chapter outline

2005 Pearson Education South Asia Pte Ltd 1

CHAPTER OUTLINE

1. Tension and Compression Test

2. Stress-Strain Diagram

3. Stress-Strain Behavior of Ductile and Brittle Materials

4. Hooke’s Law

5. Poisson’s Ratio


• The strength of a material can only be determined by experiment

• One test used by engineers is the tension or compression test

• This test is used primarily to determine the relationship between the average normal stress and average normal strain in common engineering materials, such as metals, ceramics, polymers and composites

3.1 TENSION & COMPRESSION TEST


• Before testing, 2 small punch marks identified along specimen’s length

• Measurements are taken of both specimen’s initial cross-sectional area A0 and gauge-length distance L0; between the two marks


Performing the tension or compression test

Specimen of material is made into “standard” shape and size

do=13 mm

Lo=50 mm


Performing the tension or compression test


• The machine will stretch specimen at slow constant rate until breaking point

• Seat the specimen into a testing machine shown in the figure

• At frequent intervals during test, data is recorded of the applied load P.


• Elongation δ = L − L0 is measured using either a caliper or an extensometer

• δ is used to calculate the normal strain in the specimen

• Sometimes, strain can also be read directly using an electrical-resistance strain gauge



• A stress-strain diagram is obtained by plotting the various values of the stress and corresponding strain in the specimen

Conventional stress-strain diagram• Using recorded data, we can determine nominal or

engineering stress by

3.2 STRESS-STRAIN DIAGRAM

PA0

σ =

Assumption: Stress is constant over the cross-section and throughout region between gauge points

P P

do=13 mm

Lo=50 mm


Conventional Stress-Strain Diagram

• Likewise, nominal or engineering strain is found directly from strain gauge reading, or by


δL0

=

Assumption: Strain is constant throughout region between gauge points

By plotting σ (ordinate) against (abscissa), we get a conventional stress-strain diagram


Conventional stress-strain diagram


Figure shows the characte-ristic stress-strain diagram for steel, a commonly used material for structural members and mechanical elements



Linear Elastic Behavior

• If load is removed upon reaching proportional limit, specimen will return to its original shape

• A straight line

• Stress is proportional to strain, i.e., linearly elastic

• Upper stress limit, or proportional limit; σpl



Non-linear Elastic Behavior

• If load is removed upon reaching elastic limit, specimen will return to its original shape

• A small curve line

• Stress is not proportional to strain, i.e., non-linearly elastic

• Upper stress limit is almost reaching yield stress; σy



Yielding.

• Once yield point reached, specimen continues to elongate (strain) without any increase in load

• Material deforms permanently; yielding; plastic deformation

• Yield stress, σY

• Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit

• Material is referred to as being perfectly plastic



Strain Hardening.

• Ultimate stress, σu

• While specimen is elongating, its cross-sectional area will decrease

• Decrease in area is fairly uniform over entire gauge length



Necking.

• As a result, a constriction or “neck” tends to form in this region as specimen elongates further

• Specimen finally breaks at fracture stress, σf

• At ultimate stress, cross-sectional area begins to decrease in a localized region



Specimen finally breaks at fracture stress, σf

Necking.


True stress-strain diagram


• Instead of using original cross-sectional area and length, we can use the actual cross-sectional area and length at the instant the load is measured

• Values of stress and strain thus calculated are called true stress and true strain, and a plot of their values is the true stress-strain diagram


Conventional stress-strain diagram


• In strain-hardening range, conventional σ- diagram shows specimen supporting decreasing load

• While true σ- diagram shows material to be sustaining increasing stress

True stress-strain diagram


Ductile materials

• Defined as any material that can be subjected to large strains before it ruptures, e.g., mild steel

• Such materials are used because it is capable of absorbing shock or energy. When overloaded, it will exhibit large deformation before failing

• Ductility of material is to report its percent elongation or percent reduction in area at time of fracture

3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS


Ductile materials

• Most metals do not exhibit constant yielding behavior beyond the elastic range, e.g. aluminum


• It does not have well-defined yield point, thus it is standard practice to define its yield strength using a graphical procedure called the offset method


Offset method to determine yield strength


1. Normally, a 0.2 % strain is chosen.

2. From this point on the axis, a line parallel to initial straight-line portion of stress-strain diagram is drawn.

3. The point where this line intersects the curve defines the yield strength.


Brittle Materials


• Material that exhibit little or no yielding before failure are referred to as brittle materials, e.g., gray cast iron

• Brittle materials do not have a well-defined tensile fracture stress, since appearance of initial cracks in a specimen is quite random

Tension failure ofA brittle material

Compression causesMaterial to bulge out


Brittle Materials• Instead, the average fracture stress from a set of

observed tests is generally reported


s – diagram for typical concrete mix


• E represents the constant of proportionality, also called the modulus of elasticity or Young’s modulus

3.4 HOOKE’S LAW• Most engineering materials exhibit a linear relationship

between stress and strain in the elastic region

σ = E

• Discovered by Robert Hooke in 1676 using springs, known as Hooke’s law

• E has units of stress, i.e., Pascals, MPa or GPa.


• As shown above, most grades of steel have same modulus of elasticity, Est = 200 GPa

3.4 HOOKE’S LAW

• Modulus of elasticity is a mechanical property that indicates the stiffness of a material

• Materials that are still have large E values, while spongy materials (vulcanized rubber) have low values


IMPORTANT

• Modulus of elasticity E, can be used only if a material has linear-elastic behavior.

• Also, if stress in material is greater than the proportional limit, the stress-strain diagram ceases to be a straight line and the equation is not valid

3.4 HOOKE’S LAW


• When body is subjected to axial tensile force (tension), it elongates and contracts laterally

3.5 POISSON’S RATIO

• Similarly, it will contract and its sides expand laterally when subjected to an axial compressive force

d = longitudinal elongation

d ’ = lateral contraction



long =δL

lat = ̶ δ’r

Strains of the bar for tension:

positive

negative

long = ̶δL

lat = δ’r

Strains of the bar for compression:

positive

negative


• ν is unique for homogenous and isotropic material• Why negative sign?


• Lateral strain is the same in all lateral (radial) directions• Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5

Longitudinal elongation cause lateral contraction (-ve strain) and vice versa

Poisson’s ratio, ν = −lat

long

• Early 1800s, S.D. Poisson realized that within elastic range, ration of the two strains is a constant value, since both are proportional.


EXAMPLE 3-1

Bar is made of A-36 steel and behaves elastically.Determine change in its length and change in dimensions of its cross section after load is applied.


EXAMPLE 3-1

Normal stress in the bar is

σz =PA

From tables, Est = 200 GPa, strain in z-direction is

= 16.0(106) Pa = 16 MPa

= 80(10−6) m/mz = =σz

Est

16.0(106)200(109)


EXAMPLE 3-1

Axial elongation of the bar is,

δz = zLz

Using νst = 0.32, contraction strains in both x and y directions are

x = y = −νstz

= [80(10−6)](1.5 m) = 120(10-6) m

= −0.32[80(10−6)] = −25.6(10-6) m/m


EXAMPLE 3-1

Thus changes in dimensions of cross-section are

δx = xLx

δy = yLy

= −[25.6(10−6)](0.1 m) = −2.56(10-6) m

= −[25.6(10−6)](0.05 m) = −1.28(10-6) m


3.7 SHEAR STRESS-STRAIN DIAGRAM

When an element of material issubjected to a pure shear in equilibrium, the equal shear stresses must be developed on four faces of the element.

txy = tyx

The shear stress will distort the element uniformly.

x

y gxy

2

gxy

2

2

gxy

x

y

txy

tyx


• Material will exhibit linear-elastic behavior till its proportional limit, τpl

• Strain-hardening continues till it reaches ultimate shear stress, τu

• Material loses shear strength till it fractures, at stress of τf




• Hooke’s law for shear

τ = Gγ

G is shear modulus of elasticity or modulus of rigidity

• G can be measured as slope of line on τ-γ diagram, G = τpl/ γpl

G =E

2(1 + ν)

• The three material constants E, ν, and G is related by


CHAPTER REVIEW

• Tension test is the most important test for determining material strengths. Results of normal stress and normal strain can then be plotted.

• Many engineering materials behave in a linear-elastic manner, where stress is proportional to strain, defined by Hooke’s law, σ = E. E is the modulus of elasticity, and is measured from slope of a stress-strain diagram

• When material stressed beyond yield point, permanent deformation will occur.


CHAPTER REVIEW• Strain hardening causes further yielding of material with

increasing stress

• At ultimate stress, localized region on specimen begin to constrict, and starts “necking”. Fracture occurs.

• Ductile materials exhibit both plastic and elastic behavior. Ductility specified by permanent elongation to failure or by the permanent reduction in cross-sectional area

• Brittle materials exhibit little or no yielding before failure

• Yield point for material can be increased by strain hardening, by applying load great enough to cause increase in stress causing yielding, then releasing the load. The larger stress produced becomes the new yield point for the material


CHAPTER REVIEW

• Poisson’s ratio (ν), a dimensionless property that measures the lateral strain to the longitudinal strain [0 ≤ ν ≤ 0.5]

• For shear stress vs. strain diagram: within elastic region, τ = Gγ, where G is the shearing modulus, found from the slope of the line within elastic region.

• G can also be obtained from the relationship of G = E/[2(1+ ν)]

chapter outline

Documents

stressstrain diagrampa0

stressstrain diagraml0

stressstrain diagramyielding

engineering strain

corresponding strain

stressstrain diagram

average normal strain

elasticupper stress