CE 382 L11 - Force Method
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Transcript of CE 382 L11 - Force Method
Indeterminate AnalysisForce Method1
• The force (flexibility) method• The force (flexibility) method expresses the relationships between displacements and pforces that exist in a structure.
• Primary objective of the force method is to determine the chosen set of excess unknown f d/ lforces and/or couples –redundants.
• The number of redundants is• The number of redundants is equal to the degree of static indeterminacy of the structure.
1
y1Also see pages 115 – 141 in your class notes.
Description of the Force Method ProcedureMethod Procedure
1. Determine the degree of static gindeterminacy. Number of releases* equal to the degree of static indeterminacy are applied to the structure.Released structure is referred to as the primary structure. P i b hPrimary structure must be chosen such that it is geometrically stable and statically determinateand statically determinate.Redundant forces should be carefully chosen so that the primary
2
carefully chosen so that the primary structure is easy to analyze
* Details on releases are given later in these notes.
Force Method – con’t
2. Calculate “errors” (displacements) at th i t t d d tthe primary structure redundants. These displacements are calculated using the method of virtual forces.g
3. Determine displacements in the primary structure due to unit values of redundants (method of virtual forces) These displacements areforces). These displacements are required at the same location and in the same direction as the di l t d t i d idisplacement errors determined in step 2.
3
Force Method – con’tForce Method con t
4 Calculate redundant forces to4. Calculate redundant forces to eliminate displacement errors. Use superposition equations inUse 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 thenumber of releases) that express the fact that there is zero relative displacement at each release.
4
p
Force Method – con’tThese compatibility equations guarantee a final displaced shapeguarantee a final displaced shape consistent with known support conditions, i.e., the structure fits co d o s, .e., e s uc u e stogether 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 theforces (redundants) and forces on the released structure.
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Flexibility Analysis
(1)
R1R2
=
(2)
+
D1D2
(2)
f21 (x R1)f11 (x R1)
(3)
1 (R1)
+f (x R )
6
1 (R2)f12 (x R2)
f22 (x R2)(3)
f R f R D 0+ + =11 1 12 2 1(4
f R f R D 0)
+ + =
21 1 22 2 2
(4f R f 0)
R D+ + =
Solve for R and RSolve for R1 and R2.
Using matrix methods:g
[F] {R} = -{D}
⇒ {R} = -[F]-1 {D}
7
f f⎡ ⎤[F] = 11 12
21 22
f ff f
⎡ ⎤⎢ ⎥⎣ ⎦
≡ flexibilitymatrix
[F]-1 (≡ inverse flexibility matrix)[F] (≡ inverse flexibility matrix)
22 12f f1 −⎡ ⎤22 12
21 1111 22 12 21
f f1f ff f f f
⎡ ⎤= ⎢ ⎥−− ⎣ ⎦
1D{D}
⎧ ⎫= ⎨ ⎬
≡ primary2
{D}D⎨ ⎬
⎩ ⎭ structuredisplacement
t8
vector
1R⎧ ⎫≡ redundant
f1
2
R{R}
R⎧ ⎫
= ⎨ ⎬⎩ ⎭
forcevector
1 22 1 12 2R f D f D1R f D f Ddet[F]
−⎧ ⎫ ⎧ ⎫−=⎨ ⎬ ⎨ ⎬+⎩ ⎭ ⎩ ⎭
det [F] =
2 21 1 11 2R f D f Ddet[F] − +⎩ ⎭ ⎩ ⎭
11 22 12 21f f f f−det [F]
(5)
11 22 12 21f f f f
(5)
With R1 and R2 known, remaining structure is
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remaining structure is statically determinate.
ReleasesReleases
Release is a break in the continuity ofRelease is a break in the continuity of the elastic (displacement) curve.
One release only breaks a single type of continuity.y
Figure 1 shows several types ofFigure 1 shows several types of releases.
Common release is the support reaction, particularly for continuous
10beams.
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Flexibility Equations y qPrimary structure displacements at th l l t d t ththe releases are related to the unknown redundant forces via
i ij jD f R− = (1)
displacement at release i due i jf ≡to a unit force in the direction of and at release j; flexibility coefficients
j
coefficients.
Equation 1 for the case of three redundant forces is expressed as
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redundant forces is expressed as
1 11 1 12 2 13 3D f R f R f R− = + +1 11 1 12 2 13 3
2 21 1 22 2 23 3
D f R f R f RD f R f R f R
= + +
− = + + (2a)
3 31 1 32 2 33 3D f R f R f R− = + +
Matrix form of (2a)
-{D} = [F] {R} (2b)
1D⎧ ⎫{D} = = <D1 D2 D3>T
1
2
DD
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪
= displacement vector at the3D⎪ ⎪
⎩ ⎭
13
predundant degrees of freedom
1R⎧ ⎫⎪ ⎪
{R} = = <R1 R2 R3>T1
2RR
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
= redundant force 3R⎪ ⎪
⎩ ⎭
vector
11 12 13f f f⎡ ⎤⎢ ⎥
21 22 23f f ff f f
⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
[F] =
31 32 33f f f⎢ ⎥⎣ ⎦= flexibility matrix
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y
Displacement Calculations –Displacement Calculations Method of Virtual Forces
D F d M dφli Vi ViD F d M d= + φl
subscript i direction of Ri at⇒(3)
release i
dl = differential axial displacementp
dφ = differential rotational displ
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Flexibility Coefficients –Method of Virtual ForcesMethod of Virtual Forces
a bij ij ijf f f= + (4)ij ij ijf f f+ ( )
VjFVjaij Vi
Ff F dx
EA(x)=
≡ axial deformation influence coefficient
Vjbij Vi
Mf M dx=ij Vif M dx
EI(x)≡ bending deformation
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≡ bending deformationinfluence coefficient
Force Method Examples
1 Calculate the support reactions1. Calculate the support reactions for the two-span continuous beam, EI = constant. w
L L
w
=
Primary Structure w/ Load
+
171 (x R1)Primary Structure w/ Redundant
2. Calculate the support reactions ppfor the two-span continuous beam, EI = constant.
w
L L
w
=
Primary Structure w/ Load
+
R2
+
R1
18
R2R1
Primary Structure w/ Redundant Forces
Prismatic Member Displacements
19
20
21
3. Calculate the support reactions 3 Ca cu ate t e suppo t eact o sfor the two-span continuous beam using the internal moment at B as the redundant force, IAB = 2I and IBC = I; E = constant.
PL2
Primary Structure w/ Loading22
Primary Structure w/ Loading
MB
i S / d dPrimary Structure w/ Redundant
DB = __________________
fBB = BB _________________
MB = _________________
23
4. Calculate the bar forces for th t ti ll i d t i tthe statically indeterminate truss.
St ti ll StaticallyStatically Indeterminate
Truss
Statically Determinate
Released Truss24
Truss(Redundant X)
Truss CalculationsMem L (in) F FV FVFL
192" 40 4/5 6144AB 192 40 -4/5 -6144
BC 144" 0 -3/5 0
CD 192" 0 -4/5 0
DA 144" 30 3/5 2592DA 144" 30 -3/5 -2592
AC 240" 0 1 0
BD 240" -50 1 -12000
VF FLF /f 20736/82944⎡ ⎤⎢ ⎥∑
25
VAC AC,ACF /f 20736/829.44
E A= − =⎢ ⎥
⎣ ⎦∑= 25 kips
Nonmechanical LoadingLoading
[F]{R} ({D} {D })Δ=− + (5)[F]{R} ({D} {D })= + ( )
T2 n1{D } D D DΔΔ Δ Δ= < >K2 n1{ } K
= relative dimensional change displacements calculated usingdisplacements calculated using principle of virtual forces
Displacements due to dimension changes are all relative displace-
t ll di l tments, as are all displacements corresponding to releases. They are positive when they are in the same
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positive when they are in the same vector direction as the corres-ponding release (redundant).
Structure ForcesStructure Forces
Once the redundant forces arel l d f ( ) ll hcalculated from Eq. (5), all other
support reactions and internalmember forces can be calculatedusing static equilibrium along withthe appropriate free body diagrams.
Thi i ibl i h fThis is possible since the forcemethod of analysis has been used todetermine the redundant forces orthe forces in excess of those
27required for static determinacy.
Mathematical E iExpressions
Calculation of the non-redundant f A ( t ti i t lforces Ai (support reactions, internal shears and moments, truss member forces) can be expressed usingforces) can be expressed using superposition as
RN
p
Rpi ui j ji
j 1A A (A ) R
== + ∑ (6)
piA
j 1where = desired action Ai on the primary structure due to the applied loading; = action Ai on the primary structure due to
ui j(A )
28a unit virtual force at redundant Rj and NR = number of redundants.
Example Beam Problem –Nonmechanical Loading
E 30 000 k iE = 30,000 ksiI = 288 in4
(a) Given structure
29(b) Primary structure
The interesting point of this example is that the flexibility equation will have a nonzero right hand side since the
d d t di l t i ib dredundant displacement is prescribed to equal 0.72” downward. Thus the flexibility equation isflexibility equation is
fBB RB = dB - (7)BDΔ
wheredB = prescribed displacementB p p
at redundant B= -0.72" since RB is
positive upward= -0.24"BDΔ
30relative displacementat redundant BB Bd DΔ− =
Truss Example –pNonmechanical Loading
For the truss structure on the next page compute thenext page, compute the redundant bar EC member force if the temperature in bar EF isif 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.α
31
Truss Example3 @ 20’ = 60’
CB
3 @ 20 60
DA 15’
E F
B C
A 1 D
E F
Primary Structure SubjectedPrimary Structure Subjectedto FCE = 1
32
Truss Example CalculationsMem L FV FVL
AB 240" 0 0AB 240" 0 0
AE 300" 0 0
BC 240" -4/5 -192
BE 180" -3/5 -108
BF 300" 1 300
CD 240" 0 0
CE 300" 1 300
CF 180" -3/5 -108CF 180 -3/5 -108
DF 300" 0 0
EF 240" 4/5 19233
EF 240" -4/5 -192
mCE CE Vi Vi i
1f F F L= ∑CE,CE Vi Vi ii 1
f F F LEA =
= ∑
mCE Vi iD FΔ Δ= δ∑
i 1=
T L 0 072"Δδ αΔEF EF EF
BF BF
T L 0.072"
0 3"Δ
δ = αΔ =
δ = Δ = −
Δ
BF BF 0.3δ = Δ = −
CE,CE CE CEf F D 0Δ+ =
34
Displacement C l l ti
Displacements for the statically
Calculations
Displacements for the staticallyindeterminate structure can be
l l t d i th t bcalculated using the exact memberdeformations for a truss or exactshear and moment expressionsalong with the virtual force expres-sions on the primary structure.
F t t t l l tiFor a truss structure, calculation of a joint displacement using theprinciple of virtual forces results in
Δ
35
principle of virtual forces results in
mm
Vi iF Δδ + Δ∑1 (Δ) =
i 1=∑
minti i
Vi iF LFEA
Δ⎛ ⎞+ δ + Δ⎜ ⎟
⎝ ⎠∑= (8)Vi ii
i 1EA
=
⎜ ⎟⎝ ⎠∑
ViF = primary structure member forces due to the application of ppa unit virtual force at the joint for which the displacement i d i d d i h di i
Δ
36
is desired and in the direction of Δ
ΔΔ = primary structure displaceΔ = primary structure displace-ment at desired displacement due to nonmechanical effects
iδ = exact member displacements i that are obtained for the sta-tically indeterminate structure
i th l l t d d d tusing the calculated redundant forces to determine all the member forces within the trussmember forces within the truss structure
intδintiδ = member displacements due
to nonmechanical loading on
37the member
For a frame structure, in which shear and axial deformations are ignored, the displacements are calculated asp
1( )ΔLm
iiMΔ ⎛ ⎞∑ ∫1( )Δ = intiVi i
ii 1 0
MM dxEI
Δ
=
⎛ ⎞+ κ⎜ ⎟
⎝ ⎠∑ ∫i 1 0=
(9a)Δ+ Δ
1( )θ =
Lminti
Vi iMM dxθ ⎛ ⎞
+ κ⎜ ⎟∑ ∫1( )θ = Vi ii
i 1 0
M dxEI
=
+ κ⎜ ⎟⎝ ⎠∑ ∫
Δ38(9b)Δ+ θ
M MΔ θVi ViM , MΔ θ = primary structure virtual
moments based on the desired displacement Δ ordesired displacement Δ or rotation θ
,Δ ΔΔ θ = primary structure displace-ments at Δ or rotation θ duements at Δ or rotation θ due to environmental loads or causes
intiκ = primary structure initialiκ 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 ysubjected to the external loads with the redundant forces known fromthe redundant forces known from the flexibility analysis.
Equations (8), (9a), and (9b) are cor-rect only because exact real memberrect only because exact real member forces are used in the calculation of the desired displacementsthe desired displacements.
40
Calculate the horizontal displace-ment at joint B for the staticallyment at joint B for the statically indeterminate truss.
16’40 k C
D 16
A BR1
12’
A
CD
BA 1
i S S bj d41
Primary Structure Subjected to Virtual Loading
Calculate the rotation at the center support for the two-span continuous beam, EI = constant.
w
L LR
1
R1 R2
P i St t / Vi t l
L L
Primary Structure w/ Virtual Load at Desired Displacement
Location
42
Alternatively, you can express the desired displacementthe desired displacement calculations also in matrix form following the usual superpositionfollowing the usual superposition process of the force method of analysis:analysis:
(10){ } [F ]{R} {D } {D }Δδ δ δδ = + +
where {δ} = vector of desired displacements; ={D }, {D }Δ
δ δdisplacements; vector of desired displacements for the primary structure for both
{D }, {D }δ δ
p ymechanical and non-mechanical loadings, respectively; [Fδ] =
43
g p y [ δ]
matrix of displacement influence coefficients at the desiredcoefficients at the desired displacement locations due to unit values of the redundant forcesvalues of the redundant forces {R}. Stated mathematically, the coefficients of [Fδ] arecoefficients of [Fδ] are
(11)jij i R 1Fδ == δ (11)
which simply states that the dis-
jj R 1
p yplacement influence coefficients equal the displacement at desired q pdisplacement i on the primary structure due to a unit force at
44redundant j on the primary structure.