Design of Residential Building - Islamic University of...

Design of Reinforced Concrete Structures (I)

Design of Residential Building

Prepared By:

Eng. Ibrahim Omer Abu Zuhri

2016

2016 Design of Residential Building

1 Eng. Ibrahim Omer Abu Zuhri

Design of Residential Building

The example shown in Figures 1 and 2 discusses the procedures for the design of

structural elements of a residential building.

Number of floors is 5.

Cylinder compressive strength of used concrete is 240 kg/cm2 for beams and

slabs and 280 kg/cm2 for columns and foundations.

Yield strength of used steel bars is 4200 kg/cm2.

Live load = 200 kg/cm2.

𝛾𝑠𝑜𝑖𝑙 = 1.8 𝑡/𝑚3

𝛾𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2.5 𝑡/𝑚3 .

𝛾𝑚𝑜𝑟𝑡𝑎𝑟 = 2.1 𝑡/𝑚3 .

𝛾𝑝𝑙𝑎𝑠𝑡𝑒𝑟 = 2.1 𝑡/𝑚3 .

𝑞𝑎𝑙𝑙 (𝑔𝑟𝑜𝑠𝑠) = 2.3 𝐾𝑔/𝑐𝑚2 .

𝐷𝑓 = 1.5 𝑚.

Concrete cover:

Concrete cover = 7.5 cm for the underground elements.

Concrete cover = 2.5 cm for slabs.

Concrete cover = 4 cm for other structural elements.



Figure 1: Ground floor plan



Figure 2: First floor plan

Design Procedure

1. Column distribution.

2. Beam distribution.

3. Slab thickness evaluation.

4. Slab load evaluation.

5. Rib design.

6. Beam design.

7. Column design.

8. Footing design.

9. Structural drawing preparation.



First: Column distribution

- The distribution of columns must be appropriate to the architectural space and

customer needs.

- It is advisable that, the distance between columns is limited from 4.0 to 5.0 meters.

This step reduces the cost of slab.

- The initial distribution of the columns must be on the ground floor plan.

- Calculate the total area of the first floor plan (after the ground floor), including

staircase space.

- Divide the total area of the first floor plan on 10 to get the optimum number of

permitted columns.

- Four columns must be placed for staircase.

- Distribute the other columns as much as possible on a single line - especially external

columns to facilitate implementation - , or at least put them in a projection of not more

than 10% difference between their axis.

- After completion of the distribution, move the columns from ground floor plan to first

floor plan, and make sure that there are no columns outside the external walls. See the

extended areas (Cantilever).

- Perform the previous steps on the AutoCAD program.

- Use the command (Rectangle) to draw the columns, and then press the letter (d) (then

button (Enter), then write the initial dimensions. (Preferably putting it in a separate

layer).

- Draw two intersecting lines in each column to help you identifying the center of

column.

- Use the command (Copy Object) to transfer the columns from the ground floor plan

to the first floor plan using a fixed point as a reference in two planes (Base point.)

The previous procedures are shown in Figure 3.



Figure 3: Distribution of columns (ground floor plan)

Second: Beam distribution

After the completion of the distribution of the columns on the first floor map

- Eliminate all details of the furniture and dimensions and walls, except the external

limits of the first floor.

- Draw an X on staircase space without the area of final landing. (It is part of the slab).

- Name the columns for example C1, C2, C3 and so on.

- Draw temporary lines between columns centers.



- Measure the horizontal and vertical distances between the columns c-to-c.

- Draw lines in the shortest direction for each panel to identify the direction of rip (you

will find the direction of some of the panels is horizontal and other panels is vertical).


Figure 4: Initial direction of ribs



- It’s preferable to united all the ribs in the same direction as shown in Figure 5.

Figure 5: Final direction of ribs

- Draw the initial distribution of the beams as follow:

The width of internal main beam 60-100 cm.

The width of external main beam 60-80 cm.

The width of internal secondary beam 40-50 cm.

The width of external secondary beam 40-60 cm.

Note: The main beams are perpendicular to the direction of the ribs, while secondary

beams are parallel to the direction of the ribs.

- Beams should be on the same line as much as possible.



- Draw beam with width 15-25 cm on the edges of the slab that do not contain beams,

which are called (boundary beams 'crans'). (Usually found in cantilever)

- Name the beams for example B1, B2, B3… for main beams and SB1, SB2, SB3… for

secondary beams and so on.

- Draw ribs and blocks.


Notes:

Ribs should not be less than 10 cm in width (12 is preferred to facilitate casting).

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

is equal to 40 cm).



Figure 6: Distribution of beams and ribs

Third: Slab thickness evaluation

Minimum thickness of one-way ribbed slabs was shown in Table 7.3

Minimum thickness of one-way ribbed slabs

Element Simply

supported

One end

continuous

Both ends

continuous Cantilever

One-way

ribbed slabs l/16 l/18.5 l/21 l/8

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



Notes:

The most common concrete hollow block size is 40 × 25 cm in plan and 14,

17, 20 and 24 cm in height.

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

ribs, nor less than 5.0 cm.

Tables from 1 to 3 show the evaluation of slab thickness.

Table 1: Slab (rib) thickness evaluation

Type Maximum Length (cm) Minimum Thickness (cm)

Simply supported 363.5 363.5

16= 22.72 𝑐𝑚

One end continuous 407.5 407.5

18.5= 22.03 𝑐𝑚

Both end continuous 383.5 383.5

21= 18.26 cm

Cantilever 150 150

8= 18.75 𝑐𝑚

So, the minimum thickness of rib is 24.32 cm

Table 2: Main beam thickness evaluation

So, the minimum thickness of main beam is 27.57 cm

Type Maximum Length(cm) Minimum Thickness (cm)

One end continuous 510 510

18.5= 27.57 cm

Both end continuous 548 548

21= 26.10 𝑐𝑚



Table 3: Secondary beam evaluation

Type Maximum Length (cm) Minimum Thickness (cm)

Simply supported 420 420

16= 26.25 𝑐𝑚

One end continuous 425 425

18.5= 22.97 𝑐𝑚

Both end continuous 280 280

21= 13.33 cm

Cantilever 150 150

8= 18.75 𝑐𝑚

So, the minimum thickness of secondary beam is 27.57 cm and the minimum thickness

of beams and slab = 28 cm. ( to have a uniform slab thickness with hidden beams)

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

Topping slab thickness is 28 - 20 = 8 cm.

Note:

Depth of rib is not to exceed 3.5 times the minimum web width. (3.5×12 = 42

cm).

Fourth: Slab load evaluation

The load on ribs consists of:

1. The dead load:

Own weight of the slab.

Weight of the surface finish (covering materials).

Equivalent partition load.

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



Own weight of slab

The own weight is based on the cross sections shown in Figure 7.

Figure 7: Sections in rib

Total volume (hatched) = 0.52×0.25×0.28 = 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= 0.3153 ton/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 cm 6 cm 40 cm

12 cm 12 cm 6 cm

20 cm

8 cm

52 cm

25 cm



Weight of equivalent partition load (shown in Figure 8):

- Number of concrete blocks per meter square = 12.5 blocks.

- Two layers of plaster each layer has 1.5 cm thickness.

- Height of wall = 2.95 m.

:cm thickness block 10concrete For

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

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

Length of walls 10 cm thickness = 37.6 m

Total weight of concrete blocks 10 cm thickness = 0.5546 × 37.6 = 20.85296

ton

(a)



(b)

Figure 8: (a) Section in partition; (b) Partition view (elevation)

Partial partition load calculations:

Total weight of all partition walls = 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 = own weight + covering material + equivalent partiton load

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

Total service dead load on each rib:

W service dead load,rib = 0.8672 ×40+12

100 = 0.4509 ton/m



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).

- It ranges from 200 to 250 kg/m2 in residential building.

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

Service live load on each rib:

W service live load,rib = 0.2 ×40+12

100 = 0.104 ton/m

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+12

100 = 0.7075 ton/m

Fifth: Rib design

- Multiply the total loads in 0.52 if your rib width is 12 cm and in 0.5 If your rib width

is 10 cm.

- Choose strips to represent all ribs.

- You cannot put more than two bars inside the small rib.

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 .

Factored laods and shear force diagram are shown in Figure 9.



(a)

(b)

(c)

Figure 9: (a) Factored dead load; (b) Factored live load; (c) Shear force diagram

for first rib

Vu max = 1.6 ton.

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

d= 28 − 2.5 − 0.6 −1.2

2= 24.3 cm

1.1ΦVc = 1.1Φ ×0.53× `

cf ×b×d

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

= 1975.25/1000 = 1.975 ton > 1.6 ton. OK.

Shear rainforcement is not required, but it 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 .




(a)

(b)

(c)

Figure 10: (a) Factored dead load; (b) Factored live load; (c) Shear force

diagram for second rib

Vu max = 1.8 ton.

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

D = 28 − 2.5 − 0.6 −1.2

2= 24.3 cm

1.1ΦVc = 1.1Φ ×0.53× `

cf × b×d

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

= 1975.25/1000 = 1.975 ton > 1.8 ton. OK.

Shear rainforcement is not required, but it is recommended to use Φ6 mm @ 25 cm U-

stirrups to carry the bottom flexural rainforcement.

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

or use Φ6 mm @ 20 cm

First rib:

The rib is to be designed as a rectangular section.

Factored laods and bending moment diagram are shown in Figure 11.

(a)

(b)

(c)

Figure 11: (a) Factored dead load; (b) Factored live load; (c) Bending moment

diagram for first rib

Positive moment

Mu max = 1 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

-1

-0.30

-0.90



240)3.24(129.0

)1(10353.211

4200

)240(85.02

5

req = 0.003889

𝜌𝑚𝑖𝑛 =14

yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf

1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0

4200= 01548.0 max > req OK. the section is

tension controlled.

Use = req = 0.003889

As,+ve = × b × d = 0.003889 × 12 × 24.3 = 1.13 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

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

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

n−1

S = 12−2(2.5)−2(0.6)−2(1.2)

2−1 = 3.4 > 2.5 > 1.2 OK.

Negative moment

Mu max = -1 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)3.24(129.0

)1(10353.211

4200

)240(85.02

5

req = 0.003889


yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf × 1

yf 1 = 0.85 since `

cf = 240 < 280



𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.003889

As,+ve = × b × d = 0.003889 × 12 × 24.3 = 1.13 cm2


No. bars = 1.13 / 1.13 = 1 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

Mu max = -0.30 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)3.24(129.0

)30.0(10353.211

4200

)240(85.02

5

req = 0.001133


yf=

14

4200= 00333.0 req < min Not OK.

Use = min = 0.003333

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


No. bars = 0.972 / 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.

Note:

In this case, minimum steel is used as a bottom steel.


yf=

14

4200= 00333.0

Use = min = 0.003333



As,+ve = × 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


`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)3.24(129.0

)90.0(10353.211

4200

)240(85.02

5

req = 0.003485


yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf × 1

yf 1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.003485

As,+ve = × b × d = 0.003485 × 12 × 24.3 = 1.01 cm2




n−1

S = 12−2(2.5)−2(0.6)−1(1)−1(1.2)

2−1 = 3.6 > 2.5 > 1.2 OK.

Positive moment

Mu max = 1 ton.m



`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)3.24(129.0

)1(10353.211

4200

)240(85.02

5

req = 0.003889


yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf

1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.003889

As,+ve = × b × d = 0.003889 × 12 × 24.3 = 1.13 cm2




n−1

S = 12−2(2.5)−2(0.6)−2(1)

2−1 = 3.8 > 2.5 > 1.2 OK.

Sixth: Beam design

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

Main beams: The loads that affecting on the main beam are as follow:

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

- Half of the ribs from each side.

- In case of external beam, you have to add the weight of 1 linear meter of exterior

walls.

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

weight of boundary beam.



Secondary beams: The loads that affecting on the secondary beam are as follow:

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

- In case of external beam, you have to add the weight of 1 linear meter of exterior

walls.

Note:

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

In some cases, where beam is based on another beam, take into account that you

have to calculate the loads of the first beam - whether it main or secondary - ,

and then put the reaction force as a concentrated load on the second beam.

For continuous reinforcement at the top of the beam, there must be at least:

- Two bars for beams with a width less than 60 cm.

- Four bars for beams with a width between 60 cm and 100 cm.

- Six bars for beams with a width more than 100 cm.

Stirrups for beam should be at least:

- One stirrup for beams with a width less than 60 cm.

- Two stirrups for beams with a width between 60 cm and 100 cm.

- Three stirrups for beams with a width more than 100 cm.

Design of beams for flexure:

Main beam (B1):

Factored laods and bending moment diagram are shown in Figure 12.

(a)

(b)



(c)

Figure 12: (a) Factored dead load; (b) Factored live load; (c) Bending moment

diagram for main beam (B1)

h = 28 cm (thickness of slab), and assume beam width, b of 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:

Dead load:

1. Slab load = [𝑅𝑖𝑏 𝐿𝑒𝑓𝑡

2+

𝑅𝑖𝑏 𝑟𝑖𝑔ℎ𝑡

2] × Wdead load =[

3.225

2+

0

2] × 0.8672 =

1.39836 ton/m

2. Own wieght = [(2.5 × 0.28 × 0.7) + (0.1335 × 0.7 ) + ( 0.2625 ×

0.7)] = 0.7672 ton/m

3. External wall load = 0.92335 ton/m

Service dead Load assigned to the beam = [1.39836 + 0.7672 + 0.92335] =

3.08891 ton/m

Live load:

1. Slab load = [𝑅𝑖𝑏 𝐿𝑒𝑓𝑡

2+

𝑅𝑖𝑏 𝑟𝑖𝑔ℎ𝑡

2] × Wservice load =[

3.225

2+

0

2] × 0.2 =

0.3225 ton/m

2. Beam load = 0.2 × 0.70 = 0.140 ton/m

Service live Load assigned to the beam = [0.3225 + 0.14] = 0.4625 ton/m

Toatal factored Load assigned to the beam = 1.2 [3.0889] + 1.6 [0.4625] =

4.45 ton/m



Positive moment

Mu max = 10.2 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)6.22(709.0

)3.10(10353.211

4200

)240(85.02

5

req = 0.008337


yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf 1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.008337

As,+ve = × b × d = 0.008337 × 70 × 22.6 = 13.19 cm2




n−1

S = 70−2(4)−2(0.8)−12(1.2)

12−1 = 4.18 > 2.5 > 1.2 OK.

Negative moment


`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)6.22(709.0

)3.11(10353.211

4200

)240(85.02

5

req = 0.00924




yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf

1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.00924

As,+ve = × b × d = 0.00924 × 70 × 22.6 = 14.62 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 = 1.6 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)6.22(709.0

)6.1(10353.211

4200

)240(85.02

5

req = 0.001199


yf=

14

4200= 00333.0 req < min Not OK.

Use = min = 0.00333

As,+ve = × 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

Mu max = -6.7 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)6.22(709.0

)7.6(10353.211

4200

)240(85.02

5

req = 0.00524


yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf 1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.00524

As,+ve = × b × d = 0.00524 × 70 × 22.6 = 8.29 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 = 6.3 ton.m

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

240)6.22(709.0

)3.6(10353.211

4200

)240(85.02

5

req = 0.00491




yf=

14

4200= 00333.0 req > min OK.

𝜌𝑚𝑎𝑥 =31875.0 ×

`

cf ×1

yf

1 = 0.85 since `

cf = 240 < 280

𝜌𝑚𝑎𝑥 =31875.0 × 240 × 85.0


tension controlled.

Use = req = 0.00491

As,+ve = × b × d = 0.00491 × 70 × 22.6 = 7.77 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):


(a)

(b)

(c)

Figure 13: (a) Factored dead load; (b) Factored live load; (c) Shear force

diagram for main beam (B1)



Beam width (B1) = 70 cm.

Assume Φ12 mm reinforcing bars and Φ8 mm stirrups.

d = 28 − 4 − (0.8) − (1.2/2) = 22.6 cm.

Vu max = 13.7

.74.96.22×70×240×1000

)53.0(75.0ton= CVФ

Check ductile mode of failure:

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

Vs = Vu,d −ФVC

Ф =

13.70 − 9.74

0.75 = 5.28 ton < 2.2 × 'fc × b × d

Now assume 4 legged Φ8 mm stirrups

Av = 4 ×𝜋

4× 0.82 = 2 cm2 , d = 22.6 cm , bw = 70 cm , Vs = 5.28 ton.

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

S = 22.6×2×4200

5.28×1000= 35.95 cm, Smax =

2×4200

3.5×70= 34.2 cm.

𝑑

2=

22.6

2= 11.3 𝑐𝑚.

The smallest value is 11.3 cm so that ( 𝑑

2 = max value of S)

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

Seventh: Column design

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

area method. (In this example, the area method is used)

- Draw vertical and horizontal lines between the centers of columns as shown in Figure

14.



Figure 14: Lines between centers of columns



- Draw lines that cross the center of the lines drawn in the previous step as shown in

Figure 15.

Figure 15: Lines cross the lines between centers of columns



- Extend the lines drawn in the previous step to the borders of the building as shown in

Figure 16.

Figure 16: Extended lines in cross lines



- Delete the lines drawn between the centers of columns as shown in Figure 17.

Figure 17: Columns tributary areas



- Or you can simplify it as shown in Figure 18.


- Calculate the tributary area as shown in Figure 19 and determine the loads acting on

each columns.

- Draw the final dimensions of the columns on the building plans (after final design of

columns).




Calculations of loads

Ribs area = Total area – Beam area

Own weight of one column = 𝛾𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 × (3 m × 0.2 m × 0.4 m)

Slab Load = Ribs area × (DL of Ribs) + Equivalent partition load + Covering materials

+ Beam area × (slab thickness × 𝛾𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒

)

External walls load = Length of wall 20 × Weight of wall 20

Total Dead load = Own weight + Slab load + External walls load

Total live load = Total area × Live load (ton/m2)

Pu = Number of stories × (1.2 DL + 1.6 LL)



Table 4 shows the loads acting on columns.

Table 4: Loads on columns

Design of tied column:

Columns 1 (column 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

Each column width, B = 20 cm , 𝜌𝑔 = 1%

𝐴𝑔 =𝑃𝑢

0.52[0.85𝑓𝑐` + 𝜌𝑔(𝑓𝑦 − 0.85𝑓𝑐`)]

𝐴𝑔 =78170

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

Other coulmn dimentin, ℎ =541.49

20= 27.07 𝑐𝑚

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

𝐴𝑠 = 0.01 × 541.49 = 5.4149 𝑐𝑚2

𝑢𝑠𝑒 6 ∅ 12 𝑚𝑚

Check spacing:

𝑆𝑐𝑙𝑒𝑎𝑟 =40 − 2 × 4 − 2 × 0.8 − 6 × 1.2

6 − 1= 4.64 𝑐𝑚 𝑂𝐾.

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 15.63

C2 10.11 2.24 7.871 0.59 6.5 15.87 2.02 79.36 10.11 22.28

C3 10.04 2.23 7.809 0.59 6.45 15.77 2.01 78.84 10.04 22.14

C4 6.73 2.66 4.072 0.59 5.2 11.84 1.35 59.19 6.73 16.36

C5 16.32 4.44 11.876 0.59 0 15.77 3.26 78.84 16.32 24.15

C6 16.08 4.43 11.649 0.59 0 15.56 3.22 77.79 16.08 23.82

C7 11.28 3.29 3.36 0.59 4.8 11.55 2.26 57.74 11.28 17.47

C8 13.45 4.04 9.41 0.59 4.6 17.43 2.69 87.17 13.45 25.22

C9 9.69 2.89 6.8068 0.59 4.13 13.48 1.94 67.38 9.69 19.27

C10 16.15 3.80 12.35 0.59 0 15.47 3.23 77.36 16.15 23.74

C11 15.91 3.80 12.112 0.59 0 15.27 3.18 76.33 15.91 23.41

C12 12.71 3.56 9.153 0.59 4.13 16.24 2.54 81.22 12.71 23.56

C13 5.11 2.35 2.764 0.59 4.5 9.72 1.02 48.60 5.11 13.30

C14 7.40 3.03 4.37 0.59 3.4 10.84 1.48 54.20 7.40 15.38

C15 7.29 3.03 4.261 0.59 3.35 10.70 1.46 53.50 7.29 15.17

C16 5.00 2.33 2.67 0.59 4.45 9.58 1.00 47.88 5.00 13.09



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

Use One (ties) 8 mm @ 19 cm.

Tables 5 and 6 show the detailed design and the groups of columns.

Table 5: Detailed design of columns

Table 6: Groups of columns

Eighth: Footing design

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.

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 19

C2 111.40 771.68 7.72 20 39 40 20x40 12 7 3.81 19

C3 110.68 766.67 7.67 20 39 40 20x40 12 7 3.85 19

C4 81.79 566.57 5.67 20 29 40 20x40 12 5 6.08 19

C5 120.73 836.27 8.36 20 42 45 20x45 12 7 4.15 19

C6 119.08 824.83 8.25 20 42 45 20x45 12 7 4.23 19

C7 87.34 605.00 6.05 20 31 40 20x40 12 5 5.51 19

C8 126.12 873.67 8.74 20 44 45 20x45 12 8 3.88 19

C9 96.37 667.55 6.68 20 34 40 20x40 12 6 4.75 19

C10 118.68 822.08 8.22 20 42 45 20x45 12 7 4.25 19

C11 117.06 810.86 8.11 20 41 45 20x45 12 7 4.34 19

C12 117.80 816.01 8.16 20 41 45 20x45 12 7 4.30 19

C13 66.49 460.60 4.61 20 24 40 20x40 12 4 8.30 19

C14 76.88 532.52 5.33 20 27 40 20x40 12 5 6.67 19

C15 75.86 525.47 5.25 20 27 40 20x40 12 5 6.80 19

C16 65.45 453.40 4.53 20 23 40 20x40 12 4 8.50 19



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

qall(net) = 23 − 0.45 × 2.5 − 1.05 × 1.8 =19.985 ton/m2

davg = 45 − 7.5 − 1.2= 36.3 cm

Areq = 𝑃service

𝑞𝑎𝑙𝑙,𝑛𝑒𝑡=

100.62

19.985 = 5.034 m2

Assume L= 2.25 m B= 5.034

2.25 = 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

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

= 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:

d × [1 + 2

𝛽] × 𝑏𝑜 × 'fc × 0.53 ×Ф= CVФ 1.

d × 𝑏𝑜 × 'fc ×= Ф CVФ2.

1 = 𝐶 𝑙𝑜𝑛𝑔

𝐶 𝑠ℎ𝑜𝑟𝑡=

45

25= 1.8

ΦVc, min = 0.75 × 280 × 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:

This condition must be satisfies Vu ≤ ΦVc

Long direction:

Vu = 𝑞 𝑢,𝑛𝑒𝑡 × [𝐿 × (𝐵−𝐶1

2− 𝑑𝑎𝑣𝑔)]

Vu = 24.913 × [2.25 × (2.25−0.25

2− 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 = 𝑞 𝑢,𝑛𝑒𝑡 × [𝐵 × (𝐿−𝐶2

2− 𝑑𝑎𝑣𝑔)]

Vu = 24.913 × [2.25 × (2.25−0.45

2− 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 = 𝑞 𝑢,𝑛𝑒𝑡 ×𝐵

2 × (

𝐿−𝐶2

2)2

Mu = 24.913 ×2.25

2× (

2.25−0.45

2)2 = 22.70 ton.m

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

davg = 45 − 7.5 − 1.2= 36.3 cm.

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f

280)3.36(2259.0

)7.22(10353.211

4200

)280(85.02

5

req = 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

2cm13 .1mm diameter with area of 2 Use Φ 1

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

Short direction:

Mu = 𝑞 𝑢,𝑛𝑒𝑡 ×𝐿

2 × (

𝐵−𝐶1

2)2

Mu = 24.913 ×2.25

2× (

2.25−0.25

2)2 = 28.03 ton.m

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

davg = 45 − 7.5 − 1.2= 36.3 cm.

`2

5` 10353.211

85.0

c

u

y

c

fbd

M

f

f



280)3.36(2259.0

)03.28(10353.211

4200

)280(85.02

5

req = 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

2cm13 .1mm diameter with area of 2 Use Φ 1


Table 7 shows the groups of footings.

Table 7: Groups of footings

Ninth: Structural drawings preparation

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