What is the approximate value of the tensile strength governed by yielding of the cross section of a roof truss diagonal 100 × 75 × 6 mm (Ag = 1010 mm², fy = 250 MPa) connected to the gusset plate by 4-mm welds (140 mm long at top and 310 mm long at bottom) as shown in figure? (Take partial safety factor as 1.10)
- A230 kNCorrect
- B23 kN
- C320 kN
- D32 kN
Solution
Concept
Design strength from gross section yielding is Tdg = Ag fy / γm0, as per IS 800:2007.
Formula
Tdg = Ag × fy / γm0
Step-by-step solution
1. Given Ag = 1010 mm², fy = 250 MPa, γm0 = 1.10. 2. Tdg = 1010 × 250 / 1.10 3. Tdg = 252500 / 1.10 = 229545 N 4. Tdg ≈ 229.5 kN ≈ 230 kN
Explanation
Yielding of the gross section uses only the gross area, the yield stress and γm0. The weld sizes and lengths do not affect this check. They are needed only for the weld design and block shear checks.
Why the other options are wrong
B: 23 kN is 10 times too small (decimal error). C: 320 kN has the digits mixed up and is not correct. D: 32 kN is a wrong value and also has a unit/decimal error.
Shortcut
1010 × 250 / 1.1 ≈ 1010 × 227 ≈ 230 kN
Common mistake
Correct answer: A) 230 kN
