8/6/2019 Phyc10030 Lecture 2 Deliv
1/31
Lecture 2
Stress and strain
8/6/2019 Phyc10030 Lecture 2 Deliv
2/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
3/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
4/31
Can download lab manual for free
I will post lat years lectures on blackboard in advance of
the lecture. They are broadly similar
8/6/2019 Phyc10030 Lecture 2 Deliv
5/31
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).
8/6/2019 Phyc10030 Lecture 2 Deliv
6/31
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).
8/6/2019 Phyc10030 Lecture 2 Deliv
7/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
8/31
Why can Whales grow so big?
8/6/2019 Phyc10030 Lecture 2 Deliv
9/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
10/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
11/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
12/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
13/31
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.
8/6/2019 Phyc10030 Lecture 2 Deliv
14/31
Examples: Compressive Strain in the
spine
8/6/2019 Phyc10030 Lecture 2 Deliv
15/31
Examples: Compressive and
tensile stresses in the femur
8/6/2019 Phyc10030 Lecture 2 Deliv
16/31
Examples: shear stress
8/6/2019 Phyc10030 Lecture 2 Deliv
17/31
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:
8/6/2019 Phyc10030 Lecture 2 Deliv
18/31
Example 17-4
Stretching a Bone
Example: If the mass of the weight is 21 kg,
what is the stress applied to the bone?
8/6/2019 Phyc10030 Lecture 2 Deliv
19/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
20/31
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.
8/6/2019 Phyc10030 Lecture 2 Deliv
21/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
22/31
Strain
Has units =?
0L
Lstrain
(!
What is strain when:
1.(L=0
2.(
L=L0
8/6/2019 Phyc10030 Lecture 2 Deliv
23/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
24/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
25/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
26/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
27/31
Bonds between molecules
are stretched but dont
break. Object returns to
original shape
Bonds between
molecules are beginning
to break. Object
permanently deformed
8/6/2019 Phyc10030 Lecture 2 Deliv
28/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
29/31
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
8/6/2019 Phyc10030 Lecture 2 Deliv
30/31
AY
F
strain
stressY
!
!
...
Use Youngs modulus to calculate change in length
Do in class
8/6/2019 Phyc10030 Lecture 2 Deliv
31/31
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.
Top Related