CONCEPTS OFCONCEPTS OF

FORCE-STRESSFORCE-STRESSandand

DEFORMATION- DEFORMATION- STRAINSTRAIN

For a body in equilibrium (not in motion) subjected to some external forces Pi, there are internal forces developed within the body.

P1, P2...:External forces

F : Internal forces

P1

P2

P3P4

Internal forces can Internal forces can be shown on an be shown on an imaginary cut imaginary cut section.section.

P3

P4

P1

P2

P3

P4

P3

P4

Since the body was initially in equilibrium, half of this body should also be in equilibrium.

The internal forces are derived from the equilibrium equations and can be defined as the forces to bring the body to equilibrium.

Remember force is a vectorical quantity which has a magnitude and direction.

PP11+P+P22+P+P33+P+P44 = 0 = 0 For the original body For the original body →→

ΣΣM = 0M = 0

PP11+P+P22+F = 0+F = 0 For half of the body → For half of the body →

ΣΣM = 0M = 0

F can be solved from the above set of F can be solved from the above set of equilibrium equations of the half plane.equilibrium equations of the half plane.

STRESSSTRESS

If you look at that cut section little bit closer; Force acting on an infinite small area can be shown;

Δs

ΔF

That force is called the STRESS.

In other words stress is the force intensity (force per unit area) acting on a material.

or

Stress =

Area

Force A

Fσ =

Stress Shear (τ) : acts parallel to the

area.

Normal (σ) : acts perpendicular to the area

For example; if the cut section is perpendicular to x-axis

z

y

x

ΔFΔFy

ΔFx

ΔFz

ΔA

ΔFx

σx =

ΔA

ΔFy

τy = ΔA

ΔFz

τz =

However, stresses are always represented in However, stresses are always represented in tensorial tensorial (!not vectorical!)(!not vectorical!) notation. notation.

The plane it is acting on is also presented.The plane it is acting on is also presented. Therefore, if you take an infinitesmall volume Therefore, if you take an infinitesmall volume

element you can show all of the stress element you can show all of the stress componentscomponents

The first subscript indicates the plane perpendicular to the axis and the second subscript indicates the direction of the stress component.

Stress Tensor

In tensorial notation the stress In tensorial notation the stress components are assembled in a matrix.components are assembled in a matrix.

S =

For equilibrium it can be shown that :

τij = τji for i ≠ j

τxy = τyx

τxz = τzx

τyz = τzyThis symmetry reduces the shear stress components to three.

Stresses can be grouped in several ways.

Stress

Static: A constant and continuous load causes a static stress.Dynamic: Loads having different magnitudes and at different times cause dynamic stresses.

Stress

Biaxial tension or compression

Uniaxial tension or compressionTriaxial compressionPure

shear

W

Spring is in uniaxial tension

W

Column is in uniaxial

compression

Baloon Membrane

forces

(biaxial tension)

Hydrostatic pressure

(triaxial compression)

• Simple tension: cable

Note: = M/AcR here.

Common States of StressCommon States of Stress

Ao = cross sectional

area (when unloaded)

FF

o F

A

o

FsA

M

M Ao

2R

FsAc

• Torsion (a form of shear): drive shaftSki lift (photo courtesy P.M. Anderson)

(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NM

o F

A

• Simple compression:

Note: compressivestructure member( < 0 here).(photo courtesy P.M. Anderson)

Ao

Balanced Rock, Arches National Park

Common States of StressCommon States of Stress

• Bi-axial tension: • Hydrostatic compression:

Pressurized tank

< 0h

(photo courtesyP.M. Anderson)

(photo courtesyP.M. Anderson)

Fish under water

z > 0

> 0

Common States of StressCommon States of Stress

DEFORMATIODEFORMATIONN

Deformation:Deformation: is the change in the shape or is the change in the shape or dimension of a material. In other words dimension of a material. In other words when the relative position of points within when the relative position of points within a body changes a body changes “deformation”“deformation” takes place. takes place.

a)a) Elongation:Elongation: occurs under tensile stresses. occurs under tensile stresses.

b)b) Shortening:Shortening: under compressive stresses under compressive stresses

c)c) Rotation:Rotation: due to shear stresses due to shear stresses

P

A

Δ2 B’

B

A’Δ1

Total elongation of the rod is Δ2 (cm, mm, length)

Elongation between AB is (Δ2-Δ1)

STRAINSTRAIN

Strain:Strain: represents the deformation of represents the deformation of materials per unit length and is unitless materials per unit length and is unitless (cm/cm, mm/mm) (cm/cm, mm/mm)

Strain =

DeformationOriginal

length

l-l0=ε =

Δll0 l0

εl=Δll0

εd =Δdd0

l0 l

d

d0

P

P

(+) Tensile (elongation)

(-) Shortening

When pure shear acts on an element, the When pure shear acts on an element, the element deforms into a rhombic shape.element deforms into a rhombic shape.

For convenience the element is rotated by For convenience the element is rotated by an angle an angle γγ/2 and represented as shown./2 and represented as shown.

Ξ

τγ/2

y

x

γ/2

A’A

γ

y

x

γ

D C

B B’

For small angles γ = tanγ → γ =

AA’AD

(radians)

A pure shear strain is produced in torsion.A pure shear strain is produced in torsion.

γ

B

A A’

L

θ: Angle of twist of radial line AB to position A’B

r: radius of cross-sectional area

AA’ = rθθr

γ = A

B

AA’

A’

A

γ = L

θr

Stress has units: N/m2 or kgf/cm2 or psi

Engineering StressEngineering Stress• Shear stress, :

Area, A

Ft

Ft

Fs

F

F

Fs

= Fs

Ao

• Tensile stress, :

original area before loading

Area, A

Ft

Ft

=Ft

Ao2f

m

N=

• Tensile strain: • Lateral strain:

• Shear strain:

Strain is alwaysdimensionless.

Engineering StrainEngineering Strain

90º

90º - y

x = x/y = tan

Lo

L L

wo

Adapted from Fig. 6.1 (a) and (c), Callister 7e.

/2

L/2

Lowo

Stress-Strain TestingStress-Strain Testing• Typical tensile test machine

Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)

specimenextensometer

• Typical tensile specimen

Adapted from Fig. 6.2,Callister 7e.

gauge length

Adapted from Fig. 6.11, Callister 7e.

y

Typical response of a metal

F = fracture or

ultimate

strength

Necking

TS e

ngin

eerin

g s

tres

s

engineering strain

• Maximum stress on engineering stress-strain curve.