seminar abstract non seismic walls -مقاومة قوى الزلازل بالإطارات فقط...
TRANSCRIPT
Dissertation
Effect of non Seismic Walls
On Moment Resisting Frames in buildings.
Can we neglect reinforce concrete walls like
(stairwells, elevator shafts and so forth)?
By:
Dr.Youssef Hamida
Consultant Structural Engineer
Aleppo Engineers Order
Syria - Aleppo
Web: www.dr-hamida.com
2
Seminar Abstract – Non Seismic Walls
The performance of the Dual systems (frames and walls) in re-
sisting earthquakes.
And the efficiency of neglecting the walls (Reinforce con-
crete, masonry, partition, fill) and depending Completely on
the frames in resisting the seismic loads.
and what are the Effecting factors which will occur from
neglecting these walls ?
like (Rigidity – Eccentricity – Torsion – Period - Base Shear)
On the behave of the Moment resisting Frames
Can we neglect reinforce concrete walls like stairwells, ele-
vator shafts and so forth?
*And what are the behavior of these walls during the yielding
point for the steel in work stress stage uncracked section
[Elastic Response Parameters] and after the yielding point in
Plastic stage cracked section (Ultimate strength) since
*(Plastic Hinges) will occur in the Frames during plastic
stage And the frames shall peer all the entire seismic loads
And what are these Condition and arrangements to keep
the section walls in safety during plastic stage
so they can carry just the ordinary(D+L) axial loads.
Dose reinforcement for axial ordinary loads enough for these
walls from collapsing?
All these answers you will get it when you look at the Dis-
sertation
3
2 Dissertation text
-Introductory
As it is well known to most of structural engineers who are
familiar with the types of structural systems for resisting
wind and seismic loads, they are called
Shear systems-such as:
1-Frames:
This is a frame system of rigid beams subjected to lateral
loads where the developed moments in the middle of the columns
are not existent And the shear forces will be distributed pro-
portionally with the moment of inertia of the columns and the
lateral displacements will be proportional to these forces
2-Shear walls:
These systems resist the lateral loads with the shear walls
whether these walls are separated or connected by beams.
The distribution of shear forces is proportional to the moment
of inertia of the cross sections of the walls; the displace-
ments in each floor or level are the result of the Flexural
deformations in the walls.
3-Dual systems
These systems are the result of combining the two latter
systems to resist the lateral load, in these systems the shape
of the deformations will differ from those in frames and walls
systems, where effecting interacted forces advantages of this
combination is that the frames support the walls at the top and
control their displacement. Besides, the walls support the frames
at the bottom and decrease their displacement In other words, the
shear force of the frames is bigger at the top than it is at the
bottom and it goes the other way round for the
walls occur and change the shape of shear and moment diagrams.
4
-It has been mentioned in the international and local codes
that in case we have regular frames of beams and columns along
with shear walls to resist
the lateral loads, the resistance of these members (the
frames) to the lateral
loads can be neglected, and it will be considered in the
calculations only to
resist the vertical loads, but we should conform to the
codes conditions
relating to the minimum reinforcement
and the allowed displacement of these beams and columns
-We rarely find shear systems as complete shear walls
without regular frames
(beams and columns), or absolute frames without service
walls or elevator walls
5
the purpose of our research
- And the purpose of our research is to find out if we can
neglect the presence of walls (concrete or masonry) if they
are together with the frame system,where the frame system re-
sists all the lateral shear forces, and the walls will be con-
sidered just to bear the vertical load, and what are the pro-
visions for these walls and their effect on the frames load.
-We will find the proper answer throughout our research and
experiments in this subject, and you can find the summary
of the research in the results and recommendations page
The reasons and the cause for choosing this research
and its importance, and whether it is done to fulfill an
engineering need or to solve an engineering problem that
helps in developing the engineering work.
3-Research topic Demonstration
3.1 The Dual system is the one that both shear walls and
frames participate in resisting the lateral loads resulting
from earthquakes or wind or storms, possibility to develop
plastic hinges
and the portion of the forces resisted by each one depends on
its rigidity modulus of elasticity and its ductility, and the
in its parts Knowing that the frame is a group of beams and
columns connected with each other by rigid joints that can re-
sist shear and moments, and the shear wall is considered as a
cantilever free on the top and fixed in the bottom.
55
3.2 The structural resisting system might be only shear walls
for resisting the lateral load and we can neglect the regular
frames.
3.3 The structural resisting system might be only frames
for resisting the lateral load and it is called Moment
Resisting Frames In the case of shear walls with the moment
resisting frames can we neglect the effect of these walls, and
calculate the frames to resist the whole base shear
this is the subject of our research; the existence of some
shear walls with moment resisting frames, could it be
neglected and not taken into consideration for resisting the
lateral loads, which means to calculate them only as gravity
loads resisting members, and what are structural effects and
changes resulting from that.
6
4.Purpose and benefit of the research
-Some complains and questions had come to the Engineering
Union-Civil Engineering department, about invalidating some
domestic and industrial buildings licenses where some
structural notes are present, and when there is a
contradiction between the structural requirements for the
buildings and the building manners or systems of the city mu-
nicipality where.
4.1 Elevator and service walls must be of reinforced
concrete, and that is to comply with the mechanical study,
fire resistance, and more other reasons.
4.2 Also to conform to the structural provisions of the city
municipality, the stair house walls should be of
reinforced concrete instead of using masonry walls,and be-
cause of the door openings and piers are in small
dimensions, it should be of reinforced concrete to bear the
vertical lo ads.
4.3 Since most of the industrial buildings, stores, shops
and halls or galleries are in the lower floors (base), shear
walls can’t be used to resist the seismic loads, so it is pre-
ferred to use moment resisting frames.
8
4.4 Repeating the structural study with taking into
consideration the seismic study, and including these walls in
the study will cause increment in the
base shear because of decrement in (R); the Elastic Modulus of
the structural system. In addition, the use of computers,
modeling, and advanced programs might be a little hard to pro-
cure in regular or individual offices.
4.5 That’s why we needed to make researches and investigations
about a method or a study of possibility of neglecting the
shear or masonry walls and not considering them as
participants with the frames. Also neglecting the regular
frames and not considering them as participants with the shear
walls, and considering them as gravity loads resisting members
but with other conditions that we will see in the results of
our research.
9
5-The determined basics for making the research:
To define the relationship and the effect of the shear walls
on the function of the
moment resisting frames, and to know if these walls can be
studied only for
resisting the gravity loads not the lateral load
5.1 It was necessary to go back to the theories and hypothe-
sizes of the interactive performance between frames and walls,
and we needed to seek the help of the theory of Professor
(Lain Macleod). The calculations and results have been checked
by the computer de pending on the international program
(ETABS). Due to the enormous number of analysis’s and experi-
ments according to the variety of floors and walls number and
dimensions of cross sections of the columns. We depended on a
local program (STAAD-ALARAB) for calculating the frames and
the shear walls depending on the theory and assumptions of
Professor Macleod relating to the interacted performance be-
tween the frames and the walls. The summary of this theory as
it came in the research symposium of Portland Cement institu-
tion
5.2 The dual system is the one that both frames and shear
walls contribute in resisting the lateral loads, where the
frame is a group of beams and columns connected with each oth-
er by rigid joints, and the frames bend in accordance with
(Shear Mode), whereas the deflection of the shear walls is by
a (Bending Mode) like the cantilever walls
10
5.3 As a result of the difference in deflection properties
between frames and walls, the frames will try to pull the
shear walls in the top of the building while in the bottom,
they will try to push the walls, so the frames will resist the
lateral loads in the upper part of the building, which means
an increase shear walls will resist most of the vertical loads
in the lower part of the building in the dimensions of the
cross section area of the columns in the upper part of the
frame more than what it needs to resist the gravity loads
5.4 So the distribution of the lateral loads in the top de-
pends on the rigidity of the frames where we suppose a spring
support, whose rigidity equals the rigidity of the frames in
the top, and the reaction of this spring is the share of the
frames, and the rest is the share of the walls. So, the walls
are pinned or supported by the frames at the top and fixed at
the bottom and they are resisting the seismic loads
5.5 So we need to find out the value of this reaction at the
top which equals a point load as the share of the frames ac-
cording to the (Macloed Theory).
then the share of the frames will be distributed to each frame
due to its rigidity and position relating to the center of
mass taking into consideration the torsion and shear resulting
from torsion below And that is according to the laws and rela-
tions and factors mentioned
11
6-Terminology:
Ac: cross section area of columns
B: length of the frame
C: width of column
D; depth of the beam
E: Young Modulus
Fg, Fm, Fn, FS: functions depending on the shape of the
seismic load
--------------7
H: total height of the wall
h: height of the column in every floor
Ib: moment of inertia of the beams in the nods or joints
Iw: moment of inertia of the walls
Kw: Shear rigidity of the walls
Kf: rigidity of the frames
--------- 7.1
---------- 7.2
Fs = ----------------- 7.3
W = total lateral load
∆ = total displacement in top
∆A = displacement in columns from axial load
∆B = displacement from moment
12
---------------------------- 7.4
displacement of walls ------ 7.5
∆f = displacement of Columns --- 7.6
------ rigidity of the frames
----- rigidity of the walls
--------------- 7.7
W =-------- total shear forces
P =-------- share of Frames
P1= W-P ---share of the Walls
13
7-Researches and experiments:
- The research has been done and the data has been changed to
follow the Second Static Method for calculating the Base Shear
according to the Syrian Code and The American Code: (Uniform
Building Code, UBC) according to the law
= total base shear
7.1 The local program (STAAD-ALARAB) and ETABS have been
adopted to check and calculate the internal forces resulting
from neglecting the shear walls.
7.2 Thirty five typical space and plane cases have been
chosen. In every case, the constants and dimensions are
changed to obtain the maximum stresses and forces and that is
according to the tables, calculations and the results attached
in the end of the research.
7.3 For the number of the floors, we studied from one
floor cases till twelve floor cases.
Three cases have been chosen to get maximum
torsion and its effect on the frame, and that is by
the types of the walls and their positioning with respect to
the center of mass
7.4 A case of one wall with frames different in rigidity
has been studied, with different floors for different cases
with eccentricities:
e = l / 2 e = l / 4 e = 0
14
7.5 Also the procedures are repeated for the case of two
walls for different number of floors and different examples
7.6 Also the procedures are repeated for the case of three
walls for different
number of floors, different types and rigidities
of frames and different eccentricities
7.7 Also in the research, a case of only frames without
shear walls has been studied, then we started to insert one
shear wall, then we increased the number of walls to two then
to three. Also for finding the eccentricity and increasing it
to the maximum value predicted Through the research results
and the attached tables, some important observations can be
noted, and corrections to some wrong concepts can be done for
the colleagues of engineers about the relation between the
frames and walls.
7.8 The most significant observation and the correction to
the wrong belief is that increasing the number of shear walls
and duplicating its area doesn’t duplicate the share of the
walls, but it stays almost the same as there is only one shear
wall, the increase is almost insignificant (about 15%) and it
comes from the increase of base shear resulting from the in-
crease in rigidity of the walls and the decrease of the dynam-
ic period, which means that the frames takes its share from
the base shear in the top then the rest is distributed (in the
bottom on the existed walls (one or two or three….etc.) And
that is what we observed in the research results, that is ex-
isting of a number of walls and then neglecting them is better
and gives less shear and bending stresses than the case of one
wall which is considered to be neglected
15
7.9 We found out through the results that the frame’s shear
if there is no walls (the Moment resisting frames case) is
bigger than what it is in the case of frames and neglected
walls, because the torsion shear which is caused by
theeccentricity of the neglected walls, and which is added to
the eperipheral frames’ shear is smaller than the shear that
the walls take from the peripheral frame with the maximum tor-
sion.
7.10 also noticed the increase of the base shear when
there are walls rather than the case of no walls (only frames)
and that is because of the decrease of the dynamic period for
the increase in the rigidity of the Dual system and
thedecrease of the factor(R)in the denominator of the base
shear wall
7.11 In the attached drawing of the (interactive-mutual)
performance, we noticed that in the case of only two shear
walls without a frame, the base shear is distributed equally
in the top and bottom in the walls. And if there is one frame
with those two walls, we see that the share of the walls is
maximum in the bottom, and the share of the frame is zero in
the top and the bottom but maximum at the height (0.8 H), and
that is the hypothesis of Prof. Macloed, which is to put a
constant point load equals to the maximum shear which occurs
at the height (0.8-0.95) of the total height
7.12 Referring back to the results table of the program STAAD-
ALARAB and ETABS for the internal forces or stresses and Rein-
forcement
16
7.13 we notice that it is taken into consideration:
-the dynamic period, the static period, ductility factor (Rw),
accidental eccentricity of walls positioning, ratio of frames
and walls shares, modifying the value of ductility factor (Rw)
according to that ratio, also calculation of the base consid-
ering the maximum case. shear V for every case and
7.14 we noticed from the table that the maximum share of
the frames in case of no walls (Moment resisting frames) is
larger than the case of frames with walls, for example:
8-Comparison and results:
Research example No#5:
Floor number: 7
The columns: 40 X 60
The walls: L=3,00 m
V= 61 t without walls
V= 44.46 t in case of one wall with maximum eccentricity
61 > 44.46 accepted
8.1 Also we notice from the results of ETABS:
That the shear and flexural stresses are maximum for one
shear wall, also we noticed that the minimum reinforcement for
shear and bending is adequate, so these walls can be neglected
and designed only for the gravity loads, if the number of
floors is less or equal to twelve floors.
17
8.2 We noticed when the number of the floors is more than
twelve, and the minimum reinforcement is not adequate here, we
differentiate two cases:
8.2-1 When the flexural reinforcement is not adequate, the
walls are safe, because when the walls enters the plastic
situation, the frames will intervene and take the loads from
these walls and the walls will crack but not collapse, or it
will lose its rigidity and not resist any lateral loads
8.2-2 When the shear reinforcement is not adequate these
walls will collapse by shear, and its breaking will be brittle
so it should it should be calculated, or we should neglect its
resistance to the vertical loads by putting beams over them.,
9-Results and recommendations
9.1 It is possible to neglect the walls and do not
consider them as participants with the moment resisting fames
in resisting the lateral loads and to consider that all the
lateral loads as base shear or wind loads are going to be re-
sisted only by the frames according to
9.2 The moment resisting frames is not affected by neglecting
the walls when the walls do not take part with the frames in
resisting the loads, on the contrary, the safety factor for
the frames becomes bigger and the shear resisted by the frames
is less than what it was when the frames were alone
9.3 When the number of the building’s floor is equal or less
than 12 floors, (n<12), the walls can be neglected provided
that they are qualified to bear the only vertical loads, and
the minimum reinforcement is adequate for flexural and shear
18
9.4 When the number of the building’s floor is more than
12 floors, (n>12), it is possible to neglect the participation
of the walls with the frames, provided that we should consider
finding alternative members to the walls for resisting the
vertical loads like beams or the frames itself, so when we put
beams on top of the walls or any other structural members to
transfer the vertical loads from these walls to the beams and
columns It is adequate to use the minimum reinforcement for
flexure al and shear stresses.
9.5 from(Tab-4,Tab-5)we find the ductility factor(R) decrease
to (7) instead of (8) and the base shear(v)increase (18%)
Because of affect walls stiffness.
10 conclusion
10-1 in ordinary building less than 12 story n<12 the service
walls like {stairwells, elevator shafts and so forth} can be
neglected to share frames in resisting seismic
10-2 the total base shear will take by frames after degreasing
The factor of ductility ,one degree or increasing the base
shear on frames 18%
10-3 seismic minimum reinforcement ably in both direction
As the code says.
19
REFERENCES:
- The Arabic Syrian Code and its appendixes in resisting the seismic loads.
- The American codes: UBC-ACI
- Shear wall-frame interaction
by :I.A. Macloed
-Seismic design
by :D.T. Derecho
-STAAD-ALARAB, walls and frames
-by : D r . Youssef Hamida
-Interaction of shear walls and frames
By: Khan and Sbarounis
-Concrete shear walls combined with rigid frames
By: Cardan Bernard
-Multistory frames and interconnected shear walls
American concrete institute
By Firschman prabhu
-Design of Multistory reinforced concrete building for earthquake motions
-Shear wall design philosophy
By : Portland Cement Association
20
ACI committee symposium paper by Thomas Paulay
-An inelastic approach to seismic design
by: American Society of Civil Engineering
-Response of multistory structures to lateral forces
by: Cardenas, Alex
-Symposiums and lectures of Engineers Union, Aleppo Branch
Resisting the seismic loads by walls and frames.
By: D r . Youssef Hamida
-Symposiums and lectures of Engineers Union, Damascus Branch
Analyzing and designing of buildings for Earthquakes
By: D r Karameh Baddorah - Dr. Zein-Aldeen -Dr. Alhessen
Structures dynamics and seismic engineering
By: D r Samara
21
FFrraammeess ++ oonnee wwaallll
2222
32
Etabs
Frames + two walls+ Eccentricity
42
Etabs
Frames + two walls
52
Frames only
62
Bending moment (Frames +one wall )
Etabs program
72
Shear (Frames +one wall
Etabs program
82
Bending moment (Frames +two wall) 12 stories
Etabs program
29
oriesReinforcement 12 st
Etabs program
30
staad alarab-result
dynamic
Static design
Stories
period
period
period T=
Wall from
5
DT ST 1.4*Ts<
DT ductilit
y mass center
Eccentricity
Base Shear
Frames
Columns Name
DT sT T wR a e V %
Share
60X60 Frames )x(
50.6 0.63 0.65 8 0 0 58.62 100%
wall 60.71
20x300 cm Frames
)y( 0.61 0.63 0.61 7 0 0 69.4
100%
69.4
Frames
0.5 0.42 0.5 7.5 0 0 80.8 50%
one wall 40.4
+Framese 0.47 0.42 0.47 7.5 7.5 0.71 85.87
50%
One wall 42.87
Frames
0.48 0.42 0.48 7 0 0 90.18 45%
Tow walls
41.03
Frames +e
0.45 0.42 0.45 7 7.5 1.04 96.7 45%
Tow walls
44
Story
dynamic
Static design
Wall from
5
period
period
period T=
ductility
mass center
Eccentricity
Base Shear
Frames
Columns Name DT ST
1.4*Ts<
DT wR a e V %
Share
40X60 Frames (x)
0.67 0.63 0.67 8 55.15 100%
wall 55.15
20x300 cm Frames
(y) 0.65 0.63 0.65 7 67 100%
67
Frames +
0.53 0.42 0.53 7 0 0 81.88 49%
one wall 0.284
Frames+e 0.5 0.42 0.5 7 7.5 0.79 87.26 49%
One wall 42.93
Frames+
0.51 0.42 0.51 7 0 0 85.65 44%
Tow walls 38.11
Frames +e
0.47 0.42 0.47 7 7.5 1.15 91.95 44%
Tow walls 40.92
4 – Table
staad alarab-result
3311
Story
dynamic
Static design
Wall from
5
period perio
d period
T= ductilit
y mass
center Eccentrici
ty Base Shear
Frames
Columns Name
DT ST 1.4*Ts<
DT wR a e V
% Share
30X80 Frames(x)
0.71 0.63 0.63 8 0 0 53.37 100%
wall 53.37
20x300cm Frames
(y) 0.66 0.63 0.63 7 0 - 65 100%
65.81
Frames
0.55 0.42 0.55 7 0 0 79.11 49%
one wall 38.38
Frames+e 0.51 0.42 0.51 7 7.5 0.84 84.8 49%
One wall 41.42
Frames
0.52 0.42 0.52 7 0 0 83.3 44%
Tow walls 36.6
Frames +e
0.49 0.42 0.49 7 7.5 1.21 89.51 44%
Tow walls 39.32
Stories
dynamic
Static design
Wall from
7
period perio
d period
T= ductilit
y mass
center Eccentrici
ty Base Shear
Frames
Column Name
DT ST Ts<1.4*
DT wR a e V
% Share
40X60 Frames )x(
0.87 0.81 0.87 8 61 100%
wall 61
20 x 300 cm Frames
(y) 0.82 0.81 0.82 7 74 100%
74
Frames
0.7 0.54 0.7 7.5 0 0 80.62 52%
one wall 41.82
Frames+e 0.66 0.54 0.66 7.5 7.5 .430 85.7 52%
One wall 44.46
Frames
0.69 0.54 0.69 7 0 0 87.81 49%
Tow walls 43.1
Frames +e
0.64 0.54 0.64 7 7.5 0.52 89 41%
Tow walls 36.56
5 – Table
32
staad alarab-result
Story
dynamic
atiStc design
Wall from
10
period perio
d period
T= ductilit
y mass
center Eccentrici
ty Base Shear
Frames
Columns Name DT ST
1.4*Ts<
DT wR a e V %
Share
40X60 Frames(x)
1.18 1 1.18 8 66.83 100%
wall 66.83
20x300 cm Frames
(y) 1.11 1 11.1 7 78.22 100%
78.22
Frames +
1.02 0.7 0.98 7.5 0 0 82.39 53%
one wall 44
+Framese 0.99 0.7 0.98 7.5 7.5 0.21 83.39 53%
One wall 44
Frames
0.92 0.7 0.92 7.5 0 0 88.14 52%
Tow walls 45.78
Frames +e
0.87 0.7 0.87 7 7.5 0.33 92.83 52%
Tow walls 48.2
Story
dynamic
Static
design
Wall from
12
period perio
d period
ductility
mass center
Eccentricity
Base Shear
Frames
Column Name
DT sT T wR a e V %
Share
60X60 Frames (x)
1.27 1.2 1.23 8
80.2 100%
wall 80.2
20x300 cm Frames
(y) 1.23 1.2 1.23 7
84.85 100%
84.85
Frames+
1.04 0.81 1.04 7.5 0 0 93.81 54%
one wall 50.6
+Framese 0.98 0.81 0.98 7.5 7.5 0.15 99.6 0.54%
One wall 53.4
Frames +
1.02 0.81 1.02 7.5 0 0 95.3 53%
Tow walls 50.33
Frames +e
0.96 0.81 0.96 7.5 7.5 0.24 101.4 53%
Tow walls 53.6
6 – Table
33
staad alarab-result
walls frames-Etabs
Name
Max
Shear
Max
Bending
ar She
stress
Shear
reinf Tension steel Ratio
Research ton/m
ton.m tu kg/cm2 All
Stories Typical Story
Base Story
A Frames + 38.28 202 6.38
Min Ratio
Min Ratio 0.008
Story one wall
5 Frames
+e 27 159 4.50
Min
Ratio Min Ratio
0.008
Column One wall
60X60 Frames + 30 175 5.00
Mini Ratio
Mini Ratio 0.008
wall Tow walls
20x300 cm Frames
e+ 26.61 132 4.44 Min
Ratio Min Ratio 0.008
Tow walls
walls frames-Etabs
Name
Max Shear
Max Bending
ar Shestress
Shear rein Tension steel Ratio
Research ton/m
ton.m tu kg/cm2
All Stories
Typical Story
Base Story
B Frames 44 233 7.33
Min
Ratio Min Ratio 0.008
Story one wall
5 Frames
e+ 26.41 168 4.40 Min
Ratio Min Ratio 0.008
nColum One wall
40X60 Frames 31.3 187 5.22
Min Ratio
Min Ratio 0.008 wall s Tow walls
20x300 cm
Frames e+ 25.3 125 4.22
Min Ratio
Min Ratio 0.008
Tow walls
walls frames-Etabs
Name
Max Shear
Max Bending
shear stress
ar Shereinf Tension steel Ratio
Research ton/m
ton.m tu kg/cm2
All Stories
Typical Story
Base Story
C Frames 38 212 6.33 Min
Ratio Min Ratio 0.008 Story one wall
5 Frames
+e 22
150 3.67 Min
Ratio
Min Ratio 0.008
Column One wall
30X80 Frames 28 176 4.67
Min
Ratio Min Ratio 0.008
wall Tow walls
20x300 cm
Frames
e+ 25 132 4.17 Min
Ratio Min Ratio 0.008
Tow walls
7 - Table
43
staad alarab-result
walls frames-Etabs
Name
Max Shear
Max Bending
shearstress
Shear reinf Tension steel Ratio
Research ton/m
ton.m tu kg/cm2 All Stories
Typical
Story
Base
Story
D Frames 44 226 7.33 Min Ratio Min Ratio 0.008
Stories one wall
7 Frames
+e 26.5 161 4.42 Min Ratio Min Ratio 0.008
nColum One wall
40X60 Frames 32.5 186 5.42 Min Ratio Min Ratio
0.008
wall Tow
walls
20x300 cm
Frames e+ 30 141 5.00 Min Ratio Min Ratio
0.008
Tow walls
walls frames-Etabs
Name
Max Shear
Max Bending
shear stress
r Sheareinf
Tension steel Ratio
Research ton/m
ton.m tu kg/cm2 All Stories
Typical Story
Base Story
E Frames 44.42 234 7.40 Min Ratio Min Ratio 0.008
Story one wall
10 Frames
+e 25.3 159 4.22 Min Ratio Min Ratio 0.008
Column ll One wa
40X60 Frames 32.4 191 5.40 Min Ratio Min Ratio 0.008
wall Tow
walls
20x300 cm
Frames +e 31 149 5.17 Min Ratio Min Ratio 0.008
Tow walls
walls frames-Etabs
Name
Max Shear
Max
Bending
Shear stress
Shear reinf
ension Tsteel Ratio
Research ton/m
ton.m tu kg/cm2 All Stories
Typical
Story
Base
Story
F Frames 44.5
236 7.42 Min Ratio Min Ratio 0.008
Story one wall
12 Frames
+e 31.1
188 5.18 Min Ratio Min Ratio 0.008
Column One wall
60X60 Frames 31.2 187 5.20 Min Ratio Min Ratio 0.008
wall Tow
walls
20x300 cm
Frames +e 27 153 4.50 Min Ratio Min Ratio 0.008
Tow walls
8 – Table
53
Seismic forces
yMax Eccentricit
Tx= 0.609
Ty= 0.609
Calculation
period
Ctx = 0.0731
Cty = 0.0731
Coefficient
Ftx= 0.00 Point Load
Fty = 0.00 Ft =0.07 T* V
X MR = 7.50
Y MR= 7.50
Mass Center
X CR= 7.50
Y CR= 7.50
Rigidity Center
Mtx = 45.53
Mty = 45.53
orsionT
Mx = 0.72 inelastic Drift
My = 0.72 M=0.7 R DS
9-Tab
63
Seismic forces
base Shear
Walls Share -Vy
60.71 0.00
Frames %
Frames
Share
100.00 60.71
Ft) Wx hx / s Wi hi-Fx = (V
Base Shear distribution
FX FY
20.24 20.24
16.19 16.19
12.14 12.14
8.09 8.09
4.05 4.05
60.71 60.71
10 -Tab
37