DESIGN OF SIMPLY SUPPORTED SINGLY REINFORCED BEAM


A singly supported rectangular beam is to be provided over a clear span 10 m to carry on load of 25 KN/m excluding its self-weight .design mid section using M20 grade concrete and fe 415 steel  assume width of support as 300 mm
SOLUTION

Trial section
Clear span of the baam =10m
Assume effective depth of the beam d =  span/12
                                                                    =  10000 / 12  
                                                                    =  833 mm
Say 850 mm
Assume breadth of the beam b = 0.4d
                                                       = 0.4 x 850
                                                       = 340 mm
Say 350 mm
Assume eff cover   =  50 mm
Over all depth  D  =  850 + 50 =900 mm
LOADS
Loads on the beam         =  25 KN/M
Self weight of the beam = 0.35 x 0.9 x 1 x 25
Total characteristic load = 25 + 7.875
                                           = 32.875 KN/m
Taking the partial safety factor = 1.5

Design load Wu = 1.5 x 32.875
                     Wu = 49.313 KN/m
EFFECTIVE SPAN
Effective span of the beam is least of
1. c/c distance of support = 10 + 0.3
                                             = 10.30 m
2.clear span + eff.depth   = 10 + 0.85
                                            = 10.85m
So eff span lef = 10.3 m
Design bending moment
Design bending moment at mid span mu  = WU l2/ 8
                                                                   MU = (49.313 x 10.32)/8
                                                                        = 653.95 KN.M
                                                                        = 653.95 x 106 N.MM
Effective depth requirement
For Balanced section of M20 grade of concrete
Mu lim=2.76 bd2
D2 =mulim/2.76 b
D =Mulim2.76b
D =653.95 X 1062.76 X 350
D = 822 mm
We are take already eff.depth d =850 mm
So 822 mm < 850 mm
Provided depth is more than required depth Hence ok
AREA OF STEEL REQUIREMENT
Resisting moment of under reinforced section
Mu=0.87 fy Ast [d-(fy Ast/fck b)]
Fy= tensile strength of steel
Fck=compressive strength of concrete
Ast= Area of steel
B= breadth of beam
Mu=design  bending moment
653.95 X 106 = 0.87 x 415 x Ast [850-(415 Ast/20 x 350)]
Ast = 2600 mm2
Provide 6 numbers of 25 mm dia bars (Area = 2940 mm2 )


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CRACKS IN BUILDING AND CONCRETE




CRACKS IN BUILDINGS

For cracks occur in building when the building or its components subjected to external forces higher than those which can withstand. Also cracks appear low quality construction materials used in structure and improper construction techniques etc
Cracks can be divide into following
1. Structural cracks 
2. Non structural cracks

STRUCTURAL CRACKS


Structural cracks may be occur due to rough design over loading of the structural components and similar factors structural cracks totally damage the structure because its is the cracks in main load bearing element of the building  like cracks in columns , foundation , beam or slabs etc. Some times structural cracks recover not possible.

NON STRUCTURAL CRACKS

 non structural cracks may occur due to some variations like temperature variations moisture variations and the effect of gases ad liquid cam cause internal forces developed in the building this force can make cracks non structural cracks is easily recoverable and remedial measures are taken to prevent they re occur

Non structural cracks classified as horizontal cracks vertical cracks , diagonal cracks and irregular cracks

HORIZONTAL CRACKS IN MASONRY AND PLASTER



For load bearing wall the roof slab is to be expensed in day time due to heat of sun and alternative contraction in knight time this process can make pushing out the top coarse of masonry and its cause horizontal cracks in masonry structure.

PREVENTION METHOD

Causes of cracks due to temperature variation can be minimized providing expansion and contraction joints in specified locations.
Painting top or roof with reflective finish can minimize cracks

VERTICAL CRACKS


If the building has large openings in the external walls it can be seen that portion of wall immediately below the sill is subjected to much lesser loads as compared to the portion of the wall on either side of the window openings this results in uneven loading of the wall which may cause vertical cracks .

TRANSVERSE CRACKS


Continuous structure like sunshade, long open verandah slab can expansion in day time due to heat of sun alternatively contraction in knight time it will cause transverse cracks

CRACKS IN CONCRETE

There are six type of cracks appear in concrete. They are follow as

1.thermal  contraction cracks
2.structural cracks
3.sulphate attack cracks
4.alkali aggregate reaction cracks
5.shrinkage cracks
6. Plastic cracks

THERMAL  CONTRACTION CRACKS

For horizontal cracks near window ,sunshade and sill level due to pull exerted on the wall by roof slab. Because expansion  of slab in day time and shrinkage in knight time due to climatic condition changing . This pull result in bending of wall which cause crack in weak section that is at the near of sunshade and large span rooms.

STRUCTURAL CRACKS

Cracks occur where the applied forces on the members are greater than what they could withstand the imbalance in the load causes serious strains in the structural and ultimately the sign of distress appear as cracks .

SULPHATE ATTACK CRACKS


Sulphate attack on cement i concrete and mortar of masonry in foundation and plinth would weaken these components and in course o time cause unequal settlement of foundation and cracks in the super structure if brick aggregates used in base concrete of flooring contains too much of soluble sulphates and water table is high so as to make it in damp condition the base concrete will swell and cause cracking for the concrete floor.


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HOW TO DESIGN SEPTIC TANK FOR RESIDENTIAL COLONY


Ex:1
Design a septic tank for a small residential colony having a population of 800 persons. the rate of supply is 150 litres per head per day design also the soak wall if the effluence for the septic tank is to be discharged in if.

The quantity of water supply

               =per capita demand x population

              =800 x 150

              = 120000 litres / day

Assuming that the 80 % of water supply apply appears as sewage

The quantity of sewage produced = 120000 x 0.8

                                                 = 96000 lit / day

Assuming the detention period  of 24 hours the quantity of sewage prodused during this detention period is         = (96000 x 24)/24
                      =96000 litres
Assuming the rate of deposited sludge as 40 lit/capita/year and assuming the period of cleaning is one year

The volume of sludge deposition x population x period of cleaning

               =40 x 800 x 1

              = 32000 litres

Total capacity of tank = capacity of sewage + capacity of sludge

             = 96000 + 32000

            = 128000 litres

            = 128 litres

Assuming depth of tank = 1.6 m

Area = volume of tank / depth

        = 128/1.6

        = 80 m 2

The ratio  length to width is taken as 1:2 to 1 :4

So we take L : B = 1 : 4

                       L= 4B

Area = length x  breadth

       =L XB

       = 4B x B

       = 4 B2

4B= 80 m2

B= 80/4

    = 4.5m

Length of tank  L = 4 B

                       L = 4x 4.5

                       L= 18 m

Assume free board  = 0.2 m

The over all depth    = 1.6+ 0.2
                              = 1.8m
Size of tank              =18 x 4.5 x1.8

DESIGN OF SOAK PIT
Assuming the percolating capacity of filter media of soak pit as 1260 litres /m3 / day
Volume required for soak pit

          = volume of tank / percolating capacity of filter media
          = 128 m3/1.26 m3x m3 x day
          = 101.6 m 3

If the depth of soak pit is 3.5 m

Area of soak pit   =   volume of soak pit / depth of soak pit
                         =   101.6 / 3.5
                         =  29 m2
                    D  =  6m
ALSO READ--


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