Chapter 3, Problem 3.4.15P

A uniformly tapered aluminum-alloy tube AB with a circular cross section and length L is shown in the figure. The outside diameters at the ends arc d_Aand d_B= 2d_A. A hollow section of length LB and constant thickness t = d_A/10 is cast into the tube and extends from B halfway toward A. (a)

Find the angle of twist é of the tube

when it is subjected to torques T acting at the ends. Use numerical values: d_A= 2.5 in., L = 48 in., G = 3.9 × 10⁶ psi, and T =40,000 in.-lb.

(b)

Repeat part (a) if the hollow section has con stant diameter rfj (see figure part b)

  

To determine

The angle of twist of the tube when it is subjected to torques at it ends.

Answer to Problem 3.4.15P

The angle of twist of the tube when it is subjected to torques at it end $A$is $2.793°$ .

The angle of twist of the tube when it is subjected to torques at it end $B$is $2.207°$

Explanation of Solution

Given information:

The diameter of the smaller section of the tube is $2.5\text{in}$ , the length of the tube is $48\text{in}\text{.}$ , the torque acting on the body is $40000\text{in}\cdot \text{lb}$and the diameter of the larger section of the tube is $5\text{in}$ .

Write the expression for outer diameter.

  $d_{\circ }=2dA-\frac{xdA}{L}$

Here, the elemental area of the tube is $dA$ , the length of the elemental area is $x,$and the length of the tube is $L$ .

Write the expression for inner diameter.

  $d_i=-\frac{1}{5}dA-\frac{\left(9L+5x\right)}{L}$

Write the expression for angle of twist of side $A$ .

  ${\varphi }_A=\frac{T}{G}\left(\frac{32}{\pi }\right)\left({\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{d^4{}_{\circ }-d^4{}_i}dx}+{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{d^4{}_{\circ }}dx}\right)$...... (I)

Substitute $2dA-\frac{xdA}{L}$for $d_{\circ }$and $-\frac{1}{5}dA-\frac{\left(9L+5x\right)}{L}$for $d_i$in Equation (I).

  $\begin{array}{c}{\varphi }_A=\frac{T}{G}\left(\frac{32}{\pi }\right)\left[\begin{array}{l}{\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{{\left(2dA-x\frac{dA}{L}\right)}^4-{\left(-\frac{1}{5}dA-\frac{9L+5x}{2}\right)}^4}dx}\\ +{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{{\left(2dA-x\frac{dA}{L}\right)}^4}}dx\end{array}\right]\\ =\left[\frac{T}{G}\left(\frac{32}{\pi }\right)\left(\begin{array}{l}-\frac{125L}{2}-\left(\frac{3\ln \left(2\right)+2\ln \left(7\right)-\ln \left(197\right)}{d{A}^4}\right)\\ -\frac{125L}{2}-\left(\frac{2\ln \left(19\right)+\ln \left(181\right)}{d{A}^4}\right)+\frac{19L}{81d{A}^4}\end{array}\right)\right]\end{array}$

  $\begin{array}{c}{\varphi }_A=\frac{16TL}{81G\pi d^4{}_A}\left(38+10125\ln \left(\frac{71117}{70952}\right)\right)\\ =\frac{3.868TL}{G\pi d^4{}_A}\end{array}$...... (II)

Here, the torque at the ends of the tube is $T$ , the length of the tube is $L,$and the diameter of the smaller side is $d_A$ .

Write the expression for angle of twist of side $B$ .

  $\begin{array}{l}{\varphi }_B=\frac{T}{G}\left(\frac{32}{\pi }\right)\left[\begin{array}{l}{\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{{\left(2dA-x\frac{dA}{L}\right)}^4-{\left(dA\right)}^4}dx}\\ +{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{{\left(2dA-x\frac{dA}{L}\right)}^4}}dx\end{array}\right]\\ {\varphi }_B=\frac{T}{G}\left(\frac{32}{\pi }\right)\left[\frac{1}{4}\frac{L\ln \left(5\right)+20\tan \left(\frac{3}{2}\right)}{d{A}^4}-\frac{1}{4}\frac{L\ln \left(3\right)+20\tan \left(\frac{3}{2}\right)}{d{A}^4}+\frac{19L}{81d{A}^4}\right]\\ =3.057\frac{TL}{G{d}_A{}^4}\end{array}$ ...... (III)

Calculation:

Substitute $40000\text{in}\cdot \text{lb}$for $T$ , $48\text{in}$for $L$

  $3.9\times {10}^6\text{psi}$for $G,$and $2.5\text{in}$for $d_A$ .in Equation (II).

  $\begin{array}{c}{\varphi }_A=3.868\times \frac{40000\text{in}\cdot \text{lb}\times 48\text{in}}{3.9\times {10}^6\text{psi}\times {\left(2.5\text{in}\right)}^4}\\ =3.868\times \frac{40000\text{in}\cdot \text{lb}\times 48\text{in}}{3.9\times {10}^6\text{psi}\times \left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\times {\left(2.5\text{in}\right)}^4}\\ =3.868\times \frac{1.92\times {10}^6{\text{in}}^2\cdot \text{lb}}{152.343\times {10}^6{\text{in}}^2\cdot \text{lb}}\\ =0.04875\text{radian}\end{array}$

  $\begin{array}{c}{\varphi }_A=0.04875\text{radian}\\ =0.04875\text{radian}\left(\frac{180°}{\pi \text{radian}}\right)\\ =2.793°\end{array}$

Substitute $40000\text{in}\cdot \text{lb}$for $T$ , $48\text{in}$for $L$

  $3.9\times {10}^6\text{psi}$for $G$and $5\text{in}$for $d_B$ .in Equation (III).

  $\begin{array}{c}{\varphi }_B=3.057\times \frac{40000\text{in}\cdot \text{lb}\times 48\text{in}}{3.9\times {10}^6\text{psi}\times {\left(2.5\text{in}\right)}^4}\\ =3.057\times \frac{40000\text{in}\cdot \text{lb}\times 48\text{in}}{3.9\times {10}^6\text{psi}\times \left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\times {\left(2.5\text{in}\right)}^4}\\ =3.057\times \frac{1.92\times {10}^6{\text{in}}^2\cdot \text{lb}}{152.343\times {10}^6{\text{in}}^2\cdot \text{lb}}\\ =0.03852\text{radian}\end{array}$

  $\begin{array}{c}{\varphi }_B=0.03852\text{radian}\\ =0.03852\text{radian}\left(\frac{180°}{\pi \text{radian}}\right)\\ =2.207°\end{array}$

Conclusion:

The angle of twist of the tube when it is subjected to torques at it end $A$is $2.793°$ .

The angle of twist of the tube when it is subjected to torques at it end $B$is $2.207°$

To determine

The angle of twist when the tapered shaft is hollow.

Answer to Problem 3.4.15P

The angle of twist of the tube when it is subjected to torques at it end $A$is $22.46°$ .

The angle of twist of the tube when it is subjected to torques at it end $B$is $-7.53°$ .

Explanation of Solution

Write the expression for angle of twist on side $A$ .

  ${\varphi }_A=\frac{32\left(T_{\circ }-T_A\right)}{\pi G}{\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{d_{\circ }{}^4}dx}$...... (IV)

The figure below shows the torque acting on the tube.

  

Figure-(1)

Write the expression for angle of twist on side $B$ .

  ${\varphi }_B=\frac{32\left(T_{\circ }-T_B\right)}{\pi G}{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{d_{\circ }{}^4-d_i{}^4}dx}$ ....... (V)

Write the expression for equation of compatibility.

  ${\varphi }_A+{\varphi }_B=0$...... (VI)

Substitute $\frac{32\left(T_{\circ }-T_A\right)}{\pi G}{\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{d_{\circ }{}^4}dx}$for ${\varphi }_A$and $\frac{32\left(T_{\circ }-T_B\right)}{\pi G}{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{d_{\circ }{}^4-d_i{}^4}dx}$for ${\varphi }_B$in Equation (VI).

  $\frac{32T_B}{\pi G}{\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{1}{d_{\circ }{}^4}dx-}\frac{32T_A}{\pi G}{\displaystyle \underset{L/2}{\overset{L}{\int }}\frac{1}{d_{\circ }{}^4-d_i{}^4}dx}$ ..... (VII)

Write the expression for the total torque $T_{\circ }$ .

  $T_{\circ }=T_A+T_B$...... (IX)

Here, the torque acting on the side $A$is $T_A$and the torque onthe side $B$is $T_B$ .

Calculation:

Substitute $48\text{in}$for $L$in Equation (VII).

  $\begin{array}{l}\frac{T_B}{T_A}=\frac{{\displaystyle \underset{24}{\overset{48}{\int }}\frac{{\left(48\text{in}\right)}^4}{{\left(48\text{in}+x\right)}^4-{\left(48\text{in}\right)}^4}dx}}{{\displaystyle \underset{0}{\overset{24}{\int }}\frac{{\left(48\text{in}\right)}^4}{{\left(48\text{in}+x\right)}^4}dx}}\\ \frac{T_B}{T_A}=\frac{11.2593}{3.1454}\\ T_B=3.5796T_A\end{array}$...... (VIII)

Substitute $3.5796T_A$for $T_B$in Equation (IX)

  $\begin{array}{l}{T}_A+\frac{11.2539}{3.1454}{T}_A=40000\text{in}\cdot \text{lb}\\ T_A+3.57789T_A=40000\text{in}\cdot \text{lb}\\ \text{4}{\text{.57789T}}_A=40000\text{in}\cdot \text{lb}\\ T_A=87376\text{in}\cdot \text{lb}\end{array}$

Substitute $87376\text{in}\cdot \text{lb}$for $T_A$in Equation (VIII).

  $\begin{array}{c}{T}_B=3.5796\left(87376\text{in}\cdot \text{lb}\right)\\ =312771.12\text{in}\cdot \text{lb}\end{array}$

Substitute $87376\text{in}\cdot \text{lb}$for $T_A$and $312771.12\text{in}\cdot \text{lb}$in Equation (IX).

  $\begin{array}{c}{T}_{\circ }=87376\text{in}\cdot \text{lb}+312771.12\text{in}\cdot \text{lb}\\ =400147.1296\text{in}\cdot \text{lb}\end{array}$

Substitute $400147.1296\text{in}\cdot \text{lb}$for $T_{\circ }$and $87376\text{in}\cdot \text{lb}$ for $T_A$in Equation (IV).

  $\begin{array}{c}{\varphi }_A=\frac{32\left(400147.1296\text{in}\cdot \text{lb}-87376\text{in}\cdot \text{lb}\right)}{\pi \times \left(3.9\times {10}^6\text{psi}\right)}{\displaystyle \underset{0}{\overset{24}{\int }}\frac{1}{{\left(5\text{in}\right)}^4}}\\ =\frac{32\left(3.91\times {10}^6\text{in}\cdot \text{lb}\right)}{\pi \times \left(3.9\times {10}^6\text{psi}\times \left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\right)}\times \frac{1}{625{\text{in}}^4}{\displaystyle \underset{0}{\overset{24}{\int }}1\cdot dx}\\ =10.212{\text{in}}^3\times \frac{1}{625{\text{in}}^4}\times \left[24-0\right]\\ =0.3921\text{radian}\end{array}$

  $\begin{array}{c}{\varphi }_A=0.3921\text{radian}\\ =0.3921\text{radian}\left(\frac{180°}{\pi \text{radian}}\right)\\ =22.46°\end{array}$

Substitute $400147.1296\text{in}\cdot \text{lb}$for $T_{\circ }$and $312771.12\text{in}\cdot \text{lb}$for $T_B$in Equation (V).

  $\begin{array}{c}{\varphi }_B=\frac{32\left(400147.1296\text{in}\cdot \text{lb}-312771.12\text{in}\cdot \text{lb}\right)}{\pi \left(3.2\times {10}^6\text{psi}\right)}{\displaystyle \underset{24}{\overset{48}{\int }}\frac{1}{{\left(5\text{in}\right)}^4-{\left(2.5\text{in}\right)}^4}}\\ =\frac{2.79603\times {10}^6\text{in}\cdot \text{lb}}{\pi \left(3.9\times {10}^6\text{psi}\times \left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\right)}\times \left(-0.024{\text{in}}^4\right){\displaystyle \underset{24}{\overset{48}{\int }}dx}\\ =\left(-5.47\times {10}^{-3}\right)\times \left(\text{24}\right)\\ =-0.13144\text{radian}\end{array}$

  $\begin{array}{c}{\varphi }_A=-0.13144\text{radian}\\ =-0.13144\text{radian}\left(\frac{180°}{\pi \text{radian}}\right)\\ =-7.53°\end{array}$

Conclusion:

The angle of twist of the tube when it is subjected to torques at it end $A$is $22.46°$ .

The angle of twist of the tube when it is subjected to torques at it end $B$is $-7.53°$ .