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Mechanics of Solids I Mechanics of Solids I

Stress and Strain – Axial Loading

Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses

Normal StrainNormal Strain

deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient.Normal strain

Defined as the change in length of a line segment per unit of lengthε = δ/L - dimensionless quantity

Positive = elongationnegative = contraction

strain normal

stress

==

==

L

AP

δε

σ

2

StressStress--Strain TestStrain Test

StressStress--Strain Diagram: Ductile MaterialsStrain Diagram: Ductile Materials

Conventional stress-strain diagram, σ = P/A0; ε = δ/L0

True stress-strain diagram, σ = P/A; ε = δ/L

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StressStress--Strain Diagram: Brittle Materials Strain Diagram: Brittle Materials

Ductile materials• can be subjected to large strains before it ruptures, e.g., mild steel, aluminum alloyBrittle materials• exhibit little or no yielding before failure, e.g., gray cast iron

StressStress--Strain Diagram: Strain Diagram: 00..22% Offset Method % Offset Method

o For many ductile materials, the onset of yield is not clear. The stress keeps increasing (not linear) until the ultimate strength is reach.

o Draw through the point of ε = 0.2% or 0.002 a line parallel to the initial straight-line portion of the σ-ε diagram.

o The stress σy corresponding to the point o The stress σy corresponding to the point Y is “the yield strength at 0.2% offset.”

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o Standard measure of the ductility of a material is

Measure of Ductility Measure of Ductility

• percent elongation

where subscript 0 = at initial state, B = at rupture

• percent reduction in area

0

0

-percent elongation = 100 BL LL

• percent reduction in area

0

0

-percent reduction in area = 100 BA AA

Elastic vs. Plastic BehaviorElastic vs. Plastic Behavior

o If the strain disappears when the ppstress is removed, the material is said to behave elastically.

o When the strain does not return

o The largest stress for which this occurs is called the elastic limit.

o When the strain does not return to zero after the stress is removed, the material is said to behave plastically.

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Hooke’s LawHooke’s LawMost engineering materials exhibit a linear relationship between stress and strain with the elastic regionDiscovered by Robert Hooke in 1676, known as Hooke’s law

σ = Eε

E represents the stiffness of a material, also called f i ithe modulus of elasticity or Young’s modulus

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

Most grades of steel have same modulus of elasticity, Est = 200 GPa

FatigueFatigueo Fatigue properties are shown on

S-N diagrams.

o A member may fail due to fatigueat stress levels significantly below the ultimate strength if subjected to many loading cycles.

o When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles.

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Deformations Under Axial LoadingDeformations Under Axial Loading

P• From Hooke’s Law:

PEE AEσσ ε ε= = =

• From the definition of strain:

Lδε =

• Equating and solving for the deformation,

PLAE

δ =

• With variations in loading, cross-section or material properties,

i i

i i i

P LA E

δ = ∑

Composite steel bar shown made from two segments AB and BD. Area AAB = 600 mm2 and

Example Example 22..11

g ABABD = 1200 mm2

Determine the vertical displacement of end A and displacement of B relative to C.

Ans: δA = +0.61 mm, δBC = +0.104 mm

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• The rigid bar BDE is supported by two links AB and CD

Example Example 22..22

links AB and CD.

• Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2).

• For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.

Static IndeterminacyStatic Indeterminacy• Structures for which internal forces and

reactions cannot be determined from statics alone are said to be statically indeterminate.

• Redundant reactions are replaced with unknown loads which along with the other loads

• A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.

0=+= RL δδδ

• Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.

must produce compatible deformations.

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Example Example 22..33Determine the reactions at A and B for the steel bar and loading shown assuming a close fit at bar and loading shown, assuming a close fit at both supports before the loads are applied.

Thermal StressesThermal Stresses• A temperature change results in a change in

length or thermal strain (εT).

• The thermal deformation (δT) and the deformation from the redundant support must be compatible.

coef.expansion thermal=α

( )Δ= αδ LTT( )Δ= αε ,TT

( ) 0

0

=+Δ

=+=

AEPLLT

PT

α

δδδ( )

( )TEAP

TAEPPT

Δ−==

Δ−==+=

ασ

αδδδ 0

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Example Example 22..55Knowing that a 0.5-mm gap exists when the temperature is 24 ºC

350 mm 450 mm0.5 mm

when the temperature is 24 C, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -75 MPa, (b) the corresponding exact length of the aluminum bar.

BronzeA = 1500 mm2

E = 105 GPaα = 21.6 x 10-6/ºC

AluminumA = 1800 mm2

E = 73 GPaα = 23.2 x 10-6/ºC

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

σ 0=== zyx

x Eσσσε

• The elongation in the x-direction is accompanied by a contraction in the other directions.

0≠= zy εε

• Poisson’s ratio is defined as

x

z

x

y

εε

εε

ν −=−==strain axialstrain lateral

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Generalized Hooke’s LawGeneralized Hooke’s Law• For 2D:

• For an element subjected to multi-axial loading, the normal strain components are:

νσ

EEE

EEE

EEE

zyxz

zyxy

zyxx

σνσνσε

νσσνσε

νσνσσε

+−−=

−+−=

−−+=

Shearing StrainShearing Strain• A cubic element subjected to a shear stress will

deform into a rhomboid. Shear strain is quantified deform into a rhomboid. Shear strain is quantified in terms of the change in angle between the sides.

• A plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

zxzxyzyzxyxy GGG γτγτγτ ===

where G is the modulus of rigidity or shear modulus.

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Example Example 22..66A rectangular block of material with

modulus of rigidity G = 90 ksi is bonded to modulus of rigidity G 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

SaintSaint--Venant’s PrincipleVenant’s Principle• Concentrated loads result in large

stresses in the vicinity of the load

• Saint-Venant’s Principle:Stress distribution may be assumed

• Stress and strain distributions become uniform at a relatively short distance from the load application points.

application point.

Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.

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Stress Concentration: HoleStress Concentration: Holeo Discontinuities of cross section may result in high localized or concentrated stresseslocalized or concentrated stresses.

max

ave

K σσ

=

Stress Concentration: FilletStress Concentration: Fillet