Mechanics of solids_1_lab_report

University of Engineering & Technology Peshawar,Pakistan
Department of Civil Engineering
Lab Report
CE-225L
Mechanics of Solids-1
Submitted By Hammad Shoaib
Reg No 19pwciv5288
Section M
Roll No 199
Instructor Engr.Adeel Arshad
kk21
2021

MECHANICS OF SOLIDS-1
ABOUTTHE AUTHOR
Mr. Hammad Shoaib is a regular student of Civil
Engineering Department, University of Engineering and
Technology Peshawar. The Lab Report was assigned by Engr.
Adeel Arshad in the course of Mechanics of Solid Lab.

TABLE OF CONTENTS
Experiment01.............................................................................................................................
To determine yield strength, ultimate strength, and percent elongation of a steel rebar.
Experiment 02...........................................................................................................................
To perform bend test on steel rebar.
Experiment 03...........................................................................................................................
To develop stress-strain curve for a steelrebar.
Experiment 04 ..............................................................................................................................
To experimentally determine Young's modulus and Poisson’s ratio for a concrete cylinder.
Experiment 05...........................................................................................................................
To determine experimentally the compressive strength of wood in parallel and perpendicular to the
grain directions.
Experiment 06...........................................................................................................................
To verify experimentally torsion formula by testing a member with circular cross section.
Experiment 07 ..............................................................................................................................
To verify experimentally flexure formula by testing a cantilever/simply supported beam.
Experiment 08 ..............................................................................................................................
To experimentally determine the modulus of elasticity (E) of a material by testing a given
rectangular beam.
Experiment 09 ..............................................................................................................................
To experimentally determine the modulus of rigidity (G) of a material by testing a given member
with circular cross-section in torsion.

Experiment 01
To determine yield strength, ultimate strength, and percentage
elongation of a steel rebar
OBJECTIVE
The purpose of this lab is to find the
 yield strength,ultimate strength and percentage elongation of a steel bar.
 To check weather the specimen is ductile or brittle
 Grade of steel bar.
Standard Test Method & specification:-
The following standard test methods is used to conduct this experiment;
 ASTM ( A370-20)
 ASTM ( A706/A706M)
 ASTM ( A615/A615M-20)
SCOPE:
This test method cover the procedure and definitions for the mechanical testing of steels,
stainless steels, and related alloys. The various mechanical tests herein described are used to
determine properties required in the product specifications. The scope of this experiment is to
determine the mainly tensile properties of steel rebar.
RELATED THEORY
Engineering 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. In general, this subject can
be subdivided into three branches: rigid-body mechanics, deformable-body mechanics, and
fluid mechanics. Rigid-body mechanics is essential for the design and analysis of many types
of structural members and mechanical components.
Mechanical Properties
Mechanical properties are the physical properties of the material which
describes its behaviour under the action of loads on it. Such types of properties are
independent of the amount of the material.

There are many mechanical properties of materials and some key properties among
them are given below.
I. Elasticity
This is a measure of elastic deformation of a body under stress which is recovered
when the stress is released. The ratio of stress to strain in the elastic region is known as
stiffness or modulus of elasticity (Young’s Modulus). When the stress goes beyond the
elastic limit the material will no longer return completely to its original dimension.
II. Strength
In mechanics of materials, the strength of a material is its ability to withstand the
applied load without failure or plastic deformation. It is the capacity of the material to
withstand the breaking, bowing, or deforming under the action of mechanical loads on it.
III. Ductility
Ductility is a measure of a material's ability to
undergo significant plastic deformation before rupture.
It allows the material to be deform or make into thin
wires under the action of tensile loads plastically.
Ductility of a material can be determined by
performing the tensile or bend test on the structure as the tensile test is shown in figure.
TensileTest:
We can perform tensile test on material using UTM by clamping a single piece of
material on each of its end between the upper cross head and lower cross head and pull it
apart until it breaks.
1. Stress :
“The internal force per unit area is known as stress.”
2. Strain :
“Strain is defined as the amount of deformation in the direction of the applied
force divided by the initial length of the material.”
3. Yield strength :
The magnitude of the stress at which the transition from elastic to plastic occurs is
known as the yield strength.
4. Ultimate strength :
The maximum tensile stress that a material can withstand before rupture.
5. Percentage Elongation :
“ Percent elongation (elongation to failure) is a measure of. ductility of
a material.”

MECHANICS OD SOLIDS-1
THEORATICAL EQUATION:
The effective area is given by,
Aeff = Volume/Length
The yield strength is given by;
YS = Yield load / Cross sectional area
The ultimate strength is given by;
US = Ultimate load / Cross sectional area
The % elongation is given by;
% Elongation = (Elongated Length – Original length/Original length) 100
EQUIPMENT/APPARATUS:
Universal Testing Machine
A UTM machine is used to test the tensile and
compressive strength of a material. The loads are often
applied in kilo Newton per mm2 or in Kips, the total strain
produced for each corresponding stress is measured
automatically by the advance machine itself, while in case
of mechanical machine extensiometer and
compressometer is used to determine the load.
 Test Specimen
 Vernier Caliper
 Ruler etc.
PROCEDURE:
 Prepare a test specimen of at least two feet.
 Measure caliper at least at three places and then find average.
 Insert the suitable jaws in the grip and select a suitable load scale on UTM.
 Insert the specimen in the grip by adjusting the cross heads of UTM.
 Start machine and continue applying the load.


 
 
 or determine the % elongation, difference between elongated length and original
length divided by original length multiply by 100.
 At a point when the values of the load at that point this is called yield point.
 When the specimen breaks stop the machine.
 Note the ultimate value of the load.
 Determine the yield strength and tensile strength of load dividing the yield load & ultimate load
by cross sectional area of the bar.
 For determine the % elongation, difference between elongated length and original
length divided by original length multiply by 100.
OFFSETMETHOD
If the Proportional point, elastic point , and yield point are not clear. So, 0.2% offset
method is used to elloborate these points. As below,
DATA :
Observations:
Rebar Type:Deformed
Gauge Length(mm):200
Density(kg/m3)=7850
Sample #
Nominal
Size
Length
(m) Mass (kg)
Yield
load
(MN) U.Load(MN)
Elongated
Length(mm)
1 #4 0.625 0.606 0.0453 0.0675 232
2 #5 0.615 0.598 0.0465 0.0683 226
3 #6 0.613 0.959 0.0455 0.0679 228
CALCULATION :
As we know that,
Density=Mass/volume
 Volume=Mass/density
Effective area=Volume/length
5|P a g e

 Volume=Mass/density
Effective area=Volume/length
Yield strength=yield load/Area
Ultimate strength=Ultimate load/Area
Ratio= Ultimate strength/ Yield strength
% Elongation=(Elongated Length-Original Length/ Original Length)×100
RESULT:
Sample
#
Nominal
Size Aeff (mm2
) YS(Mpa) US(Mpa) US/YD
%
Elongation
1 #8 0.000123 366.8 546.56 1.5 16
2 #8 0.000123 376.51 551.71 1.46 13
3 #6 0.000199 368.42 340.86 0.92 14
By our result, we have seen that our result requirement meet with Grade 60Rebar.
PERCENTAGE DIFFERENCE:
From percentage difference formula,
1. percentage difference of YS=Actual YS-Experimental YS/ Actual YS
=420-366.80/420
percentage difference of YS=0.12%
2. percentage difference of US=Actual US-Experimental US/ Actual US
=550-546.56/550
percentage difference of US=0.006%

CONCLUSION
From the tensile test experiment using tensile test machine, the mechanical properties of a
material can be obtained. When the material is being streched, it experiencing elastic and plastic
deformation. The strain hardening phenomenon is occur when the material is getting
strengthened until it goes to fracture. After performing the experiment, we can easily avaluate
the desired properties which is the objective of lab.
We have learned from the lab that, how to find the tensile properties of steel rebars, stress
strain curve and some basic things related to civil Engineering. We have also apply
theknowledge in daily life.

Experiment 02
TO PERFORM BEND TEST ON STEEL BAR
OBJECTIVE
 Shear strength of steel rebars.
 To determine the ultimate strength of the reinforcing steel bars when subjected to
bending load.
 To examine the reinforcing steel bars physical conditions.
STANDARD TEST METHOD AND SPECIFICATION
 ASTM ( A370-20) or ASTM (E290)
 ASTM ( A706/A706M)
 ASTM ( A615/A615M-20)
SCOPE:
These test methods cover bend testing for ductility of materials,and also find the shear
strength of stirrups and ties used in beam and columns. Simply, the bend test is one method
for evaluating ductility.
RELATED THEORY
Bend:
Bend is defined as to force (an object, especially a long or thin one) from a straight form into
a curved or angular one, or from a curved or angular form into some different form.
Ductility:
Ductility is defined by the degree to which a material can sustain plastic deformation under
tensile stress before failure.
OR
Ductility is a mechanical property commonly described as a material's amenability
to drawing (e.g. into wire).

Shear strength:
The shear strength of a material is defined as its ability to resist forces that cause the
material's internal structure to slide against itself.
EQUIPMENT/APPARATUS:
 Universal Testing Machine (UTM)
 test specimen
 bending table support pin.
PROCEDURE:
1. Take a test specimen of the steel rebar.
2. Measure the diameter of the steel rebar.Take at least 3 readings and calculate the
mean.
3. Now place the test specimen in the bending table specimen should be kept in the
bending table in such a way that the plane
4. Intersecting the longitudinal ribs is parallel to the axis of the pin.
5. Select suitable rang of scale.
6. Start the machine and start applying load continuously and uniformly throughout the
bending.
7. As the load is applied on the rod it will start bending.
8. Discontinue the application of load when the angle of bent specified in the material
specimen has been achieved before rebound.
9. Take out the specimen and examine the tension surface of the specimen for cracking.
SPECIFICATION FOR ANGLE IN BEND TEST:
Bar # 3 to Bar #11 should bend up to 180o without crack
Bar # 14 & Bar # 18 should bend upto90o without crack

Mechanics of Solid 05/02/2021
DATA:
Sample
Nominal
Size
Grade
Bend
Test
1 #4 40 OK
2 #3 40 OK
3 #4 40 OK
CONCLUSION:
It is concluded from the above experiment that Grade 60 rebars are OK and
confirmed with standard.

Experiment 03
TO DEVELOP STRESS STRAIN CURVE OF MILD STEEL
OBJECTIVE
The objective is to find
 Elastic limit,Yieldstrength,ultimatestrength,strainhardening,point of rupture and
percentage elongation of a steel bar.
 Modulus of resilience, Young Modulus and modulus of rupture.
 Grade of steel rebar.
STANDARD TEST METHOD & SPECIFICATION:
 ASTM ( A370-20)
 ASTM ( A706/A706M)
 ASTM ( A615/A615M-20)
RELATED THEORY
Stress:
“The internal force per unit area is known as stress.”
Strain :
“Strain is defined as the amount of deformation in the direction of the applied force
divided by the initial length of the material.”
In tension test of ductile metals, the properties usually determined are yield strength, ultimate tensile
strength, modulus of elasticity, percentage of elongation etc. For brittle materials only compressive
strength is determined.
The tension test is normally carried out in a Universal Testing Machine (UTM). The specimen can be
in the form of a rod or a plate. The dimensions of standard specimen can be known from accepted
specifications.
The following properties can be determined from the stress strain curve of the material:
1. Proportional limit: is that point on the stress strain curve at which the curve deviates
from linearity, i.e. from the relation

Stress = Young's modulus x strain
σ = Eε
2. Elastic limit: is the point on the stress strain curve above which plastic deformation
(that is permanent deformation) starts.
3. Yieldstrength:
It is the stress required to produce a small amount of permanent or
plastic deformation.
Universal Testing Machine
A universal testing machine (UTM), also known as a universal tester, materials testing
machine or materials test frame, is used to test the tensile strength and compressive strength
of materials. An earlier name for a tensile testing machine is a tensiometer. The "universal"
part of the name reflects that it can perform many standard tensile and compression strength
tests, shear strength testand to perform bend test on the structural material.
Components of UTM
It consists of two main components which are:
• Testing Unit/Straining Unit
• Controlling Unit
I. Testing Unit
In this unit, actual loading of the specimen takes place - consists of three cross heads
namely upper head, middle head and lower head. Using appropriate cross heads tensile,
compressive, shear, bending load with the help of different attachment can be applied.
Straining unit of a UTM consists of:
8 | P a g e
Stress = Young's modulus x strain
σ = Eε
1. Elastic limit: is the point on the stress strain curveabove which plastic deformation
(that is permanent deformation) starts.
2. Yieldstrength:
It is the stress required to producea small amount of permanent or plastic
deformation.
OffsetMethod:
In some materials such as mild steel, where there is occurrence of sharp yield point on
the stress -strain curve, the stress value at the lower yield point is taken as the yield
strength. In some materials like tor steel which do not have a sharp yield point, the
offset yield strength or proof stress is taken as the measure of the yield strength. This is
the stress at which a line drawn parallel to the initial portion of the curve, offset by a
specified strain, intersects.
The offset value is usually a strain of 0.002 (0.2% strain). The value of the yield
strength is of great importance in design calculations.

Tensilestrengthor ultimatetensilestrength(UTS):
It is the maximum load
divided by the original cross sectional area of the specimen. U.T.S. corresponds to the
peak or the highest stress value in the stress -strain curve.
PercentageElongation :
“ Percent elongation (elongation to failure) is a measure of. ductility of
a material.”
Strain Hardening:
It is region of high rate of plastic deformation after yield strength.
Modulusof Resilience:
The modulus of resilience is defined as the maximum energy that can be absorbed
per unit volume without creating a permanent distortion.
Modulusof Rupture:
Flexural strength, also known as modulus of rupture is a measure of a specimen's
strength before rupture.
Modulusof elsticity:
The modulus of elasticity of a material or Young modulus is a measure of its
stiffness. It is equal to the stress applied to it divided by the resulting elastic strain.
THEORATICAL EQUATION
The yield strength is given by;
YS = Yield load / Cross sectional area
The ultimate strength is given by;
US = Ultimate load / Cross sectional area
The % elongation is given by;
% Elongation = (Elongated Length – Original length/Original length) 100
Modulus of elasticity is;
E = Stress / Strain
Modulus of elasticity is;
E = Stress / Strain
EQUIPMENT/APPARATUS

DATA:
STRAIN(in/in) STRESS(ksi)
-0.000162862 0
-0.000325724 5.456485851
-0.000488587 6.965726618
-0.000325724 8.01058561
-0.000488587 9.171540047
-0.000651449 8.823253716
-0.000488587 9.403730934
-0.000814311 10.79687626
-0.000162862 10.33249448
-0.000162862 11.72563981
0.000325724 10.9129717
0.000325724 12.30611702
-0.000325724 13.69926235
-0.000325724 12.19002158
-0.000325724 13.69926235
-0.000162862 14.51193045
-0.000325724 15.55678945
-0.000162862 15.440694
-0.000325724 16.60164844
-0.000162862 17.53041199
-0.000162862 16.71774388
-0.000162862 17.76260288
0.000162862 19.62012997
0.000162862 18.57527098

RESULT
Stress-Strain Curve:
The result of our lab show that, the stress-strain curve is developed. But the curve is very
complicated,
CONCLUSION
From this experiment using tensile test machine, the mechanical properties of a
material can be obtained. When the material is being streched, it experiencing elastic and
plastic deformation. The strain hardening phenomenon is occur when the material is getting
strengthened until it goes to fracture.

Experiment 04
TOEXPERIMENTALLY DETERMINE THE STATIC ELASTIC MODULUS AND
POISON RATIO OF CONCRETE CYLINDER.
OBJECTIVE
 Find the modulus of Elasticity of concrete
 Find Poisson’s Ratio of Concrete
STANDARD TEST METHOD
o ASTM ( C469)
SPECIFICATION
 Correct mix
 Properly affixing strain gauges
 Properly capping the cylinder
SCOPE
This test method covers determination of chord modulus of elasticity (Young’s) and
Poisson’s ratio of molded concrete cylinders and diamond-drilled concrete cores when under
longitudinal compressive stress.
RELATED THEORY
Modulusof Elasticity
An elastic modulus is a quantity that measures an object or substance's resistance to being
deformed elastically when a stress is applied to it. The elastic modulus of an object is
defined as the slope of its stress–strain curve in the elastic deformation regionб
Math ematically E =є
Poisson Ratio:
It is defined as the ratio of lateral strain to the longitudinal strain.
EQUIPMENT/APPARATUS
Testing Machine
Extensometer
An extensometer is a device that is used to measure changes in the length of an object. It is
useful for stress-strain measurements and tensile tests. Its name comes from "extension-
meter".

Compressometer
A compressometer is a device used to determine the strain or deformation of a
specimen while measuring the compressive strength.
PROCEDURE
1. Keep the ambient temperature as steady as possible throughout the test.
2. Use two specimens to specify the compressive strength prior to the modulus of elasticity
test.
3. Attach the strain-measuring equipment to the specimen.
4. Place the specimen on the lower platen or bearing block of the testing machine.
5. Align the axis of the specimen with the center of thrust of the spherically-seated upper
bearing block.
6. Load the specimen carefully to seat the gauge and observe its performance. Any abnormal
response shall be corrected.
7. After that, apply the load continuously; record the applied load and longitudinal strain
when the longitudinal strain is 50 millionth, and the applied load is 40% of the ultimate
load.
THEORATICL EQUATION
Young’sModulus
E=(s1-s2)/(e2-.000005)
 s1=the stress corresponding to the longitudinal strain of 50 micro strain.
 s2=the stress corresponding to .4f ‘c.
 E2=the longitudinal strain corresponding to s2.
 Based on ASTM C 469
Poisson’sRatio
Calculate Poisson’s ratio, to the nearest 0.01, as follows:
μ=( e1- e2)/(e2-.000005)

where:
μ = Poisson’s ratio,
εt2 = transverse strain at mid height of the specimen pro-
duced by stress S2, and
εt1 = transverse strain at mid height of the specimen pro-
duced by stress S1.
D
DATA
Lat.strain(in/in) Long. strain (in/in) Stress (PSI)
2.87×10-06 1.38×10-05 53.3456951
2.87×10-06 1.38×10-05 53.3456951
2.88×10-06 1.38×10-05 53.3456951
2.82×10-06 1.38×10-05 53.3456951
2.88×10-06 1.38×10-05 53.3456951
CALCULATION
AS we know that poisson’s ratio is given by
V= −
𝑙𝑎𝑡𝑠𝑡𝑟𝑎𝑖𝑛
𝑙𝑜𝑛𝑔𝑠𝑡𝑟𝑎𝑖𝑛
V= - 𝟐.𝟖𝟕𝐄−𝟎𝟔
𝟏.𝟑𝟖𝐄−𝟎𝟓
Lat.strain(in/in) Long. strain(in/in) Poisson,s ratio
2.87×10-06 1.38×10-05 -2.08×10-01
2.87×10-06 1.38×10-05
-2.08×10-01
2.88×10-06 1.38×10-05
-2.08×10-01
2.82×10-06 1.38×10-05 -2. 04×10-01
2.88×10-06 1.38×10-05
-2. 09×10-01
V= -2.08E-01

RESULT
CONCLUSION
This test method provides a stress to strain ratio value and a ratio of lateral to longitudinal strain for
hardenedconcrete at whatever age and curing conditions may bedesignated.

EXPERIMENT09
TO EXPERIMENTALLY DETERMINE THE MODULUS OF RIGIDITY BY
TESTING A MEMBER WITH CIRCUAR CROSS SECTION
OBJECTIVE
 To study the torsional stress-strain relationship and determine shear modulus(G) .
 To study qualitatively the relationship between torsional load and angle of twist for a
full range of strains till failure.
 To determine the mode of failure (ductile or brittle).
LIMITATION
 Circular sections remain circular.
 The stress does not exceed the elastic limit or limit of proportionality.
 The material is elastic, following Hooke's law with shear stress proportional to shear
strain.
 The material is homogeneous, i.e. of uniform elastic properties throughout.
RELATED THEORY
Modulus of Rigidity:
Shear Modulus Of Rigidity. Shear modulus also known as Modulus of rigidity is the
measure of the rigidity of the body, given by the ratio of shear stress to shear strain. Often
denoted by G
G = 𝜏/ γ
G=shear modulus
𝜏 =shear stress
γ= shear strain
Hook’s Law:
A law stating that the strain in a solid is proportional to the applied stress within the elastic
limit of that solid.
σ= -Kε
Where,
K=proportoinality constant
σ = stress
ε =strain

Torsion :-
“Torsion is the twisting of an object caused by a moment acting about the object longitudinal axis”
Polarmomentof inertia:-
Polar Moment of Inertia is a measure of an object's capacity to oppose or resist torsion when some
amount of torque is applied to it on a specified axis. ... If the polar moment of inertia is of higher
magnitude then the torsional resistance of the object will also be greater.
Angle of twist:-
Angle of twist: For a shaft under torsional loading, the angle through which fixed end of a shaft
rotates with respect to the free end is called the angle of twist. ... As the torque is increased, the
outer region of the shaft behaves like a plastic material while the inner core is still linear elastic.
Torque:-
Torque, also called moment of a force,
“ The tendency of a force to rotate the body to which it is applied”
Unit:-
Newton meter (Nm)
Mathematical formula:-
T = F *r
EQUIPMENT/APPARATUS
 Torsion Testing Machine
 DailGuage
 Vernier Calliper
 Ruler
 Weight,etc
PROCEDURE
 Place the apparatus on a smooth horizontal surface.
 Measure the effective length of the shaft.
 Measure the diameter of the shaft.
 Adjust the zeros at 1st and 2nd measuring arms.
 Put the load in the hangers.
 Measure the 1st and 2nd angle of twist of the shaft.
 Take readings of increasing value of load and then take readings on unloading.
 Calculate the Modulus of Rigidity of the material of the shaft.

Derivation of torsion formula
As we know that,
tan = perpendicular/base
In OBB’,
𝒕𝒂𝒏= BB’/L
𝑩𝑩′
= 𝑳∅ (1)
And BB’=R∅
Modulus of Rigidity
𝐺 = τ/∅
∅ = τ/𝐺
So,
Put the value of ∅in equation (1)
𝑩𝑩′
= 𝑳(
τ
𝐺
)
And Put value of BB’= R∅ in above equation,
𝐑∅ = 𝑳(
τ
𝐺
)
By rearranging by the equation,
τ
𝑹
= 𝑮∅/𝑳
As we know that,
(2)
𝑑𝑓 = τ𝑑𝐴
Put the value 𝜏 from eq. (2),
𝒅𝒇 = (𝑮∅/𝑹)𝑹𝒅𝑨
Differential torque is given by,
𝒅𝑻 = 𝒅𝑭.𝑹 = (𝑮∅
𝑹𝟐
𝑳
) 𝒅𝑨

𝑇 = 𝐺∅/𝐿 ∫𝑅2𝑑𝐴 = (
𝐺∅
𝐿
) 𝐽
𝑻
𝑱
=
𝑮∅
𝑳
(3)
From equation (2) & (3) we get,
Data
Ange of Twist Torque(Nm)
Revs Degrees Radian 0.4% Carbon Steel
0 0 0 0
1 0.3 0.005236 0.2
2 0.6 0.010472 0.5
3 0.9 0.015708 0.6
4 1.2 0.020943 0.9
5 1.5 0.026179 1.3
6 1.8 0.031415 1.4
7 2.1 0.036651 1.7
8 2.4 0.041887 2
9 2.7 0.047123 2.2
10 3 0.052358 2.2
Length(mm) 80
Diameter(mm) 6
Radius(m) 0.003
J(m4
) 1.27×10-10
CALCULATION
As we know that,
G=TL/j∅
From the data put the value of (T/∅),
Hence,
G=45.491*80/(1.27×10-10
)1000
G=28.6028Gpa
𝑻
𝑱
=
𝑮∅
𝑳
=

𝑹

RESULT
CONCLUSION
The modulus of rigidity (G) of circular cross sectional is G=28.6028Gpa
y = 45.491x - 0.0091
R² = 0.988
0
0.5
1
1.5
2
2.5
0 0.01 0.02 0.03 0.04 0.05 0.06
Torque(Nm)
Angle of Twist (Radian)
y = 45.491x - 0.0091
R² = 0.988
0
0.5
1
1.5
2
2.5
0 0.01 0.02 0.03 0.04 0.05 0.06
Torque(Nm)
Angle of Twist (Radian)

EXPERIMENT06
TO EXPERIMENTALLY DETERMINE THE TORSION FORMULA BY TESTING
A MEMBER WITH CIRCUAR CROSS SECTION
OBJECTIVE
 To study the torsional stress-strain relationship and determine shear modulus(G) .
 To study qualitatively the relationship between torsional load and angle of twist for a
full range of strains till failure.
 To determine the mode of failure (ductile or brittle).
LIMITATION
 Circular sections remain circular.
 The stress does not exceed the elastic limit or limit of proportionality.
 The material is elastic, following Hooke's law with shear stress proportional to shear
strain.
 The material is homogeneous, i.e. of uniform elastic properties throughout.
RELATED THEORY
Modulus of Rigidity:
Shear Modulus Of Rigidity. Shear modulus also known as Modulus of rigidity is the
measure of the rigidity of the body, given by the ratio of shear stress to shear strain. Often
denoted by G
G = 𝜏/ γ
G=shear modulus
𝜏 =shear stress
γ= shear strain
Hook’s Law:
A law stating that the strain in a solid is proportional to the applied stress within the elastic
limit of that solid.
σ = -Kε
Where,
K=proportoinality constant
σ = stress
ε =strain

Torsion :-
“Torsion is the twisting of an object caused by a moment acting about the object longitudinal axis”
Polarmomentof inertia:-
Polar Moment of Inertia is a measure of an object's capacity to oppose or resist torsion when some
amount of torque is applied to it on a specified axis. ... If the polar moment of inertia is of higher
magnitude then the torsional resistance of the object will also be greater.
Angle of twist:-
Angle of twist: For a shaft under torsional loading, the angle through which fixed end of a shaft
rotates with respect to the free end is called the angle of twist. ... As the torque is increased, the
outer region of the shaft behaves like a plastic material while the inner core is still linear elastic.
Torque:-
Torque, also called moment of a force,
“ the tendency of a force to rotate the body to which it is applied”
Unit:-
Newton meter (Nm)
Mathematical formula:-
T = F *r
EQUIPMENT/APPARATUS
 Torsion Testing Machine
 DailGuage
 Vernier Calliper
 Ruler
 Weight,etc
PROCEDURE
 Place the apparatus on a smooth horizontal surface.
 Measure the effective length of the shaft.
 Measure the diameter of the shaft.
 Adjust the zeros at 1st and 2nd measuring arms.
 Put the load in the hangers.
 Measure the 1st and 2nd angle of twist of the shaft.
 Take readings of increasing value of load and then take readings on unloading.
 Calculate the Modulus of Rigidity of the material of the shaft.

Derivation of torsion formula
As we know that,
tan = perpendicular/base
In OBB’,
𝒕𝒂𝒏= BB’/L
𝑩𝑩′
= 𝑳∅ (1)
And BB’=R∅
Modulus of Rigidity
𝐺 = τ/∅
∅ = τ/𝐺
So,
Put the value of ∅in equation (1)
𝑩𝑩′
= 𝑳(
τ
𝐺
)
And Put value of BB’= R∅ in above equation,
𝐑∅ = 𝑳(
τ
𝐺
)
By rearranging by the equation,
τ
𝑹
= 𝑮∅/𝑳
As we know that,
(2)
𝑑𝑓 = τ𝑑𝐴
Put the value 𝜏 from eq. (2),
𝒅𝒇 = (𝑮∅/𝑹)𝑹𝒅𝑨
Differential torque is given by,
𝒅𝑻 = 𝒅𝑭.𝑹 = (𝑮∅
𝑹𝟐
𝑳
) 𝒅𝑨

VERIFICATION
𝑻
𝑱
=
𝑮∅
𝑳
=

𝑹
So,
𝑻
𝑱
=
𝑮∅
𝑳
… … … … … … . (𝑨)
Putting the values in Eq (A),
0.2
1.27×10−10
= (2.86028 × 10^9 ∗ 0.005236)/(80/1000)
CONCLUSION
Hence, Torsion formula is verified.
1.57×109
= 1.87×109

Experiment 05
TO DETERMINE THE COMPRESSIVE STRENGTH OF WOOD IN
PARALLEL AND PERPENDICULAR TO THE GRAIN DIRECTION.
OBJECTIVE
• To determine the compressive strength of wooden cubes in parallel and
perpendicular to the grain direction.
• To observer the anisotropic behaviors of the wood.
STANDARD TEST METHOD AND SPECIFCATION
The following test method is used for this test:
 ASTM (D3501)
RELATED THEORY
Compressive Strength
Compressive strength is the maximum compressive stress that, under a gradually
applied load, a given solid material can sustain without fracture. Compressive strength is
calculated by dividing the maximum load by the original cross-sectional area of a specimen
in a compression test.
Isotropic Material
Isotropy is a common term in materials science
that means uniform in all directions. Isotropic materials
exhibit the same material properties in all directions.
Metals and glasses tend to be isotropic
Anisotropic Material
Those materials whose material properties
changes with the change in direction is known as
anisotropic material. Common examples of anisotropic
materials are wood and composites.
Wood Grain
FIGURE 1 :- WOOD GRAIN
Wood grain is the longitudinal arrangement of
wood fibers or the pattern resulting from wood fibers.
These wood fibers arranged themselves in the form of layers over the other in the vertical
direction.

EQUIPMENT/APPARATUS
 Wooden samples
 Vernier calipers
 Universal Testing Machine
PROCEDURE
1. Take a block of wood as a specimen.
2. Measure the cross-section and length of the specimen and record the dimensions on the data sheet.
3. These measurement is taken for both cases either grains are parallel or perpendicular to load.
4. Place the specimen in the machine.
5. Apply the load continuously until the specimen fails. Record the maximum load.
OBSERVATION AND CALCULATION
Specimen L(mm) W(mm)
A = L x
W(mm2)
Parallel to
grains
57.67 43.41 2503.7
57.66 44.5 2567.23
Perpendicular
to grains
58.97 56.89 3353.36
58.9 56.29 3310.47
RESULT
S/No.
Dimension
(in×in)
Area
(in^2)
Max. Load in
parallel
direction (lb)
Max. Load in
perpendicular
direction (lb)
Comp.
strength in
parallal
direction
(psi)
Comp.
strength in
perpendicu
ar
direction
(psi)
1 2.27×1.70 3.86 22000 5699.481865
2 2.27×1.75 3.97 25800 6498.740554
3 2.31×2.23 5.15 3800 737.8640777
4 2.31×2.21 5.1 1800
 The compressive strength is very high as in case of grain parallel to direction of load.

Failure of the wooden cubes in both directions can be seen
below:
FORCE PARALLEL TO GRAINFORCE PERPENDICULAR TO GRAIN
CONCLUSION
In case, force parallel to the grain direction, each fiber is supporting the load
and act as a column.Due to which such type of failure is happens.
In case, force perpendicular to the grain direction, the shear failure of any two
consecutive layers tends to slide over the other and this will continue until the
complete failure.

EXPERIMENT08
TO EXPERIMENTALLY DETERMINE THE MODULUS OF ELASTICITY BY
TESTING A GIVEN RECTANGULAR BEAM
OBJECTIVE:
 To verify the flexure formula.
 To find the internal stresses due to external bending moment.
 To find the elasticity modulus of a materials.
LIMITATION
 Flexure formula is valid only when bending is about a principal axis and the
materials is linear elastic
 Material is homogeneous and obeys hook law.
 Beam is straight and is of constant cross-section.
 Valid for very small deflection.
 Cross-section is symmetric about x-y plane.
RELATED THEORY
Beam
A beam is a structural element that primarily resists loads applied laterally to
the beam's axis. Its mode of deflection is primarily by bending.
Bending Moment
A bending moment is the reaction induced in a structural element when an
external force or moment is applied to the element, causing the element to bend
Flexural Stresses
Stresses caused by the bending moment are known as flexural or bending stresses.
Flexural Strength
Flexural strength, also known as modulus of rupture, or bend strength, or
transverse rupture strength is a material property, defined as the stress in a material
just before it yields in a flexure test.

APPARATUS/EQUIPMENT:-
 Beam stand
 Digital Vernier caliper
 Weights
 Hanger
 Dial gauge
A dial gauge consisting of a circular graduated dial and a pointer actuated by a
member that contacts with the part being calibrated
In this experiment it is used to measure the deflection of the beam
PROCEDURE:
 Clamp the steel beam of length 0.764m in the stand and measure its length with
measuring tape
 Compute the moment of inertia of the beam by using the formula
As the beam is of rectangular cross-section so moment of inertia is given by
𝐼 =
𝑏ℎ3
12
Where “b” is the width of the beam and “h” is the height of the beam
 Now applying the load of 2.5 kgat the center of the beam so the each support reaction
must be
𝑅 =
𝑤
2
 Note the deflection of the beam with the help of dial gauge .
 Now compute the moment at the right corner of the beam which is given by
𝑀 =
𝑤𝑙
4

Derivation of flexural formula
Consider a fiber AB in the material,ata distance y from the neutral axis,when the beam bend
this will
Therefore,
Strain in fiber AB= Change in length/Original length
𝑆𝑡𝑟𝑒𝑡𝑐ℎ 𝑡𝑜 𝐴’𝐵’ = 𝐴’𝐵’ − 𝐴𝐵/𝐴𝐵 But AB=CD & CD=C’D’
So,
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐴’𝐵’ − 𝐶𝐷/𝐶𝐷
Since CD &C’D’ are on neutral axis and it is assumed that stresss on neutral axis is zero.
𝑠𝑡𝑟𝑎𝑖𝑛 = (𝑅 + 𝑦) − 𝑅/𝑅
= 𝑅 + 𝑦 − 𝑅/𝑅
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑦/𝑅
As we know that,
𝑌𝑜𝑢𝑛𝑔 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 𝐸 = 𝑠𝑡𝑟𝑒𝑠𝑠/𝑠𝑡𝑟𝑎𝑖𝑛
𝑠𝑡𝑟𝑒𝑠𝑠/𝐸 = 𝑠𝑡𝑟𝑎𝑖𝑛
Putting the value of strain
σ/E = y/R𝑜𝑟
σ
y
= E/R (1)

As we Know that,
𝑑𝐹 = σdA
Multiplying y on B.S,
dF×y = (Ey/R)ydA
∫ 𝑑𝑀 = ∫(𝐸𝑦2/𝑅)𝑑𝐴
𝑀 = 𝐸/𝑅 ∫ 𝑦2 𝑑𝐴
𝑀 = (
𝐸
𝑅
)𝐼
𝑀
𝐼
=
𝐸
𝑅
(2)
OBSERVATION & DATA
S.no Load
Defection(
m)
Moment
s(Nm)
Width
(m)
Elasticity
(Gpa)
Height
(m)
Radius(m)
Moment
of
inertia(m4
)
1 2 0.085 0.382 0.013 200 0 8.588 1.17*10-
10
2 2.5 0.012 0.477 0.013 200 0 6.901
CALCULATIONS
Now for moment,,
As we know that moment at the center of the beam is given by
𝑀 =
𝑤𝑙
4
By putting the value from the above table
𝑀 =
2.5 ∗ 0.764
4
𝑴 =0.477 N m
Now for radius,
𝑅 =
1
2𝛿
(𝛿2
+
𝑙2
4
)

𝑅 =
1
2(0.012)
(0.0122
+
0.7642
4
)
𝑹 = 𝟔. 𝟗𝟎𝒎
Now moment of inertia,
𝐼 =
𝑏ℎ3
12
Putting the value of variables from the above table
𝐼 =
0.013 ∗ 0.00483
12
𝑰 = 𝟏.𝟏𝟕 ∗ 𝟏𝟎−𝟏𝟎
m4
As we know that,
𝑀
𝐼
=
𝐸
𝑅
𝐸 = 𝑀𝑅/𝐼
Putting the values in above equation,
𝐸 =
0.477 × 6.90
1.17 ∗ 10−10
RESULT
The modulus of elasticity or young modulus is 2.81 ×1010.
CONCLUSION
It is concluded that the experimentally and theoretical value of modulus of Elasticity is
approximately equal.
𝐸 = 2.81 ×1010

EXPERIMENT07
TO VERIFY EXPERIMENTALLY FLEXURAL FORMULA BY TESTING A
CANTILIVER/SIMPLY SUPPORTED BEAM
OBJECTIVE:
 To verify the flexure formula.
 To find the internal stresses due to external bending moment.
 To find the elasticity modulus of a materials.
LIMITATION
 Flexure formula is valid only when bending is about a principal axis and the
materials is linear elastic
 Material is homogeneous and obeys hook law.
 Beam is straight and is of constant cross-section.
 Valid for very small deflection.
 Cross-section is symmetric about x-y plane.
RELATED THEORY
Beam
A beam is a structural element that primarily resists loads applied laterally to
the beam's axis. Its mode of deflection is primarily by bending.
Bending Moment
A bending moment is the reaction induced in a structural element when an external
force or moment is applied to the element, causing the element to bend
Flexural Stresses
Stresses caused by the bending moment are known as flexural or bending stresses.
Flexural Strength
Flexural strength, also known as modulus of rupture, or bend strength, or
transverse rupture strength is a material property, defined as the stress in a material
just before it yields in a flexure test.

Derivation of flexural formula
Consider a fiber AB in the material,ata distance y from the neutral axis,when the beam bend
this will
Therefore,
Strain in fiber AB= Change in length/Original length
𝑆𝑡𝑟𝑒𝑡𝑐ℎ 𝑡𝑜 𝐴’𝐵’ = 𝐴’𝐵’ − 𝐴𝐵/𝐴𝐵 But AB=CD & CD=C’D’
So,
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐴’𝐵’ − 𝐶𝐷/𝐶𝐷
Since CD &C’D’ are on neutral axis and it is assumed that stresss on neutral axis is zero.
𝑠𝑡𝑟𝑎𝑖𝑛 = (𝑅 + 𝑦) − 𝑅/𝑅
= 𝑅 + 𝑦 − 𝑅/𝑅
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑦/𝑅
As we know that,
𝑌𝑜𝑢𝑛𝑔 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 𝐸 = 𝑠𝑡𝑟𝑒𝑠𝑠/𝑠𝑡𝑟𝑎𝑖𝑛
𝑠𝑡𝑟𝑒𝑠𝑠/𝐸 = 𝑠𝑡𝑟𝑎𝑖𝑛
Putting the value of strain
σ/E = y/R𝑜𝑟
σ
y
= E/R (1)

As we Know that,
𝑑𝐹 = σdA
Multiplying y on B.S,
dF×y = (Ey/R)ydA
∫ 𝑑𝑀 = ∫(𝐸𝑦2/𝑅)𝑑𝐴
𝑀 = 𝐸/𝑅 ∫ 𝑦2 𝑑𝐴
𝑀 = (
𝐸
𝑅
)𝐼
𝑀
𝐼
=
𝐸
𝑅
(2)
Data
S.no Load
Defection
(m)
Moment
s(Nm)
Width
(m)
Elasticity
(Gpa)
Height
(m)
Radius(m)
Moment
of
inertia(m4
)
1 2 0.085 0.382 0.013 200 0 8.588 1.17*10-
10
2 2.5 0.012 0.477 0.013 200 0 6.901
SIMPLE CALCULATIONS
Now for moment,,
As we know that moment at the center of the beam is given by
𝑀 =
𝑤𝑙
4
By putting the value from the above table
𝑀 =
2.5 ∗ 0.764
4
𝑴 =0.477 N m
Now for radius,

σ
y
=
𝑀
𝐼
=
𝐸
𝑅
σ
y
=
𝑀
𝐼
=
𝐸
𝑅
σ
y
=
𝑀
𝐼
=
𝐸
𝑅
𝑅 =
1
2𝛿
(𝛿2
+
𝑙2
4
)
𝑅 =
1
2(0.012)
(0.0122
+
0.7642
4
)
𝑹 = 𝟔.𝟗𝟎𝒎
Now moment of inertia,
𝐼 =
𝑏ℎ3
12
Putting the value of variables from the above table
𝐼 =
0.013 ∗ 0.00483
12
𝑰 = 𝟏.𝟏𝟕 ∗ 𝟏𝟎−𝟏𝟎
m4
RESULT
S.No M/I E/R
1 0.326 *1010 2.328*1010
2 0.4075*1010 2.898*1010
CONCLUSION
The flexural formula is verified There is a little error due to instrumental error, experimental error
or human error.

Mechanics of solids_1_lab_report

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

Similaire à Mechanics of solids_1_lab_report

Similaire à Mechanics of solids_1_lab_report (20)

Dernier

Dernier (20)

Mechanics of solids_1_lab_report