## Unit Conversion

In Civil Engineering it may be on Construction Site while Executing any project or doing calculation in studies or in structural engineering while designing a structures or it may be in any field, it becomes problem when you are not familiar with the basics units. So In this short blog you will be able to convert any units to other unit without memorizing each one. If you know about the most basics unit then you can convert and manipulate bigger units and calculation by just using these concepts.

One thing to remember is if you convert a bigger unit to Smaller one the you have to Multiply with equivalent value and from smaller to bigger one then divide the equivalent value. The following example will clear your concepts.

### (1) Linear or One Dimensional (1D) Unit

###### Length, Distance, Displacement

1st you should know about one linear or dimensional units and these are the basics to convert to any others units if 2D (Area) & 3D(Volume) e.g foot, meters and inches ect if you don’t know about these basics then you can learn it by using a measuring tape (inchi tap). These are basics then you can convert bigger linear units e.g Km, miles etc. You can see here the feets, inches and soot (soother) on the upper part and meters, cm and mm on the lower part horizontally.

### (2) Two Dimensional(2D) Units

###### Area, surface area, cross-sectional area

So Area is a 2 Dimensional quantity and you can find these units by just squaring of the one dimensional or linear units e.g taking square of m you will got m2 and the same way all other units. It can convert from MKS (meter, kilogram,second) to FPS (Foot,pound Second).

(a)Example: Convert=>5m2 to ft3 => 20m2 = 20 (1*3.28)2 =20*10.76 ft2= 215.2ft2(here using only linear units and converted from bigger to smaller that’s why multiplied)

(b)Example: Convert=>50ft2 => m2 = 50* (1/3.28)2 =50*0.0929 ft2= 4.645m2 (here using only linear units and converted from smaller to bigger units that’s why divide)

### (3) Three Dimensional (3D)

###### Volume

Volume is a Three Dimensional Unit and you received that unit by taking cube of 1d units e.g if you take cube of ft then it will become ft3 which is 3d unit. It can be converted any 3d units of MKS(meter, kilogram, second) to FPS(foot,pound, second) with the help of linear units without memorize the exact relation between them.

(b)Example: Convert=> 5m3 to ft3 => 5m3 = 5 (1*3.28)3 =5*35.3 ft3=176.4ft3 (here using only linear units and converted from bigger to smaller that’s why multiplied )

(b)Example: Convert=>40ft3 => m3 = 40* (1/3.28)2 =40*0.028 ft3= 1.12m2 (here using only linear units and converted from smaller to bigger units that’s why divide)

Force,weight and load are same terms used in Civil Engineering and general units used for that is Newton(N), Kg, Pound (lb), KN, Kip(kilo-pound).

If you know about the most basics one e.g 1 Kg=10 N, 1 kg= 2.205 lb then you can convert any units with the help of these tow.

Examples: Convert=>

• 10 kg=> N=> 10 kg= 10 *10 N = 100 N (multiplied; Bigger=>Smaller)
• 200 lb=>KN=> suppose here we don’t know about the lb and KN relation then 1st we will convert lb to know unit that we know. So taking help from “1 kg = 10 N” & “1 kg= 2.205 lb” , 1 lb= (1/2.205) kg= 0.4535 kg , now convert the 0.4535kg into N ; 0.4535*10 N= 4.535N Now convert it to KN(divide by 1000); 0.004535 KN. So 1 lb=0.004535 KN. Now 200 lb= 200 *0.004535 KN = 0.91 KN.

The same way you can convert things without any single relations known.

### (5) Stresses (Force/Area) and Densities (Force/Volume) and Moment/Torque (Force *Distance0

So if you have learnt about how to convert forces and 1d(length), 2D(area), 3d(Volume) units then you can easily convert the units of stresses, Densities and Moments/ torques etc.

So we will solve one example for stresses

Examples: Convert=> 1Mpa to Psi

Mpa= N/mm2 and Psi= lb/in2 ; 1 N/mm2 ; So suppose here I don’t know about lb and N relation then taking help of know relations e.g 1 kg = 10 N=> 1 N= 0.1 Kg now replace the kg by pound (lb) because we know that relation; 1 N= 0.1 ( 2.205 lb)=> 1N= 0.2205 lb and the same way convert mm2 to in2 => suppose we don’t know the relations between mm to in then; 1mm2= (1/1000)2 = 1o-6 m2; converted into meters because we know the relation between meter to ft; 10-6 (3.28)2 ft2 = 0.00001076 ft2 now convert that feets to inches; 0.00001076*(12)2 in2= 0.001549 in2 .

Now put the values of 1 N= 0.2205 lb and 1mm2= 0.001549 ; 1 N/mm2(MPa)= 0.2205 lb/0.001549 in2 (Psi)=> 1 N/mm2=143 Psi but exactly it is equal to 145 Psi as I put values approximately.

so from the above example you may have cleared that only know the basics concepts you can derive any units.

Watch the Video given below I have discussed it briefly;

==========Thank You==========

## Soil Mechanics

“Soil Mechanics is the application of laws of mechanics to the Engineering Problems deals with soils e.g sediments or other unconsolidated accumulation of soil particles produce by the mechanical & chemical disintegration of rocks”

Since Soil is generally a three phase material consists of Soil Solids, Water & Air. So that’s why it exhibits different properties and behave differently under same load conditions. So the following is some of the basics properties and parameters that involves while studying Soil Mechanics and Geotechnical Engineering.

## General Parameters

1. Va=Volume of Air
2. V=Total Volume of soil mass
3. Vw=Volume of Water
4. Vs=Volume of Solids
5. Vv=Volume of voids
6. W=Weight of Soil Mass
7. Ww=Weight of Water content
8. Ws=Weight of Solids
9. Weight of Soil Mass=W=Ws+Ww,(Air weight neglected)
10. Volume of Voids=Vv=Va+Vw
11. Wsat=Weight of fully Saturated Soil
12. (Ws)sub=Submerged Weight or Buoyant weight of soil below water surface or under ground water table.

## Index Properties of Soil

1. Water Content(W)=Ww / Ws
2. Void Ratio(e)=Vv / Vs
3. Porosity (n )=Vv / V
4. Degree of Saturation(Sr)=Vw / Vv
5. Air Content(ac)=Va / Vv
6. Percentage of Air Void( a)=Va / V

## Densities

Definition: It is the Weight of any material per unit Volume of that material and Units of measurement is SI is Kg/m3 or g/cm3 .

1. Bulk Density or Bulk Unit Weight of Soil mass (r)=W / V
2. Dry Density or Dry Unit Weight of Soil(rd)=Ws / V
3. Desity of Soilds (rs)=Ws / Vs
4. Saturated Density of Soil Mass(rsat)=Wsat / V
5. Submerged Density of Soil Mass(rsub)=(Ws)sub / V

## Specific Gravities

Definition: It is the ratio of density of any material to the density of water and since it is the ratio, so this is the unitless quantity .

1. Specific gravity of Soil Solids(G)=rs / rw
2. Bulk or mass Specific gravity of Soil Solids(Gm)=r / rw

Watch the Video, these parameters have been discussed briefly.

========Thank Y0u=======

## PERT Vs CPM

### PERT

Program Evaluation & Review Technique (PERT) is a method used in Project Management to identify all the activities involved in any project and to find the unpredictable activities that upon the project is dependent and probabilistically define the time scale by the project to finish without any delayed not concern about the project cost.

### CPM

Critical Path Method (CPM) is also a Project Management technique to identify all the activities involved while preparing a schedule, in any project and based upon this method critical activities can be identify upon which whole project is dependent. The time scale and cost both has to be considered, so that project can finish with expected date and with the allocated budget.

### Basics Differences

This is the most basics differences between PERT & CPM. Watch the Video below you will make understand about how to find the Critical Path in any Project and about Early Start (ES), Early Finish (EF), Late Start(LS) & Late Finish (LF) terminologies with Example.

======Thank You======

## Weirs

Definition:“Its a low height Dam constructed across the river to increase the water level at upper stream and to control the discharge or flow of water into the ponds, irrigation land, reservoirs or to flood control, is called a weir”

## Barrages

Definition:“It is a type of low-head Dam over a river which stored water behind it at upper-stream to increase water level and consists of number of gates that regulate water to supply to irrigation land, for power generation (mechanical power to electrical), to divert water, to control flood, to maintain discharge, is called barrages.”

##### Basic Difference between Weirs & barrage
• Barrage is comparatively a big structure compared to weir.
• Barrage has regulators in the form of sluice gates which control flow of water while weirs doesn’t have any gates.
• Barrages control discharge and increase level of water by means of sluice gates (regulators) while Weirs control water flow and increase level only by its fixed super-structures (normally Concretes + Stone Masonry) upto its fixed height.
• Barrages can produce power generation by its tidal energy while weirs normally not do that.

## Irrigation Engineering and Techniques

### Irrigation Engineering

Definition:“To Irrigate Land by means of artificial techniques when rain precipitation doesn’t Satisfy its requirements is called Irrigation Engineering”

##### Techniques Used in Irrigation Engineering

The following Techniques used in Irrigation Engineering while irrigating Land:

##### (1) Surface Irrigation

Definition: “To provide water directly to soil surface from a channel located at the upper reach of the field, is called Flow Irrigation or Surface Irrigation”

Surface Irrigation further Divided into Seven (7) Types:

• Free Flooding Irrigation
• Check Flooding Irrigation
• Check Basin Irrigation
• Border Method of Irrigation
• Furrow Irrigation
• Sub-Surface Irrigation
• Sprinkler Irrigation
##### (2) Lift Irrigation

Definition: “To Lift water from under-ground surface or sub-surface by any means and provide it to irrigate land is called Liff Irrigation.

Lift Irrigation is Uses Two (2) Sources to get Water from Sub-Surface:

(1) Deep Well

(2)Open Well

• Chorus Or Mote
• Persian Wheel (Rahat)
• Door
• Archimedian Screw
• Dhenkli or Lot

Thank You

## Balanced Section, Under Reinforced Section & Over Reinforced Section

###### Parameters Used Here are:
• Ast= Area of Steel in Tension Zone
• fst or σst= Stress in Steel
• fst.u or σst.u= Ultimate Stress in Steel
• fc or σc = Stress in Concrete
• fy = Yield strength of steel
• fck = Characteristics strength of concrete
• ϵc= Strain in concrete
• ϵst = Strain in Steel
• Nc or nor Xu = Critical Neutral Axis
• N or n= Actual Neutral Axis
• D= Total Depth of Beam
• d= Effective depth of beam (from centroid of steel in tension zone to topmost fiber of concrete in compression zone).
• Note: The value of stress and strain shown in the definition below for each sections of concrete when the section is Design by limit state method (LSM), because it beyond the elast region of stress strain curve.

### Balanced Section

Definition: “The RCC Section which is reinforced with such amount of steel that when extreme fiber of compression zone of concrete reaches to its permissible allowable stress and strain value c =0.0035 and fc=0.45fck), the steel provided in tension zone also reaches to its permissible yield allowable strain and stress value st= (0.002+(0.87fy/Es)) and fs= 0.87fy) at same time then that section will be a Balanced or Critical Section.”

• The failure of such section may be due to compression or tension. So balanced section is basically the combination of both brittle and ductile section.
• The Neutral Axis of such sections lies in the middle of the section (n=nc) is called Critical Neutral Axis, normally denoted by nc or Xubal.
• The Moment of resistance (Mr) of such sections can be determined by Multiplying the lever arm of the stressed section either by Compression force or Tension force because the centroidal of both the areas (Compression and Tension zone) are almost at same distance from neutral axis.

(Moment of Resistance) Mr= Compressive Force*(d-0.42d)

or Mr= Tensile Force*(d-0.42d)

### Over-Reinforced Section

Definition: “The RCC Section which is reinforced with such amount of steel that when extreme fiber of concrete in compression zone reaches to its permissible allowable Strain and stress valuec =0.0035 and fc=0.45fck), while the allowable stresses in steel provided in tension zone doesn’t reaches to its permissible allowable yield strain and stressst= (0.002+(0.87fy/Es)) and fs= 0.87fy) then that section will be an Over-Reinforced Section.”

• It means that the percentage of reinforcement provided more than the requirements.
• That RCC element or section will fail due to brittleness which is dangerous to any section.
• The section is uneconomical due to high percentage of reinforcement provided.
• In such sections the actual Neutral Axis (NA) will move downward below Critical Neutral Axis (Nc) or n>nc.
• The Moment of resistance(M r) of such sections will be always more than the Balance section.

(Moment of resistance) Mr=b.n.(fst/2) *(d-n/3)

### Under-Reinforced Section

Definition: “The RCC Section which is reinforced with such amount of steel that when extreme fiber of concrete in compression zone doesn’t reach to its permissible allowable strain and stress valuec =0.0035 and fc=0.45fck), while the allowable stresses in steel provided in tension zone doesn’t reaches to its permissible allowable yield strain and stressst= (0.002+(0.87fy/Es)) and fs= 0.87fy) then that section will be an Under-Reinforced Section.”

This is the most desirable section(Under-Reinforced) because:

• It means that the percentage of reinforcement provided less than the requirements, so saving steel cost.
• That RCC element or section will fail in Ductile behavior which is the most desirable condition for any structure.
• Show enough warning before failure because first steel has to reach its yield stress before concrete.
• The Moment of resistance of such sections will be always less than the Balance section.
• In such sections the actual Neutral Axis (NA) will move upward above Critical Neutral Axis (Nc) or n<nc.
• Moment of resistance (Mr) of such sections can be determined by considering the stress of steel.

(Moment of Resistance) Mr=fst.Ast(d-n/3)

Note: Now You can evaluate Neutral Axis by the sections given below:

Watch the video… if still any any Query you can comment…Thank You

## Nominal Vs Mix Design of Concrete

### (1) Nominal Mix of Concrete

You may have noticed that normally on construction site, the mix ratio of concrete has been already mention in construction drawings. So you have to follow that ratio of concrete. That concrete will give you a concrete strength of ordinary strength normally from M5(1:5:10) to M25(1:1:2). The magnitude mention with the letter M mean the Compressive strength of concrete which in in N/mm2 or Megapascal(Mpa). You can see some Nominal Mix ratios in the table given below.

So this is about the nominal mix when you have to follow the already fix ratios regardless of the material specific properties. This is normally used for residential or other commercial structures where there is no need of high strength concrete.

### (2) Design Mix Ratio for Concrete

So the Design mix of concrete is done for high strength of concrete. When…

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## Nominal Vs Mix Design of Concrete

### (1) Nominal Mix of Concrete

You may have noticed that normally on construction site, the mix ratio of concrete has been already mention in construction drawings. So you have to follow that ratio of concrete. That concrete will give you a concrete strength of ordinary strength normally from M5(1:5:10) to M25(1:1:2). The magnitude mention with the letter M mean the Compressive strength of concrete which in in N/mm2 or Megapascal(Mpa). You can see some Nominal Mix ratios in the table given below.

So this is about the nominal mix when you have to follow the already fix ratios regardless of the material specific properties. This is normally used for residential or other commercial structures where there is no need of high strength concrete.

### (2) Design Mix Ratio for Concrete

So the Design mix of concrete is done for high strength of concrete. When the strength requirements beyond M25 (25 Mpa) strength of concrete and when the structures strength is the primary priority then you have to go for Mix Design of concrete.

In Mix Design of Concrete you have to check every material (Aggregates, Water, Admixtures etc) properties that you have to use in your concrete mix and upon some trial you have got the desired strength with some mix ratios. So that Mix proportion of concrete ingredients should be noted with the respective strength. For this you have to bring all the samples of your materials (Aggregates, fine Aggregates, water etc) ingredients to the Mix Ratio Laboratory of your specific area from where you considered to use the concrete ingredients Source. If you want to change the source of the concrete ingredients during construction, then again you to go for mix design. Then that will be your Design Mix ratio for concrete that you have to follow for that specific structures and specific projects. Some Examples of Mix Design ratios or given in the table below but you cannot follow it directly on your site and you to do your own MIX Design for the Concrete ingredients that you have to use in your project. So the values given in table for mix design is M30 to M70.

Thank You if you have still any doubts, you can ask in comment.

## Type of Supports in Real Structures

After reading this blog, you will be able to identify and can execute different types of support in real structure in construction site.

During Engineering you have shown just the idealized form of supports but when you go to construction site you cannot identify and execute different type support in real structures. So in this article your confusion will be vanished. Here we have discussed the most important three (3) types of supports normally used and cause confusion.

### (1) Fixed Support

Fixed support mean that it can resist and restrain all the possible movement (vertical, horizontal & bending moment) through the joint/support and the stresses generate in beam due to any type of loading can transfer from beam to column safely. Now we discuss fixed support in Reinforced Cement Concrete Structures (RCC) and Steel Structures:

##### (a) Fixed Support in RCC Structures

In RCC structure first column concrete is done upto beam bottom level and then after fixing the reinforcements of beam and slab with the column projected rebars then concrete it monolithically so that the member can act as a fixed unit. To identify that the support is fixed or hinged/pin in RCC beam-column joint or the column jointed with foundation, you must check the rebar detail in drawing and on site that if proper development length(Ld) is provided then and the desired ratio of concrete is followed at the site then the support will be consider is fixed. Because the reinforcement at the support will tell you about whether the support is fixed or hinged/pin. The proper development length (Ld) and required concrete grade will achieve sufficient bond/anchorage at that at that joint to resist all the stresses. This type of support is mainly designed to resist bending moment along with the other stresses. You can see the fixed support in the figure below:

##### (b) Fixed Support in Steel Structures

The phenomena of fixed support that it will resist the possible movements will be same as in the RCC structures. The Steel Beam-Column connections or column-footing connections will be considered as fixed if the bolts or welds around the joint (around the flange and web) are fixed/applied throughout along with gusset plates at that joint/support. If the bolts or welds are fixed/applied only at the web section then it will consider as a hinged/pinned support because it will not resist all the movements (vertical, horizontal or bending moment) except the few one. This type of support is mainly designed to resist bending moment along with the other stresses. Through this type of bolting/welding you provided sufficient anchorage to resist all the stresses. The steel has to resist all the tension not concrete. You can see the Fixed Support in Steel Structure in figure/image below:

##### (2) Hinge/Pin Support

In this type of support only the horizontal and vertical movements are restrained but cannot resist bending moment. This type of support are not designed for taking any bending moments but just take the axial stresses (axial compression and axial tension). That’s why this type of support is mainly provided only in trusses. But if the bending is not the failure criteria of any member or if the bending moment is not generate sufficiently in any support that it can fail then we provide it in RCC and Steel Frames also:

##### (a) Hinge/Pin Support in RCC-Bricks Structures

You may have seen that in residential buildings or low storey buildings in which the load bearing component is brick walls then the beams and columns provided there by the unefficient contractor at some areas or in the whole structures is which may be not needed at all and also the structure is not properly designed by the structural engineer. If you look upon the reinforcement at beam-column joints or column-footing joints then you will find that they are not provided any development length or may be not sufficient to developed the required bond/anchorage at the joint. Because the tension is resisted by the steel not concrete. The concrete is also of very low Grade, so that support will not be considered as fixed but it will be a Hinge/Pin Support. But the loads are mainly taking by brick masonry walls that’s why the structure is stable. But you cannot allow this type of supports in RCC frame structures where the load bearing element is beam-column. You can see in given figure:

##### (b) Hinge/Pin Support Steel Structures

As we have discussed earlier that This type of support are not designed for taking any bending moments but just take the axial stresses (axial compression and axial tension). That’s why this type of support is mainly provided and designed for trusses. But it can be provide in Steel frames Beam-Column and Column-Footing joints when the bending is not the failure criteria. In this type of beam-column joints or column-footing joints the bolts/welds provided only at the web section or flange section of that element which is sufficient to resist the axial stresses. You can see in given figure below:

##### (2) Roller Support

Roller support can resist only vertical forces. So it is provided only when there is only axially compressive forces are there. You have seen the simply supported beams/girders may be of Steel or RC. The idealized form of simply supported beam is one side is hinge and the other is roller (see image). We consider it because at the hinge/pin side of beam resist horizontal forces because lateral movement are not generate at the element just because of hinge/pin and the other side is roller because due to temperature changes and heavy loads there may be little lateral bit forces occur which causes expansion and bending. If we consider both side is hinged then the temperature  stresses in element will cause beam to crack and if we consider both side is roller than little component of lateral force can cause beam/girder to move horizontally which will be unstable. So when the beam which may be RCC or Steel which is lies just on both side support may be brick wall or RCC column but not casted monolithically then that will be an example of simply supported in which one side is roller and the other is hinge/pin support. You can see the roller support both in Steel and RCC structure given below:

Watch the short video given below under these discussion Topics: