- Stress is defined as the internal force per unit area within a material. It can be tensile or compressive. Common types include normal stress and shear stress. - Strain is a measure of deformation in a material under stress. Normal strain measures changes in length while shear strain measures changes in shape. - The allowable stress for a material is less than its failure stress to ensure safety under loads. Factor of safety is defined as the ratio of failure stress to allowable stress.

1. Stress and Strain
The most fundamental concepts in mechanics of material are stress and strain
Consider a prismatic bar under a axial force F (disregarding weight of bar) and
an axial force is a load directed along the axis of the member, resulting in
either tension or compression in the bar.
Stress
F F
Examples: bridge truss, connecting rods, wheel spokes and columns in buildings.
Prismatic bar: straight structural member having the same cross-sectional area A
throughout its length
Stress

2. Internal actions exposed by making imaginary cut at section m-n
F F
m
n
Continuously distributed stresses acting over entire cross section, and the axial
force, F acting at the cross section is the resultant of those stresses.
F
Resultant force
Stress has units of force per unit area and is denoted by Greek Letter σ (sigma) . In
general stresses acting on a plane may be uniform or may be vary in intensity. In
this course stresses acting on cross section assumed are uniformly distributed over
the area. Then resultant of those stress must be equal to the magnitude of the stress
times the cross-sectional area A of the bar.
Stress

3. Stress = intensity of the internal force
Generally speaking
Or symbolically,
SI units: Magnitude of stress is N/m2, called Pascal (Pa)
When bar stretched by force F, the stresses are tensile stresses
If forces are reversed than the stresses are compressive stresses
Stresses act perpendicular to the cut surface, they are also called normal stresses.
Normal stress can be either tensile or compressive
Stress

4. The bar in figure has a constant width of 35 mm and a thickness of 10 mm.
Determine The max. average normal stress in the bar when it is subjected to
the loading shown.
Examples
Determine internal normal force at section A. if rod is subject to the external
uniformly distributed loading along its length of 8KN/m
A
2m 3m
Stress

5. Rods AC and BC are used to suspend the 200-kg mass. If each rod is made of a
material for which the average normal stress can not exceed 150 MPa,
determine the minimum required diameter of each rod to the nearest mm.
Stress

6. Shear stress: If stress acts parallel or tangential to the surface of the
material.
where
τavg = Average shear stress at the section
V = Internal resultant shear force parallel to the
area
A = area at the section (or area on which it (V)
acts)
Shear stress, 𝝉 =
𝑽
𝑨
A
Stress
Greek letter τ (tau)

7. Loads are transmitted to individual members through connections
that use rivets, bolts, pins, nails, or welds
Forces applied to the bolt by the plates
These forces must be balanced
by a shear force in the bolt
This results in Shear
stress, ′𝝉’ on bolt,
𝝉 =
𝑭
𝑨
=
𝑭
𝝅𝒓 𝟐
F
F
V V=F
Single Shear
Stress

8. F
F/2 F/2
F/2=V V=F/2F
Shear stress on bolt
𝝉 =
𝑭
𝑨
=
𝑭
𝟐𝑨
Where
A is the cross sectional area of bolt
r is the radius of bolt
Double Shear
Stress

9. Bearing Stress
A normal stress that is produced by the compression of one surface against
another is called bearing stress. If the bearing stress becomes large enough then it
may deform surfaces in contact.
The bolt exerts on plate ‘A’ a force Fb equal and opposite to
the F exerted by the plate on to the bolt.
Fb represents the resultant forces distributed on the inside
surface of half cylinder of the plate.
The actual distribution of the stress is difficult to
determine. It is customary to assume that stresses are
uniformly distributed. Based on this Bearing stress can be
calculated by dividing the total bearing Force Fb by bearing
arear Ab
𝜎 𝑏 =
𝐹𝑏
𝐴 𝑏
Bearing area is projected area of the curved bearing surface
𝜎 𝑏 =
𝐹𝑏
𝑑𝑡
𝐴 𝑏 = dt
𝑤ℎ𝑒𝑟𝑒
d is diameter of bolt
t is thickness of plate A
Fb

10. Examples
Determine the average shear stress in the 20-mm diameter pin at A and the
30-mm diameter pin at B that support the beam.
Stress

11. A punch for making holes in steel plates is shown in the figure. Assume that a
punch having diameter d = 20 mm. is used to punch a hole in a 8 mm. plate, as
shown in the cross-sectional view. If a force P = 110kN is required to create the
hole, what is the average shear stress in the plate and the average compressive
stress in the punch?
Stress

12. Normal and Shear stresses on inclined sections
b
c
ϴ
A
P
2D view of the normal section
(thickness perpendicular to the
page)
2D view of the inclined section
Stress

13. The force P can be resolved into components:
Normal force N perpendicular to the inclined plane, N = P cos θ
Shear force V tangential to the inclined plane V = P sin θ

14. If we know the areas on which the forces act, we can
calculate the associated stresses.
Stress

15. A prismatic bar having cross-sectional area A=1200mm2 is compressed by an
axial load P=90KN. Determine the stress acting on an inclined section pq cut
through the bar at angle ϴ = 250
Stress
Example

16. Allowable Stress
An 80 kg lamp is supported by a single electrical copper cable of diameter d =
3.15 mm. What is the stress carried by the cable.
Example
a a
F
mg
FBD
Whether or not 80kg would be too heavy, or say 100.6MPa
stress would be too high for the wire/cable, from the safety
point of view. Indeed, stress is one of most important indicators of structural
strength.
Stress

17. When the stress (intensity of force) of an element exceeds some level, the
structure will fail. For convenience, we usually adopt allowable force or allowable
stress to measure the threshold of safety in engineering.
σ ≤ σallow
There are several reasons for this that we must take into account in engineering:
• The load for design may be different from the actual load.
• Size of structural member may not be very precise due to manufacturing and
assembly.
• Various defects in material due to manufacturing processing.
• Unknown vibrations, impact or accidental loading
To ensure the safety of a structural member it is necessary to restrict the applied
load to one that is less than the load the member can fully support.
Stress

18. One simple method to consider such uncertainties is to use a number called the
Factor of Safety, F.S. which is a ratio of failure load Ffail (found from experimental
testing) divided by the allowable Fallow
𝐹. 𝑆 =
𝐹𝑓𝑎𝑖𝑙
𝐹𝑎𝑙𝑙𝑜𝑤
If the applied load is linearly related to the stress developed in the member, as in
the case of using 𝜎 =
𝐹
𝐴
𝑎𝑛𝑑 𝜏 𝑎𝑣𝑔 =
𝑉
𝐴
, then we can define the factor of safety as a
ratio of the failure stress σfail (or to 𝜏 𝑓𝑎𝑖𝑙) to the allowable stress σallow (or 𝜏 𝑎𝑙𝑙𝑜𝑤)
𝐹. 𝑆 =
𝜎𝑓𝑎𝑖𝑙
𝜎 𝑎𝑙𝑙𝑜𝑤
𝐹. 𝑆 =
𝜏 𝑓𝑎𝑖𝑙
𝜏 𝑎𝑙𝑙𝑜𝑤
𝑂𝑅
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the
potential failure.
𝐹. 𝑆 =
𝜎𝑓𝑎𝑖𝑙
𝜎 𝑎𝑙𝑙𝑜𝑤
> 1 𝐹. 𝑆 =
𝜏 𝑓𝑎𝑖𝑙
𝜏 𝑎𝑙𝑙𝑜𝑤
> 1
𝑂𝑅

19. F. S also dependent on the specific design case. For nuclear power plant, the
factor of safety for some of its components may be as high as 3. For an
aircraft design, the higher the F.S. (safer), the heavier the structure,
therefore the higher in the operational cost. So we need to balance the
safety and cost. In case of aircraft F S may be close to one to reduce the
weight of aircraft.
Example: If the maximum allowable stress for copper is σCu,allow=50MPa.
Determine the minimum size of the wire/cable from the material strength point
of view.
Obviously, the lower the allowable stress, the bigger the cable size. Stress is
an indication of structural strength and elemental size.
Stress

20. Determine the maximum vertical force P that can be applied to the bell crank so that the
average normal stress developed in the 10-mm diameter rod CD, and the average shear
stress developed in the 6-mm diameter double sheared pin B not exceed 175 MPa and 75
MPa.
Example
Stress

21. Strain
When a force is applied to a body, it will tend to change its shape and size, in
other words the body is deformed. Strain is intensity of deformation.
Normal strain (ε): measures the change in size (elongation/contraction)
Normal Strain is a dimensionless quantity. Sometimes it is stated in terms of a ratio of
length units. Usually, for most engineering applications ε is very small, so measurements of
strain are in micrometers per meter (μm/m).

22. Shear Strain
Measures the change in shape (angle formed by the sides of a body)
θ = the angle in the deformed
state between the two initially
orthogonal reference lines
Shear strain: γ = π/ 2 -θ’ (in radians)
The shear strain is positive when the right angle decreases (θ’is smaller than π/ 2)

23. γ = tan θ = δ/ y ≅θ in radians provided that θ is very small
γ = π/ 2 - θ’ = θ (in radians)
Normal strain causes only a change in volume. on the other hand,
shear strain causes only change in shape.

24. The plate is deformed into the dashed shape as shown in figure. If in this deformed
Shape horizontal lines in the plate remain horizontal and do not change their length
Determine (a) the normal strain along the side AB and (b) the shear strain in the plate
Relative to the x and y axis.

25. The rigid beam is supported by a pin at A and wires BD and CE. If the
load P on the beam causes the end C to be displaced 10 mm downward,
determine the normal strain developed in wires CE and BD.

26. The plate is deformed uniformly into the shape shown by the dashed
lines. If at A, γxy = 0.0075 rad., while, εAB = εAF= 0 determine the average
shear strain at point G with respect to the x’ and y’ axes.