To determine The equivalent state of strain on an element. To sketch: The deformed element due to the equivalent in-plane strains within x-y plane. Explanation Given information: The state of strain at the point on the bracket has components of ε x = − 150 × 10 − 6 , ε y = 450 × 10 − 6 , and γ x y = 200 × 10 − 6 . The element oriented at an angle of θ = 30 ° counterclockwise from the original position. Calculation: Apply the strain transformation Equations as shown below. Find the normal strain component in x ′ direction as shown below. ε x ′ = ε x + ε y 2 + ε x − ε y 2 cos 2 θ + γ x y 2 sin 2 θ (1) Here, ε x is the normal strain in x direction, ε y is the normal strain in y direction, and γ x y is the shear strain of the element. Substitute − 150 × 10 − 6 for ε x , 450 × 10 − 6 for ε y , 200 × 10 − 6 for γ x y , and 30 ° for θ in Equation (1). ε x ′ = [ ( − 150 × 10 − 6 ) + ( 450 × 10 − 6 ) 2 + ( − 150 × 10 − 6 ) − ( 450 × 10 − 6 ) 2 cos 2 ( 30 ° ) + ( 200 × 10 − 6 ) 2 sin 2 ( 30 ° ) ] = ( 150 × 10 − 6 ) + ( − 300 × 10 − 6 × 0.5 ) + ( 100 × 10 − 6 × 0.866 ) = 86.6 × 10 − 6 Hence, the normal strain component in x ′ direction is ε x ′ = 86.6 × 10 − 6 _ . Find the normal strain component in y ′ direction as shown below. ε y ′ = ε x + ε y 2 − ε x − ε y 2 cos 2 θ − γ x y 2 sin 2 θ (2) Substitute − 150 × 10 − 6 for ε x , 450 × 10 − 6 for ε y , 200 × 10 − 6 for γ x y , and 30 ° for θ in Equation (2)

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ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 10.3, Problem 10.11P

To determine

The equivalent state of strain on an element.

To sketch: The deformed element due to the equivalent in-plane strains within x-y plane.

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Chapter 10.3, Problem 10.10P

Chapter 10.3, Problem 10.12P

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Mechanics Of Materials, Si Edition

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