A rolled steel joist is simply supported at its end and carries a
A rolled steel joist is simply supported at its end and carries a uniformly distributed load which causes a maximum deflection of 10 mm and slope at the ends of 0.002 radians. The length of the joist will be
A. 10 m
B. 12 m
C. 14 m
D. 16 m
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
\(\begin{array}{l} {Y_{max}} = \frac{5}{{384}}\frac{{W{L^4}}}{{EI}} = 0.010\;m\\ 0.002 = {\theta _A} = {\theta _B} = \frac{{W{L^3}}}{{24EI}}\;\\ {Y_{max}} = \frac{5}{{384}} \times 24 \times \frac{{W{L^3}}}{{24EI}} \times L\\ 0.010 = \;\frac{5}{{384}} \times 24 \times 0.002 \times L \end{array}\)
L = 16 m
Important Point:
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
