Chapter 3, Problem 3.5.13P

Two circular aluminum pipes of equal length L = 24 in. arc loaded by torsional moments T (sec figure). Pipe I has outside and inside diameters d₂= 3 in. and L₂, = 2.5 in., respectively. Pipe 2 has a constant outer diameter of d₂along its entire length L and an inner diameter of d₁but has an increased inner diameter of d₃= 2.65 in. over the middle third.

Assume that E = 10,400 ksi, u = 0.33, and allowable shear stress r_a= 6500 psi.

1. Find the maximum acceptable torques that can be applied to Pipe 1; repeat for Pipe 2.

If the maximum twist e of Pipe 2 cannot exceed 5/4 of that of Pipe 1, what is the maximum acceptable length of the middle segment? Assume both pipes have total length L and the same applied torque T.

Find the new value of inner diameter d₃of Pipe 2 if the maximum torque carried by Pipe 2 is to be 7/8 of that for Pipe L

If the maximum normal strain in each pipe is known to be_max = 811 x 10-6, what is the applied torque on each pipe? Also, what is the maximum twist of each pipe? Use the original properties and dimensions.

  

To determine

The maximum acceptable torques for pipe (1).

The maximum acceptable torques for pipe (2).

Answer to Problem 3.5.13P

Maximum acceptable torques for pipe (1) is = $17842\text{lb}\cdot \text{in}$.

Maximum acceptable torques for pipe (2) is = $13479.375\text{lb}\cdot \text{in}$.

Explanation of Solution

Given information:

The following figure shows the free body diagram of pipe 1:

       Figure-(1) shows the diagram of two pipes:

  

       Figure-(1)

The following figure shows the free body diagram of pipe 2:

  

       Figure-(2)

The length of pipe is $24\text{in}$, outside diameter of pipe (1) is $3\text{in}$, inside diameter of pipe (1) is $2.5\text{in}$, the increased diameter of the pipe (2) is $2.65\text{in}$, the young’s modulus is $10400\text{ksi}$, Poisson ratio is $0.33$and allowable shear stress is $6500\text{psi}$.

Write the expression for polar moment of inertia for pipe (1)

  $I_{p_1}=\frac{\pi }{32}\left(d_2^4-d_1^4\right)$..... (I)

Here, the polar moment of inertia is $I_P$, the outside diameter is $d_2$and inside diameter is $d_1$.

Write the expression for maximum torque for pipe (1)

  ${\tau }_a=\frac{T_a{d}_2}{2I_{p1}}$...... (II)

Here, acceptable shear stress is ${\tau }_a$, the applied torque is $T$, polar moment of inertia is $I_{p1}$and outer diameter is $d_2$.

Write the expression for polar moment of inertia for pipe (2)

  $I_{p_2}=\frac{\pi }{32}\left(d_2^4-d_3^4\right)$..... (III)

Here, the polar moment of inertia is $I_P$, the outside diameter is $d_2$and inside diameter in the middle section is $d_3$.

Write the expression for maximum torque for pipe (2)

  ${\tau }_a=\frac{T_a{d}_2}{2I_{p2}}$...... (IV)

Here, acceptable shear stress is ${\tau }_a$, the applied torque is $T$, polar moment of inertia is $I_{p1}$and outer diameter is $d_2$.

Calculation:

Substitute, $3\text{in}$for $d_2$and $2.5\text{in}$for $d_1$in equation (I)

  $\begin{array}{c}{I}_{p_1}=\frac{\pi }{32}\left({\left(3\text{in}\right)}^4-{\left(2.5\text{in}\right)}^4\right)\\ =\frac{\pi }{32}\left(81{\text{in}}^4-39.0625{\text{in}}^4\right)\\ =4.1172{\text{in}}^4\end{array}$

Substitute $4.1172{\text{in}}^4$for $I_{p_1}$, $6500\text{psi}$for ${\tau }_a$and $3\text{in}$for $d_2$in Equation (II).

  $\begin{array}{l}6500\text{psi}=\frac{T_a\left(3\text{in}\right)}{2\left(4.1172{\text{in}}^4\right)}\\ 6500\text{psi}=0.3643T_a\\ T_a=\frac{6500\text{psi}}{0.3643}\\ T_a=17842\text{lb}\cdot \text{in}\end{array}$

Substitute, $3\text{in}$for $d_2$and $2.65\text{in}$for $d_3$in equation (III)

  $\begin{array}{c}{I}_{p_2}=\frac{\pi }{32}\left({\left(3\text{in}\right)}^4-{\left(2.65\text{in}\right)}^4\right)\\ =\frac{\pi }{32}\left(81{\text{in}}^4-49.3155{\text{in}}^4\right)\\ =3.1106{\text{in}}^4\end{array}$

Substitute $3.1106{\text{in}}^4$for $I_{p_1}$, $6500\text{psi}$for ${\tau }_a$and $3\text{in}$for $d_2$in Equation (II).

  $\begin{array}{l}6500\text{psi}=\frac{T_a\left(3\text{in}\right)}{2\left(3.1106{\text{in}}^4\right)}\\ 6500\text{psi}=0.482T_a\\ T_a=\frac{6500\text{psi}}{0.482}\\ T_a=13479.375\text{lb}\cdot \text{in}\end{array}$

Conclusion:

Maximum acceptable torques for pipe (1) is = $17842\text{lb}\cdot \text{in}$.

Maximum acceptable torques for pipe (2) is = $13479.375\text{lb}\cdot \text{in}$.

To determine

The maximum acceptable length of the middle segment.

Answer to Problem 3.5.13P

The maximum acceptable length of the middle segment.

Explanation of Solution

Given information:

Maximum twist of pipe (2) cannot exceed $5/4$of pipe (1).

Write the expression for the angle of twist for pipe (1).

  ${\varphi }_1=\frac{GL}{T{I}_p}$...... (V)

Here the angle of twist is $\varphi$and shear modulus is $G$.

Write the expression for total angle of twist for pipe (2)

  ${\varphi }_2={\left({\varphi }_2\right)}_{AB}+{\left({\varphi }_2\right)}_{BC}+{\left({\varphi }_2\right)}_{CD}$...... (VI)

Here, the total angle of twist is ${\varphi }_2$, the angle of twist in the section $AB$is ${\left({\varphi }_2\right)}_{AB}$, the angle of twist in the section $BC$is ${\left({\varphi }_2\right)}_{BC}$, ${\varphi }_2$, the angle of twist in the section $CD$is ${\left({\varphi }_2\right)}_{CD}$.

Write the expression for the length of the segment $AB$and $CD$for middle segment length as $x$.

  $L_{AB}=L_{CD}=\frac{L-x}{2}$

Write expression for the angle of twist in the section $AB$.

  ${\left({\varphi }_2\right)}_{AB}=\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{AB}}$

Write expression for the angle of twist in the section $BC$.

  ${\left({\varphi }_2\right)}_{BC}=\frac{G\left(x\right)}{T{\left(I_{p2}\right)}_{BC}}$

Write expression for the angle of twist in the section $CD$.

  ${\left({\varphi }_2\right)}_{CD}=\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{CD}}$

Substitute $\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{AB}}$for ${\left({\varphi }_2\right)}_{AB}$, $\frac{G\left(x\right)}{T{\left(I_{p2}\right)}_{BC}}$for ${\left({\varphi }_2\right)}_{BC}$and $\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{CD}}$for ${\left({\varphi }_2\right)}_{CD}$in Equation (VI).

  $\begin{array}{l}{\varphi }_2=\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{AB}}+\frac{G\left(x\right)}{T{\left(I_{p2}\right)}_{BC}}+\frac{G\left(L-x\right)}{T{\left(I_{p2}\right)}_{CD}}\\ =\frac{G}{T}\left[\frac{\frac{\left(L-x\right)}{2}}{{\left(I_{p2}\right)}_{AB}}+\frac{\left(x\right)}{{\left(I_{p2}\right)}_{BC}}+\frac{\frac{\left(L-x\right)}{2}}{{\left(I_{p2}\right)}_{CD}}\right]\end{array}$...... (VII)

Write the expression for relation between ${\varphi }_1$and ${\varphi }_2$.

  ${\varphi }_2=\frac{5}{4}{\varphi }_1$..... (VIII)

Calculation:

Substitute $4.1172{\text{in}}^4$for $I_p$, $24\text{in}$for $L$in equation (V)

  ${\varphi }_1=\frac{G\left(24\text{in}\right)}{T\left(4.1172{\text{in}}^4\right)}$

Substitute $24\text{in}$for $L$

  $4.1172{\text{in}}^4$for ${\left(I_{p2}\right)}_{AB}$and $3.1106{\text{in}}^4$for ${\left(I_{p2}\right)}_{BC}$in Equation (VII).

  $\begin{array}{c}{\varphi }_2=\frac{G}{T}\left[\frac{\frac{\left(24\text{in}-x\right)}{2}}{4.1172{\text{in}}^4}+\frac{\left(x\right)}{3.1106{\text{in}}^4}+\frac{\frac{\left(24\text{in}-x\right)}{2}}{4.1172{\text{in}}^4}\right]\\ =\frac{G}{T}\left[\frac{\left(24\text{in}-x\right)}{4.1172{\text{in}}^4}+\frac{\left(x\right)}{3.1106{\text{in}}^4}\right]\end{array}$

Substitute $\frac{G}{T}\left[\frac{\left(24\text{in}-x\right)}{4.1172{\text{in}}^4}+\frac{\left(x\right)}{3.1106{\text{in}}^4}\right]$for ${\varphi }_2$and $\frac{G\left(24\text{in}\right)}{T\left(4.1172{\text{in}}^4\right)}$for ${\varphi }_1$in Equation (VIII).

  $\begin{array}{l}\frac{G}{T}\left[\frac{\left(24\text{in}-x\right)}{4.1172{\text{in}}^4}+\frac{\left(x\right)}{3.1106{\text{in}}^4}\right]=\frac{5}{4}\frac{G\left(24\text{in}\right)}{T\left(4.1172{\text{in}}^4\right)}\\ \left[\frac{\left(24\text{in}-x\right)}{4.1172{\text{in}}^4}+\frac{\left(x\right)}{3.1106{\text{in}}^4}\right]=\frac{5}{4}\frac{\left(24\text{in}\right)}{\left(4.1172{\text{in}}^4\right)}\\ \frac{x}{3.1106{\text{in}}^4}-\frac{x}{4.1172{\text{in}}^4}=\frac{\left(24\text{in}\right)}{\left(4.1172{\text{in}}^4\right)}\left(\frac{5}{4}-1\right)\\ 0.7859x=1.4573\end{array}$

  $x=18.5412\text{in}$

Conclusion:

The maximum acceptable length of the middle segment is = $18.5412\text{in}$.

To determine

The inner diameter for the given parameters.

Answer to Problem 3.5.13P

The inner diameter for the given parameters is = $2.58\text{in}$.

Explanation of Solution

Given information:

the maximum torque carried by pipe (2) is $7/8$of pipe (1).

Write the expression for allowable torque.

  $T_{allow}=\frac{7}{8}{T}_a$...... (IX)

Here, the allowable torque is $T_{allow}$

Substitute $\frac{7}{8}{T}_a$for $T_a$and $\frac{\pi }{32}\left({\left(\text{3}\text{in}\right)}^4-d_3^4\right)$for $I_{p2}$in Equation (IV).

  ${\tau }_a=\frac{\frac{7}{8}{T}_a{d}_2}{2\frac{\pi }{32}\left({\left(d_2\right)}^4-d_3^4\right)}$..... (X)

Calculation:

Substitute $6500\text{psi}$for ${\tau }_a$, $17810\text{psi}$for $T_a$and $3\text{in}$for $d_2$in Equation (X).

  $\begin{array}{l}6500\text{psi}=\frac{\frac{7}{8}\left(17810\text{psi}\right)\left(3\text{in}\right)}{2\frac{\pi }{32}\left({\left(3\text{in}\right)}^4-d_3^4\right)}\\ \left({\left(3\text{in}\right)}^4-d_3^4\right)=\frac{\frac{7}{8}\left(17810\text{psi}\right)\left(3\text{in}\right)}{2\frac{\pi }{32}6500\text{psi}}\\ \left({\left(3\text{in}\right)}^4-d_3^4\right)=36.63{\text{in}}^{\text{4}}\\ d_3^4=81{\text{in}}^4-36.63{\text{in}}^{\text{4}}\end{array}$

  $\begin{array}{l}{d}_3^4=44.369{\text{in}}^4\\ d_3=2.58\text{in}\end{array}$

Conclusion:

The inner diameter for the given parameters is $2.58\text{in}$.

To determine

Applied torque on pipe (1)

Maximum twist of pipe (1)

Answer to Problem 3.5.13P

Applied torque on pipe (1) is = $17406.56\text{lb}\cdot \text{in}$.

Applied torque on pipe (2) is = $13148.35\text{lb}\cdot \text{in}$.

Maximum twist of pipe (1) is = $1.487°$.

Explanation of Solution

Write the expression for maximum shear strain.

  ${\gamma }_{\max }=2{\epsilon }_{\max }$..... (XI)

Here, maximum shear strain is ${\gamma }_{\max }$and the allowable tensile strain is ${\epsilon }_{\max }$.

Write the expression for maximum shear stress.

  ${\tau }_{\max }=G{\gamma }_{\max }$...... (XII)

Here, the maximum shear stress is ${\tau }_{\max }$, the shear modulus of the shaft is $G$and maximum shear strain is ${\gamma }_{\max }$.

Write the expression for shear modulus of elasticity.

  $G=\frac{E}{2\left(1+v\right)}$..... (XIII)

Write the expression for maximum torque for pipe (1)

  ${\tau }_{\max }=\frac{T_{\max }{d}_2}{2I_{p1}}$...... (XIV)

Write the expression for maximum angle of twist for pipe (1)

  ${\varphi }_1=\frac{T_{\max }\times L}{G{I}_p}$...... (XV)

Maximum angle of twist in pipe (1) is ${\varphi }_1$.

Calculation:

Substitute $10400\text{ksi}$for $E$and $0.33$for $v$, $4.1172{\text{in}}^4$for $I_{p1}$and $24\text{}\text{in}$for $L$in Equation (XIII).

  $\begin{array}{l}G=\frac{10400\text{ksi}}{2\left(1+0.33\right)}\\ G=\frac{10400\text{ksi}}{2.66}\\ G=3909.77\text{psi}\end{array}$

Substitute $811\times {10}^{-6}$for ${\epsilon }_{\max }$in Equation (XI).

  $\begin{array}{c}{\gamma }_{\max }=2\left(811\times {10}^{-6}\right)\\ =1.622\times {10}^{-3}\end{array}$

Substitute $1.622\times {10}^{-3}$for ${\gamma }_{\max }$and $3909.77\text{psi}$for $G$in Equation (XII).

  $\begin{array}{c}{\tau }_{\max }=\left(3909.77\text{ksi}\right)\left(1.622\times {10}^{-3}\right)\\ =6341.65\text{psi}\end{array}$

Substitute $6341.65\text{psi}$for ${\tau }_{\max }$, $4.1172{\text{in}}^4$for $I_{p1}$, $3\text{in}$for $d_2$in equation (XIV).

  $\begin{array}{l}6341.65\text{psi}=\frac{T_{\max }\left(3\text{in}\right)}{2\left(4.1172{\text{in}}^4\right)}\\ T_{\max }=\frac{\left(6341.65\text{psi}\right)\left(2\left(4.1172{\text{in}}^4\right)\right)}{3\text{in}}\\ T_{\max }=17406.56\text{lb}\cdot \text{in}\end{array}$

Substitute $17406.56\text{lb}\cdot \text{in}$for $T_{\max }$

  $24\text{in}$for $L$, $4.1172{\text{in}}^4$for $I_p$

  $3909.77\text{psi}$for $G$in equation (XV).

  $\begin{array}{c}{\varphi }_1=\frac{\left(17406.56\text{lb}\cdot \text{in}\right)\times 24\text{in}}{\left(3909.77\text{psi}\right)\left(4.1172{\text{in}}^4\right)}\\ =\frac{427740\text{lb}\cdot {\text{in}}^2}{42818880\text{lb}\cdot {\text{in}}^2}\\ =\left(0.02594\text{rad}\right)\left(\frac{180°}{\pi \text{rad}}\right)\\ =1.487°\end{array}$

Substitute $6341.65\text{psi}$for ${\tau }_{\max }$, $3.11{\text{in}}^4$for $I_{p1}$, $3\text{in}$for $d_2$in equation (XIV).

  $\begin{array}{l}6341.65\text{psi}=\frac{T_{\max }\left(3\text{in}\right)}{2\left(3.11{\text{in}}^4\right)}\\ T_{\max }=\frac{\left(6341.65\text{psi}\right)\left(2\left(3.11{\text{in}}^4\right)\right)}{3\text{in}}\\ T_{\max }=13148.35\text{lb}\cdot \text{in}\end{array}$

Conclusion:

Applied torque on pipe (1) is = $17406.56\text{lb}\cdot \text{in}$.

Applied torque on pipe (2) is = $13148.35\text{lb}\cdot \text{in}$.

Maximum twist of pipe (1) is = $1.487°$.