STRUCTURES & MATERIALS

 

Reference:Nelson Physics VCE Units 3 & 4Chapters 16 & 17Page 433

S & M.1 Torque

A torque is a force that causes a body to rotate, it is sometimes referred to as turning moment.

Torque is a vector quantity, its direction corresponds to the direction of rotation.Clockwise rotation being defined as negative.

Torque is given by:

t=r´F

where F is the applied force and r is the perpendicular distance between the axis of rotation and the force.

The unit of torque is metre Newton (m N) and should not be confused with the unit of work and energy which is very similar.

S & M.1.1 Couples

If we have two equal and opposite forces as shown in the diagram below.

The body is free to rotate.Let us look at the effect on the body.

1.The forces are equal and opposite, so no acceleration should result.

2.The total amount of torque is:

t = r ´ F + (-r) ´ (-F)

= 2 r ´ F

Thus even though the total force is zero, a rotation results.The condition for this to occur is that the two forces do not act through the same point.

A pair of equal and opposite forces, such as these, that cause a torque is called a couple.

S & M.1.2 Centre of Gravity

When we hold an object it will rotate until it hangs at an equilibrium position.I.e.When the individual torques balance each other.If we do this at a number of different places on the object and draw a vertical line down the object, as shown:

We notice that there is a point that all of the lines go through.This is the position of the centre of gravity.For all our purposes this is the same point as the center of mass.

S & M.2 Stability

As we already know the centre of gravity of a structure is the point through which the sum of the gravitational forces acts.Therefore the centre of gravity acts as a balance point.An object will be unstable and fall over if the vertical line through its centre of gravity does not pass through its base.

Clearly the width of the base and the height of the centre of gravity affect the stability of the structure.

S & M.3 Equilibrium

An object is in equilibrium if the following two conditions are met:

i)the net force on the object is zero, and

ii)the net torque on the object is zero.

Thus when a body is in equilibrium it will stay in one position unless acted upon by other forces.

Example:

A person of mass 50 Kg stands 1 m from the end of a 3 m beam.The mass of the beam is 40 Kg.Calculate the reaction at support A and support B if the system is in equilibrium.

Note: The weight of a object such as a beam acts from the centre of mass/gravity, in this case from the centre.

Take torque about A

Anticlockwise torque = Clockwise torque

Rb´ 3 = (500 ´ 2) + (400 ´ 1.5)

Rb´ 3 = 1000 + 600

Rb´ 3 = 1600

Rb = 533.3 N

Take torque about B

Clockwise torque = Anticlockwise torque

Ra´ 3 = (500 ´ 1) + (400 ´ 1.5)

Ra´ 3 = 500 + 600

Ra´ 3 = 1100

Ra = 366.7 N

There are three possible variations within an equilibrium position.It can be either Stable, Unstable or Neutral.For stable equilibrium if the object is pushed it will return to its original position.For unstable equilibrium if the object is pushed it will not return to its original position.For neutral equilibrium if the object is pushed it may return to its original position.

Problem Set #1:TextPage 446Questions 1 – 44

S & M.4 Properties of Materials

When building a structure careful consideration has to be taken to ensure the correct materials are used.Questions such as is it strong enough?Will it wear well?Will it crack?Along with economic factors need to be considered.

A material is elastic if it returns to its original shape after being deformed (stretched or compressed).If not it is said to be plastic.

Materials that are hard resist wear.Wear occurs when surfaces rub on each other.

Tough materials can absorb a large amount of energy without breaking.They have the ability to withstand repeated stresses.A tough material is one in which the internal stresses do not build up easily.

Brittle materials tend to fracture and crumble easily.

The shaping, reinforcing and jointing used within a structure can effect its strength.

S & M.5 Forces on Structures

There are five types of forces that may affect a material.They are:

Compression Forces.These are pushing forces that try to squeeze the ends of the structure or material together.

Tension Forces.These are pulling forces that try to stretch or elongate a structure or material.

Bending Forces.These forces usually occur in beams and produce both compressive and tension forces.The bend in the beam causes compression on the top of the beam and tension on the bottom.

Torsion Forces.These are twisting forces acting about an axis. Torsion forces are produced by the action of two opposite torques acting in the same plane.

Shear Forces.These are deformation forces in which one layer of material slides (or tries to slide) over another.

S & M.6 Materials and Hooke's Law

As we have previously seen Hooke's law says:

F = -k x

However we only looked at when Hooke's law holds.Springs cannot be stretched or compressed an infinite amount.There will be a point when Hooke's law breaks down and the graph of F versus x is not straight.This point is called the Elastic Limit.Any extension past this point will cause a permanent change in the material, and it is then said to be plastic.The point at which the material becomes plastic is called the yield point.The spring constant k is also known as the Force Constant.

S & M.7 Stress

We know that a thick elastic band is harder to stretch than a thin one.The difference is the cross-sectional area.So the extension produced by a given force is inversely proportional to the cross-sectional area.

Thus

The term is called Stress.

It has the units N m-2 or Mega Pascal (M Pa)

1 Nm-2 = 1 Pa

1 M Pa = 106 Pa = 106 N m-2

The symbol used for stress is the lower case Greek letter sigma (s).

So

A stress that causes a compression is called a compressive stress; a stress that causes an extension is called a tensile stress.

Example

A force of 600 N is applied to one end of a cable of diameter 5 cm.What is the tensile stress?

Stress = Force

Area

Stress = 600

p (2.5 ´ 10-2)2

= 3.1 ´ 105 N m-2

S & M.8 Strain

We know that a long elastic band is easier to stretch than a short one.The difference is the length.So the force needed to produce a given extension is inversely proportional to the length.

Thus

Where x is change in length (DL).

The term is called Strain.It has no units.

The symbol used for stress is the lower case Greek letter epsilon (e).

So

Sometimes it is seen written as:

Example

Find the strain produced when a steel rod 2.5 m long is increased in length by 0.5 mm.

Strain = DL

L

Strain = 0.5 ´ 10-3

2.5

= 2.0 ´ 10-4

S & M.9 Young's Modulus

Since both stress and strain relate very closely to Hooke's Law a graph of Stress versus Strain will have the same shape as a graph of F vs x.

When an Engineer is designinga structure he/she will make sure that the materials used in the structure are subject to loads that fall in the lower elastic region.This will allow the structure to absorb any external forces such as those from wind, cars, etc. without breaking.

As we can see from the graph:

Stress = a constant (the gradient of the graph)

Strain

This constant is called Young's Modulus or the modulus of elasticity (E).

or

The units for Young's modulus are N m-2.

The value of Young's modulus depends on the strength of the forces between the atoms and molecules in the material.If these forces are strong then the material will resist stretching and compression.

Notes:

1.Brittle materials like glass and cast iron do not exhibit a yield point.They fracture at their elastic limit, do not show plastic behaviour,have high values for Young's modulus and have relatively low tensile strength.

2.The steeper the elastic part of a stress-strain graph, the stiffer the material.

Example.

A steel bar of cross-sectional area 16 cm2 is 2.0 m long.When a tensile force of 8 ´ 104 N is applied to it, the length increases by 0.5 mm.Find Young's modulus for this material.

F= 8 ´ 104L = 2.0DL = 5 ´ 10-4A = 1.6 ´ 10-3

Stress = FStrain = DL

AL

=8 ´ 104=5 ´ 10-4

1.6 ´ 10-32.0

=5.0 ´ 107=2.5 ´ 10-4

E=Stress

Strain

=5.0 ´ 107

2.5 ´ 10-4

=2.0 ´ 1011 N m-2

S & M.9.1 Interpreting Stress/Strain Graphs

 Brittleness

A brittle material is one which has a sudden failure soon after it has reached its elastic limit.

 Toughness

This is the amount of work required to break a material.It can be found from the area under the stress-strain graph.

 Stiffness

The stiffness of a material is how much stress is required to cause a given strain.This is measured by Young’s modulus.Material A is much stiffer than B, although B is stronger.

 Strength

A strong material requires a great deal of stress to break it.The tensile strength (TS) of a material is the greatest stress it can withstand before failure and the yield stress (YS) is the greatest stress before exceeding the elastic region.

 
Elastic

Elastic material return to their original shape once the stress has been removed.

 Plastic

Plastic materials produce a permanent deformation under stress.

 Hysteresis

This is the property of a material to partly return to its original shape after stress that exceeds its elastic limit.

S & M.10 Strain Energy

When a body is strained, work is done against the internal forces of the material.As long as the material is in the elastic region this energy can be recovered, so the strained material can be considered to possess potential energy.The work done or the energy stored in the material is known as strain energy and can be determined from the area under a Stress-Strain graph.

The area under the stress-strain graph has units J m-3. So to obtain the strain energy from the area under the stress-strain graph we must multiply the area value obtained from the graph by the volume of the material.

S & M.11 Fractures

All materials have an upper limit to the amount of stress they can with stand without breaking.This limit depends on the arrangement of theatoms and molecules in the material and the forces between them.

Under an excessive tensile stress materials snap.The minimum stress required to do this is called the tensile strength of the material.

An excessive compressive stress will crush a material.The minimum stress required to do this is called the compressive strength of the material.

Similarly an excessive shear stress will cause a fracture of the material.The minimum stress required to do this is called the shear strength of the material.

Problem Set #2:TextPage 461Questions 1 – 52

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