A cylindrical bronze sleeve (2) is held in compression against a rigid machine wall by a high-strength steel bolt (1) as shown in Figure P5.62. The steel [E = 200 GPa; α = 11.7 × 10^−6/°C] bolt has a diameter of 25 mm. The bronze [E = 105 GPa; α = 22.0 × 10^−6/°C] sleeve has an outside diameter of

Question

A cylindrical bronze sleeve (2) is held in compression against a rigid machine wall by a high-strength steel bolt (1) as shown in Figure P5.62. The steel [E = 200 GPa; α = 11.7 × [latex]10^{-6}[/latex]/°C] bolt has a diameter of 25 mm. The bronze [E = 105 GPa; α = 22.0 × [latex]10^{-6}[/latex]/°C] sleeve has an outside diameter of 75 mm, a wall thickness of 8 mm, and a length of L = 350 mm. The end of the sleeve is capped by a rigid washer with a thickness of t = 5 mm. At an initial temperature of [latex]T_{1}[/latex] = 8°C, the nut is hand-tightened on the bolt until the bolt, washers, and sleeve are just snug, meaning that all slack has been removed from the assembly but no stress has yet been induced. If the assembly is heated to [latex]T_{2}[/latex] = 80°C, calculate:
(a) the normal stress in the bronze sleeve.
(b) the normal strain in the bronze sleeve.

Accepted Answer

Equilibrium
Consider a FBD cut through the assembly. Sum forces in the horizontal direction to obtain:

[latex]\Sigma F_{x}=F_{1}+F_{2}=0 \quad herefore F_{1}=-F_{2}[/latex] (a)

Geometry-of-deformation relationship
For this configuration, the deformations of both members will be equal; therefore,

[latex]\delta_{1}=\delta_{2}[/latex] (b)

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):

[latex]\delta_{1}=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1} \quad \delta_{2}=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T L_{2}[/latex] (c)

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

[latex]\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1}=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T L_{2}[/latex] (d)

Problem Data:
For this assembly, the areas, lengths, coefficients of thermal expansion, and elastic moduli are given below:

[latex]\begin{array}{ll}A_{1}=\frac{\pi}{4}(25 mm )^{2}=490.8739 mm ^{2} & A_{2}=\frac{\pi}{4}\left[(75 mm )^{2}-(59 mm )^{2} ight]=1,683.8937 mm ^{2} \L_{1}=355 mm & L_{2}=350 mm \E_{1}=200,000 MPa & E_{2}=105,000 MPa \\alpha_{1}=11.7 imes 10^{-6} /{ }^{\circ} C & \alpha_{2}=22.0 imes 10^{-6} /{ }^{\circ} C\end{array}[/latex]

Solve the Equations
Substitute Eq. (a) into Eq. (d):

[latex]\begin{aligned}-\frac{F_{2} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1} &=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T L_{2} \F_{2}\left[\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}} ight] &=\left[\alpha_{1} L_{1}-\alpha_{2} L_{2} ight] \Delta T \F_{2} &=\frac{\alpha_{1} L_{1}-\alpha_{2} L_{2}}{\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}} \Delta T\end{aligned}[/latex]

and solve for [latex]F_{2}[/latex]:

[latex]F_{2}= \left [ \frac{\left(11.7 imes 10^{-6} /{ }^{\circ} C ight)(360 mm )-\left(22.0 imes 10^{-6} /{ }^{\circ} C ight)(350 mm )}{\frac{355 mm }{\left(490.8739 mm ^{2} ight)\left(200,000 N / mm ^{2} ight)}+\frac{350 mm }{\left(1,683.8937 mm ^{2} ight)\left(105,000 N / mm ^{2} ight)}} ight ] \left(72^{\circ} C ight)[/latex]

=-45,634.2064 N

(a) Normal stress in bronze shell.
The normal stress in the bronze shell is:

[latex]\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{-45,634.2064 N }{1,683.8937 mm ^{2}}=-27.1004 MPa =27.1 MPa ( C )[/latex]

(b) Normal strain in bronze shell.
The normal strain can easily be found from the force-deformation relationship.

[latex]\begin{aligned}\varepsilon_{2} &=\frac{\delta_{2}}{L_{2}}=\frac{F_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T \&=\frac{-45,634.2064 N }{\left(1,683.8937 mm ^{2} ight)\left(105,000 N / mm ^{2} ight)}+\left(22.0 imes 10^{-6} /{ }^{\circ} C ight)\left(72{ }^{\circ} C ight) \&=1,325.9 imes 10^{-6} mm / mm \&=1,326 \mu \varepsilon\end{aligned}[/latex]

Question 5.62: A cylindrical bronze sleeve (2) is held in compression again...

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