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7
Analysis of Stress
and Strain
Plane Stress
1200 psi
An element in plane stress is subjected to stresses sx ⫽ 4750 psi,
sy ⫽ 1200 psi, and txy ⫽ 950 psi, as shown in the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 60˚ from the
x axis, where the angle u is positive when counterclockwise. Show these stresses on a
sketch of an element oriented at the angle u.
Problem 7.2-1
950 psi
4750 psi
Solution 7.2-1
sx ⫽ 4750 psi
sy ⫽ 1200 psi
txy ⫽ 950 psi
u ⫽ 60°
sx1 ⫽
sx ⫺ sy
sx + sy
2
sx1 ⫽ 2910 psi
+
2
;
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin(2u) + txy cos(2u)
tx1y1 ⫽ ⫺2012 psi
cos(2u) + txy sin(2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 3040 psi
Problem 7.2-2 Solve the preceding problem for an element in plane stress
subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa, as shown in
the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 30˚ from
the x axis, where the angle u is positive when counterclockwise. Show these stresses
on a sketch of an element oriented at the angle u.
;
80 MPa
28 MPa
100 MPa
571
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Analysis of Stress and Strain
Solution 7.2-2
sx ⫽ 100 MPa
sy ⫽ 80 MPa
txy ⫽ 28 MPa
u ⫽ 30°
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
sx1 ⫽ 119.2 MPa
2
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin(2u) + txy cos(2u)
tx1y1 ⫽ 5.30 MPa
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 60.8 MPa
;
;
Problem 7.2-3 Solve Problem 7.2-1 for an element in plane stress subjected to
2300 psi
stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi, as shown in the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 50˚ from the x
axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch
of an element oriented at the angle u.
2500 psi
5700 psi
Solution 7.2-3
sx ⫽ ⫺5700 psi
sy ⫽ ⫺2300 psi
txy ⫽ 2500 psi
u ⫽ 50°
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
sx1 ⫽ ⫺1243 psi
2
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ 1240 psi
cos (2u) + txy sin (2u)
;
Problem 7.2-4 The stresses acting on element A in the web of
a train rail are found to be 40 MPa tension in the horizontal
direction and 160 MPa compression in the vertical direction
(see figure). Also, shear stresses of magnitude 54 MPa act in the
directions shown.
Determine the stresses acting on an element oriented at a
counterclockwise angle of 52˚ from the horizontal. Show these
stresses on a sketch of an element oriented at this angle.
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺6757 psi
;
160 MPa
A
A
Side
View
40 MPa
54 MPa
Cross
Section
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SECTION 7.2
Plane Stress
573
Solution 7.2-4
sx ⫽ 40 MPa
sy ⫽ ⫺160 MPa
txy ⫽ ⫺54 MPa
u ⫽ 52°
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
2
sx1 ⫽ ⫺136.6 MPa
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺84.0 MPa
cos(2u) + txy sin(2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 16.6 MPa
;
Problem 7.2-5 Solve the preceding problem if the normal
and shear stresses acting on element A are 6500 psi, 18,500 psi,
and 3800 psi (in the directions shown in the figure).
Determine the stresses acting on an element oriented at a
counterclockwise angle of 30˚ from the horizontal. Show these
stresses on a sketch of an element oriented at this angle.
;
18,500 psi
6500 psi
A
A
3800 psi
Side
View
Cross
Section
Solution 7.2-5
sx ⫽ 6500 psi
sy ⫽⫺18500 psi
txy ⫽ ⫺3800 psi
u ⫽ 30°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ ⫺3041 psi
;
2
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺12725 psi
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺8959 psi
Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected
to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile
stresses of magnitude 5.5 MPa in the vertical direction (see figure). Also, shear stresses
of magnitude 10.5 MPa act in the directions shown.
Determine the stresses acting on an element oriented at a clockwise angle of 35˚
from the horizontal. Show these stresses on a sketch of an element oriented at this
angle.
;
5.5 MPa
27 MPa
10.5 MPa
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Analysis of Stress and Strain
Solution 7.2-6
sx ⫽ ⫺27 MPa
sy ⫽ 5.5 MPa
txy ⫽ ⫺10.5 MPa
u ⫽ ⫺35°
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
2
sx1 ⫽ ⫺6.4 MPa
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺18.9 MPa
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx ⫹sy ⫺ sx1
sy1 ⫽ ⫺15.1 MPa
;
Problem 7.2-7 The stresses acting on element B in the
web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction (see figure). Also, shear stresses
of magnitude 3800 psi act in the directions shown.
Determine the stresses acting on an element oriented at a
counterclockwise angle of 40˚ from the horizontal. Show
these stresses on a sketch of an element oriented at this angle.
;
2600 psi
B
14,000 psi
B
3800 psi
Side
View
Cross
Section
Solution 7.2-7
sx ⫽ ⫺14000 psi
txy ⫽ ⫺3800 psi
sy ⫽ ⫺2600 psi
u ⫽ 40°
sx1 ⫽
tx1y1 ⫽ ⫺
sx ⫺ sy
2
tx1y1 ⫽ 4954 psi
sx ⫺ sy
sx + sy
2
+
sx1 ⫽ ⫺13032 psi
2
cos (2u) + txy sin (2u)
sin (2u) + txy cos (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺3568 psi
;
;
Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on
element B are 46 MPa, 13 MPa, and 21 MPa (in the directions shown in the figure) and
the angle is 42.5˚ (clockwise).
13 MPa
21 MPa
B
46 MPa
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SECTION 7.2
Plane Stress
Solution 7.2-8
sx ⫽ ⫺46 MPa
sy ⫽ ⫺13 MPa
txy ⫽ 21 MPa
u ⫽ ⫺42.5°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ ⫺51.9 MPa
tx1y1 ⫽ ⫺
sy ⫽112 psi
txy ⫽ ⫺120 psi
u ⫽ 30°
sx ⫺ sy
sx + sy
+
⫽ 187 psi
tx1y1 ⫽ ⫺
;
;
y
112 psi
30°
350 psi
O
x
120 psi
Seam
Plane stress (angle ␪ )
sx ⫽ 350 psi
2
sin (2u) + txy cos (2u)
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺7.1 MPa
;
pond is subjected to stresses sx ⫽ 350 psi, sy ⫽112 psi,
and txy ⫽ ⫺120 psi, as shown by the plane-stress
element in the first part of the figure.
Determine the normal and shear stresses acting on a
seam oriented at an angle of 30° to the element, as
shown in the second part of the figure. Show these
stresses on a sketch of an element having its sides
parallel and perpendicular to the seam.
sx1 ⫽
2
tx1y1 ⫽ ⫺14.6 MPa
cos (2u) + txy sin (2u)
Problem 7.2-9 The polyethylene liner of a settling
Solution 7.2-9
sx ⫺ sy
sx ⫺ sy
2
⫽ ⫺163 psi
2
cos 2u + txy sin 2u
;
sin 2u + txy cos 2u
;
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 275 psi
;
The normal stress on the seam equals 187 psi
;
tension.
The shear stress on the seam equals 163 psi, acting
clockwise against the seam.
;
575
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Analysis of Stress and Strain
Problem 7.2-10 Solve the preceding problem if the normal
y
and shear stresses acting on the element are sx ⫽ 2100 kPa,
sy ⫽ 300 kPa, and txy ⫽ ⫺560 kPa, and the seam is oriented
at an angle of 22.5° to the element (see figure).
300 kPa
22.5°
2100 kPa
x
O
Seam
560 kPa
Solution 7.2-10
Plane stress (angle ␪ )
sx1 ⫽
sx ⫺sy
sx + sy
+
2
⫽ 1440 kPa
tx1 y1 ⫽ ⫺
2
cos 2u + txy sin 2u
;
sx ⫺ sy
2
⫽ ⫺1030 kPa
sin 2u + txy cos 2u
;
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 960 kPa
sx ⫽ 2100 kPa
u ⫽ 22.5°
sy ⫽ 300 kPa
txy ⫽⫺560 kPa
;
The normal stress on the seam equals 1440 kPa
tension. ;
The shear stress on the seam equals 1030 kPa, acting
clockwise against the seam. ;
Problem 7.2-11 A rectangular plate of dimensions 3.0 in. * 5.0 in. is formed
by welding two triangular plates (see figure). The plate is subjected to a tensile
stress of 500 psi in the long direction and a compressive stress of 350 psi in the
short direction.
Determine the normal stress sw acting perpendicular to the line of the weld
and the shear tw acting parallel to the weld. (Assume that the normal stress sw
is positive when it acts in tension against the weld and the shear stress tw is
positive when it acts counterclockwise against the weld.)
350 psi
ld
We
3 in.
5 in.
500 psi
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SECTION 7.2
Solution 7.2-11
Plane Stress
577
Biaxial stress (welded joint)
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin 2u + txy cos 2u ⫽ ⫺375 psi
sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺125 psi
STRESSES ACTING ON THE WELD
sx ⫽ 500 psi
u ⫽ arctan
sx1 ⫽
sy ⫽ ⫺350 psi
txy ⫽ 0
sw ⫽ ⫺125 psi
3 in.
⫽ arctan 0.6 ⫽ 30.96°
5 in.
sx ⫺ sy
sx + sy
+
2
2
tw ⫽ 375 psi
;
;
cos 2u + txy sin 2u
⫽ 275 psi
12.0 MPa
Problem 7.2-12 Solve the preceding problem for a plate of dimensions
100 mm * 250 mm subjected to a compressive stress of 2.5 MPa in the
long direction and a tensile stress of 12.0 MPa in the short direction
(see figure).
Solution 7.2-12
ld
We
100 mm
250 mm
Biaxial stress (welded joint)
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin 2u + txy cos 2u ⫽ 5.0 MPa
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 10.0 MPa
STRESSES ACTING ON THE WELD
sx ⫽ ⫺2.5 MPa
u ⫽ arctan
sx1 ⫽
2
sy ⫽ 12.0 MPa
txy ⫽ 0
100 mm
⫽ arctan 0.4 ⫽ 21.80°
250 mm
sx ⫺ sy
sx + sy
+
⫽ ⫺0.5 MPa
2.5 MPa
2
cos 2u + txy sin 2u
sw ⫽ 10.0 MPa
;
tw ⫽ ⫺5.0 MPa
;
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Analysis of Stress and Strain
Problem 7.2-13 At a point on the surface of a machine the material
is in biaxial stress with sx ⫽ 3600 psi, and sy ⫽ ⫺1600 psi, as
shown in the first part of the figure. The second part of the figure
shows an inclined plane aa cut through the same point in the material
but oriented at an angle u.
Determine the value of the angle u between zero and 90° such
that no normal stress acts on plane aa. Sketch a stress element
having plane aa as one of its sides and show all stresses acting on
the element.
y
1600 psi
a
u
3600 psi
O
x
a
Solution 7.2-13
Biaxial stress
STRESS ELEMENT
sx1 ⫽ 0
u ⫽ 56.31°
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 2000 psi
sx ⫽ 3600 psi
sy ⫽ ⫺1600 psi
txy ⫽ 0
tx1y1 ⫽ ⫺
sx ⫺ sy
2
;
sin 2u + txy cos 2u
⫽ ⫺2400 psi
Find angle u for s ⫽ 0.
s ⫽ normal stress on plane a-a
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
2
cos 2u + txy sin 2u
⫽ 1000 + 2600 cos 2u(psi)
For sx1 ⫽ 0, we obtain cos 2u ⫽ ⫺
‹
2u ⫽112.62° and
1000
2600
u ⫽ 56.31°
Problem 7.2-14 Solve the preceding problem for sx ⫽ 32 MPa
y
and sy ⫽ ⫺50 MPa (see figure).
50 MPa
a
u
32 MPa
O
x
a
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SECTION 7.2
Solution 7.2-14
Plane Stress
579
Biaxial stress
STRESS ELEMENT
sx1 ⫽ 0
u ⫽ 38.66°
sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺18 MPa
sx ⫽ 32 MPa
tx1y1 ⫽ ⫺
sy ⫽ ⫺50 MPa
txy ⫽ 0
sx ⫺ sy
2
sin 2u + txy cos 2u
⫽ ⫺40 MPa
Find angles u for s ⫽ 0.
;
;
s ⫽ normal stress on plane a-a
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
cos 2u + txy sin 2u
⫽ ⫺9 + 41 cos 2u ( MPa)
For sx1 ⫽ 0, we obtain cos 2u ⫽
‹
9
41
2u ⫽ 77.32° and u ⫽ 38.66°
;
Problem 7.2-15 An element in plane stress from the frame of a racing car is
y
oriented at a known angle u (see figure). On this inclined element, the normal and
shear stresses have the magnitudes and directions shown in the figure.
Determine the normal and shear stresses acting on an element whose sides
are parallel to the xy axes, that is, determine sx, sy, and txy. Show the results on
a sketch of an element oriented at u ⫽ 0˚.
2475 psi
3950 psi
u = 40°
14,900 psi
O
Solution 7.2-15
Transform from u ⫽ 40° to u ⫽ 0°
sx ⫽ ⫺14900 psi
txy ⫽ 2475 psi
sy ⫽ ⫺3950 psi
2
sin (2u) + txy cos (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sx ⫺ sy
sx + sy
2
sx ⫺ sy
tx1y1 ⫽ ⫺4962 psi
u ⫽ ⫺40°
sx1 ⫽
tx1y1 ⫽ ⫺
+
sx1 ⫽ ⫺12813 psi
2
;
cos (2u) + txy sin (2u)
sy1 ⫽ ⫺6037 psi
;
x
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CHAPTER 7
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Analysis of Stress and Strain
Problem 7.2-16 Solve the preceding problem for the element shown in the
y
figure.
24.3 MPa
u = 55°
62.5 MPa
O
x
24.0 MPa
Solution 7.2-16
Transform from u ⫽ 55° to
sx ⫽ ⫺24.3 MPa
txy ⫽ ⫺24 MPa
u ⫽ 0°
sy ⫽ 62.5 MPa
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺32.6 MPa
u ⫽ ⫺55°
sx1 ⫽
tx1y1 ⫽ ⫺
;
sy1 ⫽ sx + sy ⫺ sx1
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ 56.5 MPa
cos (2u) + txy sin (2u)
sy1 ⫽ ⫺18.3 MPa
;
;
Problem 7.2-17 A plate in plane stress is subjected to normal stresses
y
sx and sy and shear stress txy, as shown in the figure. At counterclockwise angles
u ⫽ 35˚ and u ⫽ 75˚ from the x axis, the normal stress is 4800 psi tension.
If the stress sx equals 2200 psi tension, what are the stresses sy and txy?
sy
txy
O
sx = 2200 psi
x
Solution 7.2-17
sx ⫽ 2200 psi
sy unknown txy unknown
At u ⫽35° and
u ⫽ 75°, sx1 ⫽ 4800 psi
Find sy and txy
sx1 ⫽
sx ⫺ sy
sx + sy
2
+
2
For
u ⫽ 35°
sx1 ⫽ 4800 psi
4800 psi ⫽
cos (2u) + txy sin (2u)
2200 psi ⫺ sy
2200 psi + sy
+
2
2
* cos (70°) + txy sin (70°)
or 0.32899 sy + 0.93969 txy ⫽ 3323.8 psi
(1)
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Page 581
Plane Stress
581
0.93301sy + 0.50000 txy ⫽ 4652.6 psi
(2)
SECTION 7.2
For
u ⫽ 75°:
or
sx1 ⫽ 4800 psi
4800 psi ⫽
Solve Eqs. (1) and (2):
2200 psi ⫺ sy
2200 psi + sy
sy ⫽ 3805 psi
txy ⫽ 2205 psi
;
+
2
2
* cos (150°) + txy sin (150°)
Problem 7.2-18 The surface of an airplane wing is subjected to plane stress
y
with normal stresses sx and sy and shear stress txy, as shown in the figure. At a
counterclockwise angle u ⫽ 32˚ from the x axis, the normal stress is 37 MPa
tension, and at an angle u ⫽ 48˚, it is 12 MPa compression.
If the stress sx equals 110 MPa tension, what are the stresses sy and txy?
sy
txy
O
sx = 110 MPa
x
Solution 7.2-18
sx ⫽ 110 MPa
sy unknown
At u ⫽ 32°, sx1 ⫽ 37 MPa
sxy unknown
(tension)
At u ⫽ 48°, sx1 ⫽ ⫺12 MPa
(compression)
Find sy and txy
sx1 ⫽
For
sx ⫺sy
sx + sy
2
+
2
cos(2u) + txy sin (2u)
u ⫽ 32°
37 MPa ⫽
For
0.28081sy + 0.89879txy ⫽ ⫺42.11041 MPa (1)
u ⫽ 48°:
sx1 ⫽ ⫺12 MPa
110 MPa + sy
110 MPa ⫺ sy
⫺12 MPa ⫽
+
2
2
* cos (96°) + txy sin (96°)
or 0.55226sy + 0.99452txy ⫽ ⫺61.25093 MPa (2)
Solve Eqs. (1) and (2):
sx1 ⫽ 37 MPa
110 MPa + sy
or
sy ⫽ ⫺60.7 MPa
110 MPa ⫺ sy
+
2
2
* cos (64°) + txy sin (64°)
txy ⫽ ⫺27.9 MPa
;
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Analysis of Stress and Strain
Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are
y
sx ⫽ ⫺4100 psi, sy ⫽ 2200 psi, and txy ⫽ 2900 psi (the sign convention for these
stresses is shown in Fig. 7-1). A stress element located at the same point in the structure
(but oriented at a counterclockwise angle u1 with respect to the x axis) is subjected to
the stresses shown in the figure (sb, tb, and 1800 psi).
Assuming that the angle u1 is between zero and 90˚, calculate the normal stress sb, the
shear stress tb, and the angle u1
Solution 7.2-19
sx ⫽ ⫺4100 psi
txy ⫽ 2900 psi
For
u ⫽ u1:
Find
Stress
sy1 ⫽ sb
1800 psi u1
O
x
SOLVE NUMERICALLY:
tx1y1 ⫽ tb
2u1 ⫽ 87.32°
sb, tb, and u1
sb
u1 ⫽ 43.7°
;
Shear Stress tb
sb ⫽ sx + sy ⫺1800 psi
sb ⫽ ⫺3700 psi
;
tb ⫽ ⫺
Angle u1
sx1 ⫽
tb
1800 psi ⫽ ⫺950 psi ⫺ 3150 psi cos12u12
+ 2900 psi sin 12u12
sy ⫽ 2200 psi
sx1 ⫽ 1800 psi
sb
sx ⫺ sy
sx + sy
+
2
2
cos ( 2u) + txy sin ( 2u)
sx ⫺ sy
2
sin 12u12 + txy cos 12u12
tb ⫽ 3282 psi
;
Principal Stresses and Maximum Shear Stresses
When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane).
Problem 7.3-1 An element in plane stress is subjected to stresses sx ⫽ 4750 psi, sy ⫽ 1200 psi, and txy ⫽ 950 psi (see the
figure for Problem 7.2-1).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-1
sx ⫽ 4750 psi
sy ⫽ 1200 psi
atana
up1 ⫽
sx ⫺ sy
up1 ⫽ 14.08°
2
up2 ⫽ up1 + 90°
s1 ⫽
PRINCIPAL STRESSES
2 txy
txy ⫽ 950 psi
b
s2 ⫽
sx ⫺ sy
sx + sy
2
+
s1 ⫽ 4988 psi
s2 ⫽ 962 psi
2
sx ⫺ sy
sx + sy
2
up2 ⫽ 104.08°
+
2
;
;
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
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Page 583
SECTION 7.3
Principal Stresses and Maximum Shear Stresses
583
Problem 7.3-2 An element in plane stress is subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa (see the
figure for Problem 7.2-2).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-2
sx ⫽ 100 MPa
sy ⫽ 80 MPa
txy ⫽ 28 MPa
s1 ⫽
PRINCIPAL STRESSES
atana
up1 ⫽
2 txy
sx ⫺ sy
s2 ⫽
b
2
+
2
+
s2 ⫽ 60 MPa
up1 ⫽ 35.2°
2
sx ⫺ sy
sx + sy
s1 ⫽ 120 MPa
2
up2 ⫽ up1 + 90°
sx ⫺ sy
sx + sy
2
cos12up12 + txy sin12up12
cos12up22 + txy sin12up22
;
;
up2 ⫽ 125.17°
Problem 7.3-3 An element in plane stress is subjected to stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi
(see the figure for Problem 7.2-3).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-3
sx ⫽ ⫺5700 psi
sy ⫽ ⫺2300 psi
txy ⫽ 2500 psi
s1 ⫽
PRINCIPAL STRESSES
atana
up2 ⫽
2txy
sx ⫺ sy
s2 ⫽
b
2
+
2
+
s2 ⫽ ⫺7023 psi
up2 ⫽ ⫺27.89°
2
sx ⫺ sy
sx + sy
s1 ⫽ ⫺977 psi
2
up1 ⫽ up2 + 90°
sx ⫺ sy
sx + sy
2
cos12up12 + txy sin12up12
cos12up22 + txy sin12up22
;
;
up1 ⫽ 62.1°
Problem 7.3-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal
direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the
directions shown (see the figure for Problem 7.2-4).
Determine the principal stresses and show them on a sketch of a properly oriented element.
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Analysis of Stress and Strain
Solution 7.3-4
sx ⫽ 40 MPa
sy ⫽ ⫺160 MPa
txy ⫽ ⫺54 MPa
s1 ⫽
PRINCIPAL STRESSES
atana
up1 ⫽
2txy
sx ⫺ sy
s2 ⫽
b
2
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
s1 ⫽ 53.6 MPa
;
2
s2 ⫽ ⫺173.6 MPa
up1 ⫽ ⫺14.2°
up2 ⫽ up1 + 90°
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
;
up2 ⫽ 75.8°
Problem 7.3-5 The normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions
shown in the figure) (see the figure for Problem 7.2-5).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
Solution 7.3-5
sx ⫽ 6500 psi
sy ⫽ ⫺18500 psi
txy ⫽ ⫺3800 psi
s2 ⫽ ⫺19065 psi
PRINCIPAL ANGLES
atana
up1 ⫽
2txy
sx ⫺ sy
MAXIMUM SHEAR STRESSES
b
tmax ⫽
2
up1 ⫽ ⫺8.45°
s2 ⫽
+
2
sx ⫺ sy
sx + sy
2
up2 ⫽ 81.55°
sx ⫺ sy
sx + sy
2
sx ⫺ sy 2
a
b + txy2
A
2
us1 ⫽ up1 ⫺ 45°
up2 ⫽ up1 + 90°
s1 ⫽
s1 ⫽ 7065 psi
+
2
cos 12up12 + txy sin 12up12
saver ⫽
sx + sy
2
tmax ⫽ 13065 psi
us1 ⫽ ⫺53.4°
;
;
saver ⫽ ⫺6000 psi
;
cos 12up22 + txy sin 12up22
Problem 7.3-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of
magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction. Also,
shear stresses of magnitude 10.5 MPa act in the directions shown (see the figure for Problem 7.2-6).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
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SECTION 7.3
Principal Stresses and Maximum Shear Stresses
585
Solution 7.3-6
sx ⫽ ⫺27 MPa
sy ⫽ 5.5 MPa
txy ⫽ ⫺10.5 MPa
s2 ⫽ ⫺30.1 MPa
PRINCIPAL ANGLES
2txy
atana
b
sx ⫺ sy
up2 ⫽
2
MAXIMUM SHEAR STRESSES
tmax ⫽
up2 ⫽ 16.43°
up1 ⫽ up2 + 90°
s1 ⫽
s2 ⫽
up1 ⫽ 106.43°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
s1 ⫽ 8.6 MPa
2
cos 12up12 + txy sin 12up12
sx ⫺ sy 2
b + txy2
A
2
a
tmax ⫽ 19.3 MPa
;
us1 ⫽ up1 ⫺ 45°
saver ⫽
us1 ⫽ 61.4°
sx + sy
saver ⫽ ⫺10.8 MPa
2
;
cos 12up22 + txy sin 12up22
Problem 7.3-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression
in the horizontal direction and 2600 psi compression in the vertical direction. Also, shear stresses of magnitude 3800 psi act
in the directions shown (see the figure for Problem 7.2-7).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
Solution 7.3-7
sx ⫽ ⫺14000 psi
txy ⫽ ⫺3800 psi
sy ⫽ ⫺2600 psi
up2 ⫽
2txy
sx ⫺ sy
b
s1 ⫽
2
tmax ⫽
up1 ⫽ 106.85°
sx ⫺ sy
+
2
2
cos12up22 + txy sin12up22
MAXIMUM SHEAR STRESSES
up2 ⫽ 16.85°
sx + sy
+
s2 ⫽ ⫺15151 psi
2
up1 ⫽ up2 + 90°
sx ⫺ sy
sx + sy
s1 ⫽ ⫺1449 psi
PRINCIPAL ANGLES
atana
s2 ⫽
2
cos12up12 + txy sin12up12
sx ⫺ sy 2
a
b + txy2
A
2
us1 ⫽ up1 ⫺ 45°
saver ⫽
sx + sy
2
tmax ⫽ 6851 psi
us1 ⫽ 61.8°
saver ⫽ ⫺8300 psi
;
;
;
Problem 7.3-8 The normal and shear stresses acting on element B are sx ⫽ ⫺46 MPa, sy ⫽ ⫺13 MPa, and txy ⫽ 21 MPa
(see figure for Problem 7.2-8).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
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Page 586
Analysis of Stress and Strain
Solution 7.3-8
sx ⫽ ⫺46 MPa
sy ⫽ ⫺13 MPa
txy ⫽ 21 MPa
s2 ⫽ ⫺56.2 MPa
PRINCIPAL ANGLES
atana
up2 ⫽
2txy
sx ⫺ sy
MAXIMUM SHEAR STRESSES
b
tmax ⫽
2
up1 ⫽ up2 + 90°
s1 ⫽
s2 ⫽
up1 ⫽ 64.08°
sx ⫺ sy
+
2
2
sx ⫺ sy
sx + sy
+
2
a
sx ⫺ sy
A
2
us1 ⫽ up1 ⫺ 45°
up2 ⫽ ⫺25.92°
sx + sy
s1 ⫽ ⫺2.8 MPa
2
cos 12up12 + txy sin 12up12
saver ⫽
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 26.7 MPa
us1 ⫽ 19.08°
;
;
saver ⫽ ⫺29.5 MPa
;
cos 12up22 + txy sin 12up22
Problem 7.3-9 A shear wall in a reinforced concrete building is
subjected to a vertical uniform load of intensity q and a horizontal
force H, as shown in the first part of the figure. (The force H
represents the effects of wind and earthquake loads.) As a
consequence of these loads, the stresses at point A on the surface
of the wall have the values shown in the second part of the
figure (compressive stress equal to 1100 psi and shear stress equal
to 480 psi).
q
1100 psi
H
480 psi
A
A
(a) Determine the principal stresses and show them on a sketch
of a properly oriented element.
(b) Determine the maximum shear stresses and associated
normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-9
sx ⫽ 0
Shear wall
sy ⫽ ⫺1100 psi
txy ⫽ ⫺480 psi
2txy
sx ⫺ sy
⫽ ⫺0.87273
2up ⫽ ⫺41.11° and up ⫽ ⫺20.56°
2up ⫽ 138.89° and up ⫽ 69.44°
sx1 ⫽
sx ⫺ sy
sx + sy
2
f
s2 ⫽ ⫺1280 psi and up2 ⫽ 69.44°
(a) PRINCIPAL STRESSES
tan 2up ⫽
Therefore, s1 ⫽180 psi and up1 ⫽ ⫺20.56°
+
2
cos 2u + txy sin 2u
For 2up ⫽ ⫺41.11°:
sx1 ⫽ 180 psi
For 2up ⫽ 138.89°:
sx1 ⫽ ⫺1280 psi
;
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SECTION 7.3
587
Principal Stresses and Maximum Shear Stresses
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
sx ⫺ sy 2
b + t2xy ⫽ 730 psi
A
2
a
us1 ⫽ up1⫺ 45° ⫽ ⫺65.56° and t ⫽ 730 psi
us2 ⫽ up1+ 45° ⫽ 24.44°
saver ⫽
sx + sy
and t ⫽ ⫺730 psi
⫽ ⫺550 psi
2
f
;
;
Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust
is designed to resist a shear stress of 56 MPa and a compressive stress of 85 MPa
(see figure).
(a) Determine the principal stresses and show them on a sketch of a properly
oriented element.
(b) Determine the maximum shear stresses and associated normal stresses
and show them on a sketch of a properly oriented element.
85 MPa
56 MPa
Solution 7.3-10
sx ⫽ ⫺85 MPa
sy ⫽ 0 MPa
Txy ⫽ ⫺56 MPa
up2 ⫽
2txy
sx ⫺ sy
tmax ⫽
s2 ⫽
+
2
sx ⫺ sy
sx + sy
2
up1 ⫽ 116.4°
sx ⫺ sy
sx + sy
2
a
sx ⫺ sy
A
2
tmax ⫽ 70.3 MPa
2
up1 ⫽ up2 + 90°
+
;
(b) MAXIMUM SHEAR STRESSES
b
up2 ⫽ 26.4°
s1 ⫽
;
s2 ⫽ ⫺112.8 MPa
(a) PRINCIPAL STRESSES
atana
s1 ⫽ 27.8 MPa
2
;
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
us1 ⫽ up1 ⫺ 45°
saver ⫽
sx + sy
2
2
b + txy2
;
us1 ⫽ 71.4°
;
saver ⫽ ⫺42.5 MPa
;
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Analysis of Stress and Strain
y
sx ⫽ 2500 psi, sy ⫽ 1020 psi, txy ⫽ ⫺900 psi
Problems 7.3-11
(a) Determine the principal stresses and show them on a sketch of a properly
oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and
show them on a sketch of a properly oriented element.
sy
txy
sx
O
x
Probs. 7.3-11 through 7.3-16
Solution 7.3-11
sx ⫽ 2500 psi
sy ⫽ 1020 psi
txy ⫽ ⫺900 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
up1 ⫽
sx ⫺ sy
atan a
2txy
sx ⫺ sy
b
s1 ⫽
2
s2 ⫽
+
2
sx ⫺ sy
sx + sy
+
2
2
2
s1 ⫽ 2925 psi
For up2 ⫽ 64.7°:
s2 ⫽ 595 psi
tmax ⫽
up2 ⫽ 64.71°
sx ⫺ sy
sx + sy
For up1 ⫽ ⫺25.3°:
;
A
a
sx ⫺ sy 2
b + txy2
2
tmax ⫽ 1165 psi
us1 ⫽ ⫺70.3° and
us1 ⫽ up1 ⫺ 45°
;
t1 ⫽ 1165 psi
cos 12up12 + txy sin 12up12
us2 ⫽ up1 ⫺ 45°
t2 ⫽ ⫺1165 psi
cos 12up22 + txy sin 12up22
saver ⫽
Problems 7.3-12
;
(b) MAXIMUM SHEAR STRESSES
up1 ⫽ 25.29°
up2 ⫽ 90 ° + up1
Therefore,
sx + sy
us2 ⫽ 19.71° and
;
saver ⫽ 1760 psi
2
;
sx ⫽ 2150 kPa, sy ⫽ 375 kPa, txy ⫽ ⫺460 kPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-12
sx ⫽ 2150 kPa
sy ⫽ 375 kPa
txy ⫽ ⫺460 kPa
up1 ⫽ ⫺13.70°
up2 ⫽ 90° + up1
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
sx ⫺ sy
atana
up1 ⫽
2txy
sx ⫺sy
2
b
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
2
+
sx ⫺ sy
sx + sy
2
2
+
2
up2 ⫽ 76.30°
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
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Page 589
SECTION 7.3
tmax ⫽ 145 psi
Therefore,
For up1 ⫽ ⫺13.70°
s1 ⫽ 2262 kPa
For up2 ⫽ 76.3°
s2 ⫽ 263 kPa
us1 ⫽ ⫺58.7°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 1000 kPa
;
;
Problems 7.3-13
A
a
sx ⫺ sy
2
;
us2 ⫽ up1 + 45° us2 ⫽ 31.3° ;
and t2 ⫽⫺1000 kPa
sx + sy
saver ⫽
saver ⫽ 1263 kPa
2
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
Principal Stresses and Maximum Shear Stresses
2
b + txy2
sx ⫽ 14,500 psi, sy ⫽ 1070 psi, txy ⫽ 1900 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-13
sx ⫽ 14500 psi
sy ⫽ 1070 psi
txy ⫽ 1900 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
atana
2txy
up1 ⫽
sx ⫺ sy
2txy
sx ⫺sy
b
2
up1 ⫽ 7.90°
up2 ⫽ 90° + up1
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
2
+
2
sx ⫺ sy
sx + sy
2
up2 ⫽ 97.90°
+
2
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 7.90°
s1 ⫽ 14764 psi
;
For up2 ⫽ 97.9°
s2 ⫽ 806 psi
;
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 6979 psi
us1 ⫽ u p1 ⫺ 45°
and t1 ⫽ 6979 psi
us1 ⫽ ⫺37.1°
us2 ⫽ 52.9°
us2 ⫽ up1 + 45°
and t2 ⫽⫺6979 psi
saver ⫽
Problems 7.3-14
A
a
sx + sy
2
;
;
saver ⫽ 7785 psi
sx ⫽ 16.5 MPa, sv ⫽ ⫺91 MPa, txy ⫽ ⫺39 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
589
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CHAPTER 7
Page 590
Analysis of Stress and Strain
Solution 7.3-14
sx ⫽ 16.5 MPa
sy ⫽ ⫺91 MPa
txy ⫽ ⫺39 MPa
(b) MAXIMUM SHEAR STRESSES
(a) PRINCIPAL STRESSES
atana
2txy
tan(2up) ⫽
up1 ⫽
sx ⫺ sy
2txy
sx ⫺sy
tmax ⫽
b
s1 ⫽
us1 ⫽ ⫺63.0°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 66.4 MPa
2
s2 ⫽
+
2
2
sx ⫺ sy
sx + sy
+
2
us2 ⫽ 27.0°
us2 ⫽ up1 + 45°
and t2 ⫽ ⫺66.4 MPa
up2 ⫽72.02°
sx ⫺ sy
sx + sy
2
2
2
b + txy2
tmax ⫽ 9631.7 psi
up1 ⫽ ⫺17.98°
up2 ⫽ 90° + up1
A
sx ⫺ sy
a
cos 12up12 + txy sin 12up12
saver ⫽
sx + sy
;
;
saver ⫽ ⫺37.3 MPa
2
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ ⫺17.98°
For up2 ⫽ 72.0°
Problems 7.3-15
s1 ⫽ 29.2 MPa
s2 ⫽ ⫺103.7 MPa
sx ⫽ ⫺3300 psi, sy ⫽ ⫺11,000 psi, txy ⫽ 4500 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-15
s ⫽ ⫺3300 psi
sy ⫽⫺11000 psi
txy ⫽ 4500 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
sx ⫺ sy
atana
up1 ⫽
2txy
sx ⫺sy
b
2
up1 ⫽ 24.73°
up2 ⫽ 90° + up1
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
2
+
2
sx ⫺ sy
sx + sy
2
up2 ⫽ 114.73°
+
2
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 24.7°
s1 ⫽ ⫺1228 psi
For up2 ⫽ 114.7°
s2 ⫽ ⫺13072 psi
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
A
a
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 5922 psi
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 5922 psi
us1 ⫽ ⫺20.3°
us2 ⫽ up1 + 45°
and t2 ⫽ ⫺5922 psi
us2 ⫽ 69.7°
saver ⫽
sx + sy
2
;
;
saver ⫽ ⫺7150 psi
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Page 591
SECTION 7.3
Problems 7.3-16
Principal Stresses and Maximum Shear Stresses
591
sx ⫽ ⫺108 MPa, sy ⫽ 58 MPa, txy ⫽ ⫺58 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-16
sx ⫽ ⫺108 MPa
sy ⫽ 58 MPa
txy ⫽ ⫺58 MPa
(a) PRINCIPAL STRESSES
tan(2up) ⫽
atana
2txy
up2 ⫽
sx ⫺ sy
tmax ⫽
2txy
sx ⫺sy
b
s1 ⫽
s2 ⫽
2
+
2
sx ⫺ sy
sx + sy
2
up1 ⫽ 107.47°
sx ⫺ sy
sx + sy
+
2
A
a
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 14686.1 psi
us1 ⫽ 62.47°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 101.3 MPa
2
up2 ⫽ 17.47°
up1 ⫽ 90° + up2
(b) MAXIMUM SHEAR STRESSES
cos 12up12 + txy sin 12up12
;
us2 ⫽ 152.47° ;
us2 ⫽ up1 + 45°
and t2 ⫽⫺101.3 MPa
sx + sy
saver ⫽
saver ⫽ ⫺25.0 MPa
2
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 107.47°
s1 ⫽ 76.3 MPa
For up2 ⫽ 17.47°
s2 ⫽ ⫺126.3 MPa
;
;
Problem 7.3-17 At a point on the surface of a machine component, the stresses
y
acting on the x face of a stress element are sx ⫽ 5900 psi and txy ⫽ 1950 psi
(see figure).
What is the allowable range of values for the stress sy if the maximum shear stress
is limited to t0 ⫽ 2500 psi?
sy
txy = 1950 psi
O
sx = 5900 psi
x
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CHAPTER 7
Page 592
Analysis of Stress and Strain
Solution 7.3-17
sx ⫽ 5900 psi
sy
unknown
txy ⫽ 1950 psi
Therefore, 2771 psi … sy … 9029 psi
Find the allowable range of values for sy if the
From Eq. (1):
maximum allowable shear stresses is tmax ⫽ 2500 psi
sx ⫺ sy 2
(1)
tmax ⫽ a
b + txy2
A
2
t max (sy1) ⫽
A
a
sx ⫺ sy1
2
2
b + txy2
Solve for sy
sy ⫽ sx ⫹
J
22tmax 2 ⫺txy2
⫺ a2 2tmax2 ⫺txy2 b
K
sy ⫽ a
9029
b psi
2771
2.771 ksi
9.029 ksi
tmax(σy1)
2.5 ksi
σy1
Problem 7.3-18 At a point on the surface of a machine component the stresses
acting on the x face of a stress element are sx ⫽ 42 MPa and txy ⫽ 33 MPa (see figure).
What is the allowable range of values for the stress sy if the maximum shear
stress is limited to t0 ⫽ 35 MPa?
y
sy
txy = 33 MPa
O
sx = 42 MPa
x
Solution 7.3-18
sx ⫽ 42 MPa
sy
unknown
txy ⫽ 33 MPa
Find the allowable range of values for sy if the
maximum allowable shear stresses is tmax ⫽ 35 MPa
tmax ⫽
A
a
sx ⫺ sy
2
2
b + txy2
(1)
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Page 593
SECTION 7.3
Solve for sy
sy ⫽ sx ⫹
593
Principal Stresses and Maximum Shear Stresses
From Eq. (1):
2 2tmax 2 ⫺txy2
2
2
J ⫺ a 22tmax ⫺txy b K
65.3
sy ⫽ a
b MPa
18.7
tmax (sy1) ⫽
A
a
sx ⫺ sy1
2
2
b + txy2
Therefore, 18.7 MPa … sy … 65.3 MPa
65.3 MPa
18.7 MPa
35 MPa
tmax (σy1)
σy1
Problem 7.3-19 An element in plane stress is subjected to stresses sx ⫽ 5700 psi
and txy ⫽ ⫺2300 psi (see figure). It is known that one of the principal stresses equals
6700 psi in tension.
y
sy
(a) Determine the stress sy.
(b) Determine the other principal stress and the orientation of the principal planes,
then show the principal stresses on a sketch of a properly oriented element.
5700 psi
O
x
2300 psi
Solution 7.3-19
sx ⫽ 5700 psi
sy unknown txy ⫽ ⫺2300 psi
(a) STRESS sy
s1 ⫽ 6700 psi
s1 ⫽
2
;
(b) PRINCIPAL STRESSES
Because sy is smaller than a given principal stress,
we know that the given stress is the larger principal
stress.
sx + sy
sy ⫽ 1410 psi
Solve for sy
sx ⫺ sy 2
+
a
b + txy2
A
2
tan (2up) ⫽
2txy
sx ⫺ sy
up1 ⫽
atan a
up1 ⫽ ⫺23.50°
up2 ⫽ 90° + up1
up2 ⫽ 66.50°
2txy
sx ⫺sy
2
b
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Analysis of Stress and Strain
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
+
2
2
+ txy sin12up12
For up1 ⫽ ⫺23.5° : s1 ⫽ 6700 psi
For up2 ⫽ 66.5°
sx ⫺ sy
sx + sy
Therefore,
cos 12up12
+
2
2
+ txy sin 12up22
: s2 ⫽ 410 psi
;
;
cos 12up22
An element in plane stress is subjected to stresses sx ⫽ ⫺50 MPa
and txy ⫽ 42 MPa (see figure). It is known that one of the principal stresses equals
33 MPa in tension.
Problem 7.3-20
y
sy
(a) Determine the stress sy.
(b) Determine the other principal stress and the orientation of the principal planes,
then show the principal stresses on a sketch of a properly oriented element.
42 MPa
50 MPa
O
x
Solution 7.3-20
sx ⫽ ⫺50 MPa
sy unknown
txy ⫽ 42 MPa
up2 ⫽ ⫺26.85°
up1 ⫽ 90° + up2
(a) STRESS sy
s1 ⫽
Because sy is smaller than a given principal stress,
we know that the given stress is the larger principal
stress.
s1 ⫽ 33 MPa
s1 ⫽
sx + sy
+
2
Solve for sy
A
a
sx ⫺ sy 2
2
b +
s2 ⫽
txy2
sy ⫽ 11.7 MPa
;
tan(2up) ⫽
2txy
For up1 ⫽ 63.2°
sx ⫺ sy
up2 ⫽
atan a
sx ⫺sy
2
+
2
+ txy sin 12up12
sx ⫺ sy
sx + sy
+
2
2
+ txy sin 12up22
cos 12up12
cos 12up22
Therefore,
(b) PRINCIPAL STRESSES
2txy
sx ⫺ sy
sx + sy
2
up1 ⫽ 63.15°
: s1 ⫽ 33.0 MPa
For up2 ⫽ ⫺26.8° : s 2 ⫽ ⫺71.3 MPa
b
;
;
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SECTION 7.4
595
Mohr’s Circle
Mohr’s Circle
y
The problems for Section 7.4 are to be solved using Mohr’s circle. Consider only the
in-plane stresses (the stresses in the xy plane).
Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses
sx ⫽ 11,375 psi, as shown in the figure. Using Mohr’s circle, determine:
11,375 psi
O
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 24°
from the x axis.
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
x
Solution 7.4-1
sx ⫽ 11375 psi sy ⫽ 0 psi
(a) ELEMENT AT
u ⫽ 24°
2u ⫽ 48° R ⫽
sx
2
txy ⫽ 0 psi
;
sy1 ⫽ 1882 psi
(b) MAXIMUM SHEAR STRESSES
R ⫽ 5688 psi
sc ⫽ R sc ⫽ 5688 psi
Point C:
Point D: sx1 ⫽ R + R cos(2u)
sx1 ⫽ 9493 psi
;
;
tx1y1 ⫽ ⫺R sin (2u)
tx1y1 ⫽ ⫺4227 psi
Point S1: us1 ⫽
tmax ⫽ R
us1 ⫽⫺45°
tmax ⫽ 5688 psi
Point S2: us2 ⫽
tmax ⫽ ⫺R
;
⫺90°
2
saver ⫽ R
90°
2
;
;
us2 ⫽ 45°
;
tmax ⫽ ⫺5688 psi
;
saver ⫽ 5688 psi
;
œ
PointD : sy1 ⫽ R ⫺ R cos (2u)
y
Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses
sx ⫽ 49 MPa, as shown in the figure Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at an angle u ⫽ ⫺27° from the x
axis (minus means clockwise).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
49 MPa
O
Solution 7.4-2
sx ⫽ 49 MPa sy ⫽ 0 MPa
(a) ELEMENT AT
u ⫽ ⫺27°
2u ⫽ ⫺54.0° R ⫽
Point C:
txy ⫽ 0 MPa
sc ⫽ R
sx
2
R ⫽ 24.5 MPa
sc ⫽ 24.5 MPa
Point D:
sx1 ⫽ R + R cos (|2u|)
sx1 ⫽ 38.9 MPa
;
tx1y1 ⫽ ⫺R sin (2u)
;
tx1y1 ⫽ 19.8 MPa
œ
Point D sy1 ⫽ R ⫺ R cos ( |2u| )
;
sy1 ⫽ 10.1 MPa
x
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Analysis of Stress and Strain
(b) MAXIMUM SHEAR STRESSES
Point S1:
tmax ⫽ R
90°
2
Point S2: us2 ⫽
⫺90°
us1 ⫽
2
tmax ⫽ ⫺R
us1 ⫽ ⫺45.0°
;
tmax ⫽ 24.5 MPa
saver ⫽ R
us2 ⫽ 45.0°
;
tmax ⫽ ⫺24.5 MPa
saver ⫽ 24.5 MPa
;
;
;
Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses
of magnitude 6100 psi, as shown in the figure. Using Mohr’s circle, determine:
y
1
(a) The stresses acting on an element oriented at a slope of 1 on 2 (see figure).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
2
O
6100 psi
x
Solution 7.4-3
sx ⫽ ⫺6100 psi sy ⫽ 0 psi
txy ⫽ 0 psi
œ
Point D : sy1 ⫽ R ⫺ R cos (2u)
sy1 ⫽ ⫺1220 psi
(a) ELEMENT AT A SLOPE OF 1 ON 2
1
u ⫽ atana b
2
u ⫽ 26.565°
Point S1:
sx
2u ⫽ 53.130° R ⫽
2
Point C:
Point D:
sc ⫽ R
(b) MAXIMUM SHEAR STRESSES
;
R ⫽ ⫺3050 psi
tmax ⫽ ⫺R
sc ⫽ ⫺3050 psi
Point S2:
sx1 ⫽ R + R cos (2u)
sx1 ⫽ ⫺4880 psi
tx1y1 ⫽ ⫺R sin (2u)
;
tx1y1 ⫽ 2440 psi
us1 ⫽
;
90°
2
us1 ⫽ 45°
tmax ⫽ 3050 psi
us2 ⫽
⫺90°
2
;
;
us2 ⫽ ⫺45°
tmax ⫽ R
tmax ⫽ ⫺3050 psi
saver ⫽ R
saver ⫽ ⫺3050 psi
;
;
;
Problem 7.4-4 An element in biaxial stress is subjected to stresses sx ⫽ ⫺48 MPa and
sy ⫽ 19 MPa, as shown in the figure. Using Mohr’s circle, determine:
y
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 25°
from the x axis.
(b) The maximum shear stresses and associated normal stresses.
19 MPa
Show all results on sketches of properly oriented elements.
48 MPa
O
x
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Page 597
SECTION 7.4
597
Mohr’s Circle
Solution 7.4-4
sx ⫽ ⫺48 MPa
(a) ELEMENT AT
sy ⫽ 19 MPa
u ⫽ 25°
2u ⫽ 50.0 deg R ⫽
txy ⫽ 0 MPa
Point S1: us1 ⫽
;
|sx| + |sy|
2
Point C: sx ⫽ sx + R
(b) MAXIMUM SHEAR STRESSES
us1 ⫽ 45.0°
R ⫽ 33.5 MPa
sc ⫽ ⫺14.5 MPa
tmax ⫽ R
sx1 ⫽ ⫺36.0 MPa
;
tmax ⫽ ⫺R
;
;
tmax ⫽ ⫺33.5 MPa
saver ⫽ sc
Point D œ : sy1 ⫽ sc + R cos(2u)
;
⫺90°
2
us2 ⫽ ⫺45.0°
tx1y1 ⫽ ⫺R sin(2u)
tx1y1 ⫽ 25.7 MPa
;
tmax ⫽ 33.5 MPa
Point S2: us2 ⫽
Point D: sx1 ⫽ sc ⫺ R cos(2u)
90°
2
saver ⫽ ⫺14.5 MPa
;
;
sy1 ⫽ 7.0 MPa
Problem 7.4-5 An element in biaxial stress is subjected to stresses sx ⫽ 6250 psi
y
and sy ⫽ ⫺1750 psi, as shown in the figure. Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at a counterclockwise angle
u ⫽ 55° from the x axis.
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
1750 psi
6250 psi
O
x
Solution 7.4-5
sx ⫽ 6250 psi
(a) ELEMENT AT
sy ⫽ ⫺1750 psi
txy ⫽ 0 psi
u ⫽ 60°
2u ⫽ 120° R ⫽
|sx| + |sy|
2
Point C: sc ⫽ sx ⫺ R
Point S1: us1 ⫽
R ⫽ 4000 psi
sc ⫽ 2250 psi
Point D: sx1 ⫽ sc + R cos(2u)
sx1 ⫽ 250 psi
(b) MAXIMUM SHEAR STRESSES
us1 ⫽ ⫺45°
tmax ⫽ R
;
tx1y1 ⫽ ⫺3464 psi
Point D œ : sy1 ⫽ sc ⫺ R cos(2u)
;
sy1 ⫽ 4250 psi
;
tmax ⫽ 4000 psi
Point S2: us2 ⫽
tx1y1 ⫽ ⫺R sin (2u)
⫺90°
2
90°
2
;
us2 ⫽ 45°
;
tmax ⫽ R
tmax ⫽ 4000 psi
;
saver ⫽ sc
saver ⫽ 2250 psi
;
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Analysis of Stress and Strain
y
Problem 7.4-6 An element in biaxial stress is subjected to stresses sx ⫽ ⫺29 MPa
and sy ⫽ 57 MPa, as shown in the figure. Using Mohr’s circle, determine:
57 MPa
(a) The stresses acting on an element oriented at a slope of 1 on 2.5 (see figure).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
1
2.5
29 MPa x
O
Solution 7.4-6
sx ⫽ ⫺29 MPa
sy ⫽ 57 MPa
txy ⫽ 0 MPa
(b) MAXIMUM SHEAR STRESSES
Point S1: us1 ⫽
(a) ELEMENT AT A SLOPE OF 1 ON 2.5
u ⫽ atana
1
b
2.5
2u ⫽ 43.603° R ⫽
u ⫽ 21.801°
|sx| + |sy|
2
Point C: sc ⫽ sx + R
tmax ⫽ R
;
tmax ⫽ ⫺R
Point D: sx1 ⫽ sc ⫺ R cos (2u)
sx1 ⫽ ⫺17.1 MPa
tx1y1 ⫽ R sin (2u)
Point D
œ:
saver ⫽ sc
;
tx1y1 ⫽ 29.7 MPa
;
;
⫺90°
2
us2 ⫽ ⫺45.0°
sc ⫽ 14.0 MPa
us1 ⫽ 45.0°
tmax ⫽ 43.0 MPa
Point S2: us2 ⫽
R ⫽ 43.0 MPa
90°
2
;
tmax ⫽ ⫺43.0 MPa
saver ⫽ 14.0 MPa
;
;
;
sy1 ⫽ sc + R cos (2u)
sy1 ⫽ 45.1 MPa
Problem 7.4-7 An element in pure shear is subjected to stresses txy ⫽ 2700 psi, as shown
y
in the figure. Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 52° from
the x axis.
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
2700 psi
O
x
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Page 599
SECTION 7.4
599
Mohr’s Circle
Solution 7.4-7
sx ⫽ 0 psi
sy ⫽ 0 psi
(a) ELEMENT AT
txy ⫽ 2700 psi
Point P1: up1 ⫽
u ⫽ 52°
2u ⫽ 104.0° R ⫽ txy
(b) PRINCIPAL STRESSES
R ⫽ 2700 psi
Point D: sx1 ⫽ R cos (2u ⫺ 90°)
sx1 ⫽ 2620 psi
tx1y1 ⫽ ⫺R sin (2u ⫺ 90°)
tx1y1 ⫽ ⫺653 psi
up1 ⫽ 45°
s1 ⫽ R
Point P2: up2 ⫽
;
90°
2
;
s1 ⫽ 2700 psi
;
⫺90°
2
up2 ⫽ ⫺45°
;
;
Point D : sy1 ⫽ ⫺R cos (2u ⫺ 90°)
œ
sy1 ⫽ ⫺2620 psi
s2 ⫽ ⫺R
s2 ⫽ ⫺2700 psi
;
;
Problem 7.4-8 An element in pure shear is subjected to stresses txy ⫽ ⫺14.5 MPa, as
shown in the figure. Using Mohr’s circle, determine:
y
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 22.5°
from the x axis
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
O
x
14.5 MPa
Solution 7.4-8
sx ⫽ 0 MPa
sy ⫽ 0 MPa
(a) ELEMENT AT
txy ⫽ ⫺14.5 MPa
u ⫽ 22.5°
Point P1: up1 ⫽
2u ⫽ 45.00°
R ⫽ |txy|
270°
u ⫽ 135.0°
2 p1
s1 ⫽ R
R ⫽ 14.50 MPa
Point D: sx1 ⫽ ⫺R cos (2u ⫺ 90°)
sx1 ⫽ ⫺10.25 MPa
;
tx1y1 ⫽ R sin (2u ⫺ 90°)
Point P2: up2 ⫽
;
s1 ⫽ 14.50 MPa
⫺270°
2
up2 ⫽ ⫺135.0°
;
s2 ⫽ ⫺R
tx1y1 ⫽ ⫺10.25 MPa
;
Point D œ : sy1 ⫽ R cos (2u ⫺ 90°)
sy1 ⫽ 10.25 MPa
(b) PRINCIPAL STRESSES
;
s2 ⫽ ⫺14.50 MPa
;
;
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CHAPTER 7
Page 600
Analysis of Stress and Strain
Problem 7.4-9 An element in pure shear is subjected to stresses txy ⫽ 3750 psi, as
y
shown in the figure. Using Mohr’s circle, determine:
3
4
(a) The stresses acting on an element oriented at a slope of 3 on 4 (see figure).
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
x
O
3750 psi
Solution 7.4-9
sx ⫽ 0 psi
sy ⫽ 0 psi
txy ⫽ 3750 psi
(b) PRINCIPAL STRESSES
3
u ⫽ atana b
4
u ⫽ 36.870°
2u ⫽ 73.740°
R ⫽ txy
s1 ⫽ R
R ⫽ 3750 psi
Point D: sx1 ⫽ R cos (2u ⫺ 90°)
sx1 ⫽ 3600 psi
;
tx1y1 ⫽ ⫺R sin (2u ⫺ 90°)
tx1y1 ⫽ 1050 psi
90°
2
Poin P1: up1 ⫽
(a) ELEMENT AT A SLOPE OF 3 ON 4
Point P2: up2 ⫽
up1 ⫽ 45°
s1 ⫽ 3750 psi
;
⫺90°
2
up2 ⫽ ⫺45°
s2 ⫽ ⫺R
;
;
s2 ⫽ ⫺3750 psi
;
;
Point D sy1 ⫽ ⫺R cos (2u ⫺ 90°)
œ:
sy1 ⫽ ⫺3600 psi
;
Problem 7.4-10 sx ⫽ 27 MPa, sy ⫽ 14 MPa, txy ⫽ 6 MPa, u ⫽ 40°
y
Using Mohr’s circle, determine the stresses acting on an element oriented at an
angle u from the x axis. Show these stresses on a sketch of an element oriented at the
angle u. (Note: The angle u is positive when counterclockwise and negative when
clockwise.)
sy
txy
sx
O
Probs. 7.4-10 through 7.4-15
x
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Page 601
SECTION 7.4
Mohr’s Circle
601
Solution 7.4-10
sx ⫽ 27 MPa
sy ⫽ 14 MPa
txy ⫽ 6 MPa
sx + sy
saver ⫽ 20.50 MPa
2
R ⫽ 2(sx ⫺ saver)2 + txy 2
a ⫽ atana
b ⫽ 37.29°
Point D: sx1 ⫽ saver + R cos (b)
u ⫽ 40°
saver ⫽
b ⫽ 2u ⫺ a
txy
sx ⫺ saver
R ⫽ 8.8459 MPa
sx1 ⫽ 27.5 MPa
;
tx1y1 ⫽ ⫺R sin (b)
tx1y1 ⫽ ⫺5.36 MPa
;
Point D œ : sy1 ⫽ saver ⫺ R cos (b)
b
a ⫽ 42.71°
sy1 ⫽ 13.46 MPa
;
Problem 7.4-11 sx ⫽ 3500 psi, sy ⫽ 12,200 psi, txy ⫽ ⫺3300 psi, u ⫽ ⫺51°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a
sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-11
sx ⫽ 3500 psi
u ⫽ ⫺51°
saver ⫽
sy ⫽ 12200 psi
b ⫽ 180° + 2u ⫺ a
b ⫽ 40.82°
Point D: sx1 ⫽ saver + R cos (b)
;
sx + sy
sx1 ⫽ 11982 psi
saver ⫽ 7850 psi
2
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atana
txy ⫽ ⫺3300 psi
txy
sx ⫺ saver
b
tx1y1 ⫽ ⫺R sin (b)
R ⫽ 5460 psi
2
;
tx1y1 ⫽ ⫺3569 psi
;
Point D sy1 ⫽ saver ⫺ R cos (b)
œ:
a ⫽ 37.18°
sy1 ⫽ 3718 psi
;
Problem 7.4-12 sx ⫽ ⫺47 MPa, sy ⫽ ⫺186 MPa, txy ⫽ ⫺29 MPa, u ⫽ ⫺33°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a
sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-12
sx ⫽⫺47MPa
sy ⫽⫺186MPa
txy ⫽⫺29MPa
u ⫽ ⫺33°
saver ⫽
sx1 ⫽ ⫺61.7 MPa
sx + sy
saver ⫽ ⫺116.50 MPa
2
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atana `
Point D: sx1 ⫽ saver + R cos (b)
txy
sx ⫺ saver
b ⫽ ⫺2u ⫺ a
`b
R ⫽ 75.3077 MPa
2
a ⫽ 22.65°
b ⫽ 43.35°
;
tx1y1 ⫽⫺R sin (b)
tx1y1 ⫽⫺51.7 MPa
;
Point D œ : sy1 ⫽ saver ⫺ R cos (b)
sy1 ⫽ ⫺171.3 MPa
;
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CHAPTER 7
Page 602
Analysis of Stress and Strain
Problem 7.4-13 sx ⫽ ⫺1720 psi, sy ⫽ ⫺680 psi, txy ⫽ 320 psi, u ⫽ 14°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
Solution 7.4-13
sx ⫽ ⫺1720 psi
sy ⫽ ⫺680 psi
txy ⫽ 380 psi
Point D: sx1 ⫽ saver ⫺ R cos(b)
u ⫽ 14°
saver ⫽
sx1 ⫽ ⫺1481 psi
sx + sy
saver ⫽ ⫺1200 psi
2
tx1y1 ⫽ R sin (b) tx1y1 ⫽ 580 psi
;
Point D : sy1 ⫽ saver + R cos (b)
œ
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atana
;
txy
|sx ⫺ saver|
b ⫽ 2u + a
b
sy1 ⫽ ⫺919 psi
R ⫽ 644.0 psi
2
;
a ⫽ 36.16°
b ⫽ 64.16°
Problem 7.4-14 sx ⫽ 33 MPa, sy ⫽ ⫺9 MPa, txy ⫽ 29 MPa, u ⫽ 35°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
Solution 7.4-14
sx ⫽ 33 MPa sy ⫽ ⫺9 MPa txy ⫽ 29 MPa
Point D: sx1 ⫽ saver + R cos (b)
u ⫽ 35°
saver ⫽
sx1 ⫽ 46.4 MPa
sx + sy
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atana `
txy
sx ⫺ saver
b ⫽ 2u ⫺ a
tx1y1 ⫽ ⫺R sin (b)
saver ⫽ 12.00 MPa
2
txy2
`b
R ⫽ 35.8050 MPa
;
tx1y1 ⫽ ⫺9.81 MPa
;
Point D : sy1 ⫽ saver ⫺ R cos (b)
a ⫽ 54.09°
œ
sy1 ⫽ ⫺22.4 MPa
;
b ⫽ 15.91°
Problem 7.4-15 sx ⫽ ⫺5700 psi, sy ⫽ 950 psi, txy ⫽ ⫺2100 psi, u ⫽ 65°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
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Page 603
SECTION 7.4
603
Mohr’s Circle
Solution 7.4-15
sx ⫽ ⫺5700 psi sy ⫽ 950 psi txy ⫽ ⫺2100 psi
u ⫽ 65°
saver ⫽
Point D: sx1 ⫽ saver + R cos (b)
sx1 ⫽ ⫺1846 psi
sx + sy
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atana
tx1y1 ⫽ R sin (b)
saver ⫽ ⫺2375 psi
2
|txy|
|sx ⫺ saver|
2
b
b ⫽ 180° ⫺ 2u + a
;
tx1y1 ⫽ 3897 psi
;
Point D sy1 ⫽ saver ⫺ R cos (b)
œ:
sy1 ⫽ ⫺2904 psi
R ⫽ 3933 psi
;
a ⫽ 32.28°
b ⫽ 82.28°
Problems 7.4-16 sx ⫽ ⫺29.5 MPa, sy ⫽ 29.5 MPa, txy ⫽ 27 MPa
y
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum
shear stresses and associated normal stresses. Show all results on sketches of properly
oriented elements.
sy
txy
sx
O
x
Probs. 7.4-16 through 7.4-23
Solution 7.4-16
sx ⫽ ⫺29.5 MPa sy ⫽ 29.5 MPa txy ⫽ 27 MPa
saver ⫽
sx + sy
2
a ⫽ atana `
txy
sx ⫺ saver
`b
R ⫽ 39.9906 MPa
a ⫽ 42.47°
180° ⫺ a
2
up2 ⫽ up1 ⫺ 90°
90°⫺a
2
s2 ⫽ ⫺40.0 MPa
us1 ⫽ 23.8°
us2 ⫽ 90° + us1
us2 ⫽ 113.8°
Point S1: saver ⫽ 0 MPa
up1 ⫽ 68.8°
up2 ⫽ ⫺21.2°
tmax ⫽ R
;
;
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
up1 ⫽
s1 ⫽ 40.0 MPa
Point P2: s2 ⫽ ⫺R
saver ⫽ 0 MPa
R ⫽ 2(sx ⫺ saver)2 + txy2
Point P1: s1 ⫽ R
;
;
;
tmax ⫽ 40.0 MPa
;
;
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CHAPTER 7
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Analysis of Stress and Strain
Problems 7.4-17 sx ⫽ 7300 psi, sy ⫽ 0 psi, txy ⫽ 1300 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-17
sx ⫽ 7300 psi sy ⫽ 0 psi txy ⫽ 1300 psi
saver ⫽
sx + sy
s1 ⫽ 7525 psi
saver ⫽ 3650 psi
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atana `
Point P1: s1 ⫽ R + saver
txy
sx ⫺ saver
txy2
`b
Point P2: s2 ⫽ ⫺R + saver
a
2
up2 ⫽
a + 180°
2
s2 ⫽ ⫺225 psi
R ⫽ 3875 psi
a ⫽ 19.60°
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
up1 ⫽
up1 ⫽ 9.80°
;
⫺90° + a
2
us2 ⫽ 90° + us1
;
us1 ⫽ ⫺35.2°
us2 ⫽ 54.8°
Point S1: saver ⫽ 3650 psi
tmax ⫽ R
up2 ⫽ 99.8°
;
tmax ⫽ 3875 psi
;
Problems 7.4-18 sx ⫽ 0 MPa, sy ⫽ ⫺23.4 MPa, txy ⫽ ⫺9.6 MPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-18
sx ⫽ 0 MPa sy ⫽ ⫺23.4 MPa txy ⫽ ⫺9.6 MPa
saver ⫽
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atana `
s1 ⫽ 3.43 MPa
saver ⫽ ⫺11.70 MPa
2
txy
sx ⫺ saver
`b
Point P1: s1 ⫽ R + saver
Point P2: s2 ⫽ ⫺R + saver
s2 ⫽ ⫺26.8 MPa
R ⫽ 15.1344 MPa
(a) PRINCIPAL STRESSES
up2 ⫽ up1 + 90°
⫺90° ⫺ a
2
us2 ⫽ 90 ° + us1
up1 ⫽ ⫺19.68°
up2 ⫽ 70.32°
;
(b) MAXIMUM SHEAR STRESSES
a ⫽ 39.37°
us1 ⫽
⫺a
up1 ⫽
2
;
;
us1 ⫽ ⫺64.7°
;
us2 ⫽ 25.3°
Point S1: saver ⫽ ⫺11.70 MPa
;
tmax ⫽ R
tmax ⫽ 15.13 MPa
;
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SECTION 7.4
Mohr’s Circle
605
Problems 7.4-19 sx ⫽ 2050 psi, sy ⫽ 6100 psi, txy ⫽ 2750 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-19
sx ⫽ 2050 psi sy ⫽ 6100 psi txy ⫽ 2750 psi
saver ⫽
sx + sy
s1 ⫽ 7490 psi
saver ⫽ 4075 psi
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atana `
Point P1: s1 ⫽ R + saver
txy
sx ⫺ saver
txy2
`b
;
Point P2: s2 ⫽ ⫺R + saver
R ⫽ 3415 psi
a ⫽ 53.63°
s2 ⫽ 660 psi
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
⫺90° + a
2
us2 ⫽ 90° + us1
180° ⫺ a
up1 ⫽ 63.2° ;
2
⫺a
up2 ⫽
up2 ⫽ ⫺26.8°
;
2
up1 ⫽
us1 ⫽ ⫺18.2°
us2 ⫽ 71.8°
Point S1: saver ⫽ 4075 psi
tmax ⫽ R
;
;
tmax ⫽ 3415 psi
;
Problems 7.4-20 sx ⫽ 2900 kPa, sy ⫽ 9100 kPa, txy ⫽ ⫺3750 kPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-20
sx ⫽ 2900 kPa sy ⫽ 9100 kPa txy ⫽ ⫺3750 kPa
saver ⫽
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atana `
s1 ⫽ 10865 KPa
saver ⫽ 6000 kPa
2
txy
sx ⫺ saver
`b
Point P1: s1 ⫽ R + saver
Point P2: s2 ⫽ ⫺R + saver
s2 ⫽ 1135 kPa
R ⫽ 4865.4393 kPa
(a) PRINCIPAL STRESSES
a + 180°
2
up2 ⫽
a
2
90° + a
2
us2 ⫽ 90° + us1
up1 ⫽ 115.2°
up2 ⫽ 25.2°
;
;
(b) MAXIMUM SHEAR STRESSES
a ⫽ 50.42°
us1 ⫽
up1 ⫽
;
;
us1 ⫽ 70.2°
us2 ⫽ 160.2°
Point S1: saver ⫽ 6000 kPa
tmax ⫽ R
;
;
;
tmax ⫽ 4865 kPa
;
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CHAPTER 7
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Analysis of Stress and Strain
Problems 7.4-21 sx ⫽ ⫺11,500 psi, sy ⫽ ⫺18,250 psi, txy ⫽ ⫺7200 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-21
sx ⫽ ⫺11500 psi
sy ⫽ ⫺18250 psi
Point P1: s1 ⫽ R + saver
txy ⫽ ⫺7200 psi
saver ⫽
s1 ⫽ ⫺6923 psi
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atana `
Point P2: s2 ⫽ ⫺R + saver
saver ⫽ ⫺14875 psi
2
txy
sx ⫺ saver
`b
s2 ⫽ ⫺22827 psi
R ⫽ 7952 psi
⫺a
2
a ⫽ 64.89°
us1 ⫽
up2 ⫽
180° ⫺ a
2
270° ⫺ a
2
us2 ⫽ 90° + us1
up1 ⫽ ⫺32.4°
us1 ⫽ 102.6°
tmax ⫽ R
;
us2 ⫽ 192.6°
Point S1: saver ⫽ ⫺14875 psi
;
up2 ⫽ 57.6°
;
(b) MAXIMUM SHEAR STRESSES
(a) PRINCIPAL STRESSES
up1 ⫽
;
;
tmax ⫽ 7952 psi
;
Problems 7.4-22 sx ⫽ ⫺3.3 MPa, sy ⫽ 8.9 MPa, txy ⫽ ⫺14.1 MPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-22
sx ⫽⫺3.3 MPa
sy ⫽ 8.9 MPa
txy ⫽⫺14.1 MPa
saver ⫽
sx + sy
up1 ⫽
a + 180°
2
up2 ⫽
a
2
saver ⫽ 2.8 MPa
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atana `
(a) PRINCIPAL STRESSES
txy
sx ⫺ saver
txy2
`b
R ⫽ 15.4 MPa
a ⫽ 66.6°
Point P1:
up1 ⫽ 123.3°
up2 ⫽ 33.3°
s1 ⫽ R + saver
s1 ⫽ 18.2 MPa
Point P2: s2 ⫽ ⫺R + saver
;
;
;
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SECTION 7.4
s2 ⫽ ⫺12.6 MPa
us2 ⫽ 90° + us1
;
Point S1:
(b) MAXIMUM SHEAR STRESSES
90° + a
us1 ⫽
2
tmax ⫽ R
Mohr’s Circle
us2 ⫽ 168.3°
saver ⫽ 2.8 MPa
;
tmax ⫽ 15.4 MPa
;
us1 ⫽ 78.3°
Problems 7.4-23 sx ⫽ 800 psi, sy ⫽ ⫺2200 psi, txy ⫽ 2900 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-23
sy ⫽ ⫺2200 psi
txy ⫽ 2900 psi
sx ⫽ 800 psi
sx + sy
saver ⫽
saver ⫽ ⫺700 psi
2
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atana `
txy
sx ⫺ saver
`b
(a) PRINCIPAL STRESSES
a
up1 ⫽ 31.3°
up1 ⫽
2
up2 ⫽
180° + a
2
R ⫽ 3265 psi
a ⫽ 62.65°
Point P1: s1 ⫽ R + saver
s1 ⫽ 2565 psi
Point P2: s2 ⫽ ⫺R + saver
s2 ⫽ ⫺3965 psi
up2 ⫽ 121.3°
⫺90° + a
2
us2 ⫽ 90° + us1
us1 ⫽ ⫺13.7°
us2 ⫽ 76.3°
Point S1: saver ⫽ ⫺700 psi
;
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
;
;
tmax ⫽ R
;
;
;
tmax ⫽ 3265 psi
607
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CHAPTER 7
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Analysis of Stress and Strain
Hooke’s Law for Plane Stress
sy
When solving the problems for Section 7.5, assume that the material
is linearly elastic with modulus of elasticity E and Poisson’s ratio n.
y
Problem 7.5-1 A rectangular steel plate with thickness t 0.25 in. is
subjected to uniform normal stresses x and y, as shown in the figure.
Strain gages A and B, oriented in the x and y directions, respectively, are
attached to the plate. The gage readings give normal strains ex 0.0010
(elongation) and ey 0.0007 (shortening).
Knowing that E 30 106 psi and 0.3, determine the stresses
x and y and the change t in the thickness of the plate.
Solution 7.5-1
ây 0.0007
sx
x
Probs. 7.5-1 and 7.5-2
âz SUBSTITUTE NUMERICAL VALUES:
(s + sy) 128.5 * 106
E x
¢t âzt 32.1 * 106 in.
Eq. (7-40a):
(1 )2
O
Eq. (7-39c):
E 30 * 106 psi 0.3
sx A
Rectangular plate in biaxial stress
t 0.25 in. âx 0.0010
E
B
;
(Decrease in thickness)
(âx + ây) 26,040 psi
;
Eq. (7-40b):
sy E
(1 )2
(ây + âx) 13,190 psi
;
Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t 10 mm, the gage readings are
ex 480 106 (elongation) and ey 130 106 (elongation), the modulus is E 200 GPa, and Poisson’s ratio is 0.30.
Solution 7.5-2
Rectangular plate in biaxial stress
t 10 mm âx 480 * 106
ây 130 * 10
Eq. (7-40b):
6
sy E 200 GPa 0.3
E
(1 )2
SUBSTITUTE NUMERICAL VALUES:
Eq. (7-39c):
Eq. (7-40a):
âz sx E
(1 )2
(âx + ây) 114.1 MPa
;
(ây + âx) 60.2 MPa
(s + sy) 261.4 * 106
E x
¢t âz t 2610 * 106 mm
(Decrease in thickness)
;
;
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SECTION 7.5
Problem 7.5-3 Assume that the normal strains Px and Py for an
element in plane stress (see figure) are measured with strain gages.
Hooke’s Law for Plane Stress
y
(a) Obtain a formula for the normal strain Pz in the z direction
in terms of Px, Py, and Poisson’s ratio .
(b) Obtain a formula for the dilatation e in terms of Px, Py,
and Poisson’s ratio .
sy
txy
sx
O
x
z
Solution 7.5-3
Plane stress
Given: âx, ây, (b) DILATATION
Eq. (7-47): e (a) NORMAL STRAIN âz
Eq. (7-34c): âz Eq. (7-36a): sx Eq. (7-36b): sy (s + sy)
E x
E
(1 2)
E
(1 2)
(âx + ây)
1 2
(sx + sy)
E
Substitute sx and sy from above and simplify:
e
1 2
(â + ây)
1 x
;
(ây + âx)
Substitute sx and sy into the first equation and simplify:
âz (â + ây)
1 x
;
sy
Problem 7.5-4 A magnesium plate in biaxial stress is subjected
to tensile stresses sx 24 MPa and sy 12 MPa (see figure).
The corresponding strains in the plate are x 440 106 and
y 80 106.
Determine Poisson’s ratio and the modulus of elasticity E
for the material.
y
O
x
sx
Probs. 7.5-4 through 7.5-7
Solution 7.5-4
Biaxial stress
sx 24 MPa sy 12 MPa
âx 440 * 106 ây 80 * 106
POISSON’S RATION AND MODULUS OF ELASTICITY
Eq. (7-39a): âx 1
(s sy)
E x
Eq. (7-39b): ây 1
(s sx)
E y
Substitute numerical values:
E (440 * 106) 24 MPa (12 MPa)
E (80 * 106) 12 MPa (24 MPa)
Solve simultaneously:
0.35
E 45 GPa
;
609
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CHAPTER 7
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Analysis of Stress and Strain
Problem 7.5-5 Solve the preceding problem for a steel plate with sx 10,800 psi (tension), sy 5400 psi (compression), ex 420 106 (elongation), and ey 300 106 (shortening).
Solution 7.5-5
Biaxial stress
sx 10,800 psi sy 5400 psi
âx 420 * 10
6
ây 300 * 10
Substitute numerical values:
6
E (420 * 106) 10,800 psi (5400 psi)
E (300 * 106) 5400 psi (10,800 psi)
POISSON’S RATIO AND MODULUS OF ELASTICITY
Solve simultaneously:
1/3
1
Eq. (7-39a): âx (sx sy)
E
Eq. (7-39b): ây E 30 * 106 psi
;
1
(s sx)
E y
Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses x 90 MPa (tension) and
y 20 MPa (compression). The plate has dimensions 400 * 800 * 20 mm and is made of steel with E 200 GPa and
0.30.
(a) Determine the maximum in-plane shear strain gmax in the plate.
(b) Determine the change ¢t in the thickness of the plate.
(c) Determine the change ¢V in the volume of the plate.
Solution 7.5-6
Biaxial stress
sx 90 MPa sy 20 MPa
E 200 GPa 0.30
(b) CHANGE IN THICKNESS
Dimensions of Plate: 400 mm * 800 mm * 20 mm
Shear Modulus (Eq. 7-38):
E
G
76.923 GPa
2(1 + )
(sx + sy) 105 * 106
E
¢t âz t 2100 * 106 mm
;
(Decrease in thickness)
(c) CHANGE IN VOLUME
(a) MAXIMUM IN-PLANE SHEAR STRAIN
Principal stresses: s1 90 MPa
Eq. (7-39c): âz s2 20 MPa
Eq. (7-26): tmax s1 s2
55.0 MPa
2
Eq. (7-35): gmax tmax
715 * 106
G
From Eq. (7-47): ¢V V0 a
1 2
b(sx + sy)
E
V0 (400)(800)(20) 6.4 * 106 mm3
;
Also, a
1 2
b(sx + sy) 140 * 106
E
‹ ¢V (6.4 * 106 mm3)(140 * 106)
896 mm3
(Increase in volume)
;
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Page 611
SECTION 7.5
Hooke’s Law for Plane Stress
611
Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx 12,000 psi (tension), sy 3,000 psi
(compression), dimensions 20 30 0.5 in., E 10.5 106 psi, and 0.33.
Solution 7.5-7
Biaxial stress
sx 12,000 psi sy 3,000 psi
(b) CHANGE IN THICKNESS
E 10.5 * 10 psi 0.33
6
Eq. (7-39c): âz Dimensions of Plate: 20 in. * 30 in. * 0.5 in.
Shear Modulus (Eq. 7-38):
G
282.9 * 106
¢t âz t 141 * 106 in.
E
3.9474 * 106 psi
2(1 + )
(c) CHANGE IN VOLUME
Principal stresses: s1 12,000 psi
From Eq. (7-47): ¢V V0 a
s2 3,000 psi
Eq. (7-35): gmax
s1 s2
7,500 psi
2
tmax
1,900 * 106
G
;
(Decrease in thickness)
(a) MAXIMUM IN-PLANE SHEAR STRAIN
Eq. (7-26): tmax (s + sy)
E x
1 2
b(sx + sy)
E
V0 (20)(30)(0.5) 300 in.3
;
Also, a
1 2
b(sx + sy) 291.4 * 106
E
‹ ¢V (300 in.3)(291.4 * 106)
0.0874 in.3
;
(Increase in volume)
Problem 7.5-8 A brass cube 50 mm on each edge is compressed
in two perpendicular directions by forces P 175 kN (see figure).
Calculate the change V in the volume of the cube and the
strain energy U stored in the cube, assuming E 100 GPa and
0.34.
Solution 7.5-8
P = 175 kN
P = 175 kN
Biaxial stress-cube
sx sy P
b
2
(175 kN)
(50 mm)2
70.0 MPa
CHANGE IN VOLUME
Eq. (7-47): e 1 2
(sx + sy) 448 * 106
E
V0 b 3 (50 mm)3 125 * 103mm3
¢V eV0 56 mm3
Side b 50 mm P 175 kN
E 100 GPa 0.34 ( Brass)
(Decrease in volume)
;
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Analysis of Stress and Strain
STRAIN ENERGY
U uV0 (0.03234 MPa)(125 * 103 mm3)
Eq. (7-50): u 1
(s2 + s2y 2sxsy)
2E x
4.04 J
;
0.03234 MPa
Problem 7.5-9 A 4.0-inch cube of concrete (E 3.0 * 106 psi, 0.1)
is compressed in biaxial stress by means of a framework that is loaded as
shown in the figure.
Assuming that each load F equals 20 k, determine the change ¢V in
the volume of the cube and the strain energy U stored in the cube.
F
F
Solution 7.5-9
Biaxial stress – concrete cube
CHANGE IN VOLUME
b 4 in.
Eq. (7-47): e E 3.0 * 106 psi
V0 b 3 (4 in.)3 64 in.3
0.1
F 20 kips
1 2
(sx + sy) 0.0009429
E
¢V eV0 0.0603 in.3
;
(Decrease in volume)
STRAIN ENERGY
Joint A:
Eq. (7-50): u P F12
28.28 kips
sx sy P
1768 psi
b2
1
(s2 + s2y 2sxsy)
2E x
0.9377 psi
U uV0 60.0 in.-lb
;
Py
Problem 7.5-10 A square plate of width b and thickness t is loaded by
normal forces Px and Py, and by shear forces V, as shown in the figure.
These forces produce uniformly distributed stresses acting on the side
faces of the place.
Calculate the change V in the volume of the plate and the strain
energy U stored in the plate if the dimensions are b 600 mm and
t 40 mm, the plate is made of magnesium with E 45 GPa and
v 0.35, and the forces are Px 480 kN, Py 180 kN, and
V 120 kN.
t
V
y
Px
V
b O
x
b
V
V
Py
Probs. 7.5-10 and 7.5-11
Px
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SECTION 7.5
Solution 7.5-10
Hooke’s Law for Plane Stress
613
Square plate in plane stress
b 600 mm
t 40 mm
E 45 GPa
v 0.35 (magnesium)
Px
20.0 MPa
bt
Py
sy 7.5 MPa
bt
V
txy 5.0 MPa
bt
Px 480 kN
sx Py 180 kN
V 120 kN
CHANGE IN VOLUME
1 2
Eq. (7-47): e (sx + sy) 183.33 * 106
E
V0 b 2t 14.4 * 106 mm3
¢V eV0 2640 mm3
;
(Increase in volume)
STRAIN ENERGY
Eq. (7-50): u G
t2xy
1 2
(sx + s2y 2sxsy) +
2E
2G
E
16.667 GPa
2(1 + )
Substitute numerical values:
u 4653 Pa
U uV0 67.0 N # m 67.0 J
;
Problem 7.5-11 Solve the preceding problem for an aluminum plate with b 12 in., t 1.0 in., E 10,600 ksi, 0.33,
Px 90 k, Py 20 k, and V 15 k.
Solution 7.5-11
Square plate in plane stress
b 12.0 in.
t 1.0 in.
STRAIN ENERGY
E 10,600 ksi 0.33 (aluminum)
Px
7500 psi
bt
Py
sy 1667 psi
bt
Px 90 k
sx Py 20 k
V 15 k
txy G
u 2.591 psi
V
1250 psi
bt
U uV0 373 in.-lb
1 2
(sx + sy) 294 * 106
E
V0 b 2t 144 in.3
¢V eV0 0.0423 in.3
(Increase in volume)
E
3985 ksi
2(1 + )
Substitute numerical values:
CHANGE IN VOLUME
Eq. (7-47): e t2xy
1
2
2
Eq. (7-50): u (s + sy 2sxsy) +
2E x
2G
;
;
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Analysis of Stress and Strain
Problem 7.5-12 A circle of diameter d 200 mm is etched on a
z
brass plate (see figure). The plate has dimensions 400 400 20 mm.
Forces are applied to the plate, producing uniformly distributed normal
stresses x 42 MPa and y 14 MPa.
Calculate the following quantities: (a) the change in length ac
of diameter ac; (b) the change in length bd of diameter bd; (c) the
change t in the thickness of the plate; (d) the change V in the
volume of the plate, and (e) the strain energy U stored in the plate.
(Assume E 100 GPa and v 0.34.)
y
sy
d
sx
a
c
sx
b
x
sy
Solution 7.5-12
Plate in biaxial stress
sx 42 MPa sy 14 MPa
(c) CHANGE IN THICKNESS
(s + sy)
E x
190.4 * 106
Eq. (7-39c): âz Dimensions: 400 * 400 * 20 (mm)
Diameter of circle: d 200 mm
E 100 GPa 0.34 (Brass)
¢t âz t 0.00381 mm
(decrease)
;
(a) CHANGE IN LENGTH OF DIAMETER IN x DIRECTION
Eq. (7-39a): âx 1
(s sy) 372.4 * 106
E x
¢ac âx d 0.0745 mm
(increase)
;
(b) CHANGE IN LENGTH OF DIAMETER IN y DIRECTION
Eq. (7-39b): ây 1
(s sx) 2.80 * 106
E y
¢bd ây d
560 * 106 mm
(decrease)
;
(d) CHANGE IN VOLUME
Eq. (7-47):
e
1 2
(sx + sy) 179.2 * 106
E
V0 (400)(400)(20) 3.2 * 106 mm3
¢V eV0 573 mm3 ;
(increase)
(e) STRAIN ENERGY
Eq. (7-50): u 1 2
(s + s2y 2sx sy)
2E x
7.801 * 103 MPa
U uV0 25.0 N # m 25.0 J
;
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Page 615
SECTION 7.6
615
Triaxial Stress
Triaxial Stress
When solving the problems for Section 7.6, assume that the material is
linearly elastic with modulus of elasticity E and Poisson’s ratio n.
y
a
c
Problem 7.6-1 An element of aluminum in the form of a rectangular
parallelepiped (see figure) of dimensions a ⫽ 6.0 in., b ⫽ 4.0 in, and
c ⫽ 3.0 in. is subjected to triaxial stresses sx ⫽ 12,000 psi,
sy ⫽ ⫺4,000 psi, and sz ⫽ ⫺1,000 psi acting on the x, y, and z faces,
respectively.
Determine the following quantities: (a) the maximum shear stress tmax
in the material; (b) the changes ¢a, ¢b, and ¢c in the dimensions of the
element; (c) the change ¢V in the volume; and (d) the strain energy U
stored in the element. (Assume E ⫽ 10,400 ksi and ␯ ⫽ 0.33.)
b
O
x
z
Probs. 7.6-1 and 7.6-2
Solution 7.6-1
Triaxial stress
sx ⫽ 12,000 psi sy ⫽ ⫺4,000 psi
¢a ⫽ aâx ⫽ 0.0079 in. ( increase)
sz ⫽ ⫺1,000 psi
¢b ⫽ bây ⫽ ⫺0.0029 in. ( decrease)
a ⫽ 6.0 in. b ⫽ 4.0 in. c ⫽ 3.0 in.
E ⫽ 10,400 ksi ␯ ⫽ 0.33
( aluminum)
(a) MAXIMUM SHEAR STRESS
s1 ⫽ 12,000 psi s2 ⫽ ⫺1,000 psi
(c) CHANGE IN VOLUME
Eq. (7-56):
e⫽
s3 ⫽ ⫺4,000 psi
tmax ⫽
¢c ⫽ câz ⫽ ⫺0.0011 in. ( decrease)
s1 ⫺ s3
⫽ 8,000 psi
2
;
M
;
1 ⫺ 2␯
(sx + sy + sz) ⫽ 228.8 * 10⫺6
E
V ⫽ abc
¢V ⫽ e (abc) ⫽ 0.0165 in.3 ( increase)
;
(b) CHANGES IN DIMENSIONS
Eq. (7-53 a): âx ⫽
sx
␯
⫺ (sy + sz)
E
E
⫽1312.5 * 10⫺6
Eq. (7- 53 b): ây ⫽
sy
E
⫺
␯
(s + sx)
E z
⫽⫺733.7 * 10⫺6
Eq. (7-53 c): âz ⫽
sz
E
⫺
␯
(s + sy)
E x
⫽⫺350.0 * 10⫺6
(d) STRAIN ENERGY
Eq. (7-57a): u ⫽
1
(s â + sy ây + sz âz)
2 x x
⫽ 9.517 psi
U ⫽ u (abc) ⫽ 685 in.-lb
;
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CHAPTER 7
Page 616
Analysis of Stress and Strain
Problem 7.6-2 Solve the preceding problem if the element is steel (E = 200 GPA, ␯ ⫽ 0.30) with dimensions a = 300 mm,
b = 150 mm, and c = 150 mm and the stresses are sx ⫽ ⫺60 MPa, sy ⫽ ⫺40 MPa, and sz ⫽ ⫺40 MPa.
Solution 7.6-2
Triaxial stress
sx ⫽ ⫺60 MPa
sy ⫽ ⫺40 MPa
¢a ⫽ aâx ⫽ ⫺0.0540 mm (decrease)
sz ⫽ ⫺40 MPa
a ⫽ 300 mm
¢b ⫽ bây ⫽ ⫺0.0075 mm (decrease)
b ⫽ 150 mm
E ⫽ 200 GPa
␯ ⫽ 0.30
c ⫽ 150 mm
(steel)
(a) MAXIMUM SHEAR STRESS
s1 ⫽ ⫺40 MPa
s2 ⫽ ⫺40 MPa
s3 ⫽ ⫺60 MPa
tmax ⫽
s1 ⫺ s3
⫽ 10.0 MPa
2
Eq. (7-53 b): ây ⫽
Eq. (7-53 c): âz ⫽
sx
␯
⫺ (sy + sz) ⫽ ⫺180.0 * 10⫺6
E
E
sy
E
sz
E
(c) CHANGE IN VOLUME
Eq. (7-56):
e⫽
M
;
1 ⫺ 2␯
(sx + sy + sz) ⫽ ⫺280.0 * 10⫺6
E
V ⫽ abc
;
(b) CHANGES IN DIMENSIONS
Eq. (7-53 a): âx ⫽
¢c ⫽ câz ⫽ ⫺0.0075 mm. (decrease)
⫺
␯
(s + sx) ⫽ ⫺50.0 * 10⫺6
E z
⫺
␯
(s + sy) ⫽ ⫺50.0 * 10⫺6
E x
¢V ⫽ e(abc) ⫽ ⫺1890 mm3 (decrease)
;
(d) STRAIN ENERGY
1
(s â + sy ây + sz âz)
2 x x
⫽ 0.00740 MPa
Eq. (7-57 a): u ⫽
U ⫽ u (abc) ⫽ 50.0 N # m ⫽ 50.0 J
;
y
Problem 7.6-3 A cube of cast iron with sides of length a = 4.0 in.
(see figure) is tested in a laboratory under triaxial stress. Gages mounted
on the testing machine show that the compressive strains in the material
are Px ⫽ ⫺225 * 10⫺6 and Py ⫽ Pz ⫽ ⫺37.5 * 10⫺6.
Determine the following quantities: (a) the normal stresses sx, sy, and
sz acting on the x, y, and z faces of the cube; (b) the maximum shear
stress tmax in the material; (c) the change ¢V in the volume of the
cube; and (d) the strain energy U stored in the cube. (Assume E = 14,000
ksi and ␯ ⫽ 0.25.)
a
a
a
O
z
Probs. 7.6-3 and 7.6-4
x
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Page 617
SECTION 7.6
Solution 7.6-3
Triaxial Stress
617
Triaxial stress (cube)
âx ⫽ ⫺225 * 10⫺6
ây ⫽ ⫺37.5 * 10⫺6
(c) CHANGE IN VOLUME
âz ⫽ ⫺37.5 * 10⫺6 a ⫽ 4.0 in.
Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺0.000300
E ⫽ 14,000 ksi ␯ ⫽ 0.25
V ⫽ a3
(cast iron)
¢V ⫽ ea 3 ⫽ ⫺0.0192 in.3 ( decrease)
(a) NORMAL STRESSES
Eq. (7-54a):
(d) STRAIN ENERGY
E
[(1 ⫺ ␯)âx + ␯(ây + âz)]
sx ⫽
(1 + ␯)(1 ⫺ 2␯)
⫽ ⫺4200 psi
;
Eq. (7-57a): u ⫽
;
1
(sx âx + sy ây + sz âz)
2
⫽ 0.55125 psi
In a similar manner, Eqs. (7-54 b and c) give
sy ⫽ ⫺2100 psi sz ⫽ ⫺2100 psi ;
U ⫽ ua ⫽ 35.3 in.-lb
3
;
(b) MAXIMUM SHEAR STRESS
s1 ⫽ ⫺2100 psi s2 ⫽ ⫺2100 psi
s3 ⫽ ⫺4200 psi
tmax ⫽
s1 ⫺ s3
⫽ 1050 psi
2
;
Problem 7.6-4 Solve the preceding problem if the cube is granite (E ⫽ 60 GPa, ␯ ⫽ 0.25) with dimensions a = 75 mm and
compressive strains Px ⫽ ⫺720 * 10⫺6 and Py ⫽ Pz ⫽ ⫺270 * 10⫺6.
Solution 7.6-4
Triaxial stress (cube)
âx ⫽ ⫺720 * 10⫺6
ây ⫽ ⫺270 * 10⫺6
âz ⫽ ⫺270 * 10⫺6
a ⫽ 75 mm
␯ ⫽ 0.25
E ⫽ 60 GPa
(Granite)
;
Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺1260 * 10⫺6
V ⫽ a3
E
[(1 ⫺ ␯)âx + ␯(âx + âz)]
sx ⫽
(1 + ␯)(1 ⫺ 2␯)
;
In a similar manner, Eqs. (7-54 b and c) give
sy ⫽ ⫺43.2 MPa sz ⫽ ⫺43.2 MPa ;
(b) MAXIMUM SHEAR STESS
s1 ⫽ ⫺43.2 MPa s2 ⫽ ⫺43.2 MPa
s3 ⫽ ⫺64.8 MPa
s1 ⫺ s3
⫽ 10.8 MPa
2
(c) CHANGE IN VOLUME
(a) NORMAL STRESSES
Eq.(7-54a):
⫽ ⫺64.8 MPa
tmax ⫽
¢V ⫽ ea 3 ⫽ ⫺532 mm3 ( decrease)
;
(d) STRAIN ENERGY
1
Eq. (7-57 a): u ⫽ (sxâx + syây + szâz)
2
⫽ 0.03499 MPa ⫽ 34.99 kPa
U ⫽ ua ⫽ 14.8 N # m ⫽ 14.8 J
3
;
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CHAPTER 7
Page 618
Analysis of Stress and Strain
Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is
subjected to stresses sx ⫽ 5200 psi (tension), sy ⫽ ⫺4750 psi
(compression), and sz ⫽ ⫺3090 psi (compression). It is also known that the normal strains in the x and y directions
are Px ⫽ 7138.8 * 10⫺6 (elongation) and Py ⫽ ⫺502.3 * 10⫺6 (shortsx
ening).
What is the bulk modulus K for the aluminum?
y
sy
sz
sx
O
x
sz
sy
z
Probs. 7.6-5 and 7.6-6
Solution 7.6-5
Triaxial stress (bulk modulus)
sx ⫽ 5200 psi sy ⫽ ⫺4750 psi
sz ⫽ ⫺3090 psi âx ⫽ 713.8 * 10
ây ⫽ ⫺502.3 * 10
Substitute numerical values and rearrange:
⫺6
⫺6
(713.8 * 10⫺6) E ⫽ 5200 + 7840 ␯
(⫺502.3 * 10
) E ⫽ ⫺4750 ⫺ 2110 ␯
Find K.
Units: E = psi
sx
␯
⫺ (sy + sz)
E
E
sy
␯
Eq. (7-53 b): ây ⫽
⫺ (sx + sy)
E
E
Solve simultaneously Eqs. (1) and (2):
Eq. (7-53 a): âx ⫽
(1)
⫺6
(2)
E ⫽ 10.801 * 106 psi ␯ ⫽ 0.3202
Eq. (7-16): K ⫽
E
⫽ 10.0 * 10⫺6 psi
3(1 ⫺ 2␯)
;
Problem 7.6-6
Solve the preceding problem if the material is nylon subjected to compressive stresses
sx ⫽ ⫺4.5 MPa , sy ⫽ ⫺3.6 MPa , and sz ⫽ ⫺2.1 MPa, and the normal strains are Px ⫽ ⫺740 * 10⫺6 and
Py ⫽ ⫺320 * 10⫺6 (shortenings).
Solution 7.6-6
Triaxial stress (bulk modulus)
sx ⫽ ⫺4.5 MPa sy ⫽ ⫺3.6 MPa
sz ⫽ ⫺2.1 MPa âx ⫽ ⫺740 * 10
ây ⫽ ⫺320 * 10
Substitute numerical values and rearrange:
⫺6
⫺6
Find K.
sx
␯
⫺ (sy + sz)
E
E
sy
␯
Eq. (7-53 b): ây ⫽
⫺ (sz + sx)
E
E
Eq. (7-53 a): âx ⫽
(⫺740 * 10⫺6) E ⫽ ⫺4.5 + 5.7 ␯
(⫺320 * 10
(1)
⫺6
) E ⫽ ⫺3.6 + 6.6 ␯
(2)
Units: E = MPa
Solve simultaneously Eqs. (1) and (2):
E ⫽ 3,000 MPa ⫽ 3.0 GPa ␯ ⫽ 0.40
Eq. (7-16): K ⫽
E
⫽ 5.0 GPa
3(1 ⫺ 2␯)
;
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Page 619
SECTION 7.6
Problem 7.6-7 A rubber cylinder R of length L and cross-sectional
area A is compressed inside a steel cylinder S by a force F that
applies a uniformly distributed pressure to the rubber (see figure).
F
(a) Derive a formula for the lateral pressure p between the
rubber and the steel. (Disregard friction between the rubber
and the steel, and assume that the steel cylinder is rigid when
compared to the rubber.)
(b) Derive a formula for the shortening d of the rubber cylinder.
Solution 7.6-7
619
Triaxial Stress
F
R
S
L
S
Rubber cylinder
Solve for p: p ⫽
v
F
a b
1⫺v A
;
(b) SHORTENING
Eq. (7-53 b): ây ⫽
sx ⫽ ⫺p sy ⫽ ⫺
F
A
sz ⫽ ⫺p
âx ⫽ âz ⫽ 0
⫽⫺
E
⫺
␯
(s + sx)
E z
␯
F
⫺ (⫺2p)
EA
E
Substitute for p and simplify:
ây ⫽
F (1 + ␯)(⫺1 + 2␯)
EA
1⫺␯
(Positive ây represents an increase in strain, that is,
elongation.)
(a) LATERAL PRESSURE
Eq. (7-53 a): âx ⫽
sy
sx
␯
⫺ (sy + sz)
E
E
F
or 0 ⫽ ⫺p ⫺ ␯ a ⫺ ⫺ pb
A
d ⫽ ⫺âyL
d⫽
(1 + ␯)(1 ⫺ 2␯) FL
a
b
(1 ⫺ ␯)
EA
(Positive d represents a shortening of the rubber
cylinder.)
Problem 7.6-8 A block R of rubber is confined between plane
parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by
a force F.
(a) Derive a formula for the lateral pressure p between the
rubber and the steel. (Disregard friction between the
rubber and the steel, and assume that the steel block is
rigid when compared to the rubber.)
(b) Derive a formula for the dilatation e of the rubber.
(c) Derive a formula for the strain-energy density u of the
rubber.
;
F
F
S
S
R
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CHAPTER 7
Solution 7.6-8
Page 620
Analysis of Stress and Strain
Block of rubber
(b) DILATATION
Eq. (7-56): e ⫽
1 ⫺ 2␯
(sx + sy + sz)
E
⫽
1 ⫺ 2␯
(⫺p ⫺ p0)
E
Substitute for p:
e⫽ ⫺
sx ⫽ ⫺p
sy ⫽ ⫺p0
âx ⫽ 0
OR
ây Z 0 âz Z 0
sx
␯
⫺ (sy + sz)
E
E
0 ⫽ ⫺p ⫺ ␯ (⫺p0)
;
sz ⫽ 0
(a) LATERAL PRESSURE
Eq. (7-53 a): âx ⫽
(1 + ␯)(1 ⫺ 2␯)p0
E
(c) STRAIN ENERGY DENSITY
Eq. (7-57b):
u⫽
‹ p ⫽ ␯p0
;
1
v
(s2x + s2y + s2z ) ⫺ (sx sy + sx sz + sy sz)
2E
E
Substitute for sx, sy, sz, and p:
u⫽
(1 ⫺ ␯2)p 20
2E
;
Problem 7.6-9 A solid spherical ball of brass (E ⫽ 15 * 106 psi ␯ ⫽ 0.34) is lowered into the ocean to a depth of 10,000 ft.
The diameter of the ball is 11.0 in.
Determine the decrease ¢d in diameter, the decrease ¢V in volume, and the strain energy U of the ball.
Solution 7.6-9
E ⫽ 15 * 10
⫺6
Brass sphere
DECREASE IN VOLUME
psi ␯ ⫽ 0.34
Lowered in the ocean to depth h ⫽ 10,000 ft
Eq. (7-60): e ⫽ 3â0 ⫽ 283.6 * 10⫺6
4
4
11.0 in. 3
b ⫽ 696.9 in.3
V0 ⫽ pr 3 ⫽ (p)a
3
3
2
Diameter d ⫽ 11.0 in.
Sea water: g ⫽ 63.8 lb/ft3
Pressure: s0 ⫽ gh ⫽ 638,000 lb/ft2 ⫽ 4431 psi
DECREASE IN DIAMETER
s0
Eq. (7-59): â0 ⫽ (1 ⫺ 2␯) ⫽ 94.53 * 10⫺6
E
¢d ⫽ â0d ⫽ 1.04 * 10⫺3 in.
(decrease)
;
¢V ⫽ eV0 ⫽ 0.198 in.3
(decrease)
;
STRAIN ENERGY
Use Eq. (7-57 b) with sx ⫽ sy ⫽ sz ⫽ s0:
u⫽
3(1 ⫺ 2␯)s20
⫽ 0.6283 psi
2E
U ⫽ uV0 ⫽ 438 in.-lb
;
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Page 621
SECTION 7.6
Triaxial Stress
621
Problem 7.6-10 A solid steel sphere (E ⫽ 210 GPa, ␯ = 0.3) is subjected to hydrostatic pressure p such that its volume is
reduced by 0.4%.
(a) Calculate the pressure p.
(b) Calculate the volume modulus of elasticity K for the steel.
(c) Calculate the strain energy U stored in the sphere if its diameter is d ⫽ 150 mm
Solution 7.6-10
Steel sphere
E ⫽ 210 GPa ␯ ⫽ 0.3
(b) VOLUME MODULUS OF ELASTICITY
Hydrostatic Pressure. V0 = Initial volume
¢V ⫽ 0.004 V0
¢V
⫽ 0.004
V0
Dilatation: e ⫽
3s0(1 ⫺ 2␯)
Eq.(7-60): e ⫽
E
s0 ⫽
s0
700 MPa
⫽
⫽ 175 GPa
E
0.004
;
(c) STRAIN ENERGY (d = diameter)
d = 150 mm r = 75 mm
From Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0:
(a) PRESSURE
or
Eq. (7-63): K ⫽
Ee
⫽ 700 MPa
3(1 ⫺ 2␯)
Pressure p ⫽ s0 ⫽ 700 MPa
;
u⫽
3(1 ⫺ 2␯)s20
⫽ 1.40 MPa
2E
V0 ⫽
4pr 3
⫽ 1767 * 10⫺6 m3
3
U ⫽ uV0 ⫽ 2470 N # m ⫽ 2470 J
;
Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K ⫽ 14.5 ⫻ 106 psi) is suddenly heated around its
outer surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the center
of the sphere.
If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energy
density u at the center.
Solution 7.6-11
Bronze sphere (heated)
K ⫽ 14.5 * 10 psi
s0 ⫽ 12,000 psi (tension at the center)
6
STRAIN AT THE CENTER OF THE SPHERE
s0
Eq. (7-59): â0 ⫽ (1 ⫺ 2␯)
E
E
Eq. (7-61): K ⫽
3(1 ⫺ 2␯)
Combine the two equations:
â0 ⫽
s0
⫽ 276 * 10 ⫺6
3K
;
UNIT VOLUME CHANGE AT THE CENTER
Eq. (7-62): e ⫽
s0
⫽ 828 * 10 ⫺6
K
;
STRAIN ENERGY DENSITY AT THE CENTER
Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0:
3(1 ⫺ 2␯)s20
s02
⫽
2E
2K
u ⫽ 4.97 psi ;
u⫽
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CHAPTER 7
Page 622
Analysis of Stress and Strain
Plane Strain
y
When solving the problems for Section 7.7, consider only the in-plane strains (the
strains in the xy plane) unless stated otherwise. Use the transformation equations
of plane strain except when Mohr’s circle is specified (Problems 7.7-23 through
7.7-28).
Problem 7.7-1 A thin rectangular plate in biaxial stress is subjected to
stresses sx and sy, as shown in part (a) of the figure on the next page. The
width and height of the plate are b 8.0 in. and h 4.0 in., respectively.
Measurements show that the normal strains in the x and y directions are
Px 195 * 106 and Py 125 * 106, respectively.
With reference to part (b) of the figure, which shows a two-dimensional
view of the plate, determine the following quantities: (a) the increase ¢d in the
length of diagonal Od; (b) the change ¢f in the angle f between diagonal Od
and the x axis; and (c) the change ¢c in the angle c between diagonal Od and
the y axis.
sy
sx
h
b
x
z
(a)
y
d
c
h
f
O
x
b
(b)
Probs. 7.7-1 and 7.7-2
Solution 7.7-1
Plate in biaxial stress
For u f 26.57°, âx1 130.98 * 106
¢d âx1L d 0.00117 in.
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sinu cosu gxy sin2u
For u f 26.57°: a 128.0 * 106 rad
b 8.0 in.
h 4.0 in.
ây 125 * 106
âx 195 * 106
gxy 0
Minus sign means line Od rotates clockwise
(angle f decreases).
¢f 128 * 106 rad (decrease)
h
f arctan 26.57°
b
;
(c) CHANGE IN ANGLE c
L d 1b 2 + h2 8.944 in.
Angle c increases the same amount that f
decreases.
(a) INCREASE IN LENGTH OF DIAGONAL
¢c 128 * 106 rad (increase)
âx1 âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
sin 2u
;
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Page 623
SECTION 7.7
Plane Strain
623
Problem 7.7-2 Solve the preceding problem if b 160 mm, h 60 mm, Px 410 * 106, and Py 320 * 106.
Solution 7.7-2
Plate in biaxial stress
For u f 20.56°: âx1 319.97 * 10 6
¢d âx1 L d 0.0547 mm
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
For u f 20.56°: a 240.0 * 106 rad
b 160 mm
h 60 mm
ây 320 * 106
Minus sign means line Od rotates clockwise (angle f
decreases.)
âx 410 * 106
gxy 0
¢f 240 * 106 rad (decrease)
h
f arctan 20.56°
b
2
1
L d b + h2 170.88 mm
(c) CHANGE IN ANGLE c
Angle c increases the same amount that f
decreases.
¢c 240 * 106 rad (increase)
(a) INCREASE IN LENGTH OF DIAGONAL
âx1 âx ây
âx + ây
2
+
2
cos 2u +
;
gxy
2
;
sin 2u
Problem 7.7-3 A thin square plate in biaxial stress is subjected
to stresses sx and sy, as shown in part (a) of the figure . The
width of the plate is b 12.0 in. Measurements show that the
normal strains in the x and y directions are Px 427 * 106
and Py 113 * 106, respectively.
With reference to part (b) of the figure, which shows a
two-dimensional view of the plate, determine the following
quantities: (a) the increase ¢d in the length of diagonal Od;
(b) the change ¢f in the angle f between diagonal Od and
the x axis; and (c) the shear strain g associated with diagonals
Od and cf (that is, find the decrease in angle ced).
y
sy
y
c
sx
b
e
b
f
b
x
z
d
O
b
(b)
(a)
PROBS. 7.7-3 and 7.7-4
f
x
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CHAPTER 7
Page 624
Analysis of Stress and Strain
Solution 7.7-3
Square plate in biaxial stress
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
For u f 45°: a 157 * 106 rad
Minus sign means line Od rotates clockwise (angle f
decreases.)
¢f 157 * 106 rad (decrease)
(c) SHEAR STRAIN BETWEEN DIAGONALS
âx 427 * 106
b 12.0 in.
ây 113 * 106
Eq. (7-71b):
gxy 0
f 45°
;
gx1y1
2
âx ây
2
sin 2u +
gxy
2
cos 2u
For u f 45°: gx1y1 314 * 106 rad
L d b 12 16.97 in.
(Negative strain means angle ced increases)
(a) INCREASE IN LENGTH OF DIAGONAL
g 314 * 106 rad
âx1 âx ây
â x + ây
+
2
2
cos 2u +
gxy
2
;
sin 2u
For u f 45°: âx1 270 * 106
¢d âx1L d 0.00458 in.
;
Problem 7.7-4 Solve the preceding problem if b 225 mm, Px 845 * 106 , and Py 211 * 106.
Solution 7.7-4
Square plate in biaxial stress
(a) INCREASE IN LENGTH OF DIAGONAL
âx1 âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
sin 2u
For u f 45°: âx1 528 * 106
¢d âx1L d 0.168 mm
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
b 225 mm
ây 211 * 106
âx 845 * 106
f 45°
L d b 12 318.2 mm
gxy 0
For u f 45°: a 317 * 106 rad
Minus sign means line Od rotates clockwise (angle f
decreases.)
¢f 317 * 106 rad (decrease)
;
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Page 625
SECTION 7.7
g x1y1
2
âx ây
2
625
For u f 45°: gx1y1 634 * 106 rad
(c) SHEAR STRAIN BETWEEN DIAGONALS
Eq. (7- 71b):
Plane Strain
sin 2u +
gxy
2
cos 2u
(Negative strain means angle ced increases)
g 634 * 106 rad
y
Problem 7.7-5 An element of material subjected to plane strain (see
figure) has strains as follows: Px 220 * 106, Py 480 * 106, and
gxy 180 * 106.
Calculate the strains for an element oriented at an angle u 50° and show these
strains on a sketch of a properly oriented element.
ey
gxy
1
O
1
ex
x
Probs. 7.7-5 through 7.7-10
Solution 7.7-5
Element in plane strain
âx 220 * 106
ây 480 * 106
gxy 180 * 106
âx1 gx1 y1
2
âx ây
âx + ây
+
2
âx ây
2
2
cos 2u +
sin 2u +
g xy
2
gxy
2
sin 2u
cos 2u
ây1 âx + ây âx1
For u 50°:
âx1 461 * 106
gx1y1 225 * 106
ây1 239 * 106
Problem 7.7-6 Solve the preceding problem for the following data: Px 420 * 106, Py 170 * 106, gxy 310 * 106,
and u 37.5°.
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CHAPTER 7
Page 626
Analysis of Stress and Strain
Solution 7.7-6
Element in plane strain
âx 420 * 106
ây 170 * 106
gxy 310 * 106
âx1 gx1y1
2
âx ây
âx + ây
+
2
âx ây
2
sin 2u +
2
cos 2u +
gxy
2
gxy
2
sin 2u
cos 2u
ây1 âx + ây âx1
For u 37.5°:
âx1 351 * 106
gx1y1 490 * 106
ây1 101 * 106
Problem 7.7-7 The strains for an element of material in plane strain (see figure) are as follows: Px 480 * 106,
Py 140 * 106, and gxy 350 * 106.
Determine the principal strains and maximum shear strains, and show these strains on sketches of properly oriented
elements.
Solution 7.7-7
Element in plane strain
âx 480 * 106
ây 140 * 106
gxy 350 * 106
PRINCIPAL STRAINS
â1, 2 âx + ây
2
;
âx ây 2
gxy 2
b + a b
A
2
2
a
310 * 106 ; 244 * 106
â2 66 * 106
â1 554 * 106
gxy
1.0294
tan 2up âx ây
2up 45.8° and
up 22.9° and
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
134.2°
244 * 106
67.1°
gmax 488 * 106
For up 22.9°:
âx1 âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
us1 up1 45° 67.9° or 112.1°
sin 2u
up2 67.1°
â1 554 * 106
â2 66 * 106
;
;
;
us2 us1 + 90° 22.1°
554 * 106
‹ up1 22.9°
gmax 488 * 106
gmin 488 * 106
âaver âx + ây
2
;
310 * 106
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Page 627
SECTION 7.7
Plane Strain
Problem 7.7-8 Solve the preceding problem for the following strains: Px 120 * 106, Py 450 * 106, and
gxy 360 * 106.
Solution 7.7-8
âx 120 * 10
Element in plane strain
6
ây 450 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
gxy 360 * 106
PRINCIPAL STRAINS
â1,2 âx + ây
2
337 * 106
âx ây 2
gxy 2
;
a
b + a b
A
2
2
gmax 674 * 106
us1 up1 45° 118.9°
165 * 106 ; 377 * 106
â1 172 * 106
tan 2up gxy
âx ây
â2 502 * 106
0.6316
âaver up 163.9° and 73.9°
For up 163.9°:
âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
sin 2u
172 * 106
‹ up1 163.9°
up2 73.9°
â1 172 * 106
â2 502 * 10
;
6
;
;
us2 us1 90° 28.9°
gmin 674 * 106
2up 327.7° and 147.7°
âx1 gmax 674 * 106
âx + ây
2
;
165 * 106
627
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CHAPTER 7
Page 628
Analysis of Stress and Strain
Problem 7.7-9 A element of material in plane strain (see figure) is subjected to strains Px 480 * 106, Py 70 * 106,
and gxy 420 * 106.
Determine the following quantities: (a) the strains for an element oriented at an angle u 75°, (b) the principal strains,
and (c) the maximum shear strains. Show the results on sketches of properly oriented element.
Solution 7.7-9
Element in plane strain
âx 480 * 106
ây 70 * 106
gxy 420 * 106
âx1 gx1y1
2
âx ây
âx + ây
+
2
âx ây
2
2
cos 2u +
sin 2u +
gxy
2
gxy
2
For up 22.85°:
âx1 âx ây
âx + ây
+
2
gxy
2
sin 2u
568 * 106
sin 2u
â1 568 * 106
‹ up1 22.8°
cos 2u
2
cos 2u +
up2 112.8
°
â2 18 * 10
6
;
;
ây1 âx + ây âx1
For u 75°:
âx1 202 * 106
gx1y1 569 * 106
ây1 348 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b 293 * 106
2
A
2
2
gmax 587 * 106
us1 up1 45° 22.2° or 157.8°
gmax 587 * 106
us2 us1 + 90° 67.8°
PRINCIPAL STRAINS
â1,2 âx + ây
2
;
;
275 * 10
âx ây 2
gxy 2
b + a
b
A
2
2
6
a
; 293 * 10
6
â2 18 * 106
â1 568 * 106
gxy
1.0244
tan 2up âx ây
2up 45.69° and 225.69°
up 22.85° and 112.85°
gmin 587 * 106
âaver â x + ây
2
;
275 * 106
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Page 629
SECTION 7.7
629
Plane Strain
Problem 7.7-10 Solve the preceding problem for the following data: Px 1120 * 106, Py 430 * 106,
gxy 780 * 106, and u 45°.
Solution 7.7-10
Element in plane strain
âx 1120 * 106
gxy 780 * 10
âx1 gx1y1
2
6
+
âx ây
2
â1 254 * 106
‹ up1 65.7°
up2 155.7°
âx ây
âx + ây
2
ây 430 * 106
2
sin 2u +
cos 2u +
gxy
2
gxy
2
â2 1296 * 10
sin 2u
cos 2u
ây1 âx + ây âx1
For u 45°:
âx1 385 * 106
gx1y1 690 * 106
ây1 1165 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
521 * 106
gmax 1041 * 106
us1 up1 45° 20.7°
gmax 1041 * 106
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
â1, 2 ;
a
b + a
b
2
A
2
2
775 * 10
6
; 521 * 10
us2 us1 + 90° 110.7°
gmin 1041 * 106
6
6
â2 1296 * 10
â1 254 * 10
gxy
1.1304
tan 2up âx ây
6
2up 131.5° and 311.5°
up 65.7° and 155.7°
For up 65.7°:
âx1 âx ây
âx + ây
2
+
254 * 106
2
cos 2u +
gxy
2
;
sin 2u
âaver âx + ây
2
;
775 * 106
6
;
;
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CHAPTER 7
Page 630
Analysis of Stress and Strain
Problem 7.7-11 A steel plate with modulus of elasticity E 30 * 106 psi and
y
Poisson’s ratio 0.30 is loaded in biaxial stress by normal stresses sx and sy
(see figure). A strain gage is bonded to the plate at an angle f 30°.
If the stress sx is 18,000 psi and the strain measured by the gage is P 407 * 106,
what is the maximum in-plane shear stress (tmax)xy and shear strain (gmax)xy? What is the
maximum shear strain (gmax)xz in the xz
plane? What is the maximum shear strain (gmax)yz in the yz plane?
sy
sx
f
x
z
Probs. 7.7-11 and 7.7-12
Solution 7.7-11
Steel plate in biaxial stress
sx 18,000 psi
gxy 0
E 30 * 106 psi
sy ?
MAXIMUM IN-PLANE SHEAR STRESS
0.30
â 407 * 10
Strain gage: f 30°
(tmax)xy 6
sx sy
2
7800 psi
;
UNITS: All stresses in psi.
STRAINS FROM EQS. (1), (2), AND (3)
STRAIN IN BIAXIAL STRESS (EQS. 7-39)
âx 576 * 106
âx 1
1
(s sy) (18,000 0.3sy)
E x
30 * 106
(1)
ây 1
1
(sy sx ) (sy 5400)
E
30 * 106
(2)
âz 0.3
(s sy) (18,000sy)
E x
30 * 106
âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
âz 204 * 106
MAXIMUM SHEAR STRAINS (EQ. 7-75)
xy plane:
(3)
2
(gmax)xz
2
yz plane:
(gmax) yz
2
gyz 0
(4)
A
a
âx ây
2
gxy
2
b + a
2
(gmax) xy 676 * 10
A
a
âx âz
2
2
b + a
b
2
6
gxz
2
;
b
2
(gmax) xz 780 * 106
gxz 0
sin 2u
1
1
407 * 106 a b a
b(12,600 + 0.7sy)
2 30 * 106
1
1
+ a ba
b (23,4001.3sy) cos 60°
2 30 * 106
Solve for sy :
sy 2400 psi
(gmax) xy
gxy 0
xz plane:
STRAINS AT ANGLE f 30° (EQ. 7-71a)
âx1 ây 100 * 106
A
a
ây 2
âz 2
b + a
;
gyz 2
2
b
(gmax) yz 104 * 106
;
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Page 631
SECTION 7.7
Plane Strain
631
Problem 7.7-12 Solve the preceding problem if the plate is made of aluminum with E 72 GPa and 1/3, the stress
sx is 86.4 MPa, the angle f is 21° , and the strain P is 946 * 106.
Solution 7.7-12
sx 86.4 MPa
Aluminum plate in biaxial stress
gxy 0
E 72 GPa
sy ?
MAXIMUM IN-PLANE SHEAR STRESS
1/3
â 946 * 10
Strain gage: f 21°
(tmax) xy 6
(1)
1
1
ây (sy sx) (sy 28.8)
E
72,000
(2)
1/ 3
(86.4 sy)
âz (sx sy) E
72,000
946 * 10
2
6
2
cos 2u +
gxy
2
(gmax) xy
2
gxy 0
xz plane:
(gmax)xz
2
sin 2u
yz plane:
(gmax) yz
2
gyz 0
(4)
A
a
âx ây
2
2
b + a
gxy
2
(gmax) xy 1200 * 10
A
a
âx âz
2
2
b + a
b
2
6
gxz
2
b
;
2
(gmax) xz 1600 * 106
gxz 0
1
1
4
+ a ba
b a115.2 sy b cos 42°
2 72,000
3
sy 21.55 MPa
xy plane:
(3)
1
1
2
a ba
b a57.6 + sy b
2 72,000
3
Solve for sy:
ây 101 * 106
MAXIMUM SHEAR STRAINS (EQ. 7-75)
STRAINS AT ANGLE f 21° (EQ. 7-71a)
+
;
âz 500 * 106
1
1
1
(86.4 sy)
âx (sx sy) E
72,000
3
âx1 32.4 MPa
âx 1100 * 106
STRAIN IN BIAXIAL STRESS (EQS. 7-39)
âx ây
2
STRAINS FROM EQS. (1), (2), AND (3)
UNITS: All stresses in MPa.
âx + ây
sx sy
A
a
ây 2
âz 2
b + a
;
gyz 2
2
b
(gmax) yz 399 * 106
;
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CHAPTER 7
Page 632
Analysis of Stress and Strain
sy
Problem 7.7-13 An element in plane stress is subjected to stresses sx 8400 psi,
sy 1100 psi, and txy 1700 psi (see figure). The material is aluminum with modulus
of elasticity E 10,000 ksi and Poisson’s ratio 0.33.
Determine the following quantities: (a) the strains for an element oriented at an angle
u 30°, (b) the principal strains, and (c) the maximum shear strains. Show the results on
sketches of properly oriented elements.
txy
y
O
x
Probs. 7.7-13 and 7.7-14
Solution 7.7-13
Element in plane strain
sx 8400 psi
sy 1100 psi
txy 1700 psi
E 10,000 ksi
PRINCIPAL STRAINS
0.33
HOOKE’S LAW (EQS. 7-34 AND 7-35)
âx 1
(s sy) 876.3 * 106
E x
ây 1
(s sx) 387.2 * 106
E y
gxy txy
2txy(1 + )
G
FOR u 30°:
âx1 E
+
2
2
cos 2u +
gxy
2
2
âx ây
2
sin 2u +
gxy
2
;
2
â1 426 * 10
tan 2up 756 * 106
gx1y1
âx + ây
A
245 * 10
452.2 * 106
âx ây
âx + ây
â1,2 âx ây
a
6
2
; 671 * 10
6
gxy
âx ây
2
b + a
gxy
2
434 * 106
gx1y1 868 * 106
ây1 âx + ây âx1 267 * 106
2
6
â2 961 * 106
0.3579
2up 19.7° and 199.7°
up 9.8° and 99.8°
FOR up 9.8°:
sin 2u
âx1 âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
916 * 106
cos 2u
b
‹ up1 99.8° â1 426 * 106
up2 9.8° â2 916 * 10
6
;
;
sin 2u
sx
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Page 633
SECTION 7.7
Plane Strain
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
671 * 106
gmax 1342 * 106
us1 up1 45° 54.8°
gmax 1342 * 106
;
us2 us1 + 90° 144.8°
gmin 1342 * 106
;
âx + ây
âaver 245 * 106
2
Problem 7.7-14 Solve the preceding problem for the following data: sx 150 MPa, sy 210 MPa,
txy 16 MPa, and u 50°. The material is brass with E 100 GPa and 0.34.
Solution 7.7-14
Element in plane strain
sx 150 MPa sy 210 MPa
txy 16 MPa
E 100 GPa 0.34
HOOKE’S LAW (EQS. 7-34 AND 7-35)
âx 1
(s sy) 786 * 106
E x
ây 1
(s sx) 1590 * 106
E y
gxy txy
G
2txy(1 + )
E
429 * 106
FOR u 50°:
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
1469 * 106
gx1y1
2
âx ây
2
sin 2u +
gxy
2
cos 2u
358.5 * 106
gx1y1 717 * 106
ây1 âx + ây âx1 907 * 106
sin 2u
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
;
b + a
b
â1,2 a
2
A
2
2
1188 * 106 ; 456 * 106
â1 732 * 106 â2 1644 * 106
gxy
0.5333
tan 2up âx ây
2up 151.9° and 331.9°
up 76.0° and 166.0°
633
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CHAPTER 7
Page 634
Analysis of Stress and Strain
FOR up 76.0°:
âx1 âx ây
âx + ây
2
MAXIMUM SHEAR STRAINS
+
1644 * 10
2
cos 2u +
gxy
2
sin 2u
456 * 106
6
‹ up1 166.0° â1 732 * 106
up2 76.0° â2 1644 * 106
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
;
;
gmax 911 * 106
us1 up1 45° 121.0°
gmax 911 * 106
;
us2 us1 90° 31.0°
gmin 911 * 106
âaver âx + ây
2
;
1190 * 106
Problem 7.7-15 During a test of an airplane wing, the strain gage readings from a 45° rosette
y
(see figure) are as follows: gage A, 520 * 106; gage B, 360 * 106; and gage C, 80 * 106 .
Determine the principal strains and maximum shear strains, and show them on sketches of
properly oriented elements.
45°
B
C
45°
A
O
Probs. 7.7-15 and 7.7-16
x
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Page 635
SECTION 7.7
Solution 7.7-15
âC 80 * 10
âB 360 * 106
6
âx âA 520 * 10
6
ây âC 80 * 10
gxy 2âB âA âC 280 * 10
6
;
6
;
MAXIMUM SHEAR STRAINS
6
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
331 * 106
PRINCIPAL STRAINS
âx + ây
;
‹ up1 12.5° â1 551 * 106
up2 102.5° â2 111 * 10
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:
2
A
a
âx â y
2
2
b + a
gxy
2
b
2
220 * 106 ; 331 * 106
â1 551 * 106
â2 111 * 106
gmax 662 * 106
us1 up1 45° 32.5°or 147.5°
gmax 662 * 106
gxy
âx ây
;
us2 us1 + 90° 57.5°
gmin 662 * 106
âaver tan 2up 635
45° strain rosette
âA 520 * 106
â1,2 Plane Strain
âx + ây
2
;
220 * 106
0.4667
2up 25.0° and 205.0°
up 12.5° and 102.5°
For up 12.5°:
âx1 âx ây
âx + ây
2
+
2
cos 2u +
gxy
2
sin 2u
551 * 106
Problem 7.7-16 A 45°strain rosette (see figure) mounted on the surface of an automobile frame gives the following
readings: gage A, 310 * 106; gage B, 180 * 106; and gage C, 160 * 106.
Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.
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CHAPTER 7
Page 636
Analysis of Stress and Strain
Solution 7.7-16
45° strain rosette
âA 310 * 106
âB 180 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
âC 160 * 106
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:
âx âA 310 * 10
6
ây âC 160 * 10
gxy 2âB âA âC 210 * 10
6
2
A
a
âx â y
2
2
b + a
gxy
2
b
2
â1 332 * 106
gxy
âx â y
âaver â2 182 * 106
0.4468
2up 24.1° and 204.1°
up 12.0° and 102.0°
FOR up 12.0°:
âx + ây
âx ây
gxy
âx1 +
cos 2u +
sin 2u
2
2
2
332 * 106
‹ up1 12.0° â1 332 * 106
up2 102.0° â2 182 * 10
;
6
;
;
us2 us1 + 90° 57.0°
gmin 515 * 106
75 * 106 ; 257 * 106
tan 2up gmax 515 * 106
gmax 515 * 106
âx + ây
;
257 * 106
us1 up1 45° 33.0° or 147.0°
PRINCIPAL STRAINS
â1,2 6
âx + ây
2
;
75 * 106
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Page 637
SECTION 7.7
Problem 7.7-17 A solid circular bar of diameter d 1.5 in. is
Plane Strain
d
subjected to an axial force P and a torque T (see figure). Strain gages
A and B mounted on the surface of the bar give reading
Pa 100 * 106 and Pb 55 * 106. The bar is made of steel
having E 30 * 106 psi and 0.29.
637
T
P
C
(a) Determine the axial force P and the torque T.
(b) Determine the maximum shear strain gmax and the maximum
shear stress tmax in the bar.
B
45∞
A
C
Solution 7.7-17
Circular bar (plane stress)
Bar is subjected to a torque T and an axial force P.
E 30 * 10 psi 0.29
STRAIN AT u 45°
6
âx1 Diameter d 1.5 in.
2
2
cos 2u +
âx 100 * 106
sy 0
txy Solve for T:
pd Eâx
5300 lb
4
SHEAR STRAIN
txy
2txy(1 + )
32T(1 + )
gxy G
E
pd 3E
(T lb-in.)
T 1390 lb-in
;
Eq. (7-75):
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
111 * 106 rad
2
(0.1298 * 106)T
2u 90°
gxy (0.1298 * 106)T 180.4 * 106 rad
pd 3
ây âx 29 * 106
P
sin 2u
MAXIMUM SHEAR STRAIN AND MAXIMUM SHEAR STRESS
16T
AXIAL FORCE P
sx
4P
E
pd 2E
2
55 * 106 35.5 * 106 (0.0649 * 106)T
ELEMENT IN PLANE STRESS
P
4P
sx A
pd 2
gxy
Substitute numerical values into Eq. (1):
âA âx 100 * 106
At u 45°: âB 55 * 106
âx +
âx1 âB 55 * 106
STRAIN GAGES
At u 0°:
âx ây
âx + ây
;
gmax 222 * 106 rad
tmax Ggmax 2580 psi
;
;
(1)
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CHAPTER 7
Page 638
Analysis of Stress and Strain
Problem 7.7-18 A cantilever beam of rectangular cross
h
b
section (width b 25 mm, height h 100 mm) is loaded
by a force P that acts at the midheight of the beam and is
inclined at an angle a to the vertical (see figure). Two strain
gages are placed at point C, which also is at the midheight
of the beam. Gage A measures the strain in the horizontal
direction and gage B measures the strain at an angle b = 60°
to the horizontal. The measured strains are Pa 125 * 106
and Pb 375 * 106.
Determine the force P and the angle a, assuming the
material is steel with E 200 GPa and 1/3.
h
C
a
B
P
b
b
A
C
Probs. 7.7-18 and 7.7-19
Solution 7.7-18
Cantilever beam (plane stress)
Beam loaded by a force P acting at an angle a.
1/3
E 200 GPa
b 25 mm
h 100 mm
Axial force F P sin a
Shear force V P cos a
(At the neutral axis, the bending moment produces no
stresses.)
At u 60°:
âx sx
P sin a
E
bhE
P sin a bhEâx 62,500 N
txy
3(1 + ) P cos a
3P cos a
gxy G
2bhG
bhE
(8.0 * 109) P cos a
(1)
(2)
FOR u 60°:
STRAIN GAGES
At u 0°:
HOOKE’S LAW
âA âx 125 * 10
âB 375 * 10
6
6
ELEMENT IN PLANE STRESS
F
P sin a
sx A
bh
sy 0
âx1 âx ây
âx + ây
2
+
2
âx1 âB 375 * 106
gxy
2
sin 2u
(3)
2u 120°
Substitute into Eq. (3):
375 * 106 41.67 * 106 41.67 * 106
(3.464 * 109)P cos a
3V
3P cos a
txy 2A
2bh
or P cos a 108,260 N
âx 125 * 106
SOLVE EQS. (1) AND (4):
ây âx 41.67 * 106
cos 2u +
a 30°
tan a 0.5773
P 125 kN
(4)
;
;
Problem 7.7-19 Solve the preceding problem if the cross-sectional dimensions are b 1.0 in. and h 3.0 in., the gage
angle is b = 75°, the measure strains are Pa 171 * 106 and Pb 266 * 106, and the material is a magnesium alloy
with modulus E 6.0 * 106 psi and Poisson’s ratio 0.35.
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Page 639
Plane Strain
639
P sin a bhEâx 3078 lb
txy
3(1 + v)P cos a
3P cos a
gxy G
2bhG
bhE
(1)
SECTION 7.7
Solution 7.7-19
Cantilever beam (plane stress)
Beam loaded by a force P acting at an angle a.
E 6.0 * 10 psi
6
0.35
b 1.0 in.
h 3.0 in.
Axial force F P sin a Shear foce V P cos a
(At the neutral axis, the bending moment produces
no stresses.)
HOOKE’S LAW
âx (225.0 * 109)P cos a
STRAIN GAGES
At u 0°:
At u 75°:
sx
P sin a
E
bhE
âA âx 171 * 106
âB 266 * 10
6
ELEMENT IN PLANE STRESS
F
P sin a
sx A
bh
(3)
2u 150°
Substitute into Eq. (3):
266 * 106 55.575 * 106 99.961 * 106
3V
3P cos a
txy 2A
2bh
âx 171 * 106
FOR u 75°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âx1 âB 266 * 106
sy 0
(2)
(56.25 * 109)P cos a
ây âx 59.85 * 106
or P cos a 3939.8 lb
(4)
SOLVE EQS. (1) AND (4):
tan a 0.7813
P 5000 lb
a 38°
;
;
y
Problem 7.7-20 A 60° strain rosette, or delta rosette, consists of three
electrical-resistance strain gages arranged as shown in the figure. Gage A measures
the normal strain Pa in the direction of the x axis. Gages B and C measure the
strains Pb and Pc in the inclined directions shown.
Obtain the equations for the strains Px, Py, and gxy associated with the xy axis.
B
60°
O
60°
A
C
60°
x
Solution 7.7-20 Delta rosette (60° strain rosette)
STRAIN GAGES
Gage A at u 0°
Gage B at u 60°
Strain âA
Strain âB
Gage C at u 120°
FOR u 0°:
âx âA
Strain âC
;
FOR u 60°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âA + ây
âA â y
gxy
+
(cos 120°) + 2 (sin 120°)
âB 2
2
3ây
gxy 13
âA
âB (1)
+
+
4
4
4
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CHAPTER 7
Page 640
Analysis of Stress and Strain
FOR u 120°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âA + ây
âA â y
gxy
+
(cos 240°)
(sin 240°)
âC 2
2
2
3ây
gxy 13
âA
âC (2)
+
4
4
4
SOLVE EQS. (1) AND (2):
ây 1
(2âB + 2âC âA)
3
gxy 2
(âB âC)
13
;
;
Problem 7.7-21 On the surface of a structural component in a space vehicle, the
y
strainsare monitored by means of three strain gages arranged as shown in the figure.
During a certain maneuver, the following strains were recorded: Pa 1100 * 106,
Pb 200 * 106, and Pc 200 * 106.
Determine the principal strains and principal stresses in the material, which is a
magnesium alloy for which E 6000 ksi and 0.35. (Show the principal strains
and principal stresses on sketches of properly oriented element.)
B
30°
O
Solution 7.7-21
C
x
A
30-60-90° strain rosette
Magnesium alloy: E 6000 ksi 0.35
STRAIN GAGES
Gage A at u 0°
PA 1100 * 106
Gage B at u 90°
âB 200 * 106
Gage C at u 150°
âC 200 * 10
6
FOR u 0°:
âx âA 1100 * 106
FOR u 90°:
ây âB 200 * 106
FOR u 150°:
âx + ây
âx ây
gxy
âx1 âC +
cos 2u +
sin 2u
2
2
2
200 * 106 650 * 106 + 225 * 106
0.43301gxy
Solve for gxy: gxy 1558.9 * 106
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
â1,2 ;
a
b + a
b
2
A
2
2
650 * 106 ; 900 * 106
â1 1550 * 106
â2 250 * 106
tan 2up gxy
13 1.7321
âx ây
2up 60°
up 30°
FOR up 30°:
âx + ây
âx ây
gxy
âx1 +
cos 2u +
sin 2u
2
2
2
1550 * 106
‹ up1 30°
up2 120°
â1 1550 * 106
â2 250 * 10
6
;
;
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Page 641
SECTION 7.7
Plane Strain
641
PRINCIPAL STRESSES (see Eqs. 7-36)
s1 E
1
2
(â1 + â2)
s2 E
1 2
(â2 + â1)
Substitute numerical values:
s1 10,000 psi
s2 2,000 psi
;
y
Problem 7.7-22 The strains on the surface of an experimental device made of pure
aluminum (E 70 Gpa, 0.33) and tested in a space shuttle were measured by
means of strain gages. The gages were oriented as shown in the figure, and the measured
strains were Pa 1100 * 106, Pb 1496 * 106, and Pc 39.44 * 10 * 6.
What is the stress sx in the x direction?
B
O
Solution 7.7-22
0.33
40°
x
FOR u 140°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
STRAIN GAGES
âA 1100 * 106
Gage A at u 0°
Gage B at u 40°
âB 1496 * 10
Gage C at u 140°
Substitute âx1 âc 39.44 * 106 and
6
âC 39.44 * 10
âx 1100 * 106; then simplify and rearrange:
6
0.41318ây 0.49240gxy 684.95 * 106
âx âA 1100 * 106
SOLVE EQS. (1) AND (2):
ây 200.3 * 106
FOR u 40°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
sx 6
; then simplify and rearrange:
0.41318ây + 0.49240gxy 850.49 * 106
gxy 1559.2 * 106
HOOKE’S LAW
Substitute âx1 âB 1496 * 106 and
âx 1100 * 10
A
40-40-100° strain rosette
Pure aluminum: E 70 GPa
FOR u 0°:
40°
C
(1)
E
1 2
(âx + ây) 91.6 MPa
;
(2)
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CHAPTER 7
Page 642
Analysis of Stress and Strain
Problem 7.7-23 Solve Problem 7.7-5 by using Mohr’s circle for plane strain.
Solution 7.7-23
Element in plane strain
âx 220 * 106
gxy 180 * 106
R 2(130 * 10
158.11 * 10
a arctan
ây 480 * 106
gxy
90 * 106 u 50°
2
6 2
) + (90 * 10
6
90
34.70°
130
b 180° a 2u 45.30°
6 2
)
POINT C: âx1 350 * 106
POINT D (u 50° ):
âx1 350 * 106 + R cos b 461 * 106
gx1y1
R sin b 112.4 * 106
2
gx1y1 225 * 106
POINT D ¿ (u 140°):
âx1 350 * 106 R cos b 239 * 106
gx1y1
R sin b 112.4 * 106
2
gx1y1 225 * 106
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Page 643
SECTION 7.7
Plane Strain
Problem 7.7-24 Solve Problem 7.7-6 by using Mohr’s circle for plane strain.
Solution 7.7-24
Element in plane strain
âx 420 * 106
gxy 310 * 106
ây 170 * 106
gxy
155 * 106
u 37.5°
2
POINT C: âx1 125 * 106
POINT D (u 37.5°):
âx1 125 * 106 + R cos b 351 * 106
gx1y1
R sin b 244.8 * 106
2
gx1y1 490 * 106
POINT D œ (u 127.5°):
âx1 125 * 106 R cos b 101 * 106
gx1y1
R sin b 244.8 * 106
2
gx1y1 490 * 106
R 2(295 * 106)2 + (155 * 106)2
333.24 * 106
a arctan
155
27.72°
295
b 2u a 47.28°
643
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CHAPTER 7
Page 644
Analysis of Stress and Strain
Problem 7.7-25 Solve Problem 7.7-7 by using Mohr’s circle for plane strain.
Solution 7.7-25
Element in plane strain
âx 480 * 106
ây 140 * 106
gxy
gxy 350 * 106
175 * 106
2
MAXIMUM SHEAR STRAINS
2us2 90°a 44.17°
us2 22.1°
2us1 2us2 + 180° 224.17°
Point S1: âaver 310 * 10
6
gmax 2R 488 * 106
Point S2: âaver 310 * 106
gmin 488 * 106
R 2(175 * 10
243.98 * 10
6 2
) + (170 * 10
6 2
)
6
a arctan
175
45.83°
170
POINT C:
âx1 310 * 106
PRINCIPAL STRAINS
2up2 180° a 134.2°
up2 67.1°
2up1 2up2 + 180° 314.2° up1 157.1°
Point P1: â1 310 * 106 + R 554 * 106
Point P2: â2 310 * 106 R 66 * 106
us1 112.1°
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Page 645
SECTION 7.7
Plane Strain
Problem 7.7-26 Solve Problem 7.7-8 by using Mohr’s circle for plane strain.
Solution 7.7-26
Element in plane strain
âx 120 * 106
ây 450 * 106
gxy
gxy 360 * 106
180 * 106
2
MAXIMUM SHEAR STRAINS
2us2 90° a 57.72°
us2 28.9°
2us1 2us2 + 180° 237.72°
Point S1: âaver 165 * 10
6
gmax 2R 674 * 106
Point S2: âaver 165 * 106
R 2(285 * 10
337.08 * 10
a arctan
6 2
) + (180 * 10
6 2
)
6
180
32.28°
285
Point C: âx1 165 * 106
PRINCIPAL STRAINS
2up2 180° a 147.72°
up2 73.9°
2up1 2up2 + 180° 327.72°
Point P1: â1 R 165 * 10
6
up1 163.9°
172 * 106
Point P2: â2 165 * 106 R 502 * 106
gmin 674 * 106
us1 118.9°
645
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CHAPTER 7
Page 646
Analysis of Stress and Strain
Problem 7.7-27 Solve Problem 7.7-9 by using Mohr’s circle for plane strain.
Solution 7.7-27
âx 480 * 10
Element in plane strain
6
gxy 420 * 106
ây 70 * 106
gxy
210 * 106
2
u 75°
PRINCIPAL STRAINS
2up1 a 45.69°
up1 22.8°
2up2 2up1 + 180° 225.69°
Point P1: â1 275 * 10
6
up2 112.8°
+ R 568 * 106
Point P2: â2 275 * 106 R 18 * 106
R 2(205 * 106)2 + (210 * 106)2
293.47 * 106
a arctan
210
45.69°
205
b a + 180° 2u 75.69°
Point C: âx1 275 * 106
Point D (u 75°):
âx1 275 * 106 R cos b 202 * 106
gx1y1
R sin b 284.36 * 106
2
gx1y1 569 * 106
Point
Dœ
(u 165°):
âx1 275 * 106 + R cos b 348 * 106
gx1y1
R sin b 284.36 * 106
2
gx1y1 569 * 106
MAXIMUM SHEAR STRAINS
2us2 90° + a 135.69°
2us1 2us2 + 180° 315.69°
Point S1: âaver 275 * 10
6
gmax 2R 587 * 106
Point S2: âaver 275 * 106
gmin 587 * 106
us2 67.8°
us1 157.8°
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Page 647
SECTION 7.7
Plane Strain
647
Problem 7.7-28 Solve Problem 7.7-10 by using Mohr’s circle for plane strain.
Solution 7.7-28
Element in plane strain
âx 1120 * 106
gxy 780 * 106
ây 430 * 106
gxy
2
390 * 106
u 45°
PRINCIPAL STRAINS
2up1 180° a 131.50°
up1 65.7°
2up2 2up1 + 180° 311.50°
Point P1: â1 775 * 10
6
up2 155.7°
+ R 254 * 106
Point P2: â2 775 * 106 R 1296 * 106
R 2(345 * 106)2 + (390 * 106)2
520.70 * 106
a arctan
390
48.50°
345
b 180° a 2u 41.50°
Point C: âx1 775 * 106
MAXIMUM SHEAR STRAINS
Point D: (u 45°):
2us2 2us1 + 180° 221.50°
âx1 775 * 106 + R cos b 385 * 106
gx1y1
R sin b 345 * 106 gx1y1 690 * 106
2
Point D ¿ : (u 135°)
âx1 775 * 106 R cos b 1165 * 106
gx1y1
R sin b 345 * 106
2
gx1y1 690 * 106
2us1 90° a 41.50°
Point S1: âaver 775 * 10
us1 20.7°
6
gmax 2R 1041 * 106
Point S2: âaver 775 * 106
gmin 1041 * 106
us2 110.7°
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Page 648