Design of Residential Building

Design of residential building

Thisexample will discuss the procedures for design of each structural element of residential building.

• The number of floors = 5 floors.
• Compressive strength of used concrete = 240kg/cm2 for beams and slabs and 280kg/cm2 for columns and foundations.
• Yield strength of used steel bars 4200kg/cm2.
• Live load =200 kg/cm2.
• 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)

Procedures

1. Columnsdistribution.
2. Beamsdistribution.
3. Calculateslab thickness.
4. Calculateslab 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.

Figure3: Distribution of columns on ground floor map

Second: Beams distribution

- Determine the direction ofribs 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).

Figure4: Distribution of beams and ribs

Third:Calculate slab thickness.

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:

• So minimum thickness for rib =24.32cm

Calculate the thickness of main beam:

• So minimum thickness for main beam =27.57 cm

Calculate the thickness of secondary beam:

• So minimum thickness for secondary beam =27.57cm

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

Use block 40*25* 20 and ribsof 12 cm.

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

Note:

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

Fourth: Calculateslabloads

The load on ribs consists of:

1. The dead load:

• Own weight of the slab.
• Weight of the surface finish.
• Equivalent partition load.

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

Own weight of slab:

Figure5: Section in rib

Total volume (hatched) =0.520.250.3 = 0.0364 m³

Volume of one hollow block=0.400.250.20 = 0.020m³

Net concrete volume = 0.0364 - 0.020 = 0.0164m³

Weight of concrete =0.0164 2.5 = 0.041ton

Weight of concrete/m² =ton/m²

Weight of hollow block /m² = 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 = 0.0315ton/m2

Weight of sand = 0.07 1.8 = 0.126ton/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

Weight of equivalent partition load:

-Number of blocks per meter square =12.5 block.

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

-Height of wall =2.95m.

For block 10:

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

Weight of plaster = 20.0152100 = 63 kg/m2

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

Wu/m ' = 188*2.95 = 554.6kg/m'

Length of wall 10 cm = 37.6 m

Total weight of block 10 = 0.554637.6 = 20.85296 ton

Figure6-a:Section on partition

Figure6-b: Partition view

Partial partition load calculations:

Total weight of all partition wall = 20.85296 ton

Total Area = 162 m2

Stairs Area = 8.12m2

Equivalent partiton load (EPL) = 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 200to 250 kg/m2in 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= = 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 satisfyingthe equation 1.1ФVC>Vumax .

The shear force diagram is shown in Figure 7.

Figure 7: Shear force diagramfor first rib

Vumax=1.65ton.

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

d= 24.3cm

1.1ΦVc = 1.10.53bd

=

= 1975.25/1000 =1.975 ton1.65ton. OK.

Though shear rainforcement is not required , but is recommended to use Φ6mm@25cm 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>Vumax .

The shear force diagram is shown in Figure 8.

Figure 8: Shear force diagram for second rib

Vumax=1.78ton

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

D= 24.3cm

1.1ΦVc = 1.10.53bd

=

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

Though shear rainforcement is not required , but is recommended to use Φ6mm@25cm U-stirrups to carry the bottom flexural rainforcement.

B. Design flexural rainforcement:

Design for shrinkage reinforcement:

Use the minimum value of shrinkage reinforcementas follows:

Asmin

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

Asmin

Use Φ6mm diameter with area of 0.2826 cm2

No. bars= 1.44 /0.2826= 5.0 though use 5Φ6mm / m

So use Φ6mm @ 20cm

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

• Mumax=0.93 ton.m

ρreq = 0.003606

ρmin= = ρreqρ min OK.

1=0.85 since fc`=240 < 280

48 ρ maxρreqOK. the section is tension controlled

Use ρ = ρreq =0.003606

As,+ve = 0.0036061224.3 =1.051 cm2

Use Φ 10mm diameter with area of 0.79cm2 and Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 1.051/0.79 = 1.33 though use 1 Φ10mm and 1 Φ12 mm.

S =

S = = 3.6 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-0.87 ton.m

ρreq = 0.003365

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled.

Use ρ = ρreq =0.003365

As,+ve = 0.003365 1224.3 =0.981 cm2

Use Φ 12mm diameter with area of 1.13cm2.

No. bars= 0.981 /1.13 = 0.86 though use 1 Φ12mm.

S =

S = = 4.6 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-0.20 ton.m

ρreq = 0.00075

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

Use ρ = ρ min = 0.003333

As,+ve = 0.003333 1224.3 = 0.972 cm2

Use Φ 12mm diameter with area of 1.13 cm2.

No. bars= 0.972 / 1.13 = 0.86 though use 1 Φ12 mm.

S =

S = = 4.6 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-0.94 ton.m

ρreq = 0.003365

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled.

Use ρ = ρreq =0.003365

As,+ve = 0.003365 1224.3 = 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.

S =

S = = 3.6 > 2.5 > 1.2 OK.

Positive moment

• Mumax=0.98 ton.m

ρreq = 0.003365

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled.

Use ρ = ρreq =0.003808

As,+ve = 0.0038081224.3 =1.11 cm2

Use Φ 10 mm diameter with area of 0.79 cm2.

No. bars= 1.11/ 0.79 = 1.40 though use 2 Φ10 mm.

S =

S = = 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 = 1.2/2) = 22.6 cm

Load assigned to the beam:

1. Slab load = Wu =
2. Own wieght =
3. External wall load =

Load assigned to the beam= =

Positive moment

• Mumax=9.07 ton.m

ρreq = 0.007253

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.007253

As,+ve = 0.0072537022.6 =11.47 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 11.47/1.13 = 10.15 though use 11 Φ12mm.

S =

S = = 4.7 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-10.68 ton.m

ρreq = 0.008678

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.008678

As,+ve = 0.00867870 22.6 =13.72 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 13.72 / 1.13 = 12.14 though use 13 Φ12 mm.

S =

S = = 3.73 > 2.5 > 1.2 OK.

Positive moment

• Mumax=1.7 ton.m

ρreq = 0.001275

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

Use ρ = ρreq = 0.001275

As,+ve = 0.0033370 22.6 = 5.26 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 5.26 / 1.13 = 4.66 though use 5 Φ12 mm.

S =

S = = 13.6 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-6.68 ton.m

ρreq = 0.005224

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.005224

As,+ve = 0.00522470 22.6 = 8.26 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 8.26 / 1.13 = 7.31 though use 8 Φ12 mm.

S =

S = = 7.25 > 2.5 > 1.2 OK.

Positive moment

• Mumax=5.57 ton.m

ρreq = 0.004313

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.004313

As,+ve = 0.00431370 22.6 = 6.82 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 6.82 / 1.13 = 6.03 though use 7 Φ12 mm.

S =

S = = 8.66 > 2.5 > 1.2 OK.

Design of beams for shear

Main beam (B1):

The shear force diagram is shown in Figure 13.

Figure 13: Shear force diagram for main beam 3

Beam width (B1) = 70cm.

Assume Φ12 mm reinforcing bars and Φ8 mm stirrups.

d = 1.2/2) = 22.6 cm.

Vumax, d =12.30

ФVC =

Check ductile mode of failure:

2.2bd = 2.27022.6 10-3 = 53.91 ton.

Vs= = = 3.41ton2.2bd 

Now assume 4 legged Φ8 mm stirrups

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

1.1bd = 1.17022.610-3 = 26.959 > Vs

S = 55.67cm, Smax = 34.2cm.

The smallest value is 11.3cm so that (= 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=6.3452.517= 3.828 m2

Own weight of one column =2.50.40.2=0.59 ton

Slab Load = 3.8280.867245+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,

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

Check spacing:

Sties =Smallest of (161.2or 480.8 or 20)= 19 cm

Use One (ties) 8mm @ 19 cm

Table 2: Detailed design of columns

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 = 45cm) & Φ12 mm reinforcing bars.

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

qall(net) = qall,(gross) hcc hss

19.985 ton/m2

davg = = 36.3 cm

Areq = = 5.034 m2

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

qu,net = 24.913ton/m2

Check footing thickness for punching shear:

This condition must be satisfies Vu ΦVc

Vu =

=

= 113.70ton.

bo = 2

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

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

1.ФVC = Фd

2. ФVC = Фd

1=

ΦVc,min= 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 =

Vu= 35.71ton.

ФVC = Фd

ΦVc = 54.33 tonVu OK.

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

short direction:

Vu =

Vu= 30.10ton.

ФVC = Фd

ΦVc = 54.33 tonVu OK.

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

Design of footings for flexure:

Long direction:

Mu =

Mu=22.70 ton.m

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

davg = = 36.3 cm.

ρreq = 0.00206

As,req = 16.85 cm2

As,min 18.225cm2

As,minAs,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 Φ12mm.

Short direction:

Mu =

Mu =28.03 ton.m

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

davg = = 36.3 cm.

ρreq = 0.00256

As,req = 20.908 cm2

As,min 18.225cm2

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

Use Φ 12 mm diameter with area of 1.13 cm2

No. bars = 20.908 / 1.13= 18.5 though use 19 Φ12 mm.

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 :

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

Calculate the thickness of (rib) slab:

• So minimum thickness for rib =24.32cm

Calculate the thickness of main beam:

• So minimum thickness for main beam = 27.57 cm

Calculate the thickness of secondary beam:

• So minimum thickness for secondary beam =27.57cm

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

Use block 40*25* 20 and ribsof 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) =0.520.250.3 = 0.0364 m³

Volume of one hollow block=0.400.250.20 = 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 =ton/m²

Weight of hollow block /m² = 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 = 0.0315ton/m2

Weight of sand = 0.07 1.8 = 0.126ton/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 1.5cm thick.

-Height of wall =2.95 m.

For block 10:

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

Weight of plaster = 20.0152100 = 63 kg/m2

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

Wu/m ' = 188*2.95 = 554.6kg/m'

Length of wall 10 cm = 37.6 m

Total weight of block 10 = 0.554637.6 = 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) = 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= = 0.7075 ton/m

Check slab thickness for beam shear

Vumax= 1.78 ton

Assume Φ12 mm reinforcing bars and Φ6 mm U stirrups.

D= 24.3 cm

1.1ΦVc = 1.10.53bd

=

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

Though shear rainforcement is not required , but is recommended to use Φ6mm@25cm U-stirrups to carry the bottom flexural rainforcement.

Design of beams

Design of beams for flexure:

Main beam (B1):

kg/cm2fy = 4200 kg/cm2

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

Assume Φ12 mm reinforcing bars and Φ8 mm stirrups.

d = 1.2/2) = 22.6 cm

Load assigned to the beam:

1. Slab load = Wu =
2. Own wieght = =
3. External wall load =

Load assigned to the beam= =

Positive moment

• Mumax=9.07 ton.m

ρreq = 0.007253

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.007253

As,+ve = 0.00725370 22.6 =11.47 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 11.47 / 1.13 = 10.15 though use 11 Φ12 mm.

S =

S = = 4.7 > 2.5 > 1.2 OK.

Negative moment

• Mumax=-10.68 ton.m

ρreq = 0.008678

ρmin= = ρreq ρ min OK.

1= 0.85 since fc`= 240 < 280

48 ρ max> ρreqOK. the section is tension controlled

Use ρ = ρreq = 0.008678

As,+ve = 0.00867870 22.6 =13.72 cm2

Use Φ 12 mm diameter with area of 1.13 cm2.

No. bars= 13.72 / 1.13 = 12.14 though use 13 Φ12 mm.

S =

S = = 3.73 > 2.5 > 1.2 OK.

Design of beams for shear

Main beam (B1):

kg/cm2fy = 4200 kg/cm2

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

Assume Φ12 mm reinforcing bars and Φ8 mm stirrups.

d = 1.2/2) = 22.6 cm

Vumax, d =12.30

ФVC =

Check ductile mode of failure:

2.2bd = 2.27022.6 10-3 = 53.91 ton.

Vs= = = 3.41ton2.2bd 

Now assume 4 legged Φ8 mm stirrups

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

1.1bd = 1.17022.610-3 = 26.959 > Vs

S = 55.67 cm, Smax = 34.2 cm.

The smallest value is 11.3 cm so that (= 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=6.3452.517= 3.828 m2

Own weight of one column =2.50.40.2=0.59 ton

Slab Load = 3.8280.867245+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



kg/cm2fy = 4200 kg/cm2

B=20 cmcover = 4cm

The column dimension is So, use h=40 cm

Check spacing:

Sties =Smallest of (161.2or 480.8 or 20)= 19 cm

Use One (ties) 8mm @ 19 cm

Design of footing

Design of the first group:

kg/cm2fy = 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 cmC2 =45 cm.

qall(net) = qall,(gross) hcc hss

19.985 ton/m2

davg = = 36.3 cm

Areq = = 5.034 m2

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

qu,net = 24.913 ton/m2

Check footing thickness for punching shear:

This condition must be satisfies Vu ΦVc

Vu =

=

= 113.70 ton.

bo = 2

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

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

1.ФVC = Фd

2. ФVC = Фd

1=

ΦVc,min= 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 =

Vu= 35.71 ton.

ФVC = Фd

ΦVc = 54.33 tonVu OK.

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

short direction:

Vu =

Vu= 30.10 ton.

ФVC = Фd

ΦVc = 54.33 tonVu OK.

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

Design of footings for flexure:

Long direction:

Mu =

Mu =22.70 ton.m

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

davg = = 36.3 cm.

ρreq = 0.00206

As,req = 16.85 cm2

As,min 18.225 cm2

As,minAs,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 =

Mu =28.03 ton.m

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

davg = = 36.3 cm.

ρreq = 0.00256

As,req = 20.908 cm2

As,min 18.225 cm2

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

Use Φ 12 mm diameter with area of 1.13 cm2

No. bars = 20.908 / 1.13 = 18.5 though use 19 Φ12 mm.

Tenth: Preparation of structural drawing