indeterminate analysis force method1 -...
TRANSCRIPT
1
Indeterminate AnalysisForce Method1
• 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.
1Also see pages 10 – 43 in your class notes.
2
Description of the Force Method Procedure
1. Determine the degree of static indeterminacy.
Number of releases* equal to the degree of static indeterminacy are applied to the structure.
Released structure is referred to as the primary structure.
Primary structure must be chosen such that it is geometrically stable and statically determinate.
Redundant forces should be carefully chosen so that the primary structure is easy to analyze
* Details on releases are given later in these notes.
3
Force Method – con’t
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.
4
Force Method – con’t
4. Calculate redundant forces to eliminate displacement errors.
Use superposition equations in which the effects of the separate redundants are added to the displacements of the released structure.
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.
5
Force Method – con’t
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.
6
Flexibility Analysis
R1R2
(1)
=
1 (R1)
1 (R2)
+
+
f21 (x R1)f11 (x R1)
f12 (x R2)
f22 (x R2)
D1D2
(2)
(3)
(3)
7
11 1 12 2 1
21 1 22 2 2
(4
f R f R D 0
f R f 0
)
R D
Solve for R1 and R2.
Using matrix methods:
[F] {R} = -{D}
{R} = -[F]-1 {D}
8
[F] = 11 12
21 22
f f
f f
flexibilitymatrix
[F]-1 ( inverse flexibility matrix)
22 12
21 1111 22 12 21
f f1f ff f f f
1
2
D{D}
D
primarystructuredisplacementvector
9
det [F] =
(5)
With R1 and R2 known, remaining structure is statically determinate.
1
2
R{R}
R
redundantforcevector
1 22 1 12 2
2 21 1 11 2
R f D f D1R f D f Ddet[F]
11 22 12 21f f f f
10
Releases
Release is a break in the continuity of the elastic (displacement) curve.
One release only breaks a single type of continuity.
Figure 1 shows several types of releases.
Common release is the support reaction, particularly for continuous beams.
11
12
Flexibility Equations
Primary structure displacements at the releases are related to the unknown redundant forces via
i ij jD f R (1)
displacement at release i due to a unit force in the direction of and at release j; flexibility coefficients.
i jf
Equation 1 for the case of three redundant forces is expressed as
13
1 11 1 12 2 13 3
2 21 1 22 2 23 3
3 31 1 32 2 33 3
D f R f R f R
D f R f R f R
D f R f R f R
(2a)
Matrix form of (2a)
-{D} = [F] {R}
{D} = = <D1 D2 D3>T
= displacement vector at theredundant degrees of freedom
(2b)
1
2
3
D
D
D
14
{R} = = <R1 R2 R3>T
= redundant force vector
11 12 13
21 22 23
31 32 33
f f f
f f f
f f f
[F] =
= flexibility matrix
1
2
3
R
R
R
Displacement Calculations – Method of Virtual Forces
i i i iD p d m d v= + φ +∫ ∫ ∫ dy (3)
subscript i ⇒ direction of at release i iR
d = differential axial displacement
dφ = differential rotation displacement
dy = differential shear displacement
Flexibility Coefficients – Method of Virtual Forces
a b sij ij ij ijf f f f (4)
ja
ij ip
f p dxEA(x)
axial influence coefficient
jb
ij im
f m dxEI(x)
bending influence coefficient
js
ij is
vf v dx
GA (x)
shear flexibility influence coefficient
Nonmechanical Loading [F]{R} ({D} {D })∆= − + (5)
T1 2 n{D } D D D∆ ∆ ∆ ∆= < >… = relative dimen-
sional change displacements, calculated us-
ing principle of virtual forces
Displacements due to dimension changes are
all relative displacements, as are all displace-
ments corresponding to releases. They are
positive when they are in the same vector direc-
tion as the corresponding release.
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.
28
Mathematical Expressions
piA
Calculation of the non-redundant forces Ai (support reactions, internal shears and moments, truss member forces) can be expressed using superposition as
RNp
i ui j jij 1
A A (A ) R
where = desired action Ai on the primary structure due to the applied loading; = action
Ai on the primary structure due to a unit virtual force at redundant
Rj and NR = number of redundants.
ui j(A )
Displacement Calculations
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 dis-
placement ∆ using the principle of virtual
forces results in
1 (∆) =
m
i i
i 1
p ∆
=
δ + δ∑
=
mi i
ii
i 1
F LpEA
∆
=
+ δ∑ (6)
ip = 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 = primary structure displacement at de-
sired displacement ∆
iδ = exact member displacements that are ob-
tained for the statically indeterminate struc-
ture using the calculated redundant forces to
determine all the member forces within the
truss structure
For a frame structure, in which shear and axial
deformations are ignored, the displacements
are calculated as
1( )∆ =
Lmi
ii
i 1 0
Mm dxEI
∆ ∆
=
+ ∆∑ ∫ (7a)
1( )θ =
Lmi
ii
i 1 0
Mm dxEI
θ ∆
=
+ θ∑ ∫ (7b)
im , m∆iθ
∆
= primary structure virtual moments
based on the desired displacement ∆ or rotation
θ
,∆∆ θ = primary structure displacements at ∆
or rotation θ
In Eqs. (7a) and (7b) the moment expressions
are exact based on the statically indetermi-
nate structure subjected to the external loads
with the redundant forces known from the
flexibility analysis.
iM
Equations (6), (7a), and (7b) are correct only
because exact real member forces are used in
the calculation of the desired displacements.
17
Force Method Examples
1. Calculate the support reactions for the two-span continuous beam, EI = constant. w
L L
w
1 (x R1)
Primary Structure w/ Load
=
+
Primary Structure w/ Redundant
18
2. Calculate the support reactions for the two-span continuous beam, EI = constant.
w
L L
w
R2
Primary Structure w/ Load
=
+
R1
Primary Structure w/ Redundant Forces
19
Prismatic Member Displacements
20
21
22
3. Calculate the support reactions for the two-span continuous beam using the internal moment at B as the redundant force, IAB = 2I and IBC = I; E = constant.
Primary Structure w/ Loading
PL2
23
Primary Structure w/ Redundant
MB
DB = __________________
fBB = _________________
MB = _________________
24
4. Calculate the bar forces for the statically indeterminate truss.
Statically Indeterminate
Truss
Statically Determinate
Released Truss
(Redundant X)
25
VAC AC,AC
F FLF /f 20736/829.44
E A
= 25 kips
Mem L (in) F FV FVFL
AB 192"40 -4/5 -6144
BC 144" 0 -3/5 0
CD192" 0 -4/5 0
DA 144" 30 -4/5 -2592
AC 240" 0 1 0
BD 240" -50 1 -12000
Truss Calculations
29
Example Beam Problem –Nonmechanical Loading
E = 30,000 ksi
I = 288 in4
(a) Given structure
(b) Primary structure
30
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
fBB RB = dB - (7)
where
dB = prescribed displacementat redundant B
= -0.72" since RB is positive upward
= -0.24"relative displacementat redundant B
BD
BD
B Bd D
31
Truss Example –Nonmechanical Loading
For the truss structure on the next page, compute the redundant bar EC member force if the temperature in bar EF is increased 50 oF and member BF is fabricated 0.3 in. too short. EA = constant = 60,000 kips and
= 6x10-6 /oF.
32
Truss Example
C
D
E F
A
B
15’
3 @ 20’ = 60’
A 1
B C
D
E F
Primary Structure Subjectedto FCE = 1
33
Mem L FV FV FVL
AB 240" 0 0
AE 300" 0 0
BC 240" -4/5 153.6
BE 180" -3/5 64.8
BF 300" 1 300
CD 240" 0 0
CE 300" 1 300
CF 180" -3/5 64.8
DF 300" 0 0
EF 240" -4/5 153.6
Truss Example Calculations
34
m
CE,CE Vi Vi ii 1
1f F F L
EA
m
CE Vi ii 1
D F
CE,CE CE CEf F D 0
EF EF EF
BF BF
T L 0.072"
0.3"
41
Calculate the horizontal displace-ment at joint B for the statically indeterminate truss.
40 k
A B
CD
R1
CD
BA 1
Primary Structure Subjected to Virtual Loading
12’
16’
42
Calculate the rotation at the center support for the two-span continuous beam, EI = constant.
w
L L
Primary Structure w/ Virtual Load at Desired Displacement
Location
L L
1
R1 R2
43
Influence Lines for Statically Indeterminate
Structures
We will utilize the force
method principles developed in the
last chapter to calculate and
construct influence lines for
statically indeterminate structures.
Recall that an influence line is a
graph of a response function of a
structure as a function of position
of a downward unit load moving
across the structure.
44
Calculating the response function
values simply involves computing
the values of the desired response
function(s) for various positions
of a unit load on the structure.
Constructing the influence line(s)
simply involves plotting the
calculated values.
Earlier in this course, we produced
qualitative influence lines for
statically indeterminate structures
using the Muller-Breslau principle.
45
Influence functions for
statically indeterminate structures
are piecewise cubic for piecewise
prismatic beam members. This
is based on the solution of the
differential equation
4y
4
d uEI 0
dx (b)
where uy = transverse beam
displacement.
46
Truss structure members remain
piecewise linear since the
governing differential equation is
2x
2d u
EA 0dx
(t)
where ux = axial truss member
displacement.
47
Beam and Truss Structure Influence Lines
Consider the continuous beam in
Fig. 1(a) – calculate and draw
the influence line for the
redundant vertical reaction at B.
NOTE: Once the redundant
force(s) is (are) known, the
remaining forces can be
obtained from static equilibrium
or from superposition (Eq. 6).
Figure 1. Influence Line Construction for Single Redundant DOF Beam
The influence line for the reaction By re-
quires that we calculate By as a function of the
unit load position x. Using the force method of
analysis:
BX BB y
BXy
BB
f f B
fB
f
+ =
⇒ = −
0
(1)
where flexibility coefficient fBX denotes the de-
flection of the primary structure (beam) at B
due to a unit load at X (Fig. 1(b); and flexibility
coefficient fBB denotes the deflection at B due
to a unit value of the redundant By.
An efficient procedure for the solution of
(1) involves using Maxwell’s Law of Recipro-
cal Deflections, i.e. the deflection at B due to a
unit load at X must equal the deflection at X
due to a unit load at B. For our problem, this is
stated mathematically as fBX = fXB. Thus, (1)
can be rewritten as
XBy
BB
fB
f= − (2)
which represents the equation for the influence
line for B. We only need to determine fXB and
fBB.
Equation (2) is more convenient that (1)
since (2) shows that the unit load only needs to
be applied at B on the primary (statically de-
terminate) structure and the corresponding dis-
placement is calculated at X. Closed-formed
solutions of (b) for a number of statically de-
terminate structures subjected to point loads are
included in your notes. Also, refer to the table
solutions included in your force method of
analysis notes. The influence line of Fig. 1(d)
is obtained by plotting the solution of (2) for
various values of X; ordinates = -fXB/fBB.
The influence line equation as represented
in (2) shows the validity of Muller-Breslau’s
principle:
The influence line for a response function is given by the deflected shape of the released structure due to a unit displacement (or rota-tion) at the location and in the direction of the response function.
The ordinate of the influence line at any point
X is proportional to the deflection fXB of the
primary structure at that point due to the unit
load at B. Furthermore, this equation indicates
that the influence line for By can be obtained by
multiplying the deflected shape of the primary
structure due to the unit load at B by the scaling
factor -1/fBB.
Consider next a beam structure with multiple
degrees of freedom as shown in Fig. 2. The
procedure is the same you simply have more
redundant degrees of freedom. In Fig. 2, the
vertical reactions at B and C are taken to be re-
dundant forces leading to the equations
(3) BX BB y BC y
CX CB y CC y
f f B f C 0
f f B f C 0
+ +
+ +
=
=
Figure 2. Influence Line Construction for
Multi-Redundant DOF Beam
Using Maxwell’s reciprocal principle, (3) is
B f C 0
f f B f C 0
+ =
+ + = (4)
which facilitates the solution of the problem.
ng
rewritten as
XB Bf f+ B y BC y
XC CB y CC y
Thus, the unit load needs to be placed succes-
sively only at points B and C and the deflec-
tions fXB and fXC at a number of points X alo
the beam are computed.
Procedure for Analysis
tic indetermi-
2. sing Max-
3. ength
4. for the redun-
ce
1. Determine the degree of sta
nacy and select the redundants.
Solve the redundant equations u
well’s law of reciprocal deflections.
Select a number of points along the l
of the structure at which the numerical
values of the ordinates of the influence
lines will be evaluated.
Once the influence lines
dants have been determined, the influen
lines for other response functions can be
generated using static equilibrium.
61
Beam Example
Using the force method of analysis, calculate the ordinates for the positive internal moment at C.
(a) Beam Structure
(b)Qualitative Influence Line Diagram
EI = constant
62
This beam structure is the same as the two redundant dof qualitative example. Thus, taking By and Dyas the redundant forces (see Ghali and Neville, 2003 – “Prismatic Member Displacements”):
BB 1
2 2
f f ( 80 ', x 24 ', b 24 ', P 1)
(80 24) 242(80)(24) (24) (24)
6(80) EI
7526.4
EI
BD DB 1
2 2
f f f ( 80 ', x 56 ', b 24 ', P 1)
24(80 56)2(80) (56) (56) (24)
6(80) EI
6297.6
EI
63
DD 1
2 2
BB
f f ( 80 ', x 56 ', b 56 ', P 1)
(80 56)562(80)(56) (56) (56)
6(80)EI
7526.4f (due to symmetry)
EI
XB 1
2 2
2
f f ( 80 ', b 24 ', P 1)
(80 24) x2(80)(24) (24) x
6(80)EI
7x3264 x for x b
60EI
XB 1
2 2
2
f f ( 80 ', b 24 ', P 1)
24(80 x)2(80)x (24) x
6(80)EI
(80 x)576 160x x for x b
20EI
64
XD 1
2 2
2
f f ( 80 ', b 56 ', P 1)
(80 56) x2(80)(56) (56) x
6(80) EI
x5824 x for x b
20EI
XD 1
2 2
2
f f ( 80 ', b 56 ', P 1)
56(80 x)2(80)x (56) x
6(80) EI
7(80 x)3136 160x x
60EIfor x b
65
1
y
y
XB
XD
R [F] {D(x)}
B EI
D 16,986,931.2
f7526.4 6297.6
f6297.6 7526.4
y XB XD
XB XDy
B 4.4307 f 3.70732fEI3.70732f 4.4307 fD 10,000
y y yA
y y y y y
1M 0 E x 24 B 56 D
80
F 0 A 1 B D E
From equilibrium:
66
Influence Lines forSupport Reactions
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80
ByDyAyEy
67
Influence Line for Mc
-2
-1
0
1
2
3
4
5
6
0 8 16 24 32 40 48 56 64 72 80
Mc
(ft-
kip
s)
C y y
C y y
M 40A 16B (40 x); 0 x 40
M 40A 16B ; 40 x 80
From equilibrium:
68
Truss Example
See example 15.4 in your textbook.
1x
C D
E F
A B
15’
3 @ 20’ = 60’
69
Primary Structurew/ Loading
1x
C D
E F
A B
15’
3 @ 20’ = 60’
70
Primary Structurew/ Unit Redundant
Load
1
C D
E F
A B
15’
3 @ 20’ = 60’
1
71
Mem L A FV FVL/A
AB 240" 6 0 0
AE 300" 6 0 0
BC 240" 6 -4/5 -32
BE 180" 4 -3/5 -27
BF 300" 4 1 75
CD 240" 6 0 0
CE 300" 4 1 75
CF 180" 4 -3/5 -27
DF 300" 6 0 0
EF 240" 6 -4/5 -32
Truss Example Calculations
72
Mem FB FC
AB -8/9 -4/9
AE 10/9 5/9
BC -4/9 -8/9
BE -2/3 -1/3
BF -5/9 10/9
CD -4/9 -8/9
CE 0 0
CF 0 -1
DF 5/9 10/9
EF 8/9 4/9
73
mVi i
CE,CE Viii 1
F L1f F
E A
mB BVi iCE i
ii 1
F L1D F
E A
mC CVi iCE i
ii 1
F L1D F
E A
i iCE,CE CE CEf F D 0;
i B, C
74
Mem
AB 0 0 0
AE 0 0 0
BC 25.6 128/9 256/9
BE 16.2 162/9 81/9
BF 75 -375/9 750/9
CD 0 0 0
CE 75 0 0
CF 16.2 0 243/9
DF 0 0 0
EF 25.6 -256/9 -128/9
CCEE DB
CEE DCE,CEE f
75
CE,CE1
f (25.6 16.2 75E
16.2 25.7 6)
233.
E
5
6
BCE
1 128 162 375 256D
E 9 9 9 9
341
E
CCE
1 256 81 750 243 256D
E 9 9 9 9 9
1202
E
76
ii CECE
CE,CE
BCE
CCE
DF ; i B, C
f
F 1.46
F 5.15
77
Calculate forces in members BE, BC and EF. Note: Member forcesAB, AE, DC, and DF are statically determinate for this truss structure as are the support reactions.
1x
C D
E F
A B
15’
3 @ 20’ = 60’