PES 1110 Fall 2013, Spendier
Lecture 35/Page 1
Today:
- Elasticity (12.7)
- HW 8 due today – HW 8 solutions will be posted after class
- HW 9 assigned today (due Wednesday Dec. 4th after Thanksgiving
Note: I will hold no office hours Tuesday December 3th
- Quiz 5, Friday Nov 22nd (covers lectures 29-33, HW 8) - cover sheet with
Equations posted after class
Last time: Equilibrium

Fnet , x   Fi , x  0
i

Fnet , y   Fi , y  0
t
i
i
0
i
Stable and Unstable Equilibrium
It is hardly possible to stand a pencil on it's tip, but quite easy to stand it on it's flat end.
For example consider a cone:
Unstable equilibrium: A small force will destroy the equilibrium and send the cone into
another position.
Stable equilibrium: After a small force, the object will return to the same equilibrium
state.
An object generally topples (tips over) over if the COG is moved to outside of its base.
For the cone balanced on its tip, the COG is high so it is relatively easy to push it over
(take it out of equilibrium). But for the cone balanced on its base, the COG is lower and it
will take a large force to tilt it over.
To make something more stable:
 bigger base
 lower COG
PES 1110 Fall 2013, Spendier
Lecture 35/Page 2
Example 4:
Below you see a block of uniform mass. In which situation (a) or (b) will the block topple
over (tip to the right) and why?
Stability is vital for cars, trucks and buses. They must be designed to withstand tilt up to a
certain angle before toppling (double-decker bus). Famously a Mercedes Benz car once
failed the 'elk test' so that it toppled when swerving to avoid an elk on a test drive in
Sweden. The car had to be redesigned to be more stable.
Hint: HW 9 problem 2
A bowler holds a bowling ball (M = 7.2 kg) in the palm of his hand. His upper arm is
vertical; his lower arm (m = 1.8 kg) is horizontal. The axis of rotation is about the elbow
contact point.
a) Draw a free body diagram indicating all forces. You should have 4 individual forces.
b) What is the magnitude of the force of the biceps muscle on the lower arm?
c) What is the magnitude of the force between the bony structures at the elbow contact
point?
part a)
PES 1110 Fall 2013, Spendier
Lecture 35/Page 3
Elasticity (12.7)
Rigid bodies don’t bend, stretch or squash when forces act on them. But the rigid body is
an idealization; all real materials are elastic and do deform to some extent. Elastic
properties of materials are tremendously important. You want the wings of an airplane to
be able to bend a little, but you’d rather not have them break off. The steel frame of an
earthquake-resistant building has to be able to flex, but not too much. Many of the
necessities of everyday life, from rubber bands to suspension bridges, depend on the
elastic properties of materials. We will introduce the concepts of stress, strain, shear, and
elastic modulus
Tension: Objects that are stretched are said to be under
tension. e.g. steel is very good under tension (used for
cabling)
Compression: Objects that are pushed inwards are said to
be under compression e.g. concrete is a good material to
use for compression construction (arches)
Artificial material can be made that perform well under
both types of forces, e.g. reinforced concrete
Usually when a force is applied to an object, it at first
deforms elastically (i.e. if you remove force, the object
goes back to its usual shape). But then after some critical
threshold, the deformation is irreversible. With even more
force the material will break.
We can plot this trend of force versus deformation.
PES 1110 Fall 2013, Spendier
Lecture 35/Page 4
F
A
Force divided by area = a pressure that is called “stress” in a solid material
Here force is perpendicular to area
A = the cross-sectional area.
Stress =
L
L
Change in length divided by initial length = dimensionless measure of the effect the stress
has on the material, way to describe deformation
Strain =
In the linear region, stress is linearly related to strain so we can write:
F
L
E
( E units of [N/m2]
A
L
where the constant of proportionality, E, is called the Youngs’ modulus. E is a measure of
stiffness of an elastic material. It is determined by the tensil test
For example, for a given strain (∆L/L) a much larger stress needs to be applied to steel
than bone.
However, this doesn’t tell us about the “ultimate strength” of material, that is how much
force they can take before snapping. The Youngs’ modulus E tells us about the initial
elastic (reversible) behavior.
PES 1110 Fall 2013, Spendier
Lecture 35/Page 5
Example 2:
One end of a steel rod of radius R = 9.5 mm and length L = 81 cm is
held in a vise. A force of magnitude F = 62 kN is then applied
perpendicularly to the end face (uniformly across the area) at the other
end, pulling directly away from the vise. What are
a) the stress on the rod,
b) the elongation ∆L (E = 200 x 109 N/m2), and
c) strain of the rod?
So far we have talked about compression and tension, where forces are perpendicular to
cross-sectional area A of the material.
Shear: When forces are parallel to there areas, we can also see a deformation called
“shear”
For shearing, the force vector is in-plane but stress is still
shear stress = F/A.
Now ∆x, the change in size/shape, is perpendicular to L.
PES 1110 Fall 2013, Spendier
Lecture 35/Page 6
x
L
There is again an elastic (irreversible) region that can be described by
F
x
G
( G units of [N/m2]
A
L
where G is the shear modulus
shear strain =
Example 3:
A horizontal aluminum rod 4.8 cm in diameter projects 5.3 cm from a wall. A 1200 kg
object is suspended from the end of the rod. The shear modulus of aluminum is
3.0 x 1010 N/m2. Neglecting the rod’s mass, find
(a) the shear stress on the rod and
(b) the vertical deflection of the end of the rod.