# MCQs on Eccentric Loaded Bolted Joints

If core diameter of bolt is 13.8cm the it’s nominal diameter is given by?

a) 17.27mm

b) 15.34mm

c) 14.67mm

d) 16.34mm

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Answer: a

Explanation: D=d/0.8.

Refer to fig 1.Two plates are fastened by means of two bolts. The yield strength of bolt is 400N/mm² and factor of safety is 4. Determine the permissible shear stress in the bolts.

a) 100N/mm²

b) 50N/mm²

c) 25N/mm²

d) 75N/mm²

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Answer: b

Explanation: Permissible hear stress=0.5 x 400 /4 =50N/mm².

Refer to fig 1. Two plates are fastened by means of two bolts. The yield strength of bolt is 400N/mm² and factor of safety is 4. Determine the size of the bolts.

a) 8mm

b) 9mm

c) 10mm

d) 11mm

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Answer: a

Explanation: Permissible hear stress=0.5 x 400 /4 =50N/mm². P=2 (π xd²/4) x τ or 5000=2 x π x d² x 50/4 or dd=7.97mm.=7.9

The structure shown is subjected to an eccentric force P=5kN and eccentricity=500mm. The horizontal distance between two bolts is 200mm and vertical distance between bolts is 150mm. The yield strength of bolts is 400N/mm² and factor of safety is 3.

Referring to fig 2, Determine the primary shear force.

a) 625N

b) 1250N

c) 2500N

d) 1000N

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Answer: b

Explanation: Primary shear force P₁=P₂=P₃=P₄=P/4 =1250N.

Referring to fig 3, Determine the secondary shear force.

a) 2000N

b) 2500N

c) 1500N

d) 5000N

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Answer: d

Explanation: Secondary shear force=Moment about CG x distance from CG/sum of squares of distance of bolts from CG. F= (Pe)xr₁/(r₁²+r₂²+r₃²+r₄²). Here r₁=r₂=r₃=r₄=125mm hence F= 5000 x 500/(4×125) or F=5000N.

Referring to figure 2, determine the resultant shear force on the bolt lying left and above the CG.

a) 4068.58N

b) 4168.58N

c) 5068.58N

d) 5168.65N

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Answer: a

Explanation: After breaking shear force into primary and secondary shear force, primary acts along the vertical positively and secondary acts at an angle of 180-36.87’ ACW from the vertical. Hence net shear force= √ [5000cos36.8-1250]²+ [5000sin36.8]² =4068.58N.

In figure 2, determine the resultant shear force on the bolt lying right and above the CG.

a) 654334N b) 6047.44N c) 5047.44N d) 5989.32N

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Answer: b

Explanation: After breaking net shear force into primary and secondary shear force, primary acts along the vertical positively and secondary acts at an angle of 36.8’CW from the vertical. Hence net shear force = √ [5000cos36.8+1250]²+ [5000sin36.8]².

In figure 2, determine the size of the bolts.

a) 10.74mm b) 9.23mm c) 11.54mm d) 8.68mm

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Answer the questions 9-14 while considering figure 3

Which bolt is under maximum shear stress?

a) 1

b) 2

c) 3

d) All are under equivalent shear stress

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Answer: c

Explanation: Primary shear force acts equally on the three bolts in the vertically upward direction while the moment is CW along CG so its effect on bolts will be ACW. Hence secondary shear force acts vertically upward on bolt 3 and vertically downward on bolt 1.

Arrange the bolts in order of decreasing shear stresses.

a) 1>2>3

b) 2>1>3

c) 3>1>2

d) 3>2>1

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Answer: d

Explanation: On bolt 3,primary and secondary shear stress act in same direction, on bolt 2 there is no secondary shear stress and on bolt 1 the two act in opposite direction.

Determine the primary shear stress to which the bolts are subjected if P=3kN.

a) 3000N

b) 1000N

c) 2000N

d) None of the listed

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Answer: b

Explanation: Primary shear force=P/3.

Determine the secondary shear stress acting on the bolt 3 and its direction. The bolts are equidistant having separated by 60mm and the margin to the left and right is 25mm.Also P=5kN acts at a distance of 200mm from the channel.

a) 6100N vertically up

b) 4500N vertically down

c) 6100N vertically up

d) 4500N vertically down

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Answer: c

Explanation: Moment about CG=3000 x (75+30+200). On bolt 3 secondary shear force will be M x r₁/ (r₁²+r₂²) and will act in a direction perpendicular the line joining CG and bolt 3. As moment is CW about CG, o bolts its effect will be ACW.

Determine the size of the bolts if yield strength of bolt is 400N/mm² and factor of safety is 4. The bolts are equidistant having separated by 60mm and the margin to the left and right is 25mm. Also P=5kN acts at a distance of 200mm from the channel.

a) 14.34mm

b) 13.44mm

c) 15.44mm

d) 12.66mm

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Answer: b

Explanation: Clearly bolt 3 is under maximum shear stress. Net shear stress= Primary shear stress + Secondary shear stress or τ= (1000+6100) N or 0.5 x 400//4=7100 x 4/πd².