BPSC AE Paper 5 (Civil) 25 March 2022 Official Paper (Previous Year Paper)

Concepts:

In case of a simply supported beam subjected to a concentrated moving load:

• The maximum bending moment occurs when the load is applied at mid-span and it will occur at mid-span itself.
• The maximum shear force will occur when the load is applied at any one of the support and it will occur near the support itself.

Calculation:

Given data,

P = 8 kN

L = 15 m

For maximum bending moment load is applied at mid span 

 Mmax = PL/4 = (8 × 15)/4 = 30 kN-m

Concept:

When in addition to equilibrium equations, compatibility equations are used to evaluate the unknown reactions and internal forces in any structure, such analysis is called indeterminate analysis, and such structure is called indeterminate structure.

We have two distinct methods of analysis for such an indeterminate structure:

a. Force method of analysis

b. Displacement method of analysis

Difference between force method and Displacement method:-

Concept:

Flexibility Matrix:

• Any element ${{\rm{F}}_{{\rm{ij;}}}}$in flexibility matrix represents that the displacement (measured in the direction of force) occurs in i-th node when a unit load is applied at j-th node.
• So, for main diagonal elements ${{\rm{F}}_{{\rm{ii;}}}}$it essentially implies that displacement (measured in the direction of force) occurs in i-th node when a unit load is applied at i-th node itself.
• It is evident that if a load is applied at a certain node then the displacement at that node should be in the direction of the force. That’s why the main diagonal elements in flexibility matrix is always positive and cannot be negative at all.

Stiffness Matrix:

• Any element ${{\rm{K}}_{{\rm{ij;}}}}$in stiffness matrix represents that the action (force or moment) develops in i-th node when a unit displacement (deflection or rotation) is given at j-th node which is measured in the direction of given displacement.
• So, for main diagonal elements ${{\rm{K}}_{{\rm{ii;}}}}$it essentially implies that the action (force or moment) develops in i-th node when a unit displacement (deflection or rotation) is given at i-th node itself.
• It is evident that if displacement is given at a certain node then the action at that node should developed in the direction of the displacement. That’s why the main diagonal elements in stiffness matrix is always positive and cannot be negative at all.

In structural problems, flexibility and stiffness matrices are always square matrix.

Explanation:

Stiffness Matrix Method

1. This method is also known as displacement method or equilibrium method.
2. It is suitable if static indeterminacy (Ds) > kinematic indeterminacy (Dk­) for the structure.

Stiffness is defined as Force required to produce unit displacement, $k = \frac{P}{{\rm{\Delta }}}$

K12 → Force at 2 due to unit displacement at 1

Flexibility Matrix Method

1. This method is also known as force method or compatibility method.
2. It is suitable if kinematic indeterminacy (Dk) > static indeterminacy (Ds) for the structure.

Flexibility is defined as Displacement caused due to unit Force.

δ12 → Displacement at 2 due to Force at 1

∴ To generate a jth column of flexibility matrix a unit force is applied at coordinate j and the displacement are calculated at all coordinates.

Explanation:

True Meridian

• A true meridian is defined as the plane that passes through true north poles (geographical north pole) and true south poles (geographical South Pole) at the place of observation.
• True meridian can be established by astronomical observation as it passes through true north and south.
• The true meridians at any place are not variable and they converge to a point at the pole in the northern or southern hemisphere.
• The maps prepared by National survey departments are based on the true meridians.

Magnetic meridian

• The magnetic meridian of a place can be defined as the vertical plane which passes through the imaginary line joining the magnetic north and the south poles.
• This plane would intersect the surface of the earth in a longitude-like circle.
• A magnetic needle, which is free to swing horizontally, would then lie in the magnetic meridian and the north pole of the needle would point towards the magnetic north pole.
• Since the line joining the magnetic poles is titled with respect to the geographic axis of the earth, the magnetic meridian at a point makes an angle with the geographic meridian.
• This is the angle between the true geographic north and the north shown by a compass needle. This angle is called the magnetic declination or simply declination.

Explanation:

In general, let a frame have j joints and m members.

If m + r = 2j , then the frame is perfect frame.

If m + r < 2j , then the frame is deficient frame.

If m + r > 2j , then the frame is redundant frame.

A perfect frame can always be analyzed by the condition of equilibrium. While a redundant frame cannot be fully analyzed by the condition of equilibrium.

Hence, If in a pin-jointed plane frame (m + r) > 2 j, then the frame is stable and statically indeterminate.

Additional InformationThe static indeterminacy for frames under different conditions are:

1. Pin Jointed space Frame (3D)

Ds = m + r - 3j

2. Pin Jointed Plane Frame (2D)

Ds = m + r - 2j

 3. Rigid Jointed Plane Frame (2D):

DS = 3m + r - 3j - R;

4. Rigid Jointed Space Frame (3D):

DS = 6m + r - 6j - 3R;

Where,

m is the no of members

r is no. of support reactions

R is total no. of releases

J is the nos. of joints

Explanation:

(i) Clapeyron’s Theorem of three moment’s equation is derived using Mohr’s first and second moment theorems. This theorem can be comfortably used for simply supported and continuous beams (even when there is support settlements). It is not applicable to fixed supports.

(ii) But in case of a fixed end of a continuous beam, to apply the Clapeyron’s Theorem of Three Moments we assume the following things:

• Imaginary length (Zero length)
• Infinite stiffness
• Replacing fixed end with simply supported end.

Concept:

As per IS 456:2000, Cl.36.1

Characteristics strength of the material is defined as the strength of material below which not more than 5% of the test results are expected to fail.

Also,

Mean strength is defined as the strength below which not more than 50% of the test results are expected to fail.

fm = fck + 1.65σ

Here,

fck – characteristics strength (MPa)

fm – mean strength (MPa)

σ – standard deviation (MPa)

The values of standard deviation for different grades of concrete are given below –

Explanation:

From the strain distribution diagram in doubly reinforced beams,

The strain at the level of compression steel reinforcement ϵsc may be written as   

$\begin{array}{l} \frac{{0.0035}}{{{x_u}}} = \frac{{{\varepsilon _{sc}}}}{{{x_u} - d'}}\ {\varepsilon _{sc}} = \frac{{0.0035}}{{{x_u}}}\left( {{x_u} - d'} \right) \end{array}$

${\varepsilon _{sc}} = 0.0035\left( {1 - \frac{{d'}}{{{x_u}}}} \right)$

d' = Distance of centroid of compression steel from the extreme compression fiber of concrete.

xu = depth of neutral axis from the extreme compression fiber of concrete.

Important Points

• Maximum compressive strain in concrete at the outermost fibre (ϵ cu) is taken as 0.0035.
• Maximum strain (ϵst) in tension reinforcement at the level of centroid of reinforcement steel at the ultimate limit state shall not be less than ${\varepsilon ^*}_{st} = \frac{{0.87{f_y}}}{{{E_s}}} + 0.002$

Concept:

Diagonal tension:

Diagonal Tension is the reason for the cracks in shear failure. The diagonal tension crack originates at an angle of 45° and breaks the beam into two pieces. It is present in both the brittle and ductile material.

Shear reinforcement is required to prevent the propagation of diagonal cracks.

The diagonal tension cracks develop in the beam when the strength of the beam in diagonal tension is less than the strength in flexural tension. On this type of failure, the cracks start to develop due to the flexure at mid-span. For the prevention of diagonal tension failure, the shear link is provided.

Hence, The chances of diagonal tension cracks in RCC member reduce when axial compression and shear force act simultaneously.

According to IS 456: 200, the shear reinforcement is must be provided

1. To resist the diagonal tension or principal tension. In the beam diagonal failure occurs when the shear span is higher than three times the value of effective depth.
2. To avoid the formation of tension cracks due to any reason given below

• Due to diagonal tension
• Due to the sudden increase in stress
• Due to shrinkage and creep
• To improve the ductility of member
• To improve the rate of main reinforcement

Explanation:

High strength concrete is needed in prestressing over the low strength concrete due to following reasons,

• 1. Due to large prestressing force to be applied by tendon, large bearing stress gets developed in concrete at the ends by anchoring devices. High bearing strength can be achieved by the use of high strength concrete
• 2. Bursting stresses at the ends cannot be effectively resisted by low strength concrete.
• 3. Stresses transferred from tendons to concrete takes due to bond between concrete and steel. High strength bond can be achieved by the use of high strength concrete.
• 4. With the use of high strength concrete in prestress, Shrinkage cracks are low as compared to low strength concrete.
• 5. Use of high strength concrete reduces the size of concrete member thereby resulting in reduction of concrete and the self weight of member.

Due to any prestress loss like shrinkage and creep all the steel bars will lose their pre-tension simultaneously. To avoid that, we use few numbers of high tensile cables. This way, we have enough strain left even after some of the initial strain is lost due to prestress loss due to shrinkage, creep, elastic shortening , friction, slip etc.). Therefore, in a pre-stressed member, it is advisable to use high tension steel.

Concept :

A balanced section is one in which the area of tension steel is such that the failure of both concrete and steel occurs simultaneously.

For WSM method:

The critical depth of the neutral axis is given by –

From stress distribution diagram,

$\frac{{\frac{{{σ _{st}}}}{m}}}{{d - {X_c}}} = \frac{{{σ _{cbc}}}}{{{X_c}}}$

⇒ $\frac{{{σ _{st}}}}{{m; \times; {σ _{cbc}}}} = \frac{{d - {X_c}}}{{{X_c}}}$

Where, Xc = kd (k = Neutral axis depth factor for balanced section)

⇒ $\frac{{{σ _{st}}}}{{m; \times; {σ _{cbc}}}} = \frac{{d - kd}}{{kd}}$

⇒ $\frac{{{σ _{st}}}}{{m; \times ;{σ _{cbc}}}} = \frac{{1 - k}}{k}$

$k = \frac{{m \times {σ _{cbc}}}}{{{σ _{st}} + m \times {σ _{cbc}}}}$

Put the value of modular ratio, $m = \frac{{280}}{{3{σ _{cbc}}}}$

$k = \frac{{280}}{{3{σ _{st}} ;+; 280}}$

Thus the neutral axis factor for the balanced section depends only on σst and is independent of σcbc.

Calculation:

σ_st = 140 N / mm2

Neutral axis depth factor for balanced section

$k = \frac{{280}}{{3{σ _{st}} ;+; 280}}= \frac{{280}}{{3{\times140} ;+; 280}}$= 0.40

Hence, the depth of neutral axis for a single reinforced rectangular balanced section is 0.40d.

Explanation:

Gantry Girders: These are laterally unsupported beams provided in industrial buildings to carry cranes.

These are supposed to carry the following loads due to moving crane:

• vertical loads or (gravity load)
• lateral loads (produced due to jerk experienced while loading, unloading, and movement of the cranes)
• longitudinal loads (produced due to longitudinal movement of the crane)

Hence, Gantry girders are designed to resist lateral, longitudinal, and vertical loads.

Explanation:

Full Prestressing

When the level of prestressing is such that no tensile stress is allowed in concrete under service loads, it is called Full Prestressing (Type 1, as per IS:1343 - 1980).

Limited Prestressing

When the level of prestressing is such that the tensile stress under service loads is within the cracking stress of concrete, it is called Limited Prestressing (Type 2).

Partial Prestressing

When the level of prestressing is such that under tensile stresses due to service loads, the crack width is within the allowable limit, it is called Partial Prestressing (Type 3).

Explanation:

Durability 

• Durability to a great extent is influenced by its permeability to ingress of water, oxygen, carbon dioxide, chloride, sulphate, and other potentially deleterious substances.
• Impermeability is greatly governed by the constituents, and workmanship in making concrete.

As per IS 456:2000, the factors influencing the durability of concrete are:

• Cement content and water-cement ratio of concrete
• Cover to embedded steel
• Shape and size of member
• The Environment
• Type and Quality of constituent materials

Cement content:

• Mix must be designed to ensure cohesion and prevent segregation and bleeding. If cement is reduced, then at fixed w/c ratio the workability will be reduced leading to inadequate compaction. However, if water is added to improve workability, water / cement ratio increases and resulting in highly permeable material.
• If excess cement content is used, problems like drying shrinkage, alkali-silica reaction may occur which finally effects the durability of concrete.

Curing:

• It is very important to permit proper strength development aid moisture retention and to ensure hydration process occur completely.

Permeability:

• It is considered the most important factor for durability. It can be noticed that higher permeability is usually caused by higher porosity. Therefore, a proper curing, sufficient cement, proper compaction and suitable concrete cover could provide a low permeability concrete.

Explanation:

According to IS 456: 2000, Clause no. 12.3.1, Tolerances on Placing of Reinforcement

Unless otherwise specified by the engineer-in-charge, the reinforcement shall be placed within the following tolerances:

a) For effective depth 200 mm or less       = ± 10 mm

b) For effective depth more than 200 mm = ± 15 mm

Explanation:

Generally density of steel is 7850 kilogram per cubic metre and stainless steel generally vary in weight and density like 7480 to 8000 kilogram per cubic metre according to their composition.

Weight of Steel bar and it’s formula $D^2\over 162$

Where D is diameter of rod in mm

Unit weight of steel:- Unit Weight of steel is defined as the ratio of weight of Steel to the volume of steel, Unit Weight of steel is around 7850kg/m3 or 78.5kN/m3.

Explanation:

Short term modulus of elasticity is given by:

$E = 5000\sqrt {{f_{ck}}}$

f_ck = characteristic strength of concrete = 50 N/mm²

$E = 5000\sqrt {25} = 25000;MPa$

Which is almost equal to 25000 N/mm²

Explanation:

As per IS 456: 2000, clause 26.5.3.1,

Longitudinal reinforcement in a column:

Percentage of steel:

The cross-sectional area of longitudinal reinforcement shall be not less than 0.8 percent nor more than 6 percent of the gross cross-sectional area of the column.

Minimum diameter of longitudinal reinforcement bars:

The main longitudinal reinforcement bars used in the column shall not be less than 12 mm in diameter.

Minimum number of longitudinal bar:

The minimum number of the longitudinal bar provided in column shall be

a) Four in rectangular columns

b) Six in circular columns

Helical reinforcement:

A reinforced concrete column having helical reinforcement shall have at least six-bar of longitudinal reinforcement within the helical reinforcement.

Spacing:

The spacing of longitudinal bars measured along the periphery of the column shall not exceed 300 mm.

Explanation:

In the state of pure shear stress, the direct stress invariant is zero i.e. sum of all the principal normal stress at a point will be zero (σ₁+σ₂+σ₃ = 0).

Hence, the tension in one direction will be equal to compression in the perpendicular direction.

Explanation:

Retaining walls are constructed on the valley side of the hill roadway. These walls are constructed in stone masonry, brick masonry, or cement concrete but are usually made of dry stone masonry as it permits easy drainage of seepage water and also it is economical.

The reasons for providing retaining walls at hill roads are:

 1. It provides adequate stability to the roadway & slope. 2.  It resists lateral earth pressure.

Primary proportioning of T shaped wall:

Stem: Top width = 200 - 400 mm

Base slab width: 0.4 - 0.6 H and 0.6 - 0.75 H for surcharged wall

Base slab thickness = H/10 - H/14

Toe projection = (1/3 - 1/4) of base width

Explanation:

Surface structure: 

• It is made from a material having a very small thickness compared to its other dimensions.
• Material for surface structure may be flexible taking any form of a tent or air-inflated structure or may be made up of rigid material such as reinforced concrete.
• These structures act like cables or arches since they support loads primarily in tension or compression, with a very little bending.

∴ If the thickness of a structural member is small as compared to its length and width, it is classified as surface structures or two-dimension structures.

Concept:

The grain size distribution of fine soil is determined using sedimentation analysis. Sedimentation analysis is performed using two methods:

1. Pipette method

2. Hydrometer method.

A hydrometer is a device used to measure the specific gravity of liquids.

The following three corrections are necessary for :

1. Meniscus correction: Since the hydrometer readings increase downward on the stem, the meniscus correction (Cm) is always positive.

2. Temperature correction: If the temperature at the time of the test is more than that of calibration of the hydrometer, the observed reading will be less and the correction (Ct) would be positive and vice versa.

3. Deflocculating agent correction: The addition of the deflocculating agent increases the density of the suspension and thus a correction (Cd) is applied which is always negative.

Explanation:

Kaolinite Mineral

A basic kaolinite unit is a two-layer unit that is formed by stacking a gibbsite sheet on a silica sheet. These basic units are then stacked one on top of the other to form a lattice of the mineral. The units are held together by hydrogen bonds. The strong bonding does not permit water to enter the lattice. Thus, kaolinite minerals are stable and do not expand under saturation.

Halloysite mineral is like Kaolinite.

The basic unit is also a two-layer sheet like that of kaolinite except for the presence of water between the sheets.

Illite Minerals

Illite consists of the basic montmorillonite units but is bonded by potassium ions. There is about 20% replacement of aluminum with silicon in the gibbsite sheet due to isomorphous substitution. This mineral is very stable and does not swell or shrinks largely.

Montmorillonite Mineral

The bonding between the three-layer units is by van der Waals forces. This bonding is very weak, and water can enter easily. Thus, this mineral can imbibe a large quantity of water causing swelling. During dry weather, there will be shrinkage.

1. With reference to grain size:

Kaolinite > Illite > Montmorillonite

2. With reference to swelling and shrinkage behavior

Montmorillonite > Illite > Kaolinite

3. With reference to Plasticity Index

Montmorillonite > Illite > Kaolinite

4. With reference to Dry Strength

Montmorillonite > Illite > Kaolinite

Hence, from the above details we say that If the plasticity index of U-line lies close and above A-line of the plasticity chart, the clay mineral is montmorillonite.

Note: Illite has all the properties lies in between Montmorillonite and Kaolinite

Additional Information

Plasticity chart gives the relationship between plasticity index and liquid limit.

Equation of A-line, Ip = 0.73 ( WL - 20 ) , where Ip = plasticity index , WL = liquid limit

Explanation:

Effective stress:

• Effective stress or intergranular stress is defined as the stress caused due to soil particles coming close to each other due to the distribution of the water.
• Its function of shear strength effective stress is an abstract quantity, it cannot be measured directly in the laboratory.

Total stress (σ) = effective stress (σ’) + pore water pressure (u)

Concept:

Total stress (σ) is made up of two parts

(i) One part is due to pore water pressure (u)

(ii) The other part is due to pressure exerted by the soil skeleton which is called effective stress (σ̅ )

i.e. σ = u + σ̅ 

∴ σ̅ = σ - u

Calculation:

Given, h = 10 m, saturated soil with saturated density (γ_sat) = 2 t/m³

Total stress (σ) = γ_w × h = 10 × 10= 100 kPa

Pore water pressure (u) = γ_w × h = 10 ×  10 = 100 kPa

effective stress (σ̅ ) = σ - u = 100 - 100 = 0

Confusion Points Question is asking about above the pool level, where there is only water and no soil. So, no need of γsat in this question and total pressure will be due to water only.

Concept:

Plasticity chart gives the relationship between plasticity index and liquid limit.

Equation of A-line, Ip = 0.73 (WL - 20)

where Ip = plasticity index, WL = liquid limit

If the point gets plotted above A-line, classify soil as clay.

Further,

• If, 35 > WL ; classify as CL (low compressible clay)
• If, 35 < WL < 50 ; classify as CI (intermediate compressible clay)
• If, 50 < WL ; classify as CH (high compressible clay)

If the point gets plotted below A-line, then we need to find out whether the soil is organic or inorganic. If it's inorganic below A -line then it is silt (M)

Further,

• If, 35 > WL ; classify as ML (low compressible silt) or OL (low compressible organic)
• If, 35 < WL < 50 ; classify as MI (intermediate compressible silt) or OI (intermediate compressible organic)
• If, 50 < WL ; classify as MH (high compressible silt) or OH (high compressible organic)

Explanation:

Given,

Liquid limit, WL = 54%,

Plastic limit, WP = 27%,

Natural moisture content = 25%

Plasticity Index (Ip) = 54 - 27 = 27%

Equation of A line, Ipaline = 0.73 (WL - 20)

Ipaline = 0.73 × (54 - 20) =  24.82 <  I_p

Point will be plotted above A line, hence soil is Clay.

Concept:

Critical hydraulic gradient (ic): The quicksand / boiling condition occurs at a critical upward hydraulic gradient typically around 1.0 for many soils, when the seepage force just balances the buoyant weight of an element of soil.

At the critical conditions, the effective stress is equal to zero.

${i_c} = \frac{{G - 1}}{{1 + e}};{\rm{OR;}}\left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right)$

Calculations:

Given,

e = 0.70, G = 2.7

∵ ${i_c} = \frac{{G - 1}}{{1 + e}}$

⇒ ${i_c} = \frac{{2.7 - 1}}{{1 + 0.70}} = 1$

Concept:

$\rm{Seepage;Velocity;(V_s) = \frac{Discharge;Velocity;(V)}{Porosity;(n)}}$

$\rm{Porosity;(n)=\frac{e}{1+e}}$

Where, e = Void Ratio

Given:

Void ratio, e = 0.8, and Discharge velocity = 20 × 10-3 m/s

Calculation:

Porosity, $\rm{n= \frac{{\rm{e}}}{{1 + {\rm{e}}}} = \frac{{0.8}}{{1 + 0.8}} = 0.444}$ 

Seepage velocity, $Vs = \frac{V}{n} = \frac{{20\times {{10}^{ - 3}}}}{{0.444}} = 45 \times {10^{ - 3}}m/s$

Concept:

Weathering is the process of breaking down of rocks by mechanical or chemical process into smaller pieces.

Soil can be transported from its place of origin by wind, water, ice or any other agency and has been redeposited at another place.

Soils that had been transported by running water and deposited in the river bed and hence that will be known as alluvial soil.

Soils that have been deposited from suspension in sea water is known as Marine deposit.

Concept:

Shear stress in the direct shear test:

τ = c + σ tan ϕ

τ = shear stress

c = apparent cohesion of soil

σ = nominal stress

ϕ = angle of internal friction

Calculation:

Given:

τ = 100 kPa

σ = 100 kPa

c = 0

Shear stress in a direct shear test

τ = c + σ tan ϕ

100 = 0 + 100 tan ϕ

 tan ϕ = 1

ϕ = 45°

Explanation:

Rankine’s Active earth Pressure theory in cohesive Soil:

${P_a} = {K_a}\gamma Z - 2c'\sqrt {{K_a}}$

So, at Z =0,

${P_a} = - 2c'\sqrt {{K_a}}$

A negative sign shows that the pressure is negative.

For the stress to reverse its nature, a point will reach where the earth's pressure value will be zero,

Therefore, Pa = 0,

${K_a}\gamma Z - 2c'\sqrt {{K_a}} = 0$

${Z_c} = \frac{{2c'\sqrt {{K_a}} }}{{{K_a}\gamma }} = \frac{{2c'}}{{\gamma \sqrt {{K_a}} }}$

$K_a = {1\over N_\phi}$  

Nϕ is denotes the flow value and is it equal to the tan2[45 +(ϕ /2) ].

Z_c = $\frac{{2c'}}{\gamma }\sqrt {N_\phi}$

The above depth is known as the depth of tensile crack (Zc).

Concept:

Figure shows a clay layer, of thickness H, sandwiched between two layers of sand which serves as drainage faces.

When the layer is subjected to a pressure increment Δσ, excess hydrostatic pressure is set up in the clay layer. At the time to, the instant of pressure application, whole of the consolidating pressure Δσ is carried by the pore water so that the initial excess hydrostatic pressure u0 is equal to Δσ and is represented by a straight line u’ = Δσ on the pressure distribution diagram.

The straight line CED joining the water levels in the piezometric tubes represent this distribution. As water starts escaping into the sand, the excess hydrostatic pressure at the pervious boundaries drops to zero and remains so at all times.

After a very time tf, the whole of the excess hydrostatic pressure is dissipated so that u’=0, represented by line AGB. At an intermediate time t, the consolidating pressure Δσ is partly carried by water and partly by soil, and the following relationship is obtained.

Δσ = Δσ’ + u’

The distribution of excess hydrostatic pressure u’ at any time t is indicated by the curve AFB, joining water levels, in the piezometric tubes; this curve is known as isochrones, and number of such isochrones can be drawn at various time intervals t1,t2,t3 etc.

The slope of isochrones at any point at a given time indicates the rate of change of u’ with depth.

In a saturated clay layer consolidation with single drainage, the initial isochrones is a parabola.

Concept:

IS code of practice (IS : 2974,Part I : 1964) gives the following criteria

1. The size of foundation block (in plan) should be larger than the bed plate of the machine with a minimum allround clearance of 15 cm.
2. The width of the foundation should be at least equal to the distance of the centre of gravity of crank shaft to the bottom of foundation in all vertical machines.
3. The depth of the foundation should be such as to rest the foundation on a good bearing strata and to ensure stability against rotation in a vertical plane.
4. The combined centre of gravity of machine and the foundation block should be as much below the top of the foundation as possible.
5. Whenever possible, the operating frequency should be lower than the natural frequency of the foundation soil system and the frequency ratio should be less than 0.5. When the operating frequency is higher than the natural frequency of the foundation soil system of the machine, the frequency ratio should be more than 2 for important machines and more than 1.5 for others.

Explanation:

The ultimate load-carrying capacity(or ultimate bearing capacity or ultimate bearing resistant)** of a pile**

• It is defined as the maximum load that can be carried by a pile and at that the pile continues to sink without further increase of load.

It is determined on the basis of the following:

• Ultimate bearing capacity divided by a factor of safety
• The permissible settlement
• The overall stability of the pile foundation

The ultimate load-carrying capacity of a single pile is generally estimated by the following methods:

• Dynamic analysis
• Static analysis
• Pile load tests
• Penetration test

Concept:

Boussinesq’s equation for vertical stress

${\sigma _z} = \frac{Q}{{{Z^2}}} \times \frac{3}{{2\pi }} \times {\left[ {\frac{1}{{1 + {{\left( {\frac{r}{Z}} \right)}^2}}}} \right]^{\frac{5}{2}}}$

Where,

σZ = vertical stress in kN/m2

Q = concentrated load in kN

Z = point below the ground level where stress has to be calculated

r = radial distance

when we need to calculate the vertical stress exactly below the concentrated point load then,

r = 0

$\therefore {\sigma _z} = \frac{Q}{{{Z^2}}} \times \frac{3}{{2;\pi }} \times {\left[ {\frac{1}{{1 + {{\left( {\frac{0}{Z}} \right)}^2}}}} \right]^{\frac{5}{2}}} = \frac{Q}{{{Z^2}}} \times K_b$

Kb =  influence stress factor

Calculation:

Z = 10 m

Q = 250 kN

The vertical stress

$\therefore {\sigma _z} = \frac{Q}{{{Z^2}}} \times K_b = ;\frac{{250}}{{{10^2}}} \times 0.0179$ = 0.04475 kPa

Concept:

Negative skin friction: 

It is usually a downward shear drag acting on a pile or pile group because of the downward movement of surrounding soil relative to the piles.

Such a force can exist under one of the following conditions:

• If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will exert a downward drag force on the pile due to consolidation.
• If a fill of granular soil is placed over a layer of soft clay, it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pile.
• Lowering the water table will increase the vertical effective stress on the soil at any depth, so it can induce negative skin friction on piles.

Now the axial capacity of a pile is the summation of upward reaction due to bearing at the base and net upward skin frictional resistance. As the negative skin friction (acting downward) lowers the net skin resistance, it, in turn, reduces the axial capacity of piles.

A schematic diagram of negative skin friction on piles is shown below:

Concept:

The compatibility factor

• It is the ratio of the minimum dry density of soil to the maximum dry density of soil.

Calculation:

Given data,

The maximum dry density = 21.5 kN / m3 and

the minimum dry density = 15.5 kN / m3 

The compatibility factor is 

=  the minimum dry density of soil / the maximum dry density of soil

= 15.5 / 21.5 = 0.72

Concept:

Under-Reamed Piles

As per IS2911:1980 (part 3), cl.2.20, under-reamed piles are bored cast in situ or bored compaction concrete pile with an enlarged bulb made by either cutting or scooping out the soil or by any other suitable process.

As per IS 2911: Part III (Some other recommendations)

• The diameter of under-reamed piles may vary from 2 to 3 times the stem diameter depending upon the feasibility of construction. For Bored cast in situ under-reamed piles the bulb diameter normally be 2.5 times while for compaction piles it is 2 times.
• For piles up to 30 cm diameter, spacing of bulbs should not be greater than 1.5 times the diameter of bulb. For piles of diameter greater than 30cm, spacing can be reduced to 1.25 times the stem diameter.
• The top most bulb should be at a minimum depth of 2 times the bulb diameter. In expansive soils, it should not be less than 1.75 m below ground level. The minimum clearance below underside of pile cap embedded in ground and the bulb should be a minimum 1.5 times the bulb diameter.
• Under reamed piles with more than two bulbs are not advisable without ensuring their feasibility in strata needing stabilisation of boreholes by drilling mud. The number of bulbs in case of bored compaction piles should not exceed two in such strata.

Explanation:

Relative compaction:

It is the looseness or denseness of both cohesionless or cohesive soil. It measures the degree of compaction obtained in the field.

\({{\rm{R}}{\rm{c}}} = \frac{{{{\left( {{{\rm{\gamma }}{\rm{d}}}} \right)}{{\rm{natural;state}}}}}}{{{{\left( {{{\rm{\gamma }}{\rm{d}}}} \right)}_{{\rm{max}}}}}} \times 100\)

Relative density:

The ratio of the difference between the void ratio of soil in the loosest state and its natural void ratio to the difference between the void ratio in loosest state and fully dense state.

\({{\rm{D}}{\rm{r}}} = \frac{{{{\rm{e}}{{\rm{max}}}} - {\rm{e}}}}{{{\rm{e_{max}}} - {{\rm{e}}_{{\rm{min}}}}}} \times 100\)

The correlation between relative compaction and relative density for granular soils is

RC = 80 + 0.2Dr

RC = relative compaction and Dr = relative density

Friction circle method:

This method is also based on total stress analysis in which shearing angle of ϕ is used to analyse the stability of finite slope.

The principle of the method is that the inter granular forces are in an obliquity of ϕ to the circular surface at failure, where ϕ is the angle of the internal friction of the soil.

When the length of the arc is divided into small elements, the line of action of the inter granular forces acting on these elements can be defined by a tangent to the friction circle drawn around the centre of the sliding circle.

The radius of this friction circle is given by R sinϕ

Where,

R is the radius of the sliding circle.

Thus, the friction circle is agraphical tool that defines the line of action of the intergranularforces at any given point on the sliding circle in a convenient way.In geotechnics, the method is used primarily for slope-stability investigations in homogeneous soil when both cohesive and frictional components have to be considered.

Explanation:

• In a bending beam, a point of contraflexure is a location where the bending moment is zero (changes its sign). In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero lines.
• At the point of contraflexure, the bending moment is zero or changes its sign

For example:

A propped cantilever beam with uniformly distributed load over the entire span

The corresponding BMD is

At distance L/4 from the left support, we get point of contraflexure, as there is the change in sign. At that point, the Bending moment is zero.

Explanation:

For a three-hinged parabolic arch:

At section x - x', let Vx and Hx be the effective vertical and horizontal load acting at the section.

Then, the resultant force acting in the direction along the tangent drawn to the centerline of arch is called normal thrust and it is denoted by Nx.

Similarly, the resultant force acting the direction the normal drawn to the tangent of the centerline of the arch is called radial shear and it is denoted by Sx.

Nx = Hx cos θ + Vx sin θ

Sx = Vx cos θ - Hx sin θ

∴ The normal thrust at any section of the arch is the component of interacting forces on the section along the tangent to the centerline of the arch is correct.

∴ The radial shear at any section of the arch is the component of the interacting forces on the section along the normal to the centerline of the arch is correct.

Bending moment at any section for an arch is given by the subtraction of bending moment at section produced for the equivalent simply supported beam and the multiplication of horizontal thrust and equivalent vertical distance at the section.

\({\left. {{\rm{BM}}} \right|{{\rm{arch}}}} = {\left. {{\rm{BM}}} \right|{{\rm{S}}.{\rm{S;beam}}}} - {\rm{H}} \times {\rm{y}}\)

If bending moment for simply supported beam equals the moment from the horizontal thrust, the bending moment at any section of the arch is zero.

Such an arch is called a linear arch. E.g. two hinged and three hinged parabolic arches loaded with uniformly distributed load throughout the span.

Explanation:

Let the strength of the solid and hollow shaft is T_S and T_H 

${{\bf{T}}_{\bf{S}}} = \frac{\pi }{{16}}{{\bf{d}}^3}{τ _{per}}$, 

${{\bf{T}}_{\bf{H}}} = \frac{\pi }{{16}}{{\bf{D}}^3}\left( {1 - {{\bf{k}}^4}} \right){τ _{per}}$

where τ_per = permissible shear strength, k = Ratio of inner diameter (din) to outer diameter (D_o) i.e. $\left( \frac{d_{in}}{D_o} \right)$

The ratio of the strength of the hollow shaft to that of the solid shaft when the outer diameter of both the shafts are the same provided RPM, Material and length are the same is given by

\(\frac{{{{\bf{T}}{\bf{H}}}}}{{{{\bf{T}}{\bf{S}}}}} = 1 - {{\bf{k}}^4}\)

Here the outer diameter of both the shaft is the same 

D_o = D, d_in = $\frac{{\rm{D}}}{2}$

∴ k = $\frac{{\rm{1}}}{2}$

\(\frac{{{{\bf{T}}{\bf{H}}}}}{{{{\bf{T}}{\bf{S}}}}} = 1 - {\left( {\frac{1}{2}} \right)^4} = \frac{{15}}{{16}}\)

\(\frac{{{{\bf{T}}{\bf{S}}}}}{{{{\bf{T}}{\bf{H}}}}}; = ;\frac{{16}}{{15}}\)

Alternate Method 

This problem can also be solved by using ${M \over I} = {σ \over y}$ 

Where strength can be correlated with M (MOMENT RESISTED BY THE SECTION)$I_{Solid} = {\pi × D^4 \over 64} \ and\ I_{Hollow} = {\pi [D^4 - ({D/2})^4]\over 64}$

$Y_{Solid} = D/2\ and\ Y_{Hollow} = D/2$

Putting the above values in M/I = σ/y

We get M = (I/Y) × σ (Assuming both are having same stress), we can say that strength depends on I/Y only

Putting the value of I and Y from the above equation, we get

${{Strength \ of \ Solid} \over {Strength \ of \ hollow\ shaft} } = {I\over Y}$

${{{\pi \times D^4\over 64 } \times {D \over 2} } \over {{\pi [D^4 - {(D/2})^4]\over 64 } \times {D \over 2} }} = {1 \over (15/16)} = {16 \over 15}$

Concept:

We have the bending equation for a beam, as

$\frac{\text{ }!!\sigma!!\text{ }}{\text{y}}=\frac{\text{M}}{\text{I}}=\frac{\text{E}}{\text{R}}$

Where,

σ = Bending stress, y = distance from the neutral axis, M = Bending moment of any section, I = Moment of inertia, E = Modulus of Elasticity of material, and R = Radius of curvature.

When a beam is designed such that the extreme fibers are loaded to the maximum permissible stress ρmax by varying c/s, it will be known as the beam of uniform strength.

$\therefore \frac{\sigma }{y}=\frac{M}{I}$

$\text{ }!!\sigma!!\text{ }=\frac{\text{M}\times \text{y}}{\text{b}{{\text{d}}^{3}}/12}$

$\sigma = \frac{{M \times d \times 12}}{{{b_x}{d^3} \times 2}}$  (∵ y = d/2)

 $\sigma = \frac{{6M}}{{{b_x}{d^2}}}$

⇒ ${b_x} = \frac{{6M}}{{\sigma {d^2}}}$

b_x ∝ M

Concept:

Relation between different elastic constants E, μ, K and G are

• E = 2G (1 + μ)
• E = 3K (1 - 2μ)

Where E = Young's modulus, μ = Poisson's ratio, K = Bulk modulus, G = Modulus of rigidity

​Calculation:

Given: 

Young's modulus (E) = 120 MPa  and modulus of rigidity (G) = 48 MPa

E = 2G (1 + μ)

200 = 2 × 80(1 + μ)

1.25 = 1 + μ

μ = 0.25

Explanation:

Arches:

• Arches are structures composed of curvilinear members resting on supports. They are used for large-span structures, such as airplane hangars and long-span bridges.
• One of the main distinguishing features of an arch is the development of horizontal thrusts at the supports as well as the vertical reactions, even in the absence of a horizontal load.
• The internal forces at any section of an arch include axial compression(normal thrust), shearing force, and bending moment.
• The bending moment and shearing force at such a section of an arch are comparatively smaller than those of a beam of the same span due to the presence of the horizontal thrusts.
• The horizontal thrusts significantly reduce the moments and shear forces at any section of the arch, which results in reduced member size and a more economical design compared to other structures.

Explanation:

Muller Breslau Principle:

“If an internal stress component (shear force, bending moment, etc) or reaction component is allowed to act through a small distance thereby causing deformation of the structure, the curve of the deformed shape represents to some scale, the influence lines for that stress or the reaction component.”

Application: Muller-Breslau's Principle is applicable to all structures, determinate or indeterminate within the elastic limit.

Note:

• In the case of determinate systems, the ILD can be obtained directly by allowing unit deformation corresponding to the constraint.
• For indeterminate structure, it is applicable only when the material is within the elastic limit and obeys Hook's law so that the law of superposition holds good.

This principle gives an excellent experiment approach to determine the influence lines for indeterminate structure.

∴ Influence lines can be drawn for all structures, determinate or indeterminate.

Castigliano’s first theorem: The first partial derivative of the total internal energy (strain energy) in a structure with respect to any particular deflection component at a point is equal to the force applied at that point and in the direction corresponding to that deflection component.

This first theorem is applicable to linearly or non-linearly elastic structures in which the temperature is constant and the supports are unyielding.

Castigliano’s second theorem: The first partial derivative of the total internal energy in a structure with respect to the force applied at any point is equal to the deflection at the point of application of that force in the direction of its line of action.

The second theorem of Castigliano is applicable to linearly elastic structures with constant temperature and unyielding supports. Note that in the above statements, force may mean point force or couple (moment) and displacement may mean translation or angular rotation.

These theorems are applicable to any structure for which the force deformation relations are linear.

Explanation:

Shifting and tilting of well foundation:

• When the well is moved away horizontally from the desired position, it is called shifting of the well foundation.
• When the well is sloped against vertical alignment, it is called tilting of the well foundation.

Prevent shifting and tilting:

• The diameter of well curb should be more than the external diameter of steining. Generally, a difference of 40 to 80 mm is recommended.
• The well steining should be symmetrically placed over the curb.
• The outer surface of the well curb and steining should be smooth.
• All the sides should be uniformly dredged.
• The cutting edge should be uniformly thick and sharp.

Limitations:

• The maximum tilt allowed in case of well foundation is 1 in 60.
• The shift in well foundation should not be more than 1 % of depth of sunk.
• Beyond the above limits, well foundation is considered as dangerous and in such a case, remedial measures to rectify shifting and tilting should be followed.