Chapter 3, Problem 3.4.8P

Two sections of steel drill pipe, joined by bolted flange plates at Ä are being tested to assess the adequacy of both the pipes. In the test, the pipe structure is fixed at A, a concentrated torque of 500 kN - m is applied at x = 0.5 m, and uniformly distributed torque intensity t₁= 250 kN m/m is applied on pipe BC. Both pipes have the same inner diameter = 200 mm. Pipe AB has thickness t_AB=15 mm, while pipe BC has thickness T_BC= 12 mm. Find the maximum shear stress and maximum twist of the pipe and their locations along the pipe. Assume G = 75 GPa.

  

To determine

The maximum shear stress.

The maximum angle of twist.

Answer to Problem 3.4.8P

The maximum shear stress is $-466.2\text{MPa}$ .

The maximum angle of twist is $-5.63°$ .

Explanation of Solution

Given information:

The concentrated torque is $500\text{kN}\cdot \text{m}$ , intensity of uniformly distributed torque is $250\text{kN}\cdot \text{m}/\text{m}$ , inner diameter of both pipes is $200\text{mm}$ , thickness of AB is $15\text{mm}$ , thickness of BC is $12\text{mm}$ and the shear modulus is $75\text{GPa}$ .

Write the expression for torque at B.

  $T_B=T_i\times BC$....... (I)

Here, torque at B is $T_B$ , torque intensity is $T_i$ and length of BC is $BC$ .

Write the expression for torque at AD.

  $T_{AD}=T_B+T$........ (II)

Here, torque at AD is $T_{AD}$ and the concentrated torque is $T$ .

Write the expression for polar moment of inertia of section BC .

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

Here, polar moment of inertia is $I_{PBc}$ , diameter is $d$ and the thickness of section is $t_{BC}$ .

Write the expression for polar moment of inertia of section AB .

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

Here, polar moment of inertia is $I_{PAB}$ , diameter is $d$ and the thickness of section is $t_{AB}$ .

Write the expression for maximum shear stress for section BC .

  ${\tau }_{\max AB}=\frac{T_{\max }\left(\frac{d+2t_B}{2}\right)}{I_{PAB}}$........ (V)

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

Write the expression for maximum shear stress for section BC .

  ${\tau }_{\max BC}=\frac{T_{\max }\left(\frac{d+2t_{BC}}{2}\right)}{I_{PBC}}$........ (VI)

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

Write the expression for angle of twist.

  ${\varphi }_1\left(x\right)=\frac{T_{AD}\left(x\right)\times x}{G{I}_{PAB}}$........ (VII)

Here, function for angle of twist is ${\varphi }_1\left(x\right)$ .

Write the expression for angle of twist.

  ${\varphi }_2\left(x\right)={\varphi }_1\left(D\right)+\frac{T_{DB}\left(x\right)\times \left(x-0.5\right)}{G{I}_{PAB}}$ ......... (VIII)

Write the expression for angle of twist.

  ${\varphi }_3\left(x\right)={\varphi }_2\left(B\right)+{\displaystyle \underset{0}{\overset{x}{\int }}\frac{250\times \left(\varsigma -3.5\right)}{G{I}_{PBC}}}d\varsigma$ ....... (IX)

Here, dummy variable is $\varsigma$ ,.

Calculation:

Substitute $250\text{kN}\cdot \text{m}/\text{m}$ for $T_i$ and $1.5\text{m}$ for BC in Equation (I).

  $\begin{array}{c}{T}_B=-250\text{kN}\cdot \text{m}/\text{m}\times 1.5\text{m}\\ =-\text{375}\text{kN}\cdot \text{m}\end{array}$

Substitute $-\text{375}\text{kN}\cdot \text{m}$ for $T_B$ and $500\text{kN}\cdot \text{m}$ for $T$ in Equation (II).

  $\begin{array}{c}{T}_{AD}=-\text{375}\text{kN}\cdot \text{m}+500\text{kN}\cdot \text{m}\\ =\text{125}\text{kN}\cdot \text{m}\end{array}$

Substitute $200\text{mm}$ for

  $d$ and $12\text{mm}$ for $t_{BC}$ in Equation (III).

  $\begin{array}{c}{I}_{PBc}=\frac{\pi }{32}\left[{\left(200\text{mm}+\left(2\times 12\text{mm}\right)\right)}^4-{\left(200\text{mm}\right)}^4\right]\\ =\frac{\pi }{32}\left[917630976{\text{mm}}^{\text{4}}\right]\\ =90088210.4{\text{mm}}^{\text{4}}\end{array}$

Substitute $200\text{mm}$ for $d$ and $15\text{mm}$ for $t_{AB}$ in Equation (IV).

  $\begin{array}{c}{I}_{PAB}=\frac{\pi }{32}\left[{\left(200\text{mm}+\left(2\times 15\text{mm}\right)\right)}^4-{\left(200\text{mm}\right)}^4\right]\\ =\frac{\pi }{32}\left[1198410000{\text{mm}}^{\text{4}}\right]\\ =117653626.6{\text{mm}}^{\text{4}}\end{array}$

Substitute $-\text{375}\text{kN}\cdot \text{m}$ for $T_{\max }$ , $200\text{mm}$ for $d$ , $15\text{mm}$ for $t_{AB}$ and $117653626.6{\text{mm}}^{\text{4}}$ for $I_{PAB}$ in Equation (V).

  $\begin{array}{c}{\tau }_{\max AB}=\frac{-\text{375}\text{kN}\cdot \text{m}\times \left(\frac{200\text{mm}+2\times 15\text{mm}}{2}\right)}{117653626.6{\text{mm}}^{\text{4}}}\\ =\frac{\left(-\text{375}\text{kN}\cdot \text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\times \left(\frac{200\text{mm}+2\times 15\text{mm}}{2}\right)}{117653626.6{\text{mm}}^{\text{4}}}\\ =\left(-0.3665\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{1\text{MPa}}{{10}^{-3}\text{kN}/{\text{mm}}^{\text{2}}}\right)\\ =-366.5\text{MPa}\end{array}$

The maximum shear stress in section AB is $-366.5\text{MPa}$ .

Substitute $-\text{375}\text{kN}\cdot \text{m}$ for $T_{\max }$ , $200\text{mm}$ for $d$ , $12\text{mm}$ for $t_{AB}$ and $90088210.4{\text{mm}}^{\text{4}}$ for $I_{PAB}$ in Equation (VI).

  $\begin{array}{c}{\tau }_{\max BC}=\frac{-\text{375}\text{kN}\cdot \text{m}\times \left(\frac{200\text{mm}+2\times 12\text{mm}}{2}\right)}{90088210.4{\text{mm}}^{\text{4}}}\\ =\frac{\left(-\text{375}\text{kN}\cdot \text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\times \left(\frac{200\text{mm}+2\times 12\text{mm}}{2}\right)}{90088210.4{\text{mm}}^{\text{4}}}\\ =\left(-0.4662\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{1\text{MPa}}{{10}^{-3}\text{kN}/{\text{mm}}^{\text{2}}}\right)\\ =-466.2\text{MPa}\end{array}$

The maximum shear stress in section BC is $-466.2\text{MPa}$ .

Substitute $125\text{KN}\cdot \text{m}$ for $T_{AD}\left(x\right)$ , $117653626.6{\text{mm}}^{\text{4}}$ for $I_{PAB}$ and $75\text{GPa}$ for $G$ in Equation (VII).

  $\begin{array}{c}{\varphi }_1\left(x\right)=\frac{125\text{KN}\cdot \text{m}\times x}{75\text{GPa}\times 117653626.6{\text{mm}}^{\text{4}}}\\ =\frac{\left(125\text{KN}\cdot \text{m}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\times x}{\left(75\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\times 117653626.6{\text{mm}}^{\text{4}}}\\ =1.416\times {10}^{-5}x\end{array}$....... (X)

Substitute $0.5\text{m}$ for $x$ in Equation (X).

  $\begin{array}{c}{\varphi }_1\left(D\right)=1.416\times {10}^{-5}\times 0.5\text{m}\\ \text{=}1.416\times {10}^{-5}\times \left(0.5\text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =7.083\times {10}^{-3}\text{rad}\end{array}$

Substitute $-\text{375}\text{kN}\cdot \text{m}$ for $T_{DB}\left(x\right)$ , $117653626.6{\text{mm}}^{\text{4}}$ for $I_{PAB}$ , $7.083\times {10}^{-3}\text{rad}$ for ${\varphi }_1\left(D\right)$ and $75\text{GPa}$ for $G$ in Equation (VIII).

  $\begin{array}{c}{\varphi }_2\left(x\right)=7.083\times {10}^{-3}\text{rad}+\frac{\left(-\text{375}\text{kN}\cdot \text{m}\right)\times \left(x-0.5\right)}{\left(75\text{GPa}\right)\times 117653626.6{\text{mm}}^{\text{4}}}\\ =7.083\times {10}^{-3}\text{rad}+\frac{\left[\begin{array}{l}\left(-\text{375}\text{kN}\cdot \text{m}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ \times \left(x-0.5\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\end{array}\right]}{\left(75\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\times 117653626.6{\text{mm}}^{\text{4}}}\\ =7.083\times {10}^{-3}\text{rad}-0.04249\left(x-0.5\right)\end{array}$....... (XI)

Substitute $2\text{m}$ for $x$ in Equation (XI).

  $\begin{array}{c}{\varphi }_2\left(B\right)=7.083\times {10}^{-3}\text{rad}-0.04249\left(2-0.5\right)\\ =7.083\times {10}^{-3}\text{rad}-0.063735\\ =-0.0566\text{rad}\end{array}$

Substitute $-0.0566\text{rad}$ for ${\varphi }_2\left(B\right)$ , $90088210.4{\text{mm}}^{\text{4}}$ for $I_{PBC}$ and $75\text{GPa}$ for $G$ in Equation (IX).

  $\begin{array}{c}{\varphi }_3\left(x\right)=-0.0566\text{rad}+{\displaystyle \underset{0}{\overset{x}{\int }}\frac{250\text{kN}\cdot \text{m}/\text{m}\times \left(\varsigma -3.5\text{m}\right)}{75\text{GPa}\times 90088210.4{\text{mm}}^{\text{4}}}}d\varsigma \\ =-0.0566\text{rad}+{\displaystyle \underset{0}{\overset{x}{\int }}\frac{\left(250\text{kN}\cdot \text{m}/\text{m}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\times \left(\varsigma -\left(3.5\text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\right)}{\left(75\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^{\text{2}}}{1\text{GPa}}\right)\times 90088210.4{\text{mm}}^{\text{4}}}}d\varsigma \\ =-0.0566\text{rad}=3.7\times {10}^{-8}\left(\left(\frac{1000x^2}{2}\right)-3500\times 1000x\right)\\ -\left(\left(\frac{{2000}^2}{2}\right)-3500\left(2000\right)\right)\end{array}$ ....... (XII)

Substitute $3.5\text{m}$ for $x$ in Equation (XII).

  $\begin{array}{c}{\varphi }_3\left(C\right)=-0.0566\text{rad}+3.7\times {10}^{-8}\left(\left(\frac{1000{\left(3.5\text{m}\right)}^2}{2}\right)-3500\times 1000\times 3.5\text{m}\right)\\ -\left(\left(\frac{{2000}^2}{2}\right)-3500\left(2000\right)\right)\\ =-0.0566\text{rad}+0.154895\\ =0.098295\text{rad}\end{array}$

The maximum angle of twist occurs at C .

Convert twist angle in degrees.

  $\begin{array}{c}{\varphi }_3\left(C\right)=\left(-0.098295\text{rad}\right)\times \frac{180°}{\pi \text{rad}}\\ =-5.63°\end{array}$

Conclusion:

The maximum shear stress is $-466.2\text{MPa}$ .

The maximum angle of twist is $-5.63°$ .