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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.
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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.
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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
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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− √(
𝜎𝑥 − 𝜎𝑦
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
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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
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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]
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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.
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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)
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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,
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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,
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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,
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∆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
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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.
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(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+ √(
𝜎𝑥 − 𝜎𝑦
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
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Given:
𝜎𝑥 = 51.96 𝑁𝑚𝑚2⁄
𝜎𝑦 = 40 𝑁𝑚𝑚2⁄
𝜏𝑥𝑦 = 30 𝑁𝑚𝑚2⁄
Calculation of Shear stress:
𝜎1,2 =𝜎𝑥 + 𝜎𝑦
2± √(
𝜎𝑥 − 𝜎𝑦
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
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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:
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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
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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,
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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,
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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
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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.
Stress in z –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑧−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
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.
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For no change to occur in the volume of bar volumetric strain must be equal to zero.
𝑑𝑉
𝑉= 0
Make , Fz as unknown ,
Stress in z –direction = 𝐿𝑜𝑎𝑑 𝑖𝑛 𝑧−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
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.
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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
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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,
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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
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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
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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.
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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)
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(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
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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
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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
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Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = (10 𝑥 4) = 40kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 8 = (10 𝑥 4 𝑥 (4
2+ 1)) = 120
𝑅𝐵 = 120
8= 15𝑘𝑁
𝑅𝐴 = 40 − 15 = 25𝑘𝑁
Shear Force Diagram:
SF@B = -15kN
SF@D = -15kN
SF@C = -15 + (10 x 4) = 25kN
SF@A = -15 + (10 x 4) = 25kN
Bending Moment Diagram:
BM@B = 0
BM@D = +(15x3) = 45kN-m.
BM@C = +(15x7) -(10x4x(4/2)) = 25kN-m.
BM@A = +(15x8) -(10x4x((4/2)+1) = 0
To find the location of Maximum Bending Moment:
The maximum bending moment occurs between the points C and D, where Shear Force is zero.
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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
To find the value of Maximum Bending Moment:
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)
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4
Draw SFD and BMD and indicate the salient features of a beam loaded as shown below.
(13M) (May/June 2016) BTL3
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = (10 𝑥 9) + 15 = 105kN
Taking Moments of all forces about A, we get
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
Shear Force Diagram:
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
Bending Moment Diagram:
BM@D = 0
BM@B = -(15x1.5) = -22.5kN-m.
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BM@A = -(15x8.5)+(50.35x7) -(10x7x(7/2)) = -20.05kN-m.
BM@C = -(15x10.5)+(50.35x9) -(10x9x(9/2)) +(54.65x2) = 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 = 15 -50.35+ 10 𝑥 = 0
10x = 35.35
x = 3.535m
To find the value of Maximum Bending Moment:
The Bending Moment at any section between C and D at a distance x from D is given by,
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)
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(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
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𝜎𝑏 =𝑀 𝑦
𝐼
(7M)
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = (13.8 𝑥 4) = 55.2kN
Taking Moments of all forces about A, we get
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
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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)
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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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 8 + 10 + 15 = 33kN
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Taking Moments of all forces about A, we get
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
Shear Force Diagram:
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
Bending Moment Diagram:
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)
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(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.
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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
Reaction at Supports of the Beam:
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∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 25kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 4 = (25 𝑥 2 ) − 18.75
𝑅𝐵 = 31.25
4= 7.81 𝑘𝑁
𝑅𝐴 = 25 − 7.81 = 17.19𝑘𝑁
Shear Force Diagram:
SF@B = -7.81 𝑘𝑁
SF@C = -7.81+25 = 17.19kN
SF@A = -7.81+25 = 17.19kN
Bending Moment Diagram:
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)
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(6M)
10
Draw shear force diagram and bending moment diagram for the beam given below. (13M)
(Apr/May 2017) BTL4
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = (10 𝑥 3) + (5 𝑥 2) = 40kN
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Taking Moments of all forces about A, we get
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
Shear Force Diagram:
SF@B = -15kN
SF@D = -15+(5x2) = -5kN
SF@C = -15+(5x2) = -5kN
SF@A = -15+(5x2)+(10x3) = 25kN
Bending Moment Diagram:
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
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 = 15 -50.35+ 10 𝑥 = 0
10x = 35.35
x = 3.535m
To find the value of Maximum Bending Moment:
The Bending Moment at any section between C and D at a distance x from D is given by,
BMmax = -15(x+1.5) +50.35 x – (10 𝑥 𝑥
2)
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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.
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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
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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
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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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = (1𝑥3) + (0.5𝑥3) = 4.5kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 4 = (0.5 𝑥 3)𝑥(4 + (3
2)) − (1𝑥3)𝑥 (
3
2) = 3.75
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𝑅𝐵 = 3.75
4= 0.94 𝑘𝑁
𝑅𝐴 = 4.5 − 0.94 = 3.56𝑘𝑁
Shear Force Diagram:
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
Bending Moment Diagram:
BM@D = 0
BM@B = -(0.5x3x(3/2)) = -2.25kN-m.
BM@A = -(1x3x(3/2)) = -4.5kN-m.
BM@C = 0
(7M)
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(8M)
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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.
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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
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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
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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
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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
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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.
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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
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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
w.k.t the maximum shear stress induced as,
𝜏 =16𝑊𝑅
𝜋𝑑3
125 =16𝑥60𝑥𝑅
𝜋𝑑3
𝑅 = 0.4089𝑑3
w.k.t the stiffness of the spring as,
𝑠 =𝐺𝑑4
64𝑅3𝑛
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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
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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)
𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 (𝜋
4[𝐷0
2 − (0.6𝐷0)2]𝐿)
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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 𝐷2 𝑥 𝐿
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:
w.k.t the maximum shear stress induced as,
𝜏 =16𝑊𝑅
𝜋𝑑3
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5-64
40𝑥106 =16𝑥1000𝑥50𝑥10−3
𝜋𝑑3
𝑑3 =16𝑥1000𝑥50𝑥10−3
𝜋𝑥40𝑥106
𝑑3 = 6.366𝑥10−6
𝑑 = 0.0185𝑚
(6M)
w.k.t the stiffness of the spring as,
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,
𝑃𝑜𝑤𝑒𝑟, 𝑃 = 2𝜋𝑁𝑇
60
300𝑥103 = 2𝜋(80)𝑇
60
𝑇 = 35816.61 𝑁 − 𝑚
(1M) w.k.t the torque transmitted by the shaft, TH as
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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
𝑃𝑜𝑤𝑒𝑟, 𝑃 = 2𝜋𝑁𝑇
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
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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
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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
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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,
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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
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)
𝑊𝐻 = 𝜌 𝑥 𝑔 𝑥 (𝜋
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=
𝜌 𝑥 𝑔 𝑥 𝜋
4𝑥 𝐷2 𝑥 𝐿
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
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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,
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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
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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,
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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
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𝜏2 =126900𝑥 27.5
𝜋
32𝑥554
𝜏2 =126900𝑥 27.5𝑥32
𝜋𝑥554
𝜏2 = 3.884𝑁/𝑚𝑚2
(7M)
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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
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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.
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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.
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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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 12 + (6 𝑥 3) = 30kN
Taking Moments of all forces about A, we get
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,
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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
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(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
Bending Moment Diagram:
BM@C = 0
BM@B = -(10x2) = -20kN-m.
BM@A = -(10x4) -(4x2x(2/2)) = -48 kN-m.
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(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
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Given:
I = 78x106mm4
E= 2.1x105 N/mm2
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 50kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 6 = (50 𝑥 3 ) = 150
𝑅𝐵 = 150
6= 25𝑘𝑁
𝑅𝐴 = 50 − 25 = 25𝑘𝑁
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,
𝐸𝐼 𝑑2𝑦
𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −50(𝑥 − 3)
= 25 𝑥 ⋮ −50(𝑥 − 3)
Integrating the above equation we get,
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 25
𝑥2
2+ 𝐶1 ⋮ −50
(𝑥 − 3)2
2 → 𝐸𝑞 (1)
Integrating again we get,
𝐸𝐼 𝑦 = 25 𝑥3
6 + 𝐶1𝑥 + 𝐶2 ⋮ −25
(𝑥 − 3)3
3 → 𝐸𝑞 (2)
Where 𝐶1 𝑎𝑛𝑑 𝐶2 are constants of integration whose values can be obtained from the boundary
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conditions,
(i) at x=0, y=0 and (ii) at x=6, 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) = 25 03
6 + 𝐶10
𝐶2 = 0
Substituting condn (ii) at x=6, y=0 into complete Eq (2) (as x=6 lies in the last part of the
beam) we get,
𝐸𝐼(0) = 25 63
6 + 𝐶16 ⋮ −25
(6 − 3)3
3
𝐸𝐼(0) = 900 + 𝐶16 ⋮ −225
𝐶1 = −112.5
Substituting the values of 𝐶1 and 𝐶2 into Eq(1) and Eq(2) we get,
Slope Equation:
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 25
𝑥2
2 − 112.5 ⋮ −25(𝑥 − 3)2
Deflection Equation:
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
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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
(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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 48 + 40= 88kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 6 = (48 𝑥 1) + (40 𝑥 3 ) = 168
𝑅𝐵 = 168
6= 28𝑘𝑁
𝑅𝐴 = 168 − 28 = 140𝑘𝑁
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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,
𝐸𝐼 𝑑2𝑦
𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −48(𝑥 − 1) ⋮ −40(𝑥 − 3)
= 140 𝑥 ⋮ −48(𝑥 − 1) ⋮ −40(𝑥 − 3)
Integrating the above equation we get,
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 140
𝑥2
2+ 𝐶1 ⋮ −48
(𝑥 − 1)2
2⋮ −40
(𝑥 − 3)2
2 → 𝐸𝑞 (1)
Integrating again we get,
𝐸𝐼 𝑦 = 70 𝑥3
3 + 𝐶1𝑥 + 𝐶2 ⋮ −24
(𝑥 − 1)3
3⋮ −20
(𝑥 − 3)3
3 → 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=6, 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) = 70 03
3 + 𝐶10
𝐶2 = 0
Substituting condn (ii) at x=6, y=0 into complete Eq (2) (as x=6 lies in the last part of the
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
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:
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𝐸𝐼 𝑦 = 17.58 𝑥3
6 −
𝑥4
4− 65.82𝑥 ⋮ +
(𝑥 − 3)4
4⋮ −2
(𝑥 − 5)4
4
(7M)
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
(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
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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
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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)
Integrating the above equation we get,
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
Deflection Equation:
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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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 10 + (5𝑥2)= 20kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 8 = (10 𝑥 1) + (5 𝑥 2 𝑥(2
2+ 6 ) = 80
𝑅𝐵 = 80
8= 10𝑘𝑁
𝑅𝐴 = 20 − 10 = 10𝑘𝑁
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,
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𝐸𝐼 𝑑2𝑦
𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −10(𝑥 − 1) ⋮ −5
(𝑥 − 6)2
2
= 10 𝑥 ⋮ −10(𝑥 − 1) ⋮ −5(𝑥 − 6)2
2
Integrating the above equation we get,
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 10
𝑥2
2+ 𝐶1 ⋮ −10
(𝑥 − 1)2
2⋮ −5
(𝑥 − 6)3
6 → 𝐸𝑞 (1)
Integrating again we get,
𝐸𝐼 𝑦 = 10 𝑥3
6 + 𝐶1𝑥 + 𝐶2 ⋮ −10
(𝑥 − 1)3
6⋮ −5
(𝑥 − 6)4
24 → 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=8, 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) = 10 03
6 + 𝐶10
𝐶2 = 0
Substituting condn (ii) at x=8, y=0 into complete Eq (2) (as x=8 lies in the last part of the
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
Substituting the values of 𝐶1 and 𝐶2 into Eq(1) and Eq(2) we get,
Slope Equation:
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 10
𝑥2
2− 34.79 ⋮ −10
(𝑥 − 1)2
2⋮ −5
(𝑥 − 6)3
6
Deflection Equation:
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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
beam) we get the maximum deflcetion as,
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
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 = 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
Solving , 𝑥 = 3.979m
Hence the maximum deflection will be at a distance of 3.979m from support A.
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
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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
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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.
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 =37.5kN
Taking Moments of all forces about A, we get
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,
Bending Moment Diagram:
BM@B = 0
BM@C = +(18.75x2.9) = +54.375kN-m.
BM@A = +(18.75x5.8)-(37.5x2.9)= 0
Conjugate Beam:
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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
Taking Moments of all forces about A, we get
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)]
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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
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= −𝑤𝑥
𝑥
2= −𝑤
𝑥2
2 − − − − − −(1)
Integrating the above equation we get,
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𝐸𝐼𝑑𝑦
𝑑𝑥= −
𝑤 𝑥3
6+ 𝐶1 − − − − − −(2)
Integrating the above equation once again we get,
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
Substituting the values in eqns 2 and 3 we get,
Slope Equation:
𝐸𝐼𝑑𝑦
𝑑𝑥= −
𝑤 𝑥3
6+
𝑤 𝑙3
6
Deflection Equation:
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
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𝜃𝑓𝑟𝑒𝑒 =𝑤 𝑙3
6𝐸𝐼
To find the Deflection at free end:
Substitute x=0 in the deflection equation we get,
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
Reaction at Supports of the Beam:
∑ 𝐹𝐻 = 0
𝑅𝐴 + 𝑅𝐵 = 50kN
Taking Moments of all forces about A, we get
𝑅𝐵 𝑥 8 = (50 𝑥 3 ) + 75 = 225
𝑅𝐵 = 225
8= 28.125𝑘𝑁
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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
B.M at this section is given by,
𝐸𝐼 𝑑2𝑦
𝑑𝑥2 = 𝑅𝐴 𝑥 ⋮ −50(𝑥 − 3) ⋮ +75(𝑥 − 5)0
= 21.875 𝑥 ⋮ −50(𝑥 − 3) ⋮ +75(𝑥 − 5)0
Integrating the above equation we get,
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 21.875
𝑥2
2+ 𝐶1 ⋮ −50
(𝑥 − 3)2
2⋮ +75(𝑥 − 5) → 𝐸𝑞 (1)
Integrating again we get,
𝐸𝐼 𝑦 = 21.875 𝑥3
6 + 𝐶1𝑥 + 𝐶2 ⋮ −25
(𝑥 − 3)3
3⋮ +75
(𝑥 − 5)2
2 → 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=8, 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) = 21.875 03
6 + 𝐶10 + 𝐶2
𝐶2 = 0
Substituting condn (ii) at x=8, y=0 into complete Eq (2) (as x=6 lies in the last part of the
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
Substituting the values of 𝐶1 and 𝐶2 into Eq(1) and Eq(2) we get,
Slope Equation:
𝐸𝐼 𝑑𝑦
𝑑𝑥 = 21.875
𝑥2
2− 145.43 ⋮ −50
(𝑥 − 3)2
2⋮ +75(𝑥 − 5)
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Deflection Equation:
𝐸𝐼 𝑦 = 21.875 𝑥3
6 − 145.43𝑥 + 𝐶2 ⋮ −25
(𝑥 − 3)3
3⋮ +75
(𝑥 − 5)2
2
(5M)
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 = 21.875 𝑥2
2− 145.43 ⋮ −50
(𝑥 − 3)2
2
0 = 21.875 𝑥2
2 − 145.43 ⋮ −25(𝑥2 + 9 − 6𝑥)
Solving , 𝑥 = 3.88m
Hence the maximum deflection will be at a distance of 3.88m from support A.
Maximum Deflection:
Substituting x=3.88 into Eq (2) upto second dotted line, (as x=3.88 lies in the second part of the
beam) we get the maximum deflcetion as,
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.
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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
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,
𝐸𝐼 𝑑2𝑦
𝑑𝑥2 =
𝑤 𝐿 𝑥
2 −
𝑤. 𝑥2
2
Integrating the above equation we get,
𝐸𝐼 𝑑𝑦
𝑑𝑥 =
𝑤 𝐿 𝑥2
4 −
𝑤. 𝑥3
6+ 𝐶1 → 𝐸𝑞 (1)
Integrating again we get,
𝐸𝐼 𝑦 =𝑤 𝐿 𝑥3
12 −
𝑤. 𝑥4
24+ 𝐶1𝑥 + 𝐶2 → 𝐸𝑞 (2)
(5M)
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=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
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𝐶1 = −𝑤. 𝐿3
24
Substituting the values of 𝐶1 and 𝐶2 into Eq(1) and Eq(2) we get,
Slope Equation:
𝐸𝐼 𝑑𝑦
𝑑𝑥 =
𝑤 𝐿 𝑥2
4 −
𝑤. 𝑥3
6−
𝑤. 𝐿3
24
Deflection Equation:
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)
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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
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= −20𝑥
𝑥
2= −10𝑥2 − − − − − −(1)
Integrating the above equation we get,
𝐸𝐼𝑑𝑦
𝑑𝑥= −
10 𝑥3
3+ 𝐶1 − − − − − −(2)
Integrating the above equation once again we get,
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:
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when x=l, deflection becomes zero.
𝐸𝐼 (0) = −10 (2.3)4
12+ 40.56(2.3) + 𝐶2
𝐶2 = −69.96
Substituting the values in eqns 2 and 3 we get,
Slope Equation:
𝐸𝐼𝑑𝑦
𝑑𝑥= −
10 𝑥3
3+ 40.56
Deflection Equation:
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:
Substitute x=0 in the deflection equation we get,
𝐸𝐼𝑦𝑓𝑟𝑒𝑒 = −10 (0)4
12+ 40.56(0) − 69.96
𝑦𝑓𝑟𝑒𝑒 = −69.96
12𝑥103
𝑦𝑓𝑟𝑒𝑒 = −5.83𝑥10−3𝑚
(5M)
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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,
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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.
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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
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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
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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:
𝛿𝑑 = 𝑝𝑑2
2𝑡𝐸 (1 −
𝜇
2)
𝛿𝑑 = 1𝑥106𝑥(0.7)2
2𝑥8𝑥10−3𝑥2𝑥1011 (1 −
0.3
2)
𝛿𝑑 = 0.1301 𝑥10−3𝑚
Change in length:
𝛿𝑙 = 𝑝𝑑𝑙
2𝑡𝐸 (
1
2− 𝜇)
𝛿𝑙 = 1𝑥106𝑥0.7𝑥2.5
2𝑥8𝑥10−3𝑥2𝑥1011 (
1
2− 0.3)
𝛿𝑙 = 0.1093𝑥10−3𝑚
Change in Volume:
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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
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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
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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:
Circumferential stress,
𝜎𝑐 = 𝑝𝑑
2𝑡
𝜎𝑐 = 3𝑥106𝑥1
2𝑥1𝑥10−2
𝜎𝑐 = 1.5 𝑥 104 𝑁/𝑚2
Longitudinal stress,
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𝜎𝑙 = 𝑝𝑑
4𝑡
𝜎𝑙 = 3𝑥106𝑥1
4𝑥1𝑥10−2
𝜎𝑐 = 0.75 𝑥 104 𝑁/𝑚2
(6M) Change in diameter:
𝛿𝑑 = 𝑝𝑑2
2𝑡𝐸 (1 −
𝜇
2)
𝛿𝑑 = 3𝑥106𝑥(1)2
2𝑥1𝑥10−2𝑥2𝑥1011 (1 −
0.3
2)
𝛿𝑑 = 0.6375 𝑥10−3𝑚
Change in length:
𝛿𝑙 = 𝑝𝑑𝑙
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
Volume 𝑉 =𝜋𝑑2𝐿
4
𝑉 =𝜋𝑥12𝑥5
4= 3.926 𝑚3
𝛿𝑉 = 1.425𝑥10−3𝑥3.926
𝛿𝑉 = 5.5959𝑥10−3𝑚3
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(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
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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
𝛿𝑑 = 𝑝𝑑2
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
𝛿𝑙 = 𝑝𝑑𝑙
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
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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
The radial pressure px is given by the eqn,
𝑝𝑥 =𝑏
𝑥2− 𝑎
Apply the boundary condition to above equations,
1. At x= r1 = 200mm, 𝑝𝑥 = 8 N/mm2
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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
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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
The radial pressure px is given by the eqn,
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)
Apply the boundary condition to above equation,
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
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Now the radial pressure is given by,
𝑝𝑥 =2𝑥5000000
𝑥3− 3
Let r2=External radius of the shell.
Apply the boundary condition to above equation,
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
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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)
𝛿𝑑 = 𝑝𝑑2
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,
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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
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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
The radial pressure px is given by the eqn,
𝑝𝑥 =2𝑏
𝑥3− 𝑎
Apply the boundary condition to above equation,
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)
Apply the boundary condition to above equation,
At x= r1 = 90mm, 𝜎𝑥 = 10 N/mm2
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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.
Apply the boundary condition to above equation,
At x= r2 , 𝑝𝑥 = 0
0 =2𝑥6804000
𝑟23 − 3
𝑟23 =
13608000
3
𝑟2 = (4.536𝑥106)1
3
𝑟2 = 165.53𝑚𝑚.
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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:
Circumferential stress,
𝜎𝑐 = 𝑝𝑑
2𝑡
𝜎𝑐 = 5𝑥106𝑥1.1
2𝑥2𝑥10−2
𝜎𝑐 = 1.375 𝑥 104 𝑁/𝑚2
Longitudinal stress,
𝜎𝑙 = 𝑝𝑑
4𝑡
𝜎𝑙 = 5𝑥106𝑥1.1
4𝑥2𝑥10−2
𝜎𝑐 = 0.6875 𝑥 104 𝑁/𝑚2
(5M)
Change in diameter:
𝛿𝑑 = 𝑝𝑑2
2𝑡𝐸 (1 −
𝜇
2)
𝛿𝑑 = 5𝑥106𝑥(1.1)2
2𝑥2𝑥10−2𝑥2.1𝑥1011 (1 −
0.33
2)
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𝛿𝑑 = 0.6013 𝑥10−3𝑚
Change in length:
𝛿𝑙 = 𝑝𝑑𝑙
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
Volume 𝑉 =𝜋𝑑2𝐿
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
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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)