PLASTERING RATE PER SQ.FT

 

Cost Control Techniques in Construction

Construction is a major industry spreading rapidly with the increasing population of this world. History shows the rate of gradual increase in prices of Building materials, Skilled Labors, and  Machinery used in construction.

The concept of Supply and demand in Economics helps us to understand fluctuation and raises in the cost of Construction that cannot be  controlled as population is getting increase alarmingly.

What we can do to control the cost of construction is future cost control with present prices, so effectively we can control the cost  of construction adopting the following techniques.

1- Design of construction

Avoid over design and unnecessary use of materials and construction size ,it has been observed that mostly engineers do over designing using extra spaces and materials which causes high and unnecessary costs that is due to the lack of good and precise engineering knowledge.

2- Wastage of Labor and Machinery

It is has been observed that Managers and supervisors cannot manage scheduled usage of labor and machinery which causes low productivity and high unbearable cost burdens in construction.

Avoiding such problems Design feasible planning for labor and machinery usage and control every activity by motivation and routine supervision check and balance.

3- Appointment of skilled Team at Construction Sites

Skilled people do more productivity and precise work and also can control wastages very effectively so appoint people on a merit system.

 

4- Proper Estimation and Execution

Estimate materials and labor expenses properly and execute construction activities according to the limitations of Estimated budgets.

5-Preparation of  Detailed Drawing before Estimation.

Detailed drawing along with detailed specification should be prepared before estimation so that construction cost may be assessed accurately and also size, the number of construction elements may be controlled that will cause control of cost.

6-Maintaining Inventory very soon after Estimation.

As  we know market prices keep uplifting so purchase all material soon after the estimation and also do agreements with Labor contractors and give them time frames.

Follow times lines of project so that labor contractor can manage his teams accordingly avoiding labor and time failures.  

7-Avoid Changing in Design.

Do not change construction design when it is designed in the initial stages because when the design has changed the cost also gets the change.

  

 

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What is Organization in Management

 It is a Business group  where two or more persons work together to accomplish a target or a common purpose.

we can see many organizations around us working for some special goal or purpose as Colleges, Factories, Schools, Government Offices.

 The concept of People, Purpose, Process, and  POLCA is very important to understand the organization.

People who works in an organization

Purpose for which people work together

Process is Set of specific activities specially designed by managers to complete different tasks for the attainment of organizational goals.

In the concept of POLCA Organization Process stands on four supports of management which are as follows.

1- Planning

Planning is shaping the organizational goals and the means for achieving them in allowable Limitations.

2-Organizing

The organizing is deciding where decisions will be made, who will do what jobs, who will work for whom, and tasks in given timelines and standards.

3-Leading

Leading is inspiring and motivating workers to work hard to get organizational goals.

4-Controlling

Controlling is monitoring progress towards goal achievement and taking corrective action to fulfill deficiencies.  

In the modern era, Good managers adopt two styles of controlling.

A- assure themselves to perform these functions well.

B-  Managers are altering the way they execute these functions, thinking of themselves more like mentors, coaches, team leaders, or internal consultants. They work with anyone who can help them complete their goals rather than only following the chain of command.  

They ask others to contribute in making decisions and share information with others to maximize the progress of attaining the goals.

5-Assurance

Assurance is formulation and direction in  the quality assurance programs and policies, and also struggle hard to bring the most excellent in the organization.

 

How to Calculate shear force in Simply Supported Beam

 Shear Force is the algebraic sum of all vertical forces acting on either side of the point on the beam. Shear force is to shear off the beam along with the point where it is acting.

                                                            Shear Off Diagram

Here submitting an example of shear force acting on the simply supported beam.

Loading Diagram 

1- Calculate Reactions

we will use Minus (-) Sign for downward forces (Loads) and Positive (+) Signs for Upward Forces (Reactions or Supports)

Taking Moment at Reaction "A"

+RAx0 - (1200x60)-(3000x20)-(4000x47)+(RBx60)

-(72000)-(60000)-(188000)+RBx60

RB =  320000/60 =  5333.333 Lbs

We know that (RA+RB) = Total Load

RA+RB =  (3000+4000)+(1200x60)

RA+RB =  79000 Lbs_________Eq.1

Putting value of RB in Eq.1

RA+(5333.333) =  79000 Lbs_________Eq.1

RA = 73666.667  Lbs

Now Calculating Shear forces at different points

Shear force at point "a" due to reaction A = +73666.667 Lbs

Shear force at point "b" = 73666.667-(1200x20)-(3000) 

SF due to UDL              =73666.667-(24000)= +49666.667 Lbs

SF due to Point Load   =49666.667-(3000)= +46666.667 Lbs     

Shear force at point "c" = +46666.667-(1200x27)-(4000)

SF due to UDL               =+46666.667-(1200x27)=  +14266.667 Lbs

SF due to Point Load     =+14266.667-(4000)       = +10266.667 Lbs

Shear force at point "d" = +10266.667-(1200x13)+(5333.333)

SF due to UDL               = +10266.667-(1200x13) = - 5333.333

SF due to Reaction RB  =  -5333.333+5333.333  =0

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Function of Shear Wall

 This is a structural member of RCC structures to provide lateral stability against horizontal forces caused by Wind pressures and Earthquakes.

Shapes of Shear Wall.

Shear walls are designed in different shapes using Reinforced cement concrete materials which produce resistance against the horizontal forces that act to over-turn the structures.

Placement of Shear Walls.

Shear walls are placed at three locations

1- On the edges of the building.

2-Inner side of Building on various locations.

3-On the Center of High rise buildings.

The function of Shear Wall.

There are various useful functions of shear walls.

1-Shear Wall provides resistance against self-load of structure, movement loading, and resultant loading such as uplift loads acting on shear walls.

2-Shear Wall holds gravity lines of all areas of the building that can create extra stresses during movements and may damage nonstructural elements in the building.

3-Shear wall keeps stiffness of junctions (on column and beams)  within the design limit.

4-Shear wall provides protection in lateral displacement and sway of the building.

5-Shear wall provides safe structural movement under earthquake and wind pressures. 

The Calculation for Loading Function.

Let's have a calculation session on design of shear walls as shown in this building plan.

Suppose the each span is 20 ft and slab design load is 300 Pound/Sq.Ft.(Including weight of walls ).

Total Load of one floor = (80'x40'x300)=  960000 Pounds

Factored load                 =(960000x1.5) =1440000 Pounds

Now whole Plan will be divided 5 Parts as showing  (Area 1=800 Sft), (Area 2, =600 Sft + 600 Sft ),(Area 3,= 300 Sft+300 Sft).

Now load can be calculated for each shear force

Load on Area 1 Shear Wall =(800x1440000)/(40x80) =360000 Pounds

Load on Area 2 Shear wall = (600x1440000)/(40x80)=270000 Pounds

Load on Area 3 Shear Wall =(300x1440000)/(40x80) =135000 Pounds

 

Uniformly Distributed Load Calculation

Considering shear wall as a beam between two junctions of the slab and beams vertical height is  12 ft.

UDL  For Area 1, Shear Wall = 360000/12= 30,000 Pound/Ft

UDL  For Area 2, Shear Wall = 270000/12= 22,500  Pound/Ft

UDL  For Area 1, Shear Wall = 135000/12=  11,250 Pound/Ft

Moment of Area 1, Shear Wall=( w.L.L.x12) /8=

                                               = 30000x12x12x12/8=6,480,000 (Lbs-Inch)

Moment of Area 2, Shear Wall=( w.L.L.x12) /8=

                                               = 22500x12x12x12/8=4,860,000 (Lbs-Inch)

Moment of Area 3, Shear Wall=( w.L.L.x12) /8=

                                               = 11250x12x12x12/8=2,430,000 (Lbs-Inch)

Moment of Resistance = 250xbxd1^2   ................    EQ.1

we knows that M.r = Moment

Putting values into Eq.1 for Area 1 Shear Wall

 

Longer side in Inches =10x12= 120 Inch

Shorter side in Inches =8x12=  96  Inch.

(6,480,000/2) (Lbs-Inch) = 250x 120xd1^2

d1 = 7.34 say 7.5" Adding cover 1" on both faces = 9.5"

Repeating for Shorter Side

(6,480,000/2) (Lbs-Inch) = 250x 96 xd1^2

d1= 11.61 Say 12"  Adding cover of 1" at both faces = 14"

Maximum size of wall thickness is 14" so adopting this

Area of Steel = M/Fst.Lever Arm

6,480,000/2x20000x.75x12 =18 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(0.785x96)/18 =4.18 say 4" C/C (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2" Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force

                                          6x0.302x20000x0.75x12/(30000x12x.5x.5) =3.5" C/c

Shear Wall Area 2

Longer side in Inches =15x12= 180 Inch

 (4,860,000 (Lbs-Inch)= 250x 180xd1^2

d1 = 10.39 say 10.5" Adding cover 1" on both faces = 12.5" (Shear wall Thickness)

Area of Steel = M/Fst.Lever Arm

4,860,000 /20000x.75x10.5= 30.85 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(.785x180)/30.85 =4.5 "c/c  (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2 " Dia bar =Area of 12-Leg Stirrups x Fst x Lever Arm/Shear Force

                                      =  12x.302 x20000x0.75x10.5/(22500x12x.5) =4.25" C/c

 

 

Shear Wall Area 3

Longer side in Inches =8x12"= 96 Inch

 2,430,000 (Lbs-Inch)= 250x 96xd1^2

d1 = 10  Adding cover 1" on both faces = 12" (Shear Wall Thickness)

Area of Steel = M/Fst.Lever Arm

2,430,000 /20000x.75x10= 16.20 Sq.Inch

For Spacing Using 1 " Dia Bars = Area of Bar x width /Area of Steel

                            Spacing  =(.785x96)/16.2=4.5 "c/c  (Vertical Bars)

Now Calculating the stirrups spacing

Spacing of 1/2 " Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force

                                      =  6x.302 x20000x0.75x10/(11,250 x12x.5) =4" C/c

 

Why Curing of concrete is Important.

 

The curing process is keeping the concrete moist awaiting the safe  hydration of concrete is complete and strength is accomplished. Curing of concrete should start on soon after initial setting time of concrete and should be continued up to 28 days to attain maximum strength of concrete.

Good Curing of concrete results in the following strength.

Comparative Strength of Ordinary (Portable) Cement concrete at various ages
Curing Time	Compressive Strength
3-Days	40%
7-Days	65%
14-Days	90%
28-Days	99%
3-Months	115%
6-Months	120%
1-Year	130%

Why Curing is Important.

1-Curing is required to avoid dryness of concrete during setting and hardening of cement that is only possible with effective hydration of cement, Now how hydration of cement will be effective, it's only possible when mixing water will remain in concrete during setting time period of concrete which is 28-days.

If curing is not started after placing the concrete it will produce only 50% of the total designed compressive strength in concrete.

2-Curing ensures the safe processing for the attainment of strength in concrete structures and also helps in speedy construction practices.

For Example, Concrete Structure avails 90 % strength in 14-Days and mostly engineers remove formworks after 14-days because structures can afford to hold their self loads after 14-days. The self-load of structures remains around 50% of the total designed load so when concrete avails its 90 % strength it provides easy and safe decision to remove the formwork without any hesitation.

3-Curing can raise the strength of concrete up to 130% which is the guarantee of long life and durability of concrete in less budgeting, so it is easy to design the structures for 100 years with effective curing of concretes.

4- Curing helps to attain high and required strength in concrete which helps to avoid faulty constructions and also finally  provides a safe and reliable structures. 

Curing Methods

1- Shading.

In this method, concrete is covered to avoid evaporation from sun rays. 

2-Ponding.

This method is used only for horizontal surfaces of slabs and concrete roads but require a lot of water.

3-Spraying of Water.

This method involves the sprinkling of water on concrete surfaces to keep moist surfaces but this is difficult as requires periodic showering of water.

4-Membrane Curing.

In this method of curing a chemical coating /Using some membrane is provided on the surface of the concrete to avoid evaporation of concrete water.

5-Steam Curing/ Hot water Curing.

Curing with Steam and hot water is sometimes adopted to gain rapid development in the strength of concrete. This process of curing is best for the construction of Pre-Cast Structures.

6-Covering.

This is a broadly used technique of curing, mainly for structural concrete. The uncovered surface of the concrete is prohibited from drying out by casing it with a hessian cloth or canvas cloth.

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How to design a simply supported beam manually

 Suppose we have a beam whose span is 30 Ft , and a Point Load "F"=10000 Pounds, which includes weight of beam and imposed load.

In this case we are going to design a RCC Beam using Load Factor Theory in following steps.

1-Find out Support or Reactions

Left and Right Support = F/2= 5000  Pounds

In given symmetrical loading case both reactions are equal

2-Find out Max shear force = Load /2 = 10000/2 =5000 Pounds

Shear force at support = +5000 Pounds

Shear force at Mid      = +5000 + (-10000) =   -5000 Pounds

3-Max Bending Moment at Center of  Beam =    Reaction x L/2

Max MB=            5000x (30/2)x12= 900000 Pound-Inch--------1

4-Moment of Resistance =  K.b.d1^2----------------EQ.2

5-b (width of beam ) = (Span x12)/30 =12"

Value of  "K" = 250  in Load Factor Method using (1000-20000) Concrete

Where fc =1000 PSI  and Fs = 20000 PSI

We know that Moment of Resistance =Bending Moment  so comparing both equations EQ----1 AND EQ-----2

6-       900000 = 250x12xd1^2

d1 (Effective Depth) = 17.33 inch Say 17.5"

7-   Area of Steel in Tension zone = BM /(FstxLever Arm)

                        Ast  =900000/(20000x0.75/17.5)= 3.42 Sq.in

 No of bars according to 3/4" dia bars =  Ast/area of one bar =8 nos.

 

8-Area of  steel in compression zone =(B.M - M.r)/fst.Lever Arm

               =((900000-(250x12x17.5^2))/20000x 0.75x17.5

               =(900000-918750)/20000x0.75x17.5= - (0.07)

Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

Use  3/8" dia bar stirrups 2-Leg @ 10 " c/c

Minus sign shows that the M.r value is greater than the applied value of B.m so no need of Asc (Area of steel in compression zone ), Only two steel bars of 1/2" dia bars are enough to hold stirrups.

 

Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

9- Spacing of stirrups = Aw.Fw.Lever arm/(Shear Force)

                                    =(0.22x18000x0.75x17.5) /5000 = 10" C/c

Here Aw = Area of 2 leg stirrup bars

      Fw    =   Permissible value of Tensile stress of stirrups

Overall depth of beam = Effective depth+ 2 concrete covers

concrete cover = 1"

Concrete ratio =1:2:4

Factor of safety in Fc = 3000/1000=3 Times (considering compressive strength of concrete)

 The factor of safety in Fst= 60000/20000=3 Times (considering Tensile strength of Steel).

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