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 three-dimensional 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 x-direction

τ_xy = shear stress on yz plane in y-direction

τ_xz = shear stress on yz plane in z-direction.

 

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

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)

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 z-direction 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

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 σ₁, σ₂, and σ₃,

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

Similarly strains                

and                                      

Three-dimensional 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

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 = σ₁ = σ₂ = σ₃ = 100 MPa, hydrostatic stress

Principal strains

Change in dimensions	δ = δx = δy = δz = εa
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 z-direction. Determine change in dimensions, if E = 200 × 10₃ 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 y-direction 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 x-direction,                

Again it is deformed to A′B′CD as shown in Figure 1.7(b), strain in y-direction, .

Now the body in x-y plane is deformed to A′B′C′D, as shown in Figure 1.7(c).

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-, z-directions, 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

and shear strains are

In x-y plane, or in a two-dimensional 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-, z-directions. What is the deformed position?

Solution: Displacement components are

 

Putting the values of initial co-ordinates (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 × 1³ + 6 × 3 = 20 × 10⁻⁴

Deformed position is

            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 co-ordinates (2, 1, 2)

strain matrix                

Example 1.6 Displacement field is S = [(x² + y² + 2)i + (3x + 4y²)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 = [(x² + 2y)i + (3y + z)j + (x² + 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

 

	σxx ∝ εxx
	σxx = E εxx	(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

 

σxx	=	A11 εxx + A12 εyy + A13 εzz + A14 γxy + A15 γyz + A16 γzx
σyy	=	A21 εxx + A22 εyy + A23 εzz + A24 γxy + A25 γyz + A26 γzx
σzz	=	A31 εxx + A32 εyy + A33 εzz + A34 γxy + A35 γyz + A36 γzx	(1.2)
τxy	=	A41 εxx + A42 εyy + A43 εzz + A44 γxy + A45 γyz + A46 γzx
τyz	=	A51 εxx + A52 εyy + A53 εzz + A54 γxy + A55 γyz + A56 γzx
τxz	=	A61 εxx + A62 εyy + A63 εzz + A64 γxy + A65 γyz + A66 γzx

where A₁₁, A₁₂, …, A₆₅, A₆₆ 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 σ₁, σ₂, σ₃ and three principal strains ε₁, ε₂ and ε₃, then Hooke’s law can be written as

 

where A, B, and C are elastic constants.

In the above equation ε₁ is the longitudinal strain along σ₁ but ε₂ and ε₂ are lateral strains so, constants B = C, therefore

 

	σ1	=	A ε1 + B ε2 + C ε3
		=	A ε1 + B ε1 − B ε1 + B ε2 + B ε3
		=	(A − B)ε1 + B(ε1 + ε2 + ε3)

where ε₁ + ε₂ + ε₃ = ∆, a cubical dilatation = first invariant of strain

 

Let us denote (A − B) by 2 μ and B by λ, then

 

	σ1 = λ Δ + 2με1	(1.5)
Similarly	σ2 = λ Δ + 2με2	(1.6)
	σ3 = λ Δ + 2με3	(1.7)

λ and μ are called Lame’s coefficients.

From Eqs (1.5), (1.6), and Eqs (1.7)

 

	σ1 + σ2 + σ3	=	3λ Δ + 2μ(ε1 + ε2 + ε3)
		=	3λ Δ + 2μ Δ
		=	(3λ + 2μ) Δ

Principal stress

 

Solving Eq (1.9) for ε₁, 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/mm2
Coefficient,	λ = 1.5 μ = 120000 N/mm2

Exercise 1.6 For a concrete block E = 27.5 × 10³ MPa and Poisson’s ratio is 0.2. Determine Lame’s coefficients λ and μ.

Ans. [7639 N/mm², 11458 N/mm²].

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 σ₁ = σ₂ = σ₃ = 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
	μ	=
		=	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 y-directions.

 

Taking the summation of forces in x-direction

or

Simplifying further

Taking the summation of forces in y-directions

Simplifying further

In three-dimensional 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 two-dimensional case, equilibrium equations are

where body forces are zero.

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

 

	εxx	=	6 + x2 + y2 + x4 + y4
	εyy	=	4 + 3x2 + 3y2 + x4 + y4
	γxy	=	5 + 4xy(x2 + y4 + 2)
		=	5 + 4x3y + 4xy3 + 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 C₂ = 0

so,                    

Exercise 1.8 Given the following system of strains

 

	εxx	=	8 + x2 + 2y2
	εyy	=	6 + 3x2 + y2
	γxy	=	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

Stress,
Stress,

Shear stress, (negative, tends to rotate the body in anticlockwise direction). This represents a plane stress condition as shown in Figure 1.11.

Example 1.10 Airy’s stress function ϕ = 40 x²−30 xy + 60 y² 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 = 40x² − 30 xy + 60y²

Stress,		(i)
Shear stress,		(ii)
Moreover
Stress,		(iii)

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.

Exercise 1.9 An Airy’s stress function ϕ = 50 x² − 40 xy + 80 y² 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.

Now                    

but                       σ_yy = 0,   so constant C = D = 0

Shear stress                

but	= By + Cx
	τxy = 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 × 10⁵ 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 y³

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 y³.

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 × 10⁶ 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).

Solution:

M = −1 × 10⁶ 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

where area a = 15 × 20 = 300 mm² = 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 × 104 mm4,
F	=	1 kN = 1000 N,
τxy	=
	=	0.9375 N/mm2.

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

Resultant stress,

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

Solution: Stresses

Let us assume Airy’s stress function ϕ = C₁y³ + C₂ xy³ + C₃ xy.

Boundary conditions are

1. For τ_xy = 0, σ_yy = 0
2. For x = 0, σ_xx = 0
3. Everywhere , shear force is constant

Applying these boundary conditions

So,                    

or

Moreover

or

Putting the value of C₂D² 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 × 10³ N/mm², 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,

Since ε₃ = 0, this a plane strain condition.

Strain are

Stress tensor

Stress tensor

1. Principal stresses at a point are 120, −40,−20 MPa. What is maximum shear stress at the point?
   1. 50 MPa
   2. 70 MPa
   3. 80 MPa
   4. None of these
2. 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?
   1. −3 × 10⁻⁴ d
   2. +3 × 10⁻³ d
   3. +3 × 10⁻⁴ d
   4. None of these
3. Lame’s coefficient for a material are λ and m. What is Poisson’s ratio?
   1. 
   2. 
   3. 
   4. None of these
4. Ratio of volumetric stress/volumetric strain is known as
   1. Shear modulus
   2. Bulk modulus
   3. Young’s modulus
   4. None of these
5. Airy’s stress function is ϕ = 50x²-40xy+80y², what is normal stress syy
   1. 100 MPa
   2. 40 MPa
   3. +160 MPa
   4. 80 MPa
6. 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
   1. y = 0
   2. 
   3. 
   4. 
7. A shaft is subjected to pure twisting moment M. Surface of the shaft represents
   1. Plane strain condition
   2. Plane stress condition
   3. Both plane stress and plane strain conditions
   4. Neither plane stress nor plane strain condition
8. A thin metallic sheet is subjected to inplane shear stress, what is the state of stress of thin sheet ?
   1. a plane strain state
   2. a plane stress state
   3. a hydrostatic state of stress
   4. None of these

Answers

 

1. (c)	2. (a)	3. (b)	4. (b)	5. (a)
6. (d)	7. (c)	8. (b).

 

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 × 10⁵ MPa, μ = 0.8 × 10⁵ 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.

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 = (y² 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.