A cylindrical specimen of length L and cross - sectional area A extends by δ under a gradually applied tensile load P. The strain energy of the specimen is

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A cylindrical specimen of length L and cross - sectional area A extends by δ under a gradually applied tensile load P. The strain energy of the specimen is

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BPSC AE Paper 4 (General Engineering Science) 10 Nov 2022 Official Paper

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$\left(\frac{1}{2}\right)$P δ
$\frac{1}{2}\left(\frac{P}{A} \times \frac{\delta}{L}\right)$
(stress on the cross - section) × (strain along the axis) × (volume of specimen)
P δ

Answer 1

Explanation:

The strain energy of a cylindrical specimen subjected to a gradually applied tensile load can be calculated using the formula for strain energy in elastic materials. Strain energy ( $U$ ) is the energy stored in a body due to deformation under loading.

For a cylindrical specimen of length $L$ , cross-sectional area , extension $δ$ , and under a gradually applied tensile load , the strain energy is given by:

U=1/2×P×δ

Additional InformationThe elastic strain energy stored in a member of length (it may be curved or straight) due to axial force, bending moment, shear force and torsion are summarized below:

Axial Force, P	${U_1} = \mathop \smallint \limits_0^s \frac{{{P^2}}}{{2AE}}ds$
Bending, M	${U_2} = \mathop \smallint \limits_0^s \frac{{{M^2}}}{{2EI}}ds$
Shear Force, V	${U_3} = \mathop \smallint \limits_0^s \frac{{{V^2}}}{{2AG}}ds$
Torsion, T	${U_4} = \mathop \smallint \limits_0^s \frac{{{T^2}}}{{2GJ}}ds$

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A cylindrical specimen of length L and cross - sectional area A extends by δ under a gradually applied tensile load P. The strain energy of the specimen is

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Axial Force, P	\({U_1} = \mathop \smallint \limits_0^s \frac{{{P^2}}}{{2AE}}ds\)
Bending, M	\({U_2} = \mathop \smallint \limits_0^s \frac{{{M^2}}}{{2EI}}ds\)
Shear Force, V	\({U_3} = \mathop \smallint \limits_0^s \frac{{{V^2}}}{{2AG}}ds\)
Torsion, T	\({U_4} = \mathop \smallint \limits_0^s \frac{{{T^2}}}{{2GJ}}ds\)