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Thursday, 22 February 2018

Basic Important Question for fresher Interview

                                                 1 M Length steel Rod and its Volume
V=(Pi/4)*Dia x DiaX L=(3.14/4)x D x D X 1 (for1m length) Density of Steel=7850 kg/ cub meter
¨Weight =
¨Volume x Density=
¨(3.14/4)x D x D X1x7850 (if D is in mm ) So = ((3.14/4)x D x D X1x7850)/(1000x1000) = Dodd/162.27

                                                DESIGN MIX:M10
M10 ( 1 : 3.92 : 5.62)
¨ Cement : 210 Kg/ M 3
¨ 20 mm Jelly : 708 Kg/ M 3
¨ 12.5 mm Jelly : 472 Kg/ M 3
¨ River sand : 823 Kg/ M 3
¨ Total water : 185 Kg/ M 3
¨ Fresh concrete density: 2398 Kg/M 3
  

Isometric Rule in AutoCAD

How to Start Isometric-Go to Snap setting-Rectangular Snap-Click on Isometric Snap
-Mirror Command Does not used in same plane
-Circle Command never used
#For Circle-Go to Ellipse-axis End-Right click in Mouse-IsoCircle
-Offset Command Does not Used in Same Plan
-for Changing The plane used F5 Button.

-Ortho Always ON

Wednesday, 14 February 2018

How to convert .DWG To PDF and How to Add Material in AutoCAD 3D

For .DWG to PDF
Method-
Step 1- Ctrl+P-Then select pdf option-Paper size
Step 2-After it What to Plot-Window-Select area for Print-ok
For Material
Step 1-Mat+Enter

Step 2-For advance Material-Create New Material –Image-Trace Material

Tuesday, 13 February 2018

Some Important Tutorial Link

v AutoCAD In English 
·         AutoCAD 1st class

v Revit Architecture In Hindi

v Revit Architecture In Hindi

v Staad.Pro Tutorial In English

How to place Camera and Light in AutoCAD 3D

Method-
For Camera
Step 1-Go to visualize-Create Camera-Place Camera
Step 2-Place Camera With The help of Move command
For Light
Step 1-Go to Visualize-Light
Step 2-For Increasing intensity of Light-Select light-Right click in mouse

Step 3-Go to Properties –Intensity-Increase intensity

Monday, 12 February 2018

Basic Civil Engineering Tips you should be remembered while working on a construction site.

1. GRADE OF CONCRETE:

M5 – 1 : 4 : 8

M10 – 1 : 3 : 6

M15 – 1 : 2 : 4

M20 – 1 : 1.5 : 3

M25 – 1 : 1 : 2

2. CLEAR COVER TO MAIN REINFORCEMENT:

Footings : 50 mm

Raft Foundation (Top) : 50 mm

Raft Foundation (Bottom) : 75 mm

Raft Foundation (Side) : 75 mm

Beam : 25 mm

Strap Beam : 50 mm

Column :40 mm

Slab : 15 mm

flat Slab : 20 mm

Staircase : 15 mm

Retaining Wall : 20 – 25 mm


Some Important note for every Civil Engineer

Minimum thickness of slab is 125 mm.

Water absorption should not be more than 15 %.

Dimension tolerance for cubes + – 2 mm.

Lapping is not allowed for the bars having diameters more than 36 mm.

Chair spacing maximum spacing is 1.00 m (or) 1 No per 1m2.

Revit Lecture 5 (How to Make Roof)

For Roof by Extrusion
Method
Step 1-Command-Draw Roof

Step 2-Select Roof-Off your Define Slope-Finish Model

Sunday, 11 February 2018

Revit Lecture 3(How to Draw Floor,Roof,Trim,Offset)

For Floor
Method-
Step 1-Floor-Draw floor-Finish Mode

For Roof
Method-

Step 1-Draw roof- Select  Roof line- Close Define Slope 

How to Use Split, Paint and show shape Handel



For Split
Method-
Step 1-Go to Modify-Click on Split command.
Step 2-Click on Wall where you want split Section
For Paint
Step 1-Go to Modify-Click on Paint command.
Step 2-Click on Material –choose Material – Click On wall
For Show Shape Handel
Method-
Step 1-Create your wall with Layer-Select your wall
Step 2-Click on Create Part-select wall-Go to Properties Box –click on show shape Handel

Step 3- Trace material according to Your Requirements 

Revit Architecture Lecture 2 (Modification of Wall,Door,Window)

                                        How to Edit Wall
Method-
Step 1-Select wall-click on edit Profile
Step 2-Draw to Editing Shape according to your model
Step 3- if you want to reset your wall-click on wall-Reset Profile

                                        How to Set Door & Window in Revit
Method-
Step 1-Door command-click on edit type-Dublicate-Edit width and Height
Step 2-Then click on wall-set your Door
Step 3-For Placement of window Same Process.

Introduction About UniqueCivil And Introduction of Revit(Level,Wall,UNIT,Elevation)

                                                                Introdcution Of Revit
How to Start Work On Revit
Method-
Step 1-Set your unit
Step 2-Go to project Browers-2 times Click on any elevation-set your levels
Step 3- Click on Ground Floor-click on wall-edit type(Properties Box)-dublicate-Edit profile
           -wall Width-ok
Step 4-Draw your Wall

Wednesday, 7 February 2018

Top Basic Civil Engineering 700 MCQ

Concrete Technology

1. A higher modular ratio shows
a) higher compressive strength of con-crete
b) lower compressive strength of concrete
c) higher tensile strength of steel
d) lower tensile strength of steel
Ans:b

2. The average permissible stress in bond for plain bars in tension is
a) increased by 10% for bars in compression
b) increased by 25% for bars in compression
c) decreased by 10% for bars in compression
d) decreased by 25% for bars in com-pression
Ans:b

3. In working stress design, permissible bond stress in the case of deformed bars is more
than that in plain bars by
a) 10%
b) 20%
c) 30%
d) 40%
Ans: d

4. The main reason for providing number of reinforcing bars at a support in a simply
supported beam is to resist in that zone
a) compressive stress
b) shear stress
c) bond stress
d) tensile stress
Ans: c

5. Half of the main steel in a simply supported slab is bent up near the support at a
distance of x from the centre of slab bearing where x is equal to
a) l/3
b) l/5
c) l/7
d) l/10
where l is the span
Ans:c

6. When shear stress exceeds the permissible limit in a slab, then it is reduced by
a) increasing the depth
b) providing shear reinforcement
c) using high strength steel
www.EngineeringBooksPdf.com
d) using thinner bars but more in number
Ans: a

7. If the size of panel in a flat slab is 6m x 6m, then as per Indian Standard Code, the widths
of column strip and middle strip are
a) 3.0 m and 1.5 m
b) 1.5 m and 3.0 m
c) 3.0 m and 3.0 m
d) 1.5 m and 1.5 m
Ans:c

DEAD LOADS and LIVE LOADS

Dead loads are due to the weight of all materials that constitute a structural member. This also includes the weight of fixed equipment that are built into the structure, such as piping, ducts, air conditioning, and heating equipment. The specific or unit weights of materials are available from different sources. Dead loads are, however, expressed in terms of uniform loads on a unit area (e.g., pounds per square foot). The weights are converted to dead loads taking into account the tributary area of a member. For example, a beam section weighting 4.5 lb/ft. when spaced 16 in. (1.33 ft.) on center will have a uniform dead load of 4.5/1.33 = 3.38 psf. If the same beam section is spaced 18 in. (1.5 ft.) on center, the uniform dead load will be 4.5/1.5 = 3.5 psf. The spacing of a beam section may not be known to begin with, as this might be an objective of the design. Moreover, the estimation of dead load of a member requires knowledge as to what items and materials constitute that member. For example, a wood roof comprises roof covering, sheathing, framing, insulation, and ceiling. It is expeditious to assume a reasonable dead load for a structural member, only to be revised when found grossly out of order. The dead load of a building of light frame construction is about 10 lb/ft.2 for a flooring or roofing system without plastered ceilings and 20 lb/ft.2 with plastered ceilings. For a concrete flooring system, each 1 in. thick slab has a uniform load of about 12 psf; this is 36 psf for a 3 in. slab. To this, at least 10 psf should be added for the supporting system. Dead loads are gravity forces that act vertically downward. On a sloped roof, the dead load acts over the entire inclined length of the member.

Tuesday, 6 February 2018

SAND, SILT and CLAY

Any sample of soil is made up of a continuous array of particles
which may range from the very smallest clay particles to large
gravel. In order to describe soils scientist have established a classification
system for ranges of particle sizes; size ranges which reflect their role in many of the soil properties we are familiar with. This classification system divides the particles into four classes - gravel and stone, sand, silt,and clay. Table 1 lists the Canadian classification system for the size range of particles which may be found in a soil
sample.


CLAY
The scientists have assigned the name clay to the finest particles and
not without reason. Clay size particles are the source of most of the
chemical properties of soil.They are responsible for the retention of
many of the plant nutrients in the soil, such as calcium, magnesium,
potassium, trace elements and some of the phosphorus. Clays react with
the breakdown products of organicmatter to stabilize the humus in the
soil.A soil without clay particles can be a very infertile soil.
Clays, because of their very small size and very large surface area, are
able to retain greater amounts of water than sandy soils.On the other

Monday, 5 February 2018

Manual Design of Slab

Example 1: Design the roof slab, beam and column of house given in figure 1.
Concrete compressive strength (fc′) = 3 ksi.
Steel yield strength (fy) = 40 ksi.
Load on slab:
4″ thick mud.
2″ thick brick tile.
Live Load = 40 psf

Concrete Formwork

·        CONCRETE CONSTRUCTION
A quality reinforced concrete structure offers many advantages
over structures made with other building materials. Concrete is
a durable material that reduces building maintenance costs and
provides a longer service life. A concrete structure will reduce energy
usage because of its mass and high resistance to thermal
interchange. The use of concrete will lower insurance costs by
virtue of its high resistance to fire. Buildings made of concrete are
also more secure against theft and vandalism. Concrete floors and
walls reduce the transfer of noise, yielding a quieter environment
and happier occupants. Reinforced concrete possesses considerable
strength for resisting seismic and wind loads. These factors
and others make the selection of reinforced concrete an economical
alternative.
·        CONCRETE FORMWORK
The construction of a concrete building requires formwork to support
the slabs (horizontal formwork) as well as columns and walls
(vertical formwork). The terms concrete formwork and concrete
form carry the same meaning and are used interchangeably in this
book. Formwork is defined as a temporary structure whose purpose
is to provide support and containment for fresh concrete until
it can support itself. It molds the concrete to the desired shape

Sunday, 4 February 2018

Construction Mathematics(HOW TO USE SCIENTIFIC CALCULATOR)

Click Here for download this book
IN THIS BOOK THESE CONTENT ARE AVAILABLE.

Contents
Preface xi
About the authors xiii
Acknowledgements xv
1 Using a scientific calculator 1

STRENGTH OF MATERIALS Top 10 Numercial

Assume g = 10 m/s2 for all questions
1. a. A round steel bar with an initial diameter of 20 mm and length of 2 m is placed in tension, supporting a load of 2000 kg. If the Young’s modulus of the bar is 200 GPa what is the length of the bar when supporting the load (assuming it does not yield)?
b. If the Poisson’s ratio of the steel is 0.4, what will the diameter of the bar be when supporting the load?
Solution:                     
a. Cross-sectional area = 3.14 × 104 m2
Load = 2000 kg × g = 2 × 104 N
Stress = 2 × 104/3.14 × 104 = 6.4 × 107 Pa
Strain = 6.4 × 107/200× 109 = 3.2 × 104

New length = (3.2 × 104 × 2) + 2 = 2.00064 m
b. Lateral strain = 3.2 × 104 × 0.4 = 1.28 × 104
Change in diameter = 1.28 × 104 × 0.02 = 2.56 × 106m
New diameter = 0.02 2.56 × 106= 1.999744 × 102m = 19.99744 mm

Saturday, 3 February 2018

Useful information For Every Civil Engineer

CONCRETE GRADE:
M5 = 1:4:8
M10= 1:3:6
M15= 1:2:4
M20= 1:1.5:3
M25= 1:1:2 

CLEAR COVER TO MAIN REINFORCEMENT:
1.FOOTINGS : 50 mm
2.RAFT FOUNDATION.TOP : 50 mm
3.RAFT FOUNDATION.BOTTOM/SIDES : 75 mm
4.STRAP BEAM : 50 mm
5.GRADE SLAB : 20 mm
6.COLUMN : 40 mm
7.SHEAR WALL : 25 mm
8.BEAMS : 25 mm
9.SLABS : 15 mm
10.FLAT SLAB : 20 mm
11.STAIRCASE : 15 mm
12.RET. WALL : 20/ 25 mm on earth
13.WATER RETAINING STRUCTURES : 20/30 mm

Curing of Concrete

Curing may be defined as the process of maintaining satisfactory moisture and temperature conditions
for freshly placed concrete for some specified time for proper hardening of concrete. Curing in the early
ages of concrete is more important. Curing for 14 days is very important. Better to continue it for 7 to 14
days more. If curing is not done properly, the strength of concrete reduces. Cracks develop due shrinkage.
The durability of concrete structure reduces.
The following curing methods are employed:
(a) Spraying of water
(b) Covering the surface with wet gunny bags, straw etc.
(c) Ponding
(d) Steam curing and
(e) Application of curing compounds.

Grading of Aggregates

Aggregate comprises about 55 per cent of the volume of mortar and about 85 per cent
volume of mass concrete. Mortar contains aggregate of size of 4.75 mm and concrete contains aggregate upto a maximum size of 150 mm.

      One of the practical methods of arriving at the practical grading by trial and error method is to mix aggregates of different size fractions in different percentages and to choose the one sample which gives maximum weight or minimum voids per unit volume, out of all the alternative samples. Fractions which are actually available in the field, or which could be made available in the field including that of the fine aggregate will be used in making samples.

Sieve Analysis
This is the name given to the operation of dividing a sample of aggregate into various
fractions each consisting of particles of the same size. The sieve analysis is conducted to determine the particle size distribution in a sample
of aggregate, which we call gradation.
A convenient system of expressing the
gradation of aggregate is one which the
consecutive sieve openings are constantly doubled,
such as 10 mm, 20 mm, 40 mm etc. Under such a
system, employing a logarithmic scale, lines can be
spaced at equal intervals to represent the successive
sizes.

FIRE RESISTANT BUILDING

It is reported that in USA fire kills more people each year than all other natural disasters combined including floors, cyclones and earthquake. The fire load in a building should be kept to the minimum possible. The term fire load indicates the amount of heat liberated in kilo joules per square metre (kJ/m2) of floor area of any compartment by the combustion of the content of the building including its own combustible part. It is determined by multiplying the weights of all combustible materials by their respective calorific values and dividing that with floor area.
A building may be made more fire resistant by
1. Using suitable materials
2. Taking precautions in building construction
3. By providing fire alarm systems and fire extinguishers.

IMPROVING EARTHQUAKE RESISTANCE OF TALL BUILDINGS

Tall buildings are subjected to heavy horizontal forces due to inertia during earthquake. Hence they need shear walls. A shear wall is a R.C.C. enclosure within the building built to take shear forces. It is usually built around lift room. These shear walls must be provided evenly throughout the buildings in both directions as well as from bottom to top. Apart from providing shear walls, the following techniques are also used for making tall buildings earthquake resistant:
1. Base Isolation

2. Using Seismic Dampers.
v  Base Isolation
The idea behind base isolation is to detach (isolate) the building from the ground in such a way that earthquake motions are not transmitted up through the building, or at least greatly reduced. The concept of base isolation is explained through an example of building resting on roller [Fig. 20.3]. When the ground shakes, the roller freely roll but the building above does not move. If the gap between the building and the vertical wall of foundation pit is small, the vertical wall of the pit may

Friday, 2 February 2018

EARTHQUAKES RESISTANT BUILDINGS

An earthquake is a sudden, rapid shaking of the earth surface caused by the breaking and shifting of rocks beneath. During earthquake, ground motion occurs in a random fashion in all directions radiating from a point within earth crust, called epicentre. It causes vibrations of structures and induce inertia forces on them. As a result structure may collapse resulting into loss of property and lives. Earthquakes do not kill people, vulnerable buildings do so. Hence there is need of designing earthquake resistant buildings, especially in the earthquake prone areas.
TYPES OF EARTHQUAKES
Depending upon the possible causes, the earthquakes may be classified as:
1. Natural earthquake
2. Earthquakes due to induced activities.

BRICK MASONRY

Brick masonry is built with bricks bonded together with mortar. For temporary sheds mud mortar may be used but for all permanent buildings lime or cement mortars are used.
The various types of bonds generally used in brick masonry are
1. Stretcher bond
2. Header bond
3. English bond and
4. Flemish bond.

SOLID AND HOLLOW CONCRETE BLOCKS

Solid and hollow concrete blocks are manufactured in factories to meet the requirements of building
blocks in cities and towns. These blocks may be called as artificial stones, since they replace the stones in the masonry construction. They are manufactured with lean mixes of cement, sand and aggregates of sizes less than 12 mm. Instead of sharp edged aggregates, round aggregates are professed in the manufacture of these blocks. The properties and uses of these blocks is given in this article.

ALUMINIUM

It is present on the surface of earth crust in most of the rooks and clay. But to produce the metal bauxite
(Al2O3. 2H2O) is ideally suited ore.

Properties of Aluminum
1. It is having silver colour and bright lustre.
2. It is very light in weight.

FERROUS METALS

A ferrous material is the one in which iron is a main constituent. Iron ore is first converted into pig iron
and then pig iron is subjected to various metallurgical processes to mix different percentage of carbon
and to get the following three useful ferrous materials:
1. Cast iron—carbon content 1.7% to 4.5%
2. Wrought iron—carbon content 0.05% to 0.15%
3. Steel—carbon content 0.25% to 0.25%.
All ferrous materials contain about 0.5 to 3% silica, less than 2% manganese, 0.15% sulphur and
0.6% phosphorous.

Tests on Concrete

The following are some of the important tests conducted on concrete:
1. Slump test.
2. Compaction factor test.
3. Crushing strength test.

1. Slump Test: This test is conducted to determine the workability of concrete. It needs a slump
cone for test (Fig. 3.3). Slump cone is a vessel in the shape of a frustum of a cone with diameter at
bottom 200 mm and 50 mm at top and 300 mm high. This cone is kept over a impervious platform and is filled with concrete in four layers. Each layer is tamped with a 16 mm pointed rod for 25 times. After filling completely the cone is gently pulled up. The decrease in the height of the concrete is called slump. Higher the slump, more workable is the concrete. The desired values of slumps for

Properties of Green Concrete

1. Workability: This is defined as the ease with which concrete can be compacted fully without
seggregating and bleeding. It can also be defined as the amount of internal work required to fully
compact the concrete to optimum density. The workability depends upon the quantity of water, grading,
shape and the percentage of the aggregates present in the concrete.
Workability is measured by

(a) The slump observed when the frustum of the standard cone filled with concrete is lifted and
removed.
(b) The compaction factor determined after allowing the concrete to fall through the compaction
testing machine.
(c) The time taken in seconds for the shape of the concrete to change from cone to cylinder when
tested in Vee-Bee consistometer.
The suggested values of workability for different works are as shown in Table 3.2.