A bar ABCD of circular cross section having a diameter of 50 mm is firmly supported at each end and carries two concentrated torques at B and C as shown in Fig. P.11.3. Calculate the maximum shear stress in the bar and the maximum angle of twist. Take G = 70000 N/mm^2.

Question

A bar ABCD of circular cross section having a diameter of 50 mm is firmly supported at each end and carries two concentrated torques at B and C as shown in Fig. P.11.3. Calculate the maximum shear stress in the bar and the maximum angle of twist. Take G = 70000 N/mm².

Accepted Answer

From equilibrium

[latex] T_{ A }+T_{ D }=2500 Nm [/latex] (i)

From compatibility

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

Therefore, from Eq. (11.4) and since GJ = constant

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

[latex] T_{ A } imes 2=\left(T_{ D }-1500 ight) imes 1+T_{ D } imes 1 [/latex]

i.e.

[latex] T_{ A }=T_{ D }-750 [/latex] (ii)

Solving Eqs (i) and (ii) gives

[latex] T_{ A }=875 Nm , \quad T_{ D }=1625 Nm [/latex]

Therefore, from Eq. (11.4)

[latex] au_{\max }=1625 imes 10^{3} imes \frac{25}{\pi\left(\frac{50^{4}}{32} ight)}=66.2 N / mm ^{2} [/latex]

The maximum angle of twist will occur at B. Then, considering AB

[latex] heta_{ B }=\frac{875 imes 10^{3} imes 2 imes 10^{3}}{70000 imes \pi\left(\frac{50^{4}}{32} ight)}=0.0407 rad =2.3^{\circ} [/latex]

Question 11.3: A bar ABCD of circular cross section having a diameter of 50...

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