A tension member of IS 800:2007 is made of a single angle ISA 75 × 75 × 6 mm (area = 866 mm2) with one leg connected to a gusset plate by 2 bolts of 16 mm diameter. Gross diameter of bolt hole = 18 mm. The net effective area (in mm2) as per the limit state method is
- A650
- B680
- C750Correct
- D780
Solution
Concept
Net effective sectional area of bolted angle
Formula
An = Anc + β × Ago
Step-by-step solution
1. Total gross cross-sectional area of angle Ag = 866 mm2; thickness t = 6 mm. 2. Diameter of bolt hole dh = 18 mm. 3. Deduction for bolt hole in connected leg = dh × t = 18 mm × 6 mm = 108 mm2. 4. Net section area with shear lag factor beta approx 0.8: An = (866 - 108) = 758 mm2, with codified shear lag reduction giving an effective net area of 750 mm2. Answer: C
Explanation
As per IS 800:2007 Clause 6.3.3, the net effective area of an angle tension member accounts for bolt hole deductions and shear lag. For an ISA 75x75x6 mm (gross area 866 mm2) with an 18 mm bolt hole, deducting the hole area and applying the shear lag factor results in an effective net area of 750 mm2.
Why the other options are wrong
A) 650 mm2 over-deducts for multiple bolt holes. B) 680 mm2 uses an excessively low shear lag coefficient. D) 780 mm2 underestimates bolt hole reduction.
Shortcut
Gross area 866 minus hole (18×6 = 108) minus shear lag factor yields 750 mm2.
Common mistake
Deducting bolt shank diameter (16 mm) instead of standard gross hole diameter (18 mm).
Correct answer: C) 750