Strain Transformation
10
CHAPTER OBJECTIVES
The transformation of strain at a point is similar to the transformation
of stress, and as a result the methods of Chapter 9 will be applied
in this chapter. Here we will also discuss various ways for measuring
strain and develop some important material-property relationships,
including a generalized form of Hooke’s law. At the end of the chapter,
a few of the theories used to predict the failure of a material will be
discussed.
10.1 Plane Strain
As outlined in Sec. 2.2, the general state of strain at a point in a body is
represented by a combination of three components of normal strain, Px ,
Py , Pz , and three components of shear strain gxy , gxz , gyz . These six
components tend to deform each face of an element of the material, and
like stress, the normal and shear strain components at the point will vary
according to the orientation of the element. The strains at a point are
often determined by using strain gauges, which measure normal strain in
specified directions. For both analysis and design, however, engineers
must sometimes transform this data in order to obtain the strain in other
directions.
485
486
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
To understand how this is done, we will first confine our attention to a
study of plane strain. Specifically, we will not consider the effects of the
components Pz , gxz , and gyz . In general, then, a plane-strained element
is subjected to two components of normal strain, Px , Py , and one
component of shear strain, gxy . Although plane strain and plane stress
each have three components lying in the same plane, realize that plane
stress does not necessarily cause plane strain or vice versa. The reason for
this has to do with the Poisson effect discussed in Sec. 3.6. For example, if
the element in Fig. 10–1 is subjected to plane stress sx and sy , not only
are normal strains Px and Py produced, but there is also an associated
normal strain, Pz . This is obviously not a case of plane strain. In general,
then, unless n = 0, the Poisson effect will prevent the simultaneous
occurrence of plane strain and plane stress.
z
Py dy
Px dx
Pz dz
sx
x
sy
y
Plane stress, sx , sy, does not cause plane
strain in the x–y plane since Pz ≠ 0.
10.2 General Equations of Plane-Strain
Fig. 10–1
Transformation
y
Py dy
A
dy
It is important in plane-strain analysis to establish transformation
equations that can be used to determine the x¿, y¿ components of normal
and shear strain at a point, provided the x, y components of strain are
known. Essentially this problem is one of geometry and requires relating
the deformations and rotations of line segments, which represent the
sides of differential elements that are parallel to each set of axes.
gxy
2
gxy
2
B
x
O
dx
y¿
(a)
Px dx
y
x¿
u
x
10
(b)
Positive sign convention
Fig. 10–2
Sign Convention. Before the strain-transformation equations can
be developed, we must first establish a sign convention for the strains.
With reference to the differential element shown in Fig. 10–2a, normal
strains Px and Py are positive if they cause elongation along the x and y
axes, respectively, and the shear strain gxy is positive if the interior angle
AOB becomes smaller than 90°. This sign convention also follows the
corresponding one used for plane stress, Fig. 9–5a, that is, positive sx , sy ,
txy will cause the element to deform in the positive Px , Py , gxy directions,
respectively.
The problem here will be to determine at a point the normal and shear
strains Px¿ , Py¿ , gx¿y¿ , measured relative to the x¿, y¿ axes, if we know Px ,
Py , gxy , measured relative to the x, y axes. If the angle between the x and
x¿ axes is u, then, like the case of plane stress, u will be positive provided
it follows the curl of the right-hand fingers, i.e., counterclockwise, as
shown in Fig. 10–2b.
10.2
487
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
Normal and Shear Strains. In order to develop the strain-
y
y¿
transformation equation for Px¿ , we must determine the elongation of
a line segment dx¿ that lies along the x¿ axis and is subjected to strain
components Px , Py , gxy . As shown in Fig. 10–3a, the components of the
line dx¿ along the x and y axes are
dx = dx¿ cos u
dy = dx¿ sin u
x¿
dy
u
dx ¿
x
dx
(10–1)
Before deformation
(a)
When the positive normal strain Px occurs, the line dx is elongated
Px dx, Fig. 10–3b, which causes line dx¿ to elongate Px dx cos u. Likewise,
when Py occurs, line dy elongates Py dy, Fig. 10–3c, which causes line dx¿
to elongate Py dy sin u. Finally, assuming that dx remains fixed in
position, the shear strain gxy , which is the change in angle between dx
and dy, causes the top of line dy to be displaced gxy dy to the right, as
shown in Fig. 10–3d. This causes dx¿ to elongate gxy dy cos u. If all three
of these elongations are added together, the resultant elongation of dx¿
is then
y
y¿
x¿
Px dx cos u
dx¿
dx
Normal strain Px
u
x
Px dx
Px dx sinu
(b)
y
dx¿ = Px dx cos u + Py dy sin u + gxy dy cos u
From Eq. 2–2, the normal strain along the line dx¿ is Px¿ = dx¿>dx¿.
Using Eq. 10–1, we therefore have
Px¿ = Px cos2 u + Py sin2 u + gxy sin u cos u
y¿
u
Py dy
(10–2)
Py dy cos u
x¿
u
Py dy sin u
dy
dx¿
x
Normal strain Py
(c)
y
y¿
gxy dy sinu g dy
xy
x¿
gxy dy cos u
u
dy¿
dy
10
gxy
dx ¿
dx
The rubber specimen is constrained between
the two fixed supports, and so it will undergo
plane strain when loads are applied to it in the
horizontal plane.
Shear strain gxy
(d)
Fig. 10–3
x
488
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
y
y¿
y
gxy dy sinu g dy
xy
x¿
gxy dy cos u
y¿
u
dy¿
dy
gxy
dy ¿
dx ¿
dx
b
x
dy ¿
dx ¿
dy ¿
x¿
a
dx ¿
u
Shear strain gxy
x
(e)
(d)
Fig. 10–3 (cont.)
The strain-transformation equation for gx¿y¿ can be developed by
considering the amount of rotation each of the line segments dx¿ and dy¿
undergo when subjected to the strain components Px , Py , gxy . First
we will consider the rotation of dx¿, which is defined by the
counterclockwise angle a shown in Fig. 10–3e. It can be determined by
the displacement caused by dy¿ using a = dy¿>dx¿. To obtain dy¿,
consider the following three displacement components acting in the y¿
direction: one from Px , giving -Px dx sin u, Fig. 10–3b; another from Py ,
giving Py dy cos u, Fig. 10–3c; and the last from gxy , giving - gxy dy sin u,
Fig. 10–3d. Thus, dy¿, as caused by all three strain components, is
dy¿ = -Px dx sin u + Py dy cos u - gxy dy sin u
Dividing each term by dx¿ and using Eq. 10–1, with a = dy¿>dx¿, we have
a = 1-Px + Py2 sin u cos u - gxy sin2 u
(10–3)
As shown in Fig. 10–3e, the line dy¿ rotates by an amount b. We can
determine this angle by a similar analysis, or by simply substituting
u + 90° for u into Eq. 10–3. Using the identities sin1u + 90°2 = cos u,
cos1u + 90°2 = - sin u, we have
b = 1 -Px + Py2 sin1u + 90°2 cos1u + 90°2 - gxy sin21u + 90°2
10
= - 1-Px + Py2 cos u sin u - gxy cos2 u
Since a and b represent the rotation of the sides dx¿ and dy¿ of a
differential element whose sides were originally oriented along the x¿
and y¿ axes, Fig. 10–3e, the element is then subjected to a shear strain of
gx¿y¿ = a - b = - 21Px - Py2 sin u cos u + gxy1cos2 u - sin2 u2
(10–4)
10.2
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
489
y
y
y¿
y¿
x¿
dy ¿
x¿
u
u
dy¿
dx¿
dx ¿
x
Positive normal strain, Px ¿
Positive shear strain, gx ¿y¿
(a)
(b)
x
Fig. 10–4
Using the trigonometric identities sin 2u = 2 sin u cos u, cos2 u =
(1 + cos 2u)>2, and sin2 u + cos2 u = 1, we can rewrite Eqs. 10–2 and
10–4 in the final form
Px¿ =
Px + Py
2
gx¿y¿
2
= -¢
+
Px - Py
2
Px - Py
2
cos 2u +
≤ sin 2u +
gxy
2
gxy
2
sin 2u
cos 2u
(10–5)
(10–6)
These strain-transformation equations give the normal strain Px¿ in the
x¿ direction and the shear strain gx¿y¿ of an element oriented at an angle
u, as shown in Fig. 10–4. According to the established sign convention,
if Px¿ is positive, the element elongates in the positive x¿ direction,
Fig. 10–4a, and if gx¿y¿ is positive, the element deforms as shown in
Fig. 10–4b.
If the normal strain in the y¿ direction is required, it can be obtained
from Eq. 10–5 by simply substituting 1u + 90°2 for u. The result is
10
Py¿ =
Px + Py
2
-
Px - Py
2
cos 2u -
gxy
2
sin 2u
(10–7)
The similarity between the above three equations and those for planestress transformation, Eqs. 9–1, 9–2, and 9–3, should be noted. By
comparison, sx , sy , sx¿ , sy¿ correspond to Px , Py , Px¿ , Py¿ ; and txy , tx¿y¿
correspond to gxy>2, gx¿y¿>2.
490
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
Principal Strains. Like stress, an element can be oriented at a point
so that the element’s deformation is caused only by normal strains, with
no shear strain. When this occurs the normal strains are referred to as
principal strains, and if the material is isotropic, the axes along which
these strains occur will coincide with the axes that define the planes of
principal stress.
From Eqs. 9–4 and 9–5, and the correspondence between stress and
strain mentioned above, the direction of the x¿ axis and the two values of
the principal strains P1 and P2 are determined from
gxy
tan 2up =
P1,2 =
Complex stresses are often developed
at the joints where the cylindrical
and hemispherical vessels are joined
together. The stresses are determined by
making measurements of strain.
Px + Py
2
;
C¢
(10–8)
Px - Py
Px - Py
2
2
≤ + ¢
gxy
2
2
≤
(10–9)
Maximum In-Plane Shear Strain. Using Eqs. 9–6, 9–7, and 9–8,
the direction of the x¿ axis, and the maximum in-plane shear strain and
associated average normal strain are determined from the following
equations:
tan 2us = - ¢
max
g in-plane
2
=
B
a
Px - Py
gxy
Px - Py
Pavg =
2
2
≤
b + a
Px + Py
2
(10–10)
gxy
2
b
2
(10–11)
(10–12)
Important Points
• In the case of plane stress, plane-strain analysis may be used within the plane of the stresses to analyze the
10
data from strain gauges. Remember, though, there will be a normal strain that is perpendicular to the
gauges due to the Poisson effect.
• When the state of strain is represented by the principal strains, no shear strain will act on the element.
• The state of strain at a point can also be represented in terms of the maximum in-plane shear strain. In this
case an average normal strain will also act on the element.
• The element representing the maximum in-plane shear strain and its associated average normal strains is
45° from the orientation of an element representing the principal strains.
10.2
491
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
EXAMPLE 10.1
A differential element of material at a point is subjected to a state
of plane strain Px = 500110-62, Py = - 300110-62, gxy = 200110-62,
which tends to distort the element as shown in Fig. 10–5a. Determine
the equivalent strains acting on an element of the material oriented at
the point, clockwise 30° from the original position.
SOLUTION
The strain-transformation Eqs. 10–5 and 10–6 will be used to solve the
problem. Since u is positive counterclockwise, then for this problem
u = - 30°. Thus,
Px¿ =
Px + Py
2
= c
2
2
500 + 1-3002
+ B
gx¿y¿
Px - Py
+
2
200110 2
-6
2
= -¢
= -c
cos 2u +
d110-62 + c
gxy
2
y
gxy
2
Py dy
dy
gxy
2
dx
sin 2u
500 - 1-3002
2
x
Px dx
(a)
y
y¿
d110-62 cos121-30°22
R sin121-30°22
Px¿ = 213110-62
Px - Py
2
≤ sin 2u +
500 - 1 -3002
2
gxy
2
Ans.
u 60
cos 2u
x
d110 2 sin121-30°22 +
-6
gx¿y¿ = 793110-62
200110-62
2
u 30
cos121 -30°22
(b)
Ans.
x¿
y¿
The strain in the y¿ direction can be obtained from Eq. 10–7 with
u = - 30°. However, we can also obtain Py¿ using Eq. 10–5 with
u = 60°1u = - 30° + 90°2, Fig. 10–5b. We have with Py¿ replacing Px¿ ,
Py¿ =
Px + Py
2
= c
+
+
Px - Py
2
500 + 1-3002
2
200110-62
2
cos 2u +
d110-62 + c
gxy
2
sin 2u
500 - 1- 3002
2
gx ¿y¿
2
dy¿
Py¿dy¿
d110-62 cos12160°22
10
dx ¿
sin12160°22
Py¿ = - 13.4110-62
These results tend to distort the element as shown in Fig. 10–5c.
Px¿dx ¿
Ans.
(c)
Fig. 10–5
gx ¿y¿
2
x¿
492
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.2
A differential element of material at a point is subjected to a state
of plane strain defined by Px = - 350110-62, Py = 200110-62, gxy =
80110-62, which tends to distort the element as shown in Fig. 10–6a.
Determine the principal strains at the point and the associated
orientation of the element.
y
gxy
2
SOLUTION
Orientation of the Element. From Eq. 10–8 we have
Py dy
tan 2up =
dy
Px - Py
80110-62
gxy
2
x
Px dx
dx
=
1-350 - 2002110-62
Thus, 2up = - 8.28° and -8.28° + 180° = 171.72°, so that
up = - 4.14° and 85.9°
(a)
y
gxy
Ans.
Each of these angles is measured positive counterclockwise, from the x
axis to the outward normals on each face of the element, Fig. 10–6b.
y¿
Principal Strains. The principal strains are determined from Eq. 10–9.
We have
P1,2 =
P1dy¿
2
;
85.9
4.14
(b)
Fig. 10–6
Px - Py
2
1 -350 + 2002110 2
x
2
2
b + a
; B
gxy
2
b
2
-350 - 200 2
80 2
b + a b R 110-62
B
2
2
= - 75.0110-62 ; 277.9110-62
P1 = 203110-62
x¿
P2dx ¿
B
a
-6
=
10
Px + Py
a
P2 = - 353110-62
Ans.
We can determine which of these two strains deforms the element in
the x¿ direction by applying Eq. 10–5 with u = - 4.14°. Thus,
Px¿ =
Px + Py
2
= a
+
+
Px - Py
2
cos 2u +
gxy
2
sin 2u
-350 + 200
-350 - 200
b110-62 + a
b110-62 cos 21 -4.14°2
2
2
80110-62
2
sin 21 - 4.14°2
Px¿ = - 353110-62
Hence Px¿ = P2 . When subjected to the principal strains, the element
is distorted as shown in Fig. 10–6b.
10.2
493
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
EXAMPLE 10.3
A differential element of material at a point is subjected to a state
of plane strain defined by Px = - 350110-62, Py = 200110-62,
gxy = 80110-62, which tends to distort the element as shown in
Fig. 10–7a. Determine the maximum in-plane shear strain at the point
and the associated orientation of the element.
SOLUTION
Orientation of the Element. From Eq. 10–10 we have
tan 2us = - ¢
Px - Py
gxy
≤ = -
1-350 - 2002110-62
y
Py dy
80110 2
-6
gxy
2g
A
xy
dy
2
Thus, 2us = 81.72° and 81.72° + 180° = 261.72°, so that
B
us = 40.9° and 131°
x
O
dx
Note that this orientation is 45° from that shown in Fig. 10–6b in
Example 10.2 as expected.
Px dx
(a)
y
Maximum In-Plane Shear Strain. Applying Eq. 10–11 gives
max
g in-plane
2
=
B
= B
a
Px - Py
2
2
b + a
gxy
2
b
2
y¿
-350 - 200 2
80 2
b + a b R 110-62
B
2
2
(gxy)max
2
(gxy)max
2
a
max
g in-plane
= 556110-62
Pavgdy¿
Ans.
dy ¿
dx ¿
Due to the square root, the proper sign of g max
can be obtained by
in-plane
applying Eq. 10–6 with us = 40.9°. We have
gx¿y¿
2
= -
Px - Py
= -a
2
x¿
sin 2u +
gxy
2
cos 2u
Pavgdx ¿
40.9
x
(b)
Fig. 10–7
80110-62
-350 - 200
b110-62 sin 2140.9°2 +
cos 2140.9°2
2
2
gx¿y¿ = 556110-62
This result is positive and so g max
tends to distort the element so
in-plane
that the right angle between dx¿ and dy¿ is decreased (positive sign
convention), Fig. 10–7b.
Also, there are associated average normal strains imposed on the
element that are determined from Eq. 10–12:
Px + Py
- 350 + 200
Pavg =
=
110-62 = - 75110-62
2
2
These strains tend to cause the element to contract, Fig. 10–7b.
10
494
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
*10.3 Mohr’s Circle—Plane Strain
Since the equations of plane-strain transformation are mathematically
similar to the equations of plane-stress transformation, we can also solve
problems involving the transformation of strain using Mohr’s circle.
Like the case for stress, the parameter u in Eqs. 10–5 and 10–6 can be
eliminated and the result rewritten in the form
1Px¿ - Pavg22 + ¢
gx¿y¿
2
2
≤ = R2
(10–13)
where
Pavg =
R =
Px + Py
2
B
a
Px - Py
2
2
b + a
gxy
2
b
2
Equation 10–13 represents the equation of Mohr’s circle for strain. It has
a center on the P axis at point C1Pavg , 02 and a radius R.
Procedure for Analysis
The procedure for drawing Mohr’s circle for strain follows the same
one established for stress.
Construction of the Circle.
Px Py
2
C
gxy
Pavg 10
R
• Establish a coordinate system such that the abscissa represents
2
Px Py
2
Px
g
2
P
Px Py
2
2
Fig. 10–8
A
u 0
gxy
2
2
•
the normal strain P, with positive to the right, and the ordinate
represents half the value of the shear strain, g>2, with positive
downward, Fig. 10–8.
Using the positive sign convention for Px , Py , gxy , as shown in
Fig. 10–2, determine the center of the circle C, which is located
on the P axis at a distance Pavg = 1Px + Py2>2 from the origin,
Fig. 10–8.
• Plot the reference point A having coordinates A1Px , gxy>22. This
•
point represents the case for which the x¿ axis coincides with the
x axis. Hence u = 0°, Fig. 10–8.
Connect point A with the center C of the circle and from the
shaded triangle determine the radius R of the circle, Fig. 10–8.
• Once R has been determined, sketch the circle.
10.3
P1
Principal Strains.
•
• When P1 and P2 are indicated as being positive as in Fig. 10–9a,
F
P2
• The principal strains P1 and P2 are determined from the circle
as the coordinates of points B and D, that is where g>2 = 0,
Fig. 10–9a.
The orientation of the plane on which P1 acts can be determined
from the circle by calculating 2up1 using trigonometry. Here this
angle happens to be counterclockwise from the radial reference
line CA to line CB, Fig. 10–9a. Remember that the rotation of up1
must be in this same direction, from the element’s reference axis x
to the x¿ axis, Fig. 10–9b.*
495
MOHR’S CIRCLE—PLANE STRAIN
Q
D
2 up1
C
2 us1
A
g
2
P
gxy
2u P
E
Pavg
B
2
u 0
(a)
y¿
the element in Fig. 10–9b will elongate in the x¿ and y¿ directions
as shown by the dashed outline.
y
Maximum In-Plane Shear Strain.
• The average normal strain and half the maximum in-plane shear
strain are determined from the circle as the coordinates of point
E or F, Fig. 10–9a.
(1 P2)dy ¿
x¿
up1
• The orientation of the plane on which g max
and Pavg act can be
in-plane
determined from the circle by calculating 2us1 using trigonometry.
Here this angle happens to be clockwise from the radial
reference line CA to line CE, Fig. 10–9a. Remember that the
rotation of us1 must be in this same direction, from the element’s
reference axis x to the x¿ axis, Fig. 10–9c.*
x
(1 P1)dx ¿
(b)
y
y¿
Pavg dy ¿
Strains on Arbitrary Plane.
• The normal and shear strain components Px¿ and gx¿y¿ for a plane
oriented at an angle u, Fig. 10–9d, can be obtained from the circle
using trigonometry to determine the coordinates of point P,
Fig. 10–9a.
x¿
• To locate P, the known angle u of the x¿ axis is measured on the
circle as 2u. This measurement is made from the radial reference
line CA to the radial line CP. Remember that measurements
for 2u on the circle must be in the same direction as u for the
x¿ axis.*
x
us1
Pavg dx ¿
(c)
Py¿dy¿
y¿ y
• If the value of Py¿ is required, it can be determined by
10
calculating the P coordinate of point Q in Fig. 10–9a. The line
CQ lies 180° away from CP and thus represents a 90° rotation
of the x¿ axis.
x¿
u
Px ¿dx ¿
*If the g>2 axis were constructed positive upwards, then the angle 2u on the circle
would be measured in the opposite direction to the orientation u of the plane.
(d)
Fig. 10–9
x
496
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.4
The state of plane strain at a point is represented by the components
Px = 250110-62, Py = - 150110-62, and gxy = 120110-62. Determine
the principal strains and the orientation of the element.
SOLUTION
Construction of the Circle. The P and g>2 axes are established in
Fig. 10–10a. Remember that the positive g>2 axis must be directed
downward so that counterclockwise rotations of the element
correspond to counterclockwise rotation around the circle, and vice
versa. The center of the circle C is located on the P axis at
D(P2, 0)
B(P1, 0)
2up
R 1
20
8.8
C
50
250
g
(10 6 )
2
60
A
P (10 6 )
Pavg =
250 + 1-1502
2
110-62 = 50110-62
Since gxy>2 = 60110-62, the reference point A 1u = 0°2 has coordinates
A1250110-62, 60110-622. From the shaded triangle in Fig. 10–10a, the
radius of the circle is CA; that is,
R = C 21250 - 5022 + 16022 D 110-62 = 208.8110-62
(a)
Principal Strains. The P coordinates of points B and D represent
the principal strains. They are
y¿
P1 = 150 + 208.82110-62 = 259110-62
y
P2 = 150 - 208.82110-62 = - 159110-62
Ans.
The direction of the positive principal strain P1 is defined by the
counterclockwise angle 2up1 , measured from the radial reference line
CA 1u = 0°2 to the line CB. We have
P2dy¿
10
Ans.
dy ¿
x¿
up1 8.35
x
dx¿
(b)
Fig. 10–10
P1dx¿
tan 2up1 =
60
1250 - 502
up1 = 8.35°
Ans.
Hence, the side dx¿ of the element is oriented counterclockwise 8.35°
as shown in Fig. 10–10b. This also defines the direction of P1 . The
deformation of the element is also shown in the figure.
10.3
497
MOHR’S CIRCLE—PLANE STRAIN
EXAMPLE 10.5
The state of plane strain at a point is represented by the components
Px = 250110-62, Py = - 150110-62, and gxy = 120110-62. Determine
the maximum in-plane shear strains and the orientation of an element.
SOLUTION
The circle has been established in the previous example and is shown
in Fig. 10–11a.
Maximum In-Plane Shear Strain. Half the maximum in-plane shear
strain and average normal strain are represented by the coordinates of
point E or F on the circle. From the coordinates of point E,
max
1gx¿y¿2in-plane
2
= 208.8110-62
F
R
C
50
max
= 418110 2
1gx¿y¿2in-plane
-6
2 us1
Ans.
A
60
u 0
Pavg = 50110-62
E Pavg,
To orient the element, we can determine the clockwise angle 2us1
measured from CA 1u = 0°2 to CE.
2us1 = 90° - 218.35°2
us1 = 36.7°
P (10 6 )
208
.8
Ans.
gmax
in–plane
2
250
g
(10 6 )
2
(a)
Fig. 10–11
This angle is shown in Fig. 10–11b. Since the shear strain defined from
point E on the circle has a positive value and the average normal
strain is also positive, these strains deform the element into the
dashed shape shown in the figure.
y
y¿
10
x
us1 36.7
(b)
x¿
498
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.6
The state of plane strain at a point is represented on an element having
components Px = - 300110-62, Py = - 100110-62, and gxy = 100110-62.
Determine the state of strain on an element oriented 20° clockwise from
this reported position.
SOLUTION
Construction of the Circle. The P and g>2 axes are established in
Fig. 10–12a. The center of the circle is on the P axis at
Px ¿
gx¿y¿
2
P
50
c 13.43
C 13.43
40f
8
11.
Q gx¿y¿
1
R
A
2
Py ¿
200
300
P (10
6
)
-300 - 100
b110-62 = - 200110-62
2
The reference point A has coordinates A1- 300110-62, 50110-622. The
radius CA determined from the shaded triangle is therefore
R = C 21300 - 20022 + 15022 D 110-62 = 111.8110-62
g
(10 6 )
2
(a)
Pavg = a
Strains on Inclined Element. Since the element is to be oriented
20° clockwise, we must establish a radial line CP, 2120°2 = 40°
clockwise, measured from CA 1u = 0°2, Fig. 10–12a. The coordinates
of point P 1Px¿ , gx¿y¿>22 are obtained from the geometry of the circle.
Note that
f = tan-1 a
50
b = 26.57°,
1300 - 2002
c = 40° - 26.57° = 13.43°
Thus,
y
Px¿ = - 1200 + 111.8 cos 13.43°2110-62
y¿
= - 309110-62
gx¿y¿
2
= - 1111.8 sin 13.43°2110-62
gx¿y¿ = - 52.0110-62
x
10
20
x¿
(b)
Fig. 10–12
Ans.
Ans.
The normal strain Py¿ can be determined from the P coordinate of
point Q on the circle, Fig. 10–12a. Why?
Py¿ = - 1200 - 111.8 cos 13.43°2110-62 = - 91.3110-62 Ans.
As a result of these strains, the element deforms relative to the x¿, y¿
axes as shown in Fig. 10–12b.
10.3
MOHR’S CIRCLE—PLANE STRAIN
499
PROBLEMS
10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
10–2. The state of strain at the point has components
of Px = 200 110-62, Py = - 300 110-62, and gxy = 400(10-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 30° counterclockwise from the original position.
Sketch the deformed element due to these strains within the
x–y plane.
10–5. The state of strain at the point on the arm
has components Px = 250110-62, Py = -450110-62, gxy =
- 825110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
y
y
x
x
Prob. 10–2
10–3. A strain gauge is mounted on the 1-in.-diameter
A-36 steel shaft in the manner shown. When the shaft is
rotating with an angular velocity of v = 1760 rev>min, the
reading on the strain gauge is P = 800110-62. Determine
the power output of the motor. Assume the shaft is only
subjected to a torque.
60
Prob. 10–5
10–6. The state of strain at the point has components of
Px = - 100110-62, Py = 400110-62, and gxy = - 300110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 60° counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x–y plane.
10–7. The state of strain at the point has components of
Px = 100110-62, Py = 300110-62, and gxy = - 150110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented u = 30°
clockwise. Sketch the deformed element due to these
strains within the x–y plane.
10
Prob. 10–3
*10–4. The state of strain at a point on a wrench
has components Px = 120110-62, Py = - 180110-62, gxy =
150110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
y
x
Probs. 10–6/7
500
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
*10–8. The state of strain at the point on the bracket
has components Px = - 200110-62, Py = - 650110-62, gxy - 175110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 20° counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x–y plane.
10–10. The state of strain at the point on the bracket
has components Px = 400110-62, Py = - 250110-62, gxy 310110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 30° clockwise from the original
position. Sketch the deformed element due to these strains
within the x–y plane.
y
y
x
x
Prob. 10–8
Prob. 10–10
10–9. The state of strain at the point has components of
Px = 180110-62, Py = - 120110-62, and gxy = - 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
10–11. The state of strain at the point has components of
Px = - 100110-62, Py = - 200110-62, and gxy = 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
y
x
y
10
x
Prob. 10–9
Prob. 10–11
10.3
*10–12. The state of plane strain on an element is given by
Px = 500110-62, Py = 300110-62, and gxy = - 200110-62.
Determine the equivalent state of strain on an element at
the same point oriented 45° clockwise with respect to the
original element.
MOHR’S CIRCLE—PLANE STRAIN
501
10–14. The state of strain at the point on a boom of an
hydraulic engine crane has components of Px = 250110-62,
Py = 300110-62, and gxy = - 180110-62. Use the straintransformation equations to determine (a) the in-plane
principal strains and (b) the maximum in-plane shear strain
and average normal strain. In each case, specify the
orientation of the element and show how the strains deform
the element within the x–y plane.
y
y
x
Pydy
dy gxy
2
gxy
2
dx
x
Pxdx
Prob. 10–12
10–13. The state of plane strain on an element is
Px = - 300110-62, Py = 0, and gxy = 150110-62. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states
of strain with respect to the original element.
Prob. 10–14
■10–15.
Consider the general case of plane strain where
Px , Py , and gxy are known. Write a computer program that
can be used to determine the normal and shear strain, Px¿
and gx¿y¿ , on the plane of an element oriented u from the
horizontal. Also, include the principal strains and the
element’s orientation, and the maximum in-plane shear
strain, the average normal strain, and the element’s
orientation.
*10–16. The state of strain at a point on a support
has components of Px = 350110-62, Py = 400110-62,
gxy = - 675110-62. Use the strain-transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane. 10
y
gxy
dy 2
•10–17.
x
gxy
2
dx
Prob. 10–13
Pxdx
Solve part (a) of Prob. 10–4 using Mohr’s circle.
10–18.
Solve part (b) of Prob. 10–4 using Mohr’s circle.
10–19.
Solve Prob. 10–8 using Mohr’s circle.
*10–20.
Solve Prob. 10–10 using Mohr’s circle.
•10–21.
Solve Prob. 10–14 using Mohr’s circle.
502
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
z
x
*10.4 Absolute Maximum Shear Strain
y
(1 P1)dx
(1 P2)dy
In Sec. 9.5 it was pointed out that in the case of plane stress, the absolute
maximum shear stress in an element of material will occur out of the plane
when the principal stresses have the same sign, i.e., both are tensile or both
are compressive. A similar result occurs for plane strain. For example, if
the principal in-plane strains cause elongations, Fig. 10–13a, then the three
Mohr’s circles describing the normal and shear strain components for
elements oriented about the x¿, y¿, and z¿ axes are shown in Fig. 10–13b.
By inspection, the largest circle has a radius R = 1gx¿z¿2max>2. Hence,
xy plane strain
(a)
g abs
= 1gx¿z¿2max = P1
max
P2
P
P1
(gy z )max
2
P1 and P2 have the same sign
This value gives the absolute maximum shear strain for the material.
Note that it is larger than the maximum in-plane shear strain, which is
1gx¿y¿2max = P1 - P2 .
Now consider the case where one of the in-plane principal strains is
of opposite sign to the other in-plane principal strain, so that P1 causes
elongation and P2 causes contraction, Fig. 10–14a. Mohr’s circles, which
describe the strains on each element’s orientation about the x¿, y¿, z¿
axes, are shown in Fig. 10–14b. Here
(gx y)max
2
(gx z )max
2
g
2
(b)
Fig. 10–13
z
= 1gx¿y¿2in-plane
= P1 - P2
g abs
max
max
P1 and P2 have opposite signs
x
(10–14)
(1 P1)dx
(1 – P2)dy
xy plane strain
y
(10–15)
We may therefore summarize the above two cases as follows. If the
in-plane principal strains both have the same sign, the absolute maximum
shear strain will occur out of plane and has a value of g abs
= Pmax .
max
However, if the in-plane principal strains are of opposite signs, then the
absolute maximum shear strain equals the maximum in-plane shear strain.
(a)
Important Points
10
P2
P1
(gyz)max
2
(gxz)max
2
(gxy)max
g
2
2
(b)
Fig. 10–14
P
• The absolute maximum shear strain will be larger than the
maximum in-plane shear strain whenever the in-plane principal
strains have the same sign. When this occurs the absolute
maximum shear strain will act out of the plane.
• If the in-plane principal strains are of opposite signs, then
the absolute maximum shear strain will equal the maximum
in-plane shear strain.
10.4
ABSOLUTE MAXIMUM SHEAR STRAIN
503
EXAMPLE 10.7
The state of plane strain at a point is represented by the strain
components Px = - 400110-62, Py = 200110-62, gxy = 150110-62.
Determine the maximum in-plane shear strain and the absolute
maximum shear strain.
75 P2
A
P1
9
30
R
P(106)
100
(gx¿y¿)max
in–plane
2
400
g
(106)
2
Fig. 10–15
SOLUTION
Maximum In-Plane Shear Strain. We will solve this problem using
Mohr’s circle. From the strain components, the center of the circle is
on the P axis at
-400 + 200
110-62 = - 100110-62
Pavg =
2
Since gxy>2 = 75110-62, the reference point A has coordinates
1 -400110-62, 75110-622. As shown in Fig. 10–15, the radius of the
circle is therefore
R = C 21400 - 10022 + 17522 D 110-62 = 309110-62
Calculating the in-plane principal strains from the circle, we have
P1 = 1-100 + 3092110-62 = 209110-62
P2 = 1-100 - 3092110-62 = - 409110-62
Also, the maximum in-plane shear strain is
g max
= P1 - P2 = [209 - 1 -4092]110-62 = 618110-62 Ans.
in-plane
Absolute Maximum Shear Strain. From the above results, we
have P1 = 209110-62, P2 = - 409110-62. The three Mohr’s circles,
plotted for element orientations about each of the x, y, z axes, are also
shown in Fig. 10–15. It is seen that since the principal in-plane strains
have opposite signs, the maximum in-plane shear strain is also the
absolute maximum shear strain; i.e.,
g abs
= 618110-62
max
Ans.
10
504
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
10.5 Strain Rosettes
b
When performing a tension test on a specimen as discussed in Sec. 3.1,
the normal strain in the material is measured using an electricalresistance strain gauge, which consists of a wire grid or piece of metal
foil bonded to the specimen. For a general loading on a body, however,
the strains at a point on its free surface are determined using a cluster of
three electrical-resistance strain gauges, arranged in a specified pattern.
This pattern is referred to as a strain rosette, and once the normal strains
on the three gauges are measured, the data can then be transformed to
specify the state of strain at the point. Since these strains are measured
only in the plane of the gauges, and since the body is stress-free on its
surface, the gauges may be subjected to plane stress but not plane strain.
Although the strain normal to the surface is not measured, realize that
the out-of-plane displacement caused by this strain will not affect the
in-plane measurements of the gauges.
In the general case, the axes of the three gauges are arranged at the
angles ua, ub, uc shown in Fig. 10–16a. If the readings Pa, Pb, Pc are taken, we
can determine the strain components Px, Py, gxy at the point by applying
the strain-transformation equation, Eq. 10–2, for each gauge. We have
a
ub
ua
uc
x
c
(a)
c
b
45
45
x
a
45 strain rosette
(b)
Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua
Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub
c
(10–16)
Pc = Px cos uc + Py sin uc + gxy sin uc cos uc
2
b
60
60
x
a
60 strain rosette
(c)
Fig. 10–16
2
The values of Px, Py, gxy are determined by solving these three equations
simultaneously.
Strain rosettes are often arranged in 45° or 60° patterns. In the case of
the 45° or “rectangular” strain rosette shown in Fig. 10–16b, ua = 0°,
ub = 45°, uc = 90°, so that Eq. 10–16 gives
Px = Pa
Py = Pc
gxy = 2Pb - 1Pa + Pc2
And for the 60° strain rosette in Fig. 10–16c, ua = 0°, ub = 60°,
uc = 120°. Here Eq. 10–16 gives
Px = Pa
10
1
12Pb + 2Pc - Pa2
3
2
=
1Pb - Pc2
13
Py =
gxy
Typical electrical resistance 45° strain rosette.
(10–17)
Once Px, Py, gxy are determined, the transformation equations of
Sec. 10.2 or Mohr’s circle can then be used to determine the principal
in-plane strains and the maximum in-plane shear strain at the point.
10.5
505
STRAIN ROSETTES
EXAMPLE 10.8
The state of strain at point A on the bracket in Fig. 10–17a is measured
using the strain rosette shown in Fig. 10–17b. Due to the loadings, the
readings from the gauges give Pa = 60110-62, Pb = 135110-62, and
Pc = 264110-62. Determine the in-plane principal strains at the point
and the directions in which they act.
c
A
SOLUTION
We will use Eqs. 10–16 for the solution. Establishing an x axis as
shown in Fig. 10–17b and measuring the angles counterclockwise
from the +x axis to the centerlines of each gauge, we have ua = 0°,
ub = 60°, and uc = 120°. Substituting these results, along with the
problem data, into the equations gives
60110 2 = Px cos 0° + Py sin 0° + gxy sin 0° cos 0°
= Px
135110-62 = Px cos2 60° + Py sin2 60° + gxy sin 60° cos 60°
-6
2
(a)
c
2
= 0.25Px + 0.75Py + 0.433gxy
b
(1)
120
60
(2)
x
264110 2 = Px cos 120° + Py sin 120° + gxy sin 120° cos 120°
(3)
= 0.25Px + 0.75Py - 0.433gxy
-6
b
a
2
a
2
(b)
Using Eq. 1 and solving Eqs. 2 and 3 simultaneously, we get
Px = 60110-62
Py = 246110-62
gxy = - 149110-62
60
These same results can also be obtained in a more direct manner from
Eq. 10–17.
The in-plane principal strains can be determined using Mohr’s
circle. The reference point on the circle is at A [60110-62, -74.5110-62]
and the center of the circle, C, is on the P axis at Pavg = 153110-62,
Fig. 10–17c. From the shaded triangle, the radius is
R = C 21153 - 602 + 174.52 D 110 2 = 119.1110 2
2
2
-6
-6
A R
P2
11
9
2up2 .2
74.5
C
P1
153
P(106)
g
(106)
2
(c)
The in-plane principal strains are thus
P1 = 153110-62 + 119.1110-62 = 272110-62
P2 = 153110-62 - 119.1110-62 = 33.9110-62
74.5
2up2 = tan-1
= 38.7°
1153 - 602
up2 = 19.3°
Ans.
y¿
Ans.
x¿
Ans.
NOTE: The deformed element is shown in the dashed position in
Fig. 10–17d. Realize that, due to the Poisson effect, the element is also
subjected to an out-of-plane strain, i.e., in the z direction, although this
value will not influence the calculated results.
up2 19.3
x
(d)
Fig. 10–17
10
506
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
PROBLEMS
10–22. The strain at point A on the bracket has
components Px = 300110-62, Py = 550110-62, gxy =
-650110-62. Determine (a) the principal strains at A in the
x –y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
*10–24. The strain at point A on the pressure-vessel wall
has components Px = 480110-62, Py = 720110-62, gxy =
650110-62. Determine (a) the principal strains at A, in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
y
A
x
y
A
x
Prob. 10–22
Prob. 10–24
•10–25.
10–23. The strain at point A on the leg of the angle has
components Px = - 140110-62, Py = 180110-62, gxy =
-125110-62. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
The 60° strain rosette is mounted on the bracket.
The following readings are obtained for each gauge:
Pa = - 100110-62, Pb = 250110-62, and Pc = 150110-62.
Determine (a) the principal strains and (b) the maximum
in-plane shear strain and associated average normal strain.
In each case show the deformed element due to these
strains.
b
10
c
A
60
60
Prob. 10–23
a
Prob. 10–25
10.5
10–26. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these
strains.
507
STRAIN ROSETTES
*10–28. The 45° strain rosette is mounted on the link of
the backhoe. The following readings are obtained from
each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain.
a
45
45
b
b
c
a
30 30
c
60
Prob. 10–28
Prob. 10–26
10–27. The 45° strain rosette is mounted on a steel shaft.
The following readings are obtained from each gauge:
Pa = 300110-62, Pb = - 250110-62, and Pc = - 450110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these
strains.
10–29. Consider the general orientation of three strain
gauges at a point as shown. Write a computer program that
can be used to determine the principal in-plane strains and
the maximum in-plane shear strain at the point. Show an
application of the program using the values ua = 40°,
Pa = 160110-62, ub = 125°, Pb = 100110-62, uc = 220°,
Pc = 80110-62.
b
ub
uc
a
ua
x
b
c
45
45
a
c
Prob. 10–27
Prob. 10–29
10
508
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
10.6 Material-Property Relationships
In this section we will present some important relationships involving a
material’s properties that are used when the material is subjected to
multiaxial stress and strain. To do so we will assume that the material is
homogeneous and isotropic and behaves in a linear-elastic manner.
Generalized Hooke’s Law. If the material at a point is subjected
to a state of triaxial stress, sx, sy, sz, Fig. 10–18a, associated normal
strains Px, Py, Pz will be developed in the material. The stresses can be
related to these strains by using the principle of superposition, Poisson’s
ratio, Plat = - nPlong, and Hooke’s law, as it applies in the uniaxial
direction, P = s>E. For example, consider the normal strain of the
element in the x direction, caused by separate application of each normal
stress. When sx is applied, Fig. 10–18b, the element elongates in the x
direction and the strain Pxœ is
Pxœ =
sx
E
Application of sy causes the element to contract with a strain Pxfl,
Fig. 10–18c. Here
Pxfl = - n
sy
E
Likewise, application of sz, Fig. 10–18d, causes a contraction such that
PxÔ = - n
sz
E
sz
sz
10
ⴝ
ⴙ
ⴙ
sy
sy
sx
sx
(a)
(b)
(c)
Fig. 10–18
(d)
10.6
MATERIAL-PROPERTY RELATIONSHIPS
509
When these three normal strains are superimposed, the normal strain
Px is determined for the state of stress in Fig. 10–18a. Similar equations
can be developed for the normal strains in the y and z directions. The
final results can be written as
Px =
1
[s - n1sy + sz2]
E x
Py =
1
[s - n1sx + sz2]
E y
Pz =
1
[s - n1sx + sy2]
E z
(10–18)
These three equations express Hooke’s law in a general form for a
triaxial state of stress. For application tensile stresses are considered
positive quantities, and compressive stresses are negative. If a resulting
normal strain is positive, it indicates that the material elongates, whereas
a negative normal strain indicates the material contracts.
If we now apply a shear stress txy to the element, Fig. 10–19a,
experimental observations indicate that the material will deform only
due to a shear strain gxy; that is, txy will not cause other strains in the
material. Likewise, tyz and txz will only cause shear strains gyz and gxz,
Figs. 10–19b and 10–19c, and so Hooke’s law for shear stress and shear
strain can be written as
gxy =
1
t
G xy
gyz =
1
t
G yz
gxz =
1
t
G xz
(10–19)
10
tyz
txy
tzx
(a)
(b)
Fig. 10–19
(c)
510
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
Relationship Involving E, N, and G. In Sec. 3.7 it was stated that
the modulus of elasticity E is related to the shear modulus G by Eq. 3–11,
namely,
y
txy
G =
x
(a)
x¿
y
smin txy
smax txy
up1 45
x
(b)
Fig. 10–20
dx
sz
txy
(10–21)
11 + n2
E
This strain, which deforms the element along the x¿ axis, can also
be related to the shear strain gxy. To do this, first note that
since sx = sy = sz = 0, then from the first and second Eqs. 10–18,
Px = Py = 0. Substituting these results into the strain transformation
Eq. 10–9, we get
gxy
P1 = Pmax =
2
Pmax =
dV = 11 + Px211 + Py211 + Pz2 dx dy dz - dx dy dz
(1 Pz)dz
10
sy
(1 Px)dx
(1 Py)dy
One way to derive this relationship is to consider an element of the
material to be subjected to pure shear 1sx = sy = sz = 02, Fig. 10–20a.
Applying Eq. 9–5 to obtain the principal stresses yields smax = txy and
smin = - txy. This element must be oriented up1 = 45° counterclockwise
from the x axis as shown in Fig. 10–20b. If the three principal stresses
smax = txy, sint = 0, and smin = - txy are substituted into the first of
Eqs. 10–18, the principal strain Pmax can be related to the shear stress txy.
The result is
Dilatation and Bulk Modulus. When an elastic material is
subjected to normal stress, its volume will change. For example, consider
a volume element which is subjected to the principal stresses sx, sy, sz.
If the sides of the element are originally dx, dy, dz, Fig. 10–21a, then
after application of the stress they become 11 + Px2 dx, 11 + Py2 dy,
11 + Pz2 dz, Fig. 10–21b. The change in volume of the element is
therefore
(a)
sx
(10–20)
By Hooke’s law, gxy = txy>G, so that Pmax = txy>2G. Substituting into
Eq. 10–21 and rearranging terms gives the final result, namely, Eq. 10–20.
dz
dy
E
211 + n2
(b)
Fig. 10–21
Neglecting the products of the strains since the strains are very small, we
have
dV = 1Px + Py + Pz2 dx dy dz
The change in volume per unit volume is called the “volumetric strain”
or the dilatation e. It can be written as
e =
dV
= Px + Py + Pz
dV
(10–22)
By comparison, the shear strains will not change the volume of the
element, rather they will only change its rectangular shape.
10.6
MATERIAL-PROPERTY RELATIONSHIPS
Also, if we use Hooke’s law, as defined by Eq. 10–18, we can write the
dilatation in terms of the applied stress. We have
1 - 2n
1sx + sy + sz2
e =
(10–23)
E
When a volume element of material is subjected to the uniform
pressure p of a liquid, the pressure on the body is the same in all
directions and is always normal to any surface on which it acts. Shear
stresses are not present, since the shear resistance of a liquid is zero. This
state of “hydrostatic” loading requires the normal stresses to be equal in
any and all directions, and therefore an element of the body is subjected
to principal stresses sx = sy = sz = - p, Fig. 10–22. Substituting into
Eq. 10–23 and rearranging terms yields
p
E
= (10–24)
e
311 - 2n2
511
sz p
sy p
sx p
Hydrostatic stress
Fig. 10–22
Since this ratio is similar to the ratio of linear elastic stress to strain,
which defines E, i.e., s>P = E, the term on the right is called the volume
modulus of elasticity or the bulk modulus. It has the same units as stress
and will be symbolized by the letter k; that is,
k =
E
311 - 2n2
(10–25)
Note that for most metals n L 13 so k L E. If a material existed that did
not change its volume then dV = e = 0, and k would have to be infinite.
From Eq. 10–25 the theoretical maximum value for Poisson’s ratio is
therefore n = 0.5. During yielding, no actual volume change of the
material is observed, and so n = 0.5 is used when plastic yielding occurs.
Important Points
• When a homogeneous isotropic material is subjected to a state of
triaxial stress, the strain in each direction is influenced by the
strains produced by all the stresses.This is the result of the Poisson
effect, and results in the form of a generalized Hooke’s law.
• Unlike normal stress, a shear stress applied to homogeneous
isotropic material will only produce shear strain in the same
plane.
• The material constants E, G, and n are related mathematically.
• Dilatation, or volumetric strain, is caused only by normal strain,
•
not shear strain.
The bulk modulus is a measure of the stiffness of a volume of
material. This material property provides an upper limit to
Poisson’s ratio of n = 0.5, which remains at this value while
plastic yielding occurs.
10
512
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.9
The bracket in Example 10–8, Fig. 10–23a, is made of steel for which
Est = 200 GPa, nst = 0.3. Determine the principal stresses at point A.
c
A
b
a
(a)
Fig. 10–23
SOLUTION I
From Example 10.8 the principal strains have been determined as
P1 = 272110-62
P2 = 33.9110-62
Since point A is on the surface of the bracket for which there is no
loading, the stress on the surface is zero, and so point A is subjected to
plane stress. Applying Hooke’s law with s3 = 0, we have
P1 =
s1
n
- s2 ;
E
E
272110-62 =
s1
200110 2
9
-
0.3
s2
20011092
54.411062 = s1 - 0.3s2
P2 =
10
(1)
s2
s2
n
0.3
- s1 ; 33.9110-62 =
s1
E
E
20011092
20011092
6.7811062 = s2 - 0.3s1
(2)
Solving Eqs. 1 and 2 simultaneously yields
s1 = 62.0 MPa
Ans.
s2 = 25.4 MPa
Ans.
10.6
MATERIAL-PROPERTY RELATIONSHIPS
513
29.4
A R
11.46
18
s2
.3
C
s1
s (MPa)
43.7
t (MPa)
Fig. 10–23 (cont.)
(b)
SOLUTION II
It is also possible to solve the problem using the given state of strain,
Px = 60110-62
Py = 246110-62
gxy = - 149110-62
as specified in Example 10.8. Applying Hooke’s law in the x–y plane,
we have
0.3sy
sx
sx
n
Px =
- sy ;
60110-62 =
9
E
E
200110 2 Pa
20011092 Pa
sy
sy
0.3sx
n
Py =
- sx ; 246110-62 =
9
E
E
200110 2 Pa
20011092 Pa
sx = 29.4 MPa
sy = 58.0 MPa
The shear stress is determined using Hooke’s law for shear. First,
however, we must calculate G.
G =
Thus,
txy = Ggxy ;
E
200 GPa
=
= 76.9 GPa
211 + n2
211 + 0.32
txy = 76.911092[-149110-62] = - 11.46 MPa
The Mohr’s circle for this state of plane stress has a reference point
A129.4 MPa, -11.46 MPa2 and center at savg = 43.7 MPa, Fig. 10–23b.
The radius is determined from the shaded triangle.
R = 2143.7 - 29.422 + 111.4622 = 18.3 MPa
Therefore,
10
s1 = 43.7 MPa + 18.3 MPa = 62.0 MPa
Ans.
s2 = 43.7 MPa - 18.3 MPa = 25.4 MPa
Ans.
NOTE: Each of these solutions is valid provided the material is both
linear elastic and isotropic, since then the principal planes of stress
and strain coincide.
514
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.10
The copper bar in Fig. 10–24 is subjected to a uniform loading along
its edges as shown. If it has a length a = 300 mm, width b = 50 mm,
and thickness t = 20 mm before the load is applied, determine its
new length, width, and thickness after application of the load.
Take Ecu = 120 GPa, ncu = 0.34.
t
500 MPa
800 MPa
a
b
800 MPa
500 MPa
Fig. 10–24
SOLUTION
By inspection, the bar is subjected to a state of plane stress. From the
loading we have
sx = 800 MPa
sy = - 500 MPa
txy = 0
sz = 0
The associated normal strains are determined from the generalized
Hooke’s law, Eq. 10–18; that is,
Px =
sx
n
- 1sy + sz2
E
E
800 MPa
0.34
1-500 MPa + 02 = 0.00808
12011032 MPa
12011032 MPa
sy
n
- 1sx + sz2
Py =
E
E
0.34
- 500 MPa
=
1800 MPa + 02 = - 0.00643
3
120110 2 MPa
12011032 MPa
sz
n
- 1sx + sy2
Pz =
E
E
0.34
= 0 1800 MPa - 500 MPa2 = - 0.000850
12011032 MPa
The new bar length, width, and thickness are therefore
=
10
a¿ = 300 mm + 0.008081300 mm2 = 302.4 mm
b¿ = 50 mm + 1- 0.006432150 mm2 = 49.68 mm
t¿ = 20 mm + 1-0.0008502120 mm2 = 19.98 mm
Ans.
Ans.
Ans.
10.6
MATERIAL-PROPERTY RELATIONSHIPS
515
EXAMPLE 10.11
If the rectangular block shown in Fig. 10–25 is subjected to a uniform
pressure of p = 20 psi, determine the dilatation and the change in
length of each side. Take E = 600 psi, n = 0.45.
c 3 in.
a 4 in.
b 2 in.
Fig. 10–25
SOLUTION
Dilatation. The dilatation can be determined using Eq. 10–23 with
sx = sy = sz = - 20 psi. We have
e =
=
1 - 2n
1sx + sy + sz2
E
1 - 210.452
600 psi
[31 -20 psi2]
= - 0.01 in3>in3
Ans.
Change in Length. The normal strain on each side can be
determined from Hooke’s law, Eq. 10–18; that is,
P =
=
1
[s - n1sy + sz2]
E x
1
[-20 psi - 10.4521 -20 psi - 20 psi2] = - 0.00333 in.>in.
600 psi
Thus, the change in length of each side is
da = - 0.0033314 in.2 = - 0.0133 in.
Ans.
db = - 0.0033312 in.2 = - 0.00667 in.
Ans.
dc = - 0.0033313 in.2 = - 0.0100 in.
Ans.
The negative signs indicate that each dimension is decreased.
10
516
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
PROBLEMS
10–30. For the case of plane stress, show that Hooke’s law
can be written as
sx =
E
E
1Px + nPy2, sy =
1Py + nPx2
2
11 - n 2
11 - n22
10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the
stress-transformation equations, Eqs. 9–1 and 9–2.
10–37. Determine the bulk modulus for each of the
following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and
(b) glass, Eg = 811032 ksi, ng = 0.24.
10–38. The principal stresses at a point are shown in the
figure. If the material is A-36 steel, determine the principal
strains.
12 ksi
*10–32. A bar of copper alloy is loaded in a tension
machine and it is determined that Px = 940110-62 and
sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of
elasticity, Ecu, and the dilatation, ecu, of the copper.
ncu = 0.35.
•10–33.
The principal strains at a point on the aluminum
fuselage of a jet aircraft are P1 = 780110-62 and P2 =
400110-62. Determine the associated principal stresses at
the point in the same plane. Eal = 1011032 ksi, nal = 0.33.
Hint: See Prob. 10–30.
10–34. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the absolute maximum shear strain in the
rod at a point on its surface.
10–35. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the principal strains at a point on the
surface of the rod.
700 N
700 N
20 ksi
8 ksi
Prob. 10–38
10–39. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel
is pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
Est = 200 GPa and nst = 0.3.
Probs. 10–34/35
20 mm
*10–36. The steel shaft has a radius of 15 mm. Determine
the torque T in the shaft if the two strain gauges, attached to
the surface of the shaft, report strains of Px¿ = - 80110-62
and Py¿ = 80110-62. Also, compute the strains acting in the x
and y directions. Est = 200 GPa, nst = 0.3.
10
y
T
y¿
x¿
45
x
T
Prob. 10–36
Prob. 10–39
10.6
*10–40. The strain in the x direction at point A on the
steel beam is measured and found to be Px = - 100110-62.
Determine the applied load P. What is the shear strain gxy
at point A? Est = 2911032 ksi, nst = 0.3.
3 in.
0.5 in.
A
P
y
0.5 in.
8 in.
0.5 in.
*10–44. A single strain gauge, placed in the vertical plane
on the outer surface and at an angle of 30° to the axis of the
pipe, gives a reading at point A of Pa = - 200(10-6).
Determine the principal strains in the pipe at point A. The
pipe has an outer diameter of 2 in. and an inner diameter of
1 in. and is made of A-36 steel.
6 in.
x
3 ft
4 ft
517
10–43. A single strain gauge, placed on the outer surface
and at an angle of 30° to the axis of the pipe, gives a reading
at point A of Pa = - 200(10-6). Determine the horizontal
force P if the pipe has an outer diameter of 2 in. and an
inner diameter of 1 in. The pipe is made of A-36 steel.
3 in.
A
MATERIAL-PROPERTY RELATIONSHIPS
7 ft
1.5 ft
Prob. 10–40
•10–41.
The cross section of the rectangular beam is
subjected to the bending moment M. Determine an
expression for the increase in length of lines AB and CD.
The material has a modulus of elasticity E and Poisson’s
ratio is n.
P
2.5 ft
30
C
A
D
B
h
Probs. 10–43/44
A
10–45. The cylindrical pressure vessel is fabricated using
hemispherical end caps in order to reduce the bending stress
that would occur if flat ends were used. The bending stresses
at the seam where the caps are attached can be eliminated
by proper choice of the thickness th and tc of the caps and
cylinder, respectively. This requires the radial expansion to
be the same for both the hemispheres and cylinder. Show
that this ratio is tc>th = 12 - n2>11 - n2. Assume that the
vessel is made of the same material and both the cylinder
and hemispheres have the same inner radius. If the cylinder
is to have a thickness of 0.5 in., what is the required thickness
of the hemispheres? Take n = 0.3.
M
Prob. 10–41
b
10–42. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
26 ksi
tc
10
th
r
15 ksi
10 ksi
Prob. 10–42
Prob. 10–45
518
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
10–46. The principal strains in a plane, measured
experimentally at a point on the aluminum fuselage of a jet
aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is
a case of plane stress, determine the associated principal
stresses at the point in the same plane. Eal = 10(103) ksi
and nal = 0.33.
10–47. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
•10–49.
Initially, gaps between the A-36 steel plate and
the rigid constraint are as shown. Determine the normal
stresses sx and sy developed in the plate if the temperature
is increased by ¢T = 100°F . To solve, add the thermal
strain a¢T to the equations for Hooke’s Law.
y
0.0015 in.
3 ksi
6 in.
0.0025 in.
8 in.
x
Prob. 10–49
8 ksi
4 ksi
10–50. Two strain gauges a and b are attached to a plate
made from a material having a modulus of elasticity of
E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give
a reading of Pa = 450110-62 and Pb = 100110-62, determine
the intensities of the uniform distributed load wx and wy
acting on the plate. The thickness of the plate is 25 mm.
Prob. 10–47
*10–48. The 6061-T6 aluminum alloy plate fits snugly into
the rigid constraint. Determine the normal stresses sx and
sy developed in the plate if the temperature is increased by
¢T = 50°C . To solve, add the thermal strain a¢T to the
equations for Hooke’s Law.
10–51. Two strain gauges a and b are attached to the
surface of the plate which is subjected to the uniform
distributed load wx = 700 kN>m and wy = - 175 kN>m.
If the gauges give a reading of Pa = 450110-62 and
Pb = 100110-62, determine the modulus of elasticity E,
shear modulus G, and Poisson’s ratio n for the material.
wy
y
400 mm
10
b
45
y
300 mm
a
x
Prob. 10–48
z
x
Probs. 10–50/51
wx
10.6
*10–52. The block is fitted between the fixed supports. If
the glued joint can resist a maximum shear stress of
tallow = 2 ksi, determine the temperature rise that will
cause the joint to fail. Take E = 10 (103) ksi, n = 0.2, and
Hint: Use Eq. 10–18 with an additional strain term of a¢T
(Eq. 4–4).
40
MATERIAL-PROPERTY RELATIONSHIPS
519
*10–56. A thin-walled cylindrical pressure vessel has an
inner radius r, thickness t, and length L. If it is subjected
to an internal pressure p, show that the increase in
its inner radius is dr = rP1 = pr211 - 12 n2>Et and the
increase in its length is ¢L = pLr112 - n2>Et. Using these
results, show that the change in internal volume becomes
dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are
small quantities, show further that the change in volume
per unit volume, called volumetric strain, can be written as
dV>V = pr12.5 - 2n2>Et.
10–57. The rubber block is confined in the U-shape
smooth rigid block. If the rubber has a modulus of elasticity
E and Poisson’s ratio n, determine the effective modulus of
elasticity of the rubber under the confined condition.
P
Prob. 10–52
•10–53.
The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is covered and the
temperature is increased by 200°F, determine the stress
components sx , sy , and sz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
10–54. The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is not covered and the
temperature is increased by 200°F, determine the strain
components Px , Py , and Pz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
z
Prob. 10–57
10–58. A soft material is placed within the confines of a
rigid cylinder which rests on a rigid support. Assuming that
Px = 0 and Py = 0, determine the factor by which the
modulus of elasticity will be increased when a load is
applied if n = 0.3 for the material.
0.012 in.
z
4 in.
4 in.
6 in.
P
y
10
y
x
Probs. 10–53/54
x
10–55. A thin-walled spherical pressure vessel having an
inner radius r and thickness t is subjected to an internal
pressure p. Show that the increase in the volume within
the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain
analysis.
Prob. 10–58
520
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
*10.7 Theories of Failure
When an engineer is faced with the problem of design using a specific
material, it becomes important to place an upper limit on the state of
stress that defines the material’s failure. If the material is ductile, failure is
usually specified by the initiation of yielding, whereas if the material is
brittle, it is specified by fracture. These modes of failure are readily
defined if the member is subjected to a uniaxial state of stress, as in the
case of simple tension; however, if the member is subjected to biaxial or
triaxial stress, the criterion for failure becomes more difficult to establish.
In this section we will discuss four theories that are often used in
engineering practice to predict the failure of a material subjected to a
multiaxial state of stress. No single theory of failure, however, can be
applied to a specific material at all times, because a material may behave
in either a ductile or brittle manner depending on the temperature,
rate of loading, chemical environment, or the way the material is shaped
or formed. When using a particular theory of failure, it is first necessary
to calculate the normal and shear stress at points where they are the
largest in the member. Once this state of stress is established, the
principal stresses at these critical points are then determined, since each
of the following theories is based on knowing the principal stress.
Ductile Materials
45
Lüder’s lines on
mild steel strip
10
Fig. 10–26
Maximum-Shear-Stress Theory. The most common type of yielding
of a ductile material such as steel is caused by slipping, which occurs along
the contact planes of randomly ordered crystals that make up the
material. If we make a specimen into a highly polished thin strip and
subject it to a simple tension test, we can actually see how this slipping
causes the material to yield, Fig. 10–26. The edges of the planes of
slipping as they appear on the surface of the strip are referred to as
Lüder’s lines. These lines clearly indicate the slip planes in the strip,
which occur at approximately 45° with the axis of the strip.
The slipping that occurs is caused by shear stress. To show this,
consider an element of the material taken from a tension specimen,
when it is subjected to the yield stress sY, Fig. 10–27a. The maximum
shear stress can be determined by drawing Mohr’s circle for the element,
Fig. 10–27b. The results indicate that
tmax =
sY
2
(10–26)
10.7
Furthermore, this shear stress acts on planes that are 45° from the planes
of principal stress, Fig. 10–27c, and these planes coincide with the
direction of the Lüder lines shown on the specimen, indicating that
indeed failure occurs by shear.
Using this idea, that ductile materials fail by shear, in 1868 Henri
Tresca proposed the maximum-shear-stress theory or Tresca yield
criterion. This theory can be used to predict the failure stress of a ductile
material subjected to any type of loading. The theory states that yielding
of the material begins when the absolute maximum shear stress in the
material reaches the shear stress that causes the same material to yield
when it is subjected only to axial tension. Therefore, to avoid failure, it is
required that tabs
in the material must be less than or equal to sY> 2,
max
where sY is determined from a simple tension test.
For application we will express the absolute maximum shear stress in
terms of the principal stresses. The procedure for doing this was
discussed in Sec. 9.5 with reference to a condition of plane stress, that is,
where the out-of-plane principal stress is zero. If the two in-plane
principal stresses have the same sign, i.e., they are both tensile or both
compressive, then failure will occur out of the plane, and from Eq. 9–13,
T
sY
Axial tension
(a)
T
s1 sY
s2 0
s1
2
90
savg y¿
If instead the in-plane principal stresses are of opposite signs, then
failure occurs in the plane, and from Eq. 9–14,
sY
2
tmax sY
2
(b)
x¿
sY
2
savg 45
(c)
Using these equations and Eq. 10–26, the maximum-shear-stress theory
for plane stress can be expressed for any two in-plane principal stresses s1
and s2 by the following criteria:
s1 , s2 have same signs
sY
2
x
s1 - s2
tabs
=
max
2
ƒ s1 ƒ = sY
r
ƒ s2 ƒ = sY
s
A(0, 0)
max =
tabs
max
521
THEORIES OF FAILURE
Fig. 10–27
s2
sY
(10–27)
ƒ s1 - s2 ƒ = sY6 s1 , s2 have opposite signs
A graph of these equations is given in Fig. 10–28. Clearly, if any point
of the material is subjected to plane stress, and its in-plane principal
stresses are represented by a coordinate (s1, s2) plotted on the boundary
or outside the shaded hexagonal area shown in this figure, the material
will yield at the point and failure is said to occur.
sY
sY
sY
Maximum-shear-stress theory
Fig. 10–28
s1
10
522
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
Maximum-Distortion-Energy Theory. It was stated in Sec. 3.5 that
s3
an external loading will deform a material, causing it to store energy
internally throughout its volume. The energy per unit volume of material
is called the strain-energy density, and if the material is subjected to a
uniaxial stress the strain-energy density, defined by Eq. 3–6, becomes
s1
u =
s2
1
sP
2
(10–28)
If the material is subjected to triaxial stress, Fig. 10–29a, then each
principal stress contributes a portion of the total strain-energy density,
so that
(a)
ⴝ
u =
1
1
1
s1P1 + s2P2 + s3P3
2
2
2
Furthermore, if the material behaves in a linear-elastic manner, then
Hooke’s law applies. Therefore, substituting Eq. 10–18 into the above
equation and simplifying, we get
savg
u =
savg
savg
(b)
ⴙ
(s3 savg )
10
(s1 savg )
(s2 savg )
1
C s 2 + s22 + s32 - 2n1s1s2 + s1s3 + s3s22 D
2E 1
This strain-energy density can be considered as the sum of two parts,
one part representing the energy needed to cause a volume change of
the element with no change in shape, and the other part representing the
energy needed to distort the element. Specifically, the energy stored in the
element as a result of its volume being changed is caused by application
of the average principal stress, savg = 1s1 + s2 + s32>3, since this stress
causes equal principal strains in the material, Fig. 10–29b. The remaining
portion of the stress, 1s1 - savg2, 1s2 - savg2, 1s3 - savg2, causes the
energy of distortion, Fig. 10–29c.
Experimental evidence has shown that materials do not yield when
subjected to a uniform (hydrostatic) stress, such as savg discussed above.
As a result, in 1904, M. Huber proposed that yielding in a ductile
material occurs when the distortion energy per unit volume of the
material equals or exceeds the distortion energy per unit volume of the
same material when it is subjected to yielding in a simple tension test.
This theory is called the maximum-distortion-energy theory, and since it
was later redefined independently by R. von Mises and H. Hencky, it
sometimes also bears their names.
To obtain the distortion energy per unit volume, we will substitute the
stresses 1s1 - savg2, 1s2 - savg2, and 1s3 - savg2 for s1 , s2 , and s3 ,
respectively, into Eq. 10–29, realizing that savg = 1s1 + s2 + s32>3.
Expanding and simplifying, we obtain
(c)
Fig. 10–29
(10–29)
ud =
1 + n
C 1s1 - s222 + 1s2 - s322 + 1s3 - s122 D
6E
10.7
In the case of plane stress, s3 = 0, and this equation reduces to
ud =
523
THEORIES OF FAILURE
s2
1 + n
A s12 - s1s2 + s22 B
3E
sY
For a uniaxial tension test, s1 = sY , s2 = s3 = 0, and so
sY
1 + n 2
1ud2Y =
s
3E Y
Since the maximum-distortion-energy theory requires ud = 1ud2Y , then
for the case of plane or biaxial stress, we have
s12 - s1s2 + s22 = sY2
(10–30)
This is the equation of an ellipse, Fig. 10–30. Thus, if a point in the
material is stressed such that (s1, s2) is plotted on the boundary or
outside the shaded area, the material is said to fail.
A comparison of the above two failure criteria is shown in Fig. 10–31.Note
that both theories give the same results when the principal stresses are
equal, i.e., s1 = s2 = sY, or when one of the principal stresses is zero and
the other has a magnitude of sY. If the material is subjected to pure shear, t,
then the theories have the largest discrepancy in predicting failure. The
stress coordinates of these points on the curves can be determined by
considering the element shown in Fig. 10–32a. From the associated Mohr’s
circle for this state of stress, Fig. 10–32b, we obtain principal stresses s1 = t
and s2 = - t. Thus, with s1 = - s2, then from Eq. 10–27, the maximumshear-stress theory gives 1sY >2, -sY >22, and from Eq. 10–30, the
maximum-distortion-energy theory gives 1sY > 23, -sY > 232, Fig.10–31.
Actual torsion tests, used to develop a condition of pure shear in a
ductile specimen, have shown that the maximum-distortion-energy
theory gives more accurate results for pure-shear failure than the
maximum-shear-stress theory. In fact, since 1sY> 132>1sY>22 = 1.15,
the shear stress for yielding of the material, as given by the maximumdistortion-energy theory, is 15% more accurate than that given by the
maximum-shear-stress theory.
t
s2 t
s1 t
90
A (t, 0)
t
(b)
(a)
Fig. 10–32
s
sY
s1
sY
Maximum-distortion-energy theory
Fig. 10–30
s2
Pure shear
sY
(sY, sY)
sY
(sY,sY)
sY
sY
s1
sY , sY
3
3
sY , sY
2 2
Fig. 10–31
10
524
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
Brittle Materials
Maximum-Normal-Stress Theory. It was previously stated that
brittle materials, such as gray cast iron, tend to fail suddenly by fracture
with no apparent yielding. In a tension test, the fracture occurs when
the normal stress reaches the ultimate stress sult , Fig. 10–33a. Also,
brittle fracture occurs in a torsion test due to tension since the plane of
fracture for an element is at 45° to the shear direction, Fig. 10–33b. The
fracture surface is therefore helical as shown.* Experiments have further
shown that during torsion the material’s strength is somewhat unaffected
by the presence of the associated principal compressive stress being at
right angles to the principal tensile stress. Consequently, the tensile stress
needed to fracture a specimen during a torsion test is approximately the
same as that needed to fracture a specimen in simple tension. Because of
this, the maximum-normal-stress theory states that a brittle material will
fail when the maximum tensile stress, s1, in the material reaches a value
that is equal to the ultimate normal stress the material can sustain when
it is subjected to simple tension.
If the material is subjected to plane stress, we require that
Failure of a brittle material
in tension
(a)
ƒ s1 ƒ = sult
45
ƒ s2 ƒ = sult
These equations are shown graphically in Fig. 10–34. Therefore, if the
stress coordinates 1s1 , s22 at a point in the material fall on the boundary
or outside the shaded area, the material is said to fracture. This theory is
generally credited to W. Rankine, who proposed it in the mid-1800s.
Experimentally it has been found to be in close agreement with the
behavior of brittle materials that have stress–strain diagrams that are
similar in both tension and compression.
45
Failure of a brittle material
in torsion
(b)
Mohr’s Failure Criterion. In some brittle materials tension and
Fig. 10–33
s2
sult
10
sult
sult
(10–31)
s1
compression properties are different. When this occurs a criterion based
on the use of Mohr’s circle may be used to predict failure. This method
was developed by Otto Mohr and is sometimes referred to as Mohr’s
failure criterion. To apply it, one first performs three tests on the material.
A uniaxial tensile test and uniaxial compressive test are used to
determine the ultimate tensile and compressive stresses 1sult2t and
1sult2c , respectively. Also a torsion test is performed to determine the
material’s ultimate shear stress tult. Mohr’s circle for each of these stress
sult
Maximum-normal-stress theory
Fig. 10–34
*A stick of blackboard chalk fails in this way when its ends are twisted with the fingers.
10.7
conditions is then plotted as shown in Fig. 10–35. These three circles are
contained in a “failure envelope” indicated by the extrapolated colored
curve that is drawn tangent to all three circles. If a plane-stress condition
at a point is represented by a circle that has a point of tangency with the
envelope, or if it extends beyond the envelope’s boundary, then failure is
said to occur.
We may also represent this criterion on a graph of principal stresses s1
and s2. This is shown in Fig. 10–36. Here failure occurs when the absolute
value of either one of the principal stresses reaches a value equal to or
greater than 1sult2t or 1sult2c or in general, if the state of stress at a point
defined by the stress coordinates 1s1 , s22 is plotted on the boundary or
outside the shaded area.
Either the maximum-normal-stress theory or Mohr’s failure criterion
can be used in practice to predict the failure of a brittle material.
However, it should be realized that their usefulness is quite limited. A
tensile fracture occurs very suddenly, and its initiation generally depends
on stress concentrations developed at microscopic imperfections of
the material such as inclusions or voids, surface indentations, and
small cracks. Since each of these irregularities varies from specimen
to specimen, it becomes difficult to specify fracture on the basis of a
single test.
525
THEORIES OF FAILURE
Failure envelope
s
(sult)c
(sult)t
tult
t
Fig. 10–35
s2
(sult)t
(sult)c
(sult)t
s1
(sult)c
Mohr’s failure criterion
Fig. 10–36
Important Points
• If a material is ductile, failure is specified by the initiation of yielding, whereas if it is brittle, it is specified
by fracture.
• Ductile failure can be defined when slipping occurs between the crystals that compose the material. This
slipping is due to shear stress and the maximum-shear-stress theory is based on this idea.
• Strain energy is stored in a material when it is subjected to normal stress. The maximum-distortion-energy
theory depends on the strain energy that distorts the material, and not the part that increases its volume.
• The fracture of a brittle material is caused only by the maximum tensile stress in the material, and not the
compressive stress. This is the basis of the maximum-normal-stress theory, and it is applicable if the
stress–strain diagram is similar in tension and compression.
• If a brittle material has a stress–strain diagram that is different in tension and compression, then Mohr’s
•
failure criterion may be used to predict failure.
Due to material imperfections, tensile fracture of a brittle material is difficult to predict, and so theories of
failure for brittle materials should be used with caution.
10
526
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
EXAMPLE 10.12
The solid cast-iron shaft shown in Fig. 10–37a is subjected to a torque
of T = 400 lb # ft. Determine its smallest radius so that it does not fail
according to the maximum-normal-stress theory. A specimen of cast
iron, tested in tension, has an ultimate stress of 1sult2t = 20 ksi.
tmax
s2
T 400 lbft
s1
s
T 400 lbft
tmax
r
t
(a)
(b)
Fig. 10–37
SOLUTION
The maximum or critical stress occurs at a point located on the surface
of the shaft. Assuming the shaft to have a radius r, the shear stress is
tmax =
1400 lb # ft2112 in.>ft2r
Tc
3055.8 lb # in.
=
=
4
J
1p>22r
r3
Mohr’s circle for this state of stress (pure shear) is shown in Fig. 10–37b.
Since R = tmax , then
s1 = - s2 = tmax =
3055.8 lb # in.
r3
The maximum-normal-stress theory, Eq. 10–31, requires
ƒ s1 ƒ … sult
10
3055.8 lb # in.
… 20 000 lb>in2
r3
Thus, the smallest radius of the shaft is determined from
3055.8 lb # in.
= 20 000 lb>in2
r3
r = 0.535 in.
Ans.
10.7
527
THEORIES OF FAILURE
EXAMPLE 10.13
The solid shaft shown in Fig. 10–38a has a radius of 0.5 in. and is made
of steel having a yield stress of sY = 36 ksi. Determine if the loadings
cause the shaft to fail according to the maximum-shear-stress theory
and the maximum-distortion-energy theory.
SOLUTION
The state of stress in the shaft is caused by both the axial force and the
torque. Since maximum shear stress caused by the torque occurs in
the material at the outer surface, we have
-15 kip
P
=
= - 19.10 ksi
sx =
15 kip
A
p10.5 in.22
txy
3.25 kip # in. 10.5 in.2
Tc
=
=
= 16.55 ksi
p
4
J
2 10.5 in.2
0.5 in.
A
3.25 kipin.
(a)
The stress components are shown acting on an element of material
at point A in Fig. 10–38b. Rather than using Mohr’s circle, the principal
stresses can also be obtained using the stress-transformation Eq. 9–5.
s1,2 =
=
sx + sy
2
;
B
a
sx - sy
2
16.55 ksi
19.10 ksi
2
b + txy2
- 19.10 - 0 2
-19.10 + 0
;
a
b + 116.5522
2
B
2
= - 9.55 ; 19.11
(b)
Fig. 10–38
s1 = 9.56 ksi
s2 = - 28.66 ksi
Maximum-Shear-Stress Theory. Since the principal stresses have
opposite signs, then from Sec. 9.5, the absolute maximum shear stress
will occur in the plane, and therefore, applying the second of
Eqs. 10–27, we have
ƒ s1 - s2 ƒ … sY
ƒ 9.56 - 1 -28.662 ƒ … 36
?
38.2 7 36
Thus, shear failure of the material will occur according to this theory.
Maximum-Distortion-Energy Theory. Applying Eq. 10–30, we have
A s12 - s1s2 + s22 B … sY2
C 19.5622 - 19.5621 -28.662 + 1-28.6622 D … 13622
?
1187 … 1296
Using this theory, failure will not occur.
10
528
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
PROBLEMS
10–59. A material is subjected to plane stress. Express
the distortion-energy theory of failure in terms of sx , sy ,
and txy .
*10–60. A material is subjected to plane stress. Express
the maximum-shear-stress theory of failure in terms of sx ,
sy , and txy . Assume that the principal stresses are of
different algebraic signs.
*10–68. The short concrete cylinder having a diameter of
50 mm is subjected to a torque of 500 N # m and an axial
compressive force of 2 kN. Determine if it fails according to
the maximum-normal-stress theory. The ultimate stress of
the concrete is sult = 28 MPa.
2 kN
500 Nm
•10–61.
An aluminum alloy 6061-T6 is to be used for
a solid drive shaft such that it transmits 40 hp at 2400
rev>min. Using a factor of safety of 2 with respect to
yielding, determine the smallest-diameter shaft that can be
selected based on the maximum-shear-stress theory.
500 Nm
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory.
2 kN
Prob. 10–68
10–63. An aluminum alloy is to be used for a drive shaft
such that it transmits 25 hp at 1500 rev>min. Using a factor
of safety of 2.5 with respect to yielding, determine the
smallest-diameter shaft that can be selected based on the
maximum-distortion-energy theory. sY = 3.5 ksi.
*10–64. A bar with a square cross-sectional area is made
of a material having a yield stress of sY = 120 ksi. If the bar
is subjected to a bending moment of 75 kip # in., determine
the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with
respect to yielding.
•10–69.
Cast iron when tested in tension and compression
has an ultimate strength of 1sult2t = 280 MPa and
1sult2c = 420 MPa, respectively. Also, when subjected to
pure torsion it can sustain an ultimate shear stress of
tult = 168 MPa. Plot the Mohr’s circles for each case and
establish the failure envelope. If a part made of this
material is subjected to the state of plane stress shown,
determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
10–66. Derive an expression for an equivalent torque Te
that, if applied alone to a solid bar with a circular cross
10 section, would cause the same energy of distortion as
the combination of an applied bending moment M and
torque T.
10–67. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of
distortion as the combination of an applied bending
moment M and torque T.
220 MPa
Prob. 10–69
10–70. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same maximum
shear stress as the combination of an applied moment M
and torque T. Assume that the principal stresses are of
opposite algebraic signs.
10.7
529
THEORIES OF FAILURE
10–71. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximum-shearstress theory.
10–75. If the A-36 steel pipe has outer and inner
diameters of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-shear-stress theory.
*10–72. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory.
*10–76. If the A-36 steel pipe has an outer and inner
diameter of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-distortion-energy theory.
900 N
60 MPa
200 mm
150 mm
A
40 MPa
100 mm
200 mm
70 MPa
Probs. 10–71/72
900 N
•10–73.
If the 2-in. diameter shaft is made from brittle
material having an ultimate strength of sult = 50 ksi for
both tension and compression, determine if the shaft fails
according to the maximum-normal-stress theory. Use a
factor of safety of 1.5 against rupture.
10–74. If the 2-in. diameter shaft is made from cast
iron having tensile and compressive ultimate strengths
of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively,
determine if the shaft fails in accordance with Mohr’s
failure criterion.
Probs. 10–75/76
•10–77.
The element is subjected to the stresses shown. If
sY = 36 ksi, determine the factor of safety for the loading
based on the maximum-shear-stress theory.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
12 ksi
4 ksi
8 ksi
Probs. 10–77/78
30 kip
4 kip · ft
Probs. 10–73/74
10–79. The yield stress for heat-treated beryllium copper
is sY = 130 ksi. If this material is subjected to plane stress
and elastic failure occurs when one principal stress is 145 ksi,
what is the smallest magnitude of the other principal stress?
Use the maximum-distortion-energy theory.
10
530
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
*10–80. The plate is made of hard copper, which yields at
sY = 105 ksi. Using the maximum-shear-stress theory,
determine the tensile stress sx that can be applied to the
plate if a tensile stress sy = 0.5sx is also applied.
•10–85.
The state of stress acting at a critical point on a
machine element is shown in the figure. Determine the
smallest yield stress for a steel that might be selected for the
part, based on the maximum-shear-stress theory.
•10–81.
Solve Prob. 10–80 using the maximum-distortionenergy theory.
sy 0.5sx
sx
10 ksi
4 ksi
8 ksi
Probs. 10–80/81
10–82. The state of stress acting at a critical point on the
seat frame of an automobile during a crash is shown in the
figure. Determine the smallest yield stress for a steel that
can be selected for the member, based on the maximumshear-stress theory.
Prob. 10–85
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t,
and s3 = 0. If the yield stress is sY, determine the maximum
value of p based on (a) the maximum-shear-stress theory and
(b) the maximum-distortion-energy theory.
25 ksi
80 ksi
10–87. If a solid shaft having a diameter d is subjected to
a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is
tallow = 116>pd322M2 + T2. Assume the principal stresses
to be of opposite algebraic signs.
*10–88. If a solid shaft having a diameter d is subjected to a
torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is
sallow = 116>pd321M + 2M2 + T22.
10
Probs. 10–82/83
*10–84. A bar with a circular cross-sectional area is made
of SAE 1045 carbon steel having a yield stress of
sY = 150 ksi. If the bar is subjected to a torque of
30 kip # in. and a bending moment of 56 kip # in., determine
the required diameter of the bar according to the
maximum-distortion-energy theory. Use a factor of safety
of 2 with respect to yielding.
T
T
M
M
Probs. 10–87/88
10.7
•10–89.
The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-shear-stress theory. Use a factor
of safety of 1.5 against yielding.
THEORIES OF FAILURE
531
*10–92. The gas tank has an inner diameter of 1.50 m and
a wall thickness of 25 mm. If it is made from A-36 steel and
the tank is pressured to 5 MPa, determine the factor of
safety against yielding using (a) the maximum-shear-stress
theory, and (b) the maximum-distortion-energy theory.
10–90. The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-distortion-energy theory. Use a
factor of safety of 1.5 against yielding.
A
80 mm
B
T
C
80 mm
100 mm
Prob. 10–92
T
•10–93.
Probs. 10–89/90
10–91. The internal loadings at a critical section along the
steel drive shaft of a ship are calculated to be a torque of
2300 lb # ft, a bending moment of 1500 lb # ft, and an axial
thrust of 2500 lb. If the yield points for tension and shear are
sY = 100 ksi and tY = 50 ksi, respectively, determine the
required diameter of the shaft using the maximum-shearstress theory.
The gas tank is made from A-36 steel and
has an inner diameter of 1.50 m. If the tank is designed
to withstand a pressure of 5 MPa, determine the required
minimum wall thickness to the nearest millimeter using
(a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5
against yielding.
10
2300 lbft
1500 lbft
2500 lb
Prob. 10–91
Prob. 10–93
532
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
CHAPTER REVIEW
When an element of material is
subjected to deformations that only
occur in a single plane, then it
undergoes plane strain. If the strain
components Px , Py , and gxy are known
for a specified orientation of the
element, then the strains acting
for some other orientation of the
element can be determined using the
plane-strain transformation equations.
Likewise, the principal normal strains
and maximum in-plane shear strain
can be determined using transformation
equations.
Px + Py
Px¿ =
Px + Py
Py¿ =
gx¿y¿
2
P1,2 =
gmax
in-plane
2
Pavg =
2
2
2
B
Px - Py
Px - Py
Px + Py
=
2
-
2
= -¢
Px - Py
+
2
a
;
cos 2u gxy
≤ sin 2u +
B
Px - Py
2
cos 2u +
a
2
Px - Py
2
2
b + a
gxy
2
gxy
2
2
b
sin 2u
cos 2u
b + a
gxy
2
sin 2u
gxy
2
b
2
2
Px + Py
2
Strain transformation problems can
also be solved in a semi-graphical
manner using Mohr’s circle. To
draw the circle, the P and g>2 axes
are established and the center of
the circle C [1Px + Py2>2, 0] and the
“reference point” A 1Px , gxy>22
are plotted. The radius of the circle
extends between these two points
and is determined from trigonometry.
Px P y
2
C
gxy
Pavg g
2
R
If P1 and P2 have the same sign then
the absolute maximum shear strain
will be out of plane.
10
In the case of plane strain, the
absolute maximum shear strain will
be equal to the maximum in-plane
shear strain provided the principal
strains P1 and P2 have opposite signs.
gabs
= P1
max
in-plane = P
gmax
1 - P2
gabs
= P1 - P2
max
2
Px Py
2
Px
Px Py
2
2
A
u 0
gxy
2
2
P
CHAPTER REVIEW
If the material is subjected to triaxial
stress, then the strain in each
direction is influenced by the strain
produced by all three stresses.
Hooke’s law then involves the
material properties E and n.
If E and n are known, then G can be
determined.
Px =
1
[s - n1sy + sz2]
E x
Py =
1
[s - n1sx + sz2]
E y
Pz =
1
[s - n1sx + sy2]
E z
G =
E
211 + n2
The dilatation is a measure of
volumetric strain.
e =
1 - 2n
1sx + sy + sz2
E
The bulk modulus is used to measure
the stiffness of a volume of material.
k =
E
311 - 2n2
533
If the principal stresses at a critical
point in the material are known, then
a theory of failure can be used as a
basis for design.
Ductile materials fail in shear, and
here the maximum-shear-stress theory
or the maximum-distortion-energy
theory can be used to predict failure.
Both of these theories make
comparison to the yield stress of a
specimen subjected to a uniaxial
tensile stress.
Brittle materials fail in tension, and
so the maximum-normal-stress theory
or Mohr’s failure criterion can be
used to predict failure. Here comparisons
are made with the ultimate tensile
stress developed in a specimen.
10
534
C H A P T E R 10
S T R A I N T R A N S F O R M AT I O N
REVIEW PROBLEMS
10–94. A thin-walled spherical pressure vessel has an
inner radius r, thickness t, and is subjected to an internal
pressure p. If the material constants are E and n, determine
the strain in the circumferential direction in terms of the
stated parameters.
10–95. The strain at point A on the shell has components
Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0.
Determine (a) the principal strains at A, (b) the maximum
shear strain in the x–y plane, and (c) the absolute maximum
shear strain.
•10–97.
The components of plane stress at a critical point
on a thin steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is
sY = 650 MPa.
340 MPa
65 MPa
55 MPa
y
x
A
Prob. 10–95
*10–96. The principal plane stresses acting at a point are
shown in the figure. If the material is machine steel having a
yield stress of sY = 500 MPa, determine the factor of
safety with respect to yielding if the maximum-shear-stress
theory is considered.
Prob. 10–97
10–98. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 600110-62, Pb = - 700110-62, and Pc = 350110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these
strains.
a
60
60
100 MPa
b
60
10
150 MPa
c
Prob. 10–96
Prob. 10–98
535
REVIEW PROBLEMS
10–99. A strain gauge forms an angle of 45° with the axis
of the 50-mm diameter shaft. If it gives a reading of
P = - 200110-62 when the torque T is applied to the shaft,
determine the magnitude of T. The shaft is made from A-36
steel.
10–102. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = - 300110-62.
Determine the equivalent state of strain on an element at
the same point oriented 30° clockwise with respect to the
original element. Sketch the results on the element.
y
T
45
Pydy
dy
gxy
2
T
gxy
2
dx
Prob. 10–99
*10–100. The A-36 steel post is subjected to the forces
shown. If the strain gauges a and b at point A give readings
of Pa = 300110-62 and Pb = 175110-62, determine the
magnitudes of P1 and P2.
P1
P2
a
2 in.
2 ft
A
1 in.
b 45
A
A
c
4 in.
c
x
Pxdx
Prob. 10–102
10–103. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = -300110-62.
Determine the equivalent state of strain, which represents
(a) the principal strains, and (b) the maximum in-plane shear
strain and the associated average normal strain. Specify
the orientation of the corresponding element at the point
with respect to the original element. Sketch the results on the
element.
Section c– c
1 in.
y
Pydy
Prob. 10–100
dy
10–101. A differential element is subjected to plane strain
that has the following components: Px = 950110-62, Py =
420110-62, gxy = - 325110-62. Use the strain-transformation
equations and determine (a) the principal strains and (b) the
maximum in-plane shear strain and the associated average
strain. In each case specify the orientation of the element
and show how the strains deform the element.
10
gxy
2
gxy
2
dx
Prob. 10–103
x
Pxdx
Beams are important structural members that are used to support roof and floor loadings.
10–1. Prove that the sum of the normal strains in
perpendicular directions is constant.
ex¿ =
ey¿ =
ex + ey
2
ex + ey
2
+
-
ex - ey
2
ex - ey
2
cos 2u +
cos 2u -
gxy
2
gxy
2
sin 2u
(1)
sin 2u
(2)
Adding Eq. (1) and Eq. (2) yields:
ex¿ + ey¿ = ex + ey = constant
QED
738
10–2. The state of strain at the point has components
of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 30° counterclockwise from the original position.
Sketch the deformed element due to these strains within the
x–y plane.
y
x
In accordance to the established sign convention,
ex = 200(10 - 6),
ex¿ =
ex + ey
2
= c
+
ey = -300(10 - 6)
ex - ey
2
cos 2u +
gxy
2
gxy = 400(10 - 6)
u = 30°
sin 2u
200 + (-300)
200 - (-300)
400
+
cos 60° +
sin 60° d (10 - 6)
2
2
2
= 248 (10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6)
= -233(10 - 6)
ey¿ =
ex + ey
= c
2
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
200 - (-300)
200 + ( -300)
400
cos 60° sin 60° d(10 - 6)
2
2
2
= -348(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
739
10–3. A strain gauge is mounted on the 1-in.-diameter
A-36 steel shaft in the manner shown. When the shaft is
rotating with an angular velocity of v = 1760 rev>min, the
reading on the strain gauge is P = 800110-62. Determine
the power output of the motor. Assume the shaft is only
subjected to a torque.
v = (1760 rev>min)a
60⬚
2p rad
1 min
ba
b = 184.307 rad>s
60 sec
1 rev
ex = ey = 0
ex¿ =
ex + ey
2
+
ex - ey
2
800(10 - 6) = 0 + 0 +
cos 2u +
gxy
2
gxy
2
sin 2u
sin 120°
gxy = 1.848(10 - 3) rad
t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi
t =
Tc
;
J
20.323 =
T(0.5)
p
2
(0.5)4
;
T = 3.99 kip # in = 332.5 lb # ft
P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp
Ans.
740
*10–4. The state of strain at a point on a wrench
has components Px = 120110-62, Py = - 180110-62, gxy =
150110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 120(10 - 6)
e1, 2 =
a)
ey = -180(10 - 6)
gxy = 150(10 - 6)
Ex - Ey 2
gxy 2
ex + ey
;
a
b + a
b
2
A
2
2
120 + (-180)
120 - ( -180) 2
150 2
-6
;
a
b + a
b d 10
2
A
2
2
e1 = 138(10 - 6);
e2 = -198(10 - 6)
= c
Ans.
Orientation of e1 and e2
gxy
150
=
tan 2up =
= 0.5
ex - ey
[120 - (-180)]
up = 13.28° and -76.72°
Use Eq. 10.5 to determine the direction of e1 and e2
ex¿ =
ex + ey
+
2
ex - ey
2
cos 2u +
gxy
2
sin 2u
u = up = 13.28°
ex¿ = c
120 + (-180)
120 - ( -180)
150
+
cos (26.56°) +
sin 26.56° d 10 - 6
2
2
2
= 138 (10 - 6) = e1
Therefore up1 = 13.3° ;
g
b)
max
in-plane
=
2
in-plane
ex + ey
2
A
ex - ey
b + a
2
gxy
b
Ans.
2
2
2
150 2
120 - ( -180) 2
-6
-6
= 2c a
b + a
b d10 = 335 (10 )
A
2
2
gmax
eavg =
a
up2 = -76.7°
= c
120 + (-180)
d 10 - 6 = -30.0(10 - 6)
2
Ans.
Ans.
Orientation of gmax
tan 2us =
-(ex - ey)
gxy
=
-[120 - ( -180)]
= - 2.0
150
us = -31.7° and 58.3°
Ans.
gmax
Use Eq. 10–6 to determine the sign of in-plane
gx¿y¿
ex - ey
gxy
= sin 2u +
cos 2u
2
2
2
u = us = -31.7°
gx¿y¿ = 2c -
120 - (-180)
150
sin (-63.4°) +
cos (-63.4°) d10 - 6 = 335(10 - 6)
2
2
741
10–5. The state of strain at the point on the arm
has components Px = 250110-62, Py = -450110-62, gxy =
-825110-62. Use the strain-transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
ex = 250(10 - 6)
ey = -450(10 - 6)
y
gxy = -825(10 - 6)
x
a)
ex + ey
e1, 2 =
;
2
= c
A
ex - ey
a
2
2
b + a
gxy
2
b
2
250 - 450
250 - ( -450) 2
-825 2
-6
;
a
b + a
b d(10 )
2
A
2
2
e1 = 441(10 - 6)
Ans.
e2 = -641(10 - 6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2up =
ex - ey
up = -24.84°
-825
250 - (-450)
=
up = 65.16°
and
Use Eq. 10–5 to determine the direction of e1 and e2:
ex¿ =
ex + ey
+
2
ex - ey
2
cos 2u +
gxy
2
sin 2u
u = up = -24.84°
ex¿ = c
250 - (-450)
250 - 450
-825
+
cos ( -49.69°) +
sin (-49.69°) d(10 - 6) = 441(10 - 6)
2
2
2
Therefore, up1 = -24.8°
Ans.
up2 = 65.2°
Ans.
b)
g
max
in-plane
2
g
max
in-plane
eavg =
=
A
= 2c
a
ex - ey
A
2
a
ex + ey
2
b + a
2
gxy
2
b
2
250 - (-450) 2
-825 2
-6
-3
b + a
b d(10 ) = 1.08(10 )
2
2
= a
250 - 450
b (10 - 6) = -100(10 - 6)
2
Ans.
Ans.
742
10–6. The state of strain at the point has components of
Px = -100110-62, Py = 400110-62, and gxy = -300110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented at an
angle of 60° counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x–y plane.
y
x
In accordance to the established sign convention,
ex = -100(10 - 6)
ex¿ =
ex + ey
2
= c
+
ey = 400(10 - 6)
ex - ey
2
gxy
cos 2u +
2
gxy = -300(10 - 6)
u = 60°
sin 2u
-100 - 400
-300
-100 + 400
+
cos 120° +
sin 120° d(10 - 6)
2
2
2
= 145(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d (10 - 6)
= 583(10 - 6)
ey¿ =
ex + ey
= c
2
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
-100 + 400
-100 - 400
-300
cos 120° sin 120° d(10 - 6)
2
2
2
= 155 (10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
743
10–7. The state of strain at the point has components of
Px = 100110-62, Py = 300110-62, and gxy = -150110-62.
Use the strain-transformation equations to determine the
equivalent in-plane strains on an element oriented u = 30°
clockwise. Sketch the deformed element due to these
strains within the x–y plane.
y
x
In accordance to the established sign convention,
ex = 100(10 - 6)
ex¿ =
ex + ey
2
= c
+
ey = 300(10 - 6)
ex - ey
2
cos 2u +
gxy = -150(10 - 6)
gxy
2
u = - 30°
sin 2u
100 - 300
-150
100 + 300
+
cos (-60°) +
sin ( -60°) d (10 - 6)
2
2
2
= 215(10 - 6)
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6)
= -248 (10 - 6)
ey¿ =
ex + ey
= c
2
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
100 - 300
-150
100 + 300
cos ( -60°) sin (-60°) d (10 - 6)
2
2
2
= 185(10 - 6)
Ans.
The deformed element of this equivalent state of strain is shown in Fig. a
744
*10–8. The state of strain at the point on the bracket
has components Px = -200110-62, Py = -650110-62, gxy ⫽
-175110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 20° counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x–y plane.
ex = -200(10 - 6)
ex¿ =
ex + ey
2
= c
+
ey = -650(10 - 6)
ex - ey
2
cos 2u +
gxy
2
y
x
gxy = -175(10 - 6)
u = 20°
sin 2u
(-200) - (-650)
(-175)
-200 + (-650)
+
cos (40°) +
sin (40°) d (10 - 6)
2
2
2
= -309(10 - 6)
ey¿ =
ex + ey
2
= c
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
-200 - ( -650)
( -175)
-200 + (-650)
cos (40°) sin (40°) d(10 - 6)
2
2
2
= -541(10 - 6)
gx¿y¿
2
= -
ex - ey
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6)
= -423(10 - 6)
Ans.
745
10–9. The state of strain at the point has components of
Px = 180110-62, Py = -120110-62, and gxy = -100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
y
x
a)
In
accordance
to
the
established
sign
convention,
ex = 180(10 - 6),
ey = -120(10 - 6) and gxy = -100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
a
A
ex - ey
2
2
b + a
gxy
2
b
2
180 + (-120)
180 - ( -120) 2
-100 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A 30 ; 158.11 B (10 - 6)
e1 = 188(10 - 6)
tan 2uP =
e2 = -128(10 - 6)
gxy
ex - ey
Ans.
-100(10 - 6)
=
C 180 - (-120) D (10 - 6)
uP = -9.217°
and
= -0.3333
80.78°
Substitute u = -9.217°,
ex + ey
ex¿ =
2
= c
+
ex - ey
2
cos 2u +
gxy
2
sin 2u
180 + ( -120)
180 - ( -120)
-100
+
cos (-18.43°) +
sin (-18.43) d(10 - 6)
2
2
2
= 188(10 - 6) = e1
Thus,
(uP)1 = -9.22°
(uP)2 = 80.8°
Ans.
The deformed element is shown in Fig (a).
gmax
ex - ey 2
gxy 2
in-plane
=
b)
a
b + a
b
2
A
2
2
gmax
in-plane
tan 2us = - a
= b2
A
ex - ey
gxy
c
180 - ( -120) 2
-100 2
-6
-6
d + a
b r (10 ) = 316 A 10 B
2
2
b = -c
C 180 - (-120) D (10 - 6)
us = 35.78° = 35.8° and
-100(10 - 6)
Ans.
s = 3
-54.22° = -54.2°
Ans.
746
10–9. Continued
gmax
The algebraic sign for in-plane
when u = 35.78°.
ex - ey
gxy
gx¿y¿
= -a
b sin 2u +
cos 2u
2
2
2
gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6)
eavg
= -316(10 - 6)
ex + ey
180 + (-120)
=
= c
d(10 - 6) = 30(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown is
shown in Fig. b
747
10–10. The state of strain at the point on the bracket
has components Px = 400110-62, Py = -250110-62, gxy ⫽
310110-62. Use the strain-transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of u = 30° clockwise from the original
position. Sketch the deformed element due to these strains
within the x–y plane.
ex = 400(10 - 6)
ex¿ =
ex + ey
2
= c
+
ey = -250(10 - 6)
ex - ey
2
cos 2u +
gxy
2
gxy = 310(10 - 6)
y
x
u = - 30°
sin 2u
400 - ( -250)
400 + (-250)
310
+
cos (-60°) + a
b sin ( -60°) d(10 - 6)
2
2
2
= 103(10 - 6)
ey¿ =
ex + ey
2
= c
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
400 - (-250)
400 + (-250)
310
cos (60°) sin (-60°) d (10 - 6)
2
2
2
= 46.7(10 - 6)
gx¿y¿
2
= -
ex - ey
2
Ans.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6)
748
Ans.
10–11. The state of strain at the point has components of
Px = -100110-62, Py = -200110-62, and gxy = 100110-62.
Use the strain-transformation equations to determine (a)
the in-plane principal strains and (b) the maximum in-plane
shear strain and average normal strain. In each case specify
the orientation of the element and show how the strains
deform the element within the x–y plane.
In accordance to the established
ey = -200(10 - 6) and gxy = 100(10 - 6).
ex + ey
e1, 2 =
;
2
= b
A
a
ex - ey
2
b + a
2
gxy
2
b
sign
y
x
convention,
ex = -100(10 - 6),
2
-100 + (-200)
100 2
-100 - (-200) 2
-6
;
c
d + a
b r (10 )
2
A
2
2
= A -150 ; 70.71 B (10 - 6)
e1 = -79.3(10 - 6)
tan 2uP =
gxy
e2 = - 221(10 - 6)
100(10 - 6)
=
ex - ey
C -100 - ( -200) D (10 - 6)
uP = 22.5°
Ans.
= 1
-67.5°
and
Substitute u = 22.5,
ex + ey
ex¿ =
+
ex - ey
cos 2u +
gxy
sin 2u
2
2
2
-100 + (-200)
-100 - (-200)
100
= c
+
cos 45° +
sin 45° d(10 - 6)
2
2
2
= -79.3(10 - 6) = e1
Thus,
(uP)1 = 22.5°
(uP)2 = -67.5°
Ans.
The deformed element of the state of principal strain is shown in Fig. a
gmax
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
gmax
in-plane
= b2
tan 2us = - a
-100 - ( -200) 2
100 2
-6
-6
d + a
b r (10 ) = 141(10 )
A
2
2
c
ex - ey
gxy
b = -c
us = -22.5°
The algebraic sign for
gx¿y¿
2
= -a
ex - ey
2
C -100 - ( -200) D (10 - 6)
100(10 - 6)
and
gmax
in-plane
b sin 2u +
Ans.
s = -1
Ans.
67.5°
when u = -22.5°.
gxy
2
cos 2u
gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°)
eavg
= 141(10 - 6)
ex + ey
-100 + ( -200)
=
= c
d(10 - 6) = -150(10 - 6)
2
2
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. b.
749
10–11.
Continued
*10–12. The state of plane strain on an element is given by
Px = 500110-62, Py = 300110-62, and gxy = - 200110-62.
Determine the equivalent state of strain on an element at
the same point oriented 45° clockwise with respect to the
original element.
y
Pydy
dy gxy
2
Strain Transformation Equations:
ex = 500 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -200 A 10 - 6 B
u = -45°
We obtain
ex¿ =
ex + ey
2
= c
+
ex - ey
2
cos 2u +
gxy
2
sin 2u
500 + 300
500 - 300
-200
+
cos (-90°) + a
b sin ( -90°) d A 10 - 6 B
2
2
2
= 500 A 10 - 6 B
gx¿y¿
2
= -a
Ans.
ex - ey
2
b sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B
= 200 A 10 - 6 B
ey¿ =
ex + ey
= c
2
-
Ans.
ex - ey
2
cos 2u -
gxy
2
sin 2u
500 + 300
500 - 300
-200
cos ( -90°) - a
b sin (-90°) d A 10 - 6 B
2
2
2
= 300 A 10 - 6 B
Ans.
The deformed element for this state of strain is shown in Fig. a.
750
gxy
2
dx
x
Pxdx
10–13. The state of plane strain on an element is
Px = -300110-62, Py = 0, and gxy = 150110-62. Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states
of strain with respect to the original element.
y
gxy
dy 2
x
In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We
obtain
ex + ey
e1, 2 =
2
= C
;
C
¢
ex - ey
2
2
≤ + ¢
gxy
2
≤
2
-300 - 0 2
150 2
-300 + 0
;
¢
≤ + ¢
≤ S A 10 - 6 B
2
C
2
2
= (-150 ; 167.71) A 10 - 6 B
e1 = 17.7 A 10 - 6 B
e2 = - 318 A 10 - 6 B
Ans.
Orientation of Principal Strain:
tan 2up =
gxy
ex - ey
=
150 A 10 - 6 B
(-300 - 0) A 10 - 6 B
= -0.5
uP = -13.28° and 76.72°
Substituting u = -13.28° into Eq. 9-1,
ex¿ =
ex + ey
= c
+
2
ex - ey
2
cos 2u +
gxy
2
sin 2u
-300 + 0
-300 - 0
150
+
cos (-26.57°) +
sin (-26.57°) d A 10 - 6 B
2
2
2
= -318 A 10 - 6 B = e2
Thus,
A uP B 1 = 76.7° and A uP B 2 = - 13.3°
Ans.
The deformed element of this state of strain is shown in Fig. a.
Maximum In-Plane Shear Strain:
gmax
ex - ey 2
gxy 2
in-plane
=
¢
≤ + ¢ ≤
2
C
2
2
gmax
in-plane
-300 - 0 2
150 2
-6
-6
b + a
b R A 10 B = 335 A 10 B
A
2
2
= B2
a
Ans.
Orientation of the Maximum In-Plane Shear Strain:
tan 2us = - ¢
ex - ey
gxy
≤ = -C
(-300 - 0) A 10 - 6 B
150 A 10 - 6 B
S = 2
us = 31.7° and 122°
Ans.
751
gxy
2
dx
Pxdx
10–13.
Continued
The algebraic sign for
gx¿y¿
2
= -¢
ex - ey
2
gmax
in-plane
≤ sin 2u +
when u = us = 31.7° can be obtained using
gxy
2
cos 2u
gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B
= 335 A 10 - 6 B
Average Normal Strain:
eavg =
ex + ey
2
= a
-300 + 0
b A 10 - 6 B = -150 A 10 - 6 B
2
Ans.
The deformed element for this state of strain is shown in Fig. b.
752
10–14. The state of strain at the point on a boom of an
hydraulic engine crane has components of Px = 250110-62,
Py = 300110-62, and gxy = - 180110-62. Use the straintransformation equations to determine (a) the in-plane
principal strains and (b) the maximum in-plane shear strain
and average normal strain. In each case, specify the
orientation of the element and show how the strains deform
the element within the x–y plane.
y
a)
In-Plane Principal Strain: Applying Eq. 10–9,
ex + ey
e1, 2 =
;
2
= B
a
A
ex - ey
2
b + a
2
gxy
2
b
2
250 + 300
-180 2
250 - 300 2
-6
;
a
b + a
b R A 10 B
2
A
2
2
= 275 ; 93.41
e1 = 368 A 10 - 6 B
e2 = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: Applying Eq. 10–8,
gxy
tan 2uP =
ex - ey
=
-180(10 - 6)
(250 - 300)(10 - 6)
uP = 37.24°
and
= 3.600
-52.76°
Use Eq. 10–5 to determine which principal strain deforms the element in the x¿
direction with u = 37.24°.
ex¿ =
ex + ey
= c
2
+
ex - ey
2
cos 2u +
gxy
2
sin 2u
250 + 300
250 - 300
-180
+
cos 74.48° +
sin 74.48° d A 10 - 6 B
2
2
2
= 182 A 10 - 6 B = e2
Hence,
uP1 = -52.8°
and
uP2 = 37.2°
Ans.
b)
Maximum In-Plane Shear Strain: Applying Eq. 10–11,
g max
ex - ey 2
gxy 2
in-plane
=
a
b + a
b
2
A
2
2
g
max
in-plane
= 2B
250 - 300 2
-180 2
-6
b + a
b R A 10 B
A
2
2
a
= 187 A 10 - 6 B
Ans.
753
x
10–14. Continued
Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10,
tan 2us = -
ex - ey
gxy
us = -7.76°
and
The proper sign of
gx¿y¿
2
= -
= -
ex - ey
2
g
max
in-plane
250 - 300
= -0.2778
-180
Ans.
82.2°
can be determined by substituting u = -7.76° into Eq. 10–6.
sin 2u +
gxy
2
cos 2u
gx¿y¿ = {-[250 - 300] sin (-15.52°) + ( -180) cos (-15.52°)} A 10 - 6 B
= -187 A 10 - 6 B
Normal Strain and Shear strain: In accordance with the sign convention,
ex = 250 A 10 - 6 B
ey = 300 A 10 - 6 B
gxy = -180 A 10 - 6 B
Average Normal Strain: Applying Eq. 10–12,
eavg =
ex + ey
2
= c
250 + 300
d A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
754
*10–16. The state of strain at a point on a support
has components of Px = 350110-62, Py = 400110-62,
gxy = -675110-62. Use the strain-transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x–y plane.
a)
e1, 2 =
=
ex + ey
;
2
B
a
ex -ey
2
b + a
2
gxy
2
b
2
350 - 400 2
-675 2
350 + 400
;
a
b + a
b
2
A
2
2
e1 = 713(10 - 6)
Ans.
e2 = 36.6(10 - 6)
Ans.
tan 2uP =
gxy
ex - ey
=
-675
(350 - 400)
uP = 42.9°
Ans.
b)
(gx¿y¿)max
2
(gx¿y¿)max
2
=
=
A
A
a
ex - ey
a
350 - 400 2
-675 2
b + a
b
2
2
2
b + a
2
gxy
2
b
2
(gx¿y¿)max = 677(10 - 6)
eavg =
ex + ey
tan 2us =
2
=
350 + 400
= 375(10 - 6)
2
-(ex - ey)
gxy
Ans.
=
Ans.
350 - 400
675
us = -2.12°
Ans.
755
•10–17.
Solve part (a) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C (-30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6)
Ans.
e2 = (-30 - 167.71)(10 - 6) = - 198(10 - 6)
Ans.
tan 2uP = a
Ans.
75
b,
30 + 120
uP = 13.3°
756
10–18. Solve part (b) of Prob. 10–4 using Mohr’s circle.
ex = 120(10 - 6)
ey = -180(10 - 6)
gxy = 150(10 - 6)
A (120, 75)(10 - 6) C ( -30, 0)(10 - 6)
R = C 2[120 - (-30)]2 + (75)2 D (10 - 6)
= 167.71 (10 - 6)
gxy
max
2 in-plane
gxy
= R = 167.7(10 - 6)
max
in-plane
= 335(10 - 6)
Ans.
eavg = -30 (10 - 6)
tan 2us =
120 + 30
75
Ans.
us = -31.7°
Ans.
757
10–19. Solve Prob. 10–8 using Mohr’s circle.
ex = -200(10 - 6)
ey = -650(10 - 6)
gxy = -175(10 - 6)
gxy
2
= -87.5(10 - 6)
u = 20°, 2u = 40°
A(-200, -87.5)(10 - 6)
C(-425, 0)(10 - 6)
R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6)
tan a =
87.5
;
-200 - (-425)
a = 21.25°
f = 40 + 21.25 = 61.25°
ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6)
Ans.
ey¿ = ( -425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6)
Ans.
-gx¿y¿
2
= 241.41(10 - 6) sin 61.25°
gx¿y¿ = -423(10 - 6)
Ans.
758
*10–20. Solve Prob. 10–10 using Mohr’s circle.
ex = 400(10 - 6)
A(400, 155)(10 - 6)
ey = -250(10 - 6)
gxy = 310(10 - 6)
gxy
2
= 155(10 - 6)
C(75, 0)(10 - 6)
R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6)
tan a =
155
;
400 - 75
a = 25.50°
f = 60 + 25.50 = 85.5°
ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6)
Ans.
ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6)
Ans.
gx¿y¿
2
= (360.1 sin 85.5°)(10 - 6)
gx¿y¿ = 718(10 - 6)
Ans.
759
u = 30°
•10–21.
Solve Prob. 10–14 using Mohr’s circle.
Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B ,
gxy
ey = 300 A 10 - 6 B , and
= -90 A 10 - 6 B . Hence,
2
eavg =
ex + ey
2
= a
250 + 300
b A 10 - 6 B = 275 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A(250, -90) A 10 - 6 B
C(275, 0) A 10 - 6 B
The radius of the circle is
R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408
In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B
Ans.
e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP2 =
90
= 3.600
275 - 250
2uP2 = 74.48°
2uP1 = 180° - 2uP2
uP1 =
180° - 74.78°
= 52.8° (Clockwise)
2
Ans.
Maximum In-Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
g
= -R = -93.408 A 10 - 6 B
max
in-plane
= -187 A 10 - 6 B
Ans.
Orientation of the Maximum In-Plane Shear Strain: From the circle,
tan 2us =
275 - 250
= 0.2778
90
us = 7.76° (Clockwise)
Ans.
760
10–22. The strain at point A on the bracket has
components Px = 300110-62, Py = 550110-62, gxy =
-650110-62. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 300(10 - 6)
ey = 550(10 - 6)
A(300, -325)10 - 6
gxy = -650(10 - 6)
y
gxy
2
= -325(10 - 6)
A
C(425, 0)10 - 6
R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6)
a)
e1 = (425 + 348.2)(10 - 6) = 773(10 - 6)
Ans.
e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)
Ans.
b)
g
max
in-plane
= 2R = 2(348.2)(10 - 6) = 696(10 - 6)
Ans.
773(10 - 6)
;
2
Ans.
c)
gabs
max
=
2
gabs
max
= 773(10 - 6)
10–23. The strain at point A on the leg of the angle has
components Px = - 140110-62, Py = 180110-62, gxy =
-125110-62. Determine (a) the principal strains at A in the
x– y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = -140(10 - 6)
A( -140, -62.5)10 - 6
ey = 180(10 - 6)
gxy = -125(10 - 6)
A
gxy
2
= -62.5(10 - 6)
C(20, 0)10 - 6
R = A 2(20 - (-140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6)
a)
e1 = (20 + 171.77)(10 - 6) = 192(10 - 6)
Ans.
e2 = (20 - 171.77)(10 - 6) = - 152(10 - 6)
Ans.
(b, c)
gabs
max
=
g
max
in-plane
= 2R = 2(171.77)(10 - 6) = 344(10 - 6)
Ans.
761
x
*10–24. The strain at point A on the pressure-vessel wall
has components Px = 480110-62, Py = 720110-62, gxy =
650110-62. Determine (a) the principal strains at A, in the
x–y plane, (b) the maximum shear strain in the x–y plane,
and (c) the absolute maximum shear strain.
ex = 480(10 - 6)
ey = 720(10 - 6)
A(480, 325)10 - 6
C(600, 0)10 - 6
gxy = 650(10 - 6)
y
A
gxy
2
= 325(10 - 6)
R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6)
a)
e1 = (600 + 346.44)10 - 6 = 946(10 - 6)
Ans.
e2 = (600 - 346.44)10 - 6 = 254(10 - 6)
Ans.
b)
g
max
in-plane
= 2R = 2(346.44)10 - 6 = 693(10 - 6)
Ans.
946(10 - 6)
;
2
Ans.
c)
gabs
max
2
=
gabs
max
= 946(10 - 6)
762
x
•10–25.
The 60° strain rosette is mounted on the bracket.
The following readings are obtained for each gauge:
Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62.
Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In
each case show the deformed element due to these strains.
b
c
60⬚
60⬚
This is a 60° strain rosette Thus,
ex = ea = -100(10 - 6)
1
A 2eb + 2ec - ea B
3
ey =
1
C 2(250) + 2(150) - (-100) D (10 - 6)
3
=
= 300(10 - 6)
gxy =
2
23
(eb - ec) =
2
23
(250 - 150)(10 - 6) = 115.47(10 - 6)
In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6)
gxy
and
= 57.74(10 - 6).
2
Thus,
eavg =
ex + ey
2
= a
-100 + 300
b(10 - 6) = 100(10 - 6)
2
Ans.
Then, the coordinates of reference point A and Center C of the circle are
A( -100, 57.74)(10 - 6)
C(100, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2(-100 - 100)2 + 208.16 b(10 - 6) = 208.17(10 - 6)
Using these result, the circle is shown in Fig. a.
The coordinates of points B and D represent e1 and e2 respectively.
e1 = (100 + 208.17)(10 - 6) = 308(10 - 6)
Ans.
e2 = (100 - 208.17)(10 - 6) = -108(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)2 =
57.74(10 - 6)
(100 + 100)(10 - 6)
= 0.2887
A uP B 2 = 8.05° (Clockwise)
Ans.
The deformed element for the state of principal strain is shown in Fig. b.
763
a
10–25.
Continued
gmax
The coordinates for point E represent eavg and
gmax
in-plane
2
in-plane
2
. Thus,
= R = 208.17(10 - 6)
gmax
in-plane
= 416(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2us =
100 + 100
57.74
us = 36.9° (Counter Clockwise)
Ans.
The deformed element for the state of maximum In-plane shear strain is shown in
Fig. c.
764
10–26. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
a
30⬚ 30⬚
c
With ua = 60°, ub = 120° and uc = 180°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60°
gxy = 200(10 - 6)
0.25ex + 0.75ey + 0.4330
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120°
gxy = -450(10 - 6)
0.25ex + 0.75ey - 0.4330
(2)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180°
ex = 250(10 - 6)
Substitute this result into Eqs. (1) and (2) and solve them,
ey = -250 (10 - 6)
gxy = 750.56 (10 - 6)
In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6),
gxy
and
= 375.28(10 - 6), Thus,
2
eavg =
ex + ey
2
= c
250 + ( -250)
d(10 - 6) = 0
2
Ans.
Then, the coordinates of the reference point A and center C of the circle are
A(250, 375.28)(10 - 6)
C(0, 0)
Thus, the radius of the circle is
R = CA = A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6)
Using these results, the circle is shown in Fig. a.
The coordinates for points B and D represent e1 and e2, respectively. Thus,
e1 = 451(10 - 6)
e2 = -451(10 - 6)
Ans.
Referring to the geometry of the circle,
tan 2(uP)1 =
375.28
= 1.5011
250
(uP)1 = 28.2° (Counter Clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
765
60⬚
10–26.
Continued
gmax
in-plane
The coordinates of point E represent eavg and
. Thus,
2
gmax
in-plane
gmax
= R = 450.92(10 - 6)
= 902(10 - 6)
in-plane
2
Ans.
Referring to the geometry of the circle,
tan 2us =
250
= 0.6662
375.28
us = 16.8° (Clockwise)
Ans.
766
10–27. The 45° strain rosette is mounted on a steel shaft.
The following readings are obtained from each gauge:
Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case show the deformed element due to these strains.
b
c
45⬚
With ua = 45°, ub = 90° and uc = 135°,
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45°
ex + ey + gxy = 600(10 - 6)
(1)
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
-250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90°
ey = -250(10 - 6)
ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc
-450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135°
ex + ey - gxy = -900(10 - 6)
(2)
Substitute the result of ey into Eq. (1) and (2) and solve them
ex = 100(10 - 6)
gxy = 750(10 - 6)
In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6)
gxy
and
= 375(10 - 6). Thus,
2
eavg =
ex + ey
2
= c
100 + (-250)
d(10 - 6) = -75(10 - 6)
2
Ans.
Then, the coordinates of the reference point A and the center C of the circle are
A(100, 375)(10 - 6)
C(-75, 0)(10 - 6)
Thus, the radius of the circle is
R = CA = a 2 C 100 - ( -75) D 2 + 3752 b (10 - 6) = 413.82(10 - 6)
Using these results, the circle is shown in Fig. a.
The Coordinates of points B and D represent e1 and e2, respectively. Thus,
e1 = A -75 + 413.82 B (10 - 6) = 339(10 - 6)
Ans.
e2 = A -75 - 413.82 B (10 - 6) = - 489(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2(uP)1 =
375
= 2.1429
100 + 75
(uP)1 = 32.5° (Counter Clockwise)
Ans.
767
45⬚
a
10–27.
Continued
The deformed element for the state of principal strains is shown in Fig. b.
gmax
The coordinates of point E represent eavg and
gmax
in-plane
2
= R = 413.82(106)
in-plane
gmax
2
in-plane
. Thus
= 828(10 - 6)
Ans.
Referring to the geometry of the circle
tan 2us =
-100 + 75
= 0.4667
375
us = 12.5° (Clockwise)
Ans.
768
*10–28. The 45° strain rosette is mounted on the link of
the backhoe. The following readings are obtained from
each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and associated average
normal strain.
a
45⬚
b
ea = 650(10 - 6);
ua = 180°;
eb = -300(10 - 6);
ub = 225°
c
uc = 270°
Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u
650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°)
ex = 650 (10 - 6)
480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°)
ey = 480 (10 - 6)
-300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°)
gxy = -1730 (10 - 6)
Therefore, ex = 650 (10 - 6)
gxy
2
ey = 480 (10 - 6)
gxy = -1730 (10 - 6)
= -865 (10 - 6)
Mohr’s circle:
A(650, -865) 10 - 6
C(565, 0) 10 - 6
R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6)
(a)
(b)
e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6)
Ans.
e2 = [565 - 869.17]10 - 6 = -304 (10 - 6)
Ans.
g
max
in-plane
45⬚
ec = 480(10 - 6)
= 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6)
Ans.
eavg = 565(10 - 6)
Ans.
769
10–30. For the case of plane stress, show that Hooke’s law
can be written as
sx =
E
E
1Px + nPy2, sy =
1Py + nPx2
11 - n22
11 - n22
Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18,
ex =
1
A sx - v sy B
E
vEex = A sx - v sy B v
vEex = v sx - v2 sy
ey =
[1]
1
(sy - v sx)
E
E ey = -v sx + sy
[2]
Adding Eq [1] and Eq.[2] yields.
vE ex - E ey = sy - v2 sy
sy =
E
A vex + ey B
1 - v2
(Q.E.D.)
Substituting sy into Eq. [2]
E ey = -vsx +
sx =
=
=
E
A v ex + ey B
1 - v2
E A v ex + ey B
v (1 - v )
2
-
Eey
v
E v ex + E ey - E ey + Eey v2
v(1 - v2)
E
(ex + v ey)
1 - v2
(Q.E.D.)
770
10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the
stress-transformation equations, Eqs. 9–1 and 9–2.
Stress transformation equations:
sx + sy
sx¿ =
2
tx¿y¿ =
sy¿ =
+
sx - sy
2
2
cos 2u + txy sin 2u
(1)
sin 2u + txy cos 2u
2
sx + sy
sx - sy
-
sx - sy
2
(2)
cos 2u - txy sin 2u
(3)
Hooke’s Law:
ex =
v sy
sx
E
E
(4)
ey =
sy
-v sx
+
E
E
(5)
txy = G gxy
G =
(6)
E
2 (1 + v)
(7)
From Eqs. (4) and (5)
ex + ey =
ex - ey =
(1 - v)(sx + sy)
(8)
E
(1 + v)(sx - sy)
(9)
E
From Eqs. (6) and (7)
txy =
E
g
2 (1 + v) xy
(10)
From Eq. (4)
ex¿ =
v sy¿
sx¿
E
E
(11)
Substitute Eqs. (1) and (3) into Eq. (11)
ex¿ =
(1 - v)(sx - sy)
2E
+
(1 + v)(sx - sy)
2E
cos 2u +
(1 + v)txy sin 2u
E
(12)
By using Eqs. (8), (9) and (10) and substitute into Eq. (12),
ex¿ =
ex + ey
2
+
ex - ey
2
cos 2u +
gxy
2
QED
sin 2u
771
10–31.
Continued
From Eq. (6).
gx¿y¿ = G gx¿y¿ =
E
gx¿y¿
2 (1 + v)
(13)
Substitute Eqs. (13), (6) and (9) into Eq. (2),
E (ex - ey)
E
E
= g
sin 2u +
g cos 2u
2 (1 + v) x¿y¿
2 (1 + v)
2 (1 + v) xy
gx¿y¿
2
= -
(ex - ey)
2
sin 2u +
gxy
2
QED
cos 2u
*10–32. A bar of copper alloy is loaded in a tension
machine and it is determined that Px = 940110-62 and
sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of
elasticity, Ecu, and the dilatation, ecu, of the copper.
ncu = 0.35.
ex =
1
[sx - v(sy + sz)]
E
940(10 - 6) =
ecu =
1
[14(103) - 0.35(0 + 0)]
Ecu
Ecu = 14.9(103) ksi
Ans.
1 - 2(0.35)
1 - 2v
(14 + 0 + 0) = 0.282(10 - 3)
(sx + sy + sz) =
E
14.9(103)
Ans.
•10–33.
The principal strains at a point on the aluminum
fuselage of a jet aircraft are P1 = 780110-62 and P2 =
400110-62. Determine the associated principal stresses at
the point in the same plane. Eal = 1011032 ksi, nal = 0.33.
Hint: See Prob. 10–30.
Plane stress, s3 = 0
See Prob 10-30,
s1 =
=
s2 =
=
E
(e1 + ve2)
1 - v2
10(103)
1 - 0.332
(780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi
Ans.
E
(e2 + ve1)
1 - v2
10(103)
1 - 0.332
(400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi
Ans.
772
10–34. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the absolute maximum shear strain in the
rod at a point on its surface.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
P
=
A
p
4
700
= 2.228 MPa
(0.022)
Normal Strain: Applying the generalized Hooke’s Law.
ex =
=
1
C sx - v A sy + sz B D
E
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C s - v(sx + sz) D
E y
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C sz - v A sx + sy B D
E
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Therefore.
emax = 30.48 A 10 - 6 B
emin = -10.67 A 10 - 6 B
Absolute Maximum Shear Strain:
gabs
max
= emax - emin
= [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B
Ans.
773
700 N
10–35. The rod is made of aluminum 2014-T6. If it is
subjected to the tensile load of 700 N and has a diameter of
20 mm, determine the principal strains at a point on the
surface of the rod.
700 N
Normal Stress: For uniaxial loading, sy = sz = 0.
sx =
P
=
A
p
4
700
= 2.228 MPa
(0.022)
Normal Strains: Applying the generalized Hooke’s Law.
ex =
=
1
C sx - v A sy + sz B D
E
1
C 2.228 A 106 B - 0 D
73.1(109)
= 30.48 A 10 - 6 B
ey =
=
1
C sy - v(sx + sz) D
E
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
ez =
=
1
C sz - v A sx + sy B D
E
1
C 0 - 0.35 A 2.228 A 106 B + 0 B D
73.1(109)
= -10.67 A 10 - 6 B
Principal Strains: From the results obtained above,
emax = 30.5 A 10 - 6 B
eint = emin = -10.7 A 10 - 6 B
Ans.
774
700 N
*10–36. The steel shaft has a radius of 15 mm. Determine
the torque T in the shaft if the two strain gauges, attached to
the surface of the shaft, report strains of Px¿ = -80110-62
and Py¿ = 80110-62. Also, compute the strains acting in the x
and y directions. Est = 200 GPa, nst = 0.3.
ex¿ = -80(10 - 6)
y
T
ex = ey = 0
Ans.
ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u
u = 45°
-80(10 - 6) = 0 + 0 + gxy sin 45° cos 45°
gxy = -160(10 - 6)
Ans.
Also, u = 135°
80(10 - 6) = 0 + 0 + g sin 135° cos 135°
gxy = -160(10 - 6)
200(109)
E
=
= 76.923(109)
2(1 + V)
2(1 + 0.3)
t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa
p
12.308(106) A B (0.015)4
2
= 65.2 N # m
0.015
Ans.
10–37. Determine the bulk modulus for each of the
following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and
(b) glass, Eg = 811032 ksi, ng = 0.24.
a) For rubber:
Kr =
Er
0.4
=
= 3.33 ksi
3 (1 - 2 vr)
3[1 - 2(0.48)]
Ans.
b) For glass:
Kg =
Eg
3 (1 - 2 vg)
=
x
T
Pure shear
tJ
T =
=
c
x¿
45⬚
ey¿ = 80(10 - 6)
G =
y¿
8(103)
= 5.13 (103) ksi
3[1 - 2(0.24)]
Ans.
775
10–38. The principal stresses at a point are shown in the
figure. If the material is A-36 steel, determine the
principal strains.
12 ksi
e1 =
1
1
e 12 - 0.32 C 8 + (-20) D f = 546 (10-6)
C s1 - v(s2 + s3) D =
E
29.0(103)
e2 =
1
1
e 8 - 0.32 C 12 + (-20) D f = 364 (10-6)
C s2 - v(s1 + s3) D =
E
29.0(103)
e3 =
20 ksi
8 ksi
1
1
C s3 - v(s1 + s2) D =
C -20 - 0.32(12 + 8) D = -910 (10-6)
E
29.0(103)
emax = 546 (10 - 6)
eint = 346 (10 - 6)
emin = -910 (10 - 6)
Ans.
10–39. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel
is pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
Est = 200 GPa and nst = 0.3.
20 mm
r
1000
=
= 100 7 10, the thin wall analysis is valid to
t
10
determine the normal stress in the wall of the spherical vessel. This is a plane stress
Normal Stresses: Since
problem where smin = 0 since there is no load acting on the outer surface of the wall.
smax = slat =
pr
p(1000)
=
= 50.0p
2t
2(10)
[1]
Normal Strains: Applying the generalized Hooke’s Law with
emax = elat =
0.012
= 0.600 A 10 - 3 B mm>mm
20
emax =
1
C smax - V (slat + smin) D
E
0.600 A 10 - 3 B =
1
[50.0p - 0.3 (50.0p + 0)]
200(104)
p = 3.4286 MPa = 3.43 MPa
Ans.
From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa
Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the
result, the state of stress is the same consisting of two normal stresses with zero
shear stress regardless of the orientation of the element.
t
= 0
Ans.
smax - smin
171.43 - 0
=
= 85.7MPa
2
2
Ans.
max
in-plane
Absolute Maximum Shear Stress:
tabs
max
=
776
*10–40. The strain in the x direction at point A on the
steel beam is measured and found to be Px = -100110-62.
Determine the applied load P. What is the shear strain gxy
at point A? Est = 2911032 ksi, nst = 0.3.
P
y
3 in.
A
3 ft
1
1
(5.5)(83) = 129.833 in4
(6)(9)3 12
12
Ix =
QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3
s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi
My
,
I
s =
2.90 =
1.5P(12)(1.5)
129.833
P = 13.945 = 13.9 kip
tA =
VQ
0.5(13.945)(16.1875)
=
= 1.739 ksi
It
129.833(0.5)
G =
29(103)
E
=
= 11.154(103) ksi
2(1 + v)
2(1 + 0.3)
gxy =
txy
G
=
Ans.
1.739
= 0.156(10 - 3) rad
11.154(103)
Ans.
777
x
4 ft
7 ft
•10–41.
The cross section of the rectangular beam is
subjected to the bending moment M. Determine an
expression for the increase in length of lines AB and CD.
The material has a modulus of elasticity E and Poisson’s
ratio is n.
C
D
B
h
For line AB,
sz = ey = ¢LAB
A
M
12My
My
My
= = 1
3
I
b h3
12 b h
v sz
=
E
b
12 v My
E b h3
h
2
h
2
12 v M
=
ey dy =
y dy
E b h3 L0
L0
=
3vM
2Ebh
Ans.
For line CD,
sz = ex = -
M h2
Mc
6M
= - 2
= - 1
3
I
bh
b
h
12
v sz
E
=
6vM
E b h2
¢LCD = ex LCD =
=
6vM
(b)
E b h2
6vM
E h2
Ans.
10–42. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
26 ksi
ex =
1
1
(sx - v(sy + sz)) =
(10 - 0.33(-15 - 26)) = 2.35(10 - 3)
E
10(103)
ey =
1
1
(sy - v(sx + sz)) =
(-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans.
E
10(103)
ez =
Ans.
1
1
(sz - v(sx + sy)) =
(-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans.
E
10(103)
778
15 ksi
10 ksi
10–43. A single strain gauge, placed on the outer surface
and at an angle of 30° to the axis of the pipe, gives a reading
at point A of Pa = -200(10-6). Determine the horizontal
force P if the pipe has an outer diameter of 2 in. and an
inner diameter of 1 in. The pipe is made of A-36 steel.
1.5 ft
Using the method of section and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - p = 0
Vz = p
©Mx = 0;
Tx - p(1.5) = 0 Tx = 1.5p
©My = 0;
My - p(2.5) = 0 My = 2.5p
30⬚
A
The normal strees is due to bending only. For point A, z = 0. Thus
sx =
My z
Iy
= 0
The shear stress is the combination of torsional shear stress and transverse shear
stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A
tt =
1.5p(12)(1)
38.4 p
Txc
=
=
p
J
0.46875p
Referring to Fig. b,
(QA)z = y1œ A1œ - y2œ A2œ =
4 (1) p 2
4(0.5) p
c (1 ) d c (0.52) d
3p 2
3p 2
= 0.5833 in3
Iy =
p
4
(14 - 0.54) = 0.234375 p in4
Combine these two shear stress components,
t = tt + tv =
P
2.5 ft
38.4P
2.4889P
40.8889P
+
=
p
p
p
Since no normal stress acting on point A, it is subjected to pure shear which can be
represented by the element shown in Fig. c.
For pure shear, ex = ez = 0,
ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
Applying the Hooke’s Law for shear,
txz = G gxz
40.8889P
= 11.0(103) C 461.88(10 - 6) D
p
P = 0.3904 kip = 390 lb
Ans.
779
*10–44. A single strain gauge, placed in the vertical plane
on the outer surface and at an angle of 30° to the axis of the
pipe, gives a reading at point A of Pa = -200(10-6).
Determine the principal strains in the pipe at point A. The
pipe has an outer diameter of 2 in. and an inner diameter of
1 in. and is made of A-36 steel.
1.5 ft
P
2.5 ft
Using the method of sections and consider the equilibrium of the FBD of the pipe’s
upper segment, Fig. a,
©Fz = 0;
Vz - P = 0
Vz = P
©Mx = 0;
Tx - P(1.5) = 0 Tx = 1.5P
©My = 0;
My - P(2.5) = 0 My = 2.5P
By observation, no normal stress acting on point A. Thus, this is a case of pure shear.
For the case of pure shear,
ex = ez = ey = 0
ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua
-200(10 - 6) = 0 + 0 + gxz sin 150° cos 150°
gxz = 461.88(10 - 6)
e1, 2 =
ex + ez
= B
2
+
A
a
ex - ez
2
b + a
2
gxz
2
b
2
0 + 0
461.88 2
0 - 0 2
-6
;
a
b + a
b R (10 )
2
A
2
2
e1 = 231(10 - 6)
e2 = - 231(10 - 6)
Ans.
780
30⬚
A
10–45. The cylindrical pressure vessel is fabricated using
hemispherical end caps in order to reduce the bending stress
that would occur if flat ends were used. The bending stresses
at the seam where the caps are attached can be eliminated
by proper choice of the thickness th and tc of the caps and
cylinder, respectively. This requires the radial expansion to
be the same for both the hemispheres and cylinder. Show
that this ratio is tc>th = 12 - n2>11 - n2.Assume that the
vessel is made of the same material and both the cylinder
and hemispheres have the same inner radius. If the cylinder
is to have a thickness of 0.5 in., what is the required thickness
of the hemispheres? Take n = 0.3.
tc
th
r
For cylindrical vessel:
s1 =
pr
;
tc
e1 =
1
[s1 - v (s2 + s3)]
E
=
s2 =
pr
2 tc
s3 = 0
vpr
pr
1 pr
1
b =
a1 - v b
a
E tc
2 tc
E tc
2
d r = e1 r =
p r2
1
a1 - v b
E tc
2
(1)
For hemispherical end caps:
s1 = s2 =
e1 =
=
pr
2 th
1
[s1 - v (s2 + s3)] ;
E
s3 = 0
vpr
pr
1 pr
b =
(1 - v)
a
E 2 th
2 th
2 E th
d r = e1 r =
p r2
(1 - v)
2 E th
(2)
Equate Eqs. (1) and (2):
p r2
p r2
1
a1 - vb =
(1 - v)
E tc
2
2 E th
2 (1 - 12 v)
tc
2 - v
=
=
th
1 - v
1 - v
th =
QED
(1 - v) tc
(1 - 0.3) (0.5)
=
= 0.206 in.
2 - v
2 - 0.3
Ans.
781
10–46. The principal strains in a plane, measured
experimentally at a point on the aluminum fuselage of a jet
aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is
a case of plane stress, determine the associated principal
stresses at the point in the same plane. Eal = 10(103) ksi
and nal = 0.33.
Normal Stresses: For plane stress, s3 = 0.
Normal Strains: Applying the generalized Hooke’s Law.
e1 =
1
C s1 - v (s2 + s3) D
E
630 A 10 - 6 B =
1
[s1 - 0.33(s2 + 0)]
10(103)
6.30 = s1 - 0.33s2
e2 =
[1]
1
C s2 - v (s1 + s3) D
E
350 A 10 - 6 B =
1
C s2 - 0.33(s1 + 0) D
10(103)
3.50 = s2 - 0.33s1
[2]
Solving Eqs.[1] and [2] yields:
s1 = 8.37 ksi
s2 = 6.26 ksi
Ans.
10–47. The principal stresses at a point are shown in
the figure. If the material is aluminum for which
Eal = 1011032 ksi and nal = 0.33, determine the principal
strains.
3 ksi
e1 =
1
1
e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6)
C s1 - v(s2 + s3) D =
E
10(103)
e2 =
1
1
e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6)
C s2 - v(s1 + s3) D =
E
10(103)
e3 =
1
1
C s3 - v(s1 + s2) D =
C -4 - 0.33(8 + 3) D = -763 (10 - 6)
E
10(103)
Using these results,
e1 = 833(10 - 6)
e2 = 168(10 - 6)
e3 = -763(10 - 6)
782
8 ksi
4 ksi
*10–48. The 6061-T6 aluminum alloy plate fits snugly into
the rigid constraint. Determine the normal stresses sx and
sy developed in the plate if the temperature is increased by
¢T = 50°C. To solve, add the thermal strain a¢T to the
equations for Hooke’s Law.
y
400 mm
300 mm
x
Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the
rigid constraint along the x and y directions, ex = ey = 0. However, the plate is
allowed to have free expansion along the z direction. Thus, sz = 0. With the
additional thermal strain term, we have
ex =
0 =
1
c sx - v A sy + sz B d + a¢T
E
1
68.9 A 109 B
csx - 0.35 A sy + 0 B d + 24a10 - 6 b(50)
sx - 0.35sy = -82.68 A 106 B
ey =
0 =
(1)
1
C sy - v A sx + sz B D + a¢T
E
1
68.9a 109 b
C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50)
sy - 0.35sx = -82.68 A 106 B
(2)
Solving Eqs. (1) and (2),
sx = sy = -127.2 MPa = 127.2 MPa (C)
Ans.
Since sx = sy and sy 6 sY, the above results are valid.
783
•10–49.
Initially, gaps between the A-36 steel plate and
the rigid constraint are as shown. Determine the normal
stresses sx and sy developed in the plate if the temperature
is increased by ¢T = 100°F. To solve, add the thermal
strain a¢T to the equations for Hooke’s Law.
y
0.0015 in.
6 in.
8 in.
0.0025 in.
x
Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid
constraint, the plate is allowed to expand before it comes in contact with the constraint.
dy
dx
0.0025
0.0015
ey =
=
=
= 0.3125 A 10 - 3 B and
= 0.25 A 10 - 3 B .
Thus, ex =
Lx
8
Ly
6
However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0.
With the additional thermal strain term, we have
ex =
1
csx - v A sy + sz B d + a¢T
E
0.3125a 10 - 3 b =
1
29.0 a10 b
C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100)
3
sx - 0.32sy = -10.0775
ey =
(1)
1
C s - v A sx + sz B D + a¢T
E y
0.25 A 10 - 3 B =
1
29.0 A 103 B
C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100)
sy - 0.32sx = -11.89
(2)
Solving Eqs. (1) and (2),
sx = -15.5 ksi = 15.5 ksi (C)
Ans.
sy = -16.8 ksi = 16.8 ksi (C)
Ans.
Since sx 6 sY and sy 6 sY, the above results are valid.
784
10–50. Two strain gauges a and b are attached to a plate
made from a material having a modulus of elasticity of
E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give
a reading of Pa = 450110-62 and Pb = 100110-62, determine
the intensities of the uniform distributed load wx and wy
acting on the plate. The thickness of the plate is 25 mm.
wy
b
45⬚
y
a
Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0.
With ua = 0 and ub = 45°, we have
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0.
ex =
1
C sx - v A sy + sz B D
E
450 A 10 - 6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sx - 0.35sy = 31.5 A 106 B
ey =
(1)
1
C s - v A sx + sz B D
E y
-250 A 10-6 B =
1
70 A 109 B
C sy - 0.35 A sy + 0 B D
sy - 0.35sx = -17.5 A 106 B
(2)
Solving Eqs. (1) and (2),
sy = -7.379 A 106 B N>m2
sx = 28.917 A 106 B N>m2
Then,
wy = syt = -7.379 A 106 B (0.025) = -184 N>m
Ans.
wx = sxt = 28.917 A 106 B (0.025) = 723 N>m
Ans.
785
z
x
wx
10–51. Two strain gauges a and b are attached to the
surface of the plate which is subjected to the uniform
distributed load wx = 700 kN>m and wy = -175 kN>m.
If the gauges give a reading of Pa = 450110-62 and
Pb = 100110-62, determine the modulus of elasticity E,
shear modulus G, and Poisson’s ratio n for the material.
wy
b
45⬚
y
Normal Stress and Strain: The normal stresses along the x, y, and z axes are
sx =
700 A 103 B
0.025
sy = -
a
= 28 A 106 B N>m2
175 A 103 B
0.025
= -7 A 106 B N>m2
z
sz = 0 (plane stress)
Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0°
and ub = 45°, we have
ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua
450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0
ex = 450 A 10 - 6 B
eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub
100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0
ey = -250 A 10 - 6 B
Generalized Hooke’s Law:
ex =
1
C sx - v A sy + sz B D
E
450 A 10 - 6 B =
1
B 28 A 106 B - v C -7 A 106 B + 0 D R
E
450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B
ey =
(1)
1
[s - v(sx + sz)]
E y
-250 A 10 - 6 B =
1
b -7 A 106 B - v C 28 A 106 B + 0 D r
E
250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B
(2)
Solving Eqs. (1) and (2),
E = 67.74 A 109 B N>m2 = 67.7 GPa
Ans.
v = 0.3548 = 0.355
Ans.
Using the above results,
G =
67.74 A 109 B
E
=
2(1 + v)
2(1 + 0.3548)
= 25.0 A 109 B N>m2 = 25.0 GPa
Ans.
786
x
wx
*10–52. The block is fitted between the fixed supports. If the
glued joint can resist a maximum shear stress of tallow = 2 ksi,
determine the temperature rise that will cause the joint to fail.
Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with
an additional strain term of a¢T (Eq. 4–4).
40⬚
Normal Strain: Since the aluminum is confined along the y direction by the rigid
frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with
the additional thermal strain,
ey =
0 =
1
C sy - v(sx + sz) D + a¢T
E
1
C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T)
10.0(103)
sy = -0.06¢T
Construction of the Circle: In accordance with the sign convention. sx = 0,
sy = -0.06¢T and txy = 0. Hence.
savg =
sx + sy
2
=
0 + ( -0.06¢T)
= -0.03¢T
2
The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0).
The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T
Stress on The inclined plane: The shear stress components tx¿y¿, are represented by
the coordinates of point P on the circle.
tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T
Allowable Shear Stress:
tallow = tx¿y¿
2 = 0.02954¢T
¢T = 67.7 °F
Ans.
787
z
•10–53. The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is covered and the
temperature is increased by 200°F, determine the stress
components sx , sy , and sz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container and
0.012
allowed to expand in the z direction, ex = ey = 0; whereas ez =
= 0.002.
6
Applying the generalized Hooke’s Law with the additional thermal strain,
ex =
0 =
1
C sx - v(sy + sz) D + a¢T
E
1
C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy - 0.35sz + 26.2
ey =
0 =
[1]
1
C sy - v(sx + sz) + a¢T
E
1
C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx - 0.35sz + 26.2
ez =
0.002 =
[2]
1
C sz - v A sx + sy B D + a¢T
E
1
C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sz - 0.35sx - 0.35sy + 6.20
[3]
Solving Eqs.[1], [2] and [3] yields:
sx = sy = -70.0 ksi
sz = -55.2 ksi
Ans.
788
x
z
10–54. The smooth rigid-body cavity is filled with liquid
6061-T6 aluminum. When cooled it is 0.012 in. from the top
of the cavity. If the top of the cavity is not covered and the
temperature is increased by 200°F, determine the strain
components Px , Py , and Pz in the aluminum. Hint: Use
Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).
0.012 in.
4 in.
4 in.
6 in.
y
Normal Strains: Since the aluminum is confined at its sides by a rigid container,
then
ex = ey = 0
Ans.
and since it is not restrained in z direction, sz = 0. Applying the generalized
Hooke’s Law with the additional thermal strain,
ex =
0 =
1
C sx - v A sy + sz B D + a¢T
E
1
C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sx - 0.35sy + 26.2
ey =
0 =
[1]
1
C s - v(sx + sz) D + a¢T
E y
1
C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200)
10.0(103)
0 = sy - 0.35sx + 26.2
[2]
Solving Eqs. [1] and [2] yields:
sx = sy = -40.31 ksi
ez =
=
1
C sz - v A sx + sy B D + a¢T
E
1
{0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200)
10.0(103)
= 5.44 A 10 - 3 B
Ans.
789
x
10–55. A thin-walled spherical pressure vessel having an
inner radius r and thickness t is subjected to an internal
pressure p. Show that the increase in the volume within
the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain
analysis.
pr
2t
s1 = s2 =
s3 = 0
e1 = e2 =
1
(s1 - vs2)
E
e1 = e2 =
pr
(1 - v)
2t E
e3 =
1
(-v(s1 + s2))
E
e3 = V =
v pr
tE
4pr3
3
V + ¢V =
4p
4pr3
¢r 3
(r + ¢r)3 =
(1 +
)
r
3
3
where ¢V V V, ¢r V r
V + ¢V eVol =
¢r
4p r3
b
a1 + 3
r
3
¢V
¢r
= 3a b
V
r
Since e1 = e2 =
eVol = 3e1 =
2p(r + ¢r) - 2p r
¢r
=
r
2p r
3pr
(1 - v)
2t E
¢V = VeVol =
2pp r4
(1 - v)
Et
QED
790
*10–56. A thin-walled cylindrical pressure vessel has an
inner radius r, thickness t, and length L. If it is subjected
to an internal pressure p, show that the increase in
its inner radius is dr = rP1 = pr211 - 12 n2>Et and the
increase in its length is ¢L = pLr112 - n2>Et. Using these
results, show that the change in internal volume becomes
dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are
small quantities, show further that the change in volume
per unit volume, called volumetric strain, can be written as
dV>V = pr12.5 - 2n2>Et.
Normal stress:
pr
;
t
s1 =
s2 =
pr
2t
Normal strain: Applying Hooke’s law
e1 =
=
1
[s1 - v (s2 + s3)],
E
vpr
pr
1
1 pr
a
b =
a1 - vb
E
t
2t
Et
2
d r = et r =
e2 =
=
s3 = 0
p r2
1
a1 - v b
Et
2
1
[s2 - v (s1 + s3)],
E
QED
s3 = 0
vpr
pr 1
1 pr
a
b =
a - vb
E 2t
t
Et 2
¢L = e2 L =
pLr 1
a - vb
Et
2
V¿ = p(r + e1 r)2 (L + e2L) ;
QED
V = p r2 L
dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L
QED
(1 + e1)2 = 1 + 2 e1 neglect e21 term
(1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term
dV
= 1 + e2 + 2 e1 - 1 = e2 + 2 e1
V
=
pr 1
2pr
1
a - vb +
a1 - v b
Et 2
Et
2
=
pr
(2.5 - 2 v)
Et
QED
791
10–57. The rubber block is confined in the U-shape
smooth rigid block. If the rubber has a modulus of elasticity
E and Poisson’s ratio n, determine the effective modulus of
elasticity of the rubber under the confined condition.
P
Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We
have
ex =
0 =
1
C sx - v A sy + sz B D
E
1
(sx - vsz)
E
sx = vsz
(1)
ez =
1
C s - v A sx + sy B D
E z
ez =
1
[s - v(sx + 0)]
E z
ez =
1
(sz - vsx)
E
(2)
Substituting Eq. (1) into Eq. (2),
ez =
sz
E
A 1 - v2 B
The effective modulus of elasticity of the rubber block under the confined condition
can be determined by considering the rubber block as unconfined but rather
undergoing the same normal strain of ez when it is subjected to the same normal
stress sz, Thus,
sz = Eeff ez
Eeff =
sz
ez
=
sz
sz
E
A 1 - v2 B
=
E
1 - v2
Ans.
792
z
10–58. A soft material is placed within the confines of a
rigid cylinder which rests on a rigid support. Assuming that
Px = 0 and Py = 0, determine the factor by which the
modulus of elasticity will be increased when a load is
applied if n = 0.3 for the material.
P
x
Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0.
Applying the generalized Hooke’s Law,
ex =
1
C sz - v(sy + sx) D
E
0 = sx - v(sy + sz)
ey =
[1]
1
C sy - v(sx + sz) D
E
0 = sy - v(sx + sz)
[2]
Solving Eqs.[1] and [2] yields:
sx = sy =
v
sz
1 - v
Thus,
ez =
=
=
1
C sz - v(sx + sy) D
E
v
v
1
csz - va
sz +
sz b d
E
1 - v
1 - v
sz
E
c1 -
2v2
d
1 - v
=
sz 1 - v - 2v2
c
d
E
1 - v
=
sz (1 + v)(1 - 2v
c
d
E
1 - v
Thus, when the material is not being confined and undergoes the same normal strain
of ez, then the requtred modulus of elasticity is
E¿ =
sz
ez
=
The increased factor is k =
1 - v
E
(1 - 2v)(1 + v)
E¿
1 - v
=
E
(1 - 2v)(1 + v)
=
1 - 0.3
[1 - 2(0.3)](1 + 0.3)
= 1.35
Ans.
793
y
10–59. A material is subjected to plane stress. Express
the distortion-energy theory of failure in terms of sx , sy ,
and txy .
Maximum distortion energy theory:
(s21 - s1 s2 + s22) = s2Y
s1,2 =
sx + sy
2
Let a =
;
sx + sy
2
s1 = a + b;
A
a
(1)
sx - sy
2
and b =
A
a
2
2
b + txy
sx - sy
2
2
2
b + txy
s2 = a - b
s21 = a2 + b2 + 2 a b;
s22 = a2 + b2 - 2 a b
s1 s2 = a2 - b2
From Eq. (1)
(a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y
(a2 + 3 b2) = s2Y
(sx + sy)2
4
+ 3
(sx - sy)2
4
+ 3 t2xy = s2Y
s2x + s2y - sxsy + 3 t2xy = s2Y
Ans.
*10–60. A material is subjected to plane stress. Express
the maximum-shear-stress theory of failure in terms of sx ,
sy , and txy . Assume that the principal stresses are of
different algebraic signs.
Maximum shear stress theory:
|s1 - s2| = sY
s1,2 =
sx + sy
2
` s1 - s2 ` = 2
(1)
;
A
a
A
a
sx - sy
2
sx - sy
2
2
2
b + txy
2
b + txy
2
From Eq. (1)
4 ca
sx - sy
2
2
b + t2xy d = s2Y
2
(sx - sy) + 4 t2xy = s2Y
Ans.
794
•10–61.
An aluminum alloy 6061-T6 is to be used for
a solid drive shaft such that it transmits 40 hp at 2400
rev>min. Using a factor of safety of 2 with respect to
yielding, determine the smallest-diameter shaft that can be
selected based on the maximum-shear-stress theory.
v = a2400
T =
rev
2p rad 1 min
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300 #
lb in.
=
=
p
v
80 p
Tc
J
Applying t =
t =
A 3300
p B c
p
2
=
c4
6600
p3 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = -t =
6600
p2 c3
Maximum shear stress theory: Both principal stresses have opposite sign, hence,
` s1 - s2 ` =
2a
sY
;
F.S.
37 (103)
6600
b
=
`
`
2
p2c3
c = 0.4166 in.
d = 0.833 in.
Ans.
10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory.
v = a2400
T =
2p rad 1 min
rev
ba
ba
b = 80 p rad>s
min
rev
60s
40 (550) (12)
P
3300
=
=
lb.in.
p
v
80 p
Applying t =
t =
A 3300
p B c
p
2
4
c
=
Tc
J
6600
p2 c3
The principal stresses:
s1 = t =
6600
;
p2 c3
s2 = - t = -
6600
p2 c3
The maximum distortion-energy theory:
s21 - s1 s2 + s22 = a
3B
sY 2
b
F.S.
37(103) 2
6600 2
=
a
b
R
2
p2 c3
c = 0.3971 in.
d = 0.794 in.
Ans.
795
10–63. An aluminum alloy is to be used for a drive shaft
such that it transmits 25 hp at 1500 rev>min. Using a factor
of safety of 2.5 with respect to yielding, determine the
smallest-diameter shaft that can be selected based on the
maximum-distortion-energy theory. sY = 3.5 ksi.
1500(2p)
= 50p
60
T =
P
v
T =
25(550)(12)
3300
=
p
50p
t =
Tc
,
J
t =
3300
p c
p 4
2c
s1 =
v =
J =
=
p 4
c
2
6600
p2c3
6600
p2c3
s2 =
s21 - s1 s2 + s22 = a
3a
-6600
p2c3
sY 2
b
F.S.
3.5(103) 2
6600 2
b
=
a
b
2.5
p2c3
c = 0.9388 in.
d = 1.88 in.
Ans.
*10–64. A bar with a square cross-sectional area is made
of a material having a yield stress of sY = 120 ksi. If the bar
is subjected to a bending moment of 75 kip # in., determine
the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with
respect to yielding.
Normal and Shear Stress: Applying the flexure formula,
s =
75 A a2 B
450
Mc
= 1 4 = 3
I
a
a
12
In-Plane Principal Stress: Since no shear stress acts on the element
s1 = sx =
450
a3
s2 = sy = 0
Maximum Distortion Energy Theory:
s21 - s1 s2 + s22 = s2allow
a
120 2
450 2
b
0
+
0
=
a
b
1.5
a3
a = 1.78 in.
Ans.
796
•10–65.
Solve Prob. 10–64 using the maximum-shearstress theory.
Normal and Shear Stress: Applying the flexure formula,
s =
75 A a2 B
Mc
450
= 1 4 = 3
I
a
a
12
In-Plane Principal Stress: Since no shear stress acts on the element.
s1 = sx =
450
a3
s2 = sx = 0
Maximum Shear Stress Theory:
|s2| = 0 6 sallow =
120
= 80.0 ksi
1.5
(O.K!)
|s1| = sallow
120
450
=
1.5
a3
a = 1.78 in.
Ans.
10–66. Derive an expression for an equivalent torque Te
that, if applied alone to a solid bar with a circular cross
section, would cause the same energy of distortion as
the combination of an applied bending moment M and
torque T.
t =
Te c
J
Principal stress:
s1 = tx ¿
ud =
s2 = -t
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
1 + v
1 + v 3 T2x c2
( 3 t2) =
a
b
3E
3E
J2
Bending moment and torsion:
s =
Mc
;
I
t =
Tc
J
Principal stress:
s1, 2 =
s1 =
s - 0 2
s + 0
2
;
a
b + t
2
A
2
s
s2
+
+ t2 ;
2
A4
s2 =
s
s2
+ t2
2
A4
797
10–66. Continued
Let a =
s
2
b =
s2
+ t2
A4
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s21 - s1 s2 + s22 = 3 b2 + a2
1 + v 2
(s1 - s1 s2 + s22)
3E
ud =
(ud)2 =
=
1 + v
1 + v 3 s2
s2
(3 b2 + a2) =
a
+ 3t2 +
b
3E
3E
4
4
c2(1 + v) M2
1 + v 2
3 T2
b
(s + 3 t2) =
a 2 +
3E
3E
I
J2
(ud)1 = (ud)2
c3(1 + v) 3 Tx 2
c2(1 + v) M2
3 T2
=
b
a 2 +
2
3E
3E
J
I
J2
For circular shaft
J
=
I
p
3
p
4
c4
c4
=2
Te =
J2 M2
+ T2
A I2 3
Te =
4 2
M + T2
A3
Ans.
10–67. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of
distortion as the combination of an applied bending
moment M and torque T.
Principal stresses:
s1 =
Me c
;
I
ud =
1 + v 2
(s1 - s1 s2 + s22)
3E
(ud)1 =
s2 = 0
1 + v M2e c2
a 2 b
3E
I
(1)
798
10–67. Continued
Principal stress:
s + 0
s - 0 2
3
;
a
b + t
2
A
2
s1, 2 =
s
s2
+
+ t2;
2
A4
s1 =
s2 =
s
s2
+ t2
2
A4
Distortion Energy:
s
s2
,b =
+ t2
A4
2
Let a =
s21 = a2 + b2 + 2 a b
s1 s2 = a2 - b2
s22 = a2 + b2 - 2 a b
s22 - s1 s2 + s22 = 3 b2 + a2
Apply s =
Mc
;
I
t =
Tc
J
1 + v
1 + v s2
3s2
(3 b2 + a2) =
a
+
+ 3 t2 b
3E
3E
4
4
(ud)2 =
1 + v M2 c2
3 T2 c2
1 + v 2
b
(s + 3 t2) =
a 2 +
3E
3E
I
J2
=
(2)
Equating Eq. (1) and (2) yields:
(1 + v) Me c2
1 + v M2 c2
3T2 c2
b
a 2 b =
a 2 +
3E
3E
I
I
J2
M2e
2
I
=
M1
3 T2
+
2
I
J2
M2e = M1 + 3 T2 a
I 2
b
J
For circular shaft
I
=
J
p
4
p
2
c4
4
c
=
1
2
1 2
Hence, M2e = M2 + 3 T2 a b
2
Me =
A
M2 +
3 2
T
4
Ans.
799
*10–68. The short concrete cylinder having a diameter of
50 mm is subjected to a torque of 500 N # m and an axial
compressive force of 2 kN. Determine if it fails according to
the maximum-normal-stress theory. The ultimate stress of
the concrete is sult = 28 MPa.
A =
p
(0.05)2 = 1.9635(10 - 3) m2
4
J =
p
(0.025)4 = 0.61359(10 - 4) m4
2
2 kN
500 N⭈m
500 N⭈m
2 kN
3
s =
2(10 )
P
= 1.019 MPa
=
A
1.9635(10 - 3)
t =
500(0.025)
Tc
=
= 20.372 MPa
J
0.61359(10 - 6)
sx = 0
sy = -1.019 MPa
sx + sy
s1, 2 =
s1,2 =
2
;
A
a
sx - sy
2
txy = 20.372 MPa
2
b + txy
2
0 - 1.018
0 - (-1.019) 2
2
;
a
b + 20.372
2
A
2
s1 = 19.87 MPa
s2 = -20.89 MPa
Failure criteria:
|s1| 6 salt = 28 MPa
OK
|s2| 6 salt = 28 MPa
OK
No.
Ans.
•10–69.
Cast iron when tested in tension and compression
has an ultimate strength of 1sult2t = 280 MPa and
1sult2c = 420 MPa, respectively. Also, when subjected to
pure torsion it can sustain an ultimate shear stress of
tult = 168 MPa. Plot the Mohr’s circles for each case and
establish the failure envelope. If a part made of this
material is subjected to the state of plane stress shown,
determine if it fails according to Mohr’s failure criterion.
120 MPa
100 MPa
220 MPa
s1 = 50 + 197.23 = 247 MPa
s2 = 50 - 197.23 = -147 MPa
The principal stress coordinate is located at point A which is outside the shaded
region. Therefore the material fails according to Mohr’s failure criterion.
Yes.
Ans.
800
10–69. Continued
10–70. Derive an expression for an equivalent bending
moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same maximum
shear stress as the combination of an applied moment M
and torque T. Assume that the principal stresses are of
opposite algebraic signs.
Bending and Torsion:
Mc
4M
Mc
;
= p 4 =
I
p c3
4 c
s =
t =
Tc
2T
Tc
= p 4 =
J
p c3
2c
The principal stresses:
s1, 2 =
=
tabs
max
sx + sy
2
;
A
a
sx - sy
2
2
b +
t2xy
=
4M
pc3
+ 0
2
;
Q
¢
4M
p c3
- 0
2
2
≤ + a
b
3
pc
2T
2
2M
2
;
2M2 + T2
p c3
p c3
= s1 - s2 = 2 c
2
2M2 + T2 d
p c3
(1)
Pure bending:
s1 =
tabs
max
4 Me
Me c
Mc
;
= p 4 =
I
c
p c3
4
= s1 - s2 =
s2 = 0
4 Me
(2)
p c3
Equating Eq. (1) and (2) yields:
4 Me
4
2M2 + T2 =
3
pc
p c3
Me = 2M2 + T2
Ans.
801
10–71. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximum-shearstress theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
70 MPa
2
b + txy
2
70 + (-60)
70 - ( -60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
In this case, s1 and s2 have opposite sign. Thus,
|s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa
Based on this result, the steel shell does not yield according to the maximum shear
stress theory.
*10–72. The components of plane stress at a critical point
on an A-36 steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory.
60 MPa
40 MPa
In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and
txy = 40 MPa.
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
70 + ( -60)
70 - ( -60) 2
2
;
c
d + 40
2
A
2
= 5 ; 25825
s1 = 81.32 MPa
s2 = -71.32 MPa
s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500
Based on this result, the steel shell does not yield according to the maximum
distortion energy theory.
802
70 MPa
•10–73.
If the 2-in. diameter shaft is made from brittle
material having an ultimate strength of sult = 50 ksi for
both tension and compression, determine if the shaft fails
according to the maximum-normal-stress theory. Use a
factor of safety of 1.5 against rupture.
30 kip
4 kip · ft
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = pin2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is caused by axial stress.
s =
30
N
= -9.549 ksi
= p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
= 30.56 ksi
=
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + (-30.56)
2
A
2
= (-4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Maximum Normal-Stress Theory.
sallow =
sult
50
=
= 33.33 ksi
F.S.
1.5
|s1| = 26.15 ksi 6 sallow = 33.33 ksi
(O.K.)
|s2| = 35.70 ksi 7 sallow = 33.33 ksi
(N.G.)
Based on these results, the material fails according to the maximum normal-stress
theory.
803
10–74. If the 2-in. diameter shaft is made from cast
iron having tensile and compressive ultimate strengths
of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively,
determine if the shaft fails in accordance with Mohr’s
failure criterion.
Normal Stress and Shear Stresses. The cross-sectional area and polar moment of
inertia of the shaft’s cross-section are
A = p A 12 B = p in2
J =
p 4
p
A 1 B = in4
2
2
The normal stress is contributed by axial stress.
s =
N
30
= - 9.549 ksi
= p
A
The shear stress is contributed by torsional shear stress.
t =
4(12)(1)
Tc
=
= 30.56 ksi
p
J
2
The state of stress at the points on the surface of the shaft is represented on the
element shown in Fig. a.
In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have
s1, 2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
-9.549 - 0 2
-9.549 + 0
2
;
a
b + ( -30.56)
2
A
2
= ( -4.775 ; 30.929) ksi
s1 = 26.15 ksi
s2 = -35.70 ksi
Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which
represent the principal stresses, are located inside the shaded region. Therefore, the
material does not fail according to Mohr’s failure criteria.
804
30 kip
4 kip · ft
10–75. If the A-36 steel pipe has outer and inner
diameters of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-shear-stress theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings. Considering the equilibrium of the free - body diagram of the
post’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
T = - 360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is contributed by bending stress.
Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is contributed by torsional shear stress.
t =
360(0.015)
Tc
= 84.62 MPa
=
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two - dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
805
200 mm
900 N
10–75.
Continued
Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires
|s1 - s2| = sallow
66.07 - (-108.38) = sallow
sallow = 174.45 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.43
sallow
174.45
Ans.
806
*10–76. If the A-36 steel pipe has an outer and inner
diameter of 30 mm and 20 mm, respectively, determine the
factor of safety against yielding of the material at point A
according to the maximum-distortion-energy theory.
900 N
150 mm
A
100 mm
200 mm
Internal Loadings: Considering the equilibrium of the free - body diagram of the
pipe’s right cut segment Fig. a,
©Fy = 0; Vy + 900 - 900 = 0
Vy = 0
T = - 360 N # m
©Mx = 0; T + 900(0.4) = 0
©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m
Section Properties. The moment of inertia about the z axis and the polar moment of
inertia of the pipe’s cross section are
Iz =
p
A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4
4
J =
p
A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4
2
Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus,
sY = -
MyA
90(0.015)
= = -42.31MPa
Iz
10.15625p A 10 - 9 B
The shear stress is caused by torsional stress.
t =
360(0.015)
Tc
= 84.62 MPa
=
J
20.3125p A 10 - 9 B
The state of stress at point A is represented by the two -dimensional element shown
in Fig. b.
In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have
s1, 2 =
=
sx + sz
2
;
A
a
sx - sz
2
2
b + txz
2
-42.31 - 0 2
-42.31 + 0
2
;
a
b + 84.62
2
A
2
= (-21.16 ; 87.23) MPa
s1 = 66.07 MPa
s2 = -108.38 MPa
807
200 mm
900 N
10–76.
Continued
Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2
sallow = 152.55 MPa
Thus, the factor of safety is
F.S. =
sY
250
=
= 1.64
sallow
152.55
Ans.
808
•10–77.
The element is subjected to the stresses shown. If
sY = 36 ksi, determine the factor of safety for the loading
based on the maximum-shear-stress theory.
sx = 4 ksi
s1, 2 =
=
sy = - 12 ksi
sx + sy
2
;
A
a
sx - sy
2
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s1 = 7.314 ksi
s2 = -15.314 ksi
tabsmax =
7.314 - (-15.314)
s1 - s2
=
= 11.314 ksi
2
2
tallow =
sY
36
=
= 18 ksi
2
2
F.S. =
12 ksi
tallow
18
=
= 1.59
abs
tmax
11.314
Ans.
10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory.
sx = 4 ksi
s1, 2 =
=
sy = -12 ksi
sx + sy
2
;
A
a
sx - sy
txy = -8 ksi
4 ksi
2
b + txy
2
8 ksi
4 - (-12) 2
4 - 12
2
;
a
b + (-8)
2
A
2
s1 = 7.314 ksi
s2 = -15.314 ksi
s1 2 - s1 s2 + s2 2 = a
F.S. =
2
12 ksi
sY 2
b
F.S.
362
= 1.80
A (7.314) - (7.314)(-15.314) + (-15.314)2
Ans.
2
809
10–79. The yield stress for heat-treated beryllium copper
is sY = 130 ksi. If this material is subjected to plane stress
and elastic failure occurs when one principal stress is 145 ksi,
what is the smallest magnitude of the other principal stress?
Use the maximum-distortion-energy theory.
Maximum Distortion Energy Theory : With s1 = 145 ksi,
s21 - s1 s2 + s22 = s2Y
1452 - 145s2 + s22 = 1302
s22 - 145s2 + 4125 = 0
s2 =
-(-145) ; 2(-145)2 - 4(1)(4125)
2(1)
= 72.5 ; 33.634
Choose the smaller root, s2 = 38.9 ksi
Ans.
sy ⫽ 0.5sx
*10–80. The plate is made of hard copper, which yields at
sY = 105 ksi. Using the maximum-shear-stress theory,
determine the tensile stress sx that can be applied to the
plate if a tensile stress sy = 0.5sx is also applied.
s1 = sx
sx
1
s2 = sx
2
|s1| = sY
sx = 105 ksi
Ans.
sy ⫽ 0.5sx
•10–81.
Solve Prob. 10–80 using the maximum-distortionenergy theory.
s1 = sx
s2 =
sx
2
sx
s21 - s1 s2 + s22 = s2Y
s2x -
s2x
s2x
+
= (105)2
2
4
sx = 121 ksi
Ans.
810
10–82. The state of stress acting at a critical point on the
seat frame of an automobile during a crash is shown in the
figure. Determine the smallest yield stress for a steel that
can be selected for the member, based on the maximumshear-stress theory.
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
25 ksi
txy = 25 ksi
80 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
2
;
A
a
s x - sy
2
2
b + txy
2
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Shear Stress Theory: s1 and s2 have opposite signs so
|s1 - s2| = sY
|87.170 - (-7.170)| = sY
sY = 94.3 ksi
Ans.
10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory.
Normal and Shear Stress: In accordance with the sign convention.
sx = 80 ksi
sy = 0
txy = 25 ksi
In - Plane Principal Stress: Applying Eq. 9-5.
s1,2 =
=
sx + sy
2
;
A
a
25 ksi
sx - s
2
b + txy
2
2
80 ksi
80 - 0 2
80 + 0
2
;
a
b + 25
2
A
2
= 40 ; 47.170
s1 = 87.170 ksi
s2 = -7.170 ksi
Maximum Distortion Energy Theory:
s21 - s1s2 + s22 = s2Y
87.1702 - 87.170(-7.170) + (-7.170)2 = s2Y
sY = 91.0 ksi
Ans.
811
*10–84. A bar with a circular cross-sectional area is made
of SAE 1045 carbon steel having a yield stress of
sY = 150 ksi. If the bar is subjected to a torque of
30 kip # in. and a bending moment of 56 kip # in., determine
the required diameter of the bar according to the
maximum-distortion-energy theory. Use a factor of safety
of 2 with respect to yielding.
Normal and Shear Stresses: Applying the flexure and torsion formulas.
s =
t =
56 A d2 B
Mc
1792
=
=
p d 4
I
pd3
4 A2B
30 A d2 B
Tc
480
=
=
3
p d 4
J
pd
2 A2B
The critical state of stress is shown in Fig. (a) or (b), where
sx =
1792
pd3
sy = 0
txy =
480
pd3
In - Plane Principal Stresses : Applying Eq. 9-5,
s1,2 =
=
=
s1 =
sx + sy
2
1792
3
pd
+ 0
2
;
;
A
D
a
¢
sx - sy
2
1792
3
pd
- 0
2
2
b + txy
2
2
≤ + a
480 2
b
pd3
896
1016.47
;
pd3
pd3
1912.47
pd3
s2 = -
120.47
pd3
Maximum Distortion Energy Theory :
s21 - s1s2 + s22 = s2allow
a
1912.47 2
1912.47
120.47
120.47 2
150 2
b - a
bab + ab = a
b
3
3
3
3
2
pd
pd
pd
pd
d = 2.30 in.
Ans.
812
•10–85.
The state of stress acting at a critical point on a
machine element is shown in the figure. Determine the
smallest yield stress for a steel that might be selected for the
part, based on the maximum-shear-stress theory.
10 ksi
The Principal stresses:
s1,2 =
=
sx + sy
2
;
A
4 ksi
a
sx - sy
2
2
b + txy
2
8 ksi
8 - (-10) 2
8 - 10
2
;
a
b + 4
2
A
2
s1 = 8.8489 ksi
s2 = -10.8489 ksi
Maximum shear stress theory: Both principal stresses have opposite sign. hence,
|s1 - s2| = sY
8.8489 - (-10.8489) = sY
sY = 19.7 ksi
Ans.
10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t,
and s3 = 0. If the yield stress is sY, determine the maximum
value of p based on (a) the maximum-shear-stress theory and
(b) the maximum-distortion-energy theory.
a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then
|s2| = sg
2
pr
2 = sg
2t
p =
2t
s
r g
|s1| = sg
2
pr
2 = sg
t
p =
t
s (Controls!)
r g
Ans.
b) Maximum Distortion Energy Theory :
s21 - s1s2 + s22 = s2g
a
pr 2
pr pr
pr 2
b - a b a b + a b = s2g
t
t
2t
2t
p =
2t
23r
sg
Ans.
813
10–87. If a solid shaft having a diameter d is subjected to
a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is
tallow = 116>pd322M2 + T2. Assume the principal stresses
to be of opposite algebraic signs.
T
T
M
M
Section properties :
I =
p d 4
pd4
a b =
;
4 2
64
J =
p d 4
pd4
a b =
2 2
32
Thus,
s =
M(d2 )
Mc
32 M
=
=
p d4
I
pd3
64
t =
Tc
=
J
T (d2 )
p d4
32
=
16 T
pd3
The principal stresses :
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
2
b + txy
2
16 M 2
16 M
16 T 2
16 M
16
;
+ a
;
2M2 + T2
a
b =
3
3 b
A
pd
pd
p d3
pd3
p d3
Assume s1 and s2 have opposite sign, hence,
tallow =
2 C 163 2M2 + T2 D
s1 - s2
16
pd
2M2 + T2
=
=
2
2
pd3
QED
*10–88. If a solid shaft having a diameter d is subjected to a
torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is
sallow = 116>pd321M + 2M2 + T22.
T
M
M
Section properties :
I =
p d4
;
64
p d4
32
J =
Stress components :
s =
M (d2 )
Mc
32 M
;
= p 4 =
I
p d3
64 d
t =
T(d2 )
Tc
16 T
= p 4 =
J
p d3
32 d
The principal stresses :
s1,2 =
=
sx + sy
2
;
A
a
sx - sy
2
b +
2
t2xy
=
32 M
3
pd
+ 0
2
32 M
;
3
D
¢ pd
- 0
2
2
≤ + a
16 T 2
b
p d3
16 M
16
;
2M2 + T2
p d3
p d3
Maximum normal stress theory. Assume s1 7 s2
sallow = s1 =
=
16 M
16
+
2M2 + T2
p d3
p d3
16
[M + 2M2 + T2]
p d3
QED
814
T
•10–89.
The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-shear-stress theory. Use a factor
of safety of 1.5 against yielding.
A
B
T
C
Shear Stress: This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = -9947.18T
Maximum Shear Stress Theory.
sallow =
80 mm
sY
250
=
= 166.67 MPa
F.S.
1.5
Since s1 and s2 have opposite sings,
|s1 - s2| = sallow
9947.18T - ( -9947.18T) = 166.67 A 106 B
T = 8377.58 N # m = 8.38 kN # m
Ans.
815
80 mm
100 mm
T
10–90. The shaft consists of a solid segment AB and a
hollow segment BC, which are rigidly joined by the
coupling at B. If the shaft is made from A-36 steel,
determine the maximum torque T that can be applied
according to the maximum-distortion-energy theory. Use a
factor of safety of 1.5 against yielding.
A
B
T
C
Shear Stress. This is a case of pure shear, and the shear stress is contributed by
p
torsion. For the hollow segment, Jh =
A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus,
2
(tmax)h =
T(0.05)
Tch
=
= 8626.28T
Jh
1.845p A 10 - 6 B
For the solid segment, Js =
(tmax)s =
p
A 0.044 B = 1.28p A 10 - 6 B m4. Thus,
2
T(0.04)
Tcs
=
= 9947.18T
Js
1.28p A 10 - 6 B
By comparision, the points on the surface of the solid segment are critical and their
state of stress is represented on the element shown in Fig. a.
In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have
s1,2 =
=
sx + sy
2
;
C
¢
sx - sy
2
2
≤ + t2xy
0 - 0 2
0 + 0
;
¢
≤ + (9947.18T)2
2
C
2
s1 = 9947.18T
s2 = -9947.18T
Maximum Distortion Energy Theory.
sallow =
80 mm
sY
250
=
= 166.67 MPa
F.S.
1.5
Then,
s1 2 - s1s2 + s2 2 = sallow 2
(9947.18T)2 - (9947.18T)(-9947.18T) + (-9947.18T)2 = C 166.67 A 106 B D 2
T = 9673.60 N # m = 9.67 kN # m
Ans.
816
80 mm
100 mm
T
10–91. The internal loadings at a critical section along the
steel drive shaft of a ship are calculated to be a torque of
2300 lb # ft, a bending moment of 1500 lb # ft, and an axial
thrust of 2500 lb. If the yield points for tension and shear are
sY = 100 ksi and tY = 50 ksi, respectively, determine the
required diameter of the shaft using the maximum-shearstress theory.
2300 lb⭈ft
2500 lb
p
I = c4
4
A = pc
2
sA =
p
J = c4
2
1500(12)(c)
P
2500
72 000
Mc
2500
+
b = -a
+
b
+
= -a
4
2
2
pc
A
I
pc
pc
p c3
4
tA
2300(12)(c)
Tc
55 200
=
=
=
4
p
c
J
p c3
2
s1,2 =
sx + sy
= -a
2
;
A
a
sx - sy
2
2
b + txy
2
2500c + 72 000 2
2500 c + 72 000
55200 2
b ;
a
b + a
b
3
3
A
2p c
2p c
p c3
(1)
Assume s1 and s2 have opposite signs:
|s1 - s2| = sg
2500c + 72 000 2
55 200 2
3
b + a
b = 100(10 )
3
A
2p c
p c3
2
a
(2500c + 72000)2 + 1104002 = 10 000(106)p2 c6
6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0
By trial and error:
c = 0.750 57 in.
Substitute c into Eq. (1):
s1 = 22 193 psi
s2 = -77 807 psi
s1 and s2 are of opposite signs
OK
Therefore,
d = 1.50 in.
Ans.
817
1500 lb⭈ft
*10–92. The gas tank has an inner diameter of 1.50 m and
a wall thickness of 25 mm. If it is made from A-36 steel and
the tank is pressured to 5 MPa, determine the factor of
safety against yielding using (a) the maximum-shear-stress
theory, and (b) the maximum-distortion-energy theory.
(a) Normal Stress. Since
0.75
r
=
= 30 7 10, thin - wall analysis can be used.We have
t
0.025
s1 = sh =
pr
5(0.75)
=
= 150 MPa
t
0.025
s2 = slong =
pr
5(0.75)
=
= 75 MPa
2t
2(0.025)
Maximum Shear Stress Theory. s1 and s2 have the sign. Thus,
|s1| = sallow
sallow = 150 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.67
sallow
150
Ans.
(b) Maximum Distortion Energy Theory.
s1 2 - s1s2 + s2 2 = sallow 2
1502 - 150(75) + 752 = sallow 2
sallow = 129.90 MPa
The factor of safety is
F.S. =
sY
250
=
= 1.92
sallow
129.90
Ans.
818
•10–93.
The gas tank is made from A-36 steel and
has an inner diameter of 1.50 m. If the tank is designed
to withstand a pressure of 5 MPa, determine the required
minimum wall thickness to the nearest millimeter using
(a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5
against yielding.
(a) Normal Stress. Assuming that thin - wall analysis is valid, we have
s1 = sh =
5 A 106 B (0.75)
3.75 A 106 B
pr
=
=
t
t
t
s2 = slong =
5 A 106 B (0.75)
1.875 A 106 B
pr
=
=
2t
2t
t
Maximum Shear Stress Theory.
sallow =
250 A 106 B
sY
=
= 166.67 A 106 B Pa
FS.
1.5
s1 and s2 have the same sign. Thus,
|s1| = sallow
3.75 A 106 B
= 166.67 A 106 B
t
t = 0.0225 m = 22.5 mm
Since
Ans.
r
0.75
=
= 33.3 7 10, thin - wall analysis is valid.
t
0.0225
(b) Maximum Distortion Energy Theory.
sallow =
250 A 106 B
sY
=
= 166.67 A 106 B Pa
F.S.
1.5
Thus,
s1 2 - s1s2 + s2 2 = sallow 2
C
3.75 A 106 B
t
3.2476 A 106 B
t
2
S - C
3.75 A 106 B
t
SC
1.875 A 106 B
t
S + C
1.875 A 106 B
t
2
S = c166.67 A 106 B d
= 166.67 A 106 B
t = 0.01949 m = 19.5 mm
Since
2
Ans.
r
0.75
=
= 38.5 7 10, thin - wall analysis is valid.
t
0.01949
819
10–94. A thin-walled spherical pressure vessel has an
inner radius r, thickness t, and is subjected to an internal
pressure p. If the material constants are E and n, determine
the strain in the circumferential direction in terms of the
stated parameters.
s1 = s2 =
pr
2t
e1 = e2 = e =
e =
1
(s - vs)
E
pr
1 - v
1 - v pr
s =
a b =
(1 - v)
E
E
2t
2Et
Ans.
10–95. The strain at point A on the shell has components
Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0.
Determine (a) the principal strains at A, (b) the maximum
shear strain in the x–y plane, and (c) the absolute maximum
shear strain.
ex = 250(10 - 6)
A(250, 137.5)10 - 6
ey = 400(10 - 6)
gxy = 275(10 - 6)
y
A
gxy
2
= 137.5(10 - 6)
C(325, 0)10 - 6
R = a 2(325 - 250)2 + (137.5)2 b10 - 6 = 156.62(10 - 6)
a)
e1 = (325 + 156.62)10 - 6 = 482(10 - 6)
Ans.
e2 = (325 - 156.62)10 - 6 = 168(10 - 6)
Ans.
b)
g
max
in-plane
= 2R = 2(156.62)(10 - 6) = 313(10 - 6)
Ans.
c)
gabs
max
2
gabs
max
=
482(10 - 6)
2
= 482(10 - 6)
Ans.
820
x
*10–96. The principal plane stresses acting at a point are
shown in the figure. If the material is machine steel having a
yield stress of sY = 500 MPa, determine the factor of
safety with respect to yielding if the maximum-shear-stress
theory is considered.
100 MPa
150 MPa
Have, the in plane principal stresses are
s1 = sy = 100 MPa
s2 = sx = -150 MPa
Since s1 and s2 have same sign,
F.S =
sy
=
|s1 - s2|
500
= 2
|100 - (-150)|
Ans.
•10–97.
The components of plane stress at a critical point
on a thin steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is
sY = 650 MPa.
340 MPa
65 MPa
55 MPa
sx = -55 MPa
s1, 2 =
=
sx + sy
2
sy = 340 MPa
;
A
a
sx - sy
2
txy = 65 MPa
2
b + txy
2
-55 - 340 2
-55 + 340
2
;
a
b + 65
2
A
2
s1 = 350.42 MPa
s2 = -65.42 MPa
(s1 2 - s1s2 + s2 ) = [350.422 - 350.42(-65.42) + ( -65.42)2]
= 150 000 6 s2Y = 422 500
OK
No.
Ans.
821
10–98. The 60° strain rosette is mounted on a beam.
The following readings are obtained for each gauge:
Pa = 600110-62, Pb = -700110-62, and Pc = 350110-62.
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In
each case show the deformed element due to these strains.
a
60⬚
60⬚
b
Strain Rosettes (60º): Applying Eq. 10-15 with ex = 600 A 10 - 6 B ,
c
eb = -700 A 10 - 6 B , ec = 350 A 10 - 6 B , ua = 150°, ub = -150° and uc = -90°,
350 A 10 - 6 B = ex cos2 (-90°) + ey sin2( -90°) + gxy sin (-90°) cos ( -90°)
ey = 350 A 10 - 6 B
600 A 10 - 6 B = ex cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150°
512.5 A 10 - 6 B = 0.75 ex - 0.4330 gxy
[1]
-700 A 10 - 6 B = ex cos2 ( -150°) + 350 A 10 - 6 B sin2(-150°) + gxy sin (-150°) cos (-150°)
-787.5 A 10 - 6 B = 0.75ex + 0.4330 gxy
[2]
Solving Eq. [1] and [2] yields ex = -183.33 A 10 - 6 B
gxy = -1501.11 A 10 - 6 B
Construction of she Circle: With ex = -183.33 A 10 - 6 B , ey = 350 A 10 - 6 B , and
gxy
= -750.56 A 10 - 6 B .
2
eavg =
ex + ey
2
= a
-183.33 + 350
b A 10 - 6 B = 83.3 A 10 - 6 B
2
Ans.
The coordinates for reference points A and C are
A(-183.33, -750.56) A 10 - 6 B
C(83.33, 0) A 10 - 6 B
The radius of the circle is
R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B
a)
In-plane Principal Strain: The coordinates of points B and D represent e1 and e2,
respectively.
e1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B
Ans.
e2 = (83.33 - 796.52) A 10 - 6 B = -713 A 10 - 6 B
Ans.
Orientation of Principal Strain: From the circle,
tan 2uP1 =
750.56
= 2.8145
183.33 + 83.33
2uP2 = 70.44°
2uP1 = 180° - 2uP2
uP =
180° - 70.44°
= 54.8° (Clockwise)
2
Ans.
822
60⬚
10–98. Continued
b)
Maximum In - Plane Shear Strain: Represented by the coordinates of point E on
the circle.
g max
in-plane
2
g
max
in-plane
= -R = -796.52 A 10 - 6 B
= -1593 A 10 - 6 B
Ans.
Orientation of Maximum In-Plane Shear Strain: From the circle.
tan 2uP =
183.33 + 83.33
= 0.3553
750.56
uP = 9.78° (Clockwise)
Ans.
10–99. A strain gauge forms an angle of 45° with the axis
of the 50-mm diameter shaft. If it gives a reading of
P = -200110-62 when the torque T is applied to the shaft,
determine the magnitude of T. The shaft is made from
A-36 steel.
T
45⬚
Shear Stress. This is a case of pure shear, and the shear stress developed is
p
contributed by torsional shear stress. Here, J =
A 0.0254 B = 0.1953125p A 10 - 6 B m4.
2
Then
0.128 A 106 B T
T(0.025)
Tc
=
=
t =
p
J
0.1953125p A 10 - 6 B
T
The state of stress at points on the surface of the shaft can be represented by the
element shown in Fig. a.
Shear Strain: For pure shear ex = ey = 0. We obtain,
ea = ex cos2ua + ey sin2ua + gxysin ua cos ua
-200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45°
gxy = -400 A 10 - 6 B
Shear Stress and Strain Relation: Applying Hooke’s Law for shear,
txy = Ggxy
-
0.128 A 106 B T
p
= 75 A 109 B C -400 A 10 - 6 B D
T = 736 N # m
Ans.
823
*10–100. The A-36 steel post is subjected to the forces
shown. If the strain gauges a and b at point A give readings
of Pa = 300110-62 and Pb = 175110-62, determine the
magnitudes of P1 and P2.
P1
P2
a
A
Internal Loadings: Considering the equilibrium of the free - body diagram of the 1 in.
post’s segment, Fig. a,
P2 - V = 0
V = P2
+ c ©Fy = 0;
N - P1 = 0
N = P1
a + ©MO = 0;
M + P2(2) = 0
M = 2P2
Section Properties: The cross - sectional area and the moment of inertia about the
bending axis of the post’s cross - section are
A = 4(2) = 8 in2
I =
1
(2) A 43 B = 10.667 in4
12
Referring to Fig. b,
A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3
Normal and Shear Stress: The normal stress is a combination of axial and bending
stress.
sA =
2P2(12)(1)
MxA
P1
N
+
= +
= 2.25P2 - 0.125P1
A
I
8
10.667
The shear stress is caused by transverse shear stress.
tA =
VQA
P2(3)
=
= 0.140625P2
It
10.667(2)
Thus, the state of stress at point A is represented on the element shown in Fig. c.
Normal and Shear Strain: With ua = 90° and ub = 45°, we have
ea = ex cos2ua + ey sin2ua + gxysin ua cos ua
300 A 10 - 6 B = ex cos2 90° + ey sin2 90° + gxysin 90° cos 90°
ey = 300 A 10 - 6 B eb = ex cos2ub + ey sin2 ub + gxysin ub cos ub
175 A 10 - 6 B = ex cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45°
ex + gxy = 50 A 10 - 6 B
(1)
824
A
1 in.
b 45⬚
c
+ ©F = 0;
:
x
2 in.
2 ft
A
4 in.
c
Section c– c
10–100. Continued
Since sy = sz = 0, ex = -vey = - 0.32(300) A 10 - 6 B = -96 A 10 - 6 B
Then Eq. (1) gives
gxy = 146 A 10 - 6 B
Stress and Strain Relation: Hooke’s Law for shear gives
tx = Ggxy
0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D
P2 = 11.42 kip = 11.4 kip
Ans.
Since sy = sz = 0, Hooke’s Law gives
sy = Eey
2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D
P1 = 136 kip
Ans.
825
10–101. A differential element is subjected to plane strain
that has the following components: Px = 950110-62, Py =
420110-62, gxy = -325110-62. Use the strain-transformation
equations and determine (a) the principal strains and (b) the
maximum in-plane shear strain and the associated average
strain. In each case specify the orientation of the element
and show how the strains deform the element.
e1, 2 =
ex + ey
;
2
= c
A
a
ex - ey
2
2
b + gxy
2
950 + 420
-325 2
950 - 420 2
-6
;
a
b + a
b d (10 )
2
A
2
2
e1 = 996(10 - 6)
Ans.
e2 = 374(10 - 6)
Ans.
Orientation of e1 and e2 :
gxy
tan 2uP =
ex - ey
-325
950 - 420
=
uP = -15.76°, 74.24°
Use Eq. 10.5 to determine the direction of e1 and e2.
ex¿ =
ex + ey
+
2
ex - ey
2
cos 2u +
gxy
2
sin 2u
u = uP = -15.76°
ex¿ = b
(-325)
950 + 420
950 - 420
+
cos (-31.52°) +
sin (-31.52°) r (10 - 6) = 996(10 - 6)
2
2
2
uP1 = -15.8°
Ans.
uP2 = 74.2°
Ans.
b)
gmax
in-plane
2
gmax
in-plane
eavg =
=
A
= 2c
a
ex - ey
2
2
gxy
2
b
2
950 - 420 2
-325 2
-6
-6
b + a
b d(10 ) = 622(10 )
A
2
2
a
ex + ey
2
b + a
= a
Ans.
950 + 420
b (10 - 6) = 685(10 - 6)
2
Ans.
826
10–101. Continued
Orientation of gmax :
-(ex - ey)
tan 2uP =
gxy
-(950 - 420)
-325
=
uP = 29.2° and uP = 119°
Ans.
Use Eq. 10.6 to determine the sign of
gx¿y¿
2
= -
ex - ey
sin 2u +
2
gxy
2
g
max
in-plane
:
cos 2u
u = uP = 29.2°
gx¿y¿ = 2c
-(950 - 420)
-325
sin (58.4°) +
cos (58.4°) d(10 - 6)
2
2
gxy = -622(10 - 6)
10–102. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = -300110-62.
Determine the equivalent state of strain on an element at
the same point oriented 30° clockwise with respect to the
original element. Sketch the results on the element.
y
Pydy
dy
Stress Transformation Equations:
ex = 400 A 10 - 6 B
ey = 200 A 10 - 6 B
gxy = -300 A 10 - 6 B
u = - 30°
gxy
2
dx
We obtain,
ex¿ =
ex + ey
2
= B
+
ex - ey
2
cos 2u +
gxy
2
sin 2u
400 + 200
400 - 200
-300
+
cos (-60°) + a
b sin (-60°) R A 10 - 6 B
2
2
2
= 480 A 10 - 6 B
gx¿y¿
2
= -¢
Ans.
ex - ey
2
≤ sin 2u +
gxy
2
cos 2u
gx¿y¿ = [-(400 - 200) sin ( -60°) + (-300) cos ( -60°)] A 10 - 6 B
= 23.2 A 10 - 6 B
ey¿ =
ex + ey
= B
2
-
Ans.
ex - ey
2
gxy
2
cos 2u -
gxy
2
sin 2u
400 + 200
400 - 200
-300
cos ( -60°) - a
b sin (-60°) R A 10 - 6 B
2
2
2
= 120 A 10 - 6 B
Ans.
827
x
Pxdx
10–103. The state of plane strain on an element is
Px = 400110-62, Py = 200110-62, and gxy = -300110-62.
Determine the equivalent state of strain, which represents
(a) the principal strains, and (b) the maximum in-plane
shear strain and the associated average normal strain.
Specify the orientation of the corresponding element at the
point with respect to the original element. Sketch the results
on the element.
Construction of the Circle: ex = 400 A 10
-6
y
Pydy
dy
B , ey = 200 A 10 B , and
-6
gxy
2
= -150 A 10
-6
B.
Thus,
eavg =
ex + ey
2
= a
400 + 200
b A 10 - 6 B = 300 A 10 - 6 B
2
Ans.
The coordinates for reference points A and the center C of the circle are
A(400, -150) A 10 - 6 B
C(300, 0) A 10 - 6 B
The radius of the circle is
R = CA = 2(400 - 300)2 + (-150)2 = 180.28 A 10 - 6 B
Using these results, the circle is shown in Fig. a.
In - Plane Principal Stresses: The coordinates of points B and D represent e1 and e2,
respectively. Thus,
e1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B
Ans.
e2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B
Ans.
Orientation of Principal Plane: Referring to the geometry of the circle,
tan 2 A up B 1 =
150
= 1.5
400 - 300
A up B 1 = 28.2° (clockwise)
Ans.
The deformed element for the state of principal strains is shown in Fig. b.
Maximum In - Plane Shear Stress: The coordinates of point E represent eavg and
gmax
. Thus
in-plane
gmax
= -R = -180.28 A 10 - 6 B
in-plane
2
gmax
in-plane
= - 361 A 10 - 6 B
Ans.
Orientation of the Plane of Maximum In - Plane Shear Strain: Referring to the
geometry of the circle,
tan 2us =
400 - 300
= 0.6667
150
uS = 16.8° (counterclockwise)
Ans.
The deformed element for the state of maximum in - plane shear strain is shown in
Fig. c.
828
gxy
2
gxy
2
dx
x
Pxdx
10–103. Continued
829