1
• The force (flexibility) method expresses the relationships between displacements and forces that exist in a structure.
• Primary objective of the force method is to determine the chosen set of excess unknown forces and/or couples – redundants .
• The number of redundants is equal to the degree of static indeterminacy of the structure.
1
Also see pages 115 – 141 in your class notes.
1
1.
Determine the degree of static indeterminacy.
¾
Number of releases* equal to the degree of static indeterminacy are applied to the structure.
¾
Released structure is referred to as the primary structure .
¾
Primary structure must be chosen such that it is geometrically stable and statically determinate.
¾
Redundant forces should be carefully chosen so that the primary structure is easy to analyze
2
* Details on releases are given later in these notes.
2.
Calculate “errors” (displacements) at the primary structure redundants.
These displacements are calculated using the method of virtual forces.
3.
Determine displacements in the primary structure due to unit values of redundants (method of virtual forces). These displacements are required at the same location and in the same direction as the displacement errors determined in step 2.
3
Also see pages 115 - 141 in your class notes.
4.
Calculate redundant forces to eliminate displacement errors.
¾
Use superposition equations in which the effects of the separate redundants are added to the displacements of the released structure.
¾
Displacement superposition results in a set of n linear equations (n = number of releases) that express the fact that there is zero relative displacement at each release.
4
1

¾
These compatibility equations guarantee a final displaced shape consistent with known support conditions, i.e., the structure fits together at the n releases with no relative displacements.
5.
Hence, we find the forces on the original indeterminate structure.
They are the sum of the correction forces (redundants) and forces on the released structure.
5
(1)
Flexibility Analysis
R
1
R
2
(2)
(3)
(3)
D
1 f
11
(x R
1
)
+
D
2 f
21
(x R
1
)
1 (R
1
)
+ f
12
(x R
2
) f
22
(x R
2
)
6
1 (R
2
) f R
+ f R
2
+
D
1
=
0
(4 ) f R
+ f
22
R
2
+
D
2
=
0
Solve for R
1 and R
2
.
[F] {R} = -{D}
⇒
{R} = -[F]
-1
{D}
7
[F] =
⎡
⎢ f
11 f
12 f
21 f
22
⎤
⎦
≡ flexibility matrix
[F]
-1
(
≡ inverse flexibility matrix)
=
1 f f
− f f
⎡
⎣ f
22
− f
12
− f
21 f
11
⎤
⎦
{D}
⎧
⎩
D
D
1
2
⎫
⎭
≡ primary structure displacement vector
8
2

{R}
⎧
⎩
R
R
1
2
⎫
⎭
≡ redundant force vector
⎧
⎩
R
R
1
2
⎫
⎭
=
−
1 det[F] f D
− f D
2
⎩
− f D
+ f D
2
⎭ det [F] = f f
f f
(5)
With R
1 and R
2 known, remaining structure is statically determinate.
9
Release is a break in the continuity of the elastic (displacement) curve .
¾
One release only breaks a single type of continuity.
¾
Figure 1 shows several types of releases.
¾ Common release is the support reaction, particularly for continuous beams.
10
11
Primary structure displacements at the releases are related to the unknown redundant forces via
D i
f R j
(1) f i j
≡ displacement at release i due to a unit force in the direction of and at release j; flexibility coefficients.
Equation 1 for the case of three redundant forces is expressed as
12
3

− = + +
− = + +
− = + +
( 2a )
Matrix form of (2a)
-{D} = [F] {R} (2b)
⎧
D
1
⎫
⎨
⎩ D
3
⎬
⎭
1
D
2
D
3
> T
= displacement vector at the redundant degrees of freedom
13
⎧
⎨
R
1
2
⎫
⎬
1
⎩ R
3
⎭
R
2
R
3
> T
= redundant force vector
[F] =
⎡
⎢
⎢
⎢ f
11 f
12 f
13 f
21 f
22 f
23 f
31 f
32 f
33
⎥
⎦
⎤
⎥
= flexibility matrix
14
D i
M d
⇒ i at
(3) release i d = differential axial displacement d
φ
= differential rotational displ
15
f b ij
a ij f ij
= f ij a + f ij b
(4)
=
Vi
Vj
≡ axial deformation influence coefficient
=
M
Vi
M
Vj
EI(x) dx
≡ bending deformation influence coefficient
16
4

1. Calculate the support reactions for the two-span continuous beam,
EI = constant.
w
L L w
Primary Structure w/ Load
+
1 (x R
1
)
Primary Structure w/ Redundant
17
Prismatic Member Displacements
2. Calculate the support reactions for the two-span continuous beam,
EI = constant.
w
L L
R
1 w
Primary Structure w/ Load
+
R
2
19 20
5

21
3. Calculate the support reactions for the two-span continuous beam
I using the internal moment at B as the redundant force, I
BC
= I; E = constant.
AB
= 2I and
L
2
P
Primary Structure w/ Loading
22
M
B
Primary Structure w/ Redundant
D
B
= __________________ f
BB
= _________________
M
B
= _________________
23
4. Calculate the bar forces for the statically indeterminate truss.
Statically
Indeterminate
Truss
Statically
Determinate
Released Truss
6

Mem
AB
L (in) F F
V
F
V
FL
192" 40 -4/5 -6144
-3/5 0 BC 144" 0
CD 192" 0 -4/5 0
DA 144" 30 -3/5 -2592
AC 240" 0 1 0
BD 240" -50 1 -12000
F
AC
= −
⎡
⎢
E A
⎤
⎦
/f
AC,AC
= 25 kips
=
20736/829.44
25
Structure Forces
Once the redundant forces are calculated from Eq. (5), all other support reactions and internal member forces can be calculated using static equilibrium along with the appropriate free body diagrams.
This is possible since the force method of analysis has been used to determine the redundant forces or the forces in excess of those required for static determinacy.
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[F]{R}
{D }
D D
Δ
2
D
Δ n
T
(5)
= relative dimensional change displacements calculated using principle of virtual forces
Displacements due to dimension changes are all relative displacements , as are all displacements corresponding to releases. They are positive when they are in the same vector direction as the corres-
26 ponding release (redundant) .
Calculation of the non-redundant forces A i
(support reactions, internal shears and moments, truss member forces) can be expressed using superposition as
A i
A i p
N
R
(A ) R j
(6)
A i p desired action A i on the primary structure due to the applied loading (A ) action
A i on the primary structure due to a unit virtual force at redundant
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R j and N
R
= number of redundants .
7

(a) Given structure
(b) Primary structure
E = 30,000 ksi
I = 288 in 4
29
For the truss structure on the next page, compute the redundant bar EC member force if the temperature in bar EF is increased 50 o
F and member BF is fabricated 0.3 in. too short.
EA = constant = 60,000 kips and
= 6x10
-6
/ o
F.
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The interesting point of this example is that the flexibility equation will have a nonzero right hand side since the redundant displacement is prescribed to equal 0.72
” downward. Thus the flexibility equation is f
BB
R
B
= d
B
D
Δ
B
(7) where d
B
= prescribed displacement at redundant B
= -0.72
" since R
B is
D " positive upward d
B
Δ
B
= -0.24
−
D
Δ
B
= relative displacement
30 at redundant B
3 @ 20 ’ = 60 ’
B C
A D
15 ’
E
B
F
C
A 1 D
E F
Primary Structure Subjected to F
CE
= 1
32
8

Mem L F
V
F
V
F
V
L
AB 240" 0 0
AE 300" 0 0
BC 240"
BE 180"
BF 300"
CD 240"
CE 300"
CF 180"
DF 300"
EF 240"
-4/5
-3/5
1
0
1
-3/5
0
-4/5
153.6
64.8
300
0
300
64.8
0
153.6
33 f
CE,CE
=
1
EA m
F
Vi
F L
D
Δ
CE
m
F
Vi
i
Δ
=
δ
Δ
BF
= Δ A
BF
= −
+
Δ
CE
=
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Displacements for the statically indeterminate structure can be calculated using the exact member deformations for a truss or exact shear and moment expressions along with the virtual force expressions on the primary structure.
For a truss structure , calculation of a joint displacement using the principle of virtual forces results in
35
1 (
Δ
) = m
F
Vi
Δ
= m
F
Vi
⎛
⎝
F L i
EA i
+ δ i int
⎞
⎟ + Δ
Δ
(8)
F
Vi = primary structure member forces due to the application of a unit virtual force at the joint for which the displacement is desired and in the direction of
Δ
36
9

Δ
= primary structure displacement at desired displacement due to nonmechanical effects
i = exact member displacements that are obtained for the statically indeterminate structure using the calculated redundant forces to determine all the member forces within the truss structure
i int
= member displacements due to nonmechanical loading on the member
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For a frame structure , in which shear and axial deformations are ignored , the displacements are calculated as m
L
0
M
Δ
Vi
⎛
⎝
M
EI i i + κ i int
⎞
⎠ dx
(9a) m
L
0
M
θ
V i
⎛
⎝
M i
E I i
+ κ i int
⎞
⎠ dx
M , M
θ
Vi
= primary structure virtual moments based on the desired displacement
Δ or rotation
θ
Δ
,
Δ
= primary structure displacements at
Δ or rotation
θ due to environmental loads or causes
i int
= primary structure initial curvature strain caused by nonmechanical loading
39
In Eqs. (9a) and (9b) the moment expressions are exact based on the statically indeterminate structure subjected to the external loads with the redundant forces known from the flexibility analysis .
Equations (8), (9a), and (9b) are correct only because exact real member forces are used in the calculation of the desired displacements.
40
10

Calculate the horizontal displacement at joint B for the statically indeterminate truss.
16 ’
D
40 k C
12 ’
R
1
A B
D
C
A B
1
Primary Structure Subjected to
41 Virtual Loading
Calculate the rotation at the center support for the two-span continuous beam, EI = constant.
w
R
1
L
1
R
2
L
L L
Primary Structure w/ Virtual
Load at Desired Displacement
Location
42
Alternatively , you can express the desired displacement calculations also in matrix form following the usual superposition process of the force method of analysis:
δ =
[F ]{R}
+
{D }
+
{D } (10) where {
δ
} = vector of desired vector of desired displacements for the primary structure for both mechanical and non-mechanical loadings, respectively; [F
δ
] =
43 matrix of displacement influence coefficients at the desired displacement locations due to unit values of the redundant forces
{R}. Stated mathematically, the coefficients of [F
δ
] are
F δ ij
= δ
(11) which simply states that the displacement influence coefficients equal the displacement at desired displacement i on the primary structure due to a unit force at redundant j on the primary
44 structure.
11