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
Pa −
MA = 0
Fc L
− FB L = 0
2
2. Force-Deformation
FA = kδA
1
FB = kδB
FC = kδC
3. Compatibility
δA = uA
y
A
δ B = uA
y + L tan ϕz
δC = uA
y +
L
tan ϕA
z
2
New this lecture:
Small Angle Assumption:
tan ϕzA =
ϕA
sin ϕzA
≈ z = ϕA
z
A
cos ϕz
1
For small ϕA
z : Arc length ≈ a straight line displacemtn in y.
Rewrite compatibility.
δA = uA
y
A
δB = uA
y + Lϕz
L A
ϕ
2 z
Substitute compatibility into force-deformation
δA = uyA +
FA = kuyA
A
FB = k(uA
y + Lϕz )
2
L A
ϕ )
2 z
Substitute this result into equilibrium equations:
FC = k(uA
y +
L A
ϕ )=0
2 z
L A
L
A
A
+
ϕ
)
−L
k(u
+
Lϕ
)
=0
Pa −
k(uA
y
y
z
2 z
2
A
A
A
P − kuA
y − k(uy + Lϕz ) − k(uy +
Solve:
3
A
P = 3kuA
y + Lkϕz
2
3
5 2 A
P a = LkuA
y + L kϕz
2
4
Divide by k.
3 A
P
= 3uA
y + Lϕz
k
2
Divide by Lk/2.
2P a
5 A
= 3uA
y + Lϕz
Lk
2
So:
ϕA
z
−P
2a
=
1−
Lk
L
Substitute:
uA
y =
u(x) =
P
P
2a
+
1−
3k 2k
L
P
P
2a
P
2a
+
(1 − ) −
(1 − )x
3k 2k
L
Lk
L
UNIAXIAL LOADING
Behavior of a uniaxially loaded bar
L = LD + δ
P = kδ is a property of the bar.
STRESS =
FORCE
= σ( Like Pressure)
UNIT AREA
N
Units: 1 m
2 = 1 Pa (SI Units)
lbs
Units: 1 in2 = PSI (English Units)
σ=
P
A
σ=
P
A0
Engineering Stress
For small deformation
STRAIN =
CHANGE IN LENGTH
LENGTH
=
=
δ
L
Strain is dimensionless.
Engineering Strain (For small deformations)
=
4
δ
L0
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
=
So:
δ
L0
P
Eδ
=
A0
L0
P =
EA0
δ
L0
So:
k=
EA0
for uniaxial loading
L0
Deformation and Displacement
=
δ
(du(x) + dx) − dx)
=
L
dx
du(x)
=
dx
u(x) = axial displacement of x
du(x)
dx
σ(x)
(x) =
E
(x) =
So:
du(x)
σ(x)
P
=
=
dx
E
AE
δ
L
P
du =
dx
0
0 AE
L
δ Px
u =
AE 0
0
δ=
PL
AE
P =
AEδ
L
So:
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 = AsLEs
Al: kA = k = AALEA
Same k ⇒ As Es = AA EA
So: AA = As EEAs
So: AA ≈ 3As ⇒ the aluminum is three times bigger
6
May need to optimize weight (think about airplanes) ⇒ need to include density.
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