BAR due to point load is given as,\(δ l=\frac{PL}{AE}\)where, δl = elongation of a bar, P = load or force, L = LENGTH of a bar, A = cross-section of a bar, E = modulus of elasticitycross-section of a bar, \(A=\frac{\pi}{4}D^2\)CALCULATION:Given:δl = 1.25 mm, P = 175 kN = 175 × 103 N, L = 250 mm, \(A=\frac{\pi}{4}25^2=490.87~mm^2\)⇒ \(1.25=\frac{175~×~10^3~×~250}{490.87~\times~E}\)Therefore, E = 71301.97 MPa = 71.30 GPa

"> BAR due to point load is given as,\(δ l=\frac{PL}{AE}\)where, δl = elongation of a bar, P = load or force, L = LENGTH of a bar, A = cross-section of a bar, E = modulus of elasticitycross-section of a bar, \(A=\frac{\pi}{4}D^2\)CALCULATION:Given:δl = 1.25 mm, P = 175 kN = 175 × 103 N, L = 250 mm, \(A=\frac{\pi}{4}25^2=490.87~mm^2\)⇒ \(1.25=\frac{175~×~10^3~×~250}{490.87~\times~E}\)Therefore, E = 71301.97 MPa = 71.30 GPa

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An aluminum tensile test specimen has a diameter, do = 25 mm and a gauge length of Lo = 250 mm. If a force of 175 kN elongates the gauge length by 1.25 mm, the modulus of elasticity of the material is nearly

Materials Science Stress Strain in Materials Science . 7 months ago

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Concept:The deformation of a BAR due to point load is given as,\(δ l=\frac{PL}{AE}\)where, δl = elongation of a bar, P = load or force, L = LENGTH of a bar, A = cross-section of a bar, E = modulus of elasticitycross-section of a bar, \(A=\frac{\pi}{4}D^2\)CALCULATION:Given:δl = 1.25 mm, P = 175 kN = 175 × 103 N, L = 250 mm, \(A=\frac{\pi}{4}25^2=490.87~mm^2\)⇒ \(1.25=\frac{175~×~10^3~×~250}{490.87~\times~E}\)Therefore, E = 71301.97 MPa = 71.30 GPa

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Posted on 10 Nov 2024, this text provides information on Materials Science related to Stress Strain in Materials Science. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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