pr - shear wall analysis
TRANSCRIPT
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Shear wall structures
1. Introduction
Shear walls are often continuous down to based of building to forma vertical cantilever
Shear walls behave predominately in bending instead of shear(in spite of its name)
The floor slab usually does not have large enough out-of-planestiffness to make the walls deform as a group. As a result, each
wall bends individually with its own neutral axis.
Examples of shear walls: lift shaft, stairwell, structural wallpartition
Commonly suited for buildings up to about 35 stories
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A simple problem: Single wall
The wall is modeled as a cantilever fixed at its base, as shown in
Figure 1.
The governing differential equation is the familiar one from beambending theory:
EI
xqxy
)()(
)4( = (1)
Figure 1
QUIZ:Say whether each of the following is assumed in equation (1).
1)Linear stress-strain relationship
2)Elastic material
3)Plane-section remains plane in bending
4)Small deformation
EXERCISE:
Derive (1) from scratch.
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To make things simple, lets solve this DE for uniformly distributed
load, i.e., =)(xq constant not dependent on x .
To do that, we need to integrate (1) four times (why?). This will result
in 4 integration constants, which can be determined by requiring )(xy
to satisfy the boundary conditions of a cantilever:
a) 0)0( =y
b) 0)0(' =y
c) 0)()2( =Hy
d) 0)()3( =Hy
QUIZ:What do the boundary conditions a)-d) mean?
The resulting solution for )(xy is
1 2222 32222 45Hx
H
x
H
x
H
x
EI
qHxy
/withVariation
])(3
1
3
42[)(
m)(e.g.,ntdisplacemeofunitithquantity w
8)(
224
+= (2)
DO NOT MEMORIZE FORMULA WITHOUT UNDERSTANDING
IT
QUIZ:
1.What is the deflection at the top of the cantilever?
2.What is the highest power of Hx/ appearing in (2)?
EXERCISE:
1. Verify that )(xy given by (2) satisfies the boundary conditions
1)-4)
2. Sketch the deflection shape of the cantilever.
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The moment )(xM and shear force )(xV of a section at x from the base
are given by:
)('')( xyEIxM = (3))()(
)3( xyEIxV = (4)
QUIZ:Say whether each of the following is assumed in equations (3)
and (4).
1)Linear stress-strain relationship
2)Elastic material
3)Plane-section remains plane in bending
4)Small deformation
EXERCISE:
Sketch the moment and shear diagram.
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2. Basic considerations
When there are more than one wall acting together, the lateralload is distributed among the walls.
The amount of load shared by each individual wall depends ontheir stiffness.
Wind blows, pushes the external wall of building. Wind load is
transferred to floor slabs, then from slabs to shear walls, eventually
from shear walls to the base.
Two important phenomena:
Proportionate VS non-proportionate structures
Proportionate: the ratio of flexural stiffness among the walls is
constant with height.
Twisting vs non-twisting deformation
When either the load distribution or lateral stiffness of structure is not
symmetric in plan, twisting or torsional deformation will occur, in
addition to translational deformation.
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Non-twisting Twisting
Proportionate 1. Quantitative by handcalculation
2. Quantitative by
hand
Non-proportionate 3. Qualitative or
Quantitative by FEM(equivalent 2-D)
4. Qualitative or
3-D FEM
QUIZ:Decide whether the following structures are proportionate or
non-proportionate. Also decide whether they will twist whensubjected to the load indicated.
Figure 2
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3. Proportionate non-twisting structures
Assumptions:
1)Shear walls deform in flexural (bending).
2)Rigid floor assumption - Slab is rigid in-plane. This implies
a)If there is no twisting deformation, the lateral displacement of
all walls will be the same
b)If there is twisting, the displacement of any point on the slab can
be described in terms of a common translational and rotational
component (see later)
Equivalent 2-D models
Assumption 2a allows us to use an equivalent 2-D model to study ashear-wall structure (which is originally a 3-D problem):
OR
OR
Figure 3
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Notes to equivalent 2-D model:
1)The rigid link means that the displacement of two walls at the same
floor are the same
2)The rigid link is hinged at the two walls so it does not provide anybending resistance
3)The position of the wall is immaterial (why?)
4)Which wall the external lateral load acts on is immaterial (why?)
5)The walls bend individually, although they all have the same
displacement at any given level. This means each wall has its own
neutral axis, rather than having a common neutral axis for a groupof walls (c.f. tubular structures later). For example:
Figure 4: Wall bending with strain and neutral axis shown
QUIZ:Draw an equivalent 2-D model for the situation in Figure 4.
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Method for computing shear and moment in walls (e.g., Smith &
Coull, p.186):
Distribute shear and moment proportional to the flexural rigidity
of the wall
Why?
The following exercise helps you conclude that:
For proportionate non-twisting structures, the distribution of shear
does not depend on the level.
Note that this does not mean that the shear taken by each wall does
not depend on the level. Only the ratio among them does not.
EXERCISE: Distribution of shear based on a continuum
approach
Figure 6
Referring to the figure, we note that, from beam bending theory, for
Wall 1,
dx
xdMxV
)()( 11 =
(the minus sign is necessary but unimportant; dont let it disturb you)
and)('')()()( 111 xyxIxExM =
which means
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)]()()([)(''
1111xyxIxE
dx
dxV =
Note that )(''')()()( 111 xyxIxExV because )()( 11 xIxE is in general a function
ofx .
If we do the same thing for Wall 2, we have)]()()([)(
''
2222xyxIxE
dx
dxV =
1)Is there any relationship between )(1 xy and )(2 xy ? If so, what is it?
2)By substituting )()()()( 111 xIxEcxIxE = into the expression for )(1 xV and
)()()()( 222 xIxEcxIxE = into the expression for )(2 xV , show that
2
1
2
1
)(
)(
c
c
xV
xV=
and hence the shear distribution does not depend on the level x .3)Hence verify that
constant)()(
)()(
)(
)(
22
11
2
1
2
1 ===xIxE
xIxE
c
c
xV
xV
The previous exercise shows that the shear shared by a given wall at a
given level x is proportional to EI of the wall.
This means that for a proportionate non-twisting building, if we know
the total shear of a give level of the building, we can calculate the
amount of shear shared by the wall by just proportioning based on EI
of the wall.
The same is also true for sharing of moment among the walls (why?).
Suppose the structure is proportionate. There are n walls with EIequal to )(11 xIE , , )(xIE nn . From previous discussions, we know that
the shear force )(xVi taken by Wall i ( ni ,...,1= ) is proportional to
)()()( xIxEcxIE iii = , so we can write)()( xIEKxV
iii=
for some constant K independent of i .
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QUIZ:Does K depend on x ? Why?
Since the sum of the shear taken by the walls must balance the total
shear V at the section of the building, we have
VxVxVxVn
=+++ )()()( 21 6
VxIEKxIEKxIEK nn =+++ )()()( 2211 6
and so
=
=+++
=n
i
iinn xIE
xVxIExIExIE
xVK
1
2211 )(
)()()()(
)(6
Thus, the shear taken by Wall i is given by
1
groupby walltakensheartotal
)(
ifor Wallfactorondistributi
)(
)()(
1
xV
xIE
xIExV
n
i
ii
iii =
= 23245
The same is true for the moment taken by the wall.
YOU DONT NEED TO REMEMBER THIS FORMULA IF YOU
UNDERSTAND ITS MEANING.
The following exercise helps you illustrate that the equivalent EI of
the group of walls is equal to the sum of the EI of the walls.
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Equivalent stiffness of wall group based on a continuum approach
The following shows that the equivalent lateral stiffness of a wall
group is equal to the sum of the stiffness of all the walls.
We first write down the beam equation for each wall individually(note the indices):
)()()()(:Wall
)()()()()(:1Wall
)()()()()(:3Wall
)()()()()(:2Wall
)()()()()(:1Wall
1
)4(
12
)4(
11
32
)4(
33
21
)4(
22
1
)4(
11
xwxyxIxEn
xwxwxyxIxEn
xwxwxyxIxE
xwxwxyxIxE
xwxqxyxIxE
nnn
nnnn
=
=
===
7777
Note that the term )(xq is the external loading and the terms)(),...,(),( 121 xwxwxw n arise from the interaction between the walls.
Summing the above n equations, and noting that the terms
)(),...,(),( 121 xwxwxw n are all canceled in the summation, we obtain
)()()]()()()()()([)4(
2211 xqxyxIxExIxExIxE nn =+++ 6
that is,
=
= n
i
ii xIxE
xqxy
1
)4(
)()(
)()(
Note that this equation is identical to that of a single wall with an
equivalent stiffness equal to =
n
i
xIxE1
11 )()( , and hence we conclude that:
When multiple walls are connected through rigid links, the equivalent
lateral stiffness of the group of walls is equal to the sum of the
individual stiffness.
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The following exercise helps you get a better feel of the above result
based on what you have learnt from elementary beam formulae.
EXERCISE:
Figure 7
1)Give an expression for 1 in terms of P , Q , 1E and 1I (Hint:
look it up from a text book)
2)Give an expression for 2 in terms ofQ , 2E and 2I 3) Note that 21 = (why?). By eliminating Q in the expressions
obtained in 1) and 2), find an expression for 1 (or 2 ) in
terms of P , 1E , 1I , 2E and 2I .
By rearranging the answer for 3), you should be able to get
13
2211
stiffnessequivalent
)(3
+=
22 322 45 H
IEIEP
The middle term is the equivalent stiffness of this group of
walls. Note that if Wall 2 is absent, the stiffness is 311 /3 HIE , and
similarly, if Wall 1 is absent, the stiffness is 322 /3 HIE .
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QUIZ:Assume the walls are of the same material and thickness, and
the lateral loads are the same in both cases. Which one has a
greater horizontal displacement at point A?
(a) (b)Figure 5
QUIZ:According to our shear wall theory, assuming the walls are all
the same, arrange the following wall configurations in ascending
order of lateral stiffness.
Figure 8
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QUIZ:According to our shear wall theory, assuming the walls are all of the
same thickness and material, arrange the following wall
configurations in ascending order of lateral stiffness.
(a) (b) (c)
Figure 9
Walls acting together
When walls are connected (e.g., through concrete and reinforcement),
they act together, providing much greater lateral stiffness.
For example, in Fig. 9(b) above, when the walls are not connected,
the equivalent lateral stiffness of the wall group is just the sum of the
individual stiffnesses, i.e.,
61
6122
122)(
32333
)(9.
bt
b
ttbtbbtEI bFig
+=+= since
b
tis small
However, when the four walls are connected as shown in Fig.9(c),
they act as a section. The approximate equivalent EI should then be
calculated as
)(9.
323
)(9. )(4
6
4
2
2
12
2)(bFigcFig
EIbtb
tbbt
EI ==
+
The above suggests that a significant amount of stiffness can be
gained by couple the walls, or in general lateral systems, together so
that they deform as a whole. Later, we will see one form of structural
system, called tubular structures, which stems out from this idea.
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4. Proportionate shear wall structures with twisting
QUIZ:Decide whether the following structures are proportionate or
non-proportionate. Also decide whether they will twist whensubjected to the load indicated.
Figure 10
Two common situations where twisting will occur:
1)the load distribution is symmetric but the structure (wall
configuration) is not symmetric
2)the load is not symmetric but the structure is symmetric
In general, twisting will occur when the stiffness center does not
coincide with the section resultant center.
Strictly speaking, twisting is NOT a property of a structure. It depends
on BOTH the load pattern and the structural configuration.
However, the load distribution is symmetric in quite many situations
(e.g., unit-directional wind load), and so twisting may often beassociated with the structure, e.g., twisting structure or non-
twisting structure. It is OK to use these terms, but bear in mind that
twisting in general depends on both the loading and structure.
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Method :
E.g., see Smith & Coull, p.188-190
In what follows, we will illustrate WHY and HOW a building twists
in Case 1). Case 2) will be left as an exercise. The general case
follows from superposition.
Consider the portion of a building above a certain level. In general,
the resultant shear acting on a section at any level must balance the
load resultant (in terms of force and twisting moment).
Figure 11
The section resultant originates from the stresses at the sectionof the connecting members (i.e., columns).
The stresses are caused by deformation (strains).
Assuming the floor is rigid in-plane, the variation of thedeformation at different walls must be linear.
Just as a line bxay += can always be written as a constant b plusa linear variation xa , such linear variation of deformation can
always be decoupled into two components
1)translational (i.e., every wall moves by the same amount in
the same direction)
2)rotational (the walls move around a common point by the
same angle).
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Figure 12
Figure 13
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The lateral displacement of each wall causes stresses in thesection, which give rise to a sectional resultant of shear force.
The shear force in the walls can be considered as contributedfrom translational and rotational deformation.
The distribution of the shear forces in the shear walls must be such
that
1)the force is balanced
2)the moment is balanced
The translation contribution of shear force in the walls can be easilydetermined from 1). Since the deformation is pure translational, there
is no twisting, and so we can use the results about proportionate non-
twisting structures, which says that the shear force is distributedproportional to the EI of the wall:
V
IE
IEQ
n
j
jj
iii =
=1
twisting)(no
The less-trivial task lies in the determination of the twisting
component, which is essentially what you need to learn in this section.
First of all, we need to know where the walls rotate about. How can
we determine that? What law/principle/assumption, etc helps us find
that?
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Center of rotation
To invoke your thinking, lets look at what happens if the center of
rotation is (arbitrarily) assumed to be at the left end, that is, all the
walls rotate about the point C in Fig. 12, by an angle clockwise,
say. Then, to the first order, Wall 1 will not translate, Wall 2 willmove by 2x and Wall 3 will move by 3x . Anything wrong?
Figure 14
QUIZ:Anything wrong?
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The center of rotation has to be such that the corresponding
distribution of shear force in the walls must have a zero resultant.
So lets use this principle to find the location of the center of rotation.
In particular, suppose the center of rotation is at a distance x from the
left end, as shown in Figure 13.
Figure 15: Center of rotation.
The displacements at Wall 1, 2, , n will be given by
)()()(11
zxxzy
=,
)()()( 22 zxxzy = ,
)()()( zxxzy nn =
The corresponding shear force in Wall 1, 2, , n will be given by
[ ] [ ])('')()()()()()()( 111''
1111zzIzE
dz
dxxzyzIzE
dz
dzQ == ,
[ ] [ ])('')()()()()()()( 222''
2222zzIzE
dz
dxxzyzIzE
dz
dzQ == ,
[ ] [ ])('')()()()()()()( '' zzIzEdz
dxxzyzIzE
dz
dzQ
nnnnnnn==
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Since the structure is proportionate,
)()()()( zIzEczIzE iii =
for some )()( zIzE which does not depend on i .
Substituting into the expression for )(zVi gives
[ ] )()()('')()()()( zgxxczzIzEdz
dxxczQ iiiii ==
where [ ])('')()()( zzIzEdz
dzg = .
This means, at a given level z , the shear (due to rotation) shared by a
wall is proportional to
the distance of the wall from the center of rotation and the flexural rigidity of the wall (why?)
Summing the shear forces in all the walls and setting it to zero:
0)()(1
==
zgxxcn
i
ii
which yields (since 0)( zg )
=
=
=
= ==n
i
ii
n
i
iii
n
i
i
n
i
ii
zIzE
xzIzE
c
xc
x
1
1
1
1
)()(
)()(
If we recall the definition of the center of mass of a group of masses
1m
, 2m , ,
n
m :
=
==n
i
i
n
i
ii
m
m
xm
x
1
1
then it is natural to call x the center of rigidity.
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QUIZ:What law/principle/assumption/requirement, etc., is used to find
the location of the center of rigidity?
QUIZ:Considering the rotational component of shear forces in the
walls. Is the distribution necessarily linear among the walls?
The following exercise shows that the center of rigidity is indeed the
location where the resultant of the translational component of the
shear forces of the wall system acts.
EXERCISE:Find the location where the resultant translational component ofthe shear forces of all the walls acts and verify that it coincides
with the center of rotation.
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The walls twist about the center of rigidity of the wall system.
Now that we know center of rotation, we can pursue further to find
the amount of shear shared by the walls due to twisting action. Recall
234534523245
rigidityofcenterfrom
iWallofarmmoment
iwallofrigidity
)determinedbe(togivenforconstant
)()(component)l(rotationa
torelated
xxczgQ iii
z
=
So far )(zg is unknown, and we have to determine its value. This is
accomplished by considering the moment equilibrium of the buildingsection.
Referring to Fig. 15 showing the forces acting on a building section.
For convenience the location of the walls are measured from the
center of rotation.
The distance of the load resultant from the center of rigidity is
commonly called eccentricity, and is denoted by e here.
Recall that the center of rotation coincides with the center of rigidity,
and therefore the resultant of the translational component of wallshears passes through the center of rotation. Summing moments about
the center of rotation, we have:
345232452324522 322 45
rotationofcenterfromresultantloadofdistance
sectionatresultantload
rotationofcenterfromarmmoment
iWallofforceshearofcomponentrotational
)()()()(1
ezVxxxxczgi
n
i
ii=
=
After some algebra, we obtain
=
=
n
i
iixxc
ezVzg
1
2)(
)()(
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and so
1
2222 32222 45
factorondistributishear
total1
2
1
2 )()()(
))(()()(
)(
)()(l)(rotationa
==
=
=
n
j
jjj
iii
n
j
jj
ii
i
xxzIzE
exxzIzEzV
xxc
exxczVQ
We are almost done. The shear force taken by each wall is a sum of
the translational and rotational component, that is,
22222 322222 45222 3222 45
componenttwisting
componentnaltranslatio
1
2
1
l)(rotationaonal)(translati
)()()(
))(()(
)()()(
)()()(
(z)(z))(
==
+=
+=
n
j
jjj
iii
n
j
jj
ii
iii
xxzIzE
exxzIzE
zVzIzE
zIzEzV
QQzQ
To help you get a feel for the formula, note that
1)the shear shared by a given wall is a sum of translational and
rotational component
2)the translational component is proportional to EI of the wall
3)the rotational component is related totyeccentriciarmmoment EI
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QUIZ: In each configuration in Table 1, say which of the
following statements is true (note that the statements exhaust all
possibilities):
1)twisting must occur2)twisting must not occur
3)twisting may or may not occur, depending on the actual
dimensions
Structure
Symmetric Not symmetric
SymmetricLoad
Not symmetric
Table 1
QUIZ:
What law/principle/assumption, etc., is used to find the twistingcomponent of shear shared by each wall?
QUIZ:What law/principle/assumption, etc., is used to find the
translational component of shear shared by each wall?
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In summary, to determine the shear shared by each wall in a given
level of a proportionate shear wall structure that may twist under the
applied load:
1)determine the total shear at the level of the building
2)find the translational component of the shear shared by each wall
3)compute the location of the center of rotation, which coincides with
the location where the resultant of the translational components
acts
4)compute the twisting component of shear shared by each wall
5)sum the translational and twisting component of shear to give theshear force shared by each wall
The summary only serves to clarify what we have learnt so far, and
should not be taken as a recipe. The equations involved in the
calculations are deliberately omitted in the summary. You should
have a good idea of what they look like.
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5. Non-proportionate structures
the shear force shared by each wall is not necessarilyproportional to its rigidity, even in the absence of twisting.
the determination of shear shared by each wall requires moresophisticated analysis methods, such as finite element method.
When no twisting occurs, 2-D equivalent models may be used(which requires finite elements)
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Effect of opening at the base
Opening on the edge of wall
333
edge 1
1212
)(
=
=b
dwbdbwI
Opening in the center of wall
++
=
+
+
=
223
23
center
13114
1
12
4212
)2
(
2
b
d
b
d
b
dwb
dbdbw
dbw
I
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
bd/
12/3edge
wb
I
12/3
center
wb
I
Figure 16
As bd/ increases, edgeI decreses in a cubic manner (quite fast!) while
centerI decreases in a much slower fashion, although both correspond to
the same reduction in section shear area. This means that taking out
material in the center will have a less severe effect on the flexural
resistance than from the edge.
The ratio of centerI to edgeI is given by:
+
+=2
edge
center
/1
/131
4
1
bd
bd
I
I
Note that the ratio depends only the the ratio of d to b . As an
illustrative example, if 2/1/ =bd , then 7/ edgecenter =II , that is, opening at
the center rather than at the edge gives 6 times stiffer base!
12/3edge
wb
I
12/3center
wb
I
b
2/d 2/d
b
d
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2-D Finite Element Models
The walls are divided and modelled as beam elements (othertypes of elements are also possible)
The nodes specify the geometrical layout of the model
Elements are formed among nodes
An element is characterized by itso Element type, e.g., beam, plate, shell; specifies the
behaviour of the elemento Connectivity (which and how the nodes form the
element; determines the geometry of the element, e.g.,length, orientation)
o Material/sectional property (e.g., E, I)
Each node associated with a beam element has 3 displacementresponses, or degree-of-freedom (DOF):
o DX: horizontal displacement
o DY: vertical displacement
o DZ: rotation (how much the element has rotated at the
node)
Each node associated with a beam element has 3 elementforces:
o FX: horizontal force
o FY: vertical force
o MZ: moment
Results (displacement, internal force) are computed AT THENODES ONLY
Intuition can help understand sign convention adopted
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Illustrative example (Smith & Coull, p.192)
QUIZ:Is the shearwall structure proportionate? Will it twist?
QUIZ:Draw an equivalent 2-D model for the shearwall structure.
Finite element model for the 2-D equivalent model
Wall 1 Wall 2
1
2
3
20
21
22
23
40
41
42
43
60
1/F
2/F
3/F
20/F
Roof
1
2
3
20
19
21
22
23
40
39
41
42
43
59
60
FY1
FX1M1
FY2
FX2
M2
1st
node
2nd
node
(b) Element force definition
Wall 3
(half)
4/F
(a) Finite element model
y
x
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FEM RESULTS: NODAL DISPLACEMENTS
================================
Node DX (m) DY (m) DZ (rad)
1 2.77e-004 0.00e+000 -1.55e-004
2 1.07e-003 0.00e+000 -2.95e-004
3 2.33e-003 0.00e+000 -4.21e-004
4 4.00e-003 0.00e+000 -5.32e-004
5 6.04e-003 0.00e+000 -6.29e-004
6 8.40e-003 0.00e+000 -7.21e-004
7 1.12e-002 0.00e+000 -8.64e-004
8 1.44e-002 0.00e+000 -9.92e-004
9 1.81e-002 0.00e+000 -1.10e-003
10 2.21e-002 0.00e+000 -1.19e-003
11 2.64e-002 0.00e+000 -1.26e-003
12 3.09e-002 0.00e+000 -1.32e-003
13 3.56e-002 0.00e+000 -1.36e-003
14 4.04e-002 0.00e+000 -1.41e-003
15 4.54e-002 0.00e+000 -1.44e-003
16 5.05e-002 0.00e+000 -1.46e-003
17 5.56e-002 0.00e+000 -1.47e-003
18 6.08e-002 0.00e+000 -1.47e-003
19 6.59e-002 0.00e+000 -1.48e-003
20 7.11e-002 0.00e+000 -1.48e-003
21 2.77e-004 0.00e+000 -1.55e-00422 1.07e-003 0.00e+000 -2.95e-004
23 2.33e-003 0.00e+000 -4.21e-004
24 4.00e-003 0.00e+000 -5.31e-004
25 6.04e-003 0.00e+000 -6.33e-004
26 8.40e-003 0.00e+000 -7.06e-004
27 1.12e-002 0.00e+000 -8.69e-004
28 1.44e-002 0.00e+000 -9.90e-004
29 1.81e-002 0.00e+000 -1.10e-003
30 2.21e-002 0.00e+000 -1.19e-003
31 2.64e-002 0.00e+000 -1.26e-003
32 3.09e-002 0.00e+000 -1.32e-003
33 3.56e-002 0.00e+000 -1.36e-003
34 4.04e-002 0.00e+000 -1.41e-003
35 4.54e-002 0.00e+000 -1.44e-003
36 5.05e-002 0.00e+000 -1.46e-003
37 5.56e-002 0.00e+000 -1.47e-003
38 6.08e-002 0.00e+000 -1.47e-003
39 6.59e-002 0.00e+000 -1.48e-003
40 7.11e-002 0.00e+000 -1.48e-003
41 2.77e-004 0.00e+000 -1.55e-004
42 1.07e-003 0.00e+000 -2.95e-004
43 2.33e-003 0.00e+000 -4.21e-004
44 4.00e-003 0.00e+000 -5.32e-004
45 6.04e-003 0.00e+000 -6.30e-004
46 8.40e-003 0.00e+000 -7.18e-004
47 1.12e-002 0.00e+000 -8.65e-004
48 1.44e-002 0.00e+000 -9.91e-004
49 1.81e-002 0.00e+000 -1.10e-003
50 2.21e-002 0.00e+000 -1.19e-003
51 2.64e-002 0.00e+000 -1.26e-003
52 3.09e-002 0.00e+000 -1.32e-003
53 3.56e-002 0.00e+000 -1.36e-003
54 4.04e-002 0.00e+000 -1.41e-003
55 4.54e-002 0.00e+000 -1.44e-003
56 5.05e-002 0.00e+000 -1.46e-003
57 5.56e-002 0.00e+000 -1.47e-003
58 6.08e-002 0.00e+000 -1.47e-003
59 6.59e-002 0.00e+000 -1.48e-003
60 7.11e-002 0.00e+000 -1.48e-003
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FEM RESULTS: ELEMENT FORCE
==========================
Ele. FX1 (N) FY1 (N) MZ1 (Nm) FX2 (N) FY2 (N) MZ2 (Nm)
1 -6.91e+005 0.00e+000 2.47e+007 6.91e+005 0.00e+000 -2.23e+007
2 -6.49e+005 0.00e+000 2.23e+007 6.49e+005 0.00e+000 -2.01e+007
3 -6.38e+005 0.00e+000 2.01e+007 6.38e+005 0.00e+000 -1.78e+007
4 -5.12e+005 0.00e+000 1.78e+007 5.12e+005 0.00e+000 -1.60e+007
5 -8.12e+005 0.00e+000 1.60e+007 8.12e+005 0.00e+000 -1.32e+007
6 4.74e+005 0.00e+000 1.32e+007 -4.74e+005 0.00e+000 -1.48e+007
7 -2.37e+005 0.00e+000 1.48e+007 2.37e+005 0.00e+000 -1.40e+007
8 -6.85e+005 0.00e+000 1.40e+007 6.85e+005 0.00e+000 -1.16e+007
9 -5.07e+005 0.00e+000 1.16e+007 5.07e+005 0.00e+000 -9.85e+006
10 -4.92e+005 0.00e+000 9.85e+006 4.92e+005 0.00e+000 -8.13e+006
11 -4.50e+005 0.00e+000 8.13e+006 4.50e+005 0.00e+000 -6.55e+006
12 -3.56e+005 0.00e+000 6.55e+006 3.56e+005 0.00e+000 -5.31e+006
13 -4.94e+005 0.00e+000 5.31e+006 4.94e+005 0.00e+000 -3.58e+006
14 -3.65e+005 0.00e+000 3.58e+006 3.65e+005 0.00e+000 -2.30e+006
15 -1.77e+005 0.00e+000 2.30e+006 1.77e+005 0.00e+000 -1.68e+006
16 -1.79e+005 0.00e+000 1.68e+006 1.79e+005 0.00e+000 -1.05e+006
17 -1.30e+005 0.00e+000 1.05e+006 1.30e+005 0.00e+000 -5.99e+005
18 -9.54e+004 0.00e+000 5.99e+005 9.54e+004 0.00e+000 -2.65e+005
19 -5.67e+004 0.00e+000 2.65e+005 5.67e+004 0.00e+000 -6.64e+004
20 -1.90e+004 0.00e+000 6.64e+004 1.90e+004 0.00e+000 -2.38e-00721 -5.11e+005 0.00e+000 1.84e+007 5.11e+005 0.00e+000 -1.66e+007
22 -4.95e+005 0.00e+000 1.66e+007 4.95e+005 0.00e+000 -1.49e+007
23 -4.32e+005 0.00e+000 1.49e+007 4.32e+005 0.00e+000 -1.34e+007
24 -5.40e+005 0.00e+000 1.34e+007 5.40e+005 0.00e+000 -1.15e+007
25 -1.30e+004 0.00e+000 1.15e+007 1.30e+004 0.00e+000 -1.15e+007
26 -1.86e+006 0.00e+000 1.15e+007 1.86e+006 0.00e+000 -4.96e+006
27 -5.41e+005 0.00e+000 4.96e+006 5.41e+005 0.00e+000 -3.07e+006
28 -3.78e+004 0.00e+000 3.07e+006 3.78e+004 0.00e+000 -2.93e+006
29 -1.58e+005 0.00e+000 2.93e+006 1.58e+005 0.00e+000 -2.38e+006
30 -1.11e+005 0.00e+000 2.38e+006 1.11e+005 0.00e+000 -1.99e+006
31 -1.12e+005 0.00e+000 1.99e+006 1.12e+005 0.00e+000 -1.60e+006
32 -8.63e+004 0.00e+000 1.60e+006 8.63e+004 0.00e+000 -1.30e+006
33 -1.21e+005 0.00e+000 1.30e+006 1.21e+005 0.00e+000 -8.74e+005
34 -8.92e+004 0.00e+000 8.74e+005 8.92e+004 0.00e+000 -5.62e+005
35 -4.31e+004 0.00e+000 5.62e+005 4.31e+004 0.00e+000 -4.11e+005
36 -4.37e+004 0.00e+000 4.11e+005 4.37e+004 0.00e+000 -2.58e+005
37 -3.18e+004 0.00e+000 2.58e+005 3.18e+004 0.00e+000 -1.46e+005
38 -2.33e+004 0.00e+000 1.46e+005 2.33e+004 0.00e+000 -6.47e+004
39 -1.38e+004 0.00e+000 6.47e+004 1.38e+004 0.00e+000 -1.62e+004
40 -4.63e+003 0.00e+000 1.62e+004 4.63e+003 0.00e+000 -4.47e-008
41 -8.46e+005 0.00e+000 3.03e+007 8.46e+005 0.00e+000 -2.74e+007
42 -7.99e+005 0.00e+000 2.74e+007 7.99e+005 0.00e+000 -2.46e+007
43 -7.68e+005 0.00e+000 2.46e+007 7.68e+005 0.00e+000 -2.19e+007
44 -6.80e+005 0.00e+000 2.19e+007 6.80e+005 0.00e+000 -1.95e+007
45 -8.02e+005 0.00e+000 1.95e+007 8.02e+005 0.00e+000 -1.67e+007
46 -1.41e+005 0.00e+000 1.67e+007 1.41e+005 0.00e+000 -1.62e+007
47 -6.40e+005 0.00e+000 1.62e+007 6.40e+005 0.00e+000 -1.40e+007
48 -5.90e+005 0.00e+000 1.40e+007 5.90e+005 0.00e+000 -1.19e+007
49 -5.42e+005 0.00e+000 1.19e+007 5.42e+005 0.00e+000 -1.00e+007
50 -4.99e+005 0.00e+000 1.00e+007 4.99e+005 0.00e+000 -8.26e+006
51 -4.36e+005 0.00e+000 8.26e+006 4.36e+005 0.00e+000 -6.73e+006
52 -4.50e+005 0.00e+000 6.73e+006 4.50e+005 0.00e+000 -5.16e+006
53 -1.73e+005 0.00e+000 5.16e+006 1.73e+005 0.00e+000 -4.55e+006
54 -2.28e+005 0.00e+000 4.55e+006 2.28e+005 0.00e+000 -3.75e+006
55 -3.58e+005 0.00e+000 3.75e+006 3.58e+005 0.00e+000 -2.50e+006
56 -2.50e+005 0.00e+000 2.50e+006 2.50e+005 0.00e+000 -1.63e+006
57 -2.05e+005 0.00e+000 1.63e+006 2.05e+005 0.00e+000 -9.09e+005
58 -1.44e+005 0.00e+000 9.09e+005 1.44e+005 0.00e+000 -4.05e+005
59 -8.70e+004 0.00e+000 4.05e+005 8.70e+004 0.00e+000 -1.01e+005
60 -2.89e+004 0.00e+000 1.01e+005 2.89e+004 0.00e+000 1.91e-006
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QUIZ:Say whether each of the following should hold in the FEM
output. Verify your claim using results from the FEM output.
For every floor, sum of wall shears = external shear?
For every floor, sum of wall moments = external moment?
At the base, displacement = rotation = 0?
At every floor, all walls have the same DX?
For each element, FX1+FX2 = 0?
Top and bottom moment at each node should balance?
Top and bottom shear at each node should balance?
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-8 -6 -4 -2 0 2
4
0
2
4
6
8
1012
14
16
18
20
Wall moment kN m
Floor
Wall 1Wall 2Wall 3 (half)
External moment
Wall moment distribution
(piece-wise linear)
-500 0 500 1000 1500 2000 25000
2
4
6
8
10
12
14
16
18
20
Wall shear kN
Floor
Wall 1Wall 2Wall 3 (half)
External shear
Wall shear distribution with floor level
(note the shear at change levels 6 and 13)
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0
5
10
15
20
25
Floor
Wall 1 Wall 2 Wall 3 (half)(including external force)
Force transferred from Floor slab to walls(note that adjacent force pairs do not necessarily balance)
QUIZ:Does the floor slab forces sum to zero at each floor?
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Wall-frame structures
1. Introduction
In low rise structures (e.g., less than 20 stories), shear wallstake the majority of the lateral load
As height of structure increases, the lateral load shared by theframe increases
Economical up to 50 stories
2. Method of analysis
Case Method
Twisting 3-D FEM
Non-twisting 2-D FEM (equiv. 2-D model)
Non-twisting
(assuming simple loading and
structural property)
equiv. 2-D continuum
analytical
3. Equivalent 2-D model
Consider non-twisting cases only
Assume (In-plane) Rigid-floor
Flexural stiffness of lintel beams (that connects the walland the frame) is often (but not always) neglected.
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EXERCISE:
Draw an equivalent 2-D model for the structures in Figures 11.2
(a)-(c) of Smith & Coull (p.256).
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Coupling of wall and frame
QUIZ:
Determine in the following situations.
Figure 1(a)
Figure 1(b)
QUIZ:
Is the equivalent stiffness of a frame-wall group equal to the
sum of the stiffness of the frames and walls? Why?
In the case of proportionate shear-wall structures, the walls bend
in the same manner, or mathematically, the governing
differential equations for the deflection of the different walls are
of the same form. As a result, the stiffness of the wall-group isjust the sum of the individual stiffness, i.e., their stiffness ADD.
Frames and shear walls deform differently, and as a result their
stiffness DO NOT ADD. Thats why we need this chapter.
See Figure 11.3 of Smith & Coull (p.258).
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4. Continuum approach
Why continuum?
To gain insights about the structural behavior, which helpsinterpret the computer results and detect possible errors.
An elegant way of describing the behavior of the frame-wall system in terms of differential equations.
Idealization
Discrete Continuum
Shear wall Flexural beam
Frame Shear beam
Concentrated load
At floor level
Distributed load
along beam
Simplification (see later)
Flexural stiffness of beam (for shear wall) = constant
Shear stiffness of shear beam (for frame) = constant
Distributed load = constant
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3. Behavior of Wall (flexural beam)
The differential equation for the deflection of the wall is that of
a flexural beam:
[ ] )()()()()2(
2
2
zwzyzIzEdzd = (1)
If the rigidity )()( zIzE is constant with height, that is, does not
depend on x , then
EI
zwzy
)()(
)4( = (2)
The boundary conditions for a fixed-free situation are:
a) 0)0( =y
b) 0)0(' =y
c) 0)()2( =Hy
d) 0)()3( =Hy
4. Behavior of Frame (shear beam)
as shear beam, which deform purely in shear (no bending)
o DOES NOT mean that the columns deform in sheardeformation
ojust means that the frame as a whole deforms like a
shear beam, in the sense that the interstory drift is
approximately proportional to the shear force acting
at the story (see later).
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Figure 2: Shear beam
The shear force )(zQ on a section of a shear beam is
proportional to the slope of the beam (Figure 2(b)):
)(')( zyAGzQ = (3)
where G is the (equivalent) shear modulus and A is the
sectional area of the shear beam.
Note that when we idealize a frame as a shear beam, thevalue of G and A is NOT necessarily the corresponding
values of the columns
By static equilibrium (Figure 2(c)), the shear )(zQ is related to
the distributed load )(zw by:
)(')( zQzw = (4)
and so by combing (3) and (4) the governing differential
equation of a shear beam is given by:
AG
zwzy
)()('' = (5)
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Since (5) is a second order ODE, there should be two boundary
conditions. For a fixed-free situation, the boundary conditions
are:
a) 0)0( =y b) 0)(' =Hy
QUIZ:
Explain what the boundary conditions a) and b) mean.
QUIZ:
What is order of the differential equation for a shear beam?
5. Wall-frame system
Refer to Figure 11.5 of Smith & Coull (p.261)
For the wall, according to eq. (1),
[ ] )()()()()( )2(2
2
zuzwzyzIzEdz
d= (6)
For the frame modeled as a shear beam, from eq. (4),
)('')( zyAGzu = (7)
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Substituting (7) into (6) yields
[ ] )()('')()()('')()(2
2
zwzyzAzGzyzIzEdz
d= (6)
Boundary conditions for fixed-free situation:
a) Zero displacement at the base: 0)0( =y b) Zero slope at the based: 0)0(' =y c) Zero wall moment at the top: 0)(
)2( =Hy d) Zero shear at the top: 0)(')(
)3( = HyGAHyEI
In order to allow for analytical solution for the governingdifferential equations, we assume the properties of the walls and
frames to be constant through their height. That is,
constant)()( ==EIzIzE
andconstant)()( == GAzAzG
This assumption is often NOT met in real situations, sincegenerally the column sizes decreases up the height of the
building, due to decreasing demand in gravity load
capacity
For this reason, the results to follow is NOT expected tohelp you do the calculations for an actual wall-frame
structure, but rather to illustrate from the analytical
solution some of the important behavior.
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With the simplification mentioned, the governing differential
equation becomes
IE
zwzy
IE
AGzy
)()('')(
)4( = (8)
If we define
IE
AG=2 (9)
then (6) becomes
IE
zwzyzy
)()('')( 2)4( = (10)
For uniformly distributed load, that is, )(zq = constant, then the
solution to the ODE in (10) satisfying the boundary conditionsa)-d) is given by (see eq.(11.29) on p.268 of Smith & Coull):
123
123
factorvariation
wloadddistributeuniformlyunderbeam
cantileveraoftopat thentdisplaceme essdimensionl
1
4
)(8
)( zKEI
wHzy =
where
( )
+
+=
2
2
412
1)(sinh1cosh
cosh
1sinh8)(
H
z
H
zHzHz
H
HH
HzK
See Figure 11.8(a) on p.270 or Figure A2.1 on p.504 of Smith
& Coull for the shape of )(1 zK .
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QUIZ:
Explain, without any calculation, why 1)(1 =HK if 0= (Hint:imagine what the wall-frame model will look like if 0= ).
The reason I put the (complicated) formula for )(1 zK is notto let you memorize it, nor do I expect you to use it
frequently. It is just to let you have an idea of the form and
its complexity. It is more important to understand and
develop insights into the formula than to memorize it.
You can look up Smith and Coull for the solution for otherload distributions, such as a triangularly distributed load. It
turns out that the results do not differently qualitatively
from those for the uniformly distributed load.
6. Forces in the wall and frame
Wall moment
)(2
)(''8
)('')( 3
2
1
4
zKwH
zKHw
zyEIzMb
===
QUIZ:
Explain, without any calculation, why 1)0(3 =K if 0= .
See Figure 11.8 on p.270 or Figure A2.3 on p.505 of Smith &
Coull
wall moment becomes negative near top of wall as increases (i.e., there exist hinge-point)
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Frame moment
The frame moment can be obtained as the external moment
minus the wall moment:
)(22
)(
)(2
)()(
3
22
2
zKHwzHw
zMzHw
zMbs
=
=
Frame shear
The frame shear is equal to the external shear minus the wall
shear at that level:
)()()( zQzHwzQbs
=
QUIZ:
Can we use the following formula for the frame shear? Is it
consistent with the previous formula?
)(')( zyAGzQs =
See Figure 11.10 on p.274 of Smith & Coull
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7. Determination of parameters for continuum model
Equivalent value of IE
Just the IE of wall.
Equivalent value of AG
For a shear beam,
dz
dyAGQ =
that is,
dzdyQAG/
=
Figure 3
force needed to produce a unit slope of the shear beam
force acting needed to cause a unit slope of a story of theframe
We need to analyze the frame such as shown in Figure 3 (for
single bay) subjected to a lateral force Q and compute the
resulting lateral displacement . The resulting slope(analogous to the term dzdy/ ) of the frame is then given by h/ where h is the height of the story. The equivalent value of GA
is then obtained as
=
=
hQ
h
QGA
/
It is difficult to analyze the frame as it is a highly indeterminate
structure. The resulting expression for would be verycomplicated and not amicable for use or getting insight. Instead,
we make some assumptions to reduce the structure to a
determinate one and obtain an approximate expression for h/ .
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Simplifying assumptions
The hinges at the mid length of the columns and girders.
Neglect axial deformation
With these assumptions, the result is
gc EI
lPh
EI
Ph
)(24)(24
23
+=
and hence1
2
)(24)(24
+=
=
gc EIlh
EIhPhGA
The factor cEI)(24 comes from cEI)(122 , and 2 comes from the fact that we have got 2 columns acting.
cEI)(12 is the familiar term appearing in the lateralstiffness of a fixed-fixed end beam. Similarly for
gEI)(24 .
EXERCISE:
Derive the expression for . (Hint: virtual work method)
For multiple bays, the expression for GA is analogous:
1
)/12()/12(
+= ggc lEI
h
hEI
hGA
where c sums over all columns and g sums over allgirders in the story, where l represents the length of each girder.
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Tubular structures
1. Introduction
Lateral load predominantly resisted by perimeter rigid-jointedstructural frame
Perimeter frames act together to give substantially greaterlateral resistance than acting alone
Deep spandrel girders at the perimeter frame are needed forstrong coupling
Suitable for both steel and reinforced concrete construction
100+ stories possible (e.g., WTC in NYC: 110 stories, 411mhigh, aspect ratio 6.5:1)
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2. Basic cantilever behavior and discrepancies
cantilever in bending (plane section remains plane).
1)The elongation of a column is proportional to its
perpendicular distance from the common neutral axis of
bending
2)The perimeter frames perpendicular to the wind direction act
as the flange part of the section, and resist primary the
moment resultant of the building section. For this reason,
they are called flange frame.
3)The perimeter frames parallel to the direction of the wind act
as the web part of the section, and resist primary the shear
resultant of the building section. For this reason, they are
called web frame.
Fig. 1: Idealized behavior of tubular structure
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Separate perimeter frames
To understand the interaction between the web-frames and
flange-frames, first consider four perimeter frames without
connections, but connected by (in-plane) rigid floors only. Inthis case,
1)The frames act individually
2)The web-frames are bending in their strong direction while
the flange frames in their weak direction
3)As a result, the web-frames provide the majority of the lateral
resistance, while the lateral resistance from the flange framesis negligible.
4)The columns of the flange-frames bend with their ownneutral axis.
Fig. 2: Perimeter frames acting individually
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Perimeter frames connected together (Tubular structures)
For tubular structures, the four perimeter frames are connected
through the common columns.
1)The corner columns transmit vertical shear force from the
web-frames to the flange frames
2)The corner columns on the windward side are pulled up.
Through the spandrel girders, the inner columns of the
flange-frames on the pulled up also, but the extent depends
on the bending stiffness of the spandrel girders.
3)The phenomenon that the inner columns do not deform thesame degree as the corner columns due to flexibility of the
spandrel beams is known as shear lag.
Fig. 3: Perimeter frames acting together to form a tubular
structure
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3. Equivalent 2-D model for tubular structures
Illustrated using the tubular structure shown in Fig. 4.
Fig. 4
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Fig. 5(a): Equivalent 2-D model for Fig. 4
(based on left-right symmetry)
Notes to Fig.5(a):
1)based on considering left portion of building
2)the sign represents that the vertical displacement of the
nodes connected by the sign are constrained to be equal; it
models the connection between flange-frame and web-frame
3)apply wind load to the web-frames, but not the flange-frames
4)symmetric B.C. at mid girder of flange-frames
5)apply wind load to web-frame, but not flange-frames
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Fig. 5(b): Equivalent 2-D model for Fig. 4
(based on left-right skew symmetry of Fig. 5a)
Notes to Fig.5(b):
1)skew-symmetric B.C. at mid column imposed by
a) 2/cII= to represent half column and
b) =A (or setting 0=DY at the node)
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Fig. 6(a): Equivalent 2-D models for Fig. 4
(based on top-bottom skew symmetry)
Notes to Fig.6(a):
1)Based on considering lower half of building
2)skew-symmetrical boundary condition (B.C.) at mid columnof web-frame imposed by a) 2/cII= to represent halfcolumn and b) =A (or setting 0=DY at the node)
3)wind load to the two web-frames acts in opposite direction to
be consistent with the direction of wind load in the original 3-
D model
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Fig. 6(b): Equivalent 2-D models for Fig. 4
(based on left-right symmetry of Fig. 6a)
Notes to Fig.6(b):
1)symmetric B.C. at mid girder imposed by
a) zero horizontal displacement and
b) zero rotation
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Interpreting FEM results
The equivalent 2-D model can be analyzed using any available
frame analysis program. The results should be interpreted
carefully. The followings are noteworthy:
1)The horizontal displacement of the web-frame in the 2-D
model represents the horizontal displacement in the load
direction of the 3-D model.
2)In the 2-D model, the axial force of the columns, bending
moment and shear force of the spandrel beams represent
those of the 3-D model.
3)The horizontal displacement of the flange-frame in the 2-D
model DOES NOT represent the horizontal displacement of
the 3-D model in the load direction. Rather, the latter isassumed to be equal to the horizontal displacement of the
web-frame in the 2-D model, as a result of the rigid floor
assumption.
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QUIZ:
Decide whether the following 2-D models can be used for
studying the structure subjected to the load as shown.
Fig. 7(a)
Fig. 7(b)
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4. Reconciliation between tubular and frame model
Suppose we are given a floor plan of columns and beams. In
constructing an equivalent 2-D model, there are two natural
choices:
1)model the building as an assembly of frames connected with
rigid links
2)model the building as a tubular structure
Which one should we adopt? What factors tell us whichchoice to adopt?
Suppose spandrel girders are planned to actuate the tubular
action, then what should the equivalent 2-D model for Fig.7(b)?
Let see what goes wrong with the 2-D model in Fig.7(b). In the
2-D model for Fig.7(b),
1)frame action in subframes 1, 2, 3 and 4 are modelled
2)rigid floor assumption is enforced through the rigid links
However, the effect of the spandrel girders in promoting tubular
action is not modeled. For example, the vertical movement of
columns A, B, C and D are not related directly. In summary,
Tubular action is NOT modeled in the 2-D model of Fig.7(b)
The inner columns and spandrel girders help promote tubular
action. With tubular actions enforced, the structure shown on theleft of Fig.7(b) is often known as bundled tube structure.
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5. Equivalent 2-D model for bundled tube structure
EXERCISE:
Draw an equivalent 2-D model for the structure shown inFig.7(b), enforcing tubular action.
See also Fig.12.10 on p.304 of Smith & Coull.
6. Couple with other lateral systems
If there are other lateral structural systems in the building thatbend with their own neutral axis, they can be represented in the
equivalent 2-D model by rigid links (rigid floor assumption),
assuming no twisting.
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Miscellaneous topics
Creep/Relaxation
Creep: deformation under constant load
Relaxation: reduction of stress under constant strain
Scientific origin of creep:According to statistical/quantum mechanics atoms have a non-zero
strain rate even under constant load, at nonzero absolute temperatures
( 15.273+= CK ).
Creep in RC depends on
1.Temperature (higher temp., higher creep)
2.Steel ratio (higher steel ratio, lower creep)
3.Volume to surface area ratio (higher V/S ratio, lower creep)
4.Age, loading history, etc (higher age at loading, lower creep)
Creep effects are especially important in the design of structuresoperating at elevated temperatures, e.g., engines, furnaces, structural
members against fire.
Shear walls usually have a higher creep than columns since
1. it has a smaller steel ratio
2. it has smaller V/S ratio
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Temperature effects under normal conditions (e.g., < 100 C )
Essentially elastic under normal working conditions; may assume, e.g.,E constant
Induce deformation but no stress in determinate structures
Induce deformation and stress in indeterminate structures
Often serviceability concerns, e.g., cracks in in-filled panels, exteriorfacet.
Thermal strain is often described by
Coefficient of thermal expansion, = strain induced per C rise
In general, depends on temperature, but average values may be used
CC
C
==
/101~/107.11
/109.9 56
steel
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concrete
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Thermally-induced axial deformation
QUIZ:
Determine the strain, stress and displacement induced by thetemperature changes as shown.
T T T T
T T
(a) (b) (c)
Notes:
Strictly speaking, there is no stress associated with thermal strain (whichis obvious in unrestrained situations).
When there is mechanical restraints (e.g., axial expansion/contraction ofcolumn is restricted), however, some mechanical strain will be inducedto compensate for the thermal strain in order to satisfy the restraint.
The stress associated with such mechanical strain is often called thermalstress.
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QUIZ:
Determine the axial force and top displacement for the followings.
T T
EATP =
(a) (b)
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Equivalent load formulation
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Thermally-induced flexural deformation
Due to temperature gradient across section Temperature gradient often assumed linear across section
Flexural effect may be superimposed with axial effect
2
21TT
T+
= 1T = +
Axial Flexural
2
T
2
T
2T
Original
What is the curvature induced by the thermal gradient?
For any fibre at a distance r from the neutral
axis,
)(''/)( zyrRrr == (1)
so at face 2 where 2/br= ,
)(''2
2 zyb= (2)
On the other hand,
22
T= (3)
Combining (2) and (3) gives
b
Tzy
=
)('' (4)
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QUIZ:
Determine the moment and top lateral displacement for the following
siutations.
2
T
2
T
b
TEIM
=
(a) (b)
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Equivalent load formulation
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Temperature effects at elevated temperatures (e.g., >100 C )
A fire engineering problem
RC column behavior is considerably nonlinear. The following effectsshould also be considered:
o temperature dependence ofE,
o creep effects
o spalding (air inside concrete expand and burst off surface concrete)