# presentation ```ABSTRACT
This graduation project aims to analyze and
design a building, Northern mountain, NablusPalestine, consists of a garage, offices and
residential apartments. The structure is 14 stories
and the area of each floor is 330.11 . The total area
of the project is 4621.54
THE PROJECT CHAPTERS:
In GP1, the following have been done:
 Chapter 1: Introduction, which describes the
structure location, loads, materials, codes and
standards and the basic structural system of the
structure.
 Chapter 2: preliminary analysis and design for
slab systems-one way ribbed slab system.
 Chapter 3: preliminary analysis and design for
beams using 2D analysis on SAP.
 Chapter 4: design of columns and use live load
reduction factor.
THE PROJECT CHAPTERS:
Chapter 4: redesign of columns
 Chapter 6: Earthquake analysis.
 Chapter 7: 3D modeling for the project.
 Chapter 8: Stairs design.
 Chapter 9: Tie beams and ramp design.
 Chapter 10: Design of beams based on SAP2000
results.
 Chapter 11: Design of slab- Two way solid slab.
 Chapter 12: Analysis and design of footings.

CHAPTER FOUR COLUMN
SAP AS for columns this section in it

This table for number of bars
col #
1
2
3
4
5
6
7
8
9
10
11
As(mm2)
12780
12549
9670
57919
6899
15555
24497
9119
4225
4225
4225
bar
diameter
32
32
32
32
32
32
32
32
32
32
32
Area of bar
803.84
803.84
803.84
803.84
803.84
803.84
803.84
803.84
803.84
803.84
803.84
# of
bars
16
16
12
12
10
12
12
10
6
6
6

Section in column 11
CHAPTER FIVE: ANALYSIS FOR SHEAR
The design wind loads for buildings and other
structures
, shall be determined using one of the following
procedures:
1- Simplified Procedure for low rise
building(h&lt;18m).
2- Analytical Procedure.
3- Wind Tunnel Procedure.

DESIGN PROCEDURES:



Basic wind load and many constants shall be determined; the wind
load is obtained from the following equation:
P = qz*G*Cp
Where:
- p is the wind load(N/mm2)
- qz is the wind pressure
- G is the gust effect factor
- Cp is exposure coefficient
qz = 0.613KzKzt Kd I (N/m2);
Where:
- qz is the velocity pressure
- Kd is the wind directionality factor.
- Kz is the velocity pressure exposure coefficient.
- Kzt is the topographic factor.
- V is the wind velocity(m/s)
- I is the importance factor
FROM THE ACI CODE THE CONSTANTS ARE
CALCULATED AND FOUND TO BE AS FOLLOW:
I = 1.0 (for normal buildings).
Kz = 1.33 (box system or shear walls) from table 6.1
in ACI318-08,
KZt = 1 for the Nablus topography.
Kd=0.85 in Nablus city.
V= 35 mph= 15.65 m/s ( in Nablus city)
G=0.85 from ACI tables( in Nablus city)
0.8 for wind ward
Cp=
0.50 for lee ward
0.70 for side ward
0.70 for roof ward
FROM THE
ACI CODE THE CONSTANTS ARE
CALCULATED AND FOUND TO BE AS FOLLOW:
qz=0.613*1.13*1*0.85**1

=144.2
 P=144.2*0.85*0.8=98 N/m2
 check for Pmin
 Pmin=500/1.3∗Cp=500/1.3∗0.8 &gt; P
so, use P = 384.6 N/m2

F`call = (0.3-0.4) F`c = 105 Kg/cm2
 Story height= 3.3m
 14 stories
 Residential building
 F`c for tension= 0.10*F`c for compression= 10
Kg/cm2
 WD=8.46 KN/m2
 WL= 3KN/m2
 Load combination= 0.9D + 1.6 L

XZ-PLAIN:
CHECK THE OVERTURNING IN THE
DIRECTION:
X-
Slab weight = 8.46*330*14 = 39085.2 KN
 Walls weight = (0.3*25*46.2*28.1)= 9736.65 KN
 Walls weight= (0.2*25*46.2*10)= 2310 KN
 Wind load on the building= 0.385*46.2*10.5=
186.76 KN
 W =0.9*51131.8 = 46018.665 KN
 H =(186.76/46.5)*1.6= 8.1 KN/m
 &pound;M= -(0.5*8.1*46.2)*(2*46.2/3) + 46018.7*(35.1/2)
=801875.9 KN.m
 the building weight is sufficient to resist the
wind-load in X-direction.( the same in Ydirection and it is okay)

CHECK THE TORSIONAL EFFECT IN ( XY-PLAN)
Center of mass:
X=17 m
Y= 5.314 m
 Center of rigidity:
X= 17.11m
Y= 8.98 m

THE DIFFERENCE BETWEEN THE TWO CENTERS IN THE
X-DIRECTION
IS SMALL, BUT IN Y-DIRECTION IS LARGE, SO A SAMPLE
CALCULATION WILL BE MADE ON THE SHEAR WALL NUMBER
THE Y-DIRECTION.
2 IN
e = 8.97-5.31 = 3.66 m
 V= 0.385*10.5*46.2= 186.76 KN
 MT =V*e
 MT = 3.66*186.76= 683.55 KN.m
 J =4355 KN.m2




F= (115.36/2.31)= 49.9= 50 KN/ = 0.5 Kg/cm2 &lt; 105 Kg/cm2
F= ( 68.76/2.31) = 29.77 KN/ = 0.29 Kg/ &lt; 10.5 Kg/
the wind-load effect is very small so it can be neglected
CHAPTER SIX: ANALYSIS FOR SEISMIC
FORCE USING UBC97 CODE

Seismic Forces- Methods of analysis:
Equivalent static method:
 Dynamic analysis

Response spectrum analysis
 Time history analysis

SELECTION OF ANALYSIS METHOD
Static method: The static lateral force
procedure of Section 1630 may be used for the
following structures:
a. All structures, regular or irregular, in Seismic
Zone 1 and in Occupancy Categories 4 and 5 in
Seismic Zone 2.
b. Regular structures under 73m in height with
lateral force resistance provided by systems listed
in Table 7-E.
c. Irregular structures not more than five stories or
20m in height.

SELECTION OF ANALYSIS METHOD
Dynamic analysis: The dynamic lateral-force
procedure of Section 1631 shall be used for all other
structures, including the following:
a. Structures 240 feet (73 152 mm) or more in height.
b. Structures having a stiffness, weight or geometric
vertical irregularity of Type 1, 2 or 3,
c. Structures over five stories or 65 feet (19 812 mm) in
height in Seismic Zones 3 and 4 not having the same
structural system throughout their height
d. Structures, regular or irregular, located on Soil
Profile Type SF, that have a period greater than 0.7
second. The analysis shall include the effects of the
soils at the site.

EQUIVALENT LATERAL FORCE METHOD
(STATIC METHOD)- UBC 97

The total design base shear in a given direction shall be
determined from the following formula:
UBC97

The total design base shear need not exceed the following:
UBC97


30-4
30-5
The total design base shear shall not be less than the following:
UBC97
30-6
EQUIVALENT LATERAL FORCE METHOD
(STATIC METHOD)- UBC 97
Where:



Z= seismic zone factor, Table 7-A
I= importance factor,
R= numerical coefficient representative of the inherent over
strength and global ductility capacity of lateral- force- resisting
systems, Table 7-E

Ca= acceleration seismic coefficient, Table 7-C

Cv= velocity seismic coefficient, Table 7-D



below:
1. In storage and warehouse occupancies, a minimum of 25% of the
floor live load shall be applicable
2. Design snow loads of 1.5kN/m2 or less need not be included. Where
design snow loads exceed 1.5 kN/m2, the design snow load shall be
SOIL PROFILE TYPE

Soil Profile Types SA, SB, SC, SD and SE are
defined in Table 7-B and Soil Profile Type SF is
defined as soils requiring site-specific evaluation.
STRUCTURAL PERIOD
Method A
 The period, T is given by:




hn= height of structure in meters
Ct= factor given by:
 Ct=0.0853 for steel moment resisting frames
 Ct= 0.0731 for reinforced concrete moment resisting frames
and eccentrically braced frames
 Ct=0.0488 for all other buildings
- T= is the basic natural period of a simple one degree of
freedom system which is the time required to complete one
STRUCTURAL PERIOD

Method B:
The fundamental period T may be calculated using the
structural properties and deformational characteristics of
the resisting elements.
Where:



Mi= the mass of the building at the level i
= the deflection of the level i, calculated using the applied lateral
forces
Fi= the lateral force applied on the level i
SITE AND THE BUILDING INFORMATION:




Number of stories = 14.
Story height= 3.3 m.
Materials: concrete cylinder compressive strength at 28 days,
f’c= 28 𝑀𝑃𝑎
-Steel yield strength, 𝑓𝑦=420 𝑀𝑃𝑎

Soil: rock, Sb type in accordance with UBC 97 provisions.

All columns are square with side length equals to 650 mm.




All beams are 400 mm section width and 500mm total
thickness.
The slab is two way solid slab of 200 mm thickness.
External shear walls thickness is 300 mm, and the internal
are 200 mm wide
Shear walls weight =25 KN/m.
SITE AND THE BUILDING INFORMATION:

The live load is 3 kN/m2 – office and residential building.


The perimeter wall weight is 21 kN/m.

Importance factor (I) =1.

Zone factor (𝑍)=0.2, {2B} UBC 97

𝐶𝑎=0.20, 𝐶𝑣=0.2.

Sizes of all columns in upper floors are kept the same.

The floor diaphragms are assumed to be rigid.

Lateral-force-resisting system is: Duel systems(4.1.c)

Location of building Nablus city .


According to Uniform Building Code (UBC97), the seismic
coefficients Ca=0.20, Cv=0.20.
THE MODIFIED PLANE OF THE BUILDING
TO REDUCE THE ECCENTRICITY
WEIGHT OF THE BUILDING
Solid slab own weight= slab thickness x unit weight of concrete
𝑊𝐷=0.20*25=5 𝑘𝑁/𝑚2
𝑊𝑠𝑙𝑎𝑏=299*5=1495 𝑘𝑁/𝑚2
 - Beams weight= length of beams x cross section area x unit weight of concrete:
beams 400*500 ,Area=13.42 𝑚2
𝑊𝑏𝑒𝑎𝑚𝑠=13.42*25=335.5
Notice that the beam depth below slab is used which is 0.50-0.20= 0.30 m.
 - Columns weight= length of columns x cross section area x unit weight of
concrete: columns 650X650 at all floors ,Area=0.4225 𝑚2, 𝐼=0.014876 𝑚4
𝑊𝑐𝑜𝑙𝑢𝑚𝑛𝑠=11*0.4225*3.3*25=383.41 𝑘𝑁 ( columns at edge of walls are neglected )
 -Shear wall weight= length of wall x cross section x unit weight of concrete
Wshear walls=1295 KN
 - Parameter wall weight= length of wall* unit weight(KN/m)
Wwall=1031.5 KN
𝑊𝑠𝑢𝑝𝑒𝑟𝑖𝑚𝑝𝑜𝑠𝑒𝑑=299*3.36=1006.05 𝑘𝑁
 - Total weight of one story= 5546.46 𝑘𝑁
 - Total weight of building= 5521.46*14=77650.44 𝑘𝑁

THE STRUCTURE PERIOD
ƩIx=27.9075 m4
THE STRUCTURE PERIOD

ƩIy=6.1763 m4
BASE SHEAR IN Y-DIRECTION

R= 6.5 from UBC97 Table16-N (Duel system
4.1.c).
- The base shear, V, need not exceed
- The base shear shall not be less than
𝑉=0.11𝐶𝑎𝐼𝑊
𝑉=0.11*0.20*1*𝑊=0.022𝑊 &gt; 0.0217𝑊
So, total base shear will be 𝑉=0.022 *7756.044= 1700.6 KN
BASE SHEAR IN Y-DIRECTION

since T &gt; 0.7 sec, there is Ftop.
Ftop need not exceed 0.25V, 0.25 V= 425.15 KN
Ftop &lt; 0.25V, so Ft= 168.45 KN
and the distribution of the rest of the base shear on the stories is according to the following equation:
Where x is the number of the story.
Table 7.1 shows a summary of
distribution of forces for the stories
TABLE 7.1: DISTRIBUTION OF FORCES
TO STORIES
CENTER OF RIGIDITY
CALCULATION OF ECCENTRICITY DUE TO
SEISMIC FORCES:
center of mass(CM)
X=17.43 m
y=5.25 m

center of rigidity (CR)
X=18.187 m
Y=6.8177 m












emax= eo + e2
emin= eo - e2
eo: the real eccentricity is the distance between the CR( center of stiffness and the CM and
center of mass). 0e
eo: accidental eccentricity
e2= accidental eccentricity
e2 =0.05*L
- L: the floor dimension perpendicular to the direction of the seismic action.
eo=18.187 -17.43= 0.757 m
e2= 0.05*35.1= 1.755 m
emax=1.755 +0.757= 2.512 m
emin= -0.998 m
HORIZONTAL DISTRIBUTION OF STORY
SHEAR TO THE WALLS


The distribution of the total seismic load, to walls will be in proportion to
their rigidities.. The flexural resistance of walls with respect to their weak
axes may be neglected in lateral load analysis. Table 7.4 summarizes the
force distribution to the columns and walls in y direction.
the table 7.3: shows (Km, Kp, dm and dp) for the shear walls.
Where:
- 𝑉𝑚: The shear force of the structural walls m , parallel to the direction of the seismic action.
- 𝑉𝑝: The shear force of the structural walls p, perpendicular to the direction of the seismic
action.
- dm,dp: The distance from the centers of gravity of the structural walls m and p to the
considered center of rigidity.
- 𝑘𝑡: The torsion rigidity of the considered level.
-𝑘𝑝: The translation rigidity of the considered wall.
-K : The total translation rigidity of the considered level
THE PARALLEL WALLS ARE:
W1, W2, W3, W4,
W5, W6, W7
THE PERPENDICULAR WALLS ARE:
W4, W8, W9, W10, W11
W1, W2, W3,
TABLE 7.3: DISTRIBUTION OF FORCES TO
SHEAR WALL AT EACH STORY (V IS IN KN
TABLE 7.5: THE TOTAL FORCE ON EACH
SHEAR WALL AT EACH STORY (KN)
V=VM+VP)
(
CALCULATION OF THE BASE SHEAR IN
THE OTHER DIRECTION(X):

The same procedures as in Y-direction
The base shear, V in the y-direction will be:
R= 6.5 from UBC97 Table16-N (Duel system 4.1.c
The base shear, V, need not exceed
- The base shear shall not be less than
𝑉=0.11𝐶𝑎𝐼𝑊
𝑉=0.11*0.20*1*𝑊=0.022𝑊 &lt; 0.046 𝑊
So, total base shear will be 𝑉=0.046 *7756.044= 3576.65 KN
CALCULATION OF THE BASE SHEAR IN
THE OTHER DIRECTION(X):
since T &lt; 0.7 sec, there is no Ftop.
 The distribution of the base shear on the stories
is according to the following equation:

Where x is the number of the story. Table 7.5
shows a summary of distribution of forces for the
stories.
CALCULATION DO ECCENTRICITY DUE TO
SEISMIC FORCES:
CENTER OF MASS(CM)
center of mass(CM)
X=17.43 m
y=5.25 m

center of rigidity (CR)
X=18.187 m
Y=6.8177 m












emax= eo + e2
emin= eo - e2
eo: the real eccentricity is the distance between the CR( center of stiffness and the CM and
center of mass). 0e
eo: accidental eccentricity
e2= accidental eccentricity
e2 =0.05*L
- L: the floor dimension perpendicular to the direction of the seismic action.
eo=6.8177-5.25 =1.4617 m
e2= 0.05*10.5= 0.525 m
emax=1.4117+0.525=1.9867 m
emin= 0.8867 m
HORIZONTAL DISTRIBUTION OF STORY
SHEAR TO WALLS

Table7.7:( Km(KN), Kp(KN), dm(m), dp(m)) for
the shear walls
TABLE 7.8: DISTRIBUTION OF FORCES TO
SHEAR WALL AT EACH STORY (V IS IN KN)
TABLE 7.9: THE TOTAL FORCE ON EACH
SHEAR WALL AT EACH STORY(V=VM+VP)
(KN)
TABLES USED IN THE ANALYSIS:
TABLES USED IN THE ANALYSIS:
TABLES USED IN THE ANALYSIS:
CHAPTER 7- 3D MODELING










Structural System used:
In this project, 3D modeling will be made for the structure,
and two-way slab with beams will be used as the structural
system, since the project will be analyzed and designed to
resist the earthquakes.
No analysis will be made in this chapter, only the
modeling.
By using 2D modeling on SAP2000 and applying many
thicknesses to obtain an economical one, it was found the
following:
Slab thickness= 20 cm Beams are 40*50 cm and for the modifiers for SAP:
- Beam=0.35
- Slab=0.25
- Column=0.70
STRUCTURAL MATERIALS:
1) Concrete :
The main material used in the structural is the
concrete:
 * The cylindrical compressive strength after 28 days,
 f’c =24 MPa (B300) for slabs and beams.
* Modulus of elasticity ,E= 2.3*
f’c = 28 MPa (B350) for columns.
* Modulus of elasticity, E=2.49*
*Unit weight is 25 KN/
 2) Steel:
*The Steel yield strength used for steel reinforcement is
420 MPa
*The modulus of elasticity is 2.04*
MPa

3D-MODELING ON SAP 2000
SAP2000 RESULTS


Equilibrium check:
The final check in the model is Check local equilibrium by comparing the results of
moments for beams and slabs from structural three-dimensional model using SAP
2000 and hand calculation.

the loads from SAP as shown on the following table

Table 7.1: the loads resulted from SAP
Support reactions:
- Dead (own weight) = 49265.037 KN
- Live= 12575.64
- Parameter wall weight= 14371.363 KN
- Dead+ superimposed dead+ parameter wall= 77721.117 KN (77650.44 KN by hand calculations))
The mass and support reactions are equal to that in hand calculations
SAP2000 RESULTS
 Periods:
Table 7.2 : Structure periods
First mode- movement in y- direction
The period in Y-direction in hand calculations is 1.415seconds (UBC 97)
The period in X-direction in hand calculation= 0.665 seconds (UBC 97)
EQUIVALENT STATIC METHOD:



in this project, the base shear will be calculated by the equivalent
static method using two values for the period:
1- the period produced from the hand calculation( Tnx= 0.665 sec and Tny= 1.415
sec)
2- and the period produced by SAP 2000 (Tnx=1.533 sec and Tny=1.83 sec)
1) The first period- the manual one
Base shear resulted from
SAP 2000 using the UBC 97
code.(using the periods that
are calculated manually)
2- BASE SHEAR ACCORDING TO THE PERIOD
PRODUCED BY SAP 2000
- TNX=1.533
- TNY= 1.83
Base shear resulted from SAP
2000 using the UBC 97
code.(using the periods that are
calculated by SAP)



In order to have the same results of the hand
calculation and the equivalent static method, the
assumptions that have been made to the hand
calculation must meets the equivalent static method
conditions.
In hand calculation, we assumed that the building is
moving 100% in the X-direction to calculate the Tnx.
and the same assumption for y-direction.
but SAP 200 results show that the X-mode has 0.36
model participating mass ratio and Y-mode has 0.66
model participating mass ratio. this is opposite to the
assumption, so the equivalent static method give
results with a high percentage of error
RESPONSE SPECTRUM METHOD (UBC
97 CODE).
the main direction of the seismic force is Xdirection( U1)
the main direction of the seismic force is Ydirection (U2)
the main direction of the seismic force is Zdirection (U3)
STRUCTURE PERIODS

The modes periods and their participating mass
ratios( for 150 modes)
RESPONSE SPECTRUM METHOD (UBC
97 CODE).
the response spectrum results for the seismic force in the three directions are as shown in
the figure below:
ANALYSIS AND DESIGN OF SHEAR WALLS
S
Lw/5
3h
500 mm
ANALYSIS AND DESIGN OF SHEAR WALLS


find the concrete strength to resist shear for all shear
walls
Find the Vu from SAP, on each shear wall:
ANALYSIS AND DESIGN OF SHEAR WALLS
The shear walls have been organized in groups:

Group number one include shear walls number( 1 &amp;2 ).

Group number two include shear walls number( 3 &amp; 4 ).

Group number three include shear walls number ( 5,6,7,8 and 9).
d = min of
0.1* B =0.1*7.7 = 0.77 m
4*b = 4*0.3 = 1.2 m
The boundary reinforcement:
The web reinforcement:
DIVIDING THE SHEAR WALLS IN GROUPS
ANALYSIS AND DESIGN OF SHEAR WALLS
ANALYSIS AND DESIGN OF SHEAR WALLS
WINDOW IN THE FOLLOWING SHEAR
WALLS:(1, 2, 3, 4, 10 AND 11), SPECIAL
REINFORCEMENT SHALL BE PROVIDED AS
FOLLOW:
WINDOWS IN SHEAR WALLS
in the earlier mentioned shear walls, use the
following reinforcement:
Av= 0.0015*bw*S1
Avh=0.0025*bw*S2
S1={(d/3), 500} mm
S2={(d/3), 500} mm
TIE BEAM DESIGN

the main idea to use tie beam in structure is
decrease deferential settlement of the footing due
to loads transfer from other elements of structure
Beam ID
Tie beam
Section (mm)
500*300
Effective depth (mm)
440
(
Beam ID
(1-2)
Section(mm d(mm)
)
500*300
440
Length (m)
5.7
SAP 2000 results:
 Bending moment diagram:

Maximum positive moment = 21.3 KN.m
Use 3
Vu/
SHEAR FORCE DIAGRAM:
Design tie beam for shear:
 Ultimate shear load = 15 KN
 Ultimate Shear load at critical section = 13.9 KN
 Design shear force = Vu/
, = 0.75
 = 13.9/0.75 = 18.5 KN
 Vc = λ/6
bw*d= (1.0) (300)(440) = 116.4 KN
Vn = 18.5 KN
 Use 1
cm minimum.

RAMP 3D ANALYSIS AND DESIGN:

Ramp is a flat supporting surface tilted at an
angle, with one end higher than the other
elements
Beam
slab
Section(mm)
500*250
One way solid (h = 25cm)
MODIFICATION FACTORS FOR
STRUCTURAL ELEMENTS:
Beam modification factor
Slab modification factor
In this figure show the bending moment diagram for ramp slab.
Design ramp slab:
Slab thickness 25 cm (one way solid slab)
Design slab (A-B) for flexure:
Use area of steel As (min)
Beam Design:
Bending moment in the beam duo to
load transfer from ramp slab is very
small, so that beam design must be
depends on the minimum design
case:
Section of beam (50*25 cm)
As (min) = 0.0033*500*200 = 300 mm2
Use 3
Slab ID
Slab (A-B)
Slab (B-C)
Slab (C-D)
Slab (D-E)
Slab (E-F)
= 660 mm2
Longitudinal reinforcement
Negative moment
Positive moment
6
6
6
6
6
6
6
6
6
6
/1 m
/1 m
/1 m
/1 m
/1 m
/1 m
/1 m
/1 m
/1 m
/1 m
DESIGN BEAMS:
From SAP we design beams and take As and this picture can show that
# of
beam
And this picture can show beam number:
b1
b2
b3
b4
b5
b6
From this table we take As top and bottom in each beam:
But Asmin = .0033*400*450 =594mm =4φ14 so
we need to change every beam which is have
number of bars less than minimum to minimum.
b7
b8
b9
b1o
b11
b12
b13
b14
As of
beam
# of
bars
578
4
bottom
578
4
top
717
4
bottom
578
4
top
631
5
bottom
536
4
top
661
5
bottom
578
4
top
724
5
bottom
578
4
top
707
5
bottom
578
4
top
261
2
bottom
130
1
top
668
5
bottom
578
4
top
830
6
bottom
578
2
top
662
5
bottom
576
2
top
619
4
bottom
535
4
top
864
6
bottom
578
4
top
710
5
bottom
522
3
top
589
4
bottom
412
3
top
# of
beam
b15
b16
b17
b18
b19
b20
b21
b22
b23
b24
b25
b26
b27
As of
beam
# of
bars
top
798
6
bottom
578
4
top
574
4
bottom
283
2
top
578
4
bottom
530
4
top
578
4
bottom
415
3
top
491
4
bottom
243
2
top
578
4
bottom
278
2
top
174
2
bottom
296
2
top
199
2
bottom
336
3
top
959
7
bottom
606
2
top
611
4
bottom
552
2
top
578
4
bottom
511
4
top
428
3
bottom
212
2
top
745
5
bottom
578
4
And this is picture for As/s for shear:
We can see from this picture all values about .333
so, if we use φ10 stirrups then we need 5φ10 /m
=1φ10/20cm.
And this is beams in detail
And this cross section in beam 11:
DESIGN SLABS:
From SAP we design slabs and take As
and this pictures can show that for x
direction
bottom face
but ASmin = 360mm2/m and this more than 215mm on the
bottom face so use Asmin and this is the details in the x
direction
And this for y direction:
```