© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2=29: The block is deformed into the position shown by
the dashed lines. Determine the shear strain at corners C
and£».
15 mmL_
-—1--
r
85 mm
110 mm
100 mm
D
100 mm
Gtomttiy :
8 = 90"+Jin"' ( ^\ 97.84° « 1.70759 rad
0=jr-1.70759" 1.43401 rid
Shtar Strain :
HO(I>">
X
It
it
It
</£,)„ = J - J - J - l <
tQO mm
*&~4. f\n icsi was pcnurnrcu on a sieei speiauicil
having an original diameter of 0.503 in. and gauge length of
2.00 in. The data is listed in the table. Plot the stress-strain
diagram and determine approximately the modulus of
elasticity, the yield stress, the ultimate stress, and the rupture
stress. Use a scale of 1 in. = 20 ksi and 1 in. — 0.05 in./in.
Redraw the elastic region, using the same stress scale but a
strain scale of I in. = 0.001 in./in.
~~~
~
Load (Kip) Elongation (in.)
®^
_j ^
8.00
11.00
11.80
11.80
12.00
16.60
20.00
&(**<•')
21.50
19.50
tU>f
0
0.0005
0.0015
0.0025
0.0035
0.0050
0.0080
0.0200
0.0400
0.1000
0.2800
0.4000
0.4600
__^
y^--— o^ -J2i^L£2t
Bf
to-
L = 2.00 in.
/
1
jj
a^
f*V*()*rfrtt
7
«
- fS
40-
/
/
/
zo- //
/
/
1 tS
O
0.
o-eou
fis't
4j
£>w><» - — — = 32.0(l(r)lui
'
f~
0,10
O'Oot.
"*"!
O-IS
O-0*3
Ttr
O.£o
0-00^
_,-.
0-25
0-fOS
Am
^ /-- .
^J
76
o(ksi) e(in./in.)
0
0
7.55
0.00025
23.15 0.00075
40^6 0.00125
55.36 0.0017S
5938 00025
5938
0.0040
60.39
0.010
83J4
0.020
100.65
0.050
108.20 0.140
«13 O
820°
93W
.ao
Ji-23. The beam is supported by a pin at C and an A-36
steel guy wire AB. If the wire has a diameter of 0.2 in.,
determine how much it stretches when a distributed load of
w = 100 Ib/ft acts on the pipe. The material remains
elastic.
+ZMc =
ft
FMsn30f(lO)
- 0.1<10)(5) m O.
Fa = 1.0 kip
a * Ee;
31.83 « 29(103)£«;
0.0010981 in./in.
120
0.0010981<-^-—) = 0.152 in.
oos 30°
84
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•$-39. The rigid pipe is supported by a pin at C and an
A-36 guy wire AB, If the wire has a diameter of 0.2 in.,
determine the load P if the end B is displaced 0.10 in. to the
right. Est = 29(103) ksi.
Gtemetry :
:2ll
dm 0.05968°
96
a = 90° + 0.05968" = 90.05968°
AC = 96 lan 30" =55.4256 in
AB-.
96
cos 30°
: 110.8513 in
AB' = /962 + S5.42562 -2(96)(55.4256)cos 90.05968°
= 110.9012 in.
Normal Stress and Strain :
AB'-AB
AB
110.9012-110.8513
'
110.8513
= 0.4510(10'') in./in.
Applying Hooke's law
041 =
&H
, = 29.0( 103) 0.4510( 10'5) = 13.08 ksi
Thus.
= 410.85 Ib
Equation of Equilibrium :
410.85sin30e(8)-/'(8)=0
P*=205lb
AIM
r
44. The A-36 steel column is used to support the
symmetric loads from the two floors of a building. Determine
the vertical displacement of its top, A, if PI = 40 kip,
P2 = 62 kip, and the column has a cross-sectional area of
23.4 in2.
12ft
12ft
4o nip +O kip
Internal Forets : As shown on FBD (a) and (b)
Displacement .c _ y ^ _ -80.0(121(12)
' = ^ AE ~ 23.4(2<>.OHIO>)
= - 0.0603 in.
(-204)(I2)(12)
23.4(29.0)(10')
Ans
Negative sign indicates thai end A moves toward end C.
I
cfc;
95
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4"3S. The column is constructed from high-strength concrete and four A-36 steel reinforcing rods. If it is subjected
to an axial force of 800 kN, determine the required diameter
of each rod so that one-fourth of the load is carried by the
steel and three-fourths by the concrete. Est = 200 GPa,
Ec = 25 GPa.
Equilibrium : Require Pa = -(800) = 200 kN and
4
/>„„ = ^
4
Compatibility :
S
con = Sn
(0..3 2 -/l tt )(25.0)(IO»)
A M (200)(10»)
0.09PSI
0.09( 200)
8(600)+ 200
d- 0.03385m = 33.9mm
Ans
The assembly has the diameters and material
make-up indicated. If it fits securely between its fixed
supports when the temperature is TI = 70°F, determine
the average normal stress in each material when the
temperature reaches TI = 110°F.
2014-T6 Aluminum
304 Stainless
- C 86100 Bronze
steel
J_
D
IF. =0;
= 277.69 kip
a,, « -'— = 2.46 ksi Ana
377 fiQ
0M = -— as 22.1 ksi
AIM
137