Stress and Strain

STRESS AND STRAIN
2
Topics:
• Introduction
• Main Principles of Statics
Stress
• Normal Stress
• Shear Stress
• Bearing Stress
• Thermal Stress
Mechanics : The study of how bodies react to forces acting on them
RIGID BODIES DEFORMABLE BODIES

FLUIDS
(Things that do not change shape) (Things that do change shape)
Statics : The study of bodies

in an equilibrium
Incompressible Compressible
Dynamics :
Mechanics of Materials :
1. Kinematics – concerned
The study of the relationships
with the geometric aspects
between the external loads
of the motion
applied to a deformable body and
2. Kinetics – concerned
the intensity of internal forces
with the forces causing the
acting within the body.
motion.
1.1 Introduction 4
External Loads
Body Force Surface Forces

- developed when one body exerts a force on - caused by direct contact of one body with
another body without direct physical contact the surface of another.
between the bodies.
- e.g earth’s gravitation (weight)
concentrated force
linear distributed load, w(s)
1.2 Main Principles of Statics 5

 Axial Load
 Normal Stress
 Shear Stress
 Bearing Stress
 Allowable Stress
 Deformation of Structural under Axial Load
 Statically indeterminate problem
 Thermal Stress
 Mechanics of material is a study of the
relationship between the external loads applied
to a deformable body and the intensity of
internal forces acting within the body.
 Stress = the intensity of the internal force on a

specific plane (area) passing through a point.
 Strain = describe the deformation by changes in

length of line segments and the changes in the
angles between them
Normal Stress (σ)
- stress which acts perpendicular or normal to the
cross section of the load-carrying member.
- can be either compressive or tensile.
Shear Stress (τ)

- stress which acts tangent to the cross section of
the load-carrying member.
- refers to a cutting-like action.
1.1 Introduction 8
 Normal Stress, 
the intensity of force, or force per unit area, acting
normal to A
 = P / A
A positive sign will be used to indicate a tensile stress

(member in tension)
A negative sign will be used to indicate a compressive

stress (member in compression)
(b)
(a)
•Unit: Nm -²
Stress (  ) = Force (P) •N/mm2 or MPa
Cross Section (A) N/m2 or Pa
Assumptions :
1. Uniform deformation: Bar
remains straight before and
after load is applied, and
cross section remains flat or
plane during deformation
2. In order for uniform
deformation, force P be
applied along centroidal axis
of cross section C
1.4 Axial Loading – Normal Stress

11
  FRz  Fz ;  dF    dA
A
P  A
P

A
σ = average normal stress at any point
on cross sectional area
P = internal resultant normal force
A = cross-sectional area of the bar

12
• Use equation of σ = P/A for cross-sectional area of a member when
section subjected to internal resultant force P
Internal Loading
• Section member perpendicular to its longitudinal axis at pt
where normal stress is to be determined
• Draw free-body diagram
• Use equation of force equilibrium to obtain internal axial
force P at the section
Average Normal Stress
• Determine member’s x-sectional area at the section
• Compute average normal stress σ = P/A

13
Example 1.1:
Two solid cylindrical rods AB and BC are welded
together at B and loaded as shown. Knowing
that d1=30mm and d2=20mm, find average
normal stress at the midsection of (a) rod AB,
(b) rod BC.
Example 1.2
Two solid cylindrical roads AB and BC are welded
together at B and loaded as shown. Knowing
that d1 = 30 mm and d2 = 50 mm, find the
average normal stress in the mid section of (a)
rod AB, (b) rod BC.
 Normal strain () is the elongation or
contraction of a line segment per unit of
length  = L / Lo
L = elongation
Lo = length
* L = 
Example 1.3:
Determine the corresponding strain for a bar of
length L=0.6m and uniform cross section which
undergoes a deformation =15010-6m.
Example 1.4
A cable and strut assembly ABC supports a vertical load
P=12kN. The cable has an effective cross sectional area of
160mm², and the strut has an area of 340mm².
(a) Calculate the normal stresses in the cable and strut.
(b) If the cable elongates 1.1mm, what is the strain?
(c) If the strut shortens 0.37mm, what is the strain?

Example 1.5
The bar shown has a square cross section

(20mm x 40mm) and length, L=2.8m. If an
axial force of 70kN is applied along the
centroidal axis of the bar cross sectional area,
determine the stress and strain if the bar end
up with 4m length.
70kN 70kN
2.8m
 Tensile test is an experiment to determine
the load-deformation behavior of the
material.
 Data from tensile test can be plot into stress
and strain diagram.
 Example of test specimen
- note the dog-bone geometry
28
 Universal Testing Machine - equipment used
to subject a specimen to tension,
compression, bending, etc. loads and
measure its response
29
Stress-Strain Diagrams
A number of important mechanical properties of materials that can be

deduced from the stress-strain diagram are illustrated in figure above.
30
 Point O-A = linear relationship between stress
and strain
 Point A = proportional limit (PL)
The ratio of stress to strain in this linear region
of stress-strain diagram is called Young’s Modulus
or the Modulus of Elasticity given
 < PL
Unit: MPa
At point A-B, specimen begins yielding.

 Point B = yield point
 Point B-C = specimen continues to elongate without any increase in
stress. Its refer as perfectly plastic zone
 Point C = stress begins to increase
 Point C-D = refer as the zone of strain hardening
 Point D = ultimate stress/strength ; specimen
begins to neck-down
 Point E = fracture stress ..\stress strain diagram.doc
31
Point O to A
Point C to D
Point D to E
At point E
Normal or engineering stress can be determined by dividing the

applied load by the specimen original cross sectional area.
True stress is calculated using the actual cross sectional area at

the instant the load is measured.
31
Some of the materials like aluminum (ductile), does not have clear yield
point likes structural steel. Therefore, stress value called the offset yield
stress, YL is used in line of a yield point stress.
As illustrated, the offset yield stress is determine by;

 Drawing a straight line that best fits the data in initial (linear) portion of
the stress-strain diagram
 Second line is then drawn parallel to the original line but offset by
specified amount of strain
 The intersection of this second line with the stress-strain curve determine
the offset yield stress.
 Commonly used offset value is 0.002 or 0.2%
32
Brittle material such as ceramic and glass have low tensile
stress value but high in compressive stress. Stress-strain
diagram for brittle material.
33
 Elasticity refers to the property of a material such that
it returns to its original dimensions after unloading .
 Any material which deforms when subjected to load
and returns to its original dimensions when unloaded
is said to be elastic.
 If the stress is proportional to the strain, the material
is said to be linear elastic, otherwise it is non-linear
elastic.
 Beyond the elastic limit, some residual strain or
permanent strains will remain in the material upon
unloading .
 The residual elongation corresponding to the
permanent strain is called the permanent set .
34
• The amount of strain which is recovered upon unloading is
called the elastic recovery.
35
 When an elastic, homogenous and isotropic material is
subjected to uniform tension, it stretches axially but
contracts laterally along its entire length.
 Similarly, if the material is subjected to axial
compression, it shortens axially but bulges out laterally
(sideways).
 The ratio of lateral strain to axial strain is a constant
known as the Poisson's ratio,

v
la tera l
 a xia l
where the strains are caused by uniaxial stress only

L
 paksi @  x 
L
b d
sisi @  y   
b d 36
Example 1.7
A prismatic bar of circular cross-section is
loaded by tensile forces P = 85 kN. The bar
has length of 3 m and diameter of 30 mm.
It is made from aluminum with modulus of
elasticity of 70 GPa and poisson's ratio =
1/3. Calculate the elongation and the
decrease in diameter d.
Example 1.8
A 10 cm diameter steel rod is loaded with 862 kN by
tensile forces. Knowing that the E=207 GPa and = 0.29,
determine the deformation of rod diameter after being
loaded.
Solution
 in rod,  =
Axial strain,
Lateral strain,
38

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STRESS AND STRAIN

RIGID BODIES DEFORMABLE BODIES

Statics : The study of bodies

Body Force Surface Forces

linear distributed load, w(s)

1.2 Main Principles of Statics 5

 Stress = the intensity of the internal force on a

 Strain = describe the deformation by changes in

Shear Stress (τ)

A positive sign will be used to indicate a tensile stress

A negative sign will be used to indicate a compressive

1.4 Axial Loading – Normal Stress

1.4 Axial Loading – Normal Stress

1.4 Axial Loading – Normal Stress

(b) If the cable elongates 1.1mm, what is the strain?

(c) If the strut shortens 0.37mm, what is the strain?

The bar shown has a square cross section

A number of important mechanical properties of materials that can be

At point A-B, specimen begins yielding.

Normal or engineering stress can be determined by dividing the

True stress is calculated using the actual cross sectional area at

As illustrated, the offset yield stress is determine by;

where the strains are caused by uniaxial stress only

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