A beam AB of length L (Fig. P.13.2) is freely supported at A and at a point C which is at a distance KL from the end B. If a uniformly distributed load of intensity w per unit length acts on AC, find the value of K which will cause the upward deflection of B to equal the downward deflection midway between A and C.
Ans. 0.24.

Question

Accepted Answer

Since there are no loads on the part CB of the beam and the implication in the question is that the beam is weightless, the length CB will remain straight but will rotate about C. From Ex. 13.4 the mid-span deflection of the simply supported span AC is [latex]5 w L^{4}(1-K)^{4} / 384 E I[/latex] (see Eq. (v)).

[latex]v_{ ext {mid-span }}=-\frac{5 w L^{4}}{384 E I}[/latex] (v) from Ex. 13.4

Also, the slope of the beam at C is given by Eq. (iii)

[latex]E I \frac{ d v}{ d x}=\frac{w}{24}\left(6 L x^{2}-4 x^{3}-L^{3} ight)[/latex] (iii) from Ex. 13.4
and is [latex]w L^{3}(1-K)^{3} / 24 E I[/latex] The deflection at B is therefore

[latex]w L^{3}(1-K)^{3} K L / 24 E I[/latex]

The two deflections are numerically equal so that

[latex]\frac{5 w L^{4}(1-K)^{4}}{384 E I}=\frac{w L^{3}(1-K)^{3} K L}{24 E I}[/latex]

i.e.

[latex]\left(\frac{5}{16} ight)(1-K)=K[/latex]

from which

K = 0.24 .

Question 13.P.2: A beam AB of length L (Fig. P.13.2) is freely supported at A...

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