Question
Download Solution PDFA simply supported beam of span L and carrying a concentrated load of W at mid-span. The value of deflection at mid-span will
- \(\frac{WL^2}{48EI}\)
\(\frac{WL^3}{48EI}\)- \(\frac{WL^4}{48EI}\)
- \(\frac{WL^3}{30EI}\)
Answer (Detailed Solution Below)
\(\frac{WL^3}{48EI}\)Detailed Solution
Download Solution PDFConcept:
Moment Area Method
\(\frac{d^2y}{dx^2}~=~\int\frac{Mx}{EI_x}dx\)
\(\int_A^B\theta~=~\int_A^B\frac{Mx}{EI_x}dx\)
Mohr's Theorem No.1
\(\theta_B-\theta_A~=~[Area~of~\frac{BMD}{EI}~diagram~between~A~and~B] \)
Mohr's Theorem No.2
\(Y_B-Y_A~=~[Moment~of~Area~of~\frac{BMD}{EI}~diagram~between~B~and~A]\)
where A = Reference point where slope zero, B = Origin point where slope and deflection are to be determined.
The distance of centroid of the area should be measured from the origin.
Calculation:
As we have to find the deflection at the midpoint (C) the area of half the bending moment diagram can be considered.
Area of the bending moment diagram is given by:
\(A~=~\frac{1}{2}\times \frac{WL}{4}\times \frac{L}{2}\)
\(A~=~\frac{WL^2}{16}\)
Area of the \(\frac{{BMD}}{{EI}}\) diagram = \(\frac{{WL^2}}{{16EI}}\)
x̅ = Distance of the centroid of the Bending Moment Diagram from the origin
\(\bar x ~=~\frac{2}{3}\frac{L}{2}~=~\frac{L}{3}\)
\(Deflection~=~\frac{A\bar x}{EI}~=~\frac{WL^2}{16EI}\times \frac{L}{3}~=~\frac{WL^3}{48EI} \)
Additional Information
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
Last updated on Jul 1, 2025
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