A cantilever beam of span I and flexural rigidity EI is subjected
A. <span class="math-tex">\(\frac{{{\rm{W}}{{\rm{l}}^2}}}{{2{\rm{EI}}}}\)</span>
B. <span class="math-tex">\(\frac{{{\rm{W}}{{\rm{l}}^2}}}{{4{\rm{EI}}}}\)</span>
C. <span class="math-tex">\(\frac{{{\rm{W}}{{\rm{l}}^2}}}{{8{\rm{EI}}}}\)</span>
D. <span class="math-tex">\(\frac{{{\rm{W}}{{\rm{l}}^2}}}{{3{\rm{EI}}}}\)</span>
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Right Answer is: C
SOLUTION
Solving by Moment area method:
θA would be zero as there is fixed support at A.
\({{\rm{\theta }}_{\rm{C}}} - {{\rm{\theta }}_{\rm{A}}} = \frac{{{\rm{Area\;of\;BMD\;between\;C\;and\;A}}}}{{{\rm{EI}}}}\)
∴ \({{\rm{\theta }}_{\rm{C}}} = \frac{1}{2} \times \frac{{{\rm{WL}}}}{2} \times \frac{{\rm{L}}}{2} \times \frac{1}{{{\rm{EI}}}} = \frac{{{\rm{W}}{{\rm{L}}^2}}}{{8{\rm{EI}}}}\)
Note:
Deflection and slope of various beams are given by:
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\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
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\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
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\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
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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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\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
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\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
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\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
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\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
