Strain energy in rectangular beam, central load

Given data

$b = 25$ cm, $d = 50$ cm, $L = 6$ m $= 600$ cm, $P = 100$ kN, $E = 2\times10^6$ kg/cm$^2$, $G = 0.85\times10^6$ kg/cm$^2$.

Assumption: $1\ \text{kgf} = 9.81$ N, so $P = 100\,000/9.81 = 10\,193.7$ kgf. Shear form factor for rectangle $\alpha = 1.2$.

$$I = \frac{25\times50^3}{12} = 2.604\times10^5\ \text{cm}^4,\qquad A = 25\times50 = 1250\ \text{cm}^2$$

Strain energy due to bending

For a central load, $M = Px/2$ for $x \le L/2$:

$$U_b = 2\int_0^{L/2}\frac{(Px/2)^2}{2EI}dx = \frac{P^2L^3}{96EI}$$ $$U_b = \frac{(10193.7)^2\times(600)^3}{96\times2\times10^6\times2.604\times10^5} = 448.9\ \text{kg-cm}$$

Strain energy due to shear

$V = P/2$ throughout the span:

$$U_s = \frac{\alpha V^2 L}{2GA} = \frac{1.2\times(5096.8)^2\times600}{2\times0.85\times10^6\times1250} = 8.80\ \text{kg-cm}$$

Answer:

• Bending: $U_b = 448.9$ kg-cm ($\approx 44.0$ N·m)
• Shear: $U_s = 8.80$ kg-cm ($\approx 0.863$ N·m)

Shear energy is only about 2% of the bending energy.