The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD . When the temperature is at 30°C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130°C. Assume BF is also rigid. Prob. R9-1

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Answer 1

Textbook Question

Chapter 9, Problem 1RP

The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30°C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130°C. Assume BF is also rigid.

Chapter 9, Problem 1RP, The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the

Prob. R9-1

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

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Answer 2

Textbook Question

Chapter 9, Problem 1RP

The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30°C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130°C. Assume BF is also rigid.

Chapter 9, Problem 1RP, The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the

Prob. R9-1

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

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Answer 3

Textbook Question

Chapter 9, Problem 1RP

The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30°C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130°C. Assume BF is also rigid.

Chapter 9, Problem 1RP, The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the

Prob. R9-1

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

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Answer 4

Textbook Question

Chapter 9, Problem 1RP

The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30°C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130°C. Assume BF is also rigid.

Chapter 9, Problem 1RP, The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the

Prob. R9-1

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Answer 8

Expert Solution & Answer

To determine

Find the normal stress developed in the bolts and rod.

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$

$4.0743×10−6Fb+2.2175×10−6Fr=0.14Fb+0.5443Fr=34,361.73$

Calculate the force at the bolts AB and EF $(Fb)$ :

Substitute $2Fb$ for $Fr$ .

$Fb+0.5443(2Fb)=34,361.732.0886Fb=34,361.73Fb=34,361.732.0886Fb=16,452 N$

Calculate the force at the rod CD $(Fr)$ ;

Substitute 16,452 N for $Fb$ in Equation (1).

$Fr=2(16,452)=32,904 N$

Calculate the normal stress developed in the bolts AB and EF $(σb)$ using the formula:

$σb=FbAb$

Substitute 16,452 N for $Fb$ and $490.874 mm2$ for $Ab$ .

$σb=16,452490.874=33.5 N/mm2×1 MPa1 N/mm2=33.5 MPa$

Calculate the normal stress developed in the rod CD $(σr)$ using the formula:

$σr=FrAr$

Substitute 32,904 N for $Fb$ and $1,963.495 mm2$ for $Ar$ .

$σr=3,29041,963.495=16.8 N/mm2×1 MPa1 N/mm2=16.8 MPa$

Hence, the normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the normal stress developed in the bolts and rod.

Answer 12

Answer to Problem 1RP

The normal stress developed in the bolts and rod are $33.5 MPa_$ and $16.8 MPa_$ .

Answer 13

Explanation of Solution

Given information:

The two bolts AB and EF are made of A992 steel.

The rod CD is made of 6061-T6 aluminum.

The Young’s modulus of the steel is $(Est)$ $200×103 N/mm2$ .

The Young’s modulus of the aluminum $(Eal)$ is $68.9×103 N/mm2$ .

The coefficient of thermal expansion of the steel $(αst)$ is $12 ×10−6 m/m°C$ .

The coefficient of thermal expansion of the aluminum $(αal)$ is $24 ×10−6 m/m°C$ .

The initial temperature $(T1)$ is $30°C$ .

The finial temperature $(T2)$ is $130°C$ .

The gap between the rod and rigid member AE is 0.1 mm.

The diameter of the bolts AB and EF $(db)$ is 25 mm..

The diameter of the rod CD $(dr)$ is 50 mm.

The length of the bolts AB and EF $(Lb)$ is 400 mm..

The length of the rod CD $(Lr)$ is 300 mm.

Calculation:

Calculate the area of the bolts AB and EF $(Ab)$ using the formula:

$Ab=π4db2$

Substitute 25 mm for $db$ .

$Ab=π4×252=490.874 mm2$

Calculate the area of the rod CD $(Ar)$ using the formula:

$Ar=π4dr2$

Substitute 50 mm for $dr$ .

$Ar=π4×502=1,963.495 mm2$

Calculate the difference of temperature $(ΔT)$ using the formula:

$ΔT=T2−T1$

Substitute $30°C$ for $T1$ and $130°C$ for $T2$ .

$ΔT=130−30=100°C$

Show the free body diagram of the rigid cap as in Figure 1.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 1

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$∑Fy=0−Fb+Fr−Fb=0Fr−2Fb=0Fr=2Fb$ (1)

Here, $Fb$ is force at the bolts AB and EF and $Fr$ is force at the rod CD.

Show the initial and final position of the assembly as in Figure 2.

Statics and Mechanics of Materials (5th Edition), Chapter 9, Problem 1RP , additional homework tip 2

Here $(δT)r$ is deformation at rod due to temperature, $(δT)b$ is deformation at bolts due to temperature, $(δF)r$ is deformation at rod due to force, and $(δF)b$ is deformation at bolts due to force.

The deformation is as follows:

$(δT)r−(δF)r−0.1=(δT)b+(δF)bαalΔTLr−FrLrArEal−0.1=αstΔTLb+FbLbAbEst$

Substitute $24 ×10−6 m/m°C$ for $σal$ , $12 ×10−6 m/m°C$ for $αst$ , $100°C$ for $ΔT$ , 300 mm for $Lr$ , 400 mm for $Lb$ , $1,963.495 mm2$ for $Ar$ , $490.874 mm2$ for $Ab$ , $68.9×103 N/mm2$ for $Eal$ , and $200×103 N/mm2$ for $Est$ .

$[(24×10−6×100×300)−Fr×3001,963.495×68.9×103−0.1]=[(12×10−6×100×400)+Fb×400490.874×200×103]0.72−2.2175×10−6Fr−0.1=0.48+4.0743×10−6Fb4.0743×10−6Fb+2.2175×10−6Fr=0.72−0.1−0.48$