Explanation Given information: The state of strain at the point on the bracket has components of ε x = − 200 × 10 − 6 , ε y = − 650 × 10 − 6 , and γ x y = − 175 × 10 − 6 . The element oriented at an angle of θ = 20 ° , 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 − 200 × 10 − 6 for ε x , − 650 × 10 − 6 for ε y , − 175 × 10 − 6 for γ x y , and 20 ° for θ in Equation (1). ε x ′ = [ ( − 200 × 10 − 6 ) + ( − 650 × 10 − 6 ) 2 + ( − 200 × 10 − 6 ) − ( − 650 × 10 − 6 ) 2 cos 2 ( 20 ° ) + ( − 175 × 10 − 6 ) 2 sin 2 ( 20 ° ) ] = ( − 425 × 10 − 6 ) + ( 225 × 10 − 6 × 0.766 ) − ( 87.5 × 10 − 6 × 0.643 ) = − 309 × 10 − 6 Hence, the normal strain component in x ′ direction is ε x ′ = − 309 × 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 − 200 × 10 − 6 for ε x , − 650 × 10 − 6 for ε y , − 175 × 10 − 6 for γ x y , and 20 ° for θ in Equation (2)

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.3, Problem 10.8P

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The equivalent in-plane strains 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 10.7P

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

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

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The equivalent in-plane strains on an element.

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