Strain energy and real work deflection, overhang beam

Given data

Overhanging beam: span $AB = 5$ m (A hinge, B roller), overhang $BC = 2$ m, load $P = 20$ kN at C. Section 100 mm $\times$ 120 mm ($b = 0.1$ m, $d = 0.12$ m). $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $G = 80$ kN/mm$^2 = 8\times10^7$ kN/m$^2$.

        A            B      C
        ^------------^------+  20 kN
        |    5 m     | 2 m  v

$$I = \frac{0.1\times0.12^3}{12} = 1.44\times10^{-5}\ \text{m}^4,\quad EI = 2880\ \text{kNm}^2,\quad A = 0.012\ \text{m}^2$$

Reactions

$$R_A\times5 = 20\times2 \Rightarrow R_A = 8\ \text{kN (down)},\qquad R_B = 28\ \text{kN (up)}$$

Bending moments and shear forces (magnitudes)

Strain energy due to bending

$$U_b = \int_0^5\frac{(8x)^2}{2EI}dx + \int_0^2\frac{(20s)^2}{2EI}ds = \frac{1}{2EI}\left[\frac{64\times125}{3} + \frac{400\times8}{3}\right] = \frac{3733.33}{2\times2880}$$ $$U_b = 0.6481\ \text{kNm} = 648.1\ \text{J}$$

Strain energy due to shear

For a rectangle, the form factor is $\alpha = 1.2$:

$$U_s = \frac{\alpha}{2GA}\left[8^2\times5 + 20^2\times2\right] = \frac{1.2\times1120}{2\times8\times10^7\times0.012} = 7.0\times10^{-4}\ \text{kNm} = 0.70\ \text{J}$$

Deflection at C (real work method)

External work = total strain energy:

$$\tfrac12 P\Delta_C = U_b + U_s = 0.6481 + 0.0007 = 0.6488\ \text{kNm}$$ $$\Delta_C = \frac{2\times0.6488}{20} = 0.06488\ \text{m}$$

Answer: $U_b = 0.648$ kNm (648 J), $U_s = 0.0007$ kNm (0.70 J), and $\Delta_C = 64.9$ mm downward (bending alone gives 64.81 mm; shear adds only 0.07 mm). The shear energy is about 0.1% of the bending energy.