This set of Design of Steel Structures Interview Questions and Answers for freshers focuses on “Bolted Connections – III”.
1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1
b) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≥ 1
c) (Vsb/Vdb) + (Tsb/Tdb) ≤ 1
d) (Vsb/Vdb) + (Tsb/Tdb) ≥ 1
Explanation: Bolt required to satisfy both shear and tension at the same time should satisfy (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1 , where Vsb= factored shear force, Vdb = design shear capacity, Tsb = factored tensile force, Tdb= design tensile capacity.
2. Shear Capacity of HSFG bolts is
Explanation: Shear Capacity of HSFG bolts is μfnekhFo/γmf, where μf = coefficient of friction(0.55), ne = number of frictional interfaces offering frictional resistance to slip, kh = 1 for fasteners in clearance holes, 0.85 for fasteners in over sized and short slotted holes, γmf = 1.1 (slip resistance designed at service load), 1.25 (slip resistance designed at ultimate load), Fo = minimum bolt tension = Anbf0 , where Anb = net area of bolt, f0 = 0.7fub , fub = 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:
Explanation: 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
Explanation: 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?
Explanation: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm
e/3d0 = 33/(3×22) = 0.5, p/3d0 -0.25 = 50/(3×22) -0.25=0.5, fub /fb = 400/410=0.975
kb = minimum of (e/3d0 , p/3d0 -0.25, fub /fb, 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
Explanation: diameter of bolt = 16mm, diameter of hole =16+2 =18mm
e/3d0 = 27/(3×18) = 0.5, p/3d0 -0.25 = 40/(3×18) -0.25=0.49, fub /fb = 400/410=0.975
kb = minimum of (e/3d0, p/3d0 -0.25, fub /fb,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.
Explanation: Fo=0.7fubAnb=0.7x800x0.78x(π/4)x202x10-3=137.22 kN
Assume μf=0.5, ne=1, kh=1
Slip resistance of bolt = μf ne kh Fo/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?
Explanation: 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
Explanation: 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
Explanation: 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.
Sanfoundry Global Education & Learning Series – Design of Steel Structures.
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