Design of Machine Elements - 1 - Mechanical Engineering (MCQ) questions and answers
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1) Determine number of coils in a helical compression spring, if modulus of rigidity is 80 Gpa and spring stiffness is 50 N/ mm. Assume wire diameter and spring index as 8 mm and 5 respectively
- Published on 14 Sep 15
a. 11.8 turns
b. 12.8 turns
c. 13.3 turns
d. None of the above
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2) Which factor is used to consider the effects of direct shear stress and torsional shear stress when curvature effect stress is not considered?
- Published on 14 Sep 15
a. Shear stress concentration factor
b. Wahl shear stress concentration factor
c. Both a. and b.
d. None of the above
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3) The shear stress concentration factor (Ks) in mechanical springs is given as _____
- Published on 14 Sep 15
a. (1 + 0.5 / C)
b. 0.615 / C
c. (1 + 0.615 / C)
d. [(4C – 1) / (4C + 1)] + [0.615 / C]
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4) Which type of springs have only active coils?
- Published on 14 Sep 15
a. Helical compression springs
b. Helical tension springs
c. Both a. and b.
d. None of the above
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5) Solid length for helical compression springs having square and ground ends is given as _________.
- Published on 14 Sep 15
a. (n + 2)d
b. (n + 3)d
c. (n + 1)d
d. None of the above
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6) Which formula is used to calculate shear strength of butt weld?
- Published on 14 Sep 15
a. P = τ h l
b. P = τ h l2
c. P = τ h2 l
d. None of the above
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7) Which of the following statements is/are false?
1. Under fatigue load, casted structures are stronger than welded structures
2. Welding cannot produce complicated structures
3. Welding can join dissimilar materials
4. Rivetted joints produce light weight constructions as compared to welded joints
- Published on 14 Sep 15
a. Statement 1 and Statement 3
b. Statement 1, 2 and Statement 4
c. Statement 1, 2 and Statement 3
d. All the above statements are false
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8) If welds are designed with in-plane eccentric load, then eccentric load can be replaced by _______.
- Published on 14 Sep 15
a. parallel force
b. twisting moment
c. both a. and b.
d. none of the above
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9) In the diagram shown below la = lb = 100 mm, what is the load carrying capacity if weld size is 15 mm and shear stress for the weld is 100 MPa
- Published on 14 Sep 15
a. 200 kN
b. 310 kN
c. 212.13 kN
d. 121.13 kN
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10) Axially loaded unsymmetrical welded joint is shown below, if total length of the weld is (la + lb) then length of weld la is given as
- Published on 14 Sep 15
a. (b / (a+b)) x l
b. (a / (a+b)) x l
c. (a / (a -b)) x l
d. (b / (a-b)) x l
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