Mechanics of Materials-I

Submitted By:
Mian Ahmad Fawad
2023-ME-107

Submitted to:
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LAB LAYOUT
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Lab Session 1

Objective:
To draw the load-extension curve of a metallic wire and hence determine the modulus of elasticity
of the material of the wire.

Apparatus:
i. Young’s Modulus of Elasticity apparatus
ii. Hangers
iii. Weights
iv. Meter Rod
v. Micrometer

Figure 1. 1Young's Modulus of Elasticity

Apparatus

Modulus of Elasticity:
The Modulus of Elasticity, also known as the elastic modulus or simply modulus, is a mechanical
property of a solid material that measures its resistance to elastic deformation. It quantifies how much
a material will deform under a given stress before returning to its original shape. A higher modulus
of elasticity indicates a stiffer material, meaning it requires a greater force to produce a given amount
of deformation.

Temperature: The modulus generally increases with decreasing temperature.

Strain rate: A higher strain rate can lead to a slightly higher modulus.
Microstructure: The arrangement and properties of the material's internal structure can affect its
modulus.
Presence of defects: Defects such as cracks or voids can reduce the modulus.
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Modulus of Elasticity (GPa)

Aluminum 69
Steel (mild) 200
Stainless steel 190
Copper 120
Brass 100
Cast iron 100
Concrete 20-30
Wood (pine) 10-13
Glass 60-70
Plastic (PVC) 3-4
Nylon 3-4
Rubber 0.1-1
Titanium 110
Magnesium 45
Nickel 200
Lead 16
Zinc 100
Gold 79
Silver 100
Platinum 160

The Young’s Modulus of Elasticity apparatus consists of a wire attached to a fixed support. The
lower end of the wire is attached to the hanger with the help of a metallic plate. The extension
of the wire on loading can be measured from the scale present on metallic plate.

Stress:
Stress is a measure of the internal force per unit area within a material that arises as a result of an
external force. It is a measure of the resistance of a material to deformation.

Types of Stress:
i) Normal stress: Acts perpendicular to a cross-section of a material. It can be either tensile
(pulling) or compressive (pushing).
ii) Shear stress: Acts parallel to a cross-section of a material. It tends to cause the material to
slide or twist.

Units of Stress:
Pascal (Pa): The SI unit of stress.
Pounds per square inch (psi): A common unit used in the United States.

Strain:
Strain is a measure of the deformation of a material relative to its original dimensions. It is a
dimensionless quantity that expresses the change in length, area, or volume of a material due to
applied stress.

Types of Strain:
i) Normal strain: The change in length per unit original length.
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ii) Shear strain: The change in angle between two originally perpendicular lines.

Units of Strain:
Strain is dimensionless, but it can be expressed as a percentage or a decimal.

The Stress-Strain Relationship:

The relationship between stress and strain is often represented graphically by a stress-strain curve.
This curve shows how a material deforms under increasing stress. The shape of the curve can reveal
important properties of the material, such as its elasticity, plasticity, and ultimate strength.

Elastic Region: In this region, the material deforms elastically, meaning it returns to its original
shape when the stress is removed. The slope of the stress-strain curve in this region is called the
modulus of elasticity.
Yield Point: This is the point where the material begins to deform plastically, meaning it does not
return to its original shape when the stress is removed.
Plastic Region: In this region, the material continues to deform plastically, even as the stress remains
constant.
Ultimate Strength: This is the maximum stress a material can withstand before it fractures.

Difference b/w Stress and Pressure:

i) Direction: Stress has a direction (normal or shear), while pressure does not.
ii) Area: Stress is related to a specific area within a material, while pressure is related to a surface.
iii) Type of force: Stress can be caused by both normal and shear forces, while pressure is typically
caused by normal forces.

Derivation:
Normal stress in a solid body is defined as:
“The internal resistance force per unit area against the applied load or external force.” It is denoted
by σ. It can be tensile or compressive.
Mathematically,
Stress = Force/Area ---------- (i)
Units of stress: Newton per square meter (N/m2) = Pascal (Pa) or pounds per square inch (psi)
Normal strain in a solid body is defined as: “Change of length per Original Length.” It is
denoted by the symbol ε.
Mathematically,
Normal Strain = Change in length/Original length ---------- (ii)
Strain is measured as inch/inch.

By Hooke’s law, we know that stress is directly proportional to the strain, whenever a material is
loaded within its proportionality limit. It is denoted by E.
Mathematically,
Stress α Strain (within proportionality limit) ---------- (iii)
Units of E: Newton per square meter (N/m2) = Pascal (Pa) or pounds per square inch (psi) Consider
a body (wire) subjected to a tensile stress as shown in figure 1.1.
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Let,
P = Load or force acting on the body
L = Length of the body
A = Cross-sectional area of the body
σ = Stress induced in the body
E = Modulus of elasticity for the material of the body
ε = Strain produced in the
body δl = Deformation of the
body

From (i), (ii), and (iii)

σαε
σ=Exε
or
E = σ/ε

E = (p/ δl) (L/A)

Procedure:
1. Put the initial load to remove wrinkles in wire.
2. Measure length of wire using meter rod.
3. Measure diameter of the wire using vernier caliper.
4. Adjust main scale so that zeros of two scales coincide with each other.
5. Put a load of 0.5 lb in the hanger and measure extension.
6. Take a set of at least five readings of increasing value of load and then take readings on
unloading.
7. Check the zeros at no load.
8. Calculate the “Young’s Modulus of Elasticity (E)” of the material of the shaft.

Observations and Calculations:

Least Count of the scale of apparatus = 0.5 mm

Least Count of vernier caliper = 0.05 mm
Least Count of meter rod = 1 mm
Length of wire (L) = 895 mm
Dia of wire (d) = 1 mm
X-area of wire (A= πd2/4) =0.785 mm2
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No. of Obs. Effective Load Extension-δl P/δl Modulus of

P Elasticity
(mm) (lb/mm)
(lbs) Loading Unloading Average E=(P/δl)(L/A)
From Graph
(lb/mm2)
1. 0.0 0 0 0 0.276
2.
314.675
0.5 0 0.1 0.05
3. 1.0 0.15 0.35 0.25
4. 1.5 0.35 0.55 0.45
5. 2.0 0.6 0.75 0.675
Table 1Calculation of Modulus of Elasticity
Graph:

Load P vs Extension δl
2.5

y = 2.7603x + 0.2133
2

1.5

0.5

0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Conclusion:
Once we know the modulus of elasticity of the nylon wire, it's simple to calculate how much weight
it can hold. This calculation helps us understand the wire's strength and how much it can safely
support, which is useful for various purposes.
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Lab Session 2

Objective:
To investigate the relationship between shear stress and shear strain for rubber and to determine
the modulus of rigidity of the material.

Apparatus:
i. Modulus of rigidity of rubber apparatus
ii. Hangers
iii. Weights
iv. Steel rule
v. Dial Indicator
Rubber Block Dial

Loading Plate

Base
Hang
er
Back
Figure 2.1 Modulus of Rigidity Apparatus

A rubber block 12 x 4 x 1 inch is bonded to two aluminum alloy plates. One plate is screwed
to a wall, whilst the other has a shear load applied by a loaded weight hanger. A dial gauge
measures the deflection of the block.
This equipment is part of a range designed to both demonstrate and experimentally confirm
basic engineering principles. Great care has been given to each item so as to provide wide
experimental scope without unduly complicating or compromising the design.
Each piece of apparatus is self-contained and compact. Setting up time is minimal, and all
measurements are made with the simplest possible instrumentation, so that the student
involvement is purely with the engineering principles being taught.

Modulus of Rigidity:
The Modulus of Rigidity is a mechanical property of a solid material that measures its resistance to
shear deformation. It is also known as the shear modulus or torsional modulus.

When a material is subjected to a shear stress, it tends to deform by twisting or sliding one layer of
the material relative to another. The modulus of rigidity quantifies how much a material will deform
under a given shear stress before returning to its original shape.
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Modulus of Rigidity (GPa)

Aluminum 26
Steel (mild) 79
Stainless steel 77
Copper 42
Brass 37
Cast iron 20-30
Concrete 10-15
Wood (pine) 3-5
Glass 25-30
Plastic (PVC) 0.5-1.5
Nylon 0.5-1.5
Rubber 0.1-0.3
Titanium 44
Magnesium 17
Nickel 75
Lead 5
Zinc 37
Gold 27
Silver 30
Platinum 60

Shear Stress and Strain:

Shear stress and shear strain are fundamental concepts in mechanics, particularly in the study of
materials and structures. They describe the relationship between the forces applied parallel to a
material's surface and its resulting deformation.

Shear Stress:
Shear stress is a measure of the internal force per unit area within a material that arises as a
result of a force applied parallel to its surface. It tends to cause the material to slide or twist.

Units of Shear Stress:

Pascal (Pa): The SI unit of shear stress.
Pounds per square inch (psi): A common unit used in the United States.

Shear Strain:
Shear strain is a measure of the deformation of a material relative to its original dimensions
due to a shear stress. It is the change in angle between two originally perpendicular lines
within the material.

Units of Shear Strain:

Shear strain is dimensionless, but it can be expressed as a ratio or a percentage.
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The Shear Stress-Shear Strain Relationship:

The relationship between shear stress and shear strain is often represented graphically by a shear
stress-shear strain curve. Similar to a stress-strain curve for normal stress and strain, this curve shows
how a material deforms under increasing shear stress.

γ
Figure 2.3 Shear Stress Shear Strain Curve
-

Derivation:
The force which tends to cut off or parts off one portion of the component from the other is called
shear force. Stresses produced on the area under shear, due to shearing forces, are called shearing
stresses. Shear stress is denoted by τ.
Mathematically,
Shearing stress = Shearing force/ Area under shear ------ (i)
Units of shear stress: Newton per square meter (N/m2) = Pascal (Pa) or pounds per square inch
(psi)

Shearing strain is the angle of distortion. It can be represented by γ. ------ (ii)

The constant of proportionality relating shear stress and shear strain is modulus of rigidity. It is
represented by G.
Mathematically,
G = Shear stress/ shear strain ------ (iii)
Units of G: Newton per square meter (N/m2) = Pascal (Pa) or pounds per square inch (psi)

Let us consider the deformation of a rectangular block where the forces acting on the block are
known to be shearing stress as shown in the figure 2.2.
The change of angle at the corner of an originally rectangular element is defined as the shear strain.

Let,
Ps = Shearing load or force acting on the body
l = Length of the body
A = Area under shear = l x t
τ = Shear stress induced in the body
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G = Modulus of rigidity for the material of the body w

γ = Shear strain produced A C

δs = Deformation of the body c

l
From the figure t
B D
Cc = Dd = δs = Shear Deformation tanγ =Dd/BD = δs/w δ
d

P
For smaller angles tanγ = γ
Shear strain = δs/w Figure 2. 2 Distortion of a
Rectangular Block

From the information in (i), (ii), and (iii)

G = τ/γ
or
G = (Ps / δs) (w/ l.t)

Procedure:
1. Set the dial indicator so that its anvil rests on the top of the loading plate.
2. Set the dial indicator at zero.
3. With the hanger in position apply a load to the hanger and read the vertical displacement of
the loading plate relative to the fixing plate from the dial indicator (δs).
4. Repeat the experiment for increasing load and record the vertical displacement of the loading
plate in each case.
5. Unload and note the corresponding readings with the load decreasing.
6. Calculate the “Modulus of Rigidity (G)” of the rubber material.

Observations & Calculations:

Length of rubber block (l) = 302 mm

Width of rubber block (w) = 103 mm
Thickness of rubber block (t) = 25.6 mm
Least count of dial indicator =0.01 mm
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No. Load Shear Deformation-δs Shear Modulus of Rigidity

of Shear Strain
Obs. Ps (mm) Stress
Loading Unloading Average γ = δs / w G =τ/γ G
(lbs) τ =Ps/l .t (N/m2)
From
2
(N/m ) (N/m2) Graph
1. 0 0.00 0.00 0.000 0 0 0
2. 0.5 0.02 0.03 0.025 6.4673E-05 0.000242718 0.266452815 0.2947
3. 1.0 0.04 0.06 0.02 0.000129346 0.000485437 0.266452815
4. 1.5 0.07 0.08 0.075 0.000194019 0.000728155 0.266452815
5. 2.0 0.09 0.10 0.095 0.000258692 0.00092233 0.280476647
6. 2.5 0.11 0.11 0.11 0.000323365 0.001067961 0.302787289
Table 2Calculation of Modulus of Rigidity

Graph:

Shear stress vs Shear strain

0.00035

0.0003 y = 0.2947x - 8E-06

0.00025
Shear Stress

0.0002

0.00015

0.0001

0.00005

0
0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012
-0.00005
Shear Strain

Conclusion:
Once we know the bulk modulus of the rubber pad, we can predict how much it will compress
when something pushes on it. This information helps us understand how the material behaves under
pressure, which is important for designing things that need to withstand pressure without deforming
too much.
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Lab Session 3

Objective:
To determine the modulus of rigidity of the given material of circular shaft.

Apparatus:

i. Torsion of shaft apparatus

ii. Hangers
iii. Weights
iv. Vernier Calipers
v. Steel rule

Figure 3. 1 Torsion of shaft Apparatus

Torsion of shaft apparatus includes a shaft of circular section, two measuring scales and a pulley with
a frame.

The main purpose of the pulley with hanger is to apply some load on the circular shaft. Similarly, the
scales attached to the frame are used to measure the torsion in the circular shaft. Actually, two scales
are used, one at the front and one at the back.
The measuring arms (scales) are used to measure the magnitude of the torsion at the front and the
back of the circular shaft respectively. The front is the portion of the shaft that is near to the pulley
and the back is the portion of the shaft near the back support of the frame.
The main purpose of the frame is to support the shaft and balance the apparatus on the surface.

Torsion and its applications:

Torsion is the twisting of a solid object about an axis. It is a type of deformation that occurs when a
torque is applied to an object, causing it to rotate. This torque can be generated by forces acting in
opposite directions along different lines of action.
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Shafts: Shafts are cylindrical components that transmit rotational motion and torque. They are
commonly used in machinery, vehicles, and power generation systems. The design of shafts must
consider torsion to ensure they can withstand the twisting forces without failure.
Torsion springs: Torsion springs are used to store and release rotational energy. They are commonly
found in various applications, such as door closers, garage door openers, and tensioning mechanisms.
Automotive engineering: Torsion plays a crucial role in the design of automotive components,
including axles, steering systems, and suspension components, to ensure they can withstand the
torsional loads they experience during operation.

Polar moment of inertia:

Polar moment of inertia (also known as the polar second moment of area) is a measure of a cross-
section's resistance to torsion. It is a geometric property that quantifies how the area of a cross-section
is distributed around its centroid. For a circular cross-section with a radius r, the polar moment of
inertia J is given by:
J = πr4/ 2
Note: For other shapes, the calculation can be more complex and often involves integration.

The polar moment of inertia is crucial in understanding the torsional behavior of a shaft or beam. A
higher polar moment of inertia indicates a greater resistance to twisting. This means that for a given
applied torque, a shaft with a higher polar moment of inertia will experience a smaller angle of twist.

Angle of Twist:
Angle of twist is the angular deformation that occurs in a shaft or beam when subjected to a torsional
load. It is a measure of how much a shaft or beam rotates due to the applied torque.

Shaft Design: In designing shafts for machinery, the angle of twist is an important consideration. It
is necessary to ensure that the shaft does not twist excessively under the expected loads, as this can
affect the performance and efficiency of the machinery.
Rotor Blades: The angle of twist in helicopter rotor blades is carefully designed to achieve optimal
lift and thrust.
Joints: The angle of twist in robotic joints determines the range of motion and precision of the robot's
movements.
Prosthetic Limbs: The angle of twist in prosthetic limbs must be carefully controlled to ensure
proper functioning and comfort for the user.

Mechanical Power Transmission:

Mechanical power transmission refers to the transfer of mechanical energy from one point to another.
This energy can be in the form of rotational or linear motion. There are various methods used for
this transmission, each with its own advantages and disadvantages. In this case, we will be focusing
on rotary power transmission, such as:
Gears: Gears are toothed wheels that mesh together to transmit rotational motion and torque between
shafts. They can be used to change the speed or direction of rotation. (Spur gears, helical gears, bevel
gears, worm gears, etc)
Belts and Pulleys: Belts and pulleys are used to transmit rotational motion between shafts that are
not aligned. The belt is wrapped around the pulleys, and as one pulley rotates, it causes the other to
rotate. (Flat belts, V-belts, timing belts, etc)
Shafts and Couplings: Shafts are used to transmit rotational motion over a distance. Couplings are
used to connect shafts together. (Rigid couplings, flexible couplings, universal joints, etc)
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Derivation:
Torsion is the engineering word used to describe the process of twisting a member about its
longitudinal axis.
Consider a solid circular shaft of radius “r” and length “L” fixed at its back face as shown in figure
(b). A line AC is marked on the shaft. If a torque “T’ is applied at its free end, line AC will acquire
the shape of a helix and point A will move to A/.

L
Figure 3. 2Torsion of Shaft

Then from figure,

Angle of twist, θ = <AO A/

Now consider a longitudinal fiber at distance “ρ” from the axis of the shaft.
Deformation in longitudinal fiber, δs = AA/ = ρθ
Strain in longitudinal fiber, γ = δs /L = ρθ/L
Stress in longitudinal fiber, τ = Gθ

The shearing strain is maximum on the surface of the shaft where ρ = r.

If J is the Polar moment of inertia of the shaft, then using above information the torsional formula
for a circular shaft can be written as:

T/J = τ/r = Gθ/L

` or
G = TL/ Jθ

The torsional formula describes the relation of applied torque with the angle of twist and stresses
produced in the shafts.

Procedure:
i. Place the apparatus on a smooth horizontal surface.
ii. Measure the effective length of the shaft using steel rule.
iii. Measure the diameter of the shaft using micrometer.
iv. Adjust the Zeros at 1st and 2nd measuring arms.
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v. Put a load of ten N in the hanger.

vi. Measure the 1st and 2nd angle of twist of the shaft.
vii. Take readings from each group and take average to eliminate error for 1st and 2nd while
loading and unloading.
viii. Take a set of six readings of increasing value of load and then take readings on unloading.
ix. Calculate the “Modulus of Rigidity (G)” of the material of the shaft.

Observations & Calculations:

Effective length of shaft (L) = 249 mm

Diameter of shaft (d) = 4.075 mm
Diameter of torque pulley (D) = 127 mm
Radius of torque pulley (R=D/2) = 63.5 m
Polar Moment of Inertia of the shaft (J=πd4/32) = 27.07 mm 4

No. Load Torque Angle of twist at 1st measuring arm Modulus of Rigidity
of W WR (near end)
Obs. θ1 G=TL/Jθ
(N) (Nmm) (rad) (N/m2)
Loading Unloading Average
1. 1 63.5 0.052359878 0.052359878 0.05235988 16732.28921
2. 2 127 0.104719755 0.104719755 0.10471976 16732.28921
3. 3 190.5 0.157079633 0.157079633 0.15707963 16732.28921
4. 4 254 0.20943951 0.226892803 0.2268928 14873.14596
5. 5 317.5 0.27925268 0.27925268 0.27925268 15211.17201

No. Load Torque Angle of twist at 2nd measuring arm

of W WR (far end) Angle of twist for
Obs. θ2 effective length
(N) (Nmm) (rad) θ1 – θ2
Loading Unloading Average

1. 1 63.5 0.017453293 0.017453293 0.017453293 0.034906585

2. 2 127 0.034906585 0.034906585 0.034906585 0.06981317
3. 3 190.5 0.052359878 0.052359878 0.052359878 0.104719755
4. 4 254 0.06981317 0.06981317 0.06981317 0.157079633
5. 5 317.5 0.087266463 0.087266463 0.087266463 0.191986218
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Conclusion

In this experiment, we measured how much a shaft twisted when we applied different forces to it.
Using the modulus of rigidity, we can now predict how much the shaft will twist under different
loads. This information helps us understand the shaft’s strength and how it will react to forces.
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Lab Session 4&5

Objective:
To measure the stiffness of a compression spring and extension spring.

Apparatus:
i. Compression and Extension of Spring Apparatus
ii. Hangers
iii. Weights

Figure 8. 1 Compression spring apparatus

Figure 8. 2 Extension Spring Apparatus

Summary of Theory:

Spring:
A spring is an object that can be deformed by a force and then return to its original shape after the
force is removed. Its sole purpose is to store and utilize the energy.
Deformations in springs:
There are broadly two types of deformation:
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Elastic deformation:
When the stress is removed the material returns to the dimension it had before the load was applied.
The deformation is reversible, non-permanent.

Plastic deformation:
This occurs when a large stress is applied to a material. The stress is so large that when
removed, the material does not spring back to its previous dimension. There is a permanent,
irreversible deformation. The minimum value of the stress which produces plastic deformation
is known as the elastic limit for the material.
Any spring should be designed so that it only experience up to elastic deformation mostly for efficient
working.

Hooke’s law:
Hooke's law states that the force (F) needed to extend or compress a spring by some distance x is
proportional to that distance. That is,
F = kx
Where F is the force, x is the length of extension/compression and k is a constant of proportionality
known as the spring constant which is usually given in N/m.

Figure 8. 3 Hooke’s law visual representation

Application of Springs in different machines:

Washing Machines:
• Suspension: Springs support the washing machine drum and absorb vibrations during the
spin cycle.
• Door Latch: Springs provide tension in the door latch to keep it securely closed.

Clocks:

• Mainspring: The mainspring stores the energy that powers mechanical clocks.
• Hairspring: In balance wheel clocks, the hairspring controls the oscillation of the balance
wheel, regulating the clock's timekeeping.
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Automobiles:

• Suspension: Springs absorb shocks and vibrations from the road, ensuring a comfortable
ride.
• Engine Valves: Springs hold engine valves closed and provide the necessary force for them
to open.
• Door Hinges: Tension springs in door hinges help keep doors closed and prevent them from
sagging over time.

Calculators:

• Keyboard: Springs provide tension in the calculator keys, ensuring they return to their
original position after being pressed.

Office Equipment:

• Paper Feed Mechanisms: Springs are used in paper feed mechanisms to ensure a smooth
and consistent flow of paper.

Material of spring:
Steel alloys are the most commonly used spring materials. The most popular alloys include
high carbon (such as the music wire used for guitar strings), oil-tempered low-carbon, chrome
silicon, chrome vanadium, and stainless steel.
Other metals that are sometimes used to make springs are beryllium copper alloy, phosphor
bronze, and titanium. Rubber or urethane may be used for cylindrical, non-coil springs.
Ceramic material has been developed for coiled springs in very high-temperature
environments. One-directional glass fiber composite materials are being tested for possible use
in springs.

Types of springs according to loading conditions:

Springs can be classified depending on how the load force is applied to them:

Tension/extension spring:
The spring is designed to operate with a tension load, so the spring stretches as the load is applied to
it.

Figure 8. 4 Extension spring

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Compression spring:
It is designed to operate with a compression load, so the spring gets shorter as the load is applied
to it.

Figure 8. 5 Compression spring

Torsion spring:
The load applied to a torsion spring is a torque or twisting force, and the end of the spring rotates
through an angle as the load is applied.

Figure 8. 6 Torsion spring

Constant spring
In this the supported load will remain constant throughout the deflection of spring.

Figure 8. 7Constant spring

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Variable spring:
In this, the resistance of the coil to load varies during compression.

Figure 8. 8 Variable spring

Types of springs according to shapes:

Coil spring:
This type is made of a coil or helix of round wire.

Figure 8. 9 Coil Spring

Flat spring:
This type is made of a flat spring steel.

Figure 8. 10 Flat Spring

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Machined spring:
This type of spring is manufactured by machining bar stock with a lathe and/or milling
operation rather than a coiling operation. Since it is machined, the spring may incorporate
features in addition to the elastic element. Machined springs can be made in the typical load
cases of compression/extension, torsion, etc.

Figure 8. 11 Machined Spring

Serpentine spring:
A zig-zag of thick wire - often used in modern upholstery/furniture.

Figure 8. 12 Serpentine Spring

Derivation of stiffness formula:

A spring may be defined as an elastic member whose primary function is to deflect or distort under
the action of applied load; it recovers its original shape when load is released. Springs are energy
absorbing units whose function is to store energy and to restore it slowly or rapidly depending on the
particular application. In order to derive a necessary formula which governs the behavior of springs,
consider a closed coiled spring subjected to an axial load W.

Let,
W = axial load D = mean coil diameter
d = diameter of spring wire N = number of active coils
G = modulus of rigidity ∆ = deflection of spring
Φ = Angle of twist l = length of spring wire = πDN ---- (i)

In 1879, Alberto Castiglione’ an Italian railroad engineer, published a book in which he outlined a
method for determining the displacement / deflection & slope at a point in a body. This method which
referred to Castiglione’s Theorem is applied to the bodies, having constant temperature & material
(homogeneous) with linear elastic behavior. It states that “The derivative of the strain energy with
respect to the applied load gives the deformation corresponding to that load”.
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For a helical spring, the partial derivative of the strain energy w.r.t. the applied load gives the
deflection in the spring i.e. ∂U / ∂W = deflection. Consider a helical compression spring made up of
a circular wire and subjected to axial load W as shown in the figure above.

Strain Energy is given by: U = ½ T * Φ ----- (ii)

Whereas,
T= ½ W * D ----- (iii)
Φ = T l / JG ----- (iv)

(From Torsion formula) putting the values from eqs. # (i), (iii) & (iv) in eq. # (ii) and simplifying,
we get;
T= 4 W2D3N/d4G ----- (v)

Now applying the Castiglione’ theorem by taking the partial derivative of the strain energy with
respect to the applied load
∂U / ∂W = ∆ = 8 WD3N/d4G ----- (vi)
W / ∆ = d4G/8D3N

Stiffness = K = d4G/8D3N

Procedure:
i. Set the compression spring on apparatus.
ii. Take the reference point on reading.
iii. Put the load with the difference of 1lb.
iv. Take the deflection reading from scale on it.
v. Find the stiffness using formula.
vi. Set the extension spring on apparatus.
vii. Put the load with the difference of 1lb and take deflection.
viii. Calculate stiffness of extension spring using formula.

Observations and Calculations:

No. of Load(W) DEFLECTION Slope Theoretical Difference Percentage

obv from value error
graph
loading unloading mean
0 0 0 0 0 0.3081 0.313048 0.004948 1.580513
1 1 3 3 3 0.3081 0.313048 0.004948 1.580513
2 2 6 7 6.5 0.3081 0.313048 0.004948 1.580513
3 3 10 10 10 0.3081 0.313048 0.004948 1.580513
4 4 13 13 13 0.3081 0.313048 0.004948 1.580513
5 5 16 16 16 0.3081 0.313048 0.004948 1.580513
Table 8.1 Calculations of stiffness of a compression spring
 P a g e | 24

Compression
6

5 y = 0.3081x + 0.0097

4
Axis Title

0
0 2 4 6 8 10 12 14 16 18
Axis Title

No. of Load(W) DEFLECTION Slope Theoratical Difference Percentage

obv from value error
graph
loading unloading mean
1 1 0 0 0 0.5163 0.489615 0.026685 5.450293
2 2 1 1 1 0.5163 0.489615 0.026685 5.450293
3 3 3 3 3 0.5163 0.489615 0.026685 5.450293
4 4 5 5 5 0.5163 0.489615 0.026685 5.450293
5 5 7.5 7.5 7.5 0.5163 0.489615 0.026685 5.450293
Table 8.2 Calculations of stiffness of a extension spring

Extension
6
y = 0.5163x + 1.2962
5

4
Axis Title

0
0 1 2 3 4 5 6 7 8
Axis Title
 P a g e | 25

Conclusion:
Assessing the stiffness of compression and extension springs yields essential information about
their flexibility, facilitating a better understanding of their suitability for various tasks. This data is
critical for comprehending their performance and determining their appropriateness for specific
applications.