1. 2.001 - MECHANICS AND MATERIALS I
Lecture #8
10/4/2006
Prof. Carol Livermore
Recall from last lecture:
Find: u(x), FA , FB , FC
1. Equilibrium
Fu = 0
P − FA − F B − F C = 0
MA = 0
Fc L
Pa − − FB L = 0
2
2. Force-Deformation
FA = kδA
1

2. FB = kδB
FC = kδC
3. Compatibility
δA = uA
y
δB = uA + L tan ϕA
y z
L
δC = uA +
y tan ϕA
z
2
New this lecture:
Small Angle Assumption:
A
sin ϕz ϕA
A
tan ϕz = A
≈ z = ϕA
z
cos ϕz 1
For small ϕA : Arc length ≈ a straight line displacemtn in y.
z
Rewrite compatibility.
δA = uA
y
δB = uA + LϕA
y z
A L A
δA = uy + ϕ
2 z
Substitute compatibility into force-deformation
A
FA = kuy
FB = k(uA + LϕA )
y z
2

3. L A
FC = k(uA +
y ϕ )
2 z
Substitute this result into equilibrium equations:
L A
P − kuA − k(uA + LϕA ) − k(uA +
y y z y ϕ )=0
2 z
L L
Pa − k(uA + ϕA ) −L k(uA + LϕA ) = 0
y
2 2 z y z
Solve:
3
P = 3kuA + Lkϕz
y
A
2
3 5
P a = LkuA + L2 kϕA
y z
2 4
Divide by k.
P 3
= 3uA + LϕA
y
k 2 z
Divide by Lk/2.
2P a 5
= 3uA + LϕA
y
Lk 2 z
So:
−P 2a
ϕA =
z 1−
Lk L
Substitute:
P P 2a
uA =
y + 1−
3k 2k L
P P 2a P 2a
u(x) = + (1 − ) − (1 − )x
3k 2k L Lk L
UNIAXIAL LOADING
Behavior of a uniaxially loaded bar

4. L = LD + δ
P = kδ is a property of the bar.
FORCE
STRESS = = σ( Like Pressure)
UNIT AREA
N
Units: 1 m2 = 1 Pa (SI Units)
lbs
Units: 1 in2 = PSI (English Units)
P
σ=
A
Engineering Stress
For small deformation
P
σ=
A0
CHANGE IN LENGTH
STRAIN = LENGTH =
δ
=
L
Strain is dimensionless.
Engineering Strain (For small deformations)
δ
=
L0
4

5. Stress-Strain plot
For Uniaxial Loading:
σ=E
Material property is E, Young’s Modulus.
Note: Units of E = Pa, E 109 Pa (or GPa).
So, what is k for uniaxiail loading?
σ=E
P
σ=
A0
δ
=
L0
So: P Eδ
=
A0 L0
EA0
P = δ
L0
So:
EA0
k= for uniaxial loading
L0
Deformation and Displacement

6. δ (du(x) + dx) − dx)
= =
L dx
du(x)
=
dx
u(x) = axial displacement of x
du(x)
(x) =
dx
σ(x)
(x) =
E
So:
du(x) σ(x) P
= =
dx E AE
δ L
P
du = dx
0 0 AE
δ L
Px
u =
0 AE 0
PL
δ=
AE
So:
AEδ
P =
L
What are some typical values for E?
E
Steel 200 GPa
Aluminum 70 GPa
Polycarbonate 2.3 GPa
Titanium 150 GPa
Fiber-reinforced Composites 120 GPa
Selection of material? Optimize k for a particular A
Steel: ks = k = As Es
L
Al: kA = k = AA EA
L
Same k ⇒ As Es = AA EA
Es
So: AA = As EA
So: AA ≈ 3As ⇒ the aluminum is three times bigger
6

7. May need to optimize weight (think about airplanes) ⇒ need to include den-
sity.
What happens if you keep pulling on a material?
σy = Yield Stress
p
= Plastic Strain - Not Recoverable.
e
= Elastic Strain - Fully Recoverable.
t
= e + p = Total Strain
What about pulling on a bar in uniaxial tension?
7