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Chapter 10

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?

Step-by-Step

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

10.29