Design Of Machine Elements Set 2

Mechanical Engineering

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This set of Design of Machine Elements Multiple Choice Questions & Answers (MCQs) focuses on Design Of Machine Elements Set 2

Q1 | 1816mm and radius of centroidal axis is 100mm, then find the bending stress at inner fibre whose radius is 50mm. Area of cross section of beam is 7200mm² and the beam is loaded with 100kN of load.
  • 97.3
  • 95.8
  • 100.6
  • none of the mentioned
Answer: 100.6
Explaination:
e=100-89.816=10.8184mm, h=89.1816-50=39.1816mm, m=100 x 100 n-
Q2 | 1816/ [7200 x 10.8184 x 50] or σ=100.6N/mm².
  • 14.80mm
  • 13.95mm
  • 16.5mm
  • 17.2mm
Answer: 13.95mm
Explaination:
σ(b)=mh/ aer or σ(b)=1000x4dx(4d-0.2d-3.5d)/ 0.8d²x0.02dx3.5d , σ(b)=21428.6/d²
Q3 | A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm² and considering only direct tensile stress.
  • 4.7mm
  • 6.8mm
  • 13.95mm
  • 3.4mm
Answer: 3.4mm
Explaination:
direct tensile stress=1000/0.8d² or σ (t) =1250/d² 1250/d²=110 or d=3.4mm.
Q4 | All type of stresses vanishes after as soon as the applied load is removed.
  • true
  • false
Answer: false
Explaination:
residual stresses are independent of the load.
Q5 | Residual stresses are always added in the load stresses and hence are always harmful.
  • true
  • false
Answer: false
Explaination:
residual stresses may be beneficial when they are opposite to load stresses and hence are subtracted from load stresses.
Q6 | Which of the following is the numerator of factor safety formula?
  • working stress
  • shear stress
  • tensile stress
  • ultimate stress
Answer: ultimate stress
Explaination:
factor of safety is defined as ratio of ultimate stress and working stress. it is also called as factor of ignorance. the factor of safety is dependent on the type of load.
Q7 | Which of the following can be the factor of safety for shock loading?
  • 11
  • 13
  • 4
  • 7
Answer: 13
Explaination:
for shock loading, the range in which the factor of safety can lie is 12 to 15. therefore only 13 lies in that range and 11, 4, 7 are all values which are beyond that range.
Q8 | Factor of safety is used to find out the reliability of the design.
  • true
  • false
Answer: true
Explaination:
factor of safety tells us about how much load the material can take before it fails. basically, it states the load carrying capacity of a material beyond the actual load.
Q9 | What can understand by the factor of safety equal to one?
  • it means that the structure will fail under load
  • it means that the structure will only support the actual load
  • it means that the structure will support more than the actual load
  • there is no relation between factor safety and load application
Answer: it means that the structure will only support the actual load
Explaination:
when the factor of safety is one it means that the ultimate stress is equal to the working stress and therefore the body can only support load up to actual load and no more before failing.
Q10 | For which of the following design factor of
  • 0.1
  • 1
  • 2 d) 0.9
Answer: 2 d) 0.9
Explaination:
if the design factor of safety is not more than 1 then the design may not work and will fail under certain conditions.
Q11 | What is a safe factor of failure for a component which on failure can result in financial loss or serious injury?
  • 1
  • 2
  • 3
  • 4
Answer: 4
Explaination:
for components which on failing can be hazardous and can lead to serious injuries, death and financial loss, the factor of safety should be taken equal to or more than 4.
Q12 | Design factor for most aircraft structures is 2.
  • true
  • false
Answer: false
Explaination:
the design factor of safety for most aircraft structures or components is taken to be 1.5 and not 2. more factors affect this value, but it is most of the time taken as 1.5.
Q13 | Does ASME Standard take into consideration shock and fatigue factors?
  • yes
  • no
Answer: yes
Explaination:
moment is multiplied by a number to consider these factors while designing the shaft.
Q14 | Calculate the tension in the rope of pulley C.
  • 6778.3n and 7765.3n
  • 5948.15n and 2288.75n c) 5468.4n ad 8678.3n
  • d) none of the listed
Answer: 5948.15n and 2288.75n c) 5468.4n ad 8678.3n
Explaination:
(p₃-p₄)x150=549.3×1000 and p₃/p₄=2.6. hence p₃=5948.15n and p₄=2288.75n.
Q15 | If bending moment on point B in horizontal plate is M and in vertical plane is m, then the net bending moment at point B is?
  • m
  • m
  • m+m
  • √m²+m²
Answer: √m²+m²
Explaination:
the two moments act perpendicularly to each other.
Q16 | Calculate the shaft diameter on rigidity basis if torsional moment is 196000N-mm, length of shaft is 1000mm. Permissible angle of twist per meter is 0.5’ and take G=79300N/mm².
  • none of the listed
  • 41.2mm
  • 35.8mm
  • 38.8mm
Answer: 41.2mm
Explaination:
d⁴=584ml/gθ.
Q17 | If yielding strength=400N/mm², the find the permissible shear stress according to ASME standards.
  • 72 n/mm²
  • 76 n/mm²
  • 268 n/mm²
  • 422 n/mm²
Answer: 72 n/mm²
Explaination:
0.18×400.
Q18 | The strength of hollow shaft is more than the strength of solid shaft of same weight.
  • true
  • false
Answer: true
Explaination:
outer fibers are more effective in resisting the applied moments. in hollow shafts material is removed and spread on a larger radius.
Q19 | Solid shafts are used in epicyclic gearboxes.
  • true
  • false
Answer: false
Explaination:
in epicyclic gears, one shaft rotates inside other and hence hollow shafts are used.
Q20 | Determine the length of kennedy key required to transmit 1200N-m and allowable shear in the key is 40N/mm². The diameter of shaft and width of key can be taken as 40mm and 10mm respectively.
  • 49mm
  • 36mm
  • 46mm
  • 53mm
Answer: 53mm
Explaination:
l=m/[dbτ√2].
Q21 | 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.
  • 8mm
  • 9mm
  • 10mm
  • 11mm
Answer: 8mm
Explaination:
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.
Q22 | 8’CW from the vertical. Hence net shear force = √ [5000cos36.8+1250]²+ [5000sin36.8]².
  • 10.74mm
  • 9.23mm c) 11.54mm
  • d) 8.68mm
Answer: 10.74mm
Explaination:
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.
Q23 | Which bolt is under maximum shear stress?
  • 1
  • 2
  • 3
  • all are under equivalent shear stress
Answer: 3
Explaination:
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.
Q24 | Arrange the bolts in order of decreasing shear stresses.
  • 1>2>3
  • 2>1>3
  • 3>1>2
  • 3>2>1
Answer: 3>2>1
Explaination:
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.
Q25 | 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.
  • 14.34mm
  • 13.44mm
  • 15.44mm d) 12.66mm
Answer: 13.44mm
Explaination:
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
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