INTRODUCTION:

1.1Course Overview
1.2Mechanics of Materials
Lecture 1.3Aspects of Mechanics of Materials
1.4Methods of Analysis
1 1.5Engineering Design
1.6Review of Static Equilibrium
1.7Internal Force Resultants
1.8Problem Formulation and Solution
1.1 Course Overview:

1. Course Description: This is an introductory course to the basic concepts and principles of
Mechanics of Materials, which is considered to provide sufficient knowledge to enable any component to
be designed such that it will not fail within its service life. The course covers the following topics: stress
and strain concepts, axial load, statically indeterminate axially loaded members, thermal stress, torsion,
angle of twist, statically indeterminate torque-loaded members, bending, eccentric axial loading of beams,
transverse shear, shear flow in build-up members, curved beams, stress and strain transformation and thin-
walled pressure vessels.
2. The Course Objectives: The main objective of this course should be to develop the ability to:
1. Apply the concepts of stress, strain, torsion and bending and deflection of bar and beam in
engineering field.
2. Explain the stress, strain, torsion and bending
3. Calculate and determine the stress, strain and bending of solid body that subjected to external
and internal load
4. Use solid mechanic apparatus and analyse the experiments result
5. Work in-group that relates the basic theory with application of solid mechanics
3. Learning Outcome:
1. Define and discuss the components of stress, strain and deformation for any members subjected to
axial, torsional, moment and transverse loading.
2. Discuss and use engineer’s principal tasks, namely, the analysis and design of structures and
machines.
3. Discuss the many design considerations that an engineer should review before preparing a design.
4. Apply the various concepts and principles of Mechanics of Materials such as stress, strain, torsion
and bending and deflection of bar and beam in engineering field
5. Calculate and determine the stress, strain and bending of solid body that subjected to external
and internal load
6. Analyze a given problem in a simple and logical manner and to apply to its solution a few
fundamental and well-understood principles.
7. Use solid mechanic apparatus and analyse the experiments result.
 8. Work in-group that relates the basic theory with application of solid mechanics.
4. General Approach: In this course the study of the mechanics of materials is based on the understanding of
a few basic concepts and on the use of simplified models. This approach makes it possible to develop all
the necessary formulas in a rational and logical manner, and to indicate clearly the conditions under which
they can be safely applied to the analysis and design of actual engineering structures and machine
components.
5. Required Textbook & References:
1. Course notes and power point presentations.
2. A.C. Ugural, 2008, "Mechanics of Materials", 1st ed, Wiley& Sons.
3. T. A. Philpot, 2017, "Mechanics of Materials", 4th ed., Wiley & Sons.
1.2 Mechanics of Materials:

Engineering Mechanics: The three fundamental areas of engineering mechanics are statics, dynamics and
mechanics of materials.
 Statics and Dynamics are devoted primarily to the study of external forces and motions associated with
particles and rigid bodies (i.e. Idealized object in which any change of size or shape due to forces can

be neglected) without regard to small deformations associated with the application of load.
 Mechanics of Materials is a branch of applied mechanics that deals with the internal behaviour of
solid bodies that subjected to various types of loading. Mechanics of materials is the study of the
internal effects caused by external loads acting on real bodies that deform (meaning objects that can
stretch, bend, or twist). This course is also frequently called: "Strength of Materials", "Mechanics of
Deformable Bodies", or simply “Mechanics of Solids”. Of the states of matter, we are here concerned
only with the solid, with its ability to maintain its shape without the need of a container and to resist
continuous shear, tension, and compression.
 The “solid bodies” referred to here include shafts, bars, beams, plates, shells, and columns as well as
structures and machines that are assemblies of these components.
Mechanics of Materials Course Outlines: The Mechanics of Materials course provides engineers with the
means of analyzing and designing various machines and load-bearing structures involving the determination of
stresses and deformations. The course outlines are;
 Deals with internal behaviour of variously loaded solid bodies such as shafts, bars, beams, plates,
shells, columns, structures and machines that are assemblies of these components.
 Stress and deflection analyses, and the mechanical properties of materials are the main aspects of
mechanics of solid course.
 This course is based upon an understanding of the equilibrium of rigid bodies under the action of forces.
 This course concerned with the relationships between external loads (forces and moments) and internal
forces and deformations or displacements induced in the body.
 Stress and strain are fundamental quantities connected this subject.
 1.3Aspects of Mechanics of Materials:

 The study of Mechanics of Materials is based on upon an understanding of the equilibrium of bodies
under the action of forces by considering the relationships between external loads (forces and moments)
and internal forces and deformations or displacements included in the body. Stress and deflection
analyses, and the mechanical properties of materials are the main aspects of solid mechanics course.
1. Stress analysis: analysis of bodies under the action of external force, to determine the internal stress and
their deformation.
2. Mechanical properties of materials: consideration of such things as material strength, stability, fatigue
and brittle fracture etc.
 Regardless of the applications, a safe and successful design must address four mechanical
concerns:
a) Strength: The ability to sustain load. Is the object strong enough to carry the loads that will be applied to
it? Will it break or fracture? Will it continue to perform properly under repeated loadings?
b) Stiffness: Push per move; the ratio of deformation to associated load level. Will the object deflect or
deform so much that it cannot perform its intended function?
c) Stability: Will the object suddenly bend or buckle out of shape at some elevated load so that it can no
longer continue to perform its function? The ability of a structure to maintain position and geometry.
Instability involves collapse that is not initiated by material failure. External stability concerns the ability
of a structure’s supports to keep the structure in place; internal stability concerns a structure's ability to
maintain its shape.
d) Ductility: The amount of inelastic deformation before failure, often expressed relative to the amount of
elastic deformation.
 The main mechanical properties of materials can be briefly defined as:
 Strength: Material strength is measured by a stress level at which there is a permanent and
significant change in the material's load carrying ability. For example, the yield stress, or the
ultimate stress.
 Stiffness: Material stiffness is most commonly expressed in terms of the modulus of elasticity: the
ratio of stress to strain in the linear elastic range of material behaviour.
 Stability: As it is most commonly defined, the concept of stability applies to structural elements and
systems, but does not apply to materials, since instability is defined as a loss of load carrying ability
that is not initiated by material failure.
 Ductility: Material ductility can be measured by the amount of inelastic strain before failure
compared to the amount of elastic strain. It is commonly expressed as a ratio of the maximum strain
at failure divided by the yield strain.
1.4 Methods of Analysis:
Complete analysis of a structure under load requires the determination of stress, strain and deforation
througu the use of one of these two methods or approaches that are popular for analysis:
1.Mechanics of Materials Theory: also known as technical theory, or solid mechanics approach; is
simpler and makes assumptions that are based upon experimental evidence and the lessons of
engineering practice to makes a possible reasonable and quick solution of the basic problem i.e.;
determination of strain.
2.Theory of Elasticity: establishes every step rigorously from the mathematical point of view and
hence seeks to verify the validity of the assumptions introduced. This technique can provide ‘‘exact’’
results where configurations of loading and shape are simple. This approach yields solutions with
considerable difficulty.
Basic Principles of Analysis: generally, the basic principles of analysis through the use of the laws of
forces, the laws materials deformations, and the conditions of geometric compatibility. These
approach could be outlined in summary form as:
1.Equilibrium Conditions: The equations of static equilibrium of forces must be satisfied
throughout the member.
2.Material Behavior: The stress–strain or force–deformation relations (Hooke’s law) must apply to
the behavior of the material of which the member is made of.
3.Geometry of Deformation: The conditions of compatibility of deformations must be satisfied: that
is, each deformed portion of the member must fit together with adjacent portions.
 Boundary Conditions are used in the method of analysis to conform the loading conditions. The
stress and deformation obtained through the use of the three principles must conform to the
conditions of loading imposed at the boundaries of a member. This is known as satisfying the
boundary conditions. Applications of the preceding procedure are illustrated in the problems
presented as the subject unfolds. Note, however, that it is not always necessary to execute an analysis
in the exact order of steps listed previously.
 Instead of the equilibrium method, energy methods (based on strain energy theory) and
numerical methods (i.e.; finite element analysis) can be used as alternative methods of analysis. In
this course, you will find that your efforts are divided naturally into two parts:
1.Understanding the logical development of the concepts
2. Applying those concepts to practical situations
1.5 Engineering Design:

Engineering design: is the process of applying science and engineering methods to define a structure or
system in detail to permit its realization. The main objectives of a mechanical design process include
determination of: proper materials, dimensions, and the shape of the components of machine so that they
will support prescribed loads and perform without failure. Machine design is creating new or improved
machines to accomplish specific purposes. Design is the essence, art, and intent of engineering. A good
 design satisfies performance, cost, and safety requirements.
An optimum design is the best solution to a design problem within given restrictions. Efficiency of the
optimization may be gaged by such criteria as minimum weight or volume, optimum cost, and/or any other
standard deemed appropriate. For a design problem with many choices, a designer may often make decisions
on the basis of experience, to reduce the problem to a single variable. A solution to determine the optimum
result becomes straightforward in such a situation.
A plan for satisfying a need usually includes preparation of individual preliminary design. Each
preliminary design involves a thorough consideration of the loads and actions that the structure or machine
has to support. For each situation, an analysis is necessary. Design decisions, or choosing reasonable values
of the safety factors and material properties, are significant in the preliminary design process. We note that,
the design of numerous structures such as pressure vessels, space missiles, aircrafts, dome roofs, and bridge
decks is based upon the theories of plates and shells.
Role of Analysis in Design: This course notes provides an elementary treatment of the concept of “design to
meet strength requirements” as those requirements relate to individual machine or structural
components. That is, the geometrical configuration and material of a component are preselected and the
applied loads are specified. Then, the basic formulas for stress are employed to select members of adequate
size in each case. The role of analysis in design may be observed best in examining the phases of a design
process. The following is rational procedure in the design for strength of a load-carrying member:
1. Evaluate the most likely mode of possible failure of the member. Failure criteria that predict the
various modes of failure under anticipated conditions of service are discussed in Lecture 10.
2. Determine the expressions relating applied loading to such effects as stress, strain, and
deformation. Often, the member under consideration and conditions of loading are so significant or so
amenable to solution as to have been the subject of prior analysis. For these situations, textbooks,
handbooks, journal articles, and technical papers are good sources of information. Where the situation
is unique, a mathematical derivation specific to the case at hand is required.
3. Determine the maximum usable value of stress, strain, deformation and energy that could
conceivably cause failure. This value is obtained either by reference to compilations of material
properties or by experimental means such as simple tension test.
4. Employ this value in connection with the relationship derived in step 2 or, if required, in any of the
formulas associated with the various theories of failure.
5. Select the factor of safety as outlined in the section 2.4. The design factor of safety also reflects the
consequences of failure; i.e.; the possibility that failure will result in loss of human life or injury or in
costly repairs or danger to other components of the overall system
1.6 Review of Static Equilibrium:

a)Types of loads: External loads and Internal loads.

1.External loads: all forces acting on a body, including surface forces, body forces and reactive forces
caused by supports; are considered to be external forces.
 Surfaces forces: is of the concentrated type when it acts at a point, or a distributed load uniformly or
nonuniformly over a finite area; both acting on the surface of a body.
 Body forces: are associated with the mass rather than the surfaces of a body, and are distributed
throughout the volume of a body. Gravitational, magnetic, and inertia forces are all body forces.
They are specified in terms of force per unit volume.
 Reaction forces: caused by the supports, some of the most commonly used idealizations for 2D and
3D supports, as well as interconnections between members or elements of a structure, are illustrated
in Table 1-1.
Table 1-1 Reaction forces and support conditions in 2D or 3D static analysis
Type of support or connection Simplified sketch of support Display of restraint forces
1.Roller Support: A single reaction force R is developed and is normal to the rolling surface; force R
opposes motion into or away from the rolling surface. The rolling surface may be horizontal, vertical, or
inclined at some angle θ . If friction is present, then include a force F opposing the movement of the
support and tangential to the rolling surface. In 3D, the roller moves in the x-z plane and reaction Ry is
normal to that plane.
a.Two-dimensional roller support
(friction force F=0 for
smooth rolling surface)

Horizontal roller support [(1.1),

(1.2)]; or alternate representation
as rocker support [(1.3)] Both

Bridge with roller support (see 1.1, 1.2) downward and uplift motions are
restrained. b.Three-dimensional roller
Bridge with rocker support (see 1.3)
Vertical roller restraints support (friction force F=0
Rotated or inclined roller support for smooth rolling surface;
reaction Ry acts normal to
3D roller support
plane x-z on which roller

translates)
2.Pin Support: A single resultant force, usually shown using two rectangular components Rx and Ry in 2D
but three components in 3D, resists motion in any direction normal to the pin. The pin support cannot
resist moment, and the pin is free to rotate about the z axis. In 3D, the pin becomes a ball and-socket
joint or support.
2D pin a. Two-dimensional pin

Ball-and-socket joint model (camera

mount, 3D) support

Ball-and-socket joint model b. Three-dimensional pin

Hip prosthesis for hip replacement

support
3.Sliding Support: A support that translates without rotation is a sliding support. Examples are a collar
sliding along a sleeve or a flange moving within a slot. Reactions in 2D are a force Rx normal to the
sleeve and a moment Mz representing resistance to rotation relative to the sleeve. In 3D, the sliding
support translates on frictionless plane y-z and reaction moment components My and Mz prevent rotation
relative to that plane.
a.2D sliding support (support
translates on frictionless path
along − y or + y direction)

2D sliding support Friction F

Sliding support for column light stand opposes motion in 1y direction in
2D along sliding surface; F is
zero if smooth surface is
assumed. In 3D, add restraint
 moment Mx to prevent rotation b.Three-dimensional sliding
about x axis. support (support translates on
frictionless y-z plane)

4.Fixed Support: No translation or rotation occurs between member and support in a fixed
support. This requires three reaction components in 2D: force components Rx and Ry and
moment Mz. In 3D, three force reaction components and three moment reaction components are
required.
a. 2D fixed support

Horizontal member
Steel bollard anchored in concrete

Fixed support at base of sign post

Vertical member
b. 3D fixed support

Column bolted to footing

5.Elastic or Spring Support: In 2D, there may be a longitudinal or normal translational spring or a
combination of both. For linear springs, the support reaction at the base of the spring is the product of
 the spring constant k times the displacement δ in the direction of the spring axis. If joint A translates in
+x (δx) and +y (δy) directions, reaction forces Rx and Ry are created in -x and -y directions, respectively,
at the supports of linear translational springs. Alternatively, the support may be pinned for translation
but have moment spring kr for rotation. If joint A rotates about the +z axis ( z), reaction moment Mz is
created in the –z direction at the base of the rotational spring. In 3D, a fully elastic support consists of
three translational springs (kx, ky, kz) and three rotational springs (krx, kry, krz), and an arbitrary joint
displacement results in three reaction forces and three reaction moments. In the limit, as each spring
constant value approaches infinity, the elastic support becomes a fully fixed support.
In 3D, add spring in +z direction a. Translational spring (k) in 2D
kz with reaction force
R z=−k z δ z . In 3D, add
rotational flexural spring about
+y direction kry with reaction
moment M y =−k ry θ y and add
rotational torsional spring about
+x direction with reaction
Translational spring support for heavy
moment M x =−k rx θ x .
equipment

Torsion springs are found in

window shades and as part of the
lift mechanism in power garage
door-opening systems. b. Rotational spring (kr) in 2D
Rotational spring in a clothespin
6.Wheel on Rail Support: This support is a particular form of the 3D roller support. Now general
movement in the x-z plane is constrained by normal force Ry and lateral force Rx, both acting normal to
the rail or slot on which the wheel travels. If friction is considered, friction force F is added along the
rail in the direction opposing the wheel translation.
 Wheel rolls on rail or in slot
along z axis; friction force
opposing motion is neglected; Rx
is lateral constraint force; Ry is

normal force.
Cross section through guide
rail
7.Thrust-Bearing Support: A thrust bearing constrains translational motion along the shaft axis while
allowing rotary motion to occur about that axis.

Pillow block bearing

Thrust bearing has support reaction force
(Rx, Ry, Rz) and reaction moment
Journal bearing has no axial
components (My, Mz) & M x =0.
thrust reaction force R x =0 &
M x =0

2. Internal loads: are the forces that hold together the particles forming the body (forces resulted by the
interaction between the constituent material particles of the body). Will be discussed in details in
section 1.7.
 Unless otherwise stated, we assume in this course that body forces can be neglected and that forces
are applied steadily and slowly. The latter is referred to as static loading.
b)Conditions of Equilibrium: When a system of forces acting upon a body has zero resultant, the body is
said to be in force equilibrium. consider the equilibrium of a body in space, the equations of statics under
the equilibrium are:

∑ F x =0 ∑ F y =0 ∑ F z=0
∑ M x =0 ∑ M y =0 ∑ M z=0 … … … … … … ..(1)
In other words, for a body to be in static equilibrium, the sum of all forces acting upon a body in any
 direction is zero and also the sum of all moments taken about any axis is also zero.
c)Planar Equations of Equilibrium: For a planar body to be in equilibrium, any one of the following sets of
3 equations may be used to solve for the unknown variables.
1. ∑ F x =0 ∑ F y =0 ∑ M a=0, where the resultant moment is with respect to any axis z or any point
a in the xy-plane, or
2. ∑ F x =0 ∑ M a=0 ∑ M b=0, provided that the line connecting the points a and b is not
perpendicular to the x axis, or
3. ∑ M a =0 ∑ M b=0 ∑ M c =0 , where points a, b, and c are not collinear
d)Free Body-Diagrams:
1.Select the free body to be used.
2.Detach this body from its supports and separate it from any other bodies. (If internal force resultants are to
be found, use the method of sections).
3.Show on the sketch all of the external forces acting on the body. Location, magnitude, and direction of each
force should be marked on the sketch.
4.Label significant points and include
Table 1.1 Basic Units.
dimensions. Any other detail, however, should
be omitted.
e)Systems of Units: Both the SI and the USCS
are used in the examples and problems of
this course.
1) International System of Units (SI unit):
basic units are kg, sec, m, see Table 1.1.
2) U.S. Customary System (USCS) : basic
units are lb, sec, ft.
1.7 Internal Force Resultants:
Load: is a general term that can mean
either a force, a torque or a moment,
or any combination of them.
Depending on how it is applied, a force
can cause either an axial load or a
torsion load or a bending load or any
combinations of these loads on a
member. The resultant of the internal Figure 1.1 Method of section: (a) body acted on by external
forces, (b) internal forces acting on a plane.
forces for an axially loaded member is

normal to a section cut perpendicular to the member axis. The

body responds to the application of external forces by deforming
and by developing an internal force system that hold together the
particles forming the body.
Method of Sections: method of sections together with equations of
equilibrium are used to determine the internal axial forces. We now
deal with familiar approach could the method of sections to predict
the internal force system as shown in Figure 1.1. The steps of
method of sections are:
1.Isolate the bodies. Sketch the isolated body and show all Figure 1.2 Three-dimensional forces
and moments acting on a section of a
external forces acting on it: draw a free-body diagram. member.
2.Apply the equations of equilibrium to the diagram to determine
the unknown external forces.
3.Cut the body at a section of interest by an imaginary plane,
isolate one of the segments, and repeat step 2 for that segment.
If the entire body is in equilibrium, any part of it must be in
equilibrium. That is, there must be internal forces transmitted
across the cut sections.
Components of Internal Forces: external loads or forces are
Figure 1.3 Force and moment
balanced by internal loads or forces. Each internal force and
components in two dimensions.
moment reflects a different effect of the applied loading on
the member. These effects can be described as follows (Figure 1.2):
1.The Axial Force, Fx: acting normal to the member at the cut section and is denoted by P or N. It tends to
elongate or contract the member (termed axial tension or axial compression).
2.The Shear Force Fy and Fz: acting parallel or tangent to the cross section. This is denoted by Vy and/or Vz.
3.The Torsional or Twisting Moment, Mx: is responsible for twisting the member about its axis and is
 denoted by T.
4.The Bending Moment, M: cause the member to bend and is denoted by My and Mz.
General Sign Convention: as shown in Figures 1.2 and 1.3, the following sign convention roles are
followed through this course:
 When both the outer normal and internal force (and moment) vector component point in a positive ( +¿ + ¿¿
¿

) (or negative−¿ −¿¿

¿ ) coordinate direction, the force or moment is defined as positive (+).Hence the
tensile force at a section is positive (+). Observe that the sense of a positive twisting moment vector is
the same as that of the positive axial force vector.
 When negatively (-) directed component acts on a positive (+) or [vice versa (−¿ + ¿¿
¿ or −¿
+ ¿¿
¿ )], the force
or moment is negative. Hence the compressive force at a section is negative (-).
 Figure 1.3 represent a negative (-) shear force for a beam.
 Interestingly, the general sign convention applies to the stress components as well in the next chapters.
1.8 Problems Formulations and Solution:

The following outline may help in the formulation and solution of a problem:

1. Given: Define the problem and state briefly what is known.

2. Find: state consistently what is to be determine.
3. Assumptions: list simplifying idealization to be made.
4. Solution: apply the appropriate equations to determine the unknowns.
5. Comments: discuss the results briefly.
Significant Figures (Digits):

 Numbers beginning with 1 are recorded to 4- significant digits.

 All other numbers (that begin with 2 through 9) are recorded to 3- significant
digits.
Example 1.1:
Given: An L-shaped 3-D pipe assembly. The pipe subjected to vertical force of 50 N and couple moment of
70 N·m at end A. It is fixed to the wall at C.

Find: The resultant internal loadings acting on cross section

at O of pipe?

Assumptions: The mass of pipes AB & BC is 2.45 kg/m3.

SOLUTIONS:
W AB =2.45 ( 0.50 ) ( 9.81 )=12.02 N

W BO=2.45 ( 0.40 )( 9.81 ) =9.61 N

∑ F x =0 ⟹ F=0

Figure 1.4 (a) Pipe assembly; (b) FBD

of part ABO.

∑ F y =0⟹ V y −12.02−9.61−180=0 ⟹ V y =101.6 N

∑ F z=0 ⟹ V z =0

∑ M x =0 ⟹ T +12.02 ( 0.25 ) +80 ( 0.5 )=0 ⟹ T =−43 N . m

∑ M y =0 ⟹ M y =0

∑ M y =0 ⟹ M z−20−80 × 4−12.02 ( 0.4 )−9.6 ( 0.2 )=0 ⟹ M z =58.7 N .m

Example 1.2:
Given: A stepped circular shaft is fixed at A and has three gears that transmit the torques shown in Figure 1.4. Use
properly drawn free-body diagrams in your solution.

Find: the reaction torque at A, then find the internal torsional moments in segments AB, BC, and CD.

Assumptions: Mass of stepped circular shaft is neglected.

SOLUTION:
1)Draw the FBD of the overall shaft structure
(Figure 1.5).
2)Sum the moments about the x axis to find the
reaction moment MAx.
M AX −1900+1000+550 ⟹ M AX =350 N . m
3)Find the internal torsional moments in each
segment of the shaft:
Find the internal torque TAB (Figure 1.6a): Figure 1.4 Stepped circular shaft in torsion.
T AB=−M AX =−350 N .m
Find the internal torque TBC (Figure 1.6b):
T BC =−M AX + 1900=1550 N . m
Find the internal torque TCD (Figure 1.6c): Figure 1.5 FBD of overall shaft.
T CD =550 N . m

Example 1.2:
Given: The plane frame in Figure 1.7, is initially, member DF
has been replaced with a roller support at D. Moment MA
¿ 380 N . m is applied at pin-supported joint A, and load FB
¿ 200 N is applied at joint B. A uniform load with intensity q1
=160 N/m acts on member BC, and a linearly distributed load
with peak intensity q0 =80 N/m is applied downward
on member ED. Use
a=3 m ,b=2 m, c=6 m, d=2.5 m.
Find: the support reactions at joints A and D, then
solve for internal forces at the top of member BC.

Solution:

1) Draw the FBD of the overall frame (Figure

1.8).
2) Determine the statically equivalent concentrated forces. Distributed forces are replaced by their
statically equivalents (Fq0 and Fq1) as:
1
F q 0= ( 80 )( 6 )=240 N , F q 1=( 160 ) ( 2 )=320 N
2
3) Sum the moments about A to find reaction force Dy:

380
↺ ∑ M A =0 ⟹ 240 ( 6.5 )−D y ( 2.5 )− (320 )( 4 ) +200 ( 4/5 )( 3 ) −380 ∴ D y = =152 N
2.5

4) Sum the forces in the x and y directions to find the reaction

forces at A.

5) Find the internal forces and moment at the top of member BC.