machanics of solids

8/14/2019 Machanics of Solids

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Lecture Notes

Mechanics Of Solids

ByProf. Y. Nath


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DEFINITION


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Cordinate System

( x , y , z ) or

----

right hand tria

Displacement (u , v , w ) or

Stress = Linear , stress vector (force / area)

T Stress vector = Force / Area with outward drawn normal

Similarly for other plane


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Infinite plane exist at a point P Infinite vectors

Plane/ Area is defined by outward drawn normal.Infinite planes 3mutually outward drawn planes at point P

(x1, x2 , x3,) Planes

at P

Stressvector

State of stress at a point

State of stress at a point =


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Hypothesis

Body is continuous and remain continuous under the action of external forces -

(H1) Elastic Continuum Neighbor remain neighbor under external force No cracks

/gaps or voids(H2) There exists a unique unstressed state of the solid body to which the bodyreturns whenever all the external forces are removed.

(H3) Principle of super position holds good


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Newtons Law

....................... (1)

is mass density


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Similarly

...................... (2)

....................(3)

Euler's Equation

and Components would contribution all other components are either to x orcross it.

Neglecting Higher order terms

(No body couple)


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PABC Tetrahedron at point P formed by 3 surfaces parallel to coordinate axises with unit normalLet h is the height of the surface between P & ABC

Body forces along(x1, ,x2 x3), vol =

and

i=1,2,3 are called Cauchy formulae are sufficient to define the traction on any plane.


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NORMAL & SHEAR

STRESSES

Let and are the normal & shear comps. On the oblique plan

Substitution of

i =1,2,3 j =1,2,3


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Cross section =2 cm * 3 cm

F = 6000N

Find the normal & shear stress components on

a equally inclined/plane relative to X1,X2,X3

Normal stresscomponent


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Show that


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3 equations with 6 Unknowns

Displacement Components(u , v , w )

3 Unknowns

Stress Tensor


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6 Unknowns

6 equations

No. of Unknowns = 6 + 3 + 6 = 15

No. of equations = 3 + 6 =9

Need to have more equations for the unique solution

Simplest Relationship(LINEAR)

Superposition Principal ( the law of independence of effects of forces ( ) on the deformation (

) )

is developed in the presence of (combined

effect)

Each of stresses caused by each component of


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Hookes Law for Isotropic Material

Two material cons tants.

Isotropy No direct ional property (same in al l).

Direction of principal ( ) stress and Principal strain ( ) must

coincide


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State of stress and strain in terms of Principal components

Has same effect along 2 and 3 orthogonal

directions

Youngs Modulus E

E Modulus of elasticity

orYoungs Modulus


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Poissons Ratio

Uniaxial1D state of stress

Isotropy

Hooks Law inShear


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Cauchy Formula ( Transformation) we Know

Bulk Modulus

k = (Hydros tatic Stress ( Pressure) / Volumetric Strain


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Equation of

Equilibrium -

Isotropic Materials


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Coupled ode not easy to solve them but are cons istent and will give the Unique so lution provided , we are able to

integrate them.

Theories of failure

Ultimate Aim DESIGN

Failure material failure

Uncertainties

MaterialLoading ,Geometry

Ignorance

Mathematical limitations etc.

Simplified Criteria Theories ofyielding


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Principal stressPrincipal strain

are extremum

More useful for Brittle material ,is also called Rankine criteriafor failure.

2. Maximum-Principal Strain Theory

In case of compressive state of stress

1. Maximum-Principal Stress Theory


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3. Maximum Shear Stress Theory

is called Tresca theory useful for ductile material--- cup and cone in simple tension 450

Tresca theory fails in hydro static state of stresses

4. Total strain energy theory --( Beltrami &

Haigh)

5. Shear or Distortion Strain Energy Theory


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Energy Method

First law of thermodynamics

Adiabatic Process


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Work of Gravity


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Strain energy due to shear stress

Similarly by other components of stresses

Super position


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Ex.1 Bar in tension


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Ex.2 Torsion of circular shaft (r , L)


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Ex.3 Bending strain EnergyNormal stress


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Strain energy due to shear in Beam

Castigliano Theorem


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Castig liano Theorem

Are corresponding deflection,twists

and rotation due to

Fict itions loads are applied at the point where

there is no load and deflection is songat

at that point where there is no load

Ex.1


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Ex.2


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Ex.3


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Ex.4


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Ex.5


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Uniaxial

ENGG. Prob. Definition

A. Everyone makes / one who works makes the mistakesB. Never accept the single way solution

Equation of equilibriumNavie rs equation


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Are coupled p.d.e. and it is extremely difficult to solve

Are 3D equations

Simplifications

Most of the simple engineering problems belongs to 1D problems

(i) Uniaxial deformationBars

(ii) Torsion of shaftShaft

(iii) One axis symmetric bendingBeams

(iv) Combined state of stress problems


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