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Question

# Beam PQRS has internal hinges in spans PQ and RS as shown.The beam may be subjected to a moving distributed vertical load of maximum intensity 4 kN/m of any length anywhere on the beam.The maximum absolute value of the shear force (in kN) that can occur due to this loading just to the right of support Q shall be.

A
55
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B
30
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C
40
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D
45
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Solution

## The correct option is C 40When a cut is made just to the right of Q and displacements are given such that A'B' is parallel to B'C'.As B' is very close to Q,displacement of B' to the left will be zero and that to the right will be, 1 Hence Slope of B'C' =12 ⇒ Slope of B′A′ =120 ⇒ Ordinate at A′ =120×5=0.25 Ordinate at D′ =120×5=0.25 if udl is loading span PR,we get max.SF just to the right of Q ⇒SF=(12×0.25×10+12×20×1)×4 SF=45kN Alternative solution Rs=0 RP×5−1(5−x)=0 Rp=5−x5 RR+RQ+RP=1 RQ+RR=1−(5−x5)=5−5+x5=x5 RQ+RR=x5 RP×10−1(10−x)=RR×20 ⇒ 20×RR=5−x5×10−(10−x) /10−2x−/10+x=RR×20 −x20=RR SF to the right of Q, for x ≤5=−RR=x20 RQ+RR=1 1×(10−x)+RR×20=0 RR=x−1020 ⇒ SF to right of Q=-RR=10−x20 RQ=30−x20 SF to right of Q=(30−x)20 RQ×20+1×(x−30)=0 RQ=30−x20=−(x−30)20 SF to right of Q=−(x−30)20 SF when PR is loaded, =(12×0.25×10+12×1×20)×4 =(1.25+10)×4=40+5=45

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