Figure P.11.14 shows a thin-walled cantilever box-beam having a constant width of 50 mm and a depth which decreases linearly from 200 mm at the built-in end to 150 mm at the free end. If the beam is subjected to a torque of 1 kNm at its free end, plot the angle of twist of the beam at 500 mm

Question

Figure P.11.14 shows a thin-walled cantilever box-beam having a constant width of 50 mm and a depth which decreases linearly from 200 mm at the built-in end to 150 mm at the free end. If the beam is subjected to a torque of 1 kNm at its free end, plot the angle of twist of the beam at 500 mm intervals along its length and determine the maximum shear stress in the beam section. Take G = 25 000 N/mm².

Accepted Answer

The torque is constant along the length of the beam and is 1 kNm. Also the thickness is constant round the beam section so that the shear stress will be a maximum where the area enclosed by the mid-line of the section wall is a minimum, i.e. at the free end. Then

[latex] au_{\max }=\frac{1000 imes 10^{3}}{2 imes 50 imes 150 imes 2}=33.3 N / mm ^{2} [/latex]

The rate of twist is given by Eq. (11.25) in which [latex]\oint d s / t [/latex] varies along the length of the beam as does the area enclosed by the mid-line of the section wall. Then

[latex]\frac{ d θ}{dx} = \frac{ T}{4A^2G} \oint \frac{ d s}{t} [/latex] (11.25)

[latex] \oint \frac{ d s}{t}=\left[\frac{2 imes 50+2\left(150+\frac{50 x}{2500} ight)}{2} ight]=200+0.02 x [/latex]

where x is measured from the free end of the beam.
Also

[latex] A=50\left(150+\frac{50 x}{2500} ight)=7500+x [/latex]

Then

[latex] \frac{ d heta}{ d x}=\left(\frac{1 imes 10^{6}}{4 imes 25000} ight) \frac{(200+0.02 x)}{(7500+x)^{2}} [/latex]

or

[latex] \frac{ d heta}{ d x}=10\left[\frac{(20000+2 x)}{100(7500+x)^{2}} ight] [/latex]

i.e.

[latex] \frac{ d heta}{ d x}=0.2\left[\frac{2500} {(7500+x)^{2}}+\frac{1}{(7500+x)} ight] [/latex]

Then

[latex] heta=0.2\left[\frac{-2500}{(7500+x)}+\log _{ e }(7500+x)+ B ight] [/latex]

When [latex] x=2500 mm , heta=0 ext { so that } B =-8.96 [/latex] and

[latex] heta=0.2\left[\frac{-2500}{(7500+x)}+\log _{ e }(7500+x)-8.96 ight] rad [/latex]

When [latex] x=0, heta=4.2^{\circ} [/latex] etc.

Question 11.14: Figure P.11.14 shows a thin-walled cantilever box-beam havin...

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