To determine The maximum normal stress ( σ max ) of the column. Explanation Given information: The maximum load P is 60 kN . Calculation: Show the free body diagram of equivalent force system as shown in figure 1. Here M z , M y is moment acting about z and y direction and F is force acting downward direction. Refer to Figure 1: Apply Equations of equilibrium for horizontal forces. ∑ F x = 0 − 60 − 120 = − F − 180 = − F F = 180 kN Apply Equations of equilibrium for vertical forces. ∑ M y = ( M R ) y − 60 × 0.075 = − M y − 4.5 = − M y M y = 4.5 kN ⋅ m Take moment about z direction. ∑ M z = ( M R ) z − 120 × 0.25 = − M z − 30 = − M z M z = 30 kN ⋅ m Show the I section as shown in Figure 2. Refer Figure 3, Consider the size of big rectangle as 200 mm × 300 mm . and the size of two small rectangles as 185 mm × 270 mm . The moment of inertia of I- section is the subtraction of moment of inertia of big rectangle of 200 mm × 300 mm . and two small rectangles of 185 mm × 270 mm Determine the cross sectional area ( A ) as follows: A = b 1 d 1 − 2 ( b 2 d 2 ) (1) Here b 1 d 1 area of big rectangle and b 2 d 2 area of small rectangle. Substitute 200 mm for b 1 , 300 mm for d 1 , 92.5 mm for b 2 , and 270 mm for d 2 in Equation (1). A = 200 × 300 − 2 ( 92.5 × 270 ) = 60 , 000 − 49 , 950 = 10 , 050 mm 2 ( 1 m 2 10 6 mm 2 ) = 0.01005 m 2 Determine the moment of inertia ( I ) about z axis as shown in below: I z = b 1 d 1 3 12 − 2 ( b 2 d 2 3 ) 12 (2) Substitute 200 mm for b 1 , 300 mm for d 1 , 92.5 mm for b 2 , and 270 mm for d 2 in Equation (2). I z = 200 × 300 3 12 − 2 ( 92.5 × 270 3 ) 12 = 450 , 000 , 000 − 303 , 446 , 250 = 145 , 553 , 750 mm 4 ( 1 mm 4 10 12 m 4 ) = 0

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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 8.59P

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The maximum normal stress (σmax) of the column.

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