1
Elementary Elasticity
1.1 INTRODUCTION
An experimental stress analyst must have thorough understanding of stresses and strains developed in any structural member. Therefore, relationships between deformations, strains, and stresses developed in a body are derived in this chapter. Further generalized Hooke’s law, equilibrium equations and compatibility equations, and Airy’s stress functions for determining stresses at any point in a member subjected to known loads/stresses on its boundary form the text of this chapter. Plane stress state, plane strain state, and threedimensional displacement field along with strain matrix are also discussed in the chapter.
1.2 STRESS TENSOR
All stresses acting on a cube of infinitesimally small dimensions, i.e. ∆x = ∆y = ∆z → 0 are identified by the diagram of stresses on a cube. The first subscript on normal stress σ or shear stress τ associates the stress with the plane of the stress, i.e. subscript defines the direction of the normal to the plane, and the second subscript identifies the direction of stress as
τ_{xx} = normal stress on yz plane in xdirection
τ_{xy} = shear stress on yz plane in ydirection
τ_{xz} = shear stress on yz plane in zdirection.
Similarly the stresses on xz plane and xy planes are identified as shown in Figure 1.1. Stress tensor τ_{ij} can now be written as
Figure 1.1 Stresses on a cube element
But, generally the normal stress is designated by σ.
So, τ_{xx} = σ_{xx}, τ_{yy} = σ_{yy}, τ_{zz} = σ_{zz}
Stress tensor can now be written as
Stress tensor is symmetric, i.e. τ_{ij} = τ_{ji}
Complementary shear stresses are
Complementary shear stresses meet at diametrically opposite corners of an element to satisfy the equilibrium conditions.
Stress invariants
1.3 STRESS AT A POINT
At a point of interest within a body, the magnitude and direction of resultant stress depend on the orientation of a plane passing through the point: An infinite number of planes can pass through the point of interest so, there are infinite number of resultant stress vectors. Yet the magnitude and direction of each of these resultant stress vectors can be specified in terms of the nine stress components as shown on three faces of elemental tetrahedron (Figure 1.2)
Figure 1.2 Stresses on elemental tetrahedron
Let us take the altitude of plane abc, h → 0 and neglecting body forces, three components of σ_{r} (resultant stress) in x, y, and zdirection can be written as
From these three Cartesian components, resultant stress, σ_{r}, can be determined as follows:
The three direction cosines which define the line of action of resultant stress σ_{r} are
Normal stress σ_{n} and shear stress τ_{n} acting on the plane under consideration will be
Angle between the resultant stress vector σ_{r} and normal to the plane σ_{n} can be determined by using
The normal stress σ_{n} also can be determined by considering projections of σ_{rx}, σ_{ry} and σ_{rz} onto the normal to the plane under consideration; therefore
After the determination of normal stress σ_{n}, shear stress τ_{n} is obtained by
Example 1.1 At a point in a stressed body, the Cartesian components of stresses are σ_{xx} = 60 MPa, σ_{yy} = − 30 MPa, σ_{zz} = +30 MPa, τ_{xy} = 40 MPa, τ_{yz} = τ_{zx} = 0.
Determine normal and shear stresses on a plane whose outer normal has the direction cosines of
Solution: Let us say normal and shear stresses on plane are σ_{n} and τ_{n}, then resultant stress on plane is
For the problem, three components of σ_{r} are as follows:
σ_{rx} = σ_{xx} cos(n, x) + τ_{yx} cos(n, y)
σ_{ry} = τ_{xy} cos(n, x) + σ_{yy} cos(n, y)
σ_{rz} = σ_{zz} cos(n, z) , because τ_{yz} = τ_{yx} = 0
Putting the values of σ_{xx}, σ_{yy}, σ_{zz}, τ_{ny}, cos (n,x), cos (n, y),cos(n,z)
Normal stress σ_{n} on plane is
Shear stress,
Example 1.2 At a point in a stressed body, the Cartesian components of stress are σ_{xx} = 70 MPa, σ_{yy} = 60 MPa, σ_{zz} = 50 MPa, τ_{xy} = 20 MPa, τ_{yz} = −20 MPa, τ_{zx} = 0.
Determine the normal and shear stresses on a plane whose outer normal has directions cosines
Solution:
Exercise 1.1 Determine the normal and shear stresses on a plane whose outer normal makes equal angles with the x, y, and z axes if the Cartesian components of stress at the point are
Ans. [σ_{n} = 120 MPa, τ_{n} = 43.20 MPa].
Exercise 1.2 At a point in a stressed body, the Cartesian components of stress are σ_{xx} = 40 MPa, σ_{yy} = 60 MPa, σ_{zz} = 20 MPa, τ_{xy} = 60 MPa, τ_{yz} = 50 MPa, τ_{zx} = 40 MPa, determine (a) normal and shear stresses on a plane whose outer normal has the directions cosines
(b) The angle between σ_{r} and outer normal n.
Ans. [σ_{n} = 117.77 MPa, τ_{n} = 58.547 MPa, angle 26°35’].
1.4 PLANE STRESS CONDITION
Under plane stress condition, stresses σ_{xx}, σ_{yy}, τ_{xy} are represented in one plane, i.e. central plane of a thin element in yx plane, as shown shaded in Figure 1.3
Figure 1.3 Plane stress state
In this case stresses σ_{zz} = τ_{xz} = τ_{yz} = 0.
If E and v are Young’s modulus and Poisson’s ratio of the material, then strains are
Shear strain
But shear strain γ_{xy} is equally divided about x and y axes both. Strain tensor for plane stress condition will be as follows:
For a particular set of three orthogonal planes, where shear stresses are zero and normal stresses on these planes are termed as principal stresses σ_{1}, σ_{2}, and σ_{3},
stress tensor will be
1.5 STRAIN TENSOR
Figure 1.4 shows a body of dimensions ∆ x and ∆y subjected to shear stresses τ_{xy}, τ_{yx} shear strain develops from y to x and from x to y.
Total shear strain about xy axes is
where G is shear modulus.
This shear strain γ_{xy} is equally divided about both axes x and y. Strain tensor in a biaxial case will be
Figure 1.4 Shear stresses and shear strains on an element
Similarly strains
and
Threedimensional strain tensor will be
In the plane stress case (σ_{xx}, σ_{yy}, τ_{xy}) the strains are
Strain tensor for a plane stress state is
1.6 PLANE STRAIN CONDITION
For strain condition shown in Figure 1.5 the strain tensor is
In this, strain
Therefore, σ_{zz} = v(σ_{xx} + σ_{yy})
To satisfy the condition of plane strain
Moreover, γ_{yz} = γ_{zx} = 0
Figure 1.5 Plane strain state
Figure 1.6 Example 1.3
Example 1.3 A 100 mm cube of steel is subjected to a uniform pressure of 100 MPa acting on all faces. Determine the change in dimensions between parallel faces of cube if E = 200 GPa, v = 0.3.
Solution: σ = σ_{1} = σ_{2} = σ_{3} = 100 MPa, hydrostatic stress
Principal strains
Change in dimensions 
δ = δx = δy = δz = ε_{a} 
where 
a = side of cube 
Therefore 
= 100 × 0.4 × 10^{−3} 

= − 0.04 mm (because s is pressure) 
Exercise 1.3 A steel rectangular parallelopiped is subjected to stresses 100, − 60, + 80 MPa as shown in Figure 1.6. Dimension of the body are 150, 100, 75 mm in x, y, and zdirection. Determine change in dimensions, if E = 200 × 10_{3} MPa, v = 0.3.
Ans. [+ 0.070 mm, − 0.057 mm, + 0.028 mm].
1.7 DEFORMATIONS
There are two types of strains resulting from deformation of a body, i.e. (a) linear or extensional strains and (b) shear strains, resulting in change of shape. Let us consider an element of dimensions ∆x, ∆y, with change in x and ydirection as ∆x and ∆y as shown in Figure 1.7. A body of dimensions ∆x and ∆y, in x y plane, represented by ABCD, deforms to AB′C′D as shown in Figure 1.7(a).
Strain in xdirection,
Again it is deformed to A′B′CD as shown in Figure 1.7(b), strain in ydirection, .
Now the body in xy plane is deformed to A′B′C′D, as shown in Figure 1.7(c).
Figure 1.7 (a) Small element (b) Normal strain (c) Shear strains
Shear strain
Moreover, linear strains are on both side of coordinate axes
In order to study the deformation, or change in shape, one has to consider a displacement field(s) for a point in a body subjected to deformation.
A point P is located at position vector op = r and displaced to position vector op′ = r′. The displacement vector S = r′ = r. The displacement vector (Figure 1.8) a function of x, y, z has components u, v, w in x, y, zdirections, such that
where S is a function of initial coordinates of point P.
function of x, y, and z.
These strains at point P are defined by
Figure 1.8 Displacement vectors
and shear strains are
In xy plane, or in a twodimensional case, strains are
Example 1.4 The displacement field for a given point of a body is 9 mm
at point P (1, −2, 3) determine displacement components in x, y, zdirections. What is the deformed position?
Solution: Displacement components are
Putting the values of initial coordinates (1,−2, 3) of point P components are
u = 1 + 2(−2) + 3 = 0.0
v = 3 × 1 + 4 × (−2)^{2} = 19 × 10^{−4}
w = 2 × 1^{3} + 6 × 3 = 20 × 10^{−4}
x′ = x + u = 1 − 0 = 1
y′ = y + v = −2 + 19 × 10^{−4} = −1.9981
z′ = z + w = 3 + 20 × 10^{−4} = 3.002
Example 1.5 The displacement field for a body is given by
write down the strain components and strain matrix at point (2,1,2).
Solution: Displacement components are
Putting the values of initial coordinates (2, 1, 2)
strain matrix
Example 1.6 Displacement field is S = [(x^{2} + y^{2} + 2)i + (3x + 4y^{2})j] × 10^{−4} what is strain field at point (1, 2)?
Solution: Displacement components are
Strain components
Strain tensor
Exercise 1.4 The displacement field for a body is given by S = [(x^{2} + 2y)i + (3y + z)j + (x^{2} + z)k] × 10^{−3} what is the deformed position of point originally at (3, 1, −2)? Write down strain matrix.
Ans. [3.011, 1.001, −1.993], .
Exercise 1.5 Displacement field is
where k is a very small quantity.
What are strains at (1, −2) point?
Ans. [4k, −8k, +2k].
1.8 GENERALIZED HOOKE’S LAW
For a simple prismatic bar subjected to axial stress σ_{xx} and axial strain ε_{xx} (Figure 1.9), Hooke’s law states that stress ∝ strain

σ_{xx} ∝ ε_{xx} 


σ_{xx} = E ε_{xx} 
(1.1) 
where E is proportionality constant and E is Young’s modulus of elasticity of the material.
But for an elastic body subjected to six stress components σ_{xx}, σ_{yy}, σ_{zz}, τ_{yx}, τ_{yz}, τ_{zx} and six strain components, i.e. ε_{xx}, ε_{yy}, ε_{zz}, γ_{xy}, γ_{yz}, γ_{zx}, the generalized Hooke’s law can be expressed as
σ_{xx} 
= 
A_{11} ε_{xx} + A_{12} ε_{yy} + A_{13} ε_{zz} + A_{14} γ_{xy} + A_{15} γ_{yz} + A_{16} γ_{zx} 

σ_{yy} 
= 
A_{21} ε_{xx} + A_{22} ε_{yy} + A_{23} ε_{zz} + A_{24} γ_{xy} + A_{25} γ_{yz} + A_{26} γ_{zx} 

σ_{zz} 
= 
A_{31} ε_{xx} + A_{32} ε_{yy} + A_{33} ε_{zz} + A_{34} γ_{xy} + A_{35} γ_{yz} + A_{36} γ_{zx} 
(1.2) 
τ_{xy} 
= 
A_{41} ε_{xx} + A_{42} ε_{yy} + A_{43} ε_{zz} + A_{44} γ_{xy} + A_{45} γ_{yz} + A_{46} γ_{zx} 

τ_{yz} 
= 
A_{51} ε_{xx} + A_{52} ε_{yy} + A_{53} ε_{zz} + A_{54} γ_{xy} + A_{55} γ_{yz} + A_{56} γ_{zx} 

τ_{xz} 
= 
A_{61} ε_{xx} + A_{62} ε_{yy} + A_{63} ε_{zz} + A_{64} γ_{xy} + A_{65} γ_{yz} + A_{66} γ_{zx} 

where A_{11}, A_{12}, …, A_{65}, A_{66} are 36 elastic constants for a given material.
For homogeneous linearly elastic material, above noted six equations are known as generalized Hooke’s law. Similarly, strains can be expressed in terms of stresses as follows:
For a fully anisotropic material there are 21 constants in generalized Hooke’s law and for orthotropic materials there are nine constants in Hooke’s law.
Say for an isotropic material having the same elastic properties in all directions and material as such has no directional property, there are three principal stresses σ_{1}, σ_{2}, σ_{3} and three principal strains ε_{1}, ε_{2} and ε_{3}, then Hooke’s law can be written as
where A, B, and C are elastic constants.
Figure 1.9 Simple bar subjected to axial stress and axial strain
In the above equation ε_{1} is the longitudinal strain along σ_{1} but ε_{2} and ε_{2} are lateral strains so, constants B = C, therefore

σ_{1} 
= 
A ε_{1} + B ε_{2} + C ε_{3} 


= 
A ε_{1} + B ε_{1} − B ε_{1} + B ε_{2} + B ε_{3} 


= 
(A − B)ε_{1} + B(ε_{1} + ε_{2} + ε_{3}) 
where ε_{1} + ε_{2} + ε_{3} = ∆, a cubical dilatation = first invariant of strain
Let us denote (A − B) by 2 μ and B by λ, then

σ_{1} = λ Δ + 2με_{1} 
(1.5) 
Similarly 
σ_{2} = λ Δ + 2με_{2} 
(1.6) 

σ_{3} = λ Δ + 2με_{3} 
(1.7) 
λ and μ are called Lame’s coefficients.
From Eqs (1.5), (1.6), and Eqs (1.7)

σ_{1} + σ_{2} + σ_{3} 
= 
3λ Δ + 2μ(ε_{1} + ε_{2} + ε_{3}) 


= 
3λ Δ + 2μ Δ 


= 
(3λ + 2μ) Δ 
Principal stress
Solving Eq (1.9) for ε_{1}, we get
From elementary strength of materials
Therefore, Young’s modulus
Poisson’s ratio,
Example 1.7 For steel Young’s modulus E = 208000 MPa and Poisson’s ratio, v = 0.3. Find Lame’s coefficients λ and μ.
Solution:
Taking Eq. (i)
From Eq. (ii)
or 
λ = 0.6 λ + 0.6 μ 


λ = 1.5 μ 
(iv) 
Putting this value in Eq. (iii)
Lame’s coefficient, 
μ = 80000 N/mm^{2} 
Coefficient, 
λ = 1.5 μ = 120000 N/mm^{2} 
Exercise 1.6 For a concrete block E = 27.5 × 10^{3} MPa and Poisson’s ratio is 0.2. Determine Lame’s coefficients λ and μ.
Ans. [7639 N/mm^{2}, 11458 N/mm^{2}].
1.9 ELASTIC CONSTANTS K AND G
From the previous article 1.8, we know that
Putting the value of from Eqs (1.12) in Eqs (1.11), we get
From the elementary strength of material we know that
Therefore Lame’s coefficient μ = G, shear modulus. From Eqs (1.8) of previous article. If σ_{1} = σ_{2} = σ_{3} = p, hydrostatic stress or volumetric stress
Therefore, Bulk modulus,
Shear modulus, G = μ, Lame’s coefficient.
Example 1.8 For aluminium E = 67000 MPa, Poisson’s ratio, v = 0.33. Determine Bulk modulus and shear modulus of aluminium.
Solution:
or
From Eq. (ii)
Poisson’s ratio
or
Putting the value of λ in terms of μ in Eq. (iv)

E 
= 
2μ + μ × 0.66 = 2.66μ 








= 
Lame’s coefficient 



= 
Shear modulus 


G 
= 
25188 MPa 
(v) 

λ 
= 
1.94μ = 1.94 × 25188 = 48865 MPa 

Bulk modulus,
Exercise 1.7 For steel E = 200000 MPa and Poisson’s ratio is 0.295. Determine Lame’s coefficients and Bulk modulus K.
Ans. [λ = 111120 MPa, μ = 77220 MPa, K = 162600 MPa].
1.10 EQUILIBRIUM EQUATIONS
Consider a small infinitesimal element of a body of dimensions ∆x, ∆y and thickness t = 1 subjected to stresses varying over distances ∆x and ∆y as shown in Figure 1.10. X and Y are body forces per unit volume in x and ydirections.
Taking the summation of forces in xdirection
Figure 1.10 A small infinitesimal element in equilibrium
or
Simplifying further
Taking the summation of forces in ydirections
Simplifying further
In threedimensional case equilibrium equations can be written as
where X, Y, Z are body forces per unit volume.
In these equations, mechanical properties have not been used. So, these equations are applicable whether a material is elastic, plastic or viscoelastic.
In a twodimensional case, equilibrium equations are


We may permanently satisfy these equations by expressing stresses in terms of a function ϕ, called Airy’s stress function, as follows:
In a plane stress case a body subjected to stresses σ_{xx}, σ_{yy}, τ_{xy}, the strains are
Shear strain,
Strain compatibility equations are
Substituting the values of ε_{xx}, ε_{yy}, γ_{xy} from Eqs (1.15), (1.16), and Eq (1.18) in Eq (1.19), we will get
Putting the value of in Eq (1.20), we get
Simplifying further the equation becomes
or Δ ϕ = 0 (1.21)
Airy’s stress function ϕ chosen for any problem must satisfy the above Biharmonic equation.
Example 1.9 Following strains are given

ε_{xx} 
= 
6 + x^{2} + y^{2} + x^{4} + y^{4} 

ε_{yy} 
= 
4 + 3x^{2} + 3y^{2} + x^{4} + y^{4} 

γ_{xy} 
= 
5 + 4xy(x^{2} + y^{4} + 2) 


= 
5 + 4x^{3}y + 4xy^{3} + 8xy 
Determine whether the above strain field is possible. If it is possible determine displacement components u and v, assuming u = v = 0 at origin.
Solution: Strain compatibility condition is
Strain field is possible.
Now
Displacement component
Integrating
but v = 0 at x = 0, y = 0, constant C_{2} = 0
so,
Exercise 1.8 Given the following system of strains

ε_{xx} 
= 
8 + x^{2} + 2y^{2} 

ε_{yy} 
= 
6 + 3x^{2} + y^{2} 

γ_{xy} 
= 
10xy 
Determine whether the above strain field is possible. If possible determine displacement components u and v if u = 0, v = 0 at origin.
Ans. [strain field is possible]
1.11 SECOND DEGREE POLYNOMIAL
Let us consider an Airy’s stress function
∇^{4} ϕ = 0 satisfies the compatibility condition



Shear stress, (negative, tends to rotate the body in anticlockwise direction). This represents a plane stress condition as shown in Figure 1.11.
Figure 1.11 Plane stress condition
Example 1.10 Airy’s stress function ϕ = 40 x^{2}−30 xy + 60 y^{2} satisfies the compatibility condition ∇^{4} ϕ = 0. Determine stresses σ_{xx}, σ_{yy} and τ_{xy}, Show graphically the stress distribution. Stresses are in MPa.
Solution: ϕ = 40x^{2} − 30 xy + 60y^{2}














Note that shear stress, τ_{xy} is + ve, i.e. tending to rotate the body in a clockwise direction. Figure 1.12 shows the stress distribution which is a plane stress state. τ_{yx} are shear stresses complementary to τ_{xy}.
Figure 1.12 Stresses on element
Exercise 1.9 An Airy’s stress function ϕ = 50 x^{2} − 40 xy + 80 y^{2} satisfies the compatibility condition ∇^{4} ϕ = 0. Determine normal and shear stresses and state the type of state of stress.
Ans. [σ_{xx} = 160 MPa, σ_{yy} = 100 MPa, τ_{xy} = +40 MPa, a plane stress condition].
1.12 A BEAM SUBJECTED TO PURE BENDING
For a beam of depth d subjected to pure bending moment M and no shear force, as shown in Figure 1.13. Airy’s stress function can be a third degree polynomial
∇^{4} ϕ = 0, for this function
Taking B = 0 as stress σ_{xx} is independent of x,
y varies from − to + as shown in Figure 1.14.
Maxm. σ_{xx} in tension
where
Maxm.σ_{xx} in compression
where M = bending moment M, and
I = second moment of area (of cross section of beam) about neutral plane.
Figure 1.13 Beam subjected to pure bending
Figure 1.14 Stress distributions
Now
but σ_{yy} = 0, so constant C = D = 0
Shear stress
but 
= By + Cx 

τ_{xy} = 0, 
also because it is a case of pure bending.
So, constants B = C = 0
Finally Airy’s stress function is .
Example 1.11 A bar of circular section of diameter 30 mm is subjected to a pure bending moment of 3 × 10^{5} Nmm. What is the maximum bending stress developed in beam? What is Airy’s stress function for this case.
Solution: Moment of inertia of beam,
Bending stress
Airy’s stress function for this case
where
Airy’s stress function ϕ = 1.257 y^{3}
where y varies from − 15 to + 15 mm.
Exercise 1.10 For a beam of rectangular section B = 20 mm, D = 30 mm, Airy’s stress function ϕ = 1.6 y^{3}.
Determine the magnitude of constant BM acting on beam section.
Ans. [432 Nm].
Problem 1.1 A beam of rectangular section is subjected to shear force F = 1 kN and bending moment = −1 × 10^{6} Nmm. Section of beam is B = 20 mm, D = 60 mm. Write down stress tensor for an element located at 15 mm below the top surface (see Figure 1.15).
Figure 1.15 Problem 1.1
Solution:
M = −1 × 10^{6} Nmm (producing convexity)
I = second moment of area about neutral plane
y = 15 mm from neutral layer.
Bending stress,
To determine shear stress at y = 15 mm from neutral layer
Figure 1.16 Problem 1.1
where area a = 15 × 20 = 300 mm^{2} = area of section above the layer under consideration,
ӯ 
= 
225.,mm distance of CG of area a from neutral layer, 
b 
= 
breadth = 20 mm, 
I 
= 
36 × 10^{4} mm^{4}, 
F 
= 
1 kN = 1000 N, 




= 
0.9375 N/mm^{2}. 
Stress tensor for this state of stress, at a layer 15 mm below top surface
Problem 1.2 At a point in a stressed body the Cartesian components of stress are σ_{xx} = 80 MPa, σ_{yy} = 50 MPa, σ_{zz} = 30 MPa, τ_{xy} = 30 MPa, τ_{yz} = 20 MPa, and τ_{zx} = 40 MPa, determine (a) normal and shear stresses on a plane whose normal has the direction cosines
(b) angle between resultant stress and outer normal n.
Solution: Components of resultant stress σ_{r} are
Substituting the values as above
Normal stress, σ_{n} = σ_{rx} × cos(n, x) + σ_{ry} cos(n, y) + σ_{rz} cos(n, z)
Shear stress,
Angle between resultant stress vector σ_{r} and normal to the plane n is given by
where
So,
Problem 1.3 A cantilever beam of rectangular section B × D and of length L carries a concentrated load W at free end Figure 1.17. Consider a section at a distance x from fixed end and a layer at a distance y from neutral layer zz; derive expression for Airy’s stress functions
if 
B = 40 mm, D = 60 mm 

W = 1 kN, x = 2 m, L = 5 m 
Write stress tensor for the layer bc, if y = 20 mm
Figure 1.17 Problem 1.3
Solution: Stresses
Let us assume Airy’s stress function ϕ = C_{1}y^{3} + C_{2} xy^{3} + C_{3} xy.
 For τ_{xy} = 0, σ_{yy} = 0
 For x = 0, σ_{xx} = 0
 Everywhere , shear force is constant
Applying these boundary conditions
So,
or
Moreover
or
Putting the value of C_{2}D^{2} in Eq. (iii)
Then
Finally, Airy’s stress function is
At the layer where x = 2 m = 2000 mm, L = 5000 mm
Strees tensor
Problem 1.4 A solid circular shaft of steel is transmitting 100 hp at 200 rpm. Determine shaft diameter if the maximum shear stress in shaft is not to exceed 80 MPa. Write down stress tensor for the surface of the shaft. Show that surface of the shaft is under both plane stress and plane strain conditions E = 200 × 10^{3} N/mm^{2}, v = 0.3.
Solution: HP transmitted = 100
RPM, N = 200
angular speed,
Torque transmitted, = 20.94 rad/s
Torque transmitted,
Maxm. Shear strees,
Shaft diameter d = 60.87 mm
On the surface of the shaft, τ = ± 80 MPa
Principal stresses on the surface of the shaft are
Principal strains,
Figure 1.18 Problem 1.4
Since ε_{3} = 0, this a plane strain condition.
Strain are
Stress tensor
Stress tensor
MULTIPLE CHOICE QUESTIONS
 Principal stresses at a point are 120, −40,−20 MPa. What is maximum shear stress at the point?
 50 MPa
 70 MPa
 80 MPa
 None of these
 A bar is subjected to axial load such that its length l is increased by 0.001 l. If Poisson’s ratio is 0.3, what is change in its diameter d?
 −3 × 10^{−4} d
 +3 × 10^{−3} d
 +3 × 10^{−4} d
 None of these
 Lame’s coefficient for a material are λ and m. What is Poisson’s ratio?
 None of these
 Ratio of volumetric stress/volumetric strain is known as
 Shear modulus
 Bulk modulus
 Young’s modulus
 None of these
 Airy’s stress function is ϕ = 50x^{2}40xy+80y^{2}, what is normal stress syy
 100 MPa
 40 MPa
 +160 MPa
 80 MPa
 A rectangular section beam of breadth b, depth d is subjected to shear force F. At what depth y from top surface transverse shear stress is maximum
 y = 0
 A shaft is subjected to pure twisting moment M. Surface of the shaft represents
 Plane strain condition
 Plane stress condition
 Both plane stress and plane strain conditions
 Neither plane stress nor plane strain condition
 A thin metallic sheet is subjected to inplane shear stress, what is the state of stress of thin sheet ?
 a plane strain state
 a plane stress state
 a hydrostatic state of stress
 None of these
Answers
1. (c) 
2. (a) 
3. (b) 
4. (b) 
5. (a) 
6. (d) 
7. (c) 
8. (b). 


EXERCISE
1.1 Consider a beam of rectangular section B = 25 mm, D = 60 mm subjected to a bending moment + 1.5 × 106 Nmm. Write down (i) the stress tensor for an element located at top surface and (ii) the stress tensor for an element located in a plane 15 mm below the top surface.
Ans.
1.2 For a material, Lame’s coefficients are
λ = 1.2 × 10^{5} MPa, μ = 0.8 × 10^{5} MPa. Determine E, v, and G for the material.
Ans. [E = 208000 MPa, v = 0.3, G = 80,000 MPa].
1.3 A cantilever of rectangular section B × D is of length L as shown in Figure 1.19. Write down stress tensor to determine state of stress at section XX at a distance of x from free end and at layer at distance of y from neutral layer.
Ans.
Figure 1.19 Exercise 1.3
1.4 A cylindrical bar of length L, area of cross section A is fixed at top end. Write down Air’s stress function for stress due to self weight in bar, if w is the weight density of the bar (see Figure 1.20).
Ans.
1.5 Consider the displacement field S = (y^{2} i + 3yx j) × 10^{−2}. Find whether strain field is compatible. If yes find strain components ε_{x}, ε_{y}, at point (1, −1).
Ans.
Figure 1.20 Exercise 1.4