To determine The principal stresses at point A . The principal stresses at point B . To sketch: The results on an element at point A and point B . Explanation Given Information: Tensile force in the bolt is 40 kN. Calculation: For Point A : Sketch the free body diagram of the beam as shown in Figure 1. Sketch the internal forces and moment at point A and point B as shown in Figure 2. Refer to Figure 2. The value of internal vertical force ( V ) is 24.0 kN and the internal moment ( M ) is 2.40 kN ⋅ m . Find the moment of inertia ( I ) of the section: I = 1 12 ( 30 mm ) ( 50 mm ) 3 = 312 , 500 mm 4 Sketch the centroid distance of point A and point B as shown in Figure 3. Find the first moment of area at point A ( Q A ) using the relation: Q A = y ¯ ' A ' (1) Here, y ¯ ' is the centroid distance to point A and A ' is the area from centroid to point A . Refer Figure 3. The value of y ¯ ' is 0. Therefore Q A = 0 . Find the normal stress at point A using the relation: σ A = − M y I (2) Refer the given Figure. The value of y is 25 mm . Apply sign conventions for the internal forces. Substitute 2.40 kN ⋅ m for M , 25 mm for y , and 312 , 500 mm 4 for I in Equation (2). σ A = − ( 2.40 kN ⋅ m ) × ( 25 mm ) 312 , 500 mm 4 = − ( 2.40 kN ⋅ m × 10 3 N 1 kN × 10 3 mm 1 m ) × ( 25 mm ) 312 , 500 mm 4 = − 192 N mm 2 = − 192 MPa Find the shear stress at point A : τ A = V Q I t (3) The value of Q at point A is 0. Therefore, the value of τ A = 0 . The value of normal stresses are σ x = 0 and σ y = − 192 MPa . The value of shear stress is τ x y = 0 . Find the in-plane principal stresses ( σ 1 ) and ( σ 2 ) using the relation: σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (4) The principal stress σ 1 = σ x = 0 . The principal stress σ 2 = σ y = − 192 MPa . Therefore, the principal stresses at point A ( σ 1 ) and ( σ 2 ) are 0 _ and − 192 MPa _ . For Point B : Find the area A ' using Figure 3. A ' = 25 mm × 30 mm = 750 mm 2 The value of y ¯ ' is 12.5 mm. Substitute 12.5 mm for y ¯ ' and 750 mm 2 for A ' in Equation (1). Q B = 12.5 mm × 750 mm 2 = 9 , 375 mm 3 Find the normal stress at point B using the relation: σ B = − M y I (5) Refer Figure 3. The value of y is 0 . Therefore, the value of σ B is 0. Find the shear stress at point B : τ B = V Q I t (6) Substitute 24 kN for V , 9 , 375 mm 3 for Q , 312 , 500 mm 4 for I , and 30 mm for t in Equation (6). τ B = ( 24 kN ) ( 9 , 375 mm 3 ) ( 312 , 500 mm 4 ) ( 30 mm ) = ( 24 kN × 10 3 N 1 kN ) ( 9 , 375 mm 3 ) ( 312 , 500 mm 4 ) ( 30 mm ) = 24 N mm 2 = 24 MPa The value of normal stresses are σ x = 0 and σ y = 0

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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Chapter 9.3, Problem 9.33P

To determine

The principal stresses at point A.

The principal stresses at point B.

To sketch:

The results on an element at point A and point B.

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Chapter 9.3, Problem 9.32P

Chapter 9.3, Problem 9.34P

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The principal stresses at point A . The principal stresses at point B . To sketch: The results on an element at point A and point B .

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The principal stresses at point A. The principal stresses at point B. To sketch: The results on an element at point A and point B.

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Chapter 9 Solutions

The principal stresses at point A.

The principal stresses at point B.

To sketch:

The results on an element at point A and point B.