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Question

An overhang beam of uniform EI is loaded as shown in the figure below. the deflection at the free end will be


A
PL381EI
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B
PL327EI
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C
4PL381EI
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D
2PL327EI
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Solution

The correct option is C 4PL381EI
The over hanging part of thr beam can be replaced by moment PL3 and load P at point B. Slope at B

θB=(PL3)L3EI=PL29EI



If there was no load on overhanging part the deflection of load point would have been θB×L3. however dur to load the additional deflection P(L/3)33EI=PL381EI is caused.

Deflection under load

ΔLoad=PL29EI×L3+PL381EI=4PL381EI



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