Chapter 2, Problem 2.5.28P

Consider the sleeve made From two copper tubes joined by tin-lead solder over distance s. The sleeve has brass caps at both ends that are held in place by a steel bolt and washer with the nut turned just snug at the outset. Then, two "loadings" are applied: a = 1/2 turn applied to the nut; at the same time, the internal temperature is raised by ?T = 30°C.

(a) Find the forces in the sleeve and boll, P_sand P_B, due to both the priestess in the bolt and the temperature increase. For copper, use E_I= 120 GPa and a_c= 17 × W+C; for steel, use E, = 200 GPa and a, = 12 × 10-6/°C. The pitch of the boll threads is p = 1.0 mm. Assume s = 26 mm and bolt diameter d_B= 5 mm.

(b) Find the required length of the solder joint, s, if shear stress in the sweated joint cannot exceed the allowable shear stress t ™ 18.5 MPa.

(c) What is the final elongation of the entire assemblage due to both temperature change A T and the initial prestress in the bolt?

  

To determine

The force on the sleeve and force on the blot.

Answer to Problem 2.5.28P

The force on the sleeve is = $25.4\text{kN}$.

The force on the bolt is = $-25.4\text{kN}$.

Explanation of Solution

Given information:

The length of the copper sleeve is $40\text{mm}$, copper sleeve diameter is $25\text{mm}$, thickness of the copper sleeve is $4\text{mm}$, coefficient of thermal expansion of copper $17\times {10}^{-6}/\text{°F}$, modulus of elasticity of copper is $120\text{GPa}$, bolt diameter is $5\text{mm}$, modulus of elasticity of steel is $200\text{GPa}$, thermal expansion coefficient of steel is $12\times {10}^{-6}1/\text{°F}$, length of steel bolt is $50\text{mm}$, steel bolt diameter is $17\text{mm}$, number of turns is $\frac{1}{2}$, temperature raise is $30\text{°C}$, pitch is $1\text{mm}$and the gap is $26\text{mm}$.

Write the expression for area of bolt.

  $A_b=\frac{\pi }{4}{d}_b^2$....... (I)

Here, area of the bolt is $A_b$, diameter of the bolt is $d_b$.

Write the expression for the area of the copper sleeve.

  $A_1=\frac{\pi }{4}\left[d_1^2-d_2^2\right]$....... (II)

Here, area of the copper sleeve is $A_2$.

Write the expression for the area of steel bolt.

  $A_2=\frac{\pi }{4}\left[d_2^2-{\left(d_2-2t_2\right)}^2\right]$....... (III)

Here, area of steel bolt is $A_2$.

Write the expression for the elongation in bolt.

  ${\left({\delta }_B\right)}_1=-np+{\alpha }_s\Delta T\left(L_1+L_2-s\right)$....... (IV)

Here, elongation in bolt is $A_2$, number of turns is $n$, pitch is $p$, length of copper sleeve is $L_1$, shear bolt length is $L_2$and the gap is $s$.

Write the expression for elongation.

  ${\left({\delta }_B\right)}_2=P_B\left[\frac{L_1+L_2}{E_s{A}_b}+\frac{L_1-s}{E_c{A}_1}+\frac{L_2-s}{E_c{A}_2}+\frac{s}{E_c\left(A_1+A_2\right)}\right]$....... (V)

Here, elongation is ${\left({\delta }_B\right)}_2$, force is $P_B$, modulus of elasticity of steel is $E_s$, modulus of elasticity of copper is $E_c$.

Write the compatibility Equation for elongation.

  ${\left({\delta }_B\right)}_1={\left({\delta }_B\right)}_2$....... (VI)

Calculation:

Substitute $5\text{mm}$for $d_B$in Equation (I).

  $\begin{array}{c}{A}_b=\frac{\pi }{4}{\left(5\text{mm}\right)}^2\\ =\frac{\pi }{4}\left(25{\text{mm}}^2\right)\\ =19.635{\text{mm}}^{\text{2}}\end{array}$

Substitute $25\text{mm}$for $d_1$and $17\text{mm}$for $d_2$in Equation (II).

  $\begin{array}{c}{A}_1=\frac{\pi }{4}\left[{\left(25\text{mm}\right)}^2-{\left(17\text{mm}\right)}^2\right]\\ =\frac{\pi }{4}\left(336{\text{mm}}^{\text{2}}\right)\\ =263.893{\text{mm}}^{\text{3}}\end{array}$

Substitute $17\text{mm}$for $d_2$and $3\text{mm}$

  $t_2$in Equation (III).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}\left[{\left(17\text{mm}\right)}^2-{\left(17\text{mm}-2\left(3\text{mm}\right)\right)}^2\right]\\ =\frac{\pi }{4}\left[168{\text{mm}}^{\text{2}}\right]\\ =131.947{\text{mm}}^2\end{array}$

Substitute $\frac{1}{2}$for $n$, $1\text{mm}$for $p$, $12\times {10}^{-6}/\text{°F}$for ${\alpha }_s$, $30\text{°C}$for $\Delta T$, $40\text{mm}$for $L_1$and $50\text{mm}$for $L_2$in Equation (IV).

  $\begin{array}{c}{\left({\delta }_B\right)}_1=-\frac{1}{2}\left(1\text{mm}\right)+\left(12\times {10}^{-6}/\text{°F}\right)\left(30\text{°C}\right)\left(40\text{mm}+50\text{mm}-26\text{mm}\right)\\ =-\frac{1}{2}\text{mm}+0.02304\text{mm}\\ =-\text{0}\text{.47696}\text{mm}\end{array}$

Substitute $40\text{mm}$for $L_1$, $50\text{mm}$for $L_2$, $120\text{GPa}$for $E_c$, $200\text{GPa}$for $E_s$, $263.893{\text{mm}}^{\text{3}}$for $A_1$, $131.947{\text{mm}}^2$for $A_2$and $26\text{mm}$for $s$in Equation (V).

Substitute $1.857\times {10}^{-5}{P}_B$for ${\left({\delta }_B\right)}_2$and $-\text{0}\text{.47696}\text{mm}$for ${\left({\delta }_B\right)}_1$in Equation (VI).

  $\begin{array}{l}-\text{0}\text{.47696}\text{mm}=1.875\times {10}^{-5}{P}_B\\ P_B=\frac{-\text{0}\text{.47696}}{1.875\times {10}^{-5}}\\ P_B=\left(-25437.86\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ P_B=-25.4\text{kN}\end{array}$

Since $P_s=-P_B$, so

  $P_s=25.4\text{kN}$

Conclusion:

The force on the sleeve is $25.4\text{kN}$.

The force on the bolt is $-25.4\text{kN}$.

To determine

The required length of the joint.

Answer to Problem 2.5.28P

The required length of the joint is $25.7\text{mm}$.

Explanation of Solution

Given information:

The length of the copper sleeve is $40\text{mm}$, copper sleeve diameter is $25\text{mm}$, thickness of the copper sleeve is $4\text{mm}$, coefficient of thermal expansion of copper $17\times {10}^{-6}/\text{°F}$, modulus of elasticity of copper is $120\text{GPa}$, bolt diameter is $5\text{mm}$modulus of elasticity of steel is $200\text{GPa}$, thermal expansion coefficient of steel is $12\times {10}^{-6}/\text{°F}$, length of steel bolt is $50\text{mm}$, steel bolt diameter is $17\text{mm}$, number of turns is $\frac{1}{2}$, temperature raise is $30\text{°C}$, pitch is $1\text{mm}$, the gap is $26\text{mm}$, and the allowable shear stress is $18.5\text{MPa}$.

Write the expression for area.

  $A_s=\pi d_{2}s$....... (VII)

Here, area of sleeve is $A_s$.

Write the expression for allowable shear stress.

  $\tau =\frac{P}{A_s}$....... (VIII)

Here, allowable shear stress is $\tau$.

Calculation:

Substitute $17\text{mm}$for $d_2$in Equation (VII).

  $\begin{array}{c}{A}_s=\pi \left(17\text{mm}\right)s\\ =53.407s\end{array}$

Substitute $53.407s$for $A_s$

  $25.4\text{kN}$for $P$and $18.5\text{MPa}$for $\tau$in Equation (VIII).

  $\begin{array}{l}18.5\text{MPa}=\frac{25.4\text{kN}}{53.407s}\\ s=\frac{\left(25.4\text{kN}\right)\left(\frac{100\text{0}\text{N}}{1\text{kN}}\right)}{\left(18.5\text{MPa}\right)\left(\frac{{10}^6\text{N}/{\text{m}}^{\text{2}}}{1\text{MPa}}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(53.407\right)}\\ s=25.7\text{mm}\end{array}$

Conclusion:

The required length of joint is $25.7\text{mm}$.

To determine

The final displacement.

Answer to Problem 2.5.28P

The final displacement is $0.352\text{mm}$.

Explanation of Solution

Given information:

The length of the copper sleeve is $40\text{mm}$, copper sleeve diameter is $25\text{mm}$, thickness of the copper sleeve is $4\text{mm}$, coefficient of thermal expansion of copper $17\times {10}^{-6}/\text{°F}$, modulus of elasticity of copper is $120\text{GPa}$, bolt diameter is $5\text{mm}$modulus of elasticity of steel is $200\text{GPa}$, thermal expansion coefficient of steel is $12\times {10}^{-6}/\text{°F}$, length of steel bolt is $50\text{mm}$, steel bolt diameter is $17\text{mm}$, number of turns is $\frac{1}{2}$, temperature raise is $30\text{°C}$, pitch is $1\text{mm}$and the gap is $26\text{mm}$, allowable shear stress is $18.5\text{MPa}$.

Write the expression for displacement in bolt.

  ${\delta }_B=P_B\left[\frac{L_1+L_2-S}{E_s{A}_b}\right]$....... (IX)

Here, the displacement in bolt is ${\delta }_B$and joint length is $S$.

Write the expression for displacement in sleeve.

  ${\delta }_s=P_s\left[\frac{L_1-S}{E_c{A}_1}+\frac{L_2-S}{E_c{A}_2}+\frac{S}{E_c\left(A_1+A_2\right)}\right]$....... (X)

Here, the displacement in sleeve is ${\delta }_s$.

Write the expression for final elongation.

  ${\delta }_f={\delta }_b-{\delta }^s$....... (XI)

Here, the final elongation is ${\delta }_f$.

Calculation:

Substitute $25437.86\text{N}$for $P_B$, $200\text{GPa}$for $E_s$, $40\text{mm}$for $L_1$, $50\text{mm}$for $L_2$,and $25.7\text{mm}$for $S$in Equation (IX).

  $\begin{array}{c}{\delta }_B=25437.86\text{N}\left[\frac{40\text{mm}+50\text{mm}-25.7\text{mm}}{\left(200\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\left(19.635{\text{mm}}^{\text{2}}\right)}\right]\\ =25437.86\left(1.6373\times {10}^{-5}\text{mm}/\text{N}\right)\\ =0.416\text{mm}\end{array}$

Substitute $40\text{mm}$for $L_1$, $50\text{mm}$for $L_2$, $120\text{GPa}$for $E_c$, $200\text{GPa}$for $E_s$, $263.893{\text{mm}}^{\text{3}}$for $A_1$, $131.947{\text{mm}}^2$for $A_2$and $25.7\text{mm}$for $S$in Equation (X).

  $\begin{array}{c}{\delta }_s=-25437.86\text{N}\left[\begin{array}{l}\frac{40\text{mm}-25.7\text{mm}}{\left(120\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\left(263.893{\text{mm}}^2\right)}+\\ \frac{50\text{mm}-25.7\text{mm}}{\left(120\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\left(131.947{\text{mm}}^2\right)}\\ +\frac{25.7\text{mm}}{\left(120\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\left(263.893{\text{mm}}^2+131.947{\text{mm}}^2\right)}\end{array}\right]\\ =\left(2.515\times {10}^{-6}\text{mm}/\text{N}\right)\left(-25437.86\text{N}\right)\\ =-0.064\text{mm}\end{array}$

Substitute $-0.064\text{mm}$for ${\delta }_s$and $0.416\text{mm}$for ${\delta }_B$in Equation (XI).

  $\begin{array}{c}{\delta }_f=0.416\text{mm+}\left(-0.064\text{mm}\right)\\ =0.352\text{mm}\end{array}$

Conclusion:

The final displacement is = $0.352\text{mm}$.