Explanation Given information: The compressive load is P. Calculation : Show the free body diagram of support as shown in Figure 1. Refer to Figure 1 Consider the width of the support denoted w . Then, w = a + x Calculate the area ( A ) as follows: A = b d (1) Here b is the width of the section and d is the depth of the section. Substitute a for b and ( a + x ) for w in Equation (1). A = a ( a + x ) Calculate the moment of inertia ( I ) as follows: I = b d 3 12 (2) Substitute a for b and ( a + x ) for w in Equation (2). I = a 12 ( a + x ) 3 Find the moment ( M ) as follows: M = F × d (3) Here, F is the force and d is the distance. Substitute 0.5 x for d and P for F in Equation M = 0.5 x × P = 0.5 P x Show the internal force and moment as shown in figure 2. Express to calculate the normal stress as follows: σ = N A ± M c I (3) Here, N is the normal force, A is the area of the cross section, I is the moment of inertia, c is the centroid distance, and M moment is the moment. Refer to Figure 2. Substitute − P for N, a ( a + x ) for A , a 12 ( a + x ) 3 for I , 0.5 P x for M , and ( 1 2 ( a + x ) ) for c in Equation (3). σ = − P a ( a + x ) ± 0.5 P x ( 1 2 ( a + x ) ) a 12 ( a + x ) 3 = P a [ − 1 a + x ± 3 x ( a + x ) 2 ] Write the expression of normal stress ( σ A ) at point A as follows: σ A = − P a [ 1 a + x + 3 x ( a + x ) 2 ] = − P a [ ( a + x ) + ( 3 x ) ( a + x ) 2 ] = − P a [ 4 x + a ( a + x ) 2 ] (4) Write the expression of normal stress ( σ B ) at point B as follows: σ B = P a [ − 1 a + x + 3 x ( a + x ) 2 ] = − P a [ − ( a + x ) + ( 3 x ) ( a + x ) 2 ] = − P a [ 2 x − a ( a + x ) 2 ] (5) Calculate the maximum normal stress A as follows: Differentiate with respect to x is equal to zero. d σ A d x = 0 d σ A d x = 0 d d x [ − P a [ 4 x + a ( a + x ) 2 ] ] = 0 − P a [ ( a + x ) 2 × d d x ( 4 x + a ) − ( 4 x + a ) d d x ( a + x ) 2 ( a + x ) 4 ] = 0 − P a [ ( a + x ) 2 ( 4 ) − ( 4 x + a ) ( 2 ) ( a + x ) ( 1 ) ( a + x ) 4 ] = 0 − P ( a + x ) a ( a + x ) 4 ( 4 ( a + x ) − 2 ( 4 x + a ) ) = 0 − P a ( a + x ) 3 ( 4 a + 4 x − 8 x − 2 a ) = 0 − P a ( a + x ) 3 ( 2 a − 4 x ) = 0 Since P a ( a + x ) 3 ≠ 0 To solve the above equation to get x value

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 8.2, Problem 8.46P

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The absolute maximum normal stress (σmax) in the material.

The absolute minimum normal stress (σmin) in the material.

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Chapter 8.2, Problem 8.45P

Chapter 8.2, Problem 8.47P

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The absolute maximum normal stress ( σ max ) in the material. The absolute minimum normal stress ( σ min ) in the material.

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The absolute maximum normal stress (σmax) in the material.

The absolute minimum normal stress (σmin) in the material.