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Sunday, June 16, 2019

June 16, 2019

Calculation of Material Required For Brickwork

Calculation of Material Required For Brickwork

1. Calculation of Number of bricks required for 1 cum.

Take a wall 1and 1/2 brick with

Length of wall = 20 metre

Width of wall   = 30 cm or( 0.3 mtre)

Height of wall  = 5 m

The volume of wall = L × B × H

                                =  20 × 0.3 × 5
                                =  30 cum
Normally mortar joints are less than 1 cm, taking 1 cm mortar joints, the actual thickness of the wall be 29 cm.

Therefore actual volume = 20 × 0.29 × 5

                                        = 29 cum

The volume of 1 brick of standard size 20cm×10cm ×10cm  = 0.20 × 0.10 × 0.10

                                                                                                 = 0.002 cum

Now the number of bricks in wall  =  actual volume of wall/ volume of one brick

                                                        =  29/0.002

                                                       = 14500 Numbers

Therefore, the number of bricks per cum of nominal size = Total nos of bricks in wall/volume of wall

                                                                                            = 14500/30

                                                                                            = 484 numbers

Considering 5% breakages, wastages, etc., This may be taken 500 numbers per cubic metre.

Calculation of Quantity of mortar in Brickwork

Mortar Requirement = Total volume of brickwork minus net volume of bricks

                                 =  29 - (0.19 × 0.09 × 0.09 × 14500)

                                 = 29 - 22.315

                                 = 6.685 cum

For frog filling, for use of cut brick, for bonding, for uniform joints, wastage, etc. 15% extra mortar may be required.

Therefore, the volume of mortar = 6.685 + 6.685 × 0.15

                                                     =  7.688 cum

For dry volume increase by 1/4

Dry volume of Mortar = 7.688 + 7.688 × 0.25

                                     = 9.61 cum

For 30 cum of brickwork dry volume of mortar = 9.61 cum

For 1 cum dry volume of mortar = 9.61/30
                                                     = 0.32 cum or 11.3 cft

In practice, for cement mortar 3 cum dry mortar and for lime mortar 3.5 cum pf dry mortar are taken for 10 cum brickwork. As an approximate 30% dry mortar may be taken.
Calculation of Materials of Mortar

Approximate method to determine the quantity of materials of mortar for 1cum brickwork divide 0.3 by the sum of the numerals of the proportion of materials which gives the quantity of cement in cum.

Example - For brickwork in 1: 6 cement mortar = 0.3/ (1+6) = 0.043 cum

                                                             Sand        =  0.43 × 6 = 0.258 cum

But as the cement will go to fill up the voids in the sand, 0.045 cum of cement and 0.27 cum of sand may be taken.

Also, read - Calculation of Quantity of Material For Concrete Mix Ratio


Friday, June 14, 2019

June 14, 2019

What Is Super Elevation

What Is Super Elevation?

When a vehicle passes from a straight to a curved path or in other words when a vehicle negotiates horizontal curve following two forces act on vehicle:

1. Centrifugal Force
2. Weight of the Vehicle

1. Centrifugal Force - The centrifugal force is a function of the speed of the moving vehicle. It always acts at the center of gravity of the vehicle. It's direction always tends to outside, i.e., it always tends to push the vehicle out of the track. to counteract this tendency, the outer edge of the road is raised above the inner edge. This rise of the outer edge is called superelevation or cant or banking.
                              Thus superelevation e is the ratio of the height of the outer edge with respect to the horizontal width.
                                               e = tan θ

In practice, the value of θ is kept as 4° or a slope of 1 in 15 with horizontal.

The total height of the outer edge with respect to the inner edge

                                  E  = e × width of road    
                                       =  e B

 The centrifugal force   P   =   Wv2/gR


W = weight of the vehicle
v  =  velocity of the vehicle
R = radius of circular curve
P = centrifugal force
g = acceleration due to gravity

Effect of Centrifugal Force

1. The tendency to overturn the vehicle.
2. The tendency to skid the vehicle laterally.

Stability Condition Against Overturning

The figure shows a vehicle moving on horizontal a curve. Forces acting on the vehicle are
a.) Centrifugal force P acting outward at C.G.
b.) Weight W acting downward at C.G.

Let h be the height of C.G. of the vehicle above the road level.
The overturning moment due to centrifugal force.

                                                           = P × h

The restoring moment                       =  W × b/2

where b is the center to center distance of wheels of the vehicle.

in limiting equilibrium 
                                       Ph = Wb/2

                                        P/W = b/2h

When the centrifugal ratio, P/W is equal to B/2h there is a danger of overturning.
Thus to avoid overturning, centrifugal ratio should always be less than b/2h.
Also                       P/W = v2/gR
                               v2/gR< b/2h
Thus to avoid overturning h should be as small as possible. Only due to this reason modern passenger cars have a low center of gravity.

Stability Condition Against Skidding

The lateral thrust P = Wv2/gR is resisted by the frictional force between the tyre and pavement surface. If the lateral resisting friction is less than the centrifugal force P, then skid will occur. Thus in limiting equilibrium, P = Maximum lateral friction developed as shown in the figure.

                                            P = Fa + Fb
or                                                 P = f (RA + RB)
                                               = fW
Or                                 P/W   =  f

Thus when the centrifugal ratio attains the value equal to the lateral coefficient of friction, there is a danger of lateral skidding.

If        f < b/2h skidding would occur.

If        f > B/2h overturning at the outer edge would occur.




Friday, June 7, 2019

June 07, 2019

Material Required for Construction of WBM Road

Water Bound Macadam (WBM) 

 In the oldest type of highway pavement used in modern times is known as Macadam after the name of Johan Macadam, a Scottish Engineer. The term Macadam in the present time means the road surface and base constructed of crushed or broken aggregates cemented together by the action of rolling and water. The binding action in WBM is achieved by stone screening used as filler in the presence of water. Water bound macadam is constructed in thickness varying from 8 to 10 cm depending upon the design requirements. The surface course of WBM gets deteriorated very soon under the action of mixed traffic. so nowadays the WBM is used as a base course for the superior type of pavement such as bituminous or cement concrete surfacing. Generally, the layers in WBM are laid in 12 to 15 cm thickness. The total thickness may be up to 30 to 35 cm depending upon the design requirements. Each layer is compacted to 75 to 80% of its loose thickness by smoothed wheel roller. In WBM roads generally, camber is provided of the order of 1 in 36 to 1 in 48.

Material Required for Construction Of WBM Road

For WBM construction following materials are required.

1. Coarse aggregate
2. Screenings
3. Fillers Materials

1. Coarse Aggregates 

 The hard and durable stone should be used for coarse aggregate. traprock is one of the best stone for road metal with a specific gravity ranging from 2.8 to 3.1. hard limestone, dolomite, and granites are also satisfactory. The specific gravity of these stones varies from2.6 to 2.7. Sand stone and quartz whose specific gravity varies from 2.4 to 2.7 are some what less satisfactory especially for wearing course. The course aggregate further should be free from excess of flaky, elongated, soft and disintegrated particles and dirt. Coarse aggregate used should meet the following specifications:

Size  -  90 to 40 mm or 63 to 40 mm or 50 to 25 mm

Loss Angles abrasion value

After 500 revolution      -    40%  Max
Flakiness index             -     15%  Max


 10 to 12.5 mm size screenings may be used. They should be properly graded. The grading Requirements should be as follows:

Types of Screening
Sieve Size
% passing by weight
12.5 mm (For grading 1 or 2)
12.5 mm
10.0 mm
4.75 mm
150 micron
10 mm (For grading 2 or 3)
10 mm
4.75 mm
150 micron

3. Fillers Materials
 - To prevent raveling of WBM surface layer the filler material should contain sufficient clay content. Plasticity index of filler material may be up to 9.0.


Wednesday, June 5, 2019

June 05, 2019

Construction of WBM Road

Construction of WBM Road

Water Bound Macadam (WBM) - It is the layer of broken stone aggregates bound together by stone dust or screening material and water applied during construction, and compacted by heavy smoothed wheel roller. It is the oldest type of highway pavement used in modern times is known as Macadam after the name of Johan Macadam, A Scottish Engineer.

Construction Procedure of WBM Road

1. Preparation of Subgrade 

 The site is cleared and weak spots corrected and rolled. To check the spreading of aggregate beyond the carriageway or to provide confinement to the loose aggregate while compacting by rolling either a trench cut in the subgrade and proper camber is provided or earth or brick bunds are made. on the clayey subgrade, a blanket of granular material of a thickness of about 10 cm is provided. The subgrade should be well drained and checked for stability etc.

2. Spreading of Aggregates  

The coarse aggregate is spread on the prepared surface of subgrade evenly in layers. The thickness of each layer should be such that it gives 7.5 cm thick layer on compaction. The profile of the road is checked by placing templates across the road every 6 meters. To ensure proper camber and grade the surface is checked from time to time.

3. Dry-Rolling 

The objective of this rolling is to key the coarse aggregate thoroughly. After spreading the coarse aggregate and checking all irregularities, the rolling is done by 6 to 10 tonnes, 3 wheeled power roller. Usually rolling is started from the edges with roller running forward and backward till the aggregate is fully compacted.

4. Application of Screening 

The next step is the application and keying of screening. after the coarse aggregates are set and keyed thoroughly by the rolling, the screenings are spread uniformly and rolled. generally spreading, brooming and rolling operations are carried out simultaneously. The quantity of screening is generally used to fill about 50% of the total voids and the rest 50% voids are filled with filler materials.

5. Wet Rolling  

After spreading the screening, the surface of the layer is sprinkled with water and rolled again. The sprinkling of water and rolling is continued till all the voids are filled and a wave of grout flushed ahead of the roller. this indicates that all voids are filled with chocking material. The quantity of water and screening are generally dependent upon many factors such as size and nature of aggregate, type of surface desired, etc.

6. Application of Fillers  

The filler material is applied in two successive thin layers. the plasticity index of the filler material should not be more than 9. after the application of the filler, water is sprinkled on the surface and the slurry is allowed to fill the voids. Now rolling is done with 6 to 10-tonne roller. to wash down the binding material off the wheels of the roller water is poured on them.

7. Surface Finishing  

The section prepared as discussed above is allowed to dry overnight and then a  thin layer about 0.6 cm thick of sand or earth spread over the surface. the surface lightly sprinkled with water and rolled again.

8. Shoulders  

These are made of the same cross slope as that of pavement and are compacted by rolling. After proper drying and set of the WBM, It is opened to the traffic.


Saturday, June 1, 2019

June 01, 2019

Requirements of Good Staircase

Requirements of Good Staircase

The stair is a structure having a series of steps and affords the means of ascent and descent between the floors or landings. The enclosed room in which stair is located is known as the staircase. The opening space occupied by the stair is known as a stairway. A stair consists of treads, risers, stringers and balusters and so many other connected elements. Stair may be made from timber, stone, bricks, steel, plain cement concrete or reinforced cement concrete.

1. Location - It should be located centrally so easily accessible from every corner of the building. Light and proper ventilation should be there.

2. Width of the stair - The usually adopted average value of the stair width for public and residential building is 1.8 m and 90 cm respectively.

3. Length of flight - Number of step in flight should be restricted to a maximum of 12 and a minimum of 3.

4. Pitch of the stair - The slope of the stair should never exceed 40°C and not be flatter than 25°C.

5. HeadRoom - The headroom of the flight immediately above it should not be less than 2.14 m.

6. Material - It should be strong durable and stiff. The materials by which stair is constructed should possess fire-resisting qualities.

7. Landing - The width of the landing should not be less than the width of the stair.

8. Step Proportions- The rise and trade of each step in stair should be of uniform dimensions throughout.

9. Riser Height - The height of the riser should not be more than 20 cm.

10. Hand Rail - It should be provided with handrails.

11. Winder - Winder step as for as possible should be avoided.

Proportion of Trade and Riser

The size of trade and riser varies with the situation of the stair and the purpose for which it has to be provided. In general, the following thumb rules are considered.

1. (Going in cm) + ( 2 x Rise in cm)  =  60

2. (Going in cm) + (Rise in cm)  =   400 to 410 approx

3. Adopt the standard size going and rise as 30 cm and 14 cm respectively and for each 25 mm deducted from going and 12 to 13 mm to rise.

The results which are practically adopted in the stair constructions are

1. For Residential Building             -     25 cm x 16 cm

2. For Public Building                    -      27 cm x 15 cm to 30 cm x 13 cm


Thursday, May 30, 2019

May 30, 2019

Types of Rails

Types of Rails

Rail - Rails are steel girders placed end to end to provide a level and continuous surface for the 
movement of trains.

Function of Rail

Following are the functions of rails:

1. The rails provide s level and continuous surface for the movement of trains.
2. The rails provide a smooth pathway to trains. The pathway has very low friction. The friction between the wheels of train and steel of the rails is about 20% of the friction between the pneumatic tyres and metalled roads. 
3. The rails serve as a lateral guide for the running of wheels.
4. The rails carry the stresses developed due to vertical loads transmitted to it through axles and wheels of the rolling stocks, due to braking forces and thermal stresses etc.
5. The rail transmits the heavy load of rolling stock etc to the large area of formation through sleepers and ballast. 

Types of Rails

1. Double Headed Rail
2. Bull Headed Rail
3. Flat Footed Rail

Double Headed Rail - Originally the rails used were double headed made of I section or Dumb bell section as shown in the figure. The idea was that when the head of the rail is worn-out during the service period, the rail could be inverted and reused without increasing any extra expenditure. Such rails need to be supported on chairs that rest on sleepers. But later on, it was found that during the service the bottom table of the rail was dented by the long and continuous contact with the chair to such an extent that it was impossible to reuse it. This led to the development of bull headed rail.

Bull Headed Rail - The bull headed rail is almost similar to double head rail. The only difference between the double headed rail and Bull head rail is that in Bull headed rail more metal is added to the head to allow greater wear and tear. The lower head or table was kept of just sufficient size to be able to withstand the stress be induced by the moving loads. The rail also required chair for fixing it to the sleepers. This proved the greater drawback of this rail.

Flat Footed Rail - To remove the above drawbacks, Charles Vignoles developed an inverted T shaped section known as flat footed rail in 1936. Flat footed rail is also known as Vignole rail. It has the following advantages over double headed and bull headed rails.

Advantages of Flat Footed Rails

1. For fixing flat footed rails to sleepers, no chairs are needed. The foot of the rail may be spiked direct to the sleepers. This affects the economy to a great extent.

2. For the same weight, this rail is stronger vertically and laterally both than Bull headed rails.

3. It is cheaper than Bull headed rails.

4. It requires less fastening than Bull headed rail.

5. F.F. rail give better stability to the track as these rail distribute rolling stock load over a large number of sleepers.

6. F.F. rails develop fewer kinks and maintain a more regular top surface than B.H. rails.

7. F.F. rail give longer life to the track and reduce maintenance cost.
                                       The F.F. rails have been widely accepted throughout the world. About 90% track length of the entire world has been laid with F.F. rails. It has also been standardised on Indian Railways.

Also Read - Sleepers In Railway


Friday, May 24, 2019

May 24, 2019

What Is Shear Force and Bending Moment

What Is Shear Force and Bending Moment?

Shear Force - The shear force at any point along a loaded beam may be defined as the algebraic sum of all vertical forces acting on either side of the point on the beam. The net effect of the shear force is to shear off the beam along with the point at which it is acting. Shear force is taken +ve if it produces a clockwise moment and it is taken -ve when it produces an anticlockwise moment.

Bending Moment - Bending moment at any point along a loaded beam may be defined as the sum of the moments due to all vertical forces acting on either side of the point on the beam. The bending moment tries to bend the beam. Clockwise moments due to loads acting to the left of the section are assumed to be +ve, while anticlockwise moments are taken -ve.

Sign Convention Used For Shear Force And Bending Moment

Shear Force - Force acting in the right-hand side of the section in the upward direction is taken -ve and force in the right-hand side of the section acting in the downward direction are taken as +ve. Similarly, a force in the left-hand side of the section is taken +ve if it is acting in an upward direction and it is taken as negative if it is acting in a downward direction.

Bending Moment - First of all remove all the loads and reaction from any one side of the section. Now introduce each load and reaction one at a time and find its effect at the section. A bending moment causes concavity upwards is taken +ve and called a sagging bending moment. A bending moment which is causing convexity upwards is taken -ve and called as hogging bending moment.

Shear Force And Bending Moment Diagrams

 A shear force diagram which shows the shear force at every section of the beam due to transverse loading on it. Its baseline is equal to the span of the beam, drawn on a suitable scale. For point loads S.F. diagram has a straight horizontal line, for UDL( Uniformly Distributed Load), It has straight inclined lines, and for uniformly varying loads it has a parabolic curve.
                                                           A bending moment diagram is a diagram which shows the bending moment at every section of the beam due to transverse loading on it. in case of a simply supported beam bending moment is zero at the ends, and for a cantilever, it is zero at the free end. For point loads, B.M. diagram has straight inclined lines, for UDL, it has a parabolic curve and for the uniformly varying load, it has a cubic curve.

Important Points Must Be Kept In Mind While Drawing The Shear Force And Bending Moment Diagram

The following points must be kept in mind while drawing the shear force and bending moment diagrams-

1. First of all, consider either the left or the right-hand side of the section.

2. Add the forces( Including Reactions) normal to the beam on one of the side, if the right-hand side of the section is chosen, a force acting downwards is taken +ve while a force acting upwards is -ve.

3. The +ve values of shear force and bending moment are plotted above the baseline, and -ve values below the baseline.

4. The shear force diagram will decrease or increase suddenly shown by a vertical straight line at a section when there is a vertical point load.

5. The shear force between any two vertical loads will be constant and hence the shear force diagram between two vertical loads will be horizontal.

6. The bending moment at the two supports of a simply supported beam and also at the free end of a cantilever will be zero.