Mechanics of Materials
Mechanics of Materials
9th Edition
ISBN: 9780133254426
Author: Russell C. Hibbeler
Publisher: Prentice Hall
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Chapter 4, Problem 4.114RP

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 4, Problem 4.114RP, The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the

R4–1/2

Expert Solution & Answer
Check Mark
To determine

The normal stress developed in the bolts and rod.

Answer to Problem 4.114RP

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

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) 200×103N/mm2

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

The coefficient of thermal expansion of the steel (αst) is 12×106m/m°C

The coefficient of thermal expansion of the aluminum (αal) is 24×106m/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 0.1mm

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

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

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

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

Calculation:

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

Ab=π4db2 (1)

Substitute 25mm for db in Equation (1).

Ab=π4×252=490.874mm2

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

Ar=π4dr2 (2)

Substitute 50mm for dr in Equation (2).

Ar=π4×502=1,963.495mm2

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

ΔT=T2T1 (3)

Substitute 30°C for T1 and 130°C for T2 in Equation (3).

ΔT=13030=100°C

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

Mechanics of Materials, Chapter 4, Problem 4.114RP , additional homework tip  1

Refer Figure 1.

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

Fy=0Fb+FrFb=0Fr2Fb=0Fr=2Fb (4)

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.

Mechanics of Materials, Chapter 4, Problem 4.114RP , additional homework tip  2

Refer Figure 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)r0.1=(δT)b+(δF)bαalΔTLrFrLrArEal0.1=αstΔTLb+FbLbAbEst (5)

Substitute 24×106m/m°C for σal , 12×106m/m°C for αst , 100°C for ΔT , 300mm for Lr , 400mm for Lb , 1,963.495mm2 for Ar , 490.874mm2 for Ab , 68.9×103N/mm2 for Eal , and 200×103N/mm2 for Est in Equation (5).

[(24×106×100×300)Fr×3001,963.495×68.9×1030.1]=[(12×106×100×400)+Fb×400490.874×200×103]0.722.2175×106Fr0.1=0.48+4.0743×106Fb4.0743×106Fb+2.2175×106Fr=0.720.10.48

4.0743×106Fb+2.2175×106Fr=0.14Fb+0.5443Fr=34,361.73 (6)

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

Substitute 2Fb for Fr in Equation (6).

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

Calculate the force at the rod CD (Fr)

Substitute 16,452N for Fb in Equation (4).

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

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

σb=FbAb (7)

Substitute 16,452N for Fb and 490.874mm2 for Ab in Equation (7).

σb=16,452490.874=33.5N/mm2×1MPa1N/mm2=33.5MPa

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

σr=FrAr (8)

Substitute 32,904N for Fb and 1,963.495mm2 for Ar in Equation (8).

σr=3,29041,963.495=16.8N/mm2×1MPa1N/mm2=16.8MPa

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

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Chapter 4 Solutions

Mechanics of Materials

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