Design of concrete structure 2 project-Loay Qabaha &Basel Salame

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AN-NAJAH NATIONAL UNIVERSITY
FACULTY OF ENGINEERING
CIVIL ENGINEERING DEPARTMENT
*Prepared By: *Submitted To:
-Basel Salameh -Dr. Monther Dwakit
-Loay Qabaha -Dr. Ibrahim Irman
-Mohamad Hindi -Eng. Hanan Jafar
-Tariq Thawabi
62116112
Design of concrete structure 2 project

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*Introduction:
What is design ?
Selecting proper dimensions of reinforced concrete elements and material
used taking to account economic considerations .
The Design objectives is the designed structure should sustain all loads and
deform within limits for construction use, and should be durable. All the
above are design objectives that can be fulfilled by understanding the
strength and deformation characteristics of the materials used in the design
(concrete and steel).
Every structure has got its form, function and aesthetics. Normally, we
consider that the architects will take care of them and the structural
engineers will be solely responsible for the strength and safety of the
structure. However, the roles of architects and structural engineers are very
much interactive and a unified approach of both will only result in an
Integrated" structure.
In this project, we will design the project in two way systems, let’s to define
what the two way system?
Concrete is used mainly in slabs also in columns & beams. Slabs may be
designed with respect to loads applied over it as a solid slab either one way
or two way, ribbed slab also as one or two way, flat plate, flat slab & waffle
slab…, beams could be hidden or dropped. The main purpose is to distribute
loads and transport it to earth without causing the structure failure.
When the ratio (L/ S) is less than 2.0, it is called two-way slab, shown in.
Bending will take place in the two directions in a dish-like form. Accordingly,
main reinforcement is required in the two directions.

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Two-way Edge-supported Solid Slabs :
In this system, shown in Figure , beams provide moment interaction
with the columns especially when moment resisting frames are used to
resist lateral loads.
The slab overall thickness is greater because beams project downward,
thus the system becomes inflexible in terms of mechanical layout. The
system can be used economically for spans up to 7.0 meters.
Two-way Edge-supported Ribbed Slabs :
This system, shown in Figure 2.e, can be economically used for spans up to
7.0 meters. It is similar to the waffle slab but the voids between ribs are filled
with hollow blocks. Hidden or drop beams can be used with this system
depending on their spans.

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Waffle Slabs :
The waffle slab is capable of providing the largest spans of the conventional
concrete floor systems and can be economically used for spans up to 12.0
meters. This is attributed to the considerable reduction in dead load as
compared to other floor systems. Waffle slab construction consists of
orthogonal sets of ribs with solid parts at the columns, as shown in Figure.
The ribs are formed with fiberglass or metal dome forms. The ribs are usually
0..0to 0.90 meter on center. Shear is transferred to the columns by using
beams or shear heads.
*Problem Definition:
Given Data:
 Concrete for Beams , Columns & Slabs , 𝒇́ 𝒄 = 𝟐𝟖 𝑴𝑷𝒂 , &
 ˠ= 25 KN/m3
 Steel for Reinforcing Bars and Shear Reinforcement , Fy= 420MPa
 Floor Height = 4.7 m& number of Floors is 3.
 Superimposed dead load is the gravity load of non-structural parts of
the building .Superimposed dead load= 3.5KN/m2

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 Live loads are those loads produced by the use and occupancy of
the building or other structure and do not include construction or
environmental load such as wind load, snow load, rain load,
earthquake load, ﬂood load, or dead load. Live loads on a roof are
those produced (1) during maintenance by workers, equipment,
and materials; and (2) during the life of the structure by movable
objects such as planters and by people .Live load= 6 KN/m2.
Requirements:
Design the slab as: with full Design Detailing.
1. Two way solid slab
2. Two way ribbed slab
3. Two way voided slab
L2=frame width =10.5 m
Column strip =4 m

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Direct Design Method
The direct design method consists of certain steps for distributing moments
to slab and beam sections to satisfy safety requirements and most
serviceability requirements simultaneously. Three basic steps are involved :
a. Determination of the total factored static moment.
b. Distribution of the total factored static moment to negative and positive
Sections.
c. Distribution of negative and positive factored moments to the column
and middle strips and to the beams if any.
Limitations
The direct design method was developed from theoretical procedures for
determination of moments in slabs, requirements for simple design ,
construction procedures, and performance of existing slabs. Therefore, the
slab system, to be designed using the direct design method, should conform
to the following limitations as given by ACI Code 13.6.1 :
1. There must be three or more spans in each direction .
2. Slab panels must be rectangular with a ratio of longer to shorter span ,
center-to-center of supports, not greater than 2.0 .
3. Successive span lengths, center-to-center of supports, in each
direction must not differ by more than one-third of the longer span .
4. Columns must not be offset more than 10 % of the span in the
direction of offset from either axis between centerlines of successive
columns .
5. Loads must be due to gravity only and uniformly distributed
Over the entire panel. The live load must not exceed 2 times the dead
load.

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*Two Way Ribbed Slab System:
* Slab thickness and modification for the ribbed slab:
Assume αfm> 2:
H min. = (Ln*(0.8+fy/1400))/ (36+9β)
= (10*(0.8+420/1400))/ (36+9*1.33) = 0.23 m =230mm
Frame in south north direction (interior)
4mm8)/12=5.57*103(550*23012 =3bh=(solid)I
4mm813.5*10for the suggested cross section =(ribbed)I
(Calculated from sap frame section definition). I (ribbed) > I (solid) OK .

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To find the equivalent slab section for the ribbed slab:
13.5*108= (550*h3)/12 ….. h=308.83 mm.
* Loads and shear check:
WD (O.W) = (0.55*0.55*0.4-0.32*0.2*0.4)*25+ (0.4*0.2*0.32)*12= 2.6922 KN
WD= 2.6922/ (0.55*0.55) =8.9 KN/m2
Wu = 1.2(3.5+8.9) +1.6*6= 24.5 KN/m2
Vu= 24.5*0.55*((10.5-0.7/0.5)-0.36) = 61.17 KN
∅VC = 0.75*
𝟏
𝟔
√𝟐𝟖*1.1*150*
𝟑𝟔𝟎
𝟏𝟎𝟎𝟎
= 39.29 KN
Vs= Vu - ∅VC = 21.89 KN
(Av/s) = (21.89*1000)/ (420*360) = 0.14 mm2/mm.
(Av/s) min = max. Of {0.35*(bw/fy), 0.062*(bw/fy)*√𝒇𝒄 } =0.125 mm2/mm
so, use (Av/s) = 0.14 mm2/mm.
S max.= d/2 = 360/2 = 180 mm use 1 Ø 8mm / 180 mm
Now we have to check the assumption that αfm> 2
The suggested beams dimensions are:
B1 (700 *500) for exterior beams I (B1) =7.3*10^9 mm4
B2 (700 *800) for interior beams. I (B2) = 3*10^10 mm4
Columns (700*700) mm
αf= (I beam / (I slab* number of ribs))

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αf1=0.75 αf2 =1.44 αf3 = 0.7 αf4 = 1.2 αf5 = 1.17
All the values of αfm for all panels are less than 2.
So we have to go to 2< αfm > 0.2 case:-
the least value of αfm = 1.045
H min. = (Ln*(0.8+fy/1400))/ (36+5β*(αfm-0.2))
= ((10.5-0.7)*(0.8+420/1400))/ (36+5*1.33(1.045-0.2) = 0.26 < h=308.83 mm
…. OK
.
* Hand Calculations: analysis and design of frame 3.
A- Static moment (M0) ( for each span ) :-
M( exterior spans) = (24.5*10.5*7.3^2)/8+ (8.4*7.3^2)/8= 1770 KN. m.
M( interior span) = (24.5*10.5*8.3^2)/8+ (8.4*7.3^2)/8= 2288 KN .m
Where ,the factored own weight of beam= 8.4 KN/m .

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B- Positive and negative moments of total frame:
Refer to (Table 13.6.3.3) to find the moments in frame in end spans.
- 0.7 M (for interior negative moments).
- 0.57M (for all positive moments).
- 0.16M (for exterior negative moments).
For the interior spans:-
0.35M for positive moments
0.65 for the negative moments
C- Moments in column strip including the beam :-
( Table 13.6.4.4 for the positive moments )
L2/L1= 10.5/8=1.31 L2/L1= 10.5/9 = 1.17
αf5*( L2/L1)= 1.53>1 αf5*( L2/L1)= 1.37>1
use interpolation use interpolation
1 ………75 1 ………75
1.31…… X X=66% 1.17…..X X=70%
2 ……… 45 2………45

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(Table 13.6.4.1 for interior negative moments)
L2/L1= 10.5/8=1.31 L2/L1= 10.5/9 = 1.17
αf5*( L2/L1)= 1.53>1 αf5*( L2/L1)= 1.37>1
1 ………75 1 ………75
1.31…… X X=66% 1.17…..X X=70%
2 ……… 45 2………45
(Table 13.6.4.2 for exterior negative moments)
C = (1- 0.63*(0.5/0.7))*((0.7*0.53)/3)= 0.016 m4
Bt = C/(2*Is) = 0.016/(2*(4/0.55)* 13.5*10-4) =0.815< 2.5
Use interpolation
0 ……. 100
0.815………..X
2.5 …………..75 X= 91.85 %

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0 ……. 100
0.815………..X
2.5 …………..45
X= 82.07%
1 ……. 91.85
1.31………..X
2…………..82.07
X= 88.82%
D- Moments in beam :
αf5*( L2/L1)= 1.53>1 αf5*( L2/L1)= 1.37>1
M ( beam)= 0.85M ( C.S) M ( beam)= 0.85 M ( C.S)
E- Moments in the slab of C.S = 0.15 M ( C.s) .
********

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* Model ON SAP:
A- Slab modification:
In two way slab system
f11=f22=f12=V13=V23 = (Actual area/ Sap section area)
= (0.55*0.08+0.32*0.15)/ (0.3088*0.55) =0.54
M11=M22=M33= 0.25
Mass modifier = weight modifier = (8.9/(0.3088*25))= 1.15
B- Model Checks:
* Compatibility and Period is ok.

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* Equilibrium check.
Area of floor = 25* 38 = 950 m2
Live load = 950*6 = 5700 KN
Superimposed dead load = 950*3.5 = 3325 KN

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* Stress/Strain relationships (Moment Check):
M0 (interior span) = 2288 KN .m
From SAP: (1372+1385)/2 + 954 = 2332.5 KN .m
% of error = (2332.5-2288)/2288 = 1.94 % < 5% …. OK
Now all of the values of the moments calculated by the direct design method and
by the software (SAP) will be attached by an Excel sheet
Note: we have just five different values in the frame due to symmetry.
Note: the average width of column strip= (8+9)/4 = 4.25 m on both sides.
The width of middle strip = 10.5-4.25 = 6.25 m

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Notes:-
1- All the red cells represent that the area of steel is less than the minimum area of
steel in slab which is equal to bw*d*Ro min = 150*360*0.0033 = 180 mm2 , so this
values should be replaced by 180 mm2 .
2- The blue cells represent that the moment by draw section cut in SAP at the
exterior support in the column strip in more than the moment in the frame , so the
moment in the middle strip is negative and this region should be reinforced by the
minimum area of steel =180 mm2 .
3. The concrete zone (a) must be less than Hf= 80 mm, (As< 2493 mm2), to be able to
apply the steel percentage equation directly.

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4- The shrinkage steel in the slab:
As = 0.0018*1000*80 = 144 mm2
S max. = 5*h = 5* 80 = 400 mm
S= 1000/ (144/50) = 349 mm
so, use 1 Ø 8mm/ 350 mm in both directions.
Design of beam in frame 3 for shear and torsion:
No torsion in this beam because there is no different in span lengths and moment
values.
Vu (from SAP) = 606 KN at the interior edge of the exterior span
By hand calculation:
Wu = (8.5/2) * 24.5 = 104 KN/m ( at the top of the triangle )
W ( o.w) = 1.2*25*0.7*0.7 = 14.7 KN/m
Vu = ( M1+M2)/2 + W( o.w)*(L/2)+ 0.5*Wu*0.5*L
= (213.8+695)/2 + (14.7*8*0.5)+ ( 0.5*0.5*8*104) = 721.2 KN

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Design of beam for shear by SAP:
Av/s = 1.127 mm^2/mm
use Ø12 S = 226/1.127 = 200 mm use 1 Ø 12 / 200 mm
Design of columns: - we will use SAP to design the columns and then check one
column by hand calculation.
The framing type is sway ordinary.
A sample of steel percentage in the columns (frame 3).

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We will check the column with 2.34% steel percentage
to check if the column is braced or not, we did a second order analysis by sap and was
not there a change in the moment values, so we can assume the column is braced
Assume K=1 to be conservative:
KLu/r = 1*4700/ (0.3*700) = 22.38 so the column is short also.
Pu = 8065 KN Mu = 44 KN.m
gama=( 700 – 2*60)/ 700 = 0.83
Mu/bh^2=( (44*10^6)/ (700)^3)/7 = 0.02 Ksi
Pu/bh =( (8065*10^3)/ (700)^2)/7 = 2.35 Ksi
at gama = 0.75 Ro = 2% at gama = 0.9 Ro = 1.9 %
by interpolation Ro = 1.95 % As = 0.0195*700*700 = 9555 mm^2
Ro ( SAP) = 2.34% As= 0.0234*700*700 = 11466 mm^2 ( 14@32

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Note : there is a high difference between the hand and SAP calculations in column
steel , because the SAP designed for this moment and a minimum moment in the other
direction not only in one direction (Mu2=0 ,take Mu2 min. = 292.298 KN.m .
Ties spacing: S < 16*32 = 512 mm
S < 48*10 = 480 mm
S < 700 mm
So, use 1 Ø10 / 450 mm.

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*Details for the two way ribbed system: -
1- Cross section in beam in the middle of interior span of frame 3.
2 - Cross section in the middle strip in the middle of interior span of frame 3.
3- Cross section in the column strip slab in the middle of interior span of frame 3.
4- Cross section in column that was checked above.

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**********************************

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*Two Way Voided Slab System:
* Slab thickness and modification for the ribbed slab:
Assume αfm> 2:
H min. = (Ln*(0.8+fy/1400))/ (36+9β)
= (9.7*(0.8+420/1400))/ (36+9*1.35) = 0.221 m =221mm
Frame in south north direction (interior) (Frame 3) .
4mm8)/12=6.02*10312 =(670*2213= bh(solid)I
4mm8for the suggested cross section = 40.15*10(voided)I
(Calculated from sap frame section definition). I (voided) > I (solid) OK .

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To find the equivalent slab section for the ribbed slab:
40.15*108= (670*h3)/12 ….. h=415 mm.
Wd (O.W) = (0.67^2*0.46- 0.52^2*0.32)*25= 3 KN
Wd= 3/(0.67*0.67)=6.7 KN/m2
Wu = 1.2(3.5+6.7)+1.6*6= 21.84 KN/m2
Vu= 21.84*0.67*((10.5-0.8/0.5)-0.4) = 78 KN
∅VC = 0.75*
𝟏
𝟔
√𝟐𝟖*1.1*150*
𝟒𝟎𝟎
𝟏𝟎𝟎𝟎
= 45.6 KN
Vs= Vu - ∅VC = 32.4 KN
(Av/s) = (32.4*1000)/ (420*400) = 0.19 mm2/mm.
(Av/s) min = max. Of {0.35*(bw/fy), 0.062*(bw/fy)*√𝒇𝒄 } =0.125 mm2/mm
so, use (Av/s) = 0.19 mm2/mm.
S max.= d/2 = 400/2 = 200 mm use 1 Ø 8mm / 200 mm
The suggested beams dimensions are:
B1 (400 *800) for exterior and interior beams I (B) =1.7*10^10 mm4
Columns (800*800) mm

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αf1=0.71 αf2 =0.335 αf3 = 0.672 αf4 = 0.237 αf5 = 0.272
So we have to go to αfm< 0.2 case:-
h min. = Ln/30 = 9.7/30 = 0.32 m < h=415 mm …. OK
M( exterior spans) = (21.84*10.5*7.2^2)/8+ (8.4*7.3^2)/8= 1834.5 KN. m.
M( interior span) = (21.84*10.5*8.2^2)/8+ (8.4*7.3^2)/8= 2322 KN .m
Where :- the factored own weight of beam= 8.4 KN/m .

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( Table 13.6..4.4 for the positive moments )
L2/L1= 10.5/8=1.31 L2/L1= 10.5/9 = 1.17
αf5*( L2/L1< 1 αf5*( L2/L1)< 1
X=0.6 X=0.66
L2/L1= 10.5/8=1.31 L2/L1= 10.5/9 = 1.17
αf5*( L2/L1) < 1 αf5*( L2/L1) < 1
use interpolation ( X= 0.66) use interpolation (X=0.75)
( Table 13.6.4.2 for exterior negative moments )
C = (1- 0.63*(0.4/0.8))*((0.8*0.43)/3)= 0.01169 m4
Bt = C/(2*Is) = 0.01169/(2*((1/12)*4*0.415) =0.245< 2.5
Use interpolation
X= 0.972
D- Moments in beam :
αf5*( L2/L1) < 1 αf5*( L2/L1) <1
M ( beam)= 0.285M ( C.S) M ( beam)= 0.285 M ( C.S)
E- Moments in the slab of C.S = 0.15 M ( C.s) .
********

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* Model ON SAP:
f11=f22=f12=V13=V23 = (Actual area/ Sap section area)
= (2*0.67*0.07+0.32*0.15)/ (0.415*0.67) =0.51
M11=M22=M33= 0.25
Mass modifier = weight modifier = (6.7/(0.415*25))= 0.645

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B- Model Checks:

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Area of floor = 25* 38 = 950 m2
Live load = 950*6 = 5700 KN
M0 (interior span) = 2322 KN .m
From SAP: (8931+8011)/2 + 793= 2192 KN .m
% of error = (2322-2192)/2322 = 5.5 % < 10% …. OK

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Now all of the values of the moments calculated by the direct design method and
by the software (SAP) will be attached by an Excel sheet

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Notes:-
1- All the red cells represent that the area of steel is less than the minimum area of
steel in slab which is equal to bw*d*Ro min = 150*400*0.0033 = 198 mm2 , so this
values should be replaced by 198 mm2 .
2- The blue cells represent that the moment by draw section cut in SAP at the exterior
support in the column strip in more than the moment in the frame , so the moment in
the middle strip is negative and this region should be reinforced by the minimum area
of steel =198 mm2 .
3. The concrete zone (a) must be less than Hf= 70 mm, (As< 2657 mm2), to be able to
apply the steel percentage equation directly.
4- The shrinkage steel in the slab:
As = 0.0018*1000*70 = 126 mm2
S max. = 5*h = 5* 70 = 350 mm
S= 1000/ (144/50) = 349 mm
so, use 1 Ø 8mm/ 350 mm in both directions.

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Design of beam in frame 3 for shear and torsion:
No torsion in this beam because there is no different in span lengths and moment
values.
Wu = (8.5/2) * 21.84 = 92.82 KN/m ( at the top of the triangle )
W ( o.w) = 1.2*25*0.8*0.4 = 9.6 KN/m
Vu = ( M1+M2)/2 + W( o.w)*(L/2)+ 0.5*Wu*0.5*L
= (106.5+365.2)/2 + (9.6*8*0.5)+ ( 0.5*0.5*8*92.82) = 459 KN ( OK )

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Av/s = 0.965 mm^2/mm
use Ø12 S = 226/0.965 = 234 mm use 1 Ø 12 / 200 mm
Design of columns: - we will use SAP to design the columns and then check one
column by hand calculation.
We will check the column with 1% steel percentage
to check if the column is braced or not, we did a second order analysis by sap and was
not there a change in the moment values, so we can assume the column is braced
Pu = 2608 KN Mu = 24 KN.m
gama=( 800 – 2*60)/ 800 = 0.85
Mu/bh^2=( (24*10^6)/ (800)^3)/7 = 0.0.006 Ksi
Pu/bh =( (2608*10^3)/ (800)^2)/7 = 0.6 Ksi
at gama = 0.75 Ro = 1% at gama = 0.9 Ro = 1 %
by interpolation Ro = 1% As = 0.01*800*800 = 6400 mm^2 ( 8@32)

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S < 48*10 = 480 mm
S < 800 mm
So, use 1 Ø10 / 450 mm.
*Details for the two way ribbed system: -

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*Two Way Solid Slab System:
*. Slab thickness and modification for the ribbed slab:
Assume αfm> 2:
H min. = (Ln*(0.8+fy/1400))/ (36+9β)
Ln=max clear span =10.5-0.8=9.7 m Β=9.77.2
= (9.7*(0.8+420/1400))/ (36+9*1.34) = 0.222 m =222 mm
Try h=250 mm
To compute αf we need to compute αfm for every beam
Iedge beam =9.86*10-3 m4, L-section 800*600*400
Iinterior beam= 11.5*10-3 m4, T-section 600*400
Column 800*800

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Frame in east west direction:
I exterior = bh312 = (4*0.253)/12=5.27*10-3 m4
I interior = for the suggested cross section = 0.011m4
Frame in north south direction:
I exterior=0.0136 m4
I interior 1=0.012 m4
I interior 2=5.2*10-3
αf1=1.89 αf2 =1.05 αf3 = 1.89 αf4 = .96 αf5 = 0.85
Average for panels smallest =0.97<2
So we have to go to 2< αfm > 0.2 case:-

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H min. = (Ln*(0.8+fy/1400))/ (36+5β*(αfm-0.2))
= 0.263 try h=300mm …. OK
Wd= 0.3(25)/(3.5)=11 KN/m2
Wu = 1.2(11)+1.6*6= 22.8 KN/m2
Vu= 22.5* ((10.5-0.8/0.5)-0.26) = 104.6 KN
∅VC = 0.75*
𝟏
𝟔
√𝟐𝟖**1000*
𝟐𝟔𝟎
𝟏𝟎𝟎𝟎
= 172 KN
∅VC> Vu ok
M( exterior spans) = (22.8*10.5*8.6^2)/8= 2757.6 KN. m.
M( interior spans) = (22.8*10.5*7.6^2)/8= 1728.4 KN .m
Where :- the factored own weight of beam= 8.4 KN/m .

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Repeat αf for frame interior
I = 11.573*10-3 m
Is =0.023 m4
αf =0.5
( Table 13.6.4.4 for the positive moments )
L2/L1= 10.5/8=1.31 @L2/L1= 2
αf5*( L2/L1)= 0.85<1 αf5*( L2/L1)
use interpolation @ L2/L1=1 use interpolation
0 ………60 0 ………60
0.85…… X X=72.7% 0.85…..X X=47.2%
1 ……… 75 1………45
@ ( L2/L1)
Goal
Use interpolation
1 ………72.7
1.31…… X X=64.7%
2 ……… 47.2

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L2/L1= 10.5/8=1.31 L2/L1= 10.5/9 = 1.17
αf5*( L2/L1)= 0.85 αf5*( L2/L1)= 1.37>1
0 ………75 0 ………75
0.85…… X X=75% 0.85…..X X=49.5%
1 ……… 75 1………45
@( L2/L1)
Goal
Use interpolation
1 ………75
1.31…… X X=67%
2 ……… 49.2
(Table 13.6.4.2 for exterior negative moments)
C = (1- 0.63*(0.4/0.6))*((0.6*0.43)/3) + (1- 0.63*(0.3/0.4))*((0.4*0.333) = 8.44*10-3
m4
Bt = C/(2*Is) = 8.44*10-3 /(2*0.023) =0.183< 2.5
Use interpolation
X=100%

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D- Moments in beam:
αf5*(L2/L1) = 0.85<1
M (beam) = 0.85 *0.85 =0.72 M ( C.S)
* Model ON SAP:
f11=f22=f12=V13=V23 =1
M11=M22=M33= 0.25
Mass modifier = weight modifier = 1

47
B- Model Checks:
Area of floor = 25* 38 = 950 m2
Live load = 950*6 = 5700 KN

48
M0 (interior span) = 1728.4 KN .m
From SAP: (1132+1132)/2 + 814 = 1946 KN .m
% of error = (1946.5-1728.4)/ 1728.4 = 9 % < 10% …. OK
Now all of the values of the moments calculated by the direct design method
and by the software (SAP) will be attached by an Excel sheet

51
.Design of beam in frame 3 for shear and torsion:
No torsion in this beam because there is no different in span lengths and
moment values.
Wu = (8.5/2) * 22.8 = 96.9 KN/m ( at the top of the triangle )
W ( o.w) = 1.2*25*0.4*0.6 = 7.2 KN/m
Vu = ( M1+M2)/2 + W( o.w)*(L/2)+ 0.5*Wu*0.5*L
= (293+457)/2 + (7.2*8*0.5)+ ( 0.5*0.5*8*96.9) = 601 KN (OK) .
Av/s = 2.242 mm^2/mm
use Ø12 S = 226/2.242 = 101 mm use 1 Ø 12 / 100 mm

52
Design of columns: - we will use SAP to design the columns and then check
one column by hand calculation.
We will check the column with 1% steel percentage
to check if the column is braced or not, we did a second order analysis by
sap and was not there a change in the moment values, so we can assume the
column is braced
Pu = 2224 KN Mu = 50 KN.m
gama=( 800 – 2*60)/ 800 = 0.85
Mu/bh^2=( (50*10^6)/ (800)^3)/7 =0 Ksi
Pu/bh =( (2224*10^3)/ (800)^2)/7 = 0.49 Ksi
by interpolation Ro < 1% As = 0.01*800*800 = 6480 mm^2
( 8@32 ) .
S < 48*10 = 480 mm
S < 800 mm
so, use 1 Ø10 / 450 mm.

53
*Details for the two way solid system: -
1- Cross section in beam in the middle of interior span of frame 3.
2 - Cross section in the middle strip in the middle of interior span of frame 3.
3- Cross section in the column strip slab in the middle of interior span of
frame 3.
4- Cross section in column that was checked above.

55
*Conclusion, Recommendation:
There are differences between the hand calculation and SAP program result
but it’s Acceptable which not exceed 10%.
After analyzing each of the previous systems we have concluded that we
would use Low-rise concrete frame as our primary system. Using this we
would have a system of columns, girders, and voided box beams. This would
frame our first floor and second floor deck from there we would probably use
a steel structure to frame the second story and roof. After constructing and
executing our decision matrix we concluded that The Low-Rise Concrete
frame has the most to offer for our particular building. We used a weighted
scale to come to our conclusion. I’m sure that whether it be the owner or the
Architect they will have criteria that will have greater importance over
others. Our Group also has a good deal of previous system and we made
three floor systems to it.
The direct design method do not given accurate value so, We recommend to
use one way solid slab with girder system in order to decrease slab thickness
and to distribute the loads over coloumns and girders and increace beam
section .We recommend to use two way slab ribbed, but we offer ribbed slab
with hollow blocks to give lighter weight and good insulation too finally, We
recommend to use the most effective floor system , and the most
conservative system and low cost .
The end

Design of concrete structure 2 project-Loay Qabaha &Basel Salame

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