Explanation Given: Length of bar is 1 m . Radius of curvature of neutral axis is 0.50 m . Thickness of bar is 25 mm . Yield strength of material is 500 MPa . Elastic modulus of material is 207 GPa . Concept used: Write the expression for engineering strain in curved bars during straightening. e = t 2 ρ ............. (1) Here, e is the engineering strain in curved bars during straightening process, t is the thickness of bar and ρ is the radius of curvature of neutral axis. Write the expression for elastic limit of material. e e l = S y E ............. (2) Here, e e l is the elastic limit of material, S y is the yield strength of material and E is the elastic modulus of material. Write the expression for strain required to straighten the bar. e s t = 2 e e l ............. (3) Here, e s t is the strain required to straighten the bar. Write the expression for final length of specimen. l f = l 0 e s t + l 0 ............. (4) Here, l f is the final length of specimen and l 0 is the initial length of specimen. Calculation: Substitute 25 mm for t and 0.50 m for ρ in equation (1). e = 23 mm 2 ( 0

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ISBN: 9780134425115

Author: Schmid

Publisher: YUZU

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 2, Problem 2.69P

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The length to which the bar should be stretched so that it will be straight after unloading.

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The length to which the bar should be stretched so that it will be straight after unloading.

Solution Summary: The author explains how to write the expression for engineering strain in curved bars during straightening.

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