Transcript CONCEPTS OF FORCE, STRESS, DEFORMATION & STRAIN

CONCEPTS OF
FORCE-STRESS
and
DEFORMATION- STRAIN
For a body in
equilibrium (not in
motion) subjected to
some external forces
Pi, there are internal
forces developed
within the body.
P2
P1
P3
P4
P1, P2...:External forces
F : Internal forces
Internal forces can be
shown on an
imaginary cut section.
P4
P3
P2
P1
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.
P1+P2+P3+P4 = 0
 For the original body →
ΣM = 0
P1+P2+F = 0
 For half of the body →
ΣM = 0
 F can be solved from the above set of
equilibrium equations of the half plane.
STRESS
If you look at that cut section little bit closer; Force
acting on an infinite small area can be shown;
ΔF
s
Δs
That force is called the STRESS.
In other words stress is the force intensity (force per
unit area) acting on a material.
Stress =
Force
Area
or
σ=
F
A
Normal (σ) : acts perpendicular to the area
Stress
Shear (τ) : acts parallel to the area.
For example; if the cut section is perpendicular to xaxis
y
ΔFy
ΔFx
x
z
σx =
ΔF
ΔFz
ΔFx
ΔA
τy =
ΔFy
ΔA
τz =
ΔFz
ΔA
 However, stresses are always represented in tensorial
(!not vectorical!) notation.
 The plane it is acting on is also presented.
 Therefore, if you take an infinitesmall volume
element you can show all of the stress components
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 components
are assembled in a matrix.
For equilibrium it
can be shown that :
S=
τij = τji for i ≠ j
τxy = τyx
τxz = τzx
τyz = τzy
This symmetry reduces the shear stress components
to three.

Stresses can be grouped in several ways.
Stress
Stress
Static: A constant and continuous
load causes a static stress.
Dynamic: Loads having different
magnitudes and at different times
cause dynamic stresses.
Uniaxial tension or compression
Biaxial tension or compression
Triaxial compression
Pure shear
W
Spring is in
uniaxial tension
W
Baloon
Membrane forces
(biaxial tension)
Column is in
uniaxial
compression
Hydrostatic pressure
(triaxial compression)
Common States of Stress
• Simple tension: cable
F
F
A o = cross sectional
area (when unloaded)
F
s=
s
Ao
s
• Torsion (a form of shear): drive shaft
M
Ac
M
Fs
Ski lift
(photo courtesy
P.M. Anderson)
Ao
Fs
t =
Ao
2R
Note: t = M/AcR here.
Common States of Stress
• Simple compression:
Ao
Canyon Bridge, Los Alamos, NM
(photo courtesy P.M. Anderson)
Balanced Rock, Arches
National Park
(photo courtesy P.M. Anderson)
F
s=
Ao
Note: compressive
structure member
(s < 0 here).
Common States of Stress
• Bi-axial tension:
Pressurized tank
(photo courtesy
P.M. Anderson)
• Hydrostatic compression:
Fish under water
sq > 0
sz > 0
sh< 0
(photo courtesy
P.M. Anderson)
DEFORMATION
 Deformation: is the change in the shape or
dimension of a material. In other words
when the relative position of points within a
body changes “deformation” takes place.
Elongation: occurs under tensile stresses.
Shortening: under compressive stresses
Rotation: due to shear stresses
a)
b)
c)
Δ1
A
A’
B
B’
Δ2
P
Total elongation of the rod is Δ2 (cm, mm,
length)
Elongation between AB is (Δ2-Δ1)
STRAIN
 Strain: represents the deformation of materials
per unit length and is unitless (cm/cm,
mm/mm)
Deformation
Strain =
Original length
d
ε=
Δl
l0
l-l0
=
l0
P
εl=
d0
l
l0
εd =
P
Δl
l0
Δd
d0
(+) Tensile (elongation)
(-) Shortening
 When pure shear acts on an element, the
element deforms into a rhombic shape.
 For convenience the element is rotated by an
angle γ/2 and represented as shown.
y
γ/2
τ
Ξ
γ/2
x
y
A’
A
γ
D
 For small angles γ = tanγ → γ =
B B’
γ
C
AA’
AD
x
(radians)
 A pure shear strain is produced in torsion.
B
AA’
γ=
AB
A
L
γ
A A’
r θ
A’
AA’ = rθ
θr
γ=
L
θ: Angle of twist of radial line
AB to position A’B
r: radius of cross-sectional area
Engineering Stress
• Tensile stress, s:
• Shear stress, t:
Ft
Area, A
Area, A
Ft
Ft
Nf
= 2
s=
Ao m
original area
before loading
Ft
F
Fs
Fs
Fs
t=
Ao
F
Ft
 Stress has units:
N/m2 or kgf/cm2 or psi
Engineering Strain
• Tensile strain:
• Lateral strain:
d/2
e = d
Lo
wo
• Shear strain:
-d L
eL =
wo
Lo
dL /2
q
g = x/y = tan q
x
90º - q
y
90º
Strain is always
dimensionless.
Adapted from Fig. 6.1 (a) and (c), Callister 7e.
Stress-Strain Testing
• Typical tensile test
machine
extensometer
• Typical tensile
specimen
specimen
Adapted from
Fig. 6.2,
Callister 7e.
gauge
length
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.)
Typical response of a metal
engineering stress
• Maximum stress on engineering stress-strain curve.
Adapted from Fig. 6.11,
Callister 7e.
TS
F = fracture or
ultimate
strength
sy
Necking
engineering strain