Brick

A brick is a type of block used to build walls, pavements and other elements in masonry construction. Bricks are obtained by moulding clay in rectangular blocks of uniform size and then by drying and burning these blocks. The term brick denotes a block composed of dried clay, but is now also used informally to denote other chemically cured construction blocks. Bricks can be joined using mortar, adhesives or by interlocking them. In India, standard brick size is 190 x 90 x 90 mm as per the recommendation of BIS. With mortar thickness, the dimension of the brick becomes 200 x 100 x 100 mm which is also known as the nominal size of the modular brick. As bricks are of uniform size, they can be properly arranged, light in weight and hence bricks replace stones. Block is a similar term referring to a rectangular building unit composed of similar materials, but is usually larger than a brick. Lightweight bricks (also called lightweight blocks) are made from expanded clay aggregate.

Fig.1 Bricks

One of the oldest building material brick continues to be a most popular and leading construction material because of being cheap, durable and easy to handle. Clay bricks are used for building-up exterior and interior walls, partitions, piers, footings and other load bearing structures. A brick is rectangular in shape and of size that can be conveniently handled with one hand. Brick may be made of burnt clay or mixture of sand and lime or of Portland cement concrete. Clay bricks are commonly used since these are economical and easily available. The length, width and height of a brick are interrelated as below.

     Length of brick = 2 × Width of brick + Thickness of mortar

     Height of brick = Width of brick

Characteristics of Good Brick

The essential requirements for building bricks are sufficient strength in crushing, regularity in size, a proper suction rate and a pleasing appearance when exposed to view. The various characteristics of good brick are listed below.

1) Size and Shape

The bricks should have uniform size and plane, rectangular surfaces with parallel sides and sharp straight edges. Bricks should be table moulded, well burnt in kilns, free from cracks and with sharp and square edges.

2) Colour

The brick should have a uniform deep red or cherry colour as indicative of uniformity in chemical composition and thoroughness in the burning of the brick.

3) Texture and Compactness

The surfaces should not be too smooth to cause slipping of mortar. The brick should have pre compact and uniform texture. A fractured surface should not show fissures, holes grits or lumps of lime.

4) Hardness and Soundness

The brick should be so hard that when scratched by a finger nail no impression is made. When two bricks are struck together, a ringing sound should be produced.

5) Bricks should not absorb water more than 20 percent by weight for first class bricks and 22

percent by weight for second class bricks, when soaked in cold water for a period of 24 hours.

6) Crushing Strength should not be less than 10 N/mm².

7) Brick Earth should be free from stones, kankars, organic matter, saltpetre, etc.

8) Bricks, when soaked in water for 24 hours, should not show deposits of white salts when allowed to dry in shade.

9) Bricks when broken should show a bright homogeneous and compact structure free from voids.

10) Bricks should be low thermal conductivity and they should be sound proof.

11) Bricks should not break when dropped flat on hard ground from a height of about one meter.

Ingredients of Good Brick Earth

For the preparation of bricks, clay or other suitable earth is moulded to the desired shape after subjecting it to several processes. After drying, it should not shrink and no crack should develop. The clay used for brick making consists mainly of silica and alumina mixed in such a proportion that the clay becomes plastic when water is added to it. It also consists of small proportions of lime, iron, manganese, sulphur etc. The proportions of various ingredients are as follows.

1) Alumina

It is the chief constituent of every kind of clay. A good brick earth should contain 20 to 30 percent of alumina. This constituent imparts plasticity to earth so that it can be moulded. If alumina is present in excess, raw bricks shrink and warp during drying and burning.

2) Silica

A good brick earth should contain about 50 to 60 percent of silica. Silica exists in clay either as free or combined form. As free sand, it is mechanically mixed with clay and in combined form; it exists in chemical composition with alumina. Presence of silica prevents crackers shrinking and warping of raw bricks. It thus imparts uniform shape to the bricks. Durability of bricks depends on the proper proportion of silica in brick earth. Excess of silica destroys the cohesion between particles and bricks become brittle.

3) Lime

A small quantity of lime is desirable in finely powdered state to prevents shrinkage of raw bricks. Excess of lime causes the brick to melt and hence, its shape is last due to the splitting of bricks.

4) Oxide of Iron

A small quantity of oxide of Iron to the extent of 5 to 6 percent is desirable in good brick to imparts red colour to bricks. Excess of oxide of iron makes the bricks dark blue or blackish.

5) Magnesia

A small quantity of magnesia in brick earth imparts yellow tint to bricks and decreases shrinkage. But excess of magnesia decreases shrink leads to the decay of bricks.

The ingredients like, lime, iron pyrites, alkalies, pebbles, organic matter should not present in good brick earth.

Harmful Ingredients in Brick

1) Lime

When a desirable amount of lime is present in the clay, it results in good bricks, but if in excess, it changes the colour of the brick from red to yellow. When lime is present in lumps, it absorbs moisture, swells and causes disintegration of the bricks. Therefore, lime should be present in finely divided state and lumps, if any, should be removed in the beginning itself. Experience has shown that when lime particles smaller than 3 mm diameter hydrate and they produce only small pock mark. Particles larger than this might, if present in any quantity, cause unsightly blemishes or even severe cracking.

2) Pebbles, Gravels, Grits

It does not allow the clay to be mixed thoroughly and spoil the appearance of the brick. Bricks with pebbles and gravels may crack while working.

3) Iron Pyrites

It tends to oxidise and decompose the brick during burning. The brick may split into pieces. Pyrites decolourise the bricks.

4) Alkalis (Alkaline Salts)

It forms less than 10 per cent of the raw clay, are of great value as fluxes, especially when combined with silicates of alumina. These are mainly in the form of soda or potash. When present in excess, alkali makes the clay unsuitable for bricks. They melt the clay on burning and make the bricks unsymmetrical. When bricks come in contact with moisture, water is absorbed and crystallise. On drying, the moisture evaporates, leaving behind grey or white powder deposits on the brick which spoil the appearance. This phenomenon is called efflorescence. Efflorescence should always be dry brushed away before rendering or plastering a wall; wetting it will carry the salts back into the wall to reappear later. If bricks become saturated before the work is completed, the probability of subsequent efflorescence is increased and be protected from rain at all times. During laying, the bricks should be moistened only to the extent that is found absolutely essential to obtain adequate bond between bricks and mortar; newly built brickwork should be protected from rain.

5) Organic Matter

On burning green bricks, the organic matter gets charred and leave pores making the bricks porous; the water absorption is increased and the strength is reduced.

6) Carbonaceous Material

It is in the form of bituminous matter or carbon greatly affects the colour of raw clay. Unless proper precaution is taken to effect complete removal of such matter by oxidation, the brick is likely to have a black core.

7) Sulphur

It is usually found in clay as the sulphate of calcium, magnesium, sodium, potassium or iron, or as iron sulphide. Generally, the proportion is small. If, there is carbon in the clay and insufficient time is given during burning for proper oxidation of carbon and sulphur, the latter will cause the formation of a spongy, swollen structure in the brick and the brick will be decoloured by white blotches.

8)Water

A large proportion of free water generally causes clay to shrink considerably during drying, whereas combined water causes shrinkage during burning. The use of water containing small quantities of magnesium or calcium carbonates, together with a sulphurous fuel often causes similar effects as those by sulphur.

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Artificial Stones

Where durable natural stone is not available at reasonable cost, artificial stone, also known as cast stone becomes the choice. Artificial stone is made with cement and natural aggregates of the crushed stone and sand with desired surface finish. Suitable colouring pigments may be added. However, colouring should not exceed 15% by volume. Cement and aggregates are mixed in proportion of 1:3. Artificial stone can be moulded into the most intricate forms, cast into any size, reinforced to have higher strength, are most suitable for face work, grooves, rebates, etc., can be cast easily and are economical. Some of the artificial stones available are as follows.

1) Concrete Blocks

These are cast at site in the construction of piers or cast in moulds for steps, window sills, etc.

2) Ransom Stones

These are prepared by mixing soda silicate with cement to provide decorative flooring. These are also known as chemical stones. These have compressive strength of about 32 N/mm².

3) Victoria Stones

These are granite pieces with the surfaces hardened by keeping immersed in soda silicate for about two months.

4) Bituminous Stones

Granite and diorite are impregnated with prepared or refined tar to form bituminous stone. These are used for providing noise, wear and dust resistant stone surfaces.

5) Imperial Stones

Finely crushed granite is washed carefully and mixed with Portland cement. The mix is moulded in desired shape and then steam cured for 24 hours. The cured blocks are immersed in silicate tanks for three days. These stones are similar to Victoria stones.

6) Artificial Marbles

It can be either pre-cast or cast-in-situ. These are made from portland gypsum cement and sand. In the precast variety, the cast-stone is removed after three days. On the fifth day of casting these are treated with a solution, liquid fluorite of magnesia. It is then washed and wrapped in paper for 24 hours and then once again treated with the liquid. After one month the stone is polished by rubbing emery over the surface with a linen rag ball dipped in mixture of lime water and silicate of potash and then the process is repeated without emery. It is used for external works. Cast-in-situ variety is made by laying the mix on canvas, in thickness about 1.5 mm more than the required thickness of the stone. The surface is rubbed over and the air holes are filled with mix. Grinding is done by hand or machine. The surface is then rubbed with a polishing stone. Final rubbing is done with a ball of wool moistened with alum water dipped into a 1:3 mix of hartshorn powder and diatomite.

7) Garlic Stone

It is produced by moulding a mixture of iron slag and portland cement. These are used as flag stones, surface drains, etc.

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Tests on Building Stones

Following are different tests on building stones.

1) Acid Test

This test is carried out to understand the presence of calcium carbonate in building stone. A sample of stone weighing about 50 to 100 gm is taken. It is placed in a solution of hydrophobic acid having strength of one percent and is kept there for seven days. Solution is agitated at intervals. A good building stone maintains its sharp edges and keeps its surface free from powder at the end of this period. If the edges are broken and powder is formed on the surface, it indicates the presence of calcium carbonate and such a stone will have poor weathering quality.

2) Attrition Test

This test is done to find out the rate of wear of stones, which are used in road construction. The results of the test indicate the resisting power of stones against the grinding action under traffic. The following procedure is adopted.

• Samples of stones is broken into pieces about 60mm size.
• Such pieces, weighing 5 kg are put in both the cylinders of Deval’s attrition test machine.
• Diameter and length of cylinder are respectively 20 cm and 34 cm.
• Cylinders are closed. Their axes make an angle of 30 degree with the horizontal.
• Cylinders are rotated about the horizontal axis for 5 hours at the rate of 30 rpm.
• After this period, the contents are taken out from the cylinders and they are passed through a sieve of 1.5mm mesh.
• Quality of material which is retained on the sieve is weighed.

Percentage wear worked out as follows

3) Crushing Strength Test

Samples of stone is cut into cubes of size 40 mm and are finely dressed and finished. Maximum number of specimen to be tested is three. Such specimen should be placed in water for about 72 hours prior to test and therefore tested in saturated condition. Load bearing surface is then covered with plaster of paris of about 5mm thick plywood. Load is applied axially on the cube in a crushing test machine. Rate of loading is 140 kg/sq.cm per minute. Crushing strength of the stone per unit area is the maximum load at which the sample crushes or fails divided by the area of the bearing face of the specimen.

4) Freezing and Thawing Test

Stone specimen is kept immersed in water for 24 hours. It is then placed in a freezing machine at -12 ⁰C for 24 hours. Then it is thawed or warmed at atmospheric temperature. This should be done in shade to prevent any effect due to wind, sun rays, rain etc. This procedure is repeated several times and the behaviour of stone is carefully observed.

5) Hardness Test

For determining the hardness of a stone, the test is carried out as follows.

• A cylinder of diameter 25mm and height 25mm is taken out from the sample of stone.
• It is weighed.
• The sample is placed in Dorry’s testing machine and it is subjected to a constant pressure.
• Annular steel disc machine is then rotated at a speed of 28 rpm.
• During the rotation of the disc, coarse sand of standard specification is sprinkled on the top of disc.
• After 1000 revolutions, specimen is taken out and weighed.
• The coefficient of hardness is found out from the following equation

6) Impact Test

For determining the toughness of stone, it is subjected to impact test in an Impact Test Machine as followed.

• A cylinder of diameter 25mm and height 25mm is taken out from the sample of stones.
• It is then placed on cast iron anvil of machine.
• A steel hammer of weight 2 kg is allowed to fall axially in a vertical direction over the specimen.
• Height of first blow is 1 cm, that of second blow is 2 cm, that of third blow is 3 cm and so on.
• Blow at which specimen breaks is noted. If it is n^th blow, ‘n’ represents the toughness index of stone.

7) Microscopic Test

The sample of the test is subjected to microscopic examination. The sections of stones are taken and placed under the microscope to study the various properties such as

1. Average grain size
2. Existence of pores, fissures, veins and shakes
3. Mineral constituents
4. Nature of cementing material
5. Presence of any harmful substance
6. Texture of stones etc.

8) Smith’s Test

This test is performed to find out the presence of soluble matter in a sample of stone. Few chips or pieces of stone are taken and they are placed in a glass tube. The tube is then filled with clear water. After about an hour, the tube is vigorously stirred or shaken. Presence of earthy matter will convert the clear water into dirty water. If water remains clear, stone will be durable and free from any soluble matter.

9) Water Absorption Test

The test is carried out as follows.

• From the sample of stone, a cube weighing about 50gm is prepared. Its actual weight is recorded as W1 gm.
• Cube is then immersed in distilled water for a period of 24 hrs.
• Cube is taken out of water and surface water is wiped off with a damp cloth.
• It is weighed again. Let the weight be W2 gm.
• Cube is suspended freely in water and its weight is recorded. Let this be W3 gm.
• Water is boiled and cube is kept in boiling water for 5 hours.
• Cube is removed and surface water is wiped off with a damp cloth. Its weight is recorded. Let it be W4 gm.

From the above observations, values of the following properties of stones are obtained.

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Deterioration and Preservation of Stone Work

Deterioration of Stone Work

The various natural agents such as rain, heat, etc. and chemicals deteriorate the stones with time.

1) Rain

Rain water acts both physically and chemically on stones. The physical action is due to the erosive and transportation powers and the latter due to the decomposition, oxidation and hydration of the minerals present in the stones.

2) Physical Action

Alternate wetting by rain and drying by sun causes internal stresses in the stones and consequent disintegration.

3) Chemical Action

In industrial areas the acidic rain water reacts with the constituents of stones leading to its deterioration.

4) Decomposition

The disintegration of alkaline silicate of alumina in stones is mainly because of the action of chemically active water. The hydrated silicate and the carbonate forms of the alkaline materials are very soluble in water and are removed in solution leaving behind a hydrated silicate of alumina (Kaolinite). The decomposition of feldspar is represented as

5) Oxidation and Hydration

Rock containing iron compounds in the forms of peroxide, sulphide and carbonate are oxidised and hydrated when acted upon by acidic rain water. As an example the peroxide - FeO is converted into ferric oxide - Fe₂O₃ which combines with water to form FeO.nH₂O. This chemical change is accompanied by an increase in volume and results in a physical change manifested by the liberation of the neighbouring minerals composing the rocks. As another example iron sulphide and siderite readily oxidize to limonite and liberates sulphur, which combines with water and oxygen to form sulphuric acid and finally to sulphates.

6) Frost

In cold places frost pierces the pores of the stones where it freezes, expands and creates cracks.

7) Wind

Since wind carries dust particles, the abrasion caused by these deteriorates the stones.

8) Temperature Changes

Expansion and contraction due to frequent temperature changes cause stone to deteriorate especially if a rock is composed of several minerals with different coefficients of linear expansion.

9) Vegetable Growth

Roots of trees and weeds that grow in the masonry joints keep the stones damp and also secrete organic and acidic matters which cause the stones to deteriorate. Dust particles of organic or nonorganic origin may also settle on the surface and penetrate into the pores of stones. When these come in contact with moisture or rain water, bacteriological process starts and the resultant microorganism producing acids attack stones which cause decay.

10) Mutual Decay

When different types of stones are used together mutual decay takes place. For example, when sandstone is used under limestone, the chemicals brought down from limestone by rain water to the sandstone will deteriorate it.

11) Chemical Agent

Smokes, fumes, acids and acid fumes present in the atmosphere deteriorate the stones. Stones containing CaCO₃, MgCO₃ are affected badly.

12) Lichens

These destroy limestone but act as protective coats for other stones. Molluses gradually weaken and ultimately destroy the stone by making a series of parallel vertical holes in limestones and sand stones.

Durability of Stones

Quarrying and cutting have a great bearing on the weathering properties of stones. Stone from top ledges of limestone, granite and slate and from the exposed faces of the rock bed is likely to be less hard and durable. Highly absorbent stone should not be quarried in freezing weather since the rock is likely to split. The method of blasting and cutting also influences the strength of the stone and its resistance to freezing and temperature changes. Small, uniformly distributed charge of blasting powder has a lesser weakening effect than large concentrations of explosives.

A porous stone is less durable than a dense stone, since the former is less resistant to freezing. Also, rocks with tortuous pores and tubes are more apt to be injured by freezing than those of equal porosity having straight pores and tubes. Repeated hammering in cutting is likely to injure the stone. Polished stone is more enduring than rough surfaced work, since the rain slides off the former more easily. Stones from stratified rocks should be placed along the natural bed in order to secure maximum weathering resistance. Pyrite, magnetite and iron carbonate oxidize in weathering and cause discolouration of the stone in which they are present. Since oxidation is accompanied by a change in volume, the surrounding structure is weakened.

Preservation of Stones

Preservation of stone is essential to prevent its decay. Different types of stones require different treatments. But in general stones should be made dry with the help of blow lamp and then a coating of paraffin, linseed oil, light paint, etc. is applied over the surface. This makes a protective coating over the stone. However, this treatment is periodic and not permanent. When treatment is done with the linseed oil, it is boiled and applied in three coats over the stone. Thereafter, a coat of dilute ammonia in warm water is applied.

The structure to be preserved should be maintained by washing stones frequently with water and steam so that dirt and salts deposited are removed from time to time. However, the best way is to apply preservatives. Stones are washed with thin solution of silicate of soda or potash. Then, on drying a solution of CaCl₂ is applied over it. These two solutions called Szerelmy’s liquid, combine to form silicate of lime which fills the pores in stones. The common salt formed in this process is washed afterwards. The silicate of lime forms an insoluble film which helps to protect the stones.

Sometimes lead paint is also used to preserve the stones, but the natural colour of the stone is spoiled. Painting stone with coal tar also helps in the preservation but it spoils the beauty of the stone. Use of chemicals should be avoided as far as possible, especially the caustic alkalis. Although cleaning is easy with chemicals, there is the risk of introducing salts which may subsequently cause damage to the stone.

In industrial towns, stones are preserved by application of solution of baryta, Ba(OH)₂ — Barium hydrate. The sulphur dioxide present in acid reacts on the calcium contents of stones to form calcium sulphate. Soot and dust present in the atmosphere adhere to the calcium sulphate and form a hard skin. In due course of time, the calcium sulphate so formed flakes off and exposes fresh stone surface for further attack. This is known as sulphate attack. Baryta reacts with calcium sulphate deposited on the stones and forms insoluble barium sulphate and calcium hydroxide. The calcium hydroxide absorbs carbon dioxide from the air to form calcium carbonate.

The treatments, if carefully applied under favourable circumstances, may result in an apparent slowing down of the rate of decay. However, the rate of decay of stone is so slow that a short period experience is of very little value in establishing the effectiveness of the treatment. Also, there is some evidence that treatments which appear to be successful for few years, fail to maintain the improvement. In fact, the value of preservatives is not yet proved and they may actually be detrimental if judged over a long period.

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Couple

A pair of two equal and unlike parallel forces (i.e. forces equal in magnitude, with lines of action parallel to each other and acting in opposite directions) is known as a couple.

A couple is unable to produce any translatory motion (i.e., motion in a straight line). But it produces a motion of rotation in the body on which it acts. The simplest example of a couple is the forces applied to the key of a lock, while locking or unlocking it.

Arm of a Couple

The perpendicular distance (a), between the lines of action of the two equal and opposite parallel forces is known as arm of the couple as shown in Fig.1.

Fig. 1

Moment of a Couple

The moment of a couple is the product of the force (i.e., one of the forces of the two equal and opposite parallel forces) and the arm of the couple.

Mathematically,

                  Moment of a couple = P × a

Where,

     P = Magnitude of the force

     a = Arm of the couple

Classification of Couple

The couple may be classified into the following two categories, depending upon their direction, in which the couple tends to rotate the body, on which it acts.

1) Clockwise Couple

A couple, whose tendency is to rotate the body, on which it acts, in a clockwise direction, is known as a clockwise couple as shown in Fig. 2 (a). Such a couple is also called positive couple.

Fig. 2

2) Anticlockwise Couple

A couple, whose tendency is to rotate the body, on which it acts, in an anticlockwise direction, is known as an anticlockwise couple as shown in Fig. 2 (b). Such a couple is also called a negative couple.

Characteristics of a Couple

A couple (whether clockwise or anticlockwise) has the following characteristics.

1. The algebraic sum of the forces constituting the couple is zero.
2. The algebraic sum of the moments of the forces constituting the couple about any point is the same and equal to the moment of the couple itself.
3. A couple cannot be balanced by a single force. But it can be balanced only by a couple of opposite sense.
4. Any no. of coplanar couples can be reduced to a single couple, whose magnitude will be equal to the algebraic sum of the moments of all the couples.

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Varignon’s Principle of Moments (Law of Moments)

Varignon’s theorem states that “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”

Proof

Case (i) When the forces are concurrent

Let ‘P’ and ‘Q’ be any two forces acting at a point O along lines OX and OY respectively and let D be any point in their plane as shown in Fig. 1. Line DC is drawn parallel to OX to meet OY at B. The line OB represent the force Q in magnitude and direction and OA represent the force P in magnitude and direction.

Fig. 1

With OA and OB as the adjacent sides, parallelogram OACB is completed and OC is joined. Let ‘R’ be the resultant of forces P and Q. 

Then, according to the “Theorem of parallelogram of forces”, R is represented in magnitude and direction by the diagonal OC of the parallelogram OACB.

The point D is joined with points O and A. The moments of P, Q and R about D are given by 2 x area of ΔAOD, 2 x area of ΔOBD and 2 x area of ΔOCD respectively.

With reference to Fig1. (a), the point D is outside the <AOB and the moments of P, Q and R about D are all anti-clockwise and hence these moments are treated as positive.

Now, the algebraic sum of the moments of P and Q about

      D = 2ΔAOD + 2ΔOBD

          = 2 (ΔAOD + ΔOBD)

          = 2 (ΔAOC + ΔOBD)

          = 2 (ΔOBC + ΔOBD)

          = 2ΔOCD

          = Moment of R about D

[As AOC and AOD are on the same base and have the same altitude. ΔAOD = ΔOBC. As AOC and OBC have equal bases and equal altitudes. ΔAOC = ΔOBC]

With reference to Fig 1. (b), the point D is within the <AOB and the moments of P, Q and R about D are respectively anti-clockwise, clockwise and anti-clockwise.

Now, the algebraic sum of the forces P and Q about

      D = 2ΔAOD - 2 ΔOBD

          = 2 (ΔAOD-ΔOBD)

          = 2 (ΔAOC- ΔOBD)

          = 2(ΔOBC - ΔOBD)

          = 2ΔOCD

          = Moment of R about D

Case (ii) When the forces are parallel

Let P and Q be any two like parallel forces (i.e. the parallel forces whose lines of action are parallel and which act in the same sense) and O be any point in their plane.

Let R be the resultant of P and Q.

Then,

       R = P + Q

From O, line OACB is drawn perpendicular to the lines of action of forces P, Q and R intersecting them at A, B and C respectively as shown in Fig 2.

Fig 2

Now, algebraic sum of the moments of P and Q about O

     = P x OA + Q x OB

     = P x (OC - AC) + Q x (OC + BC)

     = P x OC – P x AC + Q x OC + Q x BC

  But P x AC = Q x BC

Algebraic sum of the moments of P and Q about O

     = P x OC + Q x OC

     = (P + Q) x OC

     = R x OC

     = Moment of R about O

In case of unlike parallel forces also it can be proved that the algebraic sum of the moments of two unlike parallel forces (i.e. the forces whose lines of action are parallel but which act in reverse senses) about any point in their plane is equal to the moment of their resultant about the same point.

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Moment of a Force

Moment is the turning effect produced by a force, on the body, on which it acts. The moment of a force is equal to the product of the force and the perpendicular distance of the point, about which the moment is required and the line of action of the force.

Fig. 1

Mathematically, moment,

                        M = P × d

  Where,

           P = Force acting on the body

           d = Perpendicular distance between the point, about which the moment is required and the

                  line of action of the force.

Fig. 2

Let a force ‘P’ act on a body which is hinged at O. Then, moment of P about the point O in the body is

                                       Moment = F x ON

Where 

           ON = perpendicular distance of O from the line of action of the force F.

Graphical Representation of Moment

Consider a force P represented in magnitude and direction, by the line AB. Let O be a point, about which the moment of this force is required to be found out, as shown in Fig. 3. From O, draw OC perpendicular to AB. Join OA and OB.

Fig. 3 Representation of Moment

Now moment of the force P about O

                       = P × OC

                       = AB × OC

But AB × OC is equal to twice the area of triangle ABO.

Thus the moment of a force about any point is equal to twice the area of the triangle, whose base is the line to some scale representing the force and whose vertex is the point about which the moment is taken.

Units of Moment

Since the moment of a force is the product of force and distance, therefore the units of the moment will depend upon the units of force and distance. Thus, if the force is in Newton and the distance is in meters, then the units of moment will be Newton-meter (N-m). Similarly, the units of moment may be kN-m (i.e. kN × m), N-mm (i.e. N × mm) etc.

Types of Moment

The moments are of two types.

1) Clockwise Moment

It is the moment of a force, whose effect is to turn or rotate the body, about the point in the same direction in which hands of a clock move as shown in Fig. 4 (a).

2) Anticlockwise Moment

It is the moment of a force, whose effect is to turn or rotate the body, about the point in the opposite direction in which the hands of a clock move as shown in Fig. 4 (b).

The general convention is to take clockwise moment as positive and anticlockwise moment as negative. But there is no hard and fast rule regarding sign convention of moments.

Fig. 4 Clockwise and Anticlockwise Moments

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Analytical Method for Finding Resultant Force - Parallelogram Law of Forces

The resultant force of a given system of forces may be found out analytically by Parallelogram law of forces.

Parallelogram Law of Forces

It states, “If two forces acting simultaneously on a particle, be represented in magnitude and direction by the two adjacent sides of a parallelogram then their resultant may be represented in magnitude and direction by the diagonal of the parallelogram, which passes through their point of intersection.”

Let forces ‘P’ and ‘Q’ acting at a point O be represented in magnitude and direction by OA and OB respectively as shown in Fig.1. Then, according to the theorem of parallelogram of forces, the diagonal OC drawn through O represents the resultant of P and Q in magnitude and direction.

Fig.1

Determination of the Resultant of Two Concurrent Forces with the Help of Law of Parallelogram of Forces

Consider two forces P and Q acting at and away from point A as shown in Fig. 2. Let the forces P and Q are represented by the two adjacent sides of a parallelogram AD and AB respectively as shown in Fig. 2. Let ‘θ’ be the angle between the force P and Q and ‘α’ be the angle between R and P. Extend line AB and drop perpendicular from point C on the extended line AB to meet at point E.

Fig. 2

Consider Right angle triangle ACE,

           AC² = AE² + CE2

                   = (AB + BE)² + CE2

                   = AB² + BE² + 2.AB.BE + CE2

                   = AB² + BE² + CE² + 2.AB.BE …………………….. (1)

Consider right angle triangle BCE,

            BC² = BE² + CE²    and     BE = BC. Cos θ

Putting BC² = BE² + CE² in equation (1), we get

            AC² = AB² + BC² + 2.AB.BE      ……………………….. (2)

Putting BE = BC. Cos θ in equation (2)

            AC² = AB² + BC² + 2.AB. BC. Cos θ

But,   AB = P, BC = Q and AC = R

               R = √ (P2 + Q2 + 2PQ Cos θ)

In triangle ACE

      

Now let us consider two forces F1 and F2 are represented by the two adjacent sides of a parallelogram

i.e. F1 and F2 = Forces whose resultant is required to be found out,

θ = Angle between the forces F1 and F2

α = Angle which the resultant force makes with one of the forces (say F1).

Then resultant force

            R= √ (F1² + F2² + 2F1 F2 Cos θ)

                  

If ‘α’ is the angle which the resultant force makes with the other force F2, then

                 

Cases

1) If θ = 0⁰ i.e., when the forces act along the same line, then

                                R_max = F1 + F2

2) If θ = 90⁰ i.e., when the forces act at right angle, then

                              R= √ (F1² + F2²)

3) If θ = 180⁰ i.e., when the forces act along the same straight line but in opposite directions, then

                               R_min= F1 – F2

In this case, the resultant force will act in the direction of the greater force.

4) If the two forces are equal i.e., when F1 = F2 = F then

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Resolution of a Given Force into Two Components in Two Assigned Direction

Let ‘P’ be the given force represented in magnitude and direction by OB as shown in Fig. 1. Also let OX and OY be two given directions along which the components of P are to be found out.

Fig. 1

Let < BOX = α and < BOY = β

From B, lines BA and BC are drawn parallel to OY and OX respectively. Then the required components of the given force P along OX and OY are represented in magnitude and direction by OA and OC respectively.

Since AB is parallel to OC, 

       <BAX = <AOC = α + β

      < AOB = 180⁰ – (α + β)

Now, in Δ OAB 

Determination of Resolved Parts of a Force

Resolved parts of a force mean components of the force along two mutually perpendicular directions. Let a force F represented in magnitude and direction by OC make an angle θ with OX. Line OY is drawn through O at right angles to OX as shown in Fig.2.

Fig. 2

Through C, lines CA and CB are drawn parallel to OY and OX respectively. Then the resolved parts of the force F along OX and OY are represented in magnitude and direction by OA and OB respectively.

Now in the right angled Δ AOC,

  

       i.e. OA = F cos θ

Since OA is parallel to BC, 

        <OCB = <AOC = θ

In the right angled Δ OBC,

 

i.e., OB = F sin θ

Thus, the resolved parts of F along OX and OY are respectively F cos θ and F sin θ.

Significance of the Resolved Parts of A Force

Fig. 3

Let 50 KN force is required to be applied to a body along a horizontal direction CD in order to move the body along the plane AB. Then it can be said that to move the body along the same plane AB, a force of 50kN is to be applied at an angle of 60° with the horizontal as CD = 50 cos60° = 25kN.

Similarly, if a force of 43.3kN is required to be applied to the body to lift it vertically upward, then the body will be lifted vertically upward if a force of 50kN is applied to the body at an angle of 60° with the horizontal, as the resolved part of 50kN along the vertical CE = 50 sin60° = 43.3kN.

Thus, the resolved part of a force in any direction represents the whole effect of the force in that direction.

Equilibriant

Equilibrant of a system of forces is a single force which will keep the given forces in equilibrium. Evidently, equilibrant is equal and opposite to the resultant of the given forces.

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Example Questions Related to Soil Phase Diagram

1) One cubic metre of wet soil weighs 19.80 kN. If the specific gravity of soil particles is 2.70 and water content is 11%, find the void ratio, dry density and degree of saturation.

Solution

Bulk unit weight, = 19.80 kN/m³

Water content, w = 11%

                                    = 0.11

Specific gravity of soil particles G = 2.70

                         

Unit weight of water, γw = 9.81 kN/m³

                       

            Void ratio, e = 0.485

 Degree of Saturation, S = 61.24 %

2) Determine the Water content, Dry density, Bulk density, Void ratio and Degree of saturation from the following data.

Sample size 3.81 cm dia. × 7.62 cm ht.

Wet weight = 1.668 N

Oven-dry weight = 1.400 N

Specific gravity = 2.7

Wet weight, W = 1.668 N

Oven-dry weight, Wd = 1.400 N

Solution

   

Specific gravity of solids, G = 2.70

3) A soil has bulk density of 20.1 kN/m³ and water content of 15%. Calculate the water content if the soil partially dries to a density of 19.4 kN/m³ and the void ratio remains unchanged.

Solution

Bulk unit weight, γ = 20.1 kN/m³

Water content, w = 15%

If the void ratio remains unchanged while drying takes place, the dry unit weight also remains unchanged since G and γw do not change.

New value of γ = 19.4 kN/m³

           

Hence the water content after partial drying = 10.86%

4) The porosity of a soil sample is 35% and the specific gravity of its particles is 2.7. Calculate its void ratio, dry density, saturated density and submerged density.

Solution

Porosity, n = 35%

         

Specific gravity of soil particles = 2.7

       

           

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