MOM_2022_23_combined.pdf

Mechanics of Materials 20ME31P
Notes by Vidyadhara C A| P a g e 1
Mechanics of Materials Lecture WEEK-01
Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of
bodies that are subjected to the action of forces.
This subject can be subdivided into three branches: rigid-body mechanics, deformable-body
mechanics, and fluid mechanics.
Rigid-body mechanics is divided into two areas; Statics and Dynamics.
 Statics deals with the equilibrium of bodies, that is, those that are either at rest or move with
a constant velocity;
 Dynamics is concerned with the accelerated motion of bodies.
ThesubjectofDynamicsmaybefurthersub-dividedintothefollowingtwobranches
1. Kinetics,and2.Kinematics.
Kinetics It is that branch of Dynamics,which deals with the bodies in motion due to the application of forces.
Kinematics It is that branch of Dynamics, which deals with the bodies in motion, without referring to the forces
which are responsible for the motion.
Force:Theforce isan agent which produces or tends to produce, destroys or tends to destroy motion.
Unit: Si unit of force is ‘newton’, abbreviated as ‘N’.
Effects of a Force
A force may produce the following effects in a body, on which in
1. It may change the motion of the body, i.e., if a body is at rest, the force moves the motion,
andifthebodyisalreadyin motion,theforcemayaccelerateit.
2. Itmayretardthemotionofabody.
3. Itmayretardtheforces,alreadyacting onabody,thusbringingittorestin
4. itmaygiverisetotheinternalstressesinthebody,onwhichitacts.
Characteristics of a force
1. Magnitudeoftheforce
2. The direction of the line,alongwhichthe force acts.It isalso known as line of action of the force.
3. Nature of the force (i.e., whether the force is push or pull). This is denoted, by placing arrow
head onthe line of action ofthe force.
The point at which (or through which) the force acts on the body.
Types of Forces
When two, or more than two, forces act on a body, they are called to form a system of forces.
Following systems of forces are important from the subject point of view:
1. Coplanar forces. The forces, whose lines of action lie on the same plane, are known as coplanar
forces.

2. Collinearforces.Theforces,whoselinesofactionlieonthesameline,-areknownascollinearforces.
3. Concurrent forces. The forces, whose meet at one point, are known as concurrent forces. The
concurrent forces may or may not be collinear.
4. Coplanar concurrent forces. The forces, which meet at one point and their lines of action also lie
on the same plane, are known as coplanar concurrent forces.
5. Coplanar non-concurrent forces. The forces which do not meet at one point, but their lines of
action lie on the same plane, are known as coplanar concurrent forces.
6. Non-coplanar concurrent forces. The forces, which meet at one point, but their lines of action
do not lie on the same plane, are known are known as non-coplanar concurrent forces.
7. Non-coplanar non-concurrent forces. The forces, which do not meet at one point and their lines of
action do not lie.on-thesame plane,are callednon-coplanar non-concurrentforces.,
Resultant Force
If a number of forces, are acting simultaneously on a particle, it is possible to find out a single force
which could replace them i.e., which would produce the same effect as produced by all the given forces. This
single force is called resultant force, and the given forces are called component forces.
Composition of forces
The process of finding out the resultant force of a number of given forces is called composition of
forcesorcomponent of forces
Parallelogram Law of forces
It states "If tit forces, acting simultaneously on a particle, be represented in magnitude and
direction by the two adjacent sides of a parallelogram; their resultant may be
represented in magnitude and direction by the diagonal of the parallelogram, which passes
through their point of intersection".
Mathematically, resultant forces:
𝑹 = √𝑭𝟏
𝟐
+ 𝑭𝟏
𝟐
+ 𝟐 𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔𝜽

Resolution of a force
Theprocessorsplittingthe,givenforceintotwoforcesalongtwoperpendiculardirectionsarecalled“resolutionofforces”.
Method of Resolution for the Resultant force
1. Fx = F cos
2. Fy = F sin
3. 𝐅 = √𝑭𝒙
𝟐
+ 𝑭𝒚
𝟐
4. 𝐭𝐚𝐧 =
𝑭𝒚
𝑭𝒙
Equilibrium: The state of a body at rest or in uniform motion, the resultant of all forces on which is
zero.
The conditions for a body to be in Static Equilibrium
1. The first condition of equilibrium states that for an object to remain in equilibrium, the net
force acting upon it in all directions must be zero.
2. The second condition of equilibrium states that the net torque acting on the object must be
zero.
Lami's Theorem:
**End**

Mechanics of Materials
Lecture
02
Force (P or F): The force is an agent which produces or tends to produce, destroys or tends to destroy
motion.
Mathematically, Force = Mass X acceleration
SI Unit: ‘newton’

By definition:
Stress is given by:
𝑆𝑡𝑟𝑒𝑠𝑠 =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
 =
𝑃
𝐴
……………………………………………………. (1)
By definition:
Strain is given by:
𝑆𝑡𝑟𝑎𝑖𝑛 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
 =
𝑙
𝑙
…………………………………………………. (2)
Applying Hooke’s Law:
Stress = Strain X Modulus of elasticity
i.e.  =  𝑋 𝐸…………………………………………….(3)
Substituting the values of Stress and strain from Eq. (1) and Eq. (2) into Eq. (3):
𝑃
𝐴
=
𝑙
𝑙
𝑋 𝐸
𝒍 =
𝑷 𝒍
𝑨 𝑬

**End**
-----****-----

Mechanics of Materials Lecture 03
Working Stress or design Stress or allowable Stress ( w / d )
The safe stress which is allowed to be undertaken by the material in designs is called ‘Working Stress or
design Stress or allowable Stress’.
Factor of Safety (FOS)
The ratio of working stress to stress elastic limit or ultimate stress is called ‘factor of safety’.
𝐅𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝐒𝐚𝐟𝐞𝐭𝐲 =
𝐒𝐭𝐫𝐞𝐬𝐬 𝐚𝐭 𝐞𝐥𝐚𝐬𝐭𝐢𝐜 𝐥𝐢𝐦𝐢𝐭 𝐨𝐫 𝐮𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐬𝐭𝐫𝐞𝐬𝐬
𝐖𝐨𝐫𝐤𝐢𝐧𝐠 𝐒𝐭𝐫𝐞𝐬𝐬
 For the structural steel work (when subjected to gradually increasing loads):
𝐒𝐭𝐫𝐞𝐬𝐬 𝐚𝐭 𝐞𝐥𝐚𝐬𝐭𝐢𝐜 𝐥𝐢𝐦𝐢𝐭
 It is taken as 2-2.5
 In case of cast iron, concrete, wood etc. (when structural steel work is subjected to suddenly
loads):
𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐬𝐭𝐫𝐞𝐬𝐬
 It is taken as 4-6

Elastic Constants
𝐋𝐢𝐧𝐞𝐚𝐫 𝐬𝐭𝐫𝐚𝐢𝐧 =
𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐭𝐡𝐞 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐫𝐜𝐞
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
Secondary or Lateral strain
It is the strain in the direction right angles to the application of force.
𝐋𝐚𝐭𝐞𝐫𝐚𝐥 𝐬𝐭𝐫𝐚𝐢𝐧 =
𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝐩𝐞𝐫𝐩𝐞𝐧𝐝𝐢𝐜𝐮𝐥𝐚𝐫 𝐭𝐨 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐫𝐜𝐞
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏
 =
𝒍
𝒍

𝐏𝐨𝐢𝐬𝐬𝐨𝐧′
𝐬 𝐑𝐚𝐭𝐢𝐨 =
𝐋𝐚𝐭𝐞𝐫𝐚𝐥 𝐬𝐭𝐫𝐚𝐢𝐧
𝐋𝐢𝐧𝐞𝐚𝐫 𝐬𝐭𝐫𝐚𝐢𝐧
𝟏
𝒎
=
𝐋𝐚𝐭𝐞𝐫𝐚𝐥 𝐬𝐭𝐫𝐚𝐢𝐧


Bulk Modulus ( K )
When a body is subjected to three mutually perpendicular stresses of equal intensity, the ratio of direct
stress to xcorresponding volumetric strain is known as ‘bulk modulus’.
𝐁𝐮𝐥𝐤 𝐌𝐨𝐝𝐮𝐥𝐮𝐬 =
𝐃𝐢𝐫𝐞𝐜𝐭 𝐬𝐭𝐫𝐞𝐬𝐬
𝐕𝐨𝐥𝐮𝐦𝐞𝐭𝐫𝐢𝐜 𝐬𝐭𝐫𝐚𝐢𝐧
Relation between Bulk Modulus and Young’s Modulus
Relation between Modulus of Elasticity and Modulus of Rigidity
Relation between Modulus of Elasticity, Modulus of Rigidity and Bulk Modulus
**End**
𝐊 =

𝒗
𝑲 =
𝐦 𝐄
𝟑(𝒎 − 𝟐)
𝑪 =
𝐦 𝐄
𝟐(𝒎 + 𝟏)
𝑬 =
𝟗 𝐂 𝐊
𝟑 𝑲 + 𝑪

Bending Moment and Shear Force in Beams
A beam is a structural member subjected to a system of external forces at right angles to its axis.
Types of Beams:
a. Cantilever beam: It is the beam that is fixed at one end and free at another.
b. Simply supported beam: It is that beam, having its ends freely resting on its supports.
c. Fixed beam: It is that beam, having its both the ends rigidly fixed or built into its supports.
d. Continuous beam: It is that beam that has got many supports.
e. Over hanging beam: It is that beam, having its both the ends extended over the supports.
The concept of Loading in Beams
The loads applied to the beam result in reaction forces at the beam's support points. The total effect of
all the forces acting on the beam is to produce shear forces and bending moments within the beams,
that in turn induce internal stresses, strains and deflections of the beam.

The Concept of Shear Force and Bending Moment in Beams
Shear Force(S.F.): The shear force at the cross-section of a beam is unbalanced vertical force to the
right or left of the section.
It tends to slide one portion of the beam, upwards or downwards with respect to the other.
It is the algebraic sum of all types of vertical forces at a particular section.
Sign Convention:
Bending Moment(B.M.): The bending moment at the cross-section of a beam is the algebraic sum of
the moments of all forces , to the right or left of the section.
It tends to bend the beam at that point to a curvature having concavity the top or convexity at the top.
It is the algebraic sum of all the moments at a particular section.
Sign Convention:

Types of Loads acting on Beams
**End**

Bending stresses in Simple Beams
The bending moment at a section beams tends to bend or deflect the beam at a section tends to bend
or deflect the beam and the internal stresses resist its bending. The process of bending stops, when
every cross- section sets up full resistance to the bending moment. The resistance offered by the
internal stresses, to the bending is called bending stresses.
This theory is called the theory of simple bending.
Assumptions in the Theory of Simple Bending
1. The beam used is of homogeneous material.
2. The beam material is stressed within the elastic limit and thus obeys Hooke’s law.
3. The transverse sections which were plane before bending, remains plane after bending.
4. Each layer of the beam is free to expand or contract, independently of the layer below or above
it.
5. Modulus of elasticity of the beam remains same for tension and compression.
6. The beam is in equilibrium.
Bending Equation
Where
M = Moment acting at the beam
I = Moment of inertia of the beam section
R = radius of curvature of the beam
 = bending stress
y = the distance of the point from the neutral axis
E = Modulus of elasticity of the beam
Neutral Axis: The line of intersection of the neutral layer, with any normal crosesw- section of a
beam is known as neutral axis of that section.
**End**
𝑀
𝐼
=

𝑦
=
𝐸
𝑅

Finite Element Methods (FEM)
Definition: The Finite Element Method (FEM) is a numerical analysis technique used by
engineers, scientists, and mathematicians to obtain solutions (approximate solutions) to the
differential equations or partial differential equations that describe, or approximately describe a
wide variety of physical and non-physical problems. Physical problems range in diversity from
solid, fluid and soil mechanics, to electromagnetism or dynamics.
BACKGROUND OF FINITE ELEMENT METHOD
Sl.No. Year
1 1942 Hrennikoff introduced the framework method, in which a plane elastic
medium was represented as collections of bars and beams.
These pioneers share one essential characteristic: mesh discretization of a
continuous domain into a set of discrete sub-domains, usually
called elements.
2 1950s 1. solution of large number of simultaneous equations became possible
because of the
digital computer.
3 1960 Ray W. Clough first published a paper using term "Finite Element Method".
4 1965 First conference on "finite elements" was held.
5 1967 the first book on the "Finite Element Method" was published by Thenkiewicz and
Chung.
6 1960-
1970
FEM was applied to a wide variety of engineering problems.
7 1970 Most commercial FEM software packages (ABAQUS, NASTR AN, ANSYS,
etc.) originated. Interactive FE programs on super Computer lead to rapid growth of
CAD systems
8 1980 Algorithm on electromagnetic applications, fluid flow and thermal analysis were
developed with the tine of FE program.
9 1990 Engineers can evaluate ways to control the vibrations and extend the use of flexible,
deployable structures in space
Trends to solve fully coupled solution of fluid flows with structural interactions, bio
mechanics related problems with a higher level of
accuracy was observed in this decade.
Basic Steps in FEM
The step by step procedure with reference to the static structural problems can be stated as follows:
1. Discretisation of the continuum: The first step in the finite element method is to divide
the given continuum into smaller regions of finite dimensions called as "Finite elements".
The original Continuum (body or structure) is then considered as an assemblage of these
elements connected at a finite number of joints called as "Nodes" or "Nodal Points". At
each node, unknown displacements are to be prescribed. The type, size, number, and
arrangement of the elements depend on the accuracy of the solution required.
2. Selection of approximating functions: Approximating functions are also known as the
displacement function or interpolation model. The displacement function may be
approximated in the form a linear function or a higher-order function. The shape or
geometry of the element may also be approximated.

3. Formation of the element stiffness matrix: After continuum is discretised with desired
element shapes, the individual element stiffness matrix is formulated.
4. Formation of overall stiffness matrix: After the element stiffness matrices in global
coordinates are formed, they are assembled to form the overall stiffness matrix. The
assembly is done through the nodes which are common to adjacent elements. The overall
stiffness matrix is symmetric and banded. Overall stiffness matrix is also known as the
global stiffness matrix.
5. Formation of the element loading matrix: The loading forms an essential parameter in
any structural engineering problem. The loading inside an element is transferred at the
nodal points and consistent element matrix is formed.
6. Formation of the overall loading matrix: the element loading matrices are assembled to
form the overall loading matrix. This matrix has one column per loading Case and it is
either a column vector or a rectangular matrix depending on the number of loading cases.
7. Formation of the overall equilibrium equation: Overall equilibrium equation is the
systematic
arrangement of the overall stiffness matrix, overall load vector and overall displacement
vector to get set of simultaneous equations.
Overall equilibrium equation can be expressed as shown below:
[k] {Q} = {F}
Where,
[k] is an overall or global stiffness matrix (Square matrix)
{Q} is an overall or global displacement vector (Column matrix)
{F) is an overall or global force vector (Column matrix)
8. Incorporation of boundary conditions: The boundary restraint conditions are to be
imposed
in the stiffness matrix to avoid the condition of singularity.
9. Calculation of unknown nodal displacements: After incorporation of boundary conditions,
elimination method or penalty methods of handling boundary condition are used to
calculate unknown nodal displacements from the equilibrium equation or simultaneous
equations.
10. Calculation of strain and stresses: Nodal displacements are utilized for the calculation of
strain and stresses using the suitable equations. This may be done for all elements of the
continuum or it may be limited to some predetermined elements
**-**
Advantages of FEM
1. The physical properties, which are intractable and complex for any closed bound solution, can
be analyzed by this method.
2. It can take care of any geometry (may be regular or irregular.
3. It can take care of any boundary conditions.
4. It can take care of any type of boundary conditions.
5. Material anisotropy and non-homogeneity can be catered without much difficulty.
6. This method is superior to other approximate methods like Galerkin and Rayleigh-Ritz methods.
7. In this method, finite element model can be altered very easily.
8. Inn this method, the nonlinear behavior existing with large deformation and nonlinear materials
can be handled easily.
9. Enable to computer programming.
Disadvantages of FEM
1. Computational time involved in the solution of the problem is high.

2. For fluid dynamics problems some other methods of analysis may prove efficient than the FEM.
3. The FEM is applied to an approximation of the mathematical model of a system (the source
so-celled inherited errors).
4. Experience and judgment are needed in order to construct a good finite element model.
5. Susceptible to user-introduced modeling errors:
i. Poor choice of element types.
ii. Distorted elements.
iii. Geometry not adequately modeled
6. Mistakes by users can be fatal.
Limitations of FEM
1. The accuracy of the obtained solution is usually a function of the mesh resolution. Special
efforts must be made to analyze suchproblems.
2. Proper engineering judgment is to be exercised to interpret results.
3. It requires large computer memory and computational time to obtain intend results.
4. There are certain categories of problems where. other methods are more effective, cage, fluid
problems having boundaries at infinity are better treated by the boundary element method.
5. For some problems, there may be a considerable amount of input data and desirable to check
it.
6. Many problems lead to round off errors as computer can handle limited no. of digits.
7. It is easy for users to make mistakes.
Applications of FEM
I. STRUCTURAL PROBLEMS
1. Stress analysis including bars. truss and frame analysis
2. Stress concentration problems typically associated geometric discontinuity (with holes, fillets or
other changes in geometry in a body).
3. Buckling Analysis:
Example: Connecting rod subjected to anal compression.
4. Vibration Analysis:
Example: A beam subjected to different types of loading.
II. NON - STRUCTURAL PROBLEMS
1. Heat Transfer analysis:
Example: Steady state thermal analysis on composite cylinder.
2. Fluid flow analysis.
Example: Fluid flow through pipes.
**-**
DISCRETISATION PROCESS
Discretization process means dividing the given continuum or problem into smaller regions called
“elements”.
The process of modeling a body by dividing it into an equivalent system of finite number of units is
called discretization and these small elements are called “finite elements”.

The consideration to be taken in the Discretization process are as follows:
1. Type of elements.
2. Size of elements.
3. Number of elements.
4. Location of nodes.
5. Node numbering scheme.
Finite Element Analysis (FEA)
It is the process of dividing or discretizing our geometry into finite nodes and elements and solving it
for stress and strains.
Meshing
It is the particular process of Discretization.
A mesh is made up of elements which contain nodes (coordinate locations in space that can vary by
element type) that represent the shape of the geometry.
Meshing is the process of turning irregular shapes into more recognizable volumes called “elements.”
In FEM, we create a mesh which splits the domain into a discrete number of elements for which the
solution can be calculated. The data is then interpolated across the whole domain.

In meshing, we divide the geometry into any one of the following shapes of elements like triangles,
quadrilaterals, tetrahedron, quadrilateral pyramids, triangular prisms, and hexahedron. This selection
of particular shape depends upon the type of analysis and shape of the geometry.
Phases of FEA
1. Pre-processing - in which the analyst develops a finite element mesh to divide the given
geometry into subdomains for mathematical analysis, and applies material properties and
boundary conditions.
2. Solution – during which the programme derives the governing matrix equations from model
and solves for primary quantities.
3. Post-processing - in which the analyst checks the validity of the solutions, examines the values
of primary quantities such as displacement and stresses and derives and examines additional
quantities such as specialized stresses and error indicators.
**-**

Stiffness Matrix of a Bar Element
The term originates from structural analysis. It is used to describe the matrix relation between force
and displacement.
Types of Stiffness Matrix
1. The elemental stiffness matrix corresponds to an individual element in a
structure/system/continuum.
2. The global stiffness matrix corresponds to the entire structure/system/continuum which is the
assemblage of /summation of all elemental stiffness matrices.
Direct method
In this method element properties are obtained by direct interpretation of physical structure by using
the basic principles of the mechanics of materials.
Analogy between Spring and a Bar element
For a spring:
𝑺𝒑𝒓𝒊𝒏𝒈 𝑺𝒕𝒊𝒇𝒇𝒏𝒆𝒔𝒔 =
𝑭𝒐𝒓𝒄𝒆
𝑫𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏
𝑲 =
𝑭
𝒖
or F = k U------------(1)
For a bar under axial loading:
𝒖 =
𝑭 𝒍
𝑨 𝑬
or
-------------(2)
From eq. 1 and 2:
𝑭 = (
𝑨 𝑬
𝒍
) 𝒖
𝒌 = (
𝑨 𝑬
𝒍
)

Stiffness Influence Coefficients kij
It is defined as the force needed to at node i to produce a unit displacement at node j while other
nodes restrained.
Stiffness influence coefficients are written in the matrix form to get over all stiffness matrix as
follows:
To derive Stiffness Matrix of a Bar Element
Let
u1 = nodal displacement at node 1
u2 = nodal displacement at node 2
If
F1 = nodal forces at node 1
F2 = nodal forces at node 2
Force due
to unit
deflection
= kij
j = 1 j = 2
i = 1
K11 = the force needed to at
node 1 to produce a unit
displacement at node 1
Node 2 fixed.
Node 1 fixed.
i= 2
Node 2 fixed.
Node 1 fixed.
𝒌 = ⌊
𝒌𝟏𝟏 𝒌𝟏𝟐
𝒌𝟐𝟏 𝒌𝟐𝟐
⌋

The equation of equilibrium for element is expressed as:
[𝒌] {𝑼} = {𝑭}
Where
[𝒌] = 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 𝒐𝒓 𝒈𝒍𝒐𝒃𝒂𝒍 𝒔𝒕𝒊𝒇𝒇𝒏𝒆𝒔𝒔 𝒎𝒂𝒕𝒓𝒊𝒙 − − − − − 𝑺𝒒𝒖𝒂𝒓𝒆 𝒎𝒂𝒕𝒓𝒊𝒙
{𝑼} = 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 𝒐𝒓 𝒈𝒍𝒐𝒃𝒂𝒍 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒗𝒆𝒄𝒕𝒐𝒓 − − − 𝑪𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒕𝒓𝒊𝒙
{𝑭} = 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 𝒐𝒓 𝒈𝒍𝒐𝒃𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 𝒗𝒆𝒄𝒕𝒐𝒓 − − − − − −𝑪𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒕𝒓𝒊𝒙
Or
⌊
𝑘11 𝑘12
𝑘21 𝑘22
⌋ {
𝑢1
𝑢2
} = {
𝐹1
𝐹2
}-------------------------------------(R1)
Where k11, k12……. = stiffness coefficients
Eq. (1), (2), (3), and (4) are combined as follows:
𝑭𝟏 = (
𝑨 𝑬
𝒍
). (𝒖𝟏 − 𝒖𝟐)_______ (4)
𝑭𝟐 = (
𝑨 𝑬
𝒍
). (−𝒖𝟏 + 𝒖𝟐)______ (5)
Eq. (4) and (5) can be written in the matrix form as follows:
(
𝑨 𝑬
𝒍
) ⌊
+1 −1
−1 +1
⌋ {
𝑢1
𝑢2
} = {
𝐹1
𝐹2
} -------(R2)
Comparing the results R1 and R2:
Displacement at node 1: Node 2 is fixed.
If F1 = force induced at node 1:
Then F1 is given by:
𝑭𝟏 = (
𝑨 𝑬
𝒍
). 𝒖𝟏-------------------(1)
And F2 = reaction at node 2:
𝑭𝟐 = − 𝑭𝟏 = − (
𝑨 𝑬
𝒍
). 𝒖𝟏 − (𝟐)
Displacement at node 2: Node 1 is fixed.
If F2 = force induced at node 2:
Then F1 is given by:
𝑭𝟐 = (
𝑨 𝑬
𝒍
) . 𝒖------------------------(3)
And F1 = reaction at node 2:
𝑭𝟏 = − 𝑭𝟐 = − (
𝑨 𝑬
𝒍
). 𝒖𝟐_____(𝟒)

⌊
𝑘11 𝑘12
𝑘21 𝑘22
⌋ = (
𝑨 𝑬
𝒍
) ⌊
+1 −1
−1 +1
⌋
or
is the required stiffness matrix for the bar
element
Strain Displacement Matrix / [B] Matrix:
Stress Matrix for 1 – D Bar Element
Global Stiffness Matrix
It is an assembly of all elemental stiffness in a systematic manner.
If there are two elements 1 and 2 in a bar element, then the Global Stiffness Matrix is given by:
[𝒌] = [𝒌𝟏] + [𝒌𝟐], where k1 and k2 are elemental stiffness matrixes of element 1 and element 2
respectively.
Properties of Stiffness Matrix
1. It is a symmetric matrix.
2. Stiffness Matrix obtained in the finite element analysis is a banded.
3. If there are n number of nodes the, global stiffness matrix is n X n, provided element is a one
dimensional and one degree of freedom at each node.
4. The main diagonal elements are always positive.
5. If rigid motion is not prevented, then stiffness matrix becomes singular.
**-**
𝒌 = (
𝑨 𝑬
𝒍
) ⌊
+1 −1
−1 +1
⌋
⌊𝑩⌋ = (
𝟏
𝒍
) [−𝟏 𝟏]
 = E ⌊𝑩⌋{𝒖}

Finite Element Method - Problems
1. The following figure shows a one –dimensional bar subjected to an axial loading. Taking it as a
single bar element, determine 1) Nodal displacement, 2) Stress in each element, 3) Reaction at the
support.
Solution:
1. Finite Element Model
Comparing both the figures:
Data:
Young’s modulus E = 2.1 X 105 N/mm2
Area of element A = 1963.5 mm2
Length of the element l = 300 mm
Load at node 2: F2 = 1500 N
2. Elemental matrix for element 1
The stiffness matrix for the 1-d bar element is given by:
For element 1:
[𝑲𝟏] = (
𝟏𝟗𝟔𝟑.𝟓 𝐗 𝟐.𝟏 𝐗 𝟏𝟎𝟓
𝟑𝟎𝟎
) ⌊
+1 −1
−1 +1
⌋ or
3. Global stiffness matrix
Since there is only one element, Global stiffness matrix = elemental stiffness matrix.
[𝑲𝟏] = 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 ⌊
+1 −1
−1 +1
⌋
[𝑲 ] = (
𝑨 𝑬
𝒍
) ⌊
+1 −1
−1 +1
⌋
⌈𝑲⌉ = [𝑲𝟏] = 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 ⌊
+1 −1
−1 +1
⌋

4. Global nodal displacement vector
The nodal displacement vector for bar is given by:
{𝑼} = {
𝒖𝟏
𝒖𝟐
}
5. Global load vector
The global load vector for bar is given by:
{𝑭} = {
𝑭𝟏
𝑭𝟐
}
= {
𝟎
𝟏𝟓𝟎𝟎
}
6. Equilibrium conditions for the given bar
The equilibrium conditions for the bar is given by:
[𝒌] {𝑼} = {𝑭}
𝟏𝟑𝟕𝟒𝟒𝟓𝟎 ⌊
+1 −1
−1 +1
⌋ {
𝒖𝟏
𝒖𝟐
} = {
𝟎
𝟏𝟓𝟎𝟎
}
7. Applying boundary conditions
Boundary conditions for node 1: u1 = 0
i.e. 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 ⌊
+1 −1
−1 +1
⌋ {
𝟎
𝒖𝟐
} = {
𝟎
𝟏𝟓𝟎𝟎
}
1374450 X [(-1X0) + (1Xu2)] = 1500
1374450 X (u2) = 1500
U2 = 0.00109 mm
The nodal displacement vector is:
{𝑼} = {
𝟎
𝟎. 𝟎𝟎𝟏𝟎𝟗
}
8. Stress in each element
Stress Matrix for 1 – D Bar Element is
Where
Strain displacement ⌊𝑩⌋ = (
𝟏
𝒍
) [−𝟏 𝟏] and {𝑼} = {
𝒖𝟏
𝒖𝟐
}
Therefore,  = (
𝟐.𝟏 𝑿 𝟏𝟎𝟓
𝟑𝟎𝟎
) [−𝟏 𝟏] {
𝟎
𝟎. 𝟎𝟎𝟏𝟎𝟗
}
= (𝟔𝟔𝟔. 𝟔𝟔) [−𝟏 𝟏] {
𝟎
𝟎. 𝟎𝟎𝟏𝟎𝟗
}
= 666.66 ((0X-1) + (1X0.00109))
= 0.7266
 = E ⌊𝑩⌋{𝒖}

Stress in each element = 0.7266 N/mm2
9. Reaction at supports
The reaction forces at supports is given by:
{𝑹} = [𝒌] {𝑼} − {𝑭}
{
𝑹𝟏
𝑹𝟐
} = [𝒌] {
𝒖𝟏
𝒖𝟐
} − {
𝑭𝟏
𝑭𝟐
}
{
𝑹𝟏
𝑹𝟐
} = 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 ⌊
+1 −1
−1 +1
⌋ {
𝟎
𝟎. 𝟎𝟎𝟏𝟎𝟗
} − {
𝟎
𝟏𝟓𝟎𝟎
}
= 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 {
(𝟏𝑿𝟎) + (−𝟏𝑿𝟎. 𝟎𝟎𝟏𝟎𝟗)
(−𝟏𝑿𝟎) + (𝟏𝑿𝟎. 𝟎𝟎𝟏𝟎𝟗)
} − {
𝟎
𝟏𝟓𝟎𝟎
}
= 𝟏𝟑𝟕𝟒𝟒𝟓𝟎 {
−𝟎. 𝟎𝟎𝟏𝟎𝟗
𝟎. 𝟎𝟎𝟏𝟎𝟗
} − {
𝟎
𝟏𝟓𝟎𝟎
}
= {
−𝟏𝟒𝟗𝟖. 𝟏𝟓
𝟏𝟒𝟗𝟖. 𝟏𝟓
} − {
𝟎
𝟏𝟓𝟎𝟎
}
= {
−𝟏𝟒𝟗𝟖. 𝟏𝟓
𝟎
} N
# The reaction at node 1: R1 = -1498.15 N = 1500 N Approximately.
# The reaction at node 2: R2 = 0
The FEM solution is:
1. The nodal displacement vector is given by:
{𝑈} = {
𝑢1
𝑢2
} = {
0
0.00109
} mm
2. Stress in each element = 0.7266 N/mm2
3. The reaction forces at supports is given by:
{
𝑅1
𝑅2
} = {
−1498.15
0
} N
**-**

Finite Element Method - Problems – Beams
Finite Element Model of a Beam Element
l = Length of the beam (span)
E = Young’s Modulus of the beam element
I = Moment of inertia of the beam =
𝒃𝒅𝟑
𝟏𝟐
Breadth of the beam section = b
Depth of the beam section = d
1. Elemental Stiffness Matrix for the Beam Element
[𝐊] =
𝐄 𝐈
𝐥𝟑
[
𝟏𝟐 𝟔𝐥 −𝟏𝟐 𝟔𝐥
𝟔𝐥 𝟒𝐥𝟐
−𝟔𝐥 𝟐𝐥𝟐
−𝟏𝟐 −𝟔𝐥 𝟏𝟐 −𝟔𝐥
𝟔𝐥 𝟐𝐥𝟐
−𝟔𝐥 𝟒𝐥𝟐
]
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
}
Where
1 = Deflection at node 1
1 = Rotation at node 1
3. Global nodal load vector

{𝑭} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
}
Where
f1 = Force atnode 1
m1 = Moment atnode 1
f2 = Force atnode 2
m2 = Moment atnode 2
4. Equilibrium conditions
[𝒌] {𝑸} = {𝑭}
**______________________________________________________________________________***
Problem- Expt. No. 9 by FEM
Solution:
Data:
Load applied node 2 = P = 10000 N
Length of the beam element = l = 5 m
Young’s modulus E = 210 X 109
N/m2
The stiffness matrix for the 1-d beam element is given by:
[𝐊] =
𝐄 𝐈
𝐥𝟑
[
𝟏𝟐 𝟔𝐥 −𝟏𝟐 𝟔𝐥
𝟔𝐥 𝟒𝐥𝟐
−𝟔𝐥 𝟐𝐥𝟐
−𝟏𝟐 −𝟔𝐥 𝟏𝟐 −𝟔𝐥
𝟔𝐥 𝟐𝐥𝟐
−𝟔𝐥 𝟒𝐥𝟐
]
=
210 X 109
X
0.2 𝑋 0.33
12
𝟓𝟑
[
𝟏𝟐 𝟔𝐗𝟓 −𝟏𝟐 𝟔𝐗𝟓
𝟔𝐗𝟓 𝟒𝐗𝟓𝟐
−𝟔𝐗𝟓 𝟐𝐗𝟓𝟐
−𝟏𝟐 −𝟔𝐗𝟓 𝟏𝟐 −𝟔𝐗𝟓
𝟔𝐗𝟓 𝟐𝐗𝟓𝟐
−𝟔𝐗𝟓 𝟒𝐗𝟓𝟐
]

[𝐊] = 𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
]
Therefore:
[𝐊] = 𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
]
The nodal displacement vector for beam is given by:
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
}
Where
load vector = { 0 0 10000 0}𝑇
The global load vector for beam is given by:
{𝑭} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
} = {
𝟎
𝟎
𝟏𝟎𝟎𝟎
𝟎
}
Where
f1 = Force atnode 1 = 0
m1 = Moment atnode 1 = 0
The equilibrium conditions for the beam is given by:
[𝒌] {𝑸} = {𝑭}

𝐢. 𝐞. 𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
] {
𝟏
𝟏
𝟐
𝟐
} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
}
Boundary conditions for node 1- Fixed
1 = Deflection at node 1 = 0
1 = Rotation at node 1 =0
i.e.
𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
] {
𝟎
𝟎
𝟐
𝟐
} = {
𝟎
𝟎
𝟏𝟎𝟎𝟎𝟎
𝟎
}
Using Elimination method for rows and columns for zeros.
𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
] {
𝟎
𝟎
𝟐
𝟐
} = {
𝟎
𝟎
𝟏𝟎𝟎𝟎𝟎
𝟎
}
The equation takes the form:
𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 −𝟑𝟎
−𝟑𝟎 𝟏𝟎𝟎
] {
𝟐
𝟐
} = {
𝟏𝟎𝟎𝟎𝟎
𝟎
}
Now we have two equations;
Eq. 1: 756000((12 X 2 ) + ( -30 x 2 ) = -10000
Eq. 2: 756000((-30 X 2 ) + ( 100 x 2 ) = 0
Solving Eq.1 and Eq.2:
2 = -0.0044 m = -4.4 mm
2 = -0.0013 rad.
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
𝟎
𝟎
−𝟎. 𝟎𝟎𝟒𝟒
−𝟎. 𝟎𝟎𝟏𝟑
}

8. Internal load vectors
After 2 and 2 have been obtained, they are substituted back into the equilibrium equation to
obtain the reaction shear force and bending moment at node 1 and node 2.
The Shear force and Bending moments are given by:
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = [𝒌] {𝑸}
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = 𝟕𝟓𝟔𝟎𝟎𝟎 [
𝟏𝟐 𝟑𝟎 −𝟏𝟐 𝟑𝟎
𝟑𝟎 𝟏𝟎𝟎 −𝟑𝟎 𝟓𝟎
−𝟏𝟐 −𝟑𝟎 𝟏𝟐 −𝟑𝟎
𝟑𝟎 𝟓𝟎 −𝟑𝟎 𝟏𝟎𝟎
] {
𝟎
𝟎
−𝟎. 𝟎𝟎𝟒𝟒
−𝟎. 𝟎𝟎𝟏𝟑
}
At node 1:
Shear force F1 = 756000 X ((12X0) + (30X0) + (-12X-0.0044) + (30X-0.0013))- = 10432 N
Bending moment M1 = 756000 X ((30X0) + (100X0) + (-30X-0.0044) + (50X-0.0013)) = 50652 N-m
At node 2:
Shear force F2 = 756000 X ((-12X0) + (-30X0) + (12X-0.0044) + (-30X-0.0013)) = -10432 N
Bending moment M2 = = 756000 X ((30X0) + (50X0) + (-30X-0.0044) + (100X-0.0013)) = -1512 N-m
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
𝟎
𝟎
−𝟎. 𝟎𝟎𝟒𝟒
−𝟎. 𝟎𝟎𝟏𝟑
}
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
2. The reaction shear force and bending moment at node 1 and node 2.
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = {
𝟏𝟎𝟒𝟑𝟐
𝟓𝟎𝟔𝟓𝟐
𝟏𝟎𝟒𝟑𝟐
−𝟏𝟓𝟏𝟐
}
𝒎
𝑵 − 𝒎
𝒎
𝑵 − 𝒎
**-**

Finite Element Method - Problems – Beams – SSB with Point Load - Experiment No: 10
AIM: Draw the shear force and bending moment diagrams for the given beam due to applied load.
Problem Description:
Compute the shear force and bending moment diagrams for the beam shown. Assume
rectangular c/s area of 0.2 m * 0.3 m, Young ‘s modulus of 210 GPa, Poisson‘s ratio 0.27.
Solution:
Data:
Load applied node 2 = P = 20000 N
N/m2
2. Elemental matrix for element 1 and element 2
The stiffness matrix for the 1-d beam element 1 is given by:
[𝑲𝟏] =
𝐄 𝐈
𝐥𝟑
[
𝟏𝟐 𝟔𝐥 −𝟏𝟐 𝟔𝐥
𝟔𝐥 𝟒𝐥𝟐
−𝟔𝐥 𝟐𝐥𝟐
−𝟏𝟐 −𝟔𝐥 𝟏𝟐 −𝟔𝐥
𝟔𝐥 𝟐𝐥𝟐
−𝟔𝐥 𝟒𝐥𝟐
]
=
210 X 109
X
0.2 𝑋 0.33
12
𝟐𝟑
[
𝟏𝟐 𝟔𝐗𝟐 −𝟏𝟐 𝟔𝐗𝟐
𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟏𝟐 −𝟔𝐗𝟐 𝟏𝟐 −𝟔𝐗𝟐
𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
]

[𝑲𝟏] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
For beam element 2:
[𝑲𝟐] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
Global stiffness matrix = stiffness matrix element 1 + stiffness matrix element 2
 [𝐊] = [𝑲𝟏] + [𝑲𝟐]
[𝑲] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎
[
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐 𝟎 𝟎
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖 𝟎 𝟎
−𝟏𝟐 −𝟏𝟐 𝟐𝟒 𝟎 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟖 𝟎 𝟑𝟐 −𝟏𝟐 𝟖
𝟎 𝟎 −𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟎 𝟎 𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔 ]
{𝑸} =
{
𝟏
𝟏
𝟐
𝟐
𝟑
𝟑}
Where
load vector = { 0 0 20000 0 0 0}𝑇
{𝒇} =
{
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
𝒇𝟑
𝒎𝟑}
=
{
𝟎
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
𝟎
𝟎 }

Where
f2 = Force atnode 2 = - 10000
[𝒌] {𝑸} = {𝒇}
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎
[
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐 𝟎 𝟎
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖 𝟎 𝟎
−𝟏𝟐 −𝟏𝟐 𝟐𝟒 𝟎 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟖 𝟎 𝟑𝟐 −𝟏𝟐 𝟖
𝟎 𝟎 −𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟎 𝟎 𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔 ] {
𝟏
𝟏
𝟐
𝟐
𝟑
𝟑}
=
{
𝟎
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
𝟎
𝟎 }
Boundary conditions for
1. node 1: Pin support -
1 = Rotation at node 1  0
2. Node 2:
2 = Deflection at node 2  0
2 = Rotation at node 2 = 0
Node 2: Roller support-
3. since the beam is symmetrical: 3 = -1
i.e.
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎
[
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐 𝟎 𝟎
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖 𝟎 𝟎
−𝟏𝟐 −𝟏𝟐 𝟐𝟒 𝟎 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟖 𝟎 𝟑𝟐 −𝟏𝟐 𝟖
𝟎 𝟎 −𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟎 𝟎 𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔 ] {
𝟎
𝟏
𝟐
𝟎
𝟎
𝟑}
=
{
𝟎
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
𝟎
𝟎 }
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎
[
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐 𝟎 𝟎
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖 𝟎 𝟎
−𝟏𝟐 −𝟏𝟐 𝟐𝟒 𝟎 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟖 𝟎 𝟑𝟐 −𝟏𝟐 𝟖
𝟎 𝟎 −𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟎 𝟎 𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔 ] {
𝟎
𝟏
𝟐
𝟎
𝟎
𝟑}
=
{
𝟎
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
𝟎
𝟎 }

𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟔 −𝟏𝟐 𝟎
−𝟏𝟐 𝟐𝟒 𝟏𝟐
𝟎 𝟏𝟐 𝟏𝟔
] {
𝟏
𝟐
𝟑
} = {
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
}
8. Solution
Eq. 1: 11812500((16 X 1 ) + ( -12 X 2 )) = 0  16 1 - 12 2 =0
Eq. 2: 11812500((-12 X 1 ) + ( 24 X 2 ) + (12 X 3 ) = -20000  -1 + 2 2 + 3 = -0.000141
Eq. 3: 11812500((12 X 2 ) + ( -12 X 3 )) = 0  12 2 + 16 3 =0
Eq.4: 3 = -1
Solving Eq.1, Eq.2, Eq.3 and Eq.4:
1 = - 0.0002115 rad
2 = -0.000282 m
2 = 0.0002115 rad
9. Reactions
The reactions are the shear force and bending moment at node 1, node 2 and node 3.
{𝑭} = [𝒌] {𝑸} − {𝒇}
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
𝑭𝟑
𝒇𝟑 }
= 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎
[
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐 𝟎 𝟎
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖 𝟎 𝟎
−𝟏𝟐 −𝟏𝟐 𝟐𝟒 𝟎 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟖 𝟎 𝟑𝟐 −𝟏𝟐 𝟖
𝟎 𝟎 −𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟎 𝟎 𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔 ] {
𝟎
− 0.0002115
−0.000282
𝟎
𝟎
0.0002115 }
−
{
𝟎
𝟎
−𝟐𝟎𝟎𝟎𝟎
𝟎
𝟎
𝟎 }
At node 1:
Shear force F1 = 11812500 X ((12X-0.0002115) + (-12X -0.000282)) = 9993.37 N
Bending moment M1 = 11812500 X ((16X-0.0002115) + (-12X-0.000282)) = 5.12X10-12
N-m
At node 2:
Shear force F2 = 11812500 X ((-12X-0.0002115) + (24X -0.000282) + (-12X-0.0002115))-(-20000) = 13.25 N
Bending moment M2 = 11812500 X ((8X-0.0002115) X (8X-0.0002115)) = 0 N-m
At node 3:
Shear force F3 = 11812500 X ((-12X-0.000282) + (-12X-0.0002115)) = 9993.37 N
Bending moment M3 = 11812500 X ((12X-0.000282) X (16X-0.0002115)) = 5.12X10-12
N-m

{𝑸} =
{
𝟎
− 0.0002115
−0.000282
𝟎
𝟎
0.0002115 }
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
2. The reaction shear force and bending moment at node 1, node 2 and node 3:
{𝑭} =
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
𝑭𝟑
𝒇𝟑 }
=
{
9993.37
5.12X10−12
13.25
0
9993.37
5.12X10−12 }
𝑵
𝑵 − 𝒎
𝑵
𝑵 − 𝒎
𝑵
𝑵 − 𝒎
**-**

Finite Element Method - Problems – SSB with UDL (Experiment No.11)
Solution:
Data:
Load applied UDL = P0 = 10000 N/m
N/m2
[𝐊] =
𝐄 𝐈
𝐥𝟑
[
𝟏𝟐 𝟔𝐥 −𝟏𝟐 𝟔𝐥
𝟔𝐥 𝟒𝐥𝟐
−𝟔𝐥 𝟐𝐥𝟐
−𝟏𝟐 −𝟔𝐥 𝟏𝟐 −𝟔𝐥
𝟔𝐥 𝟐𝐥𝟐
−𝟔𝐥 𝟒𝐥𝟐
]
=
210 X 109
X
0.2 𝑋 0.33
12
𝟐𝟑
[
𝟏𝟐 𝟔𝐗𝟐 −𝟏𝟐 𝟔𝐗𝟐
𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟏𝟐 −𝟔𝐗𝟐 𝟏𝟐 −𝟔𝐗𝟐
𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
]
[𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
Therefore:

[𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
}
Where
load vector = {
𝑃 𝑙
2
𝑝 𝑙2
12
𝑃 𝑙
2
−𝑝 𝑙2
12
}𝑇
{𝒇} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
} = {
−𝟏𝟎𝟎𝟎𝟎
−𝟑𝟑𝟑𝟑. 𝟑
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
Where
[𝒌] {𝑸} = {𝑭}
𝐢. 𝐞 [𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟏
𝟏
𝟐
𝟐
} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
}
1. node 1: Pin support -
Node 2: Roller support-

3. since the beam is symmetrical: 3 = -1
i.e.
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟎
𝟏
𝟎
𝟐
} = {
−𝟏𝟎𝟎𝟎𝟎
−𝟑𝟑𝟑𝟑. 𝟑
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟎
𝟏
𝟎
𝟐
} = {
−𝟏𝟎𝟎𝟎𝟎
−𝟑𝟑𝟑𝟑. 𝟑
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟔 𝟖
𝟖 𝟏𝟔
] {
𝟏
𝟐
} = {
−𝟑𝟑𝟑𝟑. 𝟑
𝟑𝟑𝟑𝟑. 𝟑
}
8. Solution
Eq. 1: 11812500((16 X 1 ) + ( 8 x 2 )) = -3333.3
Eq. 2: 756000((8 X 1 ) + ( 16 x 2 )) = 3333.3
1 = -3.72 X 10-6
rad
2 = 3.72 X 10-6
rad
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
0
−3.72X10−6
0
3.72X10−6
}
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
{𝑭} = {
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = [𝒌] {𝑸} - {𝒇}

{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
0
−3.72X10−6
0
3.72X10−6
} − {
−𝟏𝟎𝟎𝟎𝟎
−𝟑𝟑𝟑𝟑. 𝟑
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
At node 1:
Shear force F1 = 11812500 X ((12X0) + (12X−3.72X10−6
) + (-12X0) + (-12X3.72X10−6
))-(-10000) =10000 N
Bending moment M1 = 11812500 X ((12X0) + (16X3.72X10−6
) + (12X0) + (8X3.72X10−6
))-(-3333.3)=2980.6 N-m
At node 2:
Shear force F2 = 11812500 X ((-12X0) + (-12X−3.72X10−6
) + (12X0) + (-12X3.72X10−6
))-(-10000) =-10000 N
Bending moment M2 =11812500 X ((12X0) + (8X−3.72X10−6
)+(-12X0)+(16X3.72X10−6
))-(3333.3)=-2980.6 N-m
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
0
−3.72X10−6
0
3.72X10−6
}
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
2. The reaction shear force and bending moment at node 1 and node 2.
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = {
𝟏𝟎𝟎𝟎𝟎
𝟐𝟗𝟖𝟎. 𝟔
−𝟏𝟎𝟎𝟎𝟎
−𝟐𝟗𝟖𝟎. 𝟔
}
𝒎
𝑵 − 𝒎
𝒎
𝑵 − 𝒎
**-**

Finite Element Method - Problems – CLB with UDL (Experiment No.12)
A cantilever beam is as shown. Assume rectangular c/s area of 0.2 m X 0.3 m, Young‘s modulus of 210
GPa, Poisson‘s ratio 0.27. Solve the problem by FEM.
Solution:
Data:
Load applied UDL = P0 = 10000 N/m
N/m2
[𝐊] =
𝐄 𝐈
𝐥𝟑
[
𝟏𝟐 𝟔𝐥 −𝟏𝟐 𝟔𝐥
𝟔𝐥 𝟒𝐥𝟐
−𝟔𝐥 𝟐𝐥𝟐
−𝟏𝟐 −𝟔𝐥 𝟏𝟐 −𝟔𝐥
𝟔𝐥 𝟐𝐥𝟐
−𝟔𝐥 𝟒𝐥𝟐
]
=
210 X 109
X
0.2 𝑋 0.33
12
𝟐𝟑
[
𝟏𝟐 𝟔𝐗𝟐 −𝟏𝟐 𝟔𝐗𝟐
𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟏𝟐 −𝟔𝐗𝟐 𝟏𝟐 −𝟔𝐗𝟐
𝟔𝐗𝟐 𝟐𝐗𝟐𝟐
−𝟔𝐗𝟐 𝟒𝐗𝟐𝟐
]
[𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
Therefore:

[𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
]
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
}
Where
load vector = { 0 0
𝑃 𝑙
2
−𝑝 𝑙2
12
}𝑇
{𝒇} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
} = {
𝟎
𝟎
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
Where
f2 = Force atnode 2 = -10000 N
m2 = Moment atnode 2 = 3333.3 N-m
[𝒌] {𝑸} = {𝑭}
𝐢. 𝐞 [𝐊] = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟏
𝟏
𝟐
𝟐
} = {
𝒇𝟏
𝒎𝟏
𝒇𝟐
𝒎𝟐
}
1. node 1:
1 = Rotation at node 1 =0

Node 2:
2 = Deflection at node 2  0
i.e.
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟎
𝟎
𝟐
𝟐
} = {
𝟎
𝟎
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
𝟎
𝟎
𝟐
𝟐
} = {
𝟎
𝟎
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 −𝟏𝟐
−𝟏𝟐 𝟏𝟔
] {
𝟐
𝟐
} = {
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
8. Solution
Eq. 1: 11812500((12 X 2 ) + ( -12 x 2 )) = -10000
Eq. 2: 11812500((-12 X 2 ) + ( 16 x 2 )) = 3333.3
2 = -0.00021 m
2 = - 0.00014 rad
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
0
0
−0.00021
−0.00014
}
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
{𝑭} = {
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = [𝒌] {𝑸} - {𝒇}

{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = 𝟏𝟏𝟖𝟏𝟐𝟓𝟎𝟎 [
𝟏𝟐 𝟏𝟐 −𝟏𝟐 𝟏𝟐
𝟏𝟐 𝟏𝟔 −𝟏𝟐 𝟖
−𝟏𝟐 −𝟏𝟐 𝟏𝟐 −𝟏𝟐
𝟏𝟐 𝟖 −𝟏𝟐 𝟏𝟔
] {
0
0
−0.00021
−0.00014
} − {
𝟎
𝟎
−𝟏𝟎𝟎𝟎𝟎
𝟑𝟑𝟑𝟑. 𝟑
}
At node 1:
Shear force F1 = 11812500 X ((12X0) + (12X0) + (-12X-0.0021) + (-12X-0.00014))-(0) =10169.5 N
Bending moment M1 = 11812500 X ((12X0) + (12X0) + (-12X-0.0021) + (8X-0.00014))-(0)= 16806.8 N-m
{𝑸} = {
𝟏
𝟏
𝟐
𝟐
} = {
0
0
−0.00021
−0.00014
}
𝒎
𝒓𝒂𝒅
𝒎
𝒓𝒂𝒅
2. The reaction shear force and bending moment at node 1.
{
𝑭𝟏
𝑴𝟏
𝑭𝟐
𝑴𝟐
} = {
𝟏𝟎𝟏𝟔𝟗. 𝟓
𝟏𝟔𝟖𝟎𝟔. 𝟖
𝟎
𝟎
}
𝒎
𝑵 − 𝒎
𝒎
𝑵 − 𝒎
**-**

EXPT NO: 02 Verification of forces by Lami’s theorem Date:
Aim: Verification of forces by Lami’s theorem
Apparatus:
1. Experimental Set up
2. Scale
3. Pencil
4. A4 size paper
5. Protractor
Theory:
Equilibrium: The state of a body at rest or in uniform motion, the resultant of all forces on which is zero.
The conditions for a body to be in Static Equilibrium
1. The first condition of equilibrium states that for an object to remain in equilibrium, the net force
acting upon it in all directions must be zero.
2. The second condition of equilibrium states that the net torque acting on the object must be
zero.
Lami's Theorem:
Result: It is found that
𝑷
𝒔𝒊𝒏𝜶
=
𝑸
𝒔𝒊𝒏𝜷
=
𝑹
𝒔𝒊𝒏𝜸
Conclusion: Thus, the Lami’s theorem is verified.

Conclusion: This experiment verifies the Lami’s theorem.
Observations:
1. Whether are there three concurrent forces? ________
2. Whether are forces coplanar? ________
3. Whether is system of forces in equilibrium? ______
Tabular Column:
Trial.
No.
Forces in gm Measured Angle in Degree 𝑷
𝒔𝒊𝒏𝜶
𝑸
𝒔𝒊𝒏𝜷
𝑹
𝒔𝒊𝒏𝜸
P Q R 𝛼 𝛽 𝛾
1
2
3
Calculations:
Trial. No. 1
𝑃
𝑠𝑖𝑛𝛼
=
𝑄
𝑠𝑖𝑛𝛽
=
𝑅
𝑠𝑖𝑛𝛾
=
Trial. No. 2
𝑃
𝑠𝑖𝑛𝛼
=
𝑄
𝑠𝑖𝑛𝛽
=
𝑅
𝑠𝑖𝑛𝛾
=

Trial. No. 3
𝑃
𝑠𝑖𝑛𝛼
=
𝑄
𝑠𝑖𝑛𝛽
=
𝑅
𝑠𝑖𝑛𝛾
=
____***____

1
EXPT NO: 02 Tension Test Date:
Aim: To conduct a tensile test on a mild steel specimen and determine the following:
1. Ultimate strength.
2. Breaking stress.
3. Percentage of elongation in length.
4. Percentage of reduction in area.
5. Draw the stress – strain curve and calculate Young’s modulus from the graph.

2
Result:
1. Ultimate strength =
2. Breaking stress =
3. Percentage of elongation in length =
4. Percentage of reduction in area =
5. Young’s modulus =
Conclusion: The Young’s modulus of the material is =

3
Observations:
Initial diameter of the given specimen = d1 =
Initial area of cross section of the given specimen = a1 =
Initial gauge length of the given specimen = l1 =
Final diameter of the given specimen = d2 =
Final area of cross section of the given specimen = a2 =
Final gauge length of the given specimen = l2 =

4
Load at Upper yield point = PUYP =
Load at lower yield point = PLYP =
Ultimate load = Pu =
Breaking Load = Pb =
Tabular Column:
Sl. No. Load in Load in Change in
length
Stress Strain
Kgf newton mm N/mm2
P l 
1
2
Calculations:
1. 𝑆𝑡𝑟𝑒𝑠𝑠  =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
=
𝑃
𝑎1
=
2. 𝑆𝑡𝑟𝑎𝑖𝑛  =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
=
𝑙
𝑙1
3. Ultimate stress u = (Ultimate load/ Initial area of cross section) = (Pu/a1) =
4. Breaking stress u = (Breaking load/ Initial area of cross section) = (Pb/a1) =
5. Percentage of elongation in length = (Final gauge length- Initial gauge length) X 100/ Final gauge length
= (l2-l1) X 100/l1
=
6. Percentage of reduction in area =
(Initial area of cross section – Final area of cross section Initial gauge length) X 100/ Initial area of cross section
= (a1-a2) X 100/a1
7. Young’s modulus = Stress/strain =
***

1
EXPT NO: 04 Comression Test Date:
Aim: To conduct a compression test on a wooden block and determine the following:
1. Ultimate compressive strength.
2. Working stress.
vi) Wooden block

2
Result:
1. Ultimate compressive strength =
3. Working stress =
Conclusion: The Ultimate compressive strength of the given material is =
Observations:
Initial length of the given specimen = l=
Breadth of the given specimen = b =
Height of the given specimen = d =
Area of cross section of the given specimen = a = b X h =
Ultimate compressive Load = Pu =
Calculations:
𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐜𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐯𝐞 𝐋𝐨𝐚𝐝
𝑨𝒓𝒆𝒂 𝒐𝒇 𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏
=
2. Working stress =
𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐜𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐯𝐞 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡
𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚
=
Note: 𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚 𝒐𝒇 𝒘𝒐𝒐𝒅 = 𝟔
***

1
Tabular Column:
Sl. No. Load in Load in Change in
length
Stress Strain
Kgf newton mm N/mm2
F l  
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

1
EXPT NO: 05 Shear Test Date:
Aim: To conduct a Shear test on a given specimen under single shear action and determine the following:
1. Ultimate shear strength.
2. Working shear strength.
Result:
1. Ultimate shear strength =
2. Working shear strength =
Conclusion: The Ultimate shear strength of the given material is =
vi) Shear block

2
Observations:
1. Material of the given specimen =
2. The diameter of the given specimen = d =
3. Area of cross section of the given specimen = a =
4. Ultimate Load or Maximum = Pu =
5. Breaking Load = Pb =
Calculations:
𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐬𝐡𝐞𝐚𝐫 𝐋𝐨𝐚𝐝
𝑨𝒓𝒆𝒂 𝒐𝒇 𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏
=
𝐁𝐫𝐞𝐚𝐤𝐢𝐧𝐠 𝐥𝐨𝐚𝐝
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏
=
3. Working stress =
𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐬𝐡𝐞𝐚𝐫 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡
𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚
=
Note: 𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚 = 𝟐. 𝟓
***

Expt: No. 6 Bending Test
Aim: To study the bending or flexural behavior of the wooden beam and to determine the
Modulus of Elasticity and Modulus of Rupture of the wood.
Equipment Required
 UTM
 Deflection Gauges
 Wooden Beam
 Measuring Tape
Theory and Principle
A beam is a structural element that primarily resists loads applied laterally to the beam's
axis (an element designed to carry primarily axial load would be a strut or column). The
loads applied to the beam result in reaction forces at the beam's support points. The total
effect of all the forces acting on the beam is to produce shear forces and bending
moments within the beams, that in turn induce internal stresses, strains and deflections of
the beam. Its mode of deflection is primarily by bending. Beams are characterized by their
manner of support, profile (shape of cross-section), equilibrium conditions, length, and their
material.
Whenever a beam is loaded, it deflects from its original position. The amount by which it
deflects depends on its cross sectional area and bending moment. In modern designs, a
beam is tested for its-1. Strength 2. Stiffness
The strength criterion of the beam requires a beam should resist bending moment and
shear force. i.e. bending stresses and shear stresses.
The stiffness criterion of the beam requires a beam should resist deflection of the beam.
 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 Wmax, to the moment of its full cross-
section.
 For a simply supported beam with central loading, deflection under the load is given
by :
Write in front pages

Where,
 W =Applied load
 l = Effective span of the beam
 E = Young's Modulus of wood
 I = Moment of inertia
Modulus of Rupture, frequently abbreviated as MOR, (sometimes referred to as bending
strength), is a measure of a specimen’s strength before rupture. It is the stress at failure
in bending. It can be used to determine a wood species’ overall strength; unlike the
modulus of elasticity, which measures the wood’s deflection, but not its ultimate
strength.
Test Procedure
1. Insert the bending device in the UTM.
2. Measure the width and depth of the 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. While applying the load, the
deflection corresponding to each load is found out from the Vernier scale on the
UTM.
7. Note down the maximum deflection and the maximum load.
Results:
1. Modulus of elasticity of the beam = _______ N/mm2
2. Modulus of rupture (Flexural strength) of the beam = _______ N/mm2
3. Modulus of elasticity from the graph= ____________ N/mm2

Observation:
Effective length of the beam = l = ______mm
Breadth of the beam = b= _____________mm
Depth of the beam = d = _______________mm
Maximum Load = Wmax = _______________N
Deflection at Maximum load = max = ___________mm
Breaking Load = Wb = _______________N
Tabular Column:
Sl.No. Load in kgf Load in newton Deflection
Bending
Moment
Bending
Stress
N mm N-mm N/mm2
W δ M b
1
2
3
4
5
6
7
8
9
10
Write sketch here
To be given in the class.
Write in back pages

11
12
13
14
Calculations:
1. Bending moment = 𝑴 =
𝑾𝒍
𝟒
= ________N-mm
2. Moment of inertia = 𝑰 =
𝒃𝒅𝟑
𝟏𝟐
= ______________mm4
3. Section modulus = 𝒁 =
𝒃𝒅𝟐
𝟔
= __________mm3
4. Bending stress = b =
𝑴
𝒁
= ___________ N/mm2
5. Bending moment of Maximum load = Mmax =
𝑾𝒎𝒂𝒙 𝒍
𝟒
= ______N-mm
6. Modulus of elasticity = E =
𝑾𝒎𝒂𝒙 𝒍𝟑
𝟒𝟖 𝑰 𝜹𝒎𝒂𝒙
= _________ N/mm2
7. Bending moment of breaking load = Mb = =
𝑾𝒃 𝒍
𝟒
= ______N-mm
8. Modulus of rupture (Flexural strength) = (b)b =
𝑴𝒃
𝒁
= ______ N/mm2
9. Modulus of elasticity from the graph= E =
(𝑾𝟐−𝑾𝟏) 𝒍𝟑
𝟒𝟖 𝑰 (𝜹𝟐− 𝜹𝟏)
= ____________ N/mm2
***+++***---***

1
Expt: 7
Aim: To watch the video on the topic- Stress, Strain and Bending Stresses on Different mechanical
members and to prepare a report on the observations made.
Video 1:
https://youtu.be/MvBqCeZllpQ
Video 2:
https://youtu.be/f08Y39UiC-o
Report:

1
Experiment No: 10
rectangular c/s area of 0.2 m * 0.3 m, Young‘s modulus of 210 GPa, Poisson‘s ratio 0.27.

1
Experiment No: 11
rectangular c/s area of 0.2 m * 0.3 m, Young‘s modulus of 210 GPa, Poisson‘s ratio 0.27.

1
Experiment No: 12
Compute the shear force and bending moment diagrams for the cantilever beam shown. Assume
rectangular c/s area of 0.2 m X 0.3 m, Young‘s modulus of 210 GPa, Poisson‘s ratio 0.27.

MOM_2022_23_combined.pdf

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