A bar ABCD has a circular cross section of 75 mm diameter over half its length and 50 mm diameter over the remaining half of its length. A torque of 1 kNm is applied at C midway between B and D as shown in Fig. P.11.4. Sketch the distribution of torque along the length of the bar and calculate the

Question

A bar ABCD has a circular cross section of 75 mm diameter over half its length and 50 mm diameter over the remaining half of its length. A torque of 1 kNm is applied at C midway between B and D as shown in Fig. P.11.4. Sketch the distribution of torque along the length of the bar and calculate the maximum shear stress and the maximum angle of twist in the bar. G = 70 000 N/mm².

Accepted Answer

From equilibrium

[latex] T_{ A }+T_{ D }=1 [/latex] (i)

and from compatibility

[latex] heta_{ B ( AB )}= heta_{ B ( BC )}+ heta_{ C ( CD )} [/latex]

Then, from Eq. (11.4)

[latex] \frac{T}{J}=\frac{ au}{r}=G \frac{ heta}{L} [/latex] (11.4)

[latex] \frac{T_{ A } imes 1.0}{\pi\left(\frac{75^{4}}{32} ight)}=\frac{\left(T_{ D }-1 ight) imes 0.5}{\pi\left(\frac{50^{4}}{32} ight)}+\frac{T_{ D } imes 0.5}{\pi\left(\frac{50^{4}}{32} ight)} [/latex]

i.e.

[latex] T_{ A }=5.1 T_{ D }-2.5 [/latex] (ii)

Solving Eqs (i) and (ii)

[latex] T_{ A }=0.43 kNm , \quad T_{ D }=0.57 kNm [/latex]

Then

[latex] au_{\max }=\frac{0.57 imes 10^{6} imes 25}{\pi\left(\frac{50^{4}}{32} ight)}=23.2 N / mm ^{2} [/latex]

The maximum angle of twist will occur at C. Then

[latex] heta_{C}=\frac{0.57 imes 10^{6} imes 0.5 imes 10^{3}}{70000 imes \pi\left(\frac{50^{4}}{32} ight)}=0.00664 rad =0.38^{\circ} [/latex]

The distribution of torque along the bar is shown in Fig. S.11.4.

Question 11.4: A bar ABCD has a circular cross section of 75 mm diameter ov...

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