2.001 - MECHANICS AND MATERIALS I
Lecture #9 10/11/2006
Prof. Carol Livermore
Review of Uniaxial Loading:
δ = Change in length (Positive for extension; also called tension)
Stress (σ)
σ = P/A0
Strain () at a point
=
L − L0
δ
=
L0
L0
Stress-Strain Relationship ⇒ Material Behavior
σ = E
1
Force-Displacement Relationship
P = kδ
For a uniaxial force in a bar:
k=
EA0
L0
Deformation and Displacement
Recall from Lab:
The springs deform.
The bar is displaced.
EXAMPLE:
2
du(x)
dx
(x)dx = du
(x) =
Sign Convention
Trusses that deform
Bars pinned at the joints
How do bars deform?
How do joints displace?
EXAMPLE:
Q: Forces in bars?
How much does each bar deform?
point B displace?
Unconstrained degrees of freedom
1. uB
x
2. uB
y
3
How much does
Unknowns
1. FAB
2. FBC
This is statically determinate (Forces can be found using equilibrium)
FBD:
Fx = 0 at Pin B
−FAB − FBC sin θ = 0
Fy = 0
−P − FBC cos θ = 0 ⇒ FBC =
−FAB +
P
sin θ = 0
cos θ
So:
FAB = P tan θ
Force-Deformation Relationship
P = kδ
EA
k=
L
FAB
kAB
FBC
=
kBC
δAB =
δBC
4
−P
cos θ
AE
L sin θ
AE
=
L
kAB =
kBC
So:
P tan θ
δAB = AE
L sin θ
−P cos θ
δBC = AE
L
δAB =
δBC
P L sin θ tan θ
AE
−P L
=
AE cos θ
Check:
Compatibility
5
Algorithm to find B AB
1. If we only have uB
and δ BC would result?
x , what δ
B
AB
2. If we only have uy , what δ
and δ BC would result?
AB
BC
B
3. What is the total δ
and δ
is I have both uB
x and uy ?
B
AB
for uB
and δ BC .
x and uy from known δ
Step 1
δ BC = uB
x sin θ
LN EW
=
=
(L + δ1BC )2 + Δ2
2
2δ BC
δ BC
Δ2
L (1 + 1 + 1 2 ) + 2
L
L
L
√
LN EW
x
L
2
Δ2
δ1B C
δ1BC
=L 1+
+
+
2L2
L
2L2
1+x≈1+
2
δ1BC
→0
2L2
Δ2
→0
2L2
For δ << L:
δ BC ≈ D
So:
Lnew = L + δ1BC
6
4. Solve