Lec9 MECH ENG STRucture

Jan 19, 20110 likes277 views

The document reviews concepts of uniaxial loading, stress, strain, stress-strain relationships, force-displacement relationships, and deformation and displacement in mechanical structures. It provides an example of calculating forces and displacements in a statically determinate truss. The document outlines an algorithm to determine displacements at point B by considering compatibility of displacements for given forces.

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:

EA0
k=
L0
Deformation and Displacement
Recall from Lab:

The springs deform.
The bar is displaced.

EXAMPLE:

2

du(x)
(x) =
dx

(x)dx = du

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? How much does
point B displace?
Unconstrained degrees of freedom
1. uB
x
2. uB
y

3

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
?P ? FBC cos θ = 0 ? FBC =
cos θ
P
?FAB + sin θ = 0
cos θ
So:

FAB = P tan θ
Force-Deformation Relationship

P = kδ
EA
k=
L
FAB
δAB =
kAB
FBC
δBC =
kBC

4

AE
kAB =
L sin θ
AE
kBC =
L
So:

P tan θ
δAB =
AE
L sin θ

?P cos θ
δBC =
AE
L

P L sin θ tan θ
δAB =
AE
?P L
δBC =
AE cos θ
Check:

Compatibility

5

Algorithm to ?nd B
1. If we only have uB , what δ AB and δ BC would result?
x
2. If we only have uB , what δ AB and δ BC would result?
y
3. What is the total δ AB and δ BC is I have both uB and uy ?
x
B
4. Solve
for uB and uB from known δ AB and δ BC .
x y
Step 1

δ BC = uB sin θ
x

LN EW = BC
(L + δ1 )2 + Δ2
BC 2
2δ1 δ BC Δ2
= L (1 + + 1 2 )+ 2
L L L
√ x
1+x≈1+
L
B BC 2
δ1 C δ1 Δ2
LN EW =L 1+ + +
L 2L2 2L2
2
BC
δ1
→0
2L2
Δ2
→0
2L2
For δ << L:

δ BC ≈ D
So:

BC
Lnew = L + δ1

6

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1. 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

2. Force-Displacement Relationship P = kδ For a uniaxial force in a bar: EA0 k= L0 Deformation and Displacement Recall from Lab: The springs deform. The bar is displaced. EXAMPLE: 2

3. du(x) (x) = dx (x)dx = du 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? How much does point B displace? Unconstrained degrees of freedom 1. uB x 2. uB y 3

4. 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 ?P ? FBC cos θ = 0 ? FBC = cos θ P ?FAB + sin θ = 0 cos θ So: FAB = P tan θ Force-Deformation Relationship P = kδ EA k= L FAB δAB = kAB FBC δBC = kBC 4

5. AE kAB = L sin θ AE kBC = L So: P tan θ δAB = AE L sin θ ?P cos θ δBC = AE L P L sin θ tan θ δAB = AE ?P L δBC = AE cos θ Check: Compatibility 5

6. Algorithm to ?nd B 1. If we only have uB , what δ AB and δ BC would result? x 2. If we only have uB , what δ AB and δ BC would result? y 3. What is the total δ AB and δ BC is I have both uB and uy ? x B 4. Solve for uB and uB from known δ AB and δ BC . x y Step 1 δ BC = uB sin θ x LN EW = BC (L + δ1 )2 + Δ2 BC 2 2δ1 δ BC Δ2 = L (1 + + 1 2 )+ 2 L L L √ x 1+x≈1+ L B BC 2 δ1 C δ1 Δ2 LN EW =L 1+ + + L 2L2 2L2 2 BC δ1 →0 2L2 Δ2 →0 2L2 For δ << L: δ BC ≈ D So: BC Lnew = L + δ1 6

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Lec9 MECH ENG STRucture

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