ce8395 strength of materials for โฆ...stresses in thin cylindrical shell due to internal pressure...
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REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH/ Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0
5-1
CE8395 STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS L T P C
3 0 0 3
OBJECTIVES:
To understand the concepts of stress, strain, principal stresses and principal planes.
To study the concept of shearing force and bending moment due to external loads in
determinate beams and their effect on stresses.
To determine stresses and deformation in circular shafts and helical spring due to torsion.
To compute slopes and deflections in determinate beams by various methods.
To study the stresses and deformations induced in thin and thick shells.
UNIT I STRESS, STRAIN AND DEFORMATION OF SOLIDS 9
Rigid bodies and deformable solids โ Tension, Compression and Shear Stresses โ Deformation
of simple and compound bars โ Thermal stresses โ Elastic constants โ Volumetric strains โ
Stresses on inclined planes โ principal stresses and principal planes โ Mohrโs circle of stress.
UNIT II TRANSVERSE LOADING ON BEAMS AND STRESSES IN BEAM 9
Beams โ types transverse loading on beams โ Shear force and bending moment in beams โ
Cantilevers โ Simply supported beams and over โ hanging beams. Theory of simple bendingโ
bending stress distribution โ Load carrying capacity โ Proportioning of sections โ Flitched
beams โ Shear stress distribution.
UNIT III TORSION 9
Torsion formulation stresses and deformation in circular and hollows shafts โ Stepped shaftsโ
Deflection in shafts fixed at the both ends โ Stresses in helical springs โ Deflection of helical
springs, carriage springs.
UNIT IV DEFLECTION OF BEAMS 9
Double Integration method โ Macaulayโs method โ Area moment method for computation of
slopes and deflections in beams - Conjugate beam and strain energy โ Maxwellโs 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 โ spherical shells subjected to internal
pressure โDeformation in spherical shells โ Lameโs 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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JIT-JEPPIAAR/MECH/ Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
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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 โEngineering Mechanics of Solidsโ 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 โco. 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 โ Tension, Compression and Shear Stresses โ Deformation of simple
and compound bars โ Thermal stresses โ Elastic constants โ Volumetric strains โStresses on inclined
planes โ principal stresses and principal planes โ Mohrโs 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 Mohrโs 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
๐ธ = 3๐พ(1 โ 2๐) Where,
E = Youngโs Modulus in N/m2.
K = Bulk Modulus in N/m2.
ยต = Poissonโs Ratio.
6
Define Youngโs 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 = Youngโs modulus of the bar material.
w = weight density of the material.
Strain in the element = ๐๐ก๐๐๐ ๐
๐ธ=
๐ค๐ฅ
๐ธ
Elongation of the element = Strain x Length of the element. = ๐ค๐ฅ
๐ธ๐๐ฅ
Total Elongation = โซ๐ค ๐ฅ
๐ธ
๐ฟ
0 dx
ฮดL = WL/2E.
9 What does the radius of Mohrโs circle refer to? (Apr/May 2017) BTL1
Radius of Mohrโs Circle is equal to the maximum shear stress.
10
What is Hookeโs law? BTL1
Within the elastic limit, when a body is loaded, then stress induced is proportional to the strain.
This is called Hookeโs law.
Stress ฮฑ Strain
๐๐ก๐๐๐ ๐
๐๐ก๐๐๐๐= ๐ถ๐๐๐ ๐ก๐๐๐ก (๐๐๐ข๐๐โฒ๐ ๐๐๐๐ข๐๐ข๐ ๐ธ)
11 Define Poissonโs 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 Poissonโs ratio (ยต).
Poissonโs ratio (ยต or 1
๐) =
๐๐๐ก๐๐๐๐ ๐ ๐ก๐๐๐๐
๐๐๐๐๐๐ก๐ข๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐ ๐ ๐ก๐๐๐๐
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) = ๐ท๐๐๐๐๐ก ๐๐ก๐๐๐ ๐
๐๐๐๐ข๐๐๐ก๐๐๐ ๐ ๐ก๐๐๐๐
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
๐1 = ๐๐ฅ + ๐๐ฆ
2+ โ(
๐๐ฅ โ ๐๐ฆ
2)2 + ๐
Minor normal principal stress
๐2 = ๐๐ฅ + ๐๐ฆ
2โ โ(
๐๐ฅ โ ๐๐ฆ
2)2 + ๐
17 Give the expression for norma stress and tangential stress for a member subjected to axial
load on a oblique plane at ฮธ. BTL2
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Normal Stres, ๐๐ = ๐ ๐๐๐ 2๐
Tangential Stres, ๐๐ก = ๐
2๐ ๐๐2๐
where ๐ = ๐ฟ๐๐๐
๐ด๐๐๐
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 ) Youngโs modulus (E)
(ii) Modulus of Rigidity (or) Rigidity modulus (C, G or N)
(iii) Bulk modulus (K)
The relaionship between the three constants is
๐ธ = 3๐พ(1 โ 2๐) = 2๐บ(1 + ๐)
๐ธ = 9๐พ๐บ
3๐พ + ๐บ
Where,
E = Youngโs Modulus in N/m2.
K = Bulk Modulus in N/m2.
ยต = Poissonโs 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
๐1 ๐
๐ด1๐ธ1=
๐2 ๐
๐ด2๐ธ2
(ii) Total Load = Load in part 1 + Load in part 2
๐ = ๐1 + ๐2 ๐ = ๐1๐ด1 + ๐2๐ด2
P1, P2 - Loads in section 1 and 2
A1, A2 โ Area of section 1 and 2
E1, E2 โ Youngโs modulus of section 1 and 2
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ฯ1, ฯ 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 = ๐๐๐ก๐๐๐๐ก๐ ๐๐ก๐๐๐ ๐
๐๐๐๐๐๐๐ ๐๐ ๐ด๐๐๐๐ค๐๐๐๐ ๐๐ก๐๐๐ ๐
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, ๐ด๐ = ๐๐๐
2
4 =
๐ (32 ๐ฅ 10โ3)2
4 = 8.04๐ฅ10โ4 ๐2
Area of Aluminium tube, ๐ด๐๐ = ๐[๐ท๐๐
2 โ ๐๐๐2 ]
4 =
๐[ (45 ๐ฅ 10โ3)2โ (35 ๐ฅ 10โ3)2]
4 = 6.282๐ฅ10โ4 ๐2
Stress in Steel = Stress in Aluminium
ํ๐ = ํ๐๐ ๐๐
๐ธ๐ =
๐๐๐
๐ธ๐๐
๐๐ = 2๐ฅ1011
8๐ฅ1010 ๐๐๐
๐๐ = 2.5 ๐๐๐
(7M) Total Load = Load in steel + Load in Aluminium
๐ = ๐๐ + ๐๐๐ ๐ = ๐๐ ๐ด๐ + ๐๐๐๐ด๐๐ 300x103 = [2.5 ๐๐๐ ๐ฅ 8.04๐ฅ10โ4 ] + [๐๐๐ ๐ฅ 6.282๐ฅ10โ4]
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๐๐๐ = 113.71x106 N/m2.
๐๐ = 2.5 ๐ฅ 113.71x106
๐๐ = 284.28x106 N/m2.
(6M)
2
A point in a strained material is subjected to stress as shown below. Using Mohrโs Cricle
method, determine the normal and tangential stress across the oblique plane. Check the
answer analytically. (13M) (Apr/May 2015) BTL 5
Given:
๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ , ๐1 = 65 ๐๐๐2โ
๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ , ๐2 = 35 ๐๐๐2โ
๐โ๐๐๐ ๐๐ก๐๐๐ ๐ , ๐ = 25 ๐๐๐2โ
Angle of Oblique Plane, ๐ = 45ยฐ
Mohrโs Circle Method:
Let 1 cm = 10 ๐๐๐2โ
Then, ๐1 =65
10= 6.5๐๐
๐2 =35
10= 3.5๐๐
๐ =25
10= 2.5๐๐
(4M)
Procedure:
Take any point A and draw a horizontal line through A.
Take AB = ๐1 = 6.5 ๐๐ and AC = ๐2 = 3.5 ๐๐ towards right of A.
Draw perpendicular at B and C and cut off BF and CG equal to shear stress ๐ = 2.5 ๐๐.
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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 2ฮธ i.e., 2x45= 90ยฐ 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.
๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐ , ๐๐ = ๐ฟ๐๐๐๐กโ ๐ด๐ท ๐ฅ ๐๐๐๐๐ = 7.5 ๐ฅ 10 = 75 ๐๐๐2โ
๐๐๐๐๐๐๐ก๐๐๐ ๐๐ก๐๐๐ ๐ , ๐๐ก = ๐ฟ๐๐๐๐กโ ๐ธ๐ท ๐ฅ ๐๐๐๐๐ = 1.5 ๐ฅ 10 = 15 ๐๐๐2โ
(6M)
Analytical Method:
Normal stress ๐๐ is given by the equation,
๐๐ = ๐1 + ๐2
2+
๐1 โ ๐2
2 ๐๐๐ 2๐ + ๐ ๐ ๐๐2๐
๐๐ = 65 + 35
2+
65 โ 35
2 cos (2๐ฅ45) + 25 sin (2๐ฅ45)
๐๐ = 50 + 15 cos 90 + 25 sin 90
๐๐ = 50 + 25
๐๐ = 75 ๐๐๐2โ
Tangential stress ๐๐ก is given by the equation,
๐๐ก = ๐1 โ ๐2
2 ๐ ๐๐2๐ โ ๐ ๐๐๐ 2๐
๐๐ก = 65 โ 35
2sin(2๐ฅ45) โ 25 cos (2๐ฅ45)
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๐๐ก = 15 sin 90 โ 25 cos 90
๐๐ก = 15 โ 0
๐๐ก = 15 ๐๐๐2โ
(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 Poissonโs 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, ยต = 0.25.
Soln:
Volume V = 300x100x40 = 1200000mm3.
Stress in x โdirection = ๐ฟ๐๐๐ ๐๐ ๐ฅโ๐๐๐๐๐๐ก๐๐๐
๐ฆ ๐ง=
5000
100 ๐ฅ 40 = 1.25 N/mm2.
Stress in y โdirection = ๐ฟ๐๐๐ ๐๐ ๐ฆโ๐๐๐๐๐๐ก๐๐๐
๐ฅ ๐ง=
6000
300 ๐ฅ 40 = 0.5 N/mm2.
Stress in z โdirection = ๐ฟ๐๐๐ ๐๐ ๐งโ๐๐๐๐๐๐ก๐๐๐
๐ฅ ๐ฆ =
4000
300 ๐ฅ 100 = 0.133 N/mm2.
(6M) ๐๐
๐=
1
๐ธ(๐๐ฅ + ๐๐ฆ + ๐๐ง)(1 โ 2๐) =
1
2๐ฅ105(1.25 + 0.5 + 0.133)(1 โ 2๐ฅ0.25)
๐๐
๐=
1.883
4๐ฅ105
๐๐ = 1.883
4๐ฅ105 (1200000)
๐๐ = 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, ๐ด๐ =๐๐ท๐
2
4 =
๐ 302
4 = 706.86 mm2.
Area of Copper tube, ๐ด๐ =๐(๐ท๐,๐
2 โ ๐ท๐,๐2 )
4 =
๐ (502โ402)
4 = 706.86 mm2.
Condn (i) Strain in steel rod = Strain in copper tube
๐๐
๐ธ๐ =
๐๐
๐ธ๐
๐๐ =2.1 ๐ฅ 105
1.1 ๐ฅ 105 ๐๐
๐๐ = 1.906 ๐๐
(6M) Condn (ii) Load on steel rod + Load on copper tube = Total Load
w.k.t, Load = Stress x Area.
๐๐ ๐ด๐ + ๐๐ถ ๐ด๐ = ๐
1.906 ๐๐ ๐ฅ 706.86 + ๐๐ถ ๐ฅ 706.86 = 45000
2056.25 ๐๐ถ = 45000
๐๐ถ = 21.88 N/mm2.
๐๐ = 1.906 ๐ฅ 21.88
๐๐ = 41.77 N/mm2
Load carried by steel rod ๐๐ = ๐๐ ๐ด๐
๐๐ = 41.77 ๐ฅ 706.86 = 29525.5N
Load carried by steel rod ๐๐ = ๐๐ ๐ด๐
๐๐ = 21.88 ๐ฅ 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
ยต = 1/m = 1/3 = 0.33
Soln:
Volume, V = ๐
4๐ท2๐ฟ =
๐ 302 2000
4 = 35.343x105 mm3.
Let ๐ฟL = Change in Length,
๐ฟd = Change in Diameter,
๐ฟV = Change in Volume.
Strain of length = ๐๐ก๐๐๐ ๐
๐ธ =
๐๐
4๐2 ๐ธ
Strain of length =50000
๐
4302 2๐ฅ105
= 0.0003536
(6M)
๐น๐ณ
๐ณ= 0.0003536
ฮดL = 0.0003536 x 2000 = 1.768mm.
Poissonโs ratio = ๐ฟ๐๐ก๐๐๐๐ ๐๐ก๐๐๐๐
๐ฟ๐๐๐๐๐ ๐๐ก๐๐๐๐
Lateral Strain = 0.33 x 0.0003536 = 0.0000884
Lateral Strain = ๐น๐
๐ = 0.0000884.
ฮดd = 0.0000884 x 30 = 0.002652mm.
Volumetric Strain ๐น๐ฝ
๐ฝ=
๐น๐ณ
๐ณโ
๐๐น๐
๐ = 0.0003536 โ 2x0.0000884
๐น๐ฝ
๐ฝ= 0.0001768
ฮดV = 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 80ยฐC.
Calculate the stresses included in each metal. ฮฑ for steel = 11.2 x 10-6 per ยฐC, ฮฑ for brass =
16.5 x 10-6 per ยฐC, 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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5-13
โT = 80ยฐ C, Es=2x105N/mm2= 2x1011N/m2, Eb=1x105N/mm2= 1x1011N/m2, ฮฑs =11.2x10-6 /ยฐC, ฮฑb
=16.5x10-6 /ยฐC
Since ฮฑb is greater than ฮฑs, 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, ๐ด๐ = ๐๐๐
2
4 =
๐ (20 ๐ฅ 10โ3)2
4 = 3.141๐ฅ10โ4 ๐2
Area of Brass tube, ๐ด๐ = ๐[๐ท๐
2โ ๐๐2]
4 =
๐[ (30 ๐ฅ 10โ3)2โ (25 ๐ฅ 10โ3)2]
4 = 4.091๐ฅ10โ4 ๐2
At Equilibrium,
Tensile Load on Steel = Compressive load on Brass
๐๐ ๐ด๐ = ๐๐๐ด๐ ๐๐ ๐ฅ 3.141๐ฅ10โ4= ๐๐ ๐ฅ 4.091๐ฅ10โ4
๐๐ = 1.302 ๐๐ (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.
๐ผ๐ โ๐ ๐ฟ + ๐๐
๐ธ๐ ๐ฟ = ๐ผ๐โ๐ ๐ฟ +
๐๐
๐ธ๐ ๐ฟ
๐ผ๐ โ๐ + ๐๐
๐ธ๐ = ๐ผ๐โ๐ +
๐๐
๐ธ๐
(11.2 ๐ฅ 10โ6 ๐ฅ 80) +1.302 ๐๐
2 ๐ฅ 1011= (16.5 ๐ฅ 10โ6 ๐ฅ 80) โ
๐๐
1 ๐ฅ 1011
(0.000896) + 6.51 ๐ฅ 10โ12 ๐๐ = (0.00132) + 1 ๐ฅ 10โ11 ๐๐
๐๐ = - 121.48x106 N/m2.
๐๐ = 1.302 ๐ฅ 121.48x106
๐๐ = 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 = ๐๐๐
2
4 =
๐ (20 ๐ฅ 10โ3)2
4 = 3.141๐ฅ10โ4 ๐2
Area of steel = 2 ๐ฅ ๐๐๐
2
4 = 2 ๐ฅ
๐ (20 ๐ฅ 10โ3)2
4 = 6.282๐ฅ10โ4 ๐2
Change in length of steel = Change in length of copper.
Strain of steel x original length = Strain in copper x original length
๐๐
๐ธ๐ ๐ฟ๐ =
๐๐
๐ธ๐ ๐ฟ๐
๐๐
2.1 ๐ฅ 1011 ๐ฅ 1 =
๐๐
1.1 ๐ฅ 1011 ๐ฅ 2
๐๐ = 3.81 ๐๐ (7M)
Total Load = Load in steel + Load in Copper
๐ = ๐๐ + ๐๐ ๐ = ๐๐ ๐ด๐ + ๐๐๐ด๐ 20x103 = [3.81 ๐ฅ ๐๐ ๐ฅ 6.282๐ฅ10โ4 ] + [๐๐ ๐ฅ 3.141๐ฅ10โ4]
๐๐ = 7.386x106 N/m2.
๐๐ = 3.81 ๐ฅ 7.386x106
๐๐ = 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:
๐๐ฅ = 140 ๐๐๐2โ
๐๐ฆ = โ100 ๐๐๐2โ
๐1 = 160 ๐๐๐2โ
Calculation of Shear stress:
๐1 =๐๐ฅ + ๐๐ฆ
2+ โ(
๐๐ฅ โ ๐๐ฆ
2)
2
+ ๐๐ฅ๐ฆ2
160 =140 + (โ100)
2+ โ(
140 โ (โ100)
2)
2
+ ๐๐ฅ๐ฆ2
160 = 20 + โ(120)2 + ๐๐ฅ๐ฆ2
๐๐ฅ๐ฆ = 72.11 ๐๐๐2โ
(4M) Calculation of Maximum Shear stress:
๐๐๐๐ฅ = โ(๐๐ฅ โ ๐๐ฆ
2)
2
+ ๐๐ฅ๐ฆ2
๐๐๐๐ฅ = โ(140 โ (โ100)
2)
2
+ (72.11)2
๐๐ฅ๐ฆ = 140 ๐๐๐2โ
(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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5-16
Given:
๐๐ฅ = 51.96 ๐๐๐2โ
๐๐ฆ = 40 ๐๐๐2โ
๐๐ฅ๐ฆ = 30 ๐๐๐2โ
Calculation of Shear stress:
๐1,2 =๐๐ฅ + ๐๐ฆ
2ยฑ โ(
๐๐ฅ โ ๐๐ฆ
2)
2
+ ๐๐ฅ๐ฆ2
๐1,2 =51.96 + 40
2ยฑ โ(
51.96 โ 40
2)
2
+ (30)2
๐1,2 = 45.98 ยฑ โ(5.98)2 + (30)2
๐1,2 = 45.98 ยฑ 30.59
๐1 = 76.57 ๐๐๐2โ
๐2 = 15.39 ๐๐๐2โ
(4M) Location of Principal Plane:
๐ก๐๐2๐ = 2๐๐ฅ๐ฆ
(๐๐ฅ โ ๐๐ฆ)
๐ก๐๐2๐ = 2๐ฅ30
(51.96 โ 40)
2๐ = 78.72ยฐ
๐ = 39.36ยฐ
(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. Youngโs 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 ฯBC = 150 N/mm2 = 150x106 N/m2, P =160kN = 160
x103 N, Total extension ฮดL = 0.2mm = 0.2 x10-3 m, E=2.1x105 N/mm2 = 2.1x1011 N/m2
Stress in middle portion:
๐๐ต๐ถ = ๐ฟ๐๐๐
๐ด๐๐๐=
๐
๐๐2
4
150 ๐ฅ106 = 160๐ฅ103๐ฅ 4
๐๐2
๐ = โ160 ๐ฅ 103๐ฅ 4
๐ ๐ฅ 150๐ฅ106
๐ = 0.0368 m
(6M)
๐ด1 = ๐ด3 = ๐ (6 ๐ฅ 10โ2)2
4 = 2.826๐ฅ10โ3 ๐2
A = ๐ (3.68 ๐ฅ 10โ2)2
4 = 1.063๐ฅ10โ3 ๐2
Total Extension, ๐ฟ๐ฟ = ๐
๐ธ[
๐ฟ1
๐ด1+
๐ฟ2
๐ด2+
๐ฟ3
๐ด3]
, ๐ฟ๐ฟ = ๐
๐ธ[
0.4โ๐ฟ
๐ด1+
๐ฟ
๐ด]
0.2๐ฅ10โ3 = 160๐ฅ103
2.1๐ฅ1011[
0.4 โ ๐ฟ
2.826๐ฅ10โ3+
๐ฟ
1.063๐ฅ10โ3]
262.5 = [0.0004252 โ 1.063๐ฅ10โ3๐ฟ
2.826๐ฅ10โ3+
2.826๐ฅ10โ3๐ฟ
1.063๐ฅ10โ3]
0.000788 = [0.0004252 + 0.001763๐ฟ]
0.0003628 = [0.001763๐ฟ]
๐ฟ = 0.205๐
(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) Youngโs
modulus, (ii)Poissonโs ratio and (iii) Bulk modulus. (13M) (Apr/May 2017) BTL 5
Given:
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d =30mm= 30x10-3m, P=60kN = 60x103N, L = 200mm = 200x10-3m, ฮดL=0.1mm = 0.1x10-3m, ฮดd
=0.004mm=0.004x10-3m
Area = ๐๐2
4 =
๐ (30 ๐ฅ 10โ3)2
4 = 7.067๐ฅ10โ4 ๐2
Youngโฒs Modulus = ๐๐ก๐๐๐ ๐
๐๐ก๐๐๐๐
stress = ๐ฟ๐๐๐
๐ด๐๐๐=
60๐ฅ103
7.067๐ฅ10โ4
ฯ = 84.901x106 N/m2.
Linear strain = ๐ฟ๐ฟ
๐ฟ=
0.1๐ฅ10โ3
200๐ฅ10โ3
ฮต = 5x10-4
Youngโฒs Modulus = 84.901x 106
5x10โ4
E = 169.8 x 109 N/m2.
(6M)
๐๐๐๐ ๐ ๐๐โฒ๐ ๐ ๐๐ก๐๐ = ๐ฟ๐๐ก๐๐๐๐ ๐๐ก๐๐๐๐
๐ฟ๐๐๐๐๐ ๐๐ก๐๐๐๐
Lateral strain = ๐ฟ๐
๐=
0.004๐ฅ10โ3
30๐ฅ10โ3 = 1.33x10-4
๐๐๐๐ ๐ ๐๐โฒ๐ ๐ ๐๐ก๐๐ =1
๐ ๐๐ ๐ =
1.33x10โ4
5x10โ4
1
๐ ๐๐ ๐ = 0.266
Bulk Modulus
๐ธ = 3๐พ[1 โ2
๐]
๐พ = 169.8 ๐ฅ 109
3(1 โ 2๐ฅ0.266)
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 โWโ (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=Youngโs 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,
๐ท๐ฅ = ๐ท1 โ (๐ท1 โ ๐ท2
๐ฟ) ๐ฅ = ๐ท1 โ ๐๐ฅ
(4M)
Area of cross section of the bar at a distance โxโ from the left end,
๐ด๐ฅ = ๐๐ท๐ฅ
2
4=
๐(๐ท1 โ ๐๐ฅ)2
4
Now the stress at a distance โxโ from the left end is given by,
๐๐ฅ = ๐ฟ๐๐๐
๐ด๐ฅ=
๐๐(๐ท1โ๐๐ฅ)2
4
=4๐
๐(๐ท1 โ ๐๐ฅ)2
Now the strain in the small element of length dx is given by,
๐๐ฅ = ๐๐ก๐๐๐ ๐
๐ธ=
๐๐ฅ
๐ธ=
4๐
๐๐ธ(๐ท1 โ ๐๐ฅ)2
Extension of the small element length dx,
๐ฟ๐ฟ = ๐๐ก๐๐๐๐. ๐๐ฅ = ๐๐ฅ. ๐๐ฅ
๐ฟ๐ฟ =4๐
๐๐ธ(๐ท1 โ ๐๐ฅ)2 . ๐๐ฅ
The total extension of the bar is obtained by integrating the above equation between the limits 0
and L.
๐ฟ๐ฟ = โซ4๐. ๐๐ฅ
๐๐ธ(๐ท1 โ ๐๐ฅ)2
๐ฟ
0
The above expression can be integrated and simplified as,
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5-21
๐ฟ๐ฟ =4๐๐ฟ
๐๐ธ๐ท2
(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, ๐ด๐ =๐๐ท๐
2
4 = 4 x
๐ 102
4 = 314.16 mm2.
Area of concrete column, ๐ด๐ = (500๐ฅ500) โ ๐ด๐ = 249685.84 mm2.
Condn (i) Strain in steel rod = Strain in copper tube
๐๐
๐ธ๐ =
๐๐
๐ธ๐
๐๐ =2.1 ๐ฅ 105
1.4 ๐ฅ 105 ๐๐
๐๐ = 1.5 ๐๐
(5M) Condn (ii) Load on steel rod + Load on concrete column = Total Load
w.k.t, Load = Stress x Area.
๐๐ ๐ด๐ + ๐๐ถ ๐ด๐ = ๐
1.5 ๐๐ ๐ฅ 314.16 + ๐๐ถ ๐ฅ 249685.84 = 2x106
250156.88๐๐ถ = 2x106
(5M)
๐๐ถ = 7.99 N/mm2.
๐๐ = 1.5 ๐ฅ 7.99
๐๐ = 11.99 N/mm2
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5-22
Load carried by steel rod ๐๐ = ๐๐ ๐ด๐
๐๐ = 11.99 ๐ฅ 314.16 = 3767.5N
Load carried by steel rod ๐๐ = ๐๐ ๐ด๐
๐๐ = 7.99 ๐ฅ 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 , Poissonโs 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, ยต = 0.25.
Soln:
(i) Change in Volume:
Volume V = 250x100x50 = 1250000mm3.
Stress in x โdirection = ๐ฟ๐๐๐ ๐๐ ๐ฅโ๐๐๐๐๐๐ก๐๐๐
๐ฆ ๐ง=
500000
100 ๐ฅ 50 = 100 N/mm2.
Stress in y โdirection = ๐ฟ๐๐๐ ๐๐ ๐ฆโ๐๐๐๐๐๐ก๐๐๐
๐ฅ ๐ง=
2500000
250 ๐ฅ 50 = 200 N/mm2.
Stress in z โdirection = ๐ฟ๐๐๐ ๐๐ ๐งโ๐๐๐๐๐๐ก๐๐๐
๐ฅ ๐ฆ =
โ5000000
250 ๐ฅ 100 = -200 N/mm2.
(5M) ๐๐
๐=
1
๐ธ(๐๐ฅ + ๐๐ฆ + ๐๐ง)(1 โ 2๐) =
1
1.8๐ฅ105(100 + 200 โ 200)(1 โ 2๐ฅ0.25)
๐๐
๐=
50
1.8๐ฅ105
๐๐ = 50
1.8๐ฅ105 (1250000)
๐๐ = 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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5-23
For no change to occur in the volume of bar volumetric strain must be equal to zero.
๐๐
๐= 0
Make , Fz as unknown ,
Stress in z โdirection = ๐ฟ๐๐๐ ๐๐ ๐งโ๐๐๐๐๐๐ก๐๐๐
๐ฅ ๐ฆ =
โ๐น๐
250 ๐ฅ 100 = -4x10-5 Fz N/mm2
๐๐
๐=
1
๐ธ(๐๐ฅ + ๐๐ฆ + ๐๐ง)(1 โ 2๐)
0 = 1
1.8๐ฅ105(100 + 200 โ 4x10โ5 F๐ง)(1 โ 2๐ฅ0.25)
0 = (300 โ 4x10โ5 F๐ง)
4x10โ5 F๐ง = 300
๐น๐ง =300
4x10โ5= 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 โ types transverse loading on beams โ Shear force and bending moment in beams โ Cantilevers โ
Simply supported beams and over โ hanging beams. Theory of simple bendingโ bending stress
distribution โ Load carrying capacity โ Proportioning of sections โ Flitched beams โ 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 obeyโs hookeโs law.
The Youngโs 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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๐ = ๐
2๐ผโ๐2
4โ ๐ฆ2โ
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 = ๐ผ
๐ฆ
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 โxโ 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,
๐๐๐ฃ๐ = ๐
๐ด
๐๐๐ฃ๐ = 40 ๐ฅ 103
150 ๐ฅ 200
๐๐๐ฃ๐ = 1.33 ๐ฅ 106 ๐/๐๐2
๐๐๐๐ฅ = 1.5 ๐๐๐ฃ๐
๐๐๐๐ฅ = 1.5 ๐ฅ 1.33 ๐ฅ 106
๐๐๐๐ฅ = 2 ๐ฅ 106 ๐/๐๐2
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, ๐๐
๐๐ฅ= โ๐น
18
Calculate the sectional modulus of a circular section of diameter 200mm. (May2012) BTL3
Section Modulus,
๐ = ๐ผ
๐ฆ= (
๐
64๐4) ๐ฅ (
2
๐)
๐ = ๐
32๐3
๐ = ๐
32(200)3
๐ = 785.4 ๐ฅ 103๐๐3
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,
๐ = ๐
2๐ผโ๐ท2
4โ ๐ฆ2โ
D โ Depth
y โ 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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 8 + (3 ๐ฅ 5) = 23kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 5 = (8 ๐ฅ 7) + (3 ๐ฅ 5 ๐ฅ 5
2 ) = 93.5
๐ ๐ต = 93.5
5= 18.7๐๐
๐ ๐ด = 23 โ 18.7 = 4.3๐๐
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) โ (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 โ (3 ๐ฅ ) = 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 โ (3 ๐ฅ ๐ฅ
2)
= -8(3.56+2) + 18.7(3.56) โ 3(3.56)(3.56/2)
= -44.48 + 66.57 โ 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=ฯ x Z
Where, Z=section modulus
As all the three beams have the same alowable bending stress (ฯ), 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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๐ =๐ผ
๐ฆ=
๐๐3
12๐
2
๐ =๐ฅ. ๐ฅ3
12.2
๐ฅ
๐ =๐ฅ3
6
Section modulus of a rectangular beam
๐ =๐ผ
๐ฆ=
๐๐3
12๐
2
๐ =๐. (2๐)3
12.
2
2๐
๐ =๐. 8๐3
12.
2
2๐
๐ =2๐3
3
Section modulus of a Circular beam
๐ =๐ผ
๐ฆ=
๐๐4
64๐
2
๐ =๐๐4
64.2
๐
๐ =๐๐3
32
(6M)
Equating the section modulus of a square beam with that of a rectangular beam, we get
๐ฅ3
6=
2๐3
3
๐3 =๐ฅ3
4= 0.25๐ฅ3
๐ = 0.63๐ฅ
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Equating the section modulus of a square beam with that of circular beam, we get
๐ฅ3
6=
๐๐3
32
๐3 =32๐ฅ3
6๐= 10.18๐ฅ3
๐ = 1.1927๐ฅ
The weights of the beams are proportional to their cross-sectional areas. Hence
๐๐๐๐โ๐ก ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐๐๐๐
๐๐๐๐โ๐ก ๐๐ ๐ ๐๐ข๐๐๐ ๐๐๐๐=
๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐๐๐๐
๐ด๐๐๐ ๐๐ ๐ ๐๐ข๐๐๐ ๐๐๐๐
=๐ ๐ฅ 2๐
๐ฅ ๐ฅ ๐ฅ
=0.63๐ฅ ๐ฅ 2๐ฅ 0.63๐ฅ
๐ฅ ๐ฅ ๐ฅ
= 0.7938
๐๐๐๐โ๐ก ๐๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐๐
๐๐๐๐โ๐ก ๐๐ ๐ ๐๐ข๐๐๐ ๐๐๐๐=
๐ด๐๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐๐
๐ด๐๐๐ ๐๐ ๐ ๐๐ข๐๐๐ ๐๐๐๐
=
๐๐2
4
๐ฅ2
=๐๐2
4๐ฅ2
=๐(1.1927๐ฅ)2
4๐ฅ2
= 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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = (10 ๐ฅ 4) = 40kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 8 = (10 ๐ฅ 4 ๐ฅ (4
2+ 1)) = 120
๐ ๐ต = 120
8= 15๐๐
๐ ๐ด = 40 โ 15 = 25๐๐
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 ๐ฅ = 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) โ (10 ๐ฅ ๐ฅ
2)
= 15(1.5+3) โ 10(1.5)(1.5/2)
= 67.5 โ 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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = (10 ๐ฅ 9) + 15 = 105kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 7 = (15 ๐ฅ 8.5) + (10 ๐ฅ 7 ๐ฅ (7
2)) โ (10 ๐ฅ 2 ๐ฅ (
2
2)) = 392.5 127.5+245-20
๐ ๐ต = 352.5
7= 50.35๐๐
๐ ๐ด = 105 โ 50.35 = 54.65๐๐
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 ๐ฅ = 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 โ (10 ๐ฅ ๐ฅ
2)
= -15(3.535+1.5) +50.35x3.535 โ 10(3.535)( 3.535/2)
= -75.525 + 177.987โ62.48
= 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, ๐ค =55200
4
w = 13.8x103 N
depth, d=2b
Type of beam = simply supported beam.
Length of UDL, l=4m
Maximum bending stress, ๐๐ = 10 ๐๐๐2โ
w.k.t Bending Stress as
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5-38
๐๐ =๐ ๐ฆ
๐ผ
(7M)
Reaction at Supports of the Beam:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = (13.8 ๐ฅ 4) = 55.2kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 4 = (13.8 ๐ฅ 4 ๐ฅ (4
2))
๐ ๐ต = 110.4
4= 27.6๐๐
๐ ๐ด = 55.2 โ 27.6 = 27.6๐๐
Maximum bending moment occurs at mid-point.
Maximum bending moment,
BMmax = (27.6 x 2) โ(13.8x2x1)
BMmax = 27.6 kN-m.
Moment of Inertia of the cross section of the beam
๐ผ = ๐๐3
12 =
๐(2๐)3
12
I = 0.667b4.
10 =27.6๐ฅ 106
0.667๐4
2๐
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, ฯ2=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
๐ผ2 =๐2๐2
3
12=
10๐ฅ203
12
๐ผ2 = 6666.67๐ฅ104๐๐4
Moment of Inertia of the two steel plates about N.A
๐ผ1 = 2 ๐ฅ๐1๐1
3
12= 2 ๐ฅ
1๐ฅ203
12
๐ผ1 = 1333.33๐ฅ104๐๐4
(7M)
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Modular ratio between steel and wood is given by,
๐ =๐ธ1
๐ธ2
๐ =2๐ฅ105
1๐ฅ104= 20
The Equivalent moment of inertia I is given by
๐ผ = ๐๐ผ1 + ๐ผ2
๐ผ = 20๐ฅ1333.33๐ฅ104 + 6666.67๐ฅ104
๐ผ = 33333.2๐ฅ104
Moment of resistance of the composite section ins given by te equation,
๐ =๐2
๐ฆ๐ฅ ๐ผ
๐ =7๐ฅ104๐ฅ33333.2
10๐ฅ10
๐ = 23333.24 ๐ โ ๐
Maximum Stress in Steel
Using the equation, ๐1
๐ธ1=
๐2
๐ธ2
๐1 =๐ธ1
๐ธ2๐ฅ ๐2
๐1 = 20 ๐ฅ 7
๐1 = 140 ๐/๐๐2
(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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 8 + 10 + 15 = 33kN
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Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 5 = (8 ๐ฅ 1.5) + (10 ๐ฅ 2.5 ) + (15 ๐ฅ 4 ) = 97
๐ ๐ต = 97
5= 19.4 ๐๐
๐ ๐ด = 33 โ 19.4 = 13.6๐๐
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,
๐๐ =๐ ๐ฆ
๐ผ
The maximum bending stress occurs at the fixed end support.
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5-43
BMmax = M = 12 x 2 x 1 = 24kN-m.
M=24 x106 N-mm.
The moment of inertia of the cross section of the beam,
๐ผ = ๐๐3
12 =
230(300)3
12
I = 517.5 x106 mm4.
๐๐ =24 x106
517.5 x106 ๐ฅ
300
2
๐๐ = 6.956 ๐/๐๐2
(7M) w.k.t the maximum shear stress for rectangular cross section
๐๐๐๐ฅ = 1.5 ๐ฅ ๐๐๐ฃ
w.k.t the maximum shear force from the diagram as,
Fmax = 12 x 2 = 24kN = 24 x103N.
๐๐๐๐ฅ = 1.5 ๐ฅ 24 ๐ฅ103
230 ๐ฅ 300
๐๐๐๐ฅ = 0.5217 ๐/๐๐2
(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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5-44
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 25kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 4 = (25 ๐ฅ 2 ) โ 18.75
๐ ๐ต = 31.25
4= 7.81 ๐๐
๐ ๐ด = 25 โ 7.81 = 17.19๐๐
Shear Force Diagram:
SF@B = -7.81 ๐๐
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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = (10 ๐ฅ 3) + (5 ๐ฅ 2) = 40kN
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5-46
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 7 = (5 ๐ฅ 2 ๐ฅ (2
2+ 5)) + (10 ๐ฅ 3 ๐ฅ (
3
2)) = 105
๐ ๐ต = 105
7= 15๐๐
๐ ๐ด = 40 โ 15 = 25๐๐
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 ๐ฅ = 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 โ (10 ๐ฅ ๐ฅ
2)
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5-47
= -15(3.535+1.5) +50.35x3.535 โ 10(3.535)( 3.535/2)
= -75.525 + 177.987โ62.48
= 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, ๐๐ค๐๐ก๐๐ =๐๐2๐ฟ
4
๐๐ค๐๐ก๐๐ =๐(0.5)210
4= 1.121๐3
Weight of water = weight density of water x volume of water
Weight = 10000x1.121= 11210 N.
Volume of the pipe, ๐๐๐๐๐ =๐[๐ท2โ๐2]๐ฟ
4
๐๐ค๐๐ก๐๐ =๐[(0.54)2 โ (0.5)2]10
4= 0.3266๐3
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,
๐ =๐ค๐2
8=
3472.5(10)2
8
๐ = 43406.25 ๐ โ ๐
(5M) The maximum stress that can be obtained is given by the expression,
๐๐ =๐ ๐ฆ
๐ผ
๐ผ =๐
64[๐ท4 โ ๐4]
๐ผ =๐
64[0.544 โ 0.54]
๐ผ = 2.945 ๐ฅ 10โ3๐4
๐๐ =43406.25
2.945 ๐ฅ 10โ3 ๐ฅ
0.54
2
๐๐ = 0.9945๐ฅ106 ๐/๐๐2
(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 โyโ from the N.A. The Shear stress at this level is given by,
๐ = ๐๐ด๏ฟฝฬ ๏ฟฝ
๐ผ๐
Where, ๐ด๏ฟฝฬ ๏ฟฝ = Moment of the shaded area about N.A
= Area of triangle BEF x Distance of CG of triangle BEF from N.A
= (1
2๐ฅ 2๐ฅ ๐ฅ ๐ฅ)(
๐
2โ
2
3 ๐ฅ)
= ๐ฅ2(๐
2โ
2
3 ๐ฅ)
I = Moment of Inertia about the N.A
๐ผ = 2 ๐ฅ๐ ๐ฅ (
๐
2)3
12=
๐4
48
Substituing the values we get,
๐ = ๐๐ฅ2(
๐
2โ
2
3 ๐ฅ)
๐4
48 ๐ฅ 2๐ฅ
๐ = 4๐๐ฅ(3๐ โ 4๐ฅ)
๐4
At the top, x=0, ๐ = 0
At the N.A ๐ฅ =๐
2 ,
๐ = 4๐๐(3๐ โ 4๐ฅ)
2๐4
๐ = 2๐
๐2
(5M) Maximum Shear Stress:
Maximum shear stress wil be obtained by differetiating the above equation and equating to zero.
๐
๐๐ฅโ
4๐
๐4 (3๐๐ฅ โ 4๐ฅ2)โ = 0
4๐
๐4 (3๐ โ 8๐ฅ)= 0
4๐
๐4 ๐๐๐๐๐๐ก ๐๐ ๐๐๐ข๐๐ ๐ก๐ ๐ง๐๐๐
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5-50
(3๐ โ 8๐ฅ)= 0
๐ฅ = 3๐
8
Substituting the value of โxโ , we get the maximum shear stress as,
๐๐๐๐ฅ = 4๐
๐4๐ฅ
3๐
8(3๐ โ 4
3๐
8)
๐๐๐๐ฅ = 9๐น
4๐2
(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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = (1๐ฅ3) + (0.5๐ฅ3) = 4.5kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 4 = (0.5 ๐ฅ 3)๐ฅ(4 + (3
2)) โ (1๐ฅ3)๐ฅ (
3
2) = 3.75
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5-51
๐ ๐ต = 3.75
4= 0.94 ๐๐
๐ ๐ด = 4.5 โ 0.94 = 3.56๐๐
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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5-52
(8M)
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5-53
UNIT III - TORSION
Torsion formulation stresses and deformation in circular and hollows shafts โ Stepped shaftsโ Deflection
in shafts fixed at the both ends โ Stresses in helical springs โ 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 ๐ฟ
๐
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
๐ = ๐
16 ๐ ๐3
20 ๐ฅ 106 = ๐
16 40 ๐3
๐ = โ(8๐ฅ106)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.
๐ = ๐1 + ๐2
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.,
๐ = ๐1 + ๐2
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 ๐ป
๐ฑ=
๐ฎ๐ฝ
๐ณ=
๐
๐น
T โ Torque
J โ Polar Moment of inertia
G โModulus of rigidity
ฮธ โ Angle of twist
L โLength of shaft
ฯ โ 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 Hookeโs 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.
๐๐ = ๐ฝ
๐
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.
๐ = ๐
16 ๐ โ
๐ท4 โ ๐4
๐ทโ
16 A closed coil helical springis to carry an axial load of 500N. Its mean coil diameter is to be
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5-56
10 times its wire diameter. Calculate this diameter if the maximum shear stress in the
material is to be 80MPa. BTL3
๐ = 16 ๐ ๐
๐๐ท3
80 = 16๐ฅ500๐ฅ
10๐ท
2
๐๐ท3
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
๐ = ๐๐ = ๐
16 ๐ ๐ท3
๐ = 16 ๐ ๐
๐๐ท3
๐ฟ = 64 ๐ ๐ 3๐
๐บ๐ท4
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
๐ = 2๐๐๐
60
100 ๐ฅ 103 = 2๐๐ฅ150๐ฅ๐๐๐๐๐
60
๐๐๐๐๐ = 6.366 ๐ฅ 103 ๐ โ ๐
๐๐๐๐ฅ = 1.2 ๐ฅ 6.366 ๐ฅ 103 ๐ โ ๐
๐๐๐๐ฅ = 7.64 ๐ฅ 103 ๐ โ ๐.
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 10ยฐ Helix angle is large (> 10ยฐ)
20 Write down the expression for power transmitted by a shaft. (May2013)BTL3
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5-57
๐ = 2๐๐๐
60
N โ Speed in rpm
T โ Torque in N-m.
P โ 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,
๐ฝ๐ =๐
32[๐4]
๐ฝ๐ =๐
32[504]
๐ฝ๐ = 61.34๐ฅ104๐๐4
๐ฝ๐๐ =๐
32[804 โ 504]
๐ฝ๐๐ = 340.76๐ฅ104๐๐4
Total Torque, ๐ = ๐๐ + ๐๐๐ = 107
w.k.t, ๐๐ = ๐๐๐
๐ =๐๐ฟ
๐บ๐ฝ
Since length is same, ๐๐
๐บ๐ ๐ฝ๐ =
๐๐๐
๐บ๐๐๐ฝ๐๐
๐๐ =๐๐๐๐บ๐ ๐ฝ๐
๐บ๐๐๐ฝ๐๐
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๐๐ =๐๐๐๐ฅ40๐ฅ103๐ฅ61.34๐ฅ104
80๐ฅ103๐ฅ340.76๐ฅ104
๐๐ = 0.09๐๐๐
0.09๐๐๐ + ๐๐๐ = 107
๐๐๐ = 9.17๐ฅ106๐ โ ๐๐
๐๐ = 8.25๐ฅ105๐ โ ๐๐
(6M)
Maximum stress developed is given by the equation,
๐๐ =๐๐
๐ฝ๐
๐๐
2
๐๐ =8.25๐ฅ105
61.34๐ฅ104
50
2
๐๐ = 33.62 ๐/๐๐2
๐๐๐ =๐๐๐
๐ฝ๐๐
๐๐๐
2
๐๐๐ =9.17๐ฅ106
340.76๐ฅ104
80
2
๐๐๐ = 107.64 ๐/๐๐2
Angle of twist,
๐๐ =๐๐ ๐ฟ๐
๐บ๐ ๐ฝ๐
๐๐ =8.25๐ฅ105๐ฅ2000
40๐ฅ103๐ฅ61.34๐ฅ104
๐๐ = 0.6724 ๐๐๐๐๐๐๐
(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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To find
(i) The maximum Shear Stress Induced:
w.k.t the maximum shear stress induced as,
๐ =16๐๐
๐๐3
๐ =16๐ฅ100๐ฅ50
๐(10)3
๐ = 25.46 ๐/๐๐2
(6M)
(ii) The deflection:
w.k.t the deflection of the spring as,
๐ฟ =64๐๐ 3๐
๐บ๐4
๐ฟ =64๐ฅ100๐ฅ(50)3๐ฅ15
8.16๐ฅ104๐ฅ(10)4
๐ฟ = 14.70๐๐
(ii) The Stiffness of the spring:
w.k.t the stiffness of the spring as,
๐ =๐บ๐4
64๐ 3๐
๐ =8.16๐ฅ104๐ฅ(10)4
64๐ฅ(50)3๐ฅ15
๐ = 0.425๐ฅ108๐/๐๐
(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
๐max = 60 ๐/๐๐2
w.k.t,
๐๐๐ค๐๐, ๐ = 2๐๐๐
60
300๐ฅ103 = 2๐(200)๐
60
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5-60
๐ = 14323.94 ๐ โ ๐
63M) w.k.t the torque transmitted by the shaft, TH as
๐๐ป =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ ๐ท๐
4
๐ท0โ
14323.94 =๐
16 ๐ฅ 60๐ฅ 106 ๐ฅ โ
0.124 โ ๐ท๐4
0.12โ
โ0. 124 โ ๐ท๐4โ =
0.12 ๐ฅ 14323.94 ๐ฅ 16
๐ ๐ฅ 60 ๐ฅ 106
โ2.07๐ฅ10โ4 โ ๐ท๐4โ = 145.9๐ฅ10โ6
โ ๐ท๐4โ = 0.62๐ฅ10โ4
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.
๐max = 125 ๐/๐๐2 ๐ = 1.5 ๐/๐๐2
w.k.t the maximum shear stress induced as,
๐ =16๐๐
๐๐3
125 =16๐ฅ60๐ฅ๐
๐๐3
๐ = 0.4089๐3
w.k.t the stiffness of the spring as,
๐ =๐บ๐4
64๐ 3๐
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5-61
1.5 =4.5x104x๐4
64๐ฅ(0.4089๐3)3๐ฅ๐
1.5 =4.5x104x๐4
64๐ฅ0.06836๐9๐ฅ๐
๐5๐ฅ๐ =4.5x104
15๐ฅ64๐ฅ0.06836
๐4๐ฅ50 =4.5x104
15๐ฅ64๐ฅ0.06836
๐4๐ฅ50 = 685.70
๐4 = 13.714
d=1.924mm
(6M)
๐ = 0.4089(1.924)3
๐ = 2.912๐๐
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
ฯ = 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,
๐ =๐
16๐๐ท3=======>(1)
Torque transmitted by the hollow shaft,
๐ =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ ๐ท๐
4
๐ท0โ
๐ =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ (0.6๐ท๐)4
๐ท0โ
๐ =๐
16 ๐ฅ ๐ ๐ฅ 0.8704๐ฅ๐ท0
3 =======>(2)
(6M) Since torque transmitted by the solid shaft is equal to hollow shaft,
๐
16๐๐ท3 =
๐
16 ๐ฅ ๐ ๐ฅ 0.8704๐ฅ๐ท0
3
๐ท3 = 0.8704๐ฅ๐ท03
๐ท = 0.9547๐ฅ๐ท๐
20๐ฅ10โ3 = 0.9547๐ฅ๐ท๐
๐ท๐ = 20.94๐ฅ10โ3m
๐ท๐ = 0.6๐ฅ 20.94๐ฅ10โ3m
๐ท๐ = 12.56๐ฅ10โ3m
Weight of the solid shaft,
๐๐ = ๐ ๐ฅ ๐ ๐ฅ ๐ฃ๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐ โ๐๐๐ก
๐๐ = ๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ ๐ท2 ๐ฅ ๐ฟ
Weight of the hollow shaft,
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ ๐ฃ๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐ โ๐๐๐ก
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ (๐
4[๐ท0
2 โ ๐ท๐ผ2]๐ฟ)
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ (๐
4[๐ท0
2 โ (0.6๐ท0)2]๐ฟ)
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5-63
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ 0.64 ๐ฅ ๐ท0
2 ๐ฅ ๐ฟ
Dividing the weights of solid and hollow shaft,
๐๐
๐๐ป=
๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ ๐ท2 ๐ฅ ๐ฟ
๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ 0.64 ๐ฅ ๐ท0
2 ๐ฅ ๐ฟ
๐๐
๐๐ป=
๐ท2
0.64 ๐ฅ ๐ท02
๐๐
๐๐ป=
(0.9547๐ฅ๐ท๐)2
0.64 ๐ฅ ๐ท02
๐๐
๐๐ป=
0.9114๐ฅ ๐ท02
0.64 ๐ฅ ๐ท02
๐๐
๐๐ป= 1.424
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก =๐๐๐๐โ๐ก ๐๐ ๐๐๐๐๐ ๐ โ๐๐๐ก โ ๐๐๐๐โ๐ก ๐๐ โ๐๐๐๐๐ค ๐ โ๐๐๐ก
๐๐๐๐โ๐ก ๐๐ ๐๐๐๐๐ ๐ โ๐๐๐ก
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก =1.424 ๐๐ป โ ๐๐ป
1.424 ๐๐ป๐ฅ 100
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก = 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
๐ = 400 ๐๐/๐2 W=1000N
G=80GN/m2.
Soln:
w.k.t the maximum shear stress induced as,
๐ =16๐๐
๐๐3
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5-64
40๐ฅ106 =16๐ฅ1000๐ฅ50๐ฅ10โ3
๐๐3
๐3 =16๐ฅ1000๐ฅ50๐ฅ10โ3
๐๐ฅ40๐ฅ106
๐3 = 6.366๐ฅ10โ6
๐ = 0.0185๐
(6M)
w.k.t the stiffness of the spring as,
๐ =๐บ๐4
64๐ 3๐
๐ =๐บ๐4
64๐ 3๐
๐ =80x109x(0.0185)4
64๐ฅ(50๐ฅ10โ3)3๐ฅ20๐ฅ103
๐ =80x109x1.17x10โ7
64๐ฅ1.25x10โ4๐ฅ20๐ฅ103
๐ = 58.5 โ 59๐๐๐๐๐
(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
๐max = 60 ๐/๐๐2 Tmax=1.4 xTmean
w.k.t,
๐๐๐ค๐๐, ๐ = 2๐๐๐
60
300๐ฅ103 = 2๐(80)๐
60
๐ = 35816.61 ๐ โ ๐
(1M) w.k.t the torque transmitted by the shaft, TH as
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5-65
๐๐ป =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ ๐ท๐
4
๐ท0โ
35816.61 =๐
16 ๐ฅ 60๐ฅ 106 ๐ฅ โ
๐ท4 โ (0.6๐ท)4
๐ทโ
35816.61 =๐
16 ๐ฅ 60๐ฅ 106 ๐ฅ โ
0.8704๐ท4
๐ทโ
โ0.8704๐ท3โ = 3.0407๐ฅ10โ3
๐ท3 = 3.4934๐ฅ10โ3
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 1.5ยฐ 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
๐max = 75 ๐/๐๐2
Angle of twist, ฮธ = 1.5ยฐ, ๐ = 1.5 ๐ฅ 180
๐= 0.02617 rad
Length, L=3.3m
G= 0.8x105N/mm2
๐๐๐ค๐๐, ๐ = 2๐๐๐
60
105๐ฅ103 = 2๐(2000)๐
60
๐ = 501.33 ๐ โ ๐
๐๐๐๐ฅ = 1.36 ๐๐๐๐๐
๐๐๐๐ฅ = 1.36 ๐ฅ 501.33
๐๐๐๐ฅ = 681.81 ๐ โ ๐
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
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5-66
(6M)
Case (i): Diameter for strength:
๐ =๐
16 ๐ฅ ๐ ๐ฅ ๐ท๐
3
681.81 =๐
16 ๐ฅ 75๐ฅ 106 ๐ฅ ๐ท๐
3
๐ท๐ 3 =
16๐ฅ 681.81
๐๐ฅ 75๐ฅ106
๐ท๐ 3 = 46.29 ๐ฅ10โ6
๐ท๐ = 0.0359๐ฅ10โ3 ๐
Case (ii): Diameter for stiffness:
๐
๐ฝ=
๐บ๐
๐ฟ
681.81 ๐
32๐ท๐
4=
8๐ฅ 1010๐ฅ 0.026179
3.3
๐ท๐ 4 =
681.81๐ฅ3.3๐ฅ32
๐๐ฅ 8๐ฅ1010๐ฅ0.026179
๐ท๐ 4 = 10.94 ๐ฅ10โ6
๐ท๐ = 0.0575๐ฅ10โ3 ๐
The required shaft diameter is the greaterst of the two values,
๐ท๐ = 0.0575๐ฅ10โ3 ๐ (๐๐)57.5 ๐๐
(7M)
9
A laminated spring carries a central load of 5200N and it is made of โnโ 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
ฮด=10mm
E = 2x105N/mm2
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5-67
w.k.t stress,
๐ =3๐๐
2๐๐๐ก2
๐ =3๐ฅ5200๐ฅ500๐ฅ10โ3
2๐๐ฅ80๐ฅ10โ3(7๐ฅ10โ3)2
๐ =994.89๐ฅ106
๐
(6M) The equation for deflection is,
๐ฟ =๐๐2
4๐ธ๐ก
10๐ฅ10โ3 =994.89๐ฅ106๐ฅ(500๐ฅ10โ3)2
๐๐ฅ4๐ฅ2๐ฅ1011๐ฅ7๐ฅ10โ3
๐ =994.89๐ฅ106๐ฅ(500๐ฅ10โ3)2
10๐ฅ10โ3๐ฅ4๐ฅ2๐ฅ1011๐ฅ7๐ฅ10โ3
๐ =248.72๐ฅ106
560๐ฅ105
๐ = 4.44 โ 5 ๐๐๐๐๐
(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
ฮด=20mm
W=50N
C=8.1x104N/mm2.
w.k.t the deflection of the spring as,
๐ฟ =64๐๐ 3๐
๐บ๐4
20๐ฅ10โ3 =64๐ฅ50๐ฅ๐ 3๐ฅ ๐
8.1๐ฅ1010๐ฅ(5๐ฅ10โ3)4
๐ฟ = 14.70๐๐
Solid Length, L=nd
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5-68
2=nx5๐ฅ10โ3
n= 400
(7M)
20๐ฅ10โ3 =64๐ฅ50๐ฅ๐ 3๐ฅ 400
8.1๐ฅ1010๐ฅ(5๐ฅ10โ3)4
๐ 3 =8.1๐ฅ1010๐ฅ20๐ฅ10โ3๐ฅ(5๐ฅ10โ3)4
64๐ฅ50๐ฅ 400
๐ 3 = 7.91๐ฅ10โ6
๐ = 9.24๐๐
๐ท = 18.48๐๐
(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
ฯ = Maximum shear stress induced in each shaft.
D=20mm
Di=0.6 Do
Torque transmitted by the solid shaft,
๐ =๐
16๐๐ท3=======>(1)
Torque transmitted by the hollow shaft,
๐ =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ ๐ท๐
4
๐ท0โ
๐ =๐
16 ๐ฅ ๐ ๐ฅ โ
๐ท04 โ (0.6๐ท๐)4
๐ท0โ
๐ =๐
16 ๐ฅ ๐ ๐ฅ 0.8704๐ฅ๐ท0
3 =======>(2)
Since torque transmitted by the solid shaft is equal to hollow shaft,
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5-69
๐
16๐๐ท3 =
๐
16 ๐ฅ ๐ ๐ฅ 0.8704๐ฅ๐ท0
3
๐ท3 = 0.8704๐ฅ๐ท03
๐ท = 0.9547๐ฅ๐ท๐
20๐ฅ10โ3 = 0.9547๐ฅ๐ท๐
๐ท๐ = 20.94๐ฅ10โ3m
๐ท๐ = 0.6๐ฅ 20.94๐ฅ10โ3m
๐ท๐ = 12.56๐ฅ10โ3m
(7M) Weight of the solid shaft,
๐๐ = ๐ ๐ฅ ๐ ๐ฅ ๐ฃ๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐ โ๐๐๐ก
๐๐ = ๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ ๐ท2 ๐ฅ ๐ฟ
Weight of the hollow shaft,
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ ๐ฃ๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐ โ๐๐๐ก
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ (๐
4[๐ท0
2 โ ๐ท๐ผ2]๐ฟ)
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ (๐
4[๐ท0
2 โ (0.6๐ท0)2]๐ฟ)
๐๐ป = ๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ 0.64 ๐ฅ ๐ท0
2 ๐ฅ ๐ฟ
Dividing the weights of solid and hollow shaft,
๐๐
๐๐ป=
๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ ๐ท2 ๐ฅ ๐ฟ
๐ ๐ฅ ๐ ๐ฅ ๐
4๐ฅ 0.64 ๐ฅ ๐ท0
2 ๐ฅ ๐ฟ
๐๐
๐๐ป=
๐ท2
0.64 ๐ฅ ๐ท02
๐๐
๐๐ป=
(0.9547๐ฅ๐ท๐)2
0.64 ๐ฅ ๐ท02
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
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5-70
๐๐
๐๐ป=
0.9114๐ฅ ๐ท02
0.64 ๐ฅ ๐ท02
๐๐
๐๐ป= 1.424
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก =๐๐๐๐โ๐ก ๐๐ ๐๐๐๐๐ ๐ โ๐๐๐ก โ ๐๐๐๐โ๐ก ๐๐ โ๐๐๐๐๐ค ๐ โ๐๐๐ก
๐๐๐๐โ๐ก ๐๐ ๐๐๐๐๐ ๐ โ๐๐๐ก
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก =1.424 ๐๐ป โ ๐๐ป
1.424 ๐๐ป๐ฅ 100
% ๐๐๐ฃ๐๐๐๐ ๐๐ ๐ค๐๐๐โ๐ก = 29.77%
(8M)
2
Derive a relation for deflection of a closely coiled helical spring subjected o an axial
compressive load โWโ. (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
ฯ = Max. shear stress induced in the Wire
ฮธ = Angle oftwist in spring wire, and
ฮด = Deflection of spring due to axial load
l= Length of wire.
Now twisting moment on the wire,
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5-71
๐ = ๐๐ฅ ๐ ==> (1)
But twisting moment is also given by,
๐ =๐
16๐๐3==> (2)
Equating Eqns (1) & (2),
๐๐ฅ ๐ =๐
16๐๐3
๐ =16๐๐
๐๐3
(8M) This Equation gives the maximum shear stress induced in the wire.
Expression for deflection of the spring:
Now length of one coil = ฯD or 2ฯR
Total length of the wire = Length of one coil x No.of coils,
L =2ฯRn
Strain energy stored in the spring,
๐ =๐2
4๐ถ. ๐๐๐๐ข๐๐
๐ = (16๐๐
๐๐3)
2
๐ฅ1
4๐ถ๐ฅ(
๐
4๐22ฯRn)
๐ =32๐2๐๐
๐ถ๐4
3
Work done on the spring = Average load x deflectionm
=1
2๐๐ฟ
Equating the work done on spring to the energy stored, we get
1
2๐๐ฟ =
32๐2๐๐
๐ถ๐4
3
๐ฟ =64๐๐๐
๐ถ๐4
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.
๐
๐ฝ=
๐บ๐
๐ฟ
For the shaft AB,
๐1
๐ฝ1=
๐บ1๐1
๐ฟ1
๐1 =๐1๐ฟ1
๐บ1๐ฝ1
๐1 =๐1๐ฅ2000๐ฅ32
๐๐ฅ754๐ฅ0.3๐ฅ105
For the shaft BC,
๐2
๐ฝ2=
๐บ2๐2
๐ฟ2
๐2 =๐2๐ฟ2
๐บ2๐ฝ2
๐2 =๐2๐ฅ1000๐ฅ32
๐๐ฅ554๐ฅ0.9๐ฅ105
w.k.t
๐1 = ๐2
๐1๐ฅ2000๐ฅ32
๐๐ฅ754๐ฅ0.3๐ฅ105=
๐2๐ฅ1000๐ฅ32
๐๐ฅ554๐ฅ0.9๐ฅ105
(8M)
Solving,
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๐1 = 0.576๐2
w.k.t
0.576๐2 + ๐2 = 200000
1.576๐2 = 200000
๐2 = 200000
1.576
๐2 = 126900 ๐ โ ๐๐
๐1 + ๐2 = 200000
๐1 = 200000 โ 126900
๐1 = 73100 ๐ โ ๐๐
w.k.t
๐
๐ฝ=
๐
๐
For shaft AB
๐1
๐ฝ1=
๐1
๐ 1
๐1 =๐1๐ 1
๐ฝ1
๐1 =73100๐ฅ 37.5
๐
32๐ฅ754
๐1 =73100๐ฅ 37.5๐ฅ32
๐๐ฅ754
๐1 = 0.882๐/๐๐2
For shaft BC ๐2
๐ฝ2=
๐2
๐ 2
๐2 =๐2๐ 2
๐ฝ2
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๐2 =126900๐ฅ 27.5
๐
32๐ฅ554
๐2 =126900๐ฅ 27.5๐ฅ32
๐๐ฅ554
๐2 = 3.884๐/๐๐2
(7M)
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UNIT IV - DEFLECTION OF BEAMS
Double Integration method โ Macaulayโs method โ Area moment method for computation of slopes and
deflections in beams - Conjugate beam and strain energy โ Maxwellโs reciprocal theorems.
PART * A
Q.No. Questions
1.
What are the advantages of Macaulayโs method over other methods for the calculation of
slope and deflection? (Apr/May 2015) BTL2
In Macaulayโs 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
๐ฆ = 5๐๐ฟ3
48๐ธ๐ผ
8 = (5)(12)๐ฟ3
48๐ธ๐ผ
๐ธ๐ผ
๐ฟ3= 0.15625
Deflection at the mid span when loaded at free end as
๐ฆ = ๐๐ฟ3
24๐ธ๐ผ
๐ฆ = 7 ๐ฟ3
24๐ธ๐ผ
๐ฆ = 7
24๐ฅ0.15625
๐ฆ = 1.867 ๐๐
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,
Macaulayโs 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,
๐ = ๐ธ๐ผ๐2๐ฆ
๐๐ฅ2
Integrating the above equation,
โซ ๐ = ๐ธ๐ผ๐๐ฆ
๐๐ฅ
Again integrating the above equation once again,
โฌ ๐ = ๐ธ๐ผ ๐ฆ
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 Maxwellโs reciprocal. (May/June 2016), (Nov/Dec 2016) BTL3
Maxwell's reciprocal theorem state that โIn 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 Aโ
7
Write down the equation for the maximum deflection of a cantilever beam carrying a
central point load โWโ. (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,
๐ฆ๐๐๐ฅ = 5๐๐ฟ3
48๐ธ๐ผ
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 โlโ and carrying a point
load โWโ at the free end? BTL3
Deflection at the free end of the cantilever beam is given as,
๐ฆ๐๐๐๐ = ๐๐3
3๐ธ๐ผ
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 Mohrโs theorems. BTL3
Mohrโs 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 .
Mohrโs 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,
๐๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐ต๐๐ท ๐๐๐ก๐ค๐๐๐ ๐กโ๐๐ ๐ ๐ก๐ค๐ ๐๐๐๐๐ก๐
๐ธ๐ผ
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,
๐๐ด =๐ด๐๐๐ ๐๐ ๐ต๐๐ท ๐๐๐ก๐ค๐๐๐ ๐ด ๐๐๐ ๐ต
๐ธ๐ผ
๐๐ด =๐
๐ธ๐ผ=
0.5 ๐ฅ 2 ๐ฅ 40
12 ๐ฅ103
๐๐ด = 0.0033 ๐๐๐
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,
๐๐๐๐ฅ =๐๐2
16๐ธ๐ผ
๐๐๐๐ฅ =10๐ฅ 103๐ฅ30002
16๐ฅ2.4๐ฅ1012
๐๐๐๐ฅ = 0.00234 ๐๐๐
17
Write down the boundary conditions for a cantilever beam to find out the equations for
deflection and slope.BTL3
When x=L, ๐๐ฆ
๐๐ฅ= 0
When x=L, ๐ฆ = 0
18
State Castiglianoโs 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 โxโ 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
๐ท๐๐๐๐๐๐ก๐๐๐ = y
๐๐๐๐๐ =dy
dx
๐ต๐๐๐๐๐๐ ๐๐๐๐๐๐ก = ๐ธ๐ผ๐2๐ฆ
๐๐ฅ2
๐โ๐๐๐ ๐น๐๐๐๐ = ๐ธ๐ผ๐3๐ฆ
๐๐ฅ3
๐ฟ๐๐๐ = ๐ธ๐ผ๐4๐ฆ
๐๐ฅ4
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 Macaulayโs method. (13M) (Apr/May 2015) BTL4
Reaction at Supports of the Beam:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 12 + (6 ๐ฅ 3) = 30kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 7 = (12 ๐ฅ 5) + (6 ๐ฅ 3 ๐ฅ 3
2 ) = 87
๐ ๐ต = 87
7= 12.42๐๐
๐ ๐ด = 30 โ 12.42 = 17.58๐๐
Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
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๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 = ๐ ๐ด ๐ฅ โ 6
๐ฅ2
2โฎ +6(๐ฅ โ 3)
(๐ฅ โ 3)
2โฎ โ12(๐ฅ โ 5)
(๐ฅ โ 5)
2
= 17.58 ๐ฅ โ 3๐ฅ2 โฎ +3(๐ฅ โ 3)2 โฎ โ6(๐ฅ โ 5)2
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 17.58
๐ฅ2
2 โ 3
๐ฅ3
3+ ๐ถ1 โฎ +3
(๐ฅ โ 3)3
3โฎ โ6
(๐ฅ โ 5)3
3 โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ = 17.58 ๐ฅ3
6 โ
๐ฅ4
4+ ๐ถ1๐ฅ + ๐ถ2 โฎ +
(๐ฅ โ 3)4
4โฎ โ2
(๐ฅ โ 5)4
4 โ ๐ธ๐ (2)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = 17.58 03
6 โ
04
4+ ๐ถ10 + ๐ถ2
๐ถ2 = 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,
๐ธ๐ผ(0) = 17.58 73
6 โ
74
4+ ๐ถ17 + 0 โฎ +
(7 โ 3)4
4โฎ โ2
(7 โ 5)4
4
๐ธ๐ผ(0) = 1004.99 โ 600.25 + ๐ถ17 โฎ +64 โฎ โ8
๐ถ1 = โ65.82
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 17.58
๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ โ 3)3 โฎ โ2(๐ฅ โ 5)3
Deflection Equation:
๐ธ๐ผ ๐ฆ = 17.58 ๐ฅ3
6 โ
๐ฅ4
4โ 65.82๐ฅ โฎ +
(๐ฅ โ 3)4
4โฎ โ2
(๐ฅ โ 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 ๐๐ฆ
๐๐ฅ should be zero. Hence equating the slope equation given by Eq (1) upto second dotted
line to zero, we get
0 = 17.58 ๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ โ 3)3
0 = 17.58 ๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ3 โ 27 + 27๐ฅ โ 9๐ฅ2)
Solving , ๐ฅ = 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,
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 17.58 (3.54)3
6 โ
(3.54)4
4โ 65.82(3.54) โฎ +
(3.54 โ 3)4
4
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 129.98 โ 39.26 โ 233 + 0.021
๐ฆ ๐๐๐ฅ = โ142.26
๐ธ๐ผ
(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 AโCโ which is free at Aโ and fixed at Cโ 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 Cโ for the conjugate beam,
๐ฆ๐ = ๐ต. ๐ ๐๐ก ๐ถโฒ
๐ฆ๐ = (1
2 ๐ฅ 2 ๐ฅ
20
๐ธ๐ผ) ๐ฅ (
2
3๐ฅ 2) + ( 2 ๐ฅ
20
๐ธ๐ผ) ๐ฅ (
2
2+ 2) + (
1
3 ๐ฅ 2 ๐ฅ
28
๐ธ๐ผ) ๐ฅ (
3
4 ๐ฅ 2)
๐ฆ๐ = (20
๐ธ๐ผ) ๐ฅ (
4
3) + (
40
๐ธ๐ผ) ๐ฅ(3) + (
56
๐ธ๐ผ) ๐ฅ (
1
2)
๐ฆ๐ = (80
3๐ธ๐ผ) + (
120
๐ธ๐ผ) + (
28
๐ธ๐ผ)
๐ฆ๐ =174.67
๐ธ๐ผ
(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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 50kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 6 = (50 ๐ฅ 3 ) = 150
๐ ๐ต = 150
6= 25๐๐
๐ ๐ด = 50 โ 25 = 25๐๐
Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 = ๐ ๐ด ๐ฅ โฎ โ50(๐ฅ โ 3)
= 25 ๐ฅ โฎ โ50(๐ฅ โ 3)
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 25
๐ฅ2
2+ ๐ถ1 โฎ โ50
(๐ฅ โ 3)2
2 โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ = 25 ๐ฅ3
6 + ๐ถ1๐ฅ + ๐ถ2 โฎ โ25
(๐ฅ โ 3)3
3 โ ๐ธ๐ (2)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = 25 03
6 + ๐ถ10
๐ถ2 = 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,
๐ธ๐ผ(0) = 25 63
6 + ๐ถ16 โฎ โ25
(6 โ 3)3
3
๐ธ๐ผ(0) = 900 + ๐ถ16 โฎ โ225
๐ถ1 = โ112.5
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 25
๐ฅ2
2 โ 112.5 โฎ โ25(๐ฅ โ 3)2
Deflection Equation:
๐ธ๐ผ ๐ฆ = 25 ๐ฅ3
6 โ 112.5๐ฅ โฎ โ25
(๐ฅ โ 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,
๐ธ๐ผ๐ฆ๐๐๐๐ก๐๐๐ = 25 33
6 โ 112.5(3)
๐ธ๐ผ๐ฆ๐๐๐๐ก๐๐๐ = 112.5 โ 337.5
๐ธ๐ผ๐ฆ๐๐๐๐ก๐๐๐ = โ225
๐ฆ๐๐๐๐ก๐๐๐ =โ225๐ฅ106
2.1. ๐ฅ 1011๐ฅ78๐ฅ 10โ6
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Solving , ๐ฅ = 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,
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 17.58 (3.54)3
6 โ
(3.54)4
4โ 65.82(3.54) โฎ +
(3.54 โ 3)4
4
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 129.98 โ 39.26 โ 233 + 0.021
๐ฆ ๐๐๐ฅ = โ142.26
๐ธ๐ผ
(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 Macaulayโs
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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 48 + 40= 88kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 6 = (48 ๐ฅ 1) + (40 ๐ฅ 3 ) = 168
๐ ๐ต = 168
6= 28๐๐
๐ ๐ด = 168 โ 28 = 140๐๐
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Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 = ๐ ๐ด ๐ฅ โฎ โ48(๐ฅ โ 1) โฎ โ40(๐ฅ โ 3)
= 140 ๐ฅ โฎ โ48(๐ฅ โ 1) โฎ โ40(๐ฅ โ 3)
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 140
๐ฅ2
2+ ๐ถ1 โฎ โ48
(๐ฅ โ 1)2
2โฎ โ40
(๐ฅ โ 3)2
2 โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ = 70 ๐ฅ3
3 + ๐ถ1๐ฅ + ๐ถ2 โฎ โ24
(๐ฅ โ 1)3
3โฎ โ20
(๐ฅ โ 3)3
3 โ ๐ธ๐ (2)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = 70 03
3 + ๐ถ10
๐ถ2 = 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,
๐ธ๐ผ(0) = 70 63
3 + ๐ถ16 โฎ โ24
(6 โ 1)3
3โฎ โ20
(6 โ 3)3
3
๐ธ๐ผ(0) = 5040 + ๐ถ16 โฎ โ1000 โฎ โ180
๐ถ1 = โ643.3
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 17.58
๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ โ 3)3 โฎ โ2(๐ฅ โ 5)3
Deflection Equation:
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๐ธ๐ผ ๐ฆ = 17.58 ๐ฅ3
6 โ
๐ฅ4
4โ 65.82๐ฅ โฎ +
(๐ฅ โ 3)4
4โฎ โ2
(๐ฅ โ 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 ๐๐ฆ
๐๐ฅ should be zero. Hence equating the slope equation given by Eq (1) upto second dotted
line to zero, we get
0 = 17.58 ๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ โ 3)3
0 = 17.58 ๐ฅ2
2 โ ๐ฅ3 โ 65.82 โฎ +(๐ฅ3 โ 27 + 27๐ฅ โ 9๐ฅ2)
Solving , ๐ฅ = 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,
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 17.58 (3.54)3
6 โ
(3.54)4
4โ 65.82(3.54) โฎ +
(3.54 โ 3)4
4
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 129.98 โ 39.26 โ 233 + 0.021
๐ฆ ๐๐๐ฅ = โ142.26
๐ธ๐ผ
(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,
๐๐ต =๐๐2
2๐ธ๐ผ
๐๐ต =50000๐ฅ(2)2
2๐ฅ2๐ฅ1011๐ฅ10โ4
๐๐ต = 0.005 ๐๐๐
(4M)
Deflection at the free end is given by the equation,
๐ฆ๐ต =๐๐2
3๐ธ๐ผ+
๐๐2
2๐ธ๐ผ(๐ฟ โ ๐)
๐ฆ๐ต =50000๐ฅ(2)3
3๐ฅ2๐ฅ1011๐ฅ10โ4+
50000๐ฅ(2)2
2๐ฅ2๐ฅ1011๐ฅ10โ4(3 โ 2)
๐ฆ๐ต = 0.0067 + 0.0050
๐ฆ๐ต = 0.0117๐
(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
๐ผ =๐๐3
12
๐ผ =0.12(0.24)3
12
๐ผ = 1.3824๐ฅ10โ4 ๐4
๐ธ๐ผ = 10๐ฅ109๐ฅ1.3824๐ฅ10โ4 ๐4
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๐ธ๐ผ = 1.3824๐ฅ106 ๐ โ ๐2
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,
๐ธ๐ผ๐2๐ฆ
๐๐ฅ2= โ๐ฅ โ 2.5(๐ฅ โ 0.75) โ โ โ โ โ โ(1)
Integrating the above equation we get,
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
๐ฅ2
2โ 2.5
(๐ฅ โ 0.75)2
2+ ๐ถ1 โ โ โ โ โ โ(2)
Integrating the above equation once again we get,
๐ธ๐ผ๐ฆ = โ๐ฅ3
6โ 2.5
(๐ฅ โ 0.75)3
6+ ๐ถ1๐ฅ + ๐ถ2 โ โ โ โ โ โ(3)
Boundary Condition:
when x=2, slope becomes zero.
๐ธ๐ผ (0) = โ22
2โ 2.5
(2 โ 0.75)2
2+ ๐ถ1
๐ถ1 = 3.953
Boundary Condition:
when x=2, deflection becomes zero.
๐ธ๐ผ (0) = โ23
6โ 2.5
(2 โ 0.75)3
6+ 2๐ฅ3.953 + ๐ถ2
0 = โ1.33 โ 0.813 + 7.906 + ๐ถ2
๐ถ2 = โ5.763
Substituting the values in eqns 2 and 3 we get,
Slope Equation:
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
๐ฅ2
2โ 2.5
(๐ฅ โ 0.75)2
2+ 3.953
Deflection Equation:
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๐ธ๐ผ๐ฆ = โ๐ฅ3
6โ 2.5
(๐ฅ โ 0.75)3
6+ 3.953๐ฅ โ 5.763
(7M) To find the Deflection at free end:
Substitute x=0 in the deflection equation we get,
๐ธ๐ผ๐ฆ๐๐๐๐ = โ03
6โ 2.5
(0 โ 0.75)3
6+ 3.953(0) โ 5.763
๐ฆ๐๐๐๐ = โ5.763๐ฅ103
1.3824๐ฅ106
๐ฆ๐๐๐๐ = โ4.168๐ฅ10โ3๐
(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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 10 + (5๐ฅ2)= 20kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 8 = (10 ๐ฅ 1) + (5 ๐ฅ 2 ๐ฅ(2
2+ 6 ) = 80
๐ ๐ต = 80
8= 10๐๐
๐ ๐ด = 20 โ 10 = 10๐๐
Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
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๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 = ๐ ๐ด ๐ฅ โฎ โ10(๐ฅ โ 1) โฎ โ5
(๐ฅ โ 6)2
2
= 10 ๐ฅ โฎ โ10(๐ฅ โ 1) โฎ โ5(๐ฅ โ 6)2
2
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 10
๐ฅ2
2+ ๐ถ1 โฎ โ10
(๐ฅ โ 1)2
2โฎ โ5
(๐ฅ โ 6)3
6 โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ = 10 ๐ฅ3
6 + ๐ถ1๐ฅ + ๐ถ2 โฎ โ10
(๐ฅ โ 1)3
6โฎ โ5
(๐ฅ โ 6)4
24 โ ๐ธ๐ (2)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = 10 03
6 + ๐ถ10
๐ถ2 = 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,
๐ธ๐ผ(0) = 10 83
6 + ๐ถ18 + ๐ถ2 โฎ โ10
(8 โ 1)3
6โฎ โ5
(8 โ 6)4
24
๐ธ๐ผ(0) = 853.33 + ๐ถ18 โฎ โ571.67 โฎ โ3.33
๐ถ1 = โ34.79
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 10
๐ฅ2
2โ 34.79 โฎ โ10
(๐ฅ โ 1)2
2โฎ โ5
(๐ฅ โ 6)3
6
Deflection Equation:
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๐ธ๐ผ ๐ฆ = 10 ๐ฅ3
6โ 34.79๐ฅ โฎ โ10
(๐ฅ โ 1)3
6โฎ โ5
(๐ฅ โ 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,
๐ธ๐ผ ๐ฆ ๐๐๐ = 10 (4)3
6 โ 34.79(4) โฎ โ10
(4 โ 1)3
6
๐ธ๐ผ ๐ฆ ๐๐๐ = 106.67 โ 139.16 โ 45
๐ฆ ๐๐๐ = โ77.49
๐ธ๐ผ
Position of Maximum Deflection:
The maximum deflection is likely to happen between C and D. For the maximum deflection the
slope ๐๐ฆ
๐๐ฅ should be zero. Hence equating the slope equation given by Eq (1) upto second dotted
line to zero, we get
0 = 10 ๐ฅ2
2 โ 34.79 โฎ โ5(๐ฅ โ 1)2
0 = 5๐ฅ2 โ 34.79 โฎ โ5(๐ฅ2 + 1 โ 2๐ฅ)
0 = 5๐ฅ2 โ 34.79 โฎ โ5๐ฅ2 โ 5 + 10๐ฅ
0 = โ34.79 โ 5 + 10๐ฅ
39.79 = 10๐ฅ
Solving , ๐ฅ = 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,
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 10 (3.979)3
6 โ 34.79(3.979) โฎ โ10
(3.979 โ 1)3
6
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 104.99 โ 138.42 โ 44.06
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๐ฆ ๐๐๐ฅ = โ77.49
๐ธ๐ผ
(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 Mohrโs theorem for slope, we get
๐๐๐๐๐ ๐๐ก ๐ด = ๐ด๐๐๐ ๐๐ ๐ต.๐ ๐๐๐๐๐๐๐ ๐๐๐ก๐ค๐๐๐ ๐ด ๐๐๐ ๐ถ
๐ธ๐ผ
Area of B.M diagram between A and C = Area of triangle ACD.
=1
2๐ฅ
๐ฟ
2๐ฅ
๐๐ฟ
4โ
๐๐ฟ2
16
๐๐๐๐๐ ๐๐ก ๐ด ๐๐ ๐๐ด =๐๐ฟ2
๐ธ๐ผ
(7M) Now using Mohrโs theorem for deflection, we get
๐ฆ = ๐ด๐ฅ
๐ธ๐ผ
Where A=Area of B.M Diagram of area from A.
๐ด =๐๐ฟ2
16
x= Distance of CG of area A from A.
๐ฅ =2
3๐ฅ
๐ฟ
3=
๐ฟ
3
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๐ฆ =
๐๐ฟ2
16๐ฅ
๐ฟ
3
๐ธ๐ผ
๐ฆ =๐๐ฟ3
48๐ธ๐ผ
(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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต =37.5kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 5.8 = (37.5 ๐ฅ 2.9 ) = 108.75
๐ ๐ต = 108.75
5.8= 18.75๐๐
๐ ๐ด = 37.5 โ 18.75 = 18.75๐๐
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:
โ ๐น๐ป = 0
๐ ๐ดโฒ + ๐ ๐ต
โฒ = 2 ๐ฅ (1
2 ๐ฅ 2.9 ๐ฅ
54.38
๐ธ๐ผ)
๐ ๐ดโฒ + ๐ ๐ต
โฒ =157.4
๐ธ๐ผ
Taking Moments of all forces about A, we get
๐ ๐ตโฒ ๐ฅ 5.8 = [(
1
2 ๐ฅ 2.9 ๐ฅ
54.38
๐ธ๐ผ) ๐ฅ ((
1
3๐ฅ 2.9) + 2.9)] +[(
1
2 ๐ฅ 2.9 ๐ฅ
54.38
๐ธ๐ผ) ๐ฅ (
2
3๐ฅ 2.9)]
5.8 ๐ ๐ตโฒ =
456.54
๐ธ๐ผ
๐ ๐ตโฒ =
78.7
๐ธ๐ผ
๐ ๐ดโฒ =
157.4
๐ธ๐ผโ
78.7
๐ธ๐ผ
๐ ๐ดโฒ =
78.7
๐ธ๐ผ
(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 Cโ.
BM@Cโ = (78.71
๐ธ๐ผ ๐ฅ 2.9 ) - [(
1
2 ๐ฅ 2.9 ๐ฅ
54.38
๐ธ๐ผ) ๐ฅ (
1
3๐ฅ 2.9)]
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BM@Cโ = 228.26
๐ธ๐ผโ
76.22
๐ธ๐ผ
BM@Cโ = 152.04
๐ธ๐ผ
BM@Cโ = 152.04
4๐ฅ104
y๐ = 38.01๐ฅ10โ4 ๐ ๐๐ 3.801๐๐
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 Aโ.
SF@Aโ = โ78.7
๐ธ๐ผ
BM@Aโ = โ78.7
4๐ฅ104
ฮธ๐ด = โ19.675๐ฅ10โ4 ๐๐๐ ๐๐ โ 0.0019675 ๐๐๐๐๐๐๐
(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,
๐ธ๐ผ๐2๐ฆ
๐๐ฅ2= โ๐ค๐ฅ
๐ฅ
2= โ๐ค
๐ฅ2
2 โ โ โ โ โ โ(1)
Integrating the above equation we get,
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๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
๐ค ๐ฅ3
6+ ๐ถ1 โ โ โ โ โ โ(2)
Integrating the above equation once again we get,
๐ธ๐ผ๐ฆ = โ๐ค ๐ฅ4
24+ ๐ถ1๐ฅ + ๐ถ2 โ โ โ โ โ โ(3)
Boundary Condition:
when x=l, slope becomes zero.
๐ธ๐ผ (0) = โ๐ค (๐)3
6+ ๐ถ1
๐ถ1 =๐ค (๐)3
6
Boundary Condition:
when x=l, deflection becomes zero.
๐ธ๐ผ (0) = โ๐ค ๐4
24+
๐ค (๐)4
6+ ๐ถ2
๐ถ2 = โ๐ค ๐4
8
Substituting the values in eqns 2 and 3 we get,
Slope Equation:
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
๐ค ๐ฅ3
6+
๐ค ๐3
6
Deflection Equation:
๐ธ๐ผ๐ฆ = โ๐ค ๐ฅ4
24+
๐ค ๐3
6๐ฅ โ
๐ค ๐4
8
(7M) To find the Slope at free end:
Substitute x=0 in the slope equation we get,
๐ธ๐ผ๐๐๐๐๐ = โ๐ค 03
6+
๐ค ๐3
6
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๐๐๐๐๐ =๐ค ๐3
6๐ธ๐ผ
To find the Deflection at free end:
Substitute x=0 in the deflection equation we get,
๐ธ๐ผ๐ฆ๐๐๐๐ = โ๐ค ๐ฅ4
24+
๐ค ๐3
6๐ฅ โ
๐ค ๐4
8
๐ฆ๐๐๐๐ = โ๐ค ๐4
8๐ธ๐ผ
(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 Macaulayโs 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:
โ ๐น๐ป = 0
๐ ๐ด + ๐ ๐ต = 50kN
Taking Moments of all forces about A, we get
๐ ๐ต ๐ฅ 8 = (50 ๐ฅ 3 ) + 75 = 225
๐ ๐ต = 225
8= 28.125๐๐
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๐ ๐ด = 50 โ 28.125 = 21.875๐๐
(5M) Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 = ๐ ๐ด ๐ฅ โฎ โ50(๐ฅ โ 3) โฎ +75(๐ฅ โ 5)0
= 21.875 ๐ฅ โฎ โ50(๐ฅ โ 3) โฎ +75(๐ฅ โ 5)0
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 21.875
๐ฅ2
2+ ๐ถ1 โฎ โ50
(๐ฅ โ 3)2
2โฎ +75(๐ฅ โ 5) โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ = 21.875 ๐ฅ3
6 + ๐ถ1๐ฅ + ๐ถ2 โฎ โ25
(๐ฅ โ 3)3
3โฎ +75
(๐ฅ โ 5)2
2 โ ๐ธ๐ (2)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = 21.875 03
6 + ๐ถ10 + ๐ถ2
๐ถ2 = 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,
๐ธ๐ผ(0) = 21.875 83
6 + ๐ถ18 + ๐ถ2 โฎ โ25
(8 โ 3)3
3โฎ +75
(8 โ 5)2
2
๐ธ๐ผ(0) = 1867.67 + ๐ถ18 โฎ โ1041.67 โฎ +337.5
๐ถ1 = โ145.43
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ = 21.875
๐ฅ2
2โ 145.43 โฎ โ50
(๐ฅ โ 3)2
2โฎ +75(๐ฅ โ 5)
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Deflection Equation:
๐ธ๐ผ ๐ฆ = 21.875 ๐ฅ3
6 โ 145.43๐ฅ + ๐ถ2 โฎ โ25
(๐ฅ โ 3)3
3โฎ +75
(๐ฅ โ 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 ๐๐ฆ
๐๐ฅ should be zero. Hence equating the slope equation given by Eq (1) upto second dotted
line to zero, we get
0 = 21.875 ๐ฅ2
2โ 145.43 โฎ โ50
(๐ฅ โ 3)2
2
0 = 21.875 ๐ฅ2
2 โ 145.43 โฎ โ25(๐ฅ2 + 9 โ 6๐ฅ)
Solving , ๐ฅ = 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,
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 21.875 3.883
6 โ 145.43(3.88) โฎ โ25
(3.88โ3)3
3
๐ธ๐ผ ๐ฆ ๐๐๐ฅ = 212.95 โ 564.26 โ 5.67
๐ฆ ๐๐๐ฅ = โ356.98
๐ธ๐ผ
๐ฆ ๐๐๐ฅ = โ356.98
2๐ฅ1011๐ฅ85๐ฅ10โ6
๐ฆ ๐๐๐ฅ = โ0.02099๐
(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,
๐ ๐ด = ๐ ๐ต =๐ค๐ฅ๐ฟ
2
Consider a section X in the last part of the beam at a distance โxโ from the left most support. The
B.M at this section is given by,
๐ธ๐ผ ๐2๐ฆ
๐๐ฅ2 =
๐ค ๐ฟ ๐ฅ
2 โ
๐ค. ๐ฅ2
2
Integrating the above equation we get,
๐ธ๐ผ ๐๐ฆ
๐๐ฅ =
๐ค ๐ฟ ๐ฅ2
4 โ
๐ค. ๐ฅ3
6+ ๐ถ1 โ ๐ธ๐ (1)
Integrating again we get,
๐ธ๐ผ ๐ฆ =๐ค ๐ฟ ๐ฅ3
12 โ
๐ค. ๐ฅ4
24+ ๐ถ1๐ฅ + ๐ถ2 โ ๐ธ๐ (2)
(5M)
Where ๐ถ1 ๐๐๐ ๐ถ2 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,
๐ธ๐ผ(0) = ๐ค ๐ฟ 03
12 โ
๐ค. 04
24+ ๐ถ10 + ๐ถ2
๐ถ2 = 0
Substituting condn (ii) at x=L, y=0 into complete Eq (2) we get,
๐ธ๐ผ(0) =๐ค ๐ฟ ๐ฟ3
12 โ
๐ค. ๐ฟ4
24+ ๐ถ1๐ฟ + ๐ถ2
๐ธ๐ผ(0) = ๐ค ๐ฟ4
12 โ
๐ค. ๐ฟ4
24+ ๐ถ1๐ฟ
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๐ถ1 = โ๐ค. ๐ฟ3
24
Substituting the values of ๐ถ1 and ๐ถ2 into Eq(1) and Eq(2) we get,
Slope Equation:
๐ธ๐ผ ๐๐ฆ
๐๐ฅ =
๐ค ๐ฟ ๐ฅ2
4 โ
๐ค. ๐ฅ3
6โ
๐ค. ๐ฟ3
24
Deflection Equation:
๐ธ๐ผ ๐ฆ =๐ค ๐ฟ ๐ฅ3
12 โ
๐ค. ๐ฅ4
24โ
๐ค. ๐ฟ3
24๐ฅ
(5M)
Slope at the supports:
Let ๐๐ด = ๐๐๐๐๐ ๐๐ก ๐ ๐ข๐๐๐๐๐ก ๐ด
and ๐๐ต = ๐๐๐๐๐ ๐๐ก ๐ ๐ข๐๐๐๐๐ก ๐ต
At A, x=0 and ๐๐ฆ
๐๐ฅ= ๐๐ด
๐ธ๐ผ ๐๐ด =๐ค ๐ฟ 02
4 โ
๐ค. 03
6โ
๐ค. ๐ฟ3
24
๐ธ๐ผ ๐๐ด = โ๐ค. ๐ฟ3
24
๐๐ด = โ๐ค. ๐ฟ3
24๐ธ๐ผ
By symmetry,
๐๐ต = โ๐ค. ๐ฟ3
24๐ธ๐ผ
Maximum Deflection:
The maximum deflection is at the centre of the beam at point C, where ๐ฅ =๐ฟ
2
๐ธ๐ผ ๐ฆ ๐ ==๐ค ๐ฟ
12(
๐ฟ
2)
3
โ๐ค
24(
๐ฟ
2)
4
โ๐ค. ๐ฟ3
24(
๐ฟ
2)
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๐ธ๐ผ ๐ฆ ๐ ==๐ค ๐ฟ4
96 โ
๐ค ๐ฟ4
384โ
๐ค ๐ฟ4
48
๐ธ๐ผ ๐ฆ ๐ ==5๐ค ๐ฟ4
384
๐ฆ ๐ ==5๐ค ๐ฟ4
384๐ธ๐ผ
(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,
๐ธ๐ผ๐2๐ฆ
๐๐ฅ2= โ20๐ฅ
๐ฅ
2= โ10๐ฅ2 โ โ โ โ โ โ(1)
Integrating the above equation we get,
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
10 ๐ฅ3
3+ ๐ถ1 โ โ โ โ โ โ(2)
Integrating the above equation once again we get,
๐ธ๐ผ๐ฆ = โ10 ๐ฅ4
12+ ๐ถ1๐ฅ + ๐ถ2 โ โ โ โ โ โ(3)
(5M) Boundary Condition:
when x=l, slope becomes zero.
๐ธ๐ผ (0) = โ10 (2.3)3
3+ ๐ถ1
๐ถ1 = 40.56
Boundary Condition:
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when x=l, deflection becomes zero.
๐ธ๐ผ (0) = โ10 (2.3)4
12+ 40.56(2.3) + ๐ถ2
๐ถ2 = โ69.96
Substituting the values in eqns 2 and 3 we get,
Slope Equation:
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
10 ๐ฅ3
3+ 40.56
Deflection Equation:
๐ธ๐ผ๐ฆ = โ10 ๐ฅ4
12+ 40.56๐ฅ โ 69.96
(5M) To find the Slope at free end:
Substitute x=0 in the slope equation we get,
๐ธ๐ผ๐๐๐๐๐ = โ10 (0)3
3+ 40.56
๐๐๐๐๐ =40.56
๐ธ๐ผ
๐๐๐๐๐ =40.56
12๐ฅ103
๐๐๐๐๐ = 3.38๐ฅ10โ3๐๐๐
To find the Deflection at free end:
Substitute x=0 in the deflection equation we get,
๐ธ๐ผ๐ฆ๐๐๐๐ = โ10 (0)4
12+ 40.56(0) โ 69.96
๐ฆ๐๐๐๐ = โ69.96
12๐ฅ103
๐ฆ๐๐๐๐ = โ5.83๐ฅ10โ3๐
(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 โ spherical shells subjected to internal pressure โDeformation in
spherical shells โ Lameโs 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 Lameโs theorem for thick cylinder analysis. (Apr/May 2015)
BTL3
The following are the assumptions made in the lameโs 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
๐๐๐๐ฅ = ๐๐ โ ๐๐
2
=
๐๐
2๐กโ
๐๐
4๐ก
2
๐๐๐๐ฅ = ๐๐
8๐ก
4 Define โ 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 Lameโs equation. (May/June 2016), (Apr/May 2017) BTL2
The Lameโs equation are expressed by means of the two following equations namely,
๐๐ฅ =๐
๐ฅ2โ ๐
and
๐๐ฅ =๐
๐ฅ2+ ๐
where, โaโ and โbโ 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:
๐๐ = ๐๐
2๐ก
90 = ๐๐ฅ3000
2๐ฅ20
๐ = 1.2 ๐ฅ 106 ๐/๐2
Pressure for hoop stress:
๐๐ = ๐๐
4
60 = ๐๐ฅ3000
4๐ฅ20
๐ = 1.6 ๐ฅ 106 ๐/๐2
The safe pressure is 1.2 ๐ฅ 106 ๐/๐2
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:
๐๐ = ๐๐
4๐ก
75 = 0.5๐ฅ1000
4๐ฅ๐ก
๐ก = 1.67 ๐ฅ 10โ3 ๐
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 โpโ. BTL1
๐ฟ๐ = ๐๐2
2๐ก๐ธ (1 โ
๐
2)
๐ฟ๐ = ๐๐๐
2๐ก๐ธ (
1
2โ ๐)
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
๐๐ ๐๐๐ = ๐๐๐ฅ๐๐๐ข๐ ๐๐๐๐ ๐๐ ๐๐ก๐๐๐๐๐กโ
๐น๐๐๐ก๐๐ ๐๐ ๐ ๐๐๐๐ก๐ฆ
๐๐ = 400
5= 80 N/mm2
๐๐ = 80 = ๐๐
2๐ก๐
80 = 4๐ฅ500
2๐ฅ๐ก๐ฅ0.65
๐ก = 19.23 ๐ฅ 10โ3 ๐
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
๐๐๐๐ข๐๐๐ก๐๐๐ ๐๐ก๐๐๐๐, ๐๐ฃ = ๐ฟ๐
๐= 2๐๐ + ๐๐
๐๐ฃ = 2๐ฅ0.2๐ฅ10โ3 + 0.05๐ฅ10โ3
๐๐ฃ = 4.5๐ฅ10โ4
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
๐๐ = ๐๐
2๐ก
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๐๐ = 2 ๐ฅ 3000
4 ๐ฅ 10
๐๐ = 150 ๐/๐๐2.
17
Write down the equation for strain along the longitudinal, circumferential direction for the
thin cylinder. BTL1
๐๐ =๐ฟ๐
๐=
๐๐
2๐ก๐ธ (1 โ
๐
2)
๐๐ =๐ฟ๐
๐=
๐๐๐
2๐ก๐ธ (
1
2โ ๐)
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
๐๐ = ๐๐
2๐ก
120 = 10 ๐ฅ 750
2 ๐ฅ ๐ก
๐ก = 31.25 ๐๐ โ 32๐๐.
19
Write down the volumetric strain in a thin spherical shell subjected to internal pressure โpโ.
BTL1
๐๐ฃ = 3๐๐ =3๐๐
4๐ก๐ธ (1 โ ๐)
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 poissonโs 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
ยต=0.3
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5-108
Soln:
Circumferential stress,
๐๐ = ๐๐
2๐ก
๐๐ = 1๐ฅ106๐ฅ0.7
2๐ฅ8๐ฅ10โ3
๐๐ = 43.74 ๐ฅ 106 ๐/๐2
Longitudinal stress,
๐๐ = ๐๐
4๐ก
๐๐ = 1๐ฅ106๐ฅ0.7
4๐ฅ8๐ฅ10โ3
๐๐ = 21.88 ๐ฅ 106 ๐/๐2
Maximum Shear Stress:
๐๐๐๐ฅ = 43.74 ๐ฅ 106 โ 21.88 ๐ฅ 106
2
๐๐๐๐ฅ = 10.93 ๐ฅ 106 ๐/๐2
(6M) Change in diameter:
๐ฟ๐ = ๐๐2
2๐ก๐ธ (1 โ
๐
2)
๐ฟ๐ = 1๐ฅ106๐ฅ(0.7)2
2๐ฅ8๐ฅ10โ3๐ฅ2๐ฅ1011 (1 โ
0.3
2)
๐ฟ๐ = 0.1301 ๐ฅ10โ3๐
Change in length:
๐ฟ๐ = ๐๐๐
2๐ก๐ธ (
1
2โ ๐)
๐ฟ๐ = 1๐ฅ106๐ฅ0.7๐ฅ2.5
2๐ฅ8๐ฅ10โ3๐ฅ2๐ฅ1011 (
1
2โ 0.3)
๐ฟ๐ = 0.1093๐ฅ10โ3๐
Change in Volume:
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5-109
๐ฟ๐
๐=
๐๐
2๐ก๐ธ (
5
2โ 2๐)
๐ฟ๐
๐=
1๐ฅ106๐ฅ0.7
2๐ฅ8๐ฅ10โ3๐ฅ2๐ฅ1011 (
5
2โ 2๐ฅ0.3)
๐ฟ๐
๐= 0.21875๐ฅ10โ3
Volume ๐ =๐๐2๐ฟ
4
๐ =๐๐ฅ0.72๐ฅ2.5
4= 0.962 ๐3
๐ฟ๐ = 0.21875๐ฅ10โ3๐ฅ0.962
๐ฟ๐ = 0.2104๐ฅ10โ3๐3 (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,
๐๐ฅ =๐
๐ฅ2โ ๐
Apply the boundary condition to above equations,
1. At x= r1 = 80mm, ๐๐ฅ = 8 N/mm2
2. At x= r2 = 160mm, ๐๐ฅ = 0
8 =๐
802โ ๐ =
๐
6400โ ๐
0 =๐
1602โ ๐ =
๐
25600โ ๐
Subtracting the above equations,
8 =๐
6400โ ๐ โ
๐
25600+ ๐
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5-110
8 =4๐
25600โ
๐
25600=
3๐
25600
๐ =25600๐ฅ8
3
๐ = 68266
0 =68266
25600โ ๐
๐ =68266
25600= 2.67
(6M)
Now hoop stress at any radius x is given by,
๐๐ฅ =๐
๐ฅ2+ ๐
๐๐ฅ =68266
๐ฅ2+ 2.67
At x=80mm,
๐80 =68266
802+ 2.67 = 10.67 + 2.67
๐80 = 13.34 ๐/๐๐2
At x=160mm,
๐160 =68266
1602+ 2.67 = 2.67 + 2.67
๐160 = 5.34 ๐/๐๐2
(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, ๐๐ก=120 N/mm2
Permissible stress may be circumferential stress or longitudinal stress.
๐๐ = 90% = 0.9
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๐๐ = 40% = 0.4 Case 1:
Consider the Permissible stress as circumferential stresss ๐๐
w.k.t
๐๐ = ๐๐
2๐ก๐๐
120 = 2๐ฅ๐
2๐ฅ26๐ฅ0.9
๐ = 2808 ๐๐
(4M) Case 2:
Consider the Permissible stress as longitudinal stresss ๐๐
w.k.t
๐๐ = ๐๐
4๐ก๐๐
120 = 2๐ฅ๐
4๐ฅ26๐ฅ0.4
๐ = 2496 ๐๐
(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 poissonโs 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
ยต=0.3
Soln:
Circumferential stress,
๐๐ = ๐๐
2๐ก
๐๐ = 3๐ฅ106๐ฅ1
2๐ฅ1๐ฅ10โ2
๐๐ = 1.5 ๐ฅ 104 ๐/๐2
Longitudinal stress,
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๐๐ = ๐๐
4๐ก
๐๐ = 3๐ฅ106๐ฅ1
4๐ฅ1๐ฅ10โ2
๐๐ = 0.75 ๐ฅ 104 ๐/๐2
(6M) Change in diameter:
๐ฟ๐ = ๐๐2
2๐ก๐ธ (1 โ
๐
2)
๐ฟ๐ = 3๐ฅ106๐ฅ(1)2
2๐ฅ1๐ฅ10โ2๐ฅ2๐ฅ1011 (1 โ
0.3
2)
๐ฟ๐ = 0.6375 ๐ฅ10โ3๐
Change in length:
๐ฟ๐ = ๐๐๐
2๐ก๐ธ (
1
2โ ๐)
๐ฟ๐ = 3๐ฅ106๐ฅ1๐ฅ5
2๐ฅ1๐ฅ10โ2๐ฅ2๐ฅ1011 (
1
2โ 0.3)
๐ฟ๐ = 0.75๐ฅ10โ3๐
Change in Volume:
๐ฟ๐
๐=
๐๐
2๐ก๐ธ (
5
2โ 2๐)
๐ฟ๐
๐=
3๐ฅ106๐ฅ1
2๐ฅ1๐ฅ10โ2๐ฅ2๐ฅ1011 (
5
2โ 2๐ฅ0.3)
๐ฟ๐
๐= 1.425๐ฅ10โ3
Volume ๐ =๐๐2๐ฟ
4
๐ =๐๐ฅ12๐ฅ5
4= 3.926 ๐3
๐ฟ๐ = 1.425๐ฅ10โ3๐ฅ3.926
๐ฟ๐ = 5.5959๐ฅ10โ3๐3
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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
ฯmax=125 MN/m2=125x106 N/m2
Soln:
Maximum Shear stress,
ฯmax = ๐๐
8๐ก
125๐ฅ106 = 25๐ฅ106๐ฅ0.16
8๐ฅ๐ก
๐ก =25๐ฅ106๐ฅ0.16
8๐ฅ125๐ฅ106
๐ก = 0.004๐
(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
๐๐ = ๐๐
2๐ก
Also the longitudinal stress as
๐๐ = ๐๐
4๐ก=
๐๐
2
Let
๐ฟ๐ = ๐โ๐๐๐๐ ๐๐ ๐๐๐๐๐๐ก๐๐ ๐๐ ๐กโ๐ ๐ โ๐๐๐ ๐ฟ๐ = ๐โ๐๐๐๐ ๐๐ ๐๐๐๐๐กโ ๐๐ ๐กโ๐ ๐ โ๐๐๐ ยต=Poissonโs ratio.
Change in diameter can be found as follows,
๐ฟ๐ = ๐๐ ๐ฅ ๐
๐๐ = ๐๐
๐ธ โ ๐
๐๐
๐ธ
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5-114
๐๐ = ๐๐
๐ธ โ ๐
๐๐
2๐ธ
๐๐ = ๐๐
๐ธ (1 โ
๐
2)
๐๐ = ๐๐
2๐ก๐ธ (1 โ
๐
2)
๐ฟ๐ = ๐๐
2๐ก๐ธ (1 โ
๐
2) ๐ฅ ๐
๐ฟ๐ = ๐๐2
2๐ก๐ธ (1 โ
๐
2)
(6M) Change in length can be found as follows,
๐ฟ๐ = ๐๐ ๐ฅ ๐
๐๐ = ๐๐
๐ธ โ ๐
๐๐
๐ธ
๐๐ = ๐๐
2๐ธ โ ๐
๐๐
๐ธ
๐๐ = ๐๐
๐ธ (
1
2โ ๐)
๐๐ = ๐๐
2๐ก๐ธ (
1
2โ ๐)
๐ฟ๐ = ๐๐
2๐ก๐ธ (
1
2โ ๐) ๐ฅ ๐
๐ฟ๐ = ๐๐๐
2๐ก๐ธ (
1
2โ ๐)
Change in volume can be found as follows,
๐๐ฃ = ๐ฟ๐
๐
Change in volume ๐ฟ๐ = Final Volume โInitial Volume
Original Volume,
๐ =๐๐2๐ฟ
4
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Final Volume,
๐๐ = ๐(๐ + ๐ฟ๐)2(๐ฟ + ๐ฟ๐ฟ)
4
Expanding we get,
๐๐ = ๐(๐2๐ฟ + 2๐๐๐ฟ๐ + ๐2๐ฟ๐ฟ)
4
Now, Change in Volume,
๐ฟ๐ = ๐(๐2๐ฟ + 2๐๐๐ฟ๐ + ๐2๐ฟ๐ฟ)
4โ
๐๐2๐ฟ
4
๐ฟ๐ = ๐(2๐๐๐ฟ๐ + ๐2๐ฟ๐ฟ)
4
Volumetric Strain,
๐๐ฃ =
๐(2๐๐๐ฟ๐+๐2๐ฟ๐ฟ)
4
๐๐2๐ฟ
4
๐๐ฃ = 2๐ฟ๐
๐+
๐ฟ๐
๐
๐๐ฃ = 2๐๐ + ๐๐
๐ฟ๐ = (2๐๐ + ๐๐) ๐ฅ ๐
(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,
๐๐ฅ =๐
๐ฅ2โ ๐
Apply the boundary condition to above equations,
1. At x= r1 = 200mm, ๐๐ฅ = 8 N/mm2
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2. At x= r2 = 300mm, ๐๐ฅ = 0
8 =๐
2002โ ๐ =
๐
40000โ ๐
0 =๐
3002โ ๐ =
๐
90000โ ๐
Subtracting the above equations,
8 =๐
40000โ ๐ โ
๐
90000+ ๐
8 =2.25๐
90000โ
๐
90000=
1.25๐
90000
๐ =90000๐ฅ8
1.25
๐ = 576000
0 =576000
90000โ ๐
๐ =576000
90000= 6.4
(6M)
Now hoop stress at any radius x is given by,
๐๐ฅ =๐
๐ฅ2+ ๐
๐๐ฅ =576000
๐ฅ2+ 6.4
At x=200mm,
๐200 =576000
2002+ 6.4 = 14.4 + 6.4
๐200 = 20.8 ๐/๐๐2
At x=300mm,
๐300 =576000
3002+ 6.4 = 6.4 + 6.4
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๐300 = 12.8 ๐/๐๐2
(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, ฯx = 8N/mm2
The radial pressure px is given by the eqn,
๐๐ฅ =2๐
๐ฅ3โ ๐
Apply the boundary condition to above equation,
1. At x= r1 = 100mm, ๐๐ฅ = 7 N/mm2
7 =2๐
1003โ ๐ =
2๐
1000000โ ๐
The hoop stress ฯx is given by the eqn,
๐๐ฅ =๐
๐ฅ3+ ๐
(6M)
Apply the boundary condition to above equation,
1. At x= r1 = 100mm, ๐๐ฅ = 8 N/mm2
8 =๐
1003+ ๐ =
๐
1000000+ ๐
Adding the above equations,
15 =2๐
1000000โ ๐ +
๐
1000000+ ๐
15 =3๐
1000000
๐ =1000000๐ฅ15
8
๐ = 5000000
8 =5000000
1000000+ ๐
๐ = 8 โ 5 = 3
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5-118
Now the radial pressure is given by,
๐๐ฅ =2๐ฅ5000000
๐ฅ3โ 3
Let r2=External radius of the shell.
Apply the boundary condition to above equation,
1. At x= r2 , ๐๐ฅ = 0
0 =2๐ฅ5000000
๐23 โ 3
๐23 =
10000000
3
๐2 = (107
3)
1
3
๐2 = 149.3๐๐.
Thickness of the shell
๐ก = ๐2 โ ๐1
๐ก = 149.3 โ 100
๐ก = 49.3๐๐
(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,
ฯ= stress induced in the spherical shell.
๐ = ๐๐
4๐ก
Change in diameter:
The strain in any one direction is given by,
๐ = ๐
๐ธ โ ๐
๐
๐ธ
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5-119
๐ = ๐
๐ธ (1 โ ๐)
๐ = ๐๐
4๐ก๐ธ (1 โ ๐)
We know that the strain in any direction as,
๐ = ๐ฟ๐
๐
๐ฟ๐
๐ =
๐๐
4๐ก๐ธ (1 โ ๐)
๐ฟ๐ = ๐๐2
4๐ก๐ธ (1 โ ๐)
(6M) Change in Volume:
w.k.t the original volume of the sphere,
๐ =๐๐3
6
The final volume due to the pressure,
๐๐ = ๐(๐ + ๐ฟ๐)3
6
Change in volume can be found as follows,
๐๐ฃ = ๐ฟ๐
๐
Change in volume ๐ฟ๐ = Final Volume โInitial Volume
Now, Change in Volume,
๐ฟ๐ = ๐(๐ + ๐ฟ๐)3
6โ
๐๐3
6
๐ฟ๐ = ๐(3๐2๐ฟ๐ + ๐3)
6
Volumetric Strain,
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๐๐ฃ =
๐(3๐2๐ฟ๐+๐3)
6
๐๐3
6
๐๐ฃ = 3๐2๐ฟ๐
๐3=
3๐ฟ๐
๐
๐๐ฃ = 3๐
๐ฟ๐ = (3๐) ๐ฅ ๐
๐ฟ๐ = (3๐๐
4๐ก๐ธ (1 โ ๐)) ๐ฅ
๐๐3
6
๐ฟ๐ = ๐๐๐4
8๐ก๐ธ(1 โ ๐)
(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 ยต=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
ยต=0.3
Using the relation, ๐ฟ๐
๐ =
๐๐
4๐ก๐ธ (1 โ ๐)
๐ฟ๐
๐ =
1.4๐ฅ106๐ฅ0.9
4๐ฅ0.01๐ฅ2x1011 (1 โ 0.3)
๐ฟ๐
๐ = 11.025๐ฅ10โ5
๐ฟ๐ = 11.025๐ฅ10โ5๐ฅ๐
๐ฟ๐ = 11.025๐ฅ10โ5๐ฅ0.9
๐ฟ๐ = 9.9225๐ฅ10โ5๐
Now, Volumetric strain ๐ฟ๐
๐= 3 ๐ฅ
๐ฟ๐
๐
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5-121
๐ฟ๐
๐= 3 ๐ฅ11.025๐ฅ10โ5
๐ฟ๐
๐= 33.075๐ฅ10โ5
(6M) Volume of a sphere
๐ =๐
6๐3
๐ =๐
6(0.9)3
๐ = 0.3817๐3
Increase in volume,
๐ฟ๐ = 33.075๐ฅ10โ5๐ฅ ๐
๐ฟ๐ = 33.075๐ฅ10โ5๐ฅ 0.3817
๐ฟ๐ = 12.62๐ฅ10โ5 ๐3
(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, ฯx = 10N/mm2
The radial pressure px is given by the eqn,
๐๐ฅ =2๐
๐ฅ3โ ๐
Apply the boundary condition to above equation,
At x= r1 = 90mm, ๐๐ฅ = 4 N/mm2
4 =2๐
903โ ๐ =
2๐
729000โ ๐
The hoop stress ฯx is given by the eqn,
๐๐ฅ =๐
๐ฅ3+ ๐
(5M)
Apply the boundary condition to above equation,
At x= r1 = 90mm, ๐๐ฅ = 10 N/mm2
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH/Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0
5-122
10 =๐
903+ ๐ =
๐
729000+ ๐
Adding the above equations,
14 =2๐
729000โ ๐ +
๐
729000+ ๐
14 =3๐
1458000
๐ =1458000๐ฅ14
3
๐ = 6804000
10 =6804000
729000+ ๐
๐ = 10 โ 9 = 1
(5M)
Now the radial pressure is given by,
๐๐ฅ =2๐ฅ6804000
๐ฅ3โ 1
Let r2=External radius of the shell.
Apply the boundary condition to above equation,
At x= r2 , ๐๐ฅ = 0
0 =2๐ฅ6804000
๐23 โ 3
๐23 =
13608000
3
๐2 = (4.536๐ฅ106)1
3
๐2 = 165.53๐๐.
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH/Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0
5-123
Thickness of the shell
๐ก = ๐2 โ ๐1
๐ก = 165.53 โ 90
๐ก = 75.53๐๐
(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 poissonโs 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
ยต=0.33
Soln:
Circumferential stress,
๐๐ = ๐๐
2๐ก
๐๐ = 5๐ฅ106๐ฅ1.1
2๐ฅ2๐ฅ10โ2
๐๐ = 1.375 ๐ฅ 104 ๐/๐2
Longitudinal stress,
๐๐ = ๐๐
4๐ก
๐๐ = 5๐ฅ106๐ฅ1.1
4๐ฅ2๐ฅ10โ2
๐๐ = 0.6875 ๐ฅ 104 ๐/๐2
(5M)
Change in diameter:
๐ฟ๐ = ๐๐2
2๐ก๐ธ (1 โ
๐
2)
๐ฟ๐ = 5๐ฅ106๐ฅ(1.1)2
2๐ฅ2๐ฅ10โ2๐ฅ2.1๐ฅ1011 (1 โ
0.33
2)
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH/Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0
5-124
๐ฟ๐ = 0.6013 ๐ฅ10โ3๐
Change in length:
๐ฟ๐ = ๐๐๐
2๐ก๐ธ (
1
2โ ๐)
๐ฟ๐ = 5๐ฅ106๐ฅ1.1๐ฅ7
2๐ฅ2๐ฅ10โ2๐ฅ2.1๐ฅ1011 (
1
2โ 0.33)
๐ฟ๐ = 0.77911๐ฅ10โ3๐ (5M)
Change in Volume:
๐ฟ๐
๐=
๐๐
2๐ก๐ธ (
5
2โ 2๐)
๐ฟ๐
๐=
5๐ฅ106๐ฅ1.1
2๐ฅ2๐ฅ10โ2๐ฅ2.1๐ฅ1011 (
5
2โ 2๐ฅ0.3)
๐ฟ๐
๐= 1.2493๐ฅ10โ3
Volume ๐ =๐๐2๐ฟ
4
๐ =๐๐ฅ1.12๐ฅ7
4= 6.6523 ๐3
๐ฟ๐ = 1.2493๐ฅ10โ3๐ฅ6.6523
๐ฟ๐ = 8.3107๐ฅ10โ3๐3 (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, ๐๐ก=135 N/mm2
Permissible stress may be circumferential stress or longitudinal stress.
๐๐ = 85% = 0.85
๐๐ = 35% = 0.35 Case 1:
Consider the Permissible stress as circumferential stresss ๐๐
w.k.t
REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH/Mr.M.K.KARTHIK/IIrd Yr/SEM 04 /CE8395/STRENGTH OF MATERIALS FOR MECHANICAL
ENGINEERS/UNIT 1-5/QB+Keys/Ver1.0
5-125
๐๐ = ๐๐
2๐ก๐๐
135 = 3.5๐ฅ๐
2๐ฅ28๐ฅ0.85
๐ = 1836 ๐๐
(8M)
Case 2:
Consider the Permissible stress as longitudinal stresss ๐๐
w.k.t
๐๐ = ๐๐
4๐ก๐๐
135 = 3.5๐ฅ๐
4๐ฅ28๐ฅ0.35
๐ = 1512 ๐๐
Thus, the maximum permissible diameter of the shell, d=1512mm.
(7M)