Chapter 1
1. Consider an aluminum single crystal under a stress state,
x
= 250 psi,
y
= - 50 psi,
z
=
yz
=
zx
=
xy
= 0, where x = [100], y = [010] and z = [001].
A. What is the resolved shear stress,
nd
, on the (111) plane in the

11 0
direction? i.e. with n =
[111], d =

11 0
B. What is the resolved shear stress on the

111
plane in the [101] direction?
Solution A.
nd
=
x

xn

xd
+
y

yn

yd
= 250(1/√3)(1/√2) + (-50)(1/√3)(-1/√2) = 122.5 psi
B.
nd
=
x

xn

xd
+
y

yn

yd
= 250(1/√3)(1/√2) + (-50)(1/√3)(0/√2) = 102. psi
2. Consider the single crystal in problem 1. Now suppose that slip does occur on the (111) plane in
the

101
direction and only on that slip system. Also assume that the resulting strains are small.
Calculate the ratios of the resulting strains
x
/
x
and
z
/
x
.

x
=
nd
(

xn

xd
) =
nd
(1/√6).
x
=
nd
(

yn

yd
) = -
nd
(1/√6),
z
=
nd
(

zn

zd
) = -
nd
(0). Therefore
z
/
x
= -1 and
y
/
x
= 0.
3. A body is loaded under a stress state,
x
= 400,
y
= 100,
xy
= 120,
yz
=
zx
=
z
=0.
A. Sketch the Mohr's circle diagram.
B. Calculate the principal stresses
C. What is the largest shear stress in the body? (Do not neglect the z direction.)
Solution: A.
B.
1
,
2
= (
x
+
y
)/2± {[(
x
-
y
)/2]
2
+
xy
2
}
1/2
= 250 ± [150
2
+ 120
2
]
1/2
= 442, 58,
3
=
z
= 0.
C.
max
= (
1
-
3
)/2 = 221.
4. Three stain gauges have been pasted on the surface of a piece of steel in the pattern shown
below. While the steel is under load, these gauges indicate the strains parallel to their axes:
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Gauge A 450x10
-6
:
Gauge B 300x10
-6
: Gauge C - 150x10
-6
A. Calculate the principal strains
1
and
2
.
B. Find the angle between the 1 axis and the x-axis, where 1 is the axis of the largest principal
strain? [Hint: Let the direction of gauge B be x', write the strain transformation equation expressing
the strain e
x'
in terms of the strains along the x-y axes, solve for
xy
and finally use the Mohr's circle
equations.]
Solution: A.
B
=
A
cos
2
40 +
c
cos
2
50 +
AB
cos40cos50 =
300 = 450x0.587 –150x0.413 +
AB
0.766x0.623
AB
= 198,
1
,
2
= (
A
+
C
)/2± {[(
A
-
C
)/2]
2
+ (
xy
/2)
2
)}
1/2
= 150±[300
2
+198
2
]
1/2
= 509, -209 x10
-6
B. tan(2) =
AB
/(e
A
-
B
) = 198/600 =0.33, = atan(0.33)/2 = 9.1°
5. Consider a thin-wall tube that is 1 inch in diameter and has a 0.010-inch wall thickness. Let x, y,
and z be the axial, tangential (hoop) and radial directions respectively.
A. The tube is subjected to an axial tensile force of 80 lbs. and a torque of 100 in-lbs.
i. Sketch the Mohr's circle diagram showing stresses in the x-y plane.
ii. What is the magnitude of the largest principal stress?
iii. At what angles are the principal stress axes, 1 and 2, to the x and y directions?
B. Now let the tube be capped and let it be subject to an internal pressure of 120 psi and a torque of
100 in-lbs.
i. Sketch the Mohr's circle diagram showing stresses in the x-y plane.
ii. What is the magnitude of the largest principal stress?
iii. At what angles are the principal stress axes, 1 and 2, to the x and y directions?
Solution: A. A torque balance gives 100 in-lb =
xy
(πx1in)(0.010in)(0.5in),
xy
= 100/(0.010πx0.5) =
6366 psi. The axial stress,
x
= 80/[(πx1in)(0.010in)] = 2546 psi
i.
Figure 1.21. Arrangement of strain gauges.
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x
1
2
y
= 0
xy
xy
ii.
1
= (
x
+
y
)/2 ± {[(
x
-
y
)/2]
2
+
xy
2
}
1/2
= 2546/2 + [1273
2
+ 6366
2
]
1/2
= 7765 psi
iii. tan(2) = 6366/(2546/2) = 5.0, = (0.5)atan(5.0) = 39.3°, 129°
 Now there is a hoop stress, 
y
=Pd/(2t) = 120x1/(2x.01) = 6000 psi, an axial stress of
x
=Pd/(4t) = 3000 psi and a shear stress of 6366 psi.
i.
ii
1
= (
x
+
y
)/2+ {[(
x
-
y
)/2]
2
+
xy
2
}
1/2
= 4500 + [1500
2
+ 6366
2
]
1/2
= 11,040 psi
iii tan(2) = 6366/1500= 4.244, = (.5)atan(4.244) = 38.4°, 128.4°
6. A solid is deformed under plane-strain conditions (
z
= 0). The strains in the x-y plane are
x
=
0.010,
y
= 0.005, and
xy
= 0.007.
A. Sketch the Mohr's circle diagram.
B. Find the magnitude of
1
and
2
.
C. What is the angle between the 1 and x axes?
C. What is the largest shear strain in the body? (Do not neglect the z direction.)
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Solution: A
x
y
1
z
2

B.
1
,
2
= (+ )/2± {[(
- )/2]
2
+(/2)
2
)}
1/2
= 0.0118, 0.00320
C. tan(2/) =
xy
/(
x
-
y
) = 0.007/0.005 =1.4, = (1/2)atan(0.28) = 27.2°
D
max
= (
1
-
3
) = 
7. A grid of circles, each 10.00 mm diameter, was etched on the surface of a sheet of steel. When
the sheet was deformed the grid circles were distorted into ellipses. Measurement of one indicated
that the major and minor diameters were 11.54 and 10.70 mm respectively
A. What are the principal strains,
1
and
2
?
B. If the axis of the major diameter of the of the ellipse makes an angle of 34° to the x-direction,
what is the shear strain,
xy
?
C. Draw the Mohr's strain circle showing
1
,
2
,
x
, and
y
.
y
x
y
x
34°
Solution:
A.
1
= ln(1.154) = 0.143 
2
= ln(1.070) = 0.0676 ( or e
1
= 0.154, e
2
= .07)
B. tan(2) =
xy
/(
1
-
2
),
xy = (0.143-.067)tan(68°) = 0.188
C.
Figure 20. Circle grids printed on a metal
Sheet.
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8. Consider an aluminum single crystal under a stress state,
x
= +75 psi,
y
= +25 psi
z
=
yz
=
zx
=
xy
= 0, where x = [100], y = [010], and z = [001].
What is the resolved shear stress,
nd
, on the (111) plane and [

101
] direction?
Solution:
nd
=
x

xn
ı

xd
+
y

yn

yd
= 75(1/√6) +25(0) = 30.6 psi.
9. Consider the torsion of a rod that is 1 meter long and 50 mm in diameter.
A. If one end of the rod is twisted by 1.2° relative to the other end, what would be the largest
principal strain on the surface?
B. If the rod were extended by 1.2% and its diameter decreased by 0.4% at the same time it was
being twisted, what would be the largest principal strain?
Solution; A.
xy
= tan(1.2°) = 0.0209,
1
=
xy
/2 = .0105
B.
1
= (
x
+
y
)/2+ {[(
x
-
y
)/2]
2
+(
xy
/2)
2
]}
1/2
= (.012 - .004)/2+{[(.012 + .004)/2]
2
+.0209
2
}
1/2
= .0253
10. Two pieces of rod are glued together along a joint whose normal makes an angle, with the rod
axis, x. The joint will fail if the shear stress on the joint exceeds its shear strength,
max
. It will also
fail if the normal stress across the joint exceeds its normal strength,
max
. The shear strength,
max
, is 80% of the normal strength
max
. The rod will be loaded in uniaxial tension along its axis,
and it is desired that the rod carry as high a tensile force, F
x
as possible. The angle, , cannot exceed
65°.
A. At what angle, , should the joint be made so that a maximum force can be carried?
B. If
max
were limited to 45°, instead of 65°, how would your answer be altered? [Hint: plot
x
/
max
vs. for both failure modes.
Solution:
Let the axial stress be ,
the max normal stress be
n(max)
the max shear stress be
(max)
.
(max)
= 0.8
n(max)
n
= cos
2
. Normal failure will occur when =
n(max)/
cos
2
.
= cossin. Shear failure will occur when =
(max)
/ cossin = 0.8
n(max)/
cos
2
.
Now plotting 
n(max)
vs. , it can be seen
Figure 1.23. Glued
rod.
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A. that if ≤ 65°, the max value of
max
= 2.089
n(max)
at
= 65°
B. hat if ≤ 45°, the max value of
max
corresponds to the intersection of the two curves and
max
= 1.64
n(max)
at
= 38.65°
11. Consider a tube made by coiling and gluing a strip as show below. The diameter is 1.5 in., the
length is 6 in. and the wall thickness is 0.030 in. If a tensile force of 80 lbs. and a torque of 30 in-
lbs. are applied in the direction shown, what is the stress normal to the glued joint? [Hint: Set up a
coordinate system]
Solution: The torque, 30 in-lb = π(1.5 in)(.030in)(0.75in)
xy
.
xy
= 30/[π(1.5 in)(.030in)(0.75in)] = 283 psi ,
x
= 80lbs/[π(1.5 in)(.030 in) = 566
n
=
x
cos
2
22 -
xy
cos22cos68 = 486. Note that for the sketch shown, the torque
contributes a compressive stress on the joint.
Figure 22. Tube formed from a
coiled and glued strip.
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