PR - Shear Wall Analysis

7/31/2019 PR - Shear Wall Analysis

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CV415: Tall Buildings Lecture Notes S. K. Au

1

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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33

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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34

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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35

-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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36

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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1

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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2

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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3

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

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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5

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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6

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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7

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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8

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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9

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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12

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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13

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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14

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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1

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

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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3

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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4

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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5

3. Equivalent 2-D model for tubular structures

Illustrated using the tubular structure shown in Fig. 4.

Fig. 4


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6

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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7

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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9

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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10

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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11

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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12

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

6

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)

Documents

PR - Shear Wall Analysis