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Test: Power Screws Numericals - Mechanical Engineering MCQ


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15 Questions MCQ Test - Test: Power Screws Numericals

Test: Power Screws Numericals for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Power Screws Numericals questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Power Screws Numericals MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Power Screws Numericals below.
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Test: Power Screws Numericals - Question 1

Find the torque required to raise the load of 15kN and mean diameter of triple threaded screw being 46mm. Also given pitch=8mm and coefficient of friction is 0.15.

Detailed Solution for Test: Power Screws Numericals - Question 1

Explanation: tanἀ=l/πd or ἀ=9.429’ as l=3p. tan Ǿ=0.15 or Ǿ=8.531’. M=W x d x tan (Ǿ+ἀ)/2.

Test: Power Screws Numericals - Question 2

What is the collar friction torque if outer and inner diameters are 100mm and 65mm respectively. Coefficient of friction is 0.15 and the load acting is of 15kN. Consider uniform wear theory.

Detailed Solution for Test: Power Screws Numericals - Question 2

Explanation: M=0.15 x W(D+d)/4.

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Test: Power Screws Numericals - Question 3

For a double threaded screw, what will be the tangent of helix angle if nominal diameter and pitch are 100mm and 12mm respectively?

Detailed Solution for Test: Power Screws Numericals - Question 3

Explanation: tan ἀ= l/πd where l=2p, d=D-0.5p.

Test: Power Screws Numericals - Question 4

 If for a trapezoidal thread, angle of thread is 15’ then what will be the replacement of the coefficient of friction which is 0.15.

Detailed Solution for Test: Power Screws Numericals - Question 4

Explanation: f=0.15 x sec 15’.

Test: Power Screws Numericals - Question 5

What will be efficiency of the screw in case of raising the load when coefficient of friction is 0.1553 and tangent of helix angle is 0.0813?

Detailed Solution for Test: Power Screws Numericals - Question 5

Explanation: Efficiency= tanἀ(1-0.1553 x tan ἀ)/(0.1553+tanἀ).

Test: Power Screws Numericals - Question 6

A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the screw torque in terms of clamping force W if nominal diameter=22mm and pitch=5mm.

Detailed Solution for Test: Power Screws Numericals - Question 6

Explanation: M₁=Wd x tan(Ǿ+ἀ)/2 or M₁=W x (22-0.5×2.5)x tan(4.66’+8.531’)/2 as tan(ἀ)=l/πd and tanǾ=0.15.

Test: Power Screws Numericals - Question 7

A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the collar torque in terms of clamping force W assuming uniform wear theory if nominal diameter=22mm and pitch=5mm.

Detailed Solution for Test: Power Screws Numericals - Question 7

Explanation: M₂=0.17 x W x (55+45)/4 or M₂=4.25W N-mm.

Test: Power Screws Numericals - Question 8

A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the clamping force W if nominal diameter=22mm and pitch=5mm.

Detailed Solution for Test: Power Screws Numericals - Question 8

Explanation: Net torque=M₁+M₂ or 125 x 150=2.286W + 4.25W or W=2868.73N.

Test: Power Screws Numericals - Question 9

 A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the overall efficiency if nominal diameter=22mm and pitch=5mm.

Detailed Solution for Test: Power Screws Numericals - Question 9

Explanation: Efficiency= W l/2π(M₁+M₂) where
M₁=2.286W [M₁=Wd x tan (Ǿ+ἀ)/2 or M₁=W x (22-0.5×2.5) x tan (4.66’+8.531’)/2 as tan(ἀ)=l/πd and tan Ǿ=0.15].
M₂=4.25W [M₂=0.17 x W x (55+45)/4 or M₂=4.25W N-mm],
l=5mm,
W=2868.73N [Net torque=M₁+M₂ or 125 x 150=2.286W + 4.25W or W=2868.73N].

Test: Power Screws Numericals - Question 10

 Find the bending stress to which a screw of nominal diameter 22mm is subjected when the clamp exerts a force of 5kN acts on it. The screw is double threaded and pitch of screw is 5mm.Given: Coefficient of friction is 0.15.It is assumed operator exerts a force of 250N at the handle of length 275mm.

Detailed Solution for Test: Power Screws Numericals - Question 10

Explanation: Bending stress=32M/πdᵌ where M=250 x 275N-mm,d=22-2 x 0.5 x 5mm.

Test: Power Screws Numericals - Question 11

Find the torsional shear stress to which a screw of nominal diameter 22mm is subjected when the clamp exerts a force of 5kN on it. The screw is double threaded and pitch of screw is 5mm.Given: Coefficient of friction is 0.15.

Detailed Solution for Test: Power Screws Numericals - Question 11

Explanation: =16M/πd₂ᵌ where M=Wd₁ tan (Ǿ+ἀ)/2 where d₁=22-0.5 x 5,W=5000N,
tanἀ=2 x 5/[πx19.5] and d₂=19.5-0.5 x 5.

Test: Power Screws Numericals - Question 12

If bending stress in a screw is of magnitude 95N/mm² and torsional shear stress of magnitude 15N/mm², then what is the principal shear stress?

Detailed Solution for Test: Power Screws Numericals - Question 12

Explanation: τ(max)=√(95/2)²+15².

Test: Power Screws Numericals - Question 13

The nominal diameter of the screw is 22mm and the allowable bearing pressure for the nut is 15N//mm². Find the length of the nut if pitch of the screw is

Detailed Solution for Test: Power Screws Numericals - Question 13

Explanation: z=4W/π S(d₁²-d₂²) where S=15N/mm²,d₁=22mm and d₂=17mm. Hencez=2.2 or z=3. Length of nut=zp or l=3 x 5= 15mm.

Test: Power Screws Numericals - Question 14

 A but joint can be defined as a joint between two components lying approximately in a different plane.

Detailed Solution for Test: Power Screws Numericals - Question 14

Explanation: Butt joint is defined a joint between two plates lying in the same plane.

Test: Power Screws Numericals - Question 15

A fillet weld is an approximately triangular cross section joining two surfaces lying parallels to each other.

Detailed Solution for Test: Power Screws Numericals - Question 15

Explanation: Two surfaces lie perpendicular to each other.

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