Mechanics of Machine lab manual c 20 20ME31P

Vidya Vikas Educational Trust (R),
Vidya Vikas Polytechnic
27-128, Mysore - Bannur Road Alanahally,Alanahally Post, Mysuru, Karnataka 570028
Praperd by: Mr Thanmay J.S, H.O.D Mechanical Engineering VVETP, Mysore 1 |T o t a l P a g e ’ s ( 4 1 )
Department of Mechanical [General]
Laboratory Manual
Subject : Mechanics of Materials
Subject Code : 20ME31P
Semester : 3rd
Semester
Name of the Student: …………………………………………….
Register Number: …………………………………………….

APPENDIX 5 (Certificate issued by guide)
Name of the Institution: VIDYA VIKAS POLYTECHNIC
Address with pin code: 27-128, Mysore - Bannur Road Alanahally,Alanahally Post, Mysuru, Karnataka 570028
Department: MECHANICAL ENGINEERING (General)
CERTIFICATE
Certified that this Practical Record entitled “Mechanics of Materials 20ME31P” which is being submitted
by Mr.………………………….……………………bearing Register Number…..………………………, is a
bonafide student of Mechanical Engineering Department, studying in Third Semester in our Institution and
has fulfilment the Laboratory prescribed by Department of Technical Education, Bangalore during the year
2021-2021.
It is certified that all corrections/suggestions indicated for internal Assessment have been incorporated in the
Report.
Signature of the Staff In-charge
Signature of H.O.D
CIE (Internal Assessment Marks Obtained in Words):…………..……………………………………………
External Examiner 1:………………………………
Dept. …………………………College…….…………………
External Examiner 2:………………………………
Dept. …………………………College…….…………………

Contents
Section 01: Strength of Material
A. Force-Types of Forces-Resolution of forces (Introduction)
1) Resolution of forces- Analytical Method
2) Resolution of forces by Graphical Method
B. Verification of Forces by triangle law, parallelogram law and Lami's theorem
C. Balancing of Rotating Mass
D. Types of Loads-Tensile, Compression, Shear, Impact, Stress and Strain types, Hooks Law & Young's Modulus
E. Conduct tensile test for the given Specimen and Determine Stress-Strain-Young’s Modulus, Yield Stress-
Maximum Stress – Breaking Stress - % Elongation in Length and % Reduction in Area Also, Draw Stress - Strain
Diagram for the above Parameters.
F. Conduct Compression test for the given Specimen and Determine Stress - Strain -Young’s Modulus, Yield Stress
Maximum Stress - % Reduction in Length and % Increase in Area Also, Draw Stress- Strain Diagram for the
above Parameters.
G. Conduct Shear test for the given specimen-Problems on Members subjected to combined Stresses
H. Ansys Transient Simulation of Tensile, Compression & Shear Test
I. Conduct Bending test for the given specimen
1) Types of Beams-Types of Loads acting on Beams-Concept of Shear force-Bending moment
2) Draw Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for Cantilever subjected to Point
Load and Uniformly Distributed loads (UDL)
3) Draw Shear force Diagram(SFD) and Bending Moment Diagram(BMD) for Cantilever subjected to Point
Load and Uniformly Distributed loads (UDL)
4) Draw Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for a simply supported beam
subjected to Point Load and Uniformly Distributed loads (UDL)
5) Draw SFD and BMD for Simply supported and Cantilever beam subjected to Point Load and UDL Draw
Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for a Simply supported beam subjected
to Point Load and Uniformly Distributed loads (UDL)
6) Draw SFD and BMD for Simply supported and Cantilever beam subjected to Point Load and UDL Draw
Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for a Simply supported beam subjected
to Point Load and Uniformly Distributed loads (UDL)
J. Pure Bending-Assumptions-Neutral Axis-Bending Equation-Problems on Bending Equation
Section 02: Finite Element Methods (FEM)
A. Introduction to Finite Element Methods (FEM), Need-Back Ground
B. Methods employed in FEM- Steps in FEM
C. Advantages and Disadvantages, Limitations,
D. Applications of FEM-Concept of Discontinuity
1) Phases of FEA (Finite Element Analysis)
2) Discretization Process
3) Meshing–Element type
4) Stiffness Matrix of a Bar Element
5) Global Stiffness Matrix-Properties of stiffness matrix
6) Boundary Conditions – Methods – Types
E. Problems on 1-D elements Validate using FEM software (Ex: Ansys)
F. Problems on 2-D elements Validate using FEM software (Ex: Ansys)

Section 01: Strength of Material
Force
Force is the action of one body on another. It may be defined as an action which changes or tends to change the
state of rest or of uniform motion of body. For representing the force acting on the body, the magnitude of the force, its
point of action and direction of its action should be known. There are different types of forces such as gravitational,
frictional, magnetic, inertia or those caused by mass and acceleration.
Or
The action of one body and another, which changes or tends to change the state of rest or of uniform motion of
body is called as force.
The three requisites for representing the force acting on the body are:
• Magnitude of force
• It points of action, and
• Direction of its action
Or
Force is something which changes or tends to change the state of rest or of uniform motion of a body in a straight
line. Force is the direct or indirect action of one body on another. The bodies may be in direct contact with each other
causing direct motion or separated by distance but subjected to gravitational effects.
There are different kinds of forces such as gravitational, frictional, magnetic, inertia or those cause by mass and
acceleration. A static force is the one which is caused without relative acceleration of the bodies in question.
The force has a magnitude and direction, therefore, it is vector. While the directions of the force is measured in
absolute terms of angle relative to a co-ordinate system, the magnitude is measured in different units depending on the
situation.
Effects of a Force
A force may produce the following effects in a body, on which it acts:
1. It may change the motion of a body i.e. if a body is at rest, the force may set it in motion. And if the body is
already in motion, the force may accelerate or retard it.
2. It may retard the forces, already acting on a body, thus bringing it to rest or in equilibrium.
3. It may give rise to the internal stresses in the body, on which it acts.
Characteristics of a Force
To know the effect of force on a body, the following elements of force should be known.
1. Magnitude (i.e. 2 N, 5 kN, 10 kN etc.)
2. Direction or line of action.
3. Sense or nature (push or pull).
4. Point of application.
Representation of Force
Method Definition Representation
a) Vector
representation
Vector representation. A force can be represented
graphically by a vector (i.e, Force and Direction)
b) Bow’s notation.
Bow’s notation. It is a method of designating a force by
writing two capital letters one on either side of the force a
shown in Fig, where force P1 (200 N) is represented by AB
and force P2 (100 N) by CD.

Classification of Force
There are several ways in which forces can be classified. Some of the important classifications are given as under:
1. According to the effect produced by the force:
a) External force: When a force is applied external to a body it is called external force.
b) Internal force. The resistance to deformation, or change of shape, exerted by the material of a body is called
an internal force.
c) Active force. An active force is one which causes a body to move or change its shape.
d) Passive force. A force which prevents the motion, deformation of a body is called a passive force.
2. According to nature of the force:
a) Action and reaction. Whenever there are two bodies in contact, each exerts a force on the other. Out of these
forces one is called action and other is called reaction. Action and reaction are equal and opposite.
b) Attraction and repulsion. These are actually non-contacting forces exerted by one body or another without
any visible medium transmission such as magnetic forces.
c) Tension and thrust. When a body is dragged with a string the force communicated to the body by the string is
called the tension while, if we push the body with a rod, the force exerted on the body is called a thrust.
3. According to whether the force acts at a point or is distributed over a large area.
a) Concentrated force. The force whose point of application is so small that it may be considered as a point is
called a concentrated force.
b) Distributed force. A distributed force is one whose place of application is area.
4. According to whether the force acts at a distance or by contact.
a) Non-contacting forces or forces at a distance. Magnetic, electrical and gravitational forces are examples of
non-contacting forces or forces at a distance.
b) Contacting forces or forces by contact. The pressure of steam in a cylinder and that of the wheels of a
locomotive on the supporting rails are examples of contacting forces.
Force System
A force system is a collection of forces acting on a body in one or more planes. According to the relative positions of
the lines of action of the forces, the forces may be classified as follows:
1. Coplanar concurrent collinear force system. It is the simplest force system and includes those forces whose
vectors lie along the same straight line, as shown in the below figure.
2. Coplanar concurrent non-parallel force system. Forces whose lines of action pass through a common point
are called concurrent forces. In this system lines of action of all the forces meet at a point but have different
directions in the same plane as shown in below figure.
3. Coplanar non-concurrent parallel force system. In this system, the lines of action of all the forces lie in the
same plane and are parallel to each other but may not have same direction as shown in below figure.

4. Coplanar non-concurrent non-parallel force system. Such a system exists where the lines of action of all
forces lie in the same plane but do not pass through a common point. below figure shows such a force system
5. Non-coplanar concurrent force system. This system is evident where the lines of action of all forces do not
lie in the same plane but do pass through a common point. An example of this force system is the forces in the
legs of tripod support for camera
6. Non-coplanar non-concurrent force system. Where the lines of action of all forces do not lie in the same
plane and do not pass through a common point, a non-coplanar non-concurrent system is present.
Components of Force (Resolution of Forces)
As two forces acting simultaneously on a particle acting along directions inclined to each other can be replaced by a
single force which produces the same effect as the given force, similarly, a single force can be replaced by two forces
acting in directions which will produce the same effect as the given force. This breaking up of a force into two parts is
called the resolution of a force. The force which is broken into two parts is called the resolved force and the parts are
called component forces or the resolute. Generally, a force is resolved into the following two types of components:
1. Mutually perpendicular components
2. Non-perpendicular components.
Mutually perpendicular components:
Let the force P to be resolved is represented in magnitude and direction by oc in below figure. Let Px is the component
of force P in the direction oa making an angle α with the direction oc of the force. Complete the rectangle oacb. Then
the other component Py at right angle to Px will be represented by ob which is also equal to ac.
From the right-angled triangle oac
Non-perpendicular components.
Referring to the figure below. Let oc represents the given force P in magnitude and direction to some scale. Draw oa
and ob making angle α and β with oc. Through c draw ca parallel to ob and cb parallel to oa to complete the parallelogram
oacb. Then the vectors oa and ob represent in magnitude and direction (to the same scale) the components P1 and P2
respectively. Now from the triangle oac, by applying sine rule,

Principle of Resolved Parts
The principle of resolved parts states: “The sum of the resolved parts of two forces acting at a point in any given direction
is equal to the resolved parts of their resultant in that direction. Refer Figure below. Let the two forces P and Q be
represented by the side’s oa and ob of the parallelogram oacb and the resultant R of these two forces is given by the
diagonal oc in magnitude and direction. Let ox is the given direction. Draw bf, ae, cd and ag perpendicular to cd. Now
from the two triangles obf and acg which are same in all respects, we get
But oe, of and od represent the resolved components or parts of the forces P, Q and R respectively in the direction of
ox. It may be noted that this principle holds good for any number of forces.
Resultant of Several Coplanar Concurrent Forces
To determine the resultant of a number of coplanar concurrent forces any of the following two methods may be used:
1. Analytical method (Principle of resolved parts) or Resultant by analytical method.
2. Graphical method (Polygon law of forces) or Resultant by graphical method
Analytical method (Principle of resolved parts) or Resultant by analytical method.
It may be noted that while solving problems proper care must be taken about the signs (+ve or –ve) of the resolved
parts. Following sign conventions may be kept in view:

Graphical method (Polygon law of forces) or Resultant by graphical method
1) The below figure shows the forces P1, P2 and P3 simultaneously acting at a particle O.
2) Draw a vector ab equal to force P1 to some suitable scale and parallel to the line of action of P1.
3) From ‘b’ draw vector bc to represent force P3 in magnitude and direction.
4) Now from ‘c’ draw vector cd equal and parallel to force P3.
5) Join ad which gives the required resultant in magnitude and direction, the direction being a to d as shown in the
vector diagram.

K. Verification of Forces
Experimental verification of triangle law, parallelogram law and Lami's theorem
Two smooth small pulleys are fixed, one each at the top corners of a drawing board kept vertically on a wall as shown in
above figure. The pulleys should move freely without any friction. A light string is made to pass over both the pulleys.
Two slotted weights P and Q (of the order of 50 g) are taken and are tied to the two free ends of the string. Another short
string is tied to the center of the first string at O. A third slotted weight R is attached to the free end of the short string. The
weights P, Q and R, are adjusted such that the system is at rest.
The point O is in equilibrium under the action of the three forces P, Q and R acting along the strings. Now, a sheet
of white paper is held just behind the string without touching them. The common knot O and the directions of OA, OB and
OD are marked to represent in magnitude, the three forces P, Q and R on any convenient scale (like 50 g = 1 cm).
The experiment is repeated for different values of P, Q and R and the values are tabulated.
To verify parallelogram law
To determine the resultant of two forces P and Q, a parallelogram OACB is completed, taking OA representing P, OB
representing Q and the diagonal OC gives the resultant. The length of the diagonal OC and the angle COD are measured
and tabulated (Table).
OC is the resultant R′ of P and Q. Since O is at rest, this resultant R′ must be equal to the third force R (equilibrant)
which acts in the opposite direction. OC = OD. Also, both OC and OD are acting in the opposite direction. ∠COD must
be equal to 180º.
If OC = OD and ∠COD = 180º, one can say that parallelogram law of force is verified experimentally.
Verification of parallelogram law

To verify Triangle Law
According to triangle law of forces, the resultant of P (= OA = BC) and Q (OB) is represented in magnitude and direction
by OC which is taken in the reverse direction. Alternatively, to verify the triangle law of forces, the ratios,
𝑃
𝑂𝐴
=
𝑄
𝑂𝐵
=
𝑅
𝑂𝐶
are calculated and are tabulated. It will be found out that, all the three ratios are equal, which proves the triangle law of
forces experimentally.
To verify Lami's theorem
To verify Lami's theorem, the angles between the three forces, P, Q and R i.e, BOD = α, AOD = β and AOB = γ are
measured using protractor and tabulated, mathematically,
𝑃
sin𝛼
=
𝑄
sin 𝛽
=
𝑅
sin𝛾
are calculated and it is found that all the three ratios are equal and this verifies the Lami's theorem.
Balancing of Rotating Masses
Whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal force, whose effect is to bend the shaft
and to produce vibrations in it. In order to prevent the effect of centrifugal force, another mass is attached to the opposite
side of the shaft, at such a position so as to balance the effect of the centrifugal force of the first mass. This is done in
such a way that the centrifugal forces of both the masses are made to be equal and opposite. The process of providing
the second mass in order to counteract the effect of the centrifugal force of the first mass is called balancing of rotating
masses.
The following cases are important from the subject point of view:
1. Balancing of a single rotating mass by a single mass rotating in the same plane.
2. Balancing of different masses rotating in the same plane.
3. Balancing of different masses rotating in different planes.

Balancing of Several Masses Rotating in the Same Plane
Consider any number of masses (say four) of magnitude (m1, m2, m3 and m4) at distances of (r1, r2, r3 and r4) from
the axis of the rotating shaft. Let (θ1, θ2, θ3 and θ4) be the angles of these masses with the horizontal line OX, as
shown in below figure. Let these masses rotate about an axis through (O) and perpendicular to the plane of paper, with
a constant angular velocity of (ω) rad/s.
The magnitude and position of the balancing mass may be found out analytically or graphically as discussed
below:
1. Analytical method
Each mass produces a centrifugal force acting radially outwards from the axis of rotation. Let F be the vector sum of
these forces.
The rotor is said to be statically balanced if the vector sum F is zero.
If F is not zero, i.e., the rotor is unbalanced, then produce a counterweight (balance weight) of mass mc, at radius rc to
balance the rotor so that
The magnitude of either mc or rc may be selected and of other can be calculated. In general, if is the vector sum
of m1.r1, m2.r2, m3.r3, m4.r4, etc., then
To solve these equation by mathematically, divide each force into its x and z components,
and
and
Squaring and adding (i) and (ii),
Dividing (ii) by (i),
The signs of the numerator and denominator of this function identify the quadrant of the angle.

2. Graphical method
First of all, draw the space diagram with the positions of the several masses, and Find out the centrifugal force (or
product of the mass and radius of rotation) exerted by each mass on the rotating shaft.
Sl No Mass (m) Radius of rotation (r) (m x r)
1 m 1 r 1 (m 1 x r 1)
2 m 2 r 2 (m 2 x r 2)
3 m 3 r 3 (m 3 x r 3)
4 m 4 r 4 (m 4 x r 4)
Now draw the vector diagram with the obtained centrifugal forces (or the product of the masses and their radii of
rotation), such that ab represents the centrifugal force exerted by the mass m1 (or m1.r1) in magnitude and direction to
some suitable scale. Similarly, draw bc, cd and de to represent centrifugal forces of other masses m2, m3 and m4 (or
m2.r2, m3.r3 and m4.r4). Now, as per polygon law of forces, the closing side ae represents the resultant force in
magnitude and direction, as shown in above table. Now, as per polygon law of forces, the closing side ae represents the
resultant force in magnitude and direction The balancing force is, then, equal to resultant force, but in opposite direction.
Now to find out the magnitude of the balancing mass (m) at a given radius of rotation (r), such that
Or
(In general for graphical solution, vectors m1.r1, m2.r2, m3.r3, m4.r4, etc., are added. If they close in a loop, the system
is balanced. Otherwise, the closing vector will be giving (mc.rc). Its direction identifies the angular position of the
counter-mass relative to the other mass.)

Experimentation Problems:
Activity 01: Analytical Method
A circular disc mounted on a shaft carries three attached masses of 50 kg, 40 kg, 60 kg and 80 kg at radial
distances of 7 m, 5 m, 4m and 6 mm and at the angular positions of 45°, 60°, 140 and 210° respectively. The
angular positions are measured counterclockwise from the reference line along the x-axis. Determine the amount
of the counter-mass at a radial distance of 7.5 m required for the static balance.
Activity 02: Analytical Method
Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. The corresponding radii of
rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles between successive masses are 45°, 75°
and 135°. Find the position and magnitude of the balance mass required, if its radius of rotation is 0.2 m.

Activity 01: Graphical Method
A circular disc mounted on a shaft carries three attached masses of 50 kg, 40 kg, 60 kg and 80 kg at radial
distances of 7 m, 5 m, 4m and 6 mm and at the angular positions of 45°, 60°, 140 and 210° respectively. The
angular positions are measured counterclockwise from the reference line along the x-axis. Determine the amount
of the counter-mass at a radial distance of 7.5 m required for the static balance.
Activity 02: Graphical Method
Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. The corresponding radii of
rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles between successive masses are 45°, 75°
and 135°. Find the position and magnitude of the balance mass required, if its radius of rotation is 0.2 m.

Types of Loads-Tensile, Compression, Shear, Impact, Stress and Strain types, Hooks Law & Young's
Modulus
A load may be defined as the combined effect of external forces acting on a body.
a) The other way of classification is (Mechanical Based)
1. Tension: Two pulling (opposing) forces that stretch an object trying to pull it apart (for example,
pulling on a rope, a car towing another car with a chain – the rope and the chain are in tension or are
"being subjected to a tensile load").
2. Compression: Two pushing (opposing) forces that squeeze an object trying to compress it (for
example, standing on a soda can, squeezing a piece of wood in a vise – both the can and the wood are
in compression or are "being subjected to a compressive load").
3. Shear: Two pushing or pulling adjacent forces, acting close together but not directly opposing each
other. A shearing load cuts or rips an object by sliding its molecules apart sideways (for example,
pruning shears cutting through a branch, paper-cutter cutting paper - the branch and paper are
"subjected to a shear loading").
Another example is pulling on two pieces of glue joint or welded joints or Riveted joints etc. Can also
be called is "being subjected to a shear loading."
5. Torsion (Twisting): Created when a moment or "turning force" is applied to a structural member (or
piece of material) making it deflect at an angle (twist). A moment that causes twisting is called a
twisting or torsional moment. Torsion produces shear stresses inside the material. A beam in torsion
will fail in shear; the twisting action causes the molecules to be slid apart sideways
6. Bending Load: When a beam is subjected to a loading system or by a force couple acting on a plane
passing through the axis, then the beam deforms. In simple terms, this axial deformation is called as
bending of a beam. Due to the shear force and bending moment, the beam undergoes deformation.
These normal stress due to bending are called flexure stresses.

7. Impact Load: An impact is an extreme force or shock applied over a short time period. Load with a
sudden impact due to falling or hitting one object on another. The load produced due to these actions
is known as the Impact load. The stress produced in the machine members due to the Impact load is
known as the Impact stress. This type of acceleration or force has a greater effect than a lower force
applied over a longer period of time.
b) The loads may be classified as (Engineering Based)
1. Point or Concentrated Load: A load acting at a point of a beam is known as point or concentrated
load.
2. Uniformly Distributed Load: A load which is spread over a beam in such a way that each unit length
is loaded to the same extent, is known as uniformly distributed load.
3. Uniformly Varying Load: A load which is spread over a beam in such a way that it varies uniformly
on each unit length, is known as uniformly varying load. Sometimes, the load is zero at one end and
increases uniformly to the other. Such a load is known as Triangular Load.

Stress and Strain types, Hooks Law & Young's Modulus
a) Stress (𝝈): In mechanics, stress is defined as a force applied per unit area. It is given by the
following formula:
𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 =
𝑳𝒐𝒂𝒅
𝑨𝒓𝒆𝒂
=
𝑭
𝑨
= ⋯ 𝑁/𝑚2
.
Where, σ is the stress applied, F is the force applied and A is the area of force application.
Types of Stress
Normal Stress:
Normal stress is a stress that acts perpendicular to the area. The formula for the normal stress is given by
Tensile Stress
Tensile stress is the force applied per unit area, increasing a body’s length (or area). Objects under tensile
stress become thinner and longer. When the material is under tension, it is known as tensile. The forces that
are acting along the axis of force are responsible for the stretching of the material. The external force per unit
area of the material resulting in the stretch of the material is known as tensile stress.
Compressive Stress
Compressive stress is the force applied per unit area, which decreases the length (or area) of a body. The object
under compressive stress becomes thicker and shorter. Compressive stress is the force that is responsible for
the deformation of the material such that the volume of the material reduces. It is the stress experienced by a
material which leads to a smaller volume. High compressive stress leads to failure of the material due to
tension.
Shear Stress:
Shear stress induced in a body when it is subjected to two equal and opposite forces that acts tangential to
the area.

b) Strain
Strain is the amount of deformation experienced by the body in the direction of force applied, divided by the
initial dimensions of the body. The following equation gives the relation for deformation in terms of the length
of a solid:
𝑺𝒕𝒓𝒆𝒔𝒔 𝜺 =
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑳𝒆𝒏𝒈𝒕𝒉
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉
=
𝜹𝒍
𝒍
= ⋯ 𝑼𝒏𝒊𝒕 𝒍𝒆𝒔𝒔
Where, (𝜺) is the strain due to stress applied, (𝜹𝒍) is the
change in length and (𝒍) is the original length of the
material. In general when an external force is applied on a
body, there is some change occur in the dimension of the
body. The ratio of this change of dimension in the body to its actual dimension is called strain. The strain is a
dimensionless quantity as it just defines the relative change in shape.
Types of Strain
Strain in mechanics is of four types and these are:
1. Tensile strain: The strain produced in a body due to tensile force is called the tensile strain. The tensile
force always results in the increment of the length and decrease in the cross-section area of the body.
In this case, the ratio of the increase in length to the original length is called tensile strain.
2. Compressive strain: The strain appears due to the compressive force is called compressive strain. In
compressive force there is a decrease in the dimension of the body. So the ratio of the decrease in the
length of the body to the original length is called compressive strain.
3. Volumetric strain: The ratio of the change in the volume of a body to the original volume is called
the volumetric strain. In volumetric strain there is a change in the volume of the body due to application
of the external forces.
4. Shear strain: The strain which is produced in a body due to shear force is called shear strain.
S.no Stress Strain
1.
Stress is defined as the resisting force per unit
area.
Strain is defined as the deformation per unit
area.
2.
The ratio of resisting force (or applied load) to
the cross section area of the body is called stress.
The ratio of change in dimension of the body to
the original dimension is called strain.
3.
Stress is denoted by the Greek symbol ‘σ’
(sigma).
Strain is denoted by the symbol ‘e’.
4.
The formula of stress is given by The formula of strain is given by
5. The unit of stress is (𝑁/𝑚2
) The strain is a unit less quantity.
6. Stress can exist without strain. Strain cannot exist without stress.
7.
The various types of stress are: tensile stress,
compressive stress and shear stress.
The various types of strain are: tensile strain,
compressive strain, shear strain and volumetric
strain.

Stress-Strain Curve
When we study solids and their mechanical properties, information regarding their elastic properties is most
important. We can learn about the elastic properties of materials by studying the stress-strain relationships,
under different loads, in these materials.
The material’s stress-strain curve gives its stress-strain relationship. In a stress-strain curve, the stress and its
corresponding strain values are plotted. An example of a stress-strain curve is given below.
Explaining Stress-Strain Graph
The different regions in the stress-strain diagram are:
a) Proportional Limit: It is the region in the stress-strain curve that obeys Hooke’s Law. In this limit,
the stress-strain ratio gives us a proportionality constant known as Young’s modulus. The point OA in
the graph represents the proportional limit.
b) Elastic Limit: It is the point in the graph up to which the material returns to its original position when
the load acting on it is completely removed. Beyond this limit, the material doesn’t return to its original
position, and a plastic deformation starts to appear in it.
c) Yield Point: The yield point is defined as the point at which the material starts to deform plastically.
After the yield point is passed, permanent plastic deformation occurs. There are two yield points
(i) Upper yield point
(ii) Lower yield point.
d) Ultimate Stress Point: It is a point that represents the maximum stress that a material can endure
before failure. Beyond this point, failure occurs.
e) Fracture or Breaking Point: It is the point in the stress-strain curve at which the failure of the material
takes place.

Hooke’s Law
According to Hooke’s law, when a material is loaded within elastic limit, the stress induced in the material is
directly proportional to the strain produced. It means that the ratio of stress with the corresponding strain gives
us a constant within elastic limit. The constant is known as Modulus of Elasticity or Modulus of Rigidity or
Elastic Modulus.
In the Stress strain curve the proportionality limit indicates the Hooke’s law. The curve shows linearity within
elastic limit. The ratio of stress and corresponding strain in the stress strain curve gives us Young’s Modulus
Mathematically
Young’s Modulus or Modulus of Rigidity or Elastic Modulus.
It is defined as the ratio of stress to the strain within elastic limit. It is denoted by the letter ‘E’.
 It represents the elastic property of a material.
 The unit of Young’s modulus is N/m2
.or N/mm2
. The unit of young’s modulus is same as that of units
of stress.
Where,
σ = Stress
e = Strain
E = Proportionality constant called as Young’s Modulus

TENSION TEST (Tensile Force)
QUESTION: Conduct tensile test for the given Specimen and Determine Stress-Strain-Young’s Modulus, Yield
Stress- Maximum Stress – Breaking Stress - % Elongation in Length and % Reduction in Area Also, Draw Stress -
Strain Diagram for the above Parameters.
AIM: To study the behavior of mild steel rod under tension to find out the following material properties, and
graphs
1. Yield point stress
2. Ultimate stress
3. Percentage elongation
4. Young's modulus
5. Nominal breaking stress
6. Percentage reduction in area
7. Percentage Increase in length
Plot the Graphs:
a) Stress v/s Strain
b) Load v/s Extension
OBJECTIVE: To conduct a tensile test on a mild steel specimen and determine the following:
1. Limit of proportionality
2. Elastic limit
3. Yield strength
4. Ultimate strength
5. Young‟s modulus of elasticity
6. Percentage elongation in length
7. Percentage reduction in area.
APPARATUS:
1. Universal Testing Machine (UTM)
2. Mild steel specimens
3. Graph paper
4. Scale
5. Vernier Caliper
Universal Testing Machine

THEORY:
Tensile Testing is one of the most fundamental
mechanical tests that can be performed on a material.
A test sample is loaded in tension when it
experiences opposing forces acting upon opposite
faces both located on the same axis that attempt to
pull the specimen apart. These tests are simple to
setup and complete and reveal many characteristics
of the material that is tested. Tensile tests are
considered to be essentially the opposite of a
compression test.
Tensile Testing is a form of tension testing and is a
destructive engineering and materials science test
whereby controlled tension is applied to a sample
until it fully fails. It is used to find out how strong a material is and also how much it can be stretched before it breaks.
This test method is used to determine yield strength, ultimate tensile strength, ductility, strain hardening characteristics,
Young's modulus and Poisson's ratio.
PROCEDURE:
1. Measure the original length and diameter of the specimen. The length may either be length of gauge section which
is marked on the specimen with a preset punch or the total length of the specimen
2. Insert the specimen into grips of the test machine and attach strain-measuring device to it
3. Begin the load application and record load versus elongation data.
4. Take readings more frequently as yield point is approached.
5. Measure elongation values with the help of dividers and a ruler.
6. Continue the test till Fracture occurs.
7. By joining the two broken halves of the specimen together, measure the final length and diameter of specimen.
OBSERVATION
SPECIMEN: The specimen used for the tension test is a mild steel rod of ...........mm length and diameter, ...............mm.
a. Initial diameter of specimen d1 =
b. Initial gauge length of specimen L1 =
c. Initial cross-section area of specimen A1 =
d. Load of yield point Ft =
e. Ultimate load after specimen breaking F =
f. Final length after specimen breaking L2 =
g. Diameter of specimen at breaking place d2 =
h. Cross section area at breaking place A2 =
OBESERVATION TABLE:

CALCULATION:
1. 𝒀𝒊𝒆𝒍𝒅 𝒑𝒐𝒊𝒏𝒕 𝒔𝒕𝒓𝒆𝒔𝒔 =
𝒀𝒊𝒆𝒍𝒅 𝑳𝒐𝒂𝒅
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝑪𝒓𝒐𝒔𝒔−𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒂𝒓𝒆𝒂
= __________________𝑵/𝒎𝒎𝟐
2. 𝑼𝒍𝒕𝒊𝒎𝒂𝒕𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 =
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝑳𝒐𝒂𝒅
= __________________𝑵/𝒎𝒎𝟐
3. 𝒀𝒐𝒖𝒏𝒈′
𝒔 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 =
𝑺𝒕𝒓𝒆𝒔𝒔
𝑺𝒕𝒓𝒂𝒊𝒏
=
𝝈
𝜺
=
𝑭
𝑨
∆𝑳
𝑳
=
𝑭 𝑳
𝑨 ∆𝑳
= __________________𝑵/𝒎𝒎𝟐
4. 𝑵𝒐𝒎𝒊𝒏𝒂𝒍 𝒃𝒓𝒆𝒂𝒌𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒔𝒔 =
𝑩𝒓𝒆𝒂𝒌𝒊𝒏𝒈 𝑳𝒐𝒂𝒅
= __________________𝑵/𝒎𝒎𝟐
5. 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 =
𝑰𝒏𝒊𝒕𝒊𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉−𝑭𝒊𝒏𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉
𝑭𝒊𝒏𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉
× 𝟏𝟎𝟎 = __________________%
6. 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒓𝒆𝒅𝒖𝒄𝒕𝒊𝒐𝒏 𝒊𝒏 𝒂𝒓𝒆𝒂 =
𝑰𝒏𝒊𝒕𝒊𝒂𝒍 𝑨𝒓𝒆𝒂−𝑭𝒊𝒏𝒂𝒍 𝑨𝒓𝒆𝒂
𝑭𝒊𝒏𝒂𝒍 𝑨𝒓𝒆𝒂
× 𝟏𝟎𝟎 = __________________%
7. 𝑺𝒕𝒓𝒆𝒔𝒔 =
𝑳𝒐𝒂𝒅
𝑨𝒓𝒆𝒂
≫ 𝝈 =
𝑭
𝑨
= __________________𝑵/𝒎𝒎𝟐
8. 𝑺𝒕𝒓𝒂𝒊𝒏 =
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑳𝒆𝒏𝒈𝒕𝒉
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉
≫ 𝜺 =
∆𝑳
𝑳
= __________________(𝑵𝒐 𝑼𝒏𝒊𝒕𝒔)
RESULT:
(i) Average Breaking Stress = …………………………….
(ii) Ultimate Stress =…………………………………….
(iii) Average % Elongation = ……………………………..
VIVA-QUESTIONS
1. Which steel have you tested? What is its carbon content?
2. What general information is obtained from tensile test regarding the properties of a material?
3. Which stress have you calculated: nominal stress or true stress?
4. What kind of fracture has occurred in the tensile specimen and why?
5. Which is the most ductile metal? How much is its elongation

COMPRESSION TEST
QUESTION: Conduct Compression test for the given Specimen and Determine Stress - Strain -Young’s Modulus,
Yield Stress Maximum Stress - % Reduction in Length and % Increase in Area Also, Draw Stress- Strain Diagram for
the above Parameters.
AIM: To Perform compression test of mild steel and cast iron on UTM.
1. Stress
2. Strain
3. Percentage Reduction in length
4. Percentage Reduction in length
5. Young's modulus
6. Yield Stress
7. Maximum Stress
8. Percentage reduction in area
Plot the Graphs:
1. Stress v/s Strain
2. Load v/s Reduction
OBJECTIVES: To study the mechanical properties of Mild Steel specimen under compression.
APPARATUS: A UTM or A compression testing Machine, cylindrical or cube shaped specimen of cast iron,
Alumunium or mild steel, Vernier caliper, liner scale, dial gauge.
THEORY: Several Machines and structure components such as columns and struts are subjected to compressive load
in applications. These components are made of high compressive strength materials. Not all the materials are strong in
compression. Several materials, which are good in tension, are poor in compression. Contrary to this, many materials
poor in tension but very strong in compression. Cast iron is one such example. That is why determine of ultimate
compressive strength is essential before using a material. This strength is determined by conduct of a compression test.
Compression test is just opposite in nature to tensile test. Nature of deformation and fracture is quite different
from that in tensile test. Compressive load tends to squeeze the specimen. Brittle materials are generally weak in tension
but strong in compression. Hence this test is normally performed on cast iron, cement concrete etc. But ductile materials
like Alumunium and mild steel which are strong in tension, are also tested in compression.
MODES OF DEFORMATION IN COMPRESSION TESTING
The figure below illustrates the different modes of failure in compression testing.
a) When L/D > 5,Buckling
b) When L/D > 2.5, Shearing
c) When L/D > 2.0 and friction is present at the contact surfaces, Double barreling
d) When L/D < 2.0 and friction is present at the contact surfaces, Barreling
e) When L/D < 2.0 and no friction is present at the contact surfaces, Homogenous
compression.
f) Compressive instability due to work-softening material.
FAILURE PATTERNS:
Ductile material will have proportional limit in
compression very close to those in tension. The initial regions
of their compression stress strain diagram are very similar to
tension diagrams. When a mild steel specimen is compressed,
it begins to bulge outward on the sides and become barrel
shaped. With increasing load the specimen is flattened out, thus
offering increased resistance to further shortening.

TEST SET-UP:
A compression test can be performed on UTM by keeping the test-piece on base block and
moving down the central grip to apply load. It can also be performed on a compression testing
machine. A compression testing machine set-up is shown in fig. it has two compression
plates/heads. The upper head moveable while the lower head is stationary. One of the two heads
is equipped with a hemispherical bearing to obtain. Uniform distribution of load over the test-
piece ends. A load gauge is fitted for recording the applied load.
SPECIMEN: In cylindrical specimen, it is essential to keep
ℎ
𝑑
< 2 to avoid lateral instability due to bucking action.
𝑆𝑝𝑒𝑐𝑖𝑚𝑒𝑛 𝑠𝑖𝑧𝑒 = ℎ < 2𝑑
PROCEDURE:-
1. The original dimensions of the specimen like original dia., gauge length etc.
2. The specimen is mounted on the Universal Testing machine between the fixed and movable jaws.
3. The load range in the machine is adjusted to its maximum capacity (300 kN)
4. The dial gauge is mounted on the machine at the appropriate positions and adjusted to zero.
5. The machine is switched on and the compressive load is applied gradually.
6. For every 10 kN of load, the readings of dial gauge is noted and tabulated.
7. Remove the dial gauge at slightly below the expected load at yield point.
8. Record the load at yield point, at the yield point the pointer on load scale will remain stationary for small interval of
time and blue needle will come back by 1 or 2 divisions that point is lower yield point.
9. The specimen is loaded continuously up to the ultimate load (red needle will stops) which is to be noted.
10. The specimen is removed and final dimensions are measured.
OBSERVATION:
1. Initial length or height of specimen ho =………………… mm.
2. Initial diameter of specimen do =…………………….. mm.
3. Final length or height of specimen hf =………………… 𝑚𝑚.
4. Final diameter of specimen df =…………………….. 𝑚𝑚.
5. Original cross-section area Ao:……………………….. 𝑚𝑚2
6. Final cross-section area Af :………………………….. 𝑚𝑚2
CALCULATION:

PRECAUTIONS:
· The specimen should be prepared in proper dimensions.
· The specimen should be properly to get between the compression plates.
· Take reading carefully.
· After failed specimen stop to m/c.
RESULTS:
VIVA-QUESTIONS:
1. Compression tests are generally performed on brittles materials-why?
2. Which will have a higher strength: a small specimen or a full size member made of the same material?
3. What is column action? How does the h/d ratio of specimen affect the test result?
4. How do ductile and brittle materials in their behavior in compression test?
5. What are bi-modulus materials? Give examples.

SHEAR TEST
QUESTION: Conduct Shear test for the given specimen-Problems on Members subjected to combined Stresses
AIM: Find out the Shear strength of a given specimen using UTM.
1. To determine experimentally, the ultimate shear strength in single shear of mild steel rod.
2. To determine experimentally, the ultimate shear strength in double shear of mild steel rod.
APPARATUS: A UTM, Specimen, shearing attachment, Vernier caliper etc.
THEORY: A type of force which causes or tends to cause two contiguous parts of the body to slide relative to each
other in a direction parallel to their plane of contact is called the shear force. The stress required to produce fracture in
the plane of cross-section, acted on by the sheer force is called shear strength.
In direct shear test, the shearing stress is considered as uniformly distributed over the entire cross section. The shear
force is applied by a suitable test rig, two different cases of shearing may arise; i.e., single shear and double shear. In
single shear shearing occurs across a single surface and in double shear shearing occurs across two surfaces. Knowledge
of shear failure is important while designing any structures or machine components. Shear force causes the surface to
go out of the alignment with each other and thus the material fails.
The shearing force P in each section is P = F/2, it can be concluded that the average shearing stress is the maximum load
divided by the combined cross-sectional area of the two planes This shall be calculated from the following formula:
𝐷𝑜𝑢𝑏𝑙𝑒 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ =
𝑃
𝐴
=
𝐹
2
𝐴
=
𝐹
2 × 𝐴
Shear strength of the specimen is determined by inserting a cylindrical specimen through round holes in three hardened
steel blocks, the centre of which shall be pulled (or pushed) between the other two so as to shear the specimen on two
planes. In this test a suitable length of cylindrical specimen is subjected to double shear loading using a suitable test rig
in a testing machine under a compressive load or tensile pull and recording the maximum load P to fracture. The speed
of testing or the rate of separation of the cross-heads, at any moment during the test, shall not be greater than 10 mm/min.
Shear stress:
It is produced in a body when it is subjected to two equal and opposite forces spaced at an infinite decimal
distance or tangentially across the resisting section.
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠, 𝜎 =
𝑆ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑆𝑒𝑐𝑡𝑖𝑜𝑛
Generally it is difficult to produce conditions of pure shear as some bending effect is likely to occur due to shearing load
resulting in equal and opposite forces at a small finite distance.
TEST SPECIMEN:
Specimen shall normally be of full cross section for cylindrical metallic products up to and including 25 mm in
diameter, in the case of cylindrical products over 25 mm in diameter the samples may be turned down to a specimen of
25 mm diameter for testing. According to the Indian standard minimum length of the specimen should be twice the
diameter, the maximum length of the specimen is not specified as it has no particular effect on the test result. Shear
strengths of typical metals are of the order of about 60-80% of their ultimate strengths in tension.

PROCEDURE:
1. Insert the specimen in position and grip one end of the attachment in the upper portion and one end in the lower
position
2. Switch on the UTM
3. Bring the drag indicator in contact with the main indicator.
4. Select the suitable range of loads and space the corresponding weight in the Pendulum and balance it if necessary
with the help of small balancing weights
5. Operate (push) the button for driving the motor to drive the pump.
6. Gradually move the head control ever in left hand direction till the specimen shears.
7. Note down the load at which the specimen shears.
8. Stop the machine and remove the specimen.
9. Repeat the experiment with other specimens.
OBSERVATION:
1. Applied compressive force (F) = − − − − − − − − −𝑘𝑁.
2. Diameter of specimen = − − − − − − − − −𝑚𝑚.
3. Cross sectional area of the pin (in double shear) = (
𝜋×𝑑2
4
) − − − − − − − − − 𝑚𝑚2
4. Cross sectional area of the pin (in double shear) = (
2×𝜋×𝑑2
4
) − − − − − − − − − 𝑚𝑚2
5. Load taken by the specimen at the time of failure, W = − − − − − − − − −(𝑁)
6. Strength of the pin against shearing (τ) = (
4×𝑊
2×𝜋×𝑑2
) − − − − − − − − − 𝑁/𝑚𝑚2
PRECAUTIONS:
1. The measuring range should not be changed at any stage during the test.
2. The inner diameter of the hole in the shear stress attachment should be slightly greater than the specimen.
3. Measure the diameter of the specimen accurately.
4. The method for determining the shear strength consists of subjecting a suitable Length of steel specimen in full cross-
section to double shear, using a suitable test rig, in a testing m/c under a compressive load or tensile pull and recording
the maximum load ‘F’ to fracture.
RESULT:
Shear strength of specimen = -------------------------
VIVA-QUESTIONS:
1. Does the shear failure in wood occur along the 45° shear plane?
2. What is shear stress?
3. What is single & double shear?
4. What is finding in shear test?
5. What is unit of shear strength?

Bending test
Aim: Conduct bending test on a wooden beam and identify
a) The bending or flexural behavior of the wooden beam
b) Determine the Modulus of Elasticity
c) Modulus of Rupture of the wood.
Equipment Required
 UTM Machine
 Deflection Gauges
 Wooden Beam
 Measuring Tape
Theory and Principle
The modulus of elasticity in bending and bending strength is determined by applying a load to the center of a test
piece supported at two points. The modulus of elasticity is calculated by using the slope of the linear region of the
load-deflection curve. The bending strength of each test piece is calculated by determining the ratio of the bending
moment M, at the maximum load F (max), to the moment of its full cross-section.
If a beam is simply supported at the ends and carries a concentrated load at its center, the beam bends concave
upwards. The distance between the original position of the beams and its position after bending at different points
along the length of the beam, being maximum at the center in this case. This difference is known as ‘deflection’. In
this particular type of loading the maximum amount of deflection (δ) is given by the relation,
For a simply supported beam with central loading, deflection under the load is given by:
𝛿 =
𝑊 𝐿3
48 𝐸 𝐼
𝑜𝑟 𝐸 =
𝑊 𝐿3
48 𝛿 𝐼
Where,
𝛿 = Deflection due to Load
W =Applied load(𝑵)
L = Effective span of the beam(𝒎𝒎)
E = Young's Modulus of wood(𝑵/𝒎𝒎𝟐)
I = Second moment of area of the cross- section (i.e., moment of Inertia) of the beam, about the neutral axis, (𝒎𝒎𝟒)
Test Procedure
1) Insert the bending device in the UTM.
2) Measure the width and depth of wooden beam.
3) Adjust the support for the required distance and clamp to the lower table.
4) Fix the transverse test pan at the lower side of the lower cross head.
5) Fix it on the rollers of the transverse test brackets such that the load comes at the center and measure the length
of the span of the beam between the supports for central loading.
6) Adjust the load pointer to zero by lifting the lower table.
7) While applying the load, the deflection corresponding to each load is found out from the Vernier scale on the
UTM.
8) Note the maximum deflection and the maximum load.
9) Find the deflection (δ) in each case by subtracting the initial reading of Vernier scale.
10) Draw a graph between load (W) and deflection (δ). On the graph choose any two convenient points and between
these points find the corresponding values of W and δ by putting these in the relation δ = WL3/48EI.
11) Calculate the bending stress for different loads using relation δb = My/I.

Bending Stress:
As per bending equation,
𝑀
𝐼
=
𝜎𝑏
𝑦
Where, M = Bending moment, N-mm
I = Moment of inertia, mm4
𝜎𝑏 = Bending stress, N/mm2
and
Y = Distance of the top fiber of the beam from the neutral axis
Observation:
Total length of beam = L = ______________mm
Width of beam =b = ______________mm
Height of beam = h or t = ______________mm
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 = 𝐼 =
𝑏 𝑡3
12
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐵𝑒𝑎𝑚 𝛿 =
𝑊 𝐿3
48 𝐸 𝐼
Observation Table
Sl No Load Deflection
Young’s
Modulus
1
2
Results
• The young’s modulus for steel beam is found to be----- N/mm2.
• The young’s modulus for wooden beam is found to be----- N/mm2
Precautions:
 Make sure that beam and load are placed a proper position.
 The cross- section of the beam should be large.
 Note down the readings of the Vernier scale carefully.
 The length of the simply supported should be measured properly.
 The dial gauge spindle knob should always touch the beam at the bottom of loading point.
 Loading hanger should be placed at known distance
 Al the errors should be eliminated while taking readings.
 Beam should be positioned horizontally
Viva Questions:
1. Types of beams.
2. What is deflection?
3. Write the equation for the Slope for a cantilever beam with point load
4. Write the deflection equation for the simply supported beam with point load at the center
5. How many types of bending are there?

Ansys Transient Simulation of Compression, Tensile, Bending Test
Preprocessor
Element Type
Add/Edit/Delete
Add
Structural Solid & Brick 8 Node 185
Close
Material Props
Material Models & Structural & Linear & Elastic & Isotropic &
(Click inside the EX box) cast Iron: 2.1e11 Alluminium: 7e10
(Click inside the NUXY box) Cast Iron: 0.3 Alluminium: 0.31
Material Models & Structural & Density
(Click inside the DENS) Cast Iron: 7850 Alluminium: 2710
Close Material Models window
Modeling (Example for Compresion)
Create
Volume
Cylinder
By dimension
(Enter Outer radius = 10; z =0: 40)
Close Create window
Meshing
Mesh Tool
Volume- Hex / wedge-sweep / Sweep: click on volume
[Or Type in command bar
ESIZE, 5
VSWEEP, P]
(Click Circle Area and see in Isometric View)
Parameter (in Menu Bar top)
Functions
Define/Edit Function
Result = 10000*{time}
File
Save
(Save in suitable location with known file name ex: Load step.func)
(Exit Define/Edit Function)
Functions
Read from File
(Select location of file name ex: Load step.func and open it)
Table parameter Name; Enter name for the Table (Ex: Comp or Tens or Shear or Bend)
Close
Loads
Analysis type
New Analysis
Pick Transient
OK - Full Method - OK

Loads
Define Load
Apply
Structural
Displacement
On Nodes (or Area)
Select Entities- in menu bar
Nodes - by Location - select on Z as 0 - Plot
(Pick Box and select to All DOF)
Loads
Select Entities- in menu bar
Nodes - by Location - select on Z as 40 - Plot
Define Load
Apply
Structural
Force / Load (Or Pressure)
On Nodes (or Area)
(Pick Box and select to All Nodes)
Apply Force on Nodes in FZ- ok
(Select Existing Table (Ex: Comp or Tens or Shear or Bend))
OK
Solutions
Analysis type
Sol Control
Basic
Pick Time Increment
Time at the End of Load Step = 20
Number of Sub step = 1
Frequency select (write every sub step)
Transient
Transient effect
(Pick Ramped Loading)
Solution
Load Step Option
Read every sub step
Solution
Solve Current LS
Time History Post Processor
Add – Nodal Solution Z-Dyrection
Add – Element Solution – Von-mises Stress - Pick Area & Node
Add – Element Solution – Von-mises Elastic Strain - Pick Area & Node
Plot control – Style – Graphs – Modify Axis – Y – component – Logarithmic
Plot – Replot.
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7) Types of Beams-Types of Loads acting on Beams-Concept of Shear force-Bending moment
INTRODUCTION
Shear and bending moment diagrams are analytical tools used in conjunction with structural analysis to help
perform structural design by determining the value of shear force and bending moment at a given point of a structural
element such as a beam. These diagrams can be used to easily determine the type, size, and material of a member in a
structure so that a given set of loads can be supported without structural failure.
Another application of shear and moment diagrams is that the deflection of a beam can be easily determined
using either the moment area method or the conjugate beam method. The algebraic sum of the vertical forces at any
section of a beam to the right or left of the section is known as shear force.
It is briefly written as S.F. The algebraic sum of the moments of all the forces acting to the right or left of the
section is known as bending moment. It is written as B.M. In this chapter, the shear force and bending moment diagrams
for different types of beams (i.e., cantilevers, simply supported, fixed, overhanging etc.) for different types of loads (i.e.,
point load, uniformly distributed loads, varying loads etc.) acing on the beams, will be considered.
TYPES OF BEAMS
The following are the important types of beams:
a) Cantilever Beam.
A beam which is fixed at one of its end and
the other end is free is called a cantilever
beam. Figure below shows a cantilever beam
with one end rigidly fixed and the other end
free. The distance between fixed and free
ends is called the length of the beam.
b) Simply Supported Beam
A beam which is freely supported at both
ends is called a simply supported beam. The
term 'freely supported' implies that the end
supports exerts only the forces upon the bar but not the moments. Therefore there is no restraint offered to the angular
rotation of the ends of the bar at the supports as the bar deflects under the loads.
c) Overhanging Beam
The beam freely supported at any two points and having one or both ends projected beyond these supports is called an
overhanging beam.
d) Fixed Beams
A beam, whose both ends are fixed or built-in walls, is known as fixed beam. A fixed beam is also known as a built-in
or encased beam.
e) Continuous Beam
A beam which is provided more than two supports, is known as continuous beam.

REACTIONS AT SUPPORTS OF BEAMS
A beam is a structural member used to support loads applied at various points along its length. Beams are generally
long, straight and prismatic (i.e. of the same cross-sectional area throughout the length of the beam).
Types of Supports:
Beams are supported on roller, hinged or fixed supports as shown in Figure.
a) Simple Support:
If one end of the beam rests in a fixed support, the support is known as simple support. The
reaction of the simple support is always perpendicular to the surface of support. The beam is
free to slide and rotate at the simple support.
b) Roller Support:
Here one end of the beam is supported on a roller. The only reaction of the roller support is normal to the surface on
which the roller rolls without friction. in which four possible situations are illustrated. Examples of roller supports are
wheels of a motorcycle, or a handcart, or an over-head crane, or of a car, etc.
c) Hinged Support:
At the hinged support the beam does not move either along or normal to its axis. The beam,
however, may rotate at the hinged support. The total support reaction is R and its horizontal and
vertical components are H and V, respectively. Since the beam is free to rotate at the hinged
support, no resisting moment will exist. The hinged support behaves like the hinges provide to
doors and windows.
d) Fixed Support:
At the fixed support, the beam is not free to rotate or slide along the length of the beam or in the
direction normal to the beam. Therefore, there are three reaction components, viz., vertical
reaction component (V), horizontal reaction component (H) and the moment (M), fixed support
is also known as built-in support.
SHEAR FORCES AND BENDING MOMENT DIAMGRAMS
Definition of Shear force and bending moment
A shear force (SF) is defined as the algebraic sum of all the vertical forces, either to the left or to the right hand side of
the section.

A bending moment (BM) is defined as the algebraic sum of the moments of all the forces either to the left or to the right
of a section.
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑥 − 𝑥 = 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝐿𝑜𝑎𝑑 × 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Sign convention of SF and BM
For Shear force:
We shall remember one easy sign convention, i.e., to the right side of a section, external force acting in upward direction
is treated as negative (remember this convention as RUN —» Right side of a section Upward force is Negative). It is
automatic that a downward force acting to the right side of a section be treated as positive. Sign convention is shown in
Figure below. The signs become just reversed when we consider left side of section.
For Bending moment:
The internal resistive moment at the section that would make the beam to sag (means to sink down, droop) is treated to
be positive. A sagged beam will bend such that it exhibits concave curvature at top and convex curvature at bottom.
Positive bending moment is shown in Figure (A shear force which tends to rotate the beam in clockwise direction is
positive and vice versa)
SFD and BMD definitions
It is clear from foregone discussions that at a section taken on a loaded beam, two internal forces can be visualized,
namely, the bending moment and the shear force. It is also understood that the magnitude of bending moment and shear
force varies at different cross sections over the beam. The diagram depicting variation of bending moment and shear
force over the beam is called bending moment diagram [BMD] and shear force diagram [SFD]. Such graphic
representation is useful in determining where the maximum shearing force and bending moment occur, and we need this

information to calculate the maximum shear stress and the maximum bending stress in a beam. The moment diagram
can also be used to predict the qualitative shape of the deflected axis of a beam.
General Guidelines on Construction of SFD and BMD
Before we go on to solving problems, several standard procedures (or practices) in relation with construction of shear
force and bending moment diagrams need to be noted.
1) The load, shear and bending moment diagrams should be constructed one below the other, in that order, all with
the same horizontal scale.
2) The dimension on the beam need not be scaled but should be relative and proportionate (a 3 m span length should
not look more than 5 m length).
3) Ordinates (i.e., BM and SF values) need not be plotted to scale but should be relative. Curvature may need to be
exaggerated for clarity.
4) Principal ordinates (BM and SF values at salient points) should be labeled on both SFD and BMD.
5) A clear distinction must be made on all straight lines as to whether the line is horizontal or has a positive or negative
slope.
6) The entire diagram may be shaded or hatched for clarity, if desired.
Variation of shear force and bending moment diagrams
WORKED EXAMPLES
1) A cantilever beam of length 2 m carries the point loads as shown in Fig. Draw the shear force and B.M.
diagrams for the cantilever beam.

2) A simply supported beam of length 6 m carries point load of 3 kN and 6 kN at distances of 2 m and 4 m from
the left end. Draw the shear force and bending moment diagrams for the beam.
3) Cantilever beam carrying a distributed load with intensity varying from 𝑾𝒐 at the free end to zero at the
wall, as shown in Figure

FINITE ELEMENT ANALYSIS / FINITE ELEMENT METHOD
INTRODUCTION
Finite element analysis method is a numerical procedure that applies to many areas in engineering problems,
including structural/stress analysis, fluid flow analysis, heat transfer analysis, and electromagnetics analysis.
Finite element has several advantages and features such as the capability of solving complicated and complex
geometries, flexibility, strong mathematical foundation, and high-order approximation. Therefore, finite element
analysis (FEA) has become an important method in the design and modeling of a physical event in many engineering
disciplines.
The actual component in the FEA method is placed by a simplified model that is identified by a finite number
of elements connected at common points called nodes, with an assumed response of each element to applied loads, and
then evaluating the unknown field variable (displacement) at these nodes.
METHODS OF SOLVING ENGINEERING PROBLEMS
There are 3 common methods to solve any engineering problem:
a) Experimental method
b) Analytical method
c) Numerical method
Experimental Method
This method involves actual measurement of the system response. This method is time consuming and needs expensive
set up. This method is applicable only if physical prototype is available. The results obtained by this method cannot be
believed blindly and a minimum of 3 to 5 prototypes must be tested.
Analytical Method
This is a classic approach. This method gives closed form solutions. The results obtained with this method are accurate
within the assumptions made. This method is Applicable only for solving problems of simple geometry and loading,
like cantilever and simply supported beams, etc. Analytical methods produce exact solutions of the problem.
Numerical Method
This approximate method is resorted to when analytical method fails. This method is applicable to real-life problems of
a complex nature. Results obtained by this method cannot be believed blindly and must be carefully assessed against
experience and the judgment of the analyst. Examples of this method are finite element method, finite difference
method, moment method, etc.
PROCEDURE OF FINITE ELEMENT ANALYSIS (RELATED TO STRUCTURAL PROBLEMS)
Step (i). Discretization of the structure
This first step involves dividing the structure or domain of the problem into small divisions or elements. The analyst has
to decide about the type, size, and arrangement of the elements.

Step (ii). Selection of a proper interpolation (or displacement) model
A simple polynomial equation (linear/quadratic/cubic) describing the variation of state variable (e.g., displacement)
within an element is assumed. This model generally is the interpolation /shape function type. Certain conditions are to
be satisfied by this model so that the results are meaningful and converging.
Step (iii). Derivation of element stiffness matrices and load vectors
Response of an element to the loads can be represented by element equation of the form
[𝑲] × {𝒒} = (𝑸) 𝒐𝒓 [𝑲] × {𝑼} = |𝑭|
Where,
[k] = Element stiffness matrix,
{q} = Element response matrix or element nodal displacement vector, or nodal degree of freedom,
{Q} = Element load matrix or element nodal load vector.
Step (iv). Assembling of element equations to obtain the global equations
Element equations obtained in Step (iii) are assembled to form global equations in the form of
[𝑲] × {𝒓} = {𝑹}
Where,
[K] is the global stiffness matrix,
{r} or {U} is the vector of global nodal displacements, and
{R} is the global load vector of nodal forces for the complete structure.
The above equation describes that the behavior of entire structure.
Step (v). Solution for the unknown nodal displacements
The global equations are to be modified to account for the boundary conditions of the problem. After specifying the
boundary conditions, the equilibrium equations can be expressed as
[𝑲𝟏] × {𝒓𝟏} = {𝑹𝟏}
For linear problems, the vector {r1} can be solved very easily.
Step (vi). Computation of element strains and stresses
From the known nodal displacements {r1}, the elements strains and stresses can be computed by using predefined
equations for structure. The terminology used in the previous 6 steps has to be modified if we want to extend the concept
to other fields. For example, put the field variable in place of displacement, the characteristic matrix in place of stiffness
matrix, and the element resultants in place of element strains.
METHODS OF PRESCRIBING BOUNDARY CONDITIONS BY ELIMINATION METHOD
This method is useful when performing hand calculations. It poses difficulties in implementing in software. This method
has been used in this book for solving the problems by finite element method using hand calculations and results in

reduced sizes of matrices thus making it suitable for hand calculations. The method is explained below in brief. Consider
the following set of global equations,
Let u3 be prescribed, i.e., u3 = s=0.
This condition is imposed as follows:
1. Eliminate the row corresponding to u3 (3rd row).
2. Transfer the column corresponding to u3 (3rd column) to right-hand side after multiplying it by “s”. These steps result
in the following set of modified equations,
This set of equations now may be solved for non-trivial solution.
FINITE ELEMENT ANALYSIS SOFTWARE PACKAGE
There are 3 main steps involved in solving an engineering problem using any commercial software:
Step (i). Preprocessing
In this step, a CAD model of the system (component) is prepared and is meshed (discretized). Boundary conditions
(support conditions and loads) are applied to the meshed model.
Step (ii). Processing
In this step, the software internally calculates the elements stiffness matrices, element load vectors, global stiffness
matrix, global load vector, and solves after applying boundary conditions for primary unknowns (e.g., displacements/
temperatures etc.) and secondary unknowns (e.g., stress/strain/heat flux etc.).
Step (iii). Post-processing
Post-processing involves sorting and plotting the output to make the interpretation of results easier.
TYPES OF ELEMENTS
In general, the region in space is considered non-regular geometric. However, the FEA method divides the non-regular
geometric region to small regular geometric regions. There are 3 types of elements in finite element.
1. One-Dimensional Elements:
The objects are subdivided into short-line segments. A one-dimensional finite element expresses the object as a function
of one independent variable such as one coordinate x. Finite elements use one-dimensional elements to solve systems

that are governed by ordinary differential equations in terms of an independent variable. The number of node points in
an element can vary from 2 up to any value needed. Indeed, increasing the number of nodes for an element increases
the accuracy of the solution, but it also increases the complexity of calculations. When the elements have a polynomial
approximation higher than first order, we call that higher order elements. Figure below shows one-dimensional elements.
For example, the one-dimensional element is sufficient in dealing with heat dissipation in cooling fins.
2. Two-Dimensional Elements:
The objects can be divided into triangles, rectangles, quadrilaterals, or other suitable subregions. A two-dimensional
finite element expresses the object as a function of 2 variables such as the 2 coordinates x and y. A finite element uses
two-dimensional elements to solve systems that are governed by partial differential equations. The simplest two-
dimensional element is the triangular element. Figure below shows two-dimensional elements. For example, a two-
dimensional element is sufficient in plane stress or plane strain.
3. Three-Dimensional Elements:
The objects can be divided into tetrahedral elements, rectangular prismatic elements, pie-shaped elements, or other
suitable shapes of elements. A three-dimensional finite element expresses the object as a function of 3 variables such as
the 3 coordinates x, y, and z. A finite element uses three- dimensional elements to solve systems that are governed by
differential equations. The simplest three-dimensional element is the tetrahedral element. Figure below shows three-
dimensional elements.

FINITE ELEMENT ANALYSIS OF BEAMS
Beam is very common structure in many engineering applications because of its efficient load carrying capability. Beam
by definition is a transversely loaded structural member. Beam element is used in the analysis of beams.
This element has 2 end nodes each having 2 degrees of freedom, namely transverse displacement and slope. Beam
element gives accurate results if acted upon by nodal forces and moments. A greater number of small elements will be
necessary in the case of a beam acted upon by distributed loads in order to get good results. The interpolation equation
and element stiffness matrix for beam element are given by (𝑤 = 𝑣)
SIMPLE STEPS IN FINITE ELEMENT ANALYSIS OF BEAMS
Step 01: Discretization of Modal
Step 2: Calculate Stiffness Matrix for each
Step 3: calculate Shear force and Momentum due to Load
Step 4: Calculate Nodal Force Vector:
Step 5: Apply in Finite Element Global Equation
Step 6: Apply Boundary Conditions
Apply Vertical Deflections (𝑣 𝑜𝑟 ∆) or Slope Deflections (𝜃) to its Value or Zero (no)
Step 7: Matrices will be reduced and can be solved as equations.
Step 8: Find out the Slope and Deflections and compare with Simulations (Using Ansys)

Example 1: Simply supported Beam with Point Load
For the beam shown in Figure 5.2, determine the nodal displacements, slope, and reactions. Take E = 210 GPa
and I = 4 × 10−4 m4.
Solution
(I)Analytical method calculation
L = 10 m & P = 8 kN.
Deflection in Analytical Equation is given by,
𝜃𝐶 = 0, by symmetry.
Reaction,
(II)FEM by hand calculations
Element stiffness matrices are,

Due to symmetry,
Global equation is,
Using the elimination method for applying boundary conditions,
𝑤1 = 𝑤3 = 0.
The above matrix reduces to
By solving the above equations, we get,
𝑤2 = −0.002 𝑚 = −2 𝑚𝑚,
𝜃1 = −0.0006 𝑟𝑎𝑑, 𝜃2 = 0 𝑟𝑎𝑑, 𝑎𝑛𝑑 𝜃3 = 0.0006 𝑟𝑎𝑑.
Reaction calculation

Example 2: Simply supported Beam with Uniformly Distributed load
For the beam shown in Figure below, determine displacements, slopes, reactions, Maximum bending moment,
shear force, and maximum bending stress. 𝑻𝒂𝒌𝒆 𝑬 = 𝟐𝟏𝟎 𝑮𝑷𝒂 𝒂𝒏𝒅 𝑰 = 𝟐 × 𝟏𝟎 − 𝟒𝒎𝟒
.The beam has
rectangular cross-section of depth 𝒉 = 𝟏𝒎.
Solution
(I)Analytical method
Reaction,
𝜃𝐶 = 0, by symmetry.
Maximum bending moment,
Shear force,
Maximum bending stress,
Bending moment diagram. Shear force diagram.
(II)FEM calculations

Stiffness matrices are,
Due to symmetry,
Nodal force calculation
For element1,
Nodal forces and moments for element1is,
For element2,
Due to symmetry,

Global equation is,
𝑤1 = 𝑤3 = 0.
𝑤2 = −0.0102𝑚,
𝜃1 = −0.0036𝑟𝑎𝑑, 𝜃2 = 0𝑟𝑎𝑑, 𝑎𝑛𝑑 𝜃3 = 0.0036𝑟𝑎𝑑.
Reactions are calculated from 1st
and 5th
rows of global matrix.
Similarly from 5th
row
𝑅3𝑦 = 22.865 𝑘𝑁.

Example 3: Simply supported Beam with Uniformly Varying Load
For the beam shown in Figure 5.4, determine displacements, slopes, and reactions. 𝐓𝐚𝐤𝐞 𝐄 = 𝟐𝟎𝟎 𝐆𝐏𝐚 𝐚𝐧𝐝 𝐈 =
𝟔. 𝟐𝟓 × 𝟏𝟎 − 𝟒 𝐦𝟒
.
Solution
(I) Analytical method
Reaction,
(II) FEM calculations

Stiffness matrices are,
Due to symmetry,
Global stiffness matrix is,
Load vector,
For element 1,

For element 2,
Global load vector is,
Global equation is,
𝑤1 = 𝑤3 = 0.

Example 4: Cantilever Beam with Point Load
Beam subjected to concentrated load. For the beam shown in Figure 5.16, determine the deflections and reactions.
Let E = 210 GPa and I = 2 × 10−4 m4. Take 2 elements.
Solution
(I) FEM by hand calculations
Element stiffness matrix for element 1 is,
Element stiffness matrix for element 2 is,
Global stiffness matrix is,
The global equations are,

By using the elimination method for applying boundary conditions,
w1 = θ1 = 0.
The above matrix reduces to,
By solving the above matrix and equations, we get, Deflections and slopes as

Force Vector Calculation Formulas
Displacement Vector or Boundary Conditions

Example Problems:
8) Draw Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for Cantilever subjected to Point
Load and Uniformly Distributed loads (UDL) 𝑻𝒂𝒌𝒆 𝒀𝒐𝒖𝒏𝒈’𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝒂𝒔 𝟔𝟗 𝑮𝑷𝒂 = 𝟔𝟗 × 𝟏𝟎𝟗
𝑵/𝒎𝟐
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝑌𝑜𝑢𝑛𝑔’𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 69 × 109
𝑁/𝑚2
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐼 =
𝑏×ℎ3
12
=
0.1×0.063
12
= 1.8 × 10−6
𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝟏: 𝑵𝒐𝒅𝒆 𝟏 − 𝟐
𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑚𝑎𝑡𝑟𝑖𝑥 [𝐾1] =
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
[𝐾1] =
(69×109)(1.8×10−6)
53
[
12 6 × 5 −12 6 × 5
6 × 5 4 × 52
−6 × 5 2 × 52
−12 −6 × 5 12 −6 × 5
6 × 5 2 × 52
−6 × 5 4 × 52
]
[𝐾1] = 993.6
[
𝑉1 𝜃1 𝑉2 𝜃2
12 30 −12 30 𝑉1
30 100 −30 50 𝜃1
−12 −30 12 −30 𝑉2
30 50 −30 100 𝜃2]
=
[
𝑉1 𝜃1 𝑉2 𝜃2
11923.2 29808 −11923.2 29808 𝑉1
29808 99360 −29808 49680 𝜃1
−11923.2 −29808 11923.2 −29808 𝑉2
29808 49680 −29808 99360 𝜃2]
𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝟐: 𝑵𝒐𝒅𝒆 𝟐 − 𝟑
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
[𝐾2] =
(69×109)(1.8×10−6)
43
[
12 6 × 4 −12 6 × 4
6 × 4 4 × 42
−6 × 4 2 × 42
−12 −6 × 4 12 −6 × 4
6 × 4 2 × 42
−6 × 4 4 × 42
] = 1940.625
[
𝑽𝟐 𝜽𝟐 𝑽𝟑 𝜽𝟑
12 24 −12 24 𝑽𝟐
24 64 −24 32 𝜽𝟐
−12 −24 12 −24 𝑽𝟑
24 32 −24 64 𝜽𝟑]

[𝐾2] =
[
𝑽𝟐 𝜽𝟐 𝑽𝟑 𝜽𝟑
23287.5 46575 −23287.5 46575 𝑽𝟐
46575 124200 −46575 62100 𝜽𝟐
−23287.5 −46575 23287.5 −46575 𝑽𝟑
46575 62100 −46575 124200 𝜽𝟑]
𝑮𝒍𝒐𝒃𝒂𝒍 𝑺𝒕𝒊𝒇𝒇𝒏𝒆𝒔𝒔 𝒎𝒂𝒕𝒓𝒊𝒙
[𝐾] = [𝐾1] + [𝐾2]
=
[
𝑉1 𝜃1 𝑉2 𝜃2
11923.2 29808 −11923.2 29808 𝑉1
29808 99360 −29808 49680 𝜃1
−11923.2 −29808 11923.2 −29808 𝑉2
29808 49680 −29808 99360 𝜃2]
+
[
𝑽𝟐 𝜽𝟐 𝑽𝟑 𝜽𝟑
23287.5 46575 −23287.5 46575 𝑽𝟐
46575 124200 −46575 62100 𝜽𝟐
−23287.5 −46575 23287.5 −46575 𝑽𝟑
46575 62100 −46575 124200 𝜽𝟑]
[𝐾] =
[
𝑽𝟏 𝜽𝟏 𝑽𝟐 𝜽𝟐 𝑽𝟑 𝜽𝟑
11923.2 29808 −11923.2 29808 0 0 𝑽𝟏
29808 99360 −29808 49680 0 0 𝜽𝟏
−11923.2 −29808 11923.2 + 23287.5 −29808 + 46575 −23287.5 46575 𝑽𝟐
29808 49680 −29808 + 46575 99360 + 124200 −46575 62100 𝜽𝟐
0 0 −23287.5 −46575 23287.5 −46575 𝑽𝟑
0 0 46575 62100 −46575 124200 𝜽𝟑]
[𝐾] =
[
11923.2 29808 −11923.2 29808 0 0
29808 99360 −29808 49680 0 0
−11923.2 −29808 35210.7 16767 −23287.5 46575
29808 49680 16767 223560 −46575 62100
0 0 −23287.5 −46575 23287.5 −46575
0 0 46575 62100 −46575 124200]
𝑭𝒐𝒓𝒄𝒆 𝑽𝒆𝒄𝒕𝒐𝒓
[𝑭] = [𝑭]𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝟏 + [𝑭]𝒆𝒍𝒆𝒎𝒆𝒕 𝟐 = [
𝑭𝟏
𝑴𝟏
𝑭𝟐𝒂
𝑴𝟐𝒂
] + [
𝑭𝟐𝒃
𝑴𝟐𝒃
𝑭𝟑
𝑴𝟑
] =
[
𝑭𝟏
𝑴𝟏
𝑭𝟐𝒂 + 𝑭𝟐𝒃
𝑴𝟐𝒂 + 𝑴𝟐𝒃
𝑭𝟑
𝑴𝟑 ]
[𝑭]𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝟏 = [
𝑭𝟏
𝑴𝟏
𝑭𝟐𝒂
𝑴𝟐𝒂
] =
[
−
𝑷 × 𝒃
𝑳
−
𝑷 × 𝒂 × 𝒃𝟐
𝑳𝟐
−
𝑷 × 𝒂
𝑳
−
𝑷 × 𝒃 × 𝒂𝟐
𝑳𝟐 ]
=
[
−
𝟔 × 𝟐
𝟓
−
𝟔 × 𝟑 × 𝟐𝟐
𝟓𝟐
−
𝟔 × 𝟑
𝟓
−
𝟔 × 𝟐 × 𝟑𝟐
𝟓𝟐 ]
= [
−𝟐. 𝟒
−𝟐. 𝟖𝟖
−𝟑. 𝟔
−𝟒. 𝟑𝟐
]
[𝑭]𝒆𝒍𝒆𝒎𝒆𝒕 𝟐 = [
𝑭𝟐𝒃
𝑴𝟐𝒃
𝑭𝟑
𝑴𝟑
] =
[
−
𝑷 × 𝑳
𝟐
−
𝑷 × 𝑳𝟐
𝟏𝟐
−
𝑷 × 𝑳
𝟐
+
𝑷 × 𝑳𝟐
𝟏𝟐 ]
−
[
𝟓 × 𝟒
𝟐
𝟓 × 𝟒𝟐
𝟏𝟐
𝟓 × 𝟒
𝟐
+
𝟓 × 𝟒𝟐
𝟏𝟐 ]
= [
−𝟏𝟎
−𝟔. 𝟔𝟔
−𝟏𝟎
+𝟔. 𝟔𝟔
]

[𝑭] = [𝑭]𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝟏 + [𝑭]𝒆𝒍𝒆𝒎𝒆𝒕 𝟐 = [
𝑭𝟏
𝑴𝟏
𝑭𝟐𝒂
𝑴𝟐𝒂
] + [
𝑭𝟐𝒃
𝑴𝟐𝒃
𝑭𝟑
𝑴𝟑
] =
[
𝑭𝟏
𝑴𝟏
𝑭𝟐𝒂 + 𝑭𝟐𝒃
𝑴𝟐𝒂 + 𝑴𝟐𝒃
𝑭𝟑
𝑴𝟑 ]
=
[
−𝟐.𝟒
−𝟐. 𝟖𝟖
−𝟑. 𝟔 − 𝟏𝟎
−𝟒. 𝟑𝟐 − 𝟔. 𝟔𝟔
−𝟏𝟎
+𝟔. 𝟔𝟔 ]
=
[
−𝟐.𝟒
−𝟐. 𝟖𝟖
−𝟏𝟑.𝟔
−𝟏𝟎.𝟗𝟖
−𝟏𝟎
+𝟔. 𝟔𝟔 ]
𝑫𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝑽𝒆𝒄𝒕𝒐𝒓
[𝑼] =
[
𝑉1
𝜽𝟏
𝑉2
𝜽𝟐
𝑉3
𝜽𝟑]
=
[
0
𝟎
0
𝜽𝟐
0
𝜽𝟑]
𝒃𝒖𝒕 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝒃𝒐𝒖𝒏𝒅𝒂𝒓𝒚 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏
𝒂𝒕 𝑵𝒐𝒅𝒆 𝟏: 𝑽𝟏 = 𝟎 & 𝜽𝟏 = 𝟎 𝒇𝒊𝒙𝒆𝒅
𝑨𝒕 𝑵𝒐𝒅𝒆 𝟐: 𝑽𝟐 = 𝟎 𝒓𝒐𝒍𝒍𝒆𝒓 𝒔𝒖𝒑𝒑𝒐𝒓𝒕 𝒂𝒏𝒅 𝒂𝒕 𝑨𝒕 𝑵𝒐𝒅𝒆 𝟑: 𝑽𝟑 = 𝟎 𝒓𝒐𝒍𝒍𝒆𝒓 𝒔𝒖𝒑𝒑𝒐𝒓𝒕
[𝑲] × [𝑼] = [𝑭]
[
11923.2 29808 −11923.2 29808 0 0
29808 99360 −29808 49680 0 0
−11923.2 −29808 35210.7 16767 −23287.5 46575
29808 49680 16767 223560 −46575 62100
0 0 −23287.5 −46575 23287.5 −46575
0 0 46575 62100 −46575 124200]
×
[
0
𝟎
0
𝜽𝟐
0
𝜽𝟑]
=
[
−𝟐.𝟒
−𝟐.𝟖𝟖
−𝟏𝟑.𝟔
−𝟏𝟎.𝟗𝟖
−𝟏𝟎
+𝟔.𝟔𝟔 ]
𝑩𝒚 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒏 𝒐𝒓 𝑨𝒑𝒑𝒍𝒚𝒊𝒏𝒈 𝑩𝒐𝒖𝒏𝒅𝒂𝒓𝒚 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏𝒔 𝒘𝒆 𝒈𝒆𝒕:
[223560 62100
62100 124200
] × [𝜽𝟐
𝜽𝟑
] = [
−𝟏𝟎.𝟗𝟖
+𝟔. 𝟔𝟔
]
𝑺𝒊𝒎𝒑𝒍𝒆 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝑴𝒆𝒕𝒉𝒐𝒅
[223560 62100
62100 124200
] × [𝜽𝟐
𝜽𝟑
] = [
−𝟏𝟎.𝟗𝟖
+𝟔.𝟔𝟔
]
[
223560 × 𝜽𝟐 + 62100 × 𝜽𝟑
62100 × 𝜽𝟐 + 124200 × 𝜽𝟑
] = [
−𝟏𝟎.𝟗𝟖
+𝟔. 𝟔𝟔
]
223560 × 𝜽𝟐 + 62100 × 𝜽𝟑 = −𝟏𝟎. 𝟗𝟖… … 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏
62100 × 𝜽𝟐 + 124200 × 𝜽𝟑 = 𝟔. 𝟔𝟔… … 𝒆𝒖𝒂𝒕𝒊𝒐𝒏 𝟐
𝜽𝟐 = −0.000074333800841514726507
𝜽𝟑 = 0.000090790088826554464703

9) Draw Shear force Diagram (SFD) and Bending Moment Diagram (BMD) for Cantilever subjected to
Point Load and Uniformly Distributed loads (UDL) 𝒀𝒐𝒖𝒏𝒈’𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟐𝟎𝟎 × 𝟏𝟎𝟑
𝑵/𝒎𝟐
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝑌𝑜𝑢𝑛𝑔’𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 200 × 103
𝑁/𝑚2
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐼 =
𝑏×ℎ3
12
=
0.1×0.23
12
= 6.667 × 10−5
𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝟏: 𝑵𝒐𝒅𝒆 𝟏 − 𝟐
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
[𝐾1] =
(200×103)(6.667×10−5)
33
[
12 6 × 3 −12 6 × 3
6 × 3 4 × 32
−6 × 3 2 × 32
−12 −6 × 3 12 −6 × 3
6 × 3 2 × 32
−6 × 3 4 × 32
] = 0.4938
[
𝑽𝟏 𝜽𝟏 𝑽𝟐 𝜽𝟐
12 18 −12 18 𝑽𝟏
18 36 −18 18 𝜽𝟏
−12 −18 12 −18 𝑽𝟐
18 18 −18 36 𝜽𝟐]
[𝐾1] =
[
𝑉1 𝜃1 𝑉2 𝜃2
59256 8.8884 −59256 8.8884 𝑉1
8.8884 17.7768 −8.8884 8.8884 𝜃1
−59256 −8.8884 59256 −8.8884 𝑉2
8.8884 8.8884 −8.8884 17.7768 𝜃2]
𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝟐: 𝑵𝒐𝒅𝒆 𝟐 − 𝟑
𝐸𝐼
𝐿3
[
12 6𝐿 −12 6𝐿
6𝐿 4𝐿2
−6𝐿 2𝐿2
−12 −6𝐿 12 −6𝐿
6𝐿 2𝐿2
−6𝐿 4𝐿2
]
[𝐾2] =
(200×103)(6.667×10−5)
23
[
12 6 × 2 −12 6 × 2
6 × 2 4 × 22
−6 × 2 2 × 22
−12 −6 × 2 12 −6 × 2
6 × 2 2 × 22
−6 × 2 4 × 22
] = 1.665
[
𝑽𝟐 𝜽𝟐 𝑽𝟑 𝜽𝟑
12 12 −12 12 𝑽𝟐
12 16 −12 8 𝜽𝟐
−12 −12 12 −12 𝑽𝟑
12 8 −12 16 𝜽𝟑]

Mechanics of Machine lab manual c 20 20ME31P

Mechanics of Machine lab manual c 20 20ME31P

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