Lecture 2 METHOD OF CONSISTENT DEFORMATION Beams.ppt

LECTURE

1. Method of Consistent
Deformation
Structural Analysis
By
R. C. Hibbeler

2. Force Method of Analysis/Method of Consistent
Deformation
Consider the following beam
Number of unknown support reactions = 4
Equations of equilibrium available= 3
Degree of indeterminacy = first
So we need one more equation for solution
A B
P
Actual Beam
2

3. For this we will use principle of superposition and
consider the compatibility of displacement at one of its
support.
This is done by choosing one of its support reactions as
Redundant and temporarily removing its effect on the
beam to make it statically determinate and stable.
This beam is referred to as the Primary Structure.
3
A B
P
Actual Beam
A B
P
Primary Structure
Δ
B

4. By superposition , the unknown reaction at B, i.e., B
y,
causes the beam at B to be displaced Δ’
BB
upward.
First letter in subscript notation refers to point B where
deflection is specified and second letter refers to point B
where unknown reaction acts.
A B
P
Actual Beam
A B
P
Primary Structure
Δ
B
=
A
Δ’
BB
=B
y
f
BB
Redundant B
y
appliedB
y
4
+
B

5. Assuming positive displacements act upward, then we
can write necessary compatibility equation at the roller
as
Delta(b) and delta’(bb) must be equal
A B
P
Actual Beam
A B
P
Primary Structure
Δ
B
=
A
Δ’
BB
=B
y
f
BB
Redundant B
y
appliedB
y
5
+
B
 0
'
BBB

6. The displacement at B caused by unit load acting in the
direction of B
y
is Linear flexibility coefficient f
BB
.
f
BB is the deflection at B caused by a unit load at B
The material behaves in a linear elastic manner, a force
of B
y acting at B, instead of unit load, will cause a
proportionate increase in f
BB
.
A
f
BB
1
6
B
BByBB
fB
'

7. We can say that Linear Flexibility Coefficient f
BB is a
measure of the deflection per unit force, and its units are
m/N, ft/lb, etc.
The compatibility equation above can be written in terms
of the unknown B
y as
Once B
y
is found, the three reactions at A can be found
from equations of equilibrium.
7
BByB
fB0
BB
B
y
f
B



8. As stated previously, the choice of redundant is arbitrary.
For example, the moment at A can be determined
directly by removing the capacity of beam to support a
moment at A, by replacing the fixed support by a pin.
8
A B
Actual Beam
A B
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A
P
P

9. The rotation at A, caused by the load P is θ
A, and the
rotation at A cause by the redundant M
A at A is θ’
AA.
9
A B
Actual Beam
A B
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A
P
P

10. If we denote an angular flexibility coefficient α
AA as the
angular displacement at A cause by a unit couple
moment applied at A, then
10
A B
Actual Beam
A B
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A
P
P
AAAAA
M
'

11. The angular flexibility coefficient measures the angular
displacement per unit couple moment and has the units
of rad/N.m or rad/lb.ft, etc.
The compatibility equation for rotation at A requires
In this case,
-ve value means that M
A
acts in opposite direction
11
A B
α
AA
1
AAAA
M0
AA
A
A
M




12. Second Order Indeterminate Structures
The figure is showing a beam of second order
indeterminacy.
Two compatibility equations will be necessary for
solution.
12
A
B C
D
Actual Beam
P
1
P
2

13. We will choose the vertical forces at the roller supports B
and C, as redundants.
The resultant primary structure deflects as shown, when
the redundants are removed
13
A
B C
D Actual Beam
P
1
P
2
A
B C
D Primary
Structure
P
1
P
2
Δ
B
Δ
C

14. Each redundant force which is assumed to act
downward, deflects this beam as shown
14
A
B C
D
B
y
Δ’
BB
=B
y
f
BB
Δ’
CB
=B
y
f
CB
Redundant
B
y
Applied
A
B C
D
C
y
Δ’
BC
=B
y
f
BC
Δ’
CC=C
yf
CC
Redundant
C
y
Applied

15. 15
A
B C
D
B
y
Δ’
BB
=B
y
f
BB
Δ’
CB
=B
y
f
CB
Redundant
B
y
Applied
A
B C
D
C
y
Δ’BC=CyfBC Δ’
CC=C
yf
CC
Redundant
C
y
Applied
A
B C
D Primary
Structure
P
1
P
2
Δ
B
Δ
C
A
B C
D Actual Beam
P
1
P
2
+
+
=

16. 16
A
B C
D
1
f
BB
f
CB
A
B C
D
1

f
BC
f
CC

17. By superposition, the compatibility equations for the
deflection at B and C, respectively are
These equations may be solved simultaneously for the
two unknown forces B
y and C
y.
17



0
0
CCyCByC
BCyBByB
fCfB
fCfB

18. PROCEDURE FOR ANALYSIS
Following procedure provides a general method for
determining the reactions or internal loadings of S.I.S
using the force method.
Principle of Superposition
•Determine the number of degree of indeterminacy.
•Specify the number of redundant forces or moments
which must be removed to make the structure
determinate.
•Draw S.I.S and show it to be equal to a sequence of
corresponding S.D.S.
18

19. Principle of Superposition
•The primary structure supports the same external loads
as the S.I.S., and each of other structures added to the
primary structure shows the structure loaded with a
separate redundant force or moment.
•Sketch the elastic curve on each structure and indicate
symbolically the displacement or rotation at the point of
each redundant force or moment.
Compatibility Equations
•Write compatibility equation for the displacement or
rotation at each point where there is a redundant force
or moment.
19

20. Compatibility Equations
•These equations should be expressed in terms of the
unknown redundants and their corresponding flexibility
coefficients.
•Determine all the deflections and their corresponding
flexibility coefficients using the table on inside front
cover.
•Substitute these into the compatibility equations and
solve for the unknown redundants.
•If the numerical value for a redundant is negative, it
indicates the redundant acts opposite to its
corresponding unit force or unit couple moment.
20

21. Equilibrium Equations
•Draw a free body diagram of the structure.
•As the redundant forces have been calculated, now
calculate the remaining unknown reactions using
equations of equilibrium.
•Now draw the shear and moment diagrams.
•Also the deflection at any point can be determined using
the previous methods.
21

22. Example 1
Determine the reaction at the roller support B of the
beam in Fig. EI is constant.
A
B
C
50 KN
6 m 6 m
Actual Beam
22

23. Solution
•The beam is first degree statically indeterminate
A
B
C
50 KN
6 m 6 m
Actual Beam
23

24. Principle of Superposition
•B
y is taken as redundant
•Removal of redundant B
y requires that the roller support
in the direction of B
y be removed
•B
y
is assumed to act upward
A
B
C
50 KN
6 m 6 m
Actual Beam
=
A
B
C
50 KN
Δ
B
θ
C
Δ
C
Primary Structure
+
A
Δ’
BB
=B
y
f
BB
Redundant B
y
applied
B
y
24

25. Compatibility Equation
•Taking positive displacement as upward, we have
•Δ
B and f
BB are obtained using tables on inside front cover
of book
•Note that
A
B
C
50 KN
Δ
B
θ
C
Δ
C
Primary Structure
+
A
Δ’
BB
=B
y
f
BB
Redundant B
y
applied
B
y
(1) 0
BByB fB
m
CCB
6
25

26. Compatibility Equation
•Thus
A
B
C
50 KN
Δ
B
θ
C
Δ
C
Primary Structure
+
A
Δ’
BB
=B
y
f
BB
Redundant B
y
applied
B
y
m
CCB
6







22
)2(
3
)2(
23
L
EI
LP
EI
LP
B
m
EI
mkN
EI
mkN
B
6
2
)6)(50(
3
)6)(50(
23


EI
mkN
B
3
. 9000
26

27. Compatibility Equation
Substituting these results into Eq. (1) yields
A
B
C
50 KN
Δ
B
θ
C
Δ
C
Primary Structure
+
A
Δ’
BB
=B
y
f
BB
Redundant B
y
applied
B
y


EI
m
EI
m
EI
PL
f
BB
333
576
3
121
3







EI
B
EI
y
5769000
0
ANS 6.15kNB
y

27

28. If this reaction is placed on free body diagram of the
beam, the reactions at A can be obtained from the three
equations of equilibrium.
50 kN
6 m 6 m
15.6 kN
34.4 KN
112 kN . m
28

29. Having determined all the reactions, the moment
diagram can be constructed.
29
50 kN
6 m 6 m
15.6 kN
34.4 KN
112 kN . m
-112
3.27
6 12
x (m)
M (kN.m)
93.8

30. Example 2
Determine the moment at the fixed wall for the beam in
Fig. EI is constant.
A B
10 ft
Actual Beam
30
20 k . ft

31. Solution
•The beam is first degree statically indeterminate
A B
10 ft
Actual Beam
31
20 k . ft

32. Principle of Superposition
•M
A is taken as redundant
•The capacity of the beam to support a moment at A has
been removed
•Fixed support at A is substituted by a pin
•M
A is assumed to act counterclockwise
A B
10 ft
Actual Beam
32
20 k . ft
A B
20 k . ft
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A

33. Compatibility Equation
•Taking positive rotation as counterclockwise, we have
•θ
A and α
AA can be determined using tables on inside front
cover of book, we have
(1) 0
AAAA
M
33
EI
ftk
EI
ftftk
EI
ML
A
2
. 3.33
6
)10(. 20
6

A B
20 k . ft
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A

34. Compatibility Equation
•Taking positive rotation as counterclockwise, we have
•θ
A and α
AA can be determined using tables on inside front
cover of book, we have
(1) 0
AAAA
M
34
EI
ft
EI
ft
EI
ML
AA
.333
3
)10(1
3

A B
20 k . ft
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A

35. Compatibility Equation
Substituting these results into Eq.(1) yields
(1) 0
AAAA
M
35
ANS . 10
33.33.33
0
ftkM
EI
M
EI
A
A








A B
20 k . ft
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A

36. The negative sign indicates that M
A
acts opposite to that
shown in figure.
36
A B
20 k . ft
θ
A
Primary structure
+
A B
θ'
AA
=M
A
α
AA
Redundant M
A
applied
M
A

37. When this reaction is placed on the beam, other
reactions can be determined.
37
10 ft
3 k
A
10 k . ft
3 k
B
20 k . ft

38. The moment diagram is shown below
38
10 ft
3 k
A
10 k . ft
3 k
B
20 k . ft
x (ft)
M (k.ft)
-20
10

39. Example 3
Draw the shear and moment diagrams for the beam
shown in Fig. The support at B settles 1.5 in. Take E =
29(10
3
) ksi, I = 750 in
4
.
A C
20 k
12 ft 24 ft
Actual Beam
39
B
1.5 in
12 ft

40. Solution
Principle of Superposition
•The beam is first degree statically indeterminate.
•The centre support B is chosen as redundant, so that the
roller at B is removed.
40
A C
20 k
12 ft 24 ft
Actual Beam
B
1.5 in
12 ft

41. •B
y is assumed to act downward on the beam.
41
A C
+
A C
20 k
Actual Beam
B
1.5 in
=
B
Δ
B
Primary
Structure
20 k
A C
B
Δ’
BB
=B
y
f
BB
B
y
Redundant B
y

applied

42. Compatibility Equation
•With reference to point B, using units of ft, we require
•Use conjugate beam method to compute Δ
B
and f
BB
since
the moment diagrams consists straight line segments.
•For Δ
B
 (1)
12
5.1

BByB
fB
42
A C
B
20 k
A C
20 k
12 ft 36 ft
15 k 5 k

43. Compatibility Equation
43
A C
20 k
12 ft 36 ft
15 k 5 k
8 ft 24 ft
EI
2520
EI
1800
EI
1080
EI
3240EI
180
16 ft
conjugate beam

44. Compatibility Equation
44
8 ft 24 ft
EI
2520
EI
1800
EI
1080
EI
3240EI
180
16 ft
EI
1800
EI
1440
8 ft 16 ftV
B’
M
B’
EI
120
0M
B'

 024
1800
8
1440
' 
EIEI
M
B

EIEI
M
B
3168031680
'

45. Compatibility Equation
•For f
BB
45
A C
B
1 k
A C
24 ft 24 ft
0.5 k 0.5 k
1 k
24 ft 24 ft
conjugate beam
EI
144
EI
144
EI
288
EI
12

46. Compatibility Equation
•For f
BB
46
24 ft 24 ft
conjugate beam
EI
144
EI
144
EI
288
EI
12
EI
144
EI
144
24 ft
v
B’
m
B’
EI
12
8 ft
0M
B'

 024
144
8
144
'

EIEI
m
B

EIEI
m
B
23042304
'

47. Compatibility Equation
•Substituting these results into eq. (1), we have
•Expressing the units of E and I in terms of k and ft, we
have
 (1)
12
5.1

BByB
fB
47







EI
B
EI
.
y
230431680
12
51

      
230431680
12750121029
12
51
444422223
y
B
inftinftininkft
.








48. Equilibrium Equations
•The negative sign indicates that B
y acts upward on the
beam.
48
kB
y
56.5
A C
20 k
12 ft 24 ft
A
y
C
y
B
y
=5.56 k
12 ft

49. Equilibrium Equations
49
A C
20 k
12 ft 24 ft
A
y
=12.22 k C
y
=2.22 kB
y
=5.56 k
12 ft
0M
A
  0482456.51220 
yC
kC
y 22.2
0F
y
 022.256.520 
y
A
kA
y 22.12

50. •Using these results, shear and moment diagrams are
50
x (ft)
V (k)
-20
12.22
-7.78
-2.22
A C
20 k
12 ft 24 ft
A
y
=12.22 k C
y
=2.22 kB
y
=5.56 k
12 ft

51. •Using these results, shear and moment diagrams are
51
x (ft)
M (k.ft)
A C
20 k
12 ft 24 ft
A
y
=12.22 k C
y
=2.22 kB
y
=5.56 k
12 ft
146.7
53.3

52. Example 4
Draw the shear and moment diagrams for the beam
shown in Fig. EI is constant. Neglect the effect of axial
load.
52
A B
2 k/ft
10 ft 10 ft
Actual Beam

53. Solution
Principle of Superposition
•Since axial load is neglected the beam is second degree
statically indeterminate.
53
A B
2 k/ft
10 ft 10 ft
Actual Beam

54. Solution
Principle of Superposition
•The two end moments at A and B will be considered as
redundants.
•Beam’s capacity to resist these moments is removed by
placing a pin/hinge at A and rocker/roller at B.
54
A B
2 k/ft
10 ft 10 ft
Actual Beam

55. Solution
55
A B
2 k/ft
10 ft 10 ft
Actual Beam
=
A B
2 k/ft
Primary
structure
+
θ
A θ
B
A
B
Redundant M
A
applied
M
A
θ'
AA
=M
A
α
AA
θ‘
BA
=M
A
α
BA
+
A B
Redundant M
B
applied
M
B
θ'
AB
=M
B
α
AB
θ‘
BB
=M
B
α
BB

56. Compatibility Equation
•Reference to point A and B, requires
•The required slopes and angular flexibility coefficients
can be determined using the table on the inside front
cover. We have

(2) 0
(1) 0
BBBBAAB
ABBAAAA
MM
MM




56


57. Compatibility Equation

EIEIEI
wL
A
375
128
2023
128
3
33

57

EIEIEI
wL
B
7.291
384
2027
384
7
33


EIEIEI
ML
AA
67.6
3
201
3


EIEIEI
ML
BB
67.6
3
201
3


EIEIEI
ML
AB
33.3
6
201
6


58. Compatibility Equation
Substituting the data into Eqs. (1) and (2) yields
Cancelling EI and solving these equations simultaneously,
we have
58


























EI
M
EI
M
EI
EI
M
EI
M
EI
BA
BA
67.633.37.291
0
33.367.6375
0
ftkMftkM
BA . 8.20 . 8.45 

59. Compatibility Equation
Using these results, the end shears are calculated
59
A B
2 k/ft
10 ft 10 ft
16.25 k 3.75 k
45.8 k.ft 20.8 k.ft
x (ft)
V (k)
16.25
-3.75
8.125

60. Compatibility Equation
Using these results, the end shears are calculated
60
A B
2 k/ft
10 ft 10 ft
16.25 k 3.75 k
45.8 k.ft 20.8 k.ft
x (ft)
M (k.ft)
45.8
20.2
-20.8
8.125
3.63 14.4 20

61. Example 5
Determine the reactions at the supports for the beam
shown. EI is constant.
61
A C
120 lb/ft
12 ft 5 ft
Actual Beam
500 lb
5 ft
B

62. Solution
Principle of Superposition
•By inspection the beam is indeterminate to the first
degree.
62
A C
120 lb/ft
12 ft 5 ft
Actual Beam
500 lb
5 ft
B

63. Solution
Principle of Superposition
•We will choose the internal moment at support B as the
redundant.
•Beam is cut open and end pins or internal hinge are
placed at B to release only the capacity of beam to resist
moment at this point.
63
A C
120 lb/ft
Actual Beam
500 lb
B
θ‘
B
θ‘’
B

64. Solution
Principle of Superposition
64
A C
120 lb/ft
Primary
structure
500 lb
B
θ‘
B
θ‘’
B
A C
120 lb/ft
Actual Beam
500 lb
B
=
+
Redundant
M
B
applied
M
B
α‘
BB
M
B
α’‘
BB
M
B M
B

65. Solution
Compatibility Equations
•The relative rotation of one end of one beam with
respect to the end of other beam to be zero, that is
where
65
A C
120 lb/ft
Actual Beam
500 lb
B
θ‘
B
θ‘
BB
0
BBBB
M
'''
BBB
 
'''
BBBBBB
 

66. Solution
Compatibility Equations
•The slopes and angular flexibility coefficients can be
determined from the table on inside front cover, that is
66

EI
ftlb
EIEI
wL
θ
'
B
233
. 8640
24
12120
24


EI
ftlb
EIEI
PL
θ
'
B
222
' . 1253
16
10500
16


EI
ft
EIEI
ML
'
BB
4
3
121
3


EI
ft
EIEI
ML
'
BB
33.3
3
101
3
'


67. Solution
Compatibility Equations
•Thus
•The negative sign indicates that M
B
acts in opposite
direction.
67
0
BBBB
M
'''
BBB
 
'''
BBBBBB
 
0
33.3 4. 3125. 8640
22







EI
ft
EI
ft
M
EI
ftlb
EI
ftlb
B
ftlbM
B
. 1604

68. Solution
Compatibility Equations
68
120 lb/ft
500 lb
586 lb
854 lb 854 lb
1260 lb
410 lb 410 lb
89.6 lb
1604 lb.ft1604 lb.ft
x (ft)
V (lb)
586
-854
410
-89.64.89
12 17 22

69. Solution
Compatibility Equations
69
120 lb/ft
500 lb
586 lb
854 lb 854 lb
1260 lb
410 lb 410 lb
89.6 lb
1604 lb.ft1604 lb.ft
x (ft)
M (lb.ft)
1431
4.89
12 17
22
-1604
448