A Cantilever carries a uniformly distributed load W over its whole length and a force W acts at its free end upward. The net deflection of the free end will be:

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  1. Zero
  2. (5/24) (WL3/EI) upward
  3. (5/24) (WL3/EI) downward 
  4. None of these

Answer (Detailed Solution Below)

Option 2 : (5/24) (WL3/EI) upward
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Detailed Solution

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Calculation:

F1 Ram Ravi 12.08.21 D1

Deflection at free end = Deflection due to point load - Deflection due to UDL

\(\delta = \frac{{W{L^3}}}{{3EI}} - \frac{{w{L^4}}}{{8EI}}\)

⇒ \(\delta = \frac{{W{L^3}}}{{3EI}} - \frac{{W{L^3}}}{{8EI}}\)  (Where W = wL)

⇒ \(\delta = \frac{{W{L^3}}}{{EI}}\left( {\frac{1}{3} - \frac{1}{8}} \right) = \frac{5}{{24}}\frac{{W{L^3}}}{{EI}}\)

Important Points

 Deflection and slope of various beams are given by:

F2 A.M Madhu 09.04.20 D1

 

\({y_B} = \frac{{P{L^3}}}{{3EI}}\)

 

\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\)

F2 A.M Madhu 09.04.20 D2

 \({y_B} = \frac{{w{L^4}}}{{8EI}}\)

 

\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\)

F2 A.M Madhu 09.04.20 D3

 

\({y_B} = \frac{{M{L^2}}}{{2EI}}\)

 

\({\theta _B} = \frac{{ML}}{{EI}}\)

F2 A.M Madhu 09.04.20 D4

 

\({y_B} = \frac{{w{L^4}}}{{30EI}}\)

 

\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\)

F2 A.M Madhu 09.04.20 D5

 

\({y_c} = \frac{{P{L^3}}}{{48EI}}\)

 

\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\)


F2 A.M Madhu 09.04.20 D6

 

\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\)

 

\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\)


F2 A.M Madhu 09.04.20 D7

 

\({y_c} = 0\)

\({\theta _B} = \frac{{ML}}{{24EI}}\)

F2 A.M Madhu 09.04.20 D8

\({y_c} = \frac{{M{L^2}}}{{8EI}}\)

 

\({\theta _B} = \frac{{ML}}{{2EI}}\)


F2 A.M Madhu 09.04.20 D9

\({y_c} = \frac{{P{L^3}}}{{192EI}}\)

\({\theta _A} = {\theta _B} = {\theta _C} = 0\)

F2 A.M Madhu 09.04.20 D10

\({y_c} = \frac{{w{L^4}}}{{384EI}}\)

\({\theta _A} = {\theta _B} = {\theta _C} = 0\)

Where, y = Deflection of beam, θ = Slope of beam

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