Civil Study

Classification Of road as per IRC

noreply@blogger.com (Civil Study) — Fri, 14 Jan 2022 13:57:00 +0000

The roads can be classified in many ways. The classification based on speed and accessibility the most generic one. Note that as the accessibility of road increases, the speed reduces.Accordingly, the roads can classified as follows in the order of increased accessibility and reduced speeds.

 Freeways: Freeways are access controlled divided highways. Most freeways are four lanes, two lanes each direction, but many freeways widen to incorporate more lanes as they enter urban areas. Access is controlled through the use of interchanges, and the type of interchange depends upon the kind of intersecting road way (rural roads, another freeway etc.)

 Expressways: They are superior type of highways and are designed for high speeds(120 km/hr is common), high traffic volume and safety. They are generally provided with grade separations at intersections. Parking, loading and unloading of goods and pedestrian traffic is not allowed on expressways.

 Highways: They represent the superior type of roads in the country. Highways are of two types - rural highways and urban highways. Rural highways are those passing through rural areas (villages) and urban highways are those passing through large cities and towns, i.e. urban areas.

 Arterials: It is a general term denoting a street primarily meant for through traffic usually on a continuous route. They are generally divided highways with fully or partially controlled access. Parking, loading and unloading activities are usually restricted and regulated. Pedestrians are allowed to cross only at intersections/designated pedestrian crossings.

 Local streets: A local street is the one which is primarily intended for access to residence, business or abutting property. It does not normally carry large volume of traffic and also it allows unrestricted parking and pedestrian movements.

 Collectors streets: These are streets intended for collecting and distributing traffic to and from local streets and also for providing access to arterial streets. Normally full access is provided on these streets. There are few parking restrictions except during peak hours.

Roads can be classified based on some other criteria. They are given in detail below.

Speed vs accessibility

Based on usage

This classified is based on whether the roads can be used during different seasons of the year.

 All-weather roads: Those roads which are negotiable during all weathers, except at major river crossings where interruption of tra c is permissible up to a certain extent are called all weather roads.

 Fair-weather roads: Roads which are negotiable only during fair weather are called fair weather roads.

Based on carriage way

This classification is based on the type of the carriage way or the road pavement.

 Paved roads with hard surface : If they are provided with a hard pavement course such roads are called paved roads.(eg: stones, Water bound macadam (WBM),Bituminous macadam (BM), concrete roads) layer they is called unpaved roads. Thus earth and gravel roads come under this category.

Based on pavement surface

Based on the type of pavement surfacing provided, they are classified as surfaced and unsurfaced roads.

 Surfaced roads (BM, concrete): Roads which are provided with a bituminous or cement concreting surface are called surfaced roads.

 Unsurfaced roads (soil/gravel): Roads which are not provided with a bituminous or cement concreting surface are called unsurfaced roads.

Other criteria

Roads may also be classified based on the traffic volume in that road, load transported through that road, or location and function of that road.

 Traffic volume: Based on the traffic volume, they are classified as heavy, medium and light tra c roads. These terms are relative and so the limits under each class may be expressed as vehicles per day.

 Load transported: Based on the load carried by these roads, they can be classified as class I, class II, etc. or class A, class B etc. and the limits may be expressed as tonnes per day.

 Location and function: The classification based on location and function should be a more acceptable classification since they may be defined clearly.

Why wet process of manufacture of cement is most popular in India?

noreply@blogger.com (Civil Study) — Fri, 07 Jan 2022 15:01:00 +0000

In current world Cement plays vital role in building economic development of any country.Indian cement industry is the largest cement producing country in the world, next only to China.The first cement company became operational in Porbandar, Gujarat with a capacity of 10,000 tons in 1914. The Industry recorded an exponential growth with the introduction of partial decontrol in 1982 culminating in total decontrol in 1989 and delicensing in 1991.With the adoption of massive modernization and assimilation of state-of-the-art technology,Indian cement plants are today the most energy-efficient and environment-friendly and are comparable to the best in the world in all respects, whether it is size of the kiln, technology,energy consumption or environment-friendliness.

The cement industry contributes to environmental cleanliness by consuming hazardous wastes like Fly Ash (around 30 Mnt) from Thermal Power Plants and the entire 8 Mnt of granulated Slag produced by Steel manufacturing units and also using alternate fuels and raw materials using advanced and environment friendly technologies.

Cement plant

Production

At present, the Installed Capacity of Cement in India is 500 MTPA with production of 298 Million Tonnes per annum. Majority of the cement plants installed capacity (about 35%) is located in the states of south India. In PAT scheme, Total Installed Capacity of Cement in India is 325 MTPA which contributes to 65% coverage of total installed capacity in India.

With the increase in growth of infrastructure, the cement production in India is expected to be 500 Million Tonnes by the year 2020 and 800 Million Tonnes by 2030.

Per Capita consumption of cement

The per capita consumption of cement in India is 195 kg which is far less than world average of 500 kg and 1000 kg of China.

Type of Process

There are mainly two types of Cement Manufacturing process:

a) Wet Process b)Dry Process

Wet Process

Under wet process type, the raw materials are mixed with water so as to ensure better homogenization which leads to better quality of clinker. In the whole process, moisture content in the slurry is 35-50%. This results in higher requirement of heat which in turn require high amount of fuel. Hence in terms of energy consumption, this process consumes more amount of thermal energy. The length of Kiln in wet process is larger as compared to dry process.

Dry Process

Under dry process type, the raw materials are kept dry by utilizing the heat from Raw Mill resulting in poor homogenization which leads to inferior quality of clinker. In the whole process, moisture content in the slurry is 12%. This results in lower requirement of heat which in turn require low amount of fuel. Hence in terms of energy consumption, this process consumes less amount of thermal energy.

PAT Scheme

The Perform Achieve and Trade (PAT) scheme of Ministry of Power, Government of India has so far covered 126 numbers of cement plants in India targeting in reduce specific energy consumption since its inception from 2012 on wards.

Based on the threshold defined, 85 number of cement plants were included as DCs and their cumulative energy consumption were 15.01 million MTOE in PAT Cycle-1. Based on their specific energy consumption level, these DCs were given SEC target reduction of an average 5.43% resulting in 0.815 million TOE energy consumption reduction in absolute terms. The cement sector constituted 12.19% of the overall energy saving target under PAT Cycle 1.

The total savings achieved by Cement sector covering 75 number of designated consumers in PAT Cycle I is 1.48 million MTOE which is 0.665 million MTOE in excess of the target.

At present, the energy consumption of these cement unit as described as designated consumers is 23.246 million tonne of oil equivalent. The target given for them from PAT Cycle –II on wards is 0.94 million tonne of oil equivalent.

Global Bench-marking in terms of Energy Consumption

The table below depicts that Average Specific Thermal and Electrical energy consumption of India’s cement plants are below as compared to their global counterparts. Thus creating them as role model in World Cement Industry.

Key Stakeholders

Therefore, key stakeholders plays crucial role in development of energy efficient cement industry thereby reducing the energy consumption and mitigating the challenge of carbon reduction emissions.

The key stake holders are:-

1. The Cement Industries (Including PAT and Non-PAT)

2. The Government agencies such as Ministry of Power, Ministry of Environment and Climate Change, Bureau of Energy Efficiency, National Council of Cement and Building Materials etc.

3. The Cement Manufacturing Association.

4. Other agencies such as CII, TERI, FICCI, GIZ etc.

Importance of earthwork in construction

noreply@blogger.com (Civil Study) — Mon, 06 Dec 2021 16:01:00 +0000

For various civil engineering projects like road work, irrigation canal project, tank survey, earth moving, etc, the calculation method used are different. Some of these methods were introduced before the invention of computers and are still being continued. This document discusses about the industry practices of various methods of volume calculation so that the users can finally choose the right one for their project.

Earthworks

Earthworks are engineering works created through the moving and/or processing of massive quantities of soil or unformed rock. Earthwork is done to reconfigure the topography of a site to achieve the design levels. Earthwork involves cutting and filling to achieve the required topography.

Cutting

Cutting is the process of excavating earth material from a work location to achieve the desired topography.

Filling

Filling is the process of moving the excavated material or additional earth material to a work location to achieve the desired topography.

Applications of Earthwork

Typically, earthwork is done in the following projects:

 Road works

 Railways

 Irrigation project such as canals and dams

 Other common earthwork applications are land grading to reconfigure the topography of a site, or to stabilize slopes.

Trapezoidal Rule (End Area)

In trapezoidal method, each segment of the section is divided into various trapezoids and triangles.

Trapezoidal Area A = 1/2 X a X (b1+b2)

Trapezoidal area

Triangle area A = a * b/2

Triangle Area

Test Of Soil for road construction

noreply@blogger.com (Civil Study) — Sun, 05 Dec 2021 12:59:00 +0000

Sub grade soil is an integral part of the road pavement structure as it provides the support to the pavement from beneath. The main function of the sub grade is to give adequate support to the pavement and for this the sub grade should possess sufficient stability under adverse climatic and loading conditions. Therefore, it is very essential to evaluate the sub grade by conducting tests.

The tests used to evaluate the strength properties of soils may be broadly divided into three groups:

1. Shear tests

2. Bearing tests

3. Penetration tests

Shear tests

Shear tests are usually carried out on relatively small soil samples in the laboratory. In order to find out the strength properties of soil, a number of representative samples from different locations are tested. Some of the commonly known shear tests:

o Direct shear test,

o Triaxial compression test,

o Unconfined compression test.

California Bearing Ratio (CBR)

This test was developed by the California Division of Highway as a method of classifying and evaluating soil-sub grade and base course materials for flexible pavements.

CBR is defined as the ratio of force per unit area that is required to penetrate a soil mass with a piston of 1.25mm/min corresponding to that required for the penetration of a standard material.

CBR = Test load/Standard load * 100

Apparatus Required:

1. Steel Cutting Collar

2. Dial Gauges

3. Weights

4. IS Sieves

5. Penetration Plunger

6. Disc and loading machine

7. Cylindrical Mould and Rammers

Procedure:

The procedure for conducting the CBR test is as follows:

1. Firstly, an undisturbed sample specimen is prepared by attaching a cutting edge to the mould and pushing it gently in the mould to remove the soil on the outside.

2. The density of the sample is determined and is remoulded at Proctor’s maximum density or Optimum moisture content. The sample must be such that it must pass through a 20mm IS sieve but must be retained on a 4.75mm IS sieve.

3. About 4.5kg to 5.5 kg of the sample is mixed with water. The extension collar and base plate are then fixed to the mould and the disc is placed over the base plate.

4. The soil mix is then compacted in several layers by a rammer. Generally, for light compaction,the soil is compacted in three layers by 55 blows from a 2.6kg rammer.

5. The collar is then removed and the soil is trimmed. The mould is then turned over and the base plate and disc are removed.

6. The mould is then weighed and bulk density, as well as dry density, is determined.

7. Then the mould along with the surcharge weight is placed on the penetration test machine.

8. The load is applied to the piston such that the rate of penetration is 1.25mm/min.

9. The load readings at penetrations of 0.5,1.0,2.0,2.5,3.0,4.0,5.0,7.5,10 and 12.5 mm are recorded.

10. The mould is then detached from the loading equipment and moisture content is determined.

CBR Test

Plate Bearing Test

Plate bearing test is used to evaluate the support capability of sub-grades, bases and in some cases, complete pavement. Data from the tests are applicable for the design of both flexible and rigid pavements. In plate bearing test, a compressive stress is applied to the soil or pavement layer through rigid plates relatively large size and the deflections are measured for various stress values.

Plate load test

SPECIFIC GRAVITY AND WATER ABSORPTION OF COARSE AGGREGATES

noreply@blogger.com (Civil Study) — Sat, 04 Dec 2021 14:10:00 +0000

Theory: For design of concrete mix, information about the specific gravity and water absorption of the coarse aggregates are required. Specific gravity of aggregate provides valuable information on its quality and properties. If the specific gravity is above or below than the value normally assigned to a particular type of aggregate; it may indicate that shape and grading of aggregate has altered. It is also important in determination of moisture contact and in many concrete mix design calculations. It is also required for the calculation of volume yield of concrete.

Objective: To determine specific gravity and water absorption of coarse aggregate.

Reference: IS: 2386 (Part-3)-1963.

Apparatus: Balance of capacity 5kg weight, box wire basket 200mm in diameter, water container for immersing the wire basket, absorbent cloth for surface drying of the sample and thermos static drying oven.

Apparatus for Absorption test

Material: coarse aggregates (5 Kg),

Procedure:

1. A sample of not less than 2 Kg of the aggregate shall be thoroughly washed to remove finer particles and dust, drained and then placed in the wire basket and immersed in distilled water at a temperature between 22°C to 32°C with a cover of at least 5 cm of water above the top of the basket.

2. Immediately after immersion the entrapped air shall be removed from the sample by lifting the basket containing it 25 mm above the base of the tank and allowing it to drop 25 times at the rate of about one drop per second.

3. The basket and aggregate shall remain completely immersed during the operation and for a period of 24 ± l/2 hours afterwards.

4. The basket and the sample shall then be jolted and weighed in water at a temperature of 22°C to 32°C (weight A).

5. The basket and the aggregate shall then be removed from the water and allowed to drain for a few minutes, after which the, aggregate shall be gently emptied from the basket on to one of the dry clothes, and the empty basket shall be returned to the water and weighed in water ( weight B ).

6. The aggregate placed on the dry cloth shall be gently surface dried with the cloth, transferring it to the second dry cloth when the first will remove no further moisture. The aggregate shall then be weighed (weight C).

7. The aggregate shall then be placed in the oven in the shallow tray, at a temperature of 100 to 110°C and maintained at this temperature for 24 ± l/2 hours. It shall then be removed from the oven, cooled in the airtight container and weighed (weight D).

8. The specific gravity, apparent specific gravity and water & adsorption shall be calculated as.

Specific gravity = {𝐷⁄[𝐶 − (𝐴 − 𝐵)]} (1)

Apparent Specific gravity = 𝐷/(𝐷 − (𝐴 − 𝐵)) (2)

Water absorption (in %) = 100 × [(𝐶 − 𝐷)/𝐷] (3)

Where,

A = Weight in g of aggregate and basket in water

B = Weight in g of empty basket in water

C = Weight in g of the saturated surface - dry aggregate in air

D = Weight in g of oven dried aggregate in air.

Observations:

Weight of aggregate and basket in water (A)
Weight of empty basket in water (B)
Weight of the saturated surface - dry aggregate in air (C)
Weight of oven dried aggregate in air (D)
Specific gravity = [𝐷/(𝐶 − 𝐴 + 𝐵)]
Apparent Specific gravity = [𝐷/(𝐶 – 𝐷)]
Water absorption (in %) = 100 × [(𝐶 − 𝐷)/𝐷]

Results and discussions:

Following results are obtained for the provided coarse aggregate specimen:

a) Specific gravity : __________.

b) Apparent specific gravity : __________.

c) Water absorption : __________ %.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Different types of doors and windows in building construction

noreply@blogger.com (Civil Study) — Sun, 14 Nov 2021 06:17:00 +0000

Depending upon the manner of fixing, materials used for construction, nature of the operational movements of shutters , etc., the common varieties of windows and door used in the building can be discussed and grouped as follows:

1. Casement windows

2. Sliding windows

3. Metal windows

4. Corner windows

5. Gable windows bay windows

6. Lantern or lantern lights

7. Skylights

CASEMENT WINDOWS:

These are the windows, the shutters of which open like doors. The construction of a casement window is similar to the door construction.

Casement window

SLIDING WINDOWS:

These windows are similar to the sliding doors and the shutters moves on the roller bearings,either horizontally or vertically. Such windows are provided in trains, buses, bank counter, shops etc.

Sliding windows

METAL WINDOWS:

These are now a days widely used, especially for public building. The metal used in construction may be mild steel, bronze, or other alloys. The metal frame may be fixed direct to the wall or it may be fixed on a wooden frame.

Metal windows

CORNER WINDOWS:

These windows are provided at the corner of a room .They are placed at the corner of the room and thus they have two faces in two perpendicular direction. Due to such situation,there is entry of light and air from two direction and in many cases , the elevation of building is also improved.

GABLE WINDOWS:

These are the windows which are provided in the gable ends of a roof.

BAY WINDOWS:

These windows project outside the external wall of a room. They maybe square , splayed,circular, polygonal or of any shape. The projection of bay windows may start from floor level or sill level. These windows admit more lights, increase opening area , provide ventilation and improve the appearance of building.

LANTERNS:

These are the windows which are fixed on flat roofs to provide light to the inner portion of building where light coming from external windows are insufficient. They maybe square or rectangular or curved.

Door

A door is a moving structure used to block off, and allow access to, an entrance to or within an enclosed space, such as a building or vehicle. Similar exterior structures are called gate.Typically doors have an interior side that faces the inside of a space and an exterior side that faces the outside of that space. While in some cases the interior side of a door may match its exterior side, in other cases there are sharp contrasts between the two sides, such as in the case of the vehicle door. Doors normally consist of a panel that swings on hinges or that slides or spins inside of a space.

Panel doors:

Panel doors, also called stile and rail doors, are built with frame and panel construction.EN 12519 is describing the terms which are officially used in European Member States. The main parts are listed below:

• Stiles - Vertical boards that run the full height of a door and compose its right and left edges. The hinges are mounted to the fixed side (known as the "hanging stile"), and the handle, lock, bolt, and/or latch are mounted on the swinging side (known as the "latch stile").

• Rails- Horizontal boards at the top, bottom, and optionally in the middle of a door that join the two stiles and split the door into two or more rows of panels. The "top rail" and "bottom rail" are named for their positions. The bottom rail is also known as "kick rail".A middle rail at the height of the bolt is known as the "lock rail", other middle rails are commonly known as "cross rails".

• Mullions - Smaller optional vertical boards that run between two rails, and split the door into two or more columns of panels, the term is used sometimes for verticals in doors, but more often (UK and Australia) it refers to verticals in windows.

• Muntin - Optional vertical members that divide the door into smaller panels.

• Panels - Large, wider boards used to fill the space between the stiles, rails, and mullions.The panels typically fit into grooves in the other pieces, and help to keep the door rigid.Panels may be flat, or in raised panel designs. Can be glued in or stay as a floating panel.

• Light or Lite - a piece of glass used in place of a panel, essentially giving the door a window.

Typical door details

Compressive Strength of Concrete ||Test procedure - Civil Study

noreply@blogger.com (Civil Study) — Sun, 31 Oct 2021 12:30:00 +0000

One of the important properties of concrete is its strength in compression. The strength in compression has a definite relationship with all other properties of concrete i.e. these properties improved with the improvement in compressive strength. Thus, with this single test one judge that whether Concreting has been done properly or not. In India cubical molds of size 15 cm × 15cm × 15 cm are commonly used.The concrete is prepared with definite proportion is poured in the mold and tempered properly so as not to have any voids. After 24 hours these molds are removed and test specimens are put in water for curing. The top surface of this specimen should be made even and smooth. This is done by putting cement paste and spreading smoothly on whole area of specimen. These specimens are tested by compression testing machine after 3, 7 or 28 days curing. Load should be applied gradually at the rate of 14 N/mm2 per minute till the Specimens fails. Load at the failure divided by area of specimen gives the compressive strength of concrete. At least three specimens are tested at each selected age. The failure of the specimen is called as ‘hour glass’ type failure. This happens due to lateral restraint provided by the plates to the cubes.Practically, the compression testing system develops a complex system of stresses due to end restraints provided by steel platens of compression testing machine (CTM). Under compression loading, due to “poisons effect”, the cube specimen also undergo lateral expansion. However, the steel platens don’t undergo lateral expansion to the same extent that of concrete. Thus, there exist a differential tendency of lateral expansion between steel platens and concrete cube faces. As a result of this, tangential forces are induced between the end surfaces of the concrete specimen and the adjacent steel platens of CTM. Therefore, in addition to the applied compressive stress; lateral shearing stresses are also effective in these specimens. Effect of this shear decreases to words the center of the cube. Thus, the cube has near vertical crack at cubes center and sometimes, the cube may completely disintegrate leaving a relatively undamaged central core. Figure 8 shows, typical failure patterns of the concrete cube.

Civil Study

Objective: To determine the cube strength of the concrete of given properties.

Reference: IS: 516 - 1959, IS: 1199-1959, SP: 23-1982, IS: 10086-1982.

Apparatus: Molds for the test cubes, tamping rod, metallic sheet, Compressive testing machine.

Material: Cement, sand, aggregate and water, grease

Procedure:

1. Calculate the material required for preparing the concrete of given proportions

2. Mix them thoroughly in mechanical mixer until uniform colour of concrete is obtained

3. Pour concrete in the lightly greased cube molds.

4. Fill concrete in two layers each of approximately 75 mm and ramming each layer with 35 blows evenly distributed over the surface of layer.

5. Struck off concrete flush with the top of the molds.

6. Level the concrete at the top of the mold by means of trowel and give proper identification mark of the specimen.

7. Immediately after being made, they should be covered with wet mats.

8. Specimens are removed from the moulds after 24hrs and cured in water. Keep it for curing up to 28 days.

Testing of concrete cubes:

9. Take the cube out of water at the end of three days with dry cloth. Measure the dimensions of the surface in which the load is to be applied. Let be ‘L’ and ‘B’ respectively.

10. Place the cube in compressive testing machine and apply the load uniformly at the rate of 35N/mm2.

11. Note the load at which the cube fails. Let it be ‘P’. Also note the type of failure and appearance cracks

12. Calculate the compressive strength of the cube by using formula P/A. Where A is the area of loaded surface (i.e. L×B).

13. Repeat the same procedure (steps 9 to 12) for other two cubes.

14. Repeat the whole procedure (Step 9 to 13) to find the compressive strength of the cube at the end of 7 days and 28 days.

Observation

(a) For 3 days strength:

Sl. No.	Length (in mm)	Breadth (in mm)	Load (in N)	compressive strength in N/mm2	Remark
1
2
3

Average =

(b) For 7 days strength:

Sl. No.	Length (in mm)	Breadth (in mm)	Load (in N)	compressive strength in N/mm2	Remark
1
2
3

Average =

Sl. No.	Length (in mm)	Breadth (in mm)	Load (in N)	compressive strength in N/mm2	Remark
1
2
3

Average =

Results:

The type and grade of concrete :

The compressive strength of cement ate the end of

i) 3 days : ___________. N/mm2.

ii) 7 days : ___________. N/mm2.

iii) 28 days : ___________. N/mm2.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Fluid mechanics formulas for gate

noreply@blogger.com (Civil Study) — Sat, 23 Oct 2021 12:59:00 +0000

Some of the basic formula for fluid mechanics which can help in competitive exam.

1. 𝜏=𝜇.𝑑𝑣𝑑𝑦

2. 𝜕𝜌𝜕𝑡+𝜕𝜕𝑥(𝜌𝑣𝑥)+𝜕𝜕𝑦(𝜌𝑣𝑦)+𝜕𝜕𝑧(𝜌𝑣𝑧)=0 (continuity equation)

3. 𝐷𝜌𝐷𝑡+𝜌 ∇.𝑉=0 (continuity equation)

4. 𝐹1−𝐹2=𝑀2−𝑀1𝑡=𝑚𝑣2−𝑚𝑣1𝑡=𝜌𝑄(𝑣2−𝑣1)

5. 𝐹=𝜌𝑎(𝑣+𝑢)[(𝑣+𝑢)−𝑢]=𝜌𝑎(𝑣+𝑢)𝑣 (Jet propulsion moving with 𝑢 velocity)

6. 𝑝1𝜌𝑔+𝑣122𝑔+𝑧1=𝑝2𝜌𝑔+𝑣222𝑔+𝑧2

7. 𝜕𝜏𝜕𝑦=𝜕𝑝𝜕𝑥 and 𝜇𝜕2𝑣𝜕𝑦2=𝜕𝑝𝜕𝑥

8. 𝜏=−𝜕𝑃𝜕𝑥.𝑟2 and 𝜏=𝛾𝑤.ℎ𝐿𝐿.𝑟2

9. 𝑣=14𝜇.(−𝜕𝑝𝜕𝑥).(𝑅2−𝑟2)=𝑣𝑚𝑎𝑥.[1−(𝑟𝑅)2]

10. 𝑝1−𝑝2=32𝜇𝑉𝐿𝐷2=128𝜇𝑄𝐿𝜋𝐷4

11. 𝑓=64𝜇𝜌𝑉𝐷=64𝑅𝑒

12. 𝑣=𝑣𝑚𝑎𝑥.(1−𝑟𝑅)17⁄ (Turbulent)

13. ℎ𝐿=𝑓𝐿𝑣22𝑔𝐷

14. 𝐿𝐷5=𝐿1𝐷15+𝐿2𝐷25+𝐿3𝐷35 (Compound pipe)

15. 𝑝𝑎𝛾−ℎ =𝑝𝑠𝛾+𝑓𝑙𝑣22𝑔𝐷+0.5𝑣22𝑔+𝑣22𝑔 and 𝐻=𝑓𝐿𝑣22𝑔𝐷+1.5𝑣22𝑔 (Siphon)

16. 𝛿∗=∫(1−𝑣𝑉)∞0𝑑𝑦

17. 𝜃=∫𝑣𝑉(1−𝑣𝑉)∞0𝑑𝑦

18. 𝛿𝐸=∫𝑣𝑉(1−𝑣2𝑉2)∞0𝑑𝑦

19. Blassius boundary layer thickness 𝛿𝑥=5√𝑅𝑒𝑥

20. Displacement thickness 𝛿∗𝑥=1.729√𝑅𝑒𝑥

21. Momentum thickness 𝜃𝑥 =0.664√𝑅𝑒𝑥

22. 𝑐𝑑,𝑥=0.664√𝑅𝑒𝑥 and 𝜏𝑥=𝑐𝑑,𝑥𝜌𝑉22

23. Drag force= 𝐶𝑑𝜌𝐴𝑉22 𝑎𝑛𝑑 𝐹𝐴=𝜏0=𝐶𝑑𝜌𝑉22, 𝐶𝑑=1.328√𝑅𝑒 𝐿

24. 𝛿′=11.6 𝜗√𝜏𝑜/𝜌=11.6𝜗𝑣∗ (Laminar sub layer)

25. ℎ̅=x̅+IGAx̅ and ℎ̅=𝑥̅+𝐼𝐺𝐴𝑥̅.𝑠𝑖𝑛2𝜃

26. 𝐺𝑀=𝐼𝑉−𝐵𝐺

27. Q= 815𝐶𝑑.√2𝑔.tan𝜃/2 𝐻52⁄

28. 𝑣=√𝑔𝑥²2𝑦

29. 𝑄=Cd𝑎1𝑎2√𝑎12−𝑎22.√2gh and ℎ=𝑥(𝑠𝑚𝑠𝑓−1)

30. 𝑄=𝜓1−𝜓2

31. 𝜕∅𝜕𝑥=−𝑣𝑥 and 𝜕∅𝜕𝑦=−𝑣𝑦

32. 𝜕𝜓𝜕𝑥=𝑣𝑦 and 𝜕𝜓𝜕𝑦=−𝑣𝑥

33. (𝑅𝑒)𝑟=𝜌𝑟𝑉𝑟𝐿𝑟𝜇𝑟=1

34. (𝐹𝑟)𝑟=𝑣𝑟√𝑔𝑟𝐿𝑟

35. 𝑎𝑥=𝑣𝑥 .𝜕𝑣𝑥𝜕𝑥 +𝑣𝑦 .𝜕𝑣𝑥𝜕𝑦 +𝑣𝑧 .𝜕𝑣𝑥𝜕𝑧 +𝜕𝑣𝑥𝜕𝑡

36. 𝑎𝑦=𝑣𝑥 .𝜕𝑣𝑦𝜕𝑥 +𝑣𝑦 .𝜕𝑣𝑦𝜕𝑦 +𝑣𝑧 .𝜕𝑣𝑦𝜕𝑧 +𝜕𝑣𝑦𝜕𝑡

37. 𝑎𝑧=𝑣𝑥 .𝜕𝑣𝑧𝜕𝑥 +𝑣𝑦 .𝜕𝑣𝑧𝜕𝑦 +𝑣𝑧 .𝜕𝑣𝑧𝜕𝑧 +𝜕𝑣𝑧𝜕𝑡

38. 𝜔=12 𝑐𝑢𝑟𝑙 𝑣=12(∇×𝑣)

39. Vorticity =2𝜔

40. Circulation =𝑣𝑜𝑟𝑡𝑖𝑐𝑖𝑡𝑦×𝑎𝑟𝑒𝑎

Fluid mechanics formulas for gate

41. 𝐹=𝜌𝑎𝑣2

42. Q1=Q2[1+cosθ] 𝑎𝑛𝑑 Q2=Q2[1−cosθ] (θ is with plate)

43. Ns=N√QH3/4 ,Ns=N√PH5/4

44. 𝑄2𝑔=𝐴3𝑇

45. 𝐹=𝑣√𝑔.𝐴𝑇

46. 𝑦𝑐3=2𝑦12𝑦22𝑦1+𝑦2 and 𝐸= 𝑦12+𝑦1𝑦2+𝑦22𝑦1+𝑦2

47. 𝑑𝑦𝑑𝑥=𝑠𝑜−𝑠𝑓1−𝑄2𝑇𝑔𝐴3

48. 𝑃1+𝑀1 = 𝑃2+𝑀2 (specific force=pressure force+momentum per sec)

49. 𝑦2𝑦1[1+𝑦2𝑦1] = 2𝑞2𝑔𝑦13 (sequent depths for rectangular channel)

50. 𝐸=(𝑦2−𝑦1)34𝑦1𝑦2 (Energy loss in jump)

Splitting tensile strength test of concrete - lab Procedure

noreply@blogger.com (Civil Study) — Fri, 22 Oct 2021 14:39:00 +0000

Splitting tensile strength is generally greater than the direct tensile strength and lower than the flexural strength (modulus of rupture). Splitting tensile strength is used in the design of structural light weight concrete members to evaluate the shear resistance provided by concrete and to determine the development length of the reinforcement.

This test method consists of applying a diametrical force along the length of a cylindrical concrete at a rate that is within a prescribed range until failure. This loading induces tensile stresses on the plane containing the applied load and relatively high compressive stresses in the area immediately around the applied load. Although we are applying a compressive load but due to Poisson’s effect, tension is produced and the specimen fails in tension. Tensile failure occurs rather than compressive failure because the areas of load application are in a state of triaxial compression, thereby allowing them to withstand much higher compressive stresses than would be indicated by a uniaxial compressive strength test result. Thin, bearing strips are used to distribute the load applied along the length of the cylinder. The maximum load sustained by the specimen is divided by appropriate geometrical factors to obtain the splitting tensile strength.

Objective: To determine splitting tensile strength of cylindrical concrete specimens.

Reference: IS: 5816 - 1999, IS: 1199-1959, SP: 23-1982, IS: 10086-1982.

Apparatus: Cylindrical mould confirming to IS: 10086-1982 for splitting tensile strength,tamping rod, metallic sheet, universal testing machine.

Material: Cement, sand, aggregate and water, grease

Procedure:

1. Sampling of Materials: Samples of aggregates for each batch of concrete shall be of the desired grading and shall be in an air-dried condition. The cement samples, on arrival at the laboratory, shall be thoroughly mixed dry either by hand or in a suitable mixer in such a manner as to ensure the greatest possible blending and uniformity in the material.

2. Proportioning : The proportions of the materials, including water, in concrete mixes used for determining the suitability of the materials available, shall be similar in all respects to those to be employed in the work.

3. Weighing: The quantities of cement, each size of aggregate, and water for each batch shall be determined by weight, to an accuracy of 0.1 percent of the total weight of the batch.

4. Mixing of concrete: The concrete shall be mixed by hand, or preferably, in a laboratory batch mixer, in such a manner as to avoid loss of water or other materials. Each batch of concrete shall be of such a size as to leave about 10 percent excess after moulding the desired number of test specimens.

5. Mould: The cylindrical mould shall be of 150 mm diameter and 300 mm height conforming to IS: 10086-1982.

6. Compacting: The test specimens shall be made as soon as practicable after mixing, and in such a way as to produce full compaction of the concrete with neither segregation nor excessive laitance.

7. Curing: The test specimens shall be stored in a place, free from vibration, in moist air of at least 90 percent relative humidity and at a temperature of 27° ±2°C for 24 hours ± ½ hour from the time of addition of water to the dry ingredients.

Splitting tensile strength test of concrete - lab Procedure

8. Placing the specimen in the testing machine: The bearing surfaces of the supporting and loading rollers shall be wiped clean, and any loose sand or other material removed from the surfaces of the specimen where they are to make contact with the rollers.

9. Two bearings strips of nominal (1/8 in i.e 3.175mm) thick plywood, free of imperfections, approximately (25mm) wide, and of length equal to or slightly longer than that of the specimen should be provided for each specimen.

10. The bearing strips are placed between the specimen and both upper and lower bearing blocks of the testing machine or between the specimen and the supplemental bars or plates.

11. Draw diametric lines an each end of the specimen using a suitable device that will ensure that they are in the same axial plane. Canter one of the plywood strips along the center of the lower bearing block.

12. Place the specimen on the plywood strip and align so that the lines marked on the ends of the specimen are vertical and centred over the plywood strip.

13. Place a second plywood strip lengthwise on the cylinder, centered on the lines marked on the ends of the cylinder. Apply the load continuously and without shock, at a constant rate within, the range of 689 to 1380 kPa/min splitting tensile stress until failure of the specimen

14. Record the maximum applied load indicated by the testing machine at failure.

Note the type of failure and appearance of fracture.

Observation

• Length of Specimen (l) : _______ mm

• diameter of the specimen (d) : _______ mm

Sl. No.

Age of specimen

Maximum load

(P) in N

Spitting tensile

strength in MPa

(𝑇 = 2𝑃/𝜋 )

Average spitting

tensile strength

(MPa)

7 days

28 days

Results:

• The average 7 days tensile strength of concrete sample is : _______ MPa

• The average 28 days tensile strength of concrete sample is : _______ MPa

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Irrigation engineering and water resources valuable formulas

noreply@blogger.com (Civil Study) — Thu, 21 Oct 2021 14:59:00 +0000

I think these formulas will be very much helpful for those candidates who are preparing for competitive examinations.

1. 𝑡=𝑦𝑓log𝑒(𝑄𝑄−𝑓𝐴)

2. 𝐴𝑚𝑎𝑥=𝑄𝑓

3. 𝑆𝐴𝑅=𝑁𝑎+√𝐶𝑎+++𝑀𝑔++2

4. Δ=8.64𝐵𝐷

5. 𝐴=𝑄𝐷

6. 𝜂𝑐=𝑊𝑎𝑡𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑𝑊𝑎𝑡𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑

7. 𝜂𝑎=𝑊𝑎𝑡𝑒𝑟 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒𝑊𝑎𝑡𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑

8. 𝜂𝑠=𝑊𝑎𝑡𝑒𝑟 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒𝑊𝑎𝑡𝑒𝑟 𝑛𝑒𝑒𝑑𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒

9. 𝜂𝑑=1−𝑑𝐷, 𝐷=𝑚𝑒𝑎𝑛 𝑜𝑓 𝑑𝑒𝑝𝑡ℎ𝑠 and 𝑑=𝑚𝑒𝑎𝑛 𝑜𝑓 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑑𝑒𝑣𝑖a𝑡𝑖𝑜𝑛𝑠

10. 𝐶𝐼𝑅=𝐶𝑢−𝑅𝑒

11. 𝑁𝐼𝑅=𝐶𝐼𝑅+𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑎𝑠 𝑝𝑒𝑟𝑐𝑜𝑙𝑎𝑡𝑖𝑜𝑛

12. 𝐶𝑢=𝑘𝑝40(1.8𝑡+32)

13. 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 𝑏𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑=𝛾𝑑𝛾𝑤 𝑑(𝐹−𝑂𝑀𝐶)

14. 𝜏𝑜=𝛾𝑤𝑅𝑆

15. 𝑛=124𝑑16, 𝑑 is in meters

16. 𝑣0=0.55 𝑚 𝑦0.64

17. 𝑣=[1𝑛+(23+0.00155𝑆)1+(23+0.00155𝑆)𝑛√𝑅]√𝑅𝑆

18. 𝑣=(𝑄𝑓2140)16

19. 𝑅=52(𝑣2𝑓)

20. 𝑃=4.75√𝑄

21. 𝑆=𝑓533340𝑄16

22. 𝐷𝑠=1.35(𝑞2𝑓)13

23. 𝑣=10.8𝑅23𝑆13

24. 𝐴=𝑦2(𝜃+cot𝜃),𝑃=2𝑦(𝜃+cot𝜃) for Lined Triangular section.

25. 𝐴=𝐵𝑦+𝑦2(𝜃+cot𝜃),𝑃=𝐵+2𝑦(𝜃+cot𝜃) for Lined Trapezoidal section.

26. Launching apron scour depth, 𝐷=𝑥𝑅−𝑤𝑎𝑡𝑒𝑟 𝑑𝑒𝑝𝑡ℎ 𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑑, 𝑅=0.47(𝑄𝑓)13

27. Length of launching apron =√5 𝐷

28. 𝑡=ℎ′𝐺 , where ℎ′=𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝐻𝐺𝐿 𝑎𝑏𝑜𝑣𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟

29. 𝑡=ℎ𝐺−1 , where ℎ=𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝐻𝐺𝐿 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟

30. 𝐶=19 √𝐷𝑏′ [𝑑+𝐷𝑏], 𝑑=𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑝𝑖𝑙𝑒 𝑎𝑛𝑑 𝐷=𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑛𝑒𝑖𝑔ℎ𝑏𝑜𝑢𝑟𝑖𝑛𝑔 𝑝𝑖𝑙𝑒 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡.

31. 𝐺𝐸=𝐻𝑑 1𝜋√𝜆 , 𝜆=1+√1+𝛼22 and 𝛼=𝑏𝑑

32. ∅𝐷=1𝜋cos−1(𝜆−2𝜆) and ∅𝐸=1𝜋cos−1(𝜆−1𝜆)

33. ∅𝐶=1𝜋cos−1(𝜆2+1𝜆1),∅𝐷=1𝜋cos−1(𝜆2𝜆1) and ∅𝐸=1𝜋cos−1(𝜆2−1𝜆1) , 𝜆1=√1+𝛼12+√1+𝛼222 and 𝜆2=√1+𝛼12−√1+𝛼222 , 𝛼1=𝑏1𝑑 𝑎𝑛𝑑 𝛼2=𝑏2𝑑

34. Non-modular modules: Drowned pipe outlet, masonry sluice and wooden shoots.

35. Semi modules or flexible modules: Pipe outlet, venturi flume or Kennedy’, open flume and orifice semi module.

36. Rigid modules: Gibb’s module, Kanna’s rigid module and foote module.

37. Flexibility 𝐹=𝑑𝑞/𝑞𝑑𝑄/𝑄=𝑚𝑛.𝑦𝐻 , 𝑞=discharge in outlet and 𝑄= discharge in channel, 𝑚=outlet index and 𝑛=channel index. 𝐻=working head of outlet and 𝑦= depth of water in channel.

Irrigation engineering and water resources valuable formulas

38. Proportionality: 𝐹=1 𝑎𝑛𝑑 𝑚𝑛=𝐻𝑦

39. Setting =𝐻𝑦

40. 𝑚𝑛>𝐻𝑦 , hyper proportional outlet.

41. 𝑚𝑛<𝐻𝑦 , sub proportional outlet.

42. Sensitivity, 𝑆=𝑑𝑞/𝑞𝑑𝑦/𝑦=𝑛𝑑𝑞/𝑞𝑑𝑄/𝑄=𝑛 𝐹

43. Aqueduct: Canal over drainage with clear gap.

44. Syphon Aqueduct: Canal over drainage with syphonic action.

45. Super passage: Drain over canal with clear gap.

46. Canal Syphon: Drain over canal with syphonic action.

47. Principal stress in dam

𝜎=𝑝𝑣sec2𝛼−𝑝′tan2𝛼

where 𝑝𝑣=𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 and 𝑝′= water pressure of tail water and 𝛼=𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑎𝑚 𝑓𝑎𝑐𝑒 𝑤𝑖𝑡ℎ 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙

48. 𝐵=𝐻√𝑆𝑐−𝐶

49. 𝐻=𝑓𝛾𝑤(𝑆𝑐−𝐶+1) and 𝐻𝑚𝑎𝑥=𝑓𝛾𝑤(𝑆𝑐+1)

50. 𝑥1.85=2 𝐻𝑑0.85𝑦

Test for Bricks || civil study

noreply@blogger.com (Civil Study) — Wed, 20 Oct 2021 16:29:00 +0000

Brick is a very common construction material obtained by molding clay in rectangular blocks of uniform size and then by drying and burning them at a required temperature. Due to high strength and durability, easy availability and low cost; they are nowadays widely used for building construction.

On the basis of their size, IS 1077:1992 classifies bricks into two categories, i.e. modular and non-modular type. The sizes of modular brick are selected in conformity with the metric system considering 100 mm module as the basis of all dimensional standardization. The standard modular sizes of Indian bricks are:

Without mortar : 190 mm × 90 mm × 90 mm

With mortar : 200 mm × 100 mm × 100 mm

However, bricks of non-modular sizes are also available in India, which satisfies other requirements of the code, but not the requirements regarding dimension. The standard sizes of non-modular bricks varies region to region basis. In Odisha the standard size of non modular bricks available is 225 mm × 125 mm × 75 mm (without mortar). This size may vary slightly due to drying shrinkage. To assess the size of the brick, at least twenty numbers of whole bricks is taken at random from the stock. All blisters, loose particles of clay and small projections shall be removed. Then they shall be arranged upon a level surface successively as shown in Figure 10 in contact with each other and in a straight line. The overall length of the assembled bricks shall be measured with a steel tape or other suitable inextensible measure sufficiently long to measure the whole row at one stretch. Measurement by repeated application of short rule or measure shall not be permitted. For a good quality of brick, tolerances in dimensions are allowed within ± 3.0 %.

Bricks used in construction work should have adequate compressive strength to resist lateral and vertical loads. Ordinary bricks are designated on the basis of average compressive strength as follows:

Class

17.5

12.5

7.5

Average compressive

strength (N/mm2)

17.5

12.5

7.5

The compressive strength of any individual brick tested shall not fall below the minimum compressive strength specified for the corresponding class of brick. To access the compressive strength of the bricks, load is applied over the flat side keeping mortar filled face facing upwards.Water absorption of a brick is defined as the ratio of weight of water absorbed to the dry weight of the unit under a given method of treatment in a standard period of time. Water absorption indicates degree of porosity in a brick. Strength, stiffness, unit weight and other properties decrease with porosity. For good quality of bricks, after immersion in cold water for 24 hours, the water absorption should not be more than 20% by weight Efflorescence of bricks is usually seen as a white powder (salts of crystallization) caused by water soluble salts as Sulphates of Calcium, Magnesium, Sodium, Potassium etc. and Sodium Chloride. These salts are deposited on the surface of the bricks on the evaporation of water. Efflorescence decreases strength and stiffness of bricks. The liability to efflorescence is reported as ‘nil’, ‘slight’, ‘moderate’, ‘heavy’ or ‘serious’ in accordance with the following definitions:

Test for Bricks || civil study

Ratings of efflorescence

Efflorescence Definitions

Nil No perceptible deposit of efflorescence Slight More than ten percent of the exposed area of the brick is covered with thin deposit of salts.Moderate Heavier deposit than under “slight” and covering up to fifty percent of exposed area of the brick surface but unaccompanied by powdering or flaking of the surface.Heavy Heavy deposit of salt covering more than fifty percent or more of the exposed area of the brick surface but unaccompanied by powdering or flaking of the surface.Serious Heavy deposit of salt accompanied by powdering or flaking of the exposed surface.For good quality of bricks, the rating of efflorescence should not be more than ‘slight’.

Objective: To estimate dimension, tolerance, compressive strength, water absorption, and efflorescence of bricks.

Reference: IS 3495 (Part-1 to 4):1992.

Apparatus: (1) Dimensions and tolerance: Measuring tape, trowel and brush.

(2) Water absorption: Weighing balance, metal tray about 5 cm deep, ovens.

(3) Compressive strength: Compression testing machine

(4) Efflorescence: A shallow flat bottom dish or tray.

Material: (1) Dimensions and tolerance: Bricks (20 nos.)

(1) Water absorption: Bricks and water.

(2) Compressive strength: Bricks, cement, sand, water and two 3 mm thick plywood sheets.

(3) Efflorescence: Bricks and water.

Procedure:

For dimension and tolerance test:

1. Collect at least twenty numbers of whole bricks at random from the stock.

2. Remove all blisters, loose particles of clay and small projections from the surface of the brick.

3. Arrange the bricks upon a level surface successively as shown in Figure 1 in contact with each other and in a straight line.

4. Measure the overall length of the assembled bricks with the steel tape.

Measurement by repeated application of short rule or measure shall not be permitted. If, for any reason it is found impracticable to measure bricks in one row, then divide the samples into rows of 10 bricks each and measure them separately to the nearest millimetre. Then, add all these dimensions together.

For water absorption test:

5. Dry the specimen in a ventilated oven at a temperature of 105 to 115o C for 24 hours. Then cool it to the room temperature and determine its weight (W1).

6. Immerse the dried specimen completely in clean water at a temperature of 27 ±2 o C for 24 hours. Then remove the specimen, wiped of any traces of water and measure its weight (W2). This weighing shall be completed within three minutes after the specimen has been removed from water. Determine percentage of water absorption.

For compressive strength test:

7. Remove the unevenness observed in the bed faces to provide two smooth and parallel faces by grinding. Immerse in water at room temperature for 24 hours.Remove the specimen and drain out any surplus moisture at room temperature.

8. Fill the frog and all voids in the bed face flush with cement mortar having cement sand ratio 1:3.

9. Store under damp gunny bags for 24 hours followed by immersion in water for 3 days. Remove the bricks and wipe out traces of moisture.

10. Place the specimen with mortar filled face upward, between two 3 mm thick plywood sheets, carefully centred between platens of the testing machine.Apply load axially at a uniform rate of 14 N/mm2/min and note the maximum load at failure. Find compressive stress after dividing total load (in N) by loaded surface area (mm2)

For efflorescence test:

11. Place the end of the brick in the disk or tray. The depth of immersion in water shall be 25 mm.

12. Keep the whole arrangement at a temperature 20oC to 30oC until all the water in the dish is absorbed by the specimens and the surplus water evaporates.

13. When the water has been evaporated and the bricks appear to be dry, place a similar quantity of water in the dish and allow it to dry evaporate as before.

14. Examine the bricks for efflorescence after the second evaporation and report the results as the definition given in Table 3.

Observations:

Table 1: Observations on dimension and tolerance

Nos. of bricks tested : ________

Standard sizes of bricks taken: ______ mm × ______ mm × ______ mm.

	Length (mm)	Width (mm)	Thickness (mm)
For 20 nos. of bricks
For One bricks
Tolerance (in %)

Table 2: Observations on compressive strength test

Sl.

No.

Length

(mm)

Width (mm)

Area (mm2)

Failure

load (N)

Compressive

strength (N/mm2)

Table 3: Observations on water absorption

Sl.

No.

Dry weight of brick in

kg (W1)

Immersed weight of

brick in kg (W2)

%age of water absorption

𝑾𝟐−𝑾𝟏 /

𝑾𝟏

× 𝟏00 .

Table 4: Observations on efflorescence test

The percentage of area where efflorescence occurred is : ________ %. Thus the degree of efflorescence is ___________.

Note: Carefully observe the brick that have undergone efflorescence test. And estimate the percentage of efflorescence area.

Result:

• The dimensions of given bricks are : ______ mm × ______ mm × ______ mm

• Compressive strength for the brick specimen is : __________ N/mm2

• Percentage of water absorption for the brick specimen is : ________ % by weight.

• Degree of efflorescence for the brick specimen is found : __________

Conclusions:

• The given bricks has dimensions ______ mm × ______ mm × ______ mm and the tolerances in dimensions are bellow/ above 3.0%. Thus the bricks are of good quality/are not of good quality.

• The lowest compressive strength observed by the bricks specimen is __________N/mm2. Thus, the brick belongs to _______ class.

• The percentage of water absorption for the brick specimen is less than/ more than 20% by weight. Thus, bricks are of good quality/ are not of good quality.

• The rating of efflorescence for the brick specimen is _______. This is acceptable/not acceptable for good quality of bricks.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Transportation engineering basic formulas

noreply@blogger.com (Civil Study) — Tue, 19 Oct 2021 14:30:00 +0000

A handy reference for use in analysis and design of Transportation Engineering.

1. 𝑦=2𝑥2𝑛𝑊 (parabolic camber)

2. 𝑆𝑆𝐷 =𝑣𝑡+𝑣22𝑔(𝑓±𝑛) , 𝐼𝑆𝐷=2 ×𝑆𝑆𝐷

3. 𝑂𝑆𝐷=𝑑1+𝑑2+𝑑3 , 𝑑1=𝑣𝑏 𝑡, 𝑑2=2𝑠+𝑣𝑏𝑇 , 𝑑3=𝑣𝑇 , 𝑇=√4𝑠/𝑎

4. 𝑣2𝑅𝑔=𝑒+𝑓 , 𝑊𝑒=𝑛𝑙22𝑅+𝑉9.5√𝑅

5. 𝐿𝑠=𝑣3𝐶𝑅 (comfort condition)

6. 𝐿𝑠=𝑒𝑁(𝑊+𝑊𝑒) (rotated about inner edge)

7. 𝐿𝑠=2.7𝑉2𝑅 for plain and rolling terrain, 𝐿𝑠=𝑉2𝑅 for steep and mountainous terrain.

8. IRC recommends Spiral as transition curve.

9. 𝑆=𝐿𝑠224𝑅

10. 𝑚=𝑅(1−cos𝜃2), 𝜃=𝑆/𝑅 (single lane, 𝐿>𝑆)

11. 𝑚=𝑅(1−cos𝜃2)+𝑆−𝐿2sin𝜃2 , 𝜃=𝐿𝑅 (Single lane, 𝐿<𝑆)

12. Curve resistance =𝑇(1−cos𝛼)

13. Grade compensation =30+𝑅𝑅 or 75𝑅 whichever is less

14. 𝐿=𝑁𝑠2(√2ℎ+√2𝐻)2 when 𝐿>𝑠 (Summit curve for SSD), ℎ=0.15 𝑎𝑛𝑑 𝐻=1.2

15. 𝐿=2𝑠−(√2ℎ+√2𝐻)2𝑁 when 𝐿<𝑠 (Summit curve for SSD), ℎ=0.15 𝑎𝑛𝑑 𝐻=1.2

16. 𝐿=𝑁𝑠2(√2𝐻+√2𝐻)2 when 𝐿>𝑠 (Summit curve for OSD), 𝐻=1.2

17. 𝐿=2𝑠−(√2𝐻+√2𝐻)2𝑁 , 𝐿<𝑠 (Summit curve for OSD), 𝐻=1.2

18. 𝐿=𝑁𝑠22ℎ+2𝑠tan𝛼 when 𝐿>𝑠 (valley curve), ℎ=0.75 𝑎𝑛𝑑 𝛼=1°

19. 𝐿=2𝑠−2ℎ+2𝑠tan𝛼𝑁 , 𝐿<𝑠 (valley curve), ℎ=0.75 𝑎𝑛𝑑 𝛼=1°

20. 𝐿=2(𝑁𝑣3𝐶)12 (valley curve comfort condition)

21. 𝐶=8075+𝑉

22. 𝐹𝑙𝑎𝑘𝑖𝑛𝑒𝑠𝑠 𝐼𝑛𝑑𝑒𝑥=𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑎𝑘𝑦 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑡ℎ𝑖𝑛𝑛𝑒𝑟 𝑡ℎ𝑎𝑛 0.6 𝑑𝑚𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡×100

23. 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑑𝑒𝑥=𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑎𝑘𝑦 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑙𝑜𝑛𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 1.8 𝑑𝑚𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡×100

24. Angularity number=67−100(𝑊𝐺)𝐶, C is weight water in the cylinder, W is weight of aggregate packed in the cylinder.

25. Penetration test unit is 1/10th mm. Weight used 100 grams. Temperature 25℃.

26. 𝐺𝑎=𝑊1+𝑊2+𝑊3𝑊1𝐺1+𝑊2𝐺2+𝑊3𝐺3 and 𝐺𝑡=𝑊𝑎+𝑊𝑏𝑊𝑎𝐺𝑎+𝑊𝑏𝐺𝑏

27. 𝐺𝑚=𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑜𝑢𝑙𝑑𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑜𝑢𝑙𝑑

28. 𝑣𝑎=𝐺𝑡−𝐺𝑚𝐺𝑡 , 𝑣𝑏=𝑊𝑏𝑊𝑚×𝐺𝑚𝐺𝑏 and 𝑉𝑀𝐴=𝑣𝑎+𝑣𝑏

29. 𝑉𝐹𝐵=𝑣𝑏𝑉𝑀𝐴

30. Flow value units 1/4th mm

31. 𝑁=365𝐴[(1+𝑟)𝑛−1𝑛 ]𝑉𝐷𝐹×𝐿𝐷𝐹 and 𝐴=𝑃(1+𝑟)𝑥

32. 𝐿𝐷𝐹=0.75 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑙𝑎𝑛𝑒𝑠 𝑎𝑛𝑑 0.4 𝑓𝑜𝑟 𝑓𝑜𝑢𝑟 𝑙𝑎𝑛𝑒 (single carriageway)

33. LDF=0.75 for two lanes and 0.60 for three lane and 0.45 for four lane (dual carriageway)

34. 𝑉𝐷𝐹=(𝑃80)4, where P is in kN

35. 𝐶𝐵𝑅 𝑎𝑡 2.5 𝑚𝑚=𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑘𝑔/𝑐𝑚270=𝑙𝑜𝑎𝑑 𝑖𝑛 𝑘𝑔1370

36. 𝐶𝐵𝑅 𝑎𝑡 5 𝑚𝑚=𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑘𝑔/𝑐𝑚2105=𝑙𝑜𝑎𝑑 𝑖𝑛 𝑘𝑔2055

37. 𝐸𝑆𝑊𝐿: interpolate load for depth from line joining (𝑙𝑜𝑔𝑃,log𝑧2) and (𝑙𝑜𝑔2𝑃,log2𝑠).

38. Radius of relative stiffness 𝑙=[𝐸ℎ312𝑘(1−𝜇2)]14

39. 𝑘=𝑝 (𝑘𝑔𝑐𝑚2)0.125 𝑐𝑚 (modulus of subgrade reaction)

40. 𝑘=𝐸1.18 𝑎 (𝑎 is rigid plate radius)

41. 𝑘=𝐸1.5 𝑎 (𝑎 is flexible plate radius)

42. 𝑘1𝑎1=𝑘2𝑎2

43. 𝑅𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟=𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑡𝑦𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (below 7 𝑘𝑔/𝑐𝑚2 contact pressure is more)

44. 𝐿𝛼𝑡=𝛿2 (expansion joint)

45. 𝐿𝑐2𝑏ℎ𝛾𝑐𝑓=𝑆𝑐 𝑏ℎ

46. 𝑏ℎ𝛾𝑐𝑓=𝐴𝑠𝑡 𝑆𝑠 (tie bar area of steel per meter)

47. 𝐿2𝜋∅𝜏𝑏𝑑=𝜋∅24𝑆𝑠 (length of tie bar)

48. 𝑃(𝑥)=(𝜆𝑡)𝑥𝑒−𝜆𝑡𝑥! (Poisson distribution)

49. 𝑃(ℎ≥𝑡)=𝑒−𝜆𝑡

50. 𝑘=𝑘𝑗(1−𝑣𝑣𝑓) and 𝑣=𝑣𝑓(1−𝑘𝑘𝑗)

51. 𝑞=𝑘𝑣 and 𝑘=1000𝑠𝑝𝑎𝑐𝑖𝑛𝑔

52. 𝑃𝐻𝐹=𝑝𝑒𝑎𝑘 ℎ𝑜𝑢𝑟 𝑓𝑙𝑜𝑤4 ×𝑝𝑒𝑎𝑘 15 𝑚𝑖𝑛 𝑓𝑙𝑜𝑤 or 𝑃𝐻𝐹=30𝑡ℎ ℎ𝑜𝑢𝑟𝑙𝑦 𝑣𝑜𝑙𝑢𝑚𝑒𝐴𝐴𝐷𝑇

53. 𝑇𝑖𝑚𝑒 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑=𝑣1+𝑣2+⋯+𝑣𝑛𝑛

54. 𝑆𝑝𝑎𝑐𝑒 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑=𝑑𝑡1+𝑡2+⋯+𝑡𝑛𝑛

55. 𝑞=𝑛𝑎+𝑛𝑦𝑡𝑎+𝑡𝑤, 𝑡̅=𝑡𝑤−𝑛𝑦𝑞

56. Safe speed limit is 85th percentile speed

57. Geometric design is based on 98th percentile speed.

58. Road side facilities are based on 30th highest hourly volume.

59. 𝐶0=1.5𝐿+51−𝑌0

60. 𝑄=280𝑤 (1+𝑒𝑤)(1−𝑝3)(1+𝑤𝑙)

Transportation engineering basic formulas

61. 𝑇𝑟=𝑇𝑎+𝑇𝑚−𝑇𝑎3

62. 𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛=7 % 𝑓𝑜𝑟 300 𝑚

63. 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑎𝑖𝑟𝑝𝑜𝑟𝑡,𝑇𝑠=15−6.51000ℎ

64. Temperature correction =1% 𝑝𝑒𝑟 1℃ 𝑟𝑖𝑠𝑒 𝑜𝑓 𝑇𝑟 𝑎𝑏𝑜𝑣𝑒 𝑇𝑠

65. Gradient correction =20% 𝑝𝑒𝑟 1% 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡.

66. 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 𝑅=𝑉2125𝑓

67. 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 𝑅=0.388𝑊2𝑇2−𝑆 where 𝑠=6+𝑡𝑟𝑒𝑎𝑑2

68. Turning radius for subsonic aircraft is 120 m and for supersonic it is 180 m

69. Grade compensation for BG is 0.04%, for MG is 0.03% and for NG is 0.02% per degree of curve.

70. 𝐷=1720𝑅

71. 𝑉𝑒𝑟𝑠𝑖𝑛𝑒 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒=𝑐28𝑅

72. 𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑎𝑛𝑡=𝑉2127𝑅×𝐺

73. Theoretical cant =𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑎𝑛𝑡+𝑐𝑎𝑛𝑡 𝑑𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

74. Widening of gauge in cm, 𝑑 =13(𝐵+𝐿)2𝑅 where B is wheel base in m, lap of flange in m, 𝐿=0.02√ℎ2+𝐷ℎ , h is depth of wheel flange below rail top level, D dia of wheel in cm.

75. 𝐿𝑠=3.28𝑣3𝑅 (Transition curve)

76. 𝑦=𝑥36𝐿𝑅 (Transition curve)

77. Usually adopted transition curve for railways is cubic parabola.

Geotechnical engineering basic formulas

noreply@blogger.com (Civil Study) — Mon, 18 Oct 2021 10:28:00 +0000

A handy reference for use in geotechnical analysis and design Field Given Below :

1. 𝐼𝑓=𝑊1−𝑤2log10𝑁2/𝑁1

2. 𝐼𝑡=𝐼𝑝𝐼𝑓

3. 𝐶𝑢=𝐷60𝐷10

4. 𝐶𝑐=𝐷302𝐷60𝐷10

5. 𝐼𝑃=0.73 (𝑤𝐿−20)

6. 𝐼𝑃=𝑤𝐿−𝑤𝑃

7. 𝐼𝑠=𝑤𝑃−𝑤𝑆

8. 𝐼𝐿=𝑤−𝑤𝑃𝐼𝑃

9. 𝐼𝑐=𝑤𝐿−𝑤𝐼𝑃

10. 𝐴=𝐼𝑝𝐹 where 𝐹 is clay fraction (Activity)

11. 𝑅𝐷=𝑒𝑚𝑎𝑥−𝑒𝑒𝑚a𝑥−𝑒𝑚𝑖𝑛=1/𝛾𝑑,𝑚𝑖𝑛−1/𝛾𝑑1/𝛾𝑑,𝑚𝑖𝑛−1/𝛾𝑑,𝑚𝑎𝑥

12. 𝑘1𝑘2=tan𝛼1tan𝛼2 (non homogeneous)

13. 𝑘=𝐶𝛾𝑤𝜇𝑒31+𝑒 𝑑2

14. 𝐾=𝑘𝜇𝛾𝑤 (absolute permeability)

15. 𝑘=𝑞𝜋 ln(𝑟2/𝑟1)𝑧22−𝑧12 (Permeability in unconfined aquifer)

16. 𝑘=𝑞2𝜋𝑏 ln(𝑟2/𝑟1)𝑧2−𝑧1 (Permeability in confined aquifer)

17. 𝑘ℎ=𝑘1𝐻1+𝑘2𝐻2𝐻1+𝐻2 (effective horizontal permeability in stratified soils)

18. 𝑘𝑣=𝐻1+𝐻2𝐻1𝑘1+𝐻2𝑘2 (effective vertical permeability in stratified soils)

19. 𝑘𝑒=√𝑘ℎ𝑘𝑣 (effective permeability)

20. 𝑘=𝑎𝐿𝐴𝑡lnℎ1ℎ2 (falling head permeability test)

21. 𝑘=𝑞𝐿𝐴ℎ (constant head permeability test)

22. 𝑞=𝑘𝑒 ℎ𝑁𝑓𝑁𝑑 (seepage discharge)

23. 𝜎𝑧=3𝑄2𝜋1𝑧2 (11+(𝑟𝑧)2 )52 (Boussinesq’s formula)

24. 𝜎𝑧=𝑐𝑄2𝜋1𝑧2 (1𝑐2+(𝑟𝑧)2 )32 where 𝑐=√1−2𝜇2−2𝜇 (Wesrwegaard’s formula)

25. 𝜎𝑧=2𝑞𝜋𝑧 (11+(𝑥𝑧)2)2 (line load)

26. 𝜎𝑧=𝑞𝜋(2𝜃+sin2𝜃) where 𝜃=tan−1𝑏𝑧 (stress under centre of strip load of width 2𝑏 )

27. 𝜎𝑧=𝑞𝜋(2𝜃+sin2𝜃sin2∅) where 2𝜃=𝛽1−𝛽2 𝑎n𝑑 2∅=𝛽1+𝛽2 ( strip eccentric point)

28. 𝜎𝑧=𝑞(1−cos3𝜃) where 𝜃=tan−1𝑅𝑧 (stress under centre of circular load)

29. sin∅=𝜎1−𝜎3𝜎1+𝜎3 (for cohesion less soils)

30. sin∅=(𝜎1−𝜎3)/2𝑐cot∅+(𝜎1+𝜎3)/2 (for cohesive soils)

31. 𝜎1=2𝑐tan𝛼+𝜎3tan2𝛼 where 𝛼=45+∅2

32. tan∅=𝜏𝜎 (shear box test for cohesion less soils)

33. 𝑇=𝑐 𝜋𝐷2(𝐻2+𝐷6) (if both top and bottom surfaces contributes)

34. 𝑇=𝑐 𝜋𝐷2(𝐻2+𝐷12) (if only bottom surface contribute)

35. 𝑆𝑖=𝑞 𝐵 1−𝜇2𝐸 𝐼𝑓 (immediate settlement )

36. 𝑆𝑓=𝑆𝑝(𝐵𝑓𝐵𝑝 𝐵𝑝+0.3𝐵𝑓+0.3)2 (settlement of footing based on plate settlement)

37. Δ𝑢=𝐵(Δ𝜎𝑐)+𝐴𝐵 (Δ𝜎𝑑) (Skempton’s pore pressure parameters)

38. 𝑞=𝑚𝑃+𝑘 is stress path equation where ∅=tan−1𝑚 and 𝑐=𝑘/cos∅

39. 𝐶𝑐=Δ𝑒log10𝜎0+Δ𝜎𝜎0

40. 𝐶𝑐=0.009 (𝑤𝐿−10) (for normally consolidated soil)

41. 𝐶𝑐=0.007 (𝑤𝐿−10) (for over consolidated soil)

42. Δ𝑒1+𝑒0=Δ𝐻𝐻

43. 𝑚𝑣=Δ𝐻/𝐻Δ𝜎0

44. 𝑐𝑣=𝑘𝛾𝑤𝑚𝑣

45. 𝑇𝑣=𝑐𝑣𝑡𝑑2

46. 𝑇𝑣=𝜋4𝑈2 when 𝑈≤0.6

47. 𝑇𝑣=−0.933 log10(1−𝑈)−0.085 when 𝑈>0.6

48. 𝑆𝑓=𝐶𝑐𝐻1+𝑒0 log10𝜎0+Δ𝜎𝜎0

49. 𝑆𝑓=𝐶𝑟𝐻1+𝑒0 log10𝜎𝑐𝜎0+𝐶𝑐𝐻1+𝑒0 log10𝜎0+Δ𝜎𝜎𝑐

50. 𝐴𝑟=𝐷02−𝐷𝑖2𝐷𝑖2

51. 𝑆𝑛=𝑐𝑢𝐹𝛾𝐻

52. 𝑞𝑢=𝑐𝑁𝑐+𝑞 𝑁𝑞+0.5 𝛾 𝐵 𝑁𝛾 (Terzaghi’s strip)

53. 𝑞𝑢=1.3 𝑐𝑁𝑐+𝑞 𝑁𝑞+0.4 𝛾 𝐵 𝑁𝛾 (Terzaghi’s square)

54. 𝑞𝑢=1.3 𝑐𝑁𝑐+𝑞 𝑁𝑞+0.3 𝛾 𝐵 𝑁𝛾 (Terzaghi’s circle)

55. 𝑞𝑢=(1+0.3𝐵𝐿) 𝑐𝑁𝑐+𝑞 𝑁𝑞+(1−0.2𝐵𝐿)0.5 𝛾 𝐵 𝑁𝛾 (Terzaghi’s rectangle)

56. 𝑞𝑢=𝑐𝑁𝑐𝑆𝑐𝑑𝑐𝑖𝑐+𝑞 𝑁𝑞𝑆𝑞𝑑𝑞𝑖𝑞+0.5 𝛾 𝐵′ 𝑁𝛾 𝑆𝛾𝑑𝛾𝑖𝛾 (Meyerhof)

𝐵′=𝐵−2𝑒𝑥 and 𝐿′=𝐿−2 𝑒𝑦

57. 𝑞𝑛𝑢=𝑐𝑁𝑐 (Skempton

𝑁𝑐=5(1+0.2𝐷𝑓𝐵)(1+0.2𝐵𝐿)

Limiting value of 𝐷𝑓/𝐵 𝑖𝑠 2.5

58. 𝑄𝑢=𝑊ℎ 𝜂ℎ𝑆+𝐶 (ENR) where 𝐶=2.54 𝑐𝑚 𝑓𝑜𝑟 𝑑𝑟𝑜𝑝 ℎ𝑎𝑚𝑚𝑒𝑟 𝑎n𝑑 0.254 𝑐𝑚 𝑓𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 ℎ𝑎𝑚𝑚𝑒𝑟

Geotechnical engineering basic formulas

59. 𝑄𝑢=𝑊ℎ𝜂ℎ𝜂𝑏𝑆+𝐶2 (Hiley)

where 𝐶=𝐶1+𝐶2+𝐶3

𝐶1=9.05𝑅𝐴 with dolley and 𝐶1=1.77𝑅𝐴 without dolley and 𝐶2=0.657𝑅𝐿𝐴 𝐶3=3.55𝑅𝐴 𝐿=𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑖𝑙𝑒 𝑖𝑛 𝑚 𝑅=𝑃𝑖𝑙𝑒 𝑐𝑎𝑝𝑎𝑐𝑖t𝑦 𝑖𝑛 𝑡𝑜𝑛𝑛𝑒𝑠=0.1𝑄 𝐴=𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑖𝑙𝑒 𝑖𝑛 𝑐𝑚2

𝜂𝑏=𝑊+𝑒2𝑃𝑊+𝑃 when 𝑊>𝑃

𝜂𝑏=𝑊+𝑒2𝑃𝑊+𝑃−(𝑊−𝑒𝑃𝐸+𝑃)2 when 𝑊<𝑃𝑒

60. 𝑄𝑢=𝑊ℎ𝜂ℎ𝑆+𝑆02 (Danish) 𝑆0=√2𝜂ℎ𝑊ℎ𝐿𝐴𝐸

61. 𝑄𝑢=𝐴𝑝𝑐𝑁𝑐+𝐴𝑠𝛼 𝑐 (clays)

62. 𝑄𝑢=𝐴𝑝𝑐𝑁𝑐+𝐴𝑠 𝜆(𝜎̅+2𝑐) (clays)

63. 𝑄𝑢=𝐴𝑝𝜎̅ 𝑁𝑞+𝐴𝑠𝜎̅ 𝑘tan𝛿 (sands) 𝜎̅ 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑢𝑝𝑡𝑜 15 𝑑 𝑑𝑒𝑝𝑡ℎ

64. 𝑄𝑢=𝑁(𝐴𝑝𝑐𝑁𝑐+𝐴𝑠𝛼 𝑐) or 𝑄𝑢=(𝐴𝑔𝑝𝑐𝑁𝑐+𝐴𝑔𝑠 𝑐) (Group)

65. 𝑝𝑎=𝑘𝑎𝜎̅−2𝑐√𝑘𝑎+𝑢

66. 𝑝𝑝=𝑘𝑝𝜎̅+2𝑐√𝑘𝑝+𝑢

67. 𝑘𝑎=1−sin∅1+sin∅ and 𝑘𝑝=1+sin∅1−sin∅

68. 𝐻𝑐=2𝑐𝛾√𝐾𝑎 and unsupported vertical cut =2𝐻𝑐

69. 𝑘𝑎=sin2(𝛽+∅)sin2𝛽sin(𝛽−𝛿) (1+√sin(∅+𝛿)sin(∅−𝑖)sin(𝛽−𝛿)sin(𝛽+𝑖))2 (Coulomb’s active )

70. 𝑘𝑝=sin2(𝛽−∅)sin2𝛽sin(𝛽+𝛿) (1−√sin(∅+𝛿)sin(∅+𝑖)sin(𝛽+𝛿)sin(𝛽+𝑖))2 (Coulomb’s passive )

71. 𝑘𝑎=cos𝛽−√cos2𝛽−cos2∅cos𝛽+√cos2𝛽−cos2∅ and 𝑃𝑎=𝑘𝑎𝛾ℎ22cos𝛽 (Inclined backfill)

72. 𝑘𝑝=cos𝛽+√cos2𝛽−cos2∅cos𝛽−√cos2𝛽−cos2∅ and 𝑃𝑝=𝑘𝑝𝛾ℎ22cos𝛽 (Inclined backfill)

73. 𝑁𝑐=15+12(𝑁−15) 𝑤ℎ𝑒𝑛 𝑁>15 𝑎𝑛𝑑 𝑁𝑐=𝑁 𝑤ℎ𝑒𝑛 𝑁≤15 (dilatancy)

74. 𝑖𝑐=𝐺−11+𝑒 (Quick sand condition)

Civil Engineering Basic formula for structure analysis

noreply@blogger.com (Civil Study) — Sat, 16 Oct 2021 10:32:00 +0000

Looking at your requirement, we are sharing with you some of Important Civil Engineering Formulas & Shortcuts for Competitive Exam as well as Engineering exam.

Strength of material

1. 𝜇 = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

2. 𝑒𝑥 =𝑓𝑥𝐸 −𝜇𝑓𝑦𝐸 − 𝜇𝑓𝑧𝐸

3. 𝑒𝑦= 𝑓𝑦𝐸 − 𝜇𝑓𝑥𝐸 − 𝜇𝑓𝑧𝐸

4. 𝑒𝑧= 𝑓𝑧𝐸 − 𝜇𝑓𝑥𝐸 − 𝜇𝑓𝑦𝐸

5. 𝐺= 𝐸2(1+𝜇)

6. 𝐾=𝐸 3(1−2𝜇)

7. 𝑀𝐼=𝑓𝑦=𝐸𝑅

8. 𝜏 =VQIb

9. 𝑇𝐽 = 𝜏𝑟 = 𝐺𝜃𝐿

10. 𝑀𝑒 = 12[𝑀+√𝑀2+𝑇2]

11. 𝑇𝑒= √𝑀2+𝑇2

12. 𝜎= 3𝑤𝑙2𝑛𝑏𝑡2 and 𝛿 = 𝜎𝑙24𝐸𝑡 (Leaf spring)

13. 𝛿 = 64𝑊𝑅3𝑛𝐺𝑑4 (Helical spring)

14. 𝜎𝑛=𝜎𝑥+𝜎𝑦2+𝜎𝑥−𝜎𝑦2cos2𝜃+𝜏sin2𝜃

15. 𝜏𝑡=−(𝜎𝑥−𝜎𝑦2)sin2𝜃+𝜏cos2𝜃

16. 𝜎1=𝜎𝑥+𝜎𝑦2+√(𝜎𝑥−𝜎𝑦2)2+𝜏2

17. 𝜎3=𝜎𝑥+𝜎𝑦2−√(𝜎𝑥−𝜎𝑦2)2+𝜏2

18. 𝑃= 𝜋2𝐸𝐼𝑙2 (hinged and hinged)) ,𝑃 = 𝜋2𝐸𝐼4𝑙2 (fixed and free)

19. 𝑃 = 4𝜋2𝐸𝐼𝑙2 (Fixed and fixed) ,𝑃 = 2𝜋2𝐸𝐼𝑙2

20. 𝑃 = 𝜎𝑐.𝐴1+𝛼(𝑙𝑒𝐾)2 where 𝛼=𝜎𝑐𝜋2𝐸 is Rankine’s constant

Engineering Formula

Structural analysis

1. 𝑀̅𝐴𝐵=−𝑃𝑎𝑏2𝑙2 and 𝑀̅𝐴𝐵=𝑃𝑎2𝑏𝑙2 (Point load)

2. 𝑀̅𝐴𝐵=−𝑤𝑙212 and 𝑀̅𝐴𝐵=𝑤𝑙212 (udl)

3. 𝑀̅𝐴𝐵=−𝑤𝑙230 and 𝑀̅𝐴𝐵=𝑤𝑙220 (uvl from left to right increase)

4. 𝑀̅𝐴𝐵=𝑀𝑏𝑙2(3𝑎−𝑙) and 𝑀̅𝐴𝐵=𝑀𝑎𝑙2(3𝑏−𝑙) (clockwise moment)

5. 𝑈=𝑃2𝐿2𝐴𝐸 , 𝑈=∫𝑀22𝐸𝐼𝐿0 𝑑𝑥 and 𝑈=𝑇2𝐿2𝐺𝐽

6. 𝑈=1.2∫𝑉22𝐺𝐴𝐿0𝑑𝑥 (rectangular beam)

7. 𝐻= ∫𝑀𝑠𝑦𝐸𝐼 𝑑𝑥+∝𝑡𝑙−𝛿𝑙0∫𝑦2𝑑𝑥𝐸𝐼𝑙0 (two hinged arch)

8. H=𝑤𝑙28𝑑 (cable with udl) and Length of cable 𝐿= 𝑙+83𝑑2𝑙

9. 𝑀𝐴𝐵=𝑀̅𝐴𝐵+2𝐸𝐼𝐿(2𝜃𝐴+𝜃𝐵−3Δ𝐿) , right support sinks by Δ

10. 𝑀𝐵𝐴=𝑀̅𝐵𝐴+2𝐸𝐼𝐿(2𝜃𝐵+𝜃𝐴−3Δ𝐿), right support sinks by Δ

11. 𝑀𝐴 𝑙1+2𝑀𝐵(𝑙1+𝑙2)+𝑀𝐶𝑙2=−6𝑎1𝑥1𝑙1−6𝑎2𝑥2𝑙2

12. 𝛿=Σ𝑃𝑘𝐿𝐴𝐸

13. 𝑄=−Σ𝑃𝑘𝐿𝐴𝐸Σ𝑘2𝐿𝐴𝐸+𝐿0𝐴0𝐸0 and 𝑄=−Σ(𝑃𝐿𝐴𝐸+𝐿𝛼𝑡)𝑘Σ𝑘2𝐿𝐴𝐸+𝐿0𝐴0𝐸0

14. 𝑘𝑖𝑗 =𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑗, 𝛿𝑖𝑗=𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑗

15. Far end fixed, Transverse displacement, 𝛿=𝐿312𝐸𝐼 and 𝑘=12𝐸𝐼𝐿3

16. Far end hinged, Transverse displacement, 𝛿=𝐿33𝐸𝐼 and 𝑘=3𝐸𝐼𝐿3

RCC

1. 𝑓𝑚=𝑓𝑐𝑘 +1.65 𝜎

2. 𝐸𝑐=5000 √𝑓𝑐𝑘 and 𝑓𝑐𝑟=0.7 √𝑓𝑐𝑘

3. 𝜀𝑠𝑡=0.002+0.87 𝑓𝑦𝐸𝑠

4. 𝐸𝑠=2× 105 𝑀𝑃𝑎 , 𝑚=2803𝜎𝑐𝑏𝑐

5. 𝑏𝑓=𝑙06+𝑏𝑤+6𝐷𝑓 for T-beams.

6. 𝑏𝑓=𝑙012+𝑏𝑤+3𝐷𝑓 for L-beams.

7. 𝑙0𝑙0𝑏+4+𝑏𝑤 ≤𝑏 for isolated T-beams.

8. 0.5𝑙0𝑙0𝑏+4 +𝑏𝑤 ≤𝑏 for isolated L-beams.

9. 𝑥𝑢𝑑=0.87 𝑓𝑦. 𝐴𝑠𝑡0.36 𝑓𝑐𝑘 𝑏𝑑 and 𝑥𝑢,𝑚𝑎𝑥𝑑= 0.00350.0055+0.87 𝑓𝑦𝐸𝑠

10. 𝑀𝑢=0.87 𝑓𝑦.𝐴𝑠𝑡 (𝑑−0.42 𝑥𝑢) for 𝑥𝑢 ≤𝑥𝑢,max

11. 𝑀𝑢=0.36 𝑓𝑐𝑘.𝑏.𝑥𝑢,𝑚𝑎𝑥(𝑑−0.42 𝑥𝑢,𝑚𝑎𝑥)

12. 𝑥𝑢,𝑚𝑎𝑥𝑑= 0.87 𝑓𝑦0.36 𝑓𝑐𝑘 (𝑃𝑡,𝑙𝑖𝑚 100)

13. 𝑀𝑢=0.87 𝑓𝑦 𝐴𝑠𝑡 (𝑑−𝑓𝑦 𝐴𝑠𝑡𝑓𝑐𝑘 𝑏)

14. 𝑦𝑓=0.15 𝑥𝑢+0.65 𝐷𝑓 when 𝐷𝑓𝑑>0.2

15. 0.87 𝑓𝑦 (𝐴𝑠𝑡−𝐴𝑠𝑡,𝑙)=(𝑓𝑠𝑐−0.447 𝑓𝑐𝑘).𝐴𝑠𝑐

16. 𝑀𝑢=𝑀𝑢,𝑙 + 0.87 𝑓𝑦.(𝐴𝑠𝑡−𝐴𝑠𝑡,𝑙) (𝑑−𝑑′)

17. 𝐴𝑠𝑡 𝑏𝑑= 0.85𝑓𝑦 (minimum tension reinforcement)

18. 𝐴𝑠𝑡,𝑚𝑖𝑛=0.12 % of 𝐴𝑔 for 𝐹𝑒-415 and 0.15 % of 𝐴𝑔 for 𝐹𝑒-250. (slabs)

19. 𝐴𝑠𝑡,𝑚𝑎𝑥=4% 𝑜𝑓 𝐴𝑔 (Beams) ,𝐴𝑠𝑐,𝑚𝑎𝑥=4% 𝑜𝑓 𝐴𝑔 (Beams)

20. (𝑙𝑑)𝑚𝑎𝑥=(𝑙𝑑)𝑏𝑎𝑠𝑖𝑐𝑘𝑡 𝑘𝑐

21. (𝑙𝑑)𝑏𝑎𝑠𝑖𝑐=7 for cantilever, 20 for SS and 26 for continuous beams. If span is more than 10 m, multiply above values with 10/span for SS and continuous beams.

22. 𝐿𝑑=𝑓𝑠∅4𝜏𝑏𝑑 , multiply 𝜏𝑏𝑑 value with 1.6 for deformed bars and 1.25 for bars in compression.

23. 𝑀𝑢𝑉𝑢+𝐿0≥𝐿𝑑 and 1.3𝑀𝑢𝑉𝑢+𝐿0≥𝐿𝑑 (If confinement exists)

24. 𝜏𝑣=𝑉𝑢𝑏𝑑 , 𝑉𝑢,𝑛𝑒𝑡=𝑉𝑢+𝑀𝑑tan𝛽

25. 𝑉𝑢𝑠=0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑𝑠𝑣

26. 𝐴𝑠𝑣𝑏 𝑠𝑣≥0.40.87 𝑓𝑦 (minimum shear reinforcement)

27. 𝑠𝑣≤0.75 𝑑 𝑎𝑛𝑑 300 𝑚𝑚 for vertical stirrups.

28. 𝑉𝑒=𝑉𝑢+1.6𝑇𝑢𝑏 and 𝑀𝑒=𝑀𝑢+𝑇𝑢1.7(1+𝐷𝑏)

29. 𝑙𝑑≤35 for ss and 40 for continuous slabs for Fe-250. For Fe-415 multiply above values with 0.8

30. 𝜏𝑐2=𝑘𝑠 0.25√𝑓𝑐𝑘 , 𝑘𝑠=0.5+𝛽𝑐≤1.0 and 𝛽𝑐=𝑏𝐷

31. 𝑒𝑥,𝑚𝑖𝑛=𝑙500+𝑑30 ,20 𝑚𝑚 whichever is greater.

32. 𝑒𝑦,𝑚𝑖𝑛=𝑙500+𝑏30 ,20 𝑚𝑚 whichever is greater.

33. 𝑃𝑢=0.4 𝑓𝑐𝑘 𝐴𝑐+0.67 𝑓𝑦𝐴𝑠𝑐

34. 𝑉ℎ𝑉𝑐≥0.36(𝐴𝑔𝐴𝑐−1)𝑓𝑐𝑘𝑓𝑦 , 𝑑𝑐=𝑑−2𝑐 and 𝑑𝑚=𝑑𝑐−∅

35. 𝐶𝑟=1.25−𝑙𝑒48𝑏

36. 𝑓𝑏𝑟,𝑚𝑎𝑥=0.45 𝑓𝑐𝑘√𝐴1𝐴2 , 1<√𝐴1𝐴2≤2 , 𝐴1=Largest frustum of a pyramid with side slopes 1 in 2, 𝐴2=loaded area of column base

Engineering math formula

noreply@blogger.com (Civil Study) — Fri, 15 Oct 2021 05:42:00 +0000

Engineering Mathematics is one of the scoring section in GATE/BARC/ISRO Exam.Looking at your requirement, we are sharing with you Important Engineering Mathematics Formulas & Shortcuts for Competitive Exam as well as Engineering exam.

1. If 𝑟≠𝑟′, no solution, if 𝑟=𝑟′=𝑛, unique solution if 𝑟=𝑟′<𝑛, many solutions. (non-homogeneous)

2. If 𝑟=𝑛, trivial solution, if 𝑟<𝑛 ,then (𝑛−𝑟) linearly independent solutions. (Many solutions) and if 𝑚<𝑛, then many solutions.

3. 𝑓(𝑥+ℎ) =𝑓(𝑥) +ℎ𝑓′(𝑥) +ℎ22!𝑓′′(𝑥) +ℎ33!𝑓′′′(𝑥) + ………∞

4. If 𝑟𝑡−𝑠2>0 and 𝑟<0 𝑓(𝑥,𝑦) have maximum, if 𝑟𝑡−𝑠2>0 and 𝑟>0 𝑓(𝑥,𝑦) have minimum at(𝑎,𝑏) and if 𝑟𝑡−𝑠2<0, then saddle point. If 𝑟𝑡−𝑠2=0,𝑓𝑢𝑟𝑡ℎ𝑒𝑟 investigation is required to decide.

5. ∫𝑐(∅𝑑𝑥+𝜓𝑑𝑦)=∫∫𝐸(𝜕𝜓𝜕𝑥−𝜕∅𝜕𝑦)𝑑𝑥𝑑𝑦 (Green’s)

6. ∫𝑐𝑭.𝑑ℝ=∫𝑆𝑐𝑢𝑟𝑙𝑭.𝑁𝑑𝑠 (Stokes)

7. ∫𝑆𝑭.𝑁𝑑𝑠=∫𝐸𝑑𝑖𝑣𝑭𝑑𝑣 (Gauss)

8. 𝑦𝑒∫𝑃𝑑𝑥=∫𝑄𝑒∫𝑃𝑑𝑥𝑑𝑥+𝑐

9. If 𝑀 𝑑𝑥+𝑁 𝑑𝑦=0 be a homogeneous equation in 𝑥 and 𝑦, then 1𝑀 𝑥+𝑁 𝑦 𝑖𝑠 an integrating factor

10. If the equation of the type 𝑓1(𝑥𝑦)𝑦 𝑑𝑥+𝑓2(𝑥𝑦)𝑥 𝑑𝑦=0. If the equation 𝑀 𝑑𝑥+𝑁 𝑑𝑦=0 be of this type then 1𝑀 𝑥−𝑁 𝑦 is an integrating factor

11. If 𝜕𝑀𝜕𝑦−𝜕𝑁𝜕𝑥𝑁 be a function of x only = 𝑓(𝑥) say then 𝑒∫𝑓(𝑥)𝑑𝑥 is an integrating factor

12. If 𝜕𝑁𝜕𝑥−𝜕𝑀𝜕𝑦𝑀 be a function of y only = 𝑓(𝑦) say then 𝑒∫𝑓(𝑦)𝑑𝑦 is an integrating factor.

13. ∫𝑀𝑑𝑥𝑦=𝑐𝑜𝑛𝑡 +∫terms of N not containing x dy=c

14. 𝑃.𝐼.=1𝑓(𝐷)𝑒𝑎𝑥=1𝑓(𝑎)𝑒𝑎𝑥 , 𝑓(𝑎)≠0, if 𝑓(𝑎)=0,𝑡ℎ𝑒𝑛 𝑃.𝐼.=𝑥1𝑓′(𝑎)𝑒𝑎𝑥,𝑓′(𝑎)≠0

15. 𝑃.𝐼=1𝑓(𝐷2)sin(𝑎𝑥+𝑏)= 1𝑓(−𝑎2), 𝑓(−𝑎2)≠0, if 𝑓(−𝑎2)=0,

then 𝑃.𝐼. =𝑥1𝑓′(−𝑎2)sin(𝑎𝑥+𝑏), 𝑓′(−𝑎2)≠0

16. 𝑃.𝐼.=1𝑓(𝐷)𝑒𝑎𝑥𝑉= 𝑒𝑎𝑥1𝑓(𝐷+𝑎)𝑉

17. 𝑃.𝐼=1𝑓(𝐷)𝑥𝑚=[𝑓(𝐷)]−1𝑥𝑚,

18. (1+𝑥)−1=1−𝑥+𝑥2−⋯

19. (1−𝑥)−1=1+𝑥+𝑥2+⋯

20. 𝑥𝑛𝑑𝑛𝑦𝑑𝑥𝑛+𝑘1𝑥𝑛−1𝑑𝑛−1𝑦𝑑𝑥𝑛−1+⋯𝑘𝑛−1𝑥𝑑𝑦𝑑𝑥+𝑘𝑛𝑦=𝑋, 𝑥=𝑒𝑡 , 𝑥𝑑𝑦𝑑𝑥=𝐷𝑦, 𝑥2𝑑2𝑦𝑑𝑥2=𝐷(𝐷−1)𝑦, 𝑥3𝑑3𝑦𝑑𝑥3=D(D−1)(D−2)

21. 𝜕𝜕𝑥(∫ℎ(𝑡,𝑥)𝑑𝑡𝑔(𝑥)𝑓(𝑥))=∫𝜕𝜕𝑥ℎ(𝑡,𝑥)𝑑𝑡𝑔(𝑥)𝑓(𝑥)+𝑑𝑔𝑑𝑥 ℎ[𝑔(𝑥),𝑥]−𝑑𝑓𝑑𝑥 ℎ[𝑓(𝑥),𝑥]

22. 𝐿{𝑓(𝑡)}=∫𝑒−𝑠𝑡∞0 𝑓(𝑡)𝑑𝑡

23. 𝐿(1)=1𝑠

24. 𝐿(𝑡𝑛)=𝑛!𝑠𝑛+1

25. 𝐿(𝑒𝑎𝑡)=1𝑠−𝑎

26. 𝐿(sin𝑎𝑡)=𝑎𝑠2+𝑎2

27. 𝐿(cos𝑎𝑡)=𝑠𝑠2+𝑎2

28. 𝐿(sinh𝑎𝑡)=𝑎𝑠2−𝑎2

29. 𝐿(cosh𝑎𝑡)=𝑠𝑠2−𝑎2

30. 𝐿{𝑒𝑎𝑡𝑓(𝑡)}= 𝑓̅(𝑠−𝑎)

31. 𝑓(𝑡+𝑇)=𝑓(𝑡) then 𝐿{𝑓(𝑡)}= ∫𝑒−𝑠𝑡 𝑓(𝑡)𝑑𝑡𝑇01−𝑒−𝑠𝑇

Engineering math formula

32. 𝐿{𝑓′(𝑡)}=𝑠𝑓̅(𝑠)−𝑓(0)

33. 𝐿{𝑓𝑛(𝑡)}= 𝑠𝑛𝑓̅(𝑠)−𝑠𝑛−1𝑓(0)−𝑠𝑛−2𝑓′(0)−⋯……..𝑓𝑛−1(0)

34. 𝐿{∫𝑓(𝑥)𝑑𝑥𝑡0}= 1𝑆𝑓̅(𝑠)

35. 𝐿{𝑡𝑛𝑓(𝑡)}=(−1)𝑛𝑑𝑛𝑑𝑠𝑛.[𝑓̅(s)]

36. 𝐿{1𝑡𝑓(𝑡)}= ∫𝑓̅(s) ∞𝑆𝑑𝑠

37. 𝑓(𝑥)=𝑎02+ Σ𝑎𝑛∞𝑛=1cos𝑛𝑥+ Σ𝑏𝑛∞𝑛=1sin𝑛𝑥

38. 𝑎0=1𝜋∫𝑓(𝑥)𝑑𝑥𝛼+2𝜋𝛼, 𝑎𝑛=1𝜋∫𝑓(𝑥) cos𝑛𝑥𝑑𝑥𝛼+2𝜋𝛼, 𝑏𝑛=1𝜋∫𝑓(𝑥) sin𝑛𝑥𝑑𝑥𝛼+2𝜋𝛼

39. 𝑓(𝑥)=𝑎02+ Σ𝑎𝑛∞𝑛=1cos𝑛𝜋𝑥𝑐+ Σ𝑏𝑛∞𝑛=1sin𝑛𝜋𝑥𝑐

40. 𝑎0=1𝑐∫𝑓(𝑥)𝑑𝑥𝛼+2𝑐𝛼, 𝑎𝑛=1𝑐∫𝑓(𝑥) cos𝑛𝜋𝑥𝑐𝑑𝑥𝛼+2𝑐𝛼, 𝑏𝑛=1𝑐∫𝑓(𝑥) sin𝑛𝜋𝑥𝑐𝑑𝑥𝛼+2𝑐𝛼

41. 𝑓(𝑥)= Σ𝑏𝑛∞𝑛=1sin𝑛𝜋𝑥𝑐 , where 𝑏𝑛=2𝑐∫𝑓(𝑥) sin𝑛𝜋𝑥𝑐𝑑𝑥𝑐0

42. 𝑓(𝑥)= 𝑎02+ Σ𝑎𝑛∞𝑛=1cos𝑛𝜋𝑥𝑐 where, 𝑎0=2𝑐∫𝑓(𝑥)𝑑𝑥𝑐0, 𝑎𝑛=2𝑐∫𝑓(𝑥) cos𝑛𝜋𝑥𝑐𝑑𝑥𝑐0

43. 𝜇=Σ 𝑥𝑗𝑓(𝑥𝑗)𝑗 𝑎𝑛𝑑 𝜇=∫𝑥 𝑓(𝑥)𝑑𝑥 ∞−∞

44. 𝜎2=Σ (𝑥𝑗−𝜇)2𝑓(𝑥𝑗) 𝑗𝑎𝑛𝑑 𝜎2=∫(𝑥−𝜇)2 𝑓(𝑥)𝑑𝑥∞−∞

45. 𝑀𝑒𝑎𝑛:𝑛𝑝=𝜇 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒: 𝜎2=𝜇 (Poisson’s distribution)

46. 𝑓(𝑥)=1𝜎√2𝜋𝑒− 12(𝑥−𝜇𝜎)2 (Normal distribution)

47. 𝑦=𝑎+𝑏𝑥, Σ𝑦=𝑛𝑎+𝑏Σ𝑥, Σ𝑥𝑦=𝑎Σ𝑥+𝑏Σ𝑥2

48. 𝑦=𝑎+𝑏𝑥+𝑐𝑥2, Σ𝑦=𝑛𝑎+𝑏.Σ𝑥+𝑐.Σ𝑥2, Σ𝑥𝑦=𝑎Σ𝑥+𝑏.Σ𝑥2+𝑐 Σ𝑥3, Σ𝑥2𝑦=𝑎Σ𝑥2+𝑏.Σ𝑥3+𝑐 Σ𝑥4

49. 𝑦=𝑓(𝑥)=(𝑥−𝑥1)(𝑥−𝑥2)…(𝑥−𝑥𝑛)(𝑥0−𝑥1)(𝑥0−𝑥2)…(𝑥0−𝑥𝑛)𝑦0+(𝑥−𝑥0)(𝑥−𝑥2)…(𝑥−𝑥𝑛)(𝑥1−𝑥0)(𝑥1−𝑥2)…(𝑥1−𝑥𝑛)𝑦1+⋯+(𝑥−𝑥1)(𝑥−𝑥2)…(𝑥−𝑥𝑛−1)(𝑥𝑛−𝑥0)(𝑥𝑛−𝑥1)…(𝑥𝑛−𝑥𝑛−1)𝑦𝑛

50. (𝑑𝑦𝑑𝑥)𝑥0=1ℎ[Δ𝑦0−12Δ2𝑦0+13Δ3𝑦0−14Δ4𝑦0+⋯]

51. (𝑑𝑦𝑑𝑥)𝑥𝑛=1ℎ[∇𝑦𝑛+12∇2𝑦𝑛+13∇3𝑦𝑛+14∇4𝑦𝑛+⋯]

52. 𝑥𝑛+1=𝑥𝑛−𝑓(𝑥𝑛)𝑓′(𝑥𝑛) (Newton-Raphson)

53. ∫𝑓(𝑥)𝑑𝑥𝑥0+𝑛ℎ𝑥0=ℎ2[𝑦0+𝑦𝑛+2(𝑦1+𝑦2+⋯..+𝑦𝑛−1)] (Trapezoidal)

54. ∫𝑓(𝑥)𝑑𝑥𝑥0+𝑛ℎ𝑥0=ℎ3[(𝑦0+𝑦𝑛)+4(𝑦1+𝑦3+⋯𝑦𝑛−1)+2(𝑦2+𝑦4+⋯𝑦𝑛−2)] (Simpson’s)

55. 𝐸𝑟𝑟𝑜𝑟=−𝑏−𝑎12ℎ2 𝑓′′(𝜉)=𝑂(ℎ2) (Trapezoidal)

56. 𝐸𝑟𝑟𝑜𝑟=−𝑏−𝑎180ℎ4𝑓𝑖𝑣(𝜉)=𝑂(ℎ4) (Simpson’s)

57. 𝑦𝑘+1=𝑦𝑘+ℎ.𝑓(𝑡𝑘,𝑦𝑘) where 𝑑𝑦𝑑𝑥=𝑓(𝑡,𝑦) (Euler’s)

Test method for sieve analysis of fine and coarse aggregates

noreply@blogger.com (Civil Study) — Fri, 15 Oct 2021 03:11:00 +0000

Aggregate is the inert, inexpensive materials dispersed throughout the cement paste so as the produce a large volume of concrete. They constitute more than three quarters of volume of concrete. They provide body to the concrete, reduce shrinkage and make it durable.The aggregates are classified in two categories; fine aggregate and coarse aggregate. The size of fine aggregates is limited to a maximum of 4.75 mm, beyond which it is known as coarse aggregates. Many a time, fine aggregates are designated as coarse sand, medium sand and fine sand. These classifications do not give any precise meaning. What the supplier terms as fine sand may be really medium or even coarse sand. To avoid this ambiguity fineness modulus could be used as a yard stick to indicate the fineness of sand and in general aggregates.

Fineness modulus for a given aggregate is obtained by sieving known weight of it in a set of standard sieves and by adding the percent weight of material retained on all the sieves and dividing the total percentage by 100. It serves the purpose of comparing one aggregate with another in respect of fineness or coarseness. For classification of fine aggregates, the following limits may be taken as guidance:

Fine sand: Fineness modulus should lie in between 2.2 to 2.6

Medium sand: Fineness modulus should lie in between 2.6 to 2.9

Coarse sand: Fineness modulus should lie in between 2.9 to 3.2

Sand having a fineness modulus more than 3.2 is unsuitable for making satisfactory concrete.The coarse aggregates have fineness modulus usually more than 5.A heap of aggregate is classified as a single sized aggregate when the bulk of aggregate passes one sieve in normal concrete series and retained on next smaller size. Such aggregates are normally expressed by the maximum size of the aggregates present in considerable amount in it. For example, a heap of 20 mm size aggregate means that the heap contains maximum 20 mm size aggregate in a substantial amount.A graded aggregate comprises of a proportion of all sizes in a normal concrete series.When these sizes are so proportionate to provide a definite grading, it is known as well graded aggregate. Well graded aggregates are desirable for making concrete, as the space between larger particles is effectively filled by smaller particles to produce a well-packed structure. This minimizes the cement requirement. All-in aggregates comprise a mixture of coarse aggregate and fine aggregates. Such aggregates may directly be used for low quality concreting. But in case of good quality concreting work; necessary adjustments may be made in the grading by the addition of IS 383:1970 specifies four grading zones for fine aggregates. These four grading zones become progressively finer from Grading Zone I to Grading Zone IV (see Table). The fine aggregates within each of these grading zones are suitable for making concrete. But, the ratio of ratio of fine to coarse aggregate reduces as the fine aggregate becomes finer from Grading Zones I to IV. The grading of coarse aggregate may vary through wider limits than that of fine aggregates. However, this variation does not much affect the work ability, uniformity and finishing qualities of concrete mix. As per IS-383:1970 the grading limit of coarse aggregate,both for single size as well as graded should be as per the table given below.

Objective: To determine fineness modulus and grade of fine and coarse aggregate.

Reference: IS: 383-1970.

Apparatus: Set of sieves*; Balance; Gauging Trowel; Watch.

a) *For fine aggregates: 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron & 150 micron, pan.

b) *For coarse aggregates: 80mm, 40mm, 20mm, 10mm, 4.75mm, pan.

Material: a) Fine aggregates (1 Kg)

b) Coarse aggregates (5 Kg)

Procedure:

1. Take the aggregate from the sample by quartering.

2. Sieve the aggregate using the appropriate sieves.

3. Record the weight of aggregate retained on each sieve.

4. Calculate the cumulative weight of aggregate retained on each sieve.

5. Calculate the cumulative percentage of aggregate retained.

6. Add the cumulative weight of aggregate and calculate the fineness modulus using formula:

a. Fineness modulus for fine aggregates = Σ 𝐶 / 100

b. Fineness modulus for coarse aggregates = Σ 𝐶 / 100 + 5

Where, C denotes the cumulative percentage of mass retained in a sieve.

7. Determine the grade of aggregates from the Table 1 and the Table 2.

8. Plot the gradation curves, in a semi-log graph, between percentage of aggregates passed and size of sieve both for a) Fine aggregate and b) For coarse aggregate.

(Note: A typical grading curve for fine aggregates looks like Figure . A similar grading curve will be observed for coarse aggregates.)

A typical gradation curve for fine aggregates

Observations:

(a) For fine aggregates:

Weight of fine aggregate taken (Wf): _________Kgs

Sl. No.	Sieve size	Weight retained (in kg)	%age retained ( 𝐶3 / 𝑊𝑓×100 )	Cumulative %age retained	Percentage passed (100 –C5 )
C1	C2	C3	C4	C5	C6
1	4.75 mm
2	2.36 mm
3	1.18 mm
4	600 micron
5	300 micron
6	150 micron
7	Pan

Sum of cumulative percentage retained (excluding pan) Σ 𝐶5=

Fineness Modulus Σ 𝐶5 / 100 =

Zone to which the fine aggregate belongs:

(b) For coarse aggregates:

Weight of coarse aggregate taken (Wc): _________Kgs

Sl. No.	Sieve size	Weight retained (in kg)	%age retained ( 𝐶3 / 𝑊𝑓×100 )	Cumulative %age retained	Percentage passed (100 –C5 )
C1	C2	C3	C4	C5	C6
1	80 mm
2	40 mm
3	20 mm
4	10 mm
5	4.75 mm
6	Pan

Sum of cumulative percentage retained (excluding pan) Σ 𝐶5=

Fineness Modulus Σ 𝐶5 / 100 + 5 =

Grade to which the coarse aggregate belongs:

Results and discussions:

The fineness modulus of given samples are:

a) For fine aggregates : __________.

b) For coarse aggregates : __________.

The grading to which the given samples belong are:

a) For fine aggregates : __________.

b) For coarse aggregates : __________.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Specific gravity and water absorption test for fine aggregates

noreply@blogger.com (Civil Study) — Mon, 11 Oct 2021 14:10:00 +0000

Specific gravity of an aggregate is defined as the ratio of the mass of solid in a given volume of sample to the mass of equal volume of water at 4𝑜𝑜 𝐶𝐶. However, all rocks contain some small amount of void and the apparent specific gravity includes this voids. The specific gravity of aggregates is an indirect measure of material’s density and its quality. A low specific gravity may indicate high porosity and therefore poor durability and low strength.

Some of the pores contained by aggregates are permeable while others are impermeable.Accordingly, two types of specific gravity are defined absolute specific gravity and apparent specific gravity. If both the permeable and impermeable voids are excluded to determine the true volume of solids, the specific gravity is called true or absolute specific gravity of the aggregate. But true specific gravity has not much of practical use as volume of impermeable internal pores is too difficult to determine.In contrast, for the determination of apparent specific gravity the impermeable internal pore is added to the effective volume of the aggregates (does not include the permeable pores). Mathematically:

Apparent Specific Gravity = ( mass of aggregate / mass of water occupying the volume equal to that of solids of aggregate excluding permeable pores )

pycnometer

The apparent specific gravity is realistic one to use for concrete mix proportioning. The apparent specific gravity of most rocks lie between 2.6 to 2.7. Apparent specific gravity can be determined on the basis of surface dry condition (SSD) or oven dry condition (OD),according to the moisture condition of the aggregate.In saturated surface dry (SSD) situation, the pores of the aggregate are fully filled with water and the surface is dry. This condition can be obtained by immersing coarse aggregate in water for 24 h followed by drying of the surface with a wet cloth. When the aggregate is under the SSD condition, it will neither absorb water nor give out water during the mixing process. Hence, it is a balanced condition and is used as the standard index for concrete mix design.In contrast, the oven dry condition is obtained by keeping the aggregate in an oven at a temperature of 110o 𝐶𝐶 for 24 hrs. Due to heating, all the water from internal permeable pores gets evaporated and hence it reaches a constant weight. When the aggregate is under OD condition, it will absorb water during the concrete mixing process until the internal pores are fully filled with water.Similarly, water absorption is also provides an idea about strength of aggregate.Aggregates having more water absorption are more porous in nature and generally considered unsuitable. Usually, water absorption of coarse aggregate is about 0.5% by weight whereas water absorption of fine aggregate is about 2.0% by weight. Moreover, water absorption values are used to calculate the change in the weight of the aggregate while proportioning and mixing of concrete. Extra water is added to cater the need of water absorption

Objective: To determine specific gravity and water absorption of fine aggregate.

Reference: IS: 2386 (Part-3)-1963.

Apparatus: Pycnometer, 1000-ml measuring cylinder, thermostatically controlled oven,taping rod, filter papers and funnel.

Material: Fine aggregates (500 g)

Procedure:

1. Place 500 g of fine aggregate in a tray and cover it with distilled water at a temperature of 22 to 32°C. Remove air entrapped in or bubbles on the surface of the aggregate by gentle agitation with a rod. Keep the sample immersed under water for 24 Hrs.

2. Carefully drain the water from the sample, by decantation through a filter paper. Air dry the aggregate and solid matter retained on the filter paper, to remove the surface moisture. When the material just attains a “free-running” condition,weight the saturated and surface-dry sample (A).

3. Place the aggregate in the pycnometer and fill the remaining space by distilled water. Eliminate entrapped air by rotating the pycnometer on its side, covering the hole in the apex of the cone with a finger. Weight the pycnometer with this condition (B).

4. Empty the contents of the pycnometer into a tray. Ensure that all the aggregate is transferred. Refill the pycnometer with distilled water to the same level as before and measure the weight at this condition (C).

5. Carefully drain the water from the sample, by decantation through a filter paper.Oven-dry the aggregate in the tray at a temperature of 100 to 110oC for 24 hrs.During this period, stir the specimen occasionally to facilitate proper drying.Cool the aggregates calculate its weight (D).

6. Calculate the specific gravity, apparent specific gravity and the water absorption as follows:

Specific gravity = {𝐷/[𝐴 − (𝐵 − 𝐶)]} (1)

Apparent Specific gravity = {𝐷/[𝐷 − (𝐵 − 𝐶)]} (2)

Water absorption (in %) = 100 × [(𝐴 − 𝐷)/𝐷] (3)

Where, A = Weight in g of saturated surface-dry sample

B = Weight in g of pycnometer containing sample and filled with distilled water

C = Weight in g of pycnometer filled with distilled water only

D = Weight in g of oven dried sample only.

Observations:

Weight in g of saturated surface-dry sample (A)
Weight in g of pycnometer containing sample and filled with distilled water (B)
Weight in g of pycnometer filled with distilled water only (C)
Weight of oven dried- sample only (D)
Specific gravity = {𝐷/[𝐴 − (𝐵 − 𝐶)]}
Apparent Specific gravity = {𝐷/[𝐷 − (𝐵 − 𝐶)]}
Water absorption (in %) = 100 × [(𝐴 − 𝐷)/𝐷]

Results and discussions:

Following results are obtained for the provided fine aggregate specimen:

a) Specific gravity : __________.

b) Apparent specific gravity : __________.

c) Water absorption : __________ %.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Bulking of sand test procedure

noreply@blogger.com (Civil Study) — Sun, 10 Oct 2021 07:51:00 +0000

Theory: Increase in volume of sand due to presence of moisture is known as bulking of sand. Bulking is due to the formation of thin film of water around the sand grains and the interlocking of air in between the sand grains and the film of water. When more water is added sand particles get submerged and volume again becomes equal to dry volume of sand.

Due to the bulking, fine aggregate shows completely unrealistic volume. Therefore, it is absolutely necessary that consideration must be given to the effect of bulking in proportioning the concrete by volume. If cognizance is not given to the effect of bulking, in case of volume batching, the resulting concrete is likely to be under-sanded and harsh. It will also affect the yield of concrete for given cement content.To compensate the bulking effect extra sand is added in the concrete so that the ratio of coarse to fine aggregates will not change from the specified value. Maximum increase in volume may be 20 % to 40 % when moisture content is 5 % to 10 % by weight. Fine sands show greater percentage of bulking than coarse sands with equal percentage of moisture.

Objective: To ascertain the bulking phenomena of given sample of sand.

Reference: IS: 2386 (Part-3)-1963.

Apparatus: Beaker, 1000ml measuring jar, brush, scale, mixing tray.

Material: Fine aggregate, water.

Procedure:

1. Put sufficient quantity of dry sand into the beaker until it is about one-thirds full.

2. Level off the top of the sand and measure the height (H1) by pushing a steel rule vertically down through the sand at the middle to the bottom. Measure weight of the sand.

3. Add 2% of water; mix it thoroughly in the container. Smooth and level the top surface measure the height (H2) of soil. Find the height percentage increment.

4. Repeat the same procedure with increasing amount of water by 2% until percentage increment of sand height is reduced and attends original level.

5. Plot a graph of percentage increment of sand height against percentage of water.

(Note: A typical bulking of sand graph looks like Figure Below )

Moisture Contain

Observations:

Initial Height of sand in the Jar (H1) : _______ mm

Weight of fine aggregate : ________ g

Sl. No.

% of Water

Volume of water

(in ml)

Height of sand

(H2)

page bulking

= 𝑯𝟐−𝑯𝟏 / 𝑯𝟏

× 𝟏00

Results and discussions: From the tabulated results and the plotted graph is is observed that, the given sand specimen under goes maximum bulking at _________% of moisture contain. Maximum percentage of bulking is __________.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

Compressive strength test for cement

noreply@blogger.com (Civil Study) — Sun, 03 Oct 2021 06:45:00 +0000

Theory: Among physical properties of cement, compressive strength is the most important property. When cement is used for important structures, compressive strength test is always carried out to ascertain quality of cement. Strength test is not made on plain cement due to excess shrinkage and cracking of plain cement paste. The test is therefore carried out on blocks of mortar made of cement, sand and water.

Compressive strength of Cement test

As the quality of sand from different sources varies, it is necessary to use sand of standard quality for this test. The standard sand consists of well graded sand of quartz, light grey or whitish in colour and free of silts and organic impurities. As per IS 650: 1991, standard sand is obtained from Ennore, Tamilnadu which has the following particle size distribution:

Smaller than 2 mm but greater than 1 mm : 33.33%

Smaller than 1 mm but greater than 500 μ : 33.33%

Smaller than 500 μ but greater than 90 μ : 33.33%

In the mortar, cement and sand are used in the proportion of 1:3. Mortar cubes of size 70.6 mm are prepared, compacted and cured properly before testing under direct compression. The cubes are tested in compression testing machine at the end on three days, seven days and twenty eight days. For ordinary portland cement of 43 grade, the average compressive strength after 3, 7 and 28 days should not be less than 23 N/mm2, 33 N/mm2 and 43 N/mm2 respectively.

Objective: Determination of compressive strength of cement.

Reference: IS 4031-1988 (Part-6).

Apparatus: Vibration Machine, Poking Rod, Cube Mould of 70.6 mm size conforming to IS: 10080-1982, Balance, Gauging Trowel, Watch, Graduated Glass Cylinders, etc.

Material: Ordinary portland cement (43 grade); Water; Grease, Standard sand (IS: 650-1966).

Procedure:

Preparation of cement mortar cubes:

1. Take 200gms of cement and 600 g of standard sand (i.e. ratio of cement to sand is 1:3) in a non-porous enamel tray and mix them with a trowel for one minute.

2. Add water quantity ( p / 4 + 3.0)% of combined weight of cement and sand and mix the three ingredients thoroughly until the mixture is of uniform colour. (‘P’ is the consistency of cement). The time of mixing should be less than three minutes and not more than four minutes.

3. Immediately after mixing fill the mortar into greased cube moulds of sizes 70.6 mm.

4. Compact the mortar either by hand compaction in a standard specified manner or on the vibrating table.

5. Place the moulds in cabin at a temperature of 27º ± 2º C for 24 hours

6. Remove the specimen from the moulds and submerge them in clean water for curing.

Testing of cement mortar cubes:

7. Take the cube out of water at the end of three days with dry cloth. Measure the dimensions of the surface in which the load is to be applied. Let them be ‘L’and ‘B’ respectively.

8. Place the cube in compressive testing machine and apply the load uniformly at the rate of 14N/mm2 per minute.

9. Note the load at which the cube fails. Let it be ‘F’.

10. Calculate the compressive strength of the cube by using formula F/𝐴𝐴 . Where A is the area of loaded surface (i.e. L×B).

11. Repeat the same procedure (steps 7 to 10) for other two cubes.

12. Repeat the whole procedure (Step 7 to 11) to find the compressive strength of the cube at the end of 7 days and 28 days.

Observations:

(a) For 3 days strength:

Sl. No.	Length (L) in mm	Breadth (B) in mm	Load (F) in N	compressive strength in N/mm2
1
2
3

(b) For 7 days strength:

Sl. No.	Length (L) in mm	Breadth (B) in mm	Load (F) in N	compressive strength in N/mm2
1
2
3

Sl. No.	Length (L) in mm	Breadth (B) in mm	Load (F) in N	compressive strength in N/mm2
1
2
3

Results:

The type and grade of cement :

The compressive strength of cement ate the end of

i) 3 days : ___________. N/mm2.

ii) 7 days : ___________. N/mm2.

iii) 28 days : ___________. N/mm2.

Conclusions:

Average compressive strength of cement mortar at the end of 3 days, 7 days and 28 days are _______ N/mm2, _______ N/mm2 and _______ N/mm2, respectively. Thus the cement sample satisfies/ not satisfy the codal requirements for the given type and grade of cement.

How to find the specific gravity of cement || Le Chaterliers flask Test

noreply@blogger.com (Civil Study) — Sun, 03 Oct 2021 06:17:00 +0000

Theory: Specific gravity is defined as the ratio between weight of a given volume of material and weight of an equal volume of water. In case of cement, specific gravity is determined by use of a Le Chatelier’s flask (Figure Below). Sometimes, a specific gravity bottle may be employed to a standard Le Chatelier’s flask. To determine the specific gravity of cement, kerosene is used which does not react with cement. The specific gravity of OPC is generally around 3.15.

Le Chaterliers flask

Objective: Determination of specific gravity of cement using Le-Chatelier flash.

Reference: IS 4031 (Part-11):1988.

Apparatus: Le Chaterliers flask, weighing balance, kerosene (free from water).

Material: Ordinary Portland cement; Water; Grease

Procedure:

1. Dry the flask carefully and fill with kerosene or naphtha to a point on the stem between zero and 1 ml.

2. Record the level of the liquid in the flask as initial reading.

3. Put a weighted quantity of cement (about 60 g) into the flask so that level of kerosene rise to about 22 ml mark, care being taken to avoid splashing and to see that cement does not adhere to the sides of the above the liquid.

4. After putting all the cement to the flask, roll the flask gently in an inclined position to expel air until no further air bubble rise 3s to the surface of the liquid.

5. Note down the new liquid level as final reading.

Observations:

• Weight of cement used in g. (W1) : _______

• Initial reading of flask in ml (V1) : _______

• Final reading of flask in ml (V2) : _______

• Volume of cement particle (V2 - V1) : _______

• Weight of equal volume of water in g. (W2) : _______

• Specific gravity of cement (W1/ W2) : _______

Result and conclusion: Specific gravity of the given cement obtained as ___________.

Precautions: (Discuss about the precautions to be taken while conducting this experiment)

How to check soundness of cement || Le-Chatelier method

noreply@blogger.com (Civil Study) — Sat, 02 Oct 2021 05:25:00 +0000

Theory: The ability of cement to maintain a constant volume is known as soundness of the cement. It is essential that the cement concrete shall not undergo appreciable change in volume after setting. Unsoundness produces cracks, distortion and disintegration there by giving passage to water and atmospheric gases which may have injurious effects on concrete and reinforcement. Soundness of cement is ensured by limiting the quantities of free lime,magnesia and sulphates as these compounds undergo a large change in volume.Unsoundness in cement does not come to surface for a considerable period of time. Thus this test is designed to accelerate the hydration of free lime by the application of heat thus discovering the defects in a short time. Further, to minimize the shrinkage of cement paste,the test setup is kept immersed in water bath.This test is carried out with the help of “Le Chatelier’s apparatus” which consists of a small split cylinder of spring brass or other suitable metal of 0.5mm thickness forming a mould 30 mm internal diameter and 30mm high (Figure 2). On either side of the split mould are attached to indicators with pointed ends, the distance from these ends to the center of the cylinder being 165 mm. The mould shall be kept in good condition with the jaws not more than 50mm apart.

Le-chatlier Apparatus

Objective: Determination of soundness of cement by Le-Chatelier method.

Reference: IS 4031 (Part-3):1988.

Apparatus: Le- Chatelier apparatus conforming to IS: 5514-1969; Measuring cylinder;Gauging trowel; Balance; Water bath.

Material: Ordinary Portland cement; Water; Grease

Procedure:

1. Weigh accurately 100 g of cement to the nearest 0.15 g and add to it 0.78 times the water required to give a paste of standard consistency (i.e. 0.78×P).

2. Place the lightly grease mould on a lightly grease glass sheet and fill it with cement paste, taking care to keep the edges of the mould gently together.

3. Cover the mould with another piece of lightly grease glass sheet, place a small weight on this covering glass sheet and immediately submerge the whole assembly in water at a temperature of 27 ± 20 𝐶𝐶.

4. Keep this assembly under water for 24 hrs. After this, take the mould out of water and measure the distance between two indicators. Submerge the mould again in the water.

5. Bring the water to boiling with the mould kept submerged, and keep it boiling for 25 to 30 minutes.

Top choice

6. Remove the mould from the water allow it to cool and measure the distance between the indicator points.

7. The difference between these two measurements represents the expansion of the cement.

8. Repeat the whole procedures two more times each using fresh 100 g sample.

Observations:

Samples:

Distance between pointers before boiling (D1) in mm

Distance between pointers after boiling (D2) in mm

Expansion of the cement = E1 = (D2- D1) in mm

Average expansion of the cement in mm

Result:Average expansion of the cement is obtained is ___________ mm.

Conclusions:Average expansion of the cement as per Le- Chatelier test is less than/ more than 10 mm.Therefore the given sample of cement is found to be sound/ unsound as per IS code.

Determination of setting time of hydraulic cement

noreply@blogger.com (Civil Study) — Sat, 02 Oct 2021 02:47:00 +0000

Theory: Cement when mixed with water forms slurry which gradually becomes lesser and lesser plastic, and finally forms a hard mass. In this process a stage is obtained when the cement paste is sufficiently rigid to with stand a definite amount of pressure. The time to reach this stage is called setting time. The setting time is divided into two parts: the initial setting time and the final setting time.

Vicat apparatus

Initial set is a stage where the cement paste stiffens to such an extent that the Vicat needle is not permitted to move down through the paste within 5 ± 0.5 mm measured from the bottom of the mould. In other words, the cement paste starts losing its plasticity. The time elapsed between the moments that the water is added to the cement to the initial set is regarded as initial setting time. Any crack that may appear after initial set may not re-unite.Final setting time is the time when the paste becomes so hard that the annular attachment to the needle under standard weight only makes an impression on the hardened cement paste.In order that the concrete may be mixed, transported and placed in position conveniently,it is necessary that the initial set of cement is not too quick. But after, it has been laid; the hardening should be rapid so that the structure can be made use of as early as possible. For an ordinary Portland cement, the initial setting time should not be less than 30 minutes while the final setting time should not be more than 600 minutes.The setting time of the cement is influenced by factors such as: percentage of water,amount of kneading the paste, temperature and humidity of the environment. As per codal provisions; this test should be conducted at temperature of 27𝑜𝑜 ± 2𝑜𝑜 𝐶𝐶 and 90% humidity.Flash set and false set are two terms, which are generally encountered in relation to the setting of cements. Flash set refers to the immediate stiffening of the cement paste due to violent reaction of pure 𝐶𝐶3𝐴𝐴 with water. To prevent flash set gypsum is added to the cement clinker during grinding. A phenomenon of abnormal and premature hardening of cement within a few minutes of mixing with water is termed as false set. It differs from flash set in that no appreciable heat is evolved, and remixing the cement paste without addition of water restores plasticity of the paste and the concrete sets in normal manner without a loss of strength. This happens due to dehydration of gypsum when it comes in contact with excessively hot clinkers during grinding.

Objective: Determination of initial and final setting time of cement and determine whether the values satisfy IS standards.

Reference: IS 4031 (Part-5):1988

Apparatus: Vicat apparatus conforming to IS: 5513-1998; Weighing balance; Gauging trowel; measuring cylinder; stop watch.

Material: Ordinary Portland cement; Water.

Procedure:

1. Prepare a uniform cement paste by gauging 400 g of cement with 0.85 times the water required to give a paste of standard consistency. The procedure of mixing and filling the mould is same as standard consistency.

2. Start the stopwatch or note down the time when water is added to the cement. Determination of initial setting time:

3. Place the test block confined in the mould and resting on the non-porous plate,under the rod bearing the initial setting needle (with cross section 1 mm2);lower the needle gently until it comes in contact with the surface of the test block and quickly release, allowing it to penetrate into the test block.

4. Repeat this procedure until the needle, when brought in contact with the test block and released as described above, fails to pierce the block beyond 5.0 ±0.5 mm measured from the bottom of the mould. Note the time.

5. The difference of time between operations (2) and (4) provides the initial setting time of cement.Determination of final setting time:

6. Replace the initial setting needle of the Vicat apparatus by the needle with an annular attachment.

7. The cement shall be considered as finally set when, upon applying the needle gently to the surface of the test block, the needle makes an impression thereon,while the attachment fails to do so.

8. The interval of time between operation (2) and (7) provides the final setting time of cement.

Observations:

• Weight of given sample of cement is ___________ g.

• The normal consistency of a given sample of cement is ___________ %

• Volume of water addend (0.85 times the water required to give a paste of standard consistency) for preparation of test block ____________ ml

Time in minutes :
Height in mm fails to penetrate

Initial setting time of cement (in min):

Final setting time (in min):

Conclusion:

The given sample of cement satisfied/ does not satisfy criterion for initial setting time.

The given sample of cement satisfied/ does not satisfy criterion for final setting time.

Consistency Test of cement procedure

noreply@blogger.com (Civil Study) — Fri, 01 Oct 2021 14:55:00 +0000

Theory: This test is conducted to calculate the amount of water to be added to the cement to get a paste of standard consistency which is defined as that consistency which will permit the Vicat plunger to penetrate to a point 5 to 7 mm from the bottom of the Vicat mould. This experiment is done with the help of Vicat apparatus (Figure Below ). The time taken between adding of water to the cement and filling of mould of Vicat apparatus is called as gauging time which should be between 3 to 5 minutes.

Vicat apparatus

For finding out initial setting time, final setting time, soundness of cement and compressive strength of cement, it is necessary to fix the quantity of water to be mixed in cement in each case. Since different batches of cement differ in fineness, pastes with some water content may differ in consistency when first mixed. For this reason the consistency of the paste is standardized by varying the water content until the paste has a given resistance to penetration.

Objective: Determination of percentage of water by weight of cement required to prepare a standard acceptable (consistent) cement paste.

Reference: IS 4031 (Part-4):1988.

Apparatus: Vicat apparatus conforming to IS: 5513-1998; Weighing balance; Gauging trowel; measuring cylinder.

Material: Ordinary Portland cement; Water.

Procedure:

1. Take 400 g of cement sieved through 90 micron IS sieve and keep it on a nonporous, non-absorbent plate.

2. Add 120 ml of water (i.e. 30% by weight of cement) to the cement and mix thoroughly with two trowels for 3 to 5 minutes till a uniform cement paste is achieved.

3. Fill the past in mould and level with trowel. Shake or tap to remove air bubbles.

4. Place the nonporous plate and the mould under the plunger.

5. Release the plunger gently to touch the surface of paste. Record the initial reading.

6. Release the plunger quickly and allow penetrating into the paste. When the plunger comes to rest, note the final reading.

7. Repeat the procedure with fresh paste varying the water percentage until the plunger penetrates to a depth 5 to 7 mm from the bottom of the Vicat mould.

Sl. No.

Water added

(in ml)

Percentage

Initial

Reading

Final

Reading

Height not

penetrated (in mm)

Result:Percentage of water required to achieve normal consistency of cement paste is ___________.

Precautions: (Discuss about the precautions to be taken while conducting this experiment).

Determination of fineness of cement by dry sieving - IS 4031 (part-1) : 1988

noreply@blogger.com (Civil Study) — Sun, 26 Sep 2021 13:19:00 +0000

Theory: Cement is obtained by grinding various raw materials after calcination. The degree to which cement is ground to smaller and smaller particles is called fineness of cement. The fineness of cement has an important role on the rate of hydration and hence on the rate of gain of strength and also on the rate of evolution of heat. Finer cement offers a greater surface area for hydration and hence the faster development of strength although the ultimate strength is not affected. Fineness also provides more cohesiveness to concrete and avoid separation of water at the top of concrete (called bleeding). However, increase in fineness of cement increases the drying shrinkage and cracking of the concrete.

Fineness of cement is tested either by sieving or by determination of specific surface using air-permeability apparatus. The specific surface is defined as the total surface area of all the particles in cm2 per one gram of cement. Although determination of specific surface is more accurate to judge fineness of cement, it is rarely used except for specific purpose. In contrast sieving is most commonly used method to determine fineness of cement and is quite good for field works.

Objective: Determination of fineness of cement by dry sieving.

Reference: IS 4031 (Part-1):1988.

Apparatus: IS-90 micron sieve conforming to IS: 460 (Part 1-3)-1985; Weighing balance;Gauging trowel; Brush.

Sieve 90 Micron

Material: Ordinary Portland cement

Procedure:

1. Weigh accurately 100 g of cement to the nearest 0.01 g and place it on a standard 90 micron IS sieve.

2. Break down any air-set lumps in the cement sample with fingers.

3. Agitate the sieve by giving swirling, planetary and linear movements for a period of 10 minutes or until no more fine material passes through it.

4. Collect the residue left on the sieve, using brush if necessary, and weigh the residue.

5. Express the residue as a percentage of the quantity first placed on the sieve to the nearest 0.1 percent.

6. Repeat the whole procedures two more times each using fresh 100 g sample.

Observations:

Sl. No.	Weight of sample taken (W) (in g.)	Weight of residue (R) (in g.)	%age of residue = 𝑅 /𝑊× 100	Average % of residue
1
2
3

Result:

Percentage residue of cement sample by dry sieving is ___________ percentage.

Conclusions:

The given sample of cement contains less than/ more than 10% by weight of material coarser than 90 micron sieve. Therefore it satisfies/ not satisfies the criterion as specified by IS code.

what is gravity dam and forces acting on gravity dam

noreply@blogger.com (Civil Study) — Sun, 26 Sep 2021 07:40:00 +0000

A gravity dam is a solid concrete or masonry structure which ensures stability against all applied loads by its weight alone without depending on arch or beam action. Such dams are usually straight in plan and approximately triangular in cross-section. Gravity dams are usually classified with reference to their structural height which is the difference in elevation between the top of the dam (i.e., the crown of the roadway, or the level of the walkway if there is no roadway) and the lowest point in the excavated foundation area, exclusive of such features as narrow fault zones (1). Gravity dams up to 100 ft (30.48 m) in height are generally considered as low dams. Dams of height between 100 ft (30.48 m) and 300 ft (91.44m) are designated as medium-height dams. Dams higher than 300 ft (91.44 m) are considered as high dams.

Gravity Dam

The downstream face of a gravity dam usually has a uniform slope which, if extended, would intersect the vertical upstream face at or near the maximum water level in the reservoir.The upper portion of the dam is made thick enough to accommodate the roadway or other required access as well as to resist the shock of floating objects in the reservoir. The upstream face of a gravity dam is usually kept vertical so that most of its weight is concentrated near the upstream face to resist effectively the tensile stresses due to the reservoir water loading. The thickness of the dam provides resistance to sliding and may, therefore, dictate the slope of the downstream face which is usually in the range of 0.7 to 0.8 (H) : 1(V). The thickness in the lower part of the dam may also be increased by an upstream batter.When it is not feasible to locate the spillway in the abutment, it may be located on a portion of the dam in which case the section of the dam is modified at the top to accommodate the crest of the spillway and at the toe to accommodate the energy dissipation. The stability requirements of such overflow sections of gravity dams would be different from those of non overflow gravity dams.

FORCES ON A GRAVITY DAM

The forces commonly included in the design of a gravity dam . These are as follows (2, 3, 4):

Forces on Gravity Dam

(i) Dead Load

The dead load (Wc) includes the weight of concrete and the weight of appurtenances such as piers, gates, and bridges. All the dead load is assumed to be transmitted vertically to the foundation without transfer by shear between adjacent blocks.

(ii) Reservoir and Tail-water Loads (Ww , Ww′, W1, and W1′)

These are obtained from tail-water curves and range of water surface elevations in reservoir obtained from reservoir operation studies. These studies are based on operating and hydrologic data such as reservoir capacity, storage allocations, stream flow records, flood hydro graphs, and reservoir releases for all purposes. In case of low overflow dams, the dynamic effect of the velocity of approach may be significant and should, therefore, be considered. If gates or other control features are used on the crest, they are treated as part of the dam so far as the application of water pressure is concerned. In case of non-overflow gravity dams, the tail-water should be adjusted for any retrogression. Any increase in tail-water pressure due to curvature of flow in the downstream bucket of an overflow type gravity dam should also be considered in the design of gravity dams (4).

(iii) Uplift Forces

Uplift forces (U) occur due to internal hydraulic pressures in pores, cracks, and seams within the body of a dam, at the contact between the dam and its foundation, and within the foundation.The distribution of internal hydrostatic pressure along a horizontal section through a gravity dam is assumed to vary linearly from full reservoir pressure at the upstream face to zero or tail-water pressure at the downstream face, and to act over the entire area of the section. The pressure distribution is also adjusted depending upon the size, location, and spacing of internal drains. Experimental and analytical studies indicate that the drains set in from the upstream face at 5 per cent of the maximum reservoir depth and spaced laterally twice that distance will reduce the average pressure at the drains to approximately tail-water pressure plus one-third the difference between reservoir water and tail-water pressures It is assumed that uplift forces are not affected by earthquakes (2).

(iv) Silt Load

The construction of a dam across a river carrying sediment invariably results in reservoir sedimentation which causes an additional force (Ws) on the upstream face of the dam. The horizontal silt pressure is assumed equivalent to a hydrostatic load exerted by a fluid with a mass density of 1360 kg/m3. The vertical silt pressure is assumed equivalent to that exerted by a soil with a wet density of 1925 kg/m3.

(v) Ice Pressure

If the designer anticipates the formation of an ice sheet of appreciable thickness and its remaining on the reservoir water surface for a long duration, the ice pressures must be computed using a suitable method of their estimation. In the absence of such a method, ice pressure may be taken as 250 kPa (250 kN/m2) applied over the anticipated area of contact of ice with the face of the dam (2).

(vi) Wave Pressure

The upper portion of a dam is also subjected to the impact of waves. Wave pressure against massive dams of large height is usually of little importance. Wave pressure is related to wave height hw as follows (2):

(a) The maximum wave pressure pw (in kilopascals) occurs at 0.125 hw above the still water level and is given by the equation

pw = 24 hw

where, hw is the height of the wave in metres.

(b) The total wave force Pw (in kilonewtons) is given by

Pw = 20 hw

and acts at 0.375 hw above the still water level in the downstream direction.

hw = 0.032 VF + 0.76 – 0.27 F1/4 for F < 32 km

hw = 0.032 VF for F > 32 km

Here, V is the wind velocity in kilometres per hour and F is the fetch in kilometres.The height of the wave and the wind set-up decide the freeboard which is the vertical distance between the top of the dam and the still water level. The wind set-up S (in metres) is estimated by the Zuider Zee formula

S = V2F / 62000 D

in which, D is the average depth in metres over the fetch distance F.The minimum freeboard should be equal to wind set-up plus 4/3 times wave height above the normal pool elevation or above maximum reservoir level corresponding to the design flood, whichever gives higher crest elevation for the dam (2). The freeboard shall not, however, be less than 1.0 m above the mean water level corresponding to the design flood.

(vii) Earthquake

Gravity dams are elastic structures which may be excited to resonate by seismic disturbances. Such dams should be designed so that they remain elastic when subjected to the design earthquake. The design earthquake should be determined considering (i) historical records of earthquakes to obtain frequency of occurrence versus magnitude, (ii) useful life of the dam, and (iii) statistical approach to determine probable occurrence of earthquakes of various magnitudes during the life of the dam. A gravity dam should also be designed to withstand the maximum credible earthquake which is defined as the one having a magnitude usually larger than any historical recorded earthquake (3).Earthquakes impart random oscillations to the dam which increase the water and silt pressures acting on the dam and also the stresses within the dam. An earthquake movement may take place in any direction. Both horizontal and vertical eathquake loads should be applied in the direction which produces the most unfavourable conditions. For a gravity dam, when the reservoir is full, the most unfavourable direction of earthquake movement is upstream (so that the inertial forces acting downstream may result in resultant force intersecting the base of the dam outside middle-third of the base besides increasing the water load and, therefore, the increased overturning moment) and is downward for vertical earthquake movement as it causes the concrete, and water above the sloping faces of the dam to weigh less resulting in reduced stability of the dam. When the reservoir is empty, more unfavourable is the downstream ground motion causing inertial forces to act upstream so that the resultant may intersect the base of the dam outside middle-third of the base. The effect of earthquake forces depends on (i)their magnitude which, in turn, depends on the severity of the earthquake, (ii) the mass of the structure and its elasticity, and (iii) the earthquake effects on the water load. For estimation of earthquake load, knowledge of earthquake acceleration or intensity, usually expressed in relation to acceleration due to gravity g, is useful. This ratio of earthquake acceleration to gravitational acceleration is termed seismic coefficient and is designated as αh. The value of seismic coefficient for horizontal as well as vertical earthquake accelerations for different zones of the country are different and can be obtained from the Codes.

CAUSES OF FAILURE OF A GRAVITY DAM

A gravity dam may fail on account of overturning. For a gravity dam to be safe against overturning, the dimensions of the dam should be such that the resultant of all the forces intersects the base of the dam within its middle-third portion. Consider any horizontal section (including the base) of a gravity dam and the resultant of all the forces acting on the dam above the section. If the line of action of this resultant passes outside the downstream edge of the section, the dam would overturn. However, the section of a gravity dam is such that the line of action of the resultant force is within the upstream and downstream edges of the section and overturning would never results. But, if the line of action of the resultant passes sufficiently outside the middle-third of the horizontal section, it may cause crushing of the downstream edge of the section. This would reduce the effective width and, hence, the sliding resistance of the section and may cause the resultant to pass outside the dam section. Further, when the resultant passes downstream of the middle-third of the horizontal section, it induces tensile stresses at the upstream edge of the section. These tensile stresses may cause cracks in the dam section which would result in increased uplift pressure. The stabilising forces would, thus, be reduced. It follows, therefore, that before a gravity dam overturns bodily, other types of failures, such as crushing of toe material, sliding, cracking of the material due to tension,and increase in uplift may occur. A gravity dam is considered safe against overturning if the criteria of: (i) no tension on the upstream face, (ii) adequate resistance against sliding, and (iii) suitable quality and sufficient strength of concrete/masonry of dam and its foundation are satisfied.