Chapter 5

Chapter 5

STRESS-STRAINTEMPERATURE RELATIONSHIP BITS PILANI GOA

Contents: Introduction • Stress- Strain diagram • Tensile test Idealizations of Stress-Strain Curves Elastic Stress- Strain Relations Thermal Strain Complete Equations of Elasticity Strain Energy in an Elastic Body Criteria for Initial Yielding

σ

Stress-strain diagram for a structural steel in tension ( Not to scale)

Stress-strain diagram for a structural steel in tension ( Drawn to scale)

Typical stress- strain diagram for an aluminum alloy

Typical stress-strain diagram for brittle material

Compression stress-strain diagram for copper

Stress-strain diagram for a cast iron in tension and compression

• Elastic deformation: It is the deformation which exist when load applied and it disappears as soon as the load is released. • Plastic Deformation: when load is applied beyond elastic range, and the deformation does not disappears after the load is released. • A Ductile materials is one which the plastic deformation is much larger than elastic deformation

(a) Elastic behaviour; (b) Partially elastic behaviour

Reloading of a material and raising of the yield stress

Creep:

Creep in a bar under constant load

• Creep: Deformation increase with time under a constant load and elevated temperature i.e. paper clip deformation may be evident at 480OC and for Al components deformation may be evident at 200OC, elastic recovery and elastic after effect • Visco -elasticity: Viscoelastic behavior demonstrates a strain that is changing with time under a constant applied stress

In last class we have seen Stress- Strain diagram

Ductile material

Tension

Compression

Creep Deformation

Brittle material

Tension Compression

Failure modes • • • • • •

Excessive deformation Plastic deformation Fracture Fatigue corrosion Wear

Idealized models of material behaviour

(a) Rigid material

(b) Linearly elastic material It is useful when we are designing for small deformations or to prevent fatigue or fracture in brittle structures.

(c) Perfectly plastic material ( non- strain- hardening)

(d) Rigid– plastic material (strain- hardening)

It is useful in designing structures for their maximum loads, in studying many machining and metal forming problems.

(f) Elastic- plastic (e)Elastic- perfectly plastic material (non- strain- hardening) material (strainhardening) It is useful in designing against moderate deformations.

EX. 6

P

L

P

Geometric compatibility:

Strain- Strain Relations:

(b)

(c)

(d)

(e)

3-Dimentional State of Stress OR Triaxial State of Stress

σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z

(4.4)

A knowledge of the nine stress components is necessary in order to determine the components of the stress vector T acting on an arbitrary plane with normal n.

3-D Elastic stress-strain relations:

Assumptions: Materials are linearly elastic and Isotropic Strains are small compared to unity

Poisson’s Ratio • For a slender bar subjected to axial loading:

σx εx = σy=σz=0 E • The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),

εy = εz ≠ 0 • Poisson’s ratio is defined as

εy ε lateral strain ν= =− =− z axial strain εx εx

Initially consider an element on which only σx is acting

σx εx = E

Is their any possibility of shear strains resulting from normal stress ?

Hypothetical shear strain due to normal stress Hence γxz is zero Similarly γxy and γyz must be zero

Now in addition to σx , σy is also acting on the element

εy =

ε =ε x

z

σy E = −ν ε y

σ = −ν

y

E

Similar results are obtained for the strain due to σz

Is their any possibility of normal strains resulting from shear stress zx ?

τ

Shear Modulus : The ratio of shearing stress τ to shearing strain γ within the proportional limit of a material.

Hence for material with all stress component present, stress strain relations are

The relation between the two elastic constants is given by

E = (1 + ν ) 2G

Thermal Strain: • The strain due to temperature change in the absence of stress is called thermal strain and is denoted by the superscript t t on the strain symbol thus: ε • The following thermal strains are obtained due to change in temperature from T0 to T

ε = ε = ε = α (T − T ) γ =γ =γ =0 t

t

t

x

y

z

t

t

t

xy

yz

zx

0

For linear elastic material strains may be added

= + ε TOTAL ε ε e

t

Effect of temperature on stress- strain curve.

Complete Equations of Elasticity: The problem is to find 3-D stress-strain distribution Useful steps: Force equilibrium Geometric Compatibility Stress- Strain- Temperature relations To summarize we can write elasticity equation in 3- D for linear elastic isotropic material.

(4.14)

(4.33)

[mechanical strain, Thermal]

(5.2)

These 15 equations can be applied to deformations of isotropic, linearly elastic solids which involves small strains

Example 1 A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses σx = 12 ksi and σz = 20 ksi. For E = 10x106 psi and ν = 1/3, determine the change in: • the length of diameter AB, • the length of diameter CD, • the thickness of the plate, and • the volume of the plate.

SOLUTION: • Apply the generalized Hooke’s Law to find the three components of normal strain.

εx = +

σx E

νσ y

−

E

νσ z

−

E

1   ( ) ( ) = 12 ksi − 0 − 20 ksi  3   10 ×106 psi  1

= +0.533 ×10 −3 in./in.

νσ x

εy = −

E

+

σy E

νσ z

−

E

= −1.067 ×10 −3 in./in.

νσ x

εz = −

E

νσ y

−

σ + z E E

= +1.600 ×10 −3 in./in.

• Evaluate the deformation components.

(

)

δ B A = ε x d = + 0.533 × 10− 3 in./in. ( 9 in.) −3

(

δ B A = +4.8 × 10 in.

)

δ C D = ε z d = + 1.600 × 10− 3 in./in. ( 9 in.) −3

(

δ C D = +14.4 × 10 in.

)

δ t = ε y t = − 1.067 × 10 in./in. ( 0.75 in.) −3

δ t = −0.800 × 10− 3 in.

Determine the volumetric strain, , of a E X.2 rectangular parallelepiped of linearly elastic, isotropic material subjected to general triaxial stress. Assume that ε << 1 for all coordinate directions.

The volumetric strain is given by (1)

For uniform strain in each coordinate direction (2)

After simplification (3)

But strain in x,y and z direction in traxial stress system are

(4)

Substituting 4 in 3

In last class we have seen 3-D Elastic stress- strain relations • Normal stress produces only normal strains and no shear strains • Shear stress produces only shear strain and not normal strains Poisson’s ratio Relation between two elastic constant

E = (1 + ν ) 2G

Ex. 3

Assumptions: Uniformly distributes normal stress

σ

ε

Z

=0

y

=0

Friction force at the wall is negligible Normal stress of contact between the plate and wall is uniform

Idealized model

Equilibrium: (a)

Geometric Compatibility

(b)

Stress- Strain relations

(c) Solving a,b and c we find

(d)

Strain- Displacement Relations

(e)

In the last class we have seen Thermal strain Complete equations of elasticity. Exact solution to the elasticity problem on plate for the idealized model

Ex.4

(3.31)

Answer:

σ

x

= −137.2 MPa

δD = 0.1107 mm

EX. 5

The strains at a point on aluminium ( E=70GPa, G= 28 GPa and = 0.25) were found to be ε x = 650 micron ε y = 300 micron and γ xy = 750 micron. Determine the stresses and the strain assuming the point is in plane stress.

ν

Answer:

σ σ τ ε

x

= 54.1MPa

y

= 34.5MPa

xy z

= 21MPa = −317 microns

Strain energy in an elastic body: The energy stored in a body due to deformation is called strain energy. Strain energy density Engineering structures are designed to function without per meant deformation. For linear elastic material

For linear elastic body (5.14)

The strain energy stored in the element is (5.15)

Consider Gradual loading of element with all six component of stresses

(5.17)

Assignment: Derive equation for strain energy for isotropic material in terms of either complete stress component or strain component.

In the last class we have seen Numerical based on elasticity equations Expression for strain energy in 3-D stress system Failure modes in materials Idealization of stress-strain curve

CRITERIA FOR INITIAL YIELDING

Mises Yield Criterion Initial yielding occurs when the root mean square of the differences between the principal stresses reaches the same value that it has when yielding occurs in a tensile test 1 2 2 2   1   2     2   3     3   1   3

Let Y denote the stress at which yielding begins in simple tension test

It is also known as distortion- energy criterion for yielding

For plane stress, σ3 = 0, we will have

Tresca or maximum shear stress criterion Assumption: The yielding occurs when the absolute maximum shear stress at a point reaches the value of the maximum shear stress to cause yielding in a tensile test

Fig. 5.28 Example of biaxial stress in a thin walled cylinder

Yields according to Tresca criterion Elastic behaviour

σ2

Yields according to Mises criterion Maximum shear criterion

Mises criterion Y

σ1

Y Y

Example 5.27 A batch of 2024- T4 aluminum alloy yields in uniaxial tension at the stress σ0= 330 MN/ m2. If this material is subjected to the following state of stress, will it yield according to a) the Mises criterion, and b) the maximum shear stress criterion σx = 138 MN/ m2 σy = -69 MN/ m2 τ xy= 138 MN/m2

Answer: The given material yields according to shear stress criterion. It doesn't yield according to von- mises criterion.