Problem 6.2. Potential energy of a beam with built-in ends

Consider a beam built-in at both ends subjected to uniformly distributed load, p = 8 kN/m. Deflection and moment functions of a beam are

$v = \frac{p}{24 E I} (x^{2} L^{2} - 2 x^{3} L + x^{4}) M_{z} = \frac{p}{12} (- L^{2} + 6 x L - 6 x^{2})$

Functions are derived in Example 3.4.

Using the given functions determine the potential energy
of the beam. Length of the beam is L = 5 m, the bending stiffness is EI = 3.8 ×10¹² Nmm².

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Potential energy, π [kNm]=

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Steps

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Step 1. Determine strain energy of the beam.

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$U = \frac{1}{2} \int_{L} κ_{z} M_{z} d x = \frac{1}{2 E I} \int_{L} M_{z}^{2} d x = \frac{1}{2 E I} \int_{L} \frac{p^{2}}{12^{2}} {(- L^{2} + 6 x L - 6 x^{2})}^{2} d x = = \frac{p^{2}}{2 \times 12^{2} E I} {[L^{4} x + 36 L^{2} \frac{x^{3}}{3} + 36 \frac{x^{5}}{5} - 12 L^{3} \frac{x^{2}}{2} + 12 L^{2} \frac{x^{3}}{3} - 72 L \frac{x^{4}}{4}]}_{0}^{L} = \frac{p^{2} L^{5}}{1440 E I} = 36.55 kNm$

Eq.(6-19)

Step 2. Determine the work done by the external load.
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Eq.(6-26)

$W = - \int_{L} v p d x = - p \int_{L} \frac{p}{24 E I} (x^{2} L^{2} - 2 x^{3} L + x^{4}) d x = - \frac{p^{2}}{24 E I} {[L^{2} \frac{x^{3}}{3} - 2 L \frac{x^{4}}{4} + \frac{x^{5}}{5}]}_{0}^{L} = = - \frac{p^{2} L^{5}}{720 E I} = - 73.10 kNm$

Step 3. Give the potential energy.

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Eq.(6-25)

$π = U + W = \frac{p^{2} L^{5}}{1440 E I} - \frac{p^{2} L^{5}}{720 E I} = - \frac{p^{2} L^{5}}{1440 E I} = - 36.55 kNm$

Results

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The strain energy of the beam is

Eq.(6-19)

U = \frac{1}{2} \int_{L} κ_{z} M_{z} d x = \frac{1}{2 E I} \int_{L} M_{z}^{2} d x = \frac{1}{2 E I} \int_{L} \frac{p^{2}}{12^{2}} {(- L^{2} + 6 x L - 6 x^{2})}^{2} d x = = \frac{p^{2}}{2 \times 12^{2} E I} {[L^{4} x + 36 L^{2} \frac{x^{3}}{3} + 36 \frac{x^{5}}{5} - 12 L^{3} \frac{x^{2}}{2} + 12 L^{2} \frac{x^{3}}{3} - 72 L \frac{x^{4}}{4}]}_{0}^{L} = \frac{p^{2} L^{5}}{1440 E I} = 36.55 kNm

The work done by the external load is

Eq.(6-26)

The potential energy becomes

Eq.(6-25)

$π = U + W = \frac{p^{2} L^{5}}{1440 E I} - \frac{p^{2} L^{5}}{720 E I} = - \frac{p^{2} L^{5}}{1440 E I} = - 36.55 kNm$