Phyc10030 Lecture 2 Deliv

8/6/2019 Phyc10030 Lecture 2 Deliv

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Lecture 2

Stress and strain


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Learning objectives Define terms Stress and Strain

How can you know if one material isstronger than another e.g. is bone stronger

than steel?

Youngs Modulus

Reference text: Physics, Walker, Chap 17

sec 3


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Tutorials

Every week beginning 24th january

Tuesday 10.00 11.00

Wednesday 10.00 11.00

Thursday 10.00 11.00

Tutor: Florentina Tofoleanu

This is slightly different to what I showed on tuesday


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Can download lab manual for free

I will post lat years lectures on blackboard in advance of

the lecture. They are broadly similar


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What determines the strength of an elephants leg?

a. Area

b. Volume

c. Length

Why are there no land animals

bigger than elephants

Strength of limb w to area (A).

But mass of animal (or plant) w to volume (V).


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Strength of limb w

2.1Why are there no land animals

bigger than elephants?

Body volume :3

34

bodyrV T!

Leg area :2

legrA T!

We model the elephant as a sphere, and its legs

as cylinders.

Cross sectional area

Mass of animal (or plant) w to volume (V).


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As the animal increases in size, the typical dimensions rleg & rbodyincrease.

If all dimensions increase by 25% what is the extra increase in

volume compared to area

Do in class


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Why can Whales grow so big?


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Revisit Forces and weight

Do during class

Forces are represented as arrows

Force has a magnitude depicted by length of arrow

Force has a direction

Units: Newton

Basic Equation:

Force = mass x acceleration


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Elephant

End on

view

Have a force due

to weight of body

acting over this

area

A leg

Force dueto weight

Dimensions increase by 25%

Load has increased on legs by 95%

Consequences?Legs gets sqaushed

Maybe break?

Revisit forces on animals legs


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A leg

Force due to weight

Which leg would you expect the largest effect?

A different leg

Force due to weight

Consider same load on two different types of leg

Legs composed of same material and have equal length

F F

The effects of a force


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A leg

Force due to weight

Let S1 = Force per unit area

= F/A1

A leg

Force due to weight

Consider same load on two different legs

F F

Cross

sectional

area=A1

Cross

sectional

area=A2

S2 = Force per unit area

= F/A2

A1 > A2

Force per unit area = Stress = Force/Area

S1 < S2


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2.2 StressStress the application of a force over some

area of a body.

Stress can be compressive..

Or tensile.

Or shear

Squashing.

Stretching.

Twisting.


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Examples: Compressive Strain in the

spine


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Examples: Compressive and

tensile stresses in the femur


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Examples: shear stress


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Stress F/A units N/m2

A is the area of the material perpendicular to

the force being applied. Figure 17-10Stretching a Rod

We can calculate the stress () using the force& the area being stressed.

A

F!WStress is defined as:


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Example 17-4

Stretching a Bone

Example: If the mass of the weight is 21 kg,

what is the stress applied to the bone?


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Example 17-4

Stretching a Bone

Example: If the mass of the

weight is 21 kg,

what is the stress applied to the

bone?

Do in class


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2.3 Strain

The effect of stress = strain

The strain in an object is a result of any applied stress. It

is a measure of the deformation of the object.


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Which is easier to do

a. Stretch by 5mm a length of steel thatis 1 m long

b. Stretch by 5mm a length of steel that

is 1 mm long?

Which one is under the greaterstrain?

So strain is proportional to:

1: proportional to the amount of stretching

2. inversely proportional the original length


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Strain

Has units =?

0L

Lstrain

(!

What is strain when:

1.(L=0

2.(

L=L0


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Stress leads to strain in the material.

Strain is:

(change in length)/(original length)

0L

Lstrain

(!

Typically, the value

of the strain is

quite small.

Figure 17-13

Stress Versus Strain


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2.4 Hookes Law

What about the force involved in stretching- how does it vary

with the amount of stretch?

F w change in length

F w (Li.e. the more you stretch it the greater the force required

Hooke said

Force required to stretch (or compress) an object by (L is equalto kx (L

Why the minussign?

LkF (!

Hookes constant


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kF

!

Alternatively you can understand Hookes Law as saying

The ratio of the deformation and the force exerted is

always constant

i.e. if you double the force the object will stretch twice asmuch


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Hookes Law, which describes the behaviour of springs,

for example, talks about the elastic region of the stress

strain curve.

LkF (!Hookes Law:

What determines how much a material will

deform, i.e. what will be the strain?

This depends on thestrength of the bonds

holding the material

together.

k is the spring constant.

Beyond Hookes Law


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Bonds between molecules

are stretched but dont

break. Object returns to

original shape

Bonds between

molecules are beginning

to break. Object

permanently deformed


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kL

F!

(Hookes Law: ratio of force to change

Similar relation is called Youngs modulus

strain

stressY !

strain

stressY!

AFstress !!

0LLstrain (!

2.5 Youngs Modulus

Ratio ofstress to

strain is also

constant

0LL

AF

(!

0LL(!

W


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Youngs modulus is different

for different materials: it

indicates how easy it is tostretch a material

Young's Modulus is very

useful because:

If you know Y and the stress

you can calculate the strain

Y

stressstrain

strainstressY

!

!

Material Y (N/m2)

Steel 2.1 v 1011

Concrete(compression) 3 v 1010

Concrete (tension) 0.3 v 1010

Bone1.8 v 10

10

(Compact)

Rubber 1.0 v 10

6

Diamond 1050 x 1013

Carbon nano-tube >1000 x 1013

Spiders silk 1.3 x 1012

Oak 1.1 v 1010


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AY

F

strain

stressY

!

!

...

Use Youngs modulus to calculate change in length

Do in class


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Youngs modulus can be used to calculate by how much a given

structure will deform under stress.

YAFA

F

Y 0

0

!!

Example: assume a leg has a 1.2 m shaft of bone, with an average

cross-sectional area of 3 v 10-4 m2.What is the amount ofshortening when all of the body weight (700 N) is supported by

the leg?

mmmL

YA

FLL

15.0105.1103108.1

2.1107

4

410

2

0

!v!

vvv

vv!!

Dynamic loads may exert far higher temporary stresses, resulting

in higher strain & possibly fracture.

Phyc10030 Lecture 2 Deliv

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