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EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 2 Stress & Strain -

Axial Loading

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain - Axial Loading

Statics vs. Mechanics of Materials

Statics - Deals with un-deformable bodies (rigid bodies)

Mechanics of Materials - Deals with practical, deformable bodies

Need to calculate the stress and deformation (relative and absolute) of

a body under various loading (stress) states

Compute forces and related information (stress/strain) for certain

statically indeterminate problems


Review/Class Exercise

Write down equations that define normal stress and shearing stress

and write down the names for the terms used and their units

Normal Strain under Axial Loading

Unit for strain : dimensionless or unitless

Sometimes, people write “unit” such as mm/mm, in/in, or similar

How to measure deformation?

Example: 2 components, same materials,

same axial load, same cross-section, but

different initial length of L vs. 2L.

What about change in length in the two?

Observations:

• Same stress

• Longer initial length gives larger change in

total (or absolute) length

• Relative change in length or

(Normal) Strain ε stays the same

A

P

L

5

Common engineering questions:

• Which dimension to use?

• What material to use?

• Is the design (dimension and/or material choice) safe?

• Is the design economical?

• How much deformation (absolute/relative) would it experience?

F F

Same material

D1 D2

F F

Different material

D2 D2

Stress & Strain


Deformation in Component with Variable

Cross-section

For a component with

variable cross-sectional area,

local strain may change.

The strain at point Q is:

Total deformation (or

elongation/shrinkage)

xx

dxd

dx

d

xx

0lim

Tensile Test & Stress-Strain Curve/Diagram Metal Polymer

Load-displacement curve

F

δ

σ - ε curve useful

for obtaining &

comparing

materials’

mechanical

properties

Ceramics?

Typically NOT

by tensile test.

Instead, bending/

compression test

0L

Stress-Strain curve

Engineering Strain ε

En

gin

eerin

g S

tres

s σ

0A

P

Stress – Strain Curves for Ductile Materials

• Initial linear/elastic behavior for low stress

• Afterwards, significant non-linear/plastic

deformation before fracture or breakage

• More predictable showing signs of

distress or warning (e.g., necking) before

fracture

• For most metals and plastics

non-linear/plastic


Brittle Materials

• Small, if any, non-linear or plastic

deformation before fracture or breakage

• No obvious visual warning

• For certain metals, most glass & ceramics,

and polymer at very low temperature

Ductile vs. Brittle Fracture (or Failure)

Ductile fracture

• Shearing cone

• Significant local

narrowing or

necking

Brittle failure

• Planar fracture cross-section

• NO obvious local narrowing or “necking”

Elastic Deformation & Hooke’s Law • Elastic deformation

When stress level is low, deformation will be reversible or

elastic

• Elastic Modulus E & Hooke’s Law

For elastic deformation (i.e., low stress), in most cases, there

is a linear relationship between

normal stress and normal strain

The slope E for the linear section

is called Modulus of Elasticity

or Elastic Modulus or

Young’s Modulus

sectionlinear for Slope

E

U

E E

Elastic

deformation

Plastic Deformation & Yield Strength

• Plastic deformation

When stress increases further

(beyond elastic region),

for many materials (metals,

polymers), deformation becomes

irreversible, characterized by

non-linear - relationship

• Yield strength σY

The stress level when permanent

or plastic deformation starts to

occur (or the transition point between elastic & plastic

deformation)

Often use the value for “offset yield strength” corresponding

to 0.2% of strain (at point Y), as illustrated.


E

U

Plastic deformation

Ultimate Tensile Strength & “Necking”

• Ultimate (tensile) strength U or TS

Highest stress level after which

failure will occur


E

U

• Necking

Localized deformation

as “neck” in tensile test

beyond which absolute

load (force) required

to further deform it

starts to decrease

Often occur soon after

stress reaching U

Ductility: Elongation & Reduction in Area

• %Elongation (EL)

Normal strain (relative increase

in length) when fracture happens,

which is a measure of how

ductile a material is. Often given

by

E

U

B

100%0

0

L

LLEL B

• Reduction in area (RA)

Reduction in cross-section

area when fracture happens,

which is another measure of

how ductile a material is

1000

0

L

AARA B

Additional Concepts and Terminology

• Toughness

Energy absorbed before fracture

- The area under the - curve

• Proportional Limit

The limit, often in terms of

stress, for linear region in the

- curve

E

U

U

Proportional

limit

True Stress and True Strain

Engineering Stress = P/Ao True Stress = P/A

Ao = original area A = instantaneous area

Engineering Strain True Strain

Lo = original length L = instantaneous length

Experiences tell us material cross-section

area changes when load is applied.

For stress, which exact area A should we use? A

P

Engineering stress & strain

way more commonly used

than true stress & strain

0L

)/( LLt

0

ln

0L

L

L

dLL

L

t


Relationship between Engineering σ - ε

and True σ - ε curves

In true σ-ε curve, no decrease in stress when “necking”

occurs

Engineering σ-ε curve easier to obtain & used more widely


Class Exercise

For the - curves of different

iron and steel alloys from

pure iron to A36, A992, A709, how

do the following change?

• Elastic modulus E

No change (in the slope of

initial linear section)

• Yield strength Y

Increase (in the transition stress

between elastic and plastic regions)

• Ultimate strength U

Increase (in the highest stress

it can sustain before fracture)

• Elongation EL at fracture

Decrease (in total strain before

fracture)

Repeated Loading Unloading - Hardening

If stress increases beyond Y, after unloading, and if load again,

yielding appears at a higher stress level due to (strain)

“hardening”

Load until small plastic

deformation (beyond B to C)

and then unload

When reload:

Yielding (i.e., transition from

linear/elastic to non-linear/

plastic) appears at

stress level of ~C instead of B

Repeated Loading Unloading - Fatigue With repeated loading & unloading, but to stress level much

lower than y, failure (sudden fracture) may still eventually

happen after many cycles

Endurance Limit:

Stress level below which

fatigue failure will never

occur or the part could

be cycled indefinitely

- n curve

(i.e., stress level vs cycle life)

Deformations under Axial Loading (1)

For a homogeneous rod with cross-section area A and length L

subject to axial tensile load of P. If within elastic limit (or

proportional limit), determine the total change in length if the

elastic modulus is E

Total change in length for the rod

L

Hooke’s Law

L

E

LE

L

E

AP /

EA

PL


Class Exercise

For tensile engineering stress – strain curve for brass, estimate (a) Elastic

modulus; (b) Yield strength at “offset” strain of 0.002; (c) Maximum load

on a specimen with square cross-section with edge length of

10 mm; (d) Change in length for a 250 mm sample subject to 345 MPa

GPaMPa

E 80002.0

160~

psiMPaY

31035250

UAP 0max

mmmmL 1525006.00

Elastic modulus

(Offset) Yield strength

Max load

kNN

Pam

bAP UU

4545000

1045010100 626

2

0max

At 345 MPa, strain ε ≈ 0.06

Change in length δ


For a member with sections of different

cross-section area of A1, A2, … and each

section has length of L1, L2, … subject to

different loading of P1, P2, …, if the stress

level are all within elastic limit (or

proportion limit), the total change in length

is:

i

i i ii

ii

EA

LP

P

An axial load P applied to a member made with homogeneous materials but with variable cross-section


dxd

LL

dxExA

Pd

00 )(

dxExA

P

)(


Class Exercise

Under tensile load of 1000N, for a bar-shaped sample

with square cross-section and edge length of 1cm,

if the bar sample initial length is 50mm, please

calculate the change in sample length if the sample

material is steel with elastic modulus of 200 GPa.

Assuming it is within elastic deformation limit 1 cm 1 cm

PamNm

N

cm

N

A

F 727

24210/10

10

1000

1

1000

5

9

7

10510200

10

Pa

Pa

E

mmmmmL 5.2105.250105 35

AE

PL

Engineering stress σ

Engineering strain ε

Change in sample length δ


Class Exercise

A steel bar 100 mm long and having a square cross section

20 mm on an edge is pulled in tension with a load of 80kN,

and experiences a change in length of 0.10 mm.

Assuming that the deformation is entirely linear elastic,

please calculate the elastic modulus of the steel

GPaPam

N

mm

mN

200102001.0400

108000

)1010.0()1020(

101001080 =

9

2

9

323

33

= E

/L

AF

0/

= 2

0

=b

FL

Elastic modulus E

Statically Indeterminate Problems

A. Statically Determinate Problems:

• Problems that can be solved by Statics:

i.e. F = 0 and M = 0 & the FBD

B. Statically Indeterminate Problems:

• Problems that cannot be solved by statics alone: The

number of unknowns > the number of equations from

balance of force and moments

• Must involve considerations of deformation or strain

Example - Statically Indeterminate Problem A tube with a center rod of equal

length L is subjected to a total load

of P applied by a hard plate as

illustrated. If the cross-section

area for the center rod and tube is

A1 and A2, while the elastic

modulus for center rod and tube is

E1 and E2. Assuming it is within

the proportion limit and NO

interaction between the tube and

rod, please determine the load

borne by the rod and the tube, P1

& P2 respectively.

From Statics: balance of force

Only one equation for the balance of force along the axis

but two variables -

PPP 21

Statically Indeterminate!

Statically, balance of force

gives

One equation two variables

Example - Statically Indeterminate Problem

PPP 21

Consider geometry:

22

22

11

11

EA

LP

EA

LP

Therefore,

PEAEA

EAP

2211

111

P

EAEA

EAP

2211

222

21

Superposition Method for Solving Statically

Indeterminate Problems

1. Designate one support as redundant support

2. Remove the support from the structure & treat it as an

unknown load.

3. Superpose the displacement

Example 2.04 (1) For the structure illustrated, determine the reaction at A & B

kNRR BA 900

Assume RA and RB both up

Statically, balance of force can only give

Overall change in length for ACB is viewed as superposition of:

• Deformation for ACB hanging free with external loads (at D and K) from A with loads at D and K

• Deformation for ACB due to reaction RB

Example 2.04 (2) To determine deformation for ACB hanging free from A with loads at D and K, divide ACB into 4 parts

Part 1: No force or P1 = 0

Part 2: Tensile force of P2 = 600 kN

Part 3: Tensile force of P3 = 600 kN

Part 4: Tensile force of P4 = (300+600) = 900 kN

m 15.0

mm250mm400

4321

2

43

2

21

LLLL

AAAA

E

mN

EA

LP

i ii

ii /10125.1 9

L

Total elongation if ACB hanging free:

To determine deformation due to

reaction RB alone, divide ACB into 2 parts

Part 1 & 2: RB of compression force

Example 2.04 (3)

m 300.0

mm250mm400

21

2

2

2

1

21

LL

AA

RPP B

Geometrically (consider overall deformation):

0 RL

kNRR BA 900

Finally, remember

kNRA 323

i

B

ii

iiR

E

mR

EA

LPδ

/1095.1 3Total elongation due to support at B, RB

0

/1095.1/10125.1 39

E

mR

E

mN B

kN 577N10577 3 BR

Problem Involving Temperature Change (1)

α coefficient of thermal

expansion

The change in total

change in length due to

temperature change of T

is given by

LTT

Therefore:

= T + P = 0

For a component between two fixed walls with cross-section

area A and elastic modulus E and thermal expansion

coefficient α, if temperature increase is T, calculate the

thermally induced normal stress within the component.

Geometrically:

Change in length due to

compressive stress alone:

Change in length due to thermal expansion alone

Problem Involving Temperature Change (2)

LTT

AE

PLP

0AE

PLLTPT

TAEP TEA

P

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