250+ TOP MCQs on Design of Welds and Answers

Design of Steel Structures Multiple Choice Questions on “Design of Welds”.

1. The design nominal strength of fillet weld is given by ____________
a) f_u
b) √3 f_u
c) f_u/√3
d) f_u/(1.25 x √3)
Answer: c
Clarification: Design nominal strength of fillet weld = f_u/√3, where f_u is smaller of ultimate stress of weld or parent metal.

2. When welds are subjected to compressive or tensile or shear force alone, the stress in weld is given by :
a) P/t_tl_w
b) Pt_t/l_w
c) Pt_tl_w
d) Pl_w/t_t
Answer: a
Clarification: When welds are subjected to compressive or tensile or shear force alone, the stress in weld is given by q = P/t_tl_w , where q=shear stress in N/mm², P = force transmitted, t_t = effective throat thickness of weld in mm, l_w = effective length of weld in mm.

3. When fillet welds are subjected to combination of normal and shear stress, the equivalent stress is given by :
a) √(f_a²+q²)
b) √(f_a²+2q²)
c) √(3f_a²+q²)
d) √(f_a²+3q²)
Answer: d
Clarification: When fillet welds are subjected to combination of normal and shear stress, the equivalent stress is given by f_e = √(f_a2+3q²), where f_a = normal stresses, compression or tension, due to axial force or bending moment, q = shear stress due to shear force or tension.

4. Two plates of 12mm and 16mm thickness are to be joined by groove weld. The joint is subjected to factored tensile load of 400kN. Due to some reasons the effective length of weld that could be provided was 150mm only. What is the safety of joint if single-V groove weld is provided?
a) Safe
b) Unsafe
c) Unsafe, but adequate
d) Safe, but adequate
Answer: b
Clarification: L_w = 150mm, Throat thickness, t_e = 5×12/8 = 7.5 mm
Strength of weld = L_wt_ef_y/1.25 = 150×7.5x250x10^-3/1.25 = 225kN < 400kN. Therefore joint is unsafe and inadequate.

5. What is the effective throat thickness dimension of 10mm fillet weld made by shielded metal arc welding and submerged arc welding?
a) 4.6mm, 5mm
b) 5mm, 4.6mm
c) 8.6mm, 7mm
d) 7mm, 8.6mm
Answer: d
Clarification: a) Using shielded metal arc welding process,
effective throat thickness = 0.7a = 0.7×10 = 7mm
b) Using submerged arc welding (we get better penetration than shielded metal arc welding):
a = 10+2.4 = 12.4mm
effective throat thickness = 0.7a = 0.7×12.4 = 8.6mm.

6. What is the strength of weld per mm length used to connect two plates of 10mm thickness using a lap joint?
a) 795.36 N/mm
b) 295.5N/mm
c) 552.6 N/mm
d) 487.93 N/mm
Answer: a
Clarification: Minimum size of weld = 3mm
Maximum size of weld = 10-1.5 = 8.5mm
Assume weld size = 6mm
Effective throat thickness, t_e = 0.7 x 6 = 4.2mm
Strength of weld = t_e [f_u/(√3 x 1.25)] = 410 x 4.2 /(√3 x 1.25) = 795.36 N/mm.

7. What is the overall length of fillet weld to be provided for lap joint to transmit a factored load of 100kN? Assume site welds and width and thickness of plate as 75mm and 8mm respectively, Fe410 steel.
a) 500mm
b) 382mm
c) 201mm
d) 468mm
Answer: c
Clarification: Minimum size of weld = 3mm
Maximum size of weld = 8-1.5 = 6.5mm
Assume size of weld = 5mm
Effective throat thickness = 0.7 x 5 = 3.5mm
Strength of weld = 3.5×410/(√3 x1.5) = 552.33 N/mm
Required length of weld = 100 x 10³/552.33 = 181.05 mm
length to be provided on each side = 181/2 = 90.5mm
End return = 2×5 = 10mm
Overall length = 2 x (90.5 + 2 x 5) = 201mm.

8. The clear spacing between effective lengths of intermittent welds should not be ______
a) less than 16t in case of tension joint, where t is thickness of thinner plate
b) less than 12t in case of compression joint, where t is thickness of thinner plate
c) less than 20t in case of tension joint, where t is thickness of thinner plate
d) less than 20t in case of compression joint, where t is thickness of thinner plate
Answer: a
Clarification: The clear spacing between effective lengths of intermittent welds should not be less than 16 times and 12 times the thickness of thinner plate jointed in case of tension joint and compression joint respectively, and should never be more than 200mm.