Figure 1 uestion 6 fr 6 ft Question (2) (20%) Figure 2 below shows a front view

Question

Figure 1 uestion 6 fr 6 ft Question (2) (20%) Figure 2 below shows a front view of a lap connection between a group of 1 steel A36 steel. The connection transfer a load of P.-520 kips and it is a bearing-type connection 14" wide plates made of with A325X bolts of 0.75"-diameter each. The bolts are to be placed in two transverse lines (each has same number of bolts) as shown in the figure below. Assuming that the plates are safe a) Determine the number of bolts required to have a safe bearing-type connection. b) If the tensile dead load acting on your designed connection is Po = 200 kips, what is the maximum tensile live load Pz that can be carried by the connection. 2 Plates 14x5/8 3 Plates 14'x1/2 Pu 1/2" Figure 2 Question2 20 35 175 Page 1 of

Explanation / Answer

a )

To calculate the bearing capacity of single bolt:

t = min (2*5/8, 3*1/2) = 1.25"

Lc = 1.75" (From fig)

db = 0.75"

Fu = 36 ksi

Resistance Factor, = 0.75

Rn = min (1.5 x Lc x t x Fu, 3 x db x t x Fu) = min(1.5x1.75x1.25x36, 3x0.75x1.25x36) = 101.25 kips

Rn = 75.94kips

The no. of bolts required = Pu / Rn = 520 / 75.94 = 6.85

Since even no. of bolts are provided, No. of bolt = 8

b)

Capacity of 8 bolts = 8*75.94 = 607.5kips

Dead Load, Pd = 200kips

Therefore Live Load, Pl = 607.5-200 = 407.5kips

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