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?
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.