"പുഴയും മലയും മരങ്ങളും 

പുഴുവും പുല്ലുമതൊന്നുപോലെ കാക്കും 

വരദേ മമ കേരളാംബികേ നിൻ 

ചരണേ വീണു വണങ്ങിടുന്നിതാ ഞാൻ"

ഏവർക്കും ഹൃദയം നിറഞ്ഞ കേരളദിനാശംസകൾ.

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Shore Structure

Shoring

Shoring is the process of supporting a building, vessel, structure, or trench with shores  when in danger of collapse or during repairs or alterations. Shoring comes from shore a timber or metal prop Shoring may be vertical, angled, or horizontal.

Techniques

• Buildings

Raking Shores consist of one or more timbers sloping between the face of the structure to be supported and the ground. The most effective support is given if the raker meets the wall at an angle of 60 to 70 degrees. A wall-plate is typically used to increase the area of support.

Foundations

Shoring is commonly used when installing the foundation of a building. A shoring system such as piles and lagging or shotcrete will support the surrounding loads until the underground levels of the building are constructed.

• Trenches

During excavation, shoring systems provide safety for workers in a trench and speeds up excavation. In this case, shoring should not be confused with shielding. Shoring is designed to prevent collapse where shielding is only designed to protect workers when collapses occur. Concrete structures shoring, in this case also referred to as falsework, provides temporary support until the concrete becomes hard and achieves the desired strength to support loads.

Hydraulic Shoring

Hydraulic shoring is the use of hydraulic pistons that can be pumped outward until they press up against the trench walls. They are typically combined with steel plate or plywood, either being 1-1/8" thick plywood, or special heavy Finland Form  7/8" thick.

Beam and Plate

Beam and Plate steel I-beams are driven into the ground and steel plates are slid in amongst them. A similar method that uses wood planks is called soldier boarding. Hydraulics tend to be faster and easier; the other methods tend to be used for longer term applications or larger excavations.

Soil Nailing

Soil nailing is a technique in which soil slopes, excavations or retaining walls are reinforced by the insertion of relatively slender elements - normally steel reinforcing bars. The bars are usually installed into a pre-drilled hole and then grouted into place or drilled and grouted simultaneously. They are usually installed untensioned at a slight downward inclination. A rigid or flexible facing (often sprayed concrete) or isolated soil nail heads may be used at the surface.

• Ships

Shoring is used on board when damage has been caused to a vessel's integrity, and to hold leak-stopping devices in place to reduce or stop incoming water. Generally consists of timber 100 mm x 100 mm and used in conjunction with wedges, to further jam shoring in place, pad pieces to spread the load and dogs to secure it together. also used on board is mechanical shoring as a quick, temporary solution, however it isn't favoured due to its inability to move with the vessel.

Square Shoring

This consists of a timber member jammed on a pad piece on either the deck or deck head depending on water levels in the compartment and a strong point, this is called the proud. then the is a horizontal timber cut to size to fit between this and what it is shoring up, e.g. a splinter box, bulkhead or door. Timber wedges are then used to tighten up the structure if necessary

Vertical Shoring This is to support a hatch or splint box on the deck, consisting of a vertical timber between the deck and deck head, with to wedges used opposing each other to tighten it. pad pieces are used to spread the load on weak structures.

Gallery

• 

  Vertical or dead shore system, typically used in formwork.
 
• 

  Single steel raking shore system specifically for tilt slab shoring.
 
• 

  Angkor Wat complex, simple combination of timber raking and dead shores.  
• 

  Sketch of a timber double raking shore. Projected centre lines of floors and shores meet.
 
• 

  Carpentry detail of the joint at the top of a timber raking shore.
 
• 

  Sketch of a timber single flying shore between adjacent buildings.
 
• 

  Traditional trench shoring or Timbering.


• 

  Schematic sketch of a modern steel trench shore being lowered into a trench                                                                          

INTERIOR DESIGN

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STRUCTURAL ENGINEERING

Structural Engineering are trained to understand, predict, and calculate the stability, strength and rigidity of built structures for buildings and nonbuilding structures, to develop designs and integrate their design with that of other designers, and to supervise construction of projects on site. They can also be involved in the design of machinery, medical  equipment, vehicles etc. where structural integrity affects functioning and safety.

Structural engineering theory is based upon applied physical laws and empirical knowledge of the structural performance of different materials and geometries. Structural engineering design utilizes a number of relatively simple structural elements to build complex structural systems. Structural engineers are responsible for making creative and efficient use of funds, structural elements and materials to achieve these goals.tructures that are available to resist them. The complexity of modern structures often requires a great deal of creativity from the engineer in order to ensure the structures support and resist the loads they are subjected to. A structural engineer will typically have a four or five year undergraduate degree, followed by a minimum of three years of professional practice before being considered fully qualified. Structural engineers are licensed or accredited by different learned societies and regulatory bodies around the world. Depending on the degree course they have studied and/or the jurisdiction they are seeking licensure in, they may be accredited (or licensed) as just structural engineers, or as civil engineers, or as both civil and structural engineers. Another international organisation is IABSE (International Association for Bridge and Structural Engineering). The aim of that association is to exchange knowledge and to advance the practice of structural engineering worldwide in the service of the profession and society.

COST ESTIMATING

&

PURPOSE OF COST ESTIMATING

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Slump test

The slump test is a means of assessing the consistency of fresh concrete. It is used, indirectly, as a means of checking that the correct amount of water has been added to the mix. The test is carried out in accordance with BS EN 12350-2, Testing fresh concrete. Slump test. This replaces BS 1881: Part 102. The steel slump cone is placed on a solid, impermeable, level base and filled with the fresh concrete in three equal layers. Each layer is rodded 25 times to ensure compaction. The third layer is finished off level with the top of the cone. The cone is carefully lifted up, leaving a heap of concrete that settles or ‘slumps’ slightly. The upturned slump cone is placed on the base to act as a reference, and the difference in level between its top and the top of the concrete is measured and recorded to the nearest 5 mm to give the slump of the concrete. When the cone is removed, the slump may take one of three forms. In a true slump the concrete simply subsides, keeping more or less to shape. In a shear slump the top portion of the concrete shears off and slips sideways. In a collapse slump the concrete collapses completely. Only a true slump is of any use in the test. If a shear or collapse slump is achieved, a fresh sample should be taken and the test repeated. A collapse slump will generally mean that the mix is too wet or that it is a high workability mix, for which the flow test (see separate entry) is more appropriate. Koroth constructions. +91 8943009914

Here we presenting you detailed estimate formulas and some use ful informations.

Construction Estimating Formulas & Useful Information

Formulas, equations, tables, facts, specifications and other 
information useful in planning and estimating construction, 
decorating and other projects.

 Measure of Dimensions, Theorem ▪   Formulas, Rectangles, Squares, Triangles & Circles ▪  Board Feet Concrete Estimating ▪ Decimal and Metric Equivalents ▪  Weights, etc. ▪  Metrics ▪  Energy, Electric, Insulation ▪ Drywall Estimating,  Door Handing

Three dimensions of estimating:

Linear Measure	Square Measure	Cubic Measure
12 inches = 1 lineal ft.	12" x 12" = 144sq" = 1sq ft.	12" x 12" x 12" = 1728 cu" = 1 cubic ft.
3 lineal feet = 1 yard.	3' x 3' = 9sq feet = 1sq yd.	3' x 3' x 3' = 27 cu feet = 1 cubic yard.
5,280 lin ft = 1 mile.	43,560sq' = 1 acre.	36" x 36" x 36" = 46,656 cu" = 1 cubic yard.
Inches x .0833 = ft. Inches / 12 = ft.	Sq" x .00694 = sq ft.	Cu" x .0005787 = cu'.
Inches x .2778 = yds.	Sq" x .007716 = sq yds.	Cu' x .0002143 = cu yds.
Feet x .3333 = yds.	Sq' x .11111 = sq yds.	Cu' x .03704 = cu yds.
	100sq feet = 1 square (measure used in roofing).

Useful Mathematical Formulas:

1. One of the most useful formulas is the Pythagorean theorem.  The theorem states that the square of the hypotenuse of a right (90 degree angle) triangle is equal to the sum of the squares of its legs.

For Example:        (3' x 3' = 9') + (4' x 4' =16') = 25',  √25=5

This can be used for a wide variety of uses from squaring up a concrete form for a foundation, to attaching a new addition squarely to an existing building, or measuring a countertop -anywhere you must be certain of a right angle.  Just plug in your numbers and get a calculator that has a square root function.

                                                                                              

2.  The area of a rectangle is equal to its base x height. (A = b x h).

For example:     10 x 5 = 50

3.  The area of a square is equal to its base x height (see above).
               -or-
  The area of a square is equal to the square of one of its sides. (A = S squared).

For example:     5²=25

4.  The area of a triangle is equal to 1/2 the product of its base and height. A = 1/2 (b x h).

For example:     10 x 10 = 100 x 1/2 = 50

5.  Circles

The area of a circle is equal to 1/2 the product of its radius and circumference. (A = 1/2 (r x c). 

For example:            radius = 5, circumference = 31.42:   5 x 31.42 = 157.1 x 1/2 = 78.55

 -or-

                                A=π r², 3.1416(5 x 5) = 3.1416 x 25 = 78.54

-or-

A= d² x .7854

Find the circumference of a circle by multiplying the diameter x (π).  Find the volume of a cylinder by multiplying the height x the area of the circle.

*π = 3.1416

The circumference of a circle can be divided into equal parts with a calculator.  This will allow you to make proportioned shapes within the circle.  For example -stars, triangles, octagons, etc.

To find the divisions, multiply the diameter of the circle by the following multipliers.

Division	Multiplier		Division	Multiplier
3	.86603		8 (Octagon)	.38268
4	.70711		9	.34202
5	.58779		10	.30902
6	.50000		11	.28173
7	.43388		12	.25882

For example:

21.166" x .38268 (from above table) = 8.099"

6.  Board Feet:  A board foot is essentially a measure of wood, figured on the called size, not the actual finished size.  Therefore a 2" x 4" x 8" contains 5.33 bd. ft.  The theory is that you pay for what the mill started with before the wood is dried and planed.

    Thickness" x Width" x Length in feet, divided by 12 = board feet. T" x W" x L" ÷ 12 = Board ft.

    Thickness" x Width" x Length in inches divided by 144 = bd. ft. (T" x W" x L") ÷ 144 = Board ft.

7.  Concrete:  Concrete is usually measure in cubic yards.  A cubic yard is 3' wide x 3' long x 3' thick.  3' x 3' x 3' = 27 cubic feet.  27 cubic feet = 1 cubic yard.

The formula is:  L' x W' x T' ÷ 27 = cubic yards.
                        -or-
                        L' x W' x T" ÷ 12 ÷27 = cubic yards (where T is in inches).
                        -or-
                        L" x W" x T" ÷ 1728 ÷ 27 = cubic yards (where L, W, T, are in inches).

One cubic yard covers 324 square feet at 1" thickness.

One cubic yard is equal to approximately 40 bags of 80 pound concrete mix (+/- .675 cu' per bag).

_____________________________________________________________________________________________

The table below shows how many (approx.) 80 pound bags of concrete mix it will take to fill these tubes used to construct concrete piers, etc.

Diameter of tube	Approx. number of 80 pd. bags	To fill this many feet
6"	1 bag	3 feet
8"	1 bag	2 feet
10"	1-1/2 bags	2 feet
12"	2-1/2 bags	2 feet
14"	1-1/2 bags	1 foot
30"	7-1/2 bags	1 foot
Formula: Diameter squared x .7854 x height.  For example 30" x 30" x .7854 = 706.86 ÷144 (to convert from cubic inches to cubic feet) = 4.91 cu. ft. ÷ .675 cu. ft. per bag = 7.27 bags (rounded to 7-1/2).

One 80 pound bag of concrete mix will cover an area approximately 2' x 4' x 1'.

Decimal and Millimeter equivalents of fractions of an inch.

Inches	Inches	Mm		Inches	Inches	Mm
1/64	.01563	.397		33/64	.51563	13.097
1/32	.03125	.794		17/32	.53125	13.494
3/64	.04688	1.191		35/64	.54688	13.890
1/16	.0625	1.587		9/16	.5625	14.287
5/64	.07813	1.984		37/64	.57813	14.684
3/32	.09375	2.381		19/32	.59375	15.081
7/64	.10938	2.778		39/64	.60938	15.478
1/8	.125	3.175		5/8	.625	15.875
9/64	.14063	3.572		41/64	.64063	16.272
5/32	.15625	3.969		21/32	.65625	16.669
11/64	.17188	4.366		43/64	.67188	17.065
3/16	.1875	4.762		11/16	.6875	17.462
13/64	.20313	5.159		45/64	.70313	17.859
7/32	.21875	5.556		23/32	.71875	18.256
15/64	.23438	5.953		47/64	.73438	18.653
1/4	.250	6.350		3/4	.750	19.050
17/64	.26563	6.747		49/64	.76563	19.447
9/32	.28125	7.144		25/32	.78125	19.844
19/64	.29688	7.541		51/64	.79688	20.240
5/16	.3125	7.937		13/16	.8125	20.637
21/64	.32813	8.334		53/64	.82813	21.034
11/32	.34375	8.731		27/32	.84375	21.431
23/64	.35938	9.128		55/64	.85938	21.828
3/8	.375	9.525		7/8	.875	22.225
25/64	.39063	9.922		57/64	.89063	22.622
13/32	.40625	10.319		29/32	.90625	23.019
27/64	.42188	10.716		59/64	.92188	23.415
7/16	.4375	11.113		15/16	.9375	23.812
29/64	.45313	11.509		61/64	.95313	24.209
15/32	.46875	11.906		31/32	.96875	24.606
31/64	.48438	12.303		63/64	.98138	25.003
1/2	.500	12.700		1	1.0000	25.400

Formula: 1 divided by the fraction = thousands of an inch.  For example: 1 ÷ 4 = .250

Weights, Measures, and Miscellaneous information.

Doubling the diameter of a pipe increases its capacity four times.

A gallon of water weights 8.336 pounds at 39 degrees Fahrenheit (point of greatest density).

A gallon of water contains 231 cubic inches.

One cubic foot equals 1,728 inches.

One cubic foot of water equals 7-1/2 gallons.

_______________________________________________________________________________________

Weight of some common building materials.

Approximate weights of dry wood by cubic foot (not board foot).
Ash	3-3/4		Red  Oak	3-3/4
Birch	3-3/4		White Oak	4
Cedar	3-1/4		White Pine	2-1/3
Hemlock	2		Yellow Pine	3
Hickory	4		Poplar	2-1/2
Maple	3-1/2		Spruce	2-1/3

Metals: Approximate weights per cubic foot, In pounds.
Brass	518		Copper	550
Cast Iron	450		Lead	710
Forged Steel	490		Tin	458
Aluminum	167

Concrete with Stone Pounds per cu. ft. approx.	150	Loose Earth Pounds per cu. ft. approx.	76	Dry Sand Pounds per cu. ft. approx.	100
Frame wall with  1/2" drywall ppsf.	12	Floor sys 1/2" plywood with 3/4" flr. ppsf.	6	Roof Joist 1/2" plywood ppsf.	3
Cell Joist 1/2" drywall ppsf.	7	Glass 1/8" ppsf.	1.63	Glass 1/8" IG ppsf	3.25

Metrics:

25.4 millimeter equals 1 inch.
2.54 centimeters equals 1 inch.
1 meter equals 39.37 inches.
1.6 kilometer equals 1 mile.
100 millimeter equals 1 centimeter
100 centimeter equals 1 meter
100 meter equals 1 kilometer

Approximate Conversions:

Inches	x	25=	Mm		Mm mm	x ÷	0.04= 25.4=	Inches Inches
Inches	x	2.5=	Cm		Cm Cm	x ÷	0.4= 2.54=	Inches Inches
Feet	x	30=	Cm		Meter	x	3.3=	Feet
Yards	x	0.9=	Meter		Meter	x	1.1=	Yards
Meter	x	1.6=	Kilometer		Kilometer Kilometer	x ÷	0.6= 1.6=	Miles Miles

Temperature
Celsius	Fahrenheit
-30	-22
-20	-4
-10	14
0	32
10	50
20	68
30	86
40	104
50	122

Temperature Formula:  °Fahrenheit - 32 x 0.56 = °Celsius

                                    °Celsius x 1.8 + 32 = °Fahrenheit

Liquid:
1 ounce = 29.57 milliliters
1 quart = 0.95 liter
1 gallon = 3.79 liters

Cord Wood:

 A cord of wood measures 4' x 4' x 8'.  Stacked cord contains 128 cubic feet by measure.  Actually is 80 cubic feet -the rest is air.
Weighs 3,000 pounds when dry, 4,000 pounds when wet. Seasoned wood is defined as having 25% moisture content.  Four percent of its energy is lost in the evaporation process as it burns.  Green wood is defined as having 80% moisture content.  Fifteen percent of its energy is lost in the evaporation process.

Fuel energy BTU comparison:

Heating Oil	138,000 BTU per gallon (number 2 fuel oil)
Natural Gas	100,000 BTU per therm
L P gas	93,000 BTU per gallon
Mixed Hardwood	8,600 BTU per pound based on seasoned wood weighing +/- 3,000 pd. per cord.  24,000,000 BTU per cord (+/-)

Electrical formulas:  1.  Amps = Watts / Volts
                                     2.  Watts = Amps x Volts
                                     3.  Volts = Watts / Amps

Insulation:  The R-value of materials measures its resistance to heat flow.  The higher the R value, the better the resistance to heat flow.
                 The U-value is used to calculate heat loss.
Formula to find U-value is:  1 / R - value = U - value.

R-values of some common building materials:

1" Fiberglass batt, Insulation	3.17		1" Expanded polystyrene	4.00		1" Extruded polystyrene	5.0
1" Polyisocyanurate, foil-faced	7.20		8" concrete block	1.11		1' Poured concrete	0.08
1" Soft wood lumber	1.25		2x4 kds	4.38		2x6 kds	6.88
1/2" plywood	0.63		3/4" plywood	0.94		1/2" gypsum board	0.45
Window with  single glass	0.91		Windows with IG glass 1/4" air sp.	1.69		Windows with 1/2" Low E IG	3.13
Wood solid core 1-3/4 door	3.03		Metal ins. door no glass	15.00		1/2" - 4" air space	1.00

                      

Temperature Formula:  Degrees Fahrenheit - 32 x 0.56 = degrees Celsius

                                    Degrees Celsius x 1.8 + 32 = Degrees Fahrenheit

Liquid:
1 ounce = 29.57 milliliters
1 quart = 0.95 liter
1 gallon = 3.79 liters

Drywall board installation: Approximate quantities needed for each item per 1,000' of wall to cover.
32 pcs. 4 x 8 panels or 25 pcs. 4 x 10 or 21 pcs. 4 x 12.
138 pounds of ready mix compound.
370 lineal feet of tape.
2,000 nails or 1,275 screws.
______________________________________________________________________________

How to determine the handing of an interior passage or closet door.
Face the door on the side of the door where in order to pen the door you must pull it toward you.  If the door knob is on the right it is a right hand operating door.  If the door knob is on the left it is a left hand operating door.
Exterior doors are identified in the same way, but it is also necessary to state that the door is an inswing (swings into the building) or an outswing (swings to the outside of the building).
The active door of a Double door will also need to be identified -usually this is done by viewing the door from the outside.  It is either a left active or right active panel.
It is also wise to check with the manufacturer or sales person before ordering to make sure that they do it the same way.
_______________________________________________________________________________
Miscellaneous Math:
1.  Doubling a fraction makes it 1/2 its size.
For example:  2 x 1/16" = 1/32" and 2 x 1/4" = 1/8".
To calculate the angles of multisided objects:

Divide 360 by the number of sides and ÷ by 2.
For example:  An octagon -divide 360 by 8 (the number of sides) = 45 (the center angle) ÷ by 2 = 22-1/2 (the other 2 angles).




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                                                                                              vipin sasidharan.