Statically-Indeterminate-Beams-.pdf

beam

1. Bending Deflection –
Statically Indeterminate Beams AE1108-II: Aerospace Mechanics of Materials
Aerospace Structures
& Materials Faculty of Aerospace Engineering
Dr. Calvin Rans
Dr. Sofia Teixeira De Freitas

2. Recap I.Free Body Diagram
II.Equilibrium of Forces (and Moments)
III.Displacement Compatibility
IV.Force-Displacement (Stress-Strain) Relations
V.Answer the Question! –
Typically calculate desired internal
stresses, relevant displacements, or failure criteria
Procedure for Statically Indeterminate Problems
Solve when number of equations = number of unknowns
For bending, Force-Displacement relationships come from
Moment-Curvature relationship
(ie: use Method of Integration or Method of Superposition)

3. Statically Indeterminate Beams Many more redundancies are possible for beams:
- Draw FBD and count number of redundancies
- Each redundancy gives rise to the need for a
compatibility equation
P
AB
P
V
A
V
B
H
A
M
A
- 4 reactions
- 3 equilibrium equations
4 – 3 = 1
1
st
degree statically indeterminate

4. Statically Indeterminate Beams Many more redundancies are possible for beams:
- Draw FBD and count number of redundancies
- Each redundancy gives rise to the need for a
compatibility equation
- 6 reactions
- 3 equilibrium equations
6 – 3 = 3
3
rd
degree statically indeterminate
P
AB
P
V
A
V
B
H
A
M
A
H
B
M
B

5. Solving statically indeterminate
beams using method of integration

6. What is the difference between a
support and a force?
F
Displacement Compatibility
(support places constraint on deformation)

7. Method of Integration
P
AB
P
AB
V
B
What if we remove all redundancies
and replace with reaction forces?
If we treat V
B
as known, we
can solve!
Can integrate
to find v
2
2
dv
MEI
dz

Formulate expression for
M(z)

8. Method of Integration (cont)
P
AB
V
B

2
2
,
B
dvE
IMzV
dz




,
B
E
Iv M z V dz 


(, )
B
E
Iv M z V dz 

C
1
C
2
Determine constants of integration
from Boundary Conditions
AAt A, v = 0, θ= 0
(, )
B
vzV
How to determine V
B
?

9. Method of Integration (cont)
P
AB
V
B
(, )
B
vzV
How to determine V
B
?
P
AB
=
Compatibility BC:
At B, v = 0
V
B2
P
V
B1
V
B
changes displacement!
Must be compatible! Solve for V
B

10. Compatibility equations for beams are simply
the boundary conditions at redundant supports

11. 11
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1 Problem Statement
q
AB
Determine deflection equation for
the beam using method of
integration:
Treat reaction forces as knowns!
0
A
FH

2) Equilibrium:1) FBD:
AB
V
A
V
B
H
A
M
A
q
2
2
A
qL
LV 
Solution
AB
FVVqL



2
2
AAB
qL
MMLV


12. 12
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
4) Determine moment equation:
A
V
A
V
M
A
q
M
z
ccw
z
M
M



2
2
AA
q
M
Vz z 
Can also use step function approach

2z
qz 
A
M
A
Vz
M


0
0
A
Mz

0
A
Vz

2
0
2
q
z 

B
VzL
always off
always on

13. 13
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
23
1
26
A
A
Vq
EIv M z z z C    
5) Integrate Moment equation to get
v’ and v
2
2
AA
q
EIv M V z z  
= M(z)

= -EIθ(z)

234
12
2624
AA
MVq
EIv z z z Cz C    
= -EIv(z)
q
AB
We now have expressions for v and v’, but need to determine
constants of integration and unknown reactions
z

14. 14
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
5a) Solve for Constants of Integration
using BC’s:
= -EIθ(z)
= -EIv(z)
q
AB
z
Boundary Conditions:
At z = 0, θ= 0


 
32
1
000 0
62
A
A
V q
EI M C     
1
0 C 

0
0
Fixed support
θ= 0, v = 0
23
1
26
A
A
Vq
EIv M z z z C    
234
12
2624
AA
MVq
EIv z z z Cz C    

15. 15
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
5a) Solve for Constants of Integration
using BC’s:
= -EIθ(z)
= -EIv(z)
q
AB
z
Boundary Conditions:
0
0
Fixed support
θ= 0, v = 0
23
1
26
A
A
Vq
EIv M z z z C    
234
12
2624
AA
MVq
EIv z z z Cz C    
At z = 0, v = 0




43 2
2
0000
24 6 2
AA
VM q
EI C     
2
0 C


0

16. 16
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
5b) Solve for Reaction Forces using
BC’s (imposed by redundant support):
= -EIθ(z)
= -EIv(z)
q
AB
z
Boundary Conditions:
roller support
v = 0
23
26
A
A
Vq
EIv M z z z    
234
2624
AA
MVq
EIv z z z   
At z = L, v = 0




43 2
0
24 6 2
AA
VM q
E
ILLL    
3
4
A
A
M
qL
V
L

Recall from equilibrium:
2
2
A
A
qL
M
LV 
B
A
VqLV

17. 17
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
5b) Solve for Reaction Forces using
BC’s (imposed by redundant support):
q
AB
z
2
3
, ,
42
A
A
AABA
M qL qL
VMLVVqLV
L
  
3
8
B
VqL
5
8
A
VqL
2
8
A
qL
M
AB
V
A
V
B
M
A

18. 18
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
We were asked to determine deflection
equation:
q
AB
z

2
22
352
48
qz
vLLzz
EI

00.51
0
.4
0
.2
0
L
Max Displacement:

223
61580
48
q
vLzLzz
EI

0.5785 zL 
=0
2
(0.5785 ) 0.005416
qL
vL
E
I

234
2624
AA
MVq
EIv z z z   

19. 19
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 1
Now that the reactions are known:
q
AB
z
5
()
8
qL
Vz EIv qz   
2
()
2
AA
q
M
zEIvMVzz    
22
5
882
qL qLz qz
  
d
dz
M
4
L
2
8
qL

V
5
8
qL
3
8
qL 
5
8
L

20. Solving statically indeterminate
beams using superposition

21. Method of Superposition
P
AB
P
V
A
V
B
H
A
M
A
- 4 reactions
- 3 equilibrium equations
1
st
degree statically indeterminate
0
32
A
AB
A
BA
FH
FVVP
M
LV LP M








2) Equilibrium:
1) FBD:
2LL
Determine reaction forces:

22. Method of Superposition (cont) How do we get compatibility equation?
P
AB
P
AB
V
B
AB
δ
B1
δ
B2
Split into two statically determinate problems
3) Compatibility:
δ
B1
= δ
B2


12
0
BB
vv
Be careful! +ve v

23. Method of Superposition (cont) How do we get Force-Displacement relations?
Can integrate
to find v
2
2
dv
MEI
dz

We have been doing this in the previous lectures
Integrate Moment Curvature Relation
Standard Case Solutions
P
AB
L
2
2
B
PL
E
I


3
3
B
PL
v
E
I

+
+

24. Method of Superposition (cont) From the standard case:
P
AB
δ
B1
P
AB
L
2
2
B
PL
E
I


3
3
B
PL
v
E
I

2LL

2
2
22
2
P
PL
P
L
EI EI


1
BPP
vv L


3
14
3
P
L
EI

straight
4) Force-Displacement:

3
2
22
3
PLPL
L
EI EI


25. Method of Superposition (cont) From the standard case:
V
B
AB
δ
B2

3
3
2
39
3
BB
B
VLVL
v
EI EI


Compatibility:
12
0
BB
vv
3 3
9 14
3
B
VL PL
EI EI



14
27
B
VP 
4) Force-Displacement:
P
AB
L
2
2
B
PL
E
I


3
3
B
PL
v
E
I


26. Additional remarks about
bending deflections

27. Remarks about Beam Deflections Recall
Shear deformation
Moment deformation
+
Negligible (for long beams)
Bending Deformation = Shear Deformation + Moment Deformation

+
M M

+
V
V

V(x)
M(x)

28. Remarks about Beam Deflections
AB
C
P
P
A
P
B
V
BH
B
M
B
B
C
P
V
B
H
B
M
B
A
B
C
P
P
Axial deformation
of AB due to H
B Axial deformation
of BC due to V
B
Axial deformation << bending deformation!

29. Remark about Beam Deflections
negligible
Deformation = Axial Deformation + Shear Deformation + Moment Deformation
For bending deformation problems
A
P
B
V
BH
B
M
B
BUT!
If moment deformation is not
present, deformation is not negligible

30. 30
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2
A
q
B
D
C
Calculate reaction forces at A and D: First, can we see any simplifications?
Symmetry!
L, EI
Symmetry implies:
- Reactions at A = reactions at D
- Slope at symmetry plane = 0
- Shear force at symmetry plane = 0
We will solve using superposition
and standard cases
A
B
H
F
M
F
q

31. 31
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2 Calculate reaction forces:
A
B
H
F
M
F
q V
A
H
A
M
A
1) FBD
2) Equilibrium
A
F
FH H


2
A
qL
FV



2
8
ccw A
AFF
qL
M
MHLM

 

F
Split problem into two: beam AB and beam BF

32. 32
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2 Calculate reaction forces:
H
F
F
A
B
H
F
M
F
q
=
A B
M
F
-qL
2
/8
qL/2
+
q
B
H
F
M
F
F
M
F
-qL
2
/8 H
F
qL/2
3) Compatibility

F
0, v
B
0
(axial deformation negligible)

33. 33
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2 4a) Force-Displacement for Beam AB
H
F
A
B
M
F
-qL
2
/8
qL/2
H
F
A
B
A B
qL/2
A
B
M
F
-qL
2
/8
3
3
F
BH
L
vE
I


2
2
cwF
B
H
L
E
I



0
B
v
0
B


24
216
F
B
MLqL
v
E
IEI

 
3
8
cwF
B
MLqL
E
IEI


 
++

34. 34
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2 4a) Force-Displacement for Beam AB
H
F
A
B
M
F
-qL
2
/8
qL/2
324
3216
FF
B
HL MLqL
v
E
IEIEI


23
28
cwFF
B
HL MLqL
E
IEIEI



0, v 0
FB


Recall compatibility:
0
Still need θ
F

35. 35
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2 4b) Force-Displacement for Beam BF

3
2
6
cw
F
qL
E
I



0
F


cwF
F
M
L
E
I



q
B
H
F
M
F
F
M
F
+ qL
2
/8
H
F
qL/2
L/2
B
H
F
FB
M
F
F B
F
q
++
3
48
cwF
F
M
L qL
E
IEI



Wait! Not entirely correct!

36. 36
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2
3
48
cwF
FB
ML qL
EI EI




Previous angle relative to fixed support B
q
B
H
F
M
F
F
θ
B
2 3
2 7
48 2
cwFF
F
M
LHL qL
EI EI EI


 
0

37. 37
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2
0, v 0
FB


2 3
2 7
0
48 2
cwFF
F
ML HL qL
EI EI EI


 
Solve: Recall compatibility:
324
0
3216
FF
B
HL MLqL
v
EI EI EI


28
24
F
F
qL
H
qL
M


A
F
FH H


2
A
qL
FV



2
8
ccw A
AFF
qL
M
MHLM

 
Recall equilibrium:
2
8
2
3
A
A
qL
H
qL
M



38. 38
Aerospace Mechanics of Materials (AE1108-II) – Example Problem
Example 2
A
B
H
F
M
F
V
A
H
A
M
A
F
A
q
B
D
C
L, EI
28
24
F
F
qL
H
qL
M


2
8
2
2
3
A
A
A
qL
H
qL
V
qL
M




39. For next time