Question

A single unequal angle 100 x 75 x 10 of Fe410 grade of steel is connected to a 10mm thick gusset plate at the ends with six 16mm diameter bolts with pitch of 40mm to transfer tension. Find the tensile strength due to net section rupture if gusset is connected to 100mm leg.

(a) 526.83 kN

(b) 385.74 kN

(c) 450.98 kN

(d) 416.62 kN

I have been asked this question during an interview.

Question is from Design of Tension Members topic in section Design of Tension Members of Design of Steel Structures

Answer 1

The correct option is (d) 416.62 kN

The best explanation: dh=18 mm, fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25

Anc = (100 – 10/2 – 18) x 10 = 770 mm^2, Ag0 = (75 – 10/2) x 10 = 700 mm^2

β = 1.4 – 0.076(w/t)(fy/fu)(bs/Lc)

   = 1.4 – 0.076 [(75-5)/10] [250/410] [{(75-5)+(100-40)}/{40×5}]

  = 1.19 > 0.7 and < 1.44[(410/250)(1.1/1.25)] (=2.07)

Tdn = 0.9fuAnc /γm1 + βAg0fy /γm0

     = [0.9x410x770/1.25 + 1.19x700x250/1.1] x 10-3 = 416.62 kN.