(a) Explanation Calculate the normal stress ( σ A ) acting at point A using the relation: σ A = P A + M x y I x + M y x I y (1) Here, P is the axial load, A is the area, M x is the moment in the x -direction, y is the distance in y -direction from centroid to point A , and I x is the moment of inertia in the x -direction, M y is the moment in the y-direction, x is the distance in x -direction from centroid to point A , and I y is the moment of inertia in the y -direction. Sketch the forces and moments at point A as shown in Figure 1. Apply Equilibrium equations to find the value of forces and moment at point A . Sum of forces in x -direction at point A is equal to 0. Σ F x = 0 V x = 400 lb Sum of forces in y -direction at point A is equal to 0. Σ F y = 0 V y = 500 lb Sum of forces in z -direction at point A is equal to 0. Σ F z = 0 N z = 900 lb Sum of moments in y -direction is equal to 0. Σ M y = 0 M y = 400 × 9 in . M y = 3 , 600 lb ⋅ in . Sum of moments in x -direction is equal to 0. Σ M x = 0 M x = 400 × 9 in . M x = 4 , 500 lb ⋅ in . Find the moment of inertia of the section ( I x = I y ) using the given cross section: I x = I y = 1 12 ( 3 in . ) ( 3 in . ) 3 = 6.75 in 4 Find the area of the section ( A ) : A = ( 3 in . ) ( 3 in . ) = 9 in 2 Find the first moment of area at point A ( Q A ) x in the x -direction using the relation: ( Q A ) x = y ¯ ' A ' Here, y ¯ ' is the centroid distance to point A and A ' is the area from centroid to point A . Sketch the section and the centroid distance as shown in Figure 2. Refer Figure 2 for values of y ¯ ' . Q A = 1 in ( 1 in ) ( 3 in ) = 3 in 3 Substitute − 900 lb for P , 9 in 2 for A , 4 , 500 lb ⋅ in for M x , 1.5 in . for y , 6.75 in 4 for I x , − 3 , 600 lb ⋅ in for M y , 0.5 in . for x , 6.75 in 4 for I y in Equation (1). σ A = − 900 lb 9 in 2 + ( 4 , 500 lb ⋅ in ) ( 1.5 in . ) 6.75 in 4 − ( 3 , 600 lb ⋅ in ) ( 0.5 in . ) 6.75 in 4 = 633.33 psi Find the shear stress ( τ A ) at point A using the relation: τ z x = V Q I t Here, t is the thickness (b) To determine Find the maximum in-plane shear stress developed at point A .

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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The term stress indicates the measure of internal forces that develop due to an external force's application on an object. When a specific external force is applied to an object, internal stress generates in the material. The value of stress can obtain b…

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Chapter 14.4, Problem 75P

(a)

To determine

Find the principal stresses at point A.

(b)

To determine

Find the maximum in-plane shear stress developed at point A.

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Chapter 14.4, Problem 74P

Chapter 14.5, Problem 76P

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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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Find the principal stresses at point A .

Solution Summary: The author explains how to find the principal stresses at point A and calculate the normal stress using the relation: (sigma_A).