Chapter 1 : Slab system

Our slabs are designed as partially prestressed , that is 80% of self weight is designed as Post-tensioned flat slab and the remaining portions are designed as reinforced concrete.

A-Slab thickness

From table 9.5 (c) , the slab thickness chosen is l33 = 303.3 mm.

Thus , use 350 mm slab thickness :

B- Load calculations: 1.Dead load:

Plastering = 1510 * 21 = 31.5 Kg

Sand = 1 * 0.125 * 1800 = 225 KgMortar = 1 * 0.03 * 2100 = 63 Kg

Tile = 2010 * 20 = 40 Kg

Total DL = 360 Kg / m2 = 3.56 KN/m2

2 .Live Load: From the ACSE , the Live load is determined referring to the occupancy category of the building ( residential building ).

From the figure ; the LL is 4.79 KN¿m2

The next step is to start the design of our slabs using csi SAFE , however ; the self load multiplier of the slab own weight is set to be 0.2 since 0.8 of the slab weight will be carried by the Strands of post-tensioning , we'll start our

design by designing the slab as reinforced concrete:

1. Defining the materials , sections , spans , design strips , loads and load combinations .

The following figures will illustrates how to input the design info. Starting from the materials until being ready for running analysis and design.

The above figure illustrates splitting the Slab into design strips of width equal 10 m , this design strip is composed of a column strip of width 5 m , and two half middle strips with 2.5 width each.

Actually ; using the ACI Direct Design Method is not permitted here since the loads includes Non-gravity load ( earthquake loads and wind loads ) , however ; the csi SAFE will use a more accurate procedures of design , which is the Equivalent frame method.

Now defining the loads as follows :

Now after adding the loads , csi SAFE will automatically set the Design combinations and the design will be according to the

controlled one as showing in the figures below:

2- Design and analysis Results .

Now we are ready to the run the analysis and design of the slabs

Hint : moving the cursor at any point will give the value of the moment at that point ( eg. Max. moment is 445.4 KN/m

2).

As mentioned above , csi SAFE enable us to select the bar size that we will use in the slabs : the following figures will

illustrates the complete design of the strips:

Note : the blue line indicates the limit for the minimum reinforcement ; i.e : most slabs will require only minimum reinforcement except the exterior spans , that’s because a large quantity of the load is held by the STRANDS ( 80 % of self

weight )!!!!!!

In the N-S Direction:

In The E-W Direction:

Note here the top steel will require a design rather than minimum reinforcement :

Moving the cursor across the diagram ( reinforcement diagram ) will give the required amount of bars , that’s enable us easily for applying bar cut-off for more economic design.

Requirement of placing steel bars:

While applying cut off of bars in two way slabs , some bars must remain constant across the strips and must not be cut , the ACI

code showed this thing and recommended the following :

These figures is clear enough for applying cut-off providing that the requirement of the Development length is satisfied.

Summary of Reinforcement :

Provide minimum reinforcement bottom steel in the Design strips , That is ρmin=0 .0018 Ag uniformly distributed across the bay width , also provide the required amount of top steel as suggested by csi SAFE uniformly distributed over the effective

band with of the support.

Manual Design of post-tensioned concrete slab to balance 80% of self weight of the slab.

MATERIALS:

-f'c = 35 Mpa-ɣ for concrete = 24 KN/m.s

-fci'= 0.75f'c= 27 Mpa-fpu= 1860 Mpa , fps not exceeding 1276 Mpa

-fpy = 1690 Mpa-fpe = 1100 Mpa-Eps = 200 Gpa

-fy = 420 Mpa -Stress-relived strands ( K 270 )

-Slab thickness = 350 mm. -Unsupported length of the column ( story height ) = 4 m

-ft = 0.25√ f ' c = 1.48 Mpa. -fc = 0.45 f'c = 15.8 Mpa.

-Self weight = 24 * 0.35 = 8.4 KN/m.s-Balance ratio = 0.8 →0.8 * 8.4 = 6.72 KN/m.s

The analysis here is based on the Equivalent-frame method as recommended by the ACI Code:

As shown in the figure , the analysis here is based on an

equivalent frame taken in the N-S direction , since all spans

are same in magnitudes , also all columns are equal in dimension , analysis for one equivalent frame is enough for finding the required # of strands and non-bonded reinforcement for the whole slabs in the both directions.

Hmin= L / 30 = 10000/30 = 333.33 mm →Use h = 350 mm .Ww = 6.72 KN/m.sWu = 1.2 * 6.72 = 8.06 KN/m.s

→assume an average intensity of compressive stress due to balancing is fc = 1.17 Mpa , then unit F = 1.17 *350*1000 =

409.5 KN/m , so ; trying 12in diameter seven wires stress relived

strands ; Pe = Aps * fpe = 99*1100 = 108.9 KN , for L = 10 m ,

total Fe = 409.5 * 10 = 4095 KN .# of strands per bay is FePe =

4095/108.9 = 37 strands , and total Pe = Fe = 108.9 * 37 = 4029.3 KN , the actual unit force F = 4029.3/10 = 402.93 KN/m

, and the actual unit stress is fc = 402.9∗103

1000∗350 = 1.15 Mpa < 1.17

Mpa , which is satisfactory . Consequently , use fc = 1.15 Mpa due to load balancing , and assume a tendon profile as shown in

the figure below :

-a1 = 175+300

2 - 50 = 187.5 mm , a2 = 350 – 50 – 100 = 200 mm.

Wbal = 8F a

l2 =

8∗402.93∗0.1875102

= 6.044 KN/m.s

Wnet = 6.72 – 6.044 = 0.676 KN/m.s for the exterior span.

Wbal = 8F a

l2 =

8∗402.93∗0.2102

= 6.45 KN/m.s

Wnet = 6.72 – 6.45 = 0.27 KN/m.s for the interior span.

Equivalent frame characteristic

Kc = 4 I c Ec

l−2h =

4∗Ec∗400∗4003∗112

4000−(2∗350 ) = 2585858.586 Ec

C = ∑ (¿1−0.63 xy)( x3 y3 )¿

) = 1 -0.63 * 350400 ) * ( 3503∗400

3¿ =2565354167 mm4

Kt = ∑ 9C Ecs

l 2(1− c2l 2

)3 =

9∗2565354167∗Ecs

10,000∗(1− 40010000

)3 +

9∗2565354167∗Ecs

10,000∗(1− 40010000

)3

= 5219226.414Ecs

Kec = 1

1kc

+1k t

= 1

12585858.586

+1

5219226.414 = 1729152.367 Ecs

Ks = 4 I c Ecs

ln−C1

2 =

4∗10000∗3503∗Ecs

12∗(10000−4002 ) = 14583333.33 Ecs

Since all members are of the same concrete material , Ec cancels each other.

Exterior column :

DF = Ks

∑ K

=Ks

Kec+Ks ¿+Ks¿=

14583333.33

14583333.33+1729152.367 = 0.894

Interior columns:

DF = Ks

∑ K=

Ks

Kec+Ks ¿+Ks¿=

14583333.33

14583333.33+14583333.33+1729152.367 = 0.472

EXT. Span : FEM = w l2

12 = 0.676∗10

2

12 = 5.633 KN.m

Int . Span : FEM = w l2

12 = 0.27∗10

2

12 = 2.25 KN.m

DCBAPointDC DE

CB CD

BA BC

ABMember

0.47 0.470.47 0.470.47 0.470.894DF+2.25- 2.25

0 0+2.25- 2.25

0 0+5.6-

2.25-1.6- 1.6

-5.64.98

FEMDist.

-0.70.37 0.37

+ 2.5 0 -1.17- 0.95

-0.790.71

CODist.

5.36- 5.01

-0.734Final Mnet

Mnet = 5.36 KN.m

Slab concrete tensile stress at support.

Maximum Mnet = 5.36 – ( 0.27∗102 *

0.43 ) = 5.18 KN.m

Slab section modulus , S = b h2

6 = 10000∗350

2

6 = 204166666.7mm3.

Ftop = −P

A + MS = -1.15 +

5.18∗106

204166666.67 = -1.13 Mpa , OK.

Slab concrete tensile stress at Midspan.

Maximum Mnet = 0.27∗102

8 - 5.36 = -1.805 KN.m

Ftop = -1.15 + −1.805∗106

204166666.67 = -1.158 Mpa , OK .

Ultimate flexural strength analysis

1. Balanced Moment

-Exterior span :

FEMbal=W b l2

12 = 6.044∗102

12 = 50.4 KN.m

-Interior span:

FEMbal=W b l2

12 = 6.45∗102

12 = 53.8 KN.m

Now Running a moment distribution will determine the maximum balanced moment for the exterior column joints.

DCBAPointDC DE

CB CD

BA BC

ABMember

0.47 0.470.47 0.470.47 0.470.894DF+53.8- 53.8

0 0+53.8- 53.8

0 050.4-

53.8+1.6- 1.6

-50.444.8

FEMDist.

-0.80.37 0.37

22.4 0 -10.5- 10.5

0.80.71-

CODist.

63.9- 65.9

-5.46Final Mnet

2. Secondary Moments ( Ms ) and factored load moment ( Mu )

-Exterior span :

From tendon profile , e = 0.0 mm.

M1 = Primary moment = Pe.e = 0.0 KN.m

M bal- = 5.46 KN.m

Ms = M bal - M1 = -5.46 KN.m

Factored FEM = W u l2

12 = 8.064∗10

2

12 = 67.2 KN.m

-Interior span :

From tendon profile , e = 175-50 = 125 mm

M1 = Pe . e = 402.93 * 125 = 50.4 KN.m

M bal = 63.9 KN.m

Ms = 50.4 – 63.9 = 13.5 KN.m

Factored FEM = 67.2 KN.m

And now running a moment distribution for the factored moments , here analysis of pattern loading should be made to determine the worst case of the service and factored loads.

DCBAPointDC DE

CB CD

BA BC

ABMember

0.47 0.470.47 0.470.47 0.470.894DF+67.2- 67.2

0 067.2-

67.20 0

-67.259.81

FEMDist.

-0.80.37 0.37

29.9-14.1- 14.1

00

CODist.

82.2- 81.3

-5.46Final Mnet

3. Design Moments Mu

The design moments are the difference between the factored load moments and the secondary moments.

Exterior span:

Mu = 7.4 – 5.46 = 1.94 KN.m

Interior span BA :

Mu = 82.2 -13.5 = 68.7 KN.m

Interior span BC :

Mu = 81.3 – 15.5 = 65.8 KN.m

Moment reduction to the column face of support = V . c3

V = W u .l

2 -

∆ Mln

For the exterior span :

V = 8.064∗10

2 – 82.2−7.410 = 32.84 KN.m

V . c3

= 32.84∗0.42

= 4.4 KN.m

Centerline moment = 6.6 – 1.94 = 2.5 KN.m

Required Mn = Mu∅ =

4.70.9 = 5.2 KN.m

For the interior span BA:

V = 8.064∗10

2 – 7.4−82.210 = 47.8 KN.m

V . c3

= 47.8∗0.43

= 6.4 KN.m

Mu = 68.7 – 6.4 = 62.3 KN.m

Required Mn = Mu∅ =

62.30.9 = 69.2 KN.m

For the interior span BC:

Since the factored moment is almost the same →the required moment is smaller than the previous span , thus ignore it.

Span AB maximum positive moment

Assume point of zero shear ( or max. moment ) is at ( X ) m from face A :

X = V AB

W AB =

32.848.064 = 4.07 m

Maximum positive moment = V AB . x−W u x2

2−M u+M s

=4.07*32.84−8.06∗4.072

2−¿ 7.4 + 5.46

=64.93 KN.m

Required M n=M u

∅ = 64.930.9 = 72.14 KN.m

Span AB maximum positive moment

Unadjusted Max. +ve moment = W u .l2

8 = 8.064∗10

2

8 = 100.1 KN.m

+ve Moment = 100.1 – 82.2 + 15.5 = 33.4 KN.m

3 . Flexural strength ( Nominal moment strength )

1-Interior supports ( top face of the support )

As required by the ACI Code :

As = 0.00075 Acf

= 0.00075 * 350 * 1000 = 262.5 mm2

Thus use 5 ∅ 10 → As provided = 71*5 = 355 mm2

Spacing = 1000– (5∗10)

4 = 238 mm ≤ 300 mm , Ok.

ρp = A ps

bd =

36∗9910000∗300 = 0.0014

f pe = 1100 Mpa.

f ps =f pe + 70 + f c

'

100 ρp

= 100 + 70 + 35

100∗.0012 = 1462 Mpa.

F ps = 1462∗99∗36

10 = 521.1 KN per meter width of the slab.

F s = 420 * 355 = 149.1 KN per meter width of the slab.

Now , the total force provided by both reinforcement types

( bonded and non bonded ) is:

F t = 521.1 + 149.1 = 670.2 KN.

a=A s . f y+ Aps . f ps

0.85 f c' b

= 670.2∗103

0.85∗35∗1000 = 22.53 mm

ad p

= 22.53300

= 0.08 < 0.319 → T.C , Ok

.

M n=( A s . f y+ Aps . f ps )(d−a2 )

= 670.2 ) * 300 – 22.532 = ( 193.5 KN.m < 69.2 KN.m , Ok.

2- Mid-span section ( +ve moment )

F ps = 521.1 KN

a= 521.1∗103

0.85∗35∗1000 =17.5 mm

ad

< 0.319 , Ok ,,,,, TC!

M n=521.1∗(300−17.52 )=136.6KN >72.14KN .m

Summary :

Use 37 Strands uniformly distributed across the bay width with 5 ∅ 10 at the top fiber at the support section .