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Unit 1
Portions for CIE
Fundamental concepts: Principles of Elasticity :
Equilibrium equations,
strain displacement relationships in matrix form
Generalised Hooke’s Law, Constitutive relationships(3D)
Constitutive relationships for plane stress, plane strain , Axi-symmetric
Potential energy and equilibrium,PMPE and numericals on Springs in combination subjected to forces
Rayleigh-Ritz method applied to simple problems on axially loaded members,,with point loads and distributed loads.
Gauss elimination method and
Gaussian quadrature-1pt,2pt and 3 pt formula.
commercial packages-preprocessor, solver and post processor.
1.Explain the significance of Equilibrium equations. Derive the force equilibrium conditions
for 2D state of stress. Write equilibrium equations for 3D.
2.Sketch the variation of stresses in a 3D differential element subjected to body forces.
Establish the equations of force equilibrium in differential form
3.Write down (do not derive) the equations of equilibrium in differential form in terms of
stresses and body forces for a three dimensional element volume with the help of a neat
sketch
4.Explain principal stress, maximum stress and Von Mises stress and bring out the importance of
these in the stress analysis.
5.The stress components at a point in a body are given by x=3xy
2 z+2x ,y=5xyz+2y
,z=3xy
2 z+2x ,xy=0, yz=xz=3xy
2 z+2xy. Determine whether these components of stress
satisfy the equilibrium equations or not at the point (1,-1,2).If not then determine the suitable body force vector required at this point so that these stress components are in equilibrium.
6.Write strain displacement relations for 2D and for 3D in Cartesian coordinates.
7.Strain in a rod of length “L” fixed at one end and subjected to axial loading is given by
x=1+2x
2 .Find the tip displacement .
8.The displacement field for a body is given by u=(x2+y2)i+(3+z)j+(x2+2y)k.Find the rectangular
strain components at (3,1,-2)
9. Explain plane stress and plane strain problems. Give examples and write stiffness and
compliance matrices
10. Explain Plane stress ,Plane strain and axisymetric problems with examples.write stiffness matrix for the same
- A.Identify the idealization of 2D problems shown. State the geometric and corresponding stress/strain conditions and write stiffness matrices.
- 1.Thin Plate with Hole 2.Thin cantilever plate
11b Stress Analysis of the prototype seatbelt component shown in the figure below is to be carried out
for a tensile load. Identify the idealization of 2D problem shown. State the geometric and corresponding
stress/strain conditions and write stiffness matrices.
Given 2 D member has in plane dimensions(x,y) very large compared to Out of plane dimension(z).Since
y
x
xy xy
Plane Strain D ,
33
E
D (^) x
- The state of strain at a point is given by x=0.0015, y=-0.0025, z = yz= xz = 0, xy=
-0.004.Determine the stress tensor at the point. E =208 Gpa. and =0.28.
- Determine the strain tensor for the stress tensor at a point given by
210 140 00
[] = 140 -240 00 where E=208 GPa and =0.
00 00 00
14. Explain Axi-symmetric problems. Give examples and write stiffness and compliance **matrices.
- Explain with examples of bar and beam ,essential and Non essential boundary**
conditions
- Explain boundary value problems and Initial value problems with example. 17. Write admissible functions for i. Axial bar fixed at one end ii. Axial bar fixed at both ends
iii. Cantilever beam , iv Simply supported beam(Polynomial and trigonometric)
iv.Beam fixed at both ends
v.Propped cantilever beam
18. Using matrix notation, develop an expression for the total potential energy functional **for a 3D elastic solid subjected to body forces, surface forces and point loads.
- State principle of Minimum Potential Energy.
- Obtain displacements of the nodes in the spring system problems below using PMPE
- Explain Rayleigh-Ritz method applied to continuum. What are its disadvantages** 22. Obtain expression for displacement for uniform bar fixed at one end and loaded by a
point load at free end in tension using Rayleigh Ritz method.
23. Obtain expression for displacement for uniform bar fixed at one end and loaded by a
point load at free end in compression using Rayleigh Ritz method.
24. Obtain expression for displacement for uniform bar under uniformly distributed axial load of intensity q=C .where C is a constant using Rayleigh Ritz method .Use
second order polynomial
25. Using Rayleigh Ritz method obtain expressions for displacement and stress for
uniform bar shown in figure. Normalize the values if P=A=L=E=
26. Use Rayleigh Ritz method to find displacement of any point of the rod shown.Also
determine the displacement of the midpoint.
27. Use Rayleigh Ritz method to find displacement of any point of the rod shown above
when the bottom is not fixed. Also determine the displacement of the free end.
28. Demonstrate the importance of convergence using first order and second order polynomial in Rayleigh Ritz method to obtain expressions for displacement and stress
for uniform bar fixed at one end x=0 and subjected to linearly varying distributed axial
load of intensity q = Cx, where C is a constant.
29. Explain preprocessor,solver/processor and post processor applied to commercial
finite element package.
1. Give the principle involved in deriving Gauss quadrature formula for a simple function
F( r )= 1 + 2 r+ 3 r
2 + 4 r
3 , F( r) integrated between – 1 to +1 06
**2. Explain 1pt and 2 pt. Gaussian quadrature method
- Evaluate the integral I =** ( 1 )( 4 ) d d
2
11
2
using two point formula.
4. Evaluate I = dx x
e x
x ) ( 2 )
(^12)
(^1)
using one point and two point Gauss Quadrature.
5. Using Gaussian Quadrature formula, evaluate the following
i. ( 1 2 r 3 r 4 r ) dr
1
1
2 3 (^) 11.
1
(^1) x
dx iii. ( 2 ) d d
3
11
2
These assumptions require the material to be,
1.2 Assumptions
Continuum: The body is continuous, so displacements, Strains and stresses, can be
expressed by continuous functions in space.
Homogeneous : The body is homogeneous, i.e., the elastic properties are the same
throughout the body. Elastic constants will be independent of the location in the body.
Isotropic: The body is isotropic so that the elastic properties are the same in all directions. Thus the elastic constants will be independent of the orientation of coordinate
axes.
The two independent elastic constants are E → Young’s modulus
G → Shear modulus
Linear Elastic: The body is perfectly elastic
Obeys Hook's law of elasticity i.e linear relations between stress components and strain
components. = E. ,E → Young’s modulus = G., G → Shear modulus
Elastic(Non linear) Linear Elastic
The displacements and strains are small:
- T he displacements components of all points of the body during deformation are very
small compared with its original dimensions
- The equations of Elasticity are considerably simplified.
- Also be called small deformation theory, small displacement theory, or small displacement-gradient theory.
E
G
F(
D(
F ()
D(
1.3 Types of Forces
1.BodyForce
T x (^) z
f y
f x
f f 3 1
Acts on the volume(mass) of the body.
Dimension is Force/Volume ,
Examples :gravitational force ,Inertia
forces (in motion), Magnetic force.
Component of in X, Y, Z directions are
- Surface force
Surface force(often termed surface Traction) : Acts on the surface of the body.
Dimension is force/Area, e.x., N/m
2
Example:Contact forces ,Aerodynamic
pressures,friction hydrostatic pressure.
Component of T in X, Y, Z directions are
Forces: 3. Point load
Point load (often termed Concentrated Load ) : Idealised as acting at a point on the body.
Dimension is force, e.x., N.
Component of P in X, Y, Z directions are
T (^) 3 x 1
i 3 1
P x
Cantilever Beam Under Self-Weight Loading
Body Forces
T
x
z
T
y
T
x
T 31 T
T
x
z
P
y
P
x
Pi 31 P
In contracted notation, Stress Is written in the form of a vector as
T
6 X 1 xx yy zz xy yx zx
Development of stress concept
Sl.No Diagrams Components Definition
1.Force Force components
Push or Pull
2.Traction Force per unit area
On a surface of
specified
orientation.
(Force intensity)
3.Surface
Stress
Equilibrium SurfaceStress
components
A pair of equal and
opposite traction
on a surface of specified
orientation
4. Stress at
point
Stress
components
Surface stresses on
planes of all orientations
through a point.
A Similarities between stresses and tractions
- 1 Same dimensions (force per unit area)
- 2 The normal stress acting on a plane matches the normal traction
- B Differences between stresses and tractions
- 1 Stresses are tensor quantities and tractions are vectors.
- 2 The stress state is defined at a point using a fixed reference frame, whereas a
traction is defined on a plane with a reference frame that floats with the plane.
1.6 Equilibrium equations
- The state of stress varies from point to point in a loaded member in general.
- This variation is governed by the condition that each and every differential element considered should satisfy the conditions of equilibrium
Force equilibrium:
1.6.1 : 2 D equilibrium equations
- Consider a 2D element of size dx,dy from a loaded 2D structure.
- The element is shown under positive stress system
Let the body forces(expresses in terms of force /volume be
T
x
y
f x
f 21 f
Equilibrium Equations 2D
Fx 0 , Fy 0 Fz 0
M (^) x 0 , My 0 Mz 0
x
y
dx x
x x
xy
yx
dy y
y y
dx x
xy xy
dy y
yx yx
fx
fy
z xy yx
y
xy y y
x
x yx x
M
f x y
F
f x y
F
1.7 Displacement:
- Pattern of Deformation
- Rigid Body Motion
Two-Dimensional Example
Rigid body Rotation Rigid body Translation
Displacement vector
Displacement of a material point P
inside a body, before and after the deformation
- Initial position of the material
points of the body is described by the
Zero Strains
coordinates x, y, z of the generic point P
- Its displacement is defined by vector PP’ with components u,v,w in the reference directions x, y, z, respectively.
- The position of the point after the deformation is therefore given by the coordinates x + u ,
y + v , z + w.
- If the material is continuous before and after the deformation, the functions u ( x, y, x ),
v ( x, y, z ), w ( x, y, z ) are continuous functions of the position coordinates of the body
before the deformation, x, y, z.
- Let Intial position be
- P (x, y,z) and displaced position be , P' (x', y',z',).
- Displacement vector PP' is denoted by u.
- The displacement vector u has components Ux=u, Uy=v
and Uz=w along the x, y and z axes respectively, and can
be expressed as, u iu jv kw
2. Deformation
Biaxial stretch
Infinitesimal Strain (Small Strains)
- Reference point A is taken at location (x,y), and the displacement components of this
point are thus u(x,y) and v(x,y).
- The corresponding displacements of point B are dx x
u u
and dx x
v v
- The corresponding displacements of point D are dy y
u u
and dy y
v v
Longitudinal Strain x
1 1
1 1
x AB AB x dx x AB
AB AB
Horizontal projection of dx x
u A B dx
1 1
Vertical projection = dx x
v B B
1 1 1
2 2 2
2 2
1 12 2
x
v
x
u
x
u
dx x
v dx x
u dx
AB dx
x x
x
Assuming small deformations and strains and neglecting product of Smaller
quantities
x
u x
Similarly Longitudinal strain in Y direction
y
v
AD
A D AD
y y
1 1
Shear Strain
Shear strain measures changes in angles in terms of radians with respect to two specific
directions initially perpendicular to each other.
y
u
x
v
dx
dx y
u dx x
v
BAB DAD
xy
xy
xy
1 2
1 1 1 1
1 1 1 1
y
u
x
v
y
v
x
u
x , y xy
Extending the logic to 3D
z
u
x
w
z
v
y
w
y
u
x
v
z
w
y
v
x
u
xz
xy yz
x y z
1.9 Generalised Hooke’s Law
- The relationships between the stress and strain components are termed Constitutive
equations.
- The Constitutive equations is based on Experimental observations and established
principles.
Hooke’s Law
Robert Hooke (1635-1703) :Established tension is proportional to the stretch
- Hooke’s law established the notion of (linear) elasticity, but not yet in a way that was
expressible in terms of stress and strain.
For Linear Elastic material ,Stress is proportional to Strain and vica-versa
,GistheconstantofProportionality
,EistheconstantofProportionality
is or G
is or E
The relation between the normal stress σ and the longitudinal strain in the same
direction ε is called the longitudinal modulus of elasticity or Young’s modulus E of the material.
The relation between the Shear stress and the Shear strain is called the Shear modulus of
elasticity or Modulus of Rigidity G of the material.
Hooke's Law in Compliance Form S
E
S
(^)
E
Note G G
1.10 : 2 D Problems
By virtue of Geometry, Loading and Material Property ,certain class of problems can be
reduced in dimension from 3D to 2 D and sometimes 1 D without much loss of accuracy.
This saves considerable memory space and computational time.
Two vs Three Dimensional Problems(Sadd)
2D elastic problems
- PLANE STRESS
- PLANE STRAIN
- The basic theories of plane strain and plane stress represent the fundamental plane
problem in elasticity.
Plane Stress Problems
whose in plane dimensions(x,y)
are very large compared to
Out of plane dimension(z)
- The domain is bounded two stress free planes z
= h ,
- Since the plate is thin in the z - direction, there can be little variation in the stress
components through the thickness.
- Thus they will be approximately zero throughout the entire domain. z xz yz 0
- Under these assumptions, the stress field can be taken as
z xz yz xz yz z
x x y y xy xy
Also
x y x y x y
Plane Stress :Examples
- 1.Thin Plate with Hole 2.Thin cantilever plate
Rotating disc/Flywheel
Plane Stress D
xy
y
x
xy
y
x E
2
[D] matrix for the plane stress case is
(^332)
E
D (^) x
y
x
xy xy
z ^ x^ y
