To determine The principal stress at point A on the cross section of the arm at section a - a . The orientation of the state of stress. To sketch: The results on an element at point A . Explanation Let B x and B y be the horizontal and vertical force at point B . Let F C D be the force in the cable CD . Sketch the free body diagram of the entire arm as shown in Figure 1. Apply equations of equilibrium. Sum of moments at B is equal to 0. Σ M B = 0 F C D sin 30 ° ( 0.3 m ) − 500 N ( 0.65 m ) = 0 F C D = 2 , 166.67 N Sum of forces in the x -direction is equal to 0. Σ F x = 0 B x − F C D cos 30 ° = 0 B x − 2 , 166.67 cos 30 ° = 0 B x = 1 , 876.39 N Sum of forces in the y -direction is equal to 0. Σ F y = 0 − B y + F C D sin 30 ° − 500 = 0 − B y + 2 , 166.67 sin 30 ° − 500 = 0 B y = 583.33 N Sketch the free body diagram of the left cut segment of the arm as shown in Figure 2. Apply equations of equilibrium. Sum of forces in the x -direction is equal to 0. Σ F x = 0 1 , 876.39 N − N = 0 N = 1 , 876.39 N Sum of forces in the y -direction is equal to 0. Σ F y = 0 V − 583.33 N = 0 V = 583.33 N Sum of moments at O is equal to 0. Σ M O = 0 583.33 N ( 0.15 m ) − M = 0 M = 87.5 N ⋅ m Find the cross sectional area ( A ) about z -axis of the cross section of the arm: A = ( 20 mm × 50 mm ) − ( 12.5 mm × 35 mm ) = 562.5 mm 2 Find the moment of inertia ( I ) about z -axis of the cross section of the arm: I = 1 12 ( 20 mm ) ( 50 mm ) 3 − 1 12 ( 12.5 mm ) ( 35 mm ) 3 = 163 , 671.875 mm 4 Sketch the centroid of the I-section as shown in Figure 3. Find the first moment of area at point A ( Q A ) using Figure 3. Q A = y ¯ ' A ' = ( 0.2125 m ) ( 0.0075 m × 0.02 m ) = 3.1875 × 10 − 6 m 3 Refer the cross section, The centroid distance of the section ( y ) to point A is 0.0075 m Find the normal stress at point A ( σ A ) using the relation: σ A = N A + M y I = − 1 , 876.39 N 562.5 mm 2 + ( 87.5 N ⋅ m ) ( 0.0175 m ) 163 , 671.875 mm 4 = − 1 , 876.39 N 562.5 mm 2 + ( 87.5 N ⋅ m × 10 3 mm 1 m ) ( 0.0175 m × 10 3 mm 1 m ) 163 , 671.875 mm 4 = 6.02 MPa The thickness of the cross section is t = 0.0175 m . Find the shear stress at point A ( τ A ) using the relation: τ A = V Q A I t = ( 583.33 N ) ( 3.1875 × 10 − 6 m 3 ) ( 163 , 671.875 mm 4 ) ( 0.0075 m ) = ( 583.33 N ) ( 3.1875 × 10 − 6 m 3 × 10 9 mm 3 1 m 3 ) ( 163 , 671

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 9.3, Problem 9.31P

To determine

The principal stress at point A on the cross section of the arm at section a-a.

The orientation of the state of stress.

To sketch:

The results on an element at point A.

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Chapter 9 Solutions

Mechanics of Materials

Ch. 9.3 - In each case, the state of stress x, y, xy...Ch. 9.3 - Given the state of stress shown on the element,...Ch. 9.3 - Determine the normal stress and shear stress...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Also, find the corresponding orientation of the...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Determine the maximum principal stress at point B.Ch. 9.3 - Determine the principal stress at point C.Ch. 9.3 - Prove that the sum of the normal stresses x + y =...Ch. 9.3 - 9-2. The state of stress at a point in a member is...

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The principal stress at point A on the cross section of the arm at section a - a . The orientation of the state of stress. To sketch: The results on an element at point A .

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The principal stress at point A on the cross section of the arm at section a-a. The orientation of the state of stress. To sketch: The results on an element at point A.

Chapter 9 Solutions

The principal stress at point A on the cross section of the arm at section a-a.

The orientation of the state of stress.

To sketch:

The results on an element at point A.