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Q. 10.13

A simply supported beam of span L is loaded in such a manner that its deflection curve is a parabola ( symmetric about the middle point) with centre deflection δ. How much strain energy is stored in the beam?

Verified Solution

Let us draw the beam and the deflection curve as shown in Figure 10.29.

The deflection curve may be assumed as

y = Ax + Bx²

where y(0) = y(L) = 0 and y(L/2) = δ. Therefore,

AL+BL²=0

$A \frac{L}{2}+B \frac{L^2}{4}=\delta$

Solving the above equation, we get

$A=4\left\lgroup \frac{\delta}{L} \right\rgroup \quad \text { and } \quad B=-\frac{4 \delta}{L^2}$

Thus, the equation to the deflection curve is

$y=4\left\lgroup \frac{\delta}{L} \right\rgroup\left\lgroup x-\frac{x^2}{L} \right\rgroup$          (1)

Now, strain energy of a prismatic beam due to bending from Eq. (10.28) is

$U_{\text {bending }}=\left\lgroup \frac{E \bar{I}}{2} \right\rgroup \int_0^L\left\lgroup \frac{ d ^2 y}{ d x^2} \right\rgroup^2 d x$               (10.28)

From Eq. (1),

$\frac{ d ^2 y}{ d x^2}=4\left\lgroup \frac{\delta}{L} \right\rgroup \left[0-\frac{2}{L}\right]=-\frac{8 \delta}{L^2}$

Therefore,

$U_{\text {bending }}=\left\lgroup\frac{E \bar{I}}{2}\right\rgroup \int_0^L\left\lgroup-\frac{8}{L^2} \delta\right\rgroup d x$

$=\left\lgroup\frac{E \bar{I}}{2}\right\rgroup\left\lgroup\frac{64}{L^4}\right\rgroup \delta^2 L=\frac{32 E \bar{I} \delta^2}{L^3}$

The required strain energy stored in the beam is $U_{\text {bending }}=32 E \bar{I} \delta^2 / L^3$