Question 4.6: Two shafts (G = 28 GPa) A and B are joined and subjected to ...

Our strategy is to use the method of sections to find the internal torque in each portion of the composite shaft and then find the shear stress and angle of twist induced by this torque. First, we construct an FBD (Figure 4.37)

Equilibrium requires that 400 Nm – 1200 Nm – T_{wall} = 0.
Therefore, T_{wall} = –800 Nm (T_{wall} is clockwise, opposite from what is drawn at left.) Now use the method of sections on segments A and B (Figure 4.38)

Internal torque T_{A} = 800 Nm, and maximum shear stress occurs at c_{_A} = 0.02 m. So,

\underset{A}{\tau _{\max }}=\frac{T_A c_{_{A}}}{J_A} = \frac{T_A c_{_{A}}}{\frac{\pi}{2} c^{4}_{A}}= 63.7 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 (Figure 4.39)

Equilibrium of this section requires that the internal torque T_{B} = 400 Nm. Maximum shear stress occurs at c_{B} = 0.01 m, and

\underset{B}{\tau _{\max }}=\frac{T_B c_B}{J_B}= \frac{T_B c_B}{\frac{\pi}{2} c^{4}_{B}}= 255 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:

\phi _A+\phi _B=\phi .

Taking counterclockwise twists to be positive, as we take counterclockwise torques to be,

\phi_A=\frac{T_A L_A}{J_A G_A}=\frac{(-800  \textrm{Nm})(0.16  \textrm{m})}{(2.51\times 10^{-7}   \textrm{m}^4)(28\times 10^9  \textrm{Pa})} =-0.0182 rad (-1.04°)

\phi_B=\frac{T_B L_B}{J_B G_B}=\frac{(400   \textrm{Nm})(0.12  \textrm{m})}{(1.57\times 10^{-8}   \textrm{m}^4)(28\times 10^9  \textrm{Pa})} =0.1091 rad (6.25°).

The total angle of twist of the free end relative to the wall is then

\varphi =\varphi _{_{A}}+\varphi_{_{B}}  =–1.04°+6.25°=5.21° (CCW).