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Every object or matter occupies a space with its size and shape properties. The change in size and shape of an object through the action of external forces is called deformation. The deformation is a mechanical property of a solid which occurs due to the action of either stress or strain. Even a small force or pressure can deform any object. Deformation takes place when the intermolecular forces within an object oppose the applied external force, sometimes an object moves into a complete deformation state (permanent deformation) due to the application of higher external forces.
Key Terms: Solid Deformation, Deformations, Young’s Modulus, Bulk Modulus, Shear Modulus, Bulk Stress, Tensile Stress, Shear Stress
Types of Solid Deformation
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There are two types of solid deformations, and they are:
- Temporary Deformation
- Permanent deformation
Temporary Deformation
It is defined as short or reversible deformation where the object tends to remain deformed until external forces are applied. When the forces are eliminated, the object retains its initial state and is also known as elastic deformation.
Permanent Deformation
It can be defined as irreversible deformation where the object completely moves into a deformation state and its initial state cannot be restored. Mostly this kind of deformation occurs due to heavier external force and it can be called plastic deformation.
Stress and Strain Factor
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In the physics of solids, stress and strain are principal factors responsible for the deformation of solids.
- Stress
- Strain
Stress and Its Types
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It is defined as the restoring force per unit area whereas it explains the magnitude of forces. The SI unit of stress is N/m2 It can be mathematically written as
\(\partial = \frac{F}{A}\) (Unit: N/m2)
Where, σ - is the applied stress
F - force applied
A – area of any object
Types of Stress
- Bulk Stress: Defined as the deformation of a body due to applied external forces from all directions. Example: Pressure experienced by an ocean diver in ocean depth.
- Compressive Stress: Defined as Compression of an object due to resultant forces. Example: Formation of the mountain at convergent boundary and concrete pillar.
- Tensile Stress: Defined as the elongation of an object due to the action of forces (pull). An example of tensile stress is the stretching of a rubber band or rubber tubes.
- Shear Stress: Defined as the deformation of an object due to applied tangential force. Example: Force experienced between the ground and feet during the walk.
Strain and Its Types
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It is defined as the degree of how much an object is stretched or buckled. It can be mathematically expressed as
\(\sigma\)=\(\frac{\eth l }{L}\) (Unit: No unit - dimensionless)
Where,
- \(\sigma\) - strain due to applied stress
- δl – change in length of the material
- L – Initial or original length of the material
Types of Strain
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- Tensile Strain: It is defined as the change in area or length of a body due to applied tensile stress.
- Compressive Strain: It is defined as the change in the area or length of a body due to applied compressive strain.
Stress-Strain Curve
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The Stress-strain curve establishes a materialistic relationship between stress and strain. The curve reflects numerous properties.
Stress-Strain Curve
- Proportional Limit: The point in the graph where the material obeys Hooke’s law. Here, the ratio of stress and strain gives a proportionality constant called Young’s modulus.
- Elastic Limit: The region of limit where the material has the capability to restore its initial position after deformation.
- Yield Point: The point of limit from where the material starts to deform drastically. Beyond this limit, the material undergoes permanent deformation.
- Ultimate Stress Point: The point which defines the maximum stress tolerance capability of any materials before they undergo failure.
- Breaking Point: Beyond the ultimate stress point, the material leads to a state of failure or degradation.
Features of Solid Deformations
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- Deformation occurs when the internal intermolecular forces act against the applied external forces.
- Weak external forces may lead to negligible deformation, and acquire a new equilibrium state resulting in reformation or return to the original state.
- Heavier external forces over any object can either lead to permanent deformation or complete structural failure
- Deformation in solid takes place depending on the type of material, geometry, size, and magnitude of external forces.
Properties of Solid Deformations
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The properties of solid deformations are as follows:
Elastic Modulus & Young’s Modulus
The elastic modulus is the proportionality constant of the ratio of stress and strain. It can also be called Young' Modulus. Its SI unit is Pa. It defines the stiffness of any material whose elastic property can be understood from equation
Elastic Modulus = \( \frac{Stress}{Strain}\)
Young’s Modulus, Y = \(\frac{\sigma}{\epsilon}\)
Where,
- Y – Young’s modulus
- σ – stress
- \(\epsilon\) - strain
Shear Modulus
It is defined as the measure of the ratio of shear stress to the shear strain. The SI unit of shear modulus is Pa (pascal). It can be mathematically written as
G= \(\frac{Fl}{A\bigtriangleup x}\)
Where,
- G – Shear modulus
- F – Force on the object
- l – initial length of an object
- A – Area over which the force is acting.
- \(\bigtriangleup x\) – transverse displacement
Read More: Shear Modulus
Bulk Modulus
It is the measure of the ratio of applied pressure to the corresponding decrease in volume of any material. It can be expressed as
B = \(\frac{\bigtriangleup P}{\frac{\bigtriangleup V}{V}}\)
Where,
- B – Bulk modulus
- ΔP – Change in applied pressure
- ΔV – Change in volume of the material
- V – Initial volume of the material
Read More: Bulk Modulus
Things to Remember
- The change in size or shape of an object due to the action of applied external forces is termed deformation. It is a mechanical property and categorized into two types which are temporary and permanent deformation.
- Temporary deformation is reversible and occurs due to weak external forces whereas permanent deformation is irreversible due to large or heavy forces. Then, temporary deformation can also be called elastic deformation whereas permanent deformation is plastic deformation.
- The effect of stress and strain factors are responsible for deformation in solids. The stress is the measure of magnitude which is force per unit area, and strain is the measure of the degree of elongation or stretch.
- The stress factor exhibits four solid deformation properties such as bulk, compressive, tensile, and shear modulus whereas the strain exhibits tensile and compressive strain properties.
- The stress-strain curve responsible for solid deformations gives out five significant terms which are proportional limit, elastic limit, yield point, ultimate stress point, and breaking point.
- Finally, the type of deformation in solid occurs depending on the type of material, geometry, size, and magnitude of external forces.
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Sample Questions
Ques. A Steel wire of length 4.7m and cross-sectional area of 3.0x10-5 m2 stretched by the same amount as a copper wire of length 3.5m, and a cross-sectional area of 4.0x10-5 m2 under a given load. What is the ratio of Young's modulus to that of copper? (5 Marks)
Ans. We have numerical values for steel as: l1 = 4.7m and A1 = 3.0x10-5 m2
So, the Young modulus for steel is Y1 = F×4.73.0×10-5×?l
Similarly, for the copper we have l2 = 3.5m, and A2 = 4.0x10-5 m2
Now, the Young modulus for copper is Y2 = F×3.54.0×10-5×?l
Dividing Y1 by Y2, we get Y1Y2= 4.73.0×10-5× 4.0×10-53.5=1.79
Explanation: For the above problem, we have the initial length of both steel and copper followed by the area over with the force acting on it. So, substituting the given values for the Young modulus equation (for both steel and copper), we will get two equations as Y1 and Y2. Then, solving both equations by dividing Y1 by Y2, we get the ratio of Young’s modulus of both steel and copper. The ratio has no dimension or unit.
Ques. Read the following two statements below. And state your answer with reasons if it is true or false.
(a)The Young’s modulus of rubber is greater than that of steel.
(b)The stretching of a coil is determined by its shear modulus. (3 Marks)
Ans. (a) False
(b) True
Explanation: (a) We know that the Young modulus is the measure of the ratio of stress and strain. So, for the given material, the stretching force of elongation is higher in rubber compared to steel. Therefore, the rubber is less elastic than steel.
(b )We know that the stretching of an object is determined by the shear modulus. So, when equal and opposite forces are applied at both ends of the coil, the shape and length of the coil will change, and it involves shear modulus.
Ques. A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Shear modulus of elasticity of copper is 42 x 109 N/m2. (3 Marks)
Ans. We have the following numerical values
Cross-sectional area (A) = 15.2 x 19.1 x 10-6 m2
Tension Force (F) = 44500 N
Elasticity of copper = 42 x 109 N/m2.
So, the Strain = Stress/Elasticity
= (44500)/ (15.2 x 19.1 x 10-6)x(42 x 109)
= 3.65 x 10-3 (no unit)
Explanation: The problem has a cross-sectional area, external force, and elasticity. So, to find out the strain, we are modifying the elasticity formula which is the ratio of stress and strain. Then, make sure to convert all the measurements into SI units before putting them into the equation. Upon, solving the values, we get the resultant strain.
Ques. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 Nm-2 what is the maximum load the cable can support? (3 Marks)
Ans. We have radius of the wire (r) = 1.5 cm = 1.5 x 10-2 m
Maximum stress = 108 N/m2
We need to find the maximum load, which is max. force. So, from the stress-strain equation
F = stress x cross-sectional area
F = 108 x 22/7 (1.5 x 10-2)2
F = 7.07 x 104 N
Explanation: In this problem, we are provided with the stress and radius of the wire, and we need to calculate the maximum load. At first instant, we will find the area of the cross-section of the copper wire using its radius followed by substituting in the above equation. Here, the maximum load is nothing but the maximum force. Therefore, resolving the numerical values, we get F(maximum load) = 7.07 x 104 N
Ques. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 x 106 Pa. (5 Marks)
Ans. We have the following values,
Side of a copper cube = 10 cm
So, the volume of the copper cube is V = 10-3 m3
Applied Pressure = 7.0 x 106 Pa
Bulk modulus = 140 G Pa = 140 x 109 Pa
So, from the bulk modulus formula, we have ΔV = PV / (B)
= (7.0 x 106 x 10-3)/( 140 x 109)
= 5 x 10-8 m3 or 5 x 10-2 cm3
Explanation: First, we need to find out the volume of the copper cube followed by a change in volume. Since the unit of copper cube side is given in cm, we need to convert to SI (m) unit then to cm (in writing final solution). Then, from the bulk modulus formula, transform and modify it to find the ΔV. Finally, upon resolving all the numerical values, we get a change in volume as 5 x 10-2 cm3.
Ques. A stress of 2 N/m2 is applied to an object to produce a strain of 0.5. Find Young's modulus. (2 Marks)
Ans. Y = σ/\(\epsilon\) = 2 / 0.5 = 4 N/m2
Explanation: We have stress = 2 N/m2, and Strain = 0.5. Then, we know that the ratio of stress and strain is Young’s modulus. Therefore, dividing 2 by 0.5, we get 4 N/m2.
Ques. The Young’s modulus for material is given by 26.66 N/m2 whereas its strain is 0.15. Find the stress of the material. (2 Marks)
Ans. Stress = Young’s modulus x Strain = 26.66 x 0.15 = 4 N/m2
Explanation: We have Young's modulus and strain, we need to find out the stress. So, from the Young modulus formula, we have stress equals the product of Young's modulus and strain, substituting the value, we get Stress = 4 N/m2.
Ques. The following graph shows stress-strain for materials.
(a)Which of the materials has greater Young’s modulus?
(b)Which of the two materials is stronger? (3 Marks)
Ans. (a) Young’s modulus for A is greater than B
(b) Material A is stronger than Material B
Explanation: (a) From the above graphs, we can see that the strain for material A is greater than that of material B. So, Young's modulus for material A is greater than B. (b) Strength of any material is defined by the amount of stress required to cause failure. So, the stress equals the point of fracture. Hence, the point of fracture for material A is greater than material B (ref-figure). Therefore, material A is stronger than material B.
Ques. A Wire is suspended from a roof but no weight is attached to the wire. Is the wire under stress? (3 Marks)
Ans. Yes, the wire is under stress.
Explanation: The weight of the wire suspended from the roof itself acts as the deforming force here. The weight of the wire makes it stretch along the vertical and downward direction resulting in wire deformation.
Ques. Explain Poisson’s ratio. (3 Marks)
Ans. The ratio of lateral strain to the linear strain is called Poisson’s ratio.
Explanation: We know that the Poisson's ratio is defined as the transverse strain to longitudinal strain along the direction of stretching force. Therefore, the answer is “Ratio of transverse strain and longitudinal strain” or “Ratio of lateral strain to the linear strain”.
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