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Cantilever strain energy: bending and torsion

Short answer

For a cantilever with end load P=15kN and torque T=8kNm, length 8m, diameter 0.2m, E=200GPa, G=80GPa: bending strain energy U_b = P²L³/(6EI) = 1.222 kNm; torsional strain energy U_t = T²L/(2GJ) = 0.0204 kNm. Torsional energy is much smaller than bending energy.

A cantilever beam of length 8 m having circular cross section of diameter 20 cm is subjected to a load of 15 kN and a twisting moment of 8 kNm at its end. Calculate the strain energies due to bending and torsion. Assume necessary data.

Given data

Length L=8L = 8 m, diameter d=20d = 20 cm =0.2= 0.2 m, end load P=15P = 15 kN, twisting moment T=8T = 8 kNm.

Assumed: E=200E = 200 GPa =2×108= 2\times10^8 kN/m2^2, G=80G = 80 GPa =8×107= 8\times10^7 kN/m2^2.

Section properties

I=πd464=π(0.2)464=7.854×10−5 m4I = \frac{\pi d^4}{64} = \frac{\pi(0.2)^4}{64} = 7.854\times10^{-5}\ \text{m}^4 J=πd432=1.5708×10−4 m4J = \frac{\pi d^4}{32} = 1.5708\times10^{-4}\ \text{m}^4

Strain energy due to bending

At distance xx from the free end, M=PxM = Px:

Ub=∫0L(Px)22EIdx=P2L36EIU_b = \int_0^L \frac{(Px)^2}{2EI}dx = \frac{P^2L^3}{6EI} Ub=152×836×2×108×7.854×10−5=11520094247.8=1.2223 kNmU_b = \frac{15^2\times 8^3}{6\times 2\times10^{8}\times 7.854\times10^{-5}} = \frac{115200}{94247.8} = 1.2223\ \text{kNm}

Strain energy due to torsion

Torque is constant along the length:

Ut=T2L2GJ=82×82×8×107×1.5708×10−4=51225132.7=0.02037 kNmU_t = \frac{T^2L}{2GJ} = \frac{8^2\times 8}{2\times 8\times10^{7}\times 1.5708\times10^{-4}} = \frac{512}{25132.7} = 0.02037\ \text{kNm}

Answer: Ub=1.222U_b = 1.222 kNm (≈1222\approx 1222 J) and Ut=0.0204U_t = 0.0204 kNm (≈20.4\approx 20.4 J).

The torsional energy is small compared to the bending energy.

✓ Answer checked by Claude Opus 5.5 (high effort)

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