A bar of circular cross section is 2.5 m long (Fig. 11.8). For 2 m of its length its diameter is 200 mm while for the remaining 0.5 m its diameter is 100 mm. If the bar is firmly supported at its ends and subjected to a torque of 50 kNm applied at its change of section, calculate the maximum

Question

A bar of circular cross section is 2.5 m long (Fig. 11.8). For 2 m of its length its diameter is 200 mm while for the remaining 0.5 m its diameter is 100 mm. If the bar is firmly supported at its ends and subjected to a torque of 50 kNm applied at its change of section, calculate the maximum stress in the bar and the angle of twist at the point of application of the torque. Take G = 80000 N/mm².

Accepted Answer

In this problem Eqs (11.7) and (11.8) cannot be used directly since the bar changes section at B. Thus from equilibrium

[latex]T_{ C }=\frac{L_{ AB }}{L_{ AB }+L_{ BC }} T [/latex] (11.7)

[latex] T_{ A }=\frac{L_{ BC }}{L_{ AB }+L_{ BC }} T [/latex] (11.8)

[latex] T=T_{ A }+T_{ C } [/latex] (i)

and from the compatibility of displacement at B in the lengths AB and BC

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

or using Eq. (11.3)

[latex] T=J G \frac{ heta}{L} [/latex] (11.3)

[latex] \frac{T_{ A } L_{ AB }}{G J_{ AB }}=\frac{T_{ C } L_{ BC }}{G J_{ BC }} [/latex]

whence

[latex] T_{ A }=\frac{L_{ BC } J_{ AB }}{L_{ AB } J_{ BC }} T_{ C }[/latex] (ii)

Substituting in Eq. (i) we obtain

[latex] T=T_{ C }\left(\frac{L_{ BC }}{L_{ AB }} \frac{J_{ AB }}{J_{ BC }}+1 ight) [/latex]

or

[latex] 50=T_{ C }\left[\frac{0.5}{2.0} imes\left(\frac{200 imes 10^{-3}}{100 imes 10^{-3}} ight)^{4}+1 ight] [/latex]

from which

[latex] T_{ C }=10 kNm [/latex]

Hence, from Eq. (i)

[latex] T_{ A }=40 kNm [/latex]

Although the maximum torque occurs in the length AB, the length BC has the smaller diameter. It can be seen from Eq. (11.4) that shear stress is directly proportional to torque and inversely proportional to diameter (or radius) cubed. Therefore, we conclude that in this case the maximum shear stress occurs in the length BC of the bar and is given by

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

[latex] au_{\max }=\frac{10 imes 10^{6} imes 100 imes 32}{2 imes \pi imes 100^{4}}=50.9 N / mm ^{2} [/latex]

Also the rotation at B is given by either

[latex] heta_{ B }=\frac{T_{ A } L_{ AB }}{G J_{ AB }} ext { or } heta_{ B }=\frac{T_{ C } L_{ BC }}{G J_{ BC }} [/latex]

Using the first of these expressions we have

[latex] heta_{ B }=\frac{40 imes 10^{6} imes 2 imes 10^{3} imes 32}{80000 imes \pi imes 200^{4}}=0.0064 rad [/latex]

or

[latex] heta_{ B }=0.37^{\circ} [/latex]

Question 11.4: A bar of circular cross section is 2.5 m long (Fig. 11.8). F...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions