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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

6

2 2 2 (1,300 10 )(90 in. 0.1170 i) n.L 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

1 total 2 0.15 in. 0.1170 in. 0.0330 in.

The strain in segment (1) can now be computed:

11

1

0.0330 in.0.000825 in./in.

40 in.825 μin./in.

L Ans.




2.2 A rigid steel bar is supported by three rods, as

shown in Fig. P2.2. There is no strain in the rods before

the load P is applied. After load P is applied, the

normal strain in rod (2) is 1,080 in./in. Assume initial

rod lengths of L1 = 130 in. and L2 = 75 in. Determine:

(a) the normal strain in rods (1).

(b) the normal strain in rods (1) if there is a 1/32-in. gap

in the connections between the rigid bar and rods

(1) at joints A and C before the load is applied.

(c) the normal strain in rods (1) if there is a 1/32-in. gap

in the connection between the rigid bar and rod (2)

at joint B before the load is applied.

Fig. P2.2

Solution

(a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated:

6

2 2 2 (1,080 10 )(75 in.) 0.0810 in.L

Since rod (2) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must

move downward by an amount equal to the deformation of rod (2); therefore,

2 0.0810 in. (downward)Bv

By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods

(1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The

deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal

strain in rods (1) can now be calculated as:

11

1

0.0810 in.0.0006231 in./in.

130 in.623 μin./in.

L Ans.

(b) We can assume that the bolted connection at B is

perfect; therefore, vB = 0.0810 in. (downward).

Further, the rigid bar must remain horizontal as it

deflects downward by virtue of symmetry.

Therefore, the deflection downward of joints A and C

is still equal to 0.0810 in.

What effect is caused by the gap at A and C? When

joint A (or C) moves downward by 0.0810 in., the

first 1/32-in. of this downward movement does not

stretch rod (1)—it just closes the gap. Therefore, rod

(1) only gets elongated by the amount

1 0.03125 in.

0.0810 in. 0.03125 in.

0.04975 in.

Av




This deformation creates a strain in rod (1) of:

11

1

0.04975 in.0.0003827 in./in.

130 383 μin./in.

in.L Ans.

(c) The gap is now at joint B. We know the strain

in rod (2); hence, we know its deformation must

be 0.0810 in. However, the first 1/32-in. of

downward movement by the rigid bar does not

elongate the rod—it simply closes the gap. To

elongate rod (2) by 0.0810 in., joint B must move

down:

2 0.03125 in.

0.0810 in. 0.03125 in.

0.11225 in.

Bv

Again, since the rigid bar remains horizontal, vB =

vA = vC. The joints at A and C are assumed to be

perfect; thus,

1 0.11225 in.Av

and the normal strain in rod (1) is:

11

1

0.11225 in.0.00086346 in./in.

130 i863 μin./i

.n.

nL Ans.




2.3 A rigid steel bar is supported by three rods, as

shown in Fig. P2.3. There is no strain in the rods

before the load P is applied. After load P is

applied, the normal strain in rods (1) is 860 m/m.

Assume initial rod lengths of L1 = 2,400 mm and L2

= 1,800 mm. Determine:

(a) the normal strain in rod (2).

(b) the normal strain in rod (2) if there is a 2-

mm gap in the connections between the rigid

bar and rods (1) at joints A and C before the

load is applied.

(c) the normal strain in rod (2) if there is a 2-

mm gap in the connection between the rigid bar

and rod (2) at joint B before the load is applied. Fig. P2.3

Solution

(a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated:

6

1 1 1 (860 10 )(2,400 mm) 2.064 mmL

Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must

move downward by an amount equal to the deformation of rod (1); therefore,

1 2.064 mm (downward)A Cv v

By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod

(2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of

rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can

now be calculated as:

22

2

2.064 mm0.0011467 mm/mm 1,147 μmm/m

1,800 mm

mL Ans.

(b) We know the strain in rod (1); hence, we know its

deformation must be 2.064 mm. However, the joints

at A and C are not perfect connections. The first 2

mm of downward movement by the rigid bar does not

elongate rod (1)—it simply closes the gap. To

elongate rod (1) by 2.064 mm, joint A must move

down:

1 2 mm 2.064 mm 2 mm 4.064 mmAv

By symmetry, the rigid bar remains horizontal;

therefore, vB = vA = vC. The joint at B is assumed to be

perfect; thus, any downward movement of the rigid

bar also elongates rod (2) by the same amount:

2 4.064 mmBv

and the normal strain in rod (2) is:




22

2

4.064 mm0.0022578 mm/mm 2,260 μmm/m

1,800 mm

mL Ans.

(c) We now assume that the bolted connection at A

(or C) is perfect; therefore, the rigid bar deflection

as calculated in part (a) must be vA = 2.064 mm

(downward). Further, the rigid bar must remain

horizontal as it deflects downward by virtue of

symmetry; thus, vB = vA = vC. Therefore, the

deflection downward of joint B is vB = 2.064 mm

What effect is caused by the gap at B? When joint

B moves downward by 2.064 mm, the first 2 mm

of this downward movement does not stretch rod

(2)—it just closes the gap. Therefore, rod (2) only

gets elongated by the amount

2 2 mm

2.064 mm 2 mm

0.064 mm

Bv

This deformation creates a strain in rod (2) of:

22

2

0.064 mm0.0000356 mm/mm

1,800 m35.6 μmm/mm

mL Ans.




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 1

6( 570 10 )(900 mm)

0.5130 mm

L

Therefore, vB = 0.5130 mm (downward).

From the deformation diagram of rigid bar ABCD

240 mm (240 mm 360 mm)

600 mm(0.5130 mm) 1.2825 mm

240 mm

B C

C

v v

v

Therefore, 2 = 1.2825 mm (elongation), and thus, from the definition of strain:

62

2

2

1.2825 mm855 10 mm/mm

1,500 m855 με

mL 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

2 1 mm

1.2825 mm 1 mm

0.2825 mm

Cv




and so the strain in member (2) is

62

2

2

0.2825 mm188.3 10 mm/mm

1,500 mm188.3 με

L 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

0.5130 mm 1 mm 1.5130 mmBv

From deformation diagram of rigid bar ABCD

240 mm (240 mm 360 mm)

600 mm(1.5130 mm) 3.7825 mm

240 mm

B C

C

v v

v

and so

62

2

2

3.7825 mm2,521.7 10 mm/mm

1,5002,520 ε

μ

mmL Ans.




2.5 In Fig. P2.5, rigid bar ABC is supported by a

pin connection at B and two axial members. A slot

in member (1) allows the pin at A to slide 0.25-in.

before it contacts the axial member. If the load P

produces a compression normal strain in member

(1) of −1,300 in./in., determine the normal strain

in member (2).

Fig. P2.5

Solution

From the strain given for member (1)

1 1 1

6( 1,300 10 )(32 in.)

0.0416 in.

L

Pin A has to move 0.25 in. before it

contacts member (1); therefore, vA =

1.080 mm (downward).

0.0416 in. 0.25 in.

0.2916 in.

0.2916 in. (to the left)

Av

From the deformation diagram of

rigid bar ABC

12 in. 20 in.

20 in.(0.2916 in.) 0.4860 in. (downward)

12 in.

A C

C

v v

v

Therefore, 2 = 0.4860 in. (elongation). From the definition of strain,

62

2

2

0.4860 in.3,037.5 10 in./in.

160 in3,040 με

.L Ans.




2.6 The sanding-drum mandrel shown in Fig.

P2.6 is made for use with a hand drill. The

mandrel is made from a rubber-like material that

expands when the nut is tightened to secure the

sanding sleeve placed over the outside surface. If

the diameter D of the mandrel increases from 2.00

in. to 2.15 in. as the nut is tightened, determine

(a) the average normal strain along a diameter of

the mandrel.

(b) the circumferential strain at the outside surface

of the mandrel.

Fig. P2.6

Solution

(a) The change in strain along a diameter is found from

2.15 in. 2.00 in.

2.00 in.0.075 in./in.D

D

D Ans.

(b) Note that the circumference of a circle is given by D. The change in strain around the

circumference of the mandrel is found from

(2.15 in.) (2.00 in.)

(2.00 in.)0.075 in./in.C

C

C Ans.




2.7 The normal strain in a suspended bar of material of varying cross section due to its own weight is

given by the expression y/3E where is the specific weight of the material, y is the distance from the

free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E,

(a) the change in length of the bar due to its own weight.

(b) the average normal strain over the length L of the bar.

(c) the maximum normal strain in the bar.

Solution

(a) The strain of the suspended bar due to its own weight is given as

3

y

E

Consider a slice of the bar having length dy. In general, = L.

Applying this definition to the bar slice, the deformation of slice dy is

given by

3

yd dy dy

E

Since this strain expression varies with y, the total deformation of the bar must be found by integrating

d over the bar length:

2 2

0 063 3 2

LLL

E

y ydy

E E Ans.

(b) The average normal strain is found by dividing the expression above for by the bar length L

2

avg

/ 6

6

L E

L

L

E Ans.

(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum

value of y, that is, at y = L:

max33y L

y L

y

E

L

E Ans.




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.

6250 10 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

6 6

6 66

700 10 250 10 (2,000 ft)

700 10 250 100.225 10 /ft

2,000 ft

k

k

Substituting this value for k in the strain expression gives

6 6250 10 (0.225 10 /ft) y

(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

6 6 6250 10 (0.225 10 /ft)(1,500 ft) 587.5 1 588 0 με Ans.

(b) The total elongation is found by integrating the strain expression over the cable length; thus,

2000

6 6

0

20006

6 2

0

250 10 (0.225 10 /ft)

0.225 10(250 1

0.950 ft 11.40 in.

0 )2

dy y dy

y y

Ans.




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 rad16 mm

and thus, the shear strain in the block is

0.412 rad 412,000 μrad Ans.

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 M

62

Pa

7 kPa Ans.




2.10 A thin polymer plate PQR is

deformed such that corner Q is displaced

downward 1/16-in. to new position Q’ as

shown in Fig. P2.10. Determine the shear

strain at Q’ associated with the two edges

(PQ and QR).

Fig. P2.10

Solution

Before deformation, the angle of the plate at

Q was 90° or /2 radians. We must now

determine the plate angle at Q′ after

deformation. The difference between these

angles is the shear strain.

After point Q displaces downward by 1/16-

in., the angle ′ is

25 in. 25 in.tan

(10 in. 1 / 16-in.) 10.0625 in.

68.075215

and the angle ′ is

4 in. 4 in.tan

(10 in. 1 /16-in.) 10.0625 in.

21.678589

After deformation, the angle of the plate at Q′ is

68.075215 21.678589 89.753804 1.566499 rad

The difference in the plate angle at Q before and after deformation is the shear strain:

1.566499 rad 0.004297 ra 4,300 μrad d2

Ans.




2.11 A thin polymer plate PQR is

deformed so that corner Q is displaced

downward 1.0 mm to new position Q’ as

shown in Fig. P2.11. Determine the

shear strain at Q’ associated with the two

edges (PQ and QR).

Fig. P2.11

Solution

Before deformation, the angle of the plate

at Q was 90° or /2 radians. We must

now determine the plate angle at Q′ after

deformation. The difference between

these angles is the shear strain.

After deformation, the angle ′ is

120 mm 120 mmtan

(300 mm 1 mm) 301 mm

21.735741


750 mm 750 mmtan

(300 mm 1 mm) 301 mm

68.132764

Therefore, after deformation, the angle of the plate at Q′ is

21.735741 68.132764 89.868505 1.568501 rad

The difference in the angle at Q before and after deformation is the shear strain:

1.568501 rad 0.002295 ra 2,300 μrad d2

Ans.




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 mmtan

1,100 mm

0.072922

and the angle that side PR makes with the

vertical axis is

0.7 mmtan

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

0.002439 ra 2,440 radd μP Ans.

Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved

the extra steps involved in using the inverse tangent function. Thus,

1.4 mm

tan 0.001273 rad1,100 mm

0.7 mm

tan 0.001167 rad600 mm

0.001273 rad 0.001167 rad 0.002439 ra 2,440 μrad dP Ans.




2.13 A thin rectangular plate is uniformly

deformed as shown in Fig. P2.13.

Determine the shear strain xy at P.

Fig. P2.13

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

0.125 in.

tan 0.39788118 in.

and the angle that side PR makes with the vertical axis is

0.125 in.

tan 0.28647725 in.

By inspection, the angle of the plate at P has been increased by

and decreased by ; thus, the angle at P after deformation is:

90 90 0.397881 0.286477 90.111404

The angle in the plate at P that was initially 90° has been increased to 90.111404° = 1.572741 rad. The

shear strain in the plate at P is thus

1.572741 rad 0.001944 ra 1,944 μrd d2

aP Ans.

Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved

the extra steps involved in using the inverse tangent functions. Thus,

0.125 in.

tan 0.006944 rad18 in.

0.125 in.

tan 0.005000 rad25 in.

By inspection, the angle of the plate at P has been increased by and decreased by ; thus, the angle at

P after deformation is:

0.006944 rad 0.005000 rad 0.001944 rad2 2 2

Therefore, the shear strain in the plate at P is P = −0.001944 rad = −1,944 rad. Ans.




2.14 A thin polymer plate PQR is deformed

so that corner Q is displaced downward 1.0

mm to new position Q’ as shown in Fig.

P2.14. Determine the shear strain at Q’

associated with the two edges (PQ and QR).

Fig. P2.14

Solution

Before deformation, the angle of the plate at Q was 90° or

/2 radians. We must now determine the plate angle at Q′

after deformation. The difference between these angles is

the shear strain.

After deformation, the angle ′ is

180 mm 1.0 mm 181 mmtan

300 mm 300 mm

31.103981


500 mm 1.0 mm 499 mmtan

300 mm 300 mm

58.985614

Therefore, after deformation, the angle of the plate at Q′ is

31.103981 58.985614 90.089595 1.572360 rad

The difference in plate angle before and after deformation is the shear strain:

1.572360 rad 0.001564 ra 1d ,562

4 μrad Ans.




2.15 An airplane has a half-wingspan of 33 m. Determine the change in length of the aluminum

alloy [ A = 22.5×10-6

/°C] wing spar if the plane leaves the ground at a temperature of 15°C and

climbs to an altitude where the temperature is –55°C.

Solution

The change in temperature between the ground and the altitude in flight is

final initial 55°C 15°C 70°CT T T

The thermal strain is given by

6(22.5 10 /°C)( 70°C) 0.001575 mm/mmT T

and thus the deformation in the 33-m wing is

( 0.001575 mm/mm)(33 m) 0.0520 m 52.0 mmT L Ans.




2.16 A large cement kiln has a length of 400 ft and a diameter of 20 ft. Determine the change in

length and diameter of the structural steel [ S = 6.5×10-6

/°F] shell caused by an increase in

temperature of 350°F.

Solution

The thermal strain is given by

6(6.5 10 /°F)( 350°F) 0.002275 in./in.T T

The change in length of the 225-ft kiln length is

( 0.002275 in./in.)(400 ft) 0.9100 ft 10.92 in.TL L Ans.

The change in diameter of the 20-ft diameter is

( 0.002275 in./in.)(20 ft) 0.004550 ft 0.546 in.TD L Ans.




2.17 A cast iron pipe has an inside diameter of d = 208 mm and an outside diameter of D = 236 mm. The

length of the pipe is L = 3.0 m. The coefficient of thermal expansion for cast iron is I = 12.1×10-6

/°C.

Determine the dimension changes caused by an increase in temperature of 70°C.

Solution

The thermal strain caused by a temperature increase of 70°C is given by

6(12.1 10 /°C)(70°C) 0.000847 mm/mmT T

The dimension changes caused by a temperature increase of 70°C are

(0.000847 mm/mm)(236 mm) 0.1999 mmTD D Ans.

(0.000847 mm/mm)(208 mm) 0.1762 mmTd d Ans.

(0.000847 mm/mm)(3.0 m) 0.002541 m 2.54 mmTL L Ans.




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:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

0.08 in.

0.08 in.

0.08 in. 0.08 in.77.821°F

(12.5 10 / °F)(40 in.) (9.6 10 / °F)(55 in.)

T L T L

T L L

TL L

Since the initial temperature is 40°F, the temperature at which the gap is closed is 117.8°F. Ans.




2.19 At a temperature of 5°C, a 3-mm gap

exists between two polymer bars and a rigid

support, as shown in Fig. P2.19. Bars (1) and

(2) have coefficients of thermal expansion of

= 140 × 10−6

/°C and = 67 × 10−6

/°C,

respectively. The supports at A and C are rigid.

Determine the lowest temperature at which the

3-mm gap is closed.

Fig. P2.19

Solution

Write expressions for the temperature-induced deformations and set this equal to the 3-mm gap:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

3 mm

3 mm

3 mm 3 mm30.084°C

(140 10 / °C)(540 mm) (67 10 / °C)(360 mm)

T L T L

T L L

TL L

Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.084°C = 35.1°C. Ans.




2.20 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at

the same temperature is 5 mm longer. At what temperature will the aluminum pipe be 15 mm

longer than the steel pipe? Assume that the coefficient of thermal expansion for the aluminum is

22.5×10-6

/°C and that the coefficient of thermal expansion for the steel is 12.5×10-6

/°C.

Solution

The length of the aluminum pipe after a change in temperature can be expressed as

final initial

initial initial

A A A

AA A

L L L

L T L (a)

Similarly, the length of the steel pipe after the same change in temperature is given by

final initial

initial initial

S S S

SS S

L L L

L T L (b)

From the problem statement, we are trying to determine the temperature change that will cause the final

length of the aluminum pipe to be 15 mm longer than the steel pipe. This requirement can be expressed

as

final final 15 mm

A SL L (c)

Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship

initial initial initial initial 15 mmA SA A S S

L T L L T L

Collect the terms with T on the left-hand side of the equation

initial initial initial initial 15 mmA SA S S A

T L T L L L

Factor out T

initial initial initial initial 15 mmA SA S S A

T L L L L

and thus, T can be expressed as

initial initial

initial initial

15 mmS A

A SA S

L LT

L L

Convert all length dimensions to units of millimeters and solve

6 6

60,005 mm 60,000 mm 15 mm

(22.5 10 /°C)(60,000 mm) (12.5 10 /°C)(60,005 mm)

20 mm 20 mm

1.35 mm/°C 0.75 mm/°C 0.60 mm/°C

33.3°C

T

Initially, the pipes were at a temperature of 10°C. With the temperature change determined above, the

temperature at which the aluminum pipe is 15 mm longer than the steel pipe is

final initial 4310°C 3 .3°C3.3°CT T T Ans.




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

Steel Steel Steel (6.6 10 /°F)(60°F)(12 in.) 0.004752 in.T L

and the aluminum piece elongates by

6

Alum Alum Alum (12.5 10 /°F)(60°F)(12 in.) 0.009000 in.T L

From the deformation diagram of the pointer, the scale reading can be determined from similar triangles

pointerAlum Steel

pointer Alum Steel

1.5 in. 8.5 in.

8.5 in.5.6667 0.009000 in. 0.004752 in.

1.50.0241 in. (upward)

in.

v

v Ans.




2.22 Determine the horizontal movement of

point A of Fig. P2.22 due to a temperature

increase of 75°C. Assume that member AE

has a negligible coefficient of thermal

expansion. The coefficients of thermal

expansion are 11.9×10-6

/°C for the steel and

22.5×10-6

/°C for the aluminum alloy.

Fig. P2.22

Solution

In response to the 75°C temperature increase, the steel piece elongates by the amount

6

Steel Steel Steel (11.9 10 /°C)(75°C)(300 mm) 0.267750 mmT L

Thus, joint C moves to the right by 0.267750 mm.

Next, calculate the deformation of the aluminum piece:

6

Alum Alum Alum (22.5 10 /°C)(75°C)(300 mm) 0.50625 mmT L

Joint E moves to the right by 0.50625 mm.

Take the initial position of E as the origin. The coordinates of E in the deflected position are (0.50625

mm, 0) and the coordinates of C are (0.267750 mm, 25 mm). Use the deflected-position coordinates of

E and C to determine the slope of the pointer.

25 mm 0 25 mm

104.8218030.267750 mm 0.50625 mm 0.2385 mm

C E

C E

y yslope

x x

A general equation for the deflected pointer can be expressed as

104.821803y mx b x b




Use the known coordinates of point E to determine b:

0 104.821803(0.50625 mm)

53.066038 mm

b

b

Thus, the deflected pointer can be described by the line

104.821803 53.066038 mmy x

or rearranging to solve for x

(53.066038 mm)

104.821803

yx

At pointer tip A, y = 275 mm; therefore, the x coordinate of the pointer tip is

(53.066038 mm 275 mm) 221.933962 mm

2.11725 mm104.821803 104.82

2.118

m0

m3

2 x Ans.

The x coordinate is the same as the horizontal movement since we took the initial position of joint E as

the origin.




2.23 At a temperature of 25°C, a cold-rolled red brass [ B = 17.6×10-6

/°C] sleeve has an inside

diameter of dB = 299.75 mm and an outside diameter of DB = 310 mm. The sleeve is to be placed

on a steel [ S = 11.9×10-6

/°C] shaft with an outside diameter of DS = 300 mm. If the temperatures

of the sleeve and the shaft remain the same, determine the temperature at which the sleeve will slip

over the shaft with a gap of 0.05 mm.

Solution

The inside diameter of the brass sleeve after a change in temperature can be expressed as

final initial

initial initial

B B B

BB B

d d d

d T d (a)

Similarly, the outside diameter of the steel shaft after the same change in temperature is given by

final initial

initial initial

S S S

SS S

D D D

D T D (b)

From the problem statement, we are trying to determine the temperature change that will cause the

inside diameter of the brass sleeve to be 0.05 mm greater than the outside diameter of the steel shaft.

This requirement can be expressed as

final final 0.05 mm

B Sd D (c)

Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship

initial initial initial initial 0.05 mmB SB B S S

d T d D T D

Collect the terms with T on the left-hand side of the equation

initial initial initial initial 0.05 mmB SB S S B

T d T D D d

Factor out T

initial initial initial initial 0.05 mmB SB S S B

T d D D d

and thus, T can be expressed as

initial initial

initial initial

0.05 mmS B

B SB S

D dT

d D

Solve for the temperature change

6 6

300 mm 299.75 mm 0.05 mm

(17.6 10 /°C)(299.75 mm) (11.9 10 /°C)(300 mm)

0.30 mm 0.30 mm

0.005276 mm/°C 0.003570 mm/°C 0.001706 mm/°C

175.9°C

T

Initially, the shaft and sleeve were at a temperature of 25°C. With the temperature change determined

above, the temperature at which the sleeve fits over the shaft with a gap of 0.05 mm is

final initial 225° 01°CC 175.9°CT T T Ans.

solution - testbanktop.comtestbanktop.com/wp-content/uploads/2016/12/downloadable-solution... ·...

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