-
Flgura p.
Solution:
SOKm
105
^
- o
I OK, = 30161 + loofsr)
R, = 7 «*/ V
£P, = O
R z - 30 + 100 - 1 4 -S (> k N
Ma s * 7* ~ ' * *
= 7Vx - W
r ) = _ 5 X i 4 7* x
Ksc =
74 - s o - 10x = - ID x +
M/JC =
7VX - W X ( ~ y ) - 3 0 / X - X )
= - 5 ’x z + 4 4 x * 60
■® f w r f -ffce p o i n t o f z e r v s h e a r :
JL
-
g ~ *
14
X
4tfc
S6
=
* .«m
Beam loaded as shown in Fig. P- 406.
20 kN
40 kN
2m
le
20 kN/m
« —2 m—
u
4 m----------»
Jr,-----------
Jl,
^
F%M* m m .
*2
106
Solution:
10 Kn
■Wien
,C
1"
2 0 kM/w
A
zm *•B
4m
1
R.
= 0
H R , = 40(2? + 2 0 M + 1X0(3?
R, =
140 kl4
=0
= 4 0 + 1 0 -* 120 - 110 - 40 k N
VA6 = - X 0 -2 0 X
M a6 c - t o X - 2 0 x ( j 9 = - l ° x Z ~ * 0 *
V *: =
- 1 0 - 20X +140 = /*0 -
Mec -
- X0x - 20X f'- f- ) + 140 ( X - 2 ?
120 X - 10 X * - 290
Va> = - 2 0 - 20X + 1 4 0 -4 0 - - I O X +SO
= -2 0 X - 2 0 x ( - j - ) + l 4 0 ( x - * ? - * O ( X - 4 )
gOx - 10X * -120
10/
«M
Beam loaded as shown in Fig. P-408.
Aru. Max. M «■ 8333 kN m
«•----- 2m-----►
B
30 kN/m
15 kN/m
lii
R2
- Wr i P W l
Solution:
20*
t y £ M KjL = o
(>R,
= Id(i )- h d (s)
R, =
70 k N
108
Vju = 70-J O X
M m = 70X - 30% (-J -) = -iS * z * 70
Vg e.- 70- b O - I S ( X - t ) =■ - fcX +40
M k ,=
70x - i > o ( * - i ) - J 5. ( x - * ) * -
= —7.5 X 1 + 4 0X + 3 0
For the point o f zerv shear--.
V gc=
O
411. Cantilever beam carrying a distributed load varying from w
N/m at the free end to zero at the wall, as shown in Fig. P-411.
Arts.
H p n P-411,
Solution:
v-p
M - (w x 3/ 6 L ) - (wx2/ 2 )
109
WU
Z
K B y s im ila r- - t r ia n jf e f :
yy
L
w - y
X
V = - *YA +
n
t y - v/ -
*
W X 3ZL.
z.
ZL
-W * * -
tV X 3
,
z
413.
( 3 /
t>u
Beam loaded as shown in Fig. P-413.
25 kNm
10 kN/m
r - ± V
C
_
J
!
J
• I m 11 m
P-413.
Solution:
a Km*
A
10 Ko/ik
e
Pm* , 2»*i
* in
T Il«x
A*
IP
TTYv*— *•*
V-P
TITtttk.
(1*1)
^ jJ F "
M-P
(Kn-m)
-If
= 0
S K/
-
Z S + s o ( o . s ')
ft,
=
/o k N
tVX
110
£ F ij = O
Rz -
&
-
K>e « Jo
/O - V O k N
VBC = /o
= / OX
M sc = [ OX - 2 5
Vco = /O - I O ( X - 2 ~ ) - 3 0 - /OX
- - 5X* + 5 0 * - V 5
VPe = t o - t o ( X - 2 ) + 40 i ~ /OX + 7 0
M e? = /ox - 2 5 ~ ~ ~ ( * ~ Z ? 2 + 4 0 U - S ' )
=
—5"X2 + 7<5>X ~ 2.VS
f o r th e p o in t a f zero s h e a r ■
/O
—
=
414.
20
5TX
.
’
.
Cantilever beam carrying the loads shown in Fig. P-414.
2 kN/m
A
- — 2 m— •+«------3 m-----Flgiira P-414.
Solution:
R
V-P
Ill
£ F tf = O
*
=
**
+
= 14.5 K N
Via =
~2X
A 1A 0 =
- z x f f ) = -X *
Vac. -
-Z X - 1 *^ * 1 Z
» “
z
= —
X3
X*
- z
(f )-
zx + 1.33
420. A total distributed load o f 60 kN supported by a uniformly
distributed reaction as shown in Fig. P-420.
60 kN
n
n
-2 m -
-4 m-
-2 m - *
Flgun P-420.
Solution:
*<iti
I IT
(pO Krxfn
Im » • f ^f J'J' ^ it-L* *,C im
n ST“x
i; 7 r->T?r?r
L-P
£ /ry = 0
*PV = 4 (*>0') ; w =■ 3 0 k N / >
Vab = 3 0 *
« „
= 3 0 * (f) - / « *
112
Vbc -
6 0 - 3 0 C X - Z ) = 120 - 3 0 X
M bc = l O ( * - / ' ) - 3 o ( x - 2 ' ) ( ^ ~ j
- - / s x 2 +120X - 120
Vco =
f>0 - 120 + 3 0 (X - 6 J
M co-
L O (X -f ) - / 2 0 (X -? ? + 30
=
431.
/sx* - 2f o x
-t 1 6 0
Beam loaded as shown in Fig. P-431.
50 kN
40 k
I
2m
«
jflm
20 kN/m
10 kN/m
7m
J
*1
*2
Figure P-431.
Solution:
¥
113
= O
7R,
+ 4 0 (3 ) = 5 0 (5 )
R,
+
» 0 ( 2 ) + 1 0 0 (2 )
= 70
ZFy = O
Rz = 50 + go + 4 0 * too -7 0 = 200 k/J
432. A distributed load is supported by two distributed loads as
•.hown in Fig. P-432.
30 kN/m
3- fTTTTTT
2m
5m
w{ kN/m
Solution:
u>2 kN/m
114
0
7 -5 R ,
-
JS O
(3 .S )
R, - 70 k.N
.-.w ,
*F y
= M
= . z j . j j a-A/
-J
= 0
/fz = 150 -7 0 - 8 k M
"
435.
^
= I f
= 9 ° kN /*r>
Beam loaded and supported as shown in Fig. P-435.
20 kN 40 kN
4
1m
° m
1m ^
om
10 kN/m
1'
pi
w kN/m
Figure P—435.
Solution:
to Kn
4oKn
_y |
IO •fnlm
im
2m '
L
xm
!
*6
I t !|w *
o
-u -L
ti
f )“
i k
f
5
o
115
&
£ M «, = 0
2 y\z(s~) =
jy =
£ F ij
lb kH /»n
= O
/?, =40 ■+t o
ft, =
+
-5 2
bS k f i
439.
A beam supported on three reactions as shown in Fig.
P-439 consists o f two segments joined at a frictionlesshinge at which
the bending moment is zero.
Ans. Max. M —57.6 kN •m
40 kN
2m
Hinge
20 kN/m
1m j 1m
20 kN/m
iSJ
.......... ..... t m
~
ft,
ft,
Figure P-439.
Solution:
40Kn
2m ,
10 Xnlm
<
R,
grn
Hin6E
s
, iw ^ i"1
A &
1Rx
4«n
20 Kn/*r1
;
Ra
116
F o r 6C : < £
= O
S R 3 = Z 0 (3 )
\ R* - V-S k U
=0
R e = t O - t ? -- 3Z k N
For A B ■
=0
■9/?2 - f o ( l ) + 4 o f e )+ J Z ( * )
Rz r
70 k N
£Fy = 0
R , = 40 + 4 0 * 3 2 -7 0 - 42. AcA/
440.
A frame A B C D , with rigid comers at B and C, supports the
concentrated load as shown in Fig. P-440. (Draw shear and moment
diagrams for each o f the three parts o f the frame.)
I
D
1
L
2
Figure P-440.
Solution:
fo r
P
B
Cj
£
M
V-D
M -P
-P L
B
B C '■
117
z.
441.
A beam A B C D is supported by a hinge at A and a roller at
D. It is subjected to the loads shown in Fig. P-441, which act at the
ends o f the vertical members B E and CF. These vertical members are
rigidly attached to the beam at B and C. (Draw shear and moment
diagrams for the beam A B C D only.)
Arts.
Max. M -
5kN
Solution1
.
*5KN
A
2.rr\
B
3m
Zm
2m
- 22 kN •m
118
CQ Z M P c 0
7/?Ay
* ! * ( * )
4
s
(j
)
%Au " " ^
£ fy
s
=
s
( f ) ( 2 )
tj = & k N
* O
K p ' <°- M -- *
±F* = O
/?*<
445,
= 17 k N
Beam loaded as shown in Fig. P-445.
j4ns. Max. A f *
yvguim p-445.
-8 0 k N -m
11'
^ £ M Kz = o
(> /?, = /to U ) + — ( 3 )(g 0 ') d )
R, =
ISO k N
£Fb) - O
~ H O + n o - HD - 100 k N
For -the. p o i n t
—
r —
io
= ^ - x y , ^ x f f x ) ; x = /.22S~,
X =
1 .2 1 5 *
M
z
1 8 0 ( ¥ - . 2 Z S ) ~ U > 0 ( + . Z i S ) — ^(t.ZZS)3 ~ X
=
7t. 35 k t i
y
A -t
447.
z e r o $ h e # .r:
a
so
)
y = -^ x
J
( f t .
A )
Beam loaded as shown in Fig. P-447.
Am . Max. M — - 8 0 1 c N m
~
120
20 kN
Figure P-447.
Solution:
Q fM
R2 - O
5 R, + 2 0 ( t ) ( 3 ) ( 6 0 ) ( 3 )
R, = U
kN
£ Fy - o
Rz ~ ZD + SO +~~(5)(& 0) ~ ‘ft’ ~
For th e p o i n t o f xe.ro s h e a r :
~ ^ ( X ) ( 2 0 X ) - 4t> }
X = Z .W S m
At x =
M = ‘U , l 2 . l * S ) ~ ( 2 - 0 ) ( 2 . I ^ ) 3 T
1 ,5 .7 8 k N - m
r
121
449.
A beam carrying the triangular load shown in Fig. P-449 is
»upported on a uniformly distributed reaction.
Arts. Max. A/ — —45 kN •m
Solution:
122
450.
Beam loaded and supported as shown in Fig. P-450.
50 kN
50 kN
Solution:
50 KH
50 KM
V
im
Am
20 Kn/m
T
................I
v
im
I T|\r—
? w
- o
•f w + -
tv = 3b K N
SO + 9 0 + 5 0
k N/m
123
451.
Beam loaded as shown in Fig. P-451.
Ans. Max. M = - 4 5 kN m
Flgur* P-451.
Solution:
3*:
A
u.
3m
5m /T' 3m
U*
By C>yvYiirn&trj'/?, a /?z :
- 3CkN
124
4S3. A uniformly varying load is supported on two distributed
reactions as shown in Fig. P-4S3.
Figure P-4S3.
Solution:
= °
iR ,
/?, = 1 k N
k/, * -2- *
k N / **
125
*
« z = - t f r ) C / z ) - <7 = * 7
W 2 ^ Y =
1 3 .S k A l /bvx
F o r- -t h e * p o l r t - h c f z & n o s h & A r - C f^ - 8 ^ :
*= *x + ± X y
^
5" =
?X- f-£X (zx~)
X —
/^
2
*
M a = P . f U X O + 'k ( + ) ( * ■ ) ( $ ) = £.33
M b = o - s ( z X z ) + -L f o X z ) ( ± - t ty
=
4 ( t ) ( p. s) - ± ( , ) f e ) f l j
?
M e, = O . S ( Z ) ( 3 ) + ± ( 4 X 2 - ) \ ^ + 2 - Y + ( z ) 0 )
- i (* )(* X f )
=
!T. i 7
462. A tractor with axle loads of 4 kN and 8 k N has a wheel
base o f 3 m. Compute the maximum moment and maximum shearing
force when crossing a 6-m span.
Ans. Max. M - 12.5 kN -m ; Max. V • 10 kN
Solution:
4 km
0 KN
fel2.K N j
2m3
I
v/
- 0 '
/? = + + g = /2. k.N
R<4a = « T j ; ; <JA *
2
/iv~ A tn x . M orn e -n f~ <xh A •
M kn
,'r ( sun
A( & ------------
~ 2m —
a w— Jf*
R*
11Ri.
6/f, = /Z^Z>
t
/?, = ^ it/V
•• M a = f ( z ) - s k N - r * \
126
For- M * .x . /v1obYie.n-h
i
M j,
8 -
jP
1 8 KM
----- J ^ _ 0
B
'fKN
Zrfl
-I
0.5
1.6m
R.
'ftz.
£ /?, = /* f j
/?, = 7
____________
M 8 = 7 & 5 ) - V & ) = [/ 2 T jr Z jv -f»r
ft> r MaJC M b m e ir/ - yviHi on ly B o n "the. S f^ n
( B s h o u ld be- p o s it io n e d
Htc c.e n te r - o f -fhte. s^xin)
8 kn
c
" 1
/?, =
/?* = 4fcA/
M R =
¥ (3 )=
.
/Z k f i j - r *
M ix . V o c c u r s w / je « f i / t o i ' e r
4KN
r
8 KN
ii
=• o
C.R, - /z.
R, - 2-kAl
-•• Rz. - / z - z = /Okfi/
Arts.
464.
Thiee wheel loads roll as a unit across a 16-m span. The
loads are A — 10 kN ; B = 20 kN, 2 m to the right of A ; and C *■ 40
kN, 4 m to the right of B. Determine the maximum moment and
maximum shear in the simply supported span.
Solution:
R=7DKN
10
KM 2 0 KN ,
40WJ
2m I
4vvi
lg »W
A
=f
C
6
/? = /o + 2 0 + ■?& - 70 k N
RJa -
z o
J f,-
f
( 2. ) *■ 4 0 ( t ?
/**/- A7«X ■ Mo>nne^-h
A -■
= o
/fc /?, = 7 © ^ )
/?, = i£ .2 5 -
••■ A7„ = 2 C .Z S ( C ,) = l S 7 S k A l - m
For- M a x ■ f^lorri&trt
10KN
S :
20 KN . _
40 KN
!R
i_
A"
7m
R.
= *
/(. Rz = 7 0 (9 )
;
= 3<l.*ysk/4
■■■ R, - 70 - 31 ■37S - 30. f>2S kAl
M b = 50. t, 25(7? - /o(Z ? - m
Fc>r~ Afax- Mome-n~h tx f C -
. •* /s’ AcA/ -
128
£ M k> - o
!(. R z = 7 0 ( y )
= 3 0 .1 2 5 k N
'■ M c - 30. L 1 5 ( ? ) =
2 I 4 . 3 7 S k N -~ ,
A us.
A fo x . V o c c u r s n/hen C. is o v e r /?z :
io
20 , A
>r
’r
&
\
40
4'2tnu
I
Ra.
R.
* O
n>R, = 70 ( 2 )
R, -
8. 75
•*. R z - 7 0 - g. 75 = 6 !- 25 RrJ
AnS.
465.
A truck and trailer combination crossing a 12-m span has
axle loads o f 10 kN, 20 kN, and 30 kN separated respectively by
distances o f 3 m and 5 m. Compute the maximum moment and
maximum shear developed in the span.
Arts.
Max. M - 104 kN m; Max. V - 45 kN
Solution;
10 KN
2° KN r n . * o ^ OKN
L_
A Sm
A
&
r H
.
Sm
C
R = 10 + 20 + 30 = (,0 kfJ
RJA = 2 0 (3 ) + 3 0 (g )
For Max. M o m e n t a t A -
i.......I
*
A
R,
* __±_____
B
c*
3.5m_J
I
£ .M kz. = O
12 R , = C,C>(3.5) ;
R, - 17.5 k N
M A = I7.S (3 .5 ) - b l-2 .S k N - mn
F o r-
M a te .. M o t-v \ e -n t A rt
10 KN
8
'■
3 0 KM
2 0 KN
jR'feOKN
^ 4 /
e.
V
A
3" - t
>('
Tv
b—
^
12 R , MB -
6 0 (5 ) ;
R , = 25- k N
2 5 (5 ) - / o (s ') - <}s k N - ™
For M tK X . M o m e - n + *.t c , A / s o f f -S p a n , -th e r e fo r e
th e - r e s u lta n t -fbr 8 t C orily o n / A s sptKfj .
je /
20 KM
B
IR--50KM
ym
30KN
C
R = 50 k N
R J S * 20 C5)
t/g -
3 #■»
Mar.- M o m e n t <}J C- (& a n d C- PnUj on the. s p a n )
,
A
1
B
'------ 5>«
R,
1R*50Wl
|
'i
i'
|
---- 7)
t Rz
£ M n ,
= O
12 Rz = 5 0 (5 )
R z = 20. * 5 k N
•’• M l -
iO .«3 (5 ) =
/04 k N - ™
M m - V ( A,& And C an -the. span")
( M a x - V is o b t a in e d w h e n C is o t/e r f?z 7
I R»fo0 KN
1
\L-
V
A
£
B
T
\
\
1
r ,
1
>>
ft*
AnS.
130
£M kz = 0
IZ /?, = { ,0 (5 )
-
R ,= / S k * l
?£TA/V
/?z =
A7iVf. 1/ ( & * n d C onUi pri 1h& 9p*.n ~)
( May. V w lne*i
I
&
if o v e r R, 7
R .5 0 K M
■3m -
vB
t*
£M k, = o
12 Rz - SO (3?
R z = 1 2 . s k rJ
R , - SO - 1 2 5 - 3 7 - S k A /
5
Stresses in Beams
503. A cantilever beam, 60 mm wide by 200 inm high and 6 m
long, carries a load that varies uniformly from zero at the free end to
1000 N /m at the wall, (a ) Compute the magnitude and location o f the
maximum flexural stress, (b ) D eterm ine the type and magnitude o f the
stress in a fiber 40 mm from the top o f the beam at a section 3 m from
the free end.
Ans. (b) a = 1.13 M P a
Solution:
1
/ /
6>Om>
(».') T h e m a x i m u m S f
is a m a x i m u m -
IS
a f p o in t B
lAJhere th e
m om e-n t
- - j C t X 'o o o Y z J z (,0 0 0 N - m
Sf . kR°.iS.°.-0
T
tO IID -t
,5
b") f\+ "A - 3 ( o f . E~)
Me - - j ( 3)(*C>oXj-)z 750
f\f
‘fO m m
5 f --
-from top
My
I/VA
iso (o-oe.')
' "
Sf -
foX
e
/■ 115 M Pa
( 'fencile.■)
- bO mr-n
132
507. In a laboratory test of a beam loaded by end couples, the
fibers at layer A B in Fig. P-507 are found to increase 30 X 10-3 mm
while those at C D decrease 90 X 10~3 mm in the 200-m m -gage length.
Using E = 100 GPa, determine the flexural stress in the top and bottom
fibers.
■*-200 mm *-
_L
50 mm)
r
100 mm\
_L
75 mmj
Figure P-507.
Solution:
S ab - 30 X IO~yrnm c
S CP -
% I0~3 m n 1
<5AB _
ij
-
& CP
o.i - y
30 x / O' 3
C^ec.reeise.')
10 X !0~3
1 -
O.I
pKOfnn
& - PLAB
- SL
'
£
SB
S
-
3 0 X I 0 -6CH>0 * 109)
o .z
S top
O- 0 7 5
S +op --
_
-S'/is
O. O Z S
3 S nB
--
O . IS
3 ( I S M P a ')
= IS M P a .
133
S b o it o m
~
S b c ^ --
^
^ Ag
L (IS ) *
10 M r* .
508.
Determine the minimum width b of the beam shown in Fig.
P-508 if the flexural stress is not to exceed 10 MPa.
Ans. b — 75.0 mm
5000 N
200 mm
■
■
Fl^ur* P-508.
Solution:
5 0 00 H
: ..............
/ /
ZOO mm
ZOOO ti/xn
A
In '
Ki.
*»
bh*
I n*
b (o - z )3
000
T a IK
e>
TUrt;500
V- P
I* * -
(N )
-
2
-jfooo
(.000
=
12-
0
3 R i = 5000C O
+ 3 0 0 0 (2 )
R, = TOOO N
L F y -O
/?2 = 5000 ■+ SOOO - 7 0 0 0 - 6 0 0 0 Al
67 * /<3"* b
134
~To <j£rt&rm/n& the.
cim n
n
&
m
/>»<2
m o m e n t m i n e s &.+ t h e p o i n t s o f 2ie .ro s h e a r :
M f, -
- ' — (> )( 2 0 0 0 ) z -1000 N - m
M b z
- iO O O + - £ ( 5 0 0 0 -t 1000) ( 2) = 5 0 0 0 N - a t j
Max
Me.
5f
M o m e n t - 5000 N - m
I hJA
'
fOOO ( 0 . 1)
IO X l O 6
b
6 .6 7 X I O - * b
— 0 .0 7 5 m
— 75" m m
509.
A box section used in aircraft is constructed of tubes
connected by thin webs as shown in Fig. P-509. Each tube has a
cross-sectional area o f 130 mm2. If the average stress in the tubes is not
to exceed 70 M Pa. determine the total uniformly distributed load that
can be supported on a simple span 4 m long. N eglect the effects of the
webs.
0(V, mm
___
zUU
100 mm
© ------ <6>
100 mm
<§>------©
150 mm
<?>----- ©
© ------©
Figure P-509.
Solution:
100 mm
W kM
NA.
C y - Q r ------- 0 - 0
w
Xna
to
V-P
Ik
2rr\
B y S y m m e f-ri*
_
*
i M *
z
g ( n o X \D'b)(o .0 1 s ) 2
Ina -• s i 5 *
2. m
W
z
_
1-SOmiY)
/
4m
R,
ZDOmin __ lODrvtrn
0 - 0 --------- Q - Q
*v
135
I~he
rn o m & n f oec-o/rs a.j- /"/j-e pnin /
o f z & r o sh&Ar-.
., M aA M , J - ( 2 ) ( J £ ^ J £ kl\l - m
Mtz.
WA
t>
7 O X. I O 6 -
*v
=
~T X I 0 3 ( 0
07S)
X t o '*
10.92
k M
512. The circular bar 20 mm in diameter shown in Fig. P—512 is
bent into a semicircle with a mean radius of 600 mm. If P = 2000 N
and F — 1000 N , compute the maximum flexural stress developed in
•ection a -a . Neglect the deformation o f the bar.
Ans. a = 331 MPa
Solution-.
B(t.2l=
1.2 B
F fin 60° ( 0 . 6 ) + P i i n 3 0 ° ( O . 6 )
- I O O O i i n l> 0 °(0 .t,l * 2000 S i n 3 0°(0.6')
B : 133 N
136
=
H * -a .
(O * ) ' IO O O e * i b 0 ° ( 0 . £ ' ' )
= ZbO A/- m
c
Sf
-
Me,
-‘-NA
For «
a r c -H le ir- b a r .
T.
T fr*
J-NA ~
-
^jT ( ° O lO ) +
J v/) = 7. £ 5 x IO
5f *
= 33l M P a
7.0f x / ° " 9
514. The right-angled frame shown in Fig. P-514 carries a uni­
formly distributed loading equivalent to 200 N for each horizontal
projected meter o f the frame; that is, the total load is 1000 N. Compute
the maximum flexural stress at section a - a if the cross section is 50 mm
square.
Ans. a = 30.0 MPa
Figures P-514 and P-S1S.
Solution;
50 vn m
V
50 vmy\
\ 2-
(0 .0 5 ) 4
:
iz
I « A = 5 . 208 X io '7™4
Since ike- hota.I horizon fat /oaei is 200 x S= /OOO tJ,
R a = R e - SOO N.
P( - J ios
- 3 ( - £ - ) - Z. t m
- R * (* ■ + ') ' £■* ( z o o ) 0 .2 )
- SOO ( Z. f ) - 2 . * (20t$(l.z)
-
b Z 0- N - m
137
5f
=
Ifi/A
bZ V (0 .0 2 5 )
s .z o e , x 10-7
S f = 24. I S M P a .
520. A beam with a W200 x 27 section (see Appendix B) is used
cantilever beam 6 m long. Find the maximum uniformly distributed
i<>ad which can be applied over the entire length o f the beam, in
addition to the weight o f the beam, if the flexural stress is not to exceed
140 M N / m 2.
m i
Solution:
Theotien is a IV2 0 0 x 27 Seotion. From
Appendix. 3 , we
g e t th e -following cJat*
m a S S / l- - 26.6 kg /nrt
X x _ x = 25.8 x 10>
Depth ■= 0.207 yyi
ji
For the- c^tntileVer t>e.avyi, th e m a x i mu m m onJent
a t J>f- A , therefore,
M a * (b w )(i) = 16w N -rr1
Me.
In a
S f-
18 W (o . 1035)
net
IH-O X lO ^
2 5 .8 X I O - *
VM -
>939 N / r »
H> -
1439 -
'•‘ t/m eter t»r the- beam
-
1939 -
2 6 .6 ( 1 - 8 )
16 78
N /m
An i­
525.
In Problem 524, if the load on the overhang is 10 kN /m
and the overhang is x meters long, find the maximum values of P and x
that can be used simultaneously.
138
\
:
w N/m
3m
3m
i ‘ 2m
R2
ft,
Figure* P-524 and P-525.
Solution:
ID kN/m
3m
/“
Rz
/0X
\
(1
®k
V-D
maximum values o f P a n d X fhat £^2n
For
be- used si Mix l+amous/y / A t - Az
M A
kA/ - /W
3 R,
z
MB - ~ [('0* ) ( X? - 5X*
'*1
l5° m>?
r
bh3
S 's
- na
12.
I» A --
*3 7 S x ,0
0 .1 5 (0 .3 )3
300
m m
-A
Me.
Sf -
X ha
t X to 6 -
3R,
(O. I S ) X t o 3
3 . 3 7 5 X / O '*
R, = booo H = 6 KN
iince. A, - A 2.
3R , * S x 23 (6 ? .
■J¥
n
139
= o
6 R, - PCs) + 10 (
z
/1 °
t ( i ) - 3 P + S ( i . S 1 7 ) 2- = O
P-- IS krJ
526.
A rectangular beam, 120 mm wide by 400 mm deep, is
li >aded as shown in Fig. P-526. I f w = 3 kN/m , find P to cause a
maximum bending stress o f 10 MPa.
Ans. P = 32.5 kN
Solution:
lrr\
4m
3 kfi/n
!20mm
f-ir'S /
m e- t h e i/alue^ e>f
V - O
If/A :
(o. iz')(o. + ')J
IZ
■NA
b. 4 x \0~* m +
KN/ m
for*
vtncJ& r' P .
JTa7 fh & a b o v e she*ar- d i a g r a m , t h e yv%ax/rvi h rr\
W o m e n f o e c-u rs a / fhe^ p o i n t o f zBT'o sh& *r~ a n d
/£ &aiA<*/ ’A>:
Una*
c
5 f
*
( + ) ( 4 w ') z 8 vu k A! - rrt
Me.
—
±NA
- —
JO x I O * !A>
M
-
& M/ x i o 3 Co. z o )
t . f x to -*
¥ k N / nn
140
Therefore., sinee- H iegive-n
$hear cjiagranr) is as -follows:
-5”f t, R,
£Fy
■+ 1 5 ( 2 . 5 )
P ( 0
---j- + 7 . 5
=
w - 3 k N / m , the ct>rrec+-
kA/
O
Rz - P + 15 - C~^~•* 1, 5 ')
+ 7.S kM
Rz
The- m a x im u m m<yme-ni occurs a f the- point
o f zer~e> she-ar~.
tyiax. M
7.S
O
~ ¥.5]
_
Me.
S f -- —
NA
10*10* -
& ) (0 .Z ) K !Os
C . f
P -
A
IO ~ *
32. S k N
529. A 10-m beam simply supported at the ends carries a uni­
form ly distributed load o f 16 kN /m over its entire length. What is the
lightest W shape beam that will not exceed a flexural stress o f 120 MPa?
What is the actual stress in the beam selected?
Arts. W610 X 82; 113 M Pa
141
'.olution:
l i kN Irv\
S f < n o m Poe
10m
V
symmetry ;
I?( - R2
/k
R,
i(,, o o o (io ■) _ eo
z
4 0 ,0 0 0 N
Sm
SSSNVSSSs\
V-D
5m
-20,000 H
Them om ent occurs at the point o f ze-r&
shear- a t th& carrier:
M ax
M - - ~ ( S ') ( 8 0 , o o o ) - 2 0 0 ,0 0 0 A ! - m
M
w h e re Z is the- s e c tio n m o d u lu s
Sf
2.00, 000
120 x 10 *
7
I
-
1666. 67 X /O'6 m *
1 666. 67 x /0s m m 3
F ro rn +he- ta b le o f p r o p e r tie s o-f W ih a p es a n d s ta rtin g
a t the- b o t t o m r w e f i n d t h a t th e f i r s t b e a m w hoses e c t i o n m o d u lu s ('S / n t a b le " ) is g r e a t e r th a n /666.67
X /0 3 m m 3 is IV 2.SO X w e w ith Z - 18^0 x !0 3 m m 3. I n
t h e W 3/0 g ro u p , w e f i n d a W3lO X H 8 w ith Z -1750X ID1m m 3,
w hich is also s a tiifa c-to ry / as w ell as lig h te r. I n th e W 560
g r o u p w e f i n d a W 360 x 101 w ith Z - /6^0 X /03m m 3, which
is also s a tis fa c to ry , as w e ll as be/hg lig h te r th a n iYSIOx H8.
T h e fo llo w m a S u m m a riz e s th e l i s t o f choices ■
‘'
Z , IP3mm *
Desig nation
W 250 x /V<?
18+0
W3/0 X US
W 360 * 101
nso
1610
W tIO x 100
I ‘120
1*460 X 81
W 530 X 8Z
W6/0 X 82
1810
1770
1870
T h e b e * t c h o ic e is W6/P x 82 b e ca u s e the lig h te s t S u ita ­
ble beam in th e r e m a in in g g ro u p s have a m ass p er m e t e r
W hich is g r e a te r th a n &/• 1 k g / m , th e m a s s p e r m e t e r o f
th e W 6I0 x 82 b e a m .
142
The- W e i g h t o f the. be-^m
/c SI. f C 1). S ) l ^ / m = %D2.t>2. ^ / m
whicM <5 in s u fficie n t to
location o f the m a x im u m
yyir pv\ent re s u ltiru j -from the. c o m b in e d liv e a n d d e a d loads.
Tlenc-C, th e d ea d lo a d m o m e n t M p r tf ,X - 5 m is co m p u te d
as -falloivs ■
Kj= 4013.1 N
H-013 .1H
= ~ ( ‘tP IZ .O C s ) - I0,03Z.7S A/-*n
Z
IDP3Z.7S
U p x 10 *
Sf
2 p -
- 85 hi X 10"** yy\Z
8 3 . b ! x IO ^ yy> m *
1~C> che-c-k :
Z G
> Z L -> Z p
1 0 70 n /0s i
ib b h . <*7 X /O3 + 85. *>/ x 10 s
T h e r e jo r e ^
I V (,/D x 8 2 is S a t is fa c t o r y .
WC,/P x 82.
f ln s
T h e a c iiA a i t e a m s tre s s is d e t e r m in e d fr o m the r a tio
o f the beann m o d u lu s , to th e design m o d u lu s . Therefore.,
S
=
M
7. + Z„
IbbC,. 67 X 10s + S3. t>! x to *
120 x !Ob
Sa =
/870 X IO 3
H Z .5 2 M P a
A ns.
O b serva tion o f choices in d ica te- t h u f
W 53D x 8 Z.
IS also g o o d , w h ic h is o n ly s lig h tly h e a v ie r a t 82. V k j
531. A concentrated load of 90 kN is applied at the center of a
simply supported beam 8 m long. Select the lightest suitable W shape
section using an allowable stress o f 120 M N / m 2.
Ans. W530 x 74
Ill
Solution:
‘fokH
Avy\
S f < n o wPa.
4hn
%
(?2 = 4 5 k tf
+5, OOOM
4m
v-p
■fm
-45,000
Max
M = *6-000(4') = HD,OOON-rvs
M
_
/fg,
ISO, 000 .
6
1500 x / 0 ' 6 am ■*
120 x !Dk
1500 X /0 s yyim 3
>
Z, IOSfnrr,3
itic n
Des.
WZ5D x 13/
W3/0 x 107
W3t>0 x •?/
WV/0 x ?5
Ibto
1510
1510
15/0
tbtO
X 82
1550
x 7?
mo
^6/0 x ^2
The- b est c-hoice- is W530 x 7t/ which is the- tightest.
*The iveicjhf o f tftr beam is 1¥. 7 kQ /tv.i x f. 8 - 732. 0(>N/m
which is jnmffic-ie-n-f■ io c-hanye- the- location o f maxi rnum m om & nt. /fence-, the- dead load rytotnevit M p
a t X - 1 w\ is
W 530
7 HI. O b N / m
&m
292 S. 24 N
Zp
=
¥ 8 .8 0 X / 0 3tr,ty,JI
n D = -j(i< tZ t.2 ¥ ')C ¥ ')
M p = 5856. t S M - m
144
cA e-c-k. ■
To
Z 6 > Z L + Z„
X 10s >
ISSO
choose-
1500
X J O 3 ■+ U . 80 X 103
Ans.
W 530 X Tf
533. A beam simply supported on a 12-m span carries a uni­
formly distributed load o f 30 kN /m over the middle 6 m. Using an
allowable stress o f 140 MPa, determine the lightest suitable W shape
beam. What is the actual maximum stress in the selected beam?
Arts. W610 X 125; 133 MPa
Solution;
6m
<,0, 000n
3o M / m
tr
l«2.
?0,OOO
Max M = 9 0 ,0 0 0 (3 )
t — (lP , 0 0 0 )(3 )
■10,000
*
M
jo sooo
Sf
M O X 106
m
- ifps, POD N - m
2812 . 8k x 10-
Z L > 2812. U x 10* mrrj3
Dtsiq n a tion
'Z , /t>s m rvi*
y\(3/0 x 102
3050
W 3tO x
3120
3080
3 ! tO
1120
3500
/7?
y v th O x i z i
W53D x 138
W6/0 x U S
W 6 10 x [25
T h e bes t c-ho)ce- is
W blO
X 125.
change* the location o f m a x i m u m nnom&mt.
g
(l,22S.‘tS)CH)
*1 ~ *2 " ------ J--------
1215.18 * / m
12m
**
R , - R t = 7355. 88 N
rsss-tSN
V- D
6
6 *r\
- 735*. 98 hi
141
M p
* —
(73SS. se) ( ( , ) = 220t>7'
ZZ067. 6H-
Mp
-
M O X to *
•Sf
< / -m
157. i>3 x IP
Z„ = >57. 65 x 103 n ,n ,s
T o ch e c-k ZB
>
3220
Z L + Zp
X I D3
>
Z 8 1 2 -8i> X I 0 3 t I S 7 ■ t>3 X I Os
WlblO X
fjoftz ■ Wb^P X
125
h ea vier-
b tj
A n s .
1ZG
can also be suitable0.1
kty />^>
-than
w h ich IS only
Wi>IO
x
IZG .
+ Zi
Sf
ZS IZ.Sh x
t o 3 + 157-65 x
to3
3 2 20
I V O x I Ot
Sa
=
Ans.
132 . (.3 M f a .
536. A simply supported beam 10 m long carries a uniformly
distributed load o f 20 kN /m over its entire length and a concentrated
load o f 40 kN at midspan. If the allowable stress is 120 M Pa, determine
the lightest W shape beam which can be used.
Ans. W610 x 125
Solution:
40KH
Str\
S vy\
/'
s
10 m
R , = /?*
R.
Wert,
120, 000 H
Sm
m
=-
ZO, 00C'] (5 )
ODON
V-P
20,000O o ) + 40,000
R , = R2
20 k.H/m
Sm
- ID, oooti
-120,000 N
3 5 0 ,0 0 0
Z > ^ L
%
ZL
-
2.91C,. !o7 X / o ' 6 m 3
UO, X 10*
>
29M>.(>7 X IOSMY,nn3
146
De-uqn a fion
Z , 10 s b u m 3
W3/0 x 202
W3i,0 x /7?
3120
3080
305-0
P V V tt) X
/V4
W530 x
/5<?
3 ! HO
W 6/0 x
/.Z5
3220
X
125
3500
T h e btzsl choice- is
W t l p X 125. T h e -w e ig h i is
I2.S.I X *1■£ - 12-2-5. *fK N/ tm w h ic h m/it/ n o / 'the lo c a tio n o f m a x i m u m m om e-n t-
*1 ‘ *z /T
/Ow
*
Kl
M p =—
z
/ 2 2 5 . <78 {7<?>
2.
b U I . I A/
( e , u ^ . ‘f ) ( s )
= ]5321-. 75 N - m
=
2. p
-
Zi
* 2P
/27. 71 X IO ~u m 3
IZ7.7lx lO3 m m
3
To c-he-ck ■
Zs
>
3 1 2 0 X ID3 > 2 < m > .(,J x /0J +
'• c h o o s e -
WHO
X
/ZS
/27-7/ X ID 3
A rts .
538.
Floor joists 50 mm wide by 200 mm high, simply supported
on a 4-m span, carry a floor loaded at 5 kN /m 2. Compute the centerline spacing between joists to develop a bending stress o f 8 MPa. What
safe floor load could be carried on a center-line spacing of 0.40 m?
Arts. 0.267 m; 3.34 kN /m 2
Solution:
w - f>4 N/m
4 vn
Ri
R, * R2 --
147
M:-jr(2f>A)(l)
2 f a ll
M - zpa rJ-m
2m
M - 1 (Sooo)a. U- m
Zm
M r lo, ooo a. f i- m
-2 fa hi
bhi
12.
"y
I ma
0.05 (0-Z~)3
= 553 x I O - s m *
12.
_
a. -
ID, 000a. (O .t)
5.35 x to- s
0. 26.7 m
If
al = O. V m
p'a.' = pa~
p'(D.4)= sdod (o.zt.v')
p ' = 5553.55 N/rrt 2-
Ans.
539.
Timbers 300 X 300 mm, spaced 0.90 m apart on centers, are
driven into the ground and act as cantilever beams to back up the sheet
piling of a coffer dam. What is the maximum safe height of water
behind the dam if the density o f water is 1000 kg/m3 and the bending
Arts, h - 2.90 m
•tress is limited to 8 M N /m2?
Solution:
W.= 1000 Kg/m3
S f < 8 MPa.
0,9 m
Whole prt-iture cf tva.ter on
wa.H -
>vA A
300 X 3oo m»n
= O' 45 w h 2 M t'm g <*t A-
height h/j above Hie.
ground-
M . /><£)
M = 0 -?5 w h * (-y )
M -
0. IS w h 3 -
148
£ x to*’ -
0. 15(1000 x q. g) h s (0. IS)
b.75 x 10-*
A
" 2.
Z .‘lO
W t mm
Ams.
=
540.
Timbers 200 mm wide by 300 mm deep and 5 m long,
sirroly supported at top and bottom, back up a dam restraining watei
3 m deep. The density o f water is 1000 kg/m 3, (a ) Determine the center
line spacing o f the timbers if the flexural stress is 8 MPa. (b ) W ill this
spacing be safe if the maximum flexural stress is limited to 12 M Pa and
the water reaches its maximum depth o f 5 m?
Ans. (a) 0.939 m
Solution:
SH-o] ( a}
= W
200 fatn
ry = o
R, + Rz ~ W = O
R, = *>/s
;£ > Z M Rl = 0
w(*') - Rz (*) = 0
R z -~ w
B y ra tio a n d p ro p o rtio n
Of in t e n s it y ,
./
2 V = J*. - -i-
-j/
•
* o
u = o. z q g jv
c/
8
e/ = (/TT
= /••?■? 1
71
fn
M
=
VJ- f t h A
0.571 W
= (tODO X ? . s ) ( t - ) ( J x
AJ
=
0 . 5 7 ^ ( ‘f4 l O O a . ') -
s ~
. —
My
sf
I W
WA
A ,
a .) = H>DO eu
2 5 5 3 3 . ^ a-
V . / . I C/ ’ O / ^
± ± f ± I > l s f.5x / O 'W
/VA
n
2S533.^a- (0 .1 5 )
w1
A
+ .5 * /0*f
=
/4ns.
i7. 9 4 m
I
<=/
Rz =
zy = o
Rt + R i * W
'/ *
A t z e -ro s h e a r ■■
J f - - L C a l ) ( i ') = 0
JV_
zs
z
3
J z =
J
=
25
3
2.81
L/ = 0. 25/ W
150
Max M = ~ ( 2 . S f ) ~ y ( 2. 8‘l')(£ > .2 3 / lv )(-^ P -)
M ux M = 0.0,12- W
tV = uvhA = C/ooo x
tV= I22SD0 a.
W = 122500
id = ns, is o aJ
M a x M - D . t 1 Z ( n s , i S o ) - 73110,.3 N - m
*
- - ?
■
T h e re . f>re- , ~fhe. spacking /S tioj- safe-.
a n d m u s t be- re.c/ u ced to
Arts■
) = ° -+ * * -r r \
543. A portion of the floor plan o f a certain building is shown in
Fig. P-543. The total loading (including live and dead loads) in each
bay is as shown. Select the lightest suitable W shape beams if the
allowable flexural stress is 120 MPa. Assume the beams are adequately
braced.
Wall
2.5 m-
-2.5 m-
-2.5 m-
-2.5 m-
-2.5 m-
-2.5 m
15 kN/m2
15 kN/m2
9 kN/m2
9 kN/m2
15 kN/m2
15 kN/m2
I
(B-3)
7m
(B -l)
(B-2)
(G-l)
Stair well
(B-2)
r
2m
____ L
Wall
Figure P-543.
(B-3)
151
Solution:
Wz s J ^ L ( 2.G)
yrtz
w= 22.S' M / m
R.
R i=
Rz
,
=
l l ‘ SO°
S i,Z S O
AI
5b 250 N
Z.Svn
V- V
2 5m
- 56250•
M
=
= - - C 2 5 0 ( - ' S^
2.
70312. . 5
, I S ’. 2. ! 2- - 5
7
S f
-n n
= S - g G . ^ q - x. i o - 6 m 3
/ Z O X IO ‘
x /O3 m ry .3
Z. = S'S S'.
WWO
u se.
N
Z
(&-D
X v?f W ith
-
A ns.
to * m m 3
x
S62-SOfil
M z 2 8 / 2 5 (2 .S ')
/js 2.5m
'.,5m
M = 7A5/2 . S A l - m
2.5m
Z.5m
IMShl
2212SM
Z -
21IZSH
Sf
7<93/-2 ■S'
2.5m
V -D
z =• /2<9 x /O -
15m
-zins n
Z - S 8 5 . ? f X I O - &ry,3
Z = 5”4?5*.
U se
W*i-IO X 3 9
w ith
Z
-
x
X /O3 rrtrr} J
to 3 m m *
2*125 N
c& -o
-
w *
K"
IM KN
w ,^ r
/v
<Lrn
A
S tti
R.
Rz
W, =
*
z
jv2 = /s'
x -2 * tri-
= /<?. 7S' M ljm
V *
x « = .
152
Z M k, = O
. !S,150(z')CO -I- 2.S/2 S ( 2 ) +3 0 0 0 0 f s ) ( 4 , s ) z/= Q
*z - 1°* *2 t-t/i A/
R ,+
Rz -
O )-0
I S ,7 5 0 ( 2 ) - 2 8 1 2 5 - 3 0 0 0 0 ( 5 ) = O
K, = I0SSO 3.S? N
J:
105803. 57 N
109821 ■ ^ 3 a/
30 0 0 0 -N/m
6 S 3 0 S .S 7 *
d -
3. 66 rn
- I098Z1.43
M c lx
M = 2M *
A7a_x M =
= — ( 3 . £ t ) ( I 0 9 S 2 . / . V3)
Z O l , OI Z . 4 4 N - m
ZOIOI2.H-4
S
=
120 X /06
Z =
X lO3 m m 3
Z =
| £/se
tyg/O X
wi+h
/675 x t O - L m *
Z = 1%7Q x I Q ^ m r r i 3
(B -3 )
is kN
mz
iV - ----- — X 2 .5 m
w
A
7<«
K;
Ans.
w = 3 7 .5
KM /m
M a x M = — (131, Z S O )(3 .e ) - Z2<i(>87.S A1 - m
2.
z = s
« -iililZif
120 X /0£
= iw .o tm o -* ",*
Z =■ l ‘J I 4 - . 0 & x l O 3 m m 3
Use-
W S>IO X 92
* v if h
Z = ^ /Vi? x / O ^ m n j3
A ns.
153
M9. Determine the maximum tensile and compressive bending
i. -.c* developed in the beam shown in Fig. P-549. The cross section
i* Uic given properties.
Arts, a, = 20.0 MPa; ac = 10.0 M Pa
C
~~j 60 mm
200 mm
_ A
Rj
^na= 40 X 106 mm4
Flgur* P-549.
'.olution:
5kN
—*—
to mm
T
200mm
_ L
I f,/. - fO X IOkntm 4
^ £ M Ki = O
SR,
=S(z)
+5
R., = 3 k M
£Fjr =■ o
R2 *S--3 -- 2kfiJ
Ai f>+- A ■
M a
= -Z k .ti-m
■■ The- b ea m as ben t c.on c.a ve cjovunujar'd.
T
Xfc = o. 06 m
C
/c - O. 2 m
154
Z x l o 3 X o . o t , ')
fo x fO-e
•Z X /O3 ('o. 2}
M*c.
T-n /\
■f-O x
=
/£>-
/o M Pa.
<" /A/VS.}
^>/. s :
= jk N -r r ,
The. b e a m is i e n t co n Q a v e . upn/ard '.
< ± )
^
/t - * . 2
T
bf i =
5 fc =
I X /o3 ( o z )
J.O x / o -«
=
20 MPc\
CAns.)
f x /o* fo. Ofcj
fo
X
-
/ o -«
6
M Pa.
555.
A cast-iron beam carries the loads shown in Fig. P-555. If
a, < 20 M N / m 2 and ac < 80 M N / m 2, compute the permissible limit*
o f th* overhang.
Ans_ * = 2.0 to 2.5 m
4kN
4kfJ
bm
Ifc k N
'<■ K
/*'
R|
X '
------------------ 7 c....... —
1
fh
F'9u r» P -S S S .
Solution:
ZOmm
_
-
/
/
/
Ld
NA
ZOO mm
* $0 x 10 6mm 4
155
A t jx > in i A •
- - 4*
kfij - t~r~t
...
y«
>fi
-
0.08
yc
.- o . £ .
Myt
--
■T A<4
IX
20 X to 6
X / o * (Q
O S ')
<fO K t o - e
Sfc
I ha
80 x
/O* = . ^ X C/OSl ( ° - ^ l
4o x / O -*
} * *
i'Orr,
A t pt. B :
A7S = -4 X + - C S X * ) =' - + *■ ■+
•■•
-m
y& = o/o8
_ Wife
■HA
2.0 X I O 6
(~4x+ iz . ') ( i o %')(o . z . ')
= ---------- — -------- - — r ----------■fO X / o -
;
X
-
2 .0
C~4x + J2 ) ( / o a) ( o . o 8 )
j X - -7 m
4 0 x / o -c
• 2 < X £ 2-S556.
A T beam supports the three concentrated loads shown in
Fig. P-556. Show that the N A is 70 mm from the top and that
15.52 X 106 mm4. Then use these values to determine the maxi­
mum value of P so that a. < 30 M Pa and ac < 70 MPa.
Ans. P = 1.41 kN
80 mm
4P
------------ 1
v m
L
20
1m |
f
ft,
R2
Figure P-556.
160 mm
I
I I
156
fP
Solution: S
' ^______ 2<n
z»7
7\--- ------------------/m
1Im
Rz
\/-D
-P
-zP L
%0 mrv\
•sr
_ go(20)(no) + 2o(/6o')CSo')
~r~
^— 3
So C*o) + 2o(t<zo-)
70mm
=.
//O mr
Cfhorr7 /Ae &orthsm
or* 70 m m fhorr-j +htc_-
IbOmm
HO m m
20 mm
- I t + A ,d ,z + I z + A2d2.
I ha
( j z- )t +
80 ( a e » J , * - +
__ &_(«,> * + so(30) a 0y
/.Z
T
A-f poirt+
-
v- 160 ( 20) J -
< * ° 0 * o £ + 2o(/f.o)(so)3
A2-
K
15. S 2 X A3 6 rrtr-n 4
fy :
M A =
~P
f j - An
yt
= 0 .0 7 rn
yc
-
0.11 n -,
157
M yt
%
=
30
x to
P X to 3 Co. 0 7 )
/S.52. y. /O'®
p
S f,
=
G .G 57
^/V
=
70 x
P X /O3 Co. n~)
/o« =
=
P
X / 0 '&
/S. 52.
?. 87C. k r j
A i f>o in j- 3 ■'
A7B =
-P
+
4P
=
.3/=>
yt -- 0.07 m
yt
O./i m
=.
T
M yt
3 ft
X /O 3 C o . i O
/S'. 5-JL X / O '-*
Z7 = A 4-I I
k Aj
M yc
-Sic
«
7o x / o ‘
=
■5^X
,5\ / 7 3
P-
558.
maximum.
/ 0 3 (0 .0 7 ~ )
/S. 5 2
X /O - e
4 a/
I . 4 H k/J
In Problem 557, find the values o f x and w so that w is a
Ans. x ** 1.59 m; w = 3.16 kN /m
■10 m
I
w N/m
’T
Ri
T
n a J
T
180 'mm
____ l _
/NA ' 36 X 106 mm1
Figure* P-5S7 and P-SSS.
158
T
10m
Solution;
w
180 mm
x
NR
---- *d__
50 mm
V 7 V Z 7 \ -jk-
I/ka ' 36 x iO*mm4
Due h> S y m m e try :
K, -- «t --
-- £M
A i p o in -f A '■
M a = -~ C w x )C x ) * -
WX ;
2
r
yt = O. IS fr,
y t = O. OS" m
s ft =
M yt
M x/0‘ =
“ /x„* ( 0./8;>
z ________
3S
'S/a =
X
*VX2* ? o o o - 0
/ O -C
M Sc
2 C o - O S ')
JO X /O'
2..(.30, X / © " « >
<4^ pom-f- 3 ■
- w
1
V - X . ) 2- * /
-------+ (_--------Ms
)
M/X2 - //£-, 200
_ w (iz ^.s -s x )N
Eve-n i f we. do ncrf knouJ uJ a n d X , ~fi~om Hie~
load d ia g r a m , w / e cart c.on<z/ude- +haj~ a t o/. B ,
f A e , t>e.am is besrf- canc-ai/e- u p w a rd Yc
T
_
Sft
M ift
=
0 .1 8 m
y^=
O. O S 'm
159
/„e
W (It * - S X ) ( o . o S )
20 *
-------
w(l2.5
S*.
- S'*)
=
0
—
14,400
M yc
■NA
so x
M/O z .s - s x .)
jc, * to -
10 * .
w(/2.5 - S X )
=
/£,COO
For m a x i m u m
<S>
x
a n d m/, equate- Q> at,<J ( x )
PVC ! 2 - S
8000
w -
WX s - S0OO
5X?
=
X-z
14-, 4-00
14-4-o q
/Z.S - s x
gooo
_
X*
/4 >4oo
J 2 .S -& X
/4 , 4 o o X 2 + 4 0 . 0 0 0 x - 1 0 0 , 0 0 0 z 0
X
z I.& 3 nn
•'• v S n b s f i i u f / n ^
*
Jn
(j)
^Jo °s °T *
~~
N/ tn
559.
A beam is composed o f 6 planks, each 100 mm wife ancj 20
mm thick, piled loosely on each other to an overall dimension o f 100
mm wide by 120 mm high, (a ) Compare the strength of such a beam
with that o f a solid beam of equal overall dimensions, (b) ty},at woujcj
be the ratio if the built-up beam consisted of 12 planks each 100 mm
wide by 10 mm thick?
Ans. (a ) 1 to 6; (b) 1 to 12
Solution:
~ z b ® ,o° mrr?
120mm
X 2 onint
110mm
loo
100mm
160
M
•S
}
=
F o r Sctr-n^.
M\ e*ncj
S
;
dorr}pare- Z. 's
r
NAi
-
_o_// o . o 2y 3 l
12.
L
5
Z « -
J
O. o/
-
4 x / o -5 m'-
o . / ( p . IZ ~ )S
1-NA
12.
o . oc
~
/
&
■? X /<=>
o . OOOZ.-4-
©
a>;
>(J) too nim x lomm
IZOmrrt
<00mm
=■ O . t3000Z_
Zc -
io
X
> JVX2* ffOOO -CD
=
3C
ZB
X ZO -6
O. 00002O . 0002.f
/2-
A n j.
560. The wide-flange beam shown in Fig. P-560 is strengthened
by riveting two cover plates 160 mni by 20 mm to the top and bottom
flanges. I f the maximum flexure stress is 110 MPa, compute the total
force (a ) in each cover plate and (b ) in each flange. Neglect the
weakening effect o f the rivet holes.
Arts, (a ) 336 k N ; (b ) 304 kN
J-«—360mm-*j
1n
Figure P-560.
H i— Xmm
-‘Fh1 20 n
]—j 20n
Tm m
Ito
A/A
s f, ------ 2_
S
l
At th e neutral axis
S f s O and at the. tr>p , S f is
triAX/munn • Sinc.e. M a n d J arc constant th rou gh ou t
the cross - section , the. fle-xurml stress varies linearly
with the ejistanc-e 14 'from fhi*._ n eu tra l a x is atven
Ho MPa_ „
Ay
1 O .tZ tn
Sf
y
'* SO° V
(a
A i ■the b o tto m o f th e . c o v e r -p la t e . o r to p o f th e .
•flange. , y * O.ZO , a n d
S p - SOO ( o . 2 j - / o o M P a T h e - -force, in t h e co ve Jr' p la te , is
< 6 .* x
Pc
(b )
Pc. = A S^-Ve ~
x «.«]
3 3 6, 00O A/
Ans.
At the bottom o f th e -flange. , i^ - O.lS m/ a n d
S f - 500 ( 0 . 19 ) = ‘t o M P a .
T h e -fo rc e a t t h e f l a n g e , is Pp . r IO O + 9 o
( o . / L X o . o z ) \ ------------ X IO* J
PiT
3 0 4 , OOP N |
Ans.
561. A T section has the dimensions given in Fig. P-561. Show
that the neutral axis is 60 nun below the top and that /NA * 26.67 X l ( f
mm4. If the tensile stress at the bottom of the flange is 10 M N / m \
determine (a) the total tensile force in the flange and (b ) the total
compressive force in the cross section. Also determine (c) the moment
of the total compressive force and (d) the moment of the total tensile
force about the N A . (e) How does the sum of (c) and (d ) compare with
the total applied bending moment as computed from the flexure for­
mula?
Ans.
(a) 96.0 k N ; (b) 98.0 kN; (c) 9.15 k N • m; (d) 4.19 kN • m
162
Flgur* P-561.
Solution:
IT160 mm
/4£)wm
- A/ . A .
i
1
Z0»irrt
T ^ ik ih ln e . ba.se. <zts ~th& r e
_ _ 2,4*/
y
_ O- cz ( o . !G>) (o . 0 8 ) + O. I2.(o.Q 4-)(o. /*)
O.OZCOslG') + 0.12. (0.04-')
0 .14 m
=
fro m
-hhe- b a s e , or~
|Q. O Q> rn fr o m
- NA
Ig\Z<s.1j
s
-the -hop
Ans.
+ A^ )
o, o z ( o - /fe)3
■+ o. o z (o.i&~) (o . o&')2-
12-
o . 12. (a . o 4 ) 3 ,
„ .
*■ o . 12. (o
( o . o4-~)Co. 0 4 ) 2.
12.
--------- ---------- —
26 . (,7 X IO - f m *
-AM
Since-. S f = o a-t the n e u tra l ax/S; M/e hav^ th is
fle x u r a l s t r e s s d ia g r a m b a s ed o n
_
•Sr =
/o M P a .
I
y
IM
Soo ,
W h ere- y
M fV
is th e d is ta n c e . -prom t h e n e u t r a l an is ■
T h e r e f o r e , n t the- fop a f t h e f la n g e . ,
S f = 30 W Pa-
lj
■6 0
( a ) T he- -frofa.1 t e n s ile
Pf
fo rc -e in t h e -fla n g e , is
„ r s o •+ /o
A S al/tL = C0. /2 x o . o 4 ) ^ ----- ------ x
and
=
,")
9C>,ooo AJ
Pf =
(b ~ ) A t th e n e -u tra l a x i s ,
S f - O a n d a t th e b o t t o m ,
-the c o m ^ r e s s / V e - f o r c e is m a x i m un-i w ith
ij - O . l j m
S f - S o o (O ■1 4 ) - 70 M P a . . T h e to ta l c o m p r e s s 1de­
and
f o r c e m th& c r o s s - s e c t i o n /-S'
{0
Pc - -
Ig o o o N
(c~)Thm om ent- af
/S
M c
-
P„d
Me
-
-
a vera ge
r
"\ r>
,1
-----X
Ans.
th e- f o f a l
c o m p r e s s i v e . -far-c&
9 8 , 0 0 0 ( - j - X o ./ 4 ~ )
9/ 4 (> . 6 7
T h e
2.
+ 70
A S ave = o . o z ( o . i 4 - ) [ —
Po --
M - m
c o m p r e s ^ /v e -
-f o r c e , a c t s
-
- j 'C o . 14) f r o m -f-be: n & u tra l ax/s, b&oau>s<^
1~he c o m p r e s s iy ^ . f o r c e d 'iajrX am /<s
f r i a ncj I a r~/ z & r o c\J- fbe. r->&u -hr&l czX/s
&ind rn a x i m u m ctt the. b o t fn m .
a f a. p o m T
ma
^
10 m £
(d ~ )
—)
0.0+M
HI
o.oZm
flt
end
L—
ha.
fh e- j u n c t i o n o f th e io e b a n d f l a n g e , y z a . o z m
S p r to M P a
R t th e n e u t r a l a x is , j/ « O
s i/e forc^. be/ouj th e j u n c d /o n
Pt /
=
A S aV4t =
=
and
S f = O . The ten -
o f th e u'f’ A a n d fla n g e -
2
C o . o z ) < ' o . o . y '[
Z D O O A/
X
/ o «J
164
m c im j a t
0 - 02- / z m
fro m
th e . n e u t r a l a x is .
■~~]—ijzagy 7~h+ ’tests'lle -forc.e on the flange- sh o u ld 6*
|
Tfy r e s o lv e d in o r d e r to fa c ilita te C*>mputn. tio n
0.0*
* I /
th e m o m e n t - C o n jm a lly t r a p e z o id a l)
tom>»1 /
Pt l = A S
4 8 0 0 0 a/ acting a t o.e>4- m from the.
n eutral axis.
=
t-n
( 0 . 0 4 - ') Co. u ) ( / o x / o c ")
-
Pt% = ^ sA„a
Pi3 = ( o . o * ) f o . i 2) [ -2t>± °
X 106J
o. o4
- 4t,000 N acting at — ~ —
tort
from the. •Ayo.
~The f o f a ! m o m e n t o f th e -tensile, forces is
Mt
= JEP* d -
z o o o (^ f ^ +
000 ( 0 . 0 4 ') + 48000 Co. Of, M t *
4173.33
A/ - m
Ans.
(* * )
Mt
M
f
Mc
-
4173.33 + 9 1 46 - 67
=
1 3 3 2 0 N - rrt
SJ
_ 3 0 X /Q *CZC . & 7 X / Q - t )
vj
O .o c
70 x /o_« _C2_& . & 7 x /o_cp
A/ =
13333. 33
T h e a n s w e r s a re .
Aj - m
a l m o s t e<^ 4a !
Ans.
563. A box beam is made from 50-mm by 150-mm piece*
screwed together as shown in Fig. P-563. If the maximum flexure stress
is 8 MPa, determine the force acting on the shaded piece and the
moment of this force about the N A . (Hint: Use the results of Problem
562.)
Ans. F *■ 45.0 kN; M ■» 3.50 kN • m
n gura* P -M 3 and P-SM.
Ul
Solution:
P
-
-
F -
-* ■*
4 50 0 0 N
.
I
s I
/0 * (o.ts x o.os)(o.o7s>
Ans
0 .IS (0 .0 S )3
+
' ----------- /2
- ' ' "" +
o. /£T^5-o g ) ( ° 7 S ) * = 4.jr7^ X /o~e m +
« X » < V f.S
f WJl x / e - 0
= --------- ------------------—
O. /
_
A7 - — I
3
A? =
3 ,SOO k] - rrt j
Ans.
561. A timber beam 90 mm wide by 160 mm high is subjected to
• vertical shear V - 20 kN. Determine the shearing stress developed at
layers 20 mm apart from top to bottom of the section.
Solution:
mm
If
40
777T. to T
is
N.M.—
HO mm
1
kh3
_
0 -0 9 ( o . /ft3J
/Z
"
U
-H A
S
=
3 . o j z y io~*m*
^
Xb
lb
_> L ,
X/>
J L
J 07£ * I 0 ' s ( 0 . 0 9 )
m*
s , - 7z . 3f x t o ’ c a ')
< *- A y
0
O
20 mm
(o.oz)Co.e> 9)C o .o 7)
4o mm
Co. 04-)C0.09) CO. 0*1
/.St A/Pa­
&Omm
(o -o t )C o .a 9 )C o .o s )
r is
80 mm
( 0. o t ) ( o . 09) Co.04)
2 Of MP a .
lOOmm
Co- i) C o . o g )C 0. 0s )
/■IS MPk.
/20mm
Co-/z)Co. o j ) ( o . 02)
/ - « A1Fk_
Momm
Co. /4~)C0. o 9)C o . o r )
0.9// M P a .
/60 mm
0
O
0.91/ M P a .
M Pa .
166
571. The T section shown in Fig. P -5 7 1 is the cross section of a
beam formed by joining two rectangular pieces o f wood together. I !i*
beam is subjected to a maximum shearing force o f 60 kN. Show that (It*
N A is 34 mm from the top and that /NA = 10.57 X 106 mm4. Using
these values, determine the shearing stress (a) at the neutral axis ami
(b ) at the junction between the two pieces o f wood.
(a) 3.28 M Pa; (b ) 3.18 MPa, 31.8 Ml*.
Arts,
-200 mm-
I-
-1 u
20 mm
Figure P—571.
Solution:
200m m
Z4mnt
'— NA
10b
IDOmm
mm
[A
10mm
_
y,
y
A, + A2.
lj -
-Ena ~
+ A2 y i
(o . 2 ') ( o . o 4-)(o . Q 2l ) + Q- / ( o . o z ) C o . 09)
0 . 2 (0 . 0 4 - ) + ( o . i ) C o . o z . )
0.034 l'vl ~ 3 4 m r*-i frorri "the.
iz>f>
I + Aid,2+ i z +
= \~ir~), A>d' +(ir )z +
+ 0.1 ( 0. 04X 0 . 0,4 ) * +
a
jz.
+
-Nt\
/. O S-73 X
o . 02. ( 0. i ~ ) ( o . o s ra .')2-
I O --S
12.
^
r
u./
cd~
N A
1,0 X I0 S ( 0 - Z
X 0 . 0 3 + ') ( 0 . 0 l 7 ')
^5 - — —
I.O S 7 3
lb
=
3 .2 8
X
I O - s C o -2 -')
MPa~
f t i -th e . J u n c . t i o n ■.
J u s t
a b o v e - th e - j u n c - t i o n ,
VAy
_
x o .o 4)C o -o i 4 ')
/.OS73 x io's (a.z')
lb
J T u s t’
lO 3 (o .z o
(, o x
Ss =
b - O .Z O
Ss =
-3- I S M P ^
b e -lo w
th e - ju n c - t io n
(gO X
■Sc =
I Q 3 C °-
,
b -
C o - o * > C&
o .o z .
0 /4 }
/. OS73 X /0~-S’ (O.O Z~)
Ss =
-3 / -S
M P a -
573.
The cross section o f a beam is an isosceles triangle with
vertex uppermost, o f altitude h and base b. If V is the vertical shear,
show that the maximum shearing stress is 3V/bh located at the mid­
point o f the altitude.
Solution:
HA
X b
V (- k
[%
h -
- y 2]
" s ’ n~\
168
To -fin d -the. Irvtaximurvy s h e + r m a s t r e s s , d if f e r e n t m t e Ss w i t h r e s p e c t to ij a n d e^udrtc. t h e r e s u l t i n g
e<^*attor\ t o z e r o .
-~ £ .
- - ^ — \h - 2 » 1
3J»* L
*0
1
■■■' h -~ 2 y
y
=—
2.
HA
b u t-
X ma
V(iz)h-
••• s
3
S*
2>M
3C.
4 bh*
m
5 V
bh
574.
In the beam section shown in Fig. P-574, prove that the
maximum horizontal shearing stress occurs at a layer A/8 above or
below the N A .
169
Z » a I>
c
v (-b x y )(-z
At = "
----------------------
i
-
i
t
l
, - 0-
M i.
*
-
*
’ }
f —^ - — h 1
»i«» L 1
Js
•'■ J b _
_
_£ . U
i
"
3 “
y
r - J- ^
J
= «
J
frory, +hm- -k>f>
Ho¥umve4~ fr w n A M » / A e m « X
^
___ £>. - —
h
-
Sj
«
- 3h
A rectangular beam of width A meters and height A meters
carries a central concentrated load P on a simply supported span of
length L meters. Express the maximum r in terms of the maximum of.
Ans. r “ t>jh/2L
Solution;
r'
1
1
T,
,
*1 **1 • * £
*•/*
1
M *x
' *
V = P/z -
3V
SV
2A
*■ 5 i » H
170
I
V
^ bh
M ay.
M
f- Ss b h L
3
,
M a x
M
f V* ) = " I T
= PA
= ~J~Ss b h L
. _M y
x
M ax
A7
S f 51
b h 3
12G>
<ZM
Af<ax -S f =
C3 S s b h ) L-
b h *■
b h 2-
S fh
Arxs.
ZL.
583. A wide-flange section having the dimensions shown in Fig.
P-583 supports a central concentrated load P on a simple span L
meters long. Determine the ratio o f the maximum flexure stress to the
maximum shear stress.
\
UZZZZZZZZZA'.
/
_L
25 mm
250 mm
► — 25 mm
T7777W777m
[—200
mm-»-J f
25 mm
Figures P-SS3 and P-5S4
Solution:
L
E27:
iff?
26 nini
m a.
2 5 m m -is (■
EZ±iX L
L/z
V*
I
■%
-to o
mm
; an m m
171
M ax
P/ 2-
1/ =
M a x M = P/z- ( LA ) =
02 . (o - 3 > 3
12-
2fc/,3
NA ~
12.
T na -
M
«X
S f
2 .2 2 / 3 5 "
My
=
X
( o . o s - r s ') ( o - 2. s ')^
12-
IO -4 m -
Co / s )
_
I
= 168. 82- PL.
2 . 22/35T x
V 2 .A u
I O -*
% . [ ] 0 .0 2 S (0 . 2 ')(0 . 1 'S 7 S ') + 0 .0 2 s C o /2 .s)(biaiZ5)\
Xb
Z. 2 2 /3 5 ' X !0 ~ *(o.ozs)
P/z (% . Z 2 S 1 2 X
/o~ *>
2 .2 Z 1 S S X 1 0 -4 ( 0 . 0 2 5 }
M ax Ss =
79.4 S P
/£■*. ffz /*/_
55
2. 124 L
Arts.
7 ?- 48 P
584. A built-up timber beam having the same cross section as
that in Problem 583 is used to support a concentrated load P on a
Mmply supported span 8 m long. Determine P and its location that
would cause simultaneously a maximum flexural stress o f 8 M Pa and a
maximum shearing stress o f 1.2 MPa.
Solution:
t m
r
=
o
Px - RzCs') = O
8m
Rz f
sy=
o
R,
Rz. = p i
„
Ss Xb
S s = ---- 2 - • M a x V = — -—
Ifc '
Ay
Px
= p (• ~ T >
M ax M - Vx
g -x
VAm
k,
r*
t
1- 2. x/ 06 X Z.22I35 X I0 -4 X o .o z c
= ------------------------------ -----------------
M e JX V r
8.82812. X 10-4
7 S 4 8 . 6 6 >V
172
^
1
tj
0 .15
Mm
M » /1SM7.2 N
Max M * Vx » IIS47.Z * 7 S 4 t £ i x
X 8 /.S7»,~]
V = p (t -
t
)
7 S 4 t.it = P ( I P =
93*/ . g A/ |
9t& A simply supported beam L meters long carries a uniformly
distributed load of 16 kN/m over its entire length and has the cross
section shown in Fig. P-585. Find L to cause a maximum flexural stress
of 40 MPa. What maximum shearing stress is then developed?
Ans. L - 1.77 m; r - 5.55 MPa
|"»----- 140 mm — *|
7
160 m m
J
L
20 mm
P
L
$*•01
- u 3iA y s e>.0i f a l 6 ) ( 0. 09) * O.iA-(o o i)C o n ) ^
SA
O . O Z (O . f ) * 0 .1 4 (0 .0 *.)
Stmm
173
y ;
O. l i t m
I MA =
a bove. M®. kasc-
+ 0. 02.( 0 ./6) ( 0. 0* Z ) * +
II
0. 14. ( 0. 02) ^ + a / 4 ( o . o i 'X o . o + t ') * II
I
MA
=.
x
1 .^ 0 1 5 9
IO ~ g r * *
tOOO L
Max V > *000 iV -0
Max Sf
-M0ML
« 2000 L *
y k governs »
o*
m
2000L x ( O - V t t )
40 x 10
l.lo is n A
»
/ O -*
<4n(.
iz n
S€ *
VAg
■'
Xb
**t" /#>*. neutral a x is
(tODO x ! - 7M > L o . H (0 .0 t )(0 .0 4 * ) +• O .O 3t(o.0Z)(o.0H X\
M O I S3 x /<?-*■
St a *5~.Sg
|
<4nf.
SIT. Hie wide flange beam shown in Fig. P-587 supports die
concentrated load W and a total uniformly distributed load of 2 W.
Determine the maximum safe value of W if «y < 10 MPa and r < I.4
AnS. W — 2.62 kN
MPa.
20 mm
zzzzzz&zzzzzzi
IT
2m
2W
1m
|
—
2m
20 mm
Plpm PM M r.
200 m m —
1
174
Solution:
vJ
2m
Z W
im
2O m m
2 >n
~i / / / '/ ' A ~£bmm
t
*»
<
/O A7A3*.
'V/»
cf. 2 (o.ZZ~)3
12
O. /gCo /s)3
/■4 MPa.
/2z m
Ri = O
=. g-WfftC, x /O~s m 4
2 I ' V ( i ') + w ( s ) - R z ( s ' ) = O
/?2 =
W
2lV
2 >w
✓ -P .
M ax
zm
1/ = 2
>V
■w
Jw
2 IV
Max M - Z w
M-D
M i
to x /oc =
-
w = 4030 ■s
Ss = — —
I b
aJ
; 1-4 x /o 4 = 2 Jvf’
’
x
42 x e>. 10 +
-1
O - O q X D .o z
X O .P 4 -S J
g.<W 8l>L X Z O ' s X O .O Z .
IV r
26/<?. /S'A/
/In s.
588. The distributed load shown in Fig. P-S88 is supported by a
box beam with the given dimensions. Determine the maximum value of
w that will not exceed a flexural stress o f 14 M N / m 2 or a shearing stress
o f 1.2 M N / m 3.
I
175
|*— 300 mm — **j
w N/m
1m
i-
" ■ "2 ,,,
250 mm
150 m;
1m
,t
'
200 mm
Solution;
\ b 3 0 D m m —3|
£
14 M P a .
;
S s
<
1
L .
1 -2
= °
M P a .
UNA
3W(2..S') = 3 R z
R2 = Z . S
r
/ 25Omm
jfc-
^ !tn
Zm
1
S f
t
/50mm
kv w/m
In1
0 .3 ( 0. 2s ) 3
o . 2. ( o . i s ) 3
12
12.
X MA = 3.JV375-X / o -+ m 4
iv
s y -- o
R , -* R2 - 3m/ = o
R,
=
o.S IV
►
V
j.5W____ 05 iv
~ N
V-0
By ratio a n d p ro p o rtio n
t a e / point- o f -^e-r o
shne-i
e-a r
2 -^
o .G hJ
! ' s **
d
2 - d
-/•5 IV
V
/W ax
A 1 = O . S h) ( i ') - t - - £ ( o . & ) ( o - s w )
M a x
M
c
=
-
/ .5 w
d
M *X
0. ( , z s
Af
W
N
-
n r,
-
0 ■b Z S w ( p . I 2 S )
14 X ! 0 k -
jv =
■ 3 .3 4 3 7 s X 1 0 -4 -
S ^ lz o
r i/ m
-
o .S m
176
* Ag
16
E
,
l.SW 0-3 X O.125 X
t.2 x to **3
M* s
.........
... - .- ■
J . 3437S
X
O.Z X 0.07S X
J
-------
-- - — .
tO ~ * X o . t o
Is, o n . S 4- d / n
A rts .
M > A channel section carries two concentrated load* W and a
total uniformly distributed load of 8 W , as shown in Fig. P-589. Verify
that the NA is SO nun above the bottom and that /NA — 13.96 >
10^ nun4. Then use these values to determine the maximum value of W
that will not exceed allowable stresses in tension of 30 MPa, in comprusion of 70 MPa, or in shear of 20 MPa.
A m . W - 3.19 kN
Solution:
J*— * —j| jf—xflmrn
w
I
IKf
r
gw
/mf
__
N -*-
f.ffw
lm
fm
X
120mm
K,t5w
— _ 2. At) _ O- I S Co. l 4 ) ( o - Q 7 } - Q - I 4 Co. l z ) C O . Q 8 )
0 .1 8 (0 . !+ )- 0.14 (0.12")
*A
I iJ a
X ma s
j MA =
OS
m
a b o v e - 4-ha. b * s c -
-«• A J a>)
0 ‘ 8 ( 0:.'* 2 L
+ o .i* ( 0 . 14 X 0 . 0^
A rts.
17'
— ‘4 ( 0 'J * 2 *
12.
-
~
0 . 1 4 ( 0 . 12 ) ( 0 . 0 3 ) *
/■59C> x /o~s m ‘1
LV/1
My
, \ y , = o. 03
X = / m • M - - w v^V
y c = o. os' ►»»
>St -
X * <5 n-> ,
iv = S 3 2 0 A/
i-S^e. x /o'-5
tv f o.oer)
7o x !<><■
vSc =
o f
Ml (o.Q 3 )
x /o6 =
»V= 2 2 ,3 4 4 a/
i.sqe x 10 ~s
M
=
ik
= 0 .0 3 ^
y £ = O'OS'm
s t = 3 0 X /O6 * v5H' r ° — • ;• W = vj, /S.2 A/
f
I.5 3 C X IO~s
>3IV To. 09 )
s c = 70 x /o*= y
M ax
=
IV =
V =
y/> w
“
*
^ (V [o .w x o .o ? x < 5 .045J ^
: 2 0 x to — ---------------
lb
• •-
-f /5 7 . Tff A/
VV =
/9, 7 0 3 . 70
>V =
■31-3 2. Aj
A/
Ar,S.
_- -------------
178
592.
A wide-flange section is formed by bolting together thre«
planks, each 80 mm by 200 mm, arranged as shown in Fig. P-592. If
each bolt can withstand a shearing force of 8 kN, determine the pitch if
the beam is loaded so as to cause a maximum shearing stress of
1.2 MPa.
Ans. e = 98.2 mm
T
- 80 mm x 200 mm
Flgure P-592.
Solution;
|«_ 0.2 m
It
o.ot,
II
It
11 ?
II O.tfn
I;
o.ot
RI
V « ,f
i
N.A.
R =
5., =
Ve
I
V *!)
VG>,
Xfe
lb
R X 6.
Ib S s fy
R6
S s b<£)f
e =
e =
^oooLo z (p.os) + 0-14- + O.0s(t>.o{o.os )3
1-2 X !0 b (0.08 ) ( o . OZ X O.OS X O.I4-')
o. o is z
Ans.
593.
A box beam, built up as shown in Fig. P-593. is secured b
screws spaced 100 mm apart. The beam supports a concentrated load P
at the third point o f a simply supported span 3 m long. Determine the
maximum value of P that will not exceed a shearing stress o f 800 kPa in
the beam or a total shearing force o f 1200 N in the screws. What is the
maximum flexural stress in the beam?
179
200 mm
Figure P-593.
Solution:
ZMR, = o
Irrt
I
r
2m
pen -R z (s ) = O
r2 -T P
.
sy = o
R, * *2 - P -- °
R .- j- P
O. 16 C0 .2~)3
o. iz ( o - i e -')*
J2.
12.
-NA
= 6.S707 x 10's
_2
3 H
2m
Max
V = -J- P
M a x
M =
V -D
It n
. P
3
<5
P
3
2 „ _ l2 Q o (b .S 7 Q 7
v -- A L
—
•3
X to "*}
O . t ( O . 12 X O. 0 2
X O■O S )
p = *S4 7 5 .s e , A/
'S f J b
Z p
? 0 0 ,0 0 0 (b S 7 0 7 X /O ~s ) ( 0
O V)
1/ —------- ’
\ <5r ‘ — O./b X 0.1 X o . O S - 0 1 2 X D. OS X 0 .0 4 TA i
Pz.
7 S S / . S 8 Aj
P
z S 4 7 S . S 6 A/
M
Q .S 7 0 7 X / O '
- S.S(> MPa-
Arts.
\
180
594.
A distributed load o f w N / m is applied over the entire
length o f a simply supported beam 4 m long. The beam section is that of
Problem 593, but used here so that the 160-mm dimension is vertical
Determine the maximum value o f w if af < 10 MPa, r < 800 kPa, and
the screws have a shear strength o f 800 N and a pitch o f 50 mm.
Ans.
w ■< 2.05 kN/in
Solution:
Q.-
De*ferm'me^ th *. yrtaxtnnurr. Value o f M/ i f '■
Of
i IOMP a , ;
T i SOO kPa. j
RSire** - SOON) e. z SOmm
4- J.200mm—^
tv
'
fw
Rl = 2 w
M a x . Shear
AU x
Kx-Zy!
-
M
2 W
S
Ib 0 (l2 0 )3
o o (tb o ’) 3
/z
Q fS A ~
(200 x
= zw
S
lz
M
- (ib o X i P
X
a 1 = 200 X 20 X 70 =
Shear (wood) -
=
V=
4S.Z x 10^ m m *
X 30) - 3 S Z x
2X0 X 103 yv\rv\-3
—
„ .
(SDO X /DSXVS-Z X /D-t ) ( 0 . D t ) .
Z W * ....................... .. ................ ................ /
352
Shear (screws)
X
/ o -6
^ _ HI
.
(Sl'e.
Z (g Q o )(*S -Z X IQ-*-)
2\M a
(Z 80 x i o - (-)( . o . o s o ' )
IV a 2.S'* k N / m
F le x u r e . :
ilV -
to-3 m m 3
<ri
A ? a --------
(IP x lo6)(4S.Z X /O -*?
,
w - 2. o s kN m
INI
2.S. 3 kN /n
H/ =
M ax
Vs/
Ans.
2 . OS k f t / m
=
595.
A concentrated load P is carried at midspari o f a simply
•imported beam 6 m long. The beam is made o f 50-mm by 150-mm
l«ieces of wood, screwed together as shown in Fig. P-595. I f the
maximum flexural stress developed is 9 M N / m 2, find the pitch o f the
•i rews if each screw can resist 800 N .
Figure P-595.
Solution:
5 0 * 150mrv)
f
T
ZOO mm
■:h : .
i!
THz '-
r , s pA
I
— loo— %
S f 4 <JMP a .
R=
■ k r -n ,
SOON
M a.x
V =
P/z
M ax
M
p/ z ( 3 ) = Z
=
-
± Z (0 .Z )3
_
e/ 2,
± j(Q ;j2 1
"*
12
sf =
,.2 S
^ P (o .io )
.
7/Sooti
/. 2S x IO-*
^
Rz
e
x
'■*.
12
=
goo x 2
sc.re.ujs =■ l,6 DOfi
R J
( l b O O ) ( t . Z S X 10- 6 )
VQ
( 7S& °/ l)(o ./ S x O- os X O. 07S)
0 . O 9 V-S ry.
A n s.
182
596.
Three planks 100 mm by 150 mm, arranged as shown m
Fig. P-59 6 and supported by bolts 0.4 m apart, are used to support a
concentrated load P at the center o f a simply supported span 6 m long,
I f P causes a maximum flexural stress o f 12 M Pa, determine the boll
diameters, assuming the shear between the planks is transmitted by
friction only. The bolts are tightened to a tension o f 140 M Pa, and the
coefficient o f friction t tween the planks is 0.40.
Arts, d = 19.1 mm
300 m m
Solution-.
= 12. M P a -
S f
L
7 la,
- A
R1--P
M ax V =
P/z
Max M -
(3 ) -
-J.37S X i O ' * m *
.
12
V-V
My .
^
(0.IS-')
O.IG(O-S)*
12-
\0Qvnm
f> z ]&,OOOfi
Sontn*
N. A
jo n m
—h
lOOvnm
I,
(tsooo \
Ve
n ■*. — <5 = -i
R
I
4)(p.is xo.l X O -0
- ----------------------
'
3 .37 ,5' X /O-*
R = IbOPO M
R-. f N
■
tb,000 -- 0.4-N i
N - 10,000 rJ
Al z A S t } 40.W 0 -- ~ o '2 (>*o x
J :
O- 01^07 m
cJ -
19. 07 rnrr\
Iteams Deflections
606. Determine the maximum deflection 5 in a simply supported
i r.im of length L carrying a uniformly distributed load o f w N /m
applied over its entire length.
Ans. S - (5/^84\ w L * / E I) - (5/384)( W L?/ E l)
<3- Pe/«r-m/ne-
tx . d & flcction S
Solution:
w ■*/«
A
L
R| "
wL
wL
7
Q zm 8 = o
R .(L -) - r t ( L ) ( - j - ) = 0
TFv
-
W L.
+ Rz - hVL. - O
WL
«* iv ix
z
wx*-
WL-X2-
yvx 3
12.
WL.
*
= M
2.
6
tv L x 3
O
wx 4
z-f
+ c,
^/X + C 2.
184
<xt f>b B
o =
ij = o , x = L — >
©
“12L- 4 - ~ * 4
-WJL.*
c,
H
w lx 3
WX4
" 2.4
)2
V Y L 3X
2 -4
'
a t m i d s p a n ■ ( m a x . d e-fl& ch & rt o e ic .u rs ')
X =
E lZ
WL- 4
3<£
L.
w
/ L_~) 1 _
jz-f \ * /
W L.3 f L \
2-4
{ 2 /
38^
Ans •
£
=
^ lY L 3
3 S4 E l
■SfVLl384
608.
Find the equation o f the elastic curve for the cantilever
beam shown in Fig. P-608; it carries a load that varies from zero at the
wall to w N / m at the free end. Take the origin at the wall.
L ----
Flgu«» M O I.
<5 ' F m d equation o f &-Iashc~ c^urve .
SOLUTION:
(p
2L F v = O
WNf „
IMS
WL.
2.
M
/ r\ „
®
^
©
e
i j 7
„
_
_
m^ x *
—
r
_
t v i* x _
** « - c '
w l zx *
w lx *
E l y --------------------—
<sf y>A. ^ •
tv x * +
ij z o , X = O
w xs
- - ^ 2 +Ci X ^ C ^
- — > etjn. ( 5 )
.-. C 2 = 0
o f p f. A ■ d~L
- O /
X = o
----> e^ n . ( e )
dX
E ^n «
&/as iic* cziAr\/& •
*
-
- ^ s r -
2^
-
■
&
«
«
*
.
609.
As shown in Fig. P-609, a simply supported beam carries
two symmetrically placed concentrated loads. Compute the maximum
deflection 0 and compare one-half this result with the midspan 5 to case
7, Table 6-2, page 270. Check your answer by letting a - L / 2 and
comparing it with the answer to Problem 60S.
Ans. 9 - (P a /2 4 E I)(3 L 3 - 4a2)
186
f
P
n 0 «m P-60*.
C3 : Comt>le-i&-
Solution:
maX/ m u m d e -flec^han £
„
/\
~P\
e
->
R,
■Sijmme-irioally lo a d ed b e a m
,y~'ay
wr/le* Ihe- m om eJrh e-ouai~/or\ o j -
oaj®. h a ir o f
the. beam, b ecau se. i r i s su m m e t/rie*/ .
Us/ntj th e l e l f h a l f only @
H-T
=
a x 2-
©
E ly
Px. -
A-
U ° < X <
= M
P < X - # )
- -?
=
cy/- p t-
(.X -a y + C ,X * C i
!j = O / x = O — => ( ^ ) (neglec-hrit^ Values
.\ C z = O
or/ m / d s P G n ■■
-
-E L ?
S
ft, - f?z = P
° / 'A' /<5SS ™ a/7 * 0
----->
= O
,
X
-
L/ z
+
P L
4
i
-
P 2a- +
S T
(£ )
/ l
2.
PL
P.
7 *
8
C-.
£ = ( Pa-/z4 E l ) ( j L 2 - I * . * )
\So -for case. T , Table. C - 2. f one- half 1c>f -this result
with the- rn ids pan £ aiVes E I S i f <K = L/ 2PL
P SL
B IS
- £
- L ‘)
in th is yDrob!e>ry\
P -
E lg -
24
ZP
whicJi att/es the sarrtc~
( 2P ^ L '
48
r e s u ltIS
p r o h le J n £ o S ,
612. Compute the midspan value o f EIS for the beam loaded as
ihown in Fig. P-612.
Ans. EIS — 657 N - m 3
2m
300 N/m
1.5 m
0.5 m
4m
£
Figure P-612.
(3. •
. Compute- m id sp a n value
E I .£ ,
Solution:
2. W
A
300 N/m
1.5 ►
05w
/r
Ri
R,
^
2M b = O
/?, ( 4 ) - 3 0 0 C 2 ) / - S = O
R,
= 2 2 5
N
B
188
SFv- = o ]
M -
Rz = S 7 S M
e i
i f
^ x - j . sy
E I~ ~
= 2Z5X -150< X -I.S )£ *■150(X - 3 . 5 } *
Q )
E I-^ 2 -
= ~ X 2 - S 0(x - i.s y a+ 5 0 (X - s . s y + C,
©
E IH =
X3 ~ ^ ( x - I . s ) 4 + ^ ( X - J . 5 y * c , x * C Z
at p t A
■■■ y = o ,
at p t 6 ■■
x = o •— >• ( c )
y =o ,
* : 4
0 s ” i ( 4y - 22. (z .s v + m o . s ) < + 4 t ,
b
4
4
C, -- - m
N ■m*-
£ i > = ~ T X* ' f < * ~ I S/ + J ( x - r f - *7*X
M idspon value- o f E I £
X = 2*r>
E l y -- - Z f W 3- f
= -(.5 7 N
•• E I S
(0 .5 )1 - 4 7 S (Z )
r**
= C 5 7 M .n ,*
615. Compute the value o f E ly midway between the supports for
the overhanging beam shown in Fig. P-615.
Ans.
4 kN
1r
2 kN/m
2m
2 kN/m
2m
‘i
R2
2m
E ly — — 4.66 k N -m3
189
Solution;
Q-- C om pute- the- value, o f E I *
S upp o r k .
J
m id *)a u b e t wetn
J
SDL U T 10N --SkN
/
2 kN/m
A
Zm
2-m
2kM/m
B
.... '/^
2m
Rz
R,
6 ) Z M e = o j R, ( 4) - 2 ( 2) 3 - 4 ( 2) + 2 ( 2)1 = o
R, *
£ F v
=
4 - 2 (2 )-4
o ]
Rz
M
*
4 kN
E l
-
+
Rz
-
2 (e )
,
o
S k rt
4 X -Z ¥ -'K X -2 ) +
2 < X - £ > 2-
2
+ 8{ X - ^ ) r ~ - ^ ^
@
E l H L J I.
d x 2-
©
E I~ ~ = 2 X * - T - £ < * - * > * ■ *<x-f>*
=
4 X - X z- - i < X - 2 > + < X - 2 > z + g < X - 4 ' > - < X - 4 ) * -
< x - i >3 + c .
< X - 4 > 4‘
12
at pt. A ■- y z o f
x * o
----- >
(2 )
?
(& )
= o
a t p f. 6 '-
ij - o /
X r
190
t - j w
- -c j r - i ( * r +
C, =
- <f.33
<J£ + c . « ) ~ o
k f i l - m 2-
aj- a pfr m id w a y beta/ten su p p e rh ft ^ 3
X - Ztn
E Ip j
- ~ ( 2 ) 3-
E ly
-
-
~ 4 -3 3 (Z ?
AA/.m3
620. Find the midspan deflection 5 for the beam shown in Fig
P-620, carrying two triangularly distributed loads. ( H in t: For con
venience, select the origin o f the axes at the midspan position o f the
elastic curve.)
Arts. S = (9/1920Xh,L 4/£7)
Figure P-620.
Solution:
a= F in d m i d s p a n d e f l e c t i o n S '
191
At/X 3
W L z
3Z_
E l^ 2 - =
d*
fj.©
•
/- 2 L
H ^ fv T 2 _
.
f i t f>t. / ) :
*± 1
Z1
y
= o j
-------»
dx
(2 )
—o
At p i R : --- ^ =r £> ; x = o
'
'
+ C ,X + C,z,
X - <?
.*.
+ r
'
®
••• c , = o
F7
£
~
E h
626.
ML*- f L >*
1* ( z ) ~
fV
v j )
3W L *
~ iT z o
Beam loaded as shown in Fig. P-626.
/4/ty.
(area)^B • xB =* 8.25 kN m3
400 N/m
i 1m
3m
1 mi
Figure P-626.
<3 -- C o m p u te - ~fhe- w o o w i f
jr a tn
/Ac A1 d fa -
b e -tn / e c n f h » C m a c -iio n s a b o u t b o th
+ h » le -f t a n d r i g h t r & a c h o n .
192
Solution:
4 00
N/m
ft
A
!^
1m
3 m
B ’
Im
R z.
tyZM B * o
R, C s ) - 4 0 0 ( s ) ( z . s ) = O
R,
S; Fv =
=
0
Ri
ts >
-(m
)M
=
400( 3)
6< 7 < ?/ V
. «
■ i ^ a l l (d f
=• O
193
=
4 X 0 0 ( 7/ s )
+ zoo -
t s o o ( 7/ i )
= * 1 5 0 AJ - m 3
=
S. 2 5
k M - m 3
- ^ f & l j w + r]
= I2g00 + ^ 0 0 - 5850
= $2.50 M - m 3
S . 25 k M - m s
628. Beam loaded with a uniformly varying load and a couple as
*hown in Fig. P-628.
Ans. (area)/1B■xA = 2.13 kN-m3
M= iKN-m
Figures P -62 8 and P-629.
Solution:
z
\
3 "curve ^
\
N
\
\
1-5
2.
194
Q 2 M s = O
R, (S ') - — 0 ) 3 ( 3 ) + 2 = 0
R , = ~ r ~ = O .s K N
JEFv- -
o
O.S - -^ -(l )3 + Rz = O
Rz = / k/U
M= R , ( s )
r O.S ( 3 )
1 1-5 k H - m
-
1
.1
t
z
- 1 . 5 k tl-m
M = ZRz
- 2KN -hi
■ XA -- j 0 . 5 ) ( 3 ) ( z ) , j ( z ) U ) ( 3 + j ) - ^ C O s.s
- j ( ‘- s) 3 ( !
2 .1 3 3
KAJ - m 3
■ 5
)
A
631. Determine the value of P for the beam loaded as shown in
Fig. P-631 so that the moment of area, about A of the M diagram
between A and B will be zero. What is the physical significance of this
result?
Figure P-631.
Solution:
£m a
-P
= o
+ goo (3 )1 -5
3 Rz
- 3RZ
3kOO - P
=
Rz = 1200 -
\rr\
M=
M = -«0 0(3 )(l.5 )
£00 N/rn
3m
r 3600 - P
( Area.
-
L
(J J
— -
(>
+
A6 ) ■ X A - O
( f
)
5 4 0 0
+
( u
o
-
1 -5 P
/. t>7 P
p
o
(m e a n s th a t the- ta n g e n t lin e
pt- A h horizonta.1)
) 3
- P
-
( ! 2
2 7 0 0
_
J
-
( U
0 0 )
o
- 2700
=
Ib Z O
N
A ns.
, ( L
. }
=
0
196
636. The cantilever beam shown in Fig. P-636 has a rectangulai
cross section 50 mm wide by h mm high. Find the height h if thr
maximum deflection is not to exceed 10 mm. Use E = 10 X 10® N/m 2.
Arts, h = 328 mm
Flgura P-636.
Q- F m J ihe. height
Solution:
50vr\W
\
i
S max. I
E = ID X \o9
Mux & = Io vnm
T -
hh 3
12.
Q o s H hs )
12.
t
A/ c
-
( a re a ) AC
- ^
=
-
m .7 S
= - p. o/ m
XA
( 4
x
• « ) /o3
E 1 S = t n/c.
IO K / O 9
X
jz
0 .0 5 (h )3
( o . O l ) - 14. 75 X IO3
197
h 3 =. O. OSSV m 3
=
h
32g
mm
|
Here :
t A/c- - De.vta.iion o f A -from a -tangent
drawn at C.
640. Compute the value of $ at the concentrated load in Problem
619. Is the deflection upward or downward?
Am. 8 = 46.9 mm; downward
Q : Co/nj^leie 1toe vo.!iae* o f* & £*t co n ce n ir~{\-fed lo a d
Solution:
z°uAf\ie
-
12.
2 • e-urve
—
Z fV = o
VA + 4 - 3 ( 4 ) = o
VA = Skfil
^ 2 M a = o
Ma +
M A
i
b/ a
4 (5 ) - S O t )( z ) = o
=
= (a r & o O eA
12 k r J ■nr,
■ x B
= i M M ( T 3) - . ( l z * sX i ' s) - ( i x , 3 -5X i ' 0
= - 2*. 1 2 5
k f J ■y>~t 3
B I S = t t/ A
(/OX. i c V ( 6 0 X IO-t) S
- 28./ZS X IOJ
198
£
=
875
Ans.
642. Find the maximum deflection for the cantilever beam
loaded as shown in Fig. P-642 if the cross section is 50 mm wide by 150
mm high. Use E = 69 G N /m2.
Ans. S — 28.0 mm
4
2m
kN/m
1m
Figu re P -6 4 2 .
<2 -• Find max. Reflection
(^M = 10 kfj-m
50vnm
15 Ovnm
Solution
x _ SOfiSO)3
12.
E; t n
= - ( u M O - 1 0 (3 )0 * )-
S =
(IS - 4S - ~t )
1 0 *'
X tp *
& =■
28. 12. m m
A ns.
1o 3
199
645. Compute the deflection and slope at a section 2 m from the
•'«ll for the beam shown in Fig. P-645. Assume that E = 10 GN/m 2
*m(I / «= 30 X 106 mm4.
Ans. S = 16.4 mm: 9 = 0.739°
<3
Co m Mie-
de-fle^hon and slope-
ext a s& ch or\
2. m
-f r o m
W a ll.
Solution;
3°^urve
Z Fv = o
6,
6>
~~ - J - MOO
= °
V -j 0 * O O )(s ) = O
v = /Soo M
M = 3600 N
n r,
* M 6 = ^ ( S o o ) ( z ) ( j z J = S33.33
E
= IO G p *.
I
= 30 X t o 6 m m 4
E -IO b /a = (areA^nB
200
E l t
(* )
E ie
b/ a
~ Care,o.~)0/\ ■ y B
B/a = ( ~ X it O O x z ) - ( z X 3 l,D o )- (- ~ X Z. X £ 3 3.3 3 ^
=
& bJa =
© b/ a =
(b )
- 38(,(>. (>i-s
38(.C.f>LS ___________
(■/ox lo*)(3o Xto 6)
-
o. 0 /2 ? x i s o °
E I t z/A
O. 73 9/ '
Tf
=
r*J .
(<Jt'oton ft> r~
-ihjh O
( (\retI e/\') ■ X 6
= ( i X 3U>0 X 2. ) ( j
= - f^ 0 7
o. o i z q
t ) - (H D 0 X z X i ■z ) - ( i X « 5 x
rJ-yyi3
4907_________
(/O X t o 9) ( 3 0 X 10 ~ * )
£ =
647.
-647.
/b. 4 /*7«7
Find the maximum- value of EIS for the beam shown in Fig.
Ans. EIS = (121/1920)wL*
Aa
L 3 2.
Z\
2.4
201
M =-
W L.
V
WL 224
t * /e = r= T
-
>
MlZ ~
E s*
1 ~ f
a s *
=
EI£
i
'
/ M
^
2.
\3/~z4 " L
,
L
'
W L-*-
L 7 ~ T ~ ^
L
I
/ L \
'1 'J l T )
w l *
Vo x «
WL<
112.0
653. Compute the midspan value of £/6 for the beam shown in
l-ig. P-653. {Mint: Draw the M diagram by parts, starting from mid­
ipan toward the ends. Also take advantage of symmetry to note that the
tangent drawn to the elastic curve at midspan is horizontal.)
Arts. EIS “ 3350 N-m3
600 N/m
600 N/m
2m
1m
2m
'
R2
Figure P-853.
Q.:
M idi par) Value-
B I$>
R, , Hz c 1200 N
Solution:
6 0 0 N/m
lm
Z.m
^*■
\
\
R,
\
^
^
f
B
600 N/m
2m
^
2 02
M = 1200 N - m
600 tl/vr
I20OH
V= o
+ 12OO
N-m
1° Z 1
-12 00
N-wi
Because o f sum m e.trij , th e ta n g e n t draw n to the^elasho curve- tv maid span is h o r iz o n b jE I & = E - I t c/ B = (area.')ca • * c
=
(U P O
X 2 -S )C 2 .S
j
) - ( J X
1200 K 2 ~ ) ( j X 2 )
E I S = 3350 r i m * Cpi- & is above ia n je n t a f B )
656. Find the value of EIS at the point of application of the 100
N-m couple in Fig. P-656.
Arts. EIS -> 342 N m3
500 N
M= 100 N-m
2m
~* || 1 m
■■
Figure P-658.
Q.: Value o f E I S a t p t- o f a p p lica tio n o f
/Oo f i ■m ' couple.
M = 10 0 N • *n
500 u
I
/+)£M c - O
R/4-) -5 o o (5 ) -IOO * 0
R, =
400
N
■£Fv = 0
400 - 5DO + R 1 - 0
R2 - 100N
1600
3
3i
4
= IZOON-wi
I
150 0
3
15 00
IO O O N •rr)
N-m
100
N-m
— (artx~)cn ■ *c-
- /X too X z
!‘
H>7
BI
BP =
t B/ a
- ~
(Area.~) B A ■ X B
1
204
( i x ,DO° x zX i * * ) ]
= T j [ ( i x 1200*
t %
-
//3% z
E IS
=
E l [ B P -
=
1 *7 5 -
t B/a]
//33
f\ ns-
663. Determine the maximum deflection of the beam carrying a
uniformly distributed load over the middle portion, as shown in Fig.
P-663. Check your answer by letting 2b *= L and comparing with case
8 in Table 6-2.
Ans. EIS - (wb/24)(L3 - 2Lb2 + 63)
2b
iv N / m
R2
Flgur* P-663.
<3:
M a x . d e fle c .iio n o f 1be.amr>
Solution:
2b
w N/m
/IK
«—
L
R,
M
s'.
B'
Ri = Rz
wb
Syinm®Vic^l - » wax. Iothk occurs
l/ £ ^VnuJspan)
a IS =
E l t A / b ' = (a r e a )A B '■ X A
- t t i 1
C IS . ( " l / n ) ( L ’ - 2 L t '+ l , ’ )
* £ -J £
= -± __
A / o / c C a s e 8 in ta ble.
*v7.*
& - 2 , j> a ^e -27/
666. Determine the value of EIS at the right end of the overhang­
ing beam shown in Fig. P-666.
Aits. EIS — (wb3/24)(4a + 34)
w N/m
Flgur* P-666.
a-
va lu e o f E I S At r ‘j h t e *id
Solution:
tV / 3
(*re.a )A B ■ xA
206
2.
t c/ e
( * '* « ■ ) C6 . Xc.
_
y vb *
8EI
(~~) S i i n d i c a t e s that f>h ■ Pi $ C. lie, below the-hanger f Hue draoOt^ a { B .
d y s im ila r irianyle^ :
DE.
Ii A/ B I
b
(A.
■■■pe
i
= PE
+
| t C/B I
E I S = -^ ( — * b * j +
668. For the beam shown in Fig. P-668, compute the value of P
that will cause the tangent to the elastic curve over support R 2 to be
horizontal.
Aru. P - 1350 N
300 N
P
Figure P- m i .
207
Solution:
Q: C om pu te -fhe value- o f P th a t w ill cause the.
t a n g e n t fa th e e J a s h c c u rv e over Support Rz
-fa be h o riz on ta.1.
SCON
P
If
yV
&
a.m
Rt
R,
\
0 2 M c. = o
R, (55 - P fO - 3 0 0 ( 2 ) = o
2-P
R, = - T -
- 200
SPv = O
-— • - 200 - P + (?2 - 3 0 0
R- .
■
—
+ 500
2 P - 6>00
£ i l t A /c - (a rea .) AC ■ X f\ = O
( - j ) ( s ) ( z p - M o ) ( f x j ) ~ ( j - ) ( 2 ) ( 2 P ) ( ' + f X 2) » ^
/=> r
13SO N
GJO. Determine the value of £IS at the left end of the overhang­
ing beam shown in Fig. P-670.
Ans. EIS - 428 N -m3 down
208
Q
Pe-fe^m in e- the- va lu e o f E I S
a!
the. Ie.fi -end
Solution:
M - 600 N •m
ZFv = O
650 - - £ ( 9 0 0 ) ( $ ) + R 2
-6 00
R . C O -■^ ( 9 0 o ) ( 0 ( s ) s O
R2 = 700 M
R, = 6 5 0 N
1950
/
- 'to o
E l t c//3 = ( ar&a_)c-e, ■ X «=
=
E H
-1 3 5 0
( ~ J X 3 X l ‘3 5 C > ' ) ( l ' ) - ( 3 x . b O o ) ( - ^ ) - ( ^ X I 3 S O x z ) ( 3/ s ' )
- 383
N ■r * 3
A/& = (area . ~ ) / i s
• X A
= (-6 0 0 X / ) ( - £ - )
=
-300
N -
AM3
S / j n in d ic -a ie s A
C
3 ij s t m i !a .r f r i a n g /<£s :
are. beJous the. t a n je * if Line, a t B .
209
E I &
-
~ (3 S 5 ) + 30 0
E IS
=
428 Al- m 3 J o k i h
675. Repeat Problem 674 for the overhanging beam shown in
I ig. P-675.
Arts. 8 — wa2L 2/32EI
w N/m
IS
Figure P-675.
a ■
■ F /n J
rte
s u p p o rts
d e fle .c.fio n
m id w a y b e tw e e n the-
show n.
Solution:
W
2S
N/ w
7\_
o o
/777 ~77~T
c [~
/y/
B
S7/ '77
R, = WA
f?2. - W« -
M =^
2.
W
N.
<
II
J
>
^ ------------ --------------- ---------------
wa.
VI/A*2.
2.
•2<S
s•
£
ty *
w
- I•VA.2- L Z
ItrE l
210
2. S =
vva.1 L 21C, E l
S =
fv<a2 L z
32 E l
679. Determine the midspan value of EIS for the beam shown in
Fig. P-679 which carries a uniformly varying load over part of the span
Arts. EIS - 2940 N m '
900 N/m
f
3m
2m 1m I
--------6 m-------
Ri
R2
Figure P-679.
Q:
i
Deter mi ne. the m id span sains. o f E I S .
Solution:
Considering F B D o f left
h a lf o f tne- beam .
900 N m
IB
oo
N•*>i
H-- 1 o o ( 3 ) . 1 0 0 ( j )
= 1S00 N • >r>
-Goo N■m
211
Z S = i A/a
t * /b = —
i »/B
=
fartmJ/tB ■ Xfl
— : [ ( isoo x z ) ( / + - ^ ( z) ) + ( j x
IX w ) ( j X
i)
~ ( j - X bOO X 2 ^ / t z x j J ]
± A/ e -- 5 W 0 / e r
-2 5 -
* %
fir s
r
=
sxso j E1
yv- m 3
681. Show that the midspan value of £76 is (wb/48)(L3 — 2L62
+ fc3) for the beam in part (a) of Fig. P-681. Then use this result to find
the midspan EIS of the loading in part (b) by assuming the loading to
extend over two separate intervals that start from midspan and adding
the results.
Arts. EIS — 9280 N -m3
2
^
0
►
w N/m
800 N/m
3m
2m
R,
*2
(a)
(b )
Figure P-0*1
fl. F \ n d m id s p a n
va.!ue- o f £ T S
Solution:
-A
w nyn
MX
N
<■
lm
—
L
-4
W N/nr
r~
zs
4
-
B
-
'
Rz.
212
wbL
t a/b - ~ ~ (ftre x )A 6 • xA
- t t ■*?■>)(t - P J ]
wb
2S
=
[}* - ZLb* +hi]
t *U
wb
18 El
(b)
[ l s - 2 L b z + b3J
Mfiny results o f
gOO N/m
A
-Iwi---,13m
2m
3m
L = I
l /z
2m
-f 5 + / = 6
=
3
Using formula, in part (a )
F o r -the. ritfjhi p o r t io n :
F o r the, l e f t p o r t io n
b is
- — t 2 ^
£IS=
b = 2b - /
3 - 2 d X 0 z + ( ‘ ) 5J
213
687. Determine the midspan deflection of the beam shown in
I ik P-687 if E = 10 X 109 N/m 2 and / = 20 X 106 mm4.
Ans. 25.8 mm
2kN
1m
3m
|
1 kN/m
-4 m -
Figure P-SS7.
<5
De
fh tz m i t i s p a n d e f l e c t i o n o f t h e . b e a m .
Solution:
G iven -
£ = to x JO 3 N/ .m 2
I
= 20 X ! 0 6 m m *
For the. u n ifo rm ly cjis fri buie4 load ./ sie, - c_____
U
ase.
Midspan £
a ■
3
'
.,
pV = I M / im j
S
w L
/ -4 m
*-
3 g tf E l
S (/ Q O O )(_
3&+EI
S3 33
E l
F o r the, c o n c e n t r a t e d l o a d :
U s e c a se J a. = 3 nn j
b - tm,
M id s p a n £ - ~
*
T
■(-3 L 2 - i - b %)
®
L - 4 nr,
)
a* >
b
214
2 000 ( , )
~7TFI
- 4 (0 * ]
1 S33
£ 1
3333
+
S I 66
I
E l
E l
=
O. 0258 rn
^ T ~ (l O X I 0 a) ( 2 0 X lO~b)
&T
689. The beam shown in Fig. P-689 has a rectangular cross
section 100 mm wide by 200 mm deep. Compute the value of P that will
limit the midspan deflection to 40 mm. Use E = 10 X 109 N/m 2.
Arts. P - 7.07 kN
4 kN/m
6m
lmj
f
Ri
Figure P-689.
<3: C o n n the- v a l u e o f
P-
Solution:
G t v e -t i :
M id -s p o r t d & f lc c .'f 't o n
-
“f O m m
E = /o x t o 6 a/y'/100mm
r 2.00mm
4 « n/ „
I V'V)
&m
T‘ r".
12.
12
: 15
'f+4C>Oo)H
(P-t- 4, 000) N
_______
4 KN/ r
“7v
Q>m
Ri
R,
M om ent -
2 0 0 0 ■+ P
F o r L s tn i-fb rrY ily c j i s t r t o u t e d
c<we
U s e .
s --
-
8
W
-
^ -O O O
lo a d -
/
L
=
s3&+ E J
3S* E l
For-
&
SS.^f x / o J
E l
•'
£/re c a s e
II cin<d 12. -
M
- 2000 +
P }
L - 6
*■ * r e - v
_z_ r f i o o o -)£ IL
S
m id s p a n
U S ±2 ? 1 =
J
/£
-
O- OV-O -
lif.
£2
----------- —--------------------„ f o . S - 4-5^1
(to x
7 x t o ~ <,) L
J
P = 7070 /V
p =.
7 .0 7
kAl
691. Determine the midspan deflection for the beam shown in
Fig. P-691. {Hint: Apply case 7 and integrate.)
Ans. EIS = (wa2/48X3L2 - 2a2)
u> N/m
m N/m
a —«-|
|— a
------------L ----------R*
Rv
Figure P-691.
<5 1 D e te rm in e - the. M id s p a n cJe.flec.tion :
Solution:
w
N/m
— i
W N/r»
Af>p ly,ng c-asc. 7 ( 2 . 7 / )
e
i £
Pb
= £
-
~
(3 L * ftff
a ( w dx ) X
2.
r
2
#v r
2.W
4 1
x■
18 L
696.
In Fig. P-696, determine the value of P for which the
deflection under P will be zero.
Arts. P = 400 N
800 N/m
1.5 m
j
3m
R,
*1
Ftgura P-S98
Q. ■' De-iermine- -the- VAlue- £>f P )
if
Solution:
gOO N /m
1.5 m
£ -
PL*
■3 E l
3 yy\
W L '
M L
Z t E I
3 E l
21
ip -
/*-5*
if
I S
r
1r
r
j v t *
E l l
_
*-+
M L . *] _
PL3
3
3 E l
J
1.^ \ 800 (s
(3')-3
)-*
E
l l
O
p j) ^
U - sS P
3 - ]] _
2 -4
J
3
P =■
P (i s ) 3
400 /J
697.
Two identical cantilever beams in contact at their ends
<upport a distributed load over one of them as shown in Fig. P-697.
I >etermine the restraining moment at each wall.
Arts. M a = — 3 wL 2/16; M b = — 5wL 2/16
&■ P e f e -rm in e a t c a c-h
Hie. r e s t r a i n i n g m o m e -n j-
Wad-
Solution:
•1/1/ N/ m
ti
A
7 / e—
l
—
L
8
----- j
yA
i
j
PL*
3
218
EIS,
PL*
J
V .
E Igx
tv L 4
&
p -
a t p o in t
=
PL3
3
^ w L/6
A ■
.
P
.
i n
k
U6 -i/6
°
»w _
- ^
,
Mg =
PL
-
a t p o in t B
M g -------- w l *■
iv L *
z
""*=
!(,
7
Restrained Beams
704. Compute the reaction R and sketch the shear and moment
ilmgrams for the propped beam shown in Fig. P-704.
Ans. R «■ w L/10
V \|
|| w N / m
1
{
L
Figure P-704.
Q- Find R a n d J r * * / th e sh ea f a n J
m o m e n t d ia g ra m *.
Solution:
220
A t pt. A
S, -
Si
RL
LJ
J
h/L ;
L
S
ML
R =
lO
Max. m o m e n t =
z
wL
L
3
to
ST
IV L.
X
is fT
706. A couple M is applied at the propped end o f the beam
shown in Fig. P-706. Compute R at the propped end and also the wall
restraining moment.
Arts. R - 3M/2L
M
Figure P-706.
<5 -■
C o m p u te . R a t th e . p ro p p e d e n d a n d a U o
wall restraining rno.me.nf.
Solution:
ZEIS = O
M L2
M
th e.
iil
Determine the end moments for the restrained beam nI i hwi i
P-709.
Ans. M a = - wL2/30; M B - >
w l ‘ / 20
709.
Ih l i g .
Q.- D efe^rm inc
e-nJ m o m e n ts ~for'
the- r e s tr a in e d beam s h o w n Solution:
Mi
M r
c
a t -the- l e f t e-nJ ( p t. A )
E 1 6
r
( ^
X
-
O
t 1) * ^ L ( i ) ~ m> l ■ 0
- * Tf
*
' »* - °
©
222
E l S - O
WL 2
WL z
Solving
's
v»
©
Vm L
a
and
r-r
- —
J ML
........,’
20
WL:
30
b
£imnlh<neously,
^M A ~
^
30
do WL) L -
yvL-
WL •
20
713. Determine the end moment and midspan value o f EIS for
the restrained beam shown in Fig. P-713. (Hint: Because o f symmetry,
the end shears are equal and the slope is zero at midspan. Let the
redundant be the moment at midspan.)
Arts. M - - (2/3)/>a; EIS - (5/24)Pai
r
r
LJL
Figure P-713.
<3:
Solution;
Determine' the. end m om en t a nJ midspan
value- o f E I S .
223
tJ- m i d span :
EI 0 = O
(M + Pck
3V * ) ~ P*- ( V i ) = °
M = - 2/ j Pa.
a t m id s p a n :
A
3
= -
5/ u
Pa3
E I S = S/z 4 Pa3 downwards
715. Determine the end moment and maximum E l i tor the
restrained beam shown in Fig. P-71S. (Hint: Let the redundant! be the
shear and moment at midspan. Also note that the midspan shear is zero.
Why?)
1900 N/m
2m
4m
Pl9Mm P-715.
Solution:
Free Body diagram
of left h*lf
224
Z E IQ
= O
at m id spay,
~ l&° ° (~ J ~ ) + ( m a + !B 00)+ = O
M A - -15 00
M A +
IZ O O
' Z E I S *
= 300
-/SOO
( T o checJc)
) -2.(l200 * j ) + 300
~ |-woo rt- *n3 |
MiJsrpar, d e fla t io n is d o w n w a rd a s indicate*/
by
+he ( - )
720. Compute the reaction'/? and sketch shear and moment
diagrams for the propped beam shown in Fig. P-720.
Figure P-720.
Q : Compute the reaotion
and dram the shetn
and m om ent diagram s. Assume unye-i/din*
Supports.
'
Solution;
^ -''"'1 /
' //?! T
225
= (a r e a ') A B • KA
E H
a/ b
= O
e h
*/b =(- j - l - RL) ( , ^ r ) ' (~ 4 'L
WL
&
R L 3 _ W L* = 0
6
30
1VL
£ =
/O
724. The beam shown in Fig. P-724 is only partially r e s t r a i n e d
ut the wall so that, after the uniformly distributed load is applied, th e
slope at, the wall is wL3/48E l upward to the r ig h t . If t h e s u p p o r t s
remain at the same level, determine R.
Ans. R - (7/ 16)wL
w N/m
R
Figures P-724 and P-725.
Q ■ Peter m i n e R J s lo p e A / w a l l a f 't e r lo a d is
W L 3/ 4-2 E I npn/4 r J Jo -tha- r i g h i .
Solution:
G&= WL}
n ti
W "/m
t b/6 =
( a r e a ') A B ■ XA *■ O b *
226
WL 4
■fS
RL3
WL ■
WL 4
48
R =
7wL
16
728.
Determ ine the end moment and maximum deflection for n
perfectly restrained beam loaded as shown in Fig. P-728.
Q-
De-tejrminc- the. m-nd m o m e -n t a n d
mecAimunn d cfla c-fio n -
Solution:
E I 9 As - (area')A3 = O
(*!k)+( L . ±
\ Z /
\Z
Z
. ” L l ) - ( l . ± . » L ? - ) =0
8
/ If
z
2.4 /
F
227
wL 3 _ wL_ 3
ML
3Z
t
/> //}
=
-
192
S
- O;
( a r e a ) 0 A - X0
-
I j. *v i J
■ (-£ « ' X f . y + i f
-
/?2
E IS
= -1 —
nzo
IVL 4
3840
731. The beam shown in Fig. P-731 is connected to a vertical
rod. If the beam is horizontal at a certain temperature, determine the
increase in stress in the rod if the temperature of the rod drops 50°C.
Both the beam and the rod are made of steel with E = 200 X 109
N/m2. For the beam, use / ■» 60 X 106 mm4.
Ans. a = 95.8 MN/m2
Q : DeiervY\ 'me^
£ r
monzASaz. in si-rezz in the. rod -
200 X /O &
le tA M
-
&°
*
A T = S o ° c.
Solution:
s /s s s /y
P,
L
=
IO
A * SOmmz
p o< = U - 7 M r n / ( m . ac ' )
228
r~—
)
\ / \ E / rod
______________ P ( i o ' )
(S O
.
X IO -b }(2 0 0
=
+ (~ ^ f)-
\ J E I J bbam
(2 L a t ) ROD
P U ) 3
3(200 x / o a ) ( 6 o X / O ' * )
X l°2 )
(V /.7 X / 0 - b ) ( l o ) ( 5 c >)
= -7790 A/
A ^' -
A
- ■■
s o x r o "6
change, in stress -
-
3 S'. & x I O 6 N / m z-
7
35- & MPa-
737. In the perfectly restrained beam shown in Fig. P-737.
support B has settled a distance A below support ^ . Show that h i. - Ma - 6£/A/LJ.
1 Ai
§ t
in
^
Wa
P1«ura P -m .
(3-- Show fh*S- Mg = - Mn ~ & E I & / i-z
Solution:
B I O aa = (a re a )A e = ^
i. Va l ) + ( m a L ) =
O
229
-t-
M a L
= o
©
E l t 3/ a - (a r e a ) b a ■ x g - - E J A
(■*£-')(-+
Solving G^)n < (*)
-eiA
®
(T ) 3imultanc^>us.ly
12 E l A
V/>
=
Ma -
L3
6aiA
L*-
M B = I/aL + M a
12 E l A
&E l A
Lz
C .E IA
741. Compute the moment at the restrained end of the propped
beam shown in Fig. P-741.
Figure P-741.
Q. - Compute, the m om ent a t the restrained er»d
of
iha propped b e a m .
Solution:
Mft
2.H)
R2 = &DON
'F
Am
I
E li
= (a re a )A B
*
*
W L* _
JfC
£
R*
■ XA = o
( ¥ ) ■
°
w/--» + M g t a =
_ O
/2.0
.J
=
--{T o
w L*
743. Compute the end moments in the restrained beam described
in Problem 712 (page 292).
<2•' Compuie the cnA moments
2K»J/m
Va
£ k .y m
z-m
3m
- 3ooo (4- m
Ma.
E I O ac
- (a rea lf\c
£ - 5-
=
<?
-3 •30oo)*(f-M A)
7500 - 2 2 5 0 + 2.5 M
E l *■ c / a
-
+ 2 . 5 Me.
a
(a r& a ~ ) Cf\ ■ X C
7500 ( j - 5 ) - 2250 ( i
+ 2 , 5 M e . (’
J
Ma
=
a.
5 )
and
-57(>ti-yn ;
°
©
= O
- ? ) + 2 .5 M b ( - J • S )
=
12500 -1350 + S . 3 3 3 3 M A
S o lv in g et^n'j
-
°
+ 4-I&7 M e
= O
b
S/Mulfotneauslij
M e
= - 1524 N m
(£)
232
751.
A timber beam with a square cross section supports
loads shown in Fig. P—751. Determine the cross-sectional dimension*
the allowable flexural stress is 10 M N/m 2. What is the maximu
shearing stress developed in the beam?
2 kN
.,
1m
4 kN
U
2m
3 kN/m
-4 m-
m
Figure P-751.
Ol ■ Determine- the c-rosi - a c tio n a I dim ensions for*.
Square beam i f the allowable fltf.itr a ! stress is lOMPa..
Solution:
W L 2- _ P, abz
Ml -. M
l
-
12.
:
MX =
,
(3 ) ( 4 ) * _ 2 Q ) ( 3 )
12.
(4 ? z
(4 )2 -
- w L * _ P, Azb _
-
IZ
Lz
3 (4 )* - _
2. ( l ) 2- ( i )
12.
Pia*b
L 2_
c4 ) *
4 ( Z ) Z(~Z-)"
C 4 ) 2-
■=
-&.38kfi-m
&M
gA l
( f o r square.)
'
~bT2 '
b3
h= b
^
*
L2
-7-13 k il-m ( Use ibis')
Mi =
M r -
_ Pzabz.
L*
iF V P r _
h = V <T
3A
(7-13 x /Oi )
J
,o x /O*
I
233
h =
0, I b Z
y z
Ik Z nr\ m
-
m
i>.3& = - 7 - ! 3 + + V l
VL - 1 .6 9 x
MAX
T
=
-
/ o s N (m a x im u m sh e a r)
V
-J _£
2
- Z(3) - 4 ( Z ) - 3(4)Z
A
d - w * le>s)
( D. l£>2) *
5S4- x
/oJ
N/<
752.
Using Table 7-2, check the values o f end moment and
midspan deflection for the restrained beam in Problem 713 (page 293).
Q_: D e t e r m i n e - f h e e*id m o m & n f c a n d
m id span d eflection .
Solution-.
P
P
M
a
.......
1
{ — >
A
Va
=
m a
.
m b
l v B =
P
z
=
P a (ia )z
_
'
^
-
_
1 J *y ~
^
Ma
= Me
=
E1S
=
Z-
E IZ
=
_
2*
(J * ? *
ZPa
—J ~
f t 2-
S Pa. -
P (2 »)«?
(S L
- 4b)
p
799. An S130 x 22 steel beam 4 m long carries a load varying
uniformly from zero at the left end to 15 kN/m at the right end. Th*
beam is perfectly restrained against rotation at its ends, but the right
end settles 10 mm relative to the left end. Determine the ratio of th*
maximum flexural stress to the flexural stress if no settlement had
occurred. Use E - 200 GPa.
Ans. 1(to
(2-- D e t e r m in e - fhe. r a t i o o f m a X / m u m -fle xu ra l
'S ire a t u f U x u r a ! stre ss i f
SettI*,m e n i
had occurredSolution:
Given -
E
= 200 GPfy
S 130 A 2-Z beam :
I
=
6 .S 3 x
/ O * rrim
4 ( f ™ ni -fable.)
11= I7ll /.
M
_
- ( is X ! Q 3 ') ( 4 - ') :l
_ 6 (Z O O X I 0 9)((,.5 3 X lO (’) ( - O l ')
3o
42
- iz .7 5 kA/- m
&E T A
L2
m b
=
~(IS X
I O * ) ( 4 ' ) 2-
---------- — ---------- *
M b
=
6 ( 2 0 0 X ! 0 9)(&.5S X / 0 ~ * ) ( 0 . 0 / ' )
------------------- ^ -----------------------
- 7 . 2 5 k rf-y y ,
W ithout se-ttlc.me.nt:
20
M
= Mb =
~ /2..00 kfi ■m
8
('ontinuous Beams
■•3.
See Fig. P-803.
Arts.
55wL3
Solution:
B y sy m m etry :
R, = x *
=
=
WL
¥
6A b
6A a
-
~
*
b
[~A/L *»
L
l_ 3 2 .
SW L
32
_
W t. * 1
I
qz. J
236
7. See Fig. P-807. Solve by combining the results for Problem*
SOS and 806.
6 00 N/m -
Solution.
M’ fOON-m
A
.
too #/m
low rift
+) Z M a
50 0 -
K0 ( s )
s 0
/OO/V
4m
1 J 1*
+ 7“ 27 f^y
— Q
too
■=. o
1
Ra
=
-I [ i
_
zizoo
£ ± I
=
A .
, . -r
£>Ab
s_.
b *3
L
O
Rp s
m » won- «
IN
=
(i) -
=
/ o o /*
"X t)]
N -m *
For
L
I ~s-co ( H ) ( - £ ) - - t ( 5 ) ( S 0 0 ) ( - § - ) j
for 1 * 1.
2500
N - m 7-
Spo
_ay
L
+Ht
Sw
-s o o
237
r
Rp- 720 H
R a= ISON
- q oou-m
£)-ZM a
=
<9
(3 )(tO O )(z
+ -f* % 3 )
Kp
+ T Z F !f
>?A
=
-
/ T p fs )
=
O
7-2.0 N
=
+
7Z<?
-
Ra
=
/ fid
O
*
£r S- ' - f J j f s V ^ X i *
SH 3 6
JL
^
=
f
N - / H *
jjr C s X -’ * * ) ( - t *
6/4 6
L
Hom
N ■ &>"*•
Com bining
the.
r e s u l t s , W<2 ft a v fi .'
=
Z 2 £50 + S H 3 b
=.
76J6
A/ /n*
=
Z 3 0 0
=
(63/^
/y • /n-*
+
HotH-
238
814.
See Fig. P-814.
Arts.
600 N
lm t
M 2 - - 262 N
800 N
2m
700 N
|r l m
|
Flgur*
Solution:
000fJ
bOON
-4 -
SfWtlrZrW I
400 K
Spa* z= 4»l
the. t r ia n g u la r l o a d i n g -the in t e n s it y i
a t s u p p o r t R t is : o b t a i n e d k>y r a tio a n d
p r o p o r t io n .
i
_
/2 g o
3
S'
f
(* X
M3 = - l o o
=
7 2 0 A%
72 o )(-j
( i)
X j )
-7 0 0
n
=
-10 8 0 N-nj
•m
w ritin g t h e t h r e e - m o m e n t- ec^uat-ion b&ttoe&n
s p a n s / a/7«y z , v je h a v e .
M , i- , +
m t
( l , t LZ )
t
Lz
+ ■
+
*- #
bz
=
- r*)
Z
lo a d in g in &pan I , w e <ca/> rA&of\/e
it in to a u n ifo rm lo a d in g o-f 7 7.0 N/m a n d a
tr ia n g u la r lo a d in g .
2J9
1 2 0 0 MM
Ito %
720 m/mI
“480 n/m
j
L,
~
H
4,0
_
7ZC>(ZY
~
H
60
z ) 3
G/\ /
=
~ T T "
t q s -z
tA z k z ,
N ■m
=
( Lz -b z )
&Az t’z —
— 6 (5 o
£ut>sti-t-uh'ng
tf ■rr>
fhe.&e
v a lv e s
i'n+o
( Z)
ZM z f 2 f ^ ) -
w ill
give. :
-
1 09 0
+
Mz
*
€> I S O
=
—
O
\- Z & I . e 3
N •m
817. See Fig. P-817.
Flguraa P-S17 and P -«1 8 .
TOO ( h ) + n s z
240
Solution:
m - izoo N-m
1400N
t---------
000%
3m
'
2rrt ' v IW
f= 4 m
r
"V l j im
'
Rr
R.
^
--------— Span 2*3m— }|
F ro m th e
d e f in it io n o f b e n d in g m o rn e n tit Ms
= O . W ritin g +h& -three. - mom-e-nt- etfr cation b e tw e e n s p a n s / a n d -z , w e h a ve :
M,L,
t 2 M Z ( L, + L i )
- *? b *
=
+
V 3 LZ
+ ■? * ' 3 >
*■/
+
o - ( a )
L -2.
M,
-
-600 ( i )
Fo r- s p a n
s
-11.00
aj - t n
I :
J
*,
i m
=
£>
Y ( * ) ( * * < > ) ( I + T * 3)
-
/?a
+
=
R i
= “100 N
t
=
o
=
6
-
-300 N
R,
t r 1- “ - $ [ k ( • * ' * « » ( *
(/200 ) ( I +
■6^ ' a>.
<oA1 t>:
=
5-0/0 N - m ' 2-
o
X s jj
241
( f ]
* t>'z, —
— z + o o N m ’z
L-i
S ubs-hhu i-in g
■M<es.<fi va lu e s ,
info
E t f * .(d ) u*M
give- •'
— 1 1 o o ( H ) + 2 M * ( H * s ) + £TO>o - z HOO = O
Mz
IS & .H 3
—
N -m
820. Solve Problem 819 if the concentrated load is replaced by a
uniformly distributed load o fw N / m over the first span.
Ans.
= —
2 (a
h ’L 2
+ P)
4
4(1 + « ) ( a + f i ) — a 2
wL2
a
4
4(1 + a )(a + f i) — a2
\f m
m + _____ __________________________
Solution:
w K/m ]
1V-
L
— & v a - # fc - pL
*Kx
I?.
iRs
*4
fc.Sp3n1 -^)Spem2|^-5pah3 —>j
virirtnq -the. -th re e m o m en t- e q u a tio n b e -tu j<een
Spans. I a n d 2 , anti b e t iv e e n spans, z and 3
w e
/7a v e :
A f,*.,
y- 2 M j ( L, + L z ' ) +
M 3 ^z
' 3 ,_ -fLi
M ji;
■/• Z M 3 ( ^ 2 * £ 3 ) +• A f * Z - 3
+ — A * a * ■+
t-1
- * A3 *3
^3
=
(b)
Prom th e d e f-inih'on o f bending m o m e n t-, A7,=
Atf « O- He.nc& e q u a tio n s m W-i a n d Mz t iohioh
ca n b e &o!ved uShen t t a va lve s of- S A a / l .
242
a nd &A ~E / l ■f'or' th e g iv e n lo a d in g s a re k no con
tf&inq Ta-hl-e. 0 - 1 , u>e oom ptd-e -+h&s.e v a lu e r ,
H e n c e , a q u a tio n s ( a ) a n d ( b ) ca n be a p a i r
o f s i m u lt a n e o u s e ^ u tt-lio n s in m z a n d m $
w h ic h c a n b e s o lv e d w h a n th e v a lv e s o f'
6 A a / l a n d 6 A b / t f o r the g iv e n lo a d in g s
a r e A no tun&A, a,
& A x b-i.
<*>L 3
__
i-z.
S u b s titu tin g
o>e have. :
_
6A-Z a -i
_
6>A 3 t >3
~
L-Z
~
ZTi
th e values.
m+-o Er^ -S -. ( a ) a n d ( b )
Z M : ( L + o ( L ) + M s ( c C l.')-—
M2 ( < L )
+ -ZfAz C < L + ^ L . )
P ivid in g eufroah'ons ( c ) a n d ( d )
•Z M i
( I + «C ) +
M% «C
OCM-z -h Z M 3 C *
^
-
O — C c)
=
0 -
(d )
by L. , ooe. h a v e :
j / 7L
t — —----
=
0 — (6 )
+ Is >
=
O - (f)
F rom a j* . ( f )
m ,
-
Substituting
~ * A1g
Z (<*.+?)
Into B ’c^- ( e )
Z M z ( I + oC) + t< (
VM z f
Mz jj/
/ * * ) r * + ? ') -
MZ 0
«*
tOL
+
H
+ -* > ± L .
=
MZ +
(* )
ft<. + p )
=
( / + « . )(< * . + ? ) - o i j j
<5
0
= = J £ t l (* )
(oC + ( * ')
24.1
Mz
and
-U i L .1
—
2. ( oC + P )
V ( / + oC.)C< + f i ) - d 2
Ms ■
Ml *
-
-
Ms
=
+ <0
M3
OUL'
s
829. A uniform load is carried over three equal spans as shown
in Fig. P-829.
wj 2
Ans.
M - M3-
-----jQ -; R t = R4 - 0.4wL; R2 - jR3 * 1.1wL
u/N/m
L
1
L
/^
x>
'
» ■ *
t
*3
*1
Figure P-829.
Solution:
w
"•
L
A
N/m
L
"
1
R.
R,
Rz.
|^— 5pa« 1 - > fc - Span 2 —
R*
Sp^nS ->|
Simpif beam
regciiori
O.ftfL O.fwL
Coufie reactin')
0
MWI.
D.W* 0-9*)L
Vferti»l sh«r
0.5U)1 0.5h)L
0
0-iwL
O.fki
o.5*JL H W t
OjH
W riting
-three - mom&n-h e q u a tio n
spans r
w e /?ave
2 , s/jc^
:
spans
b&f-uoee/r
2
3,
244
M ,i,
+
2M*
( L, + L i )
=
M * /■a ■+
<? -
ZM-& ( +
6/1 3 ^ 3
+
Lz
g/>, a ,
+
r-a )
L -g ') +■ yitf. Z-3 V-
=
0 -
(!> -)
F ro m +h& d-ef-mifaon o f t>-&nding m o m & n -h , M i
- Aitf - o • &ot>stihrtin^ in fo f ^ - s . ( a ) a n d { £ ,)
u>& h a v e :
c> y
2M Z ( l + l )
f-
3l
m
+
v- -A^ 3 *r
t
MZL
r
a -
z m 3 ( l. + l )
r*)
+ o
+ -^ 3
*
_**L I
r * ;
Ht/!z.L
M z*.
M 3 i.
tit)! 3
*• — ------
WM3 L
+
+
+
2
=.
(5 — f '. g )
=
*
AJi. 3
-
r/ >
£ & N in g ^ ig 'S . ( e ) a n d ( f ) sim oU-an& ousjij -fior
-the. +tue> unknow ns. M-z a n d M 3 0 /‘v«fts -*
M fc
= — 0 . / *AZ. *■
M 3
=
M^
L-
—■<?./ a) L 2Ms
0 , / 60L *' -
(5
O- t "> L t -
O
n;
=
K'3
— ----------—-----------
M an
V
-
O ■ (? <*>L
*, =
O- H toL
=
(. I 101—
£?. I coL
= . £ > ■ ! t oL
245
To d e t-e .rm in e -the- m a x im u m m o m e n t- M -Pnom
fh& s h e a r d ia g r a m , toe. have. ( ra h o a n d p r o ­
p o rtio n ) .
O. Co UlL
X
I X
W&x
M
=
X
o. -y L
= - i- ( O H L ) ( o . ‘t-L ') -
O. oe> coL 2
844. In the propped beam shown in Fig. P-844, determine the
prop reaction.
6 kN
Figure P-844.
Solution:
7 ??e
p&H>ec.Htj f in & d e n d is c o n s id e r e d &cfoi—
v a le n t- to im a q in a rx ] s p a n z. . tonitin^ th e
th r e e - m om ent- e<fruatiZm b e tto e e h s p a n s 2
a n d z / U3& have. :
M> L, +
ZM -z ( L , + L ? ) +■ M 3 Lz. t
<2>Az. b-z
In a p p ly in g th is
te rm
re fe r rin g
M ,
m
6 -
&A, a,
(9 )
e«jj-o a b 'o r) , U3e n e g l-e c -t any
to -the im a g in a r y s p a n .
~ <Z (
!■ s )
=
— °!
kN - m
246
± •
F o r sp a n
+ )
Z
M
,
r
^
4KM/W
- /?2 ( s )
=
O
( ~ T * 3 ) + lf
o
=
Rz •=
R i= 4 K *
R z =2KN
£.f=y
*-f
Z
-
2. k N
O
R,
+
)
/?,
-^Krwir
=
=
<5
^Ar/V
|OKk- «
zni
-fcKN-
jrrf dtqrte, <Uirye.
-
~Th&r-e.fors
(s )
,
+
s ubxhf-uHng
2 M
z
t
( j ) f
O
^
=
3 . £ > f c ^
M2
-
Z M L
- V -
£
+
R,
s
k N • m *•
into Eif.. (a ') usil! give :
2
M
=
/
& .3 6
S
. 3 6
+
<3
=r
k N ■m
- ~ r ( 3 ) ( ¥ ) ( Z + -J -X
s )
£>
3 )
/?,(*)
= ’ iz. srsze
kN
848. Determine the support moments and reactions for the beam
shown in Fig. P-848.
9kN
Ftgur* P-848.
24/
Solution:
(5p*« o)
fcf&n 1)
(fpw z)
(5 p n i )
(.imaginary)
R4
(iwa^ironj)
T h e p e r f e c t ly -f ix e d ends, a r e c o n s id e r e d
ecysivale-rrt- to im a g in a ry
spans o
a n d 3 . lOn'i~ing -the. -three - m o m e n t- e<froah'on -for spans,
o
a n d i , f o r s p a n s ± a n d z , a n d -For- sp a n s. -Z
9h d 3 , u>e h a v e •
V 0L0 +
=
hi, L , +
2M-2 ( L , + l z )
&A -Z. ^*2
A72 Z.z
6 A o
2M , ( L 0 f L , ) + M z L , +
+■ Z M s
O -
( L2 f
(a ~ )
+■ M s i z
=
o -
To
+ - ~ y — — t-
(t o
Z-3 > V- M ^ i - 3
i-Z
6
(c)
L 3.
In a p p ly in g t t e s e etfroahoos. , u>e netgleot- ant)
t e r m r e f e r r i n g to t h e im a g in a r y s p a n s . U s in g
T & b te 0 - / , uxs n e x t- c o m p u te , t h e f o llo w in g
• v a lu e s :
6 A , a,
i;
7
j&A z
^ 2.
3z
L-z
&
6,0
"&cT
m L 3
_
7
60
kAI- m 7/2 . &
/cA/-
■
3
IOL 3
^
6
60
w L3
*
248
( 3* - Z l )
g A z t?2
L7
_
~
3
.
^
L
h
/ / 2 _ fc-2 ) —
1
'
/)
$*. * £ * .
=
3 7 .5 -
Z/^3)3 ,
( 3 * - ! * )
W . m Z
l~2
tn -the
Sut>&tiH>Hnq ^ f t e s e
v a lu e s ,
e n t e tfr o a h 'o n s ^ / v < a s ;
Fnsm
E % -( 3 )
Ftr>m
e - f . ( k>) M , ( s )
-H ir e e - m o m ­
+ \?.6>
Z M , f 3 ) + M z /”3j)
=
o
-(d )
+ Z M 2 (''fc )
-A M s
r 3 _ ) •#•
y- 3 7 S '
=r 0
- ^e)
/■V ^
P ro /7 7
+
) M z f 3 )
+- V 3 . S = 0
2 M 3 ('S ')
- ( f )
&o/ving
g iv & s ■
&c^oa H'ons*
s im u !/-a ne.ous>ltj
M,
~
-
0 -7 7 5 *
A :A/ -
A^2
—
—
Z .
A A/ • m
/*75
=
—
S ’.
-Hres-e.
6 5 "
ZS *
/cA/ ■ m
Jm
5»mple
heart
reaction
2.0
r e a c tio n
- o - fc z y
i.j? r
(fcuple
\krti <M
_ ✓
/?,
s h «a r
=.
Z
65-
3m
4 .0
-
fc.O
«?.0
o ,(.a f
-1.632
4.C-Z5
W ot
1-097.
I0.C& 2
<3 .7 7 5 -
---------------------------ST. 4 z S *
-
2 . £ s-S "
*z
.-. /?,
/.
/?z
=
/?3
=1
3
7 5 “
^
^ _
...
=
0. fc z r
ita»j
=
/ .fl3 Z
Ar#
* a;
6 zs- +
10.03
fn
Z
H . 10 Q
=
3. S ’3 2
A rt
854. Solve for the moments over thr •u|>|mui« m ili»
•• shown in Fig. P-854.
1
w L2
A/, - M 4 - + - j j ~
An*.
p
2^
w l.‘
- M , ---------^
:
2
2+ „
w N/m
1>
aL
V
aL
L
1«
J,
FlgurM P -M 4 and P-S55.
S o lu tio n :
W riting t h e
A flve.M 0 L0
+
-Hirae - m o m e n t-
Z M ,(t-o
+ L , )
6>A , t>,
i- ,
M ,L,
+
6A / 3/
S ut*s.Hfwh'ng in to
0 +
2M, ( 0 +
o r ••
2M ,
and
M , (r t i.
_
td L "3
=
_ _ _
£ < y -.s .
*
('a
L )
-Z M 2
( & ') ,
-
M
M z
A
( b ) , u>&
have :
)
+ tCL)
—
3
_ a
M i o ii.
( L
—+
6 A 1 t>i
-fM * t(.L. + O +
eOL3
Fro m
-h —
(b ')
and M z
*•
+
(a )
o -
s y m m e t r y ,M , —
G A z t>z
-
+■ - $ A f 3o +
V- L - z ) + M 3 L z
6 A z b>z
By
+ M z L
o
2M 2 f L ,
Hons. , w e
«= o
O
-- ( c )
+ O
+
M z *
O
— ( d )
~ 2W
«=0
+
250
Sufr&H+v-ting
M , e(.L
M,
mho
Btp ( d ) , to o g e -f" ;
COL3
-
( L. + o U ) - 2 M , L. -h ■
ZVM L .
=
*
1
/ Z
and:
Mz
— o
2
fo C
C O L '* -
Mz
17.
IVL*
/z
-
—
2
y- t f.
2
2
«C
860. Determine the value of £75 at the end of the overhang ami
midway between the supports for the beam shown in Fig. P-860.
w N/m
Jsr
Figure P-860.
Solution:
N/«
k -----a. -------- — b -*-^J
K
Irz
ZTMc
= o
Mb ( ' z ~ ) ~ * ‘ ( a ) ~
R,
=
°
23
za
M-
+he
Z-z
=
ub ^■
end
o f - +b&
■za + b
7.3
o\e-r-h-a.n^ :
W riting -the. ttir+ & - m o m e n t
u»een a , C a n d P , utc. h a v e ;
nation
berh-
251
M A L\
+
Z M
( L,
t
-r L z " ) +
Mp Lz
+
+
e«r-z7 - -it-)
#3 d e f - i m 'H o n , M a
— °
« * / *#
_
*
ZAi bi
_
0>L*
Lz
Me
-
—
hp —
Sp
h fi, =
» b 3
H
oob
—
~
r-f) ■
H
U )b ^
_
O
J h e r a -fo re . , s v b & ,t i H j h n q
0 + 2
L , «= a , L z = £ .
snd
( ~ ^
■) ( a
in to
+ b~) + 0
w ill g iv e ■
3
+ 0
+
+
3 b )
6 £ I f ‘, - 4 r )
—
u>abz —
<W- :
u b 3
ETSp
M id usay
=
<u£>3
Z*
t?&+u>&en t h e
(
s u p p o r t s -:
W ritin g the. t h r e e — m o m -e rrt ezfroab'on
w ee n A , B a n d C , toe. h a v e :
_
6EI
*
*•*
=
>
II
M AL, * Z M b (!L , * L i ‘) + M C L Z
> ,c
m
8 *
6 A , A ,
A
M e
=
2 M
Z M
=
2
-
l
l
=
=
-
IV *1
w bz
2 a
4
vvb z
uvb z
2 a
6Aibi
b e t­
252
*5H b s + i + v f i n g i n t o
0 + 0 -
Ee^- ( t> ) W 9 w ill h a v e
-S e
6EX
E IS b
a/ z ■)
J =
(■ a/z.
a
LOQz 1>'Z
'
3Z
. Determine the midspan value of EIS for the beam shown in
866
Fig. P-866.
Figure P-866.
Solution:
A l * 4 f r 0 OH-m
1200^
m
‘■K“ ir^ 3m
&
Ri » ? ooK
Hi* ioooii
j*------SfBK 1-^rt)------^ -----Sfiati
-
veoo
Rz
=
+*[*
- h - L ( 3 ) ( i 2 . o o ' ) ( s - + -z) -
lO O O
N
=. o
( 3 V I ■zoo') +
R,
Ifn ------
=
8 0 0
N
to GO
«=• O
rz
(e ^ -o
I
n i
W ritin g t h e -tb ra a - m o m e n t e q u a t io n
A, 6 a n d C , tots h a v e :
2 M g ( L , + L 2. ) +■
Ma L , +
6 £ I
^•2
F^*/" s p a ^
+■
>
tm|
= eoo
l -l
=
/-Z «
/%
=
O
<e>A . a
-
J
±
. ( 3
-
^600
—
^
a
*
-
L . 1 )
Q*
Li
_
<SAX Sk
—
6
0
0
N
- m
F&r span 2 :
Me *0
^ S -O
JEF^y
*«■<> -
+ 4 £- ')
i- J /
— - u t o o si-m
h <soo
VrT7
Z
(S A ^ 3^
=
6 A , g,
1 :
Me
/?3
b e tw e e n
=
2 f
/V
(3
11.06) +
R+
-
O
25 4
-
I 3 -5"O hi
6A
£ A _ z _ t z ;_ _
s ^ e s ’ N m z
Lz
=
A*
=
Si
S u b s.ti-t-u b n q r t e s e va/u-e-s thbo +h& -tbr&& —
m o m & n 't S ^ u & b 'o n , toe. A s v e :
O +
Z ( - IHOO } ( h + *•) + 6
51& S-
-
+ ( -
‘+ 6 0 0 ')
+
SEX
(
E I $
b
x=
-
7^<3 S '
£ IS b
-
-TOOS’
A/-/TJ3
N - m 3 ( v p ")
869. Find the value of £76 at the center of the first span of the
continuous beam in Fig. P-869 if it is known that M 2 = — 2040 N • m
and A/3 - —2810 N •m.
Ans. EI8 = 120 N m3 up
6kN 3 kN
Figure P-S89.
Solution:
<-«N JKN
lr»i
A /|i Nim ’ zm /|>C
' T
'
Rt
0PM/m
5m
R* Mjs -lO'IO N'ffl
Mj *'2810
R*
255
v jr iu n q the. r t r e e - m om ent- e q u a t io n
A, B a n d c ., u>e. h a ve •
M+L,
+
2 M S ( L, + / - z ) +
b e tw e e n
M c.L z. +
+l-ti
Me
=
M z
-2 0 *0 =
«=
K, ( * )
( z ) (
R,
—
-3 -3 0
Me =
=
39<J
MA
=
O
=.
ZAf ^
- ^ ■ ( 2 ) ( ‘T O O ) ( 2 +
9 0 0 ) (
Z. X - j - )
N
(2.)
-
-
6 )0 0
l&o
N -m
__
(j>A ,a ,
l
=
N - m 7L z
=
— he.
S u b s h 'fv + t'h g
h & se .:
0 + 2 .
-
N -m
K, ( i ) - ~ ( Z ) ( 3 6 0 ' ) f-| 9
=
M e
L,
- -ZOH-O
rs Z m
—$ B
th & s e
w a iv e s
( ! & o ) ( 2 + 2 .) -
ETSb
—
£XSb
—
-1 2 0
in fo
E e f .(a )
2.0*f-0 ( - z )
+ <=0,0 +
N- m s
12.0 N - m *
( vp)
Loe
9
Combined Stresses
905. A wooden beam 100 mm by 200 mm, supported as shown in
I ig. P-905, carries a load P. What is the iargest safe value of P if the
maximum stress is not to exceed 10 MPa?
^ L
200 mm
Figure P-905.
0.6P
L-P
0-0 P
M-O.iPfp.l)
M= o.ogp
3R, + y P(<>- l ) - -p-P-0
f
R,
= <5.17 3 3 P
ZFH
«
<5
O .l 7 ? P
M -0
I
S
r?2f]
=
o . h z -i P
CV)
2 P x ST 6
ff2x = -J-P
JwA -
0-1 f<?-2)3
/2
258
O-H Z IP
K
o . z h -i r
I na =
6.67 X
m*
w\-D
(N-rft)
A = o.i (o .z ) -
m2
T h e m a x im u m sfress oc.ex>rs jc i& t t o +H&
r i g h t o-f -the. a p p lie d lo a d .
a x ia l lo a d
O . & P u>hic.h is a c o m p r e s s i v e t o a d .
T h e re -t-o re . , th e . rnBK ■ s t r e s s , i s :
is.
Me.
— Pa
^hlA
P
O.&P
o . o -z
=
7
_
0 . H Z 1 P ('O .l')
& . 47 X to
IcN
907. Determine the largest load P that can be supported by
platform o f the cast-iron bracket shown in Fig. P-907 if a, <
MN/m * and oc < 70 M N/m 2.
Arts. P - 32.9
Solution:
JUJltilUilJltl
S /t
—
30 M Pa
Sj c
£z
70 M Pa
— Z0 x l o 6 mm v
Sin e a
-Hie m o m e n t- a t A - B i s n e g a t iv e , the.
b e a m i s b e n t- l i k e th is r((| , - t h e r e f o r e A
/* /Vj t e n s io n e n d 8 i s in c o m p r e s s i o n . P a i s
t e n s i l e a n d is e ^ u a ! to P.
St A •
SA =
Pa
A
£ js f
—
- I O * IG fc _
25*>
20 x to
=
p
=
at
£>. 3 e ( o . os')
e a o o x /o *
s '* . 3
20
X
JO - *
kN
B:
•s B =
7(5 X /O6 =
Pa
A
E na
O ■3 P ( 0. i s )
p
8 000 x /<j-fc
P
=
3 Z . “J
P
a
3 2 . <? « N
2 -0 X
1 6 - 6
kN
915. A concrete dam has the profile shown in Fig. P-915. If the
density o f concrete is 2400 kg/m3 and that of water is 1000 kg/m3,
determine the maximum compressive stress on section m - h if the depth
of the water behind the dam is h ■» 15 m.
Solution:
NA.
,
6
X
2s ) ( l )
260
i ay
=
=
*>,
/2
/. 7 6 ¥ - K IO £ AJ
u)z_ =r
I WA =. &0-lSrrj‘i-
■ZHOO ( 1 8 ) ( 3 X
* * ) ( ! )
A
= / f*j)
Ulz_ =
I. 7 6 ^ y- IC>& N
Tr
Pave
=
A
R a ve
=
±-e3h;*--TP3h
\ F
=
~
F
s
( 1 0 0 0 ) ( <r.a)
( i s ) ( IS
=
x 1)
/. t o - z s x
ia & N
- It/, ( O S ) + U)z ( 3 ) -
F ( s)
5= - l . t k H
3 ( /-7 **
(O S ) x /0*4X
I O * > ) — S ( I. I O Z S -
% to *>)
M
=
x 10 *
-i.iozs-
N -m
~Th& n e g a t i v e S ig n in d i c.at&S. +b31- + fre b& Q m
is bsrt-f- tik e . + h is t ( ( |
, +h e .re .f-o re m_ is in + & n s io n a n d Jt_ is in e-am pr& xs i o n .
Pa
=
<*>, + u-z
=
«
( i ~ti>H-
x i<54 ) = .
S . S 2 © X IO &
( c .»m y r -& s ,& iy a ')
Sc
-
-
Pa
A
3.s z e
Me.
I HA
k i«‘
I. l o z s r X I O b ( H . s )
e>o. is S,
=
- t+7 H k P a
917. The structure shown in Fig. P-917 is hinged to fixed sup­
ports at A and E. Compute the maximum compressive stress developed
in bar BDE if its cross section is 200 mm square. Neglect the weights of
the members.
Solution:
+) z; m £
=
o
- 6 0 ( 2.) + & A y + 6
Ay
ZFg
= O
(" t o ')
O
=z
O
Ay + E (j =■ 60
&.y
=. 60 k N
7 ) JTM g
=
0
E x f 3 ) + 60 ( z ) - 60 O )
= O
Hx
g x
s
4-0 ArAJ
=-
0
=
H O kN
TF -y
—
O
<oO k N
T h « fo rce s.
a c tin g
on
Ey
-f B y - 6 0
By
= o
BD£ a re :
® O
262
M b x • M oc-c.urs
M -
2V f z . s )
— 60
m ax .
D :
■&+■
kN
S
6>0 k N -
m
Jv&h b e fo re . pt-.
p from
£ ( &>mpre.&sive. & > rce )
-
~. Fa-------------f>
Z na
p> =. ( o . z . ) * o . o n ­ to •
( 0.2.)**
bh*
~~Tz~ =
Xn* ~
max.
•L - /- z
"
= - . -6 a ' g o °
_
=
M Pa
d ■OH-
-
-V6..7/
=
1-333
X to
.11
- 6 ° . * . “ >3 ( 0 - 0 .
X
1.33 3
IO ~ ^
919. From the data in Problem 918, what additional load applied
at the centroid is necessary so that no tensile stress exists anywhere on
the cross section?
Arts. 112 kN
Solution:
B
l. * fM Pa
St
*|;
St
11
fa •= H&omm
+
- p
A
+
( f*e x ) X
X *
st
- t> +
A
p
f\
e o , OOO
o. z ( o .
M x(lT )
bh*
IZ
6 My
bh*
hb 3
/Z
6Mx
hb*
261
fe ( * 0 ,0 0 0
_
~Th*
"
^ MPa
<3Z ( O ' 4* } *
6M X
bfc2
6
r go< ^ og> * o -o fe )
O.H (o-T.)' *-
I . & M Pg
A
s
C
1. 0 0
— 1.00
- I - 00
-
+ o.bo
+ O-bO
-Q .bO
- o.t>o
St
p
A
-
6Mi)
t/7Z
x 0 . 0+ )
<SMx
P
1. 00
-
1.90
+
I- 80
+ 1-80
— /. 0O
-
2.20
+ I. H-0
+ 0.-Z.0
-
3 . HO
ra tio a n d p r o p o r t io n o-f- st/-&ss + ria n g le stoe. h ave :
3 ,4 M Pa
P
/• ¥
<?>" X
=
2. Z
0. Z
0.Z -X
i/
=5 /.
M at .
M Pa
T h e r e fo r e > +h e
S t — - I t o M Pa
a nyu ih erc. o v e r
Pa
Si
Fa
=
or y
0 • <37 7 0 m
3. *
0. - Z. - JJ
=z
a-^ c o r n e r
o. o i i i
m
&
a d d itio n a l l o a d Pa mus.t- caose
to e lim in a t e t e n s il e s t r e s s
+he c.ros& - s e c t i o n ■
Si A
—
I- **o x IO 6 ( o .- z X i . » )
(1 2 , OOO H
928. A small block is 50 mm long, 30 mm high, and 10 mm thick.
The block is subjected to uniformly distributed tensile forces having the
resultant values shown in Fig. P-928. Determine the stress components
developed along the diagonal AB.
Ans. o = 83.9 MPa; r ------39.7 MPa
264
tan 6
=
&
=
4£T
SO
—
0 .6
3 0 .1 6 °
•V 2 0 k N
Solution:
<
_
‘tS ’OQ0
* ~ o . o s (o . o i )
=s i s o b\pa
= f.o s fo ^ o i)
"The
»
c o o r d in a t e s o f M o h r 's
c-irc-te
A ( is o , 0 ) on -the. x - -f-aoe
B (
6o, o )
on t h e
are. :
y - /ace
J o in in g -the +too p o in t s coil/ grive th e d ia m e —
•ter of- th e M o h r 's cj'ro le w h o s e o e -n te r is a t
£ ( IO S , o ) . T h e r a d iu s BC. is to S - 60 , o r M S
MPa ■ P o in t D r e p r e s e n t s the. s t r e s s co m p o ne n +s a lo n g d ia g o n a l A B , w h o s e n o r m a l is
a t t 3 o .<=1 6 0 to -the y - f a o e . The a n g le on the
c ,ir d e is p l o t t e d c-ou nt e.r cloc-k uj is.& f r o m -the
y — f a t e a n d is d o u b le t h e a n g le b e h u j& e rj
t h e a o tu a t p la n e s . T h e r e f o r e . ,
Sn — I O S 5S =
- h-s
*+ S
«/n
c-os
61.1 Z
6i.q 2*
=
9 3 . 8 7 . MPa
- 3 7 . 70 M Pa
265
934.
If an element is subjected to the state o f stress shown in Fig.
I' -934, find the principal stresses and the maximum shearing stresses.
Also determine the stress components on planes whose normals are at
45* and 135° with the X axis. Show all results on complete sketches of
the appropriate elements.
Solution:
&o MF^
ho MPa
<2»<5MPa----- x
40 MPa
M o h r ' s c-irc-te f o r th&
S t a t e , o f s t r e s s g iv e n
is shou>n b>&!ouj.
ASs
-Sx
- 6 0 M Pa
Syg
t *to MPa
Sf,
Syx
- H-o M P b
__ F
: = 7z.
E .■
/e>o IP
y S<?
Si>
Min Sr> = 7Z./IM y 33 (NT*/
75/np
Kr
' Sri
Max Sn = 72 /1 MfJj
x y-axis
The. s/’r « s s « s o n t h e X - f a o B a r e g i v e n k>y
p o i n t A cuhic-h h a s a n a t > s c i s s a o f — 60
C c o m p r e s s / v e ) a n d a n o r d i n a t e o f H o . S*<j
is <^ons- i d e r e r i p o s i t i v e , b e -c .a u s e its m o m ­
e n t - 5 <s/is<5 is tJo tzk u tis te a b o u t- t h e o e n t e r
266
o f -the e l e m e n t ■ T h e s t r e s s e s on t h e y - f a c e
3 r<s g iv e n b y p o in t- B , ushi'cJt h & s a n a b s c i ­
s s a o f <oO a n d a n o r d i n a t e o f - h o ( c o u n t e r c l o c k u i s e ) . C o in in g ft a n d s g iv e s H ie d i a ­
m e t e r o f -th e M o h r I &. c i r c le , its c e n t e r b e ­
in g m id co -a y b e to o e e n H i e a b s c is s a e , o f A
a n d B> o r a t- t h e o r ig in ■ H e n c e , t h e r a d io s
is c o m p u t e d f r o m -the r ig h t - t r i a n g l e w h o se sid e s, a re . q-o a n d &<£> • t h e r a d i u s i s 7 2 . H
M P a ■ T h e p r in c ip a l s t r e s s e s a r e g iv e n b y
p o in t s p a n d £ , i n h e r e t h e s h e a r i n g s t r e s s ­
e s a r e ze ro .
Th e re fo re ,
M ax
Sn
ss.
M in .
5n
-
7 Z .//
M Pa
j - 7 2 . ;/ M P a
T h e ra d iu s to D m a k e s a c o u n t e r c lo c k w is e
a n g le 2 e m e a s u r e d f r o m -th e r a d i o s C-8
to h ic h d e n o t e s t h e q - a x / s . f ^ r o m t h e . c i r ­
c l e , f a n ze> = t*°/6>o o . 6 G > T , a n d hence.
Z & = 3 3 . 7 a n d & — I G . 0! 0. T h e p r i n c i p a l
S + r e s . s e s a n d p l a n e s a n e a s s h o w n o n th e
d i f f e r e n t i a l & l e m ent~ ( n o s h e a r i n g s + n s s s
on t h e p n 'n c ip a / p l a n e ) .
*The s t r e s s e s on -th e p l a n e s of- m & x im o m
s h e a r i n g s t r e s s a r e g i v e n b y p o in t s p a n d
& , th e. v a lu e s b e in g M a x . £ s = 72.// M P a ,
& n d M in . S s = - 7 z . 11 M P a . T h e n o r m a l stnsss
on e a c h p la n e , is z e r o a s ca n b e s e e n fro m
th e M o h r ‘s c i r c l e a s t h e v e r t i c J e l i n e —
t h r o u g h t h e c e n t e r c o i n c i d e s ip ith t h e £ $ ~
a x is . T h e ra d iu s . c F is l o ° c o o n -h e rc lo c Jz —
w is e , -fro m C D . s o fhsit~ th e . n o r m a l t o t h e
p l a n e o f m a x im u m s h e a r i n g s t r e s s i s
t s ° c o u n t e r c l o c k t o is A f r o m t h e m a x im u m
p r i n c i p a l p l a n e , o r a t - ttS '° + !£>.
= 61.1°
utifh t h e y - a x i s . T h e p l a n e of- m a x im u m
.'ft /
eh & an 'nq *tr& & s is shocun on ttiA
J i f f mtmnh.-tl
e l& m & n t.
Cf- a x is
7 2 . II M P a
- 7 Z . i l M f^
------ - x- ams
—72. li MPa
7Z. II MPa
Te> c o m p l& r& H i a s o lu tio n t t h e s t r e s s e s o n
fhts. p l a n e r u>hos>& n o r m a l is. a t t H S *
OJith t h e x - a x fs a r e r e p r e s e n t e d
p o ir ft
H , lo c -a te c j a t the. in tA r-s e -c.H 'o n o f th-e. /-a <af/'«s <£/* u jit h M o h r ’ s zj'rcJtS. . The. a n q t e
b & tio e & n the. n o r m a ls -b a r x j -to** fa cies. is
la id off - d o u b le . s/"s«s <3/7
<zincJe. . H e n c e .,
anql-e. A C H = Z X 4fcfir* zz ° lo ° , a n d a n q le A C S
ca. <fo« - 3 3 .7 / * = £r^. 2 f » T h e r & f o n s the. c o ­
o r d i n a t e s of- p o i n t H a r & •
ST, =s - 7 2 . iJ <S£5S S-6.2*7a
=
Sfc
=: I - . S ’?. ^*7MPa
= - 72.. n sm
S b .'li 0
s m a s s is s on the. p i a n a s u>ho&e. n o r m a l
Is a t + /3 S,<J tu/V-b t h e
a r e r& p n e s rs n t
Ad by p o i n t X . Paints It e n d I a re - IQO° a p a r t
on th e cir& le. s in c e , t h e p l a n t s th e y ne.pn esertf
a re . a c -tu a U y ^ o ° a p a r t . The. c o o r d in a t e s a r e :
So
=
72.//
e o s £~6. Z ^ °
=
St
=
7 2. II
S in
=s
S6>. 7.°!°
|
02 M Pa 1
5*7. <?>?
935.
For the element shown in Fig. P-935, determine the values
of ox and oy if the principal stresses are known to be 20 MPa and —80
MPa.
268
Solution:
S x ^ - -S O M P a
£jyx — + 30 M P^
3 o MPa
-30Mfgi
"7*7^ p r i n c i p a l s f r e £ & & s . h a v e c o o r d in a t e s .
f \ ( -8 C >, o ) a n d & ( 2 0 , o ) t iohicJi a n s t h e & n d p o in t& o f th e d i a m e t e r o f t h e M o h r ' s cSncJei.
The. c e n t e r c. i s m id tva t/ b -e to o ^& n t h e a b s ­
c is s a e . o f ft a n d B , o r a t - 3 0 M P a f r o m the
o r ig in • H e n c e ., the- r a d i u s i s e o m p o -h e d +t> b &
the. le n g t h o f FkC o r
, w h i c h is £~o M P a Sxtf & gN e -n to t>e. — S O M P # & n d th e r e are.
two s o lu tio n s , one. a t p o i n t & a n d the- o t h e r
a t p o in t B . T h e re , f o r a ,
Sx, =x r ~ cr=
Sjr,
=
JSXz =
s-o -
=
l+O
so - J s o * - 3d*
=
/o
- 30 _
-
— 3 0 - n/s<3 z - 3 0 *
—
- 30 —
=
—
70
*0
~P?<s c o r r e s p o n d i n g
Sy, —
% z
Sy
values, ana :
—
-2 0 -H O -
~ 30 - C &
r ~
=
=
C -f=
SO
— H o s*
- - 70M F&
IO M P a
T h e re fo re . :
Sx
—
IO MPa.
1
and S% -
- l o MPa
,
-
*=■
70 M P a
to MP&
940. The state of stress at a point is the result of three separate
actions that produce the three states of stress shown in Fig. P-940.
Determine the principal stresses and principal planes caused by the
•uperposidon of these three stress states.
Ans. Max. a “ 37.8 MPa at 9 — 26.2°
20 MPa
40 MPa
FI0UM P-M O .
Solution:
F o r s ta te
(a ) ,
Sx
=■ "30 MPa
Sy
=
S
x
u
«
o
=
0
270
T h e M o h r ‘s. c i r c l e f o r t h e s t a i e . o f s t r e s s
(b " ) is shocon . T h e s t r e s s e s o n t h e f a c e a t
- 6 0 ° uji+h t h e x - fa ce- a r e r e p r e s e n t e d by
p o i n t fit , uJhich h a s an a b s c i s s a <jf - Ho (c o m }'
nas*z-'ve ) a n d an o r d in a t e of- z e r o . T h e s t r e s s ­
on -Hie f a c e a s + so® noi+h th e x - ax/s
g iv e n b y p a in t & , u>hich has &n a b s c is s a
o f o and an o r d in a te o f o . J o in in g fit a n d B
g iv e s -the d ia m e t e r - o f M o h r 's c i r c l e , its
c e n t e r b e in g m id toay b etto e& n t h e a b s c i s ­
s a e of- fit and B j o r a t - 2 0 M P a fr v m th e o r ­
igin. H en ce., t h e radios. is 2 0 MP& . T h e * r a c e is t & o° u> 'rf-h t h e f a c e h a vin g c o o r d in 3 t e s fit ( - ho , o ) . T h e s t r e s s e s on t h e x face. ca n b e d e t e r m i n e d f r o m th e. M o h r 's
c ir c le by ro ta tin g t t z o 0 fro m po\nt fit, a nd
g iv e n by -the c o o rd in a te s , o f p o in t P - The.
S tresses
or? -the y - fa c e . are. g iv e n b y the.
c o o r d i n a t e s o f p o in t - B . T h e r e f o r e . ,
es
are
Sx =
- 20 + r c o s
Sg
- r -
=
e>o*
r c o s 60"
—
-t o
MPa
s
- -20 - z o c o s <hoa
=.
-3 0
MPa
Sx« — — r s in <bo 0
— - z o s m <£>s° -
&yx ~
— ‘z- ° * ,n & c° — ' 7 - 3 2 .
r sln
600
- f7. s z
M Pa
M Pa
T h e M o h r 's c ir c ie f o r t h e s t a t e o f s t r e s s
( c ) is s h o w n . T h e s t r e s s e s on the. fa ce. + 6 0 °
cuith t h e x - fac&. are: r e p r e s e n t e d by poirrt A
a n d t h e s t r e s s e s on th e f a c e a t - 3 0 0 u>Hti
■the x - f a c e are. r e p r e s e n t e d b y p o in t B T h e c e n t e r c is rn id tva y b & lt o e e n th e a b s c i -
271
% , o M P a *r>m + * * o n / „-a*is
3£_/
g in •
. * *
rad,u&
jc /O M P a . T h e X - fa c e
i s a t - 6 0 ° to i+ h th A —
h a vin g c& ordin —
face:
afAS- ft ( t o .
and
2 tr,s 5 s e s a r e g/van
p o /> 7*
P . a#* -
^ v?" 7
point- & - The nh-e&sAS.
on the. t j -
9r& ~
g N & n t>U +he c o o r d ,.—
ft&tes. <?f point- 3 •
„
77, e r e
<
Sx =
;© - ; o
* » ' *
ffM P a
, „ ♦ , « « s to- =
s„
=
. *
s «x
=
6,n
* > '
=
=
- a
0 ■
“
M Ps
u>e. h-axe i
s « r e r F ° * " Z * " • * ’ rS * s te ,B S ’ "
s „ = 3 o - l o t s ’ = 2 ^ M Pa
Sa =
» . 3 » H r
= -,* -M P a
s *9
= i - n.3Z -*■** = - 2 6
MPa
69X
= o +
* *
“
M PS
•
& or fh<s r & s u t t a n i- A t r M o h r '* * i r * 'S
* ,*
co& SSS& i s 9 've n
f a d ia m & f& r .
o r d i n a t e * * e n d pom +*
A ( 2 5 -.-Z 6 )
g fw ff.z * )
272
ss
The c e n t e r c
is m id to ay b e t w e e n -the a b s c i
s s .s e o f A a n d B , o r a t -t S MF& fr o m t h e o r ig in . H e n c e t t h e ra d iu s i s r is c o m p u t e d
from t h e rig h t- tr ia n g le w h o s e s i d e s a r e 20
a n d zt> •, -the radius, i s Z 2. .& M P a . The. p r i n ­
c ip a l s t r e s s e s a re. r e p r e s e n t e d t>y p o i n t s
P a n d E , w h e r e t h e s h e a r in g s t r e s s ezoor—
d/'n&tTes a r e Ztsrw . F ro m t h e g e o m e t r y of- the.
c irc le , w e h a \e ,
M a x . S n sr O P
— S' + 3 2 . &
=
=.
( 2 *>. 2 0 )
37. e M P S
- 2 7 . 9 M Pa
( 1 1 6 . Z. • )
M in .
Sn -
&& = - 3 2 . 0 + S
The
w is e
r a d io s
a n g le
to o
m a k e s a c o u n te r c lo c k —
z& m e a s u r e d f r o m t h e ra d io s
, w h ic h d e n o t e s
cS rd e , -tan -z. »
th e
x - a xis. . F r o m t h e
= z */z£> »
/. 3 , a n d hence.
— S ”Z . lt e’ a n d & = Z £ > - 2 ° . T h e p r in c ip a l
s t r e s s e s a n d p la n e s , a r e a s s h o w n on th e
d if f e r e n t ia l e le m e n t26
273
946. Explain why the stress trajectories in Fig. 9-22 tend to
Income horizontal as they approach the wall. Where are they exactly
horizontal? What are the stress trajectories for axial tension or compres­
sion?
Solution:
S in cm . & i/9 n e s. uJi+h -Hi ■& ra tio s V k / co here
t a n -Z& = ' z S z / s f , a + -the. w a l l w h e r e ,
is m a x i m u m , © is a m in im u m . T h e a n g le &
ufi!l a p p r o a c h -z&re> a s S f b & c o m & s muc.h tar" ~
a e r -than £ s ■ S + re s s t r a j « c . t v n e s a r& e x a c -i-Iq
h o r i z o n t a l a t- t h e u p p & r a s coelt a s t h e tocoe-r e d g e s o f H ie b e a m ■ P o r a x i a l t e n s i o n
o r t^ o m p r<2s s io n , -the s t r e s s . tra j& c .fo ri< e s
w i l l t>e p a r a l l e l to t h e l o n g i t u d i n a l a x i s .
947. The solid shaft in a small hydraulic turbine is 100 mm in
dkmeter and supports an axial compressive load of 140ir kN. De­
termine the maximum power that be developed at 4 r/s without
exceeding a maximum shearing stress of 70 M N/m 2 or a maximum
normal stress o f 90 M N/m 2.
Arts. 273 kW
Solution:
d
— O. im
p
s=
/?<? rr k N
/ =. ¥ Vs
Ss £
to
Sh *
^<3 M P 9
For
C
- ____ r _ _
S ~
A “
ItO .O O C tT T '
T T (a. o s ) z
=. se> M P a
M Pa
A SS
Sn £. °10
M3X.
274
— ze - r
sz -
r
F=or
S & £.
-JO M P a
rer
70 M P a
a VZMP&
In e>rd&r nof- -to e x <2e e d
ar>y
s+rass, r =
6 2 M Pa • T h e . r e f -o r e ,
r 2=
z e z + 5t 2 =
St
s s . 32
-
-r-_ _
S* 7Pr3 _
z
~
7<r
=
/ tf,S62
P=
27T/Te
P -s .
2 7 -S fcfe/
62*
5TS-. 3 Z X IO *> X 7P X O.GS*
5
*
.0 6
/V w
2 1 T (h ) ( t o e 6 2 . 0 6 ) =
=3
272,<?^3
UJ
948.
A solid shaft 100 mm in diameter is subjected simulta­
neously to an axial compressive force of 600 kN and to a torque which
twists the shaft through an angle of 1.5° in a length of 8 m. If
G •= 80 X 109 N/m 2, compute the maximum normal and shearing
stresses in the shaft.
Ans. r = 40.4 MN/m
Solution:
r — o. o s m
Pa =
- 6 0 0 , OOO N
e
=
/. S’ *
L
=
S m
&
=
00 x to *
“
-P a
A
OrG
St
—
__
l
r Qr
=
- 7 6 . 3<?
N/ m 'z
- <£>00, 0 0 0
r r ( o . o s ) 2-
_
~
e =
(l. S x
/3.0*? M P a
) ( a . o s ) ( a o y /a**)
f to r s io n a l
s tt-ts s & )
/k
76. 3 ? MPa
nk
I3.o i
MPa
275
/N
M ax. 5JS = J ( 3 0 . /‘T'S' ) 2 -h ( 1 3 - 0 9 ) ^
V<5.38 M P a
Man.. S t
M in.
St
=
o p = - K s . /^s"
-
2 . /<=? M P a
= < ? £ ■ = • - 3 0 . /‘I S ’ =
- IS .
ho
. se
HO. 3 8
5 7 S' M P a
955.
A closed-end cylindrical pressure vessel has an external
diameter of 300 mm and is fabricated from 10-mm steel plate. If the
lank is subjected to an internal pressure of 6 M N/m 2, find the maxi­
mum torque which can also be applied if the normal stress in the walls
of the vessel is limited to 100 M N/m 2. Neglect the possibility of
buckling.
Solution:
D
-
O.Zm
b =
0. o i m
p -
6 MPa
S/7 —
t o o MPa
r
=
R - t
-
J
=
J L ( R * i - r n-) ^ J L ( o ., s + -
o. /sr - o . o / =• o . i H / n .
=
o.i h h )
/. °iie> x t o ~ ^ m *
( lo n g it u d in a l]
q o MPa
( -tange-nHa! )
276
r
2 2
.
S
2
M ax. 5n =
/tfo M P j
e v i r + /- =
/d o
r
^ 2 2-S2 +
_
J
=
-
32
S
2 3 . A^ipa
=
f 23.
=
-z ^ ^ l
-
3 0
r
T
S'
32.5"
S^=
&J~ _
f-
HS
8
1
S'
X JO 6 ) ( t - 1 1 &
<?./s-
N
-
X /P ~ ¥ )
m
k N ■m
956.
Compute the principal stresses and maximum shearing
stress at point A in Fig. P-956 at the section x — 250 mm. The beam is
rectangular, 20 mm wide by 120 mm deep, and point A is 20 mm above
the centerline of the beam. Assume the 50-kN load acts at the centroid
of the cross section. Show answers on complete sketches of appropriate
differential elements. (Hint: Be sure to include the shearing stress
caused by the applied load.)
Ans. Max. a =* 72.6 MPa at 0 =* — 12.9°
277
M
30°
50 kN
i
Flguraa P-956 and P 957.
Solution:
o.ozm
X = o. -zsrm
T
o. o&nj
<
Mr
O- OZni
\
>
0. CWn
;
^ E
^ ----- N-A.
XT 30 °
O.obm
u
S O kM
ZFx
=
0
s r o .o o o ( o . & ) -
Pa =
ZFy
£) ZM
=
O
V
=
=
O
r „ A.
_
— O
H - o . o o o N (+&nsit&)
V - S o , o o o ( o . <5?) =
O
N ( p o s -‘h'v <2 S h e a r )
30, o o o
S O , O O O f O. & ) ( O- Z S ) Mr =
Pat
M r
■=■ O
7, STOO N ■ m
frh ? _
O. O Z C o . t - 2 . ) 3
IZ
l-Z
-
z.& &
x
ro~*
m *
firT A ;
S t = J3i_ +
14
,
A
"
T
X
_
<+0,000
T jJ S O O
=
Q . O Z X o. »z
2.08
Sfc c
6 0 . 7S-
M Pa
_
_
VA g
_
3<9. OOO f t o .g -2
5 “
it
“
( 0 . 02. )
X
X O . O H - ) ( Q.OT.
2 .e g
* *
0
f^Tdzl
/O-A
x to-4
278
%
=
16. 6 7
M Pa
2
I------------------^
_____ v '
< > © . 7 s - M P a ------- x
16 . 6 t
7ft<9 M o h r ' s c ir c le , -dor H ie
a b o v e is g iv e n b-etou) .
MPa
sh a be o f- s H -e s s
TfriS o e n t e r is. b e tu ie & n th e a b s i / s s a ^
A
an<^ 3 < o r a t + 3 H . 3 8 MPa fro m th e o rig in .
H e n c e , t h e ra d io s is tzomputts, d fro m Hie- rig h rt
-triangle, w h o s e s id e s a r e 3<+. 3 6 a n d i£>. & T i
■Hie. ra d iu s is 3&. - Z/ MP& .~The p r in c ip a l s t r e s s ­
e s a r e g iv e n b y p o in t s p e n d £ , tuh c-re th e
s h e a r in g s t r e s s e s , a r e -ze.ro . Th e re C o n s . , M &t.
S n = O P — 3H-. 3& + Z & . - z l = 72. srg M Pa ; M in.
S n = O E = - 3 & . - Z 1 +■ 3 H - . S & — - 5. S 3 MF% .T h e
ra d io s to p m a k e s a doc.ku>is *2 a n g le %& #
m e a ^ u r & d f r o m H ie r a d iu s C A Ufhich d e n o H s s
■Hie x - ax/s . P r a m t h e c-frcJe , t a n
=
!6>.<s>7/ 3 ^ 3 S
o . i+ e s - a n d h& hce. -T.& =•
2LS". &e>° a n d Q — / 2 , ? 3 4 . T S « p r in c ip a l s t r e s ­
s e s a n d p la n e s are. as. s to w n
tfje. afr'ffe r& n ti'a l e le m e n t -.
M /n
/
s n
= - 3 . S 3
M a x . S ri
=■
M Pa
7 2 . £"*7
T h e s t r e s s e s c>n th e p la n e s of- m a x irrrvm
S h e a r i n g s ri-ia ss a r e g i v e n t>g p o in t s p=
pnc/ <5-,
v a lu e s , b e -in g M &x. S s = 3 a . 2/
17V
M P a , a n d M in 5>s =■ —3 0 . "2./ M P a J f/i<s n o r m ­
a l s/ ra sft ort « a < ^ p l a n s /s. + 3 4 ■ Z & M P a .
"The. r a d io s . c - F is. ° i o ° ^ o im te .r 'o J o c J z u jr s .e
fro m
C -P , s o t h e n o r m a l t e f h c p la n a , o f - m a x t~
m u m s h e a rin g stress is t5°Counterc/cxzk wise. fro m the maxi­
m um
p r i n c i p a l p la n e . , o n
a t"
* t s ° — /z ■c1 3 a *=•
3 2 . 07 ° uii+h -the. x - axis: . T b e r e s u l t s ® n s
s-h o w n on th e e l & m ozt/" E>&Jolo .
I*
X
967. A state o f strain is defined by tx = — 400 X 10 6, fy — 200
10~6, and
= 800 X IO-6. I f E = 200 GPa and »- - 0.30, de­
termine the principal stresses and stress components on the face whose
normal is at + 4 0 ° from the X axis.
A ns.
Max. a = 48.3 M P a ; min. a = — 106 M Pa;
a = - 97.2 M P a ; t = - 34.8 M P a
Solution:
= “ H -o o x
io ~ 6>
& y
=
Z
<5 X
IO ~ *
Y iy
~
& oo X
/o “ 6
£:
=.
z o o <5/^
ir
=■
o.so
x
M o h r >s c -ird e . f o r tbs. q iv & n s ta h e . orf- s t r a in
is &hou>n b e lo w . -The.
f a c t o r t o " 6* t>e.ing o m i t ­
t e d , Th-e. c o o r d in a te s
o f- p o i n t f* s n s i£x =
- H - O O a n d ~ I f y y — *+00 /
c o o rd in ­
a te s o f B a re
= Z O O a n d -£r Y*y
-*+66
The- x - axis, is r e p r » s . e n t - ^ d b y r a d iu s C .',
a n d -Hie y - a x i s bej th e r a d iu s . £. .
280
=
S OO
X
16
OCe =s -/<3<3 X IO—Co
Zfi>
ta n
=
.^ g g -
& s
J
Z<?
~
S 3 . 1 3 °
'z e ,.s < i,G ‘>
«s
=
** T ^ p r
£>s
—
y o °
i£> ~ 63 ( 2 00 x / o '* )
*
1+0.3
Rs.
f?s -
7 6 . <?z
r ««js =
r
Ex l p - 6 ( - zoo X 10* )
t - <3.3
f<?£.)s = f<?<=)s =
M o h r 's
MPa
- Z S . S 7
c-ir-cJtS.
M^a
fo r-
th e
s h -e x te s
/s q iv & n
bta/ouJ.
Max.. S» =
- 2 a . s-r + 76- <rz =
Mm. Sft =
-
M Pa
Z 0 . 5-7 -
76. <72 =
■/ o s . m
MPa
On
Sn
+n<s
=
h o ° -
f a c js :
- -2e . £ 7
-
7 6 • “f z
-
Ss
=
-7 6 . ‘JZ s/n
Z 6 . £ ^65- a =
?7. 3 7
Z fe .
M Pa 1
5-6-5-a = ; [ 3 » . • »
M P
31
10
Reinforced Beams
1002. A timber beam is reinforced with steel plates rigidly
attached at the top and bottom as shown in Fig. P-1002. By what
■mount is the moment increased by the reinforcement if n =» IS and the
allowable stresses in the wood and steel are 8 MPa and 120 MPa,
re«pectively?
Ans. 52.2 kN-m
mm n
10 mm
300 mm
10 m m
Figures P-1002, P-1003, P-1004.
Solution:
l£- X
/ S 'C —
2 Z S O mm
ta mm
3 0 0 mrr)
io mm
Suj
S S
S e fo r< s .
8 M Pa
^
f2 -0 M Pa
r e - ir fo r-c & m & rrf-■'
282
Q. -2.ST ( Q . 3 ) 3
m n -a.
—
*
-
/Z
~rh& r& s is .H n q m o m & n t
mt/m u>ood s t r e s s is :
^
w
_
“
S u, X
y
__ ( 8
=
=
S'. e>~zs x to
m t e r m s &P the. m a x i -
X. to 6 ) ( £. &■2 S *
o.is
I O~H )
3 0 , 00(5 /V- >n
fcvV* re in -forcc-tne -n t- :
•N-A.
st S’. 6ZS X /<3
+
X N.A. =
Z. Z S
/. 6 ^
p * . Z S r ■\(O.0i)'3
L
iz.
z
f o . O l ) f 0. I S 5 ) 2J
X lo - 3 m H
( i-z o x /a & ) ( ’/■ &■*+*+ x io ~ 3~)
o. /<a
Mu
__
=.
ez, zoo
_
re
—
3 7, 6So
SuiX
y
-th e re fi ons, m =
Ov&i~s.tr&&£, a n y
AM
=
sz.zo )
n ■m
X I Q * ) ( !■ &HH- X <0~3)
O. I S/V- AW
0 2 , -zo o tn o r d e r - no t- to
m a t e r ia / .
-
so,e>oc> =
s z , z o o hi m
1005. A timber beam 150 mm by 250 mm is reinforced at the
bottom only by a steel plate, as shown in Fig. P-1005. Determine the
concentrated load that can be applied at the center of a simply sup­
ported span 6 m long if n — 20, o, < 120 MPa, and ow < 8 MPa. Show
that the neutral axis is 170.2 mm below the top and that /NA — 416 x
10s mm
Ans. P - 13.1 kN
260 mm
ngurM M 0 H , K lO O t, P-1007.
^ i o o _ ^ j— T
10 mm
Solution:
T
—
r
no.-z
mm
Z s ’o mm
loj w in
T F
e ° l- e m m
2.0 x loo - -zooa mm ■)J-
Ss
^
t Z O MPa
S ia
—
S M Pa
loi+h Hi4& r e f e r fip c e
ax / s 3+" fh is
b o tf& m , y~ —
ZLAy
ZA
'
—_
a - i r C o . z s ' i C o . o i +■ o . i z s ) + z Co-Pi")
o. i s ( O . Z S ) +
y ~
2.
f O. 0 0 5 ^
(o-oi)
u =s
.00^8 m
-N-A. =
2:
— 6 9 . 3 m m -from +tis bart+om
an t i o - Z m m from +b& -hap
bh3
Ad* )
IZ.
P ./ S ’ ( Q .- Z S ) 3
1-2.
-
O.t-ZS ) *
(<9.0 B ^ e
r w.A. =
-f
O .IS ( o . - z . s ) ( o . n o -z
+ ,2 ( d g 0 1 " " +
- <5.
)2
*!>■/<£> X I O - * m *
-
H-16, X. to ** mm *
Max M -
% (3 ) =
3m
3m
f*. = %
/-7
sx
m
=
__
fe
Muj
=
----------------- M 7 0 2 ----------------------
=
/ ef S S 3
x io6 )(
. HT
2 Co o l )
x
N m
i o - * )
284
7 2 tf X /P
—) { + .!6 >
20
X I O '* )
o ■ o a <?s
=£ 2 7 , 7*? S’ . I O N m
• -M a x
M
=.
M S G S . * !
2
=
n s s -z
P
=
/3 0 3 B . 6-S*N
3P
N m
. HT
1012. An aluminum beam having the properties of a W200 X 4f.
section is reinforced by bolting 6-mm steel plates 203 mm wide to the
flanges, as shown in Fig. P-1012. The allowable stresses in the steel and
aluminum are 140 MPa and 100 MPa, respectively, and the ratio
E J E„ ■* 3. Determine (a) the percentage change in the strength of the
original unreinforced aluminum section and (b) the percentage changr
in the flexural rigidity E l.
Ans. (a) +21.4%; (b) + 175%
— 203 mm —-~»|
I -
J - L
------f
—
6 mm
cQ - .- t - :------FM L 6 mm
~
f
Figure P-1012.
Solution:
k— 3
X 203
m m -------- i
=
A\V\\\\\\\\\VV\\\\\V
2d %
ss lo l.S 'm m
W 8 X io*6m3
L ____
AWWWVVaWWWW
Ss
B s.
Ez
(a)
^
M Pa
=
i
Sg
*£
/oo M P ’a
3
U n r -A i n f -o r c . e d
ftlu m in u m
beam *’
285
<
—
^
X
Ma =
“
H Sa
R e in fo rc e d
e
«
(•+•+& x lo - ‘> )( too x 10* )
=
*ineoo
A lu m in u m
beam :
,- 6 + ^
r«.A .=
-m
n
vs-.s- x i o ~ » + 1
/"<?•
fg - ^ g )3
/z
,
+ o. 6 0 9 ( 0 .0 0 6,') ( 0.101 S’
TN.fi -
0.OO3)*J
t z s . s s * io-*> m +
' “ i* ) < ■ < « . 3 3 »
/Ws “
( O. I O I S
.0Z
;<>-«)
+■ 0 . 0 0 6 )
=
S’* ,
N- m
_
( /<?0 x W 6 )C /Z£\ 33 X /0 - 6 )
o.iots
a
=
/2L 3 , ¥ 7 7 . 8 3
Ai m
.’. ys<2 M =
S ¥ , ¥ 0 6 . 0 2 M-n?
%
/« s+retngth =
/ n c re a s^
SH
. 82 - W&co ,
-------------------------- ------- ’
¥¥800
Z /. ¥ H e/»
(b ) F o r t/nrein f a r c e d P tlu m in u m
=
For
E l
H a ( h -s . s
x tc>~6>')
r e i n - f o r c e d A lu m in u m
= £a
% /pch sas£
beam :
beam:
f /ZS-. 3 3
X to-* )
/Z5". 3 3 X l o ~ 6 — HS15
s* ------------------^ ^----------
//>
X IQ " 6
I6-&
'
=
/ 7S". VS*
%
286
1014. A rectangular section 150 mm wide by 250 mm deep resists
a bending moment of 140 kN -m. The material of the beam is noniso­
tropic with the tensile modulus of elasticity twice the compressive
modulus. Determine the maximum tensile and compressive stresses in
the section.
Ans. a, = 108 MPa; ac = 76.5 MPa
Solution:
M
— l*+o
=
STa«
kN- m
O
Z ff*
o .ts ( a . - z s r - a
)
-
e - s y f- f)
( o.zs--7j) 2 s
o . o e - z s - O.Sef
y *
5
a.-2-s- - y
■A/A
~
O - / S ' c o . I H- 6H- ) ' 3
--------- i i
f 0. /¥6^^- Z
(
UA/A
S
^
j
L
+
2y z
-
t- H '1
=
a. /(53 4> m
— d . /^6 ‘f m
,
+
01S
<5.3 ( 0. I 0 3 6 ) 3
------- ------------------ +
=
2 . £>3 / X 1 0 " ^ m *
=
z&a.i
_
Me f
*
about- Hie
A/./J.
io*> m m *
X
_
o o o (o .m m -)
2.6>&1 X / fl-¥
“
7 6 . ¥£T M Pa
287
st « 25; =
St
-
j
L
1 * 0 , O O P ( O. 1 03 6 ,)
7
2 6 0 / X 10-+ J
/ o s . -z MPa
1018. Compute the allowable total vertical shear for a beam
having the same cross section as in Problem 1005, if n = 20 and the
maximum shear stress is 800 kN/m2.
Ans. V = 23.0 kN
Solution:
Ss ^
800
11
—
(/sing -the.
kN/rn*.
20
r«-£ tV fe
erf- P - l o o s '
V A cL
V d}
xb
$s _ ( e g o x i q 3 ) ( H .m> x fo-4)
v$>
“
( a. i s x 0 . 1 7 0 Z )
(o . is )
/7*J2j
(/'O.
o . I7<
1 2
V
=
Z 2 <? 77 • O S N
V
=
z z . <?e kN
1027. Determine the maximum stresses produced in the concrete
and steel of a reinforced beam by a bending moment of 70 kN-m if
b = 300 mm, d *= 500 mm, As — 1200 mm2, and n = 8.
Ans. f c - 6.91 M N/m 2; fs = 130MN/m2
Solution:
b = o.zm
T MM
*kci
o.svn
_
NA.
o . s - kd
J.
jTTtrft M =
nA s
= af/zoo)
70 k N - m
— <?&oo /nnr2 = o. 0 0 9 6 rT)%
1
288
z .a y =
O
£>.0096 ( o .s r -k d )
=
kd
o. is m
kd = O. S’ - — (o . IS ) = O. •VStv
fc
( bkd) ( jd )
d-
Jd
Me -
=
70 X /6 3 — -£-./* f o . s x 0. /S-jf O. VS")
/c — 6. «?/ MPa
Ms =
70 X /O5 =
=
/s
0 '^ )
j s ( tz o o y. J0“ 6
*£■)
12*?. £ 3 MPa
1030. In a reinforced beam, b = 250 mm, d = 450 mm,
1400 mm2, and n -- 8. Determine the safe uniformly distributed load
that can be carried on a simply supported span 4 m long if fc < 12 MPn
and fs < 140 MPa. Assume 50 mm o f concrete below the reinforcing
steel, and include the weight o f the beam. The density o f concrete is
2400 kg/m3.
Solution:
b -
o .z s m
•3j£---------- N-A •
O .m r-K A
a = o.asm
nA$ -
R
nAs
/
& ( iH -o o )
/s
(3 :
=
2
±
1-2. M Pa
*
/*<? M Pa
^
0
0
K g / m
3
— I I 2 OO m m 2
= n ." z x, i o - 3 m '2
ay
o. z s
—
o
( k d ) (-
=
kd
=
\
/;
Z
X
IO ~ 3 ( o . ^ s - k d )
0 . 161 m
2H9
jd
=
d ~ 4 rk d
J
~
O .H S — ~ - ( O . l k l ) =
3
O.S<*6>
M
m
Mc = -j-A r (b k d )C jd )
= -~ -( 17. X lO£>) ( O- -ZS- X O. IG>1)(o.Z^Ca)
Me
-
^ S C -,3 H
N - rrt
MS =
/ s A s jd
=
IH-& X IOa ( m o o
x
/o- * )
f<?. 3 < ? G )
7 7 61 6 ,
Ms y s e
M
N ■m
=
7 7 .
6 /6
N- m
Max . M - -L (z to )(-z )
/ v ja jf.
M
=
2 0 ) =.
zw
776/6
/V
/77
2io
For concrete ,
-
(?C X A
=•
2.3 B Z O - ~ r ( O . - Z S X O.S-O)
Wc =
7here tore t Hies sa-As value, of- u> =
30 8 0 0
- 2 9 ^ 0
s
3 S S fe a
N/m
1041. A simply supported beam 6 m long is designed to carry a
concentrated load o f 80 k N at midspan. Compute b and A s for a depth
d = 600 mm using balanced-stress reinforcement with f c = S M N / m 2,
fs = 120 M N / m 2, and n = 10. A llo w 50 mm o f concrete below the steel
and include the weight o f the beam, assuming the density o f concrete is
2400 kg/m 3. (H in t: Assume an initial weight per meter fo r the beam
and check this assumption after you have found dimensions.)
Arts,
b = 278 mm; A s — 2220 mm2
290
Solution:
P - 8 0 kN
w%
&m
R «= -£ + 3 «o
t^ - ~ + 3 u i
d
=
5 .6 m
A
-
& M Pa
/s
=
1 2 0 MPa
n
=s
10
^
s
O. OB' m
=
2 *0 0
*^ / m 3
=
Z3 S 2 0
H/m 3
M ax M
± (* H i
■ 3 * > + Z .)
A 7ax
M
ss ^ . ( p
+ 3oa)
Max M ~ -J-p v -|-w
M
-
I -2 0 , 0 0 0
+ J -L uj
N -m
z
r
A = &MF&
T
_
<o/r?
Kd
I
.
N.A..
»/ts
k d
^5-
~ * - £ ~ n r
k d ___________________________________________________
e x to
f a x w ^
Ar«rf s
o . z v m
jd
d
=
A ssum e
_
kd
to =
-
„
„
o. ( b
-
/z x
—
H-ooo ^ /m
o .z *
=. <2. s-2 m
291
M
n
cf
-h - s - ( H -OO O ) =
iza.aoo
l 3 Q , 6 t 30
z
1 3 9 ,0 0 0
b
=( 0
X.
A/-/W
I 0 & ) ( b
y 0 . z ‘t ) ( a . £ - 2 )
0 . 2 . 7 6> m
=
ch&c King the.
w e ig h t:
a ;
z i . f c . h -
Z 3 S 2 0
CO
=
^ 2 / “? . ^
| j5 . 2 7 g
( O. O S ' + 0 6 ) j
'V / tj
77ji’s ;s h i g h e r t h a n -th*s a & s c / m e d u». c h o o s e .
& va lu e - h i g h - e r th a .n f h & c a l c u l a t e d v a lu e
Crf- CO ■
f rs s .u m &
M
co
-
t+ 'Z S £ ~ f^/tn
12.0,000
=
-t
( H 2 S S )
•=■
1 3 ^ 1 ^ 7 . 6 ’
z
b
=
»
=
N -m
(J & U 2 L * . )-
0 -2 7 6
I
0 . 2 .76
/3S OOO
f
/
< 3 . 2 7 S 3 )
=
o. 2 7 © 3 m
H2S'H-.<2>S'
^/m
uJhiezh /'s approximately -the. s a m e a s t t e
a s h a m e d value. :
"T h e r e P o n s . ,
b
=
O- 2 7 & 3 nr
-
■
2- 7£3 . 3 mm
S i n c e th e c o m p r ness/ve iv r c .^ in -the- c o n —
Cnste e ^ y a /s +h& tc.ns.He fo rc e , in t h e s t e e l
cu& o b t a in ,
~2^c
• ~ C 0
X
— ^S -6;
I O * ) ( o . 2.70 3
A<s =
Z.ZZ6
X
0.2.H-')
/ JO~3
2 2 Z 6. ^
mm2
C 120
~
As.
=.
-ZZZ<J . ? mn»*
X /0*)
292
1043. Design a reinforced concrete beam with balanced-stre.su
reinforcement to carry a live load o f 80 k N / m over a simple span 4 m
long. Use f c — 12 M N / m 2 and f s = 140 M N / m 2, and n = 8. Allow 50
mm of concrete below the steel and include the weight o f the beam,
assuming the density of concrete is 2400 kg/m3. Assume b = 200 mm
(Use the hint in Problem 1041.)
Solution:
UiVm - 80,000%, f
h
=
h
ss /VO Mfj?
n
=
e
a
=
<3. OS" /7?
^•m
= 160,COO -f-
z
■ZuJp
12 Mfb
f c -
U.A.
V
/2
=
ZH OO
*9 / m s
=
-ZZS2C
*//n*
b
—
O -'Z m
3
4
Max. M =
= i| jt
I B O , ooo + 2uit>') =
s=
1 6 0 , OOO
+
■Z.LOa M • tn
£/
=
17. S ’ X. IO & +
K
=
o.H-o-Tm
j
-
Kd
kd
hs& um a
M
=
to
=
-
/- - . y 7 =
3000
OOO
/Z X. IO 6
•#- 2
N/m
(3 0 0 0 ) -
1 6 ,6 > ,0 0 0 f\lm
M e = - j- fe. ( b k d ) C j d )
tie.Ce , O O O =
-J - /
' X /O 6
O- Z
X O. HO~T X ( J )
■
293
(O. 06+ d )
O. 6-Z7 m
d -
C Jie& tong th «- u>&>'<gtrt:
U) =
"T h is
is
XA =
Z3S2.0 £b-Z ( 0.6Z7 +O.0£?)J
s lig h tly
b i g g & r -th a n th e . a & s v m t a d
a v a lu e , o f 10 h i g h e r t h a n
3000 M/m ■ O h o o s e
■Hie c a l c u la t e d
v a lv e .
P r& s u m e
3 110 N/m
to
=»
M
=
160.000
Me
= - j - f c
+
2 ( 3 1 ^ 0)
e
(b > k d )(jd )
~ - j - ( 12 X /O6 ) ( 0 - 2
166360
166330 H m
*O .H -O 7d )
( 0. & 6 ‘t d ' )
d
—
cJ*ecJiin<%
O. 6 Z Q m
-the.
u j& ig h t:
Z3S-20 [ 0 . 2 ( o . 623 + O . O S )J
Ui vs f t x A =
s
w h ie -h is.
3/5*7. 3
a p p r o x / m a f -& U j
th e
m
sa m e
<5»S fh A .
& s-
som&d valve..
T h e re fo re ,
-£*/<=■
-J" f
d
=•
=
c. 6 2 3
m
A s/S
x /O 6 )(< ? . Z x O- * 0-7 X 0. 6 Z S ) =
/?S
=
As =
2 . J tJ I K l O ' s r n ' z
2 /*? / mm *
A i
11
Columns
1103. An aluminum strut 2 m long has a rectangular section 20
mm by SO mm. A bolt through each end secures the strut so that it acts
ms a hinged column about an axis perpendicular to the 50-mm dimen­
sion and as a fixed-ended column about an axis perpendicular to the
20-mm dimension. Determine the safe central load, using a factor of
•afety of 2.5 and E ~ 70 GPa.
Arts. P - 9.24 kN
Solution:
r — r a d io s , o f
t ^ r r n in e -fir s tt a b I is. f o r +H &
2 0 mm
in e r tia
on
g y r -a tio n - A s F r o m th e
m o m & n + s . o f-
p ■ 6>oe
f +h&
r a d iu s o f g y r a tio n o f &
r & c -fa n g le . is r =
h
.
■ J=_
'
For
p in n & d
L.
r
Fs>r f i x e d
L
~F
For
p in n e d
P_
A
ends
r
~
h
:
7 .0 0 0 \J/Z
en d s. :
2.0 06 7 /Z
2.0
e .n d & •
n o x i o * Crr 2 )
------- ( 13* T*-----
3 S B MPa
PZot-
-f-in & d
Gnds :
P
HIT'
( L/ r V
A
Th& -formula for fixed ends is four" timesthe formula -for pinned &nds. because. Frvm
■the. -table on page HH1 the fix e d - ended
column is four times, -the strength of- a pinned
ended column o r hinged column .
.
P
9 IT*
(T O
X to* ) _
=
(S H C > ) z
Z 3 . / M Pa
~Ti> d e t e r m i n e rthe. s a fe central load use -the
fcooer P/A as Hie •value of- the s fr e s s , hehce
S sa 2 3. / MPa.
AS _ i z 3 I x la 6 ) ( o .< s z')(o .a s ')
P ■=
-Z .S
FS
P =
1109. Select the lightest W shape that will act as a column 12 m
long with fixed ends and support an axial load o f 700 kN with a factor
of safety of 2.0. Assume that the proportional limit is 200 M Pa and
E - 200 GPa.
Solution:
F o r fin e d
Le
ends. , t h e
-
~
z L
= ~
PL z ( F . j i . )
eir z
z- ~
e ffe c tiv e
_
7 0 0
=
ZS. 5 X \o-*> m
4 \-r
-
1 00
•
r ^
^
y
tmimS
lo o
^
=
6
X
10 3 ( 6 ) x
(z )
'zoo x to** rr 'z
Z
..
z
length.
(gi OO O
too
ZG. S’ % to * mm **
V*
J ^
— 6 0 mm
F rom -Hie. apf.>endiK on u >id e - F l a n g e s e c t i o n s
■the u ) s h a p e . b e a m ttia f has. a n 1 > 2.S. s* X
IO & m m *t a n d i~ — *0 m m i s to -z o o x 0 6 . F o r
th is s e c t i o n
297
I
=
31. v
x
r
=.
5"3. 3
mm
A
ss
//, /<3<3
m m 2
...
f*
S o lv in g
_
=
/£> * mm *
i £ £ l
.5-3.3
TT2
f
=
I 'M - ) 2
=
AS
P =
„ 3
-For +h& s^-r&s-s ;
_
•
=
7TZ
f ZOO X
T '- s T 5
( lit t o o
FS.
“
B<oO
KN
/o'7 )
X
-
I O - 6 ) ( !S £ -
M Fa
X
/a6 )
Z
~th&+ w e a n s
ba&m is g a p a & e o f ca rryin g
P s= g s o AcAJ (ohfc-h is. g r e a t e r -than +h& g iven
lo a d of- ~joc> ArAJ a/»<^ rf- is +b& re fare, a c ^ e -p l'abl& • tfow & ve-r if- H is s e i e o t f o n is b a s e d on
jfoe f>t~oporb'an&.l Hm t-h, aso ly e fo r +fr&
~
U\r&d ar&a an rf m a fa s
&&l&c.fr'on.
P ( P. &■)
R&^t/ir&d fit =
S fi.
_ ~7£>c> *
K eefru ired
fr
,o Z
~
7 .0 0 Y
lO *
s=
7 X i &~ 3
-
In m a king ■a c-hotc^.
r a. 6 <7 m m
•, /) 2
~IOOO m m z
y o o o m m *■ 3-/W
Prom
Hus. a p p e n d i x . o n CO s h a p e s . , f r y id
7.50 x 7 3 w/Vrt A =s <?za<? m m 2 snc/ r =
mm.
■Sinc^ met &re
-Hie. Ifglrh esf' b e a m s e t tion , use Ui z s o X 7 3 .
COT.SO X 7 3
1113. Using A IS C column specifications, determine the maxi­
mum length o f a W360 X 122 section if it is used as a hinged-end
column to support a load o f 1200 kN. Use ow = 450 MPa.
Arts. L ■“ 7.25 m
298
F 'r a m
x
+h&
a p p & n d i'y ,
(2 2 , r
<3 c .3 m m
„
_
/ 2 1 T *E
V
on
#
s
_
I-Z 1 T * f 2 0 0
£
S ug pp,
V
V
ii ff S 'O
=
X
— C-c.
~
w o r k in g
^ 3 . 7
-
TT2- e
_
7 7 . H M Pa
=
lu<s.
/n m , «t/<s
& tre .& £ >,
; z 7r * f 2 0 0
Z3
2 3
n~r
X fO * )
/O *
/O 3
o .o t s s
L//,f
f o r th e
s to —
X
*7 3 . 7
1 2 .0 0
P
/>
Sfc,
CO s h a p e s f o r UJ3G>o
i s , s a c * m m '2’.
X
/o^ )
f ^ 3 .7 )
M Pa
•Sinc,&
S cu = t i 7 M P a i s g r e a t e r t h a n
M P a , ~Th<2 E u l e r ' s , f o r m o l - a a p p l i e s . .
-
S to
i-z i r r
—
b
—
23
2 3
-F
S/W<s<s
r =
i- =
77 V
(* )■
=
6 3 mrr>
//S' ^<5. 0 6 3 ;
=.
7 .2 5 -/ n
1116.
A W360 X 134 section is used as a column with hinged
ends. Using A IS C specifications, determine the maximum load which
can be applied if (a ) L = 9 m; (b ) L = 15 m. Use
= 290 MPa.
Arts, (a ) 1740 k N ; (b) 687 kN
Solution:
P=rom 7 Table. B - 2
mm z and r -
.
_
Cc
— 4
j
, A p p e n d i x . B , A ~ I T , IO O
m m f o r W36c X 1 3 + ■
2 IT 2 E ~
_
s^p
~
=
I 27T* {-Z O O
X /O ^T
2 “TO X to *
4
11 7
299
(a) if Le =. °im
° ie > o o
Le
r
T h is
va lu e,
=
*?S\7
I AS.js. +ti&n C-c. =
is
I I 7 /rim + ft«n s •
:
=
p. s.
S3
6
f : s. =
X C c .*
e ( in)*
I . ‘t o
(L e / r ) 2 7
=
=•
[ ' - ■
Z
r
r i s - . T ) 2"1
2 r //7? 2J
-
S to
=
1 0 2 .
P
=
A Sco
P
=
j /7 ¥
ft>7
0
X
3 (q s -.i)
& X /17
■ h *
S io
(L e / ,,)*
3 ( L* / r )
P . s.
J
2 <70 x. 1 6 *
I. «?
M Pa
f/ O Z
=
X
/
KN
ls~m
/ -«
IS 'G O O
—
r
Le
S i'o c e .
—
>
Cc. -
/Z 7 T * B
23
H
l£>0
H7
_
2
/Z7T*
“
1^200
2 3
X /<3^)
C / fe O ) 2
?0
Z
S<t» =
M Pa
. P
=
A S 14)
P
=
f o .p m
) ( h- o ■?. x
p
=
~ fc g 7
fcN ]
to *)
1117.
A W200 X 100 section is used as a column 9 m long with
built-in ends, (a) Using A IS C specifications, compute the safe load that
can be applied if the effective length is three-quarters the given length,
(b ) What is the safe load if the column is also braced at its midpoint?
Use oyp = 380 MPa.
Ans. (a) 837 kN ; (b ) 2400 kN
300
F rv m
B- 2
T& M &
,
f* p p * n d ix
lo o , f) = 1 2 , T O O mm z
r
=
I
41
Z 7 r* g
-
n/
/z7T^
s/
=
I)
L b
=
.
Le
S m cA >
0.75* 1.
ss
fc75T0
r
“
£T3.e
Cc
=
/O Z
~
65.
P
=
A S 44,
lo z a o
X
X /o**f
=
6.75-/77
IZ S ~
“
X lO * * )
-zs ( i z e y
M Pa
-7
— f o . 0 12.7) ( t>S. *7 X / O 6*)
S 3 7 fcA/
Sra&Ad a f mid point- :
Le
= 0.7 ( o . T £■ x
Le
—
z . 3 & 3
. Le
>- _
2 3 C .3
F.S. =
F .S =
=
5 (ll
=
5w =
10
= 0.7 (o.-rs~ x
m
S ’S . S
15in<z& J=£L
Cc. —
-4 -
3
+
102 .
■ (■ fr j
_ ± z £
0 Y .C e.
& X Ce. *
3 / * 3 . «T)
8 ( 102)
r
a ( I OZ) *
/• 8 2
r, m z i
F5
L
J8 Q
x /o^
' 02
X
IO *
I2 .1 T 2 ( 2 0 0
--------^
• z.e^g
uj
fz o p
3 0 0
o . 7 S - f <?)
=
Z3
S
, f o r
/D 2 «m
fZ T T *e
^
Suj -
B
S/7c/ /-= &3 8 mm.
2 X
J
r
_ / ^ 3 .- ? )g
l_ '
* ( 1 0 2 )2
l
J
301
—
/ s*?
M Pa
P
=
A S to
=
P
=
( 0 . 012.7 ) ( l & l X / o 6 )
2 .H -0 0
kN
1121. In the bridge truss shown in Fig. P-1121, the end chord
AB is composed o f two C230 X 30 channels latticed together so that the
fabricated section has equal moments o f inertia about the axes o f
symmetry. If the safe load P on the truss is governed by the strength o f
member A B , determine P using A IS C specifications with aw — 290
MPa.
Ans. J » - 2 1 0 k N
B
Flfura P-1121.
Solution:
&
f?A =
/- S P
=
R
P r o m T -a b le . s - H- , A p p e n d i x B , f o r £ x
2 3 o X s o , I t = z s . S * IO * m m V , A = 3 0 0 0
m m 2 a n d r x — 0 1 . * m m . S f n o c H ie h o o
c-h&nnels- & r e l a t t i c e d + e > g e .th e r s o
I s
e q u a l a b o u t -th e a x e s
s y m m e try ,
I * <=■ X y = 2 S . S ~ X /o6 /n/T7
. . I
=
2. ( -z s . S’ x l o b ) -
b
=
2 (3 3 0 0 )
=
Tof =
mm
rx
=
01.
S '! x i o & m m **
7 6 )0 0
m m *■
TZ> d e t & f m in e .
fo r o e a o h h g o n A R , ta k e
j o i n t A a n d s o lv e f o r P # b i
302
ZFy
.
R » = /. S P
S
/ 2 7 7 -2 E
Pa b
= z .s -p
i-A8
—
\/
II-
=
/zz
01.
l b
=
/ZZ
> C*
Ther'e.-For-e. , E u l e r 's .
S iv —
/O ffl
Z<7£3
=.
lO O O O
^~g
/-
S in c e .
Pab( - t ) =
[. Z7T2 f 2 0 0
_
** V
*
O
=
s
X
X /<?**)
/<9 6
//v
<s^ua+ton g o v e r n s ..
/2 7T 2 £
2 3 (’i<W
)
2
/Z 7T z f Z<30 X lO * 1 )
23 ( / 2 Z ) z
Sw
=
6<7. 2
M Pa
*'• PAB -
-4S ov
=
r AB =
5 "Z 6
p
_
Pab
2 . 5*
p
=
Z IO
76 00
X /o - 6 ( 6<?. Z
X /O6 )
£ -2 6
2 .5 *AJ
1126. The connecting rod o f an engine has a cross section with
the following properties: area = 300 mm2; rx — 3.00 mm; and
1.40 mm. The wrist pin and crank pin cause the rod to act as a hinge*I
column about the X axis and as a fixed-ended column about the Y axu
Applying the concept of effective length, find the safe load for a length
o f 250 mm, using the Rankine-Gordon equation, Eq. (11-9).
Ans.
P « 25.8 kN
Solution:
F o r f h e x - a x i s , -th e c o lu m n «
L e — L. , -H i& re F o r e ,
J -e
Oc
_
■zsro
=. © 3 . 3
h in g e d s o
303
F o r t h e y - a*/"s , t f i e c o l u m n i s
e d , L e - Vz. L. , t h e r e f o r e . ,
Le
_
_
=
'/ « ( z s o )
------ ----- ---------
a
fix e d — e n d ­
______
e -7 .3
Choose
t h e l o n g e r - - - - s ir t c .e th th e R a n ­
kin**. - C o r d o n e q u a t i o n , - y r i s a t th e d e ­
n o m i n a t o r s o w e n e e d a b ic /g & r value, -to
p r o d u c e a / « o e r w o r k i n g s -t r e s s . .
IZ +
-------------
~
18 X IO 3
m
/ Z 4'
+
7
8
T
« ~ ' s ,| ;| Z
s «j
=
& S - . ‘f
M Pa
.
P
—
PtSvo
=
300 X / o - 6 ( e s . q x / O * )
P
=
Z S .
8
> cN
1134. A W360 X 122 section is used as a column with an effec­
tive length of 10 m. Determine the maximum load that can be carried at
an eccentricity of 300 mm. Should the load be placed on the X or Y
axis? Assume o ^ = 290 MPa.
Am . P - 190 kN
Solution:
F r o m the. T a b l e B - Z , A p p e n d i x S f o r hi 36,0
x / z z , A = i s , BOO m m * , le a s t- r = 6 3 m m
5 x =■ 2 0 / 0 m m 3 (
_ < j7 g m m 3
C 5 is. t h e % .e c.h o n m o d u lu s )
Sir>c.e
^
Sx
>
/ F* s h o u ld b e
_
I 2 TT^ e
~
\J
_
Syp
Le
—
s / n c fi
osed .
..
lo.ooo
--------- 6 3 ----^
/ z 7 T 2 f Z<9<9 X / O ^ T
V
r=
on t h e y - a x i s
Z<?0
X /O6
117
=
ss |i*7 / e u l e r ' s
e q u a tio n
is
304
/Z7Tz E
Su) =
I ZTTz
Z .S
23
(t r
*to. 7 M P a
Sco —
M
c
( 2 0 0 X IQ * 1 )
( ISH ) *
-
P<e s=
_
40 ”
P ,
"A
P ( o . s ) .N - m
M
”
O. S P
<3. O I S S
P
=
2 0 / 0
X
/O ~ 6
/*TC> AcAJ
1137. A steel pipe 2.5 m long, built in at its lower end and free ;ii
its upper end, supports a sign whose center of gravity is 0.6 m from the
axis o f the pipe. Applying the concept o f effective length, determine the
maximum weight of the sign. The outside diameter of the pipe ii
140 mm, its area is 2800 mm2, and its moment of inertia is 6.32 X 10*
mm4. Use 0yp = 250 M N / m 2.
Solution:
F ro m p.
, i f on*?- & n d
o-hbfzr- € -nd is -finse ■ ~Th*s-
is. f i x e d A n d it*s
le n g th L a
— \zl. •=■ •z ('Z .s ') = S'mz & oo
£
/ z ?r2 e~
^ ~ \l
•Syp
Cc
Le.
S in c.&
L&
r
£
j£
3
+
- g
“ 3
,
=
‘
2.so x io £
s~ooa
*7 - S'
PS
I 2 7 r z ( 7.00 x i o * T
12.(3
-
PS
__
~
ss
/.<=»/
=
Cc
/^ S ‘
=
/Z£>
- 1 - L s J . _______ L H J —
6 Cc
&Cc 3
3 O Q g?
0(1-26)
-0 ( /Z&)
3
305
S io
=
FS
2^ z J
L
25~Q
X IQ 6
!■**!
5to
=
5 to
—
#5\
L
_ ( I P S ) 2- ~J
7. (I7.6>)* J
0£r. H- MPa
P
Me
A
■£#/»
P =
id
M =
o.6io
vo
,
H X /0
N -m
~
.
2 .8 0 0 x /0"fo
/ 'O P T )
X
<<? " fo
o.o,co
32
12
Riveted and Welded Connections
1204. The longitudinal seam o f a boiler is a triple-riveted lap
joint, with the pitch in the outer rows equal to 140 mm and that in the
middle row equal to 70 mm. Diameter of the rivet holes is 23.5 mm;
thickness of the plate is 12 mm. Determine the strength of a repeating
section and the efficiency.
Solution:
P
I
",
s z
( ==>
<==>
n-r =
*
Methods o f -PaHt/re :
3 )
Ps
=
—
•
—
b -)
pb
d*
SS
y
N r
C o .O Z S S -)^
=
/O H -
=
td
X IO * ) ( + )
\
kN
S t,
(t o
X
N r
308
=. ( o . o i - z . ) ( o . o ‘z s s - ) ( i 3 C
pt,
x
io 6 ) ^
)
—
/¥7
kN
=
( p
=
( o - ih- — rx o . o z 3 s r ) ( o . on?)(&e>x
- n , d )
t
St
/£>*>)
P tz
—
H Z.
kN
—
( P -
n -z d ) t S t
-f R R , x n ,
— (o .m — z x a. o-zss-) ( o ■on. )
=:
( QO
X
IIS ’
AcA/
3 ) SH'sncjth o f
IO ¥
IO 6 )
a re-peah'nq
s .tr -e .n q fh
0+
a
R*
a
& < a h 'd p / a h a
x
pt
=
&exhort
—
S ._____
X
/oo
/oo
/<=>*/•, 0 0 0 c t o o ')
(.! • + ■ )( o . O I-Z -)( 8 0
=
X t
kN
Et t
f/ y
IOH- X I O 3
+
7 7 .
X IO * )
%
1208. In a double-riveted butt joint o f the pressure type in which
the upper cover plate extends over only the inner rows o f rivets, the
thickness o f the main plates is 14 mm, that o f the shorter upper cover
plate is 6 mm, that o f the longer lower cover plate is 10 mm. Diameter
o f rivet holes is 20.5 mm; long pitch is 100 mm; short pitch is 50 mm.
Compute the strength o f a repeating section.
Ans. P = 75.5 kN
309
Solution:
T
<
=
=
>
/7Z
=
2
===>
PraJirn ina rzj
а)
t
Single
Ps=
=
б
c-'a/col&fi'ans :
s / ie a r
a r>ve.f
~ d ^ S & = J L (o .o z o s )-*
( i 60
X IO 6 )
shear a
W va/*:
/<7. e
) £>ouk?te
/cA/
/ R s'=
2 (!<!.& )■=.
c ) TZro*,h
in th e
ss.
to ' Si.
=
~
3 <1.6. kfij
main p/a/e ;
( # . OI H- ) ( & ■ O Z O S - )
( 130
X IO 6 )
3 7 .3 1
kN
d ) &ru£.h in the
p/ai-e.
py ^ ~
t '“
upper
<5o v e r
d £b
=. ( o. o ec,) ( e . o z . o s ') ( i s o x i o * )
—
I S . **9
e ) Cru&t7 in -the
^
kN
lo w e r
cover plafs:
=
K *
=z
( O .O i')( O . O Z O S ) (
=
2 6 . fcs-
M e t h o d s o -f -fa ilu r e :
3.) fftve-h C a p &=-i-hj ■
&
kN
130
X / < ? *)
310
pR
r r , x n,
=
— m<n {
+ RRZ x
nz
Pt>m /
P ^ lJ
Ps
,
RR, =
min { 19-B , 31-31 , - Z6. f>s} = l<f-6kN
RRg —
mm f
i Pbm
Pbc =
m in { Ps.
, P b c*}
=
IS ■3<3
.'. R R Z =
T e a r in g
l<3 . e
r
m in { Ps ,
3 S . 79
AcAJ
= ^-8 fO
*
o -f
m a in
,Z S
*S.7<3 f - z )
. 71 ]
=
9 / . 3 S A/sJ
p ta + & :
P-tm, -
( P ~ n , d ")
t St
=
( o .i — / x
o- o z o s r ) ( o ■o i + )
( s&
=
x
/<?*)
© «» • 0 V
frAs/
= ( p ~~ nz d ) t St
^trrif
=
f
<? . JO -
2
8 S-.&&
e ) F a ilu re
P*&tA —
=
=
m
RR i x
O.
nt
O lt)
19 8 x /^3 f/)
kN
+h&cos/dr p!a>te&
( P - r>i d )
/ 1, St
("o . i -
-z x 0 . o i o s ^ C o - o o e > )
(& o
x
io * )
Z
k/v
Z S
. 3
= ( p -
n ; d ) t L Si
= ( 0.1 -
* x o . o z o s ~) ( o. o t o ' ) f 8 o y /o*J>
= + 7 . 7.0
kN
=
^6tA
+
Pci.
Pen =
m in
{
Pt £ U
Pe.
+
X <?. 0 2 O S ) (
( &o x /£>*? +
=
Pbeuj
=5 3 S’. 79 kN
min { 3*7- 6> , 37.31
=
FR
+
/ Pbc. J
,
=- min { 2&. 3 Z
Pbn
,
X z
,
Ps
x z
}
z x is. 9 * , z x 19. 0 }
311
ss
The
kN
28 .3 2
=
m in
! Picu
=
f77/'n
f
=
H I-
-Z.O
Pe.
■=
ZS.3Z.
+ H -7 . -ZO
Pc
=
-1 5 .S -2 .
kN
P
=
3X
,3 V PS
2 6 .6 5 ",
3
7
X
l<7 9 J
kN
& f r - e . n g H 7r e p e a t i n g
■/h<s s m a ll e s t -
» 3 X
H 1 .-Z O ,
s e c .f io n
/.s
P.
7 5 *. S Z
A tN
1211. A quadruple-riveted butt joint like that in Fig. 12-2d has a
long pitch of 430 mm. Diameter of rivet holes is 32.5 mm; thickness of
main plate is 20 mm, and of each cover plate, 14 mm. Compute the
strength o f a repeating section, using a factor o f safety of 4, based on
the ultimate stresses of t = 300 M Pa in single shear and r = 520 M Pa
in double shear, ob — 660 MPa, and a, = 400 MPa. If this joint is the
longitudinal seam o f a boiler carrying an internal pressure o f 1.8 MPa,
and the girth joint is 50% efficient, what is the maximum allowable
boiler diameter?
Ans. D = 2.03 m
Solution:
n, = /
nz = z
nz = H
nk -
p
v
M-
312
P*
—
+
€>Z.Z2. ( l )
f>Z .Z Z (Z ~ ) -t
IO~I. ■*(•*■)
+ l o -l . 3 (tf-')
=
/oq-S
J>) ~Tearing
of
p-bm, ~
kN
/nait)
p la te :
( p -n .d ')
£Si
=.
( £>. ^3 -
—
7 ^ s- fcN
=
( p - n - t d ) t St
=
( o.*#s - 2 x o . o s z s - ) ( o . a z ') ( i o o x t o t )
I X 0 - 0 3 - Z S - X o . OZ~) ( / OO x /06.)
+ 6Z.ZZ ( t )
=
pim s ~
=
cr) F a ilu re ,
P fc i =
x
R R , X n,
/03
7<3Z.7. kn
( P~"sdl
( £>■ H Z
+
=
+
—
V
786. 7
0 - C 2 . ') ( l 0 0
0 . 0 3 2 S 1
X /o'3 ( l )’
6Z. 2. 2.
in
t s t + RR, y. n, + R R 2 x /7*
■V
+
X
£ ,2 .-Z Z (-z )
ArN
+h&
c ts v a r p la c e s :
C p - n,d')
t ‘ St
( O . H S ~ <i Xo- C Z Z S T ^ C
O.OIH)
( io o x /© * )
=
-V 2 0
fc/si
+■
=
m in
{ P tcj ,
—
mi n
f HZ O , e x 62 z ,
=r
*iZO
k.N
JT7>>» [
=S
m in
Ss
=
si
=
Ptci ,
fnzo,
sa
& x P& , &
»
X
,
n x *z-z
fc N
-----
=
74- M R *
=
'3 0
M Pa
x Pb }
q *
-ja r. o e J
II X P b }
, // x 7 s r.o & J
/ £ * ,)
313
5b
=
=
/6S -
MPa
Si
— — e^ >-
-
IO O
MPa
P r & h 'm in a r y
8>
ps
=
<£alc.vlations.:
—■ ( o . o z z s ^
P
s -
c ) Pbm =
=
( - 7S X t o * ) = 6 2 ,-ZZkH
( O
. 032£- ) * (130 X fO6')- IO
-J.&kN
t d 5/, s
1
07 . 3
( o . o z ) ( <?. o 3 z s ) { / 1 £ - x / o * )
*N
d') Pkfc. — ± 'd sy — ( o. o t* t) ( o . o 3 z .s " ) ( ie ,s x /o*^
=
7S~. 0 3
kN
-to t a l
Pbc.
= 2 x m/n { Ps ,
=
=
pt<,
2 X fnin jf 62.2 2
Z X
« . Z 2
y
, 7S - .o e j
St
/(A/
M eth o d s
o f -f a i lu r e :
a") P*
RR, x n, + PR? x n? + PP^ % r?s + PR H
=
x
/7*
P P S =. Ftp H
s=
C h o o s .<s
P
7o
VZO
=
rrjifi { ps , p t m ,
=•
tjiin f f>Z.-2.-Z , /<T7. 3 , 7S". *3g J
=
6 Z . 22.
S
m in { P i
=
m in f 101. a ,
s
/<?7 . 3
-f V 2 0
Me
;
y»y-a/
lo~i. 3 , i- z h .+ h J
es S n O
kN
s m a lle s t- p
=
=
se»/vtf
,
y
7© 6. 7
KM
-tfie b o ile r - d ia m & t & r :
L o n g itu d in a l -It.
"Zp
sa
■z(-ret.!* t o *)
D
=
p DL
— C / .0 K i o * ) ( o ) ( o . 9 s )
2 . 0 3 /fJ
314
G i r t h Joint' ;
e U
£ • # X P sr
p
x ir D t s t
-
<3. 0 2
p *
X / < ? *)
(o .S ')(lO O
D
%
= -2 .2 2
x P
=
^
( / . & r. IO6)
m
C h o o £ .< a
th e
s m a l l e r D k*e.c,au& &
fy p r o p e r ft'o n a l h i f h & s l r e & s . .
p
2.03
=
P
is d i r c c . t -
m
1213. Double-riveted lap joint in Problem 1203: load — 350 kN
per meter o f length.
Solution:
m . o f / ertgfit
=
n,
I
/
/. p
P = 3 sva x
/» =s
_g_
_
^
_
AcN
2 0 X AO*
2g
X /<SS
2 (o .o iz . ) ( o . o n s )
A t,
Z0
0.O6
S=
<6.7A/Pa
2 6 X /O *
jjp. 0 6 — <9. 0/7S"J 0. 0/2
37. 3
M P*
Ss
=
5-0. z
A'fPa
sb
=
66- 7
M Pa
St
=
3T7. 3
M pa
1217. Triple-riveted butt joint in Problem 1209: load on a repeat'
ing section — 200 kN.
315
Solution:
. t
1
#L
O
• •
p
Sh ear
n,
■= t
nt
—
2
w*
»
2
( single s h e a r )
(
(
Povble
sh e a t-)
Doob!&
sh ea r)
/ * v X z = <7 s h e a r
a rea s . -
a rea s
p
=
p
kN
7.00
=
■
Shear
a rea
Ss
200
=
22. Z Z
ss
P?
2 ? . ZZ-
*A»
= 2 2 .
ZZ
X
Z
=
2 2 . Z Z X K>3
=
kh)
.=.
H O .S M F a
7 '*
. **H X i<9 *
Y<J. ©/<. ’) ( o OZ65-)
== / o v e
MPa
Ma>n p / a t e ;
S t,
=
=
S f2
_
=
S i3
s
C over
zo o
x t o * _____________
C O Z - i X 6.0 2 6 S ')( 0 0 / 6 )
Pt
72
M Pa
P-r - P. * n,
_
( p -n ^ d ) t
~ ( o . - Z - Z X <JO 2<>S ) ( O.OI6)
2 Q O »M ?
- 22 . ZZ f l )
7 S’. 6
p !a + e & :
( at- H ie
in n e r m o s t -
ro w )
316
n o te
•
The
to w e r-
cover
p la te ,
has.
£■ s h e a r
a re a s .
w A ereas
th e
u p p e r c o w e r p la te
h a s t+ s h e a r a r e a s . - T h e r e f o r e -the
lo t u & r c o v e r p l a t e
c a r r ie s , t h e q re a -h e r
lo a d .
PLow er
-S i=
C -P
_
( p - ni d ) t l
5 tc
=
6 ?
( 2 2 . 2 2 ) X IO %
( e > i - z x o .0 7 .(> s ')(.o i-z )
M ra
Ss =
*tO ■3
M Pa
=
lo w . a
M Pa
•=*
-7S\ 6
m f >3
Sf
S
122S. Find the sale load on the lap connection shown in Fig
P-1225 if the rivets are o f 19-mm diameter and the plates are 8 mm
thick. Use allowable stresses o f r «■ 95 MPa, a, - 140 MPa, and ak 220 M Pa.
Ans. P - 242 kN
Solution:
317
c/K
Ps
=
-
/9 m m
df, •=
iq + 3
-f"
S s x <?
=
zz
mm
=. -J - (O.0I9)'1 (a s- X lO*)(<3) Pb
=
ia f*
St
2+2. HkH
-X 9
= ( o . o o 8 } ( o.O!<3)f z z o * tof>)(g)si 301 kN
Pi, =
( p - n , dh~) t St
= f o-zs- - / x o .o z z ") ( o - o o & ') ( /+o x to6)
-H
sr
Z S S .+
=
( p - n ^ d t,) t St
Pt2 =■
B
kN
C o- Z S
-
-z X f . OZZ
0 - 00 & ) ( 1 + 0 X
/ o*)
Z S+ .
± p is =
Pt}
=
kH
( p - n-z dh ) t St
~(o.zs
-
3 x o-oz-z)( o o o a )
( I + 0 X /O * )
\
P
s
3 09.1
kN
-
Z+Z . +
kN
1227. Two plates 250 mm wide and 20 nun thick are to be
connected by a riveted butt joint with two strap plates using rivets of
22-nun nominal diameter. The axial tensile load on the joint is 400 kN.
If the allowable stresses are r — 70 M Pa, 0 ,-1 1 0 M Pa, and ot — 130
MPa, determine: (a ) the minimum number of rivets; (b ) the minimum
rows of rivets and the best distribution of rivets in each row; (c ) the
minimum thickness of each strap plate, using the distribution in part
(b).
Arts, (c ) / — 10.4 nim
318
Solution:
dK
—
2 2
d h
-
z i + 3
mm
s
2S
mm
P ro b le m : P e s / q n a b u tt Joint far P = t+oo
Ac*/
using pla tes, zsro m m w id e and
zo mm
■ th ic k .
a) T o d e t e r m in e
the.
minimum
num ber o f
rivets. :
sh ear :
D o u b le
z x ~
p -
_
HOOXIO*
fly =
£s x n T
152
Z Y J L (O .O Z ‘2 .')3' (-TO Y- I O6 ) n T
riwehs
B e a r in g ■
P
t d K Sb
~
x
/£>3
( o- oz)Co .ozz) ( 1 3 o
y
to*)nT
&.<*<) 3
nT m in
b") To
=
X /7t
no.
of
d eterm in e,
r iv e ts
—
th e
7 . S~Z
m ini mom
rou>s of rivets :
r Q r | $> |
n u m b e r ofi
R o */ l :
P
s
t+oo x
( p - n, d h ) t S t
/a*
=
( t>. zs - ( HO
n, =
n, x o . o z s ~ ) ( o o -z )
X /<?*)
2 .7 3
S in c e
in s o l v i n g f o r
S t , n , i s in th e
d e n o m in ­
a t o r a n d -the d e n o m i n a t o r h a s to b e
b ig g e r,
n, - t h e r e f o r e
m u s t b e r o u n d e d o f f to t h e
l o w e r va lv e..
P,
x
2. r i v e t *
319
R ow Z :
-|• p
=
( P - nT d h')
t St
£ (* o o x /o») = (o. z s ( no
nz =
£ > m c&
10 * )
h . ss
F o r th e
sam e re a s o n
b e r o u n d e d o f f t o -th e
nz =
*
n* x o . o z s ) ( o . o i )
h
-th e rm
as
tn
lo w e r -
ro w I , nz
n u m b e r-.
m u s t-
«v fi/ s
a re
z
m o re
r iv e ts .
to b e
d is tr ib ­
u te d
a n d ro u i t c a n h a v e a m a x im u m o f z r iv e t s
a n d r o w z c a n h a v e a m a x i m u m o f H r i v e t s o n ly
■Hie m in im u m n o . o f r o w s is & .
m in . n o . o f
ro w s
-
The possible
distributions are. .-
o
o
o
o
C) o
o
s
o
o
C)
q>
o
C)
0 o
0
q>
o
C)
o
( b )
Ca )
o
o cd
o
( c )
The f i r s t - d is trib u tio n ( a ) is not- a g o o d choice
b e c a u s e /h r o w 3 , t h e r e a r e H r i v e t s tuhieh
ivill m ake -the p i-a te l e s s r e s i s t s rrt to t e a r ing . T h e s a m e thing is t r u e w ith distribu tion
( c ) . The four h o le s m the zn d r o w cutll m ake
t h e plat-e l o s s r e s i s t a n t -to te a rin g . T h e r e fore- d is trib u tio n ( b~) is a g o o d c h o ic e .
e ) To
**’ *• m in im u m
p la ts : .
d e te rm in e
ih e
% trap
P
( p - nidh' ). t ‘
=
*toe>
X /0 s
ssr
th ic k n e s s c f
* Z
C O. ZS - 3 X O O Z S " ) t ‘
320
(n o
t
—
O - o i o H- rrt
x
s
) ( -z)
io ■
mm
1235. Rivets 22 mm in diameter are used in the connection
shown in Fig. P-1235. If P -» 90 kN, what thickness o f plate is required
so as not to exceed a bearing stress o f 140 MPa?
Arts, t — 9.90 mm
Solution:
=
90
~
=
Pe
qo.ooo ( o.oe.j
=
7. Z k N - m
+ Xy*
kN
=
6 ( C O S )1
4- v ( o . o e ) *
sx * . 0 6
Sb -
l+°
X » 6 ' Z <7»*
321
A s t/ io w n in
diagram, the m
ost heavi­
ly l o a d e d rivets are A a n d F
=a n d since
-th
e
y are equidistant t'rom&, 4f,eu have-foe
same Pt .
Pdx
r*
6
=
Hx
90
6
_
“
IS
=
kN
7. 2 X I O 3 ( O . 0 8 ~)
V. <96> X /£> - z
* £y £
/v.
— o
T>
7- 2 X I O *
S ix ’1 + Z y *
=
X
/O ~ z
+
^ y) 2
kW
=
y r^ X
+■ P</X ) * +
f P*«,
=
J C tH .W
+ / S ') 2
( 0 . e . 7 ) z
=
3 < 3 - 5T
ffc
•*•
0 .8 7
( Q O S )
0 6)
^?
=
=
Ar/\y
td S
\,
3 0 . s- x ;<s 3
*
+
=■ t
( 'o
o z z ') (
9 .9 x i o ~ 3 m
=
m o
x io 6 )
*?. 9 m m
1239. For the connection shown in Fig. P-1239, determine the
shearing stress in the most heavily loaded of the three 22-mm rivets.
j4ns. t «■ 159 MPa.
80 mm
80 mm
60 kN
4
Figure P-1239.
322
Solution:
'
T
=
t i e f a j s " ) + 36> ( . o e )
T
=
<?. / z
5 Ix z
kN — m
+ Zy*
=
/ (0 -O 6 )2
+ Z ( o . o * ' ) ' * + -z f . o B ) 2
+ "Z -ij1
5Txz
=
10. 2 X /(?”
mr*r
F ro m
is
th e
£
A g i/ r e
th e
m ost
h e a v ily
b eca u se
H ie
x
cf c o m p o n e n t a
and
lo a d e d
B d d iH v e .
Pdx
_
3
36
Pdtj
n*
3
-
=
ft,
j+e_
_
=
-
=
I6>
kN
—
/Z
kN
r*
Zoc* + Slg*
/ S ’. 0 3
-
x IQ 3
18 ■ Z
x
( O O Z ')
/0
kN
T x __________
_
2 x z + r</z “
9 d > .c 9
9/2
kN
9- IZ
X
/O*
f O. O g )
/0 • 2. x 10 - 3
r r 'v a re
3123
= J (P d x
pk
^ f ( I (,
-
60 6
5s
=
Ss
=
+
P tx Y
IS . 03
+
+
(
+
f/ Z
) 2
+
H -O
01 ) '
fcAJ
&/i
. is.
b
o ■
<*>
fr
—
/ 5 -< ?
v
x
/r>
/o 3
( o. o-zz ) *
M P a
1244. A ISO X 100 X 13 mm angle is to be welded to a gusset
plate with the 150-mm leg against the plate. If the angle carries a
centroidal load o f 400 kN, what lengths of 8-mm side fillet welds will be
required along the toe and heel o f the angle. Assume the allowable
shearing stress through the throats o f the welds is 145 MPa.
Ans. 161 mm and 324 mm
Solution:
to o • I mm
p = ^ oc, kNj
4-^*1 mm
IS O
K
\0O
X
IS
Sr?g/<£
The. d a n fe r a f g ra vity of- t h e
i s o x to o x /3
m m d n g l e c a n b e o b t a i n e d -Prom -Hie a p p e n d ix
p- 6 s o ( P r o p e r t i e s o f u n e q u a l a n g l e s s c .t t 'o n s .)
"E-Mp^
"EM p,
t
“
= o
= o
—
ISO P,
s
HOO
f/O O .l)
P,
=
Z67
W
i s o p^
=
*+ oo
f y q .q ')
Pz
-
13 3
kN
to s t
— 10 3 ( e ) -
9 z v N/ mm
^
324
L
’
L
2
= J -L
f-
2fe7 X ! Q %
—
&Z.H
- J-Z - =
733 * /<j3
f
e z - t*
16 1
mm
1250. A plate is lapped over and welded to a gusset plate as
shown in Fig. P-1250. Determine the size o f fillet welds to be specified
to the nearest millimeter. What maximum value o f P can be applied
vertically as shown if 8-mm welds are used? Use r = 145 M Pa through
the throats of the welds.
P - 90 kN
U-150 mm-»
t
150
mm
t
F
B
mm 1 150 mm-»-
Figure* P -1 2 5 0 , P -1 2 5 1 , and P -1 2 5 2 .
Solution:
P =
+■ a
Z ( ISO)
2 . "2S ’ y. t o *
mm
3
O kM
3
qe> X /O*
P
f d y = 2 .L
f d * ~~
T
-
Max
~~
*0 ( t o o + 7 5 - )
Pe =
/ y .^ iT x i*
z .z s - x
<**
=
J*.
,s
f
=
/ ffrfx
^
=
\ / (,£ ’ Z 5 ‘ ) Z
j
=
-
~
< ,7 S
15
*
' ° * < 7 S '>... „
+-
f
3 00
/ < 5 3 fc
*
=
-2 | § -
a
9 .5 -m m
to mm
tf 0 - mm
io& lds.
a r&
usexJ t
"> *
=
res
Ncrn* :
x /<?6
S J±_
mm
+
S - 2 S - )2
%
=
-T h e re fo re . ;
s r z «r _N_
mm
tftx ) 2 + (ftd^ + ft 3 ) ' 2
+
^
-
mm
/5' . 7S"v i o %
kN — m m
=
z.-zs
~Ther^ flora , u s e
f
~
A o r E a s show n in -the fic ji/r e .
at
f . ^ 2 ± =
l>r*
J
0*9
N
~3c>o
t
P =
w e .I d s .
C a ) =■ e z + N / mm
-§ -§ ■— (*7o j
78
=
7£-.8
L o a d P /* p rop& rft'onal "to ^7<s
S fr-«e .n g M of- -the. ■fiide.i- oo& ld .
kN
V' ?
"'ViV
-
.
-
13
Special Topics
1304. As shown in Fig. P-1304, two aluminum rods A B and BC,
hinged to rigid supports, are pinned together at B to carry a vertical
load o f 20 kN. I f each rod has a cross-sectional area o f 400 mm2 and
E - 70 GPa, compute the horizontal and vertical deflections of point B.
Assume a = 30° and 9 * 30°.
Am.
- 0.412 mm; Sv = 3.57 mm
Q:
C o m p u te ,
the. h o r i z o n t a l
d e fle .c .t io n s . o f p o i n t 0 .
Solution:
v e r tic a l
Z<3kU
/. O k N
/A
s
=
i
»
£>r
S no rm al.
A
f°r
& H o R rz o N T * t,
' •
/ X !C -
^NORMAL.
= ( Ho o
io - 6 ) ( - to
k
J jra f / ) 3 - 2 0
- • S nokmal
~
^
^
x ia * )
*
( '- / ) £ ]
3 S’ 7
r»7
3 . ^ 7 /77/7T
fi HOmiorrrM.
=• (
X to - 6 ) ( 7 0
^ z o (o- s - n ) z
X /£• *• j *
— -z a (.S 7 v )
zTJ
0. H U . m m
1307. A circular bar is bent into the shape o f a half-ring and
supported in a vertical plane as shown in Fig. P-1307. Determine the
horizontal movement o f point C and the vertical movement o f point B.
.
nS'
,
s
P R 3it
c* "
2E l ’ *• ”
F lg u rM P -1 3 0 7 , P -1 3 C 8 , P -1 3 0 9 .
PR3
2£7
329
3
D e te r m in e
-the. f r e r iT o r r t a l m o vem e n t- ofpoint- C ' a n d th e vartica! m o vem e n t- erf
point 8 .
Solution:
M om a-n t-
=
P
( R s in
a)
Mm d s
/■
h o w zo n ta l-
Hom-zoN-rAi-
EX
J
“
m
R sin
zs.
&
/ P*
ej J
( PR sin 9 ~) R sin & (Rd6)
in teg ra tin g
and evaluating :
ie . horizontal
zz-
P R * TT
2 EX
S B VW T1M I
=
-rz: /
/
r **
going -to the. righ t
( P R *m » ) ( - R cos s ') Rd6
e x ;v*
in te g r a tin g
and
eva lu a tin g :
' 8 VKKTICAL
—
z ex
PR*
down w A rd s
1312. The rigid frame shown in Fig. P—1312 is supported by a
hinge at A and a roller at D and carries the triangularly distributed load.
Assuming constant E l, compute EJS at the roller D . Neglect axial
deformation.
Ans. EI8 “ 2970 N •m3
330
Solution:
Com pv+a
E/s
9 t.
ro lle r -
%>
c
4m
Mon).
j loon
N
o . N & q /e.c.f- avia/
331
=
erxst,
• z ( a r e a ) ac t ^ ( a r e a )B c • *c
r i3 5 fl
zrxs* =
IS S -O
X 3
X 3 1
^----------------*----J *
SJS-O
X 3
/ I
„ , "N
/ ? S-O
> 3
* [*
I? * )]
3
E T £p.
=
1316. A 6-kg mass falls 0.8 m and strikes the head o f the steel
bolt shown in Fig. P-1316. Assuming all the energy is absorbed by the
bolt, compute the required thickness t of its head if the shearing stress
on the cylindrical surface through the head is not to exceed 80 M N / m 2.
Assume £ — 200 G N / m 1.
Arts, t — 12.5 mm
1.5 m
20-mm dia
steel bolt—
0.8 m
t _ 3 ---- t
Figure F-131*.
d : G om puhe.
h ea d i f th e
th e r e q u ir e d
th ic k n e s s ,
S t r e a r i n q s t r e s s , on t h e
cat
s u rfa c e
th ro u g h
80
MN / m * -
H ie
head
is
t o f it s
c y tin d r i -
n o t to
exceed
Solution:
& St e e l .
s
steel.
—
—
_
a e
/• h o s
( & x 9- e p
rr ,
v
—
f o o z)
x
to ~ b m
i - s ■_________
,
( z o o x /o *»)
332
-
J
J
/ +
•* r e « .
t
t
/
Zh
+
&S.T
I +
2 C ° &s>
/. t+ O S
X IO-&
I ObB
£ ftT SB U
P
-
w
®s r
A
=
*
=
t
-
-
I o (,&
( <1.01
P =
6Z..8&
kN
P
TT dt
-
X
*7*
fei-86
x
it ^ 0 -0 2 ) (
r < y ‘ Tx
"
t l . S’
x 10
6)
to 3
x /<>•)
tn
i z . S ’ mm
1318. Compute the impact factor for a simply supported beam 3
m long subject to an impact by a 900-kg mass that drops from a height
o f 2.5 m onto a point 1.0 m from one end. Assume the beam section is
rectangular, 40 mm wide and 90 mm deep, and E = 200 X 109 N / m J.
Ans. 25.9
Neglect the mass o f the beam.
Solution:
C?ompvt~e t h e i m p a c t f a t h e r -fo r a s im p ly s u p ­
p o rte d b ea m
3 m. lo n g
s u b j e c t ho art im p a c t
by a
900 - k g . m a s s fh a + d ro p s , -fr o m a
h e ig h t o f z . s m
o n to 9 p o in t t . o m f r o m
o n e e .n d . A s s u m e t h e b e a m n e c .h o n is r e c t ­
a n g u la r , v o m m u .iid * a n d
mm d e e p , and
B = -ZOO x 10 * N / m z . N e q l e c . t t h e m a s s 0+
H ie b e a m .
X
~
b h *
J2
I
=
2 .^ 3
-
HO
= /2
x/ d > * m m **•
3
333
F ro m
^ase
(
L
=
s
_
4T “
7
7^W «
3 tn
,
X =
f ^<90
3 =
6 -2
Im
x q.ei ) (
, b
___________
6 ^ 3 ) -ZOO X /O'* ('■Z.‘*S X
S 5T =
0 .07*
6
&ST
=
/ +
=■
/
i
*
X
=■ -Zm )
/O '*)
/0 - 3 /77
/
y
S sr
8.<37^ x 1 0 - * m
z s : . ‘jo s -
S ST
1323. Determine the position o f the shear center for a section
composed o f a thin-walled cylinder o f thickness t and mean radius r
which is split along one longitudinal element, as in Fig. 1 3 -17c.
Ans. e *= 2r measured along the axis o f symmetry
from the center of the cylinder in a direction
opposite to the split element.
Solution:
P & f’K .rmine. ■/*« p o s it i o n o f the. s h e a r c.& nhsr
f o r a s&c-hdn c o m p o s e d o f a t h i n - eoall& d c y l ­
i n d e r o f H iic k n e s ^ t
a n d m e-B n r a d iu s r
w h ich is s p lit- a lo n g one. lo n g itu d in & l gjeme,rrh.
UM o
Ve
=
O
J C fd U ) r
334
Ve
=
We
=
e
e
PZtT
V 't r ■*
K
*
°
/>zrr
V r« t
X
I
r‘
X
[V
+
—
—
,
( t - c o s . ? ) Crd<j>) t-
r * t
X
*TT
r
-Jo
[z r r ]
bo-h:
I
a
.. e
e -
rr r * t
r
f- 2.Tr)
V r-3 t
F z"
|
1325. The thin-walled section shown in Fig. P-1325 consists of a
semicircular ring o f mean radius r and two straight pieces o f length r
Show that the shear center is e — (tr * / I\-n + 3) to the left of O, and
hence for r = 50 mm and t = 2.5 mm, that e = 86.0 mm. Need thr
value o f t be specified?
Solution:
shau> that -Hie shear- can+re is & ■=.
/ t ")
(n + z ) -ho Hie left- of c , j hence for t~ — SO
mm and t = 2.sr mm +h a +e = &<,. o mm.
Need -the. valve o f t bo specified. **
335
V ( t r ') C r ')
V<3
fA
=•
I
~
X
T& —
f A
+
=-
vfe f
1
_v_ Pt
( *rd < f> ) (
x «/*
—
- iitltli. ( ,
2 «o
=
°
e =
+ * , „ 4 'i
^ T f A
V t r-
L
r =
ir tr 5
x =
t r *
(
z
+ -| 1 )
c7 T + 3 )
Z.
+
!T /2
5~Q ( i r + 3 )
2- V- ffy 8 6
r f * )
( i + s;/? <(> ) ( rd<t>') r
7r + 3
X
r
r cos. 4 )
;
/ " Y ^
IT
V e-
X
va
~ x “
.
?B —
Ve -
V t r *
~
0 mm
+
V f-r *
336
1334. A beam simply supported at the ends has the cross section
and is loaded with a concentrated load P as shown in Fig. P-1334. If
the maximum bending stress is not to exceed 120 MPa, determine the
maximum safe value of P.
Ans. P — 36.2 kN
r
2m
L
Figure P-1334.
Solution:
P e te rm m e
m a x-im t/m
x—
of-
P.
- K 1
'2 .0 m m
■• .
T
Kz
t « .
s a f e v a lu e
ISO mm
s :m 2
<=* 0
= %
/wax- A*3
"2.0 mm I
•E-Fy = £
=
n . .. isro m m -}
IS O
( /<?o ) 3
__
/2.
rx =
X
2.
/ /■ 3 5
X
/-Z.
IS O »
^
/
*
20 * J
/O6 /nm *
X J-STO *
x
/
(
) +
{ IS°
i d 6 /nm ^
337
Mx
=
Mx
=
ZP
O- S' 7 7
■ZP
M tj
My
=
18
=
so*
P
S in
0.333
(5
o - s *' 7 7
cos.
N ■ rn
36
P
N - m
M *£f
M yx
Xx
xv,
/ -z o x. io *
P ( ''a .
X
P
0 . 3 3
/ 0 _<s
3
//. 3 S T
=
/ Z O
X
=
3 6 .Z
P
X
07S )
1 0 -6
/O 6
kW
1345.
The cross section o f a ring is the T section shown in Fig.
P - 1345. The inside diameter o f the ring is 366 mm. Determine the value
of P that will cause a maximum stress of 120 M N / m 2.
Arts. P = 80.2 kN
20 mm
100
Section AB
Figure P -1 3 4 5 .
338
Solution:
p e t& rm in e . t h e values, o f P t h a t u>HI ca-u&e a
mert-imum stro & s. o f I'Z-O M Pa-
A x
— Y. ax.
(2 0 0 0
+
X *
=
Xy
=
3000) x
=.
( -Z.OOO x ic>) ( s o o o
x
=
6>I m m
4.
/w *
r
:
p a o j
+
)
z o a o (s -O ^ J
3 000 ( 3 9 ) ^
+
^
=
M
—
tA
-
m . 3 6>
P
x
/rwr» ^
f
<?. z w
P n ■m
3 <b<Z>
t-
R
6
c
/© 6
e>\
/
R
c.
F ro m
T a b le
K0
-
13 - z .
6 06
on
;
x. 9S ')
p . &£&
K,
=
/. Z.S"
+•
339
p o irj+ A :
(TA
=
— /Z<?
--T -
X
x.\
/O 6
—
—
s ro o o
fo . Z W
P )
/^ . 3 6
P
=
r g -
x /o
—
/• Z S "
X
0 . 06/
X
ZO -6
S o . z ArN
/c<s Me.
sroao x /o —2 V" <3. d6 X
f p .
z » »
p
m. 36
use
)
( o . i o q )
x io -*>
P
=
86. z
kN
P
ss
SO.
ArAi
Z
1346.
The cylinder for a hydraulic press has an inside diameter
o f 300 mm. Determine the wall thickness required if the cylinder is to
withstand an internal pressure o f 40 M Pa without exceeding a shearing
stress o f 80 MPa.
Ans. I -> 62.1 mm
Solution:
D e te r m in e
lic
d(
—
fh-& t o s l l +hickn&s.<s.
O tjlin d & r - q i v c . n :
3 oo
=
M a x . *7'
e<5 X /o 6
mm ,
BO
-
P =■ H-O M Pg
t*P a
h>i
„
b* - a*
1
J l* _ L Z £ fJ L L
( t +■ is ro
S&tv/ng -f-crr t '■
•fc
>
=
£ 2 . / mm
of a
,
h y d ra u ­
340
1349. A thick-walled cylinder is built up by shrinking an
aluminum tube 20 mm thick upon a hollow aluminum cylinder having
an outside diameter of 150 mm and an inside diameter o f 100 mm,
thereby causing a contact pressure o f 20 MPa. What is the largest
internal pressure that can be applied to the assembly without exceeding
a tangential stress o f 100 M Pa at the inner surface? Ans. p — 97.4 MPa
Solution:
D e t e r m in e .
Hie. l a r g e s t - i n t e r n a l p r & s s c / r e
g iv e n :
co n ta ct- p re s s u re .
-2 .0 WP&
6y
<=
to o
2^ _
~ t , _ a x
P
=-
_ ^ £ Z £ 1 L
-7 S *- s a *
~
-1 -2 .0
M Pa
bg
b *
a*
a *
O 'T
*v
IS *
<T-r
t
-t
so*
-
S -0 Z
/■ -r&£>
*T
M Pa
( du&
c o f'1'ac*~
p re s s u re )
r * o )
'
Pc
p i
f> ;
F in a l
s tr& £ ^ : —
lo o
=.
I 76>6>
pi
=
9 7 . V- M P a
i . i t e , f>(
p / - 7 2 .(5
-
7 - 2 . <5
14
Inelastic Action
1401.
A solid circular shaft 80 mm in diameter is subjected to a
torque T. If the yield stress tw — 140 M N / m J, determine the maximum
elastic torque and the limit torque. I f T ~ 16 kN-m , to what radius
Arts, r, ■* 33.5 mm
does the elastic action extend?
Solution:
P e fm n m in e .
m a x im u m
t i o r t ju e .
and
limi-t- -fz>nqv& g iven :
a )
b)
F^or
d
p o l l 'd
c j'r c . u la ir
cTyp
=
1*0
rf
=
It,
T
=
*T’j p
=
. 07
lirnrfr
fro ra ves :
71
T
kN
—
BO
mm.
• m
(m o
T yp
7i
S h a f+
M Pa
kN ■ m
^
y 3
'T u p
=
%
( IH.07)
=
IB .17
kN - m
x /o 6 )
( max. elastic.
■berque )
342
S o / v/ r7 ^
-fo r
r ,' ;
Kr
/
—
3 U r *
if T
=
16
*r
=
»?■
=
3 3 .
ri
=
3 3 • S*
-
s ! Wr
kN
6T
V "T y p
■m
v ( o o it ' ) 3
S’
x
6 (
*
to 3 )
1T (
/H6 x
to 6 )
f77
mm
1402.
Determine the ratio o f the limit torque to the yield torque
in a hollow circular shaft whose outer radius is twice the inner radius.
Ans. 1.24
Solution:
p e .* -& r m in e .
tH e. r a t i o o +
l i m i t - - t o r q u e . tz>
t h e y i& / d + o r q u &
/« a
h o llo w
c .ir o u l-a r s h a ffw hose.
o u -h & r r a d i o s , is., f u j i ^ e .
4 b e i n n e r ra d *
it/s.
‘Typ
_
*r%p
=
Tl
=
J P ( '7'H P d
Tl
=
<ryp
T,
=
Tgp
—
f^ n p
3Z
27T
Tl
T ijP
_
5~6
^
v r* )
<yp
p e
IS/
S3
32
TL
£ *\
/&>
Tl
V tP
"TV p
343
1403.
A t what fraction o f the maximum elastic torque will the
elastic region extend to three-quarters o f the outer radius in a solid
circular shaft?
Ans. 1.19
Solution:
M -
u jh a -h
+ o rq u &
f r a c - H 'o n
u ji/ I
q u a rte rs ,
+h&
o f
o f
m a x. im v r n
oJ a s
th e
o o f& r
r e q io n
to
r a d iu s
in
e la s tic .
th r e e
a
—
s o lid
c-irc-otar shaft 7
Tu
=
%L
=
TI
=
% p
=
ru
“
IT r
TTr 3
6
■ZZ3
'T y p
*7"w >
T tr-
e7 y t> IT r
3
6
X
X
TV r
6 *t
(
z z j )
< T tfp
V
r *
z
_
TL
"*!)P
“
229
/32
1404.
Verify the ratios M L/ M yp specified in Table 14-1 for the
solid circle and thin-walled tube.
Solution:
\ ]A riftj the- r a t i o s
M I~/m *ip
£>p&&6'&d in T a b le
I H - I "fo r the. s o lid c .ir d e
a n d th in — t v »lle d
■ fvb e.
Thin — tra iled tube. :
I
=
1fr3 t
Mi
=
Z 6 yp Q
Mu
«
Z6yF (ir r t) f j~ )
Ml
=
H b r z 6yp
344
” ir
=
M ijp
=
- f
=
n t-r
M
l
~u
/
S'W
r iftP
up
2
6^
( ~
)
6 ljF
_z.
~
IT
I—
L/ Z7 3 Z
Ml
“
z 6 tfp Q.
=
*
=
-±
II
S o ! id C-ir^./’e :
61
6 yp
r*
e.
M »JP
~
6 HP -% • r 3
Ml
Mtfp
«/3
”/ h
Ml
/. 6 9 7 6
1405. Compute the ratio o f the limit moment to the yield mo­
ment for a W200 x 100 beam.
,4/u. 1.15
Solution:
C o m p u te .
2 .1 0
M!/ m u p
&>r
a
w
rnm
S =
Mu
—
K to o B
2 6 y p S ' <3
X /a 3 rnm'
345
II
Aft.
6cjp S
_
■ZTLGl
M y?
XQ
S
2 3 .7
= ( 210 X Z S . ? ) ( 11H- S
4 fy o . 8 x
ZG
S i\
=
Mt,
Mojp
Mu
Mup
x
io 3 m m 3
2 fs-7/ )
96*?
/. / . S * 7
1409. A rectangular beam 50 mm wide and 160 mm deep is
made o f an elastic-perfectly plastic material for which an ™ 300 MPa.
Compute the bending moment that will cause one-half o f the section to
be in the plastic range.
Ans. M «■ 88.0 lcN * m
Solution:
Com pute, -the. bonding m o m e n t "th&t will caose.
Y-z e rf Hie. s e c t i o n
-fa be. in + h c
p la & H c . r a n q e
given :
=
5"o mnt
i CoO m m
Sc5
HO
M
z c o MPa
yj
«
ss
l ^ 'i
*
2£* )
r;
-
Zi
=
Q
=
( o . O* x 0 . O s ) ( 0 -06)
<5
=
/. z
*fO m m
bh3
<3- OS'
12
(<?- 0 8 ) 3
/z
2. 1 33
x
lo ~ 6 m *
H-o
Mo
HO
x/<3'^m3
346
M
=
33 x 10
3 0 0 X IO
OH
+
-Z 11.2.
X IO
('■
fi g . O K N • M
1411. If Oyp - 270 MPa, compute the limit moment for the
section shown in Fig. P-1411.
Ans. M L — 177 kN-m
80 mm
Solution:
G i v e n 6yP — 2 70 M P j t c o m p u t e
m om en t- -tor -Hie. g i v e n t e c - h o n .
-th&
lim it'
ftO fnrrj
C?
=
£3
Q,
= 3 20
=3 2 0
Mt =
=r
( b o x i o o ) ( s o ') — ( t o n
a f
**o) ( s o )
x /C-* mm3
x (O
m*
x t o * ') ( 3Z0 x /o
I y y 1 7 .0
N
)
347
1416. The bar shown in Fig. P-1416 is firmly attached to rigid
supports. The yield strengths for steel and aluminum are, respectively,
290 and 330 MPa. Determine the residual stresses if the limit load is
applied at P and then removed.
-300 mm-
—500 mm -
Aluminum
Steel
£ = 70 GPa
A = 600 mm2
E - 200 GPa
A • 900 mm2
Figure P-1410 and p-1417.
Solution:
d e t e r m in e .
r e s id u a l s tr e s s e s
rfi
a t P a n d thes? r e m o v e d .
/is a p p lie d
S' o o m m
lim i-t loa d
-i:
<r—
A lorn
E -
70 OPa
A
6 0 0 i»m l
=
S te e f
E - -2.0 0 erf’s
h
T g p = 33oMPa
T .F
=
O
Ps
( 6 y p A ) »/
*■' Pu
=
330
*
m m '2-
^ 0 0
+ P's
= P
+ ( 6 q p fit) s t
X (O *
pl
Z^O
= ( 6ejp A ) a !
— 198
A’A/( T )
■=■ ( g y p A ) s t
=
“Z 61
kM ( c )
=
£> a
AI5-<?
(& o o
■=
IO ~ 6 ) t
io*> ( 960
PL
=
Typ - Z ‘JO MPa
X
X
x to - * )
tN
Feur a lu m in u m '.
L im i t lo a d
For
s te e l :
L im it lo a d
F o r- e la s tic ,
u n lo a d in g :
[ * * ] * ■ =
&s
ST
348
_
P * I (3 0 0 )
6 00 ( 70 )
Pal
w
Pal
+ Psh
S a v in g
p s+ f c o o )
<tOO ( ’2 .0 0 )
~
£>• 3 a e *
-
Pst
***<1
s im v l+ a h e o v s lij
Pal
-
R e s id u a l
iz e .s k N
( c)
j
P sf -
Z 3 0 . s '( '- r )
loads. ;
12 0 . s-
Pal
-
/*>&
Ps-h
=
3 3 0 . s- -
~
R e s id u a l
s
-ze>i
&q s -
k N (~ r)
k N C~r)
=
•’
^
■ -a f -
r . o W f -j
^
--££-■
-&
&
&
- 1» *
-•
EEE
1418. A sandwich beam is made by bonding 4-mm strips of
2024-T3 aluminum alloy between layers of foam plastic to form the
section shown in Fig. P - 1418. The foam plastic acts only to separate the
aluminum strips; its effect on bending resistance is negligible. A positive
bending moment o f 16 k N -m is applied and then removed. Determine
the residual stresses if oyp = 300 MPa.
Arts. + 20 M P a on outer strips; ± 40 M Pa on inner strips
50 mm
Figure P-141S.
Solution:
P e t e r m in e r e s id u a l s ^ re & s e s . -for ih e giver/
s a n d w ic h b e a m if- M = + it, k N - m a p p lie d
a nd tb&n r e m o v e d . A n d G y p — 3 o o M P a .
349
2 j f - z oo)
*N-A. e
— s- o
Y ie ld in g
+
x.ff'zoe>)fsro)^j
to 6 mm 4t
k
art- s t r i p s
” yp
6 l
_
e
~
=
For
(
b e g in s
3 00
X IQ *
( S ’ X !Q ~ *)
o. too
IS .O
xN
tf, =
300
M Pa ( c )
6't
/ S 'o
M P a (c ')
=
a t -:
• m
m o m A n t- a p p lie d
m e x c e s s &+■
/ KN • m
w h ie J i i s
c s r t i ’s d b y
Ih e
— /«*
=■ 16
4 tte
=
6 -tr ip s , .
M
U s,in ^
/ X lO 3
2.Pd
s
2 fZ O ^ > V
=
JSTO
6*
=
if 2
=
S + re s .s e s
( n&qa-h've
ZO~*jJ
|c?. ^■S'<o]
M Pa ( c )
v- <5*
2 <00
=
/s-o + s »
M P a (*£-')
<y^c /t> s / a s tic .
u n lo a d in q :
be-ndinc) m o m e trf-')
6,
=
6,
--$ ■
'
sr
in n e r
I
^
* > ° 3) - £ d Z L
s-.o x/o -*
320
s
/^<5
=
3ZO
(
t)
M / s fr ;
350
R e s id u a l
s i-r e s s e s :
—
R es.
—300
3 z o
■ZO M P a
6, R es
=
6 z KBS
*z
+
- -z o o
-
R es
+ /6>o
HO M P a
1426. A sheet o f steel 10 mm thick is bent over a 90° arc of a
circular die 100 mm in radius. If
= 270 M Pa and E = 200 GPa,
determine the residual radius o f curvature.
Ans. R -» 110 mm
Solution:
D a t e r m in «
+ h e r e s id u a l
r a d io s , c f c jtr v a + o r e .
f o r a s h e d - o f- S t e e l to m m -f*>ie.fc b e n - f - a v e r
a
9 o ° a r e . arf- a c i r c u l a r d i e
t o o m m th r a d ­
ius .
tjp
=s
es
-zoo &Pa
Ro
~
too +
Ro
=
IOS
mm
=
//
' // R
ko
o
—
~
l/L
-
f
270
M Pa
IO
>/
'
3
3 fo r/ E -ff
_
3 f Z7Q X IQ*>)
1 0 0 x to * ( o . oro}
'/K f.
=
<?.//<?
Rf
=
II O mm
1436.
Determine the limit load for the propped beam loaded as
M
shown in Fig. P-1436.
Ans. P . = ~ t ( L + a)
ab
Figure P-1436.
351
Solution:
D e t e r m i n e the.
beam s h o w n .
timH~ lo a d
-fior -Hie p r o p p e d
p
K
Plastic,
hinges
toill -form &t pts. / and z.
M,
=
[x m ] u
Mu
-
Ra
MZ
=
[iM ] t
-
Ml
S o lvin g
=
(a )
K l. -
e q u a tio n s . (
Pk>
(b )
and (b )
by
eliniin-arting Ft.
p ° ~ % r ( L * 31
1437.
Tw o cantilever beams are separated by a roller as shown in
Fig. P-1437. If both'beams have the same limit moment M L, determine
the load P at which collapse impends.
Ans. P = 2M L/a
P
Solution:
D e t e r m i n e -Hie lo a d P ef~ w h ic h c o l l a p s e
p e n d s , o n -the <sa n ti le v e * b e & m s shocon -
im ­
352
P ’o r £<3S£ a
P / a s fc h in g e s
Ma
=
:
R
P
—
in ih&
to
hing-& s.
Me
-
( ^ M ) t-j
M p
—
(X M )
P
t>s.e
=
R
2
a
P
=■
i_-
m
*
and
D
3fa /?
-
.
—Mt
— 3a R - Pa
equation
2 Mz.
^isx' < za s *
ptos.H a .
E 'U m in a H n cj
a
&t A
(Z M )«
E ’/imin ahna
c .s s .e .
af- c
and D
M t.
=
) — M l.
~
Z9K.
3 a l? -
Pa
H ie & qtsa-tions :
Ml
g ML
/ la r g e r Hian /n )
I
ca&e. a
/
353
1443. Determine the load P in terms o f the limit moment at
which collapse will impend for the restrained beam loaded as shown in
Arts. P = 2 M l /3
Fig. P-1443.
P.
2P
Solution:
ih e load P in +&nr)s, o f -the hmi-h
mom&rrt- a t (ufn'c-h <zoltaps.e. w il l i m p e n d f o r
■the r& s + r& ln e c f
b e a m 'Z.hovon.
D e te r m in e
2.P
M'
..w..
O
£>V
case (a ).
■v
B
c a s e ( t>) •'
l=c>r b e rth
=( t z m )
M p
-M
l
c -a s e s . ■
-
Va
ra se
(a )
Mb
Mu
-
=
l
VA ( b ) -
+
ZP ( h) -
P fi)
A 5- P
: P / a s tic . h in g e s .
—( z m ) l ■
a t p is .
Mu
=-
Ml
V*
= Mt
/ .S P
- Ml
A , B ^ p
-h
VA ( 2. )
354
R
tza&a ( b ) : P fa s tis
h>nq&s.
= ( i M ) l.
Me
P
-
=
;
Ml
at- A ,£ ^ P .
Mu
- M l + Y A (s ? ) - 2 P fa )
2Ml
=
s (+ r )
-
P
=*
1.333
M l.
£>P
o.e.e>e>7 Ml
1446. Determine the load P in terms of the limit moment for the
continuous beam loaded as shown in Fig. P - 1446.
Ans.
1
P = 2.5M l /a
1
2a — ^*-a *|"2a —^
r2
r3
Figure P-1440.
Solution:
C>ate.rm in& H ie l o a d P m te rm s .
m o m e n t- -f o r t h e c o n tin u o u s , b e.am
S h ow n .
case- (t>) '/y
y
l r<z ’ v
-the iim i+
lo a d e d a s
355
P o r ca&ts
( a ) : Pla&tic.
h in g e s . a~h A , B . C .
M b
=
( X M ) u
j
M l
=
-
M l
+
Va ( a )
Mc
=
( I M j
l.
;
- Ml -
-
M l
+
Va (" 3 a )
- P ( za)
S olving th e
z
ecfVBhons
P =
f -o r
simuIf-ane.ous.lu :
3 M l
ta s f i
( b ) • Plastic, hinges.
M0
=
(J M ) *
j
Ml
-
*3 f 23)
Me.
=
C s : M ) «
; -Mi.
=
K 3 ( 3a ) ~ P (a )
His.
z
e q u a t io n s
P
»
^
=
z.. S' Mi.
S o lv in g
us. a
P
S Ml
31 -
£ and O
s im u H 'a n e a u s .lq :
357
Appendix A
Board Problems
£h- E. Board Ex a m . Jan.,
/.
, July, M7I
f\ pressure i/esfeJ i> -ft- in Jiam&f&rand 0. 7 5 " thick-fo r ih long'itudina.1 j o i n t A q u a d ru ­
ple - rit/e-ted b u t t jo i n t~ /W>/'<s/j is c^tou farted ~tb f a i l o f
700, OOP# total loa d fo r a repeating section / t " /onq. Theeffic-iency o f the* ji r t h j o / n t IS
°/> a n d th e- unintate
Sfrenath <*f the- pfa-te- /s £of ooopa- Using a factor o f safe.-tijc f'S , -fin d -the Safe, interna/ freisurez.
a-nain ftate- has
S O L U T IO N For the. to n j ittid in a ( jo i n t " .
7^
,
ry
■T
,
,
D
2P- F = f P L
j
2 (140, OP#) = p((>x IzJClSj
p
-
ZfC p s i
For the j i r t h J o in t" ■
£ff*P = F
pi /C r )
f
£ff*TTPtS= p * ? £ p * („ - n X e . K } ^ i ) - p * « £ & )
j?
or
-
2 .1 0 p s /
u fina
■flt x i f f
60, O O P
s
p (b
'
X !Z ~ )
# x d . J s T b - ¥ 2 ')
P rr Z / O
358
2-
C-h. B. Board £xa*»i . A p ril, 1477
ft Short cylindrical post is haded ecientricallu
as shewn ■
*
lO O /O O O *
T - .'3. M r *
Z=
l.ltC z -)*
X=
/Z.57-
Calculate- theoompressiue- stress.
>1 = 77VZ
= 7T(2)z
--------- N . A .
A = 12.57 ;*
C -
This is a problem
2 "
com bined s1re.sse.s.
Mo
Sr. =
*NA
P - f00, 00O*Ccompressn/e- load}
M = /00 , 0 0 0 ( 2 ) = Z O O , O P O i n - *
100,000
_
12.5-7
Sc -
2 0 0 , 0 0 0 (Z~)
71777
~ 3 ^,77 7 p si
(T h e ncjatU'e- i'u)n irnditates compression.')
Sc - 3 1 -7 7 1 psi
3.
Ch. E n yy - Board E<am . Oct- I el7g
A ) Draw the' Shear a n d rnonne-nf diagrams,
b ) Calculate, -the m a xim u m shear a n d
m a xim u m mome-nt.
1Ooo"
100 * / ■ ? +
S
Nr >
r Je „X X
Je > £_s :L_!
■ 5 ' ---
* M Kz = o
ID R 1 - !0 0 ( 1 o)C Io l
5
-K
Rx
m
R,
~ 1000
Rz -
IPOD + IOOO - IOOD ~ IDOO
-\ ooo
4.
Ch. £ . Board
Oct■ I t 7t>
fj /s - ft- beam is freely Supported a t one e,rtd
and
5 ft- frent the, other estd. Determine* the
ma.Kimum M>mPrcSiil/e and tensile, bendnna
stresses o f the- team i f the only toad on the beam
ii Us own vueiqht, ‘which is umformly A *0 lbs/ftthrvuahouf its length- The beam has a. T cross Section j The horizontal position of the T being
JO inchei wide, and Z inches deep, a n d the vertical
portion of the T bema /D inches
Z 'inches wide--
<**d
l O O ’f r / f f
5'
^
\O '
) \
360
* 0
to R ,
-
to o (1 5 ) (7 . 5 )
K , -
H 2 5 *
Rz -
is o o - IIZS
=• 3 1 S *
£.£5
*
V -P
i°-x
~
-3 7 5
-5 0 0
To solve- -for X :
-£
£ • - - «x£ • ;
/0-X
X. J.W
M a - - j ^ ( S o o ) U ') - - 1250 f t - *
M b = -f (3- Ts)(3T5) = -JOS. /5 fA- #
^ = io (n ) +
„ ,
----- —--------- — = £ /*WM /*<5
"
+ ^
bottom
4" I
NA:
I ma - i , + AfJ, z +• Za + A J *
10"
p
£
2"
M
J/»>l=
1253.33 in
A:
-
- 12SO f t - $ (T h e negAiiw siqn indicates
cun/a.ture dousrm/Ar/J■ ihere-fbre ( y/j«-
/s
c
__^y±
Sf* -
At
,
_
M ijc-
*
“
7 7 a"
B
.
-
*/t - j f " a„ j
l2 S P (l2 )(* y
/ iT J T ir
1250 (8 )(lZ )
'
« » • »
* 7- ' 7
_
lt>
'
M b - 703 .13 f t
The positive- siyn
indicATe< c-urvAture- w,yH/ard ■ Therefore
■the bea.hn /s bent
&nd
y* -S i n
And Ve- - 1 "
_ 9»
361
Sft =
703. ! 3 (l Z ')(g ')
1253. 33
S3. Si, p ii
5ft -
703-13
1253.33
2 6 .1 3 psi
M a x im u m
Sft =
S3. SO, pst
Sfc = V - 76 psi
S.
Ch .B .
Board Exam. Jan.
R reinforced concrete column is to be built 14 inches
in dia.yne.ler- and 15 f t high. The concrete h ft> havean ultimate- compressive strength o f 5OOOpsi. Thecolum n is 1b be reinforced yvHh vertical rods.
30, 000,000 psi ■ E c ~ 3,500, OOO psi ; Ss -10,000psiFind the n u m ber o f vertical / - inch iauare steel barf
which m u ff be used for the column to Safely carry
230, OOO tbs.
NOTE-. Since- th e stresses art9:2 1 0 ,0 0 0 *
given , use- have -to fin d fir s t ihcgot/ernine stress by using th e ir
de-farm&Tion s. yVhen a. load n
Applied o v e r a. colum n the d e for­
m a tion fhould be- the Same f » r
fa /"
both the steel rods and concrete.
P<
30 x 106
3.5 % IOI
30
J f Sc. s 5000 p ii
S i » - j | (5000) = < fl,g S 7 p s'
362
T he value- -for Sc can't be used because fh e-
Allowable- Ss - 10,000 psi
•■- 5 j = 10,000 /t -thegoverning stress and fhe valueo f Sc- is obfained from -fhe relation.
3 .5
Oo, o o o )
30
5c = 3-S
30
5c =
*
m > 7 f si
From fhe equation o f St*.fics i
Ps f Pc -
2 3 0,0 0 0
n.jcAs Sf + Ac.Sc = 230,000
n ( ) * / ) ( w,OOO) +
Z- * 0 * 1 ) ] llt>7 - 230,000
n = 5.7
vi = (, bars
M E . board Exam
ft cylindrical funk., hav'mg a ve-rticoJ axis, is 6 ft.
in diameter- a n d lOft. h iy h . Zff sides owe- held in
position by means of two Steel hoops, Pvte a t the
-tvp And one <*/ the bottom- Determ i ne the diam eters
o f these hoops when fhe tank is -filled w ith water.
Assume, the -tensile strength o f stee l is 15,000 psi.
StLUTtON ■ The maximum -force occurs
a t the bottom o f the tank an d -there-fare.
-the bottom hoop governs. The force a c tm a at the bottom is equal to fhe weight
o f water- in the tank. ^
p
F = W in t o = PV =
■fp
P - ( AS) hoor
2P = F
r
2 (A & h o^r = DAh
2 dif
05,000') -
t>2. V ( L ) z (lO )
^Uoop = O. g65 '»>->.
363
E> ca.rd E x a w \
R IO - f t sphericalpressure t/esseJ if rna.de by betting
two hern iSpherical flanged Sections together-. How
mmnig / inch -diam eter- bolts Are re a uired id Secure,
■the two htAJt/es, i f the pressure tuifM/n the tsesseJ ts
XOpsi ? Hie. a. workina Stresi o f 15,000 psi And croffSectional areA. f f O.Zb! in z -for the bolts ■
M E
From the figure* it is obviously shown -thud- the force within
the i/esse*l is count er-axded by the- bolts placed Along th e
flanged sections- Therefore a.t the- girth j o i n t :
P - F
n x ( A S )m + = p A
IT*
** * CAS)ix>t{- = f> x ^ ^vetse/
n x (0.5St)(isJ300) = 2 0 x
(/ o x t z ) x
n = 27. 4 bolts
n -
S.
C. E
And
Z.& b&l-ts
Ch- E- ■ Board E xam s
ft vertical wooden m a s t S O ft- high tm.pe.rs linearly
from *? - inch diam eter' At the base % 4 - inch d ia m e fer- A t the b p - (a .) A t w h o f point w ill the rrtAst break
u n d e r a horiz.ontn.1 toAd P Applied At the fop ? fZo/veby m ax im *. A n d m in im *. i ) ( b ) 2 f the ultim ate strength
/H fh e w ood is S’OOO p si, calculate- the magnitude- o f fne
force- which w ill cause' failure.
364
^7 = P tj
ft '#
= U Py
in - &
The serfion m odulus (z.~) is :
„ _ ffir* _ T f x i
z - - ----------- —
.
Fret*i +he fUtitre- stress formula^ ;
S =
M
UPH
TfX3
Z
=
HSPy
Trx.3
Uith^ ratio andproportion:
x-x
j) . Jf--Z
SO
4.5-2
X
= ^
X
_ JC - i
Z.S
50
, * *-J L
20
=
20
S u b stitu tin g in-k>
$=11?*
'ifx 3
4gp y
~ If
. 384> 000 P ■
7f
y (* o l3
(HO* 9 »
^
( t j ■* i o Y
'Toget the- point -from the- top where, the m ast k>ill break,
differentiate- 5 with respect 1t> t j And etjpmie- 1o xen>.
d S _ 38H,00Q P \ (y + W ) 3 ~ 3 y (* 4 * V ^ a ~|
dy
7T
L
( y ■* + O} *
J
365
(y+Ho)3 =
y
3 y (y + 4 e )x
t HO -
3 LJ
2 y s
HO
*j (b )
If
20 ft. fre>m the top
5 - $000psi ,
t j - 2o f t
Front the. equation:
ig v, ooo p
it
tt
{tj * HO)3
5-ODD -
M 'O M P x
**
V
(t o )3
P*
5000 C r ) ( t o ) 3
P *
H V1#
*0(3*4,000)
M £ iSoard - A u g . /^75
A I ooo A/s vertical lo a d /s Supported bg a vertical
bar a n d but two tan a / (in si^e) in clin e d bars m akin g
f5 ° With fne horizontal f ir m in g an in ve rte d n/ye-,
A nd fixe d a t the ends. The length o f the bars are.
eaual a n d AS /ooc^n- b u t the cress - section*/ area
a j the Vertical b a r ft e-c^uat to Z tim es the area, o f
the in d in e d bars. D eterm ine tbe forces th a t are.
Jictinei on the bans.
SOLUTION ■
©
^Fv = 0
2 200 lbs - Pv + 2-Pi Si* HS°
©
S i sin HS° - i v
S v :« V .
SL
Assuming bars o f th e same m ater/aj
P i ( I ) Sin $5°
_ Pv (/ sin H S °)
AL
2 Ai
366
Pi
in
10.
Q
= 0.5 Pv
17.00 -
pv *
P v ( - f z ')
pv -
u s e - . 73 I * .
PL --
b f t . 3b5 tbs.
3 o a rJ - A pril, /777
The single Support o f fhe Spherica.1 water ienk
/s made o f a s ite ! pipe. with an outside dian»et*ro f /£ in. and inside.1diay*ie.ter o f /5 in f a centerSolid cast iron rod o f 3 ir>. diar»e tef~ and the space
between them with concrete. Jhe steel pipe, has
* n £s -- Z f x / o 6psi. the concrete has £ c - 2 x/O^psi.
and the. &as+ iron has E ci - /S x /O *p s i • I t is
fou n d that the unit stress fo r the steel pipe- t's
!&, /O O p si, fhe unit stress for the cast iron rod is
7,Boo psi and the unit stress fo r concrete is /o&opsiHow rnantj Cathie, m aters 0f w ater /S in fhe tank?
CH£
S O L U T IO N ■
- ~ \ > 2 - IS *]
Ast
= 2H.3H7342. i n *
Ac.I. = - f
-
(3 )*
7. 0(>&5g }„ *
b to n c *^ *
- 7.0bgSS
= Ib l.b t b /’« 4
G iv e n :
5st -
If, I OP f i t
E i t = 24
* >Ob pii
S c . i . - 1SD0 f f i
B e.V- M
* IOh pii
S cD nc-
Eccnc.- 2
X
10^0 Pfl
L-ii
5«t
-
.
L c. i .
-
L oey,t .
5 c .'l .
.
S con c.
21
15
Governing
:
5 ft -
IfOSO psi )
ID6 p ii
2.
Sc.i - 7X00 fit )
Scotic.- 10*^0 pfi
Pit - (2 4 .1 4 7 3 )(l5 0 S 0 )
=
3 b7 .J5 7 .2 g Hi.
ft.l-- (7 0 b iS t)(7 S 0 0 ) -- 55,134.124 lbs.
pc<,^.z ( i b ‘).b 'tb ')iio > io ') - n b . m i . & i
P totai.
-
v .
516,724. 04
(> 2 .4
V *
27/. 10711 « , ^
ibi.
M S ,7 2 4 .0 * Ibi.
1 5 1 4 . 136$ f t *
C £
& o*.rd - Feb., 1171
A slecl angle, attacheJ to the fla n ge o f a s i eel " H " colum n by means o f f S
/ i " rivets supports a trianguUfr load o f S300 lbs. Tine following permissible stresses
for s i eel are as follows:
Shearing Stress ‘ ti, 0O0 psi
Bearing Stress = 32,000 psi
Tensile stress
=
20, OOO psi
Determine, the actual Jen site stress o f the m o s t
-stressed rivet an d actual m a xim u m bearing stress of
the steel angle i f a 6 " x 3 '/ * ” x / ” /s used. I s the
angular 'Support adequate ?
SOLUTION:
Allowable- toAdi -
Ps -
7O bf.5 8 2 4 Ibi.
Pb = 24,000 lbs.
For tensile-:
PT -
9835. 722 lbs.
M a x i m u m allowable, load without exceeding any o f
~thegiven p erm iiiibl* stresses P = 706g. 5 *2 4
l/smg
P -
5300
368
Actual ie*tsi/-e siren
5300
ST - 5300/ i f ( l _ \ x
ST =
St
‘t
.
'
m , 1 2 6 .™
H
- U ,W t ,.V H psi
Act- Maximum Searing Strati:
Sy - ------ --------m
12. CB Board -
-
7 Obb.blpH
<
'
M a y I9SI
A f X 8 tim b e r beam ( *
1600p s i , Fj, ItOO pst,) is reinforced at its bottom with a fla t jfe e t
plate ■?" x ‘/ z " (F t - 2 0 , ooo p si, Es - 30 x /o * p s i).
The beam spans /£ f t - be.tuse.en -simple- -Supp>orfsaJ F in d the fata] u n iform , lo a d that the beam
can carry.
b ) D e te rm in e the length o f the span oi/tr which
the -steel reinforcingplate, is required.
For Simply Supported beam with uniform loading -.
tA/L. ^ /
M ax . Moment - / 8 ) L = /(, '
= 5Z »V
Steel ple\te,
E i * 5 0 x l0 spsi
E Z Z Z ]/ .
Fs = m ° 00pS'1
4
20,000 = ~
««/ --
Ms z 3331.33 " # i
H 7 3 3 .3 ? "# V
A7»v = V777. 7 r ' # J
Ms r z n . l S ' ^ ' s
S Z iv = H i l l . ~lg’ #•$
3Zw-
ISS. 5S5&Z ,h/ f t
w
hlT
=
WT
= /^V. 211,21 V «
2 77. I f
g.6SO(> '/f t ,
I55.55LZ + g , L t O b
369
/3. C E Board - Fe-b-, /976
I t is proposed to strtnafhen a f " x to " h ruber beam
by rein forcing it With a s u e ! p U te f"x f t " thick u>nn ecfaJ
h the- potbm surface- o f the lim ber beam. The beam has ajpan c f H> f t find the new load is <350 p/f. Investigate the*
proposed desijr btj chcchnj the- in*xim i*m flexural stress
in the, h m b er and the steel plate i f Ew * 2. f * / o 6psc.
E s z 3 0 x lo b p u. What is the shear stress « / the linear
-Surface o f the tim ber and the steeifiv - tfOD psi .
f s - IS, ooo psi.
I
A W
1
J.
. V - —
2.
Mm . M •
= * f p ° ±?
7-
8
r
8
Z S O O lbs.
.
U .to o ff-n ..
6 (n ,z o o ') x >2.
ft
fs -
J.
7*v =
fiv =
> H O O O psi
3 2 2 5 boo psi
( NOT SAFE)
(, ( I I , 2 0 0 ) X 12-
------------------H (io )z -
2016. psi >
15 0 0 psi (N O T S A F E )
WOO D
STEEL
cv ^ ^v; W
4"
370
m.
CB
Board -
A n j-, /16S
A (>" X I®" tim ber beam it rennjorc^d by the addi­
tion o f two
lo in. / S . 31 lb. channe-Js, proper/g belted
fo the* / o " vT/V^r o f the
Mem ber. What at the b ttijmq sfre-nafh o f the reinforced beam cowpared wifh fhe!
o f the
H r»b ir be*™ ? Utt n o m in * ! d,mWStons.
All ortabter stresses are so, ooo pit. f o r steel and tf foopsi
-for Wood.
io "
,__ ORieiNAuuy
IO->5.3#'s
ft = —
6"xlO"
—
-
to"- 15.5#'s
ISOO
M = iso, OOO i~ - tbs.
)V/M -two channels :
fo r
/ o " x S".3 i t ' f
5 - /3.S /« 3
I - b~7. y
fs z
I3.S
n fo r i n
«
- 20, ooo
1
z 70, OOO
- !t>s
M f c r i t •■ s q o .o o o in -lb s
fiendina strength o f the -two channel Added A the
origin*! section is g r e y e r since, d cauld cerry ck y rea fer
lead es seen in }h larger- M om ent capacity-
371
15.
HELICAL
SPRINGS
C £ H o a r d - Alov. t M g I
f) load Pis supported by two concen­
tric steel Springs arranged as fhousn.
The inner- spring consists o f 2*t turns of
% -in-di<* kvtre on a mean drameter-of
6 i n - t h e outer spring has 20 turns of
l-in d ia . mire- on a mean diamet-er- of
&in. Compute -the maximum lead that
mill not exceed a shearing stress of 2P,t>00
psi //7 either spring ■Use 6 - 1 2 x 10 ^psi
s o l u t io n -
For the- inner spring , n, - 2 4 ]
d, - 3/if."■ - 3 "
For iha- outer spring, n2 - 2 0 ; d2 - / " ; P-z - t "
The- condition o f static- equilibrium git/es
p -
p.
+ rx
where- f, and Pz are- the loads Supported by the inner and
outer spr'mas, respectively- $mce- no other e-aueition o f
ifa.HC' e-^uilibrium Is a vAitable fo indicate- ^ruhat proporfTon o f ‘'the load is Supported by each spring , we- consider
the deformation ■ The- deflection o f each sprina isaiven bu
r
d r
&J*-
I t is evident th a t the load causes the springs to deform
er a„ y t * * * ’
s,
-- &z
61P, R-fn,
(,‘t-Pj.Rjyiz
Gtd?
G ^d z
Since- the two springs ared the Same m aterial} /-«■• > Steel,
<?/ - Gz . Substituting the values of R ,n and d and
Simplifying , we get
P, - */t Pj_
This is the- governing retoifFon between the loads Supported
by the tw o springs. The shearing Stress fo rm u la for~
helical springs is
c
_ /£ PK r
d x
J
“ ir js
-9KJ
I f the inner spring is stressed to its lim it o f 2 0 , 0 0 0 p s i ,
the bad it can support is obtained as follows ■■
2 0 ,0 0 0
P,
tb P, (3?
' ------T T (3 /^
= — .
=
s
(/ +
5 1 ^ . 7 5 lbs.
3/4
* x 3 '
N
)
372
I f the o u t e r spring is S tressed to its l i m i t o f 20,000p s it
fhe load it can carry is o b ta in ed as fo llo w s :
TTCO3
Pz -
921. 00 Ibi
I f the i n n e r sp rm g is stressed tb ih lim it o f 20,000p s i,
the Corresponding loa d in the o u te r spring from theg o v e m m g relation m ill b e Pz = ~
P,
z ~
( S / 9 - 75)= S31. 60 Ibi
I n o th e r w ords, the Outer Spring to n /d n o t he siressed
fo ih l i m i t o f 2o, 000p s i w ith o u t overs fre-ss ina the.
i n n e r spring- The Actual loads are th ereb y d eterm in ed
tb b e P, - 5 / 9 . 7 5 Ib i a n d P* - 8 3 l.e > p lb s. These
values can th e n be su b s titu te d into t h e equation o f
S i*tic e^tiiH iriu m fo g e t
P - P, + Pz
lb.
-
S /9. 75 f
8 3 / . 6 o r 1351. 5 5 !bs\ Ans.
M D V /N & L O A P S
C E B o a rd - Feb ■ >96 9
Determ ine- the m a x im u m bending *nom ent a n d
M a x i m u m sh ea rin g three Set up bg -the truck w ith
axle loads A , B , a n d C as show n wJien p a ssin g
Over- dimply Supported span 20 f t Iona, (/ton 5T
. 7#
>
/
>c
2, OOO Ibi)
11 tons
9T
lOtt
■7\
2 0 ft
r2
R.
SOLUTI ONI f the three loads are on th e span, -the re s u lta n t is
R = 2 5 tons. The position o f th e resu lta n t can be obtained
b y tfangnon's th e o r e m ■ Thus
of the components cnoit ^-25tons iit
2 M a Of ih t resultant
1 ( 1 ) + / '( I ? ) = 2 5 J
d - 1 0 f t to the right o f A
Maxim um
(a )
B e n d in g
M o m e n t-:
Three loads on t h e span -
373
R=25T
ct,
c:r
*-
s'
IIT
T
S' ,
-
(j*> l
<?
©
Th& position o f th e loads th a t w ill cause- the ben d ­
in g m o m e n t to b e m a x im u m u n d e r A is shown above,
where -fhe cen te r line- o f the beam is m id w a y b etw een
A a n d R • H ow ever, load C com es o f f the spant which
.is contrary to the a ssum p tion that all t h r e e toads are
On the spanThis indicates the p o ssib ility o f a m a x im u m fe n d ­
ing m o m e n t u n d e r A when only toads A a n d B are.
On the span - /-Ve n e x t c o n s id e r th e figure- be/ous
Where- -the toads a r e so lo ca ted th a t th e cen te r tine
o f th.e. beam is m id way betw een B and R . Taking the.
m om en t Summ ation about R x to b e z ero , w e h a ve( x m „ 2 = 0 ) 2DK, - 2S ( S . 5 ) r O
R,
5T
=
to . 6 2 5
i~0*7S
9T
4t
I IT
r<=2ST
i.s'
A
I
T h e re fo re ., th e b en d in g m o m e n t A j- 3 is
\M -
( M g ) L] M
= IO.b2S(f.5)-S(7)
- SS. 5125 ft - tons.
We now f>o(ition -the loads SO thut the centerline is between
C and R i in order to have the maximum bending m o m e n t
under C . We find that A comes off the span, whie-h is
contrary to the assumption that All three- loads Are on
the Span. This indic-A-tes the possibility o f a m a x im u m
lending m om e-nf u n d e r C- when only loads fS A n d CAre on the span .
374
(b )
A and
B only on SpAn ■
K'=14T
When only lends A a n d & are
ST
5T
On the. span t "their resultant is
/?' - lb- tm s . The position p f
the resultant is obtained * s
'225
-4.5
folIOH/i : (E M a) W -d - ^ ( 7 )
(5)
d - 15 tb the.
©
right o f A
The position oj the loads to cause M axim um bending
m o m e n t under' A is $hou>n in which the centerline- o f
the beam is m idw ay between A and R ' .
Taking moments about Rz.
et^ualto zero, we find R, % be
R: I4T
(Z M ^ -O )
R '(2 0 )-lt(lD -2 .z fy 0
Rt r 5. tfZ5 tons
M - - ( L M a ) l -- S. H25 ( 7.75 )
* ‘t2 .0 W 7 5 ft -tons.
I f H/e now position the loads so that the centerline- of
the beam is midway between 8 and R , in order A
have- a maximum bending m om ent under &Takma a summation o f
<IT
>
hnomenfi about Rz , we g e t
t '
'T
M
L>is
R, (20 )-14(11.25) r o
W
4.5'
R, = 7. *75 tons
(f 5
M -- ( S M 8)tr 7-875(H -2 5 )~ 5 (7 )
-- 53. 513 75 f t - tons.
I <?I
10 '
..’
Rz
R,
( C)
8 and C only on the- Span-.
iMhen o n ly loads 6 a n d C are On th e span, t h e i r r e s u l­
t a n t it R " = 20 t o n s . The p o s itio n o f t h e r e s u lta n t is
O b ta in e d as fo llo w s :
C S M fs )
,R=20T
20 d - II (1 0 )
9T
IIT
J = 5.5 ft to -the
right o f B
®
5.5'
1 4 .5 ©
375
I f me now position the
loads so thAt the centerlineof the beam if m id way
b e tw e e n B a n d R , irt ord e r
tv h a v e t\. m a x im u m ben d ­
in g m o m e n t u n d e r- C- / we.
n o te -th a t L o a d A is on th e
Span w h izh is ten tra ry to
t h e a s s u m p tio n t h a t o n ly
B a n d C are o n f h e SpanW e n e x t c o n s id e r- t h e
R = 20T
■figure w h e re the loads a r e
h c-a te d so t h a t th e c e n te rlin e
o f th e beam /s m id w a y b e t­
w een C a n d /?• S e ttin g t h e
y n o m e n ts a b o u t R/ to ze ro ,
toe h a v e
( X M Ki = 0 )
2 0 ( ! D - Z . Z S ) - Ri.(2D')--0
R z = 1.15 ft -t o n s
and the bendma moment at C is
M = C Z M c ) k - 7.75(10- 2. Z5 ) ~ bO.DbZ5 ft-tons.
A comparison o f the results short that the m o st d a nge­
occurmg
rous bendmt* m o m e n t is S O ■O 6 2 5 f t - tons **r.unnsi
K/hen only loads 6 a n d C- are on -fhe span u n d e r load C .
Ans.
This is also e^ual tv I U ) , I Z 5 f t - lbs
M a x i m 1* m
S h etAv-irict
force :
I f All the' three, loads are on fhe span, the resultant R
is 7m from R z when C is over R ^ ■ it is 10 m from R,
when A is over /?/- The m a n i m unr> shearing stress is a t R^
Since it is nearer- the resultant load. There/pre u/e have.
( Z M R' = 0 )
Z S ( l S ) - R z ( 2 0 ) - - °0
R z - lb . 25 tons
The condition w h e n Only loads A and & are an th e span
need n o t b e considered because the resu lta n t o f I 1
/ - tens
IS less fhan -the reaction -foundabove-.
Whe.n loads 3 and C are on the span 1 i~he m axim um
reaction is at Rt when 3 is over R, , a n d the resultantload R " - ZO tons is S’. S m -from R , . Therefore ,
-- <>)
2 0 ( 1+ .$ ) - R 2 (20~) - 0
R.2 * I t s -tons
There Lore, the. m a x i m u m shear/na force. /
lb. 25 tons or 32-j S’OO lbs.
Ans.
S H EA R F\NP M O M E N T DIAGRAMS
C E Board - AiAcy- I ‘MS
Fihd the value o f L so that it will be in eamiibriutn ■
Draw the- shear a n d m o m e n t diagrams ana label all
coordinated and location o f the M a x im u m shear and
mom&nt. The beam rem ains to be horizontal■
1---IS'--- 1
K— IS'~H
15 Kips M
.........
■
IS Kips./f t
-'B
X
,0
C,
1
W K ip s /-f +
I--------------------------
X
E
---------------------------1
SOLUTI ON" .
Applying the conditions for e^ui tihrium / we have.
+ 1 (SFc, = 0)
15(15)+ 15(15)- ur£ = O
w£ = 4 50
£)
(Z M a
o)
-
0)
I S ( 1 5 ) ( 1 5 / 2 ) + / 5 ( i s ) ( i f 5 + 15/z)
- ty£ ft/ *-) - O
Z 7, 0 0 0
w ll
Dividing Et^- ( l )
I
^ bOft
and
From Efy ( ' ) ,
fa
by Ee^. ( O , w e g e t
X
- J£ - bD - 0
we g e t w - ^ 5 0 /(,o - 7 . 5 kips/ft
The equiva lent toad diagram for the given beam is
Shown be/otts af w ell as the shear a n d m o m e n t
d iagram s .
7.B * f %
.5 '
3C '
'
..................
■"
"J
7.5 K ips/f+
«
k'l%
"
15'
112.5
Max V = 112.5 lbs
V-D,
is '
*s '
/
IS '
15'
Man M - I6S7-5 fi-lbs
377
l&.
V e tlG N
FOR FLEXURE A N D S H E A R
CB Board - Feb. H lS
An S " x 12" X I S ' Simple b e a m carries a concentrated
load ofr b, 000 lbs a t fh e center'. Is if a d eat*ate. 1o curry
the load i f th e m a x i m u m fle x u r a l stress is /, 2 0 0 psi
and the- m a x im u m shearing Stress is 120 psi ? I f not,
■find fh e Saje load that it could carry at the center.
Hey l e d fhe. deflectionSOLUTION-
By symmetry, R, ~ Rj_ - 3 0 0 0 lbs.
n eutral axis /s
The m o m e n f o f inertia, ahput the- /
± h l
NA
.
a
in
12
12
i3 -= n s z
P - 6000*
7 .5 '
,,
7 .5 '
r 2= 3 0 0 0 *
From the shear an d m o m e n t diaaram s, fh e m a x im u m
shear force due fo fhe b,000/b lo ed is II = 3000 lbs a n d
+he m axim um bending m o m e n t is M - 1 2 15 0 0 ft - lbs. To
deterrrtine-wkether the beam is adeaiiate o r not, use can
com pute the actual flexural a n d horizontal shearing
Stresses due to the* b,000 lb b a d a n d then compare, them
with the m a x im u m stresses that fhe- beam can hat/e.
3000#
V -D ...
////
- 3000*
2Z S O O f + - lb s
M - D ----
The m axim u m flexural stress for agn /en b e n d m y m o m e n t is g ive n byS
Sf
-.
- Mp f
-.
22500 Cb)
nsz
H7.2 psi
I
378
The m axim um horizontal shearing Stress o f a g i v e n shear
force it yii/en -by.
5* -- ~
8 ^ n & i c r 11 z V6^ r
$inc-e the m axim um , -flexuralstress and the max'/m um
horizontal shearing Stress in du ced by the. b, ODD lb /oad
d id h o t e x c e e d the m ax im u n n that the beam & tn hatfe,
it is th erefore adequate to carry th e load- ■
Ans.
H.
TORSION
M E 3 o « r d - Oct. 1181
A steel shaft is transm itting SO hp at 12.00 rpm. The
nature o f the load and the type o f the sen/ice. o f the
Shaft is such that the alleu/able shearing Stress based
on p ure torsion is VOOO psi. Use & = 12 X 10^ psi.
(a . ) Compute the d ia m eter o f the shaft tb the
n e a re st com m ercial size o f shafting ■
( b ) Calculate the h rsionaldeflection o f the shaft
in ia I in degrees p e r foot of the shaft.
SOLUTION :
( a ) To determine the toraue transmitted, w e have
r = h3 00Q hr
N
S’n
T - blDOO
IZOO
= 2bZS i n -l b
To satisfy the condition o f strength, u/e apply the torsion
formula., and g e t
. Tr- _ T C d / i )
!(, r
s "
J
-Z-JV3Z
~ rrJ3
.......
/b ( 2 b2 S)
fOOO - -------——
7r d 3
d3=
3.31
d
I.
-
" o r / 2_ " comm ercial size
Ans.
( b ) The angle o f tw is t relation is given b y Q -T L ./J &
a n d the deflection p er fo o f o f the s h a f t is g ive n
by &/L ~ T/<J<S. With d = t’ S " / n/e have
9
T
—-----jpJ&
ZbZS
...
-4
- — ------- -------------- = f . ^ x / O
rad
J T ( / . S ) (lZ X IO&)
Converting into d e g r e e s , w e g e t O . O Z S ° p e r foot
o f the s h a f t . Ans.
379
APPENDIX
B
The International System
of Units (SI)
THE INTERNATIONAL SY STEM OF UNITS
E -l INTRODUCTION The importance of the regulation of weights and
measures was recognized as early as 1787 when Article 1, Section 8 , of the United
States Constitution was written. The metric system was legalized In the United
Stales in 1866; in 1893, the international metre and kilogram became the funda­
mental standards of length and mass both for metric and customary weights and
measures. International standardization began with an International Mctric Con­
vention in 1375, which established a permanent International Bureau of Weights
and Measures. The National Bureau of Standards represents the United States in
this international body.
The original metric system provided a set of units for the measurement of length,
area, volume, capacity, and mass based on two fundamental units: the metre and
the kilogram. With the addition of a unit of time, practical measurements began to
be based on the meire-kilogram-second (MKS) system. In i960, the Eleventh
General Conference .on Weights and Measures formally adopted the International
System of Units, for which the abbreviation is SI in all languages, as the interna­
tional standard. Thirty-six countries, including the United States, participated in
this conference.
E-2 SI UNITS The International System of Units adopted by the conference
includes three classes of units: (1) base units, (2) supplementary units, and (3)
derived units. The system is founded on the seven base units listed in Tible E-l.
Table E -l
Quantity
Name of
Base SI Unit
Length
Mass
Time
Electric current
Thermodynamic temperature
Amount of substance
Luminous intensity
metre
kilogram
second
ampere
kelvin
mole
candela
Symbol
m
kg
s
A
K
mol
cd
380
Certain units of the international system have not been classified under
either base units or derived units. These units, listed in Table E-2, are
called supplementary units and may be regarded either as base units or as
derived units.
Table E -2
Quantity
Name of
Supplementary SI Unit
Symbol
Plane angle
Solid angle
radian
steradian
rad
sr
Derived units are expressed algebraically in terms o f base units and/or
supplementary units. Their symbols are obtained by means o f the
mathematical signs of multiplication and division. For example, the SI unii
for velocity is metre per second (m /s) and the SI unit for angular velocity
is radian per second (rad/ s). For some o f the derived units, special names
and symbols exist; those o f interest in mechanics are listed in Table E-3:
E-3 M ULTIPLES OF SI U N ITS Prefixes are used to form names and
symbols of multiples (decimal multiples and sub-multiples) of SI units. The choice
of the appropriate multiple is governed by convenience and should usually
chosen so that the numerical values will be between 0.1 and 1000. Only one prefix
should be used in forming a multiple of a compound SI unit, and prefixes in the
denominator should be avoided. Approved prefixes with their names and symbols
are listed in Table E -4 .
be
Table E-3
Quantity
Special Name Symbol
Derived SI Unit
—
Area
square metre
m2
Volume
cubic metre
m3
Linear velocity
metre per second
m/s
Angular velocity radian per second
—
rad/s
Linear acceleration metre per second squared
m/sJ
Frequency
(cycle) per second
hertz
Hz
Density
kilogramper cubic metre
kg/m3
Force
kilogram-metre per second newton
N
squared
—
Moment of force newton•metre
N •m
Pressure
newton per metre squared
pascal
Pa
Stress
newton per metre squared
pascal
Pa or N/mJ
Work, energy
newton metre
joule
J
Power
joule per second
watt
W
—
—
—
—
381
Table E -4
Prefix
Factor by which
Unit is Multiplied
Name
Symbol
1012
tera
T
10»
106
giga
mega
G
M
103
10*
10
kilo
hecto*
deca*
k
h
da
10"1 .
deci*
10-2
centi*
d
c
10- J
milli
m
10'6
10"9
micro
f*
n
10- 12
pico
1 0 ,s
femto
P
f
10- 1
8
atto
a
nar.o
*T o be avoided where possible.
E-4 CONVERSION BETWEEN THE SI AND GRAVITATIONAL ENG­
LISH SYSTEMS. As the use of SI becomes more commonplace in the United
States, engineers will be required to be familiar with both SI and the English
system in common use today. As an aid to interpreting the physical significance of
answers in SI units for those more accustomed to the English system, the following
conversion factors are provided:
CONVERSION FACTORS
English to SI
SI to English
length:
Area:
Volume:
1 tn. = 25.40 mm
I ft-0.3048 m
1 m —39.37 in.
1 m —3.281 ft
1 in.2“ 6.452 cm2
1cm2-0.1550 in.2
1 ft2- 0.092 90 m2
1 m2- 10.76 ft2
1 in.3- 16.39 cm3
1 ft3-0.028 32 m3
1 cm3- 0.06102 in.3
1 m3- 35.31 ft3
Moment of Inertia:
1 in."*—41.62 cm4
1 cm4-0.024 03 in.4
Force;
1 lb -4.448 N
1 N - 0.2248 lb
Distributed Load:
1 lb/ft - 14.59 N/m
1 k N / m -68.53 lb/ft
Pressure or Stress:
1 psi —6.895 kPa
1 k s i - 6.895 MPa
1 MPa - 145.0 psi
1 MPa - 145.0 psi
Bending Moment or Torque:
1 ft-lb-1.356 N m
1 N m —0.7376 ft lb
Work or Energy:
1 ft l b - 1.356 J
1 J - 0.7376 ft lb
Power:
1 hp-745 7 W
1 k W - 1.341 hp
382
C o n v e rs io n F a c to rs
C la s s ifie d
L is t O l U n it s
T o convert from
(Symbols o f SI units given in parentheses)
to
Multiply by
metre/second1 (m /s’ ) . . . . ................
metre/second' ( m / s ') .......................
metre/second’ (m /s’ ) .......................
metre/second’ (m /s’ ) .......................
3.048 000*1 0 1
9.806 650*1 i <*i
1.000 000*1 O'
2.540 000*1 (>.'
a c r e ......................................................... metre’ (m ’ ) ........................................
barn
..................................................... metre’ (m 2) ........................................
circular mih ........................................... metre’ (m ’ ) ........................................
fo o t’ ....................................................... m etre’ (m 1) .....................................
m etre1 (m ’ ) .......................................
inch’ ......................................................
m ile’ (U .S . s t a t u t e ).............................
metre’ (m 2) .......................................
section ................................................... metre’ (m ’ ) .......................................
tow nship................................................. metre’ (m 1) .......................................
yard1 ......................................................
metre’ (m ’ j .......................................
4 046 856 I * (I*
1.000 000*1
5.067 075 I
10
9.290 304*1
01
6.451 600*1IH
2.589 988 !•*<>»•
2.589 988 E <Ilf.
9.323 957 E t(H
8.361 274 I 01
ACCELERATION
foot/second' .........................................
free fall, standard .................................
gal ( g a l i l e o ) .........................................
inch/second1 .........................................
AREA
BEND ING M O M E NT OR TORQUE
dyne-tentim ctre.....................................
kilogram-force-metre ...........................
ounce-force-inch ...................................
peund-force-inch...................................
pou n d-force-foot...................................
newton-metre ( N m ) .......................
newton-metre ( N - m ) .......................
newton-metre( N - m ) .........................
newton-metre( N - m ) .........................
newton-metre ( N - m ) .......................
1.000 000*b ()(
9.806 650*1- t <*>
7 061 552 E 01
1.129 848 E
01
1.355 8 18 E^Otl
ENERGY / AREA TIM E
Btu (thermochemical )/ foo t’ -second . . . watt/metre’ (W /m "|
Btu (therm ochem ical)/foot’ -minute .
. watt/metre’ (W / m ’ )
Btu (therm ochem ical)/foot’ -h our.......... watt/metre’ (W / m ’ )
Btu (thermochemical)/inch’ -second .. . watt/metre’ (W / m ')
........................
........................
........................
........................
1.134 893 E + 04
1.891 489 E-tOl
3.152 481 E-MHI
1.634 246 E + 06
calorie , (thermochemical )/centimetre’ minute ......................................... watt/metre’ (W / m ’ ) ..................................
6.973 333 E + 0J
erg/centimetre’ -second........................... watt/metre’ (W / m ’ ) ........................
1.000 000*E
watt/centimetre’ .................................... watt/metre’ (W / m 1) .........................
1.000 000* E-t 04
01
A n asterisk ( * ) after the sixth decimal
place indicates that the conversion factor is exact
and that all subsequent digits are zero. A ll other
conversion factors have been rounded.
Reprinted by permission
Copyright the American Society
for Testing and Materials.
383
To'convert from
to
Multiply by
.............. .......................... .........................
ENERGY (INCLUDES W O RK)
British thermal unit (International
T a b le )" ............................................. joule (J ) ......................... .................
British thermal unit (m e a n ).................
joule (J ) ......................... .................
British thermal unit (therm ochem ical). joule (J ) ..........i ............. ..................
British thermal unit (39 F ) .................
joule ( J ) ......................... .................
joule (J ) ......................... .................
British thermal unit (60 F ) .................
1.055 056 E + 03
1.055 87 E + 03
1.054 350 E + 03
1.059 67 E + 03
1.054 68 E + 03
calorie (International T a b l e )...............
calorie (m e a n ).......................................
caloric (th erm ochem ical)...................
calorie {15 C ) .......................................
calorie (2 0 C ) .......................................
calorie (kg. International T ab le) ........
calorie (kg, m ean) ...............................
calorie (kg, thermochemical) .............
electron v o l t .......................................
" g ........................................................
joule (J ) ......................... ..................
joule (J ) ......................... .................
joule (J ) ......................... .................
joule (J ) ......................... ..................
joule (J ) ......................... .................
joule ( J ) ...................
.................
joule (J ) .1..................... ..................
joule ( J ) ......................... .................
joule ( J ) ......................... ..................
joule (J ) ......................... ..................
4186 800*E + 00
4 190 02 E +00
4 184 000* E * 00
4.185 80 E +00
4.181 90 E +00
4.186 800* E + 03
4.190 02 E+03
4.184 000*E + 03
1.602 19 E - 19
1.000'000* E - 07
fo o t-p o u n d T o rc t...................................
foot-poundal .........................................
joule (J ) ......................... ..................
joule (J ) ......................... ..................
1.355 818 E +00
4.214 O il E 02
joule, international U.S. (Jim--us)*
..............
joule, U.S. legal 1948
joule (J ) ......................... .................
joule (J ) ......................... .................
1.000 182 E +00
1.000 017 E +00
kilocalorie (International T a b le) .......
kilocalorie (m ean) ...............................
kilocalorie (therm ochem ical) ..............
kilow att-hour.........................................
kilowatt-hour, international U.S.
(kWhlKT-U*)® .......................
kilowatt-hour, U.S. legal 1948
(k W h ,» . .) .......................................
joule (J )
joule (J )
joule (J )
joule (J )
..................
.................
..................
.................
4.186 800*E+03
4 190 02 E+03
4.184 000»E+03
3.600 000*E+06
joule ( J ) ......................... ..................
3 600 655 E + 06
joule ( J ) ........................... ..................
3.600 061
E +06
ton (nuclear equivalent o f T N T )
........
joule (J ) ......................... .................
4.20
E+09
w a tt-h d u r...............................................
w att-secon d ...........................................
jdult (J ) ......................... ..................
joufc (J ) ......................... ..................
3.600 090* E + 03
1.000 000*E+00
.........................
.........................
.........................
.........................
FORCE
dyne .......................................................
newton ( N ) ................... ...............
1 000 000 *E - 05
kilopond ...............................................
kip
...................................................
.................
newton ( N ) ................... .................
newton ( N )
.................. .................
9.806 650*E + 00
9.806 650* E +00
4.448 222 E + 03
ounce-force (avoirdupois) ...................
newton ( N )
................... .................
2.780 139 E -01
pound-force (Ib f avoirdupois) O ............
newton ( N ) ................... .................
newton ( N ) ................. ..................
4.448 222 E +00
1.382 550 E -01
pound.tl
.................................................
■ This value was adopted in l956.-$ome of the older International Tables use the value 1.055 04 E+03.
The'exact conversion factor is 1.055 055 852 62*E+03.
. • In 1948 a new international agreement was reached on absolute electrical units, which changed the value of the volt
*scd in this country by about 300 parts per million. Again in 1969 a new base of reference was internationally adopted
making a further change of 8.4 parts per million. These changes (and also changes in ampere, joule, watt, coulomb) re­
quire careful terminology and conversion factors for exact use of old information. Terms used in this guide are
volt as used prior to January 1948—volt, international U.S. (V INr.vv)
volt as used between January 1948 and January 1969—volt, U.S. legal 1948 (VUM.«a)
volt as used since January 1969—volt (V)
Identical treatment is given the ampere, coulomb, watt, and joule. Since the henry, farad, and ohm' were not changed in
1969, only conversions as listed first above are shown.