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# Test: Power Screws Numericals

## 15 Questions MCQ Test Machine Design | Test: Power Screws Numericals

Description
This mock test of Test: Power Screws Numericals for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 15 Multiple Choice Questions for Mechanical Engineering Test: Power Screws Numericals (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Power Screws Numericals quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Test: Power Screws Numericals exercise for a better result in the exam. You can find other Test: Power Screws Numericals extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
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.

Solution:

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.

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.

Solution:

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

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?

Solution:

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

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.

Solution:

Explanation: f=0.15 x sec 15’.

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?

Solution:

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

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.

Solution:

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.

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.

Solution:

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

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.

Solution:

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

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.

Solution:

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].

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.

Solution:

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

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.

Solution:

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.

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?

Solution:

Explanation: τ(max)=√(95/2)²+15².

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

Solution:

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.

QUESTION: 14

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

Solution:

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

QUESTION: 15

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

Solution:

Explanation: Two surfaces lie perpendicular to each other.