Chapter 11 Deformation of Solids

CHAPTER 11 DEFORMATION OF SOLIDS prepared by Yew Sze Ling@Fiona, KML

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Chapter 11 Deformation of Solids

Curriculum Specification Remarks

Before After Revision

11.1 Stress and Strain

a) Distinguish between stress and strain for tensile and

compression force. (C1, C2, C3, C4)

b) Discuss the graph of stress and strain for a metal under

tension. (C1, C2)

c) Discuss elastic and plastic deformation. (C1, C2)

d) Discuss graph of force – elongation, f-e for brittle and

ductile materials. (C1, C2)

11.2 Young’s Modulus

a) Define Young’s modulus. (C1, C2)

b) Discuss strain energy from force – elongation graph

(C1, C2)

c) Discuss strain energy per unit volume from stress – strain

graph. (C1, C2)

d) Solve problems related to the Young’s Modulus.

(C3, C4)


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11.1 Stress and Strain

The elastic property of a material is an important criterion to determine the strength of the

material.

The elastic behaviour of the material measures the tendency of the material to return to its

original shape when the stress on the material is removed.

Stress

Stress, σ is defined as the force acting per unit cross sectional area.

The basic unit for stress is kg m-1 s-2 or normally known as Nm-2

or Pa (Pascal)

Stress exists within material.

The rod shown in (a) is said to be under tension or tensile stress. This tensile stress exists throughout the material.

Compressive stress is the exact opposite of tensile stress. Instead

of being stretched, the material is compressed: the forces act inwardly on the object as shown in (b).

Strain

Strain, ε is defined as the extension per unit length.

There is no unit for strain.

It is a measure of how much the object has been deformed.

Stress is applied to the material by external agents, whereas strain is the material’s response to

the stress.

The Graph of Stress – Strain

𝜎 =𝐹

𝐴

𝜀 =𝑒

𝑙0=∆𝑙

𝑙0

A : Limit of proportionality

B : Yield point

D : Breaking stress

E : Object Breaks

OB : Elastic deformation

BE : Plastic deformation


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Proportionality limit (Point A) :

Maximum stress for which Hooke’s law is still valid.

OA: When the force is small, the extension is directly proportional to the force. The object

obeys Hooke’s law.

When the force is removed, the object returns to its original length. The object shows elastic behaviour in this region.

Yield point (Point B) :

The yield point marked a change in the internal structure of the object.

This is the elastic limit of the object. The object returns to its original length along path

BAO when the force is removed.

Plastic deformation start beyond point B. Beyond this point, the object is permanently

stretched and will never regain its original shape and length.

OB region is known as elastic deformation.

If the force on the object is removed at C, the wire has a permanent extension of OP.

The planes of atoms slide across each other. When a force on the wire is further increased, extension increases rapidly.

Breaking stress (Point D) :

The force or stress on the object is at maximum. The stress at D is

known as the breaking stress. After point D, the wire thin and ‘necks’

are formed.

Object Breaks(Point E) :

The object breaks or fractures.

BE region is known as plastic deformation.

Deformation

Forces not only cause acceleration, but can also

cause changes in the inner structure of solids

(externally changes shape or size). Such changes are

called “deformations”.

ELASTIC DEFORMATION PLASTIC DEFORMATION

Extension does not exceed the elastic limit of

material.

Extension exceeds the elastic limit of material.

When stretching force is removed, the object

returns to its original length.

When stretching force is removed, the object

does NOT return to its original length. The object is permanently lengthened.

Obeys Hooke’s law Does NOT obeys Hooke’s law

Does NOT change the internal structure of a

wire.

Change the internal structure of a wire.

When Hooke’s law is obeyed, the energy store

during elastic deformation is fully recovered when the force is removed.

During plastic deformation, energy is converted

into heat.


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Brittle vs Ductile Materials

Brittle material: No significant plastic deformation prior fracture.

Ductile material: Sustain significant plastic

deformation prior fracture

Young’s Modulus

Young’s modulus measures the resistance of a solid to a change in its length.

If the stress is not high & the limit of proportionality has not been exceeded, Young’s Modulus, Y of a material (such as wire) can be defined as the ratio of tensile stress to strain.

SI unit for Y : N m-2 or Pa (Pascal)

Young’s modulus does not depend to the length of the wire but it depends to the material made

the wire.

Greater Y, stiffer the materials.

Stiffer means stronger inter atomic bonding. Greater force

must be applied to separate the atoms in turn to stretch the

material.

Stiffness is a measure of a material's resistance to deformation.

A stiff material does not easily bend or changed in shape.

Strain Energy

When an elastic object is stretched by a force, the work done on the object is transformed into strain

energy. The strain energy, which is also a form of potential energy, is stored in the atoms or

molecules due to their displacement from their mean position.

𝑌 =stress

strain=𝜎

𝜀


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Strain Energy from Force – Elongation Graph, F-e

Thus, the strain energy stored in an object if the elastic limit is not exceeded:

𝑊 = 𝐹𝑥 𝐹 = 𝑘𝑥 and

𝑑𝑊 = 𝐹𝑑𝑥

𝑊 = ∫𝐹 𝑑𝑥

𝑊 = ∫𝑘𝑥 𝑑𝑥

𝑊 = [𝑘𝑥2

2]0

𝑒

𝑊 = [𝐹𝑥

2]0

𝑒

𝑊 =1

2𝐹𝑒 Area under F-e graph

𝑼 =𝟏

𝟐𝑭𝒆

U = Area under F-e graph


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Strain Energy per Unit Volume from Stress – Strain Graph, σ-ε

Thus, strain energy per unit volume:

Unit for strain energy per volume is J m-3.

𝜎 =𝐹

𝐴

𝜀 =𝑒

𝑙0

𝐹 = 𝜎𝐴

𝑒 = 𝜀𝑙0

𝑈 =1

2𝐹𝑒

𝑈 =1

2(𝜎𝐴)(𝜀𝑙0)

𝑈

𝐴𝑙0=1

2𝜎𝜀 𝐴𝑙0 = 𝑉 and

𝑼

𝑽=𝟏

𝟐𝝈𝜺 Area under σ-ε graph


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Exercise

Stress and Strain

1. Two stretched cables both experience the same stress. The first cable has a radius of

3.5×10-3 m and is subject to a stretching force of 270 N. The radius of the second cable is

5.1×10-3 m. Determine the stretching force acting on the second cable.

2. A wire originally 2 m long suffers a 0.1% strain. Calculate its stretched length.

Young’s Modulus

1. A tow truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has

a radius of 0.50 cm. When the car just begins to move, the tension in the cable is 890 N.

Given the Young’s modulus of the steel cable is 2.0×1011 N m-2, how much has the cable stretched?

2. Is the Young’s modulus for a bungee cord smaller or larger than that for an ordinary rope?

Explain your answer.

3. A marble column of cross-sectional area 1.4 m2 supports a mass of 25,000 kg. Given the

Young’s modulus of the marble is 50×109 N m-2.

a) What is the stress within the column?

b) What is the strain?

4 A piece of bone is under tension. The graph below shows variations between stress and

strain for the bone which is under tension. Calculate the Young’s modulus for the bone.

5. A wire X has length 100 cm and diameter 2.0 mm. A wire Y has length 200 cm and diameter

4.0 mm. Forces of same magnitude pull both wires and the extensions of X and Y are 0.10

mm and 0.04 mm respectively. Determine the ratio of the Young’s modulus for wire X to that for wire Y.

6. An object of mass 6.0×103 kg is suspended from a steel cable of length 50 m and cross-

sectional area 2.0×10–4 m2. Determine the extension of the cable when the mass-cable

system

a) at rest,

b) moves upwards at constant speed,

c) moves upwards at constant acceleration 2.0 m s–2.

Assume the cross-sectional area to be constant.

(Young’s modulus for steel = 2.0 ×1011 Pa)


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7.

The force F against elongation e graphs for wire X and wire Y as shown in figure above. The wires have the same original length and same cross sectional area of 5.0 mm2.

a) Calculate the strain energy stored because of extensions for wire X and wire Y by 1.5 mm.

State which wire is more rigid.

b) From the graphs above, calculate the ratio of the Young’ modulus of wire X to the Young’ modulus of wire Y.

8. A metal bar that has a cross-sectional area of 4.0 cm2 and length of 50.0 cm is stretched by a

force of 1.0×104 N.

a) Find the stress on the wire.

b) If the bar elongates by 0.5 cm, what is the strain in the wire?

c) Determine the Young’s Modulus of the wire.

d) Calculate the strain energy stored in the wire.

e) Calculate the strain energy per unit volume of the wire.

9. The force constant of the copper wire and steel wire are 2.04×105 N m-1 and 3.39×105 N m-1

respectively. If both undergoes the same elongation, which of the wire has more strain

energy?

10. Two steel wires P and Q are suspended from the ceiling and each wire is extended by a

weight W. If the original length of P and Q are 2 cm and 4 cm respectively, and the

extension produced on the two wires is same, what is the ratio of diameter of wire P to

diameter of wire Q?

Chapter 11 Deformation of Solids

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