2.1 When an axial load is applied to the
ends of the bar shown in Fig. P2.1, the total
elongation of the bar between joints A and C
is 0.15 in. In segment (2), the normal strain
is measured as 1,300 in./in. Determine:
(a) the elongation of segment (2).
(b) the normal strain in segment (1) of
the bar.
Fig. P2.1
Solution
(a) From the definition of normal strain, the elongation in segment (2) can be computed as
(1,300 10 6 )(90 in.) 0.1170 in.
2
2 L2
Ans.
(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that
occurs in segment (1) must be
0.15 in. 0.1170 in. 0.0330 in.
1
total
2
The strain in segment (1) can now be computed:
0.0330 in.
1
0.000825 in./in. 825 μin./in.
1
L1
40 in.
Ans.
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2.4 A rigid bar ABCD is supported by two bars as
shown in Fig. P2.4. There is no strain in the
vertical bars before load P is applied. After load P
is applied, the normal strain in rod (1) is −570
m/m. Determine:
(a) the normal strain in rod (2).
(b) the normal strain in rod (2) if there is a 1-mm
gap in the connection at pin C before the load is
applied.
(c) the normal strain in rod (2) if there is a 1-mm
gap in the connection at pin B before the load is
applied.
Fig. P2.4
Solution
(a) From the strain given for rod (1)
1
1 L1
( 570 10 6 )(900 mm)
0.5130 mm
Therefore, vB = 0.5130 mm (downward).
From the deformation diagram of rigid bar ABCD
vB
vC
240 mm (240 mm 360 mm)
600 mm
vC
(0.5130 mm) 1.2825 mm
240 mm
Therefore,
2
2
2
L2
= 1.2825 mm (elongation), and thus, from the definition of strain:
1.2825 mm
855 10 6 mm/mm 855 με
1,500 mm
Ans.
(b) The 1-mm gap at C doesn’t affect rod (1);
therefore, 1 = −0.5130 mm. The rigid bar
deformation diagram is unaffected; thus, vB =
0.5130 mm (downward) and vC = 1.2825 mm
(downward).
The rigid bar must move downward 1 mm at C
before it begins to elongate member (2).
Therefore, the elongation of member (2) is
vC 1 mm
2
1.2825 mm 1 mm
0.2825 mm
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and so the strain in member (2) is
0.2825 mm
2
188.3 10
2
L2
1,500 mm
6
mm/mm
188.3 με
Ans.
(c) From the strain given for rod (1), 1 =
−0.5130 mm. In order to contract rod (1) by this
amount, the rigid bar must move downward at B
by
vB 0.5130 mm 1 mm 1.5130 mm
From deformation diagram of rigid bar ABCD
vB
vC
240 mm (240 mm 360 mm)
600 mm
vC
(1.5130 mm) 3.7825 mm
240 mm
and so
3.7825 mm
2
2,521.7 10 6 mm/mm
2
L2
1,500 mm
2,520 με
Ans.
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2.8 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A
uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point,
the weight of the cable produces an additional normal strain that is proportional to the length of the cable
below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700
in./in., determine
(a) the strain in the cable at a depth of 500 ft.
(b) the total elongation of the cable.
Solution
Call the vertical coordinate y and establish the origin of the y
axis at the lower end of the steel cable. The strain in the cable
has a constant term (i.e., = 250 ) and a term (we will call it
k) that varies with the vertical coordinate y.
250 10 6 k y
The problem states that the normal strain at the cable drum (i.e.,
y = 2,000 ft) is 700 in./in. Knowing this value, the constant k
can be determined
700 10 6 250 10 6 k (2, 000 ft)
700 10 6 250 10 6
0.225 10 6 /ft
2, 000 ft
Substituting this value for k in the strain expression gives
250 10 6 (0.225 10 6 /ft) y
k
(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft
is
Ans.
250 10 6 (0.225 10 6 /ft)(1,500 ft) 587.5 10 6 588 με
(b) The total elongation is found by integrating the strain expression over the cable length; thus,
2000
dy
250 10
6
(0.225 10 6 /ft) y dy
0
6
(250 10 ) y
0.225 10
2
0.950 ft 11.40 in.
2000
6
y
2
0
Ans.
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2.9 The 16 × 22 × 25-mm rubber blocks shown
in Fig. P2.9 are used in a double U shear mount
to isolate the vibration of a machine from its
supports. An applied load of P = 690 N causes
the upper frame to be deflected downward by 7
mm. Determine the average shear strain and the
shear stress in the rubber blocks.
Fig. P2.9
Solution
Consider the deformation of one block. After a downward
deflection of 7 mm:
7 mm
tan
0.4375
0.412410 rad
16 mm
and thus, the shear strain in the block is
Ans.
0.412 rad 412,000 μrad
Note that the small angle approximation most definitely does not
apply here!
The applied load of 690 N is
carried by two blocks; therefore,
the shear force applied to one
block is V = 345 N. The area
subjected to shear stress is the area
that is parallel to the direction of
the shear force; that is, the 22 mm
by 25 mm surface of the block.
The shear stress is
345 N
(22 mm)(25 mm)
0.627 MPa
627 kPa
Ans.
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2.12 A thin rectangular plate is uniformly
deformed as shown in Fig. P2.12.
Determine the shear strain xy at P.
Fig. P2.12
Solution
Before deformation, the angle of the plate at P
was 90° or /2 radians. We must now
determine the plate angle at P after deformation.
The difference between these angles is the shear
strain.
After deformation, the angle that side PQ makes
with the horizontal axis is
1.4 mm
tan
1,100 mm
0.072922
and the angle that side PR makes with the
vertical axis is
0.7 mm
tan
600 mm
0.066845
Therefore, the angle of the plate at P that was initially 90° has been reduced by the sum of and :
0.072922 0.066845 0.139767 0.002439 rad
Since shear strain is defined as the change in angle between two initially perpendicular lines, the shear
strain in the plate at P is
Ans.
0.002439 rad 2,440 μrad
P
Note: Since these angles are small, we could have just as well used tan ≈ and tan
the extra steps involved in using the inverse tangent function. Thus,
1.4 mm
tan
0.001273 rad
1,100 mm
0.7 mm
tan
0.001167 rad
600 mm
0.001273 rad 0.001167 rad 0.002439 rad 2,440 μrad
P
≈
and saved
Ans.
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2.18 At a temperature of 40°F, a 0.08-in.
gap exists between the ends of the two
bars shown in Fig. P2.18. Bar (1) is an
aluminum alloy [ = 12.5 × 10−6/°F] and
bar (2) is stainless steel [ = 9.6 ×
10−6/°F]. The supports at A and C are
rigid. Determine the lowest temperature
at which the two bars contact each other.
Fig. P2.18
Solution
Write expressions for the temperature-induced deformations and set this equal to the 0.08-in. gap:
0.08 in.
1 T L1
2 T L2
T
L
1 1
2
L2
0.08 in.
0.08 in.
0.08 in.
77.821°F
-6
(12.5 10 / °F)(40 in.) (9.6 10-6 / °F)(55 in.)
1 L1
2 L2
Since the initial temperature is 40°F, the temperature at which the gap is closed is 117.8°F.
T
Ans.
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2.21 Determine the movement of the pointer
of Fig. P2.21 with respect to the scale zero in
response to a temperature increase of 60°F.
The coefficients of thermal expansion are
6.6×10-6/°F for the steel and 12.5×10-6/°F for
the aluminum.
Fig. P2.21
Solution
In response to the 60°F temperature increase, the steel pieces elongate by the amount
(6.6 10 6 /°F)(60°F)(12 in.) 0.004752 in.
Steel
Steel T LSteel
and the aluminum piece elongates by
(12.5 10 6 /°F)(60°F)(12 in.) 0.009000 in.
Alum
Alum T LAlum
From the deformation diagram of the pointer, the scale reading can be determined from similar triangles
vpointer
Alum
Steel
1.5 in.
vpointer
8.5 in.
8.5 in.
Alum
1.5 in.
Steel
5.6667 0.009000 in. 0.004752 in.
0.0241 in. (upward)
Ans.
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