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) (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

View Answer

Explanation: 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_{f}n_{e}k_{h}F_{o}

b) μ_{f}n_{e}k_{h}F_{o}γ_{mf}

c) μ_{f}n_{e}k_{ho}γ_{mf}

d) μ_{f}n_{e}k_{h}F_{o}/γ_{mf}

View Answer

Explanation: Shear Capacity of HSFG bolts is μ

_{f}n

_{e}k

_{h}F

_{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

_{nb}f

_{0}, where A

_{nb}= net area of bolt, f

_{0}= 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

View Answer

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

View Answer

Explanation: Bolts will be in single shear. Diameter of bolt = 16mm. Net area = 0.78x(π/4)x16

^{2}=156.83mm

^{2}.

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

View Answer

Explanation: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm

e=1.5×22=33mm, p=2.5×20=50mm

e/3d

_{0}= 33/(3×22) = 0.5, p/3d

_{0}-0.25 = 50/(3×22) -0.25=0.5, f

_{ub}/f

_{b}= 400/410=0.975

k

_{b}= minimum of (e/3d

_{0}, p/3d

_{0}-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

View Answer

Explanation: diameter of bolt = 16mm, diameter of hole =16+2 =18mm

e=1.5×18=27mm, p=2.5×16=40mm

e/3d

_{0}= 27/(3×18) = 0.5, p/3d

_{0}-0.25 = 40/(3×18) -0.25=0.49, f

_{ub}/f

_{b}= 400/410=0.975

k

_{b}= minimum of (e/3d

_{0}, p/3d

_{0}-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

View Answer

Explanation: F

_{o}=0.7f

_{ub}A

_{nb}=0.7x800x0.78x(π/4)x20

^{2}x10

^{-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%

View Answer

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

View Answer

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

View Answer

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.

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