Mechanics of Solids

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Mechanics Of Solids

Suraj kr. Ray (surajjj2445@gmail.com) Department of Civil Engineering

Mechanics of Solids

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Mechanics of Solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading.

Other names for this field of study are strength of materials and mechanics of deformable bodies.

Introduction:-

A force exerted on a body can cause a change in either the shape or the motion of the body. The unit of force is the Newton, N.

No solid body is perfectly rigid and when forces are applied to it, changes in dimensions occur. Such changes are not always perceptible to the human eye since they are so small. For example, the span of a bridge will sag under the weight of a vehicle and a spanner will bend slightly when tightening a nut. It is important for engineers and designers to appreciate the effects of forces on materials, together with their mechanical properties. The three main types of mechanical force that can act on a body are: (i) tensile, (ii) compressive, and (iii) Shear.

1. Tensile force:

When two equal magnitude force occurred on a body in such a way that their senses (Directions) are opposite and outward from the CG of the body, is known as tensile force. Tension is a force which tends to stretch a body.

Examples:

A rope or Cable of crane carrying load is in tension.

When nut is tightened, bolts are in tension.

2. Compressive force: When two equal magnitude force

occurred on a body in such a way that their senses (Directions) are opposite and toward the CG of the body, is known as compressive force. Compression is a force which tends to squeeze or crush a body.

Examples:

A pillar or column supporting bridge or slab is in compression.

The sole of a shoe is in compression.

3. Shear force:

When two equal magnitude forces occurred on a body in such a way that their senses (Directions) are opposite and line of actions are also different, is known as shear force.

Shear is a force that tends to slide one face of the material over an adjacent face.

Example:

A rivet holding two plates together is in shear stress if tensile force is applied between the plates.

A horizontal beam is subjected to shear force.

Mechanics of Solids

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Try Yourself:

Stress: Stress is nothing but only a ratio of

applied force (say F) to the cross-sectional area (say A).

Types of stress Symbol used

Tensile stress σ(Greek word sigma)

Compressive stress

σ

Shear stress τ (Greek word tau)

From definition:

S.I. unit of stress:

So, Pascal is the S.I. unit of stress. Problem: A rectangular bar having a cross-sectional area of 75 mm2 has a tensile force of 15 KN applied to it. Determine the stress in the bar. Given data:

A =cross-sectional area = 75 mm2

=75 x 10-6 m2

F = 15 KN = 15 x 103 N

Stress in bar,

Problem: A circular wire has a tensile force of 60 N applied to it and this force produces a stress of 3.06 MPa in the wire. Determine the diameter of the wire. Given data: F = 60 N σ = 3.06 MPa = 3.06 x 106 Pa d = ? (say d is dia. of bar) Since,

i.e. diameter of wire, d = 5.0 mm Sign convention for stress:

Tensile stress or tension is taken as or with positive (+ve) sign.

Compressive stress or compression is taken as or, with (-ve) sign.

Problem:

In Shown fig. state the types of

Solution:

A is supporting member, is in

Compression.

B is Horizontal beam, is in Shear.

C is holding load, is in Tension.

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Let’s try it:

1. In given fig. if, Exerted force = 900 KN, dia. of rope =10 mm. find stress in rope.

2. A hollow circular column supporting a building is attached to a metal plate and bolted into the concrete foundation as shown in fig. The column outside diameter is 100 mm and an inside diameter is 75 mm. the metal plate dimensions are 200 mm x 200 mm x 10 mm. the load P is estimated at 800 KN. Determine the compressive stress in the column.

3. A square sectioned support of side 12 mm is loaded with a compressive force of 10 KN. Determine the compressive stress in the support. [69.44MPa]

4. A tube of outside diameter 60 mm

and inside diameter 40 mm is subjected to a load of 60 KN. Determine the stress in the tube.

[38.2 MPa]

5. A rectangular bar having a cross-sectional area of 80 mm2 has a tensile force of 20 KN applied to it. Determine the stress in the bar. [250 MPa]

Strain:

Strain is a ratio of change in length (say δ) to the original length (say L).

Types of strain Symbol used

Tensile strain ε (Greek word epsilon)

Compressive strain ε

Shear strain γ(Greek word gamma)

% strain,

S.I. unit of strain:

Problem: A bar 1.60 m long contracts axially by 0.1 mm when a compressive load is applied to it. Determine the strain and the percentage strain. Given data: L = 1.60 m = δ = 0.1 mm Since,

% strain,

Problem: A wire of length 2.50 m has a percentage strain of 0.012% when loaded with a tensile force. Determine the extension of the wire. Given data: L = 2.50 m = 2.50 x 103 mm

% = 0.012%, = ?

Since,

Mechanics of Solids

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Problem: If in shown fig. inner & outer dia. of a hollow circular tube are respectively d1 =10 mm, d2 = 10.5 mm, and the shortening due to the load is 0.012 mm. determine the compressive stress and strain.

Given data: L = 40 mm P = 100 N d1 = 10 mm d2 = 10.5 mm δ = 0.012 mm Cross-sectional area of the tube:

A =

=

2

Compressive stress in the tube:

2

Compressive strain in the tube:

-4

Let’s try it:

1. A pipe has an outside diameter of 25 mm, an inside diameter of 15 mm and length 0.40 m and it supported a compressive load of 40 KN. The pipe shortens by 0.5 mm when the load is applied. Determine (a) the compressive stress, (b) the compressive strain in the pipe when supporting the load. [127.3 MPa, 0.00125]

2. A circular hole of diameter 50 mm is to be

punched out of a 2 mm thick metal plate. The shear stress needed to cause fracture is 500 MPa. Determine (a) the minimum force to be applied to the punch, and (b) the compressive stress in the punch at this value. [157.1 KN, 80.03 MPa]

Mechanical properties of materials:

1. Elasticity: It is the property of material by virtue of which body regain its original shape and size after removal of external forces.

2. Plasticity: It is the property of

material by virtue of which a body doesn’t return to the original shape and size after removal of forces too.

Hooke’s law: Within the limit of proportionality, the extension of a material is proportional to the applied force.

In other words,

Within the limit of proportionality of a material, the strain produced is directly proportional to the stress producing it.

Mathematically it can be written as,

Young’s modulus of elasticity:

Within the limit of proportionality,

Hence,

This constant of proportionality is called young’s modulus of elasticity and is given the symbol E. So,

Its unit is same as unit of stress, i.e. Pascal. Stiffness: it is the ratio of

Force (F) to the change in length or extension (δ).

Value of stiffness is said to be high if have a large value of Young’s modulus. From definition,

Mechanics of Solids

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Since,

Then,

i.e.

Problem: A wire is stretched 2 mm by a

force of 250 N. Determine the force that would

stretch the wire 5 mm, assuming that the limit

of proportionality is not exceeded.

When δ = 2 mm, F = 250 N

From Hook’s law,

2 = k x 250

From which constant

Again when, When δ = 5 mm, then 5=kF

i.e.

From which, Force F = 5 (125) = 625 N Hence, to stretch the wire 5 mm a force of 625 N

is required.

Problem: A copper rod of diameter 20 mm

and length 2.0 m has a tensile force of 5 kN

applied to it. Determine (a) the stress in the

rod, (b) by how much the rod extends when

the load is applied. Take the modulus of

elasticity for copper as 96 GPa.

Given data: d = 20 mm, L = 2.0 m, F = 5

KN, E = 96 GPa.

(a) Stress,

(b) Extension ,

i.e. extension of rod is 0.332 mm.

Problem A bar of thickness 15 mm and having a

rectangular cross-section carries a load of 120 kN.

Determine the minimum width of the bar to limit

the maximum stress to 200 MPa. The bar, which is

1.0 m long, extends by 2.5 mm when carrying a

load of 120 kN. Determine the modulus of

elasticity of the material of the bar.

Given data:

t = 15 mm, thickness of bar.

F = 120 kN, load.

b = ?, width of the bar.

σ = 200 MPa stress

L = 1.0 m length of the bar.

δ = 2.5 mm extension.

E = ? modulus of elasticity.

Since,

Again since,

Modulus of elasticity,

1. A brass tube has an internal diameter of

120 mm and an outside diameter of 150

mm and is used to support a load of 5 kN.

The tube is 500 mm long before the load

is applied. Determine by how much the

tube contracts when loaded, taking the

modulus of elasticity for brass as 90 GPa.

[4.37µm]

2. In an experiment to determine the

modulus of elasticity of a sample of mild

steel, a wire is loaded and the

corresponding extension noted. The

results of the experiment are as shown.

Draw the load/extension graph.

The mean diameter of the wire is 1.3 mm

and its length is 8.0 m. Determine the

modulus of elasticity E of the sample, and

the stress at the limit of proportionality.

Load (N) 0 40 110 160 200 250 290 340

Extension

(mm)

0 1.2 3.3 4.8 6.0 7.5 10.0 16.2

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