The mass moment of inertia of two rotors in a two-rotor system are 100 kg m2 and 10 kg m2. The length of the shaft of uniform diameter between the rotor is 110 cm.The distance of node from the rotor of lower moment of inertia is ________
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Solution
Given: Mass moment of inertia of rotor (1)
= I1= 100 kgm2 Mass moment of inertia of rotor(2)
= I2= 10 kgm2
l1+ l2 = 110 cm
Now,l1I1 = l2I2
l1 × 100 = l2 × 10
l2= 10 l1
l1+ 10I1= 110
11l1= 110
⇒l1 = 10cm; l2 = 100 cm
A hollow shaft of outer diameter 40 mm and inner diameter of 20 mm is to be replaced by a solid shaft to transmit the same torque at the same maximum stress.Then the diameter of solid shaft will be ________
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Solution
For hollow shaft,
ζ (Shear stress) =\(\frac{TdD}{\frac{\pi(d_{0}^{4}-d_{i}^{4})}{16}}\)
or T=\(\frac{\pi(d_{0}^{4}-d_{i}^{4})\zeta }{16}\)
where, T= Torque, do= outer diameter, di= inner diameter
ζ= shear stress Given : do= 40 mm, di= 20 mm
Now,torque and shear stress will be same,
So,\(\frac{\pi D^{3}\zeta }{16}=\frac{\pi(d_{0}^{4}-d_{i}^{4})\zeta }{16d_{0}}\)
\(D^{3}=\frac{d_{0}^{4}-d_{i}^{4}}{d_{0}}=\frac{(40)^{4}-(20)^{4}}{40}\)
=\(\frac{256\times 10^{4}-16\times 10^{4}}{40}\)
\(D^{3}=\frac{240\times 10^{4}}{40}= 6 \times 10^{4}\)
D=\(\sqrt[3]{6\times 10^{4}}\)= 39.15 mm
The maximum distorsion energy theory of failure is suitable to predict the failure of:
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Solution
The maximum distorsion energy theory of failure is suitable to predict the failure of ductile materials.
From a tensile test, the yield strength of steel is found to be 200 N/mm2 using a factor of safety of 2 and maximum principal stress theory of failure, the permissible stress in the steel shaft subjected to torque will be ________
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Solution
Given: yield strength (σyt)= 200 N/mm2
Factor of safety (FOS) = 2
Now,permissible stress in steel shaft (σPerm)
\(\frac{\sigma _{yt}}{FOS}=\frac{200}{2}\)= 100 N/mm2
Which one of the following statements is correct in the case of screw dislocation ?
(\(\underset{b}{\rightarrow}\)= Burger’s vector,\(\underset{t}{\rightarrow}\)= Imaginary vector)
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Solution
In screw dislocations, Burger’s vector (\(\underset{b}{\rightarrow}\)
) is parallel to imaginory vector (\(\underset{t}{\rightarrow}\)
).
Which one of the following welding processes consists of minimum heat affected zone (HAZ)?
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Solution
Laser Beam welding has minimum HAZ.
A shaft of diameter 27 mm and a bearing are assembled along with a clearance fit. Allowances and tolerances are given as :
Allowance = 0.005 mm, Tolerance of hole = 0.009 mm,Tolerance on shaft = 0.003 mm. Then the limit of size for the hole and the shaft will be ……………….
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Solution
As we know that, minimum clearance is equal to the allowance.
Now,using Hole Basis system,Hole⇒maximum = 27.009 mm
Minimum = 27.000 mm
Shaft basis, maximum = 26.995 mm
Minimum = 26.992 mm
The correct sequence of above 4-cycles of T-s figure is:
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Solution
The correct sequence will be :
\(\underset{(1)}{Rankine\, cycl}\rightarrow \underset{(2)}{otto\, cycl}\rightarrow \underset{(3)}{Diesel\, cycl}\rightarrow \underset{(4)}{carnot\, cycl}\)
In a population queuing model, arrivals follow a Poisson distribution with mean = 3/hour. The service times are exponential and equal to 3 minutes. The expected length of queue will be ………………..
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Solution
Given : l = 3/hour = 3/60 =1⁄20/min.
p=3⁄20= 0.15
Expected length of queue is :
\(L=\frac{p^{2}}{1-p}=\frac{(0.15)^{2}}{10.15}=\frac{0.0225}{0.85}= 0.0265\)
For inventory control, the following data is given :
Annual requirement : ₹18,000 units
Preparation cost :₹18/order
Inventory holding cost :₹4/unit/year
Production rate : 100 units/day
Working days: 250/year
Then, the number of production runs and the total incremental cost will be ………………..
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Solution
Using the formula,
Build up EOQ model, Q=\(\sqrt{\frac{2A.D}{h(1-d/p)}}\)
where,
A= ordering cost/order
D= Annual demand
P= Production rate
d= depletion rate
h= holding cost
Given : D= 18,000 units, A= Rs 18; h = Rs 4P= 100 units, d=\(\frac{1800}{250}=75\)
Q =\(\sqrt{\frac{2\times 18\times 18,000}{4\left ( 1-\frac{72}{100} \right )}}=\sqrt{\frac{648000}{1.12}}=\sqrt{540000}\)
= 734.85 units
Now, number of production run =\(\frac{D}{Q}=\frac{18000}{734.85}=24.5\)
Total incrementel cost =\(\sqrt{2ADh(1-d/p)}\)
\(\sqrt{648000\times 4\left ( 1-\frac{72}{100} \right )}=\sqrt{725760}\)
= ₹851.92