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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Important Relationships Based on Soil Phase Diagram

A number of useful relationships may be derived based on the foregoing definitions and the soil-phase diagram.

1) Relationships Involving Porosity, Void Ratio, Degree of Saturation, Water Content, Percent Air Voids and Air Content

This may provide a practical approach to the determination of n.

This may provide a practical approach to the determination of e.

                                 

From the above equation, we can derive

These interrelationships between n and e facilitate computation of one if the other is known.

              

By multiplying both of these equations

By definition,

                      

By multiplying the equation of S and e

This equation is valid even if both w and S are expressed as percentages.

For saturated condition,

S = 1.

We know that,

But

2) Relationships Involving Unit Weights, Grain Specific Gravity, Void Ratio and Degree of Saturation

But

                 

              w G = S e

This is a general equation from which the unit weights corresponding to the saturated and dry states of soil may be got by substituting S = 1 and S = 0 respectively.

The submerged unit weight γ′ may be written as

                   

On solving

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Composition of Soil – Three Phase Diagram

Soil is a complex physical system. A mass of soil includes accumulated solid particles or soil grains and the void spaces that exist between the particles. The void spaces may be partially or completely filled with water or some other liquid. Void spaces not occupied by water or any other liquid are filled with air or some other gas. ‘Phase’ means any homogeneous part of the system different from other parts of the system and separated from them by abrupt transition. In other words, each physically or chemically different, homogeneous and mechanically separable part of a system constitutes a distinct phase. A system consisting of more than one phase is said to be heterogeneous.

Since the volume occupied by a soil mass may generally be expected to include material in all the three states of matter - solid, liquid and gas. Soil is referred to as a “three-phase system”. A soil mass as it exists in nature is a more or less random accumulation of soil particles, water and air-filled spaces as shown in Fig. 1 (a). For purposes of analysis it is convenient to represent this soil mass by a block diagram, called ‘Phase-diagram’, as shown in Fig. 1 (b). It may be noted that the separation of solids from voids can only be imagined. The phase-diagram provides a convenient means of developing the weight-volume relationship for a soil.

Fig. 1 (a) Actual Soil Mass (b) Representation of Soil Mass by Phase Diagram

When the soil voids are completely filled with water, the gaseous phase being absent, it is said to be ‘fully saturated’ or merely ‘saturated’. When there is no water at all in the voids, the voids will be full of air, the liquid phase being absent; the soil is said to be dry. (It may be noted that the dry condition is rare in nature and may be achieved in the laboratory through oven drying). In both these cases, the soil system reduces to a ‘two-phase’ one as shown in Fig. 2 (a) and (b). These are merely special cases of the three-phase system.

Fig. 2 (a) Saturated Soil (b) Dry Soil Represented as Two-Phase System

Basic Terminology – Weight Volume Relationship

The general three-phase diagram for soil will help in understanding the terminology and also in the development of more useful relationships between the various quantities. Conventionally, the volumes of the phases are represented on the left side of the phase diagram, while weights are represented on the right side as shown in Fig. 3.

Fig. 3 Soil-Phase Diagram

Va = Volume of air                                Wa = Weight of air (negligible or zero)

Vw = Volume of water                        Ww = Weight of water

Vv = Volume of voids                           Wv = Weight of material occupying void space

Vs = Volume of solids                           Ws = Weight of solids

V = Total volume of soil mass            W = Total weight of solid mass

        Wv = Ww

1) Porosity (n)

Porosity of a soil mass is the ratio of the volume of voids to the total volume of the soil mass. It is denoted by the letter symbol ‘n’ and is commonly expressed as a percentage. Porosity is also known as percentage voids. 

Here       Vv = Va + Vw

               V = Va + Vw + Vs

2) Void Ratio (e)

Void ratio of a soil mass is defined as the ratio of the volume of voids to the volume of solids in the soil mass. It is denoted by the letter symbol ‘e’ and is generally expressed as a decimal fraction.

Both porosity and void ratio are measures of the denseness (or looseness) of soils. As the soil becomes more and more dense, their values decrease. The term porosity is more commonly used in other disciplines such as agricultural engineering. In soil engineering, the term void ratio is more popular. It is more convenient to use void ratio than porosity. When the volume of a soil mass changes, only the numerator (i.e. Vv) in the void ratio changes and the denominator (i.e. Vs) remains constant. However, if the term porosity is used, both the numeration and the denominator change and it will become inconvenient.

3) Degree of Saturation (S)

Degree of saturation of a soil mass is defined as the ratio of the volume of water in the voids to the volume of voids. It is designated by the letter symbol ‘S’ and is commonly expressed as a percentage.

For a fully saturated soil mass, Vw = Vv

Therefore, for a saturated soil mass S = 100%.

For a dry soil mass, Vw is zero.

Therefore, for a perfectly dry soil sample S is zero.

In both these conditions, the soil is considered to be a two-phase system. The degree of saturation is between zero and 100%, the soil mass being said to be ‘partially saturated’ and is the most common condition in nature.

4) Percent Air Voids (na)

Percent air voids of a soil mass is defined as the ratio of the volume of air voids to the total volume of the soil mass. It is denoted by the letter symbol ‘na’ and is commonly expressed in percentage.

5) Air Content (ac)

Air content of a soil mass is defined as the ratio of the volume of air voids to the total volume of voids. It is designated by the letter symbol ‘ac’ and is commonly expressed as a percentage.

6) Water Content/Moisture Content (w)

Water content or Moisture content of a soil mass is defined as the ratio of the weight of water to the weight of solids (dry weight) of the soil mass. It is denoted by the letter symbol 'w' and is commonly expressed as a percentage. 

7) Bulk Unit Weight/Mass Unit Weight (𝛾)

Bulk unit weight or Mass unit weight of a soil mass is defined as the weight per unit volume of the soil mass. It is denoted by the letter symbol 'γ'. Hence, 

Here,         W = Ww + Ws

         and    V = Va + Vw + Vs

The term ‘density’ is used for ‘unit weight’ in soil mechanics, although density means the mass per unit volume and not weight.

8) Unit Weight of Solids (γs)

Unit weight of solids is the weight of soil solids per unit volume of solids alone. It is also sometimes called the ‘absolute unit weight’ of a soil. It is denoted by the letter symbol 'γs'.

9) Unit Weight of Water (γw)

Unit weight of water is the weight per unit volume of water. It is denoted by the letter symbol 'γw'.

It should be noted that the unit weight of water varies in a small range with temperature. It has a convenient value at 4°C, which is the standard temperature for this purpose. γo is the symbol used to denote the unit weight of water at 4°C. The value of γo is 1g/cm³ or 1000 kg/m³ or 9.81 kN/m³.

10) Saturated Unit Weight (γsat)

The saturated unit weight is defined as the bulk unit weight of the soil mass in the saturated condition. This is denoted by the letter symbol γsat.

11) Submerged Unit Weight/Buoyant Unit Weight (γ′)

The submerged unit weight or buoyant unit weight of a soil is its unit weight in the submerged condition. In other words, it is the submerged weight of soil solids (Ws)sub per unit of total volume, V of the soil. It is denoted by the letter symbol γ′.

(Ws)sub is equal to the weight of solids in air minus the weight of water displaced by the solids. Hence

                                                (Ws)sub = Ws – (Vs . γw)

Since the soil is submerged, the voids must be full of water.

The total volume V must be equal to (Vs + Vw) . (Ws)sub may now be written as,

                         (Ws)sub = W – Ww – Vs . γw

                                           = W – Vw . γw – Vs . γw

                                           = W – γw (Vw + Vs)

                                           = W – V . γw

Dividing throughout by V, the total volume,

Or

γ′ = γsat – γw

It may be noted that a submerged soil is invariably saturated, while a saturated soil need not be submerged. This equation may be written as a direct consequence of Archimedes’ Principle which states that the apparent loss of weight of a substance when weighed in water is equal to the weight of water displaced by it. Thus,

γ ′ = γsat – γw

12) Dry Unit Weight (γd)

The dry unit weight is defined as the weight of soil solids per unit of total volume, the former is obtained by drying the soil, while the latter would be got prior to drying. The dry unit weight is denoted by the letter symbol 'γd' and is given by

13) Mass Specific Gravity (Gm)

The mass specific gravity of a soil may be defined as the ratio of mass or bulk unit weight of soil to the unit weight of water at the standard temperature (4°C). This is denoted by the letter symbol Gm and is given by

This is also referred to as ‘bulk specific gravity’ or ‘apparent specific gravity’.

14) Specific Gravity of Solids (G)

The specific gravity of soil solids is defined as the ratio of the unit weight of solids (absolute unit weight of soil) to the unit weight of water at the standard temperature (4°C). This is denoted by the letter symbol G and is given by

This is also known as ‘Absolute specific gravity’ and ‘Grain Specific Gravity’.

15) Specific Gravity of Water (Gw)

Specific gravity of water is defined as the ratio of the unit weight of water to the unit weight of water at the standard temperature (4°C). It is denoted by the letter symbol, Gw and is given by

Since the variation of the unit weight of water with temperature is small, this value is very nearly unity and in practice is taken as such.

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