BPSC AE Paper 5 (Civil) 14 Oct 2022 Official Paper (Previous Year Paper)

Concept:

Section Modulus(z):

• Section modulus is defined as the ratio of moment of inertia of a beam about its C.G to the maximum distance of extreme section of the beam (Ymax).
• The section modulus (Z) of the cross-sectional shape is significant in designing beams.
• It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.
• The unit of the Section modulus is mm3

Section modulus, $Z = \frac{I}{y_{max}}$

For rectangular section:

Moment of inertia and the maximum distance of neutral axis from top fiber for a rectangular beam is given by,

I = $\frac{bh^3}{12}$ and ymax = h/2

Hence, Z = $\frac{\frac{bh^3}{12}}{\frac{h}{2}}$ = $\frac{bh^2}{6}$

Concept-

When a structure is subjected to a system of loads, it is stable and hence functional until a sufficient number of plastic hinges have been formed to render the structure unstable. As soon as the structure reaches an unstable condition, it is considered to have been failed. The segments of the beam between the plastic hinges are able to move without an increase of load. This condition in a member is called mechanism.

If an indeterminate structure has redundancy r, the insertion of r plastic hinges makes it statically determinate. Any further hinge converts this statically determinate structure into mechanism. Hence for collapse the number of plastic hinges required are r + 1.

Number of independent mechanism-

The number of independent mechanism can be determined as follows,

Let N = Number of possible plastic hinges, r = Number of redundancies

Then number of possible independent mechanism will be N - r.

Given data and Calculation-

Possible location of plastic hinges → 4. (A, B, C, D).

At E, No plastic hinge will form as it is already a hinge joint.

Redundancy of the structure = 2.

No. of independent mechanism = 4 – 2 = 2.

(Beam mechanism & Sway mechanism)

Concept:

The shape factor of some cross-section is given below:

Rectangular section - 1.5

Triangular section - 2.34

Thin Hollow tube - 1.27

Solid circular section - 1.7

${\rm{Shape;Factor}},{\rm{;SF}} = \frac{{{\rm{Plastic;Moment}}}}{{{\rm{Yield;Moment}}}}$

Plastic Moment, Mp = Zp fy

Yield Moment, Me = Ze fy

Where,

Zp is plastic section modulus

Ze i the elastic section modulus

fy is yield stress.

∴ \({\rm{SF}} = \frac{{{{\rm{Z}}{\rm{p}}}{\rm{;}}}}{{{{\rm{Z}}{\rm{e}}}}}\)

i.e. Shape factor is the ratio of Plastic Section Modulus to the Elastic section Modulus.

Shape factor of every type cross-section is always greater than 1 because it represents the reserve strength of material beyond yield point to reach plastic stage.

More shape factor means more reserves strength beyond yield point to reach plastic stage.

Explanation:

The beam under the given loading is

Taking moment about the point A, we get

M0 - RB × L = 0

RB = M0/L (↑)

Taking, Σ V = 0

RA + RB = 0 (As there is no vertical loading)

RA =  M0/L (↓ )

So, the FBD of the above beam can be drawn as

So, the SFD will be

From, the SFD we can say that the maximum shear force will be M0/L throughout the beam, including the midsection.

BMD

The maximum bending moment in the beam is

= M/2 or Mo/2

Explanation:

a) In a bending beam, a point of contra flexure 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 contra flexure, the bending moment is zero.

​b) The rate of change of bending moment at any section is equal to the shear force at that section.

• $\frac{{dM}}{{dx}} = F \Rightarrow dM = Fdx \Rightarrow M = \smallint Fdx$

c) The rate of change of shear force at any section is equal to load intensity at that section.

• $\frac{{dF}}{{dx}} = \omega \Rightarrow dF = \omega dx \Rightarrow F = \smallint \omega dx$

d) For bending moment to be maximum or minimum, shear force should change its sign.

${d {\ [a (x - {x^3 \over L^2}) } ]\over dx } = V$

${ {\ a (1 - {3x^2 \over L^2}) } } = 0$

x = L/√3

Explanation:

Link support

It is a support that allows rotation and translation perpendicular to the direction of the link only.

It does not allow translation in the direction of the link.

It has a single linear resultant force component in the direction of the link which can be resolved into vertical and horizontal components.

There will always be only 1 unknown if link support is used and that is the force acting along the axis of the link.

Explanation:

1. When a truss member is in compression:

• The member tends to push outward against the joints to which it is connected.
• Compression members resist shortening, and their internal force is directed inward (to counteract the external load), but the reaction force they apply to the joints is outward.

2. This principle is in accordance with Newton's Third Law of Motion:

• Action and reaction forces are equal and opposite.
• If the internal force in the member is directed inward (compression), the force exerted on the joints is outward.

There is load on the beam and hence there will be no reaction

Explanation:

According to the Müller-Breslau Principle, to construct an Influence Line Diagram (ILD) for a reaction or internal force, the structure is modified to show the deflected shape under the respective influence.

• When constructing the ILD for a vertical reaction at a pin joint, the Müller-Breslau Principle dictates that the pin support should be replaced by a roller guide
  
• This roller guide will allow the support to move freely in the horizontal direction while still resisting vertical forces, which helps in visualizing how vertical reactions change as a load moves across the structure.

This transformation models the system’s behavior under varying loads and accurately reflects the influence on the vertical reaction at the pin joint.

Explanation:

A statically indeterminate structure has more redundancies and can redistribute the internal forces when parts of the structure experience damage or failure. This redistribution capacity makes such structures more resilient and better at withstanding dynamic loads like those caused by earthquakes.

• Statically determinate structures are less flexible in redistributing loads when subjected to unexpected forces or moments, and any failure in one element may lead to catastrophic collapse.
  

• In contrast, statically indeterminate structures have multiple load paths, which helps them handle varying forces more effectively, especially in an earthquake where dynamic and unexpected forces act on the structure.

Therefore, statically indeterminate structures tend to perform better during an earthquake.

Concept:

Sign convention for rotation:

Explanation:

When right support (i.e. B) settles by δ, the following figure shows the induced rotations -

It is clear from the above figure -

The rotation at both ends is clockwise i.e. (+ive).

• Rotations are always considered from the horizontal axis.

Explanation:

Transverse reinforcements, also known as shear reinforcements, are provided in reinforced concrete to serve multiple purposes:

• Distribute the load: Transverse reinforcements help in distributing the applied load across the structure and prevent localized stress concentrations.
  
• Resist temperature stresses: These reinforcements help in handling the stresses caused due to temperature changes, which can lead to expansion or contraction in the concrete.
  
• Resist shrinkage stress: They also help in resisting shrinkage stresses that occur when concrete dries and shrinks, reducing the chances of cracking.

Hence, transverse reinforcements contribute to all the aspects listed above.

Explanation:

Losses of prestressing:

From IS 456:2000, 26.5.3.2 (c)

The spacing of transverse reinforcement or pitch of column should not be more than the least of the following:

• The least lateral dimension of the compression member (In case of rectangular column least cross-sectional dimension and in case of circular column diameter of that column)
• Sixteen times the smallest diameter of the longitudinal reinforcement
• 300 mm

Note:

If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of the tie are effectively tied in two directions. additional longitudinal bars in between these bars need to be tied in one direction by open ties.

Explanation:

Lap splice:

• It is done by overlapping the bars over a certain length, thereby making possible the transfer of axial force from the terminating bar to the connecting bar by the mechanism of anchoring (development) bond with the surrounding concrete.
• According to 3rd amendment of IS 456: 2000 in 2007, the value was changed from 36 mm to 32 mm.
• So, 36 mm is an old provision and 32 mm is a new provision
• 32 is not given in the given options, so we have to go with the old provision ie 36 mm

Additional Information

• However, where welding is not practicable, CI. 26.2.5.1(a) of IS 456:2000 recommends the use of additional spirals around the lapped bars.
• It is always desirable to bend the bars slightly (especially for large diameter bars) at the location of the splice in order to have a collinear force transfer without any eccentricity.
• As per IS 456, the straight length of the lap should not be less than 15ϕ or 200 mm (whichever is more).
• The lap length must be at least equal to the development length.
• In no case, the lap length should be less than 30ϕ for flexural and direct tension and 24ϕ for compression.
• When two different diameter bars meet at the splice location then lap length should be calculated based on the smaller bar diameter.
• Splices in tension members should be enclosed by spirals made up of bars of diameter not less than 6 mm along with pitch not exceeding 100 mm

Concept: 

Moment of resistance for balanced section,

${M_{u,lim}} = Q{f_{ck}}b{d^2}$

Force from the compression side,

C = 0.36 × f_ck × b × x_u

Force from the tension side,

T = 0.87 × f_y × A_st

To determine the A_st, we have to equate the compressive and tensile forces 

 0.36 × fck × b × Xu = 0.87 × fy × Ast

0.36 × 20 × 300 × X_u = 0.87 × 415 × 4 × π × 20²/4

X_U = 209.9 ≈ 210 mm

Concept: 

Moment of resistance for balanced section,

${M_{u,lim}} = Q{f_{ck}}b{d^2}$

Force from the compression side,

C = 0.36 × fck × b × xu

Force from the tension side,

T = 0.87 × fy × Ast

Calculation:

Given, B = 350 mm, d = 600 mm, fck = 20 MPa and fy = 415 MPa.

For a balanced section, neutral axis depth, x_u = x_u,lim

For Fe 415, xu (lim) = 0.48 × d = 0.48 × 600 = 288 mm

From the static equilibrium condition, equating the compressive force (C) due to concrete and tensile force (T) due to steel along the neutral axis:

C = T

C = 0.36 fck× B × xu, lim 

T = 0.87 fy × Ast, lim

\({{\rm{A}}{{\rm{st}},{\rm{lim}}}} = \frac{{0.36 × {{\rm{f}}{{\rm{ck}}}} × {\rm{;B;}} × {\rm{;}}{{\rm{x}}{{{\rm{u}}{{\rm{lim}}}}}}}}{{{{\rm{f}}_{{\rm{ds}}}}}}\)

${{\rm{A}}_{{\rm{st}},{\rm{lim}}}} = \frac{{0.36{\rm{;}} × {\rm{;}}20{\rm{;}} × {\rm{;}}350{\rm{;}} × {\rm{;}}288}}{{0.87{\rm{;}} × {\rm{;}}415}}$

∴ A_{st lim} = 2010 mm²    ........ (1)

Now we have to calculate the actual area to be provided, for which we have to add the area of tensile and compressive reinforcement

A_{st pro} = A_st1 + A_st2

= 6 × π × 25² /4 +  4 × π × 202/4

A_{st pro} = 4201 mm² ........ (2)

From the above 2 equations, we can say that area of reinforcement provided in greater than limiting reinforcement, so the given section is over reinforced section

Concept:

Where

bf = effective width of flange

lo = distance between points of zero moments in the beam

bw = breadth of the web

df = thickness of the flange, and

b = actual width of the flange

${b_f} = {b_w} + \frac{{{l_o}}}{6} + 6{d_f}$

l₀ = 12000 mm, d_f = 100 mm, b_w = 350 mm

${b_f} = {350} + \frac{{{12000}}}{6} + 6\times{100}$

b_f = 2950 mm

Explanation:

Three hinge arch is a statically determinate structure.

As we know, 

No. of support reactions, R = 4

Equations of static equilibrium, re = 3

Additional equation due to released reaction at hinge at crown (∑ M = 0), rr= 1

Degree of static determinacy, Ds = R - (re + rr) = 4 - (3 + 1) = 0.

Explanation:

A fillet weld can have different profiles based on how the weld bead is shaped. The options provided describe different possible shapes of a fillet weld:

1. Mitre weld: A fillet weld with an approximately 45-degree angle where the two surfaces meet.
   
2. Concave weld: A fillet weld where the face of the weld is slightly indented or curved inward.
   

3**. Convex weld:** A fillet weld where the face of the weld is slightly bulged or curved outward..

In the case of torsion box type sections are preferred.

Such sections are used as they are symmetrical and the polar moment of inertia of these sections are higher than other sections.

This due to larger area lying away from the axis of the section.

The value of the polar moment of inertia of the channel section is smaller than that of the box section as a larger area lies closer to the axis of the section.

Concepts:

A minimum effective length of weld is four times the size of weld.

When placing a fillet weld, though the welder tries to build up the weld to its full dimension from the beginning, there is always a slight tapering off where the weld starts and where it ends. The width of this tapering is called end returns and is equal to two times the size of weld on either side of welding. This tapered portion of weld is not able resist or carry any load hence cannot be the part of weld length. Therefore, the length of weld which is actually effective to resist loads, called effective length, is taken as overall length of weld minus two times the size of weld in case fillet weld as per IS 800:2007 codal provisions.

Note:

1. The effective length of weld is taken equal to overall length of weld minus twice the size of weld. This is done to avoid stress concentration at ends.
2. To avoid stress concentration at edges of plate, the length of longitudinal (side) fillet weld shall not be less than the width of plate.
3. The perpendicular distance between longitudinal fillet welds is limited to 16 times the thickness of the thinner plate jointed.
4. In case of Groove weld in butt joints, the effective length is taken as the length of continues full size weld, but in no case, it should not be less than the four times the size of weld.

Concept:

Purlins:

• Beams that carry the roof loads in trusses are called purlins. These are placed in a tilted position over the principal rafters of the trusses.
• Purlins may be designed as simple, continuous or cantilever beams. the simple beam design yields the greatest moments and deflections.
• Purlins have a tendency to sag in the direction of sloping roof and sag rods are sometimes provided midway or at third points, between roof trusses, to take up the sag. these rods reduce the moment and result in a smaller purlin section. So generally purlins are placed at the panel points so as to avoid bending moment in rafters.

Additional Information

• If purlins are assumed to be simply supported the moment will be $wl^2 \over 8$.
• If purlins are assumed to be continuous the moment will be $wl^2 \over 10$.

Concept: -

As per IS 800 : 2007, cl. 3.7.

Plate elements of a cross-section may buckle locally due to compressive stresses. The local buckling can be avoided by before limit state is achieved by limiting the width to thickness ration of teach element of a cross-section subjected to axial force, moment, or shear.

As per IS 875, the design wind speed at any height z is given by:

${V_z} = {V_b}{k_1}{k_2}{k_3}$

Where,

Vb = Basic Maximum wind speed

K1 = Probability factor or risk coefficient

K2 = Terrain, height and structure size factor

K3 = topography factor

Risk coefficient is dependent on:

a) Mean probable design life of structures

b) Basic wind speed

Example: For Important buildings and structures such as hospitals, communication buildings/towers for 100 years of design life of structure and wind speed of 33 m/s, Risk coefficient is 1.05.

Concept:

Net Sectional area:

When tension members are attached to the gusset plate, some material of plate is removed due to bolt/rivet hole. The net area at any section is equal to the gross area minus the deductions for holes at that section.

As per IS 800,

i. For a rivet upto 25 mm diameter:

Dia of rivet hole = Rivet Dia + 1.5 mm

ii. For rivet diameter > 25 mm

Dia of rivet hole = (d + 2) mm

Given:

d = 20 mm, B = 300 mm, t = 10 mm

Net sectional area = (b – nd0) x t

Where,

n = Number of rivet holes, d0 = Diameter of one rivet hole, t = Thickness of plate, and b = Width of plate.

Here, d0 = (20 + 1.5) = 21.5mm

∴ Net Sectional Area = (300 – 21.5) × 10 = 2785 mm2 ≈ 2800 mm²

Important Point:

Consideration of Net Sectional area makes conservative design.

Dia of holes in case of:

Concept:

Column base:

• Column base transmits the column load to the concrete or masonry foundation blocks.
• The column base spreads the load on a wider area so that the intensity of bearing pressure on the foundation block is within the bearing strength.

There are two types of column bases commonly used in practice:

• Slab base
• Gusseted base

Slab Base:

• These are used in columns carrying small loads.
• In this type, the column is directly connected to the base plate by welding or through cleat angles.
• The load is transferred to the base plate through bearing.

Gusseted Base:

• For columns carrying heavy loads, gusseted bases are used. In a gusseted base, the column is connected to the base plate through gussets.
• The load is transferred to the base partly through bearing and partly through the gusset.
• The thickness of the gusset plate in any case should not be less than 12mm.

Concept:

For a fixed-fixed beam,

For each end, the no. unknowns are 2, so total 4 unknowns.

We have 2 available equation of equilibrium which are

\({\rm{;}}\sum {{\rm{F}}{\rm{x}}} = 0,\sum {{\rm{F}}{\rm{y}}} = 0,{\rm{and;}}\sum {{\rm{M}}_{\rm{z}}} = 0{\rm{;}}\)

Static Indeterminacy = No. of unknowns – No. of equilibrium equations

Static Indeterminacy = 4 – 2 = 2.

Explanation:

Concept:

The influence line for a function (like bending moment) at a specific point on a structure shows how that function varies as a unit load moves across the beam.

For a simply supported beam:

• The influence line for the bending moment at any section is a triangle.
• It is zero at the supports and reaches a maximum at the point of interest.

Mathematically:

For a beam of length $L$, and a point C at distances $a$ from the left and $b$ from the right:

$Ordinate = \frac{ab}{L}$

This creates a triangular shape with linear rise and fall on either side of point C.

Concept:

The fixed end moments generated due to the application of Moment M on a beam fixed at both ends are shown below:

⇒ MAB = $\frac{{\rm{M}}}{{{{\rm{L}}^{2{\rm{}}}}}} \times \left( {3{\rm{a}} - {\rm{L}}} \right) \times {\rm{b}}$

When the moment is at the center, a = b = L/2

⇒ MBA = $\frac{{\rm{M}}}{{{{\rm{L}}^{2{\rm{}}}}}} \times \left( {3{\rm{L/2}} - {\rm{L}}} \right) \times {\rm{L/2}}$

M_BA = M_AB = M/4

Mistake Points

Here Moment is in Anticlockwise Direction.

MAB = $- \frac{{\rm{M}}}{{{{\rm{L}}^{2{\rm{}}}}}} \times \left( {3{\rm{a}} - {\rm{L}}} \right) \times {\rm{b}}$

MBA = $- \frac{{\rm{M}}}{{{{\rm{L}}^{2{\rm{}}}}}} \times \left( {3{\rm{b}} - {\rm{L}}} \right) \times {\rm{a}}$

Note: MAB and MBA must have the same sign. a and b, will always be positive which means to have the same sign-on MAB and MBA (3a – L) and (3b – L) must have the same sign.

Fully saturated soil is said to be a two-phase system with soil and water as shown in the figure below:

Concept:

Coefficient of uniformity $= \frac{{{D_{60}}}}{{{D_{10}}}}$

Coefficient of curvature $= \frac{{D_{30}^2}}{{{D_{60}}; \times ;{D_{10}}}}$

Calculation:

CU = 4, Cc = 1

${C_U} = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow 4 = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow {D_{60}} = 4;{D_{10}}$

${C_c} = 1.6 \Rightarrow \frac{{D_{30}^2}}{{{D_{60}}; \times ;{D_{10}}}} = 1 \Rightarrow \frac{{D_{30}^2}}{{4;{D_{10}}; \times ;{D_{10}}}} = 1$

$\Rightarrow \frac{{D_{30}^2}}{{D_{10}^2}} = 4 \Rightarrow \frac{{{D_{30}}}}{{{D_{10}}}} = \sqrt {4} = 2$

Concept:

Void ratio (e):

Void ratio (e) is the ratio of the volume of voids (Vv) to the volume of soil solids (Vs), and is expressed as a decimal.

e = Vv/Vs

Porosity (n):

Porosity (n) is the ratio of the volume of voids to the total volume of soil (V), and is expressed as a percentage.

n = (Vv/V) × 100

Relation between void ratio and porosity

n = e/(1+e)

Calculation:

Given,

Vv = Vs

∴ e = Vv/Vs

e = Vv/Vv = 1

∵ n = e/(1 + e)

n = 1/(1 + 1) = 1/2 = 0.5

Explanation:

• When flow takes place in an upward direction, seepage pressure (ps) also acts in an upward direction and effective stress is reduced.
• If seepage pressure equals the submerged weight of soil mass then effective stress reduces to zero. In such a case, cohesionless soil loses all its shear strength and has the tendency to flow along with water. This phenomenon is known as the quicksand condition.

For quicksand / Piping condition:

Effective stress = γ'z - Ps = 0

γ'z = Ps = i × z × γw

Where Ps = Seepage pressure = i z γw

γ = i × γw

i = $\left ( \frac{\gamma '}{\gamma _{w}} \right )=\frac{G-1}{1+e}.\gamma _{w}$

ic = $\frac{G-1}{1+e}$

Note:

The hydraulic gradient at which quicksand condition occurs is termed as the critical hydraulic gradient piping gradient, floating gradient, or bursting gradient.

For piping, the hydraulic gradient should be more than or equals to the critical hydraulic gradient.  ​

Important Points

• For most of the soils, the critical hydraulic gradient is close to unity. So when the hydraulic gradient is nearly unity piping will occur.

Concept

The time required for any degree of consolidation can be obtained from Taylor's formula, it is a dimensionless parameter

Tv = Cv × t /d2

Where,

Cv = Coefficient of consolidation is constant for the same type of soils,

Tv = Time factor will also remain constant for the same degree of consolidation.

Thus, Time Factor depends on Coefficient of consolidation, Drainage path & Time.

Important Points

The time factor is given by,

For consolidation less than 60 %

${T_v} = \frac{\pi }{4} \times {\left( {\frac{U}{{100}}} \right)^2}$

For consolidation greater than 60 %

${T_v} = 1.781 - 0.933{\log _{10}}\left( {100 - U\% } \right)$

In the consolidated drained (CD) test on a saturated soil sample, drainage is permitted throughout the test, during the application of both normal as well as shear stress so that full consolidation occurs and no excess pore pressure is set up at any stage of the test. Generally, it takes 2 to 5 days.

Important Points:

In the unconsolidated-undrained (UU) test, no drainage is permitted either during the consolidation stage or the shear stage. Therefore, there is no dissipation of pore pressure during the test and test can be conducted in a few minutes (5 minutes to 10 minutes).

In the consolidated-undrained (CU) test, drainage is permitted under the initially applied normal stress only and fully primary consolidation is allowed to take place and no drainage is allowed in the shear stage. Generally, it takes 24 hours.

Hence, pore water pressure is zero during both consolidation and shearing stage.

Given,

Degree of consolidation = 50%

U = 50%

Time factor

$T_v = \frac \pi 4(U^2)$ when U ≤ 0.6

Tv = 1.781 - 0.933 log10 (100 - U%)

U > 0.6

Than $T_v = \frac \pi 4 \left(\frac {50}{100}\right)^2$

$= \frac \pi 4 \left(\frac 1 2 \right)^2$

Tv = 0.196

So, Tv ≃ 0.2

Explanation:

There are two main types of a soil sample that is collected for the study of the properties of soils:

1. Disturbed Soil Samples:

• During the sampling process of the soil sample If the natural structure of the soil gets disturbed, then this type of soil sample is called a Disturbed Soil Sample.
• The disturbed Soil Samples can be used for the determination of the grain size, plasticity characteristics, and specific gravity of the soil.
• These samples are collected by different methods such as Auger Boring, Wash Boring, Rotary Drilling, and Percussion Drilling.

1. Undisturbed Soil Samples:

• During the sampling process of the soil sample, if the natural structure of the soil and water content does not disturb, that means the soil retained its natural structure and water content, then these types of soil samples are called undisturbed soil samples.
• The undisturbed soil samples are used for the determination of engineering properties of soils such as shear strength, consolidation test, permeability, and compressibility.
• The undisturbed soil samples are collected by Thin-walled samplers.

Concept:

Lateral earth pressure is the pressure that soil exerts against a structure in a sideways, mainly horizontal direction. Since most open cuts are excavated in stages within the boundaries of sheet pile walls or walls consisting of soldier piles and laggings and since struts are inserted progressively as the excavation proceeds, the walls are likely to deform (as shown in the figure).

Little inward movement can occur at the top of the cut after the first strut is inserted. As the earth pressure magnitude increases due to an increase in effective stress while moving from top to bottom, so deflection of a sheet pile in a braced cut increases from top to bottom.

The soils formed at a place may be transported to other places by agents of transportation, such as water, wind, ice, and gravity.

1) Water transported Soils: Flowing water is one of the most important agents of the transportation of soils. Swift running water carries a large quantity of soil either in suspension or by rolling along the bed.

The size of the soil particles carried by water depends upon the velocity.

All types of soils carried and deposited by water are known as alluvial deposits. Deposits made in lakes are called lacustrine deposits. Such deposits are laminated or varved in layers. Marine deposits are formed when the flowing water carries soils to the ocean or sea.

2) Wind transported Soils: Soil particles are transported by winds. The particle size of the soil depends upon the velocity of the wind. Soils deposited by wind are known as aeolian deposits.

Loess is a silt deposit made by wind. These deposits have low density and high compressibility. The bearing capacity of such soils is very low.

3) Glacier-Deposited Soils: Glaciers are large masses of ice formed by the compaction of snow. As the glaciers grow and move, they carry with them soils varying in size from fine-grained to huge boulders.

Drift is a general term used for the deposits made by glaciers directly or indirectly. Deposits directly made by the melting of glaciers are called till. The soil carried by the melting water from the front of a glacier is termed out-wash.

4) Gravity deposited soil: These are soils transported through short distances under the action of gravity. Colluvial soils such as talus have been deposited by the gravity. Talus consists of irregular, coarse particles. It is a good source of broken rock pieces and coarse-grained soils for many engineering works.

Skempton’s Pore Pressure coefficients are A and B,

Where A → depends upon the degree of saturation and over consolidation ratio

B → depends upon the degree of saturation of soil

For over-consolidated soil, the value of A is negative.

For fully saturated soil, the value of B is 1

 Note: If there is no description about saturation then 

  A < 1 and B < 1

Explanation :

• The angle of friction between soil particles depends upon the shape, size, roughness, grading, and packaging of particles.
• For Sand, particle size and shape roughly remain the same and thus it cannot be used to estimate the angle of friction of sandy soil.
• However, the angle of internal friction would be very high if the sand particle is compactly dense which is indicated by the value of the density index.
• The density index is the ratio of the difference between the void ratios of a cohesionless soil in its loosest state and existing natural state to the difference between its void ratio in the loosest and densest states.
• The higher value of the relative index indicates a higher angle of internal friction.

So we can say that parameter that can be used as a rough guide to determine the type of failure in the soil is the Angle of internal friction which is indicated by the value of the density index.

Sheet pile:

A sheet pile wall consists of a series of sheet piles driven side by side into the ground, thus forming a continuous vertical wall for the purpose of retaining an earth bank.

They are commonly used for waterfront structures temporary construction and lightweight construction where sub-soil is poor for supporting a retaining wall.

A sheet pile wall may be of three types.

1. Cantilever sheet piling
2. Anchored sheet pilling
3. Braced sheeting.

Certain configurations of sheet pilling are termed as bulkheads or cofferdams. A bulkhead is a sheet pile retaining wall of water front, backed up ground.

A cantilever sheet pile wall or bulkhead derives its stability entirely from the lateral resistance of the soil into which it is driven.

The bulkhead is adequately embedded into the soil below the dredge into so that a driven line of sheeting acts as wide cantilever beam in resisting the lateral earth pressures developed above the dredge line. A cantilever bulkhead is used for moderate height only.

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:

Critical hydraulic gradient

 n = 1/3, G = 2.5

 G - 1 = 1.5

\(\begin{array}{*{20}{c}} {{i_c} = \left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right) = 1.5 \times \left( {1 - \frac{1}{3}} \right) = 1} \end{array}\)

According to Terzaghi and Peck the allowable soil pressure in cohesionless soils is directly proportional to allowable settlement. So to double the allowable soil pressure, permissible settlement should be doubled.

This concept comes from settlement calculations in foundation engineering. By allowing a greater amount of settlement, the soil's bearing capacity is effectively increased because the structure is designed to handle the increased settlement without adverse effects. This leads to an increase in allowable bearing pressure.

Concept:

Bearing Capacity can be calculated based on the following field test:

(a) Standard Penetration Test (S.P.T)

(b) Plate Load test

(c) Static cone penetration test/Static Cone resistance test.

Standard Penetration Test (S.P.T)-

S.P.T. Test procedure

• Suitable for Granular soil.
• Split spoon samples are used in the borehole.
• The borehole is advanced to a depth at which N-value is to be calculated.
• The split-spoon sampler is allowed to penetrate the soil by applying an impact load of 65 Kg having a free fall of 75 cm.
• The sampler is allowed to penetrate for 150 mm depth, but reading is not noted i.e.(No. of blows required for 150 mm penetration is not Noted).
• Then the sampler is allowed to penetrate, further for 300 mm and No. of blows required to penetrate the sampler to this 300 mm is the SPT N-value
• The next test is carried out at level 750 mm below the previous test reference level.
• If borehole depth is large, then the interval of the next test is taken at a depth of 2.5 to 2 m or the change of strata.

Calculation of S.P.T N value:

• Bearing capacity of Granular soil can be calculated by using corrected S.P.T N-values.
• The values of N – is determined at Number of Selected boreholes and the average value of corrected (N) is calculated.
• While calculating the average value of (N), any value greater than 50% of the average value is discarded.

e.g. if $\left[ {\frac{{{N_1} + {N_2} + {N_3} + {N_4}}}{4}} \right] = 30$ and if N3 = 40

Then (N3) is accepted as [30 × 1.5 = 45] and (40 < 45) and if N2 = 48 then (N2) is discarded as 48 is > 45

• New average is taken as $= \left[ {\frac{{{N_1} + {N_3} + {N_4}}}{3}} \right]$
• Minimum of all the avg. values of various boreholes are used in the design of foundations.

Important Points

The following corrections are applied to the N-values obtained above:-

1. Overburden Correction

• It is necessary because the N-value will affect it due to the confinement of soil at various depth.
• Two granular soils possessing the same relative density but having different confining pressures are tested, the one with higher confining pressure will give higher N value.
• Since the confining pressure increases with depth, the N values at shallow depths are underestimated and the N values at larger depths are overestimated.
• If N0 = observed S.P.T. value

${N_1} = {N_0} \times \frac{{350}}{{\left( {\bar \sigma + 70} \right)}}$

Where, σ̅ = Effective stress at the level of the test (KN/m2)

N1 = Corrected N-value of overburden.

• Overburden correction will not be applied if σ̅ > 280 KN/m2

2. Dilatancy Correction

• It is applied to the already corrected N-values for overburden pressure. Dilatancy correction is required only if N1 > 15 in saturated fine sand and silt
• (N1 > 15) represents the Dense sand which will tend to dilate under rapid loading (undrained condition) and (-VA) pore water pressure will develop. Hence, observed (N) value will be more because shear resistance will increase.

Corrected N value after Dilatancy Correction $:{N_2} = 15 + \frac{1}{2}\left( {{N_1} - 15} \right)$

The line drawn tangent to all the Mohr’s circles at failure is called as the failure envelope as shown in the below figure.

A stress point above the failure envelope does not show any stress condition and does not exist.

Or we can say that the condition of the stress of a point above Mohr’s envelope indicates the imaginary condition.

Explanation:

Plate load test: 

• The plate load test is a field test, which is performed to determine the ultimate bearing capacity of the soil and the probable settlement under a given load.
• This test is very popular for the selection and design of the shallow foundation.

Procedure:

• Excavate test pit up to the desired depth. The pit size should be at least 5 times the size of the test plate (Bp).
• At the center of the pit, a small hole or depression is created. The size of the hole is the same as the size of the steel plate.
• A mild steel plate is used as a load-bearing plate whose thickness should be at least 25 mm thickness and size may vary from 300 mm to 750 mm. The plate can be square or circular. Generally, a square plate is used for square footing and a circular plate is used for circular footing.
• A column is placed at the center of the plate. The load is transferred to the plate through the centrally placed column.
• The load can be transferred to the column either by the gravity loading method or by the truss method.
• At least two dial gauges should be placed at diagonal corners of the plate to record the settlement. The gauges are placed on a platform so that it does not settle with the plate.
• Apply seating load of 0.7 t/m2 = 70 g/cm2 and release before the actual loading starts.
• The initial readings are noted.
• The load is then applied through the hydraulic jack and increased gradually. The increment is generally one-fifth of the expected safe bearing capacity or one-tenth of the ultimate bearing capacity or any other smaller value. The applied load is noted from the pressure gauge.
• The settlement is observed for each increment and from the dial gauge. After increasing the load-settlement should be observed after 1, 4, 10, 20, 40, and 60 minutes and then at hourly intervals until the rate of settlement is less than .02 mm per hour. The readings are noted in tabular form.
• After completing the collection of data for a particular loading, the next load increment is applied and readings are noted under new load. This increment and data collection is repeated until the maximum load is applied. The maximum load is generally 1.5 times the expected ultimate load or 3 times of the expected allowable bearing pressure.

Additional Information

Plate Load Test Apparatus / Equipment: 

• Test plate
• Hydraulic jack & pump
• Reaction beam or reaction truss
• Dial gauges
• Pressure gauge
• Loading columns
• Necessary equipment for the loading platform.
• Tripod, Plumb bob, spirit level, etc.

Concept:

The allowable settlement is defined as the acceptable amount of settlement of the structure and it usually includes a factor of safety.

The allowable settlement depends on many factors, including the following:

Type of Construction

Use of the Structure

Presence of Sensitive Finishes

The maximum permissible settlement as per IS-1904(1966) for isolated foundations are 40 mm on sandy soil and 65 mm in clayey soils.

The permissible settlement for the raft foundation on clay soil is 65-100 mm and for sandy soil, it is 65 mm.

Differential settlement:

Foundation on clay soil = 40 mm.

The foundation on Sandy soil = 25 mm.

The value of factor of safety adopted in foundation design is 2 - 3