Question
Download Solution PDFA 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:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
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:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
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\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
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\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
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\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
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\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
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\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
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\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
Last updated on Jun 4, 2025
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