Question 5.3: Two shafts (G = 28 GPa) A and B are joined and subjected to ......
Two shafts (G = 28 GPa) A and B are joined and subjected to the torques shown. Section A has a solid circular cross section with diameter 40 mm, and is 160 mm long; B has a solid circular cross section with diameter 20 mm, and is 120 mm long.
Find (a) the maximum shear stress in sections A and B; and (b) the angle of twist of the rightmost end of B relative to the wall.
Given: Dimensions and properties of composite shaft in torsion.
Find: Shear stresses, angle of twist of free end.
Assume: Hooke’s law applies; stress–concentration at the step in the composite shaft is not included.
Our strategy is to use the method of sections to find the internal torque in each portion of the composite shaft, then find the shear stress and angle of twist induced by this torque.
First, we construct an FBD:
Equilibrium requires that 400 N · m − 1200 N · m − T_{wall} = 0.
Therefore, T_{wall} = −800 N · m (T_{wall} is clockwise, opposite from what is drawn). Now use the method of sections on segments A and B:
Internal torque T_A = 800 N · m, and maximum shear stress occurs at c_A = 0.02 m. So:
\tau_{\max , A}=\frac{T_A c_A}{J_A}=\frac{T_A c_A}{\frac{\pi}{2} c_A^4}=63.7 \mathrm{MPa} .
To find the internal resisting torque in section B, we must look at the whole shaft from the wall to our imaginary section cut:
Equilibrium of this section requires that the internal torque T_B = 400 N · m. Maximum shear stress occurs at c_B = 0.01 m, and
\tau_{\max , B}=\frac{T_B c_B}{J_B}=\frac{T_B c_B}{\frac{\pi}{2} c_B^4}=255 \mathrm{MPa} .
Next, we will calculate the angles of twist of both A and B, and then find the resultant twist of the free end with respect to the wall: ΦA + ΦB = Φ.
Taking counterclockwise twists to be positive, as we have taken counterclockwise torques to be
\begin{aligned} & \phi_A=\frac{T_A L_A}{J_A G_A}=\frac{(-800 \mathrm{~N} \cdot \mathrm{m})(0.16 \mathrm{~m})}{\left(2.51 \times 10^{-7} \mathrm{~m}^4\right)\left(28 \times 10^9 \mathrm{~Pa}\right)}=-0.018 \mathrm{rad}\left(-1.0^{\circ}\right), \\ & \phi_B=\frac{T_B L_B}{J_B G_B}=\frac{(400 \mathrm{~N} \cdot \mathrm{m})(0.12 \mathrm{~m})}{\left(1.57 \times 10^{-8} \mathrm{~m}^4\right)\left(28 \times 10^9 \mathrm{~Pa}\right)}=0.109 \mathrm{rad}\left(6.2^{\circ}\right) . \end{aligned}The total angle of twist of the free end relative to the wall is then
\phi=\phi_A+\phi_B=-1.0^{\circ}+6.2^{\circ}=5.2^{\circ} \text { (counterclockwise) }
