Chapter 3, Problem 3.8.9P

Two sections of steel drill pipe, joined by bolted flange plates at B, arc subjected to a concentrated torque 4000 kip-in. at x = 3 ft, and a uniformly distributed torque t₀= 50 kip-ft/ft is applied on pipe BC. Let G = 11,800 ksi and assume that pipes AB and BC have the same inner diameter, d = 12 in. Pipe AB has a thickness t_AB= 3/4 in., and pipe BC has a thickness t_BC= 5/8 in. Find the reactive torques at A and C and the maximum shear stresses in each segment.

  

To determine

The reaction torque at A and the reaction torque at B .

The twist angle at the center of the bar.

Answer to Problem 3.8.9P

The reaction torque at A is $\text{225}\text{N}\cdot \text{m}$ and the reaction torque at B is $\text{225}\text{N}\cdot \text{m}$ .

The twist angle at the center of the bar is $-4.34°$ .

Explanation of Solution

Given information:

The concentrated torque is $400\text{kip}\cdot \text{ft}/\text{ft}$ , diameter of shaft is $12\text{in}$ , diameter of the right shaft is $25\text{mm}$ , length is $\text{3}\text{m}$ and the distributed torque is $150\text{kip}\cdot \text{ft}/\text{ft}$ , thickness of left pipe is $\frac{3}{4}\text{in}$ and the thickness of right pipe is $\frac{5}{8}\text{in}$ .

Write the expression for polar moment of inertia of left bar.

  $I_{PAB}=\frac{\pi }{32}\left[{\left(d+2t_{AB}\right)}^4-d^4\right]$........ (I)

Here, polar moment of inertia of the left bar is $I_{PAB}$ and the thickness of the left bar is $t_{AB}$ .

Write the expression for polar moment of inertia of right bar.

  $I_{PBC}=\frac{\pi }{32}\left[{\left(d+2t_{BC}\right)}^4-d^4\right]$ ........ (II)

Here, polar moment of inertia of the right bar is $I_{PBC}$ and the thickness of the right bar is $t_{BC}$ .

Write the equilibrium Equation for torques.

  $-T_A-T_C-50L_{BC}+4000=0$ ........ (III)

Here, torque at A is $T_A$ , torque at C is $T_C$ and the length between B and C is $L_{BC}$ .

Write equilibrium equation for AD .

  $\begin{array}{l}-T_A+T_{AD}=0\\ T_D=T_A\end{array}$......... (IV)

Here, torque at D is $T_D$ .

Write equilibrium equation for DB .

  $\begin{array}{l}-T_A+4000+T_{DB}=0\\ T_{DB}=T_A-4000\end{array}$........ (V)

Write the equilibrium Equation for BC .

  $-T_A+4000-50\left(L-x\right)+T_{BC}=0$........ (VI)

Here, torque in segment BC is $T_{BC}$ , length is $L$ and the distance from left end is $x$ .

Write the expression for angle of twist for AD .

  ${\varphi }_D=\frac{T_A{L}_{AD}}{G{I}_{PAB}}$ ....... (VII)

Here, angle of twist is ${\varphi }_D$ , modulus of rigidity is $G$ and length of section AD is $L_{AD}$ .

Write the expression for angle of twist for B .

  ${\varphi }_B=\frac{T_{DB}{L}_{DB}}{G{I}_{PAB}}$ ....... (VIII)

Here, angle of twist is ${\varphi }_B$ , modulus of rigidity is $G$ and length of section DB is $L_{DB}$ .

Substitute $T_A-4000$ for $T_{DB}$ in Equation (VIII).

  ${\varphi }_B=\frac{\left(T_A-4000\right)L_{DB}}{G{I}_{PAB}}$ ....... (IX)

Write the expression for angle of twist for BC .

  $\begin{array}{c}{\varphi }_{BC}\left(x\right)={\displaystyle \underset{L_{AB}}{\overset{x}{\int }}\frac{-T_A-7600+50\varsigma }{G{I}_{PBC}}}d\varsigma \\ =\frac{1}{G{I}_{PBC}}{\left[-T_A-7600\varsigma +25{\varsigma }^2\right]}_{L_{AB}}^x\\ =\frac{1}{G{I}_{PBC}}\left(\left(T_{A}x-7600x+25x^2\right)-\left(T_A{L}_{AB}-7600L_{AB}+25L_{AB}^2\right)\right)\end{array}$ ....... (X)

Write the compatibility equation for angle of twist.

  ${\varphi }_D+{\varphi }_B+{\varphi }_C=0$…... (XI)

Here, angle of twist for C is ${\varphi }_C$ .

Write the expression for maximum shear stress in AB .

  ${\tau }_{\max ,AB}=\frac{T_{\max ,AB}}{I_{PAB}}\times \frac{d+2t_{AB}}{2}$........ (XII)

Here, maximum shear stress is ${\tau }_{\max ,AB}$ .

Write the expression for maximum shear stress in BC .

  ${\tau }_{\max ,BC}=\frac{T_{\max ,BC}}{I_{PBC}}\times \frac{d+2t_{BC}}{2}$ ........ (XIII)

Here, maximum shear stress is ${\tau }_{\max ,BC}$.

Calculation:

Substitute $12\text{in}$ for $d$ and $\frac{3}{4}\text{in}$ for $t_{AB}$ in Equation (I).

  $\begin{array}{c}{I}_{PAB}=\frac{\pi }{32}\left[{\left(12\text{in}+2\times \frac{3}{4}\text{in}\right)}^4-{\left(12\text{in}\right)}^4\right]\\ =\frac{\pi }{32}\left(12479.0625{\text{in}}^4\right)\\ =1225.13{\text{in}}^4\end{array}$

Substitute $12\text{in}$ for $d$ and $\frac{5}{8}\text{in}$ for $t_{BC}$ in Equation (II).

  $\begin{array}{c}{I}_{PBC}=\frac{\pi }{32}\left[{\left(12\text{in}+2\times \frac{5}{8}\text{in}\right)}^4-{\left(12\text{in}\right)}^4\right]\\ =\frac{\pi }{32}\left(10086.19141{\text{in}}^4\right)\\ =990.21{\text{in}}^4\end{array}$

Substitute $4\text{f}t$ for $L_{BC}$ in Equation (III).

  $\begin{array}{l}-T_A-T_C-50\times 4\text{ft}+4000=0\\ T_A+T_C=1600\end{array}$........ (XIV)

Substitute $6\text{ft}$ for $L$ in Equation (VI).

  $\begin{array}{l}-T_A+4000-50\left(\left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)-x\right)+T_{BC}=0\\ T_{BC}=T_A-7600+50x\end{array}$......... (XV)

Substitute $3\text{ft}$ for $L_{AD}$ , $11800\text{ksi}$ for $G$ and $1225.13{\text{in}}^4$ for $I_{PAB}$ in Equation (VII).

  $\begin{array}{c}{\varphi }_D=\frac{T_A\times 3\text{ft}}{11800\text{ksi}\times 1225.13{\text{in}}^4}\\ =\frac{T_A\times \left(3\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{11800\text{ksi}\times 1225.13{\text{in}}^4}\\ =2.49\times {10}^{-6}{T}_A\end{array}$

Substitute $3\text{ft}$ for $L_{DB}$ , $11800\text{ksi}$ for $G$ and $1225.13{\text{in}}^4$ for $I_{PAB}$ in Equation (VIII).

  $\begin{array}{c}{\varphi }_B=\frac{\left(T_A-4000\right)\times 3\text{ft}}{11800\text{ksi}\times 1225.13{\text{in}}^4}\\ =\frac{\left(T_A-4000\right)\times \left(3\text{ft}\right)\left(\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}{11800\text{ksi}\times 1225.13{\text{in}}^4}\\ =2.49\times {10}^{-6}\left(T_A-4000\right)\end{array}$

Substitute $10\text{ft}$ for $x$ , $6\text{ft}$ for $L_{AB}$ , $11800\text{ksi}$ for $G$ and $990.21{\text{in}}^4$ for $I_{PBC}$ in Equation (X).

  $\begin{array}{c}{\varphi }_{BC}\left(C\right)=\frac{1}{11800\text{ksi}\left(990.21{\text{in}}^4\right)}\left(\left[\begin{array}{l}\left(T_A\times 10\text{ft}-7600\times 10\text{ft}+25{\left(\left(10\text{ft}\right)\right)}^2\right)\\ -\left(T_A\times \left(6\text{ft}\right)-7600\times \left(6\text{ft}\right)+25{\left(\left(6\text{ft}\right)\right)}^2\right)\end{array}\right]\right)\\ =\frac{1}{11800\text{ksi}\left(990.21{\text{in}}^4\right)}\left(\left[\begin{array}{l}\left(\begin{array}{l}{T}_A\times 10\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)-7600\times 10\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\\ +25{\left(\left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2\end{array}\right)\\ -\left(\begin{array}{l}{T}_A\times \left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)-7600\times \left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ +25{\left(\left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2\end{array}\right)\end{array}\right]\right)\\ =8.558\times {10}^{-8}\left(48T_A-134400\right)\end{array}$

Substitute $2.49\times {10}^{-6}{T}_A$ for ${\varphi }_D$ , $2.49\times {10}^{-6}\left(T_A-4000\right)$ for ${\varphi }_B$ and $8.558\times {10}^{-8}\left(48T_A-134400\right)$ for ${\varphi }_C$ in Equation (XI).

  $\begin{array}{l}2.49\times {10}^{-6}{T}_A+2.49\times {10}^{-6}\left(T_A-4000\right)+8.558\times {10}^{-8}\left(48T_A-134400\right)=0\\ T_A=\left(2361.61\text{kip}\cdot \text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ T_A=196.80\text{kip}\cdot \text{ft}\end{array}$

The reactive torque at A is $-196.80\text{kip}\cdot \text{ft}$ .

Negative sign is due to selected coordinate.

Substitute $2361.61\text{kip}\cdot \text{in}$ for $T_A$ in Equation (IV).

  $\begin{array}{l}2361.61\text{kip}\cdot \text{in}+T_C=1600\\ T_C=\left(-761.61\text{kip}\cdot \text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ T_C=-63.47\text{kip}\cdot \text{ft}\end{array}$

The obtained torque is negative so reverse the direction of torque.

The reactive torque at C is $63.47\text{kip}\cdot \text{ft}$ .

Substitute $12\text{in}$ for $d$ and $\frac{3}{4}\text{in}$ for $t_{AB}$ , $2361.61\text{kip}\cdot \text{in}$ for $T_{\max ,AB}$ and $1225.13{\text{in}}^4$ for $I_{PAB}$ in Equation (XII).

  $\begin{array}{c}{\tau }_{\max ,AB}=\frac{2361.61\text{kip}\cdot \text{in}}{1225.13{\text{in}}^4}\times \frac{12\text{in}+2\times \frac{3}{4}\text{in}}{2}\\ =\frac{2361.61\text{kip}\cdot \text{in}}{1225.13{\text{in}}^4}\times 6.75\text{in}\\ =\text{13}\text{.01}\text{ksi}\end{array}$

The maximum shear stress in segment AB is $\text{13}\text{.01}\text{ksi}$ .

Substitute $12\text{in}$ for $d$ and $\frac{5}{8}\text{in}$ for $t_{BC}$ , $-1638.39\text{kip}\cdot \text{in}$ for $T_{\max ,BC}$ and $990.21{\text{in}}^4$ for $I_{PBC}$ in Equation (XIII).

  $\begin{array}{c}{\tau }_{\max ,BC}=\frac{-1638.39\text{kip}\cdot \text{in}}{990.21{\text{in}}^4}\times \frac{12\text{in}+2\times \frac{5}{8}\text{in}}{2}\\ =\frac{-1638.39\text{kip}\cdot \text{in}}{990.21{\text{in}}^4}\times 6.625\text{in}\\ =-\text{10}\text{.96}\text{ksi}\end{array}$

The maximum shear stress in the segment BC is $-\text{10}\text{.96}\text{ksi}$ .

Conclusion:

The reactive torque at A is $-196.80\text{kip}\cdot \text{ft}$ .

The reactive torque at C is $63.47\text{kip}\cdot \text{ft}$ .

The maximum shear stress in segment AB is $\text{13}\text{.01}\text{ksi}$ .

The maximum shear stress in the segment BC is $-\text{10}\text{.96}\text{ksi}$ .