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Introduction to StaticallyIndeterminate Analysisndeterminate Analysisu
forces of statically determinate
using only the equations of
equilibrium. However, theanalysis of statically indeter-minate structures requires
the geometry of deformation of
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Additional equations come fromcompatibility relationships,
displacements throughout the
structure. The remaininequations are constructed frommember constitutive equations,.e., re at ons ps etween
stresses and strains and the
over the cross section.
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Design of an indeterminates ruc ure s carr e ou n aniterative manner, whereby the
members are initially assumed
and used to analyze the structure.Based on the computed results(displacements and internalmem er orces , e mem er
sizes are adjusted to meet.iteration process continues untilthe member sizes based on the
results of an analysis are close tothose assumed for that analysis.
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Another consequence of
structures is that the relativevariation of member sizesinfluences the magnitudes ofthe forces that the member
.another way, stiffness (largemember size and/or hi hmodulus materials) attracts
force.
Despite these difficulties withstatically indeterminate
structures, an overwhelmingmajority of structures being
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u o ay are s a ca yindeterminate.
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Advantages Statically
n e erm na e ruc ures
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Statically indeterminate
structures typically result insmaller stresses and greater
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s ness sma er e ec onsas illustrated for this beam.
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u
if middle support is removed
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ystructures introduce redundancy,
one part of the structure will notresult in catastrophic or collapse
8failure of the structure.
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Disadvantages of
a ca y n e erm na eStructures
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Statically indeterminate structure-
settlement, which producesstresses as illustrated above.
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Statically indeterminate struc-
tures are also self-strained dueto temperature changes and
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a r ca on errors.
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Indeterminate Structures:
n uence nesInfluence lines for staticall
indeterminate structuresprovide the same informationas influence lines for staticallydeterminate structures, i.e. it
response function at aarticular location on the
structure as a unit load movesacross the structure.
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Our goals in this chapter are:
1.To become familiar with theshape of influence lines for thesupport react ons an nternaforces in continuous beams
.2.To develop an ability to sketch
influence functions for
indeterminate beams andframes.
3.To establish how to positiondistributed live loads oncontinuous structures to
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values.
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Qualitative Influence
nes or a ca y n e-terminate Structures:
-
In many practical applications, itis usually sufficient to draw onlythe qualitative influence lines toec e w ere o p ace e ve
loads to maximize the response.Muller-Breslau Principle pro-vides a convenient mechanism
to construct the qualitativeinfluence lines, which is stated
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:
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The influence line for a force (ormoment res onse function is
given by the deflected shape of
the released structure by
remov ng e sp acemenconstraint corresponding to the
from the original structure and
giving a unit displacement (orrotation) at the location and in
the direction of the response
unc on.
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Procedure for constructing
qualitative influence lines forindeterminate structures is: (1)remove rom e s ruc ure erestraint corresponding to the
,apply a unit displacement orrotation to the released structureat the release in the desired
response function direction, andw u v
shape of the released structure
support and continuityconditions.
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Notice that this rocedure is
identical to the one discussed forstatically determinate structures.
However, unlike staticallydeterminate structures, theinfluence lines for staticallyindeterminate structures arey y u v .
Placement of the live loads tomaximize the desired response
function is obtained from thequalitative ILD.
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Uniforml distributed live
loads are placed over thepositive areas of the ILD tomaximize the drawn responsefunction values. Because the
diminish rapidly with distancefrom the res onse functionlocation, live loads placed more
than three span lengths awaycan e gnore . nce t e veload pattern is known, an
structure can be performed todetermine the maximum value of
18the response function.
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19QILD for RA
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20QILDs for RC and VB
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QILDs for (MC)-,M + and R
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Live Load Pattern to
Maximize Forces inMultistory Buildings
Building codes specify thatmembers of multistorybuildings be designed tosupport a uniformly distributed
load of the structure. Deadn liv l r n rm ll
considered separately sincethe dead load is fixed inposition whereas the live loadmust be varied to maximize a
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of the structure. Such
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maximum forces are
typically produced bypatterned loading.
Qualitative Influence Lines:
. displacement at the desiredres onse function location.
2. Sketch the displacement
dia ram alon the beam orcolumn line (axial force incolumn) appropriate for theun sp acemen an
assume zero axial
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.
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3. Axial column force (do notconsider axial force in beams):
(a) Sketch the beam linequalitative displacementdiagrams.
b Sketch the column linequalitative displacement
diagrams maintaining equalityo e connec on geome rybefore and after deformation.
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4. Beam force:
(a) Sketch the beam lineualitative dis lacement
diagram for which the releasehas been introduced.
(b) Sketch all column linequalitative displacement
connection geometry before.
the column line qualitativedisplacement diagrams from
the beam line diagram of (a).
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c Sketch remainin beam
line qualitative displacementdiagrams maintaining con-nection geometry before andafter deformation.
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VerticalReaction F
Load Pattern toMaximize F
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M
Maximize M
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QILD and Load Pattern for
Center Beam Moment M
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M
QILD and Load Pattern for
End Beam Moment M
Ex anded Detailfor Beam EndMoment
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Envelope Curves
Design engineers often usen uence nes o cons ruc s earand moment envelope curves for
for bridge girders. An envelopecurve defines the extremeboundary values of shear or
bending moment along the beamue o cr ca p acemen s o
design live loads. For example,-
continuous beam.
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Qualitative influence lines for
positive moments are given,shear influence lines arepresented later. Based on thequalitative influence lines, critical
determined and a structuralanal sis com uter ro ram canbe used to calculate the member
end shear and moment valuesor t e ea oa case an t ecritical live load cases.
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a b c ed
1 2 3 4
-
a b c ed
QILD for (Ma)+
a b c ed
1 2 3 4
+
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a b c ed
1 2 3 4
+ c
a b c ed
QILD for (M )+
a b c ed
1 2 3 4
34
e
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a b c ed
1 2 3 4
Critical Live Load Placement
+ a
a b c ed
1 2 3 4
Critical Live Load Placementfor (Ma)-
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a b c ed
1 2 3 4
Critical Live Load Placement
+ b
a b c ed
1 2 3 4
Critical Live Load Placementfor (Mb)-
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a b c ed
1 2 3 4
Critical Live Load Placement
+ c
a b c ed
1 2 3 4
Critical Live Load Placementfor (Mc)-
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a b c ed
1 2 3 4
Critical Live Load Placement
+ d
a b c ed
1 2 3 4
Critical Live Load Placementfor (Md)-
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a b c ed
1 2 3 4
Critical Live Load Placement
+ e
a b c ed
1 2 3 4
Critical Live Load Placementfor (Me)-
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Calculate the moment envelope
curve for the three-spancontinuous beam.
a b c ed
1 2 3 4
L L L
L = 20 = 240,
A = 60 in2
= 4
wDL = 1.2 k/ft dead load
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LL = .
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Shear and Moment Equations
for a Loaded SpanqMi
xi
ie
V = V x
Viie
Mie = -Mi + Vi xi 0.5q (xi)
2
Shear and Moment Equationsfor an Unloaded Span
=
Vie = Vi
41Mie = -Mi + Vi xi
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Load Cases
wDL
a b c ed
wLL wLL
a b c ed
wLL
a b c ed
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wLL
a b c ed
LC4
a b c ed
wLL
1 2 3 4wLLLC5
a b c ed
LL
1 2 3 4
w
LC6
a b c ed
431 2 3 4LC7
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A summar of the results from
the statically indeterminate beamanalysis for each of the sevenload cases are given in yourclass notes.
----- RESULTS FOR LOAD SET: 1***** M E M B E R F O R C E S *****
MEMBER AXIAL SHEAR BENDINGMEMBER NODE FORCE FORCE MOMENT
(kip) (kip) (ft-k)
1 1 0.00 9.60 0.002 -0.00 14.40 -48.00
. . .
3 -0.00 12.00 -48.00
3 3 0.00 14.40 48.00
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4 -0.00 9.60 0.00
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The equations for the internal
shear and bending moments foreach span and each load caseare:
Load Case 1
V = 9.6 1.2xM12 = 9.6x1 0.6(x1)
2
= M23 = -48 + 12x2 0.6(x2)
2
34 . . 3
M34 = -48 + 14.4x3 0.6(x3)2
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Load Case 2
12 = . . x1M12 = 43.2x1 2.4(x1)2
V23 = 0M23 = -96
V34 = 52.8 4.8x3M34 = -96 + 52.8x3 2.4(x3)
2
Load Case 3
-12 .M12 = -4.8x1
23 = . 2M23 = -96 + 48x2 2.4(x2)2
4634 = .
M34 = -96 + 4.8x3
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Load Case 4
12 = . . x1M12 = 41.6x1 2.4(x1)2
V23 = 8M23 = -128 + 8x2
V34 = -1.60M34 = 32 - 1.6x3
Load Case 5
12 .M12 = 1.6x1
23 = -
M23 = 32 - 8x2
4734 = . . x3
M34 = -128 + 54.4x3 2.4(x3)2
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Load Case 6
V12 = 36.8 4.8x1M12 = 36.8x1 2.4(x1)2
V23 = 56 4.8x2M23 = -224 + 56x2 2.4(x2)
2
V34 = 3.2M34 = -64 + 3.2x3
Load Case 7
-12 .M12 = -3.2x1
23 = . 2M23 = -64 + 40x2 2.4(x2)2
4834 = . . x3
M34 = -224 + 59.2x3 2.4(x3)2
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Bending Moment Diagram LC1
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Live Load E-Mom (+) Live Load E-Mom (-)
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A spreadsheet program listing is
gives the moment values alongthe span lengths and is used tograph the moment envelopecurves.
In the spreadsheet:
ve oa - om +
= max (LC2 through LC7)- -= min (LC2 through LC7)
- =
+ Live Load E-Mom (+)Total Load E-Mom - = LC1
54+ Live Load E-Mom (-)
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Total Load E-Mom (+) Total Load E-Mom (-)
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Construction of the shearenve ope curve o ows e sameprocedure. However,just as is
envelope, a complete analysisshould also load increasing/decreasing fractions of the span
where shear is being considered.
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1
1 2 3 4
QILD (V1)+
a b c ed
-1
QILD (V2L
)+
a b c ed
1 2 3 4
ILD V R +
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a b c ed
1 2 3 4
-1 ILD V L +
1
a c e
1 2 3 4
QILD (V3R)+
a b c ed
1 2 3 4
-1QILD (V )+
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Shear ILD Notation:Su erscri t L = ust to the left of
the subscript point
Su erscri t R = ust to the ri htof the subscript point
To obtain the negative shear
qualitative influence line dia-grams s mp y p t e rawnpositive qualitative influence line
.
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,
exact shear envelope is usuallyunnecessary since an approximateenvelope obtained by connectingthe maximum possible shear at the
w x upossible value at the center of the
.course, the dead load shear must
be added to the live load shearenvelope.