Chapter 2, Problem 2.5.21P

Pipe 2 has been inserted snugly into Pipe I. but the holes Tor a connecting pin do not line up; there is a gap s. The user decides to apply either force P_:lo Pipe I or force P-, to Pipe 2, whichever is smaller. Determine the following using the numerical properties in the box.

(a) If only P_{is applied, find P_t{tips} required to close gap s; if a pin is then inserted and P_tremoved, what are reaction forces R_Aand R_Bfor this load case?

(b) If only P₂is applied, find P₂{kips) required to close gap a; if a pin is inserted and P₂removed, what are reaction forces R^ and R_Bfor this load case?

(c) What is the maximum shear stress in the pipes, for the loads in parts (a) and (b)?

(d) If a temperature increase IT is to be applied to the entire structure to close gaps{instead of applying forces P_tand P₂), find the AT required to close the gap. If a pin is inserted after the gaphas closed, what are reaction forces .''.', and R_Bfor this case? (e) Finally, if the structure (with pin inserted) then cools to the original ambient temperature, what are reaction forces R_tand P

To determine

The reactions at A and B.

Answer to Problem 2.5.21P

The reaction at A is = $-55.24\text{kips}$.

The reaction at B is = $55.24\text{kips}$.

Explanation of Solution

Given information:

The gap between the pipes is $s$and the force $P_1$is applied, modulus of elasticity of pipe 1 is $30000\text{ksi}$, modulus of elasticity of pipe 2 is $14000\text{ksi}$, the gap is $0.05\text{in}$, length of pipe 1 is $56\text{in}$, length of pipe 2 is $36\text{in}$, diameter of pipe 1 is $6\text{in}$, diameter of pipe 2 is $5\text{in}$, thickness of pipe 1 is $0.5\text{in}$thickness of pipe 2 is $0.25\text{in}$, area of pipe 1 is $8.64{\text{in}}^2$and the area of pipe 2 is $3.73{\text{in}}^2$.

Write the expression for elongation in pipe 1.

  $s=\frac{P_1{L}_1}{E_1{A}_1}$....... (I)

Here, length of pipe 1 is $L_1$, modulus of elasticity of pipe 1 is $E_1$and the area is $A_1$.

Write the expression for elongation at point B.

  ${\left({\delta }_B\right)}_2=R_B\left(\frac{L_1}{A_1{E}_1}+\frac{L_2}{A_2{E}_2}\right)$........ (II)

Here, elongation at point B is ${\left({\delta }_B\right)}_2$, length of pipe 2 is $L_2$, modulus of elasticity of pipe 2 is $E_2$, and area of pipe 2 is $A_2$and the reaction at B is $R_B$.

Write the expression for net elongation at B.

  ${\left({\delta }_B\right)}_1+{\left({\delta }_B\right)}_2=0$........ (III)

Here, elongation at point 1 is ${\left({\delta }_B\right)}_1$.

Substitute $-s$for ${\left({\delta }_B\right)}_1$and $R_B\left(\frac{L_1}{A_1{E}_1}+\frac{L_2}{A_2{E}_2}\right)$for ${\left({\delta }_B\right)}_2$in Equation (III).

  $\begin{array}{l}-s+R_B\left(\frac{L_1}{A_1{E}_1}+\frac{L_2}{A_2{E}_2}\right)=0\\ R_B=\frac{s}{\frac{L_1}{A_1{E}_1}+\frac{L_2}{A_2{E}_2}}\end{array}$....... (IV)

Write the reaction at point A.

  $R_A=-R_B$........ (V)

Here, reaction at A is $R_A$.

Calculation:

Substitute $0.05\text{in}$for $s$, $56\text{in}$for $L_1$, $30000\text{ksi}$for $E_1$and $8.64{\text{in}}^2$for $A_1$in Equation (I).

  $\begin{array}{l}0.05\text{in}=\frac{P_1\left(56\text{in}\right)}{\left(30000\text{ksi}\right)\left(8.64{\text{in}}^2\right)}\\ P_1=\frac{0.05\text{in}\times 30000\text{ksi}\times 8.64{\text{in}}^2}{56\text{in}}\\ P_1=231.43\text{kips}\end{array}$

The force required to close the gap is $231.43\text{kips}$.

Substitute $0.05\text{in}$for $s$, $56\text{in}$for $L_1$, $30000\text{ksi}$for $E_1$and $8.64{\text{in}}^2$for $A_1$, $36\text{in}$for $L_2$, $3.73{\text{in}}^2$for $A_2$, $14000\text{ksi}$for $E_2$in and $231.43\text{kips}$for $P_1$Equation (IV).

  $\begin{array}{c}{R}_B=\frac{0.05\text{in}}{\frac{\left(56\text{in}\right)}{\left(30000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(8.64{\text{in}}^2\right)}+\frac{\left(36\text{in}\right)}{\left(14000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(3.73{\text{in}}^2\right)}}\\ =\frac{0.05\text{in}}{2.16\times {10}^{-7}\text{psi}\cdot {\text{in}}^2+6.89\times {10}^{-7}\text{psi}\cdot {\text{in}}^2}\\ =\left(55.24\times {10}^3\text{psi}\cdot {\text{in}}^2\right)\left(\frac{1\text{kips}}{1000\text{psi}\cdot {\text{in}}^2}\right)\\ =55.24\text{kips}\end{array}$

Substitute $55.24\text{kips}$for $R_B$in Equation (V).

  $R_A=-55.24\text{kips}$

Conclusion:

The reaction at A is $-55.24\text{kips}$.

The reaction at B is $55.24\text{kips}$.

To determine

The reaction at A is $-55.24\text{kips}$.

The reaction at B is $55.24\text{kips}$.

Answer to Problem 2.5.21P

The reaction at A is $-55.24\text{kips}$.

The reaction at B is $55.24\text{kips}$.

Explanation of Solution

Given information:

The gap between the pipes is $s$, modulus of elasticity of pipe 1 is $30000\text{ksi}$, modulus of elasticity of pipe 2 is $14000\text{ksi}$, the gap is $0.05\text{in}$, length of pipe 1 is $56\text{in}$, length of pipe 2 is $36\text{in}$, diameter of pipe 1 is $6\text{in}$, diameter of pipe 2 is $5\text{in}$, thickness of pipe 1 is $0.5\text{in}$, thickness of pipe 2 is $0.25\text{in}$, area of pipe 1 is $8.64{\text{in}}^2$and the area of pipe 2 is $3.73{\text{in}}^2$.

Write the expression for force applied at pipe 2.

  $P_2=2\left(\frac{E_2{A}_2}{L_2}\right)s$…... (VI)

Here, force applied at pipe 2 is $P_2$.

Calculation:

Substitute $36\text{in}$for $L_2$, $3.73{\text{in}}^2$for $A_2$, $14000\text{ksi}$for $E_2$and $0.05\text{in}$for $s$in Equation (VI).

  $\begin{array}{c}{P}_2=2\left(\frac{\left(14000\text{ksi}\right)\left(3.73{\text{in}}^2\right)}{36\text{in}}\right)0.05\text{in}\\ =\frac{\left(0.1\text{in}\right)\left(14000\text{ksi}\right)\left(3.73{\text{in}}^2\right)}{36\text{in}}\\ =145.05\text{kips}\end{array}$

Substitute $0.05\text{in}$for $s$, $56\text{in}$for $L_1$, $30000\text{ksi}$for $E_1$and $8.64{\text{in}}^2$for $A_1$, $36\text{in}$for $L_2$, $3.73{\text{in}}^2$for $A_2$, $14000\text{ksi}$for $E_2$in Equation (IV).

  $\begin{array}{c}{R}_B=\frac{0.05\text{in}}{\frac{\left(56\text{in}\right)}{\left(30000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(8.64{\text{in}}^2\right)}+\frac{\left(36\text{in}\right)}{\left(14000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(3.73{\text{in}}^2\right)}}\\ =\frac{0.05\text{in}}{2.16\times {10}^{-7}\text{psi}\cdot {\text{in}}^2+6.89\times {10}^{-7}\text{psi}\cdot {\text{in}}^2}\\ =\left(55.24\times {10}^3\text{psi}\cdot {\text{in}}^2\right)\left(\frac{1\text{kips}}{1000\text{psi}\cdot {\text{in}}^2}\right)\\ =55.24\text{kips}\end{array}$

Substitute $55.24\text{kips}$for $R_B$in Equation (V).

  $R_A=-55.24\text{kips}$

Conclusion:

The reaction at A is $-55.24\text{kips}$.

The reaction at B is $55.24\text{kips}$.

To determine

The maximum shear stress in pipe 1 and pipe 2.

.

Answer to Problem 2.5.21P

The maximum shear stress in pipe 1 is $13.9\text{ksi}$.

The maximum shear stress in pipe 2 is $19.44\text{ksi}$.

Explanation of Solution

Given information:

The gap between the pipes is $s$and the force $P_1$is applied, modulus of elasticity of pipe 1 is $30000\text{ksi}$, modulus of elasticity of pipe 2 is $14000\text{ksi}$, the gap is $0.05\text{in}$, length of pipe 1 is $56\text{in}$, length of pipe 2 is $36\text{in}$, diameter of pipe 1 is $6\text{in}$, diameter of pipe 2 is $5\text{in}$, thickness of pipe 1 is $0.5\text{in}$, thickness of pipe 2 is $0.25\text{in}$, area of pipe 1 is $8.64{\text{in}}^2$and the area of pipe 2 is $3.73{\text{in}}^2$.

Write the expression for maximum shear stress in pipe 1.

  ${\tau }_{\max 1}=\frac{\frac{P_1}{A_1}}{2}$........ (VII)

Here, maximum shear stress in pipe 1 is ${\tau }_{\max 1}$.

Write the expression for maximum shear stress in pipe.

  ${\tau }_{\max 2}=\frac{\frac{P_2}{A_2}}{2}$........ (VIII)

Here, maximum shear stress in pipe 2 is ${\tau }_{\max 2}$.

Calculation:

Substitute $231.43\text{kips}$for $P_1$and $8.64{\text{in}}^2$for $A_1$in Equation (VII).

  $\begin{array}{c}{\tau }_{\max 1}=\frac{\frac{231.43\text{kips}}{8.64{\text{in}}^2}}{2}\\ =\frac{26.7858\text{kips}/{\text{in}}^2}{2}\\ =\left(13.39\text{kips}/{\text{in}}^2\right)\left(\frac{1\text{ksi}}{1\text{kips}/{\text{in}}^2}\right)\\ =1.39\text{ksi}\end{array}$

The maximum shear stress in pipe 1 is = $1.39\text{ksi}$.

Substitute $145.05\text{kips}$for $P_2$

  $3.73{\text{in}}^2$for $A_2$in Equation (VIII).

  $\begin{array}{c}{\tau }_{\max 2}=\frac{\frac{145.05\text{kips}}{3.73{\text{in}}^2}}{2}\\ =\frac{38.8873\text{kips}/{\text{in}}^2}{2}\\ =\left(19.44\text{kips}/{\text{in}}^2\right)\left(\frac{1\text{ksi}}{1\text{kips}/{\text{in}}^2}\right)\\ =19.44\text{ksi}\end{array}$

Conclusion:

The maximum shear stress in pipe 1 is = $13.9\text{ksi}$.

The maximum shear stress in pipe 2 is = $19.44\text{ksi}$.

To determine

The rise in temperature required to close the gap.

The reactions.

Answer to Problem 2.5.21P

The rise in temperature required to close the gap is $65.8\text{°F}$.

The reactions are $0$.

Explanation of Solution

Given information:

The gap between the pipes is $s$and the force $P_1$is applied, modulus of elasticity of pipe 1 is $30000\text{ksi}$, modulus of elasticity of pipe 2 is $14000\text{ksi}$, the gap is $0.05\text{in}$, length of pipe 1 is $56\text{in}$, length of pipe 2 is $36\text{in}$, diameter of pipe 1 is $6\text{in}$, diameter of pipe 2 is $5\text{in}$, thickness of pipe 1 is $0.5\text{in}$, thickness of pipe 2 is $0.25\text{in}$, area of pipe 1 is $8.64{\text{in}}^2$and the area of pipe 2 is $3.73{\text{in}}^2$.

Write the expression for temperature raise.

  $\Delta T=\frac{s}{{\alpha }_1{L}_1+{\alpha }_2{L}_2}$........ (IX)

Here, raise in temperature is $\Delta T$, coefficient of thermal expansion for pipe 1 is ${\alpha }_1$and the coefficient of thermal expansion for pipe 2 is ${\alpha }_2$.

Calculation:

Substitute $0.05\text{in}$for $s$, $56\text{in}$for $L_1$, $36\text{in}$for $L_2$, $6.56\times {10}^{-6}/\text{°F}$for ${\alpha }_1$and $11\times {10}^{-6}/\text{°F}$for ${\alpha }_2$in Equation (IX).

  $\begin{array}{c}\Delta T=\frac{0.05\text{in}}{\left(6.5\times {10}^{-6}/\text{°F}\right)\left(56\text{in}\right)+\left(11\times {10}^{-6}/\text{°F}\right)\left(36\text{in}\right)}\\ =\frac{0.05\text{in}}{760\times {10}^{-6}\text{in}/\text{°F}}\\ =65.8\text{°F}\end{array}$

Since the temperature remains constant, so the reactions are zero.

Conclusion:

The temperature raise required to close the gap is $65.8\text{°F}$.

The reactions are $0$.

To determine

The reaction at A.

The reaction at B.

Answer to Problem 2.5.21P

The reaction at A is $-55.24\text{kips}$.

The reaction at B is $55.24\text{kips}$.

Explanation of Solution

Given information:

The gap between the pipes is $s$and the force $P_1$is applied, modulus of elasticity of pipe 1 is $30000\text{ksi}$, modulus of elasticity of pipe 2 is $14000\text{ksi}$, the gap is $0.05\text{in}$, length of pipe 1 is $56\text{in}$, length of pipe 2 is $36\text{in}$, diameter of pipe 1 is $6\text{in}$, diameter of pipe 2 is $5\text{in}$, thickness of pipe 1 is $0.5\text{in}$, thickness of pipe 2 is $0.25\text{in}$, area of pipe 1 is $8.64{\text{in}}^2$and the area of pipe 2 is $3.73{\text{in}}^2$.

Calculation:

Substitute $0.05\text{in}$for $s$, $56\text{in}$for $L_1$, $30000\text{ksi}$for $E_1$and $8.64{\text{in}}^2$for $A_1$, $36\text{in}$for $L_2$, $3.73{\text{in}}^2$for $A_2$, $14000\text{ksi}$for $E_2$in and $231.43\text{kips}$for $P_1$Equation (IV).

  $\begin{array}{c}{R}_B=\frac{0.05\text{in}}{\frac{\left(56\text{in}\right)}{\left(30000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(8.64{\text{in}}^2\right)}+\frac{\left(36\text{in}\right)}{\left(14000\text{ksi}\right)\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\left(3.73{\text{in}}^2\right)}}\\ =\frac{0.05\text{in}}{2.16\times {10}^{-7}\text{psi}\cdot {\text{in}}^2+6.89\times {10}^{-7}\text{psi}\cdot {\text{in}}^2}\\ =\left(55.24\times {10}^3\text{psi}\cdot {\text{in}}^2\right)\left(\frac{1\text{kips}}{1000\text{psi}\cdot {\text{in}}^2}\right)\\ =55.24\text{kips}\end{array}$

Substitute $55.24\text{kips}$for $R_B$in Equation (V).

  $R_A=-55.24\text{kips}$

Conclusion:

The reaction at A is = $-55.24\text{kips}$.

The reaction at B is = $55.24\text{kips}$.