Question 11.SP.2: (a) Taking into account only the effect of normal stresses d......

MODELING:
Bending Moment. Using the free-body diagram of the entire beam (Fig. 1), determine the reactions

R _A=\frac{P b}{L} \uparrow \quad R _B=\frac{P a}{L} \uparrow

Using the free-body diagram in Fig. 2, the bending moment for portion AD of the beam is

M_1=\frac{P b}{L} x

Similarly using the free-body diagram in Fig. 3, the bending moment for portion DB at a distance ν from end B is

M_2=\frac{P a}{L} ν

ANALYSIS:
a. Strain Energy. Since strain energy is a scalar quantity, add the strain energy of segment AD to that of DB to obtain the total strain energy of the beam. Using Eq. (11.15),

\begin{aligned} U & =U_{A D}+U_{D B} \\ & =\int_0^a \frac{M_1^2}{2 E I} d x+\int_0^b \frac{M_2^2}{2 E I} d ν \\ & =\frac{1}{2 E I} \int_0^a\left(\frac{P b}{L} x\right)^2 d x+\frac{1}{2 E I} \int_0^b\left(\frac{P a}{L} ν\right)^2 d ν \\ & =\frac{1}{2 E I} \frac{P^2}{L^2}\left(\frac{b^2 a^3}{3}+\frac{a^2 b^3}{3}\right)=\frac{P^2 a^2 b^2}{6 E I L^2}(a+b) \end{aligned}

or since (a + b) = L,        U=\frac{P^2 a^2 b^2}{6 E I L}

b. Evaluation of the Strain Energy. The moment of inertia of a W10 × 45 rolled-steel shape is obtained from Appendix E, and the given data is restated using units of kips and inches.

P = 40 kips                                                   L = 12 ft = 144 in.
a = 3 ft = 36 in.                                            b = 9 ft = 108 in.
E = 29 × 10^6 psi = 29 × 10³ ksi                I = 248 in^4

Substituting into the expression for U,

U=\frac{(40 \text {  kips })^2(36 \text {  in. })^2(108 \text {  in. })^2}{6\left(29 \times 10^3  ksi \right)\left(248 \text {  in }^4\right)(144 \text {  in. })}

U = 3.89 in·kips