Problem 2 The tension member shown below consists of a pair

Problem 2. The tension member shown below consists of a pair of L8x6x1/2 angles with the long legs bolted to a gusset plate as shown. Determine if the design strength of the tension member (neglecting block shear) is sufficient for a P, of 298 kips. The angle is made of A36 steel and the bolts are 7/8 in. in diameter. (20 points) s-2.75 in., Le-1.5 in., g13 in, g-3 in Section 1-1

Solution

steel type = A36 STEEL

assume Ae = 0.75 An

Pny = Fy*Ag = 36*(2(2.75*3))*(1/2) =36*8.25 =297kip

Pn = Fu*Ae

An = Ag-Ah = 8.25-((7/8+1/8)(1/2)(2)) =7.25in2

Ae = 0.75An =0.75*7.25 =5.4375

Pnf = Fu*Ae = 58*5.4375 =315.375kips

a36 steel Fy=36ksi Fu=58 ksi

(DS)y = t*Pny = 0.9*297 = 267.3 kips

(DS)f = t*Pnf = 0.75*315.375 =236.531 kips

(DS)final = 2*236.531 = 473.0625 kips for double angle

ASD:

Pny/ty = 0.6*Fy*Ag =0.6*36*8.25 =178.2 kips

Pnf/tf = 0.5*Fu*Ae = 0.5*58*5.4375 = 157.6875 kips

dobule angle = 2*157.6875 =315.375