MECHANICAL PROPERTIES OF SOLIDS
Deforming Force & Restoring Force
the body may get deformed. Then
the force developed inside the body,
which try to bring the body back to
its original shape and size is called
restoring force.
Elasticity
Plasticity
9.2 ELASTIC BEHAVIOUR OF SOLIDS
Spring-ball model for the illustration of elastic behavior of solids
9.3 STRESS AND STRAIN
Stress�The restoring force or deforming force experienced by a unit area is called stress.
�
S.I unit = Nm−2
Dimension is [ML-1T-2]
9.3 STRESS AND STRAIN
A] Normal Stress:
When the elastic restoring force or deforming force acts perpendicular to the area, the stress is called normal stress. Normal stress can be sub-divided into the following categories
1) Tensile Stress 2) Compressive Stress
3) Volume Stress
9.3 STRESS AND STRAIN
B] Tangential or Shearing Stress:
When the elastic restoring force or deforming force acts parallel to the surface area, the stress is called tangential stress
9.3 STRESS AND STRAIN
Strain
Types of Strain
9.3 STRESS AND STRAIN
Longitudinal Strain
9.3 STRESS AND STRAIN
Volumetric Strain:
9.3 STRESS AND STRAIN
Shearing Strain
Cause of Elasticity
Intermolecular Forces: (Click in black portion)
9.4 HOOKE’S LAW
Hooke’s law states that within the elastic limit, stress is directly proportional to strain.
Stress α Strain
stress = k × strain
where k is known as Modulus of elasticity
9.5 STRESS-STRAIN CURVE
9.5 STRESS-STRAIN CURVE
Part OA
This part OA obeys Hooke’s law. The point ‘O’ is called elastic limit or yield point. Corresponding stress is called yield strength. In this region the material behaves as elastic material.
Part AB
In this region a small increase in stress produce a large change in strain. At any point between AB,if the deforming force is removed, the body will never return to the original length. But results a permanent change in length (Eg: OO’). This is known as permanent set.
9.5 STRESS-STRAIN CURVE
Part BC :
The stress corresponding to the point ‘C’ is called Ultimate tensile strength (Su). This is the maximum stress that can be applied to a wire.
Part CD :
Beyond the point C, additional strain is produced even by a reduced stress and the wire breaks at D
Ductile Solids
Materials have a large CD region is called ductile solids. So these materials can be drawn into thin wires eg: copper , Aluminum
Brittle Materials
If C and D are very close, the material is said to be brittle. It suddenly breaks as soon as the ultimate strength (C) is crossed Eg : Glass.
Elastomer
Substances which can stretch to large values of strain are called elastomer.
These materials does not obey Hooke’s law
e.g.: Rubber band, Aorta
Elastic Modulii
According to Hooke’s law, within elastic limit,
�Stress ∝ Strain�Stress = k × Strain�
It is known as modulus of elasticity
Types of modulus of elasticity
Young’s Modulus (Y)
�
�
∴
�
Where,�F - Force applied�r - Radius of the wire�l - Original length�Δl - Change in length�Unit → Nm−2 or Pascal (denoted by Pa)
Bulk Modulus (B)
If P is the increase in pressure applied on the spherical body, then�P = F/A
B =
�
∴ B =
�V - Original volume�ΔV - Change in volume�Unit → Nm−2 or Pascal�
Compressibility (k) −
Reciprocal of bulk modulus of elasticity (B)
i.e., k = 1/B
Rigidity Modulus (G)
G =
G =
Where,
F - Force applied�a - Area�L- Original length�ΔL - Change in length�Units → Nm−2 or Pascal
Q.1Which one is more elastic Rubber or Iron?
Answer: Those materials for which large amount of stress causes less amount of strain are more elastic. When iron and rubber are subjected to some stress(pressure) then rubber can be stretched easily but iron cannot. Hence iron is more elastic than rubber.
Q.2Why the springs are made up of iron and not of copper?
Answer: A good spring will be that in which a large restoring force is developed on being deformed. This in turn depends on the elasticity of the material of the spring. Since Young's modulus for steel is more than that of copper, therefore the springs made of steel and not of copper.
What we know?
Stress, Strain & Modulii of Elasticity
Types of Stress | Stress | Strain | Change in | Elastic Modulus | Name of Modulus | State of Matter | |
Shape | Volume | ||||||
Tensile or Compressive | Two equal and opposite forces perpendicular to opposite faces(σ=F/A | Elongation or compression parallel to force direction(∆L/L)(Longitudinal Strain | Yes | No | Y=(FxL)/Ax∆L) | Young’s Modulus | Solid |
Shearing | Two equal and opposite forces parallel to opposite surfaces (forces in each case such that total force & total torque on the body vanishes (σs=F/A) | Pure Shear,Ѳ | Yes | No | G=(F x Ѳ)/A | Shear modulus | Solid |
Hydraulic | Forces Perpendicular everywhere to the surface. Force per unit area is same everywhere | Volume change (Compression or elongation(∆V/V) | No | Yes | B=-P/(∆V/V) | Bulk Modulus | Solid ,Liquid & Gas |
Applications of Elastic Behavior of Materials
Construction of a Beam
Consider a bar of length l, breadth b, and depth (height) ‘d ‘ . Let a load ‘W’ be applied at the mid point of the bar. The mid-point will sag by an
amount ‘δ’ is given by
[Given; the beam has to support a maximum load ‘W’ and Beam length or span is ‘l’]
Applications of Elastic Behaviour of Materials
From the expression it is clear that, to reduce bending of a Steel Bar; Use material of large Young’s modulus (Y) . It is better to increase the
depth (δ α 1/d3) rather than the breadth (δ α 1/b).
Activity: Try to bend your scale by holding it vertically & Horizontally ,then see and share the difference
Applications of Elastic Behaviour of Materials
As depth increases the chance of buckling also increases.
So to avoid buckling and bending I section beams are used.
This shape reduces the weight of the beam without sacrificing the
strength and hence reduces the cost.
Height of a Mountain
At the base of the mountain pressure exerted is= (hρg)
Where h=height of mountain,
ρ= density of rocks (3×103) kgm−3.
But the elastic limit of typical rock at the bottom of mountain is=30 × 107 Nm-2 .
Equating; hρg = 30 × 107 Nm-2
h = 10km
From the calculations it is clear that maximum height of mountain must be less than 10 km.
h
The maximum height of mountain on earth depends upon shear modulus of rock. At the base of the mountain, the stress due to all the rock on the top should be less than the critical shear stress at which the rock begins to flow.
9.6.5 Poisson’s Ratio
lateral strain
lateral stain is directly proportional to the longitudinal strain.
longitudinal strain in a stretched wire
is called Poisson’s ratio
Q.3 Why Bridges are declared unsafe after many years of use?
Ans. Due to the repeated stress and strain, the material used in the bridges loses elastic strength and ultimately may be collapsed. Hence bridges are declared unsafe after long use.
9.6.6 Elastic Potential Energy in a Stretched Wire
When a wire is put under a tensile stress, work is done against the inter-atomic forces.
This work is stored in the wire in the form of elastic potential energy
F = YA (l/L).
work done dW = F dl or YAld l /L
W = ½ × Young’s modulus strain2 volume of the wire
= ½ × stress strain volume of the wire
This work is stored in the wire in the form of elastic potential energy (U). Therefore the elastic potential energy per unit volume of the wire (u) is