September 9, 2011Page 1

MECH 321 -Solid Mechanics II

Week 1, Lecture 2

Review of CIVL 220 (cont’d)

September 9, 2011Page 2

The Stress–Strain Diagram

Conventional Stress

Conventional Stress

–

–

Strain Diagram

Strain Diagram

•Nominal or engineering stress is obtained by dividing

the applied load P by the specimen’s original cross-

sectional area.

•Nominal or engineering strain is obtained by dividing the

change in the specimen’s gauge length by the specimen’s

original gauge length.

0

A

P

=

σ

0

L

δ

ε

=

September 9, 2011Page 3

The Stress–Strain Diagram

Conventional Stress

Conventional Stress

–

–

Strain Diagram

Strain Diagram

•Elastic Behaviour

Stress is proportional to the strain.

Material is said to be

linearly elastic.

{

Proportional, linearly

elastic range

September 9, 2011Page 4

The Stress–Strain Diagram

•Yielding

Increase in stress above elastic limit will cause material

to deform permanently.

September 9, 2011Page 5

The Stress–Strain Diagram

Strain Hardening.

After yielding a further load will reaches a ultimate stress.

September 9, 2011Page 6

The Stress–Strain Diagram

•Necking

At ultimate stress, cross-sectional area begins to decrease

in a localized region of the specimen.

–Specimen breaks at the fracture stress.

September 9, 2011Page 7

True Stress–Strain

The values of stress and strain computed from these

measurements are called true stress

and true strain

.

Use this diagram since most engineering design is done within

the elastic range.

September 9, 2011Page 8

Stress vs. Strain –Ductile Material

Ductile material –Low Carbon Steel.

σu

= Ultimate Strength

(450 MPa)

σY

= Yield Strength

(265 MPa)

Strain hardening -

stress increases

Necking –stress decreases

September 9, 2011Page 9

Ductile Material -Aluminum

Length increases linearly at a slow rate.

σY

–yield stress point occurs after a large amount of

deformation largely due to shear stresses (internal slippage

along grains).

Rupture

σB

σY

σu

(450 MPa)

300

0.002

Necking

0.2

Aluminum Alloy

ε

September 9, 2011Page 10

Brittle Material

σu

= σB

Rupture

ε

Cast Iron

September 9, 2011Page 11

Hooke’s Law

σ= E εor E = σ/ ε

E= modulus of elasticity (same units as stress, σ)

σ

ε

Tempered steel

High-C steel

Low-C steel

Iron

Different: σY, σU, σB, εB

Same: E

September 9, 2011Page 12

Hooke’s Law

For the straight section of stress-strain (σ–ε) curve, stress

is directly proportional to strain.

Proportional Limit: limit of linear stress-strain (σ–ε)

relationship.

Alloys of the same material will have different properties

(strength, ductility, corrosion resistance, etc), but will often

have the same E. Same stiffness within their linear range.

σ= E εor E = σ/ ε

September 9, 2011Page 13

Hooke’s Law

Strain Hardening

Strain Hardening

•When ductile material is loaded into the plastic region and then

unloaded, elastic strain is recovered.

•The plastic strain remainsand material is subjected to a permanent

set.

September 9, 2011Page 14

Strain Energy (new)

•When material is deformed by external loading, it will store

energy internally throughout its volume.

•Energy is related to the strains called strain energy.

Modulus of Resilience

Modulus of Resilience

•When stress reaches the

proportional limit (σpl), the

strain-energy density is the

modulus of resilience, ur.

E

u

pl

plplr

2

2

1

2

1

σ

εσ

==

September 9, 2011Page 15

Strain Energy

Modulus of Toughness

Modulus of Toughness

•Modulus of toughness, ut, represents the entire area

under the stress–strain diagram.

•It indicates the strain-energy density of the material just

before it fractures.

September 9, 2011Page 16

Example 3.2

The stress–strain diagram for an aluminum alloy that is used for making aircraft

parts is shown. When material is stressed to 600 MPa, find the permanent strain

that remains in the specimen when load is released. Also, compute the modulus of

resilience both before and after the load application.

September 9, 2011Page 17

Poisson’s Ratio

•Poisson’s ratio, v(nu), states that in the elastic range, the

ratio of these strains is a constantsince the deformations

are proportional.

•Negative sign since longitudinal elongation (positive strain)

causes lateral contraction (negative strain), and vice versa.

long

lat

v

ε

ε

−=

Poisson’s ratio is dimensionless.

Typical values are 1/3 or 1/4.

September 9, 2011Page 18

Poisson’s Ratio

L

long

δ

ε

=

r

lat

δ

ε

′

=

long

lat

ε

ε

υ

=

September 9, 2011Page 19

Poisson’s Ratio

Unloaded Loaded

September 9, 2011Page 20

Example 3.4

A bar made of A-36 steel has the dimensions shown. If an axial force of is applied to

the bar, determine the change in its length and the change in the dimensions of its

cross section after applying the load. The material behaves elastically.

September 9, 2011Page 21

The Shear Stress–Strain Diagram

•For pure shear, equilibrium

requires equal shear stresses

on each face of the element.

•When material is

homogeneous and isotropic,

shear stress will distort the

element uniformly.

September 9, 2011Page 22

The Shear Stress–Strain Diagram

•For most engineering materials the elastic

behaviour is linear, so Hooke’s Law for shear

applies.

•3 material constants, E, and G are actually

related by the equation

γ

τ

G

=

G= shear modulus of elasticity

or the modulus of rigidity

()

v

E

G

+

=

12

September 9, 2011Page 23

Example 3.5

A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram

is shown. Find the shear modulus G, the proportional limit, and the ultimate shear

stress. Also, findthe maximum distance d that the top of a block of this material

could be displaced horizontally if the material behaves elastically when acted upon

by a shear force V. What is the magnitude of V necessary to cause this

displacement?

September 9, 2011Page 24

Next Time

Axial Loading (review) and the

Principle of Superposition

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