Saturday, May 7, 2022

How to calculate storm water runoff volume


There are a few complicated experimental formulae for calculating Run-Offs of catchments which are not strictly accurate and give unreliable results. The rational method has been evolved in the form of the following general equation for calculating the flood discharge of Run-Off an area.

Q=  (R.A.P)/36

Where Q is total Run-Off in Cu.Metrs/Second

R is Intensity of Maximum rainfall in Centimeter /Hour.

A   is drainage area in hectares contributing to Run-Off

P is Factor of imperviousness

Example

Suppose length of catchment is 1900 ft having width of 40 ft for expected Run-Off in drainage area ,considering rainfall depth as given by weather department is 5cm in peak hour of rain intensity.   

  Length of drainage Area= 1900        

Width of strip causing Run-Off = 40 Ft.

Area of Run-Off = 1900 x 40 = 76000 Sft

1 Hectares = 107639 Sft

A= Converting Sft Area into Hectares = 0.70 Hectares

R= Max. Intensity of Rain fall depth = 05 Cm/Hour

P= This value varies according to nature of Ground surface.

We have following factors and we must choose one as per condition of our area, Factor no. 4 matches with our case.

1- Steep bare rock = 0.90

2- Rock steep but wooded = 0.80

3- Plateaus lightly covered, ordinary ground bare = 0.70

4-Densely built-up areas of cities with metallic Roads

and paths = 0.70 to 0.90

5- Residential areas not densely built up with metallic roads =0.50 to 0.70

6- Residential areas not densely built up with Un-metallic roads =0.20 to 0.50

 

4-Densely built-up areas of cities with metallic Roads

and paths = 0.70 to 0.90

P= Average value (0.80)

Q= (R.A.P)/36

Putting values in above formula

Q= (5 Cm/Hour x 0.70 Hectares x 0.70)/36 = 0.068Cubic. Meter

So, we have Discharge of 0.10 Cum = 2.4 Cubic Feet /Sec which is also equal to (0.068x1000) = 68 Liters/Second


Note: Rainfall data will be taken from local departments as per acctual locations in the world.

Wednesday, March 16, 2022

Lintel Beam Design Example

 

In Lintel beams there are three type of loading.

1) Masonry load (Triangular Loading  concept is used as an angle of 60 degrees develops due to brick masonry bonding but in case of block masonry total load of block wall over the lintel will be considered).

2) Concentrated Load may be point load or may be the uniformly distributed load in case of slab load (Slab loads should be considered in case of bigger sizes of window lintels).

3) Self Weight of Lintel .





Example : Design RCC Lintel beam for window opening size of 6 Ft and brick wall over the Lintel is 6 Ft high. The width of wall is 9 inches and the Ratio of concrete is (1:2:4) having 3000 Psi  compressive strength.

Here in this case the triangular loading concept will be used as  three sided equilateral triangle may be assumed but when brick work is well bonded and extends to each side of the opening a distance equal to at least half the span of the lintel beam. The usual practice is to take a load Uniformly distributed equivalent to the weight of the brick work contained in an equilateral triangle. This takes account of the fact that if the lintel were to fail, only a triangular block (roughly) of brick work would drop out because of the arch effect exerted by the bonding. The area of the equilateral triangle is 0.433 x (Length of side )^2 . It will be assumed that the lintel may appear to be fixed at its supports, the amount of actual fixing is not great ,so that the bending moment may , therefore, be assumed to be Wl/8 .(The actual bending moment for triangular load is Wl/6 but as mentioned above , in case of lintels, it is quite usual to consider the load as being uniformly distributed.

Coming to the example the wall bearing will be considered as width of wall so the side length of equilateral triangle will be 6.75 Ft

Weight of brick work = 0.433 (6.75X6.75) X0.75X 120 = 1775 Pounds .

Assuming size of beam ( width must be equal to width of beam and depth must be equal to number of brick courses).

So Width = 9" and depth may be assumed 1 inch for each ft of span = 6.75 x1         = 6.75 Inches so the size is increasing from two brick courses of 6Inches therefore we must exceed size up to 9".

Now calculating weight of beam = 6.75 x0.75x0.75x150 = 570 Pounds

Total Load = 1775+ 570 = 2345 Pounds.

M= Wl/8

M= Bending moment in pound-inch

l = effective span in ft.

M= 2345 x  6.75 x12 /8 = 23743 Pounds-Inch

Moment of Resistance = 250 . b.d1.d1

we know that Moment of resistance = Bending moment

Putting values in above equation

23743 = 250.(9).(d1.d1)

d1= 3.25"

Adding 1" cover on top and bottom

Overall size = 3.25 + 1+1= 5.25

now here we must consider the depth of brick courses which is 6" for two courses so the overall depth will be 6".

and d1(Effective depth will be 4" instead of 3.25 inches.

Area of Steel in tension zone = Bending moment / Fst.Lever Arm

= 23743/20,000x0.75x4  =0.396 Sq.Inch

No of 1/2 Inch Dia bars = 0.396 / 0.196 =2 Nos

Spacing of Stirrups  = AW.FW.Lever Arm/ Shear Force

Aw= Area of Legs of stirrups

Fw. Permissble tensile strength of steel bars (20000 Psi ) considering factor of safety of 3 times.

Lever arm  in load factor method = 0.75 .d1

Shear Force =  W/2 = 2345 /2 = 1173 Pounds

so trying stirrups of 3/8" dia bars

Spacing = (0.11x2)x 20000x( 0.75x4) x 1/ 1173 = 11.25 Inch  Center / Center

So the results are as under

Width of beam = 9 Inches

Depth of beam  = 6"

2 Nos of 1/2 " dia bars at top and 2 No of supporting bars at top and 3/8" dia bar stirrups @ 11.25 Inch Center to Center.

 







Tuesday, March 15, 2022

how to calculate weight steel bars in kg

 

The basic formula to calculate weight of steel bars is as follows

Weight = Volume of bar x Density of Mild steel

Volume of bar = (π/4) x d2  X Length of Bar

Density of Mild Steel is 490 Pounds per Cft

So calculating weight of bars in Kgs/ft

 

2/8" dia bar weight in Kgs/ft = 0.075 Kgs/Ft

3/8" dia bar weight in Kgs/ft = 0.17 Kgs/Ft

4/8" dia bar weight in Kgs/ft = 0.302Kgs/Ft

5/8" dia bar weight in Kgs/ft = 0.473 Kgs/Ft

6/8" dia bar weight in Kgs/ft = 0.681 Kgs/Ft

7/8" dia bar weight in Kgs/ft = 0.927 Kgs/Ft

8/8 " dia bar weight in Kgs/ft = 1.21  Kgs/Ft

Tuesday, February 15, 2022

Axially Loaded Column Design

 

Axially Loaded Column Design

Axially loaded meaning is that the line of the resultant force of the loads supported by the column is coincident with the center of gravity line of column .

Axially loaded column is also known as short column. A short column may be considered to be short when its effective height does not exceeds 15 times its least breadth.

Example

Suppose a column is subjected to carry a load of 150 Ton having concrete of (1:2:4) and we have to design it as short column and size of column is also required. The effective height of column is 15 ft.

First of all we will put formulas of short column designs

1) Longitudinal steel = The cross sectional area of steel must not be less than 0.8% of gross cross sectional area of column and also not more than 8 % of gross  cross sectional area of column.

2) Concrete cover = The cover to outside of longitudinal bars must be a minimum of 1.5 inches or the diameter of the bar whichever is greater. In case of column minimum dimension of 7.5" or under, whose bars do not exceed 1/2" in diameter ,1" cover may be used.

3) Splices = The length of lap of bars in compression should not be less than =

The bar diameter x The compressive stress in bar/ Five times the permissible average bond stress

OR 24 bar diameter whichever is greater

For example if the compressive stress in the bars is 18000 Lbs per sq.in and the permissible average bond stress is 120 Lb. per sq.in. (nominal 1:2:4 mix ) the length of lap will be calculated as follows

Bar Diameter x 18000/(5x120)= Bar diameter x 30

4) Lateral Ties= The diameter of transverse reinforcement should not be less than one quarter the diameter of the main rods and in no case less than 3/16 in.

5) Pitch.  The distance apart of the ties must not be more than the least of following

a. least lateral dimension of column

b.Twelve times diameter of the smallest longitudinal bars.

c. Twelve inches

6)  P =  fcc.Ac+fsc.Asc

      P= fcc (Ag-Asc)+ fsc.Asc

fcc = permissible stress of concrete in direct compression (760 Lbs./sq.in for 1:2:4 mix)

fsc = permissible compressive stress for column bars (18000 Lbs./sq.in for Grade 60 steel)

Ac = cross sectional area of concrete excluding any finishing material and reinforcement steel i.e

Ac= Ag- Asc where

Ag= Gross cross sectional area of column

Asc= Cross sectional area of longitudinal bars

Now Putting values

Considering given load we will put values in formula no 6 , and will assume 2% of steel (As we can use range of steel from 0.8 % to 8% of gross cross sectional area of column.

      P= fcc (Ag-Asc)+ fsc.Asc

(150X2240) = 760( Ag - 2/100.Ag)+18000x2/100Ag)

336000 = 760Ag-15.2Ag + 360.Ag

336000= 1104.8 Ag

Ag= 304.127 Sq.inch

Considering Rectangular column having 12 inches as Least dimension of column does not exceeds 15 times of effective height of column = 1x15 =15 so it will be short column

Longer side of column = 304.127/12 = 25.34 say 26 inches so size of column will be 24"x12"

Longitudinal bars of columns = 24x12x 2/100= 5.76 Sq.In

Using 3/4" dia bars

No of bars = Asc /Area of bars = 5.76/0.44= 13 Nos

The length of lap of bars in compression should not be less than =

The bar diameter x The compressive stress in bar/ Five times the permissible average bond stress

OR 24 bar diameter whichever is greater

Lets try above rule to find out lap length

The bar diameterx18000/5X120 =30 Times of Diameter of bars

or 24 bar diameter

so we will use lap length = 30 times of diameter of bars

=30x 3/4" = 22.5 Inches

Pitch.  The distance apart of the ties must not be more than the least of following

a. least lateral dimension of column

b.Twelve times diameter of the smallest longitudinal bars.

c. Twelve inches

Pitch will be used as Twelve times of diameter of longitudinal bars = 3/4"x12=9" c/c

 

 

Monday, February 14, 2022

Double Bituminous Surface Treatment

 

DESCRIPTION OF DST/DOUBLE SURFACE TREATMENT

Double bituminous surface treatment is a term used to explain a common type of pavement surfacing construction usually 3/4" thick layer which involves two applications of asphalt binder material (bitumen) and mineral aggregate. The bitumen is applied by a pressure distributor, followed immediately by an application of mineral aggregate, and finished by rolling.

Double surface treatment are generally necessary on untreated water-bound macadam roads subject to heavy traffic .A semi grout with seal coat or thin carpet should be preferred to two coats of surface dressing if cost is not prohibitive. Where the road is rutted or there are depressions more than 6 mm deep ,two coat dressing should be done.

The second coat may be applied 24 hours after the first coat or an interval of several months may elapse. The advantage of allowing an interval between each coat is that the first coat will have time to settle down under traffic and show up depressions which can be made up in the second coat, thus ensuring an even surface. When an interval is to elapse between each application ,the second method of construction is exactly the same as in the single coat work. 

Materials and Procedures

 After completion of Water-bound layer construction of DST consists on following steps.

 1) Cleaning of top surface of water-bound and removal of loose particles of stone dust (used to fill the voids of water-bound macadam layer).The cleaning should provide a clear sight of stone's top surface not having dust or stone pieces.

2)After cleaning water-bound surface we will apply emulsion of prime coat @ 0.09  kgs / sft of surface area of water-bound layer

3) In First Coat 12MM OR 1/2 " SIZE crush stone will be used @0.05 cu.ft/ Sft of Water-Bound, Including shower of bitumen @.09 Kgs/Sft  covering  80 % height of 1/2" sized crush stones and this will be the first layer of DST.

4) In Second Coat/Renewal Coat 8 to 9 MM OR 3/8" SIZE crush stone will be used @0.03 cu.ft/ Sft of Water-Bound, Including shower of bitumen @.11 Kgs/Sft  covering  80 % height of 1/2" sized crush stones and this will be the first layer of DST.

 

Seal Coat

Surface dressing (gritting)or seal coat is required over open textured carpets. Premixed  seal using medium course sand or fine grit 32-Cft per every 1000 Sq.Ft and Bitumen 98 Kgs Per every 1000 Sq.ft

Sunday, February 13, 2022

One Way Slab Design

 

One way Slab Design


Any slab having Longer and Shorter span ratio more than 1.5 will be considered as One way slab which means Shorter will carry the loading shorter side bars will contribute as per their specific bending moments and longer side bars will be used as distribution bars or as temperature bars.

Example

Suppose we have a slab panel sizing 20 ft x 10 ft  having imposed UDL of 150 Lbs/Ft.

Finding Longer and shorter span ratios (20/10 =2 so it is greater value of 1.5 so we will design it as One Way slab)

1) First we will find the roughly Estimated thickness of slab =span/30

= 10/30= 0.333 Ft x12 =  4 Inches

 

So self load of slab = Unit Volume of slab x density of RCC

=  1x1x4/12x150 = 50 Lbs/Sft

Total UDL =      Imposed Load+ Self Load Using estimated Thickness

150+ 50 = 200 Pounds per Sq.Ft

Formulas for Bending Moment Mx= ax.w.lx.lx

Max Bending Moment (Simply supoorted span) = w.L.L./8

 

L =Shorter span

w= UDL (Uniformly distributed Load

So Putting values in Bending Moment formulas 

Max Bending Moment = 200x10x10x1/8x 12 = 30000 Lbs-Inch

In above equation 12 is used to convert Lbs-Ft to Lbs-Inch

As per Elastic Theory Method (Using 1:2:4) concrete

Moment of resistance = Bending Moment

Moment of Resistance in Elastic theory method 

= 184.b.d1.d1

where b is unit width of slab

d1 = Effective depth of slab (depth without covers)

Putting values in above equation 

30000= 184 x 12 x d1 x d1

d1 = 3.72  inch say 3.75

So adding 1/2" concrete cover on top and bottom overall depth will be 4.75 Inches

 

Ast = Bending Moment /Fst x Lever Arm

where Ast = Area of steel in tension zone

Fst = Factored Tensile stress of steel (which is 20,000 Psi) 

Factor of safety is 3 

Lever arm = 0.857 x Effective Depth

Putting Values in Above equation 

Ast = 30000/ 20000x0.857x3.75  = 0.466 Sq.Inch

Spacing of Bars = (Area of bar x 12) /Ast

Using 1/2 inch dia bar 

Spacing = (0.196 x 12 )/0.466= 5.04 say 5 Inch c/c 

So 1/2" dia bar will be used @ 5 inch center to center on shorter side  and slab thickness will be 4.75  inch using (1:2:4) concrete.

Now Designing Distribution bar

Area of Distribution bars = 0 .2 percent of unit cross sectional area of slab

           = 0.20/100 x12x4.75 = 0.114 Sq.Inch

Spacing of Distribution bar = Area bar x 12 /Area of Distribution bars

Using 3/8 inch dia bars spacing = 0.11x12/.114= 11.57 Say 11.5 inch center to center

As per BS Code spacing should not be more than 5 times of effective depth (d1)

= 3.75 x5 = 18.75  so We can use 3/8" diameter bars @ 11.5 inch center to center

 

quality of coarse aggregate

 

Quality of  Coarse Aggregate




Since characteristics of concrete are directly related to those of its constituent aggregates, aggregates for load bearing concrete should be hard ,strong, non-porous ,free from friable ,elongated and laminated particles, and should be suitable for the purposes required. Stones absorbing more than 10 per cent of their weight of water after 24 Hours immersion in water are considered porous. Porous materials corrode reinforcement.

A friable aggregate will produce a concrete of similar nature. Elongated or laminated particles are weak in shear. Stones having mica inclusion should be avoided. Stones of the varieties of granite, quartzite , trap and basalt, and those with rough non-glossy surface are considered best.

All sand-stones tend to be porous. Soft varieties of sand stones make poor concrete and also produce shrinkage cracks. Limestone is quite good provided it is hard ,crystalline and entirely free from dust. Limestone should not be used in works subject to excessive heat. Both lime and sand-stones and other porous stones are not suitable for structures retaining water, or existing in the atmosphere near to sea sides.

Size of Aggregates

The material retained on a 4.75 mm IS sieve is classified as coarse aggregate and below that  size as fine aggregate or sand. The material passing a 75-micron IS sieve is generally considered to be clay, fine silt or fine dust in an aggregate.

Coarse aggregates should be ordered in separate sizes and recommended in proper proportions while batching.  40 mm to 20 mm , 20 mm to 10 mm and 10 mm to 4.75 mm. A 20 mm nominal maximum size will be ordered in two sizes--- 20 mm to 10 mm , and 10 to 4.75 mm. Separate stockpiles should be maintained for the different sizes.

Angular and roughly cubical particles are ideal. River gravel makes the best coarse aggregate.

Storing or Stockpiling of Aggregate

During storing or handling of aggregate it is of utmost importance to see that there is no segregation i.e., separation of the various sizes of particles. Stockpiling segregation does take place if successive consign- me it's are dropped on the same place each time and it forms a pyramid like heap, as the coarser materials roll down the sides of the pile while the finer particles stay on in the centre of the pile on concentration at the top. Therefore, all the material should not be piled at the same place but should be placed in individual units side by side not larger than a truck load and should not be thrown from a height as this will also result in segregation by the winds. Each separate size of the coarse aggregate and that of sand should be stacked separately ,large in area and low in height 1 to 1.5 meters. All washed aggregate should be stacked for draining at least 12 hours before being batched.

 

Maximum Aggregate Size.

The maximum size of aggregate is governed by the type of the work to be built. The bigger the maximum size of the aggregate, less than voids when the aggregate is graded. A coarse aggregate which has less void content is economical and will give higher strength for the same amount of cement in the concrete. The maximum size of aggregate may be up to 160 mm for mass concrete, but size up to 225 mm has been used in dams. Aggregates of this size require careful mix design to avoid segregation and it is probably wise to limit maximum size to 80 mm. Large stones which are embedded in mass concrete works are called "plumbs". Plumbs should be sound and hard and should not be placed nearer than 160 mm to one another or to an exposed surface.

Testing of Aggregates

In order to decide the suitability of the aggregate for use in pavement construction, following tests are carried out:

1-Crushing test.

2-Abrasion test.

3-Impact test.

4-Soundness test.

5-Shape test.

6-Specific gravity and water absorption test.

7-Bitumen adhesion test.

Thursday, February 10, 2022

Design of Two Way Slab by Coefficient Method

 

Any slab having Longer and Shorter span ratio less than 1.5 will be considered as Two way slab which means both spans (Longer and Shorter) will carry the loading and both side bars will contribute as per their specific bending moments.

Example

Suppose we have a slab panel sizing 20 ft x 20ft having imposed UDL of 150 Lbs/Ft.

1) First we will find the rough thickness of slab =span/35

= 20/35= 0.571 Ft =  6.85 Inches say 7"

 

So self load of slab = Unit Volume of slab x density of RCC

=  1x1x7/12x150= 87.5 Lbs/Sft

Total UDL =      150+ 87.5 = 238 Pounds per Sq.Ft

Formulas for Bending Moment at Longer span Ly and Shorter Span Lx is as follows.

Mx= ax.w.lx.lx

My= ay.w.ly.ly

Mx= Maximum Bending Moment at Middle strip of shorter span

My= Maximum Bending Moment at Middle strip of Longer span

ax= Shorter span

ay =Longer span

ax= Co-efficient of shorter span

ay= Co-efficient of longer span

Following coefficient table may be used as per Ly/Lx Ratios.



so in our case span Ratio = 20/20 = 1 

So ax= 0.062

     ay= 0.062

So Putting values in Bending Moment formulas 

Mx= ax.w.lx.lx

0.062 x 238 x 20 x 20 x 12 = 70828 Pounds-Inch 

In above equation 12 is in inches 

My= ay.w.ly.ly

0.062 x 238 x 20 x 20 x 12 = 70828 Pounds-Inch

As per Elastic Theory Method (Using 1:2:4) concrete

Moment of resistance = Bending Moment

Moment of Resistance in Elastic theory method 

= 184.b.d1.d1

where b is unit width of slab

d1 = Effective depth of slab (depth without covers)

Putting values in above equation 

70828= 184 x 12 x d1 xd1

d1 = 5.75 inch for Mx and My 

So adding 1/2" concrete cover on top and bottom overall depth will be 6.75 Inches


Ast = Bending Moment /Fst x Lever Arm

where Ast = Area of steel in tension zone

Fst = Factored Tensile stress of steel (which is 20,000 Psi) 

Factor of safety is 3 

Lever arm = 0.857 x Effective Depth

Putting Values in Above equation 

Ast = 70828/ 20000x0.857x5.75  = 0.718 Sq.Inch

Spacing of Bars = (Area of bar x 12) /0.718 

Using 1/2 inch dia bar 

Spacing = (0.196 x 12 )/0.718 = 3.27 say 3.25 Inch c/c 

So 1/2" dia bar will be used @ 3.25 inch center to center both ways and slab thickness will be 6.75 inch using (1:2:4) concrete.



Saturday, September 25, 2021

Types of Pile Foundation.

 

What is Pile?

An constructional element placed vertically or nearly so in the ground to increase the bearing capacity of soil or to resist the lateral load is known as pile.


A pile foundation is usually provided where the soil material under the base of a structure has deficient bearing power to take the load of the structure, and the soil near the ground surface is also incapable of supporting a raft foundation .Piles are used for erecting foundations of structures in order to transmit load to the underlying soil and to improve the load bearing capacity of low bearing capacity soils.

Piles are classified into two categories

1) Piles As per functionality

2) Piles As per material of construction

 1) Piles as per function  

Piles are classified as per their function as discussed below.

a)Bearing Piles.

The piles which rest on hard strata and act as columns to bear the load of the structure are known as Bearing piles. These piles are used to bear vertical loads .They take and transfer the load to the hard stratum lying underneath.

These piles are driven through soft strata and go deep to rest on hard surface and support the load by the resistance developed at their points by end bearing, and act as  long columns. In these piles the cross section should be comparatively greater to resist the buckling effect.

 

 b) Friction Piles or Floating Piles.

The piles which do not rest on hard strata and bear the loads on account of frictional resistance between their outer surface and the soil in contact are called Friction piles.

These piles are used when the soil is soft and there is no hard strata available up to a considerable depth. These are generally long in length. Frictional resistance can be increased by making the surface of these piles rough or by increasing their surface area likely to come in contact with soil.

c ) Friction Cum -bearing Piles.

The piles which rest on a hard strata and resist the structural load partly by bearing and partly by their skin friction are known as friction-cum bearing piles. These piles are used when the bearing capacity of soil strata lying under them is not sufficient to resist load of structure.

These piles are driven into hard strata and their bearing capacity is provided chiefly by friction of their side surface against the soil. This friction is called skin friction.

Here are some significant frictional resistance offered by various soil materials to the pile surface.

 

- Sand and Gravel          (4.9 to 8.8 ) Tonnes/Sq.m

-Stiff Clay                         (3.9 to 5.9 ) Tonnes/Sq.m

-Clay and sand mixed     (1.9 to 9.9 ) Tonnes/Sq.m

-Dried and compact silt  (1.0 to 1.5 ) Tonnes/Sq.m

-Silt and soft clay             (0.25 to 0.50 ) Tonnes /Sq.m

 

d) Batter piles

The piles driven at an inclination to resist inclined loads are known as batter piles.

These piles are used generally to resist lateral forces in case of retaining walls ,abutments etc.

e) Guide Piles

These piles are mainly used in the formation of cofferdams which are temporarily constructed to provide foundations under water.

f) Sheet Piles

The piles which consist of thin steel sheets, driven in the ground to enclose an area are known as sheet piles.

These piles are used to enclose soil so as to prevent the leakage of water and to enclose soft material. They are used in the construction of cofferdams. Sheet piles are not required to carry any load but should be strong enough to take the lateral pressure of earth filling ,water etc.

2) Piles As per material of construction

a) Timber Piles

The piles made of wood are called timber or Wooden Piles.

The timber to be used for their construction should be free from detects, decay etc and it should be well seasoned. These piles are circular (20 cm to 50 cm  ) in diameter or square (15 cm to 50 cm side) in cross-section. Length of these piles is generally 20 times of their diameter of side length. Top of these piles is provided with an iron ring to prevent it from splitting under blows of the hammer. The bottom is fitted with an iron shoe to facilitate sinking of the piles. These piles are driven by blows of drop hammer of a pile driving machine.

Timber piles are broadly used as they have advantage of flexibility and weightlessness, and in many places they are cheaper than other materials. Their disadvantages is lack of durability in certain conditions.

strength depends on the type of wood ,its moisture content ,and its position. In general timber piles are durable in permanently wet or permanently dry positions, but not where they are alternatively wet and dry or where the moisture content is widely variable.

Timber piles if used longer than 30 year or thereabouts and such are usually preferred for temporary works and also for semi permanent marine structures. Timber piles should be impregnated with solignum , creosote or treated with some such anti rot compounds to enhance their service life.

preservative treatment may not be necessary for piles which will be completely and permanently submerged in water-logged ground because in this case seasoning is not necessary and piles may be stored in water prior to use.

Suitability

These piles are generally used for buildings ,bridges and cofferdams but their use is not recommended in sea water.

b) Pre-Cast Concrete Piles

Reinforced concrete piles are widely used in construction practice with and without pre-stressed reinforcement. They are usually of square or rectangular cross-section, as they are easier to cast than a circular section. The usual size is 15 to 60 cm but piles have been made up to 90 cm size with cylindrical holes inside. Hollow piles have advantages where exceptional lengths are required because they provide stiffness and large perimeter with lesser weight than solid piles.

The piles having larger than 36 x 36 cm an octagonal section is preferable to a square section . Square piles should have chamfered corners. The maximum lengths of piles are usually 12 m for 30 cm square, 15 m for 36 cm square,18 m for 40 cm square, 21 m for 46 cm square. It is preferable to keep the length less than 4o times the side for friction piles and less than 20 times the side for bearing piles.

Where the piles are considered to act as column, stresses should be calculated as for ordinary columns. To prevent damage to the head of pile, the top edges should be chamfered liberally and additional lateral reinforcement provided and kept back from the head about 50 to 75 mm according to the diameter. Concrete piles should be cured at least a month. Lifting holes should be made at one-fourth to one-fifth the length of the pile from each lifting hole. 2.5 cm diameter gas pipe ferrules may be fixed in the holes. Concrete piles are generally designed for 70 to 80 tonnes.



c ) Cast-in-situ Piles.

These piles are made with driven hollow tubes or by using heavy steel pipe castings and then withdrawing them or by boring and filling the holes formed with concrete. The tube is placed at the top of loose cast iron point before driving into the ground and is slowly and steadily drawn out of the ground as concrete is filled in. Piles are also formed by driving in a steel shell ,leaving it permanently there and filling it with concrete. The shells should be strong enough so that they are not distorted by soil pressure or the driving of adjacent piles. These piles are built in diameter of 0.4 to 2 m and in lengths of up to 50 m. The bearing capacity may be as high as 600 Ton per pile. These piles can also be made with bulb toes giving greater bearing value.

There may be many patented processes for these piles such as Franki, Simplex, Vibro.

No driving pile should be withdrawn until all piles within 3m radius have been driven. Built in place piles are made without vibrating the soil, which is particularly important for construction work near or inside existing buildings and installations. As these piles are  not intended to carry impact loads under normal conditions no reinforcement is necessary and where required it is placed in the upper part only. Any reinforcement used should be made up into cages sufficiently well wired because the bars should be openly spaced , and the lateral ties should not be closer than 15 cm center to center.

The reinforcement should be exposed for a sufficient distance to permit it to be adequately bonded into the pile cap. In bored piles care should be taken to prevent influx of soil into the castings during boring. Before placing the concrete the holes should be inspected by lowering a light for any undesirable materials or water. Placing of concrete should not be started until all the shells in a group has been completed. Bored piles unless sunk into hard and compact ground should be test loaded. Steel shells which are to be filled with concrete should be coated externally with bituminous composition or tar ,etc, before they are driven . In other cases all surfaces should be coated if tar is employed , it should be neutralized with slaked lime.

 

Reinforcement and concrete materials for the Piles

The area of the main reinforcement of longitudinal bars may be 1.5% of gross cross sectional area of the pile if length of pile is 30 times of the least width of pile , and 2 % if the lengths are 30 to 40 times , which may be increased up to 3 % for longer lengths. The links usually are 6 mm diameter up to 12 m and 10 mm diameter above 12 m are spaced 50 to 75 mm for lengths up to 3 times the side at each end of the pile, lengthening to 15 to 20 cm at the center.

Reinforced concrete piles should be of 1:1.5:3 or richer mix,  with well graded aggregate of maximum size limited to 12 mm, and a slump of about 40 mm.



Wednesday, September 22, 2021

Sound absorbing materials for Office and Home

 

The materials used for reducing the reflection of sound waves in buildings are called Sound absorbing materials.

There are many materials which may be used for reducing the sounds effects and also  in buildings  to reduce noise pollution.

Examples of these materials are following

1) Perforated card boards

2) Porous materials

3) Heavy curtains

4) Maps

5) Pictures put upon the walls

6) Carpets on the floors

7) Wood-wool Acoustic Panel

8) Moss Panels

9) Fabric Acoustic Panels

10) Cloth papers

11) Ceiling Baffle

12) Printed Acoustic Panels

13) Fiber boards

14) Porous plaster

15) Oil Cloth

16) Glass Silk

17) Porous Plastics

18) Sound Absorbing Foam (Pro Studio Acoustics Tiles)

19) Acoustic Panels (ATS Acoustics)

20) Acoustic Curtains (Utopia Thermal Blackout Curtains)

21) Moving Blankets (Sure Max Heavy Duty).

22) Door Sealing Gasket & Sweep Kit.

 

These materials suitably applied to walls ,ceiling and floor of a room or Hall to reduce reflection of sound waves from their surfaces. Sound absorbing materials should be porous ,inelastic ally flexible, compressible or they may be having combination of two or more of these properties.

The absorbent co-efficient of commercial sound absorbing materials can be increased by various means such as perforating ,slotting or by providing small apertures into the body of materials , by doing so energy centre of sound dispersed sideways.

Architectural Soundproofing . This  includes everything used in the structure of a building, such as soundproof windows, soundproof walls, doors, and decoupling products used to install them.



Auralex Acoustics Studio foam Wedgies Acoustic Absorption Foam, 2" x 12" x 12", 24-Panels, Charcoal

 


Pro Studio Acoustics - Blue/Charcoal - 12"x12"x2" Acoustic Wedge Foam Absorption Soundproofing Tiles - 12 Pack

 

       Acoustic Panels (ATS Acoustics)



            Acoustic Curtains


               Perforated card boards



          Wood-wool Acoustic Panel


     ceiling baffles sound absorption



              Printed acoustic panels

        Fiber boards for sound proofing





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