To determine the safe number of bolts and plate thickness, we need to calculate the required area of steel and the available area of steel.
Required Area of Steel:
Total Load = 200 kips
Shear Force per bolt = (200 kips) / (2 bolts) = 100 kips
Nominal shear strength of one bolt = 0.50FuAb
where Fu = 65 ksi and Ab = (π/4)*(6/8)^2 = 0.265 in^2 (for 6/8 in bolts)
Nominal shear strength per bolt = 0.50(65 ksi)(0.265 in^2) = 8.47 kips
Required number of bolts = Shear force / Nominal shear strength per bolt
= 100 kips / 8.47 kips = 12 bolts
Available Area of Steel:
Thickness of outer plates = 6 in
Thickness of center plate = 11 in
Diameter of bolt hole = 1.125 in (for 6/8 in bolts)
Net area of one A992 plate = Ag - An
where Ag = Thickness * Width = 6 in * 12.25 in = 73.5 in^2
and An = Number of bolt holes * Area of one bolt hole
= 4 * (π/4)*(1.125 in)^2 = 3.98 in^2
Net area of one A992 plate = 73.5 in^2 - 3.98 in^2 = 69.52 in^2
Total available net area of steel = 2 * (69.52 in^2) + 69.52 in^2 = 208.56 in^2
Required area of steel = Number of bolts * Area of one bolt = 12 * 0.265 in^2 = 3.18 in^2
Safe Number of Bolts and Plate Thickness:
We can use the equation for bolt spacing to determine the required layout of bolts:
Bolt spacing = 1.13d = 1.13*(0.75 in) = 0.85 in (for 6/8 in bolts)
The distance between bolts is limited by the plate thickness, so we need to use the larger of the two values:
Bolt spacing = MAX(0.85 in, 6 in/12 - 1.125 in) = 0.725 in
Therefore, the safe layout of bolts is 12 bolts spaced 0.725 in apart. We can use two outer plates with a thickness of 0.5 in each, and a center plate with a thickness of 1 in. This gives a total available net area of steel of 210 in^2, which is greater than the required area of steel of 3.18 in^2.
Connection as in Fig. 11.6 carries a total load of 200 kips. Plates of A992 with Fy=50ksi, Fu=65 ksi are attached with 6/8 in A325 bolts, double shear, threaded. If the outer plates are 6 in, center plate 11 in, what is the safe # and layout of bolts and plate thickness?
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