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The structure shown is subjected to an eccentric force P=5kN and eccentricity=500mm. The horizontal distance between two bolts is 200mm and vertical distance between bolts is 150mm. The yield strength of bolts is 400N/mm² and factor of safety is 3. Determine the secondary shear force in the below figure.

The structure shown is subjected to an eccentric force P=5kN and eccentricity=500mm. The horizontal distance between two bolts is 200mm and vertical distance between bolts is 150mm. The yield strength of bolts is 400N/mm² and factor of safety is 3. Determine the secondary shear force in the below figure. Correct Answer 5000N

Secondary shear force=Moment about CG x distance from CG/sum of squares of distance of bolts from CG. F= (Pe)xr₁/(r₁²+r₂²+r₃²+r₄²). Here r₁=r₂=r₃=r₄=125mm hence F= 5000 x 500/(4×125) or F=5000N.

Related Questions

The structure shown is subjected to an eccentric force P=5kN and eccentricity=500mm. The horizontal distance between two bolts is 200mm and vertical distance between bolts is 150mm. The yield strength of bolts is 400N/mm² and factor of safety is 3. Determine the primary shear force from the following figure.
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