The beam shown in Fig. P.13.26 is simply supported at each end and is provided with an additional support at mid-span. If the beam carries a uniformly distributed load of intensity w and has a flexural rigidity EI, use the principle of superposition to determine the reactions in the supports.
Ans. 5wL/4 (central support), 3wL/8 (outside supports).

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Accepted Answer

Release the beam at the mid-span point. The mid-span deflection of the resulting simply supported beam is given by Eq. (v) of Ex. 13.4

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

and is [latex]5 w(2 L)^{4} / 384 E I[/latex]. Suppose that the mid-span support reaction is R. The deflection of a simply supported beam carrying a central concentrated load is, from Eq. (v) of Ex. 13.5,

[latex]v_{ ext {mid-span }}=-\frac{W L^{3}}{48 E I}[/latex] Eq. (v) from Ex.13.5

[latex]R(2 L)^{3} / 48 E I[/latex]. From the principle of superposition the two deflections must be equal but opposite in direction. Equating them gives R = 5wL/4.
From equilibrium and symmetry the reactions at the outside supports are

[latex]\frac{w(2 L)-\frac{5 w L}{4}}{2}=\frac{3 w L}{8}[/latex].

Question 13.P.26: The beam shown in Fig. P.13.26 is simply supported at each e...

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