CE8395 STRENGTH OF MATERIALS FOR …...Stresses in thin cylindrical shell due to internal pressure circumferential and longitudinal stresses and deformation in thin and thick cylinders

REGULATION : 2017 ACADEMIC YEAR : 2018-2019

JIT-JEPPIAAR/MECH/ Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL

ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0

5-1

CE8395 STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS L T P C

3 0 0 3

OBJECTIVES:

To understand the concepts of stress, strain, principal stresses and principal planes.

To study the concept of shearing force and bending moment due to external loads in

determinate beams and their effect on stresses.

To determine stresses and deformation in circular shafts and helical spring due to torsion.

To compute slopes and deflections in determinate beams by various methods.

To study the stresses and deformations induced in thin and thick shells.

UNIT I STRESS, STRAIN AND DEFORMATION OF SOLIDS 9

Rigid bodies and deformable solids – Tension, Compression and Shear Stresses – Deformation

of simple and compound bars – Thermal stresses – Elastic constants – Volumetric strains –

Stresses on inclined planes – principal stresses and principal planes – Mohr’s circle of stress.

UNIT II TRANSVERSE LOADING ON BEAMS AND STRESSES IN BEAM 9

Beams – types transverse loading on beams – Shear force and bending moment in beams –

Cantilevers – Simply supported beams and over – hanging beams. Theory of simple bending–

bending stress distribution – Load carrying capacity – Proportioning of sections – Flitched

beams – Shear stress distribution.

UNIT III TORSION 9

Torsion formulation stresses and deformation in circular and hollows shafts – Stepped shafts–

Deflection in shafts fixed at the both ends – Stresses in helical springs – Deflection of helical

springs, carriage springs.

UNIT IV DEFLECTION OF BEAMS 9

Double Integration method – Macaulay’s method – Area moment method for computation of

slopes and deflections in beams - Conjugate beam and strain energy – Maxwell’s reciprocal

theorems.

UNIT V THIN CYLINDERS, SPHERES AND THICK CYLINDERS 9

Stresses in thin cylindrical shell due to internal pressure circumferential and longitudinal

stresses and deformation in thin and thick cylinders – spherical shells subjected to internal

pressure –Deformation in spherical shells – Lame’s theorem.

TOTAL: 45 PERIODS

OUTCOMES Students will be able to

Understand the concepts of stress and strain in simple and compound bars, the importance of

principal stresses and principal planes.

Understand the load transferring mechanism in beams and stress distribution due to shearing

force and bending moment.

Apply basic equation of simple torsion in designing of shafts and helical spring

Calculate the slope and deflection in beams using different methods.

Analyze and design thin and thick shells for the applied internal and external pressures.




5-2

TEXT BOOKS:

1. Bansal, R.K., "Strength of Materials", Laxmi Publications (P) Ltd., 2016

2. Jindal U.C., "Strength of Materials", Asian Books Pvt. Ltd., New Delhi, 2009

REFERENCES: 1. Egor. P.Popov “Engineering Mechanics of Solids” Prentice Hall of India, New Delhi, 2002

2. Ferdinand P. Been, Russell Johnson, J.r. and John J. Dewole "Mechanics of Materials", Tata

McGraw Hill Publishing ‘co. Ltd., New Delhi, 2005.

3. Hibbeler, R.C., "Mechanics of Materials", Pearson Education, Low Price Edition, 2013

4. Subramanian R., "Strength of Materials", Oxford University Press, Oxford Higher Education

Series, 2010.




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Subject Code: CE8395 Year/Semester: II /04

Subject Name: STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

Subject Handler: Mr.M.K.Karthik

UNIT I - STRESS, STRAIN AND DEFORMATION OF SOLIDS

Rigid bodies and deformable solids – Tension, Compression and Shear Stresses – Deformation of simple

and compound bars – Thermal stresses – Elastic constants – Volumetric strains –Stresses on inclined

planes – principal stresses and principal planes – Mohr’s circle of stress.

PART * A

Q.No. Questions

1.

What do you mean by thermal stress? (Apr/May 2015) BTL2

Thermal stresses are the stresses induced in a body due to change in temperature. Thermal stresses

are set up in a body, when the temperature of the body is raised or lowered and the body is not

allowed to expand or contract freely.

2

Draw the Mohr’s circle for the state of pure shear in a strained body and mark all salient

points in it. (Apr/May 2015) BTL3

3

Differentiate Elasticity and elastic limit. (Nov/Dec 2015) BTL1

The property by virtue of which certain materials return back to their original position after the

removal of the external force is called as elasticity.

Thers is limiting value of force up to and within which, the deformation completely disappears on

the removal of force. The value of stress corresponding to this limiting force is called as elastic

limit.




5-4

4

What is principle of super position? (Nov/Dec 2015) BTL3

The principle of superposition simply states that on a linear elastic structure, the combined effect

of several loads acting simultaneously is equal to the algebraic sum of the effects of each load

acting individually.

5

Obtain the relation between E and K. (May/June 2016) BTL3

𝐸 = 3𝐾(1 − 2𝜇) Where,

E = Young’s Modulus in N/m2.

K = Bulk Modulus in N/m2.

µ = Poisson’s Ratio.

6

Define Young’s modulus. (Nov/Dec 2016) BTL1

Young's modulus is a mechanical property that measures the stiffness of a solid material. It defines

the relationship between stress (force per unit area) and strain (proportional deformation) in a

material in the linear elasticity regime of a uniaxial deformation.

7

What do you mean by principal planes and principal stress? (May/June 2016) , (Nov/Dec

2016), (Nov/Dec 2017) BTL2

The plane where the maximum normal stress exist and the value of shear stress is zero is called

principal plane and these maximum positive and maximum negative value of normal stresses are

known as principal stress.

8

Derive a relation for change in length of a bar hanging freely under its

own weight. (Apr/May 2017) BTL3

Let,

L = length of the bar.

A = Area of cross section.

E = Young’s modulus of the bar material.

w = weight density of the material.

Strain in the element = 𝑆𝑡𝑟𝑒𝑠𝑠

𝐸=

𝑤𝑥

𝐸

Elongation of the element = Strain x Length of the element. = 𝑤𝑥

𝐸𝑑𝑥

Total Elongation = ∫𝑤 𝑥

𝐸

𝐿

0 dx

δL = WL/2E.

9 What does the radius of Mohr’s circle refer to? (Apr/May 2017) BTL1

Radius of Mohr’s Circle is equal to the maximum shear stress.

10

What is Hooke’s law? BTL1

Within the elastic limit, when a body is loaded, then stress induced is proportional to the strain.

This is called Hooke’s law.

Stress α Strain

𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑌𝑜𝑢𝑛𝑔′𝑠𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸)

11 Define Poisson’s Ratio. BTL2




5-5

When a member is stresses within elastic limit, the ratio of lateral strain to its corresponding

linear strain remains constant throughout loading. This constant is called Poisson’s ratio (µ).

Poisson’s ratio (µ or 1

𝑚) =

𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑜𝑟 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛

12

Define Shear stress and Shear strain. BTL2

When two equal and opposite force act tangential on any cross section plane of a body tending to

slide on part of the body over the other part, the stress induced is called shear stress and the

corresponding strain is known as shear strain.

13

What is a compound bar? BTL3

A bar made up of two or more different materials, joined together is called a compound bar or

composite bar. The bars are joined in such a manner, that the system extends or contracts as a single

unit, equally when subjected to tension or compression.

14

What is a bulk modulus? (Nov/Dec2017) BTL2

When a body is stressed, the ratio of direct stress to the correspoding volumetric strain is constant

within elastic limit. This constant is called as Bulk modulus. It is denoted by K.

Bulk Moduus (K) = 𝐷𝑖𝑟𝑒𝑐𝑡 𝑆𝑡𝑟𝑒𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛

15

What is meant by strain energy and proof resilience?. BTL2

Strain energy: During deformation of a body some work is done by the internal resistance

developed in the body which is stored in the form of energy. This energy is known as strain

energy. Unit is N-m.

Proof resilience: The maximum strain energy that can be stored in a material within the elastic

limit is known as proof resilience.

16

Give the expressions for normal stresses on member subjected to like stresses and a shear

stress. BTL2

Major normal principal stress

𝜎1 = 𝜎𝑥 + 𝜎𝑦

2+ √(

𝜎𝑥 − 𝜎𝑦

2)2 + 𝜏

Minor normal principal stress

𝜎2 = 𝜎𝑥 + 𝜎𝑦

2− √(


2)2 + 𝜏

17 Give the expression for norma stress and tangential stress for a member subjected to axial

load on a oblique plane at θ. BTL2




5-6

Normal Stres, 𝜎𝑛 = 𝜎 𝑐𝑜𝑠2𝜃

Tangential Stres, 𝜎𝑡 = 𝜎

2𝑠𝑖𝑛2𝜃

where 𝜎 = 𝐿𝑜𝑎𝑑

𝐴𝑟𝑒𝑎

18

What are the different types of elastic constants and give their inter relationship. BTL2

The types of elastic constants are,

(i) Modulus of Elasticity (or ) Young’s modulus (E)

(ii) Modulus of Rigidity (or) Rigidity modulus (C, G or N)

(iii) Bulk modulus (K)

The relaionship between the three constants is

𝐸 = 3𝐾(1 − 2𝜇) = 2𝐺(1 + 𝜇)

𝐸 = 9𝐾𝐺

3𝐾 + 𝐺

Where,

E = Young’s Modulus in N/m2.

K = Bulk Modulus in N/m2.

µ = Poisson’s Ratio.

19

What are the governing equations of compound bar? BTL2

The governing equations of compound bar of two material are

(i) Elongation in part 1 = Elongation in part 2

𝑃1 𝑙

𝐴1𝐸1=

𝑃2 𝑙

𝐴2𝐸2

(ii) Total Load = Load in part 1 + Load in part 2

𝑃 = 𝑃1 + 𝑃2 𝑃 = 𝜎1𝐴1 + 𝜎2𝐴2

P1, P2 - Loads in section 1 and 2

A1, A2 – Area of section 1 and 2

E1, E2 – Young’s modulus of section 1 and 2




5-7

σ1, σ 2 - Stresses in section 1 and 2

20

Define Factor of safety. BTL2

Factor of Safety is defined as the ratio of ultimate stress to the working stress or permissible

stress or allowable stress.

Factor of Safety = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

𝑊𝑜𝑟𝑘𝑖𝑛𝑔 𝑜𝑟 𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

PART * B

1

A steel rod of diameter 32mm and length 500mm is placed inside an aluminium tube of

internal diameter 35mm and external diameter 45mm which is 1mm longer than the steel

rod. A load of 300kN is placed on the assembly through the rigid collar. Find the stress

induced in steel rod and aluminum tube. Take the modulus of elasticity of steel as 200GPa

and that of aluminium as 80GPa. (13M) (Apr/May 2015) BTL 3

Given:

ds = 32mm=32x10-3m, L=500mm = 500x10-3m, Dal=45mm=45x10-3m, dal=35mm=35x10-3m,

P=300kN=300x103N, Es=200Gpa= 2x1011N/m2, Eal=80Gpa= 8x1010N/m2.

Area of Steel rod, 𝐴𝑠 = 𝜋𝑑𝑠

2

4 =

𝜋 (32 𝑥 10−3)2

4 = 8.04𝑥10−4 𝑚2

Area of Aluminium tube, 𝐴𝑎𝑙 = 𝜋[𝐷𝑎𝑙

2 − 𝑑𝑎𝑙2 ]

4 =

𝜋[ (45 𝑥 10−3)2− (35 𝑥 10−3)2]

4 = 6.282𝑥10−4 𝑚2

Stress in Steel = Stress in Aluminium

휀𝑠 = 휀𝑎𝑙 𝜎𝑠

𝐸𝑠=

𝜎𝑎𝑙

𝐸𝑎𝑙

𝜎𝑠 = 2𝑥1011

8𝑥1010 𝜎𝑎𝑙

𝜎𝑠 = 2.5 𝜎𝑎𝑙

(7M) Total Load = Load in steel + Load in Aluminium

𝑃 = 𝑃𝑠 + 𝑃𝑎𝑙 𝑃 = 𝜎𝑠𝐴𝑠 + 𝜎𝑎𝑙𝐴𝑎𝑙 300x103 = [2.5 𝜎𝑎𝑙 𝑥 8.04𝑥10−4 ] + [𝜎𝑎𝑙 𝑥 6.282𝑥10−4]




5-8

𝜎𝑎𝑙 = 113.71x106 N/m2.

𝜎𝑠 = 2.5 𝑥 113.71x106

𝜎𝑠 = 284.28x106 N/m2.

(6M)

2

A point in a strained material is subjected to stress as shown below. Using Mohr’s Cricle

method, determine the normal and tangential stress across the oblique plane. Check the

answer analytically. (13M) (Apr/May 2015) BTL 5

Given:

𝑀𝑎𝑗𝑜𝑟 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜎1 = 65 𝑁𝑚𝑚2⁄

𝑀𝑖𝑛𝑜𝑟 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜎2 = 35 𝑁𝑚𝑚2⁄

𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜏 = 25 𝑁𝑚𝑚2⁄

Angle of Oblique Plane, 𝜃 = 45°

Mohr’s Circle Method:

Let 1 cm = 10 𝑁𝑚𝑚2⁄

Then, 𝜎1 =65

10= 6.5𝑐𝑚

𝜎2 =35

10= 3.5𝑐𝑚

𝜏 =25

10= 2.5𝑐𝑚

(4M)

Procedure:

Take any point A and draw a horizontal line through A.

Take AB = 𝜎1 = 6.5 𝑐𝑚 and AC = 𝜎2 = 3.5 𝑐𝑚 towards right of A.

Draw perpendicular at B and C and cut off BF and CG equal to shear stress 𝜏 = 2.5 𝑐𝑚.




5-9

Bisect BC at O. Now with O as centre and radius equal to OF or OG draw a circle.

Through O draw a line OE making an angle of 2θ i.e., 2x45= 90° with OF.

From E, draw ED perpendicular to AB produced. Join AE.

Then length AD represents the normal stress and length ED represents the shear stress.

By measurements,

Length AD=7.5cm and Length ED=1.5cm.

𝑁𝑜𝑟𝑚𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜎𝑛 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝐷 𝑥 𝑆𝑐𝑎𝑙𝑒 = 7.5 𝑥 10 = 75 𝑁𝑚𝑚2⁄

𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜎𝑡 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝐸𝐷 𝑥 𝑆𝑐𝑎𝑙𝑒 = 1.5 𝑥 10 = 15 𝑁𝑚𝑚2⁄

(6M)

Analytical Method:

Normal stress 𝜎𝑛 is given by the equation,

𝜎𝑛 = 𝜎1 + 𝜎2

2+

𝜎1 − 𝜎2

2 𝑐𝑜𝑠2𝜃 + 𝜏 𝑠𝑖𝑛2𝜃

𝜎𝑛 = 65 + 35

2+

65 − 35

2 cos (2𝑥45) + 25 sin (2𝑥45)

𝜎𝑛 = 50 + 15 cos 90 + 25 sin 90

𝜎𝑛 = 50 + 25

𝜎𝑛 = 75 𝑁𝑚𝑚2⁄

Tangential stress 𝜎𝑡 is given by the equation,

𝜎𝑡 = 𝜎1 − 𝜎2

2 𝑠𝑖𝑛2𝜃 − 𝜏 𝑐𝑜𝑠2𝜃

𝜎𝑡 = 65 − 35

2sin(2𝑥45) − 25 cos (2𝑥45)




5-10

𝜎𝑡 = 15 sin 90 − 25 cos 90

𝜎𝑡 = 15 − 0

𝜎𝑡 = 15 𝑁𝑚𝑚2⁄

(3M)

,

3

A metallic bar 300mm x 100mm x 40mm is subjected to a force 5kN(tensile), 6kN(tensile)

and 4kN(tensile) along x, y and z directions respectively. Determine the change in the

volume of the block. Take E = 2x105N/mm2 and Poisson’s ratio = 0.25. (13M) (Nov/Dec

2015) BTL 3

Given:

x = 300mm, y = 100mm, z = 40mm,

Fx=5kN= 5000N, Fy=6kN=6000N, Fz=4kN=4000N,

E=2x105N/mm2, µ = 0.25.

Soln:

Volume V = 300x100x40 = 1200000mm3.

Stress in x –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑥−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑦 𝑧=

5000

100 𝑥 40 = 1.25 N/mm2.

Stress in y –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑦−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑥 𝑧=

6000

300 𝑥 40 = 0.5 N/mm2.

Stress in z –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑧−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑥 𝑦 =

4000

300 𝑥 100 = 0.133 N/mm2.

(6M) 𝑑𝑉

𝑉=

1

𝐸(𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧)(1 − 2𝜇) =

1

2𝑥105(1.25 + 0.5 + 0.133)(1 − 2𝑥0.25)

𝑑𝑉

𝑉=

1.883

4𝑥105

𝑑𝑉 = 1.883

4𝑥105 (1200000)

𝑑𝑉 = 5.649 mm3.

(7M)

4

A steel rod of 3cm diameter is enclosed centrally in a hollow copper tube of external diameter

5 cm and internal diamter 4cm. The composite bar is then subjected to axial pull of 45000N.

If the length of each bar is equal to 15cm, determine: (i) the stress in the rod and tube, (ii)

load carried by each bar. Take E for steel = 2.1x105N/mm2 and for copper = 1.1x105N/mm2.

(13M) (Nov/Dec 2015), (Nov/Dec 2016) BTL 5

Given:

Ds = 3cm =30mm, Do,c = 5cm =50mm, Di,c = 4cm =40mm

Axial pull, P =45kN=450000N,




5-11

Length of each bar = L = 15cm =150mm,

Es=2.1x105N/mm2, Ec=1.1x105N/mm2.

Soln:

Area of steel bar, 𝐴𝑠 =𝜋𝐷𝑠

2

4 =

𝜋 302

4 = 706.86 mm2.

Area of Copper tube, 𝐴𝑐 =𝜋(𝐷𝑜,𝑐

2 − 𝐷𝑖,𝑐2 )

4 =

𝜋 (502−402)

4 = 706.86 mm2.

Condn (i) Strain in steel rod = Strain in copper tube

𝜎𝑠

𝐸𝑠=

𝜎𝑐

𝐸𝑐

𝜎𝑠 =2.1 𝑥 105

1.1 𝑥 105 𝜎𝑐

𝜎𝑠 = 1.906 𝜎𝑐

(6M) Condn (ii) Load on steel rod + Load on copper tube = Total Load

w.k.t, Load = Stress x Area.

𝜎𝑠 𝐴𝑠 + 𝜎𝐶 𝐴𝑐 = 𝑃

1.906 𝜎𝑐 𝑥 706.86 + 𝜎𝐶 𝑥 706.86 = 45000

2056.25 𝜎𝐶 = 45000

𝜎𝐶 = 21.88 N/mm2.

𝜎𝑠 = 1.906 𝑥 21.88

𝜎𝑠 = 41.77 N/mm2

Load carried by steel rod 𝑃𝑠 = 𝜎𝑠 𝐴𝑠

𝑃𝑠 = 41.77 𝑥 706.86 = 29525.5N

Load carried by steel rod 𝑃𝑐 = 𝜎𝑐 𝐴𝑐

𝑃𝑐 = 21.88 𝑥 706.86 = 15474.5N

(7M)

5

A steel bar 20mm in diameter, 2m long is subjected to axial pull of 50kN. If E = 2x105N/mm2

and m=3. Calculate the change in the (1)length, (2)diameter and (3)volume. (13M) (May/June

2016) BTL 3

Given:

Length L = 2m = 2000mm,




5-12

Diameter d = 20mm,

Tensile Load = 50kN = 50000N,

E = 2x105N/mm2

µ = 1/m = 1/3 = 0.33

Soln:

Volume, V = 𝜋

4𝐷2𝐿 =

𝜋 302 2000

4 = 35.343x105 mm3.

Let 𝛿L = Change in Length,

𝛿d = Change in Diameter,

𝛿V = Change in Volume.

Strain of length = 𝑆𝑡𝑟𝑒𝑠𝑠

𝐸 =

𝑃𝜋

4𝑑2 𝐸

Strain of length =50000

𝜋

4302 2𝑥105

= 0.0003536

(6M)

𝜹𝑳

𝑳= 0.0003536

δL = 0.0003536 x 2000 = 1.768mm.

Poisson’s ratio = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛

𝐿𝑖𝑛𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛

Lateral Strain = 0.33 x 0.0003536 = 0.0000884

Lateral Strain = 𝜹𝒅

𝒅 = 0.0000884.

δd = 0.0000884 x 30 = 0.002652mm.

Volumetric Strain 𝜹𝑽

𝑽=

𝜹𝑳

𝑳−

𝟐𝜹𝒅

𝒅 = 0.0003536 – 2x0.0000884

𝜹𝑽

𝑽= 0.0001768

δV = 0.0001768 x 35.343x105 = 624.86 mm3

(7M)

6

A mild steel bar 20mm in diameter and 40cm long is encased in a brass tube whose external

diameter is 30mm and internal diameter is 25mm. The composite bar is heated through 80°C.

Calculate the stresses included in each metal. α for steel = 11.2 x 10-6 per °C, α for brass =

16.5 x 10-6 per °C, E for steel = 2x105N/mm2 and E for brass = 1x105N/mm2 (13M) (May/June

2016), (Nov/Dec 2016) BTL 5

Given:

ds = 20mm=20x10-3m, L=40cm = 40x10-2m, Db=30mm=30x10-3m, db=25mm=25x10-3m,




5-13

∆T = 80° C, Es=2x105N/mm2= 2x1011N/m2, Eb=1x105N/mm2= 1x1011N/m2, αs =11.2x10-6 /°C, αb

=16.5x10-6 /°C

Since αb is greater than αs, brass will expand more than steel. But both are clamped. Hence brass

tube will be subjected to compressive stress and steel rod wil be subjected to tensile stress.

Area of Steel rod, 𝐴𝑠 = 𝜋𝑑𝑠

2

4 =

𝜋 (20 𝑥 10−3)2

4 = 3.141𝑥10−4 𝑚2

Area of Brass tube, 𝐴𝑏 = 𝜋[𝐷𝑏

2− 𝑑𝑏2]

4 =

𝜋[ (30 𝑥 10−3)2− (25 𝑥 10−3)2]

4 = 4.091𝑥10−4 𝑚2

At Equilibrium,

Tensile Load on Steel = Compressive load on Brass

𝜎𝑠𝐴𝑠 = 𝜎𝑏𝐴𝑏 𝜎𝑠 𝑥 3.141𝑥10−4= 𝜎𝑏 𝑥 4.091𝑥10−4

𝜎𝑠 = 1.302 𝜎𝑏 (7M)

Actual Expansion of steel = Actual Expansion of Brass

Free expansion of steel + Tensile stress expansion of steel =

Free expansion of brass - Compressive stress compression of brass.

𝛼𝑠∆𝑇 𝐿 + 𝜎𝑠

𝐸𝑠 𝐿 = 𝛼𝑏∆𝑇 𝐿 +

𝜎𝑏

𝐸𝑏 𝐿

𝛼𝑠∆𝑇 + 𝜎𝑠

𝐸𝑠 = 𝛼𝑏∆𝑇 +

𝜎𝑏

𝐸𝑏

(11.2 𝑥 10−6 𝑥 80) +1.302 𝜎𝑏

2 𝑥 1011= (16.5 𝑥 10−6 𝑥 80) −

𝜎𝑏

1 𝑥 1011

(0.000896) + 6.51 𝑥 10−12 𝜎𝑏 = (0.00132) + 1 𝑥 10−11 𝜎𝑏

𝜎𝑏 = - 121.48x106 N/m2.

𝜎𝑠 = 1.302 𝑥 121.48x106

𝜎𝑠 = 158.17x106 N/m2.

(6M)

7 Two steel rods and one copper rod, each of 20mm diameter, together support a load of 20kN

as shown below. Find the stresses in the rods. Take E for steel = 210kN/mm2 and E for copper

= 110kN/mm2 (13M) (May/June 2016) BTL 4




5-14

Given: P = 20kN =20x103 N, ds = dc = 20mm=20x10-3m, Ls=1m, Lc=2m, Es=210kN/mm2=

2.1x1011N/m2, Ec=110kN/mm2= 1.1x1011N/m2.

Area of copper = 𝜋𝑑𝑐

2

4 =

𝜋 (20 𝑥 10−3)2

4 = 3.141𝑥10−4 𝑚2

Area of steel = 2 𝑥 𝜋𝑑𝑠

2

4 = 2 𝑥

𝜋 (20 𝑥 10−3)2

4 = 6.282𝑥10−4 𝑚2

Change in length of steel = Change in length of copper.

Strain of steel x original length = Strain in copper x original length

𝜎𝑠

𝐸𝑠 𝐿𝑠 =

𝜎𝑐

𝐸𝑐 𝐿𝑐

𝜎𝑠

2.1 𝑥 1011 𝑥 1 =

𝜎𝑐

1.1 𝑥 1011 𝑥 2

𝜎𝑠 = 3.81 𝜎𝑐 (7M)

Total Load = Load in steel + Load in Copper

𝑃 = 𝑃𝑠 + 𝑃𝑐 𝑃 = 𝜎𝑠𝐴𝑠 + 𝜎𝑐𝐴𝑐 20x103 = [3.81 𝑥 𝜎𝑐 𝑥 6.282𝑥10−4 ] + [𝜎𝑐 𝑥 3.141𝑥10−4]

𝜎𝑐 = 7.386x106 N/m2.

𝜎𝑠 = 3.81 𝑥 7.386x106

𝜎𝑠 = 28.14x106 N/m2.

(6M)

8 Direct stresses of 140N/mm2 tensile and 100 N/mm2 compression exist on two perpendicular

planes at a certain point in a body. They are also accompanied by shear stress on the planes.

The greatest principal stress at the point due to these is 160 N/mm2.




5-15

(1) What must be the magnitude of the shear stresses on the two planes?

(2) What will be the maximum shear stress at the point? (8M) (May/June 2016) BTL 4

Given:

𝜎𝑥 = 140 𝑁𝑚𝑚2⁄

𝜎𝑦 = −100 𝑁𝑚𝑚2⁄

𝜎1 = 160 𝑁𝑚𝑚2⁄

Calculation of Shear stress:

𝜎1 =𝜎𝑥 + 𝜎𝑦

2+ √(


2)

2

+ 𝜏𝑥𝑦2

160 =140 + (−100)

2+ √(

140 − (−100)

2)

2

+ 𝜏𝑥𝑦2

160 = 20 + √(120)2 + 𝜏𝑥𝑦2

𝜏𝑥𝑦 = 72.11 𝑁𝑚𝑚2⁄

(4M) Calculation of Maximum Shear stress:

𝜏𝑚𝑎𝑥 = √(𝜎𝑥 − 𝜎𝑦

2)

2

+ 𝜏𝑥𝑦2

𝜏𝑚𝑎𝑥 = √(140 − (−100)

2)

2

+ (72.11)2

𝜏𝑥𝑦 = 140 𝑁𝑚𝑚2⁄

(4M)

9

A point in a strained material is subjected to the stresses as shown below. Locate the principle

plane and find the principle stresses. (7M) (Nov/Dec 2016) BTL 5




5-16

Given:

𝜎𝑥 = 51.96 𝑁𝑚𝑚2⁄

𝜎𝑦 = 40 𝑁𝑚𝑚2⁄

𝜏𝑥𝑦 = 30 𝑁𝑚𝑚2⁄

Calculation of Shear stress:

𝜎1,2 =𝜎𝑥 + 𝜎𝑦

2± √(


2)

2

+ 𝜏𝑥𝑦2

𝜎1,2 =51.96 + 40

2± √(

51.96 − 40

2)

2

+ (30)2

𝜎1,2 = 45.98 ± √(5.98)2 + (30)2

𝜎1,2 = 45.98 ± 30.59

𝜎1 = 76.57 𝑁𝑚𝑚2⁄

𝜎2 = 15.39 𝑁𝑚𝑚2⁄

(4M) Location of Principal Plane:

𝑡𝑎𝑛2𝜃 = 2𝜏𝑥𝑦

(𝜎𝑥 − 𝜎𝑦)

𝑡𝑎𝑛2𝜃 = 2𝑥30

(51.96 − 40)

2𝜃 = 78.72°

𝜃 = 39.36°

(3M)

10

The bar shown below is subjected to tensile load of 160kN. If the stress in middle portion is

limited to 150 N/mm2, determine the diameter of the middle portion. Find also the length of

the middle portion if the total elongation of the bar is to be 0.2mm. Young’s modulus is 2.1x105

N/mm2. (13M) (Apr/May 2017) BTL 3




5-17

Given:

L = 400mm = 400x10-3m, Diameter of AB & CD = d1=6cm =6x10-2m, Diameter of BC = d,

Length of BC = L, stress in middle portion σBC = 150 N/mm2 = 150x106 N/m2, P =160kN = 160

x103 N, Total extension δL = 0.2mm = 0.2 x10-3 m, E=2.1x105 N/mm2 = 2.1x1011 N/m2

Stress in middle portion:

𝜎𝐵𝐶 = 𝐿𝑜𝑎𝑑

𝐴𝑟𝑒𝑎=

𝑝

𝜋𝑑2

4

150 𝑥106 = 160𝑥103𝑥 4

𝜋𝑑2

𝑑 = √160 𝑥 103𝑥 4

𝜋 𝑥 150𝑥106

𝑑 = 0.0368 m

(6M)

𝐴1 = 𝐴3 = 𝜋 (6 𝑥 10−2)2

4 = 2.826𝑥10−3 𝑚2

A = 𝜋 (3.68 𝑥 10−2)2

4 = 1.063𝑥10−3 𝑚2

Total Extension, 𝛿𝐿 = 𝑃

𝐸[

𝐿1

𝐴1+

𝐿2

𝐴2+

𝐿3

𝐴3]

, 𝛿𝐿 = 𝑃

𝐸[

0.4−𝐿

𝐴1+

𝐿

𝐴]

0.2𝑥10−3 = 160𝑥103

2.1𝑥1011[

0.4 − 𝐿

2.826𝑥10−3+

𝐿

1.063𝑥10−3]

262.5 = [0.0004252 − 1.063𝑥10−3𝐿

2.826𝑥10−3+

2.826𝑥10−3𝐿

1.063𝑥10−3]

0.000788 = [0.0004252 + 0.001763𝐿]

0.0003628 = [0.001763𝐿]

𝐿 = 0.205𝑚

(7M)

11

A bar of 30mm diameter is subjected to a pull of 60kN. The measured extension on gauge

length of 200mm is 0.1mm and change in diameter is 0.004mm. Calculate (i) Young’s

modulus, (ii)Poisson’s ratio and (iii) Bulk modulus. (13M) (Apr/May 2017) BTL 5

Given:




5-18

d =30mm= 30x10-3m, P=60kN = 60x103N, L = 200mm = 200x10-3m, δL=0.1mm = 0.1x10-3m, δd

=0.004mm=0.004x10-3m

Area = 𝜋𝑑2

4 =

𝜋 (30 𝑥 10−3)2

4 = 7.067𝑥10−4 𝑚2

Young′s Modulus = 𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛

stress = 𝐿𝑜𝑎𝑑

𝐴𝑟𝑒𝑎=

60𝑥103

7.067𝑥10−4

σ = 84.901x106 N/m2.

Linear strain = 𝛿𝐿

𝐿=

0.1𝑥10−3

200𝑥10−3

ε = 5x10-4

Young′s Modulus = 84.901x 106

5x10−4

E = 169.8 x 109 N/m2.

(6M)

𝑃𝑜𝑖𝑠𝑠𝑜𝑛′𝑠 𝑅𝑎𝑡𝑖𝑜 = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛

𝐿𝑖𝑛𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛

Lateral strain = 𝛿𝑑

𝑑=

0.004𝑥10−3

30𝑥10−3 = 1.33x10-4

𝑃𝑜𝑖𝑠𝑠𝑜𝑛′𝑠 𝑅𝑎𝑡𝑖𝑜 =1

𝑚 𝑜𝑟 𝜇 =

1.33x10−4

5x10−4

1

𝑚 𝑜𝑟 𝜇 = 0.266

Bulk Modulus

𝐸 = 3𝐾[1 −2

𝑚]

𝐾 = 169.8 𝑥 109

3(1 − 2𝑥0.266)

K = 120.9 x 109 N/m2.

(7M)

PART * C




5-19

1

(i) Draw stress strain curve for mild steel and explain the salient points on it. (7M) (Apr/May

2017) BTL 2

If tensile force is applied to a steel bar it will have some extension. If the force is small the ratio of

the stress and strain will remain proportional. And the graph will be a straight line ( up to point A)

So the 0 to point A is the limit of proportionality.

If the force is considerably large the material will experience elastic deformation but the ratio of

stress and strain will not be proportional. (point A to B). This is the elastic limit. Beyond that point

the material will experience plastic deformation. The point where plastic deformations starts is the

yield point which is show in the figure as point B.

0 B is the upper yield point. Resulting graph will not be straight line anymore. C is the lower yield

point. D is the maximum ultimate stress. E is the breaking stress.

(7M)

(ii) Derive a relation for change in length of a circular bar with uniformly varying diameter,

subjected to an axial tensile load ‘W’ (8M) (Apr/May 2017) BTL 4

Consider a bar uniformly tapering from a diameter D1 at one end to a diameter D2 at the other end

as shown below,




5-20

Let,

P =Axial tensile load on the bar.

L=Total Length of the bar.

E=Young’s modulus.

Consider a small element of length dx of the bar at a distance x

from the left end. Let the diameter of the bar be Dx at a distance

x from the left end.

Then,

𝐷𝑥 = 𝐷1 − (𝐷1 − 𝐷2

𝐿) 𝑥 = 𝐷1 − 𝑘𝑥

(4M)

Area of cross section of the bar at a distance ‘x’ from the left end,

𝐴𝑥 = 𝜋𝐷𝑥

2

4=

𝜋(𝐷1 − 𝑘𝑥)2

4

Now the stress at a distance ‘x’ from the left end is given by,

𝜎𝑥 = 𝐿𝑜𝑎𝑑

𝐴𝑥=

𝑃𝜋(𝐷1−𝑘𝑥)2

4

=4𝑃

𝜋(𝐷1 − 𝑘𝑥)2

Now the strain in the small element of length dx is given by,

𝜖𝑥 = 𝑆𝑡𝑟𝑒𝑠𝑠

𝐸=

𝜎𝑥

𝐸=

4𝑃

𝜋𝐸(𝐷1 − 𝑘𝑥)2

Extension of the small element length dx,

𝛿𝐿 = 𝑆𝑡𝑟𝑎𝑖𝑛. 𝑑𝑥 = 𝜖𝑥. 𝑑𝑥

𝛿𝐿 =4𝑃

𝜋𝐸(𝐷1 − 𝑘𝑥)2 . 𝑑𝑥

The total extension of the bar is obtained by integrating the above equation between the limits 0

and L.

𝛿𝐿 = ∫4𝑃. 𝑑𝑥

𝜋𝐸(𝐷1 − 𝑘𝑥)2

𝐿

0

The above expression can be integrated and simplified as,




5-21

𝛿𝐿 =4𝑃𝐿

𝜋𝐸𝐷2

(4M)

2

A load of 2MN is applied on a short concrete column 500mm x 500mm. The column is

reinforced with four steel bars of 10mm diamter, one in each corner. Find the stresses in

concrete and the steel bars. Take E for steel as 2.1x105 N/mm2 and for concrete as

1.4x105N/mm2. (15M) (Nov/Dec 2017) BTL 4

Given:

Ds = 10mm ,

No.of Steel bars = 4

Axial pull, P =2MN=2x106N,

Area of concrete column, Ac = 500mmx500mm,

Es=2.1x105N/mm2, Ec=1.4x105N/mm2.

Soln:

Area of steel bar, 𝐴𝑠 =𝜋𝐷𝑠

2

4 = 4 x

𝜋 102

4 = 314.16 mm2.

Area of concrete column, 𝐴𝑐 = (500𝑥500) − 𝐴𝑠 = 249685.84 mm2.

Condn (i) Strain in steel rod = Strain in copper tube

𝜎𝑠

𝐸𝑠=

𝜎𝑐

𝐸𝑐

𝜎𝑠 =2.1 𝑥 105

1.4 𝑥 105 𝜎𝑐

𝜎𝑠 = 1.5 𝜎𝑐

(5M) Condn (ii) Load on steel rod + Load on concrete column = Total Load

w.k.t, Load = Stress x Area.

𝜎𝑠 𝐴𝑠 + 𝜎𝐶 𝐴𝑐 = 𝑃

1.5 𝜎𝑐 𝑥 314.16 + 𝜎𝐶 𝑥 249685.84 = 2x106

250156.88𝜎𝐶 = 2x106

(5M)

𝜎𝐶 = 7.99 N/mm2.

𝜎𝑠 = 1.5 𝑥 7.99

𝜎𝑠 = 11.99 N/mm2




5-22

Load carried by steel rod 𝑃𝑠 = 𝜎𝑠 𝐴𝑠

𝑃𝑠 = 11.99 𝑥 314.16 = 3767.5N

Load carried by steel rod 𝑃𝑐 = 𝜎𝑐 𝐴𝑐

𝑃𝑐 = 7.99 𝑥 249685.84 = 1994989.5N

(5M)

3

A bar 250mm long, cross-sectional area 100mm x 50mm, carries a tensile load of 500kN

along lengthwise, a compressive load of 5000kN on its 100mm x 250mm faces and a tensile

load of 2500kN on its 50mm x 250mm faces. Calculate i) the change in volume, ii) what

changes must be made in the 5000kN load so that no change in the volume of the bar occurs.

Take E = 1.8x105 N/mm2 , Poisson’s ratio=0.25. (15M) (Apr/May 2018) BTL 4

Given:

x = 250mm, y = 100mm, z = 50mm,

Fx=500kN= 500000N, Fy=2500kN=2500000N, Fz=-5000kN=-5000000N,

E=1.8x105N/mm2, µ = 0.25.

Soln:

(i) Change in Volume:

Volume V = 250x100x50 = 1250000mm3.

Stress in x –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑥−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑦 𝑧=

500000

100 𝑥 50 = 100 N/mm2.

Stress in y –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑦−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑥 𝑧=

2500000

250 𝑥 50 = 200 N/mm2.


𝑥 𝑦 =

−5000000

250 𝑥 100 = -200 N/mm2.

(5M) 𝑑𝑉

𝑉=

1

𝐸(𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧)(1 − 2𝜇) =

1

1.8𝑥105(100 + 200 − 200)(1 − 2𝑥0.25)

𝑑𝑉

𝑉=

50

1.8𝑥105

𝑑𝑉 = 50

1.8𝑥105 (1250000)

𝑑𝑉 = 347.22 mm3.

(5M) (ii) What changes must be made in the 5000kN load so that no change in the volume of the bar

occurs.




5-23

For no change to occur in the volume of bar volumetric strain must be equal to zero.

𝑑𝑉

𝑉= 0

Make , Fz as unknown ,


𝑥 𝑦 =

−𝐹𝑍

250 𝑥 100 = -4x10-5 Fz N/mm2

𝑑𝑉

𝑉=

1

𝐸(𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧)(1 − 2𝜇)

0 = 1

1.8𝑥105(100 + 200 − 4x10−5 F𝑧)(1 − 2𝑥0.25)

0 = (300 − 4x10−5 F𝑧)

4x10−5 F𝑧 = 300

𝐹𝑧 =300

4x10−5= 7500000

(5M) Therefore for the bar to have no change in any volume,

The load applied on the bar should be 7500kN which means an additional compressive load of

2500kN along with the existing 5000kN should be applied.


JIT-JEPPIAAR/MECH/Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL


5-24

UNIT II - TRANSVERSE LOADING ON BEAMS AND STRESSES IN BEAM

Beams – types transverse loading on beams – Shear force and bending moment in beams – Cantilevers –

Simply supported beams and over – hanging beams. Theory of simple bending– bending stress

distribution – Load carrying capacity – Proportioning of sections – Flitched beams – Shear stress

distribution.

PART * A

Q.No. Questions

1.

Define: (a) Shearing force, (b) Bending moment. (Apr/May 2015) BTL2

(a) Shear force: Shear force at a cross section is defined as the algebraic sum of all the forces

acting on either side of the beam.

(b) Bending moment: Bending moment at a cross section is the algebraic sum of the moments

of all the forces which are placed either side from that point.

2

What is neutral axis of a beam section? How do you locate it when a beam is under simple

bending? (Apr/May 2015) BTL3

It is the axis or layer of the beam where the bending stress is zero. It is normally located at the

centre of gravity point of the cross section of the beam.

3

Write the assumption in the theory of simple bending? (Nov/Dec 2015), (Nov/Dec 2016)

BTL1

The material is perfectly homogeneous and isotropic and obey’s hooke’s law.

The Young’s modulus is same in tension and well as compression.

Transverse section which are plane before bending remains plane after bending.

Raduis of curvature of the beam is very large compared to the cross section.

The resultant force on a transverse section of the beam is zero.

4 What are the types of beams? (Nov/Dec 2015) BTL3

Beams are classified based upon supports as: (a) cantilever beam, (b) simply supported beam, (c)

overhanging beam, (d) fixed beam, (e) continuous beam, (f) propped cantilever beam.

5

Discuss the fixed and hinged support. (May/June 2016) BTL2

Fixed Support: When a beam is completely fixed or built in wall, the support is called as fixed or

built in support. It has horizontal reaction, vertical reaction as well as moment.

Hinged support: A body is said to be hinged or pin joined when connected to rotating body. The

reaction at the hinged support may be either vertical or horizontal or inclined based on the

loading. But the ends are not restained against rotation at the support.

6

What are the advantages of flitched beams? (May/June 2016) BTL3

A composite section beam may be defined as a beam made up of two or more different materials

joined together in such a manner that they behave like a single piece and material bends to the

same radius of curvature.

7 Draw the SFD and BMD for the cantilever beam carries uniformly varying load of zero

intensity at the free end w kN/m at the fixed end. (Nov/Dec 2016) BTL2




5-25

8

Draw shear force diagram for a simply supported beam of length 4m carrying a central

point load of 4kN. (Apr/May 2017) BTL3

9 Prove that the shear stress distribution over a rectangular section due to shear force is

parabolic. (Apr/May 2017) BTL1

The Shear stress distribution for a rectangular section is given by the expression,




5-26

𝜏 = 𝑉

2𝐼⌊𝑑2

4− 𝑦2⌋

10

State the theory of simple bending. BTL1

When a beam is subjected to bending load, the bottom most layer is subjected to tensile stress and

top most layer is subjected to compressive stress. The neutral layer is subjected to neither

compressive stress nor tensile stress. The stress at a point in the section of the beam is directly

proportional to its distance from the neutral axis.

11

What is meant by section modulus? (May2012) BTL2

It is defined as the ratio of moment of inertia of the section to the distance of the extreme layer

from the neutral axis. It is denoted by Z.

Z = 𝐼

𝑦

12 Define point of contraflexure. (May2013) BTL1

It is defined as the point at which bending moment changes to zero. It occurs mostly in

overhanging beam.

13

In a simply supported beam how will you locate point of maximum bending moment. BTL2

The bending moment is maximum when shear force is zero. Write SF equation at that point and

equating to zero we can find out the distances ‘x’ from one end. Then find maximum bending

moment at that point by taking all moment on right or left hand side of the beam.

14

A rectangular beam 150mm wide and 200mm deep is subjected to shear force of 40kN.

Determine the average shear stress and maximum shear stress. BTL2

Avergae shear stress,

𝜏𝑎𝑣𝑔 = 𝑉

𝐴

𝜏𝑎𝑣𝑔 = 40 𝑥 103

150 𝑥 200

𝜏𝑎𝑣𝑔 = 1.33 𝑥 106 𝑁/𝑚𝑚2

𝜏𝑚𝑎𝑥 = 1.5 𝜏𝑎𝑣𝑔

𝜏𝑚𝑎𝑥 = 1.5 𝑥 1.33 𝑥 106

𝜏𝑚𝑎𝑥 = 2 𝑥 106 𝑁/𝑚𝑚2

15 Write down any four types of beam. (May2013) BTL2

Cantilever beam

Simply supported beam




5-27

Fixed beam

Continuous beam

Overhanging beam.

16

A simply supported beam of length 5m span is subjected to a concentrated load of 10kN at a

distance of 3m from the left support. Draw the bending moment diagram. BTL3

17

Write the relationship between bending moment and shear force. (May2012) BTL3

The rate of change of bending moment is equal to the shear force at that section and can be

expressed as, 𝑑𝑀

𝑑𝑥= −𝐹

18

Calculate the sectional modulus of a circular section of diameter 200mm. (May2012) BTL3

Section Modulus,

𝑍 = 𝐼

𝑦= (

𝜋

64𝑑4) 𝑥 (

2

𝑑)

𝑍 = 𝜋

32𝑑3

𝑍 = 𝜋

32(200)3

𝑍 = 785.4 𝑥 103𝑚𝑚3

19 Define Shear center. (May2012) BTL3




5-28

It is the point of intersection of the bending axis and th plane of transverse section. It is also

called as the center of twist.

20

What is the shear stress distribution value of flange portion of the I-section?. (May2012)

BTL3

The shear stress distribution for the flange portion of the I-section is given by,

𝜏 = 𝑉

2𝐼⌊𝐷2

4− 𝑦2⌋

D – Depth

y – Distance from the neutral axis.

PART * B

1

An overhanging beam ABC of length 7m is simply supported at A and B over a span of 5m

and the portion BC overhangs by 2m. Draw the shearing force and bending moment

diagrams and determine the point of contra-flexure if it is subjected to uniformly

distributed loads of 3kN/m over the portion AB and a concentrated load of 8kN at C. (13M)

(Apr/May 2015) BTL4

Reaction at Supports of the Beam:

∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 8 + (3 𝑥 5) = 23kN

Taking Moments of all forces about A, we get

𝑅𝐵 𝑥 5 = (8 𝑥 7) + (3 𝑥 5 𝑥 5

2 ) = 93.5

𝑅𝐵 = 93.5

5= 18.7𝑘𝑁

𝑅𝐴 = 23 − 18.7 = 4.3𝑘𝑁

Shear Force Diagram:

SF@C = +8kN

SF@B = +8-18.7 = -10.7kN

SF@A = -10.7+(3x5) = +4.3kN

Bending Moment Diagram:

BM@C = 0

BM@B = -(8x2) = -16kN-m.




5-29

BM@A = -(8x2) + (10.7x5) – (3x5x5

2)= 0

To find the location of Maximum Bending Moment:

The maximum bending moment occurs between the points A and B, where Shear Force is zero.

The Shear force at any section between A and B at a distance x from B is given by,

SF@X = -8 + 18.7 – (3 𝑥 ) = 0

3x = 10.7

x = 3.56m

To find the value of Maximum Bending Moment:

The Bending Moment at any section between A and B at a distance x from B is given by,

BMmax = -8(x+2) + 18.7x – (3 𝑥 𝑥

2)

= -8(3.56+2) + 18.7(3.56) – 3(3.56)(3.56/2)

= -44.48 + 66.57 – 19.01

= 3.08 kN-m

(7M)




5-30

(6M)

2

Three beams have the same length, the same allowable stress and the same bending moment.

The cross-section of the beams are a square, a rectangle with depth twice the width and a

circle. Find the ratios of weights of circular and the rectangular beams with respect to the

square beam. (13M) Apr/May 2015) BTL3

Given:

Let

x=side of a square beam.

b=width of rectangular beam.

2b=depth of rectangular beam.

d= diameter of a circular section.

The moment of resistance of a beam is given by,

M=σ x Z

Where, Z=section modulus

As all the three beams have the same alowable bending stress (σ), and the same bending moment

(M), therefore the section modulus (Z) of the three beams must be equal.

Section modulus of a square beam




5-31

𝑍 =𝐼

𝑦=

𝑏𝑑3

12𝑑

2

𝑍 =𝑥. 𝑥3

12.2

𝑥

𝑍 =𝑥3

6

Section modulus of a rectangular beam

𝑍 =𝐼

𝑦=

𝑏𝑑3

12𝑑

2

𝑍 =𝑏. (2𝑏)3

12.

2

2𝑏

𝑍 =𝑏. 8𝑏3

12.

2

2𝑏

𝑍 =2𝑏3

3

Section modulus of a Circular beam

𝑍 =𝐼

𝑦=

𝜋𝑑4

64𝑑

2

𝑍 =𝜋𝑑4

64.2

𝑑

𝑍 =𝜋𝑑3

32

(6M)

Equating the section modulus of a square beam with that of a rectangular beam, we get

𝑥3

6=

2𝑏3

3

𝑏3 =𝑥3

4= 0.25𝑥3

𝑏 = 0.63𝑥




5-32

Equating the section modulus of a square beam with that of circular beam, we get

𝑥3

6=

𝜋𝑑3

32

𝑑3 =32𝑥3

6𝜋= 10.18𝑥3

𝑑 = 1.1927𝑥

The weights of the beams are proportional to their cross-sectional areas. Hence

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑒𝑎𝑚

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑒𝑎𝑚=

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑒𝑎𝑚

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑒𝑎𝑚

=𝑏 𝑥 2𝑏

𝑥 𝑥 𝑥

=0.63𝑥 𝑥 2𝑥 0.63𝑥

𝑥 𝑥 𝑥

= 0.7938

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑏𝑒𝑎𝑚

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑒𝑎𝑚=

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑏𝑒𝑎𝑚

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑒𝑎𝑚

=

𝜋𝑑2

4

𝑥2

=𝜋𝑑2

4𝑥2

=𝜋(1.1927𝑥)2

4𝑥2

= 1.1172

(7M)

3 Draw the shear force and B.M diagrams for a simply supported beam of length 8m and

carrying a uniformly distributed load of 10kN/m for a distance of 4m as shown below.

(13M) (Nov/Dec 2015) BTL4




5-33


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = (10 𝑥 4) = 40kN


𝑅𝐵 𝑥 8 = (10 𝑥 4 𝑥 (4

2+ 1)) = 120

𝑅𝐵 = 120

8= 15𝑘𝑁

𝑅𝐴 = 40 − 15 = 25𝑘𝑁


SF@B = -15kN

SF@D = -15kN

SF@C = -15 + (10 x 4) = 25kN

SF@A = -15 + (10 x 4) = 25kN


BM@B = 0

BM@D = +(15x3) = 45kN-m.

BM@C = +(15x7) -(10x4x(4/2)) = 25kN-m.

BM@A = +(15x8) -(10x4x((4/2)+1) = 0


The maximum bending moment occurs between the points C and D, where Shear Force is zero.




5-34

The Shear force at any section between C and D at a distance x from D is given by,

SF@X = -15 + 10 𝑥 = 0

10x = 15

x = 1.5m


The Bending Moment at any section between C and D at a distance x from D is given by,

BMmax = 15(x+3) – (10 𝑥 𝑥

2)

= 15(1.5+3) – 10(1.5)(1.5/2)

= 67.5 – 11.25

= 56.25 kN-m

(7M)

(6M)




5-35

4

Draw SFD and BMD and indicate the salient features of a beam loaded as shown below.

(13M) (May/June 2016) BTL3


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = (10 𝑥 9) + 15 = 105kN


𝑅𝐵 𝑥 7 = (15 𝑥 8.5) + (10 𝑥 7 𝑥 (7

2)) − (10 𝑥 2 𝑥 (

2

2)) = 392.5 127.5+245-20

𝑅𝐵 = 352.5

7= 50.35𝑘𝑁

𝑅𝐴 = 105 − 50.35 = 54.65𝑘𝑁


SF@D = 15kN

SF@B = 15-50.35 = -35.35kN

SF@ARHS = 15-50.35+(10 x 7) = 34.65kN

SF@ALHS = 15-50.35+(10 x 7)-54.65 = -20kN

SF@C = 15-50.35+(10 x 9)-54.65= 0


BM@D = 0

BM@B = -(15x1.5) = -22.5kN-m.




5-36

BM@A = -(15x8.5)+(50.35x7) -(10x7x(7/2)) = -20.05kN-m.

BM@C = -(15x10.5)+(50.35x9) -(10x9x(9/2)) +(54.65x2) = 0




SF@X = 15 -50.35+ 10 𝑥 = 0

10x = 35.35

x = 3.535m



BMmax = -15(x+1.5) +50.35 x – (10 𝑥 𝑥

2)

= -15(3.535+1.5) +50.35x3.535 – 10(3.535)( 3.535/2)

= -75.525 + 177.987–62.48

= 40.01 kN-m

(7M)




5-37

(6M)

5

Find the dimensions of a timber joist, span 4m to carry a brick wall 230mm thick and 3m

high if the unit weight of brickwork is 20kN/m3. Permissible bending stress in timber is

10N/mm2. The depth of the joist is twice the width. (13M) (May/June 2016) BTL3

Given:

Span of timber beam = 4m

Thickness of the brickwall = 230mm = 0.23m

Height of the brick wall=3m

Volume of the brick wall, Vb= 4x3x0.23 = 2.76m3.

Weight of the brick wall, W=2.76x20x103 = 55.2x103 N

udl, 𝑤 =55200

4

w = 13.8x103 N

depth, d=2b

Type of beam = simply supported beam.

Length of UDL, l=4m

Maximum bending stress, 𝜎𝑏 = 10 𝑁𝑚𝑚2⁄

w.k.t Bending Stress as




5-38

𝜎𝑏 =𝑀 𝑦

𝐼

(7M)


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = (13.8 𝑥 4) = 55.2kN


𝑅𝐵 𝑥 4 = (13.8 𝑥 4 𝑥 (4

2))

𝑅𝐵 = 110.4

4= 27.6𝑘𝑁

𝑅𝐴 = 55.2 − 27.6 = 27.6𝑘𝑁

Maximum bending moment occurs at mid-point.

Maximum bending moment,

BMmax = (27.6 x 2) –(13.8x2x1)

BMmax = 27.6 kN-m.

Moment of Inertia of the cross section of the beam

𝐼 = 𝑏𝑑3

12 =

𝑏(2𝑏)3

12

I = 0.667b4.

10 =27.6𝑥 106

0.667𝑏4

2𝑏

2

b3 = 4.13x106

b = 160.44mm




5-39

d=2b = 2x160.44mm

d=320.88mm.

(6M)

6

A flitched beam consists of wooden joist 10cm wide and 20 cm deep strengthened by two

steel plates 10mm thick and 20cm deep as shown below. If the maximum stress in the

wooden joist is 7 N/mm2, find the corresponding maximum stress attained in steel. Find also

the moment of resistance of the composite section. Assume Es = 2 x 105 N/mm2, Ew = 1 x 104

N/mm2 (13M) (May/June 2016) BTL3

Width of wooden joist, b2=10cm

Depth of wooden joist, d2=20cm

Width of one steel plate, b1=10cm

Depth of one steel plate, d1=10cm

Number of steel plates, n=2

Max stress in wood, σ2=7 N/mm2

E1 = 2 x 105 N/mm2,

E2 = 1 x 104 N/mm2

Moment of Inertia of the wooden joist about N.A

𝐼2 =𝑏2𝑑2

3

12=

10𝑥203

12

𝐼2 = 6666.67𝑥104𝑚𝑚4

Moment of Inertia of the two steel plates about N.A

𝐼1 = 2 𝑥𝑏1𝑑1

3

12= 2 𝑥

1𝑥203

12

𝐼1 = 1333.33𝑥104𝑚𝑚4

(7M)




5-40

Modular ratio between steel and wood is given by,

𝑚 =𝐸1

𝐸2

𝑚 =2𝑥105

1𝑥104= 20

The Equivalent moment of inertia I is given by

𝐼 = 𝑚𝐼1 + 𝐼2

𝐼 = 20𝑥1333.33𝑥104 + 6666.67𝑥104

𝐼 = 33333.2𝑥104

Moment of resistance of the composite section ins given by te equation,

𝑀 =𝜎2

𝑦𝑥 𝐼

𝑀 =7𝑥104𝑥33333.2

10𝑥10

𝑀 = 23333.24 𝑁 − 𝑚

Maximum Stress in Steel

Using the equation, 𝜎1

𝐸1=

𝜎2

𝐸2

𝜎1 =𝐸1

𝐸2𝑥 𝜎2

𝜎1 = 20 𝑥 7

𝜎1 = 140 𝑁/𝑚𝑚2

(6M)

7

A simply supported beam AB of length 5m carries point loads of 8kN, 10kN and 15kN at

1.5m, 2.5m and 4m respectively from left hand support. Draw shear force diagram and

bending moment diagram. (13M) (Nov/Dec 2016) BTL4


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 8 + 10 + 15 = 33kN




5-41


𝑅𝐵 𝑥 5 = (8 𝑥 1.5) + (10 𝑥 2.5 ) + (15 𝑥 4 ) = 97

𝑅𝐵 = 97

5= 19.4 𝑘𝑁

𝑅𝐴 = 33 − 19.4 = 13.6𝑘𝑁


SF@B = -19.4kN

SF@E = -19.4kN

SF@D = -19.4+15 = -4.4kN

SF@C = -19.4+15+10 = 5.6kN

SF@B = -19.4+15+10+8 = 13.6kN

SF@A = -19.4+15+10+8 = 13.6kN


BM@B = 0

BM@E = (19.4x1) = 19.4kN-m.

BM@D = (19.4x2.5)-(15x1.5) = 26kN-m.

BM@C = (19.4x3.5)-(15x2.5)-(10x1) = 20.4kN-m.

BM@A = (19.4x5)-(15x4)-(10x2.5)-(8x1.5) = 0

(7M)




5-42

(6M)

8

A cantilever beam AB of length 2m carries a uniform distributed load of 12kN/m over the

entire length. Find the shear stress and bending stress, of the size of the beam is 230mm x

300mm. (8M) (Nov/Dec 2016) BTL3

Given:

Span of beam, l=2m

Udl, w=12kN/m

Breadth, b=230mm,

Depth, d=300mm

w.k.t the bending stress as,

𝜎𝑏 =𝑀 𝑦

𝐼

The maximum bending stress occurs at the fixed end support.




5-43

BMmax = M = 12 x 2 x 1 = 24kN-m.

M=24 x106 N-mm.

The moment of inertia of the cross section of the beam,

𝐼 = 𝑏𝑑3

12 =

230(300)3

12

I = 517.5 x106 mm4.

𝜎𝑏 =24 x106

517.5 x106 𝑥

300

2

𝜎𝑏 = 6.956 𝑁/𝑚𝑚2

(7M) w.k.t the maximum shear stress for rectangular cross section

𝜏𝑚𝑎𝑥 = 1.5 𝑥 𝜏𝑎𝑣

w.k.t the maximum shear force from the diagram as,

Fmax = 12 x 2 = 24kN = 24 x103N.

𝜏𝑚𝑎𝑥 = 1.5 𝑥 24 𝑥103

230 𝑥 300

𝜏𝑚𝑎𝑥 = 0.5217 𝑁/𝑚𝑚2

(6M)

9

Construct the SFD and BMD for the beam shown below. (13M) (Nov/Dec 2016) BTL5





5-44

∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 25kN


𝑅𝐵 𝑥 4 = (25 𝑥 2 ) − 18.75

𝑅𝐵 = 31.25

4= 7.81 𝑘𝑁

𝑅𝐴 = 25 − 7.81 = 17.19𝑘𝑁


SF@B = -7.81 𝑘𝑁

SF@C = -7.81+25 = 17.19kN

SF@A = -7.81+25 = 17.19kN


BM@B = 0

BM@CRHS = (7.81x2) = 15.63kN-m.

BM@CLHS = (7.81x2)+18.75 = 34.37kN-m

BM@A = (7.81x4)+18.75-(25x2) = 0

(7M)




5-45

(6M)

10

Draw shear force diagram and bending moment diagram for the beam given below. (13M)

(Apr/May 2017) BTL4


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = (10 𝑥 3) + (5 𝑥 2) = 40kN




5-46


𝑅𝐵 𝑥 7 = (5 𝑥 2 𝑥 (2

2+ 5)) + (10 𝑥 3 𝑥 (

3

2)) = 105

𝑅𝐵 = 105

7= 15𝑘𝑁

𝑅𝐴 = 40 − 15 = 25𝑘𝑁


SF@B = -15kN

SF@D = -15+(5x2) = -5kN

SF@C = -15+(5x2) = -5kN

SF@A = -15+(5x2)+(10x3) = 25kN


BM@B = 0

BM@D = (15x2)- (5x2x(2/2)) = 20kN-m.

BM@C = (15x4)- (5x2x((2/2)+2)) = 30kN-m.

BM@A = (15x7)- (5x2x((2/2)+5)) - (10x3x(3/2)) = 0




SF@X = 15 -50.35+ 10 𝑥 = 0

10x = 35.35

x = 3.535m



BMmax = -15(x+1.5) +50.35 x – (10 𝑥 𝑥

2)




5-47

= -15(3.535+1.5) +50.35x3.535 – 10(3.535)( 3.535/2)

= -75.525 + 177.987–62.48

= 40.01 kN-m

(7M)

(6M)

PART * C

1

A water main of 500mm internal diameter and 20mm thick is full. The water main is of cast

iron and is supported at two points 10m apart. Find the maximum stress in the metal. The

cast iron and water weigh 72000N/m3 and 10000N/m3 respectively. (15M)BTL6

Soln:

Internal diameter, d= 500mm,

Thicknss, t=20mm,

Outer Diameter, D=500+(2x20)=540mm.

Length, L=10m.

Weigth density, Wci=72000N/m3.




5-48

Weight density, Ww=10000 N/m3.

Volume of the water, 𝑉𝑤𝑎𝑡𝑒𝑟 =𝜋𝑑2𝐿

4

𝑉𝑤𝑎𝑡𝑒𝑟 =𝜋(0.5)210

4= 1.121𝑚3

Weight of water = weight density of water x volume of water

Weight = 10000x1.121= 11210 N.

Volume of the pipe, 𝑉𝑝𝑖𝑝𝑒 =𝜋[𝐷2−𝑑2]𝐿

4

𝑉𝑤𝑎𝑡𝑒𝑟 =𝜋[(0.54)2 − (0.5)2]10

4= 0.3266𝑚3

Weight of water = weight density of water x volume of water

Weight = 72000x0.3266= 23515 N.

Total Weight = 11210+23515=34725N.

Weight per unit length , w =34725/10 = 3472.5 N/m (or) 3.4725 kN/m.

(5M) We know that the maximum bending moment for a simply supported beam loaded with udl

throughout the entire length as,

𝑀 =𝑤𝑙2

8=

3472.5(10)2

8

𝑀 = 43406.25 𝑁 − 𝑚

(5M) The maximum stress that can be obtained is given by the expression,

𝜎𝑏 =𝑀 𝑦

𝐼

𝐼 =𝜋

64[𝐷4 − 𝑑4]

𝐼 =𝜋

64[0.544 − 0.54]

𝐼 = 2.945 𝑥 10−3𝑚4

𝜎𝑏 =43406.25

2.945 𝑥 10−3 𝑥

0.54

2

𝜎𝑏 = 0.9945𝑥106 𝑁/𝑚𝑚2

(5M)

2

A beam of square section is used as a beam with one diagonal horizontal. The beam is

subjected to a shear force F, at a section. Find the maximum shear in the cross section of the

beam and draw shear stress distibution diagram for the section. (15M) (Apr/May 2017)

BTL4




5-49

Let b= Length of diagonal Ac. This is also length of diagonal BD.

The Neutal Axis of the beam passes through the diagonal AC.

Consider a level EF at a distance ‘y’ from the N.A. The Shear stress at this level is given by,

𝜏 = 𝑉𝐴�̅�

𝐼𝑏

Where, 𝐴�̅� = Moment of the shaded area about N.A

= Area of triangle BEF x Distance of CG of triangle BEF from N.A

= (1

2𝑥 2𝑥 𝑥 𝑥)(

𝑏

2−

2

3 𝑥)

= 𝑥2(𝑏

2−

2

3 𝑥)

I = Moment of Inertia about the N.A

𝐼 = 2 𝑥𝑏 𝑥 (

𝑏

2)3

12=

𝑏4

48

Substituing the values we get,

𝜏 = 𝑉𝑥2(

𝑏

2−

2

3 𝑥)

𝑏4

48 𝑥 2𝑥

𝜏 = 4𝑉𝑥(3𝑏 − 4𝑥)

𝑏4

At the top, x=0, 𝜏 = 0

At the N.A 𝑥 =𝑏

2 ,

𝜏 = 4𝑉𝑏(3𝑏 − 4𝑥)

2𝑏4

𝜏 = 2𝑉

𝑏2

(5M) Maximum Shear Stress:

Maximum shear stress wil be obtained by differetiating the above equation and equating to zero.

𝑑

𝑑𝑥⌈

4𝑉

𝑏4 (3𝑏𝑥 − 4𝑥2)⌉ = 0

4𝑉

𝑏4 (3𝑏 − 8𝑥)= 0

4𝑉

𝑏4 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑧𝑒𝑟𝑜




5-50

(3𝑏 − 8𝑥)= 0

𝑥 = 3𝑏

8

Substituting the value of ‘x’ , we get the maximum shear stress as,

𝜏𝑚𝑎𝑥 = 4𝑉

𝑏4𝑥

3𝑏

8(3𝑏 − 4

3𝑏

8)

𝜏𝑚𝑎𝑥 = 9𝐹

4𝑏2

(5M)

(5M)

3

A beam ABCD, 10m long is simply supported at B and C which are 4m apart, and

overhangs the support by 3m. The overhanging part AB carries UDL of 1kN/m and the part

CD carries UDL of 0.5kN/m. Calculate the position and magnitude of the least value of the

bending moment between the supports. Draw the S.F and B.M diagrams. (15M) BTL5


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = (1𝑥3) + (0.5𝑥3) = 4.5kN


𝑅𝐵 𝑥 4 = (0.5 𝑥 3)𝑥(4 + (3

2)) − (1𝑥3)𝑥 (

3

2) = 3.75




5-51

𝑅𝐵 = 3.75

4= 0.94 𝑘𝑁

𝑅𝐴 = 4.5 − 0.94 = 3.56𝑘𝑁


SF@D = 0

SF@BRHS = (0.5x3) = 1.5kN

SF@BLHS = (0.5x3) -0.94 = 0.56kN

SF@ARHS = (0.5x3) -0.94 = 0.56kN

SF@ALHS = (0.5x3) -0.94 -3.56 = -3kN

SF@C = (0.5x3) -0.94 -3.56+(1x3) = 0


BM@D = 0

BM@B = -(0.5x3x(3/2)) = -2.25kN-m.

BM@A = -(1x3x(3/2)) = -4.5kN-m.

BM@C = 0

(7M)




5-52

(8M)




5-53

UNIT III - TORSION

Torsion formulation stresses and deformation in circular and hollows shafts – Stepped shafts– Deflection

in shafts fixed at the both ends – Stresses in helical springs – Deflection of helical springs, carriage

springs.

PART * A

Q.No. Questions

1.

What is meant by torsional stiffness? (or) Define torsional rigidity. (Apr/May 2015)

(Nov/Dec 2016) BTL2

Torsional rigidity or stiffness of the shaft is defined as the product of modulus of rigidity G and

polar moment of inertia of the shaft.

Torsional rigidity = GJ =T 𝐿

𝜃

2

What are the uses of helical springs?(Apr/May 2015) BTL3

Railway Industry

Car suspension system

Watches.

3

The shearing stress in a solid shaft is not to exceed 40N/mm2 when the torque transmitted is

20kN-m. Determine the minimum diameter of the shaft. (Nov/Dec 2015) BTL1

𝑇 = 𝜋

16 𝜏 𝑑3

20 𝑥 106 = 𝜋

16 40 𝑑3

𝑑 = √(8𝑥106)3

d = 200mm

4

What are the various types of springs?(Nov/Dec 2015) BTL3

Helical springs

Spiral springs

Leaf springs

Disc spring or Belleville springs

5

Draw and discuss the shafts in series and parallel. (May/June 2016) BTL2

When two shafts of different diameters are connected together to form one shaft, it is then known

as composite shaft.

Series Shaft:

If the driving torque is applied at one end and the resisting torque at the other end, then the shafts

are said to be connected in series.




5-54

In such cases, each shaft transmits the same torque and the total angle of twist is equal to the sum

of the angle of twists of the two shafts.

𝜃 = 𝜃1 + 𝜃2

Parallel Shaft:

When the driving torque (T) is applied at the junction of the two shafts, and the resisting torques

T1 and T2 at the other ends of the shafts, then the shafts are said to be connected in parallel

In such cases, the angle of twist is same for both the shafts and the total torque is equal to the

sum of the torques of the two shafts.,

𝑇 = 𝑇1 + 𝑇2

6 List out the stresses induced in the helical and carriage springs.(May/June 2016) BTL3

Shear Stress

Bending Stress.

7

Define Torque. (May 2011) BTL2

When a pair of forces of equal magnitude but oppposite directions acting on a body, it tends to

twist the body. It is known as twisting moment or torsion moment or simply as torque.

Torque is equal to the product of the force applied and the distance between the point of

application of the force and the axis of the shaft.

8

What is a spring? Name the two important types of springs.(Nov/Dec 2016) BTL2

A spring is an elastic member which deflects or distorts under the action of load and regains its

original shape after the load is removed.

The two important types of springs are,

Helical springs.

Leaf springs.

9 Draw shear stress distribution of a circular section due to torque.(Apr/May 2017) BTL1




5-55

10 What is meant by spring constant? (Apr/May 2017) BTL3

Spring constant is defined as the ratio of mean or pitch diameter to the diameter of wire for the

spring.

11

What is composite shaft?BTL2

Sometimes a shaft is made up of composite section i.e., one type of shaft is sleeved over other

types of shaft. At the time of sleeving, the two shafts are joined together, that the composite shaft

behaves likes a single shaft.

12

Give the torsion formula. (May 2013) BTL1 𝑻

𝑱=

𝑮𝜽

𝑳=

𝝉

𝑹

T – Torque

J – Polar Moment of inertia

G –Modulus of rigidity

θ – Angle of twist

L –Length of shaft

τ – Shear Stress

R - Radius

13

State any four assumptions involved in simple theory of torsion. (Dec 2010) BTL2

The material of the shaft is homogeneous, perfectly elastic and obeys Hooke’s law.

Twist is uniform along the length of the shaft.

The stress does not exceed the limit of proportionality

The shaft circular in section remains circular after loading.

Strain and deformations are small.

14

Define polar modulus or torsional sectional modulus of a section. BTL3

It is the ratio between polar moment of inertia and radius of the shaft. It is denoted by Zp.

𝑍𝑝 = 𝐽

𝑅

15

Write down the expression for torque transmitted by hollow shaft. BTL3

It is the ratio between polar moment of inertia and radius of the shaft. It is denoted by Zp.

𝑇 = 𝜋

16 𝜏 ⌈

𝐷4 − 𝑑4

𝐷⌉

16 A closed coil helical springis to carry an axial load of 500N. Its mean coil diameter is to be




5-56

10 times its wire diameter. Calculate this diameter if the maximum shear stress in the

material is to be 80MPa. BTL3

𝜏 = 16 𝑊 𝑅

𝜋𝐷3

80 = 16𝑥500𝑥

10𝐷

2

𝜋𝐷3

D =12.6mm

17

State the expression for maximum shear stress and deflection of close coiled helical spring

when subjected to axial load W. BTL3

𝑇 = 𝑊𝑅 = 𝜋

16 𝜏 𝐷3

𝜏 = 16 𝑊 𝑅

𝜋𝐷3

𝛿 = 64 𝑊 𝑅3𝑛

𝐺𝐷4

18

Determine the maximum torque developed in a shaft transmitting a power of 100kW

running at 150rpm. The maximum torque is 20% more than the mean torque. BTL3

𝑃 = 2𝜋𝑁𝑇

60

100 𝑥 103 = 2𝜋𝑥150𝑥𝑇𝑚𝑒𝑎𝑛

60

𝑇𝑚𝑒𝑎𝑛 = 6.366 𝑥 103 𝑁 − 𝑚

𝑇𝑚𝑎𝑥 = 1.2 𝑥 6.366 𝑥 103 𝑁 − 𝑚

𝑇𝑚𝑎𝑥 = 7.64 𝑥 103 𝑁 − 𝑚.

19

Differentiate between close coiled and open coiled helical spring , and state the type of stress

induced in each spring due to an axial load.(May 2013) BTL1

Closed Coiled Helical Spring Open Coiled Helical Spring

The Spring wires are coiled very closely,

eaach turn is nearly at right angles to the

axis of helix.

The wires are coiled such that there is a gap

between the two consecutive turns.

Helix angle is less than 10° Helix angle is large (> 10°)

20 Write down the expression for power transmitted by a shaft. (May2013)BTL3




5-57

𝑃 = 2𝜋𝑁𝑇

60

N – Speed in rpm

T – Torque in N-m.

P – Power Transmitted in Watts.

PART * B

1

A brass tube of external diameter 80mm and internal diameter 50mm is closely fitted to a

steel rod of 50mm diameter to form a composite shaft. If a torque of 10kNm is to be resisted

by this shaft, find the maximum stresses developed in each material and the angle of twist in

2m length. Take modulus of rigidity of brass and steel as 40x103N/mm2 and 80x103N/mm2

respectively. (13M) (Apr/May 2015) BTL4

Soln:

d=50mm

D0=80mm

Di=50mm

T=10000N-m=107N-mm.

L=2m=2000mm.

Gs=40x103N/mm2

Gbr=80x103N/mm2

Polar moment of inertia of steel rod,

𝐽𝑆 =𝜋

32[𝑑4]

𝐽𝑆 =𝜋

32[504]

𝐽𝑆 = 61.34𝑥104𝑚𝑚4

𝐽𝑏𝑟 =𝜋

32[804 − 504]

𝐽𝑏𝑟 = 340.76𝑥104𝑚𝑚4

Total Torque, 𝑇 = 𝑇𝑠 + 𝑇𝑏𝑟 = 107

w.k.t, 𝜃𝑠 = 𝜃𝑏𝑟

𝜃 =𝑇𝐿

𝐺𝐽

Since length is same, 𝑇𝑠

𝐺𝑠𝐽𝑠=

𝑇𝑏𝑟

𝐺𝑏𝑟𝐽𝑏𝑟

𝑇𝑠 =𝑇𝑏𝑟𝐺𝑠𝐽𝑠

𝐺𝑏𝑟𝐽𝑏𝑟




5-58

𝑇𝑠 =𝑇𝑏𝑟𝑥40𝑥103𝑥61.34𝑥104

80𝑥103𝑥340.76𝑥104

𝑇𝑠 = 0.09𝑇𝑏𝑟

0.09𝑇𝑏𝑟 + 𝑇𝑏𝑟 = 107

𝑇𝑏𝑟 = 9.17𝑥106𝑁 − 𝑚𝑚

𝑇𝑠 = 8.25𝑥105𝑁 − 𝑚𝑚

(6M)

Maximum stress developed is given by the equation,

𝜏𝑠 =𝑇𝑠

𝐽𝑠

𝑑𝑠

2

𝜏𝑠 =8.25𝑥105

61.34𝑥104

50

2

𝜏𝑠 = 33.62 𝑁/𝑚𝑚2

𝜏𝑏𝑟 =𝑇𝑏𝑟

𝐽𝑏𝑟

𝑑𝑏𝑟

2

𝜏𝑏𝑟 =9.17𝑥106

340.76𝑥104

80

2

𝜏𝑏𝑟 = 107.64 𝑁/𝑚𝑚2

Angle of twist,

𝜃𝑠 =𝑇𝑠𝐿𝑠

𝐺𝑠𝐽𝑠

𝜃𝑠 =8.25𝑥105𝑥2000

40𝑥103𝑥61.34𝑥104

𝜃𝑠 = 0.6724 𝑟𝑎𝑑𝑖𝑎𝑛𝑠

(7M)

2

A close coiled helical spring made up of 10mm diameter steel wire has 15 coils of 100mm

mean diameter. The spring is subjected to an axial load of 100N. Calculate: (i) the maximum

shear stress induced, (ii)the deflection, and (iii) stiffness of the spring. Take modulus of

rigidity of the material of the spring as 8.16x104 N/mm2. (13M) (Apr/May 2015) BTL3

Given:

d=10mm

n=15 coils

D=100mm

W=100N

G=8.16x104 N/mm2.




5-59

To find

(i) The maximum Shear Stress Induced:

w.k.t the maximum shear stress induced as,

𝜏 =16𝑊𝑅

𝜋𝑑3

𝜏 =16𝑥100𝑥50

𝜋(10)3

𝜏 = 25.46 𝑁/𝑚𝑚2

(6M)

(ii) The deflection:

w.k.t the deflection of the spring as,

𝛿 =64𝑊𝑅3𝑛

𝐺𝑑4

𝛿 =64𝑥100𝑥(50)3𝑥15

8.16𝑥104𝑥(10)4

𝛿 = 14.70𝑚𝑚

(ii) The Stiffness of the spring:

w.k.t the stiffness of the spring as,

𝑠 =𝐺𝑑4

64𝑅3𝑛

𝑠 =8.16𝑥104𝑥(10)4

64𝑥(50)3𝑥15

𝑠 = 0.425𝑥108𝑁/𝑚𝑚

(7M)

3

A hollow shaft of external diameter 120mm transmitts 300kW power at 200rpm. Determine

the maximum internal diameter if the maximum stress in the shaft is not to exceed

60N/mm2 (13M) (Nov/Dec 2015) BTL4 Given:

External diameter, D0=120mm,

Power, P=300kW

Speed, N=200rpm

𝜏max = 60 𝑁/𝑚𝑚2

w.k.t,

𝑃𝑜𝑤𝑒𝑟, 𝑃 = 2𝜋𝑁𝑇

60

300𝑥103 = 2𝜋(200)𝑇

60




5-60

𝑇 = 14323.94 𝑁 − 𝑚

63M) w.k.t the torque transmitted by the shaft, TH as

𝑇𝐻 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − 𝐷𝑖

4

𝐷0⌉

14323.94 =𝜋

16 𝑥 60𝑥 106 𝑥 ⌈

0.124 − 𝐷𝑖4

0.12⌉

⌊0. 124 − 𝐷𝑖4⌋ =

0.12 𝑥 14323.94 𝑥 16

𝜋 𝑥 60 𝑥 106

⌊2.07𝑥10−4 − 𝐷𝑖4⌋ = 145.9𝑥10−6

⌊ 𝐷𝑖4⌋ = 0.62𝑥10−4

Di = 88.73mm

(7M)

4

A closely coiled helical spring is to have a stiffness of 1.5N/mm of compression under a

maximum load of 60N. The maximum shear stress produced in the wire of the spring is 125

N/mm2. The Solid length of the spring is 50mm. Find the diameter of coil, diameter of wire

and number of coils.Take C=4.5x104N/mm2 (13M) (Apr/May 2018) BTL5 Given:

W=60N

G=4.5x104 N/mm2.

Solid length= nd =50mm.

𝜏max = 125 𝑁/𝑚𝑚2 𝑠 = 1.5 𝑁/𝑚𝑚2


𝜏 =16𝑊𝑅

𝜋𝑑3

125 =16𝑥60𝑥𝑅

𝜋𝑑3

𝑅 = 0.4089𝑑3


𝑠 =𝐺𝑑4

64𝑅3𝑛




5-61

1.5 =4.5x104x𝑑4

64𝑥(0.4089𝑑3)3𝑥𝑛

1.5 =4.5x104x𝑑4

64𝑥0.06836𝑑9𝑥𝑛

𝑑5𝑥𝑛 =4.5x104

15𝑥64𝑥0.06836

𝑑4𝑥50 =4.5x104

15𝑥64𝑥0.06836

𝑑4𝑥50 = 685.70

𝑑4 = 13.714

d=1.924mm

(6M)

𝑅 = 0.4089(1.924)3

𝑅 = 2.912𝑚𝑚

Solid length nd=50mm

n=50/d

n=50/1.924

n=26 coils

(7M)

5

A solid circular shaft 20mm in diameter is to be replaced by a hollow shaft the ratio of

external diameter to internal diameter being 5:3. Determine the size of the hollow shaft if

maximum shear stress is to be the same as that of a solid shaft. Also find the percentage

savings in mass. (13M) (May/June 2016) BTL3

Soln:

Let

D0 = Outer diameter of the hollow shaft

Di = Inside diameter of the hollow shaft

D'= Diameter of the solid shaft

P= Power transmitted by hollow shaft. or by solid shaft

N= Speed of each shaft

τ = Maximum shear stress induced in each shaft.

D=20mm




5-62

Di=0.6 Do

Torque transmitted by the solid shaft,

𝑇 =𝜋

16𝜏𝐷3=======>(1)

Torque transmitted by the hollow shaft,

𝑇 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − 𝐷𝑖

4

𝐷0⌉

𝑇 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − (0.6𝐷𝑜)4

𝐷0⌉

𝑇 =𝜋

16 𝑥 𝜏 𝑥 0.8704𝑥𝐷0

3 =======>(2)

(6M) Since torque transmitted by the solid shaft is equal to hollow shaft,

𝜋

16𝜏𝐷3 =

𝜋

16 𝑥 𝜏 𝑥 0.8704𝑥𝐷0

3

𝐷3 = 0.8704𝑥𝐷03

𝐷 = 0.9547𝑥𝐷𝑜

20𝑥10−3 = 0.9547𝑥𝐷𝑜

𝐷𝑜 = 20.94𝑥10−3m

𝐷𝑖 = 0.6𝑥 20.94𝑥10−3m

𝐷𝑖 = 12.56𝑥10−3m

Weight of the solid shaft,

𝑊𝑠 = 𝜌 𝑥 𝑔 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡

𝑊𝑠 = 𝜌 𝑥 𝑔 𝑥 𝜋

4𝑥 𝐷2 𝑥 𝐿

Weight of the hollow shaft,

𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡

𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 (𝜋

4[𝐷0

2 − 𝐷𝐼2]𝐿)


4[𝐷0

2 − (0.6𝐷0)2]𝐿)




5-63

𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 𝜋

4𝑥 0.64 𝑥 𝐷0

2 𝑥 𝐿

Dividing the weights of solid and hollow shaft,

𝑊𝑠

𝑊𝐻=

𝜌 𝑥 𝑔 𝑥 𝜋



4𝑥 0.64 𝑥 𝐷0

2 𝑥 𝐿

𝑊𝑠

𝑊𝐻=

𝐷2

0.64 𝑥 𝐷02

𝑊𝑠

𝑊𝐻=

(0.9547𝑥𝐷𝑜)2

0.64 𝑥 𝐷02

𝑊𝑠

𝑊𝐻=

0.9114𝑥 𝐷02

0.64 𝑥 𝐷02

𝑊𝑠

𝑊𝐻= 1.424

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 =𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡 − 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 ℎ𝑜𝑙𝑙𝑜𝑤 𝑠ℎ𝑎𝑓𝑡

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 =1.424 𝑊𝐻 − 𝑊𝐻

1.424 𝑊𝐻𝑥 100

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 = 29.77%

(7M)

6

A closely coiled helical spring made from round steel rod is required to carry a load of

1000N for a stress of 400 MN/m2, the spring stiffness being 20N/mm. The diameter of the

helix is 100mm and G for the material is 80GN/m2. Calculate (1) the diamter of the wire and

(2) the number of turns required for the spring. (13M) (May/June 2016) BTL3

Given:

s=20N/mm

D=100mm

𝜏 = 400 𝑀𝑁/𝑚2 W=1000N

G=80GN/m2.

Soln:


𝜏 =16𝑊𝑅

𝜋𝑑3




5-64

40𝑥106 =16𝑥1000𝑥50𝑥10−3

𝜋𝑑3

𝑑3 =16𝑥1000𝑥50𝑥10−3

𝜋𝑥40𝑥106

𝑑3 = 6.366𝑥10−6

𝑑 = 0.0185𝑚

(6M)


𝑠 =𝐺𝑑4

64𝑅3𝑛

𝑛 =𝐺𝑑4

64𝑅3𝑠

𝑛 =80x109x(0.0185)4

64𝑥(50𝑥10−3)3𝑥20𝑥103

𝑛 =80x109x1.17x10−7

64𝑥1.25x10−4𝑥20𝑥103

𝑛 = 58.5 ≅ 59𝑐𝑜𝑖𝑙𝑠

(7M)

7

A hollow shaft is to transmit 300kW power at 80rpm. If the shear stress is not to exceed

60N/mm2 and the internal diameter is 0.6 of the external diameter. Find the external and

internal diameters assuming that the maximum torque is 1.4 times the mean. (13M)

(Nov/Dec 2017) BTL5

Internal diameter, d=0.6D,

Power, P=300kW

Speed, N=80rpm

𝜏max = 60 𝑁/𝑚𝑚2 Tmax=1.4 xTmean

w.k.t,


60

300𝑥103 = 2𝜋(80)𝑇

60

𝑇 = 35816.61 𝑁 − 𝑚

(1M) w.k.t the torque transmitted by the shaft, TH as




5-65

𝑇𝐻 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − 𝐷𝑖

4

𝐷0⌉

35816.61 =𝜋

16 𝑥 60𝑥 106 𝑥 ⌈

𝐷4 − (0.6𝐷)4

𝐷⌉

35816.61 =𝜋

16 𝑥 60𝑥 106 𝑥 ⌈

0.8704𝐷4

𝐷⌉

⌊0.8704𝐷3⌋ = 3.0407𝑥10−3

𝐷3 = 3.4934𝑥10−3

D = 0.1517m (or) 151.7mm

d=0.6D = 0.6x0.1517=0.09102

d = 0.09102m (or) 91.02mm

(7M)

8

A solid shaft has to transmit the power 105kW at 2000 rpm. The maximum torque

transmitted in each revolution exceeds the mean by 36%. Find the suitable diamter of the

shaft, if the shear stress is not to exceed 75N/mm2 and maximum angle of twist is 1.5° in a

length of 3.3m and G= 0.8x105N/mm2 (13M)(Nov/Dec 2016) BTL4

Given:

Power, P=105kW

Speed, N=2000rpm

Tmax=1.36 Tmean

𝜏max = 75 𝑁/𝑚𝑚2

Angle of twist, θ = 1.5°, 𝜃 = 1.5 𝑥 180

𝜋= 0.02617 rad

Length, L=3.3m

G= 0.8x105N/mm2


60

105𝑥103 = 2𝜋(2000)𝑇

60

𝑇 = 501.33 𝑁 − 𝑚

𝑇𝑚𝑎𝑥 = 1.36 𝑇𝑚𝑒𝑎𝑛

𝑇𝑚𝑎𝑥 = 1.36 𝑥 501.33

𝑇𝑚𝑎𝑥 = 681.81 𝑁 − 𝑚




5-66

(6M)

Case (i): Diameter for strength:

𝑇 =𝜋

16 𝑥 𝜏 𝑥 𝐷𝑠

3

681.81 =𝜋

16 𝑥 75𝑥 106 𝑥 𝐷𝑠

3

𝐷𝑠3 =

16𝑥 681.81

𝜋𝑥 75𝑥106

𝐷𝑠3 = 46.29 𝑥10−6

𝐷𝑠 = 0.0359𝑥10−3 𝑚

Case (ii): Diameter for stiffness:

𝑇

𝐽=

𝐺𝜃

𝐿

681.81 𝜋

32𝐷𝑠

4=

8𝑥 1010𝑥 0.026179

3.3

𝐷𝑠4 =

681.81𝑥3.3𝑥32

𝜋𝑥 8𝑥1010𝑥0.026179

𝐷𝑠4 = 10.94 𝑥10−6

𝐷𝑠 = 0.0575𝑥10−3 𝑚

The required shaft diameter is the greaterst of the two values,

𝐷𝑠 = 0.0575𝑥10−3 𝑚 (𝑜𝑟)57.5 𝑚𝑚

(7M)

9

A laminated spring carries a central load of 5200N and it is made of ‘n’ number of plates,

80mm wide, 7mm thick and length 500mm. Find the number of plates, if the maximum

deflection is 10mm. Let E = 2x105N/mm2 (13M) (Nov/Dec 2016) BTL3

Given:

W=5200N

b=80mm

t=7mm

L=500mm

δ=10mm

E = 2x105N/mm2




5-67

w.k.t stress,

𝜎 =3𝑊𝑙

2𝑛𝑏𝑡2

𝜎 =3𝑥5200𝑥500𝑥10−3

2𝑛𝑥80𝑥10−3(7𝑥10−3)2

𝜎 =994.89𝑥106

𝑛

(6M) The equation for deflection is,

𝛿 =𝜎𝑙2

4𝐸𝑡

10𝑥10−3 =994.89𝑥106𝑥(500𝑥10−3)2

𝑛𝑥4𝑥2𝑥1011𝑥7𝑥10−3

𝑛 =994.89𝑥106𝑥(500𝑥10−3)2

10𝑥10−3𝑥4𝑥2𝑥1011𝑥7𝑥10−3

𝑛 =248.72𝑥106

560𝑥105

𝑛 = 4.44 ≅ 5 𝑐𝑜𝑖𝑙𝑠

(7M)

10

A closed coiled helical spring is to be made out of 5mm diameter wire 2m long so that it

deflects by 20mm under an axial load of 50N. Determine the mean diameter of the coil.

Take C = 8.1x104N/mm2 (13M)(Nov/Dec 2016) BTL4

Given:

d=5mm

L=2m

δ=20mm

W=50N

C=8.1x104N/mm2.

w.k.t the deflection of the spring as,

𝛿 =64𝑊𝑅3𝑛

𝐺𝑑4

20𝑥10−3 =64𝑥50𝑥𝑅3𝑥 𝑛

8.1𝑥1010𝑥(5𝑥10−3)4

𝛿 = 14.70𝑚𝑚

Solid Length, L=nd




5-68

2=nx5𝑥10−3

n= 400

(7M)

20𝑥10−3 =64𝑥50𝑥𝑅3𝑥 400

8.1𝑥1010𝑥(5𝑥10−3)4

𝑅3 =8.1𝑥1010𝑥20𝑥10−3𝑥(5𝑥10−3)4

64𝑥50𝑥 400

𝑅3 = 7.91𝑥10−6

𝑅 = 9.24𝑚𝑚

𝐷 = 18.48𝑚𝑚

(6M)

PART * C

1

A hollow shaft having an inside diameter 60% of its outer diameter, is to replace a solid

shaft transmitting in the same power at the same speed. Calculate percentage saving in

material, if the material to be is also the same. (15M) (Apr/May 2017) BTL4

Soln:

Let

D0 = Outer diameter of the hollow shaft

Di = Inside diameter of the hollow shaft

D'= Diameter of the solid shaft

P= Power transmitted by hollow shaft. or by solid shaft

N= Speed of each shaft

τ = Maximum shear stress induced in each shaft.

D=20mm

Di=0.6 Do

Torque transmitted by the solid shaft,

𝑇 =𝜋

16𝜏𝐷3=======>(1)

Torque transmitted by the hollow shaft,

𝑇 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − 𝐷𝑖

4

𝐷0⌉

𝑇 =𝜋

16 𝑥 𝜏 𝑥 ⌈

𝐷04 − (0.6𝐷𝑜)4

𝐷0⌉

𝑇 =𝜋

16 𝑥 𝜏 𝑥 0.8704𝑥𝐷0

3 =======>(2)

Since torque transmitted by the solid shaft is equal to hollow shaft,




5-69

𝜋

16𝜏𝐷3 =

𝜋

16 𝑥 𝜏 𝑥 0.8704𝑥𝐷0

3

𝐷3 = 0.8704𝑥𝐷03

𝐷 = 0.9547𝑥𝐷𝑜

20𝑥10−3 = 0.9547𝑥𝐷𝑜

𝐷𝑜 = 20.94𝑥10−3m

𝐷𝑖 = 0.6𝑥 20.94𝑥10−3m

𝐷𝑖 = 12.56𝑥10−3m

(7M) Weight of the solid shaft,

𝑊𝑠 = 𝜌 𝑥 𝑔 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡

𝑊𝑠 = 𝜌 𝑥 𝑔 𝑥 𝜋


Weight of the hollow shaft,

𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡


4[𝐷0

2 − 𝐷𝐼2]𝐿)


4[𝐷0

2 − (0.6𝐷0)2]𝐿)

𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 𝜋

4𝑥 0.64 𝑥 𝐷0

2 𝑥 𝐿

Dividing the weights of solid and hollow shaft,

𝑊𝑠

𝑊𝐻=




4𝑥 0.64 𝑥 𝐷0

2 𝑥 𝐿

𝑊𝑠

𝑊𝐻=

𝐷2

0.64 𝑥 𝐷02

𝑊𝑠

𝑊𝐻=

(0.9547𝑥𝐷𝑜)2

0.64 𝑥 𝐷02




5-70

𝑊𝑠

𝑊𝐻=

0.9114𝑥 𝐷02

0.64 𝑥 𝐷02

𝑊𝑠

𝑊𝐻= 1.424

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 =𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡 − 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 ℎ𝑜𝑙𝑙𝑜𝑤 𝑠ℎ𝑎𝑓𝑡

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 =1.424 𝑊𝐻 − 𝑊𝐻

1.424 𝑊𝐻𝑥 100

% 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 = 29.77%

(8M)

2

Derive a relation for deflection of a closely coiled helical spring subjected o an axial

compressive load ‘W’. (15M) (Apr/May 2017) BTL4

Close-coiled helical springs are the springs in which helix angle is very small or in other words

the pitch between two adjacent turns is small. A close-coiled helical spring carrying an axial load

is shown below. As the helix angle in case of close-ooiled helical springs are small, hence the

bending effect on the spring is ignored and we assume that the coils of a close-coiled helical

springs are to stand purely torsional stresses.

Let

d = Diameter of spring wire

p = Pitch ofths helical spring

n = Number of coils

R = Mean radius of spring ooil

W = Axial load on spring

C = Modulus of rigidity

τ = Max. shear stress induced in the Wire

θ = Angle oftwist in spring wire, and

δ = Deflection of spring due to axial load

l= Length of wire.

Now twisting moment on the wire,




5-71

𝑇 = 𝑊𝑥 𝑅 ==> (1)

But twisting moment is also given by,

𝑇 =𝜋

16𝜏𝑑3==> (2)

Equating Eqns (1) & (2),

𝑊𝑥 𝑅 =𝜋

16𝜏𝑑3

𝜏 =16𝑊𝑅

𝜋𝑑3

(8M) This Equation gives the maximum shear stress induced in the wire.

Expression for deflection of the spring:

Now length of one coil = πD or 2πR

Total length of the wire = Length of one coil x No.of coils,

L =2πRn

Strain energy stored in the spring,

𝑈 =𝜏2

4𝐶. 𝑉𝑜𝑙𝑢𝑚𝑒

𝑈 = (16𝑊𝑅

𝜋𝑑3)

2

𝑥1

4𝐶𝑥(

𝜋

4𝑑22πRn)

𝑈 =32𝑊2𝑛𝑅

𝐶𝑑4

3

Work done on the spring = Average load x deflectionm

=1

2𝑊𝛿

Equating the work done on spring to the energy stored, we get

1

2𝑊𝛿 =

32𝑊2𝑛𝑅

𝐶𝑑4

3

𝛿 =64𝑊𝑛𝑅

𝐶𝑑4

3

(7M)

3 Two solid shafts AB and BC of aluminium and steel respectively are rigidly astened

together at B and attached to two rigid supports at A and C. Shaft AB is 7.5cm in diameter




5-72

and 2m in length. Shaft BC is 5.5cm in diameter and 1m in length. A torque of 2000N-cm is

applied at the junction B. Compute the maximum shearing stress in each material. What is

the angle of twist at the junction? Take modulus of rigidity of the materials as

Gal=0.3x105N/mm2 and Gst=0.9x105N/mm2 (15M) BTL4

Given:

L1=2m

D1=7.5cm

G1=0.3x105N/mm2 L2=1m

D2=5.5cm

G2=0.9x105N/mm2 T=20000N-cm

The torque is appled at junction B, hence angle of twist in shaft AB and in shaft BC will be same.

Using the equation.

𝑇

𝐽=

𝐺𝜃

𝐿

For the shaft AB,

𝑇1

𝐽1=

𝐺1𝜃1

𝐿1

𝜃1 =𝑇1𝐿1

𝐺1𝐽1

𝜃1 =𝑇1𝑥2000𝑥32

𝜋𝑥754𝑥0.3𝑥105

For the shaft BC,

𝑇2

𝐽2=

𝐺2𝜃2

𝐿2

𝜃2 =𝑇2𝐿2

𝐺2𝐽2

𝜃2 =𝑇2𝑥1000𝑥32

𝜋𝑥554𝑥0.9𝑥105

w.k.t

𝜃1 = 𝜃2

𝑇1𝑥2000𝑥32

𝜋𝑥754𝑥0.3𝑥105=

𝑇2𝑥1000𝑥32

𝜋𝑥554𝑥0.9𝑥105

(8M)

Solving,




5-73

𝑇1 = 0.576𝑇2

w.k.t

0.576𝑇2 + 𝑇2 = 200000

1.576𝑇2 = 200000

𝑇2 = 200000

1.576

𝑇2 = 126900 𝑁 − 𝑚𝑚

𝑇1 + 𝑇2 = 200000

𝑇1 = 200000 − 126900

𝑇1 = 73100 𝑁 − 𝑚𝑚

w.k.t

𝑇

𝐽=

𝜏

𝑅

For shaft AB

𝑇1

𝐽1=

𝜏1

𝑅1

𝜏1 =𝑇1𝑅1

𝐽1

𝜏1 =73100𝑥 37.5

𝜋

32𝑥754

𝜏1 =73100𝑥 37.5𝑥32

𝜋𝑥754

𝜏1 = 0.882𝑁/𝑚𝑚2

For shaft BC 𝑇2

𝐽2=

𝜏2

𝑅2

𝜏2 =𝑇2𝑅2

𝐽2




5-74

𝜏2 =126900𝑥 27.5

𝜋

32𝑥554

𝜏2 =126900𝑥 27.5𝑥32

𝜋𝑥554

𝜏2 = 3.884𝑁/𝑚𝑚2

(7M)




5-75

UNIT IV - DEFLECTION OF BEAMS

Double Integration method – Macaulay’s method – Area moment method for computation of slopes and

deflections in beams - Conjugate beam and strain energy – Maxwell’s reciprocal theorems.

PART * A

Q.No. Questions

1.

What are the advantages of Macaulay’s method over other methods for the calculation of

slope and deflection? (Apr/May 2015) BTL2

In Macaulay’s method a single equation is formed for all loading on a beam, the equation is

constructed in such a way that the constant of integration apply to all portions of the beam. This

method is also called as method of singularity functions.

2

In a cantilever beam, the measured deflection at free end was 8mm when a concentrated

load of 12kN was applied at its mid-span. What will be the defection at mid-span when the

same beam carries a concentrated load of 7kN at the free end? (Apr/May 2015) BTL3

w.k.t Deflection at the free end when loaded at mid span as

𝑦 = 5𝑊𝐿3

48𝐸𝐼

8 = (5)(12)𝐿3

48𝐸𝐼

𝐸𝐼

𝐿3= 0.15625

Deflection at the mid span when loaded at free end as

𝑦 = 𝑊𝐿3

24𝐸𝐼

𝑦 = 7 𝐿3

24𝐸𝐼

𝑦 = 7

24𝑥0.15625

𝑦 = 1.867 𝑚𝑚

3

What are the methods of determining slope and deflection at a section in a loaded beam?

(Nov/Dec 2015), (Nov/Dec 2016)BTL1

Double integration method,

Moment area method,

Macaulay’s method,

Conjugate beam method.

4 What is the equation used in the case of double integration method?(Nov/Dec 2015) BTL3




5-76

The Bending moment at a point,

𝑀 = 𝐸𝐼𝑑2𝑦

𝑑𝑥2

Integrating the above equation,

∫ 𝑀 = 𝐸𝐼𝑑𝑦

𝑑𝑥

Again integrating the above equation once again,

∬ 𝑀 = 𝐸𝐼 𝑦

5

How the deflection and slope is calculated for the cantilever beam by conjugate beam

method? (May/June 2016) BTL2

The slope at any section of a cantilever beam, relative to the original axis of the beam is equal to

the shear in the conjugate beam at the corresponding section.

The deflection at any given section of a cantilever beam, relative to the original axis of the beam

is equal to the bending moment in the conjugate beam at the corresponding section.

6

State the Maxwell’s reciprocal. (May/June 2016), (Nov/Dec 2016) BTL3

Maxwell's reciprocal theorem state that “In a linearly elastic structure, the deflection at any point.

A due to a load applied at some other point B will be equal to the deflection at B when the same

load is applied at A”

7

Write down the equation for the maximum deflection of a cantilever beam carrying a

central point load ‘W’. (Apr/May 2017) BTL1

The maximum deflection of the cantilever beam happens at the free end when the beam is loaded

at the mid-span and is given as,

𝑦𝑚𝑎𝑥 = 5𝑊𝐿3

48𝐸𝐼

8 Draw conjugate beam for a double side over hanging beam. (Apr/May 2017) BTL3

9

Explain the theorem for conjugate beam method. BTL3

Theorem I: The slope at any section of a loaded beam, relative to the original axis of the beam is

equal to the shear in the conjugate beam at the corresponding section.

Theorem II: The deflection at any given section of a loaded beam, relative to the original axis of

the beam is equal to the bending moment in the conjugate beam at the corresponding section.

10

What is the deflection at the free end of a cantilever beam of span ‘l’ and carrying a point

load ‘W’ at the free end? BTL3

Deflection at the free end of the cantilever beam is given as,

𝑦𝑓𝑟𝑒𝑒 = 𝑊𝑙3

3𝐸𝐼

11 What is deflection of beam? BTL3

Deflection of the beam is the vertical distance at a point between the elastic curve of the deflected

beam to the unloaded neutral axis of the actual beam.

12

State Mohr’s theorems. BTL3

Mohr’s Theorem I: The change of slope between any two points is equal to the net area of the

bending moment diagram between these points divided by EI .

Mohr’s Theorem II: The total deflection between any two points is equal to the moment of the

area of the bending moment diagram between these two points about the reference line divided by

EI.




5-77

13

Write down the formula used to find the deflection of the beam by moment area

method.(May 2010)BTL3

The change in deflection of the beam at between two sections is expresseed as,

𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐵𝑀𝐷 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒𝑠𝑒 𝑡𝑤𝑜 𝑝𝑜𝑖𝑛𝑡𝑠

𝐸𝐼

14

A cantilever beam of span 2m is carrying a point load of 20kN at free end. Calculate the

slope at the free end. Assume EI = 12x 103 kN-m2. BTL3

By Area moment method:

Slope at support A,

𝜃𝐴 =𝐴𝑟𝑒𝑎 𝑜𝑓 𝐵𝑀𝐷 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵

𝐸𝐼

𝜃𝐴 =𝑎

𝐸𝐼=

0.5 𝑥 2 𝑥 40

12 𝑥103

𝜃𝐴 = 0.0033 𝑟𝑎𝑑

15 What is slope of beam? BTL3

Slope of the beam is the angle between deflected beam to the actual beam at the same point.

16

A simply supported beam of span 3m is subjected to a central load of 10kN. Find the

maximum slope and deflection of the beam. Take I = 12x 106 mm4 and E=200Gpa.

The maximum slope of the simply supported beam carrying a central load is given by,

𝜃𝑚𝑎𝑥 =𝑊𝑙2

16𝐸𝐼

𝜃𝑚𝑎𝑥 =10𝑥 103𝑥30002

16𝑥2.4𝑥1012

𝜃𝑚𝑎𝑥 = 0.00234 𝑟𝑎𝑑

17

Write down the boundary conditions for a cantilever beam to find out the equations for

deflection and slope.BTL3

When x=L, 𝑑𝑦

𝑑𝑥= 0

When x=L, 𝑦 = 0

18

State Castigliano’s theorem.BTL3

In any beam or truss subjected to any load system, the deflection at any point r is given by the

partial differential coefficient of the total strain energy stored with respect to force Pr acting at

that point r in the direction in which the deflection is desired.

19

How do you determine the maximum deflection in a simply supported beam?

(May2012)BTL3

The maximum deflection occurs when the slope is zero. The position of the maximum deflection

is found out by equating the slope equation to zero. Then the value of ‘x’ is substituted in the

deflection equation to calculate the maximum deflection.




5-78

20

Give the relationship between intensity of load, shear force, bending moment, slope and

deflection in a beam.BTL3

𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 = y

𝑆𝑙𝑜𝑝𝑒 =dy

dx

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 = 𝐸𝐼𝑑2𝑦

𝑑𝑥2

𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒 = 𝐸𝐼𝑑3𝑦

𝑑𝑥3

𝐿𝑜𝑎𝑑 = 𝐸𝐼𝑑4𝑦

𝑑𝑥4

PART * B

1

A horizontal beam of uniform section and 7m long is simply supported at its ends. The

beam is subjected to uniformly distributed load of 6kN/m over a length of 3m from the left

end and a concentrated load of 12kN at 5m from the left end. Find the maximum deflection

in the beam using Macaulay’s method. (13M) (Apr/May 2015) BTL4


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 12 + (6 𝑥 3) = 30kN


𝑅𝐵 𝑥 7 = (12 𝑥 5) + (6 𝑥 3 𝑥 3

2 ) = 87

𝑅𝐵 = 87

7= 12.42𝑘𝑁

𝑅𝐴 = 30 − 12.42 = 17.58𝑘𝑁

Consider a section X in the last part of the beam at a distance ‘x’ from the left most support. The

B.M at this section is given by,




5-79

𝐸𝐼 𝑑2𝑦

𝑑𝑥2 = 𝑅𝐴 𝑥 − 6

𝑥2

2⋮ +6(𝑥 − 3)

(𝑥 − 3)

2⋮ −12(𝑥 − 5)

(𝑥 − 5)

2

= 17.58 𝑥 − 3𝑥2 ⋮ +3(𝑥 − 3)2 ⋮ −6(𝑥 − 5)2

Integrating the above equation we get,

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 17.58

𝑥2

2 − 3

𝑥3

3+ 𝐶1 ⋮ +3

(𝑥 − 3)3

3⋮ −6

(𝑥 − 5)3

3 → 𝐸𝑞 (1)

Integrating again we get,

𝐸𝐼 𝑦 = 17.58 𝑥3

6 −

𝑥4

4+ 𝐶1𝑥 + 𝐶2 ⋮ +

(𝑥 − 3)4

4⋮ −2

(𝑥 − 5)4

4 → 𝐸𝑞 (2)

Where 𝐶1 𝑎𝑛𝑑 𝐶2 are constants of integration whose values can be obtained from the boundary

conditions,

(i) at x=0, y=0 and (ii) at x=7, y=0

Substituting condn (i) at x=0, y=0 into Eq (2) upto first dotted line(as x=0 lies in the first

part of the beam) we get,

𝐸𝐼(0) = 17.58 03

6 −

04

4+ 𝐶10 + 𝐶2

𝐶2 = 0

Substituting condn (ii) at x=7, y=0 into complete Eq (2) (as x=7 lies in the last part of the

beam) we get,

𝐸𝐼(0) = 17.58 73

6 −

74

4+ 𝐶17 + 0 ⋮ +

(7 − 3)4

4⋮ −2

(7 − 5)4

4

𝐸𝐼(0) = 1004.99 − 600.25 + 𝐶17 ⋮ +64 ⋮ −8

𝐶1 = −65.82

Substituting the values of 𝐶1 and 𝐶2 into Eq(1) and Eq(2) we get,

Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 17.58

𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥 − 3)3 ⋮ −2(𝑥 − 5)3

Deflection Equation:

𝐸𝐼 𝑦 = 17.58 𝑥3

6 −

𝑥4

4− 65.82𝑥 ⋮ +

(𝑥 − 3)4

4⋮ −2

(𝑥 − 5)4

4




5-80

(6M)

Position of Maximum Deflection:

The maximum deflection is likely to happen between C and D. For the maximum deflection the

slope 𝑑𝑦

𝑑𝑥 should be zero. Hence equating the slope equation given by Eq (1) upto second dotted

line to zero, we get

0 = 17.58 𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥 − 3)3

0 = 17.58 𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥3 − 27 + 27𝑥 − 9𝑥2)

Solving , 𝑥 = 3.5476m

Hence the maximum deflection will be at a distance of 3.54m from support A.

Maximum Deflection:

Substituting x=3.54 into Eq (2) upto second dotted line, (as x=3.54 lies in the second part of the

beam) we get the maximum deflcetion as,

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 17.58 (3.54)3

6 −

(3.54)4

4− 65.82(3.54) ⋮ +

(3.54 − 3)4

4

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 129.98 − 39.26 − 233 + 0.021

𝑦 𝑚𝑎𝑥 = −142.26

𝐸𝐼

(7M)

2

A cantilever of span 4m carries a uniformly distributed load of 4kN/m over a length of 2m

from the fixed end and a concentrated load of 10kN at the free end. Determine the slope and

deflection of the cantilever at the free end using conjugate beam method. Assume EI is

uniform throughout. (13M) (Apr/May 2015) BTL3


BM@C = 0

BM@B = -(10x2) = -20kN-m.

BM@A = -(10x4) -(4x2x(2/2)) = -48 kN-m.




5-81

(6M)

To Draw the Conjugate Beam:

Now construct conjugate beam A’C’ which is free at A’ and fixed at C’ by dividing the B.M at

any section by EI. The loading on the conjugate beam will be negative.

Then, according to the conjugate beam theorem,

Deflection at the free end = B.M at C’ for the conjugate beam,

𝑦𝑐 = 𝐵. 𝑀 𝑎𝑡 𝐶′

𝑦𝑐 = (1

2 𝑥 2 𝑥

20

𝐸𝐼) 𝑥 (

2

3𝑥 2) + ( 2 𝑥

20

𝐸𝐼) 𝑥 (

2

2+ 2) + (

1

3 𝑥 2 𝑥

28

𝐸𝐼) 𝑥 (

3

4 𝑥 2)

𝑦𝑐 = (20

𝐸𝐼) 𝑥 (

4

3) + (

40

𝐸𝐼) 𝑥(3) + (

56

𝐸𝐼) 𝑥 (

1

2)

𝑦𝑐 = (80

3𝐸𝐼) + (

120

𝐸𝐼) + (

28

𝐸𝐼)

𝑦𝑐 =174.67

𝐸𝐼

(7M)

3

A beam 6 long, simply supported at its ends, is carrying a point load of 50kN at its centre.

The moment of inertia of the beam is given as equal to78x106mm4. If E for the material of

the beam = 2.1x105 N/mm2, calculate: (i) deflection at the centre of the beam and (ii) slope

at the supports. (13M) (Nov/Dec 2015) BTL4




5-82

Given:

I = 78x106mm4

E= 2.1x105 N/mm2


∑ 𝐹𝐻 = 0



𝑅𝐵 𝑥 6 = (50 𝑥 3 ) = 150

𝑅𝐵 = 150

6= 25𝑘𝑁

𝑅𝐴 = 50 − 25 = 25𝑘𝑁



𝐸𝐼 𝑑2𝑦

𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −50(𝑥 − 3)

= 25 𝑥 ⋮ −50(𝑥 − 3)


𝐸𝐼 𝑑𝑦

𝑑𝑥 = 25

𝑥2

2+ 𝐶1 ⋮ −50

(𝑥 − 3)2

2 → 𝐸𝑞 (1)


𝐸𝐼 𝑦 = 25 𝑥3

6 + 𝐶1𝑥 + 𝐶2 ⋮ −25

(𝑥 − 3)3

3 → 𝐸𝑞 (2)





5-83

conditions,

(i) at x=0, y=0 and (ii) at x=6, y=0



𝐸𝐼(0) = 25 03

6 + 𝐶10

𝐶2 = 0


beam) we get,

𝐸𝐼(0) = 25 63

6 + 𝐶16 ⋮ −25

(6 − 3)3

3

𝐸𝐼(0) = 900 + 𝐶16 ⋮ −225

𝐶1 = −112.5


Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 25

𝑥2

2 − 112.5 ⋮ −25(𝑥 − 3)2


𝐸𝐼 𝑦 = 25 𝑥3

6 − 112.5𝑥 ⋮ −25

(𝑥 − 3)3

3

(7M)

Deflection at the centre of the beam:

Substituting x=3 into Eq (2) upto first dotted line, (as x=3 lies in the first part of the beam) we

get the deflcetion at the centre as,

𝐸𝐼𝑦𝑐𝑒𝑛𝑡𝑟𝑎𝑙 = 25 33

6 − 112.5(3)

𝐸𝐼𝑦𝑐𝑒𝑛𝑡𝑟𝑎𝑙 = 112.5 − 337.5

𝐸𝐼𝑦𝑐𝑒𝑛𝑡𝑟𝑎𝑙 = −225

𝑦𝑐𝑒𝑛𝑡𝑟𝑎𝑙 =−225𝑥106

2.1. 𝑥 1011𝑥78𝑥 10−6




5-84



Maximum Deflection:



𝐸𝐼 𝑦 𝑚𝑎𝑥 = 17.58 (3.54)3

6 −

(3.54)4

4− 65.82(3.54) ⋮ +

(3.54 − 3)4

4

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 129.98 − 39.26 − 233 + 0.021

𝑦 𝑚𝑎𝑥 = −142.26

𝐸𝐼

(6M)

4

A beam of length 6m is simply supported at its ends and carries two point loads of 48kN

and 40kN at a distance of 1m and 3m respectively from the left support. Using Macaulay’s

method find: (i) deflection under the load, (ii) maximum deflection, (iii)the point at which

maximum deflection occurs. Given E = 2x105 N/mm2, I =85x106mm4. (13M) (Nov/Dec 2015)

BTL5


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 48 + 40= 88kN


𝑅𝐵 𝑥 6 = (48 𝑥 1) + (40 𝑥 3 ) = 168

𝑅𝐵 = 168

6= 28𝑘𝑁

𝑅𝐴 = 168 − 28 = 140𝑘𝑁




5-85



𝐸𝐼 𝑑2𝑦

𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −48(𝑥 − 1) ⋮ −40(𝑥 − 3)

= 140 𝑥 ⋮ −48(𝑥 − 1) ⋮ −40(𝑥 − 3)


𝐸𝐼 𝑑𝑦

𝑑𝑥 = 140

𝑥2

2+ 𝐶1 ⋮ −48

(𝑥 − 1)2

2⋮ −40

(𝑥 − 3)2

2 → 𝐸𝑞 (1)


𝐸𝐼 𝑦 = 70 𝑥3

3 + 𝐶1𝑥 + 𝐶2 ⋮ −24

(𝑥 − 1)3

3⋮ −20

(𝑥 − 3)3

3 → 𝐸𝑞 (2)


conditions,

(i) at x=0, y=0 and (ii) at x=6, y=0



𝐸𝐼(0) = 70 03

3 + 𝐶10

𝐶2 = 0


beam) we get,

𝐸𝐼(0) = 70 63

3 + 𝐶16 ⋮ −24

(6 − 1)3

3⋮ −20

(6 − 3)3

3

𝐸𝐼(0) = 5040 + 𝐶16 ⋮ −1000 ⋮ −180

𝐶1 = −643.3


Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 17.58

𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥 − 3)3 ⋮ −2(𝑥 − 5)3





5-86

𝐸𝐼 𝑦 = 17.58 𝑥3

6 −

𝑥4

4− 65.82𝑥 ⋮ +

(𝑥 − 3)4

4⋮ −2

(𝑥 − 5)4

4

(7M)



slope 𝑑𝑦



0 = 17.58 𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥 − 3)3

0 = 17.58 𝑥2

2 − 𝑥3 − 65.82 ⋮ +(𝑥3 − 27 + 27𝑥 − 9𝑥2)



Maximum Deflection:



𝐸𝐼 𝑦 𝑚𝑎𝑥 = 17.58 (3.54)3

6 −

(3.54)4

4− 65.82(3.54) ⋮ +

(3.54 − 3)4

4

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 129.98 − 39.26 − 233 + 0.021

𝑦 𝑚𝑎𝑥 = −142.26

𝐸𝐼

(6M)

5

A cantilever of length 3m is carrying a point load of 50kN at a distance of 2m from the

fixed end. If I=108mm4 and E=2x105N/mm2, find i)Slope at the free end and ii)Deflection at

the free end. (8M) (Nov/Dec 2017) BTL3

Given:

L=3m,

W=50kN

I=108mm4




5-87

E=2x105N/mm2

Slope at the free end is given by the equation,

𝜃𝐵 =𝑊𝑎2

2𝐸𝐼

𝜃𝐵 =50000𝑥(2)2

2𝑥2𝑥1011𝑥10−4

𝜃𝐵 = 0.005 𝑟𝑎𝑑

(4M)

Deflection at the free end is given by the equation,

𝑦𝐵 =𝑊𝑎2

3𝐸𝐼+

𝑊𝑎2

2𝐸𝐼(𝐿 − 𝑎)

𝑦𝐵 =50000𝑥(2)3

3𝑥2𝑥1011𝑥10−4+

50000𝑥(2)2

2𝑥2𝑥1011𝑥10−4(3 − 2)

𝑦𝐵 = 0.0067 + 0.0050

𝑦𝐵 = 0.0117𝑚

(4M)

6

A cantilever 2m long is of rectangular section 120mm wide and 240mm deep. It carries a

uniformly distributed load of 2.5kNper metre length for a length of 1.25m from the fixed

end and a point load of 1kN at the free end. Find the deflectiion at the free end. Take

E=10GN/m2. (13M) (Apr/May 2018) BTL3

Moment of Inertia

𝐼 =𝑏𝑑3

12

𝐼 =0.12(0.24)3

12

𝐼 = 1.3824𝑥10−4 𝑚4

𝐸𝐼 = 10𝑥109𝑥1.3824𝑥10−4 𝑚4




5-88

𝐸𝐼 = 1.3824𝑥106 𝑁 − 𝑚2

Double Integration Method:

Taking A as origin and using double integration method, the bending moment at any section X at

a distance of x from A,

𝐸𝐼𝑑2𝑦

𝑑𝑥2= −𝑥 − 2.5(𝑥 − 0.75) − − − − − −(1)


𝐸𝐼𝑑𝑦

𝑑𝑥= −

𝑥2

2− 2.5

(𝑥 − 0.75)2

2+ 𝐶1 − − − − − −(2)

Integrating the above equation once again we get,

𝐸𝐼𝑦 = −𝑥3

6− 2.5

(𝑥 − 0.75)3

6+ 𝐶1𝑥 + 𝐶2 − − − − − −(3)

Boundary Condition:

when x=2, slope becomes zero.

𝐸𝐼 (0) = −22

2− 2.5

(2 − 0.75)2

2+ 𝐶1

𝐶1 = 3.953

Boundary Condition:

when x=2, deflection becomes zero.

𝐸𝐼 (0) = −23

6− 2.5

(2 − 0.75)3

6+ 2𝑥3.953 + 𝐶2

0 = −1.33 − 0.813 + 7.906 + 𝐶2

𝐶2 = −5.763

Substituting the values in eqns 2 and 3 we get,

Slope Equation:

𝐸𝐼𝑑𝑦

𝑑𝑥= −

𝑥2

2− 2.5

(𝑥 − 0.75)2

2+ 3.953





5-89

𝐸𝐼𝑦 = −𝑥3

6− 2.5

(𝑥 − 0.75)3

6+ 3.953𝑥 − 5.763

(7M) To find the Deflection at free end:

Substitute x=0 in the deflection equation we get,

𝐸𝐼𝑦𝑓𝑟𝑒𝑒 = −03

6− 2.5

(0 − 0.75)3

6+ 3.953(0) − 5.763

𝑦𝑓𝑟𝑒𝑒 = −5.763𝑥103

1.3824𝑥106

𝑦𝑓𝑟𝑒𝑒 = −4.168𝑥10−3𝑚

(6M)

7

A beam AB of 8m span is simply supported at the ends. It carries a point load of 10kN at a

distance of 1m from the end A and a uniformly distributed load of 5kN/m for a length of 2m

from the endB. If I=10x10-6m4, Determine i)Deflection at mid-span, ii)Maximum deflection,

and iii)Slope at the end A. (13M) (Apr/May 2018) BTL3


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 = 10 + (5𝑥2)= 20kN


𝑅𝐵 𝑥 8 = (10 𝑥 1) + (5 𝑥 2 𝑥(2

2+ 6 ) = 80

𝑅𝐵 = 80

8= 10𝑘𝑁

𝑅𝐴 = 20 − 10 = 10𝑘𝑁






5-90

𝐸𝐼 𝑑2𝑦

𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −10(𝑥 − 1) ⋮ −5

(𝑥 − 6)2

2

= 10 𝑥 ⋮ −10(𝑥 − 1) ⋮ −5(𝑥 − 6)2

2


𝐸𝐼 𝑑𝑦

𝑑𝑥 = 10

𝑥2

2+ 𝐶1 ⋮ −10

(𝑥 − 1)2

2⋮ −5

(𝑥 − 6)3

6 → 𝐸𝑞 (1)


𝐸𝐼 𝑦 = 10 𝑥3

6 + 𝐶1𝑥 + 𝐶2 ⋮ −10

(𝑥 − 1)3

6⋮ −5

(𝑥 − 6)4

24 → 𝐸𝑞 (2)


conditions,

(i) at x=0, y=0 and (ii) at x=8, y=0



𝐸𝐼(0) = 10 03

6 + 𝐶10

𝐶2 = 0


beam) we get,

𝐸𝐼(0) = 10 83

6 + 𝐶18 + 𝐶2 ⋮ −10

(8 − 1)3

6⋮ −5

(8 − 6)4

24

𝐸𝐼(0) = 853.33 + 𝐶18 ⋮ −571.67 ⋮ −3.33

𝐶1 = −34.79


Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 10

𝑥2

2− 34.79 ⋮ −10

(𝑥 − 1)2

2⋮ −5

(𝑥 − 6)3

6





5-91

𝐸𝐼 𝑦 = 10 𝑥3

6− 34.79𝑥 ⋮ −10

(𝑥 − 1)3

6⋮ −5

(𝑥 − 6)4

24

(7M)

Deflection at mid-span:

Substituting x=4 into deflection upto second dotted line, (as x=4 lies in the second part of the


𝐸𝐼 𝑦 𝑚𝑖𝑑 = 10 (4)3

6 − 34.79(4) ⋮ −10

(4 − 1)3

6

𝐸𝐼 𝑦 𝑚𝑖𝑑 = 106.67 − 139.16 − 45

𝑦 𝑚𝑖𝑑 = −77.49

𝐸𝐼



slope 𝑑𝑦



0 = 10 𝑥2

2 − 34.79 ⋮ −5(𝑥 − 1)2

0 = 5𝑥2 − 34.79 ⋮ −5(𝑥2 + 1 − 2𝑥)

0 = 5𝑥2 − 34.79 ⋮ −5𝑥2 − 5 + 10𝑥

0 = −34.79 − 5 + 10𝑥

39.79 = 10𝑥



Maximum Deflection:

Substituting x=3.979 into Eq (2) upto second dotted line, (as x=3.979 lies in the second part of

the beam) we get the maximum deflcetion as,

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 10 (3.979)3

6 − 34.79(3.979) ⋮ −10

(3.979 − 1)3

6

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 104.99 − 138.42 − 44.06




5-92

𝑦 𝑚𝑎𝑥 = −77.49

𝐸𝐼

(6M)

8

Derive the formula to find the deflection of a simply supported beam with the point load W

at the centre by moment area method. (13M) (Nov/Dec 2016) BTL4

Consider a simply supported beam AB of length L and carrying a point load W at the centre of

the beam at point C. This is a symmetrical loading, hence slope is zero at the centre at point C.

The bending moment diagram is shown below,

Now using Mohr’s theorem for slope, we get

𝑆𝑙𝑜𝑝𝑒 𝑎𝑡 𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐵.𝑀 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐶

𝐸𝐼

Area of B.M diagram between A and C = Area of triangle ACD.

=1

2𝑥

𝐿

2𝑥

𝑊𝐿

4−

𝑊𝐿2

16

𝑆𝑙𝑜𝑝𝑒 𝑎𝑡 𝐴 𝑜𝑟 𝜃𝐴 =𝑊𝐿2

𝐸𝐼

(7M) Now using Mohr’s theorem for deflection, we get

𝑦 = 𝐴𝑥

𝐸𝐼

Where A=Area of B.M Diagram of area from A.

𝐴 =𝑊𝐿2

16

x= Distance of CG of area A from A.

𝑥 =2

3𝑥

𝐿

3=

𝐿

3




5-93

𝑦 =

𝑊𝐿2

16𝑥

𝐿

3

𝐸𝐼

𝑦 =𝑊𝐿3

48𝐸𝐼

(6M)

9

A simply supported beam of span 5.8m carries a central point load of 37.5kN, find the

maximum slope and deflection, let EI = 40000 kN-m2. Use conjugate beam method.(13M)

(Nov/Dec 2016) BTL3

Given : EI = 4x104 kN-m2.


∑ 𝐹𝐻 = 0

𝑅𝐴 + 𝑅𝐵 =37.5kN


𝑅𝐵 𝑥 5.8 = (37.5 𝑥 2.9 ) = 108.75

𝑅𝐵 = 108.75

5.8= 18.75𝑘𝑁

𝑅𝐴 = 37.5 − 18.75 = 18.75𝑘𝑁

Constructing the bending moment diagram we get,


BM@B = 0

BM@C = +(18.75x2.9) = +54.375kN-m.

BM@A = +(18.75x5.8)-(37.5x2.9)= 0

Conjugate Beam:




5-94

Reaction at Supports of the Conjugate Beam:

∑ 𝐹𝐻 = 0

𝑅𝐴′ + 𝑅𝐵

′ = 2 𝑥 (1

2 𝑥 2.9 𝑥

54.38

𝐸𝐼)

𝑅𝐴′ + 𝑅𝐵

′ =157.4

𝐸𝐼


𝑅𝐵′ 𝑥 5.8 = [(

1

2 𝑥 2.9 𝑥

54.38

𝐸𝐼) 𝑥 ((

1

3𝑥 2.9) + 2.9)] +[(

1

2 𝑥 2.9 𝑥

54.38

𝐸𝐼) 𝑥 (

2

3𝑥 2.9)]

5.8 𝑅𝐵′ =

456.54

𝐸𝐼

𝑅𝐵′ =

78.7

𝐸𝐼

𝑅𝐴′ =

157.4

𝐸𝐼−

78.7

𝐸𝐼

𝑅𝐴′ =

78.7

𝐸𝐼

(7M) To find maximum deflection:

The maximum deflection occurs at the mid-span i.e., Point C

By Conjugate beam therorem,

Defeclection at C = Bending moment BM at C’.

BM@C’ = (78.71

𝐸𝐼 𝑥 2.9 ) - [(

1

2 𝑥 2.9 𝑥

54.38

𝐸𝐼) 𝑥 (

1

3𝑥 2.9)]




5-95

BM@C’ = 228.26

𝐸𝐼−

76.22

𝐸𝐼

BM@C’ = 152.04

𝐸𝐼

BM@C’ = 152.04

4𝑥104

y𝑐 = 38.01𝑥10−4 𝑚 𝑜𝑟 3.801𝑚𝑚

To find maximum slope:

The maximum slope occurs at the end points. Point A and B.

By Conjugate beam therorem,

Slope at A = Shear Force SF at A’.

SF@A’ = −78.7

𝐸𝐼

BM@A’ = −78.7

4𝑥104

θ𝐴 = −19.675𝑥10−4 𝑟𝑎𝑑 𝑜𝑟 − 0.0019675 𝑟𝑎𝑑𝑖𝑎𝑛𝑠

(6M)

10

A Cantilever of length l carrying uniformly distributed load w kN per unit run over whole

length.Derive the formula to find the slope and deflection at the free end by double

integration method. (13M) BTL5




𝐸𝐼𝑑2𝑦

𝑑𝑥2= −𝑤𝑥

𝑥

2= −𝑤

𝑥2

2 − − − − − −(1)





5-96

𝐸𝐼𝑑𝑦

𝑑𝑥= −

𝑤 𝑥3

6+ 𝐶1 − − − − − −(2)


𝐸𝐼𝑦 = −𝑤 𝑥4

24+ 𝐶1𝑥 + 𝐶2 − − − − − −(3)

Boundary Condition:

when x=l, slope becomes zero.

𝐸𝐼 (0) = −𝑤 (𝑙)3

6+ 𝐶1

𝐶1 =𝑤 (𝑙)3

6

Boundary Condition:

when x=l, deflection becomes zero.

𝐸𝐼 (0) = −𝑤 𝑙4

24+

𝑤 (𝑙)4

6+ 𝐶2

𝐶2 = −𝑤 𝑙4

8


Slope Equation:

𝐸𝐼𝑑𝑦

𝑑𝑥= −

𝑤 𝑥3

6+

𝑤 𝑙3

6


𝐸𝐼𝑦 = −𝑤 𝑥4

24+

𝑤 𝑙3

6𝑥 −

𝑤 𝑙4

8

(7M) To find the Slope at free end:

Substitute x=0 in the slope equation we get,

𝐸𝐼𝜃𝑓𝑟𝑒𝑒 = −𝑤 03

6+

𝑤 𝑙3

6




5-97

𝜃𝑓𝑟𝑒𝑒 =𝑤 𝑙3

6𝐸𝐼

To find the Deflection at free end:


𝐸𝐼𝑦𝑓𝑟𝑒𝑒 = −𝑤 𝑥4

24+

𝑤 𝑙3

6𝑥 −

𝑤 𝑙4

8

𝑦𝑓𝑟𝑒𝑒 = −𝑤 𝑙4

8𝐸𝐼

(6M)

PART * C

1

A beam of length 8m is simply supported at its ends and carries point load of 50kN at a

distance of 3m from the left support and a moment of 75kN-m clock-wise at the distance of

6m from the left support. Using Macaulay’s method find: (i) maximum deflection, (ii)the

point at which maximum deflection occurs. Given E = 2x105 N/mm2, I =85x106mm4. (15M)

BTL5


∑ 𝐹𝐻 = 0



𝑅𝐵 𝑥 8 = (50 𝑥 3 ) + 75 = 225

𝑅𝐵 = 225

8= 28.125𝑘𝑁




5-98

𝑅𝐴 = 50 − 28.125 = 21.875𝑘𝑁

(5M) Consider a section X in the last part of the beam at a distance ‘x’ from the left most support. The


𝐸𝐼 𝑑2𝑦

𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −50(𝑥 − 3) ⋮ +75(𝑥 − 5)0

= 21.875 𝑥 ⋮ −50(𝑥 − 3) ⋮ +75(𝑥 − 5)0


𝐸𝐼 𝑑𝑦

𝑑𝑥 = 21.875

𝑥2

2+ 𝐶1 ⋮ −50

(𝑥 − 3)2

2⋮ +75(𝑥 − 5) → 𝐸𝑞 (1)


𝐸𝐼 𝑦 = 21.875 𝑥3

6 + 𝐶1𝑥 + 𝐶2 ⋮ −25

(𝑥 − 3)3

3⋮ +75

(𝑥 − 5)2

2 → 𝐸𝑞 (2)


conditions,

(i) at x=0, y=0 and (ii) at x=8, y=0



𝐸𝐼(0) = 21.875 03

6 + 𝐶10 + 𝐶2

𝐶2 = 0


beam) we get,

𝐸𝐼(0) = 21.875 83

6 + 𝐶18 + 𝐶2 ⋮ −25

(8 − 3)3

3⋮ +75

(8 − 5)2

2

𝐸𝐼(0) = 1867.67 + 𝐶18 ⋮ −1041.67 ⋮ +337.5

𝐶1 = −145.43


Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 = 21.875

𝑥2

2− 145.43 ⋮ −50

(𝑥 − 3)2

2⋮ +75(𝑥 − 5)




5-99


𝐸𝐼 𝑦 = 21.875 𝑥3

6 − 145.43𝑥 + 𝐶2 ⋮ −25

(𝑥 − 3)3

3⋮ +75

(𝑥 − 5)2

2

(5M)



slope 𝑑𝑦



0 = 21.875 𝑥2

2− 145.43 ⋮ −50

(𝑥 − 3)2

2

0 = 21.875 𝑥2

2 − 145.43 ⋮ −25(𝑥2 + 9 − 6𝑥)



Maximum Deflection:



𝐸𝐼 𝑦 𝑚𝑎𝑥 = 21.875 3.883

6 − 145.43(3.88) ⋮ −25

(3.88−3)3

3

𝐸𝐼 𝑦 𝑚𝑎𝑥 = 212.95 − 564.26 − 5.67

𝑦 𝑚𝑎𝑥 = −356.98

𝐸𝐼

𝑦 𝑚𝑎𝑥 = −356.98

2𝑥1011𝑥85𝑥10−6

𝑦 𝑚𝑎𝑥 = −0.02099𝑚

(5M)

2

Derive the formula to find the deflection of a simply supported beam with the uniformly

distributed load w throughout the entire length. (15M) BTL4

A simply supported beam Ab of length L and carrying a uniformly distributed load of w per unit

length over the entire length is shown below.




5-100

The reactions at A and B will be equal. Also the maximum delection will be at the centre. Each

vertical reaction,

𝑅𝐴 = 𝑅𝐵 =𝑤𝑥𝐿

2



𝐸𝐼 𝑑2𝑦

𝑑𝑥2 =

𝑤 𝐿 𝑥

2 −

𝑤. 𝑥2

2


𝐸𝐼 𝑑𝑦

𝑑𝑥 =

𝑤 𝐿 𝑥2

4 −

𝑤. 𝑥3

6+ 𝐶1 → 𝐸𝑞 (1)


𝐸𝐼 𝑦 =𝑤 𝐿 𝑥3

12 −

𝑤. 𝑥4

24+ 𝐶1𝑥 + 𝐶2 → 𝐸𝑞 (2)

(5M)


conditions,

(i) at x=0, y=0 and (ii) at x=L, y=0

Substituting condn (i) at x=0, y=0 into Eq (2) we get,

𝐸𝐼(0) = 𝑤 𝐿 03

12 −

𝑤. 04

24+ 𝐶10 + 𝐶2

𝐶2 = 0

Substituting condn (ii) at x=L, y=0 into complete Eq (2) we get,

𝐸𝐼(0) =𝑤 𝐿 𝐿3

12 −

𝑤. 𝐿4

24+ 𝐶1𝐿 + 𝐶2

𝐸𝐼(0) = 𝑤 𝐿4

12 −

𝑤. 𝐿4

24+ 𝐶1𝐿




5-101

𝐶1 = −𝑤. 𝐿3

24


Slope Equation:

𝐸𝐼 𝑑𝑦

𝑑𝑥 =

𝑤 𝐿 𝑥2

4 −

𝑤. 𝑥3

6−

𝑤. 𝐿3

24


𝐸𝐼 𝑦 =𝑤 𝐿 𝑥3

12 −

𝑤. 𝑥4

24−

𝑤. 𝐿3

24𝑥

(5M)

Slope at the supports:

Let 𝜃𝐴 = 𝑆𝑙𝑜𝑝𝑒 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝐴

and 𝜃𝐵 = 𝑆𝑙𝑜𝑝𝑒 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝐵

At A, x=0 and 𝑑𝑦

𝑑𝑥= 𝜃𝐴

𝐸𝐼 𝜃𝐴 =𝑤 𝐿 02

4 −

𝑤. 03

6−

𝑤. 𝐿3

24

𝐸𝐼 𝜃𝐴 = −𝑤. 𝐿3

24

𝜃𝐴 = −𝑤. 𝐿3

24𝐸𝐼

By symmetry,

𝜃𝐵 = −𝑤. 𝐿3

24𝐸𝐼

Maximum Deflection:

The maximum deflection is at the centre of the beam at point C, where 𝑥 =𝐿

2

𝐸𝐼 𝑦 𝑐 ==𝑤 𝐿

12(

𝐿

2)

3

−𝑤

24(

𝐿

2)

4

−𝑤. 𝐿3

24(

𝐿

2)




5-102

𝐸𝐼 𝑦 𝑐 ==𝑤 𝐿4

96 −

𝑤 𝐿4

384−

𝑤 𝐿4

48

𝐸𝐼 𝑦 𝑐 ==5𝑤 𝐿4

384

𝑦 𝑐 ==5𝑤 𝐿4

384𝐸𝐼

(5M)

3

Catilever of length l carrying uniformly distributed load w kN per unit run over whole

length.Derive the formula to find the slope and deflection at the free end by double

integration method. Calculate the deflcetion, if w=20kN/m, l =2.3m, and EI = 12000 kN-m2.

(15M) (Nov/Dec 2016) BTL5




𝐸𝐼𝑑2𝑦

𝑑𝑥2= −20𝑥

𝑥

2= −10𝑥2 − − − − − −(1)


𝐸𝐼𝑑𝑦

𝑑𝑥= −

10 𝑥3

3+ 𝐶1 − − − − − −(2)


𝐸𝐼𝑦 = −10 𝑥4

12+ 𝐶1𝑥 + 𝐶2 − − − − − −(3)

(5M) Boundary Condition:

when x=l, slope becomes zero.

𝐸𝐼 (0) = −10 (2.3)3

3+ 𝐶1

𝐶1 = 40.56

Boundary Condition:




5-103

when x=l, deflection becomes zero.

𝐸𝐼 (0) = −10 (2.3)4

12+ 40.56(2.3) + 𝐶2

𝐶2 = −69.96


Slope Equation:

𝐸𝐼𝑑𝑦

𝑑𝑥= −

10 𝑥3

3+ 40.56


𝐸𝐼𝑦 = −10 𝑥4

12+ 40.56𝑥 − 69.96

(5M) To find the Slope at free end:

Substitute x=0 in the slope equation we get,

𝐸𝐼𝜃𝑓𝑟𝑒𝑒 = −10 (0)3

3+ 40.56

𝜃𝑓𝑟𝑒𝑒 =40.56

𝐸𝐼

𝜃𝑓𝑟𝑒𝑒 =40.56

12𝑥103

𝜃𝑓𝑟𝑒𝑒 = 3.38𝑥10−3𝑟𝑎𝑑

To find the Deflection at free end:


𝐸𝐼𝑦𝑓𝑟𝑒𝑒 = −10 (0)4

12+ 40.56(0) − 69.96

𝑦𝑓𝑟𝑒𝑒 = −69.96

12𝑥103

𝑦𝑓𝑟𝑒𝑒 = −5.83𝑥10−3𝑚

(5M)




5-104

UNIT V - THIN CYLINDERS, SPHERES AND THICK CYLINDERS

Stresses in thin cylindrical shell due to internal pressure circumferential and longitudinal stresses and

deformation in thin and thick cylinders – spherical shells subjected to internal pressure –Deformation in

spherical shells – Lame’s theorem.

PART * A

Q.No. Questions

1.

Distinguish between thin and thick shells. (Apr/May 2015), (May/June 2016) BTL2

Thin Shell Thick Shell

The ratio of wall thickness to the diameter of

the cylinder is less than 1/20

The ratio of wall thickness to the diameter of the

cylinder is more than 1/20

Circumferential stress is assumed to be

constant throughout wall thickness.

Circumferential stress varies from inner to outer

wall thickness.

2

State the assumptions made in Lame’s theorem for thick cylinder analysis. (Apr/May 2015)

BTL3

The following are the assumptions made in the lame’s theorem used for thick cylinder analysis,

The Material of the cylinder is homogeneous and isotropic.

Plane sections of the cylinder perpendicular to the longitudinal axis remain plane under

the pressure.

3

State the expression for maximum shear stress in a cylindrical shell. (Nov/Dec 2015)BTL1

𝜏𝑚𝑎𝑥 = 𝜎𝑐 − 𝜎𝑙

2

=

𝑝𝑑

2𝑡−

𝑝𝑑

4𝑡

2

𝜏𝑚𝑎𝑥 = 𝑝𝑑

8𝑡

4 Define – hoop stress and longitudinal stress. (Nov/Dec 2015) BTL3

The stress acting along the circumference of the cylinder is called circumferential stress or hoop

stress whereas the stress acting along the length of the cylinder is known as longitudinal stress.

5

State Lame’s equation. (May/June 2016), (Apr/May 2017) BTL2

The Lame’s equation are expressed by means of the two following equations namely,

𝑝𝑥 =𝑏

𝑥2− 𝑎

and

𝜎𝑥 =𝑏

𝑥2+ 𝑎

where, ‘a’ and ‘b’ are constants which can be determined from the boundary conditions.

6 Name the stresses develop in the cylinder.(Nov/Dec 2016) BTL3

The stresses developed in the thin cylinder subjected to internal fluid pressure are,

Circumferential Stress,




5-105

Longitudinal Stress.

7 Define radial pressure in thin cylinder. (Nov/Dec 2016) BTL2

Radial stress is defined as the stress in directions coplanar with but perpendicular to the symmetry

axis. Most often the thin sections cylinders have negligibly small radial stress,

8

How does a thin cylinder fail due to internal fluid pressure? (Apr/May 2017) BTL1

If the strees induced in the cylinders exceeds the permissibe limit, the cylinder is likely to fail in

any one of the following two ways,

It may split into two troughs and

It may split up into two cylindres.

9

A cylinder air receiver for a compressor is 3m in internal diameter and made of plates of

20mm thick. If the hoop stress is not to exceed 90N/mm2 and the axial stress is not to exceed

60N/mm2, find the maximum safe air pressure. BTL3

Pressure for circumferential stress:

𝜎𝑐 = 𝑝𝑑

2𝑡

90 = 𝑝𝑥3000

2𝑥20

𝑝 = 1.2 𝑥 106 𝑁/𝑚2

Pressure for hoop stress:

𝜎𝑙 = 𝑝𝑑

4

60 = 𝑝𝑥3000

4𝑥20

𝑝 = 1.6 𝑥 106 𝑁/𝑚2

The safe pressure is 1.2 𝑥 106 𝑁/𝑚2

10

A spherical shell of 1m diameter is subjected to an internal pressure of 0.5N/mm2. Find the

thickness if the allowable stress in the material of the shell is 75N/mm2. BTL3

Pressure for circumferential stress:

𝜎𝑐 = 𝑝𝑑

4𝑡

75 = 0.5𝑥1000

4𝑥𝑡

𝑡 = 1.67 𝑥 10−3 𝑚

11

What do you understand by the term wire winding of thin cylinder? BTL3

The thin cylinders are sometimes pre-stressed by winding with steel wire under tension in order

to increase the tensile strength of the thin cylinders to withstand high internal pressure without

excessive increase in wall thickness.

https://en.wikipedia.org/wiki/Radial_stress




5-106

12

Write down the expression for the change in diameter and change in length of a thin

cylindrical shell when subjected to an internal pressure ‘p’. BTL1

𝛿𝑑 = 𝑝𝑑2

2𝑡𝐸 (1 −

𝜇

2)

𝛿𝑙 = 𝑝𝑑𝑙

2𝑡𝐸 (

1

2− 𝜇)

13

What is the effect of riveting a thin cylindrical shell? BTL3

It reduces the area offering the resistance. Due to this, the circumferential and longitudinal

stresses are more. It reduces the pressure carrying capacity of the shell.

14

A cylindrical shell of 500mm diameter is required to withstand an internal pressure of

4MPa. Find the minimum thickness of the shell, if maximum tensile strength in the material

is 400N/mm2 and the efficiency of the joint is 65%. Take factor of safety as 5. BTL4

𝜎𝑠𝑎𝑓𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑛𝑠𝑖𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ

𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦

𝜎𝑐 = 400

5= 80 N/mm2

𝜎𝑐 = 80 = 𝑝𝑑

2𝑡𝜂

80 = 4𝑥500

2𝑥𝑡𝑥0.65

𝑡 = 19.23 𝑥 10−3 𝑚

15

In a thin cylindrical shell, if hoop strain is 0.2x10-3 and longitudinal strain is 0.05x10-3 , find

out the volumetric strain. BTL3

𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑆𝑡𝑟𝑎𝑖𝑛, 𝑒𝑣 = 𝛿𝑉

𝑉= 2𝑒𝑐 + 𝑒𝑙

𝑒𝑣 = 2𝑥0.2𝑥10−3 + 0.05𝑥10−3

𝑒𝑣 = 4.5𝑥10−4

16

A thin spherical shell of 3m inner diameter and 10mm thickness is subjected to an internal

pressure of 2MPa. What is the maximum principal stress? BTL1

𝜎𝑐 = 𝑝𝑑

2𝑡




5-107

𝜎𝑐 = 2 𝑥 3000

4 𝑥 10

𝜎𝑐 = 150 𝑁/𝑚𝑚2.

17

Write down the equation for strain along the longitudinal, circumferential direction for the

thin cylinder. BTL1

𝑒𝑐 =𝛿𝑑

𝑑=

𝑝𝑑

2𝑡𝐸 (1 −

𝜇

2)

𝑒𝑙 =𝛿𝑙

𝑙=

𝑝𝑑𝑙

2𝑡𝐸 (

1

2− 𝜇)

18

For the thickness of the pipe due to an internal pressure 10N/mm2 if the permissible stress is

120N/mm2. The diameter of the pipe is 750mm. BTL1

𝜎𝑐 = 𝑝𝑑

2𝑡

120 = 10 𝑥 750

2 𝑥 𝑡

𝑡 = 31.25 𝑚𝑚 ≅ 32𝑚𝑚.

19

Write down the volumetric strain in a thin spherical shell subjected to internal pressure ‘p’.

BTL1

𝑒𝑣 = 3𝑒𝑐 =3𝑝𝑑

4𝑡𝐸 (1 − 𝜇)

20 Will the radial stress vary over the thickness of the wall in a thin cylinder? BTL1

No. The radial stress developed in its wall is assumed to be constant since the wall thickness is

very small compared to the cylinder diameter in thin cylinders.

PART * B

1

A thin cylindrical shell, 2.5m long has 700mm internal diameter and 8mm thickness. If the

shell is subjected to an internal pressure of 1Mpa , find (i) the hoop and longitudinal stress

developed, (ii) maximum shear stress induced and (iii) the changes in diameter length and

volume. Take modulus of elasticity of the wall material as 200GPa and poisson’s ratio as

0.3. (13M) (Apr/May 2015) BTL4

Given:

L=2.5m

d=700mm = 0.7m

t=8mm = 8x10-3 m

p=1MPa=1x106 N/m2

E=200GPa=2x1011 N/m2

µ=0.3




5-108

Soln:

Circumferential stress,

𝜎𝑐 = 𝑝𝑑

2𝑡

𝜎𝑐 = 1𝑥106𝑥0.7

2𝑥8𝑥10−3

𝜎𝑐 = 43.74 𝑥 106 𝑁/𝑚2

Longitudinal stress,

𝜎𝑙 = 𝑝𝑑

4𝑡

𝜎𝑙 = 1𝑥106𝑥0.7

4𝑥8𝑥10−3

𝜎𝑐 = 21.88 𝑥 106 𝑁/𝑚2

Maximum Shear Stress:

𝜏𝑚𝑎𝑥 = 43.74 𝑥 106 − 21.88 𝑥 106

2

𝜏𝑚𝑎𝑥 = 10.93 𝑥 106 𝑁/𝑚2

(6M) Change in diameter:


2𝑡𝐸 (1 −

𝜇

2)

𝛿𝑑 = 1𝑥106𝑥(0.7)2

2𝑥8𝑥10−3𝑥2𝑥1011 (1 −

0.3

2)

𝛿𝑑 = 0.1301 𝑥10−3𝑚

Change in length:


2𝑡𝐸 (

1

2− 𝜇)

𝛿𝑙 = 1𝑥106𝑥0.7𝑥2.5

2𝑥8𝑥10−3𝑥2𝑥1011 (

1

2− 0.3)

𝛿𝑙 = 0.1093𝑥10−3𝑚

Change in Volume:




5-109

𝛿𝑉

𝑉=

𝑝𝑑

2𝑡𝐸 (

5

2− 2𝜇)

𝛿𝑉

𝑉=

1𝑥106𝑥0.7

2𝑥8𝑥10−3𝑥2𝑥1011 (

5

2− 2𝑥0.3)

𝛿𝑉

𝑉= 0.21875𝑥10−3

Volume 𝑉 =𝜋𝑑2𝐿

4

𝑉 =𝜋𝑥0.72𝑥2.5

4= 0.962 𝑚3

𝛿𝑉 = 0.21875𝑥10−3𝑥0.962

𝛿𝑉 = 0.2104𝑥10−3𝑚3 (7M)

2

A thick cylinder with external diameter 320mm and internal diameter 160mm is subjected

to an internal pressure of 8 N/mm2.Draw the variation of radial and hoop stress in the

cylinder wall. Also determine the maximum shear stress in the cylinder wall. (13M)

(Apr/May 2015) BTL3

Given:

Internal diameter =160mm

Internal radius r1 = 80mm.

External diameter =320mm

External radius r2 = 160mm.

Fluid pressure P0 = 8 N/mm2

The radial pressure px is given by the eqn,

𝑝𝑥 =𝑏

𝑥2− 𝑎

Apply the boundary condition to above equations,

1. At x= r1 = 80mm, 𝑝𝑥 = 8 N/mm2

2. At x= r2 = 160mm, 𝑝𝑥 = 0

8 =𝑏

802− 𝑎 =

𝑏

6400− 𝑎

0 =𝑏

1602− 𝑎 =

𝑏

25600− 𝑎

Subtracting the above equations,

8 =𝑏

6400− 𝑎 −

𝑏

25600+ 𝑎




5-110

8 =4𝑏

25600−

𝑏

25600=

3𝑏

25600

𝑏 =25600𝑥8

3

𝑏 = 68266

0 =68266

25600− 𝑎

𝑎 =68266

25600= 2.67

(6M)

Now hoop stress at any radius x is given by,

𝜎𝑥 =𝑏

𝑥2+ 𝑎

𝜎𝑥 =68266

𝑥2+ 2.67

At x=80mm,

𝜎80 =68266

802+ 2.67 = 10.67 + 2.67

𝜎80 = 13.34 𝑁/𝑚𝑚2

At x=160mm,

𝜎160 =68266

1602+ 2.67 = 2.67 + 2.67

𝜎160 = 5.34 𝑁/𝑚𝑚2

(7M)

3

A boiler is subjected to an internal steam pressure of 2 N/mm2. The thickness of boiler plate

is 2.6 cm and permissible tensile stress is 120 N/mm2. Find the maximum diameter, when

efficiency of longitudinal joint is 90% and that of circumference point is 40%.(8M)

(Nov/Dec 2015) BTL4 Given:

t=2.6cm

Permissible stress, 𝜎𝑡=120 N/mm2

Permissible stress may be circumferential stress or longitudinal stress.

𝜂𝑙 = 90% = 0.9




5-111

𝜂𝑐 = 40% = 0.4 Case 1:

Consider the Permissible stress as circumferential stresss 𝜎𝑐

w.k.t

𝜎𝑐 = 𝑝𝑑

2𝑡𝜂𝑙

120 = 2𝑥𝑑

2𝑥26𝑥0.9

𝑑 = 2808 𝑚𝑚

(4M) Case 2:

Consider the Permissible stress as longitudinal stresss 𝜎𝑙

w.k.t

𝜎𝑙 = 𝑝𝑑

4𝑡𝜂𝑐

120 = 2𝑥𝑑

4𝑥26𝑥0.4

𝑑 = 2496 𝑚𝑚

(4M) Thus, the maximum permissible diameter of the shell, d=2496mm.

4

Calculate: (i) the change in diameter, (ii) change in length and (iii) change in volume of a

thin cylindrical shell 100cm diameter, 1cm thick and 5m long when subjected to internal

pressure of 3 N/mm2. Take the value of E = 2x105 N/mm2 and poisson’s ratio as 0.3. (13M)

(Nov/Dec 2015), (Nov/Dec 2016), (Nov/Dec 2017) BTL5 Given:

L=5m

d=100cm = 1m

t=1cm = 1x10-2 m

p=3 N/mm2=3x106 N/m2

E=2x105 N/mm2=2x1011 N/m2

µ=0.3

Soln:


𝜎𝑐 = 𝑝𝑑

2𝑡

𝜎𝑐 = 3𝑥106𝑥1

2𝑥1𝑥10−2

𝜎𝑐 = 1.5 𝑥 104 𝑁/𝑚2





5-112

𝜎𝑙 = 𝑝𝑑

4𝑡

𝜎𝑙 = 3𝑥106𝑥1

4𝑥1𝑥10−2

𝜎𝑐 = 0.75 𝑥 104 𝑁/𝑚2

(6M) Change in diameter:


2𝑡𝐸 (1 −

𝜇

2)

𝛿𝑑 = 3𝑥106𝑥(1)2

2𝑥1𝑥10−2𝑥2𝑥1011 (1 −

0.3

2)

𝛿𝑑 = 0.6375 𝑥10−3𝑚

Change in length:


2𝑡𝐸 (

1

2− 𝜇)

𝛿𝑙 = 3𝑥106𝑥1𝑥5

2𝑥1𝑥10−2𝑥2𝑥1011 (

1

2− 0.3)

𝛿𝑙 = 0.75𝑥10−3𝑚

Change in Volume:

𝛿𝑉

𝑉=

𝑝𝑑

2𝑡𝐸 (

5

2− 2𝜇)

𝛿𝑉

𝑉=

3𝑥106𝑥1

2𝑥1𝑥10−2𝑥2𝑥1011 (

5

2− 2𝑥0.3)

𝛿𝑉

𝑉= 1.425𝑥10−3


4

𝑉 =𝜋𝑥12𝑥5

4= 3.926 𝑚3

𝛿𝑉 = 1.425𝑥10−3𝑥3.926

𝛿𝑉 = 5.5959𝑥10−3𝑚3




5-113

(7M)

5

Calculate the thickness of metal necessary for a cylindrical shell of internal diameter 160

mm to with stand an internal pressure of 25 MN/m2, if maximum permissible shear stress is

125 MN/m2. (8M) (Nov/Dec 2016) BTL3

Given:

d=160mm = 0.16m

p=25 MN/m2=25x106 N/m2

τmax=125 MN/m2=125x106 N/m2

Soln:

Maximum Shear stress,

τmax = 𝑝𝑑

8𝑡

125𝑥106 = 25𝑥106𝑥0.16

8𝑥𝑡

𝑡 =25𝑥106𝑥0.16

8𝑥125𝑥106

𝑡 = 0.004𝑚

(8M)

6

Derive a relation for change in volume of a thin cylinder subjected to internal fluid

pressure. (13M) (Apr/May 2017) BTL4

Let,

L= length of the shell

D=diameter of the shell

t= thickness of the shell

p=intensity of internal pressure

w.k.t

circumferential stress as

𝜎𝑐 = 𝑝𝑑

2𝑡

Also the longitudinal stress as

𝜎𝑙 = 𝑝𝑑

4𝑡=

𝜎𝑐

2

Let

𝛿𝑑 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑒𝑙𝑙 𝛿𝑙 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑒𝑙𝑙 µ=Poisson’s ratio.

Change in diameter can be found as follows,

𝛿𝑑 = 𝜖𝑐 𝑥 𝑑

𝜖𝑐 = 𝜎𝑐

𝐸 − 𝜇

𝜎𝑙

𝐸




5-114

𝜖𝑐 = 𝜎𝑐

𝐸 − 𝜇

𝜎𝑐

2𝐸

𝜖𝑐 = 𝜎𝑐

𝐸 (1 −

𝜇

2)

𝜖𝑐 = 𝑝𝑑

2𝑡𝐸 (1 −

𝜇

2)

𝛿𝑑 = 𝑝𝑑

2𝑡𝐸 (1 −

𝜇

2) 𝑥 𝑑


2𝑡𝐸 (1 −

𝜇

2)

(6M) Change in length can be found as follows,

𝛿𝑙 = 𝜖𝑙 𝑥 𝑙

𝜖𝑙 = 𝜎𝑙

𝐸 − 𝜇

𝜎𝑐

𝐸

𝜖𝑙 = 𝜎𝑐

2𝐸 − 𝜇

𝜎𝑐

𝐸

𝜖𝑙 = 𝜎𝑐

𝐸 (

1

2− 𝜇)

𝜖𝑙 = 𝑝𝑑

2𝑡𝐸 (

1

2− 𝜇)

𝛿𝑙 = 𝑝𝑑

2𝑡𝐸 (

1

2− 𝜇) 𝑥 𝑙


2𝑡𝐸 (

1

2− 𝜇)

Change in volume can be found as follows,

𝜖𝑣 = 𝛿𝑉

𝑉

Change in volume 𝛿𝑉 = Final Volume –Initial Volume

Original Volume,

𝑉 =𝜋𝑑2𝐿

4




5-115

Final Volume,

𝑉𝑓 = 𝜋(𝑑 + 𝛿𝑑)2(𝐿 + 𝛿𝐿)

4

Expanding we get,

𝑉𝑓 = 𝜋(𝑑2𝐿 + 2𝑑𝑙𝛿𝑑 + 𝑑2𝛿𝐿)

4

Now, Change in Volume,

𝛿𝑉 = 𝜋(𝑑2𝐿 + 2𝑑𝑙𝛿𝑑 + 𝑑2𝛿𝐿)

4−

𝜋𝑑2𝐿

4

𝛿𝑉 = 𝜋(2𝑑𝑙𝛿𝑑 + 𝑑2𝛿𝐿)

4

Volumetric Strain,

𝜖𝑣 =

𝜋(2𝑑𝑙𝛿𝑑+𝑑2𝛿𝐿)

4

𝜋𝑑2𝐿

4

𝜖𝑣 = 2𝛿𝑑

𝑑+

𝛿𝑙

𝑙

𝜖𝑣 = 2𝜖𝑐 + 𝜖𝑙

𝛿𝑉 = (2𝜖𝑐 + 𝜖𝑙) 𝑥 𝑉

(7M)

7

Determine the maximum and minimum hoop stress across the section of a pipe of 400mm

internal diameter and 100mm thick, when the pipe contains a fluid at a pressure of 8N/mm2.

Also sketch the radial pressure distribution and hoop stress distribution across the section.

(13M) (Apr/May 2017), (Nov/Dec 2017) BTL4

Given:

Internal diameter =400mm

Internal radius r1 = 200mm.

External diameter =600mm

External radius r2 = 300mm.

Fluid pressure P0 = 8 N/mm2


𝑝𝑥 =𝑏

𝑥2− 𝑎

Apply the boundary condition to above equations,

1. At x= r1 = 200mm, 𝑝𝑥 = 8 N/mm2




5-116

2. At x= r2 = 300mm, 𝑝𝑥 = 0

8 =𝑏

2002− 𝑎 =

𝑏

40000− 𝑎

0 =𝑏

3002− 𝑎 =

𝑏

90000− 𝑎

Subtracting the above equations,

8 =𝑏

40000− 𝑎 −

𝑏

90000+ 𝑎

8 =2.25𝑏

90000−

𝑏

90000=

1.25𝑏

90000

𝑏 =90000𝑥8

1.25

𝑏 = 576000

0 =576000

90000− 𝑎

𝑎 =576000

90000= 6.4

(6M)

Now hoop stress at any radius x is given by,

𝜎𝑥 =𝑏

𝑥2+ 𝑎

𝜎𝑥 =576000

𝑥2+ 6.4

At x=200mm,

𝜎200 =576000

2002+ 6.4 = 14.4 + 6.4

𝜎200 = 20.8 𝑁/𝑚𝑚2

At x=300mm,

𝜎300 =576000

3002+ 6.4 = 6.4 + 6.4




5-117

𝜎300 = 12.8 𝑁/𝑚𝑚2

(7M)

8

A thick spherical shell of 200mm internal diameter is subjected to an internal fluid pressure

of 7N/mm2. If the permissible stress in the shell material is 8N/mm2, find the thickness of the

shell. (13M) BTL6

Given:

Internal dia = 200mm,

Internal radius, r1=100mm.

Fluid pressure, p=7N/mm2.

Permissible tensile stress, σx = 8N/mm2


𝑝𝑥 =2𝑏

𝑥3− 𝑎

Apply the boundary condition to above equation,

1. At x= r1 = 100mm, 𝑝𝑥 = 7 N/mm2

7 =2𝑏

1003− 𝑎 =

2𝑏

1000000− 𝑎

The hoop stress σx is given by the eqn,

𝜎𝑥 =𝑏

𝑥3+ 𝑎

(6M)


1. At x= r1 = 100mm, 𝜎𝑥 = 8 N/mm2

8 =𝑏

1003+ 𝑎 =

𝑏

1000000+ 𝑎

Adding the above equations,

15 =2𝑏

1000000− 𝑎 +

𝑏

1000000+ 𝑎

15 =3𝑏

1000000

𝑏 =1000000𝑥15

8

𝑏 = 5000000

8 =5000000

1000000+ 𝑎

𝑎 = 8 − 5 = 3




5-118

Now the radial pressure is given by,

𝑝𝑥 =2𝑥5000000

𝑥3− 3

Let r2=External radius of the shell.


1. At x= r2 , 𝑝𝑥 = 0

0 =2𝑥5000000

𝑟23 − 3

𝑟23 =

10000000

3

𝑟2 = (107

3)

1

3

𝑟2 = 149.3𝑚𝑚.

Thickness of the shell

𝑡 = 𝑟2 − 𝑟1

𝑡 = 149.3 − 100

𝑡 = 49.3𝑚𝑚

(7M)

9

Derive the expression for the change in diameter and for the change in volume of a volume

of a thin spherical shell when it is subjected to an internal fluid pressure. (13M)BTL4

There is no shear stress at any point in thin spherical shells.

Let,

σ= stress induced in the spherical shell.

𝜎 = 𝑝𝑑

4𝑡

Change in diameter:

The strain in any one direction is given by,

𝜖 = 𝜎

𝐸 − 𝜇

𝜎

𝐸




5-119

𝜖 = 𝜎

𝐸 (1 − 𝜇)

𝜖 = 𝑝𝑑

4𝑡𝐸 (1 − 𝜇)

We know that the strain in any direction as,

𝜖 = 𝛿𝑑

𝑑

𝛿𝑑

𝑑 =

𝑝𝑑

4𝑡𝐸 (1 − 𝜇)


4𝑡𝐸 (1 − 𝜇)

(6M) Change in Volume:

w.k.t the original volume of the sphere,

𝑉 =𝜋𝑑3

6

The final volume due to the pressure,

𝑉𝑓 = 𝜋(𝑑 + 𝛿𝑑)3

6

Change in volume can be found as follows,

𝜖𝑣 = 𝛿𝑉

𝑉

Change in volume 𝛿𝑉 = Final Volume –Initial Volume

Now, Change in Volume,

𝛿𝑉 = 𝜋(𝑑 + 𝛿𝑑)3

6−

𝜋𝑑3

6

𝛿𝑉 = 𝜋(3𝑑2𝛿𝑑 + 𝑑3)

6

Volumetric Strain,




5-120

𝜖𝑣 =

𝜋(3𝑑2𝛿𝑑+𝑑3)

6

𝜋𝑑3

6

𝜖𝑣 = 3𝑑2𝛿𝑑

𝑑3=

3𝛿𝑑

𝑑

𝜖𝑣 = 3𝜖

𝛿𝑉 = (3𝜖) 𝑥 𝑉

𝛿𝑉 = (3𝑝𝑑

4𝑡𝐸 (1 − 𝜇)) 𝑥

𝜋𝑑3

6

𝛿𝑉 = 𝜋𝑝𝑑4

8𝑡𝐸(1 − 𝜇)

(7M)

10

A spherical shell of internal diameter 0.9m and of thickness 10mm is subjected to an

internal pressure of 1.4N/mm2. Determine the increase in diameter and increase in volume.

Take E = 2x105 N/mm2 and µ=0.3. (13M) BTL6

Given:

Internal diameter, d=0.9m

Thickness, t=10mm=0.01m

Internal Pressure, p=1.4N/mm2=1.4x106N/m2

E = 2x105 N/mm2=2x1011 N/m2

µ=0.3

Using the relation, 𝛿𝑑

𝑑 =

𝑝𝑑

4𝑡𝐸 (1 − 𝜇)

𝛿𝑑

𝑑 =

1.4𝑥106𝑥0.9

4𝑥0.01𝑥2x1011 (1 − 0.3)

𝛿𝑑

𝑑 = 11.025𝑥10−5

𝛿𝑑 = 11.025𝑥10−5𝑥𝑑

𝛿𝑑 = 11.025𝑥10−5𝑥0.9

𝛿𝑑 = 9.9225𝑥10−5𝑚

Now, Volumetric strain 𝛿𝑉

𝑉= 3 𝑥

𝛿𝑑

𝑑




5-121

𝛿𝑉

𝑉= 3 𝑥11.025𝑥10−5

𝛿𝑉

𝑉= 33.075𝑥10−5

(6M) Volume of a sphere

𝑉 =𝜋

6𝑑3

𝑉 =𝜋

6(0.9)3

𝑉 = 0.3817𝑚3

Increase in volume,

𝛿𝑉 = 33.075𝑥10−5𝑥 𝑉

𝛿𝑉 = 33.075𝑥10−5𝑥 0.3817

𝛿𝑉 = 12.62𝑥10−5 𝑚3

(7M)

PART * C

1

A thick spherical shell of 180mm internal diameter is subjected to an internal fluid pressure

of 4N/mm2. If the permissible stress in the shell material is 10N/mm2, find the thickness of

the shell. (15M) BTL5

Given:

Internal dia = 180mm,

Internal radius, r1=90mm.

Fluid pressure, px=4N/mm2.

Permissible tensile stress, σx = 10N/mm2


𝑝𝑥 =2𝑏

𝑥3− 𝑎


At x= r1 = 90mm, 𝑝𝑥 = 4 N/mm2

4 =2𝑏

903− 𝑎 =

2𝑏

729000− 𝑎

The hoop stress σx is given by the eqn,

𝜎𝑥 =𝑏

𝑥3+ 𝑎

(5M)


At x= r1 = 90mm, 𝜎𝑥 = 10 N/mm2




5-122

10 =𝑏

903+ 𝑎 =

𝑏

729000+ 𝑎

Adding the above equations,

14 =2𝑏

729000− 𝑎 +

𝑏

729000+ 𝑎

14 =3𝑏

1458000

𝑏 =1458000𝑥14

3

𝑏 = 6804000

10 =6804000

729000+ 𝑎

𝑎 = 10 − 9 = 1

(5M)

Now the radial pressure is given by,

𝑝𝑥 =2𝑥6804000

𝑥3− 1

Let r2=External radius of the shell.


At x= r2 , 𝑝𝑥 = 0

0 =2𝑥6804000

𝑟23 − 3

𝑟23 =

13608000

3

𝑟2 = (4.536𝑥106)1

3

𝑟2 = 165.53𝑚𝑚.




5-123

Thickness of the shell

𝑡 = 𝑟2 − 𝑟1

𝑡 = 165.53 − 90

𝑡 = 75.53𝑚𝑚

(5M)

2

Calculate: (i) the change in diameter, (ii) change in length and (iii) change in volume of a

thin cylindrical shell 110cm diameter, 2cm thick and 7m long when subjected to internal

pressure of 5 N/mm2. Take the value of E = 2.1x105 N/mm2 and poisson’s ratio as 0.33.

(15M) (Nov/Dec 2015), (Nov/Dec 2016), (Nov/Dec 2017) BTL5 Given:

L=7m

d=110cm = 1.1m

t=2cm = 2x10-2 m

p=5 N/mm2=5x106 N/m2

E=2.1x105 N/mm2=2.1x1011 N/m2

µ=0.33

Soln:


𝜎𝑐 = 𝑝𝑑

2𝑡

𝜎𝑐 = 5𝑥106𝑥1.1

2𝑥2𝑥10−2

𝜎𝑐 = 1.375 𝑥 104 𝑁/𝑚2


𝜎𝑙 = 𝑝𝑑

4𝑡

𝜎𝑙 = 5𝑥106𝑥1.1

4𝑥2𝑥10−2

𝜎𝑐 = 0.6875 𝑥 104 𝑁/𝑚2

(5M)

Change in diameter:


2𝑡𝐸 (1 −

𝜇

2)

𝛿𝑑 = 5𝑥106𝑥(1.1)2

2𝑥2𝑥10−2𝑥2.1𝑥1011 (1 −

0.33

2)




5-124

𝛿𝑑 = 0.6013 𝑥10−3𝑚

Change in length:


2𝑡𝐸 (

1

2− 𝜇)

𝛿𝑙 = 5𝑥106𝑥1.1𝑥7

2𝑥2𝑥10−2𝑥2.1𝑥1011 (

1

2− 0.33)

𝛿𝑙 = 0.77911𝑥10−3𝑚 (5M)

Change in Volume:

𝛿𝑉

𝑉=

𝑝𝑑

2𝑡𝐸 (

5

2− 2𝜇)

𝛿𝑉

𝑉=

5𝑥106𝑥1.1

2𝑥2𝑥10−2𝑥2.1𝑥1011 (

5

2− 2𝑥0.3)

𝛿𝑉

𝑉= 1.2493𝑥10−3


4

𝑉 =𝜋𝑥1.12𝑥7

4= 6.6523 𝑚3

𝛿𝑉 = 1.2493𝑥10−3𝑥6.6523

𝛿𝑉 = 8.3107𝑥10−3𝑚3 (5M)

3

A boiler is subjected to an internal steam pressure of 3.5 N/mm2. The thickness of boiler

plate is 2.8 cm and permissible tensile stress is 135 N/mm2. Find the maximum diameter,

when efficiency of longitudinal joint is 85% and that of circumference point is 35%. (15M)

(Nov/Dec 2015) BTL4 Given:

t=2.8cm

pressure, p=3.5 N/mm2

Permissible stress, 𝜎𝑡=135 N/mm2

Permissible stress may be circumferential stress or longitudinal stress.

𝜂𝑙 = 85% = 0.85

𝜂𝑐 = 35% = 0.35 Case 1:

Consider the Permissible stress as circumferential stresss 𝜎𝑐

w.k.t




5-125

𝜎𝑐 = 𝑝𝑑

2𝑡𝜂𝑙

135 = 3.5𝑥𝑑

2𝑥28𝑥0.85

𝑑 = 1836 𝑚𝑚

(8M)

Case 2:

Consider the Permissible stress as longitudinal stresss 𝜎𝑙

w.k.t

𝜎𝑙 = 𝑝𝑑

4𝑡𝜂𝑐

135 = 3.5𝑥𝑑

4𝑥28𝑥0.35

𝑑 = 1512 𝑚𝑚

Thus, the maximum permissible diameter of the shell, d=1512mm.

(7M)

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CE8395 STRENGTH OF MATERIALS FOR …...Stresses in thin cylindrical shell due to internal pressure circumferential and longitudinal stresses and deformation in thin and thick cylinders