Chapter 3, Problem 3.4.22P

-22 Two tubes (AB, BC) of the same material arc connected by three pins (pin diameter = d ) just left of B as shown in the figure. Properties and dimensions for each tube are given in the figure. Torque 2ris applied at x = 2L/5 and uniformly distributed torque intensity t_Q= 37/L is applied on tube BC. (a)

Find the maximum value of load variable T(N m) based on allowable shear (t_x) and bearing(c_ha ) stresses in the three pins which connect the two tubes at B.

Use the following numerical properties:

L = 1.5m, E = 74GPa, v = 0.33, d_p= 18mm, t_a=45MPa, =90 MPa, d_i=85 mm, di = T$ mm, and d₃— 60 mm.

(b)

What is the maximum shear stress in the tubes

for the applied torque in part (a)?

  

To determine

The maximum value of load variable for the given parameters.

Answer to Problem 3.4.22P

The maximum value of load variable for the given parameters is = $875306\text{N}\cdot \text{mm}$.

Explanation of Solution

Given information:

The total length of the tube is $1.5\text{m}$, young’s modulus of elasticity is $74\text{GPa}$, poison ratio is $0.33$, diameter of the pin is $18\text{mm}$, shear stress at point (A) is ${\tau }_a$and the bending stress is $90\text{MPa}$, the outer diameter of the first tube is $85\text{mm}$, the outer diameter of the second tube is $73\text{mm}$and the inner diameter of the second tube is $60\text{mm}$.

       Figure (1) shows the segment from point (A) to $x\le \frac{2L}{5}$.

  

       Figure-(1)

Write the expression for moment about $x$axis for the system.

  $R_A+2T_o-t_o\left(\frac{2L}{5}\right)=0$...... (I)

Here, the reaction force at point (A) is $R_A$, the length of the beam tube is $L$and the uniform distributed torque intensity is $t_o$.

Substitute $\frac{3T_o}{5}$for $t_o$in Equation (I)

  $\begin{array}{l}{R}_A+2T_o-\frac{3T_o}{5}\left(\frac{2L}{5}\right)=0\\ R_A=\frac{-4T_o}{5}\end{array}$...... (II)

Here, $T\left(x\right)=-R_A$.

Substitute $-T\left(x\right)$for $R_A$in Equation (II).

  $T(x)=\frac{4T_o}{5}$

       Figure-(2) shows the segment from point $A$to point between $\frac{2L}{5}<x\le \frac{3L}{5}$.

  

       Figure-(2)

Write the expression for moment about $x$axis for the system.

  $\begin{array}{l}{T}_2\left(x\right)=\frac{4}{5}{T}_o-2T_o\\ T_2\left(x\right)=-\frac{6}{5}{T}_o\end{array}$...... (III)

Write the expression for the moment about $x$axis for the element $\frac{3}{5}L\le x\le L$.

  $T_3\left(x\right)=\frac{4}{5}{T}_o-2T_o+t_o\left(x-\frac{3}{5}L\right)$...... (IV)

Substitute $\frac{3L}{5}$for $x$in Equation (IV)

  $\begin{array}{c}{T}_3\left(x\right)=\frac{4}{5}{T}_o-2T_o+t_o\left(\frac{3L}{5}-\frac{3}{5}L\right)\\ =-\frac{6}{5}{T}_o\end{array}$

Substitute $L$for $x$in Equation (IV)

  $\begin{array}{c}{T}_3\left(x\right)=\frac{4}{5}{T}_o-2T_o+\frac{3T_o}{L}\left(L-\frac{3}{5}L\right)\\ =0\end{array}$

Maximum torsional moment in both $AB$and $BC$is same

  $T_{\max }=\frac{6T}{5}$...... (V)

Write the expression for torque.

  $T=F\times r$...... (VI)

Here, torque is $T$, force on the pin is $F$and the radius is $r$.

Write the expression for force on the pin.

  $F=\frac{\frac{6T}{5}}{3\left(\frac{d_1}{2}-\frac{t_1}{2}\right)}$...... (VII)

Here, the force on the pin is $F$, the diameter of the tube $AB$is $d_1$and the thickness of the tube is $t_1$.

Write the expression for bearing stress on the pin due to the torque.

  ${\sigma }_{b1}=\frac{F}{d_p{t}_1}$...... (VIII)

Here, the bearing stress on the pin is ${\sigma }_{b1}$and the diameter of the pin is $d_p$.

Substitute $\frac{\frac{6T}{5}}{3\left(\frac{d_1}{2}-\frac{t_1}{2}\right)}$for $F$in Equation (VIII).

  $\begin{array}{l}{\sigma }_{b1}=\frac{\frac{6T}{5}}{3\left(\frac{d_1}{2}-\frac{t_1}{2}\right)}\times \left(\frac{1}{d_p{t}_1}\right)\\ T=\left({\sigma }_{b1}\times d_p{t}_1\right)\frac{3}{2}\left(d_1-t_1\right)\frac{5}{6}\end{array}$...... (IX)

Calculation:

Substitute $90\text{MPa}$for ${\sigma }_{b1}$, $18\text{mm}$for $d_p$, $6\text{mm}$for $t_1$and $85\text{mm}$for $d_1$in Equation (IX).

  $\begin{array}{c}T=\left(\left(90\text{MPa}\right)\times \left(18\text{mm}\right)\left(6\text{mm}\right)\right)\frac{3}{2}\left(\left(85\text{mm}\right)-6\text{mm}\right)\frac{5}{6}\\ =\left(\left(90\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)\times \left(18\text{mm}\right)\left(6\text{mm}\right)\right)\frac{3}{2}\left(\left(85\text{mm}\right)-6\text{mm}\right)\frac{5}{6}\\ =972\times {10}^7\times \frac{3}{2}\left(\left(85\text{mm}\right)-6\text{mm}\right)\frac{5}{6}\\ =959850\text{N}\cdot \text{mm}\end{array}$

Substitute $90\text{MPa}$for ${\sigma }_{b1}$, $18\text{mm}$for $d_p$, $6.5\text{mm}$for $t_1$and $73\text{mm}$for $d_1$in Equation (IX).

  $\begin{array}{c}T=\left(\left(90\text{MPa}\right)\times \left(18\text{mm}\right)\left(6.5\text{mm}\right)\right)\frac{3}{2}\left(\left(73\text{mm}\right)-6.5\text{mm}\right)\frac{5}{6}\\ =\left(\left(90\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)\times \left(18\text{mm}\right)\left(6.5\text{mm}\right)\right)\frac{3}{2}\left(\left(73\text{mm}\right)-6.5\text{mm}\right)\frac{5}{6}\\ =1.053\times {10}^{10}\times \frac{3}{2}\left(\left(73\text{mm}\right)-6.5\text{mm}\right)\frac{5}{6}\\ =875306\text{N}\cdot \text{mm}\end{array}$

Conclusion:

The maximum value of load variable for the given parameters is = $875306\text{N}\cdot \text{mm}$.

To determine

The maximum shear stress in the tubes.

Answer to Problem 3.4.22P

The maximum shear stress in the tubes is = $25.28\text{MPa}$.

Explanation of Solution

Write the expression for maximum shear stress for the applied torque

  ${\tau }_{\max }=\frac{T_{\max }\times \left(\frac{d}{2}\right)}{I_p}$...... (X)

Here, the maximum shear stress is ${\tau }_{\max }$and the polar moment of inertia is $I_p$

Write the expression for maximum torque.

  $T_{\max }=\frac{6}{5}T$...... (XI)

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

  $I_{p1}=\frac{\pi }{32}\left(d_1^4-d_2^4\right)$..... (XII)

Here, the polar moment of inertia is $I_{p1}$.

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

  $I_{p2}=\frac{\pi }{32}\left(d_2^4-d_3^4\right)$..... (XII)

Here, the polar moment of inertia is $I_{p2}$.

Calculation:

Substitute $85\text{mm}$for $d_1$and $73\text{mm}$for $d_2$in Equation (XII)

  $\begin{array}{c}{I}_{p1}=\frac{\pi }{32}\left({\left(85\text{mm}\right)}^4-{\left(73\text{mm}\right)}^4\right)\\ =0.098\left({\left(85\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4-{\left(73\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4\right)\\ =2.3368\times {10}^{-6}{\text{m}}^4\end{array}$

Substitute $73\text{mm}$for $d_1$and $60\text{mm}$for $d_2$in Equation (XII).

  $\begin{array}{c}{I}_{p2}=\frac{\pi }{32}\left({\left(73\text{mm}\right)}^4-{\left(60\text{mm}\right)}^4\right)\\ =0.098\left({\left(73\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4-{\left(60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4\right)\\ =1.516\times {10}^{-6}{\text{m}}^4\end{array}$

Substitute $875306\text{N}\cdot \text{mm}$for $T$in Equation ()

  $\begin{array}{l}{T}_{\max }=\frac{6}{5}\left(875306\text{N}\cdot \text{mm}\right)\\ =1050367.2\text{N}\cdot \text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =1050\text{N}\cdot \text{m}\end{array}$

Substitute $1050\text{N}\cdot \text{m}$for $T_{\max }$

  $2.3368\times {10}^{-6}{\text{m}}^4$for $I_p$and $85\text{mm}$for $d$in Equation (X).

  $\begin{array}{c}{\tau }_{\max }=\frac{1050\text{N}\cdot \text{m}\times \left(\frac{85\text{mm}}{2}\right)}{2.3368\times {10}^{-6}{\text{m}}^4}\\ =\frac{1050\text{N}\cdot \text{m}\times \left(\frac{85\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}{2.3368\times {10}^{-6}{\text{m}}^4}\\ =19.1\text{MPa}\end{array}$

Substitute $1050\text{N}\cdot \text{m}$for $T_{\max }$

  $1.516\times {10}^{-6}{\text{m}}^4$for $I_{p2}$and $73\text{mm}$for $d$in Equation (X).

  $\begin{array}{c}{\tau }_{\max }=\frac{1050\text{N}\cdot \text{m}\times \left(\frac{73\text{mm}}{2}\right)}{1.516\times {10}^{-6}{\text{m}}^4}\\ =\frac{1050\text{N}\cdot \text{m}\times \left(\frac{73\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right)}{1.516\times {10}^{-6}{\text{m}}^4}\\ =25.28\text{MPa}\end{array}$

Conclusion:

The maximum shear stress in the tubes is $25.28\text{MPa}$.