CE6402 SOM- By Easy Engineering.net 1 PDF

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Chapter No. 1

Title

ENERGY PRINCIPLES 1.1 Strain Energy

1 1

1.2

Proof Stress

1

1.3 1.4

Resilience Proof Resilience

2 2

1.5

Modulus Of Resilience

2

1.6

Castigliano€s first theorem second Theorem. 1.7 Principle Of Virtual Work 1.8 Strain Energy Stored in a Rod of Length L and Axial Rigidity AE To an Axial Force P 1.9 State the Various Methods for Computing the Joint Deflection of a Perfect Frame 1.10 State the Deflection of the Joint Due To Linear Deformation

2

Page No.

3 13 13 13 13

1.

14

1. 1.

14 27

1.

27

IN 2.

29 29

2.

29

2.3

State the Degree Of Indeterminacy in A Fixed Beam

29

2.4 2.5

State the Degree Of Indeterminacy in The Given Beam State the Degree Of Indeterminacy in The Given Beam

29 30

State the Methods Available for Analyzing Statically Indeterminate Structures 2.7 Write the Expression Fixed End Moments and Deflection for a Fixed Beam Carrying Point Load at Centre 2.8 Write the Expression Fixed End Moments and Deflection for a Fixed Beam Carrying Eccentric Point Load 2.9 Write the Expression Fixed End Moments for a Fixed Due To Sinking of Support 2.10 State the Theorem of Three Moments

30

2.11 Effect Of Settlement Of Supports In A Continuous Beam

35

2.6

3

COLUMNS 3.1

Columns

30 30 30 34 52 52

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3.2

Struts

52

3.3

Mention the Stresses Which are Responsible for Column Failure.

52

3.4

End Conditions of Columns

52

3.5 3.6

Explain the Failure of Long Column State the Assumptions Made in the Euler€s Column theory and Explain the Sign Conventions Considered In Columns. 3.7 Derive the Expression for Crippling Load When the Both Ends of the Column are Hinged 3.8 Derive the Expression for Buckling Load (Or) Crippling Load When Both Ends of the Column are Fixed 3.9 Derive the Expression For Crippling Load When Column With One End Fixed and Other End Hinged 3.10 Derive the Expression for Buckling Load for the Column With One End Fixed and Other End Free 3.11 Expression For Crippling Load

53 54

3.12 Expression for Buckling Load (Or) Crippling Load

62

3.13 Expression For Crippling Load When Column With One End

62

3.

Fixed

62

3. 3.

Which

62 62

3.

nd

63

54 56 58 59 61

3.

67

3.19 Derive The Expression For Core Of A Rectangular Section 3.20 Derive The Expression For Core Of A Solid Circular Section Of Diameter D

67 68

4 4.1

STATE OF STRESS IN THREE DIMENSIONS Stress

92 92

4.2

Principal Planes

92

4.3

Spherical Tensor

92

4.4 4.5

Deviator Stress Tensor Stress Components At A Point

92 93

4.6

95

4.7

The Energy Of Distortion ( Shear Strain Energy ) And Dilatation State The Principal Theories Of Failure

101

4.8

Limitations Of Maximum Principal Stress Theory

102

4.9

Maximum Principal Stress Theory

102

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4.10 4.11

5

Maximum Shear Stress Theory Limitations Of Maximum Shear Stress Theory

102 102

4.12 Shear Strain Energy Theory

102

4.13 Limitations Of Distortion Energy Theory 4.14 Maximum Principal Strain Theory

102 102

4.15 Limitations In Maximum Principal Strain Theory

102

4.16 Stress Tensor In Cartesian Components

102

4.17 Three Stress Invariants 4.18 Two Types Of Strain Energy

103 104

4.19 The Maximum Principal Stress

104

4.20 Explain The Maximum Shear Stress (Or) Stress Difference Theory 4.21 Explain The Shear Strain Energy Theory

105

4.22

107

Explain The Maximum Principal Strain Theory

106

4.23 Explain The Strain Energy Theory 4.

109 110

A 5. 5.

119 119 119

5.

119

5.

119

5. 5. 5.7

ction Derive The Equation Of Shear Center For Unequa-Lsei Ction

5.8

120 120 121

Derive The Stresses In Curved Bars Us Ing Winkl–Erbach Theory 5.9 State The Parallel Axes And Principal Moment Of Inertia 5.10 Stress Concentration

122

5.11 Stress Concentration Factor

135

5.12 Fatigue Stress Concentration Factor 5.13 Shear Flow

135 135

5.14 Explain The Position Of Shear Centre In Various Sections

136

5.15

136

State The Principles Involved In Locating The Shear Centre

135 135

5.16 State The Stresses Due To Unsymmetrical Bending 5.17 Fatigue

136 136

5.18 Types Of Fatigue Stress

136

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5.19 State The Reasons For Stress Concentration

137

5.20

137

Creep

CE6402

STRENGTH OF: MATERIALS Downloaded From www.EasyEngineering.net

LT P C 3104

OBJECTIVES: To know the method of finding slope and deflection of beams and trusses using energy theorems and to know the concept of analysing indeterminate beam. To estimate the load carrying capacity of columns, stresses due to unsymmetrical bending and various theories for failure of material. UNIT I ENERGY PRINCIPLES 9 Strain energy and strain energy density – strain energy due to axial load, shear, flexure and torsion – Castigliano‟s theorems – Maxwell ‟s reciprocal theorems - Principle of virtual work – application of energy theorems for computing deflections in beams and trusses - Williot Mohr's Diagram. UNIT II INDETERMINATE BEAMS 9 Concept of Analysis - Propped cantilever and fixed beams-fixed end moments and reactions – Theorem of three moments – analysis of continuous beams – shear force and bending moment diagrams. UNIT III COLUMNS AND CYLINDER 9 Eulers theory of long columns – critical loads for prismatic columns with different end conditions; Rankine-Gordon formula for eccentrically loaded columns – Eccentrically loaded short columns – middle third rule – core section – Thick cylinders – Compound cylinders. UNIT IV STATE OF STRESS IN THREE DIMENSIONS 9 Determination of principal stresses and principal planes – Volumetric strain –Theories of failure – Principal stress - Principal strain – shear stress – Strain energy and distortion energy theories – application in analysis of stress, load carrying capacity. UNIT V AD Unsymmetrical be curved beams – W

9 – Shear Centre -

T:15): 60 PERIODS OUTCOMES: Students w energy me

eams and use of ses.

They will materials.

ms and failure of

TEXT BOOKS: 1.

Rajput R.K. "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2010.

2.

Egor P Popov, “Engineering Mechanics of Solids”, 2nd edition, PHI Learning Pvt. Ltd., New Delhi, 2012.

REFERENCES: 1.

Kazimi S.M.A, “Solid Mechanics”, Tata McGraw-Hill Publishing Co., New Delhi, 2003.

2.

William A .Nash, “Theory and Problems of Strength of Materials”, Schaum ‟s Outline Series,Tata McGraw Hill Publishing company, 2007.

3.

Punmia B.C."Theory of Structures" (SMTS) Vol 1&II, Laxmi Publishing Pvt Ltd, New Delhi 2004.

4.

Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi,2011.

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CE 6402

STRENGTH OF MATERIALS

CHAPTER - I ENERGY PRINCIPLES Strain energy and strain energy density- strain energy in traction, shear in flexure and torsion- Castigliano’s theorem – Principle of virtual work – application of energy theorems for computing deflections in beams and trusses – Maxwell’s reciprocal theorem.

1.1 STRAIN ENERGY Whenever a body is strained, the energy is absorbed in the body. The energy which is absorbed in the body due to straining effect is known as strain energy. The strain energy stored in the body is equal to the work done by the applied load in stretching the body. 1.2 PROOF STRESS The stress in its proof stress.

rgy is termed as

Derive the expression f Axial loading (ii) Flexu (i)Axial Loading

cases. (i)

Let us consid load P is appl as strain ener deformation ∆. Thus

rea A. If an axial rk done, stored plied by the

U = ½ P. ∆ But ∆ = PL / AE

U = ½ P. PL/AE = P2 L / 2AE ---------- (i) If, however the bar has variable area of cross section, consider a small of length dx and area of cross section Ax. The strain energy dU stored in this small element of length dx will be, from equation (i) P2 dx dU = --------2Ax E The total strain energy U can be obtained by integrating the above expression over the length of the bar. L

U=

 0

SCE

P 2 dx 2 Ax E

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CE 6402

STRENGTH OF MATERIALS

(ii) Flexural Loading (Moment or couple ) Let us now consider a member of length L subjected to uniform bending moment M. Consider an element of length dx and let di be the change in the slope of the element due to applied moment M. If M is applied gradually, the strain energy stored in the small element will be dU = ½ Mdi But di d ------ = ----- (dy/dx) = d2y/d2x = M/EI dx dx M di = ------- dx EI Hence dU = ½ M (M/EI) dx = Integrating U =

1.3 RESILIENC The resilience is defined as the capacity of a strained body for doing work on the removal of the straining force. The total strain energy stored in a body is commonly known as resilience. 1.4 PROOF RESILIENCE The proof resilience is defined as the quantity of strain energy stored in a body when strained up to elastic limit. The maximum strain energy stored in a body is known as proof resilience. 1.5MODULUS OF RESILIENCE It is defined as the proof resilience of a material per unit volume. Proof resilience Modulus of resilience = ------------------Volume of the body Two methods for analyzing the statically indeterminate structures.

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CE 6402

STRENGTH OF MATERIALS

a.Displacement method (equilibrium method (or) stiffness coefficient method b.Force method (compatibility method (or) flexibility coefficient method) 1.6 Castigliano’s first theorem second Theorem. First Theorem. It states that the deflection caused by any external force is equal to the partial derivative of the strain energy with respect to that force. Second Theorem It states that “If U is the total strain energy stored up in a frame work in equilibrium under an external force; its magnitude is always a minimum.

Castigliano’s first theorem: It states that the deflection caused by any external force is equal to the partial derivative of the strain energy with respect to that force. A generalized statement of the theorem is as follows: “ If there is any elastic system in equilibri W 1 , W 2, W 3 of moments the partial de moments tak actions.”

a set of a forces ……. δn and a set 3,…….. Φn , then of the forces or n its direction of

Expressed U  1 W 1 U  1 M 1

------------- (i) -------------

(ii)

Proof: Consider an elastic body as show in fig subjected to loads W 1, W 2, W 3 ………etc. each applied independently. Let the body be supported at A, B etc. The reactions RA ,RB etc do not work while the body deforms because the hinge reaction is fixed and cannot move (and therefore the work done is zero) and the roller reaction is perpendicular to the displacements of the roller. Assuming that the material follows the Hooke‟s law, the displacements of the points of loading will be linear functions of the loads and the principles of superposition will hold. Let δ1, δ2, δ3……… etc be the deflections of points 1, 2, 3, etc in the direction of the loads at these points. The total strain energy U is then given by

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CE 6402

U = ½ (W 1δ1 + W 2 δ2 + ……….)

STRENGTH OF MATERIALS

--------- (iii)

Let the load W 1 be increased by an amount dW 1, after the loads have been applied. Due to this, there will be small changes in the deformation of the body, and the strain energy will be increased slightly by an amount dU. expressing this small increase as the rate of change of U with respect to W 1 times dW 1, the new strain energy will be U+

U xdW1 W1

---------

(iv)

On the assumption that the principle of superposition applies, the final strain energy does not depend upon the order in which the forces are applied. Hence assuming that dW 1 is acting on the body, prior to the application of W1, W 2, W 3 ………etc, the deflections will be infinitely small and the corresponding strain energy of the second order can be neglected. Now when W1, W 2, W 3 ………etc, are applied (with dW1 still acting initially), the points 1, 2, 3 etc will move through δ1, δ2, δ3……… etc. in the direction of these forces and the strain energy will be given as above. Due to the application of W 1, rides through a distance δ1 and produces the external work increment dU = dW 1 . δ1. Hence the strain energy, when the loads are applied is U Since the fina U

Which proves the proportion. Similarly it can be proved that Φ1=

U . M 1

Deflection of beams by castigliano’s first theorem: If a member carries an axial force the energies stored is given by L P 2 dx U=  0 2Ax E In the above expression, P is the axial force in the member and is the function of external load W 1, W 2,W3 etc. To compute the deflection δ1 in the direction of W 1 L U P p dx = δ 1= W1 0 AE W1 If the strain energy is due to bending and not due to axial load L M 2 dx U=  2 EI 0

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CE 6402

STRENGTH OF MATERIALS

L

U M dx = M δ 1= W 1 0 W1 EI

If no load is acting at the point where deflection is desired, fictitious load W is applied at the point in the direction where the deflection is required. Then after differentiating but before integrating the fictitious load is set to zero. This method is sometimes known as the fictitious load method. If the rotation Φ1 is required in the direction of M1.

Φ1=

U L M dx = M M 1 0 M 1 EI

Calculate the central deflection and the slope at ends of a simply supported beam carrying a UDL w/ unit length over the whole span. Solution: a) Central defl Since no fictitious load W

d, apply the

δc

Consider Bending moment at x, wx 2  wL W   x  M=  2 2  2 M x  2 x

c 

2 EI

l 2

 0

  wL W  wx 2  x  dx  x     2 2 2  2 

Putting W=0, 2 c  EI

l 2



0

  wL  wx 2  x  dx   x  2  2  2  l

=

SCE

2   wLx 3 wx 4  2    16  0 EI   12

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CE 6402

c 

STRENGTH OF MATERIALS

5 wl 4 384 EI

b) Slope at ends To obtain the slope at the end A, say apply a frictions moment A as shown in fig. The  wl m   wl m  reactions at A and B will be    and    l   2 l   2 Measuring x from b, we get

 A=

u  m

1 l

EI  Mx

Mx .Dx -------------------------------- 2 M

0

Where Mx is the moment at a point distant x from the origin (ie, B) is a function of M. Wx 2  wl m  Mx =    x  2 Mx x  m l l w A= 1   EI 0  2

Putting M=0

a 

1 wl WX 2 x x dx  2 l Ei 0 2

L

3 wx 4  1  wx  A    EI  6 8L  0 wL3 A  24 EI

State and prove the Castigliano’s second Theorem. Castigliano’s second theorem: It states that the strain energy of a linearly elastic system that is initially unstrained will have less strain energy stored in it when subjected to a total load system than it would have if it were self-strained.

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CE 6402

STRENGTH OF MATERIALS

u =0 t For example, if  is small strain (or) displacement, within the elastic limit in the direction of the redundant force T, u = t  =0 when the redundant supports do not yield (or) when there is no initial lack of fit in the redundant members.

Proof: Consider a redundant frame as shown in fig.in which Fc is a redundant member of geometrical length L.Let the actual length of the member Fc be (L-  ),  being the initial lack of fit.F2 C represents thus the actual length (L-  ) of the member. When it is fitted to the truss, the member will have to be pulled such that F2 and F coincide.

According to Ho F2 F1 = D Where T is the fo Hence FF

=

TL ------------------------------------ ( i ) AE

Let the member Fc be removed and consider a tensile force T applied at the corners F and C.

FF1 = relative deflection of F and C u1 ------------------------------------------ ( ii ) = T According to castigliano‟s first theorem where U1 is the strain energy of the whole frame except that of the member Fc. Equating (i) and (ii) we get TL u 1 =  -AE T (or)

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CE 6402

STRENGTH OF MATERIALS

TL u 1 + =  ----------------------- ( iii ) T AE

To strain energy stored in the member Fc due to a force T is UFC = ½ T.

TL T 2L = AE 2AE

TL U FC  AE T

Substitute the value of

TL in (iii) we get AE

U u' U FC     (or) T T T

When U= U1 + U

Note: i) Castigliano indeterminate bea ii) If the degr distribution meth

lysis of statically an two. d or the moment

A beam AB of span 3mis fixed at both the ends and carries a point load of 9 KN at C distant 1m from A. The M.O.I. of the...