mechanics of materials solutions ch05

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5.44 A 20-mm-diameter by 3.5-m-long steel rod (1) is stress free after being attached to rigid supports, as shown in Fig. P5.44. At A, a 25-mm-diameter bolt is used to connect the rod to the support. Determine the normal stress in steel rod (1) and the shear stress in bolt A after the temperature drops 60°C. Use E = 200 GPa and α = 11.9 × 10−6/°C. Fig. P5.44

Solution Section properties: For the 20-mm-diameter rod, the cross-sectional area is:

2 21 (20 mm) 314.1593 mm

4A π= =

Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is:

1 11 1 1

1 1

F Le T LA E

α= + Δ

Since the rod is attached to rigid supports, e1 = 0.

1 11 1

1 1

0F L T LA E

α+ Δ =

Thus, the force produced in the rod by the temperature drop is:

1 11 1 1 1 1 1

16 2 2(11.9 10 / C)( 60 C)(314.1593 mm )(200,000 N/mm )

44,861.95 N

A EF T L T A EL

α α

−

= − Δ = − Δ

= − × ° − °=

The normal stress in the steel rod is:

1 2

44,861.95 N 142.8 MPa (T)314.1593 mm

σ = = Ans.

The 25-mm-diameter bolt has a cross-sectional area of:

2 2bolt (25 mm) 490.8739 mm

4A π= =

Since the bolt is loaded in double shear, the shear stress in the bolt is

bolt 2

44,861.95 N 45.7 MPa2(490.8439 mm )

τ = = Ans.


5.45 A 1.5-in.-diameter by 30-ft-long steel rod (1) is stress free after being attached to rigid supports. A clevis-and-bolt connection, as shown in Fig. P5.45, connects the rod with the support at A. The normal stress in the steel rod must be limited to 24 ksi, and the shear stress in the bolt must be limited to 42 ksi. Assume E = 29,000 ksi and α = 6.6 × 10−6/°F and determine: (a) the temperature decrease that can be safely accommodated by rod (1) based on the allowable normal stress. (b) the minimum required diameter D for the bolt at A for the temperature decrease found in part (a). Fig. P5.45

Solution Section properties: For the 1.5-in.-diameter rod, the cross-sectional area is:

2 21 (1.5 in.) 1.7671 in.

4A π= =

Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is:

1 11 1 1

1 1

F Le T LA E

α= + Δ

Since the rod is attached to rigid supports, e1 = 0.

1 11 1

1 1

0F L T LA E

α+ Δ =

which can also be expressed in terms of the rod normal stress:

11 1 1

1

0L T LE

σ α+ Δ =

Solve for ΔT corresponding to a 24 ksi normal stress in the steel rod:

1 11

1 1 1 1 1

6

1

24 ksi 125.4 F(6.6 10 / F)(29,000 ksi)

LTE L E

σσα α

−

Δ = − = −

= − = − °× °

Ans.

The normal force in the steel rod is: 2

1 (24 ksi)(1.7671 in. ) 42.4104 kipsF = = If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear bolt is

2 2bolt

bolt

42.4104 kips2 1.009771 in.4 42 ksi

0.802 in.

D

D

π⎡ ⎤ ≥ =⎢ ⎥⎣ ⎦

∴ ≥ Ans.


5.46 A steel [E = 200 GPa and α = 11.9×10-6/°C] rod containing a turnbuckle has its ends attached to rigid walls. During the summer when the temperature is 28°C, the turnbuckle is tightened to produce a stress in the rod of 25 MPa. Determine the stress in the rod in the winter when the temperature is –30°C[tap3].

Solution The normal strain in the rod can be expressed as:

TEσε α= + Δ

Since the rod is attached to rigid walls, the rod strain after the temperature change is ε = 0.

0TEσε α= + Δ =

The change in temperature between the summer and winter is winter summer 30 C 28 C 58 CT T TΔ = − = − ° − ° = − ° Solve for the stress increase created by the 58°C drop in temperature.

6(11.9 10 / C)( 58 C)(200,000 MPa) 138.04 MPa (T)T Eσ α

−

= − Δ= − × ° − ° =

In the summer, the rod had a tension normal stress of 25 MPa; therefore, the rod stress in the winter is: winter 25 MPa 138.04 MPa 163.0 MPa (T)σ = + = Ans.


5.47 A high-density polyethylene [E = 120 ksi and α = 78 × 10−6/°F] block (1) is positioned in a fixture as shown in Fig. P5.47. The block is 2-in. by 2-in. by 32-in.-long. At room temperature, a gap of 0.10 in. exists between the block and the rigid support at B. Determine: (a) the normal stress in the block caused by a temperature increase of 100°F. (b) the normal strain in block (1) at the increased temperature.

Fig. P5.47

Solution If the polyethylene block were completely free to elongate, a temperature change of 100°F would cause an elongation of 6

1 1 1 (78 10 / F)(100 F)(32 in.) 0.2496 in.e T Lα −= Δ = × ° ° = Since this elongation is greater than the 0.10-in. gap, the temperature change will cause the polyethylene block to contact the support at B, which will create normal stress in the block. Force-Temperature-Deformation Relationship The relationship between the internal force, temperature change, and the deformation of an axial member is:

1 11 1 1 1

1 1

F Le T LA E

α= + Δ

In this situation, the elongation of member (1) equals the 0.10-in. gap:

1 11 1 1

1 1

0.10 in.F L T LA E

α+ Δ =

This relationship can be stated in terms of normal stress as

1 11 1 1

1

0.10 in.L T LE

σ α+ Δ =

(a) Normal stress: The normal stress in the block due to the 100°F temperature increase is:

[ ] 11 1 1 1

1

6

0.10 in.

120 ksi0.10 in. (78 10 / F)(100 F)(32 in.) 0.561 ksi 561 psi (C)32 in.

ET LL

σ α

−

= − Δ

⎡ ⎤= − × ° ° = − =⎣ ⎦ Ans.

(b) Normal strain: The normal strain in the polyethylene block is:

11

1

0.10 in. 0.003125 in./in. 3,125 με32 in.

eL

ε = = = = Ans.


5.48 The assembly shown in Fig. P5.48 consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 15,000 ksi, α = 9.8 × 10−6/°F] has an outside diameter of 2.00 in. and an inside diameter of 1.25 in. The ceramic core [E = 42,000 ksi, α = 1.7 × 10−6/°F] has a diameter of 1.25 in. At a temperature of 60°F, the assembly is unstressed. Determine the largest temperature increase that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 21 ksi.

Fig. P5.48

Solution Section properties: The cross-sectional areas of brass shell (1) and ceramic core (2) are:

2 2 2 2 21 2(2.00 in.) (1.25 in.) 1.9144 in. (1.25 in.) 1.2272 in.

4 4A Aπ π⎡ ⎤= − = = =⎣ ⎦

Equilibrium Consider a FBD cut through the assembly. Sum forces in the horizontal direction to obtain: 1 2 2 10xF F F F FΣ = − − = ∴ = − (a)

Force-Temperature-Deformation Relationships The relationship between internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2):

1 1 2 21 1 1 2 2 2

1 1 2 2

F L F Le T L e T LA E A E

α α= + Δ = + Δ (b)

Geometry of Deformations Relationship For this configuration, the elongations of both members will be equal; therefore, 1 2e e= (c) Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation:

1 1 2 21 1 2 2

1 1 2 2

F L F LT L T LA E A E

α α+ Δ = + Δ (d)

Solve the Equations Since a limiting stress is specified for brass shell (1), express Eq. (d) in terms of normal stress:

1 21 1 1 2 2 2

1 2

L LT L T LE E

σ α σ α+ Δ = + Δ (e)

Based on Eq. (a), the normal stress σ2 can be expressed in terms of σ1 as:


2 1 1 1 12 1

2 2 2 1 2

F F F A AA A A A A

σ σ= = − = − = −

Substitute this expression into Eq. (e) to obtain

1 1 21 1 1 1 2 2

1 2 2

L A LT L T LE A E

σ α σ α+ Δ = − + Δ

Rearrange terms

1 1 21 2 2 1 1 2 2 1 1

1 2 2

( )L A L T L T L L L TE A E

σ α α α α⎡ ⎤

+ = Δ − Δ = − Δ⎢ ⎥⎣ ⎦

and solve for ΔT, recognizing that both the shell and the core have the same length:

1 1 2 11 1

1 2 2 1 2 2

2 2 1 1 2 1

1 1L A L AE A E E A E

TL L

σ σ

α α α α

⎡ ⎤ ⎡ ⎤+ +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦Δ = =− −

(f)

Substitute the problem data into Eq. (f) to compute ΔT that will produce a normal stress of 21 ksi in the brass shell:

2

2

6 6

1 1.9144 in. 1( 21 ksi)15,000 ksi 1.2272 in. 42,000 ksi

1.7 10 / F 9.8 10 / F269.1319 F

T − −

⎡ ⎤± +⎢ ⎥

⎣ ⎦Δ =× ° − × °

= °∓ Since the problem asks for the largest temperature increase, max 269 FTΔ = ° Ans.


5.49 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown in Fig. P5.49. Bar (1) is an aluminum alloy [E = 10,000 ksi, ν = 0.32, α = 12.5 × 10−6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi, ν = 0.12, α = 9.6 × 10−6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine: (a) the lowest temperature at which the two bars contact each other. (b) the normal stress in the two bars at a temperature of 250°F. (c) the normal strain in the two bars at 250°F. (d) the change in width of the aluminum bar at a temperature of 250°F.

Fig. P5.49

Solution (a) Lowest Contact Temperature Before the gap is closed, only thermal strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 0.04-in. gap:

[ ]1 1 2 2

1 1 2 2

-6 -61 1 2 2

0.04 in.0.04 in.

0.04 in. 0.04 in. 48.6381°F(12.5 10 / °F)(32 in.) (9.6 10 / °F)(44 in.)

T L T LT L L

TL L

α αα α

α α

Δ + Δ =Δ + =

∴Δ = = =+ × + ×

Since the initial temperature is 60°F, the temperature at which the gap is closed is 108.6°F. Ans. (b) Equilibrium From the results obtained for part (a), we know that the gap will be closed at 250°F, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. 1 2 0xF F FΣ = − + = (a) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap:


1 2 0.04 in.e e+ = (c) Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2

0.04 in.F L F LT L T LA E A E

α α+ Δ + + Δ = (d)

Solve the Equations This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:

1 1 1 21 1 1 2 2 2

1 1 2 2

1 1 1 21 1 1 2 2 2

1 1 2 2

1 21 1 1 1 2 2 2

1 1 2 2

1 1 1 2 2 21

1 2

1 1 2 2

0.04 in.

0.04 in.

0.04 in.

0.04 in.


F L F L T L T LA E A E

L LF T L T LA E A E

T L T LFL L

A E A E

α α

α α

α α

α α

+ Δ + + Δ =

+ = − Δ − Δ

⎡ ⎤+ = − Δ − Δ⎢ ⎥

⎣ ⎦− Δ − Δ=⎡ ⎤

+⎢ ⎥⎣ ⎦

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, ΔT1 = ΔT2 = ΔT = 190°F:

1 22 2

1 2

1 26 6

1 2

32 in. 44 in.

(3 in.)(0.75 in.) 2.25 in. (2 in.)(0.75 in.) 1.50 in.10,000 ksi 28,000 ksi

12.5 10 / °F 9.6 10 / °F

L L

A AE E

α α− −

= =

= = = == =

= × = ×

Substitute these values into Eq. (e) and calculate F1 = −47.0702 kips. Backsubstitute into Eq. (a) to calculate F2 = −47.0702 kips. Note that the internal forces are compression, as expected. Normal Stresses The normal stresses in each axial member can now be calculated:

11 2

1

22 2

2

47.0702 kips 20.9201 ksi 20.9 ksi (C)2.25 in.

47.0702 kips 31.3801 ksi 31.4 ksi (C)1.50 in.

FAFA

σ

σ

−= = = − =

−= = = − = Ans.

(c) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = e/L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths:


1 21 1 1 2 2 2

1 1 2 2

F FT TA E A E

ε α ε α= + Δ = + Δ (f)

Substitute the appropriate values to calculate the normal strains in each member:

11 1 1

1 1

62

47.0702 kips (12.5 10 / °F)(190 F)(2.25 in. )(10,000 ksi)

0.000283 in./in. 283 με

F TA E

ε α

−

= + Δ

−= + × °

= = Ans.

22 2 2

2 2

62

47.0702 kips (9.6 10 / °F)(190 F)(1.50 in. )(28,000 ksi)

0.000703 in./in. 703 με

F TA E

ε α

−

= + Δ

−= + × °

= = Ans. (d) The change in width of the aluminum bar (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in the aluminum bar caused by the internal force F1 is:

61long, 2

1 1

47.0702 kips 2,092.01 10 in./in.(2.25 in. )(10,000 ksi)

FA Eσε −−= = = − ×

The accompanying lateral strain due to the Poisson effect is thus 6 6

lat, long, (0.32)( 2,092.01 10 in./in.) 669.44 10 in./in.σ σε νε − −= − = − − × = × The lateral strain caused by the temperature change is 6 6

lat, 1 1 (12.5 10 / °F)(190 F) 2,375 10 in./in.T Tε α − −= Δ = × ° = × Therefore, the total lateral strain in aluminum bar (1) is

lat lat, lat,

6 6 6669.44 10 in./in. 2,375 10 in./in. 3,044.44 10 in./in.Tσε ε ε

− − −

= +

= × + × = × The change is width of the aluminum bar is thus 6

latwidth (width) (3,044.44 10 in./in.)(3 in.) 0.00913 in.ε −Δ = = × = Ans.


5.50 At a temperature of 5°C, a 3-mm gap exists between two polymer bars and a rigid support, as shown in Fig. P5.50. Bar (1) is 50-mm wide and 20-mm thick [E = 800 MPa, α = 140 × 10−6/°C]. Bar (2) is 75mm-wide and 25-mm thick [E = 2.7 GPa, α = 67 × 10−6/°C] bar. The supports at A and C are rigid. Determine: (a) the lowest temperature at which the 3-mm gap is closed. (b) the normal stress in the two bars at a temperature of 60°C. (c) the normal strain in the two bars at 60°C.

Fig. P5.50

Solution (a) Before the gap is closed, only thermal strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 3-mm gap:

[ ]1 1 2 2

1 1 2 2

-6 -61 1 2 2

3 mm3 mm

3 mm 3 mm 30.99°C(140 10 / °C)(500 mm) (67 10 / °C)(400 mm)

T L T LT L L

TL L

α αα α

α α

Δ + Δ =Δ + =

∴Δ = = =+ × + ×

Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.99°C = 36.0°C. Ans. Equilibrium (b) From the results obtained for part (a), we know that the gap will be closed at 60°C, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. 1 2 0xF F FΣ = − + = (a) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap: 1 2 3 mme e+ = (c) Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation:


1 1 2 21 1 1 2 2 2

1 1 2 2

3 mmF L F LT L T LA E A E

α α+ Δ + + Δ = (d)

Solve the Equations This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:

1 1 1 21 1 1 2 2 2

1 1 2 2

1 1 1 21 1 1 2 2 2

1 1 2 2

1 21 1 1 1 2 2 2

1 1 2 2

1 1 1 2 2 21

1 2

1 1 2 2

3 mm

3 mm

3 mm

3 mm


F L F LT L T L

A E A E

L LF T L T L

A E A ET L T LF

L LA E A E

α α

α α

α α

α α

+ Δ + + Δ =

+ = − Δ − Δ

⎡ ⎤+ = − Δ − Δ⎢ ⎥

⎣ ⎦− Δ − Δ

=⎡ ⎤

+⎢ ⎥⎣ ⎦

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, ΔT1 = ΔT2 = ΔT = 55°C:

1 22 2

1 2

1 26 6

1 2

500 mm 400 mm

(50 mm)(20 mm) 1,000 mm (75 mm)(25 mm) 1,875 mm800 MPa 2,700 MPa

140 10 / °C 67 10 / °C

L L

A AE E

α α− −

= =

= = = == =

= × = ×

Substitute these values into Eq. (e) and calculate F1 = −3.30108 kN. Backsubstitute into Eq. (a) to calculate F2 = −3.30108 kN. Note that the internal forces are compression, as expected. Normal Stresses The normal stresses in each axial member can now be calculated:

11 2

1

22 2

2

( 3.30108 kN)(1,000 N/kN) 3.30 MPa 3.30 MPa (C)1,000 mm

( 3.30108 kN)(1,000 N/kN) 1.761 MPa 1.761 MPa (C)1,875 mm

FAFA

σ

σ

−= = = − =

−= = = − = Ans.

Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = e/L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT TA E A E

ε α ε α= + Δ = + Δ (f)



11 1 1

1 1

62 2

( 3.30108 kN)(1,000 N/kN) (140 10 / °C)(55 C)(1,000 mm )(800 N/mm )

0.00357365 mm/mm 3,570 με

F TA E

ε α

−

= + Δ

−= + × °

= = Ans.

22 2 2

2 2

62 2

( 3.30108 kN)(1,000 N/kN) (67 10 / °C)(55 C)(1,875 mm )(2,700 N/mm )

0.00303294 mm/mm 3,030 με

F TA E

ε α

−

= + Δ

−= + × °

= = Ans.


5.51 At a temperature of 5°C, a 3-mm gap exists between the ends of the two polymer bars shown in Fig. P5.51. Bar (1) is 50-mm wide and 20-mm thick [E = 800 MPa, α = 140 × 10−6/°C]. Bar (2) is 75-mm wide and 25-mm thick [E = 2.7 GPa, α = 67 × 10−6/°C] bar. The supports at A and C are rigid. Determine: (a) the temperature at which the normal stress in bar (2) will be equal to −3.0 MPa. (b) the lengths of the two polymer bars at this temperature.

Fig. P5.51

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. 1 2 0xF F FΣ = − + = (a) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap: 1 2 3 mme e+ = (c) Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2

3 mmF L F LT L T LA E A E

α α+ Δ + + Δ = (d)

(a) Temperature that produces −3 MPa in bar (2) Since a limiting stress is specified for bar (2), express Eq. (d) in terms of normal stress:

1 21 1 1 2 2 2

1 2

3 mmL LT L T LE E

σ α σ α+ Δ + + Δ = (e)

Based on Eq. (a), the normal stress σ1 can be expressed in terms of σ2 as:

1 2 2 2 21 2

1 1 1 2 1

F F F A AA A A A A

σ σ= = = =

Substitute this expression into Eq. (e) to obtain


2 1 22 1 1 2 2 2

1 1 2

3 mmA L LT L T LA E E

σ α σ α+ Δ + + Δ =

Rearrange terms

2 1 21 1 2 2 2 2

1 1 2

3 mm A L LT L T LA E E

α α σ σΔ + Δ = − −

and solve for ΔT:

2 1 22

1 1 2

1 1 2 2

3 mm A L LA E E

TL L

σ

α α

⎡ ⎤− +⎢ ⎥

⎣ ⎦Δ =+

(f)

For this structure, the lengths, areas, and elastic moduli are given below:

1 22 2

1 2

1 26 6

1 2

500 mm 400 mm

(50 mm)(20 mm) 1,000 mm (75 mm)(25 mm) 1,875 mm800 MPa 2,700 MPa

140 10 / °C 67 10 / °C

L L

A AE E

α α− −

= =

= = = == =

= × = ×

Substitute these values into Eq. (f) and calculate ΔT:

2

2

6 6

1,875 mm 500 mm 400 mm3 mm ( 3 MPa)1,000 mm 800 MPa 2,700 MPa

71.9015 C(140 10 / C)(500 mm) (67 10 / C)(400 mm)

T − −

⎡ ⎤− − +⎢ ⎥

⎣ ⎦Δ = = °× ° + × °

Since the bars were initially at a temperature of 5°C, the temperature at which the normal stress in bar (2) reaches −3 MPa is 5°C 71.9015 C 76.9 CT = + ° = ° Ans. (b) Lengths of bars (1) and (2) The corresponding normal stress in bar (1) is:

2

21 2 2

1

1,875 mm( 3 MPa) 5.625 MPa1,000 mm

AA

σ σ= = − = −

Thus, the elongation of bar (1) is

61 11 1 1 1

1

( 5.625 MPa)(500 mm) (140 10 / °C)(71.9015°C)(500 mm) 1.5175 mm800 MPa

Le T LE

σ α −−= + Δ = + × =

The elongation of bar (2) is

62 22 2 2 2

2

( 3 MPa)(400 mm) (67 10 / °C)(71.9015°C)(400 mm) 1.4825 mm2,700 MPa

Le T LE

σ α −−= + Δ = + × =

The lengths of bars (1) and (2) are thus: 1 500 mm 1.5175 mm 501.52 mmL = + = Ans.

2 400 mm 1.4825 mm 401.48 mmL = + = Ans.


5.52 The axial assembly shown in Fig. P5.52 consists of a solid 1in.diameter aluminum alloy rod (1) [E = 10,000 ksi, ν = 0.32, α = 12.5 × 10−6/°F] and a solid 1.5in.diameter bronze rod (2) [E = 15,000 ksi, ν = 0.15, α = 9.4 × 10−-6/°F]. If the supports at A and C are rigid and the assembly is stress free at 0°F, determine: (a) the normal stress in both rods at 160°F. (b) the displacement of flange B. (c) the change in diameter of the aluminum rod.

Fig. P5.52

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. 1 2 1 20xF F F F FΣ = − + = ∴ = (a)

Force-Temperature-Deformation Relationships

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations Relationship 1 2 0e e+ = (c) Compatibility Equation

1 1 2 21 1 1 2 2 2

1 1 2 2

0F L F LT L T LA E A E

α α+ Δ + + Δ = (d)

Solve the Equations From Eq. (a), F1 = F2. The temperature change is the same for both members; therefore, ΔT1 = ΔT2 = ΔT. Eq. (d) then can be written as:

1 1 1 21 1 2 2

1 1 2 2


α α+ Δ + + Δ =

Solving for F1:

[ ]

[ ]

1 1 1 21 1 2 2

1 1 2 2

1 21 1 1 2 2

1 1 2 2

1 1 2 21

1 2

1 1 2 2


L LF T L LA E A E

T L LF

L LA E A E

α α

α α

α α

+ = − Δ − Δ

⎡ ⎤+ = −Δ +⎢ ⎥

⎣ ⎦Δ +

= −⎡ ⎤

+⎢ ⎥⎣ ⎦

(e)


For this structure, the lengths, areas, and elastic moduli are given below.

1 2

2 2 2 21 2

1 26 6

1 2

15 in. 22 in.

(1 in.) 0.7854 in. (1.5 in.) 1.7671 in.4 410,000 ksi 15,000 ksi

12.5 10 / °F 9.4 10 / °F

L L

A A

E E

π π

α α− −

= =

= = = =

= =

= × = ×

Substitute these values along with ΔT = +160°F into Eq. (e) and calculate F1 = −23.0263 kips. From Eq. (a), F2 = F1 = −23.0263 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated:

11 2

1

23.0263 kips 29.318 ksi 29.3 ksi (C)0.7854 in.

FA

σ −= = = − = Ans.

22 2

2

23.0263 kips 13.030 ksi 13.03 ksi (C)1.7671 in.

FA

σ −= = = − = Ans.

(b) Displacement of Flange B The displacement of flange B is equal to the elongation (i.e., contraction in this instance) of rod (1). The elongation of rod (1) is given by:

1 11 1 1 1

1 1

62

( 23.0263 kips)(15 in.) (12.5 10 / °F)(160°F)(15 in.) 0.013977 in.(0.7854 in. )(10,000 ksi)

F Le T LA E

α

−

= + Δ

−= + × = −

The displacement of flange B is thus: 1 0.01398 in.Bu e= = ← (c) Change in diameter of the aluminum rod The change in diameter of aluminum rod (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in the aluminum rod caused by the internal force F1 is:

61long, 2

1 1

23.0263 kips 2,931.78 10 in./in.(0.7854 in. )(10,000 ksi)

FA Eσε −−= = = − ×

The accompanying lateral strain due to the Poisson effect is thus 6 6

lat, long, (0.32)( 2,931.78 10 in./in.) 938.17 10 in./in.σ σε νε − −= − = − − × = × The lateral strain caused by the temperature change is 6 6

lat, 1 1 (12.5 10 / °F)(160 F) 2,000 10 in./in.T Tε α − −= Δ = × ° = × Therefore, the total lateral strain in aluminum rod (1) is

lat lat, lat,

6 6 6938.17 10 in./in. 2,000 10 in./in. 2,938.17 10 in./in.Tσε ε ε

− − −

= +

= × + × = × The change in diameter of the aluminum rod is thus 6

1 lat 1 (2,938.17 10 in./in.)(1 in.) 0.00294 in.D Dε −Δ = = × = Ans.


5.53 A load of P = 170 kN is supported by a structure consisting of rigid bar ABC, two identical solid bronze [E = 100 GPa, α = 16.9 × 10−6/°C] rods, and a solid steel [E = 200 GPa, α = 11.9×10−6/°C] rod, as shown in Fig. P5.53. The bronze rods (1) each have a diameter of 20 mm and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 24 mm. The bars are unstressed when the structure is assembled at 40°C. When the temperature decreases to –10°C, determine: (a) the normal stresses in the bronze and steel rods. (b) the normal strains in the bronze and steel rods. (c) the downward deflection of rigid bar ABC.

Fig. P5.53

Solution Equilibrium: By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: 1 22 0yF F F PΣ = + − = (a) Force-Temperature-Deformation Relationships: The relationship between internal force, temperature change, and deformation can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: A B Cv v v= = (c) All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical elongation of rods (1): 1Av e= (d) and the rigid bar deflection vB will cause an identical elongation of rod (2): 2Bv e= (e) Substitute Eqs. (d) and (e) into Eq. (c) to obtain the geometry of deformation equation: 1 2e e= (f) Compatibility Equation: Substitute the force-deformation relationships (b) into the geometry of deformation relationship (f) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2


α α+ Δ = + Δ (g)

Solve the Equations: For this situation, ΔT1 = ΔT2 = ΔT. Solve Eq. (g) for F2:


[ ]

1 1 2 22 1 1 1 2 2 2

1 1 2

1 2 2 2 21 1 1 2 2

2 1 1 2

F L A EF T L T LA E L

L A E A EF T L LL A E L

α α

α α

⎡ ⎤= + Δ − Δ⎢ ⎥⎣ ⎦

= + Δ −

Substitute this expression into Eq. (a)

[ ]1 2 2 2 21 1 1 1 2 2

2 1 1 2

2 L A E A EF F T L L PL A E L

α α+ + Δ − =

and derive an expression for F1:

[ ]

[ ]

1 2 2 2 21 1 1 2 2

2 1 1 2

2 21 1 2 2

21

1 2 2

2 1 1

2

2

L A E A EF P T L LL A E L

A EP T L LLF L A E

L A E

α α

α α

⎡ ⎤+ = − Δ −⎢ ⎥

⎣ ⎦

− Δ −=

+ (h)

For this structure, P = 170 kN = 170,000 N, and the lengths, areas, and elastic moduli are given below:

1 2

2 2 2 21 2

1 26 6

1 2

2, 400 mm 1,800 mm

(20 mm) 314.1593 mm (24 mm) 452.3893 mm4 4100,000 MPa 200,000 MPa

16.9 10 / °C 11.9 10 / °C

L L

A A

E E

π π

α α− −

= =

= = = =

= =

= × = ×

Substitute these values along with ΔT = −50°C into Eq. (h) and calculate F1:

[ ] 2 21 1 2 2

21

1 2 2

2 1 12 2

6 6

2

2

2

(452.3893 mm )(200,000 N/mm )(170,000 N) ( 50 C) (16.9 10 )(2,400) (11.9 10 )(1,800)1,800 mm

2,400 mm 452.3893 mm 200,000 MPa21,800 mm 314.1593 mm 1

A EP T L LLF L A E

L A E

α α

− −

− Δ −=

+

⎡ ⎤− − ° × − ×⎣ ⎦=

⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 00,000 MPa

37,346.59 N

⎛ ⎞⎜ ⎟⎝ ⎠

=Backsubstitute this result into Eq. (a) to calculate F2 = 95,306.83 N. (a) Normal Stresses: The normal stresses in each rod can now be calculated:

11 2

1

37,346.59 N 118.9 MPa314.1593 mm

FA

σ = = = Ans.

22 2

2

95,306.83 N 211 MPa452.3893 mm

FA

σ = = = Ans.


(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = e/L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT TA E A E

ε α ε α= + Δ = + Δ (f)


11 1 1

1 1

62 2

37,346.59 N (16.9 10 / °C)( 50 C)(314.1593 mm )(100,000 N/mm )

0.0003438 mm/mm 344 με

F TA E

ε α

−

= + Δ

= + × − °

= = Ans.

22 2 2

2 2

62 2

95,306.83 N (11.9 10 / °C)( 50 C)(452.3893 mm )(200,000 N/mm )

0.0004584 mm/mm 458 με

F TA E

ε α

−

= + Δ

= + × − °

= = Ans. (c) Rigid bar deflection: The downward deflection of the rigid bar can be determined from the elongation of rods (1):

1 11 1 1 1

1 1

62

(37,346.59 N)(2,400 mm) (16.9 10 / °C)( 50°C)(2,400 mm)(314.1593 mm )(100,000 MPa)

0.825 mm

BF Lv e T LA E

α

−

= = + Δ

= + × −

= ↓ Ans.


5.54 A solid aluminum [E = 70 GPa, α = 22.5 × 10−6/°C] rod (1) is connected to a solid bronze [E = 100 GPa, α = 16.9 × 10−6/°C] rod at flange B, as shown in Fig. P5.54. Aluminum rod (1) has an outside diameter of 40 mm and bronze rod (2) has an outside diameter of 120 mm. The bars are unstressed when the structure is assembled at 30°C. After the 300-kN load is applied to flange B, the temperature increases to 45°C. Determine: (a) the normal stresses in rods (1) and (2). (b) the deflection of flange B.

Fig. P5.54

Solution Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: 1 2 300 kN 0xF F FΣ = − + + = (a) Force-Deformation Relationships:

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations: 1 2 0e e+ = (c) Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2


α α+ Δ + + Δ = (d)

Solve the Equations: For this situation, ΔT1 = ΔT2 = ΔT. Solve Eq. (d) for F1:

[ ]2 1 1 1 11 2 1 1 2 2

1 2 2 1

L A E A EF F T L LL A E L

α α= − − Δ +

Substitute this expression into Eq. (a):

[ ]2 1 1 1 12 1 1 2 2 2

1 2 2 1

300 kNL A E A EF T L L FL A E L

α α+ Δ + + = −

and solve for F2:

[ ]

[ ]

2 1 1 1 12 1 1 2 2

1 2 2 1

1 11 1 2 2

12

2 1 1

1 2 2

1 300 kN

300 kN

1

L A E A EF T L LL A E L

A ET L LLF L A E

L A E

α α

α α

⎡ ⎤+ = − − Δ +⎢ ⎥

⎣ ⎦

+ Δ += −

+ (f)


Member lengths, areas, elastic moduli, and coefficients of thermal expansion are given below:

1 2

2 2 2 21 2

1 26 6

1 2

2,600 mm 1,000 mm

(40 mm) 1,256.637 mm (120 mm) 11,309.734 mm4 470,000 MPa 100,000 MPa

22.5 10 / °C 16.9 10 / °C

L L

A A

E E

π π

α α− −

= =

= = = =

= =

= × = ×

Substitute these values along with ΔT = +15°C into Eq. (f) and compute F2 = −328.4395 kN. Backsubstitute this result into Eq. (a) to find F1 = −28.4395 kN. (a) Normal Stresses: The normal stresses in each rod can now be calculated:

11 2

1

28,439.5 N 22.6 MPa (C)1,256.637 mm

FA

σ −= = = Ans.

22 2

2

328,439.5 N 29.0 MPa (C)11,309.734 mm

FA

σ −= = = Ans.

(b) Deflection of Flange B: The deflection of flange B can be determined from the elongation of rod (1):

1 11 1 1 1

1 1

62

( 28, 439.5 N)(2,600 mm) (22.5 10 / °C)(15°C)(2,600 mm)(1,256.637 mm )(70,000 MPa)

0.0369 mm

BF Lu e T LA E

α

−

= = + Δ

−= + ×

= → Ans.


5.55 A steel [E = 30,000 ksi, α = 6.6 × 10−6/°F] pipe column (1) with a cross-sectional area of A1 = 5.60 in.2 is connected at flange B to an aluminum alloy [E = 10,000 ksi, α = 12.5 × 10−6/°F] pipe (2) with a cross-sectional area of A2 = 4.40 in.2. The assembly (shown in Fig. P5.55) is connected to rigid supports at A and C. It is initially unstressed at a temperature of 90°F. (a) At what temperature will the normal stress in steel pipe (1) be reduced to zero? (b) Determine the normal stresses in steel pipe (1) and aluminum pipe (2) when the temperature reaches –10°F.

Fig. P5.55

Solution Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: 1 2 60 kips 0xF F FΣ = − + − = (a) Force-Temperature-Deformation Relationships:

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations: 1 2 0e e+ = (c) Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2


α α+ Δ + + Δ = (d)

Solve the Equations: Set F1 = 0 and solve Eq (a) to find F2 = 60 kips. Substitute these values for F1 and F2 into Eq. (d) along with the observation that the temperature change for both axial members is the same (i.e., ΔT1 = ΔT2 = ΔT) and solve for ΔT:

1 1 2 22

1 1 2 26 6

1 1 2 2

(60 kips)(144 in.)0(4.40 in. )(10,000 ksi) 75.758 F

(6.6 10 / F)(120 in.) (12.5 10 / F)(144 in.)

F L F LA E A ET

L Lα α − −

− − −Δ = = = − °

+ × ° + × °

Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel pipe (1) is reduced to zero is 90 F 75.748 F 14.24 FT = ° − ° = ° Ans. (b) Solve Eq. (a) for F2 to obtain 2 1 60 kipsF F= + (e)


When the temperature reaches −10°F, the total change in temperature is ΔT = −100°F. Substitute this value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for F1:

[ ]

[ ]

[ ]

1 1 1 21 1 2 2

1 1 2 2

1 2 21 1 1 2 2

1 1 2 2 2 2

1 2 21 1 1 2 2

1 1 2 2 2 2

21 1 2 2

2 21

1 2

1 1 2 2

( 60 kips)

(60 kips)

(60 kips)

(60 kips)


L L LF T L LA E A E A E

L L LF T L LA E A E A E

LT L LA EF

L LA E A E

α α

α α

α α

α α

++ = − Δ − Δ

⎡ ⎤+ + = −Δ +⎢ ⎥

⎣ ⎦⎡ ⎤

+ = −Δ + −⎢ ⎥⎣ ⎦

−Δ + −=

+

and compute F1:

6 62

1

2 2

6

(60 kips)(144 in.)( 100 F) (6.6 10 )(120 in.) (12.5 10 )(144 in.)(4.40 in. )(10,000 ksi)

120 in. 144 in.(5.60 in. )(30,000 ksi) (4.40 in. )(10,000 ksi)

0.259200 in. 0.196364 in.714.2857 10 i

F

− −

−

⎡ ⎤− − ° × + × −⎣ ⎦=

+

−=× 6 6

0.062836 in.n./kip 3,272.7273 10 in./kip 3,987.0130 10 in./kip

15.7602 kips

− −=+ × ×

=

From Eq. (a), F2 has a value of 2 1 60 kips 15.7602 kips 60 kips 75.7602 kipsF F= + = + = (a) Normal Stresses: The normal stresses in each axial member can now be calculated:

11 2

1

15.7602 kips 2.8143 ksi 2.81 ksi (T)5.60 in.

FA

σ = = = = Ans.

22 2

2

75.7602 kips 17.2182 ksi 17.22 ksi (T)4.40 in.

FA

σ = = = = Ans.


5.56 A load P will be supported by a structure consisting of a rigid bar ABCD, a polymer [E = 2,300 ksi, α = 2.9 × 10−6/°F] bar (1) and an aluminum alloy [E = 10,000 ksi, α = 12.5 × 10−6/°F] bar (2), as shown in Fig. P5.56. Each bar has a cross-sectional area of 2.00 in.2. The bars are unstressed when the structure is assembled at 30°F. After a concentrated load of P = 26 kips is applied and the temperature is increased to 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the vertical deflection of point D.

Fig. P5.56

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: 1 2(30 in.) (84 in.) (66 in.) 0AM F F PΣ = + − = (a) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles:

30 in. 84 in.

B Dv v= (c)

Since there are no gaps, clearances, or other misfits at pins B and D, the elongation of member (1) will equal the deflection of the rigid bar at B and the elongation of member (2) will equal the deflection of the rigid bar at D. Therefore, Eq. (c) can be rewritten in terms of the member elongations as:

1 2

30 in. 84 in.e e= (d)


Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2

1 130 in. 84 in.


α α⎡ ⎤ ⎡ ⎤

+ Δ = + Δ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(e)

Solve the Equations Note that the temperature change for both axial members is the same (i.e., ΔT1 = ΔT2 = ΔT). Solve Eq. (e) for F1:

1 1 2 22 2 1 1

1 1 2 2

2 1 1 1 11 2 2 2 1 1 1

1 2 2 1

30 in.84 in.

30 in. 30 in.84 in. 84 in.


L A E A EF F T L T A EL A E L

α α

α α

⎡ ⎤= + Δ − Δ⎢ ⎥

⎣ ⎦

= + Δ − Δ (f)

Substitute this expression into equilibrium equation (a):

2 1 1 1 12 2 2 1 1 1 2

1 2 2 1

30 in. 30 in.(30 in.) (84 in.) (66 in.) 084 in. 84 in.

L A E A EF T L T A E F PL A E L

α α⎡ ⎤

+ Δ − Δ + − =⎢ ⎥⎣ ⎦

and solve for F2:

21 1

2 2 1 1 11

2 22 1 1

1 2 2

(30 in.)(66 in.) (30 in.)84 in.(30 in.) 84 in.

84 in.

A EP T L T A ELF

L A EL A E

α α− Δ + Δ=

+ (g)

For this structure, P = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below:

1 22 2

1 2

1 26 6

1 2

72 in. 96 in.

2.00 in. 2.00 in.2,300 ksi 10,000 ksi

2.9 10 / °F 12.5 10 / °F

L L

A AE E

α α− −

= =

= == =

= × = ×

Substitute these values along with ΔT = 70°F into Eq. (g) and calculate F2 = 19.3218 kips. Backsubstitute into Eq. (f) to calculate F1 = 3.0991 kips. Normal Stresses The normal stresses in each axial member can now be calculated:

11 2

1

3.0991 kips 1.550 ksi (T)2.00 in.

FA

σ = = = Ans.

22 2

2

19.3218 kips 9.66 ksi (T)2.00 in.

FA

σ = = = Ans.


Deflections of the rigid bar Calculate the elongation of member (2):

62 22 2 2 2 2

2 2

(19.3218 kips)(96 in.) (12.5 10 / °F)(70°F)(96 in.) 0.1767 in.(2.00 in. )(10,000 ksi)

F Le T LA E

α −= + Δ = + × = (h)

Since there are no gaps at pin D, the rigid bar deflection at D is equal to the elongation of member (2); therefore: 2 0.1767 in.Dv e= = ↓ Ans.


5.57 Rigid bar ABCD is loaded and supported as shown in Fig. P5.57. Bar (1) is made of bronze [E= 100 GPa, α = 16.9 × 10−6/°C] and has a cross-sectional area of 400 mm2. Bar (2) is made of aluminum [E = 70 GPa, α = 22.5 × 10−6/°C] and has a cross-sectional area of 600 mm2. Bars (1) and (2) are initially unstressed. After the temperature has increased by 40°C, determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A.

Fig. P5.57

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin D gives the best information for this situation: 1 2 2 1(3 m) (1 m) 0 3DM F F F FΣ = − = ∴ = (a) Force-Temperature-Deformation Relationships

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)


4 m 3 m 1 m

CA B vv v= = (c)

There are no gaps, clearances, or other misfits at pins B and C; therefore, Eq. (c) can be rewritten in terms of the member elongations as:

1 21 23

3 m 1 me e e e− = ∴ = − (d)

Note: To understand the negative sign associated with e1, see Section 5-5 for discussion of statically indeterminate rigid bar configurations with opposing members.


Compatibility Equation

1 1 2 21 1 1 2 2 2

1 1 2 2


α α⎡ ⎤

+ Δ = − + Δ⎢ ⎥⎣ ⎦

(e)

Solve the Equations For this situation, ΔT1 = ΔT2 = ΔT = 40°C. Substitute Eq. (a) into Eq. (e):

( )1 21 11 1 2 2

1 1 2 2

33

F LF L T L T LA E A E

α α⎡ ⎤

+ Δ = − + Δ⎢ ⎥⎣ ⎦

and solve for F1:

( )2 2 1 11

1 2

1 1 2 2

6 6

2 2 2 2

39

(40 C) 3(22.5 10 / C)(920 mm) (16.9 10 / C)(840 mm)840 mm 9(920 mm)

(400 mm )(100,000 N/mm ) (600 mm )(70,000 N/mm )13,990 N 13.990 kN

T L LF L L

A E A E

α α

− −

Δ += −

+

⎡ ⎤° × ° + × °⎣ ⎦= −+

= − = − Backsubstitute into Eq. (a) to find F2 = −41.970 kN. (a) Normal Stresses The normal stresses in each axial member can now be calculated:

11 2

1

13,990 N 35.0 MPa (C)400 mm

FA

σ −= = = Ans.

22 2

2

41,970 N 70.0 MPa (C)600 mm

FA

σ −= = = Ans.

(b) Deflection of the rigid bar at A Calculate the elongation of one of the axial members, say member (1):

1 11 1 1 1

1 1

62 2

( 13,990 N)(840 mm) (16.9 10 / °C)(40 C)(840 mm)(400 mm )(100,000 N/mm )0.27405 mm

F Le T LA E

α

−

= + Δ

−= + × °

= Since there are no gaps at pin B, the rigid bar deflection at B is equal to the elongation of member (1); therefore, vB = e1 = 0.27405 mm (upward). From similar triangles, the deflection of the rigid bar at A is related to vB by:

4 m 3 m

A Bv v=

The deflection of the rigid bar at A is thus:

4 m 4 m (0.27405 mm) 0.365 mm3 m 3 mA Bv v= = = ↑ Ans.


5.58 Rigid bar ABCD in Fig. P5.58 is supported by a pin connection at A and by two axial bars (1) and (2). Bar (1) is a 30-in.-long bronze [E = 15,000 ksi, α = 9.4 × 10−6/°F] bar with a cross-sectional area of 1.25 in.2. Bar (2) is a 40-in.-long aluminum alloy [E = 10,000 ksi, α = 12.5 × 10−6/°F] bar with a cross-sectional area of 2.00 in.2. Both bars are unstressed before the load P is applied. If a concentrated load of P = 27 kips is applied to the rigid bar at D and the temperature is decreased by 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the normal strains in bars (1) and (2). (c) the deflection of the rigid bar at point D.

Fig. P5.58

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: 1 2(36 in.) (84 in.) (98 in.)(27 kips) 0AM F FΣ = − + − = (a) Force-Deformation Relationships

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)


36 in. 84 in.

CB vv = (c)

There are no gaps, clearances, or other misfits at pins B and C; therefore, Eq. (c) can be rewritten in terms of the member elongations as:

1 2

36 in. 84 in.e e− = (d)




1 1 2 21 1 1 2 2 2

1 1 2 2

1 136 in. 84 in.


α α⎡ ⎤ ⎡ ⎤

− + Δ = + Δ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(e)

Solve the Equations Solve Eq. (e) for F1:

2 2 1 11 2 2 2 1 1 1

2 2 1

2 2 1 12 2 2 1 1 1

2 2 1

36 in.84 in.

3 37 7


F L A ET L T LA E L

α α

α α

⎡ ⎤⎛ ⎞= − + Δ − Δ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎡ ⎤

= − + Δ + Δ⎢ ⎥⎣ ⎦

Substitute this expression into equilibrium equation (a) and solve for F2 using ΔT1 = ΔT2 = −100°F:

1 2

2 2 1 12 2 2 1 1 1 2

2 2 1

(36 in.) (84 in.) (98 in.)(27 kips)

3 3(36 in.) (84 in.) (98 in.)(27 kips)7 7

F F

F L A ET L T L FA E L

α α

− + =

⎡ ⎤⎛ ⎞− − + Δ + Δ + =⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

2 1 12

1 2 2

1 1 1 12 2 2 1 1 1

1 1

2

22

3 (36 in.) 84 in. (98 in.)(27 kips)7

3(36 in.) (36 in.)7

3 40 in. 1.25 in. 15,000 ksi (36 in.) 84 in. (98 in.)(27 kips)7 30 in. 2.00 in. 10,000 ksi

(36 i

L A E FL A E

A E A ET L T LL L

F

α α

⎡ ⎤+ =⎢ ⎥

⎣ ⎦

− Δ − Δ

⎡ ⎤+ =⎢ ⎥

⎣ ⎦

−2

6

26

3 (1.25 in. )(15,000 ksi)n.) (12.5 10 )( 100 F)(40 in.)7 30 in.

(1.25 in. )(15,000 ksi)(36 in.)(9.4 10 )( 100 F)(30 in.)30 in.

−

−

× − °

− × − °

22,646 kip-in. 482.1429 kip-in. 634.5000 kip-in. 3,762.6429 kip-in.

19.2857 in. 84 in. 103.2857 in.36.4295 kips

F + += =+

= Backsubstitute into Eq. (a) to find F1:

1(84 in.)(36.4295 kips) (98 in.)(27 kips) 11.5022 kips

36 in.F −= =

Normal Stresses The normal stresses in each axial member can now be calculated:


11 2

1

11.5022 kips 9.20 ksi (T)1.25 in.

FA

σ = = = Ans.

22 2

2

36.4295 kips 18.21 ksi (T)2.00 in.

FA

σ = = = Ans.

(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = e/L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT TA E A E

ε α ε α= + Δ = + Δ


11 1 1

1 1

62

6

11.5022 kips (9.4 10 / °F)( 100 F)(1.25 in. )(15,000 ksi)

326.55 10 in./in. 327 με

F TA E

ε α

−

−

= + Δ

= + × − °

= − × = − Ans.

22 2 2

2 2

62

6

36.4295 kips (12.5 10 / °F)( 100 F)(2.00 in. )(10,000 ksi)

571.48 10 in./in. 571 με

F TA E

ε α

−

−

= + Δ

= + × − °

= × = Ans. (c) Deflection of the rigid bar at D Calculate the elongation of one of the axial members, say member (2):

62 22 2 2 2 2

2 2

(36.4295 kips)(40 in.) (12.5 10 / °F)( 100 F)(40 in.)(2.00 in. )(10,000 ksi)

0.022859 in.

F Le T LA E

α −= + Δ = + × − °

= This elongation can also be determined from the strain in member (2): 6

2 2 2 (571.48 10 in./in.)(40 in.) 0.022859 in.e Lε −= = × = Since there are no gaps at pin C, the rigid bar deflection at C is equal to the elongation of member (2); therefore, vC = e2 = 0.022859 in. (downward). From similar triangles, the deflection of the rigid bar at D is related to vC by:

84 in. 98 in.

C Dv v=

The deflection of the rigid bar at D is thus:

98 in. 98 in. (0.022859 in.) 0.0267 in.84 in. 84 in.D Cv v= = = ↓ Ans.


5.59 The pin-connected structure shown in Fig. P5.59 consists of a rigid bar ABC, a steel bar (1), and a steel rod (2). The cross-sectional area of bar (1) is 1.5 in.2 and the diameter of rod (2) is 0.75 in. Assume E = 30,000 ksi and α = 6.6 × 10−6/°F for both axial members. The bars are unstressed when the structure is assembled at 70°F. After application of a concentrated force of P = 20 kips, the temperature is decreased to 30°F. Determine: (a) the normal stresses in bar (1) and rod (2). (b) the normal strains in bar (1) and rod (2). (c) the deflection of pin C from its original position.

Fig. P5.59

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin B gives the best information for this situation:

1 2(12 in.) (20 in.)

(15 in.)(20 kips) 0BM F FΣ = − +

− = (a) Force-Temperature-Deformation Relationships

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2


α α= + Δ = + Δ (b)


12 in. 20 in.

CA vv = (c)

There are no gaps, clearances, or other misfits at pins A and C; therefore, Eq. (c) can be rewritten in terms of the member elongations as:

1 2

12 in. 20 in.e e− = (d)




1 1 2 21 1 1 2 2 2

1 1 2 2

1 112 in. 20 in.


α α⎡ ⎤ ⎡ ⎤

− + Δ = + Δ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(e)

Solve the Equations Solve Eq. (e) for F1:

2 2 1 11 2 2 2 1 1 1

2 2 1

2 2 1 12 2 2 1 1 1

2 2 1

12 in.20 in.

12 1220 20


F L A ET L T LA E L

α α

α α

⎡ ⎤⎛ ⎞= − + Δ − Δ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎡ ⎤

= − + Δ + Δ⎢ ⎥⎣ ⎦

Substitute this expression into equilibrium equation (a) and solve for F2 using ΔT1 = ΔT2 = −40°F:

1 2

2 2 1 12 2 2 1 1 1 2

2 2 1

(12 in.) (20 in.) (15 in.)(20 kips)

12 12(12 in.) (20 in.) (15 in.)(20 kips)20 20

F F

F L A ET L T L FA E L

α α

− + =

⎡ ⎤⎛ ⎞− − + Δ + Δ + =⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

Note that the area of rod (2) is A2 = π/4(0.75 in.)2 = 0.4418 in.2.

2 1 12

1 2 2

1 1 1 12 2 2 1 1 1

1 1

2

22

12 (12 in.) 20 in. (15 in.)(20 kips)20

12(12 in.) (12 in.)20

12 80 in. 1.5 in. 30,000 ksi (12 in.) 20 in. (15 in.)(20 kips20 32 in. 0.4418 in. 30,000 ksi

L A E FL A E

A E A ET L T LL L

F

α α

⎡ ⎤+ =⎢ ⎥

⎣ ⎦

− Δ − Δ

⎡ ⎤+ =⎢ ⎥

⎣ ⎦2

6

26

)

12 (1.5 in. )(30,000 ksi)(12 in.) (6.6 10 )( 40 F)(80 in.)20 32 in.

(1.5 in. )(30,000 ksi)(12 in.)(6.6 10 )( 40 F)(32 in.)32 in.

−

−

− × − °

− × − °

2300 kip-in. 213.84 kip-in. 142.56 kip-in. 656.40 kip-in.

61.113626 in. 20 in. 81.1136267 in.8.092 kips

F + += =+

= Backsubstitute into Eq. (a) to find F1:

1(20 in.)(8.092 kips) (15 in.)(20 kips) 11.513 kips

12 in.F −= = −


(a) Normal Stresses The normal stresses in each axial member can now be calculated:

11 2

1

11.513 kips 7.68 ksi (C)1.5 in.

FA

σ −= = = Ans.

22 2

2

8.092 kips 18.32 ksi (T)0.4418 in.

FA

σ = = = Ans.

(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = e/L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT TA E A E

ε α ε α= + Δ = + Δ


11 1 1

1 1

62

6

11.513 kips (6.6 10 / °F)( 40 F)(1.5 in. )(30,000 ksi)

519.8 10 in./in. 520 με

F TA E

ε α

−

−

= + Δ

−= + × − °

= − × = − Ans.

22 2 2

2 2

62

6

8.092 kips (6.6 10 / °F)( 40 F)(0.4418 in. )(30,000 ksi)

346.5 10 in./in. 347 με

F TA E

ε α

−

−

= + Δ

= + × − °

= × = Ans. (c) Deflection of the rigid bar at C Calculate the elongation of one of the axial members, say member (2):

62 22 2 2 2 2

2 2

(8.092 kips)(80 in.) (6.6 10 / °F)( 40 F)(80 in.)(0.4418 in. )(30,000 ksi)

0.027723 in.

F Le T LA E

α −= + Δ = + × − °

= This elongation can also be determined from the strain in member (2): 6

2 2 2 (346.5 10 in./in.)(80 in.) 0.027723 in.e Lε −= = × = Since there are no gaps at pin C, the rigid bar deflection at C is equal to the elongation of member (2); therefore: 2 0.0277 in.Cv e= = ↓ Ans.


5.60 Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Fig. P5.60. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 70°F. After the temperature has been increased to 250°F, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B and C relative to rigid support A.

Fig. P5.60

Aluminum (1) L1 = 10 in. A1 = 0.8 in.2 E1 = 10,000 ksi α1 = 12.5×10-6/°F

Cast Iron (2) L2 = 5 in. A2 = 1.8 in.2 E2 = 22,500 ksi α2 = 7.5×10-6/°F

Bronze (3) L3 = 7 in. A3 = 0.6 in.2 E3 = 15,000 ksi α3 = 9.4×10-6/°F

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. 1 2 1 20xF F F F FΣ = − + = ∴ = (a) Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension. 2 3 3 20xF F F F FΣ = − + = ∴ = (b)


3 31 1 2 21 1 1 1 2 2 2 2 3 3 3 3

1 1 2 2 3 3

F LF L F Le T L e T L e T LA E A E A E

α α α= + Δ = + Δ = + Δ (c)

Geometry of Deformations Relationship 1 2 3 0e e e+ + = (d) Compatibility Equation

3 31 1 2 21 1 1 2 2 2 3 3 3

1 1 2 2 3 3

0F LF L F LT L T L T LA E A E A E

α α α+ Δ + + Δ + + Δ = (e)

Solve the Equations From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members; therefore, ΔT1 = ΔT2 = ΔT3 = ΔT. Eq. (e) then can be written as:


( ) ( )2 1 2 32 21 1 2 2 3 3

1 1 2 2 3 3

0F L F LF LT L T L T LA E A E A E

α α α+ Δ + + Δ + + Δ =

Solving for F2:

[ ]

[ ]

3 32 1 2 21 1 2 2 3 3

1 1 2 2 3 3

31 22 1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 32

31 2

1 1 2 2 3 3

F LF L F L T L T L T LA E A E A E

LL LF T L L LA E A E A E

T L L LF LL L

A E A E A E

α α α

α α α

α α α

+ + = − Δ − Δ − Δ

⎡ ⎤+ + = −Δ + +⎢ ⎥

⎣ ⎦Δ + +

= −+ +

(f)

Substitute the problem data along with ΔT = +180°F into Eq. (f) and calculate F1 = −19.1025 kips. From Eq. (a), F1 = −19.1025 kips and from Eq. (b), F3 = −19.1025 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated:

11 2

1

19.1025 kips 23.878 ksi 23.9 ksi (C)0.8 in.

FA

σ −= = = − = Ans.

22 2

2

19.1025 kips 10.613 ksi 10.61 ksi (C)1.8 in.

FA

σ −= = = − = Ans.

33 2

3

19.1025 kips 31.838 ksi 31.8 ksi (C)0.6 in.

FA

σ −= = = − = Ans.

(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force: 19.10 kipsA DR R= = Ans. (c) Deflection of Joints B and C The deflection of joint B is equal to the elongation (i.e., contraction in this instance) of rod (1). The elongation of rod (1) is given by:

1 11 1 1 1

1 1

62


F Le T LA E

α

−

= + Δ

−= + × = −

The deflection of joint B is thus: 1 0.001378 in.Bu e= = ← Ans. The elongation of rod (2) is given by:


2 22 2 2 2

2 2

62


F Le T LA E

α

−

= + Δ

−= + × =

The deflection of joint C is: 2 0.001378 in. 0.004392 in. 0.00301 in.C Bu u e= + = − + = → Ans.


5.61 Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Fig. P5.61. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 20°C. After the temperature has been increased to 100°C, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B and C relative to rigid support A.

Fig. P5.61

Aluminum (1) L1 = 440 mm A1 = 1,200 mm2 E1 = 70 GPa α1 = 22.5×10-6/°C

Cast Iron (2) L2 = 200 mm A2 = 2,800 mm2 E2 = 155 GPa α2 = 13.5×10-6/°C

Bronze (3) L3 = 320 mm A3 = 800 mm2 E3 = 100 GPa α3 = 17.0×10-6/°C

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. 1 2 1 20xF F F F FΣ = − + = ∴ = (a) Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension. 2 3 3 20xF F F F FΣ = − + = ∴ = (b)


3 31 1 2 21 1 1 1 2 2 2 2 3 3 3 3

1 1 2 2 3 3

F LF L F Le T L e T L e T LA E A E A E

α α α= + Δ = + Δ = + Δ (c)

Geometry of Deformations Relationship 1 2 3 0e e e+ + = (d) Compatibility Equation

3 31 1 2 21 1 1 2 2 2 3 3 3

1 1 2 2 3 3

0F LF L F LT L T L T LA E A E A E

α α α+ Δ + + Δ + + Δ = (e)

Solve the Equations From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members; therefore, ΔT1 = ΔT2 = ΔT3 = ΔT. Eq. (e) then can be written as:

( ) ( )2 1 2 32 21 1 2 2 3 3

1 1 2 2 3 3

0F L F LF LT L T L T LA E A E A E

α α α+ Δ + + Δ + + Δ =


Solving for F2:

[ ]

[ ]

3 32 1 2 21 1 2 2 3 3

1 1 2 2 3 3

31 22 1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 32

31 2

1 1 2 2 3 3

F LF L F L T L T L T LA E A E A E

LL LF T L L LA E A E A E

T L L LF LL L

A E A E A E

α α α

α α α

α α α

+ + = − Δ − Δ − Δ

⎡ ⎤+ + = −Δ + +⎢ ⎥

⎣ ⎦Δ + +

= −+ +

(f)

Substitute the problem data along with ΔT = +80°C into Eq. (f) and calculate F1 = −148.80 kN. From Eq. (a), F1 = −148.80 kN and from Eq. (b), F3 = −148.80 kN. (a) Normal Stresses The normal stresses in each rod can now be calculated:

11 2

1

148,800 N 124.00 MPa 124.0 MPa (C)1,200 mm

FA

σ −= = = − = Ans.

22 2

2

148,800 N 53.143 MPa 53.1 MPa (C)2,800 mm

FA

σ −= = = − = Ans.

33 2

3

148,800 N 186.0 MPa 186.0 MPa (C)800 mm

FA

σ −= = = − = Ans.

(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force: 148.8 kNA DR R= = Ans. (c) Deflection of Joints B and C The deflection of joint B is equal to the elongation (i.e., contraction in this instance) of rod (1). The elongation of rod (1) is given by:

1 11 1 1 1

1 1

62 2

( 148,800 N)(440 mm) (22.5 10 / °C)(80°C)(440 mm) 0.01257 mm(1, 200 mm )(70,000 N/mm )

F Le T LA E

α

−

= + Δ

−= + × =

The deflection of joint B is thus: 1 0.01257 mmBu e= = → Ans. The elongation of rod (2) is given by:

2 22 2 2 2

2 2

62 2

( 148,800 N)(200 mm) (13.5 10 / °C)(80°C)(200 mm) 0.14743 mm(2,800 mm )(155,000 N/mm )

F Le T LA E

α

−

= + Δ

−= + × =

The deflection of joint C is: 2 0.01257 mm 0.14743 mm 0.1600 mmC Bu u e= + = + = → Ans.

mechanics of materials solutions ch05

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