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What is the value of kb in nominal bearing strength for a bolt of 20mm diameter of grade 4.6?

a) 0.5
b) 1
c) 0.97
d) 2

Answer: a
Explanation: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm
e=1.5×22=33mm, p=2.5×20=50mm
e/3d0 = 33/(3×22) = 0.5, p/3d0 -0.25 = 50/(3×22) -0.25=0.5, fub /fb = 400/410=0.975
kb = minimum of (e/3d0 , p/3d0 -0.25, fub /fb, 1) = 0.5.

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