A beam has a second moment of area I for the central half of its span and I/2 for the outer quarters. If the beam carries a central concentrated load W, find the deflection at mid-span if the beam is simply supported and also the fixed-end moments when both ends of the beam are built-in.
Ans. $3 W L^{3} / 128 E I, 5 W L / 48$ .

Question

A beam has a second moment of area I for the central half of its span and I/2 for the outer quarters. If the beam carries a central concentrated load W, find the deflection at mid-span if the beam is simply supported and also the fixed-end moments when both ends of the beam are built-in.
Ans. [latex]3 W L^{3} / 128 E I, 5 W L / 48[/latex].

Accepted Answer

The moment-area method may be used to solve this problem. For the case of the simply supported beam and taking the origin of axes at the left hand support Eq. (13.10)

[latex]x_{ B }\left(\frac{ d v}{ d x} ight)_{ B }-x_{ A }\left(\frac{ d v}{ d x} ight)_{ A }-\left(v_{ B }-v_{ A } ight)=\int_{ A }^{ B } \frac{M}{E I} x d x[/latex] (13.10)

becomes

[latex]v_{ ext {mid-span }}=\int_{0}^{L / 4}\left(\frac{W x^{2}}{2\left(\frac{E I}{2} ight)} ight) d x+\int_{L / 4}^{L / 2}\left[\frac{W x^{2}}{2 E I} ight] d x[/latex]

which gives

[latex]v_{ ext {mid-span }}=\frac{3 W L^{3}}{128 E I}[/latex]

When the beam is built in at both ends the slope of the beam at both ends and at mid-span is zero. Suppose that the fixed-end moments are M. Then, from Eq. (13.7)

[latex]\left(\frac{ d v}{ d x} ight)_{ B }-\left(\frac{ d v}{ d x} ight)_{ A }=\int_{ A }^{ B } \frac{M}{E I} d x[/latex] (13.7)

[latex]0=\int_{0}^{L / 4}\left(\frac{2}{E I} ight)\left(\frac{W x}{2}-M ight) d x+\int_{L / 4}^{L / 2}\left(\frac{1}{E I} ight)\left(\frac{W x}{2}-M ight) d x[/latex]

from which

[latex]M=\frac{5 W L}{48}[/latex].

Question 13.P.28: A beam has a second moment of area I for the central half of...

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