PROBLEM 7.2
For the given state of stress, determine the normal and shearing stresses
exerted on the oblique face of the shaded triangular element shown. Use a
method of analysis based on the equilibrium of that element, as was done in the
derivations of Sec. 7.2.
SOLUT!ON
80 y'Pq
Stresses
.;,
Areas
Forces
LF = 0:
oA
-
80A cos55o cos55o + 40A sin 55o sin 55o = 0
o = 80 cos255o -
i'r IF
40sin2 55.
o = -0.521MPa
{
= 0:
rA
-
SOAcos 55o sin 55o
r = 120 cos 55o sin 55o
-
4OA sin 55ocos 55o
r = 56.4 MPa {
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PROBLEM 7.9
For the given state of stress, determine (c) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
d,
(a)
tan 20, =
or-o,
2t xy
--
=
-60 MPa
-60 + 4o
(2X3s)
Oy
=-40MPa
T,y
=35 MPa
=0.2857
9. = 8.0o,
20- =15.95"
98.0' <
(b)
-eo:
(c)
o' = ouu"
qo)'z+
(3s)2
ox+ov
=--4=
22
-60-40
c*
=36.4 MPa
{
o'=-50.0 MPa {
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reproduced, or distributed in any form or by any means, without the prior written
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to teachers and educators permitted by McGraw-Hill
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PROBLEM 7.13
i |::i
For the given state of stress, determine the normal and shearing stresses after the element
shown has been rotated through (a) 25" clockwise, (b) l0' counterclockwise.
,*
1
!
:
SOLUTION
Or=0 Oy=8kSi 7'r=5kSi
o'-oY
ox+ov
=4
.1
L'
(a) 0=-25"
ksi
=-4
ox+ov o -6.
oo
=T*-
,ro,
=-1J
o,,
=
o*-o,
sin20
ksi
cos20 +
-
f
sin20
* r,, cos20
ox+ov o -o
T
t,
"os
20
- t,,
sin 20
20=-50"
o,,--4-4cos(-50")+5sin(-50")
r,,r,=4 sin (-50')+5 cos (-50')
ot=-2'40ksi {
t,'r'=0'15 ksi
<(
o*' =7'95 ksi
{
or,--4+4cos(-50')-5sin(-50)oi=10'40ksi{
(b)
0
=10"
20
=20"
(20')
tr,r, = 4sin (20") + 5 cos (20')
o,, = 4- 4 cos (20") + 5 sin
ou, = 4+ 4 cos (20")
-
5 cos
(20')
ttv' = 6'07 ksi 1
or' = 6'05 ksi
{
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reproduced, or distributed in any form or by any means, without the prior written
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!
ii;\1i't,,
PROBLEM 7.17
t
,,
I
t,,.t,,.,,.
,, , , \li,r
itt
T
The grain of a wooden member forms an angle of l5o with the vertical. For the
state-of stress shown, determine (a) the in-plane shearing stress parallel to the
T":
'',.'"
;
SOLUTION
6" = -4MPa dy = -1.6 MPa
0 = -15" 20 = -30"
o--o
(a\ r*, -- -ai!
- - -4 -
sin20
(-1'6)
sin
* t,
T"y = 0
cos20
(-30.) + 0
riy, = -0.600 MPa
<(
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prior written permission of the publisher, or used beyond the limited
reproduced, or distributed in any form or by any means, without the
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PROBLEM 7.19
steel pipe of 12-in. outer diameter is fabricated from {-in.-thick plate by
welding along a helix that forms an angle of 22.5" with a plane perpendicular
to the ixis of the pipe. Knowing that a 40-kip axial force P and an 80-kip ' in.
A
+: I,
. --;-*
I
+tl+I I
1l
II
1tl:::::::i
.:.lii:,..]
torque T, each directed as shown, are applied to the pipe, determine
in directions, respectively, normal and tangential to the weld.
Si-l
ii, : ,\\:.,1,1"t
'"1.. .
..
.r.r
o
and
t
:c
iA
:::.:-: $Y l
SOLUTION
y,
lg
d,/ =12 in., z =!d, = 6 in., t = 0.25 in.
", 2_
=c2-t=5.75in.
n(62 -5.7s2) =
A= n(c) -
r =+(r: -ri)=tro-
LSocZ
lc"
i
l[__il_.
c1
"?)=
1 1-33411a;
9.2284 in2
-s.7s4)=378.67 ina
I
Stresses:
P
Q =--
=
A
40
- 9.2284 = -4.3344 ksi
Tc-,
,\
J
_ (80x6) = 1.5063 ksi
318.67
6r=0,
Choose the
x'and y'
Then
oy=-4.3344ksi,
?ry =1.5063 ksi
axes, respectively, tangential and normal to the weld.
o*=oy' and t*=Tiy'
ox+ov or-6
o,,
=
-
7
fros2?
0=22'5"
- r,r sin20
_ (-4.3344) _L-e$344)lcos
45o _ 1.5063 sin 45o
o,=-436ksi {
= -4.76 ksi
o..
-
o..
^), -4sin20
2
r,,,, =
=
t-(-4.3344)l
--srn 2
= -0.467 ksi
+ t,rcos 20
.
45o + I .5063 cos 45o
r,=4.467 ksi {
pRopRIETARy MATERIAL. O 2012 The McGraw-Hiil companies, Inc. All rights reserved. No part of this Manual may be displayed,
publisher, or used beyond the limited
reproduced, or distributed in any form or by any means, without the prior wriften permission of the
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A student using this manual is using it
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individual
educators permitted by McGraw-Hill for their
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