Explanation Given information: The load at point C is F C = 150 kN . Load intensity at B D is w = 20 kN/m . The allowable bearing stress for the plates is σ b a l l o w = 1.8 MPa . Free body diagram of beam: For the beam to be in equilibrium, ∑ F V = 0 ∑ M A = 0 Bearing stress in plate is given by using the formula of bearing stress. σ C = F b e a r i n g A b a s e p l a t e Base plate area is given by: A b a s e p l a t e = a 2 b a s e p l a t e Calculation: For the vertical equilibrium of beam, ∑ F V = 0 R A − w × B D − F C + R B = 0 R A + R B = w × A D + F C R A + R B = 20 kN/m × 2 m + 150 kN R A + R B = 190 kN ............ 1 For the moment equilibrium about A , ∑ M A = 0 w × B D × A B + B D 2 + F C × A C − R B × A B = 0 20 kN/m × 2 m × 4 + 2 2 m + 150 kN × 2 m − R B × 4 m = 0 200 kN-m + 300 kN-m = R B × 4 m R B = 500 kN-m 4 m R B = 125 kN .............. 2 From (1) and (2), R A + R B = 190 kN R A + 125 kN = 190 kN R A = 65 kN For the base plate at A ' : σ C = F b e a r i n g A b a s e p l a t e 1

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ISBN: 9780137605521

Author: HIBBELER

Publisher: RENT PEARS

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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 1.7, Problem 83P

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The size of square bearing plates A' and B' .

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Chapter 1.7, Problem 82P

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The size of square bearing plates A ' and B ' .

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