ENGR 225Section 3.1 - 3.8

Load increase –> stress increase.

Deformation increase –> strain increase.

Relation between Stress and Strain.

Tension Test

Standard test specimen

Universal Testing Machine (UTM)

Tension Test

http://www.youtube.com/watch?v=9suShuEwc7I&feature=related

Nominal Stress: σ = P/A0

Where A0 is original cross section

Nominal Strain: ε = δ/L0

Where L0 is the original gauge length and δ is the change in gauge length

No two stress-strain diagrams for a particular material will be exactly the same since the results depend on:

material’s compositionmicroscopic imperfections,way material is manufacturedrate of loadingtemperature

Stress-Strain Diagrams

Stress-Strain Diagrams

Stress is proportional to strain

(within proportional limit)

in most Solids

Not so for fluids !!

Mild Steel

y = 25 pl

Ductility

• The extent of plastic deformation that a material undergoes before fracture.

Ductile Material– A material that can be subjected to large strains before it ruptures.

– Ductility can be measured by percent elongation or percent reduction in area at the time of fracture.

– Mild Steel : 38%

– Mild Steel : 60%

%)100(ElongationPercent o

of

L

LL

%)100(Areain Reduction Percent o

of

A

AA

Ductility• Why use ductile materials?

– Capable of absorbing shock or energy– When overloaded, usually exhibit large deformation before failing.

Aluminum

Yield Strength• Yield Strength is not a physical property of the material.

• We will use approach that – Yield strength– Yield point– Elastic limit– Proportional limit

• All coincide unless otherwise stated

Natural Rubber(Nonlinear Elastic Behavior)

Brittle Material

– A material that exhibits little of no yielding before failure. It fractures suddenly under tension.

Grey Cast Iron

Concrete

Modulus of Elasticity

E = Stress / Strain

Hooke’s Law

= E

Modulus of Elasticity• Modulus of Elasticity (E) indicates stiffness of a material.

If material is stiff, E is large

(for steel, E = 200 GPa)

If material is spongy, E is small

(for vulcanized rubber, E = 0.70 MPa)

• Strength• Ductility• Brittleness• Stiffness• Resilience• Toughness• Endurance• Rigidity

Material Properties

Strain Hardening

• Strain hardening is used to establish a higher yield point for a material

• The modulus of elasticity stays the same.

• The ductility decreases.

Chapter 3 Lecture Example 1

A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the graph the ultimate stress and the fracture stress.

Strain Energy

• Energy stored in a material due to deformation

Modulus of Resilience

Chapter 3 Lecture Example 2

The stress-strain diagram for an aluminum alloy that is used for making aircraft parts is shown. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application.

Toughness

• The area under the stress-strain curve

• The amount of energy per unit volume that the material dissipates prior to fracture.

Modulus of Toughness

Strength, Toughness and Ductility

• Strength ~ related to the height of the curve

• Ductility ~ related to the width of the curve

• Toughness ~ related to area under the curve

Poisson’s Ratio

Poisson’s Ratio

Poisson’s Ratio

• When a deformable body is subjected to a tensile force, not only does it elongate, but it also contracts.

• Longitudinal is in the direction of the tensile force.

long

lat

French mathematician Simeon Denis Poisson

• Value of is positive

• Value of same in tension and compression

• Range of 0.25 to 0.35

• constant only in the elastic range

• Maximum possible value of is 0.5 (Section 10.6)

• Only Longitudinal force is acting to cause the lateral strain.

Poisson’s Ratio

Shear Stress –Strain Diagram

Shear Stress –Strain Diagram

G

Modulus of Rigidity

G)1(2 v

EG

Derivation of this equation in section 10.6

Chapter 3 Lecture Example 3

This is a titanium alloy. Determine the shear modulus, G, the proportional limit, and the ultimate shear stress. Find the maximum d where the material behaves elastically. What is the magnitude of V to cause d?

Failure of turbine blade due to creep

Failure of steam pipe due to creep

Creep• Creep is the time-related deformation of a

material for which temperature and stress play an important role.

• Creep results from sustained loading below the measured yield point.

• Members are designed to resist the effects of creep based on their creep strength, which is the highest amount of stress a member can withstand during a specified amount of time without experiencing creep strain.

Creep

Fatigue

• Fatigue occurs in metals when stress or strain is cycled. It causes a brittle fracture to occur.

• Members are design to resist fatigue by ensuring that the stress in the member does not exceed its endurance limit.

• This the maximum stress member can resist when enduring a specified number of cycles.

Concept Questions

• How would you explain strain to one of your engineering classmates?

– Strain is a linear deformation due to stress– Percent increase in length under tension or

compression

Concept Questions

• How would you characterize a ductile material? Give several examples.

– A material that can be subjected to large strains before it ruptures.

– Steel, wood, natural rubber, brass, copper, gold, aluminum

Concept Questions

• How would you characterize a brittle material? Give several examples.

– A material that exhibits little or no yielding before failure.

– Cast iron, concrete, glass, ceramics

Concept Questions

• What is the difference between engineering stress and true stress?

– Engineering stress assumes a constant cross sectional area during elongation.

A tale of two cities• How can you explain Elasticity?

– Property of material by which it returns to original dimensions on unloading.

• How can you explain Plasticity?

– Characteristic of material by which it cannot return to original dimensions on unloading and undergoes permanent deformation.

Strength – Capacity to resist loads – yield stress. Higher y, higher strength.

Resilience – Measure of energy absorbed without permanent damage .

Toughness – Measure of energy absorbed before fracturing.

Ductility – Property of material which allows large deformation before fracture.

Brittleness – Property of material which allows little or no yielding before fracture.

Endurance – Ability to sustain cyclic loads.

Stiffness – Mechanical property indicated by E. Higher E means stiff material. Steeper slope in the Stress – Strain diagram.

Rigidity - Mechanical property of material indicated by G.

Material Properties

3.7 (a) A structural member in a nuclear reactor is made of Zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine the required cross sectional area. Yield stress for Zirconium is 57.5 ksi. Factor of safety 3.

(b) What value of load would cause an elongation of 0.02 inch in this member if length of member is 3 ft. Ezr = 14 x 103 ksi.

3.26 A short cylindrical block of 2014-T6 Aluminum having an original diameter of 0.5 in and an original length of 1.5 in is placed in the smooth jaws of a vise and a compressive force of 800 lb is applied. Determine

(a) The decrease in length.(b) The new diameter.