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2073 Bhadra · 8 marks

Cantilever strain energy: bending, shear, torsion

Short answer

For a cantilever with end load 10kN and torque 5kNm, length 4m, diameter 0.15m, E=200GPa, G=80GPa: bending strain energy U_b = P²L³/(6EI) = 0.2146 kNm; shear strain energy U_s = αP²L/(2GA) = 1.57×10⁻⁴ kNm; torsional strain energy U_t = T²L/(2GJ) = 0.01258 kNm. Shear energy is negligible.

A cantilever beam of length 4 m and having circular cross section of diameter 15 cm is subjected to a concentrated load of 10 kN and a twisting moment 5 kNm at its end. Calculate the strain energies due to bending, shear and torsion. $E = 200\ \text{kN/mm}^2$ , $G = 80\ \text{kN/mm}^2$ .

Given data

$L = 4$ m, $d = 15$ cm $= 0.15$ m, $P = 10$ kN, $T = 5$ kNm, $E = 200$ kN/mm $^2 = 2\times10^8$ kN/m $^2$ , $G = 80$ kN/mm $^2 = 8\times10^7$ kN/m $^2$ .

Section properties

\begin{aligned} A &= \frac{\pi d^2}{4} = 0.017671\ \text{m}^2\\ I &= \frac{\pi d^4}{64} = 2.4850\times10^{-5}\ \text{m}^4\\ J &= \frac{\pi d^4}{32} = 4.9701\times10^{-5}\ \text{m}^4 \end{aligned}

Bending energy

$M = Px$ , so

U_b = \frac{P^2L^3}{6EI} = \frac{10^2\times 4^3}{6\times2\times10^8\times2.4850\times10^{-5}} = \frac{6400}{29820.6} = 0.2146\ \text{kNm}

Shear energy

$V = P$ constant. Shape factor for a solid circle $\alpha = 10/9$ :

U_s = \frac{\alpha P^2 L}{2GA} = \frac{(10/9)\times 100\times 4}{2\times 8\times10^{7}\times 0.017671} = \frac{444.44}{2827433} = 1.572\times10^{-4}\ \text{kNm}

Torsion energy

U_t = \frac{T^2L}{2GJ} = \frac{5^2\times4}{2\times8\times10^7\times4.9701\times10^{-5}} = \frac{100}{7952.2} = 0.01258\ \text{kNm}

Answer:

Energy	Value
Bending $U_b$	0.2146 kNm (214.6 J)
Shear $U_s$	$1.57\times10^{-4}$ kNm (0.157 J)
Torsion $U_t$	0.01258 kNm (12.58 J)

Shear energy is negligible compared with bending energy.

✓ Answer checked by Claude Opus 5.5 (high effort)

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