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

 

 

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

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.

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.