Explanation The expression to find the normal stress ( σ x ' ) using the stress transformation equation is shown below: σ x ' = σ x + σ y 2 + σ x − σ y 2 cos 2 θ + τ x y sin 2 θ (1) Here, σ x is the normal stress in the x -direction, σ y is the normal stress in the y -direction, θ is the angle, and τ x y is the shear stress in the xy plane. The expression to find the shear stress ( τ x ' y ' ) using the stress transformation equation is shown below: τ x ' y ' = − σ x − σ y 2 sin 2 θ + τ x y cos 2 θ (2) The procedure to construct Mohr’s circle is shown below. The coordinate systems are the horizontal axis represents the normal stress with positive towards right and the vertical axis represents shear stress with positive downwards. Plot the center of the circle C which is located σ avg = σ x + σ y 2 from the origin. Plot the reference points A having coordinates A ( σ x , τ x y ) . This point represents that the x ′ axis coincides with the x axis. Hence, θ = 0 ° . Plot the reference points B having coordinates B ( σ y , − τ x y ) . Connect the point A with the centre C and find CA by trigonometry which represents the radius of the circle R . Sketch the circle. Compare the procedure and the given Figure to find the reference points. The coordinates of the reference point C is ( ( σ x + σ y 2 ) , 0 ) . The coordinates of the reference point A is ( σ x , τ x y ) . The coordinates of the reference point C is ( σ y , − τ x y ) . Find the radius of the circle using the relation: R = ( σ x − σ avg ) 2 + ( τ x y ) 2 Substitute σ x + σ y 2 for σ avg . R = ( σ x − ( σ x + σ y 2 ) ) 2 + ( τ x y ) 2 = ( σ x − σ y 2 ) 2 + ( τ x y ) 2 The expression for normal stress in the x ' plane ( σ x ' ) using the Mohr circle is shown below: σ x ' = σ x + σ y 2 + ( σ x − σ y 2 ) 2 + ( τ x y ) 2 cos θ ' (3) Refer the given Figure. θ ' = 2 θ P − 2 θ (4) cos 2 θ p = σ x − σ x + σ y 2 ( σ x + σ y 2 ) 2 + τ x y 2 (5) sin 2 θ p = τ x y ( σ x + σ y 2 ) 2 + τ x y 2 (6) The expansion of cos for the function ( 2 θ P − 2 θ ) is shown below: cos ( 2 θ P − 2 θ ) = cos 2 θ p cos 2 θ + sin 2 θ p sin 2 θ (7) Substitute Equation (7) and Equation (5), Equation (6) in Equation (3)

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.4, Problem 49P

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Show the coordinates of point P (σx', x'y') gives the same value as the stress transformation equations.

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Chapter 14.4, Problem 48P

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Chapter 14 Solutions

Statics and Mechanics of Materials (5th Edition)

Ch. 14.3 - In each ease, the state of stress x, y, xy...Ch. 14.3 - Given the state of stress shown on the element,...Ch. 14.3 - Determine the normal stress and shear stress...Ch. 14.3 - Prob. 2FP Ch. 14.3 - Determine the equivalent state of stress on an...Ch. 14.3 - Prob. 4FP Ch. 14.3 - The beam is subjected to the load at its end....Ch. 14.3 - Prob. 6FP Ch. 14.3 - Prove that the sum of the normal stresses x+y=x+y...Ch. 14.3 - Determine the stress components acting on the...

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Show the coordinates of point P ( σ x ' , x ' y ' ) gives the same value as the stress transformation equations.

Solution Summary: The author explains that the coordinates of point P left (sigma) are the same as the stress transformation equations. The expression to find the normal stress is shown below.

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Show the coordinates of point P (σx', x'y') gives the same value as the stress transformation equations.

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Chapter 14 Solutions