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926

October 2019

34742 - Mechanics of Materials

Question Paper Code: 926

October 2019

PART – A (1)Question No 8 in PART-A is compulsory

(2) Answer any FOUR Questions

(3) Each Question carries 2 marks

1. Define elasticity and plasticity. (Each carries one mark)

Elasticity:

The elasticity is defined as its ability of material to come back to its original

position after deformation when the stress or load is removed.

Plasticity:

The plasticity of a material is its ability to undergo some permanent

deformation without rupture or failure.

Materials such as clay, lead etc. are plastic at room temperature. Steel is plastic

when at bright red heat.

2. Define shear stress and shear strain. (Each carries one mark)

Shear stress: Shear stress is the amount of force per unit area perpendicular to the axle of the

member.

Shear strain: Shear strain is defined as the length of deformation divided by the

perpendicular length in the plane of the force applied.

3. Define centroidal axis. (2 Marks)

Centroidal axis:

A line passing through the centroid is known as centroidal axis.

4. Define thick and thin cylindrical shell.

Thick cylindrical shell: (Each carries one mark)

If the thickness of the cylindrical shell is greater than 1/20 times of its

diameter, then it is called thick cylindrical shell. t≥d/20

Thin cylindrical shell:

If the thickness of the cylindrical shell is less than 1/20 times of its diameter,

then it is called thin cylindrical shell. t≤d/20

5. Write a short note on torsional rigidity. (2 Marks)

Torsional rigidity:

Torsional rigidity is the product of modulus of rigidity and polar moment of

inertia of the shaft.

6. Give the applications of worm and worm wheel gears.(any two points)

Applications of worm and worm wheel gears: (Each carries one mark)

They are used in the indexing heads.

They are used in the final drive and steering gear box of automobiles.

They are used in the wind shield wiper.

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7. Define the angle of lap in flat belt drive. (2 Marks)

Angle of lap:

The angle through which the belt is contact with the pulley is called angle of

contact or angle of lap.

8. What is meant by tension spring? Give examples. (2 Marks)

Tension spring:

A helical spring is said to be a tension spring, when it is subjected to axial

load, it open out and it closes when the load is released.

Examples: Spring balances and cycle stand.

PART – B (1)Question No 16 in PART-B is compulsory

(2) Answer any FOUR Questions

(3) Each Question carries 3 marks

9. How is percentage elongation and percentage reduction in area calculated in tensile test?

Calculation of Percentage elongation in tensile test: (Each carries 1½ Marks)

The ratio of extension on the specimen after fracture to the gauge length,

expressed as a percentage is called as percentage of elongation.

Percentage of elongation = (Increase in length /Original length) x 100

Percentage elongation= (δl/l) x 100

Calculation of Percentage reduction in area:

The ratio of maximum reduction in cross sectional area of the fractured test

piece to the original area of cross section, expressed as a percentage is called percentage of

reduction in area.

Percentage reduction in area = [(Original area of cross section – Area of neck after

fracture) / Original area of cross section] x 100

10. Describe about the radius of gyration.

Radius of gyration: (Explanation – 2 Marks; Diagram – 1 Mark)

It may be defined as the uniform distance from the reference axis at which the

entire area is assumed to be distributed.

Consider a plane area A. The moment of inertia of this area about a

Reference axis PQ, as shown in fig.

IPQ = ∑ da.r2

IPQ = ∑ da x k2 = Ak

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k = √IPQ / A

11. Explain failure of thin cylindrical shell subjected to internal pressure. (3 Marks)

Failure of thin cylindrical shell subjected to internal pressure:

When a thin cylindrical shell is subjected to an internal pressure its wall is

subjected to the following stresses.

Hoop or circumferential stress.

Longitudinal stress

When these stresses exceed the permissible limit, the cylinder is likely to fail

in the following two ways.

1. Split up into two troughs (semicircular halves)

2. Split up into two short cylinders.

12. Explain velocity ratio of gear drive. (3 Marks)

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Velocity ratio of gear drive:

It is the ratio between follower speed to driver speed.

n1 = speed of the driver in rpm

T1 = number of teeth on the driver

D1 = pitch circle diameter of driver

Similarly, n2, T2 & D2 are speed, number of teeth & pitch circle dia of follower

for proper meshing of gear pairs,

pitch of driver = pitch of follower

p = (∏D1) / T1 = (∏D2) / T2

(D1 / D2) = (T1 / T2) ……………….(1)

linear velocity of the driver = linear velocity of thefollower

[(∏D1n1) / 60] = [(∏D2n2) / 60]

Velocity ratio (n2/n1) = (D1/D2) …………..(2)

Add the equation (1) and (2)

(n2 / n1) = (D1 / D2) = (T1 / T2)

13. Explain moment of resistance. (Explanation – 2 Marks; Diagram – 1 Mark)

Moment of resistance:

When a beam is subjected to an external bending moment it is resisted by an

internal moment set up inside the beam section. This internal moment is known as moment

of resistance.

14. Explain the power transmitted by the shaft. (3 Marks)

Power transmitted by the shaft:

Consider a shaft rotating at speed N rpm, transmitting a power P.

Let Tmean = mean torque of the shaft in N-m.

Workdone per second = Mean torque x angle turned per second

= Tmean x (N/60)x2∏ N-m/sec

Power transmitted, P = (2∏NTmean) / 60 N-m/sec (or) watts

or T = [(60P) / (2∏N)] N-m = [(30P) / (∏N)] N-m

for designing the shaft, we have to take

T = Tmax = 130 to 140% of Tmean

15. Explain briefly with sketch how flat belt connects perpendicular shafts.

Flat belts connecting perpendicular shafts: (3 Marks)

M f E

--- = --- = ---

I y R

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16. Explain types of loading on beams. (3 Marks)

Types of loading on beams:

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PART – C Each question Carries -10 Marks

17. (a) The rod of a hydraulic lift is 1.2m long and 32mm in diameter. It is attached to a

plunger of 100mm in diameter under a pressure of 8N/mm2. If E = 2 x 10

5 N/mm

2, find

the change in the length of the rod.

Given data:

Length, l = 1.2m 1200mm

Diameter of rod, d = 32mm

Diameter of plunger, D = 100mm

Pressure, p = 8N/mm2

Young’s modulus, E = 2 x 105 N/mm

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To find:

Change in length, (δl) (10 Marks)

Solution:

Force on plunger and on rod = Area of plunger x pressure

P = Area x pressure

= (∏ /4) x D2 x p

Force, P = (∏ / 4) x 1002 x 8

= 62831.85N

Area of rod, A = (∏ /4) x d2

= (∏ /4) x 322 = 804.248mm

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Change in length, δl = Pl / AE = (62831.82 x 1200) / (804.248 x 2 x 105)

= 0.4687mm

Change in length, δl = 0.4687mm

(Or)

(b) A steel specimen 150mm2 cross section stretches by 0.05mm over a 50mm gauge

length under an axial load of 30kN. Calculate the strain energy stored in the specimen at

this stage. If the load at the elastic limit for the specimen is 50kN, calculate the elongation

at elastic limit and proof resilience.

Given data:

Area of the specimen, A = 150mm2

Length of the specimen, l = 50mm

Load , P = 30kN = 30 x 1000 N

Change in length, δl = 0.05mm

To find:

i. Strain energy stored below elastic limit (U) (4 Marks)

ii. Elongation at the elastic limit (δl) (4 Marks)

iii. Proof resilience (Umax) (2 Marks)

Solution:

i.Strain energy stored below elastic limit (U):

P = P/A = (30 x 103) / 150 = 200 N/mm

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E = p.l / δl = (200 x 50) / 0.05 = 2 x 105 N/mm

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U = (p2 / 2E) x A x l = [(200

2 / 2 x 2 x 10

5) x 150 x 50] = 750 N-mm

ii.Elongation at the elastic limit (δl):

Stress at elastic limit, pmax = elastic limit load / area

= (50 x 1000) /150 = 333.33 N/mm2

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Elongation at elastic limit = (pmax x l ) / E

= (333.33 x 50) / 2 x 105 = 0.0833mm

iii.Proof resilience (Umax):

Proof resilience = Maximum energy stored upto elastic limit

= (p2

max/2E) x A x l = [(333.332 / 2 x 2 x 10

5) x 150 x 50]

Umax = 2083.33 N-mm

18. (a) Determine the moment of inertia of channel section having 300 x 100 x 25mm about

XX axis. (Diagram – 2 Marks; x – 3 Marks; Ixx – 5 Marks)

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(Or)

(b) Calculate the increase in volume of a spherical shell 1m in diameter and 10mm thick,

when it is subjected to an internal pressure of 1.2N/mm2. Take E = 200kN/mm

2 and

1/m = 0.3. (10 Marks)

19. (a) A simply supported beam of span 6m carries a UDL of 20kN/m through its length and

a point load of 30kN at a distance of 2m from the right support. Draw SFD and BMD and

find the position and magnitude of maximum bending moment.

Solution: (SFD – 3 Marks; BMD – 3 Marks; Calculation – 4 Marks)

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(Or)

(b) A timber beam is freely supported on supports 6m apart. It carries a uniformly

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distributed load of 12kN/m run and a concentrated load of 9kN at 2.5m from the left

support. If the stress in timber is not to exceed 8N/mm2. Design a suitable section making

the depth twice the width. (Diagram – 3 Marks; Calculation – 7 Marks)

9kN

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20. (a) A solid shaft 100mm diameter transmits 100kW at 180 rpm. Calculate the maximum

intensity of shear stress induced and angle of twist in degree in a length of 10m.

Take N=0.8 x 105 N/mm

2. (fs – 5 Marks; θ – 5 Marks)

Given data:

D = 100mm; P = 100kW; n = 180rpm

l = 10m = 10 x 1000mm; N = 0.8 x 105 N/mm

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(Or)

(b) A truck weighing 30kN and moving at 5kmph has to be brought to rest by a buffer. Find

how many springs each of 18 coils will be required to store the energy of motion during

a compression of 200mm. The spring is made of 25mm diameter steel rod, coiled to a

mean diameter of 240mm, N=0.084x106 N/mm

2. (10 Marks)

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21. (a) (i) List the merits and demerits of flat belt drive.

Merits of Flat belt drive: (3 Marks)

The belt is in simple construction.

The cost of the belt is less.

It runs smoothly.

It is suitable for transmitting power between two shafts arranged in any

position to each other.

It transmits power over considerable distance between driver and

driven.

Demerits: (2 Marks)

Slip of the belt may occur.

It requires comparatively large size.

The life of the belt is short.

(ii) List the merits and demerits of ‘V’ belt drive.

Merits of V Belt drive: (3 Marks)

They are suitable for transmission of power for short centre distance

between two pulleys.

The space required for the drive is less.

Less vibration and noise.

They transmit more power.

Slip of the belt will not occur.

Replacement and maintenance is easy.

V-belts are made endless. Hence there is no joint trouble.

Demerits of V Belt drive: (3 Marks)

The cost of the belt is more.

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The V-belts are not durable as flat belts.

The construction of pulleys for v-belts is more complicated than flat

belts.

V-belts cannot be used to transmit power for long distance.

The belt wears out quickly.

(Or)

(b) Sketch and explain the nomenclature of gear.

Nomenclature of gear: (Diagram – 5 Marks; Explanation – 5 Marks)

Addendum: The radial distance between the Pitch Circle and the top of the teeth.

Arc of Action: Is the arc of the Pitch Circle between the beginning and the end of the engagement

of a given pair of teeth.

Arc of Approach: Is the arc of the Pitch Circle between the first point of contact of the gear teeth

and the Pitch Point.

Arc of Recession: That arc of the Pitch Circle between the Pitch Point and the last point of

contact of the gear teeth.

Backlash: Play between mating teeth.

Base Circle: The circle from which is generated the involute curve upon which the tooth profile is

based.

Center Distance: The distance between centers of two gears.

Chordal Addendum: The distance between a chord, passing through the points where the Pitch

Circle crosses the tooth profile, and the tooth top.

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Chordal Thickness: The thickness of the tooth measured along a chord passing through the

points where the Pitch Circle crosses the tooth profile.

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Circular Pitch: Millimeter of Pitch Circle circumference per tooth.

Circular Thickness: The thickness of the tooth measured along an arc following the Pitch Circle

Clearance: The distance between the top of a tooth and the bottom of the space into which it fits

on the meshing gear.

Contact Ratio: The ratio of the length of the Arc of Action to the Circular Pitch.

Dedendum: The radial distance between the bottom of the tooth to pitch circle.

Diametral Pitch: Teeth per mm of diameter.

Face: The working surface of a gear tooth, located between the pitch diameter and the top of the

tooth.

Face Width: The width of the tooth measured parallel to the gear axis.

Flank: The working surface of a gear tooth, located between the pitch diameter and the bottom of

the teeth

Wheel:Larger of the two meshing gears is called wheel..

Pinion: The smaller of the two meshing gears is called pinion.

Land: The top surface of the tooth.

Line of Action: That line along which the point of contact between gear teeth travels, between the

first point of contact and the last.

Module: Ratio of Pitch Diameter to the number of teeth..

Pitch Circle: The circle, the radius of which is equal to the distance from the center of the gear to

the pitch point.

Diametral pitch: Ratio of the number of teeth to the of pitch circle diameter.

Pitch Point: The point of tangency of the pitch circles of two meshing gears, where the Line of

Centers crosses the pitch circles.

Prepared by:

K.Nandhakumar,

Staff ID: 42222003

222, Arulmigu Palaniandavar Polytechnic College,

Palani.