This document provides an introduction and overview of mechanics of materials. It defines key terms like stress, strain, normal stress, shear stress, factor of safety, and allowable stress. It also gives examples of calculating stresses in structural members subjected to various loads. The document is an introductory reading for a mechanics of materials course that will analyze the relationship between external forces and internal stresses and strains in structural elements.

1. Assistant Professor
Department of Mechanical Engineering
National University of Science and Technology
College of E M E
ME211-Mechanics of Materials 1
Contact: naveeddin@ceme.nust.edu.pk

2. Introduction to mechanics of materials
Mechanics of material is a study of the relationship between the external
loads applied to the body and the stress and strain caused by the internal
loads within the body.
In this course we shall be concerned with what might be called the internal
effects of forces acting on a body. The bodies themselves will no longer be
considered to be perfectly rigid as was assumed in statics; instead, the
calculation of the deformations of various bodies under a variety of loads will
be one of our primary concerns.
Basic Definitions
Equilibrium: A state of no acceleration, in either translational or rotational
senses.
Scalar: A quantity which has only magnitude. Examples include mass and
area.
Vector: A quantity which has both magnitude and direction e.g. displacement
force.

3. Force: The interaction between bodies which gives rise to an acceleration or
to the deformation of the body.
Moment: The product of the magnitude of a force and the perpendicular
distance of its line of action from a particular point. (Also a vector.)
Couple: It consists of two forces equal in magnitude but opposite in direction
whose line of action are parallel but no collinear.
1. Surface forces:
(a) caused by the direct contact of one body with the surface of another.
Forces are distributed over the area of contact between bodies. If
contacted area is small as compared to the area of the body then surface
force can be idealised as concentrated force (or point load).e.g. bicycle
wheel contact with surface.
(b) If load is linearly distributed along a narrow area over a specified length.
To deal with distributed loads, the resultant force is equivalent to the
area under the distributed loading curve and act through the centroid of
that area
A body is subject to only two types of external loadings
Introduction to mechanics of materials

4. 2. Body Forces: No direct contact e.g. effects caused by earth’s gravitation
The surface force that develop at the supports or point of contact between bodies
are called reactions. If the support prevents motion (translation, rotation) in a
given direction , then a force or a moment must be developed on the member in
that direction.
Support Reactions
Introduction to mechanics of materials

5. Type of Supports
Introduction to mechanics of materials

6. Equation of equilibrium
Equilibrium of a body requires both a balance of forces, to prevent the body from
translating Or having accelerated motion along a straight or curved path, and a
balance of moments to Prevent body from rotating.
Sum of all forces acting on the body and sum
of the moments of all forces about any point
O either on or off the body must be zero
𝐹 = 0 𝑀 𝑜 = 0
For a 3-D body at rest the coordinate system used is the x-y-z Cartesian system, in
which the definition of positive moments is given by the right hand rule that states
that moments are positive is their sense is counterclockwise as shown in Figure 1
3-D positive axis system (Right-Hand) 2-D positive axis system
Figure 1 Figure 2
Introduction to mechanics of materials

7. Force equilibrium equation is given as:
and Moment equilibrium as
The 2D x & y -axis system looks like Fig 2: For a two dimensional body in the xy-axis
system, the 3D equilibrium Eqs. simplify to:
Introduction to mechanics of materials

8. Free- body diagram
A sketch of the outlines shape (or simplified line sketch of the structure) of the
body isolated from its surrounding. On this sketch all forces and couple moments
that the surrounding exert on the body together with any support reactions must be
shown correctly. Only then applying equilibrium equations will be useful.
Internal loadings
These internal loading acting on a specific region within the body can be attained
by the Method of Section.
Method of Section: Imaginary cut is made through the body in the region where
the internal loading is to be determined. The two parts are separated and a free
body diagram of one of the parts is drawn. Only then applying equilibrium would
enable us to relate the resultant internal force and moment to the external
forces. Point O is often chosen as the
centroid of the sectioned area
Apply Equilibrium at this stage
Introduction to mechanics of materials

9. Three Dimensional Loading:
Normal force, N: This force act perpendicular to the area.
Shear Force, V: This force lies in the plane of the area (parallel)
Torsional Moment or Torque, T: This torque is developed when the external loads
tend to twist one segment of the body with respect to the other
Bending Moment, M: This moment is developed when the external loads tend to
bend the body.
Four types of internal loadings can be defined:
Introduction to mechanics of materials

10. If the body is subjected to a coplanar system of forces then only
normal force N, shear force V, and bending moment Mo
components will exist at the section.
Coplanar Loading
Introduction to mechanics of materials

11. • After sectioning, decide which segment of the body will be studied. If this
segment has a support or connection than a free body diagram for the entire
body must be done first to calculate the reactions of these supports.
• Pass an imaginary section through the body at the point where the resultant
internal loadings are to be determined and put the three unknowns (V, Mo,
N) at the cut section. Then apply equilibrium.
• Moments should be summed at the cut section. This will eliminates
unknown forces V and N and solve directly for Mo.
Procedure of Analysis

12. A
E
B
C
D
1.5m
1m 1m 1m
500Kg
Determine the resultant internal loading
acting on the cross section of the boom at
point E
Introduction to mechanics of materials
Determine the reactions at A and B
Examples

13. Introduction to mechanics of materials
Determine the resultant internal loading acting on the cross section of the boom at
point C
The shaft is supported by a smooth thrust bearing at A and a smooth
journal bearing at B. Determine the resultant internal loadings acting on
the cross section at C.

14. 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

15. Internal actions exposed by making imaginary cut at section mn
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

16. 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

17. 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 road is subject to the external
uniformly distributed loading along its length of 8KN/m
A
2m 3m

18. 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.

19. 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)

20. 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

21. 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

22. Determine the largest internal shear
force resisted by the bolt.
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.

23. 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?

24. 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

25. 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

26. 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

27. 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