design of concrete structure 2 project-loay qabaha &basel salame

1

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

62\11\6112

Design of concrete structure 2 project

3

*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.

4

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.

5

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..0 to 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 , �́�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

6

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

7

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.

8

*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)

4mm 8)/12=5.57*103(550*23012 =\3bh =(solid)I

4mm 813.5*10for the suggested cross section = (ribbed)I

(Calculated from sap frame section definition). I (ribbed) > I (solid) OK .

9

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))

11

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 .

11

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

12

(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 %

13

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) .

********

14

* 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.

15

* Equilibrium check.

Area of floor = 25* 38 = 950 m2

Live load = 950*6 = 5700 KN

Superimposed dead load = 950*3.5 = 3325 KN

16

* 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

18

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.

19

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

21

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).

21

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

22

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.

23

*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.

25

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

26

*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) .

4mm 8)/12=6.02*10312 =(670*221\3= bh(solid)I

4mm 8for the suggested cross section = 40.15*10 (voided)I

(Calculated from sap frame section definition). I (voided) > I (solid) OK .

27

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


28

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 .






29





( 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) .

********

31

* 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

31

B- Model Checks:


32


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

33

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




34

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.

35

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 )


36

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

Assume K=1 to be conservative:


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)

37


S < 48*10 = 480 mm

S < 800 mm

So, use 1 Ø10 / 450 mm.

*Details for the two way ribbed system: -

41

*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=9.7\7.2

= (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

42

Frame in east west direction:

I exterior = bh3\12 = (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:-


43

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 .







44



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

45


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*0.33\3) = 8.44*10-3

m4

Bt = C/(2*Is) = 8.44*10-3 /(2*0.023) =0.183< 2.5

Use interpolation

X=100%

46

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 Assume K=1 to be conservative:


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

Engineering