Title: CONCEPTS OF FORCESTRESS and DEFORMATION STRAIN
1CONCEPTS OF FORCE-STRESSand DEFORMATION-
STRAIN
2For 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
3- Internal forces can be shown on an imaginary cut
section.
4- 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.
5- P1P2P3P4
0 - For the original body ?
- SM 0
- P1P2F
0 - For half of the body ?
- SM 0
- F can be solved from the above set of equilibrium
equations of the half plane.
6 7- If you look at that cut section little bit
closer Force acting on an infinite small area
can be shown
- That force is called the STRESS.
- In other words stress is the force intensity
(force per unit area) acting on a material.
8For example if the cut section is perpendicular
to x-axis
9- 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
10- In tensorial notation the stress components are
assembled in a matrix.
- For equilibrium it can be shown that
- tij tji for i ? j
- txy tyx
- txz tzx
- tyz tzy
- This symmetry reduces the shear stress components
to three.
11- Stresses can be grouped in several ways.
12Column is in uniaxial compression
Hydrostatic pressure (triaxial compression)
Membrane forces (biaxial tension)
Baloon
13Common States of Stress
Simple tension cable
Ski lift (photo courtesy P.M. Anderson)
Torsion (a form of shear) drive shaft
Note t M/AcR here.
14Common States of Stress
Simple compression
Note compressive structure member (s lt 0 here).
(photo courtesy P.M. Anderson)
15Common States of Stress
Bi-axial tension
Hydrostatic compression
Pressurized tank
(photo courtesy P.M. Anderson)
(photo courtesy P.M. Anderson)
16 17- 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
Total elongation of the rod is ?2 (cm, mm,
length) Elongation between AB is (?2-?1)
P
18 19- Strain represents the deformation of materials
per unit length and is unitless (cm/cm, mm/mm)
() Tensile (elongation)
(-) Shortening
20- When pure shear acts on an element, the element
deforms into a rhombic shape. - For convenience the element is rotated by an
angle ?/2 and represented as shown.
?
- For small angles ? tan? ? ?
21- A pure shear strain is produced in torsion.
AA r?
?
r
? Angle of twist of radial line AB to position
AB r radius of cross-sectional area
22Engineering Stress
Tensile stress, s
F
t
Area, A
F
t
F
N
t
s
f
2
A
m
o
original area
? Stress has units N/m2 or kgf/cm2 or psi
before loading
23Engineering Strain
Tensile strain
Lateral strain
d
/2
-
d
L
o
w
o
Shear strain
q
?x
y
90º - q
Strain is always dimensionless.
90º
Adapted from Fig. 6.1 (a) and (c), Callister 7e.
24Stress-Strain Testing
Typical tensile test machine
specimen
extensometer
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.)
25Typical response of a metal
Maximum stress on engineering stress-strain
curve.
Adapted from Fig. 6.11, Callister 7e.
TS
F fracture or ultimate
strength Necking
?y
engineering stress
engineering strain