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Design of residential buildingThis example will discuss the procedures for design of each structural element of

residential building.

The number of floors = 5 floors.

Compressive strength of used concrete = 240 kg/cm2 for beams and slabs and

280 kg/cm2 for columns and foundations.

Yield strength of used steel bars 4200 kg/cm2.

Live load = 200 kg/cm2.

γsoil=1.8 t /m3

γ Reinforced Concrete=2.5 t /m3 .

γmorter=2.1 t /m3 .

γ plaster=2.1t /m3 .

qall(gross)=2.3 Kg /c m2 .

Df =1.5 m.

Concrete cover:

Concrete cover = 7.5 cm for the underground element.

Concrete cover = 2.5 cm for slab.

Concrete cover = 4 cm for other elements.

Figure 1: Ground floor map (plan)

Figure 2: First floor map (plan)

Procedures1. Columns distribution.

2. Beams distribution.

3. Calculate slab thickness.

4. Calculate slab loads.

5. Design of ribs.

6. Design of beams.

7. Design of columns.

8. Design of footing.

9. Preparation of calculation sheet.

10. Preparation of structural drawing.

First: Columns distribution

- The distribution of columns must be appropriate to the architecture plan.

Figure 3: Distribution of columns on ground floor map

Second: Beams distribution

- Determine the direction of ribs in the slab.

- Draw the initial distribution of the beams.

- Draw ribs and blocks.

Notes:

The main beams is the beam perpendicular to the direction of ribs, secondary

beams is parallel to the direction of the ribs.

Ribs are not to be less than 10 cm in width (prefer 12 to facilate steel and

casting works).

Clear spacing between ribs is not to exceed 75.0 cm (usually used width of

block which equal 40 cm).

Figure 4: Distribution of beams and ribs

Third: Calculate slab thickness.

Minimum thickness of one-way ribbed slabs

ElementSimply

supportedOne end

continuousBoth ends continuous

Cantilever

One-way ribbed slabs

l/16l/18.5l/21l/8

where l is the span length in the direction of bending and measure center to center.

Notes:

The most common concrete hollow block sizes are 40 × 25 cm in plan and

heights of 14, 17, 20 and 24 cm.

Topping slab thickness is not to be less than 1/12 the clear distance between

ribs, nor less than 5.0 cm.

Calculate the thickness of (rib) slab:

Member Type Maximum Length Minimum Thickness (cm)

- Simply supported 363.5 363.516

=22.72 cm

- One end continuous 407.5407.518.5

=22.03 cm

- Both end continuous 383.5 383.521

=18.26 cm

- Cantilever 1501508

=18.75 cm

So minimum thickness for rib = 24.32 cm

Calculate the thickness of main beam:


- Simply supported - -

C1C7 One end continuous 51051018.5

=27.57cm

C5C10 Both end continuous 54854821

=26.10 cm

- Cantilever - -

So minimum thickness for main beam = 27.57 cm

Calculate the thickness of secondary beam:


C1C2 Simply supported 42042016

=26.25 cm


=22.97 cm


=13.33cm

Cantilever Cantilever 1501508

=18.75 cm

So minimum thickness for secondary beam = 27.57 cm

So the minimum thickness for beam & slab = 28 cm.

Use block 40*25* 20 and ribs of 12 cm.

So the topping slab thickness = 28 - 20 = 8 cm.

Note:

Depth of rib not more than 3.5 times the minimum web width.

Fourth: Calculate slab loads

The load on ribs consists of:

1. The dead load:

Own weight of the slab.

Weight of the surface finish.

Equivalent partition load.

2. The live load is dependant on the intended use of the building.

Own weight of slab:

Figure 5: Section in rib

Total volume (hatched) = 0.52×0.25×0.3 = 0.0364 m³

Volume of one hollow block = 0.40×0.25×0.20 = 0.020 m³

Net concrete volume = 0.0364 - 0.020 = 0.0164 m³

Weight of concrete = 0.0164 ×2.5 = 0.041 ton

Weight of concrete /m² =0.041

0.52× 0.25=0.3153ton/m²

Weight of hollow block /m² = 20 /1000

0.52× 0.25=¿ 0.1538 ton/m²

Own weight = 0.3153 + 0.1538 = 0.4692 ton/m²

Weight of slab covering materials:

This weight per unit area depends on the type of finishing which is usually made of

Weight of layer = thickness × unit weight ton/m2

Weight of plaster = 0.015 × 2.1 = 0.0315 ton/m2

Weight of sand = 0.07× 1.8 = 0.126 ton/m2

Weight of mortar = 0.025 × 2.1 = 0.0525 ton/m2

Weight of tiles = 0.025 × 2.1 = 0.052 ton/m2

Total weight = 0.2625 ton/m2

6 6 cm

40

12 cm

12 6

24

6 cm

Weight of equivalent partition load:

- Number of blocks per meter square = 12.5 block.

- Two layers of plaster each layer have 1.5 cm thick.

- Height of wall = 2.95 m.

For block 10:

Weight of blocks = 12.5 × 10 = 125 kg/m2

Weight of plaster = 2×0.015×2100 = 63 kg/m2

Total weight / m2 = 125 + 63 = 188 kg/m2

Wu/m ' = 188*2.95 = 554.6 kg/m'

Length of wall 10 cm = 37.6 m

Total weight of block 10 = 0.5546 × 37.6 = 20.85296 ton

Figure 6-a: Section on partition

Figure 6-b: Partition view

Partial partition load calculations:

Total weight of all partition wall = 20.85296 ton

Total Area = 162 m2

Stairs Area = 8.12 m2

Equivalent partiton load (EPL) =20.85

162−8.12=0.1355 ton/m2

Total Service Dead Load = O.W +C.M + EPL

= 0.4692 + 0.2625 + 0.1355= 0.8672 t/m2.

Live Load:

- It depends on the purpose for which the floor is constructed. Table 7.2 shows typical

values used by the Uniform Building Code (UBC).

- Its equal 200 to 250 kg/m2 in residential building.

Live Load (L.L) = 0.20 ton /m2 (In this example).

Total factored load per meter square of the slab:

Wu= 1.2(0.8672) + 1.6(0.20) =1.3606 ton/m2

Total factored load on each rib:

Wu,rib= 1.3606 × 40+12100 = 0.7075 ton/m

Fifth: Design of ribs

- Take several strips of persistent rip representing all the ribs in the slab.

A. Check rib width for beam shear:

First rib:

The rib width must be adequate to resist beam shear by satisfying the equation

1.1ФVC>Vu max .

The shear force diagram is shown in Figure 7.

Figure 7: Shear force diagram for first rib

Vu max = 1.65 ton.

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

d= 28−2.5−0.6−1.22

=¿ 24.3 cm

1.1ΦVc = 1.1Φ×0.53×√F c×b×d

= 1.1 × (0.75 ) ×(0.53)× 12× 24.3×√240

= 1975.25/1000 = 1.975 ton > 1.65 ton. OK.

Though shear rainforcement is not required , but is recommended to use Φ6 mm @

25 cm U-stirrups to carry the bottom flexural rainforcement.

Second rib:

The rib width must be adequate to resist beam shear by satisfying the equation

1.1ФVC>Vu max .


Figure 8: Shear force diagram for second rib

Vu max= 1.78 ton


D = 28−2.5−0.6−1.22

=¿ 24.3 cm

1.1ΦVc = 1.1Φ×0.53×√F c×b×d

= 1.1 × (0.75 ) ×(0.53)× 12× 24.3×√240

= 1975.25/1000 = 1.975 ton > 1.78 ton. OK.



B. Design flexural rainforcement:

Design for shrinkage reinforcement:

Use the minimum value of shrinkage reinforcement as follows:

As min ¿0.0018 ×b× h

calculate for astrip of 1m wide (b =100 cm) , htopping = 8 cm.

As min ¿0.0018 ×100 ×8=1.44 cm2

Use Φ6 mm diameter with area of 0.2826 cm2

No. bars = 1.44 / 0.2826 = 5.0 though use 5Φ6 mm / m

So use Φ6 mm @ 20 cm

First rib:

The rib is designed as a rectangular section

The bending moment diagram is shown in Figure 9.

Figure 9: Bending moment diagram in first rib

Positive moment

Mu max = 0.93 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105( 0. 93 )

0 . 9×12×(24 . 3 )2×240 ]ρreq = 0.003606

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85

4200=0.01548 ρ max > ρ req OK. the section is tension

controlled

Use ρ = ρ req = 0.003606

As,+ve ¿ ρ ×b× d = 0.003606×12× 24.3 = 1.051 cm2

Use Φ 10 mm diameter with area of 0.79 cm2 and Φ 12 mm diameter with area of

1.13 cm2.

No. bars = 1.051 / 0.79 = 1.33 though use 1 Φ10 mm and 1 Φ12 mm.

S = b−2 (concrete cover )−2 (φstirrup )−n(φbar)

n−1

S = 12−2 (2.5 )−2 (0.6 )−(1)−(1.2)

2−1 = 3.6 > 2.5 > 1.2 OK.

Negative moment

Mu max = -0.87 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105( 0. 87 )

0 . 9×12×(24 . 3 )2×240 ]ρreq = 0.003365

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled.

Use ρ = ρ req =0.003365

As,+ve ¿ ρ ×b× d = 0.003365 ×12× 24.3 = 0.981 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars = 0.981 / 1.13 = 0.86 though use 1 Φ12 mm.


n−1

S = 12−2 (2.5 )−2 (0.6 )−1(1.2)

2−1 = 4.6 > 2.5 > 1.2 OK.

Negative moment


ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(240 )4200 [1−√1− 2 .353×105 (0. 20 )

0 . 9×12×(24 . 3 )2×240 ]ρreq = 0.00075

ρmin = 14fy

= 144200 = 0.00333 ρ req < ρ min OK.

1= 0.85 since fc`= 240 < 280

Use ρ = ρ min = 0.003333

As,+ve ¿ ρ ×b× d = 0.003333 ×12× 24.3 = 0.972 cm2




n−1

S = 12−2 (2.5 )−2 (0.6 )−1(1.2)

2−1 = 4.6 > 2.5 > 1.2 OK.

Negative moment


ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105( 0. 87 )

0 . 9×12×(24 . 3 )2×240 ]ρreq = 0.003365

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled.

Use ρ = ρ req =0.003365

As,+ve ¿ ρ ×b× d = 0.003365 ×12× 24.3 = 0.981 cm2

Use Φ 10 mm diameter with area of 0.79 cm2 and Φ 12 mm diameter with area of

1.13 cm2.

No. bars = 0.981 / 0.79 = 1.24 though use 1 Φ10 mm and 1 Φ12 mm.


n−1

S = 12−2 (2.5 )−2 (0.6 )−(1)−(1.2)

2−1 = 3.6 > 2.5 > 1.2 OK.

Positive moment

Mu max = 0.98 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(240 )4200 [1−√1− 2 .353×105( 0. 98 )

0 . 9×12×(24 . 3 )2×240 ]ρreq = 0.003365

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled.

Use ρ = ρ req =0.003808

As,+ve ¿ ρ ×b× d = 0.003808 ×12× 24.3 = 1.11 cm2




n−1

S = 12−2 (2.5 )−2 (0.6 )−2(1)

2−1 = 3.8 > 2.5 > 1 OK.

Sixth: Design of beams

There are two types of beams, main and secondary beams.

Main beams: the following loads affect it:

- Self weight = width × thickness × 2500 kg/m'

- Half of the ribs from each side.

- If the external beam, add weight of 1 linear meters of exterior walls.

- If one end of the beam is cantilever, take the all distance not half of it, and add the

weight of Crans.

Secondary beams: the following loads affect it:

- Self weight = width × thickness × 2500 kg/m'

- If the external beam, add weight of 1 linear meters of exterior walls.

Note:

If the live load is maximum, use the envelope moment in the anylsis of beams.

Design of beams for flexure:

Main beam (B1):

The bending moment diagram is shown Figure 11.

Figure 11: Bending moment diagram for main beam 1

h = 28 cm (thickness of slab) b = 70 cm

Assume Φ16 mm reinforcing bars and Φ8 mm stirrups.

d = 28−4−(0.8 )−¿1.2/2) = 22.6 cm

Load assigned to the beam:

1. Slab load = [Rib ¿

2+Rib¿

2]× Wu =[3.225

2+ 0

2 ]× 1.3606=2.1939 ton /m

2. Own wieght =

1.2 [ (0.7 ×0.28 ×2.5 )+(0.1335+0.2625)]+1.6 [ 0.7 ×0.2 ]=1.2872 ton /m

3. External wall load =1.2 ×0.92335=1.01802ton /m

Load assigned to the beam = [2.1939+1.2872+1.01802] =4.45 ton /m

Positive moment

Mu max = 9.07 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2. 353×105(9 .07 )

0 .9×70×(22 .6)2×240 ]ρreq = 0.007253

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.007253

As,+ve ¿ ρ ×b× d = 0.007253×70× 22.6 = 11.47 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−11(1.2)

11−1 = 4.7 > 2.5 > 1.2 OK.

Negative moment


ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105 (10 .68 )

0 .9×70×(22 .6)2×240 ]

ρreq = 0.008678

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.008678

As,+ve ¿ ρ ×b× d = 0.008678×70× 22.6 = 13.72 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−13(1.2)

13−1 = 3.73 > 2.5 > 1.2 OK.

Positive moment

Mu max = 1.7 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105 (1.7)

0 .9×70×(22 .6)2×240 ]ρreq = 0.001275

ρmin = 14fy

= 144200 = 0.00333 ρ req < ρ min OK.

1= 0.85 since fc`= 240 < 280

Use ρ = ρ req = 0.001275

As,+ve ¿ ρ ×b× d = 0.00333×70× 22.6 = 5.26 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−5(1.2)

5−1 = 13.6 > 2.5 > 1.2 OK.

Negative moment


ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105(6 .68 )

0 .9×70×(22 .6)2×240 ]ρreq = 0.005224

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.005224

As,+ve ¿ ρ ×b× d = 0.005224 ×70× 22.6 = 8.26 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−8(1.2)

8−1 = 7.25 > 2.5 > 1.2 OK.

Positive moment

Mu max = 5.57 ton.m

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105 (5 .57 )

0 .9×70×(22 .6)2×240 ]

ρreq = 0.004313

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.004313

As,+ve ¿ ρ ×b× d = 0.004313 ×70× 22.6 = 6.82 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−7(1.2)

7−1 = 8.66 > 2.5 > 1.2 OK.

Design of beams for shear

Main beam (B1):


Figure 13: Shear force diagram for main beam 3

Beam width (B1) = 70 cm.


d = 28−4−(0.8 )−¿1.2/2) = 22.6 cm.

Vu max, d =12.30

ФVC= 0 .75(0 . 53)1000

×√240×70×22 .6=9.74 ton .

Check ductile mode of failure:

2.2×√ fc ' ×b×d = 2.2×√240×70×22.6× 10-3 = 53.91 ton.

Vs= V u, d−Ф V C

Ф = 12.30−9.74

0.75 = 3.41 ton < 2.2×√ fc ' ×b×d C

Now assume 4 legged Φ8 mm stirrups

Av = 4 × π4

× 0.82 = 2 cm2 , d = 22.6 cm , bw = 70 cm , Vs = 3.41 ton.

1.1×√ fc ' ×b×d = 1.1×√240×70×22.6×10-3 = 26.959 > Vs

S = 22.6 ×2 ×4200

3.41 ×1000=¿55.67 cm, Smax =

2× 42003.5 ×70

=¿ 34.2 cm.

d2=22.6

2=11.3cm.

The smallest value is 11.3 cm so that (d2 = max value of S)

The value of S = 11 cm will be taken for the whole sections of B1.

Seventh: Design of columns

- The loads of columns can be determined by two methods reaction method and area

method. (In this example area method are used)

- Calculate the bounded area as show in Figure 17.

- After finished the design of columns, draw the final dimensions of columns on the

maps.

Figure 17: Area of columns

Calculation of loads

Table 1: Loads on columns

Design of tied column:

Columns 1, Columns Group (3):

Ribs area = 6.345 −¿ 2.517 = 3.828 m2

Own weight of one column =2.5×0.4×0.2×2.95 = 0.59 ton

Slab Load = 3.828 ×0.867245 + 2.517 × (0.28× 2.5 + 0.1355 + 0.2625) = 6.083 ton

External walls load = 5.05 × 0.92335 = 4.6629 ton

Total Dead load = 0.59 + 6.083 + 4.6629 =11.336 ton

Total live load = 0.20 ×6.345 = 1.269 ton

Pu = 5 × (1.2 × 11.336 +1.6 ×1.269 ) = 78.17 ton

B = 20 cm , ρg=1 %

Ag=Pu

0.52 [0.85 fc +ρg ( fy−0.85 fc ) ]

Ag=78170

0.52[0.85× 280+0.01 (4200−0.85 ×280 )]=541.49 cm2

h=541.4920

=27.07 cm

The minimum column dimension is 20 cm× 40 cm So, use h = 40 cm

A s=0.01 ×541.49=5.4149 cm2

use6∅ 12mm

Check spacing:

Col no. Total area Beam area Ribs area Own wt. External wall length Total DL Total L.L DL service LL service Pu C1 6.35 2.52 3.828 0.59 5.05 11.34 1.27 56.68 6.35 15.63C2 10.11 2.24 7.871 0.59 6.5 15.87 2.02 79.36 10.11 22.28C3 10.04 2.23 7.809 0.59 6.45 15.77 2.01 78.84 10.04 22.14C4 6.73 2.66 4.072 0.59 5.2 11.84 1.35 59.19 6.73 16.36C5 16.32 4.44 11.876 0.59 0 15.77 3.26 78.84 16.32 24.15C6 16.08 4.43 11.649 0.59 0 15.56 3.22 77.79 16.08 23.82C7 11.28 3.29 3.36 0.59 4.8 11.55 2.26 57.74 11.28 17.47C8 13.45 4.04 9.41 0.59 4.6 17.43 2.69 87.17 13.45 25.22C9 9.69 2.89 6.8068 0.59 4.13 13.48 1.94 67.38 9.69 19.27C10 16.15 3.80 12.35 0.59 0 15.47 3.23 77.36 16.15 23.74C11 15.91 3.80 12.112 0.59 0 15.27 3.18 76.33 15.91 23.41C12 12.71 3.56 9.153 0.59 4.13 16.24 2.54 81.22 12.71 23.56C13 5.11 2.35 2.764 0.59 4.5 9.72 1.02 48.60 5.11 13.30C14 7.40 3.03 4.37 0.59 3.4 10.84 1.48 54.20 7.40 15.38C15 7.29 3.03 4.261 0.59 3.35 10.70 1.46 53.50 7.29 15.17C16 5.00 2.33 2.67 0.59 4.45 9.58 1.00 47.88 5.00 13.09

Sclear=40−2× 4−2×0.8−6 ×1.2

6−1=4.64 cmOK .

Sties = Smallest of (16×1.2 or 48×0.8 or 20) = 19 cm

Use One (ties) 8 mm @ 19 cm

Table 2: Detailed design of columns

Col no. Pu Factor Ag Req. As b h required h actual Dimenition ф # of Bars Space Smax of ties C1 78.17 541.49 5.41 20 28 40 20x40 12 5 6.50 19C2 111.40 771.68 7.72 20 39 40 20x40 12 7 3.81 19C3 110.68 766.67 7.67 20 39 40 20x40 12 7 3.85 19C4 81.79 566.57 5.67 20 29 40 20x40 12 5 6.08 19C5 120.73 836.27 8.36 20 42 45 20x45 12 7 4.15 19C6 119.08 824.83 8.25 20 42 45 20x45 12 7 4.23 19C7 87.34 605.00 6.05 20 31 40 20x40 12 5 5.51 19C8 126.12 873.67 8.74 20 44 45 20x45 12 8 3.88 19C9 96.37 667.55 6.68 20 34 40 20x40 12 6 4.75 19C10 118.68 822.08 8.22 20 42 45 20x45 12 7 4.25 19C11 117.06 810.86 8.11 20 41 45 20x45 12 7 4.34 19C12 117.80 816.01 8.16 20 41 45 20x45 12 7 4.30 19C13 66.49 460.60 4.61 20 24 40 20x40 12 4 8.30 19C14 76.88 532.52 5.33 20 27 40 20x40 12 5 6.67 19C15 75.86 525.47 5.25 20 27 40 20x40 12 5 6.80 19C16 65.45 453.40 4.53 20 23 40 20x40 12 4 8.50 19

Table 3: Groups of columns

Eighth: Design of footing

There are three groups of footings according the groups of columns.

Design of the first group:

The unfactored load of column (Pservice) = 100.62 ton.

The factored load of column (Pu) = 126.125 ton.

Assume footing thickness (h = 45 cm) & Φ12 mm reinforcing bars.

hc= 45 cm , hs = 150 - 45 = 105 cm , C1=25 cm , C2 =45 cm.

qall(net) = qall, (gross) −¿hc c −¿ hs s

q all(net )=20−0.45× 2.5−1.05×1.8=¿19.985 ton/m2

davg = 45−7.5−1.2= 36.3 cm

Areq = P service

qall , net=100.62

19.985 = 5.034 m2

Assume L= 2.25 m B= 5.0342.25 = 2.237 ~ 2.25 m.

qu,net = Pu

Anew= 126.1252

2.25 × 2.25=¿ 24.913 ton/m2

Check footing thickness for punching shear:

This condition must be satisfies Vu≤ ΦVc

Vu = qu ,net ×[ (B × L )−(C1+davg ) (C2+davg )]

= 24.913 ×[(2.25 ×2.25 )−(0.25+0.363 ) (0.45+0.363 ) ]

= 113.70 ton.

bo = 2 (C1+davg )+2 (C2+davg )

= 2 (0.25+0.363) +2(0.45+0.363) = 2.852 m

The value of ΦVc is taken as the smaller of the following tow equations:

1. ФVC = Ф× 0.53× √ fc ' ×[1+ 2β]× bo ×d

2 .ФVC = Ф× √ fc ' × bo ×d

1=C longC short

=4525

=1.8

ΦVc, min¿0.75 ×× 285.2× 36.3/1000= 129.926 > Vu OK.

i.e. The footing thickness is aduqate for resisting punching shear.

Check footing thickness for beam shear:


Long direction:

Vu =qu ,net ×[ L×( B−C 1

2−davg)]

Vu = 24.913 ×[2.25 ×( 2.25−0.252

−0.363)]=¿ 35.71 ton.

ФVC = Ф× 0.53× √ fc ' × bo ×d

ΦVc = 0.75 ×0.53 ×√280 × 225× 36.3× 10−3=¿ 54.33 ton > Vu OK.

i.e. The footing thickness is aduqate for resisting beam shear in the long direction.

short direction:

Vu =qu ,net ×[B ×( L−C 2

2−davg)]

Vu = 24.913 ×[2.25 ×( 2.25−0.452

−0.363)]=¿ 30.10 ton.

ФVC = Ф× 0.53 ×√ fc ' × bo ×d

ΦVc = 0.75 ×0.53 ×√280 × 225× 36.3× 10−3=¿ 54.33 ton > Vu OK.

i.e. The footing thickness is aduqate for resisting beam shear in the short direction.

Design of footings for flexure:

Long direction:

Mu = qu ,net ×B2

× ¿

Mu =24.913 × 2.252

× ¿ 22.70 ton.m

Now design the footing as arectangular section with b = 225 cm.

davg = 45−7.5−1.2= 36.3 cm.

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(280 )4200 [1−√1− 2 .353×105 (22. 7 )

0 . 9×225×(36 . 3 )2×280 ]ρreq = 0.00206

As,req ¿0.00206 ×225 ×36.3 = 16.85 cm2

As,min ¿0.0018 ×b× h=0.0018× 225× 45=¿ 18.225 cm2

As,min >As,req so ues As,min = 18.225 cm2

Use Φ 12 mm diameter with area of 1.13 cm2

No. bars = 18.225 / 1.13= 16.12 though use 17 Φ12 mm.

Short direction:

Mu = qu ,net ×L2

× ¿

Mu =24.913 × 2.252

× ¿ 28.03 ton.m


davg = 45−7.5−1.2= 36.3 cm.

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(280 )4200 [1−√1− 2 .353×105 (28. 03 )

0 . 9×225×(36 . 3 )2×280 ]ρreq = 0.00256

As,req ¿0.00256 ×225 ×36.3 = 20.908 cm2

As,min ¿0.0018 ×b× h=0.0018× 225× 45=¿ 18.225 cm2

As,min < As,req so ues As,req = 20.908 cm2



Table 4: Groups of footings

Ninth: Preparation of calculation sheet

Calculation Sheet

Design of Multi Story Building Using ACI 318-08 Code

Determination of slab thickness

To control deflection the min. thickness required must be calculated from this table :

ElementSimply

supportedOne end

continuousBoth ends continuous

Cantilever

One-way ribbed slabs

l/16l/18.5l/21l/8

where l is the span length in the direction of bending and measure center to center.

Calculate the thickness of (rib) slab:


- Simply supported 363.5 363.516

=22.72 cm

- One end continuous 407.5407.518.5

=22.03 cm

- Both end continuous 383.5 383.521

=18.26 cm

- Cantilever 1501508

=18.75 cm

So minimum thickness for rib = 24.32 cm

Calculate the thickness of main beam:


- Simply supported - -


=27.57cm


=26.10 cm

- Cantilever - -

So minimum thickness for main beam = 27.57 cm

Calculate the thickness of secondary beam:


C1C2 Simply supported 42042016

=26.25 cm


=22.97 cm


=13.33cm

Cantilever Cantilever 1501508

=18.75 cm

So minimum thickness for secondary beam = 27.57 cm

So the minimum thickness for beam & slab = 28 cm.

Use block 40*25* 20 and ribs of 12 cm.

So the topping slab thickness = 28 - 20 = 8 cm.

Load calculation for 1 m2 of slab

Own weight of slab:

Total volume (hatched) = 0.52×0.25×0.3 = 0.0364 m³

Volume of one hollow block = 0.40×0.25×0.20 = 0.020 m³

Net concrete volume = 0.0364 - 0.020 = 0.0164 m³

Weight of concrete = 0.0164 ×2.5 = 0.041 ton

Weight of concrete /m² =0.041

0.52× 0.25=0.3153ton/m²

6 6 cm

40

12 cm

12 6

24

6 cm

Weight of hollow block /m² = 20 /1000

0.52× 0.25=¿ 0.1538 ton/m²

Own weight = 0.3153 + 0.1538 = 0.4692 ton/m²

Weight of slab covering materials:

Weight of layer = thickness × unit weight ton/m2

Weight of plaster = 0.015 × 2.1 = 0.0315 ton/m2

Weight of sand = 0.07× 1.8 = 0.126 ton/m2

Weight of mortar = 0.025 × 2.1 = 0.0525 ton/m2

Weight of tiles = 0.025 × 2.1 = 0.052 ton/m2

Total covering material weight = 0.2625 ton/m2

Weight of equivalent partition load:

- Number of blocks per meter square = 12.5 block.

- Two layers of plaster each layer have 1.5 cm thick.

- Height of wall = 2.95 m.

For block 10:

Weight of blocks = 12.5 × 10 = 125 kg/m2

Weight of plaster = 2×0.015×2100 = 63 kg/m2

Total weight / m2 = 125 + 63 = 188 kg/m2

Wu/m ' = 188*2.95 = 554.6 kg/m'

Length of wall 10 cm = 37.6 m

Total weight of block 10 = 0.5546 × 37.6 = 20.85296 ton

Partial partition load calculations:

Total weight of all partition wall = 20.85296 ton

Total Area = 162 m2

Stairs Area = 8.12 m2

Equivalent partiton load (EPL) =20.85

162−8.12=0.1355 ton/m2

Total Service Dead Load = O.W +C.M + EPL

Total Service Dead Load = 0.4692 + 0.2625 + 0.1355= 0.8672 t/m2.

Live load:

Live load (L.L) = 0.20 ton /m2.

Total factored load per meter square of the slab:

Wu= 1.2(0.8672) + 1.6(0.20) =1.3606 ton/m2

Total factored load on each rib:

Wu,rib= 1.3606 × 40+12100 = 0.7075 ton/m

Check slab thickness for beam shear

Vu max= 1.78 ton


D = 28−2.5−0.6−1.22

=¿ 24.3 cm

1.1ΦVc = 1.1Φ×0.53×√F c×b×d

= 1.1 × (0.75 ) ×(0.53)× 12× 24.3×√240

= 1975.25/1000 = 1.975 ton > 1.78 ton. OK.



Design of beams

Design of beams for flexure:

Main beam (B1):

f c=240 kg/cm2 fy = 4200 kg/cm2

h = 28 cm (thickness of slab) b = 70 cm cover = 4 cm


d = 28−4−(0.8 )−¿1.2/2) = 22.6 cm

Load assigned to the beam:

1. Slab load = [Rib ¿

2+Rib¿

2]× Wu =[3.225

2+ 0

2 ]× 1.3606=2.1939 ton /m

2. Own wieght = =

1.2 [ (0.7 ×0.28 ×2.5 )+(0.1335+0.2625) ]+1.6 [ 0.7 ×0.2 ]=1.2872 ton /m

3. External wall load =1.2 ×0.92335=1.01802ton /m

Load assigned to the beam = [2.1939+1.2872+1.01802] =4.45 ton /m

Positive moment

Mu max = 9.07 ton.m

ρ=0 . 85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2. 353×105(9 .07 )

0 .9×70×(22 .6)2×240 ]ρreq = 0.007253

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.007253

As,+ve ¿ ρ ×b× d = 0.007253×70× 22.6 = 11.47 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−11(1.2)

11−1 = 4.7 > 2.5 > 1.2 OK.

Negative moment


ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 .85(240 )4200 [1−√1− 2 .353×105 (10 .68 )

0 .9×70×(22 .6)2×240 ]ρreq = 0.008678

ρmin = 14fy

= 144200 = 0.00333 ρ req > ρ min OK.

ρmax=0.31875 ×fc × 1

fy 1= 0.85 since fc`= 240 < 280

ρmax=0.31875 ×240 ×0.85


controlled

Use ρ = ρ req = 0.008678

As,+ve ¿ ρ ×b× d = 0.008678×70× 22.6 = 13.72 cm2




n−1

S = 70−2 (4 )−2 (0.8 )−13(1.2)

13−1 = 3.73 > 2.5 > 1.2 OK.

Design of beams for shear

Main beam (B1):

f c=240 kg/cm2 fy = 4200 kg/cm2

h = 28 cm (thickness of slab) b = 70 cm cover = 4 cm


d = 28−4−(0.8 )−¿1.2/2) = 22.6 cm

Vu max, d =12.30

ФVC= 0 .75(0 . 53)1000

×√240×70×22 .6=9.74 ton .

Check ductile mode of failure:

2.2×√ fc ' ×b×d = 2.2×√240×70×22.6× 10-3 = 53.91 ton.

Vs= V u, d−Ф V C

Ф = 12.30−9.74

0.75 = 3.41 ton < 2.2×√ fc ' ×b×d C

Now assume 4 legged Φ8 mm stirrups

Av = 4 × π4

× 0.82 = 2 cm2 , d = 22.6 cm , bw = 70 cm , Vs = 3.41 ton.

1.1×√ fc ' ×b×d = 1.1×√240×70×22.6×10-3 = 26.959 > Vs

S = 22.6 ×2 ×4200

3.41 ×1000=¿55.67 cm, Smax =

2× 42003.5 ×70

=¿ 34.2 cm.

d2=22.6

2=11.3cm.

The smallest value is 11.3 cm so that (d2 = max value of S)

The value of S = 11 cm will be taken for the whole sections of B1.

Design of short columns

Design of tied column:

Columns 1, Columns Group (3):

Ribs area = 6.345 −¿ 2.517 = 3.828 m2

Own weight of one column =2.5×0.4×0.2×2.95 = 0.59 ton

Slab Load = 3.828 ×0.867245 + 2.517 × (0.28× 2.5 + 0.1355 + 0.2625) = 6.083 ton

External walls load = 5.05 × 0.92335 = 4.6629 ton

Total Dead load = 0.59 + 6.083 + 4.6629 =11.336 ton

Total live load = 0.20 ×6.345 = 1.269 ton

Number of stories = 5

Pu = 5 × (1.2 × 11.336 +1.6 ×1.269 ) = 78.17 ton

f c=240 kg/cm2 fy = 4200 kg/cm2

B = 20 cm ρg=1% cover = 4 cm

Ag=Pu

0.52 [0.85 fc +ρg ( fy−0.85 fc ) ]

Ag=78170

0.52[0.85× 280+0.01 (4200−0.85 ×280 )]=541.49 cm2

h=541.4920

=27.07 cm

The column dimension is 20cm× 40 cm So, use h = 40 cm

A s=0.01 ×541.49=5.4149 cm2

use6∅ 12mm

Check spacing:

Sclear=40−2× 4−2×0.8−6 ×1.2

6−1=4.64 cmOK .

Sties = Smallest of (16×1.2 or 48×0.8 or 20) = 19 cm

Use One (ties) 8 mm @ 19 cm

Design of footing

Design of the first group:

f c=240 kg/cm2 fy = 4200 kg/cm2

c= 2.5 ton /m3 s = 1.8 ton/m3

The unfactored load of column (Pservice) = 100.62 ton.

The factored load of column (Pu) = 126.125 ton.

Assume footing thickness (h = 45 cm) & Φ12 mm reinforcing bars.

hc= 45 cm hs = 150 - 45 = 105 cm

C1=25 cm C2 =45 cm.

qall(net) = qall, (gross) −¿hc c −¿ hs s

q all(net )=20−0.45× 2.5−1.05×1.8=¿19.985 ton/m2

davg = 45−7.5−1.2= 36.3 cm

Areq = P service

qall , net=100.62

19.985 = 5.034 m2

Assume L= 2.25 m B= 5.0342.25 = 2.237 ~ 2.25 m.

qu,net = Pu

Anew= 126.1252

2.25 × 2.25=¿ 24.913 ton/m2

Check footing thickness for punching shear:


Vu = qu ,net ×[ (B × L )−(C1+davg ) (C2+davg )]

= 24.913 ×[(2.25 ×2.25 )−(0.25+0.363 ) (0.45+0.363 ) ]

= 113.70 ton.

bo = 2 (C1+davg )+2 (C2+davg )

= 2 (0.25+0.363) +2(0.45+0.363) = 2.852 m

The value of ΦVc is taken as the smaller of the following tow equations:

1. ФVC = Ф× 0.53× √ fc ' ×[1+ 2β]× bo ×d

2 .ФVC = Ф× √ fc ' × bo ×d

1=C longC short

=4525

=1.8

ΦVc, min¿0.75 ×× 285.2× 36.3/1000= 129.926 > Vu OK.

i.e. The footing thickness is aduqate for resisting punching shear.

Check footing thickness for beam shear:


Long direction:

Vu =qu ,net ×[ L×( B−C 1

2−davg)]

Vu = 24.913 ×[2.25 ×( 2.25−0.252

−0.363)]=¿ 35.71 ton.

ФVC = Ф× 0.53× √ fc ' × bo ×d

ΦVc = 0.75 ×0.53 ×√280 × 225× 36.3× 10−3=¿ 54.33 ton > Vu OK.

i.e. The footing thickness is aduqate for resisting beam shear in the long direction.

short direction:

Vu =qu ,net ×[B ×( L−C 2

2−davg)]

Vu = 24.913 ×[2.25 ×( 2.25−0.452

−0.363)]=¿ 30.10 ton.

ФVC = Ф× 0.53 ×√ fc ' × bo ×d

ΦVc = 0.75 ×0.53 ×√280 × 225× 36.3× 10−3=¿ 54.33 ton > Vu OK.

i.e. The footing thickness is aduqate for resisting beam shear in the short direction.

Design of footings for flexure:

Long direction:

Mu = qu ,net ×B2

× ¿

Mu =24.913 × 2.252

× ¿ 22.70 ton.m


davg = 45−7.5−1.2= 36.3 cm.

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(280 )4200 [1−√1− 2 .353×105 (22. 7 )

0 . 9×225×(36 . 3 )2×280 ]ρreq = 0.00206

As,req ¿0.00206 ×225 ×36.3 = 16.85 cm2

As,min ¿0.0018 ×b× h=0.0018× 225× 45=¿ 18.225 cm2

As,min >As,req so ues As,min = 18.225 cm2


No. bars = 18.225 / 1.13= 16.12 though use 17 Φ12 mm.

Short direction:

Mu = qu ,net ×L2

× ¿

Mu =24.913 × 2.252

× ¿ 28.03 ton.m


davg = 45−7.5−1.2= 36.3 cm.

ρ=0 .85 f c

f y [1−√1−2 .353×105 M u

φ bd2 f c]

ρ=0 . 85(280 )4200 [1−√1− 2 .353×105 (28. 03 )

0 . 9×225×(36 . 3 )2×280 ]ρreq = 0.00256

As,req ¿0.00256 ×225 ×36.3 = 20.908 cm2

As,min ¿0.0018 ×b× h=0.0018× 225× 45=¿ 18.225 cm2

As,min < As,req so ues As,req = 20.908 cm2



Tenth: Preparation of structural drawing

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