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Question

A bar of circular cross section is clamped at ends P and Q as shown in the figure. A torsional moment T = 150 Nm is applied at a distance of 100 mm from end P. The torsional reactions (TP,TQ) in Nm at the ends P and Q respectively are

A
(50, 100)
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B
(75, 75)
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C
(100, 50)
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D
(120, 30)
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Solution

The correct option is C (100, 50)
TP+TQ=T
θPR+θRQ=0
TPLGIP+(TPT)2LGIP=0
TP+2TP=2T
TP=2T3=23×150=100Nm
From equation (i),
TQ=T3=1503=50Nm

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