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Chapter 10

Q. 10.11

A simple beam with an overhang supports a load P at the free end of the beam as shown in Figure 10.27.
Calculate the strain energy of the beam.

10.27

Step-by-Step

Verified Solution

From Eq. (10.22) for prismatic beam, we get

U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}_{z z}} \right\rgroup \int_0^L M^2 d x                (10.22)

U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup \int_0^L M_x^2 d x

Now, from Figure 10.28, we observe the following:

M_x=\left\{\begin{array}{lr} -R_{ A } x ; & 0 \leq x \leq L \\ -R_{ A } x+R_{ B }(x-L) ; & L \leq x \leq(L+a) \end{array}\right\}

where M_x is the bending moment at any section at a distance x from end A. Therefore,

U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup\left[\int_0^L M_x^2 d x+\int_L^{L+a} M_x^2 d x\right]              (1)

Now, putting the value of R_{ A } , we get

\int_0^L M_x^2 d x=R_{ A }^2 \int_0^L x^2 d x=\frac{R_{ A }^2 L^3}{3}=\frac{P^2 a^2 L}{3}

and            \int_L^{L+a} M_x^2 d x=\int_L^{L+a}\left[\left(R_{ B }-R_{ A }\right) x-L R_{ B }\right]^2 d x

=\frac{\left(R_{ B }-R_{ A }\right)^2}{3}\left[(L+a)^3-L^3\right]+\left(L^2 R_{ B }^2\right)[(L+a)-L]

-L R_{ B }\left(R_{ B }-R_{ A }\right)\left[(L+a)^2-L^2\right]

=\frac{P^2}{3}\left(3 L^2 a+3 a L^2+a^3\right) a+(L+a)^2 P^2-(L+a) P^2 \cdot\left(2 L a+a^2\right)

=P^2\left[\left\lgroup L^2 a+a^2 L+\frac{a^3}{3} \right\rgroup+\left(a L^2+a^3+2 L a^2\right)-\left(2 L^2 a+a^2 L+2 L a^2+a^3\right)\right]

=\frac{P^2 a^3}{3}

From Eq. (1),

U_{\text {bending }}=P^2\left[\frac{a^2 L}{3}+\frac{a^3}{3}\right]\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup

The required strain energy of the beam is

U_{\text {bending }}=\frac{P^2 a^2(a+L)}{6 E \bar{I}}

10.28