250+ TOP MCQs on Bolted Connections – III and Answers

Design of Steel Structures Interview Questions and Answers for freshers on “Bolted Connections – III”.

1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1
b) (V_sb/V_db)2 + (T_sb/T_db)2 ≥ 1
c) (V_sb/V_db) + (T_sb/T_db) ≤ 1
d) (V_sb/V_db) + (T_sb/T_db) ≥ 1
Answer: a
Clarification: Bolt required to satisfy both shear and tension at the same time should satisfy (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1 , where V_sb= factored shear force, V_db = design shear capacity, T_sb = factored tensile force, T_db= design tensile capacity.

2. Shear Capacity of HSFG bolts is
a) μ_fn_ek_hF_o
b) μ_fn_ek_hF_oγ_mf
c) μ_fn_ek_h? /F_oγ_mf
d) μ_fn_ek_hF_o/γ_mf
Answer: d
Clarification: Shear Capacity of HSFG bolts is μ_fn_ek_hF_o/γ_mf, where μ_f = coefficient of friction(0.55), n_e = number of frictional interfaces offering frictional resistance to slip, k_h = 1 for fasteners in clearance holes, 0.85 for fasteners in over sized and short slotted holes, γ_mf = 1.1 (slip resistance designed at service load), 1.25 (slip resistance designed at ultimate load), F_o = minimum bolt tension = A_nbf₀ , where A_nb = net area of bolt, f₀ = 0.7fub , f_ub = ultimate tensile stress of bolt.

3. The maximum number of bolts of diameter 25mm that can be accomodated in one row in 200mm wide flat are:
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Minimum end distance = 2.5×25 = 62.5mm
Number of bolts that can be accommodated = (200-2×62.5)/25 = 3 bolts.

4. Calculate strength in shear of 16mm diameter of bolt of grade 4.6 for lap joint
a) 50 kN
b) 40 kN
c) 29 kN
d) 59 kN
Answer: c
Clarification: Bolts will be in single shear. Diameter of bolt = 16mm. Net area = 0.78x(π/4)x16²=156.83mm².
Strength of bolt in shear = Anbfub/(√3 x 1.25) = 156.83x400x10^-3/1.25x√3 = 28.97kN.

5. What is the value of kb in nominal bearing strength for a bolt of 20mm diameter of grade 4.6?
a) 0.5
b) 1
c) 0.97
d) 2
Answer: a
Clarification: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm
e=1.5×22=33mm, p=2.5×20=50mm
e/3d₀ = 33/(3×22) = 0.5, p/3d₀ -0.25 = 50/(3×22) -0.25=0.5, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀ , p/3d₀ -0.25, f_ub /f_b, 1) = 0.5.

6. Calculate bearing strength of 20mm diameter bolt of grade 4.6 for joining main plates of 10mm thick using cover plate of 8mm thick using single cover butt joint.
a) 70.26 kN
b) 109.82 kN
c) 50.18 kN
d) 29.56 kN
Answer: c
Clarification: diameter of bolt = 16mm, diameter of hole =16+2 =18mm
e=1.5×18=27mm, p=2.5×16=40mm
e/3d₀ = 27/(3×18) = 0.5, p/3d₀ -0.25 = 40/(3×18) -0.25=0.49, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀, p/3d₀ -0.25, f_ub /f_b,1) = 0.49
bearing strength = 2.5kbdtfu/1.25 = 2.5×0.49x16x8x400x10-3/1.25 = 50.18 kN.

7. Find the number of HSFG bolts of diameter 20mm, grade 88 for connection of member carrying factored tensile load of 200kN when no slip is permitted.
a) 5
b) 4
c) 3
d) 2
Answer: b
Clarification: F_o=0.7f_ubA_nb=0.7x800x0.78x(π/4)x20²x10^-3=137.22 kN
Assume μ_f=0.5, n_e=1, k_h=1
Slip resistance of bolt = μ_f n_e k_h F_o/1.25 = 0.5x1x1x137.22/1.25 =54.88 kN
Number of bolts required = 200/54.88 = 3.64 = 4(approximately).

8. What is the efficiency of joint when strength of bolt per pitch length is 60kN and strength of plate per pitch length is 150kN?
a) 25%
b) 30%
c) 35%
d) 40%
Answer: d
Clarification: Efficiency = (strength of bolt per pitch length/ strength of plate per pitch length)x100 = 60×100/150 = 40%.

9. Strength of bolt is
a) minimum of shear strength and bearing capacity of bolt
b) maximum of shear strength and bearing capacity of bolt
c) shear strength of bolt
d) bearing capacity of bolt
Answer: a
Clarification: Strength of bolt is minimum of shear strength and bearing capacity of bolt. Design shear strength = nominal shear capacity/1.25, Design bearing strength = nominal bearing capacity/1.25.

10. Prying forces are
a) friction forces
b) shear forced
c) tensile forces
d) bending forces
Answer: c
Clarification: In connections subjected to tensile stresses, the flexibility of connected parts can lead to deformations that increases tension applied to bolts. This additional tension is called prying force.