Machine Design Questions and Answers – Eccentric Loaded Bolted Joints

«
»

This set of Machine Design Multiple Choice Questions & Answers (MCQs) focuses on “Eccentric Loaded Bolted Joints”.

1. 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
View Answer

Answer: a
Explanation: D=d/0.8.
advertisement

2. In the following figure, 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.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q2
a) 100N/mm²
b) 50N/mm²
c) 25N/mm²
d) 75N/mm²
View Answer

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

3. In the following figure, 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.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q2
a) 8mm
b) 9mm
c) 10mm
d) 11mm
View Answer

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 d=7.97mm.
advertisement
advertisement

4. 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. Determine the primary shear force from the following figure.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q4
a) 625N
b) 1250N
c) 2500N
d) 1000N
View Answer

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

5. 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. Determine the secondary shear force in the below figure.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q4
a) 2000N
b) 2500N
c) 1500N
d) 5000N
View Answer

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.
advertisement

6. 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. Determine the resultant shear force on the bolt lying left and above the CG.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q4
a) 4068.58N
b) 4168.58N
c) 5068.58N
d) 5168.65N
View Answer

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.

7. 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. Determine the resultant shear force on the bolt lying right and above the CG.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q4
a) 654334N
b) 6047.44N
c) 5047.44N
d) 5989.32N
View Answer

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.
advertisement

8. 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. Determine the size of the bolts.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q4
a) 10.74mm
b) 9.23mm
c) 11.54mm
d) 8.68mm
View Answer

Answer: a
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.

9. Which bolt is under maximum shear stress in the following figure?
machine-design-questions-answers-eccentric-loaded-bolted-joints-q9
a) 1
b) 2
c) 3
d) All are under equivalent shear stress
View Answer

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.
advertisement

10. Arrange the bolts in order of decreasing shear stresses.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q9
a) 1>2>3
b) 2>1>3
c) 3>1>2
d) 3>2>1
View Answer

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.

11. Determine the primary shear stress to which the bolts are subjected if P=3kN.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q9
a) 3000N
b) 1000N
c) 2000N
d) None of the listed
View Answer

Answer: b
Explanation: Primary shear force=P/3.

12. 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.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q9
a) 6100N vertically up
b) 4500N vertically down
c) 6100N vertically up
d) 4500N vertically down
View Answer

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.

13. 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.
machine-design-questions-answers-eccentric-loaded-bolted-joints-q9
a) 14.34mm
b) 13.44mm
c) 15.44mm
d) 12.66mm
View Answer

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.

Sanfoundry Global Education & Learning Series – Machine Design.
To practice all areas of Machine Design, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter