Explanation Given information: The normal strain along x-axis is − 1120 × 10 − 6 , normal strain along y-axis is − 430 × 10 − 6 ,and shear strain along xy-plane is 780 × 10 − 6 and orientation of element is 45 ° . Explanation: The following are the steps to draw the Mohr’s circle Draw the x-axis as normal strain and y-axis as shear strain and indicate the origin as point O ( 0 , 0 ) . Write the expression to locate the center point C of the circle from the origin O(ie., OC). OC = ε a v g = ε x + ε y 2 .....(I) Here, the normal strain along x-axis is ε x , normal strain along y-axis is ε y , and average strain is ε a v g . Mark the center point O: ( ε a v g , 0 ) . Write the expression for the radius of Mohr’s circle. R = ( ε x − ε a v g ) 2 + ( γ x y 2 ) 2 .....(II) Here, the radius of the Mohr’s circle is R and shear strain along xy-plane is γ x y . Write the expression for principle angle. α = tan − 1 ( γ x y 2 ε x − ε a v g ) .....(III) Here, principle plane angle is α . Write the expression for angle of oriented strain plane. β = 180 ° − 2 θ − α .....(IV) Here, angle of oriented strain plane is β , and θ is the orientation of element. Write the expression for the strain at point D. ( ε x ) 1 = ε a v g + R cos β .....(V) Here, the strain at point D is ( ε x ) 1 Write the expression for the strain at point D ′ . ( ε x ) 2 = ε a v g − R cos β ...... (VI) Here, the strain at point D ′ is ( ε x ) 2 . Write the expression for the shear strain at point D. ( γ x y ) 1 2 = R sin β ...... (VII) Here, the shear strain at point D ( γ x y ) 1 . Write the expression for the shear strain at point D ′ ( γ x y ) 2 2 = − R sin β ...... (VIII) Here, the shear strain at point D ′ ( γ x y ) 2 . Calculation: Substitute − 1120 × 10 − 6 for ε x , and − 430 × 10 − 6 for ε y in Equation (I). ε a v g = ( − 1120 × 10 − 6 ) + ( − 430 × 10 − 6 ) 2 = − 1550 × 10 − 6 2 = − 775 × 10 − 6 Substitute − 1120 × 10 − 6 for ε x , − 775 × 10 − 6 for ε a v g and 780 × 10 − 6 for γ x y in Equation (II). R = ( ( − 1120 × 10 − 6 ) − ( − 775 × 10 − 6 ) ) 2 + ( 780 × 10 − 6 2 ) 2 = ( 345 × 10 − 6 ) 2 + ( 390 × 10 − 6 ) 2 = 520

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ISBN: 9781337093354

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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Chapter 7, Problem 7.7.32P

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The principle strains and maximum shear strain.

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