- Introduction

A simple multi-storey residential unit at Suresh Nagar, Gwalior in India had been chosen. It had been designed by the team of design campus comprising of architects, engineers and designers.

Figure 1: Front-side view of the multi-storey building

The total area of the ground floor plan is 2400 sq. ft. It is a G+3 structure i.e. it consists of four floors including the ground floor. The structure is a reinforced framed-structure supported on the column-beam frame. The lowermost floor that is partially under the ground level has been allotted for parking of vehicles.

The area under consideration is covered by alluvial soil . The city spans over an area of about 500sq/km and is found to lie between the longitudes 78°0' to 78°15' E and the latitudes 26°0' to 26°15’ N. The average monthly temperature is found to vary from 28℃ to 46℃. Besides alluvium, the area is also covered by sandstone, Quartzite, dolerite and shale. The permeability index (PI) of the groundwater in the area ranges between the values 0.425 and 199.65 epm .

Characteristics of the soil are useful determinants of its performance under loading and hence predicts the stability of the structure. The property exercising maximum influence on the on the physical characteristics is the particle-size distribution. Besides this, the surface texture, moisture content, density and the chemical composition are other significant parameters in ascertaining the soil behaviors. Under ordinary circumstances, the properties of the soils composed primarily of coarse materials are controlled by the properties of its particles. However, in the case of clayey or colloidal soils, the moisture content plays the most important role. Furthermore, some soils and ground water can also have corrosive action on the metals that are used for building construction such as cast iron. It may also bring about the deterioration of cement concrete. The deteriorating action can be attributed to several external agencies such as industrial wastes, sea water or other saline waters, sulphates originating in clay soils and the acidicity .

With the help of this data, it can be observed that the soil is acidic in nature with a pH value ranging from 8.3 to 8.7 which is quite above the neutral value. The mineral content is within the permissible ranges, therefore no special treatment is required for them. However, the amount of sodium is slightly lesser.

## Proposal

The various structural elements of the multi-storey building such as columns, beams, slabs and staircase need to be designed taking the strength and stability factors into consideration. The building needs to be designed to follow the framed-structure type i.e. the beams, columns and slabs will be connected to each other rigidly and the beams and columns will form a grid-like pattern. For the design of the structural members, limit state design method has been applied. In limit state design, the ultimate loading conditions are used for designing. The plans and structural details of the building have been given in the next few pages.

Figure 2: Ground floor plan of the multi-storey building

Figure 3: Typical floor plan of the multi-storey building

Figure 4: Parking floor plan of the multi-storey building provided at the basement

Figure 5: Plan showing the door & window schedule of the multi-storey building

- Foundations

2.1 Introduction

That part of the structure which is responsible for transmitting the load of the structure to the soil underneath is known as foundation. Different soils show different behavior and in turn affect the structure in various ways. The aim of providing a foundation to a structure is to distribute its load on the bearing soil in a way such that the load carrying capacity of the soil is not exceeded and such that the settlement is within reasonable limits. The foundation of a structure carries the entire weight of the structure i.e. the dead load .

When a structure gets constructed, it has to additionally bear the live load as well. Live load or imposed load comprises of the weight due to the users, wind and slow load, seismic loads etc. However, most of settlement already takes place when the structure gets built and before any live load can be experienced. Furthermore, as the compactness of the soil increases, its yielding reduces proportionally with the increase in load. As the load is applied, the settlement of the soil does not take place instantly. Its full impact can be felt only when it is applied over a considerable period of time. Also, there is uneven settlement under live load compared to that due to dead load. To minimize this effect, a method of proportioning the areas of footings is based on the principle that the area of a footing should be designed primarily for dead load plus a fraction of live load with a reduced unit pressure so that it can take full load if necessary but with partial loading, the unit pressure under all columns is the same .

The foundation should be provided in a manner such that the structure does not undergo any tilting. If the centre of the gravity of loads does not coincide with the centre of gravity of the bearing area, the bearing pressures will become non-uniform and a much higher pressure zone will be generated at the edges of the footing nearer to the centre of gravity of the loads. This will result in higher settlement of the ground towards a single edge causing the structure to lean in that direction. Therefore, the centre of gravity of the loads should be made to coincide with the centre of gravity of the bearing area. Furthermore, if the foundation of a building is standing partly upon rock and partly upon the earth, then it is bound to cause unequal settlements. In such a situation, it is advisable to isolate the two sections of the building and allow thus them to settle independently .

The foundations are taken sufficiently below the ground level so as to provide them the stability to resist the actions of wind and water and also to check the heaving up of soil on the sides due to internal failure. The depth of the foundation below the ground level can be calculated using the Rankine’s formula, which is given by the following expression.

h=pw1-sinØ1+sinØ2

Here h, p and w represent the depth of the foundation below the ground level, pressure of the soil and the unit weight of the soil respectively. Ø stands for the angle of repose .

The study of soil and its behavior under a foundation is called soil mechanics. When load is applied upon the earth, it gets compressed. The amount of settlement depends upon the intensity of the load, the nature of soil and the depth below ground level. The settlement of different parts of the same structure should be the same so that no secondary stresses develop since no allowance is made for them in the design. To ensure equal settlement, it is necessary that the bearing area is so provided that the load on the soil per unit area under all footings is the same. Furthermore, it must be ensured that each footing is designed for its own load .

As the flexibility of the foundation base increases, the distribution of load under the base of foundation tends to be uniform. In the case of cohesive soils, the pressure under the edges of rigid bases tends to be infinite but under flexible bases, it is nearly uniform. In the practical situations, in order to be able to assume uniform distribution of pressure under foundation, it is necessary to make the base of the foundation as flexible as possible. By reducing the thickness of footing’s concrete towards the free edges lowers concrete consumption as well as provides flexibility for a uniform load distribution on the soil. The actual distribution of pressure depends upon the nature of soil and the extent of the flexibility of the base .

- Load calculations

- Dead load for each floor in the residential unit can be taken to be equal to 200kg/sq. m. Therefore, the total dead load due to all the four floors of the structure will become equal to 800kg/ sq. m.

- Load due to stairs and landing is 300 kg/sq. m.

- Superimposed or live load is contributed by the people or users and the immovable objects brought inside the building. The average weight of a man is 68kg. 5 men standing on a space of 1 sq. yard will exercise a weight of 0.84 sq. m. Since, the building is a multistory, it can be expected that it will be used by many people at a given point of time. The total live load can be taken as 400 kg/ sq. m.

- Superimposed loads can also be attributed to wind and seismic loads. Another 100 kg/ sq. m. should be added to counter-balance these effects.

Thus, the total amount of load that the structure needs to carry is 1600 kg/ sq. m.

Depth of foundation=120 cm. in the case of multi-storey building on a firm soil.

- Design of the R.C.C. footing

## The type of foundation to be provided here is spread foundation.

- There are eleven F1-Columns. The length and breadth of the footings for these columns will be equal to 6’0”. The area of the footing then becomes equal to 36 sq. ft.

- Again, there are six F2-Columns. For these 9 columns, the footing size shall be taken as 25 sq. ft., with the length and breadth carrying a value of 5’0” each.

- The footing size for each of the five F3-Columns can be taken as 16 sq. ft., with the size of length and breadth equal to 4’0” each.

Figure 6: Column Layout and Footing Plan of the multi-storey building

- Columns

- Introduction

In a building, columns play extremely significant role. Columns are the vertical support members to which the other elements such as beams, slabs and walls are rigidly connected. Failure of the column can lead to the collapse of the entire structure. In a framed structure, where the columns are rigidly connected to other structural elements, besides the direct loads large bending moments are imposed on the columns. Reinforced columns are reinforced with the help of longitudinal bars are meant to carry the tensile stresses besides sharing the compressive forces with the concrete. Any bending moment taking place due to the accidental eccentricity of load on the column can be counter-balanced by providing adequate reinforcement. The columns can be reinforced with longitudinal bards and closely wound spiral reinforcement round the vertical bars. The spiral being wound closely has a confining effect on concrete within it. It is brought in tension and has got the ability to check the expanding tendency of concrete under loading.

The load bearing capacity of a column is given by the expression, P=σcAc+m σsAsc. Here, σc and σs represent the working stresses in concrete and steel. The stress allowed in steel is about 1.5 to 2 times the value of mσc.

stress in concrete=σc=Ece ; stress in steel=σs=Ese

modular ratio, m=σsσc

It has been observed that when the column is gradually loaded, the stresses in steel and concrete will be in the proportion of their modulus of elasticity as long as the stresses are within the elastic limits.

The ultimate load carrying capacity of a column is given by the factor (ασcuAs+ σsyAs), where σsy represents the yield point compressive stress in steel, σcu is the ultimate compressive strength of concrete cubes and α is a factor less than unity.

Figure 6: Column Layout Plan of the multi-storey building

- Design of columns

## Step 1: Geometrical properties of column

Ag(gross area of column)=length * breadth

Asc=nπr2

Area of concrete, Ac=Ag-Asc

## Step 2: Slenderness ratio

γ=Effective lengthLeast lateral diameter

## Step 3: Evaluate the strength of column

Ultimate load carrying capacity of columnMinimum eccentricty of column

Ultimate load carrying capacity of column=Ag[0.4fck1-p100+p100*0.67fy

emin=l500-D30

Strength of column, Pcu can be calculated using the expression,

Pcu=[0.4fckAc+0.67fyAsc]

## Step 4: Calculation of longitudinal reinforcement

Asc=p.Ag

A∅=π4*Diameter of reinforcement bars22

## Step 5: Calculation of transverse reinforcement

Pitch of the lateral ties can be any one of the following depending upon the conditions,

- Least dimension of the column

- 16 times the diameter of the longitudinal bar

- 48 times the diameter of the lateral ties

Pitch of the spiral can be taken as 1/6th of the core diameter or smaller than or equal to 75mm, whichever holds a lesser value.

- Beams

- Introduction

Beams are horizontal structural members constructed below the slab and are connected to the columns. These members span along the length of a given edge. Beams are responsible for carrying the load of the structural elements built above and then transmitting the load to the columns connected to it. Beams can be constructed using wood, steel, concrete, reinforced cement concrete and even plastic. The construction of beams can be carried out in two ways i.e. cast-in situ and laying of prefabricated beams. Beams are constructed at an average distance of 3m centre to centre. By suitably fixing the length and depth of the beam, the requirement of the number of beams can be reduced and hence the overall construction cost can be controlled.

- Design of beams

When a beam is subjected to a bending moment, compressive and tensile stresses are set up in its fibres. The intensity of stress in any fibre is proportional to its distance from the neutral axis. As long as the stresses in the concrete are within their elastic limits, the stresses in steel and concrete are in their modular ratio and the beam behaves homogeneously. The tensile stress in concrete surrounding the steel is equal to σst'm, where σst' is the tensile stress at any stage of loading .

The equations used for the design of reinforced cement concrete beams are given below .

stress in concrete=σc=Ece ; stress in steel=σs=Ese

modular ratio, m=σsσc

Further, Ac=net sectional area of concrete

Asc=net sectional area of steel embedded in concrete

The total load on concrete=σcAc

Total load on steel=m σsAsc

Therefore, the final load, P=σcAc+m σsAsc

stress in concrete, σc=P(Ac+mAsc)

Load on concrete, Pc=PAc(Ac+mAsc)

Load on steel, Ps=PAsc(Ac+mAsc)

The factor (Ac+mAsc) is called the equivalent area of the section.

When the depth and the breadth of a beam needs to follow certain restrictions due to various reasons such as enhancing the appearance, it becomes essential to provide steel on the compression side of the beam as well. If the permissible dimension given by the expression, M=Kbd2, are greater than the restricted dimension then the moment of resistance with respect to compression with smaller than the applied bending moment. Steel provided on the compressive side increases the capacity of the beam to resist compressive forces without increasing its size. Beams of such kind are known as doubly reinforced beams. Let Asc and At be the areas of steel in the compression and tension zones of the section respectively. Since plane sections of the beam remain plane after bending, we have:-

σcbσcb1=NdNd-ad=NN-a

Where, N, a and d represent the neutral axis coefficient, width and depth of the beam section while σcb and σcb1 are the compressive stresses in the concrete of beam. The value of the neutral axis coefficient can be calculated using the following equation.

N=11+σst'mσcb'

## Steps in beam-design are given as follows:

- Calculation of the minimum depth required with respect to the safe compression for the given bending moment

M=Bdsσcb2Nd-ds2Ndjd

Here B stands for the effective length, dsfor the thickness of slab and σcb represents the stress in concrete. Also, N is the co-efficient of neutral axis and d is the depth of the beam.

- Calculation of the minimum depth required for the upper limit of shear stress

- Calculation of the economical depth

d=M.r.σst.j.br

## Here ‘r’ represents the ratio of the cost of steel to that of concrete.

- Calculation of the tensile steel required

- Provide shear reinforcement if the shear stress exceeds the safe value for concrete

Figure 7: Plinth Beam Plan of the multi-storey building

Figure 8: Roof Beam Plan of the multi-storey building

- Design of slabs

Reinforced concrete slabs are the horizontal members acting as a top covering in a building and spans to the entire length of the floor. In the ground floor, a thick concrete slab supported on foundations or constructed directly on the subsoil is used. However, in the case of high rise buildings or skyscrapers the slabs are connected to the steel frames to create the floors and ceilings at each level. These slabs are thinner and are pre-cast concrete slabs. Cast-in situ slabs are constructed in buildings that are of much smaller scale in comparison.

In a rectangular slab with uniformly distributed load there is only one plane of bending and the load is transferred to these two supports. However, if a slab is supported on all the four edges, the load is transferred to all these supports. Thus the bending moments and the deflection is reduced considerably thereby reducing the need for thicker slabs. The load is carried in two directions to the respective supports, and the bending moment is much less than what it would have been if the load was carried in a single direction only.

The slabs are designed for the bending moments WB. B2/8 and WL.L2/8 per unit width along the short and long spans respectively. The depth is calculated for the larger of the two values. Since the moment in the short span is larger therefore the effective depth needs to be higher in this case. Therefore, the reinforcement parallel to short span should be placed below and parallel to the long span. Hence, the available effective depth of the slab for long span will be smaller than that for the shorter span.

Figure 9: Slab layout Plan of the multi-storey building

In the given project, to design the slab the following calculations need to be carried out.

Let the thickness of the slab be 15 cm or 150 mm.

Weight of the slab per sq. m. = 15*24= 360 kg

Total load=1600kg on all four floors

Total load on a single floor=400kg

Dimensions of slab-1= 5.5m x 10m

The effective spans=5.86m and10.36 m

r=10.365.86=1.77

wB=wr41+r4=400*9.811+9.81=363.185kg

wL=w-wB=400-363.185=36.815kg

Maximum bending moment along the short span=

363.185*5.528=1373.293

Maximum bending moment along the long span=

36.815*1028=460.1875

Depth of slab required=MbK=156.95=12.53

Therefore, an effective depth of 13 cm needs to be provided.

Area of steel required along the short span=

Mσstjd=9.43

Use 12mm dia. steel bars 10.0 cm apart from centre to centre. As the steel bars for long span will be placed above these bars for short span, effective depth for long span will be 11.4 cm.

Area of steel required for long span=

Mσstjd=6.43

Use 10mm dia. bars at 12.0 cm centre to centre distance.

Shear force on short edges = 1/2 wB= 1100 kg

Shear force on long edges =

wBr2+r=1032.89kg

Shear stress at long edges =

=Qbjd

= 0.993 kg/ sq cm

Shear stress at short edges = 1.21 kg/ sq cm

Therefore, the slab is safe at shear. Alternate bars can be bent up in each span without exceeding the permissible limits. The bending of bars can be done at 1/7th span from centre of supports. The ends of the bars must be provided with hooks.

- Design of Staircase

There are several arrangements that can be adopted to build reinforced concrete staircases. The simplest type of staircase is the one in which individual R.C. steps can project out of the wall as cantilevers. If a flat ceiling is required, an inclined R.C. slab cantilevering from the wall and supporting the steps can be built. The usual practice in the proportioning of staircase is to keep the rise equal to 15 cm to 20 cm and the tread equal to 23 cm to 25 cm. exclusive of the nosing that can be about 2 cm.

Figure 10: Section of the multi-storey building showing the staircase

Figure 11: Detailed Section specifically showing the staircase of the multi-storey building

Mbk

D=16.2 cm

Reinforcement, At =

Mbkdeff

At = 11.5 sq cm

Using 12 mm dia. bars, number of bars required in 1.25 m width =

1.25*11.51.13=12.7

Spacing of 12 mm steel bars = 125/13 = 9.6 cm.

- Final Proposal

The building comprises of four floors but follows the same plan at all the levels. The lowest floor has been allotted for parking of two wheelers and four wheelers. The land is covered by alluvial soil and hence it is favorable for construction. Furthermore, no harmful chemical or mineral is found in any significant number. The ground water conditions have also been found to be satisfactory. Therefore, no pre-processing of the ground is required prior to the construction.

The type of foundation selected for the building is the shallow foundation and all the members are designed according to the limit state design method. The collapse conditions as well as the serviceability requirements are significant aspects of the limit state design. Limit state design is an empirical method that takes into account the probabilities of collapse load and the possibility of reduction in building strength with increased load. The design of the building follows a simple rectangular form and it is a framed structure. The total load considered for design purposes consists of the self weight of the structure i.e. the dead load and the superimposed loads comprising of the live load, wind load, seismic loads etc. The load on the topmost floor is transferred to the floor beneath it through beams onto the columns. This load gets continuously transferred to the preceding floors and finally the columns transfer the total load to the footing. Ultimately, the entire load is transmitted to the soil by the footings. However, unequal loads at different point of the structure can cause differential settlement of the soil. This is turn can prove to be very dangerous for the durability of the structure. Thus, the designing has been carried out in a manner such that the safety parameters are appropriately calculated and incorporated.

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