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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Test: Eccentric Loaded Bolted Joints - Mechanical Engineering MCQ

Test: Eccentric Loaded Bolted Joints - Mechanical Engineering MCQ


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10 Questions MCQ Test - Test: Eccentric Loaded Bolted Joints

Test: Eccentric Loaded Bolted Joints for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Eccentric Loaded Bolted Joints questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Eccentric Loaded Bolted Joints MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Eccentric Loaded Bolted Joints below.
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Test: Eccentric Loaded Bolted Joints - Question 1
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 If core diameter of bolt is 13.8cm the it’s nominal diameter is given by?

Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 1

Explanation: D=d/0.8.

Test: Eccentric Loaded Bolted Joints - Question 2
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Refer to fig 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.

Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 2

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

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Test: Eccentric Loaded Bolted Joints - Question 3
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Refer to fig 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.

Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 3

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 d=7.97mm.

Test: Eccentric Loaded Bolted Joints - Question 4
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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 , Determine the primary shear force.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 4

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

Test: Eccentric Loaded Bolted Joints - Question 5
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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 , Determine the secondary shear force.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 5

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.

Test: Eccentric Loaded Bolted Joints - Question 6
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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 figure , determine the resultant shear force on the bolt lying left and above the CG.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 6

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.

Test: Eccentric Loaded Bolted Joints - Question 7
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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.

In figure , determine the resultant shear force on the bolt lying right and above the CG.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 7

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.

Test: Eccentric Loaded Bolted Joints - Question 8
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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.

In figure , determine the size of the bolts.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 8

Explanation: The maximum shear force to which any bolt is subjected is 6047.44N. Hence 0.5 x 400/3= 4 x 6047.44/πd² or d=10.74mm.

Test: Eccentric Loaded Bolted Joints - Question 9
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Which bolt is under maximum shear stress?
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 9

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.

Test: Eccentric Loaded Bolted Joints - Question 10
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Arrange the bolts in order of decreasing shear stresses.
Detailed Solution for Test: Eccentric Loaded Bolted Joints - Question 10

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.

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