BPSC AE Paper 5 (Civil) 11 Nov 2022 Official Paper (Previous Year Paper)

Concept:

Shape Factor:

• It represents the reserve of the strength of a section beyond the yield point.
• It doesn't show the reserve of the strength of the complete structure.
• It depends upon the geometry of the cross-section only.
• It is defined as the ratio of the plastic moment capacity of a beam section to its yield moment capacity.

Mathematically,

\({\rm{S}}{\rm{.F = }}\dfrac{{{{\rm{M}}{\rm{p}}}}}{{{{\rm{M}}{\rm{y}}}}}{\rm{ = }}\dfrac{{{{\rm{f}}{\rm{y}}}{{\rm{Z}}{\rm{p}}}}}{{{{\rm{f}}{\rm{y}}}{\rm{Z}}}}{\rm{ = }}\dfrac{{{{\rm{Z}}{\rm{p}}}}}{{\rm{Z}}}\)

Where,

fy = Yield stress.

Zp = Plastic modulus of a section.

Z = Elastic modulus of a section.

Zp is calculated about the equal area axis and Z is calculated about the centroidal axis.

It can also be defined in terms of the moment of resistance under plastic and elastic conditions:

Shape Factor = (MOB)p/(MOB)e

(MOB​)p​ = Plastic moment of resistance

(MOB​)e​ = Elastic moment of resistance

Additional Information

Shape Factor for the different section:

Type of section	Shape factor
I – Section	1.12 - 1.14
Rectangular section	1.5
Hollow Circular section	1.27
Circular section	1.7
T-section	1.8
Diamond Section	2
Triangular Section	2.34

Concept-

Conjugate Beam Theorem-

Theorem 1: The slope or rotation at a point in the real beam is numerically equal to the shear force at the corresponding point in the conjugate beam.

Theorem 2: The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam.

Explanation:

The number of zero force members is 2.

Additional InformationTo identify the zero force members in truss, the following rules are required to be used:

1. If at a joint three members are meeting, two of which are collinear then force in third member is zero provide that no external load is acting on joint.
2. If two non-collinear members are meeting at joint and no external load is acting on that joint, then force in both members is zero.

Explanation:

Degree of indeterminacy is given by

D_s = 3 x number of cuts required to make tree like structure fixed - number of reaction required to make joint

Ds = 3 x 4 - 2 = 10

Concept:

To analyze a continuous beam using the slope deflection method, specific boundary conditions and equilibrium equations must be considered at each support. The conditions will differ based on the type of support and load applied to each span.

Given Beam:

• Support A is fixed, implying $\theta_A = 0$.
• Support B and Support C are simply supported, meaning no moment resistance at C ($M_{CB} = 0$).
• Span AB has a uniformly distributed load $w$ per meter run.
• Span BC has a point load $W$ at its mid-span.
• The beam has a uniform cross-section but different span lengths for AB and BC.

Applying the Slope Deflection Method:

• Fixed Support A: Since support A is fixed, the rotation $\theta_A$ is zero.
• Moment at Joint B: The sum of moments at B due to adjacent spans should be zero for equilibrium, giving $M_{BA} + M_{BC} = 0$.
• Simply Supported Support C: The moment at support C is zero ($M_{CB} = 0$) since it is simply supported.

Conditions for Analysis:

$\theta_A = 0$

$M_{BA} + M_{BC} = 0$

$M_{CB} = 0$

Concept:

Stiffness of a member is a measure of its resistance to deformation under load. The stiffness of a member in terms of flexural rigidity (EIEIEIEI

) and the length (ℓℓℓℓ

) is calculated using standard formulas for different boundary conditions and configurations.

The distribution factor (DF) is a coefficient used in the moment distribution method to distribute moments to various members connected at a joint based on their relative stiffness. It is calculated using the stiffness of each member.

Stiffness for Different Members:

$\text{Stiffness of member EB} = \frac{4EI}{\ell}$

$= \frac{4E(2I)}{6} = \frac{4}{3}EI$

$\text{Stiffness of member ED} = \frac{4EI}{3}$

$\text{Stiffness of member EF} = \frac{4EI}{4} = EI$

Distribution Factors for Different Members:

$(DF)_{EB} = \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI}$ $= \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI} = \frac{4}{11}$

$(DF)_{ED} = \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI}$ $= \frac{4}{11}$

$(DF)_{EF} = \frac{EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI}$ $= \frac{3}{11}$

concept:

According to 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.

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.

Concept:

To determine the location of the Plastic Neutral Axis (PNA) of an I-section beam, we need to find the point where the areas of the cross-section above and below this axis are equal. This is achieved by calculating the areas and balancing them around the PNA.

Given Data:

• Width of top flange ($b_t$) = 250 mm
• Thickness of top flange ($t_t$) = 50 mm
• Width of bottom flange ($b_b$) = 400 mm
• Thickness of bottom flange ($t_b$) = 50 mm
• Width of web ($b_w$) = 50 mm
• Depth of web ($d_w$) = 200 mm

Calculation:

1. Calculate the Area of Each Section:

• Top flange area ($A_t$): $250 , \text{mm} \times 50 , \text{mm} = 12500 , \text{mm}^2$
• Bottom flange area ($A_b$): $400 , \text{mm} \times 50 , \text{mm} = 20000 , \text{mm}^2$
• Web area ($A_w$): $50 , \text{mm} \times 200 , \text{mm} = 10000 , \text{mm}^2$

2. Calculate the Total Area and Half Area:

$A_{\text{total}} = A_t + A_b + A_w = 12500 + 20000 + 10000 = 42500 , \text{mm}^2$

$\frac{A_{\text{total}}}{2} = \frac{42500}{2} = 21250 , \text{mm}^2$

3. Determine the Distance from the Top Fiber:

• Area of top flange: $12500 , \text{mm}^2$
• Remaining area needed: $21250 - 12500 = 8750 , \text{mm}^2$
• Web area per mm depth: $50 , \text{mm}$
• Depth required in web: $\frac{8750}{50} = 175 , \text{mm}$

The plastic neutral axis is located at a distance of $50 , \text{mm} + 175 , \text{mm} = 225 , \text{mm}$ from the topmost fiber.

concept:

• Static Theorem (also known as the Lower Bound Theorem of Plasticity):
• The static theorem states that for a given structure and loading, if there exists a statically admissible distribution of bending moments that is both safe (does not exceed the plastic moment capacity) and in equilibrium with the applied loads, then the applied load WW W

W must be less than or equal to the collapse load WcWc W_c

• In simpler terms, if you can find a moment distribution that satisfies equilibrium and does not exceed the material's yield criteria, then the structure will not collapse under that loading.

 Additional Information

• Kinematic Theorem (Upper Bound Theorem):
• This theorem states that for any assumed failure mechanism, the load calculated by equating the internal and external work done during the mechanism is an upper bound to the actual collapse load.
• Uniqueness Theorem:
• The uniqueness theorem asserts that if both a statically admissible stress distribution and a kinematically admissible mechanism predict the same collapse load, then that load is the true collapse load for the structure.

concept:

• Plastic Moment Capacity:
• The plastic moment capacity is the maximum moment that a cross-section can resist once it has fully yielded.
• At this moment, the entire cross-section has reached the yield stress, and the section is in a fully plastic state.
• This is an important concept in plastic analysis and design, where the strength of a structure is determined based on its ability to undergo large deformations before collapsing.

Additional Information

• 1. Moment of Resistance:
• The moment of resistance is the moment at which the internal stresses within a structural member balance the external forces. It can vary depending on whether the section is elastic, partially plastic, or fully plastic.
• 2. Yield Moment:
• The yield moment is the moment at which the first fiber of the cross-section reaches the yield stress. It is the moment that marks the transition from elastic behavior to plastic behavior, but not all fibers in the section are yielding yet.
• 3. Flexural Rigidity:
• Flexural rigidity is the product of the modulus of elasticity E the moment of inertia I of a section. It represents the resistance of a section to bending under an applied moment but is not directly related to the moment at which yielding occurs.

Explanation:

If we denote stiffness matrix as M and flexibility matrix as Δ

The flexibility matrix (F) is the inverse of the stiffness Matrix.

then flexibility matrix is: Δ = M-1

Calculation:

Δ = \frac{1}{{\left( {12\times5 - 4\times4} \right)}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\ { - 4}&{12} \end{array}} \right]

∴ Δ = \frac{1}{{44}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\ { - 4}&{12} \end{array}} \right]

Explanation:

Shape factor

• It is defined as the ratio of the plastic moment to yield moment.
• It is a function of cross-section or shape and is represented by ' f '.
• $f = \frac{{{M_P}}}{{{M_y}}} = \frac{{{f_y}{Z_{Pz}}}}{{{f_y}{Z_{ez}}}} = \frac{{{Z_{Pz}}}}{{{Z_{ez}}}}$

Shape factor for different cross-sectional shapes:

Explanation:

Given Data:

Mp = 150 kN-m

Me = 100 kN-m

Collapse load for the each span of the given beam is

Wu = 6Mp/L1

Total Collapse load for the given beam is

Wu = (6Mp/L1)+(6Mp/L2) = 12Mp/5 = 12×150/5 = 360 kN

Working load on the given beam is 

Ww = 30 + 30 = 60 kN

Load Factor is 

= Wu/Ww = 360/60 = 6

Shape factor = Mp/Me = 150/100 = 1.5

FOS = Load factor/Shape factor = 6/1.5 = 4

concept:

As per the relevant IS code, for a built-up beam, the maximum pitch of rivets in the tension flange should not be more than lesser of 200 mm and 16t, where t is the thickness of the thinner outside plate.

This is applicable to any two adjacent rivets and rivets lying in a line parallel to the force in the member.

The minimum pitch should be 2.5 times the nominal diameter of the rivet

Maximum values of Effective Slenderness Ratios

Explanation:

King Post Truss - Span length up to 5 meter (i.e ≤ 5 meter )

Fink Truss - Span length is in between 5 to 10 meter

Queen Post Truss - Span length is in between 8 to 12 meter.

Howe Truss - Span length is in between 6 to 30 meter.

Concept:

The correct expression as per IS 456: 2000 to find Target mean strength is:

fcm = fck + (1.65 × σ)

fcm = Target mean strength

fck = Characteristic strength

σ = Standard deviation

1.65 is generally a risk factor

Calculations:

fck = 20 N/mm2

σ = 4.0 N/mm2

Target mean strength is:

fcm = fck + (1.65 × σ)

fcm = 20 + (1.65 × 4)

fcm = 26.6 N/mm2

Explanation

These are the assumptions taken in the limit state of collapse: flexure-

1. Plane section before bending remains plane even after bending
2. The maximum strain in concrete at the outermost compression fiber is taken as 0.0035 in bending.
3. Relationship between compressive stress distribution in concrete and strain in concrete may be assumed to be rectangular, trapezoidal, parabolic or any other shape which results in prediction of strength in substantial agreement with the result of test.
4. The tensile strength of concrete is ignored
5. The partial factor of safety for steel is taken as 1.15 and the stress in the steel are derived from stress strain curve.

Explanation:

Stress and Strain Diagram of a Singly Reinforced Beam in Limit State Method:

Where Xu = Actual depth of neutral axis

fck = Characteristic compressive strength of concrete

fy = Yield stress of steel

The centroid of the compressive force is at a distance of 0.42Xu from the extreme compression fibre.

Concept:

The ratio of limiting depth of neutral axis to the effective depth of the beam is given by

$\frac{{{x_u}}}{d} = \frac{{{\rm{Max;strain;concrete}}}}{{\frac{{{\rm{Grade;of;steel}}}}{{{\rm{Modulus;of;Elasticity;of;steel}} \times {\rm{F}}.{\rm{O}}.{\rm{S}}}} + 0.002 + {\rm{Max;strain;concrete}}}}$

Note:

Without getting into the tedious calculation,

We know the standard values of the ratio of limiting depth of neutral axis to the effective depth (k) of the beam for different steel sections as following:

Concept:

As per IS 456: 2000, clause 26.5.3.1,

Longitudinal reinforcement in a column:

Spacing:

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

Additional Information

Percentage of steel:

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

The minimum diameter of longitudinal reinforcement bars:

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

The minimum number of longitudinal bar:

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

• four in rectangular columns
• 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.

Explanation:

As per IS: 800 – 2007

For column, effectively restrained against translation at both the ends but restrained against rotation at one end only

Effective length is 0.8L

Design value of effective length factor for various combination are given below:

Concept:

As per IS456: 2000, cl. 26.3.2,

The horizontal distance between two parallel main reinforcing bars shall usually be not less than the greatest of the following –

1. The diameter of the bar if the diameters are equal.
2. The diameter of the larger bar if the diameters are unequal. 3. 5 mm more than the nominal maximum size of coarse aggregate.

Explanation:

Dia of main reinforcing bar = 16 mm

Size of coarse aggregate = 20 mm

The horizontal distance between two parallel reinforcing bars should not be less than greatest in the following:

1. Diameter of the bar = 16 mm 2. 5 mm more than the nominal maximum size of coarse aggregate = 20 + 5 = 25 mm

So, the horizontal distance between two parallel reinforcing bars should not be less than 25 mm.

Area of steel reinforcement is regulated by the following norms:

i) Minimum area of steel $\left( {{A_{s{t_{min}}}}} \right) = \frac{{0.85}}{{{f_y}}} \times b \times d$

Minimum area of steel is to prevent any sudden failure and to provide minimum required ductility to the section.

ii) Maximum area of steel $\left( {{A_{s{t_{max}}}}} \right) = 4\%$ of gross area

Maximum area of steel is restricted, so as to avoid the problem of compaction while concreting.

Concept:

In the case of one-way slab, the slab bends primarily in one direction (shorter span), and reinforcement is provided accordingly.

According to IS 456:2000, a slab is considered a one-way slab if:

Longer span/Shorter span ​≥ 2

In this question:

7/3 ​= 2.33 ≥ 2 ⇒ One-way slab

Reinforcement Rule for One-Way Slab:

• Main reinforcement is provided in the shorter direction (i.e., to resist bending in the longer span).
• Distribution steel (secondary reinforcement) is provided in the longer direction to take care of temperature, shrinkage, and crack control.

🔹 But note: the slab bends along the longer direction (i.e., reinforcement is placed perpendicular to the direction of bending – hence in the shorter span).

For slab of 7 m × 3 m:

• Main reinforcement = in 3 m direction (i.e., along the shorter span)
• Distribution steel = in 7 m direction

Concept:

To calculate the design two-way shear (punching shear) for a circular column and footing, we use the formula based on the areas of the column and the footing.

$\text{Design Punching Shear} = q_0 \pi (R^2 - r_0^2)$

Given Data:

• Radius of the circular column, r0=200mm=0.2mr0=200mm=0.2m
• Radius of the circular footing, R=1.25mR=1.25m
• Depth of the footing, d=600mm=0.6md=600mm=0.6m
• Factored earth pressure, ququ

Calculation:

1. Substitute the given values into the formula:

$\text{Design Punching Shear} = q_u \pi (1.25^2 - 0.2^2)$

2. Calculate the squared values:

$1.25^2 = 1.5625$ and $0.2^2 = 0.04$

3. Substitute these values back into the formula:

$\text{Design Punching Shear} = q_u \pi (1.5625 - 0.04)$

$\text{Design Punching Shear} = q_u \pi (1.5225)$

The design two-way shear is approximately 1.5225πqu1.5225πqu

.

Concept:

• IS 2250:1981 specifies the code of practice for the preparation and use of masonry mortars. It outlines the guidelines for selecting the proportions, mixing, and application of mortars in masonry construction, ensuring the strength and durability of brick masonry work.

Additional Information

1. IS 456:2000: This code of practice is for plain and reinforced concrete, providing guidelines for concrete mix design, construction, and maintenance. It does not directly relate to mortars used in brick masonry.
2. IS 1332:1982: This standard specifies the composition and properties of phenolic resins for use in general-purpose molding powders. It is not related to masonry mortars.
3. IS 2116:1980: This standard deals with sand for masonry mortars and plastering, specifying the requirements for sand used in these applications.

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:

1. 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.
2. 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.
3. Lowering of 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:

∴ Negative skin friction on a pile under vertical compressive load acts downwards and reduce the load-carrying capacity.

Explanation:

Compaction Curve

• A compaction curve is plotted between the water content as abscissa and the corresponding dry density as ordinate.
• It is observed that the dry density initially increases with an increase in water content till the maximum density is attained. With a further increase in water content, the dry density decreases.
• The water content corresponding to the maximum dry density is known as the optimum water content (O.W.C.) or the optimum moisture content (O.M.C.).

Important Points

• The maximum dry density obtained increases with the increase in the compaction effort. The higher the compaction energy applied, the closer will be the packing of the materials leading to high dry density and low Optimum moisture content.
• Since IS heavy compaction test (modified proctor test) imparts more compaction energy than the IS light compaction test (standard proctor test),
• Fine soils have relatively more OMC but less maximum dry density compared to coarser soil because due to the high specific surface is more water is required for lubrication, hence showing a higher OMC.
• Fine-grained soils have a higher void ratio, occupying more volume, and hence show a lesser value of maximum dry density compared to coarser soil.
• The zero air voids line is the line that indicates an imaginary situation where there is no air in the soil and hence the air content in the soil is 0.

As Sr (Degree of saturation) + Ac (Air content) = 1, So the degree of saturation will be 1 for zero air content.

Explanation

According to IS 2911: Part III 1973, the ratio of bearing resistance for the double under-reamed pile to that of a single under-reamed pile is 1.5 for sandy and clayey soils including black cotton soils.

Note:-

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

1. For deep deposits of expansive soils the minimum length of piles, irrespective of any other considerations shall be 3.5 m below the ground level. For recently filled-up grounds or other strata or poor bearing, the piles should pass through them and rest in good-bearing strata.
2. 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.
3. For piles up to 30 cm in diameter, the spacing of bulbs should not be greater than 1.5 times the diameter of the bulb. For piles of diameter greater than 30 cm, spacing can be reduced to 1.25 times the stem diameter.
4. The topmost 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 the underside of the pile cap is embedded in the ground and the bulb should be a minimum of 1.5 times the bulb diameter.
5. Under-reamed piles with more than two bulbs are not advisable without ensuring their feasibility in strata needing stabilization of boreholes by drilling mud. The number of bulbs in case of bored compaction piles should not exceed tow in such strata.

Explanation:

Stokes Law

Assumptions

• It is assumed that particles show discrete settling (i.e. grains of different sizes fall through a liquid at different velocities).
• Particles assumed spherical in shape.
• Medium is assumed infinite.
• Particles size range 0.2 mm to 0.0002 mm.

Settling velocity,

${V_s} = {\frac{{(G - 1){γ _w}d}}{{18μ }}^2}$

where,

G = specific gravity, γw = unit weight of water, d = diameter of particle, μ = dynamic viscosity

Additional Information

Limitations in the use of Stoke's Law in Sedimentation analysis:

• Stoke's law is applicable for spherical particles only. The fine clay particles are not spherical in shape. While applying Stoke's law, the concept of equivalent diameter is used. The equivalent diameter of a soil particle is defined as the diameter of an imaginary sphere that has the same specific gravity as the soil particle and settles with the same terminal velocity as that of the soil particle.
• It is assumed that every particle settles independently without interference from other particles as well as from the sides of the jar. To minimize error due to this assumption, it is recommended that not more than 50 gms of soil particles be taken in 1000 ml of soil suspension.
• The soil particles in the soil suspension may have different values of specific gravity. But in the computations, an average value of G is used.
• The lower limit of particle size for the validity of Stoke's Law is 0.0002 mm. However, the upper limit for the same is 0.2 mm. For particles of size less than 0.0002 mm, Brownian movement affects their settlement and in the case of particles larger than 0.2 mm, turbulence affects the settlement.
• Stokes's law is valid upto a maximum Reynolds number of 1.

$Consistency;index,;{I_C} = \frac{{{w_L} - w}}{{{w_L} - {w_P}}}$

If ${I_C} = 0; \Rightarrow w = {w_L};i.e;the;soil;is;in;liquid;limit.$

Concept:

Shear strength of soil is the capacity of soil to resist shearing stresses.  It can be defined as the maximum value of shear stress that can be mobilized within a soil mass. If this value exceeded by the shear stress on any plane or a surface at a point, failure will occur in soil because of movement of a portion of the soil mass along that particular plane. The soil is then said to be fail in shear.  Therefore, shear strength generally represented the strength of soil.

As per Mohr- Coulomb criteria, Shear strength of soil is given as:

τ = C + σ tanϕ

Where,

C is the Cohesion

σ is normal stress

ϕ is the friction angle

From above, it can be concluded that

a. It increases with an increase in the normal stress.

b. It is proportional to the cohesion of the soil i.e. increases with increase in cohesion.

c. it increases with increases in friction angle.

Explanation:

Types of Triaxial Test on the basis of Drainage:

(i) Unconsolidated Un-drained (UU) Test:

• In this test expulsion of pore water is not permitted in both stages. It is used for clays in short-term analysis for clays under un-drained conditions at a fast loading rate
• Due to undrained conditions in both the stages, the ϕ does not get mobilized and gives 0 value, which can be verified by drawing Mohr failure envelope for UU test.

From the above graph, we can conclude the ϕ is 0, for UU. As ϕ is defined as the slope of failure envelope and here the slope of failure envelope is 0

(ii) Consolidated Drained (CD) Test:  

• In this test expulsion of pore water is permitted in both stages.
• It is used for short-term and long-term stability analysis in saturated sands and long-term stability analysis in clays.

(iii) Consolidated Un-drained (CU) Test:

• In this test, the expulsion of pore water is permitted in 1^st stage but not in the second stage.
• It is used for the investigation of the safety of the earthen dam which may occur due to the sudden drawdown of the water table.

Explanation:

Given data,  coefficient of active earth pressure, K_a= 1/3

We know that, $K_a \times K_p = 1$

from the above,

$K_p = \frac{1}{\frac{1}{3}} = 3$

The coefficient of passive pressure Kp is 3.

Additional InformationActive earth pressure:

It is the minimum pressure acting on the wall when the wall moves away from, the backfill.

Passive earth pressure:

It is the maximum pressure acting on the wall when the wall moves towards the backfill.

Explanation:

As per Cl. 7.2, IS:1904,

All foundations shall extend to a depth of at least 50 cm below the natural ground level.

The minimum depth of footing on sandy soil is 0.8 - 1 m, for rocky soil, it is 0.05 - 0.5 m and for clayey soil, it is 0.9 - 1.6 m.

The minimum depth of foundation for the load-bearing wall is 900 mm.

Important Points

The design of building foundations is performed based on prescribed serviceability and strength criteria laid down by various standard codes adopted by different countries.

Generally, the total settlement of a foundation is described under the serviceability criterion; whereas a strength criterion is described by bearing capacity of the soil or rock underlying the foundation.

As per Rankine formula, minimum depth of foundation is given as

${{\rm{D}}_{\rm{f}}} = \frac{{\rm{q}}}{{\rm{\gamma }}} \times {\left( {\frac{{1 - {\rm{sin}}\phi }}{{1 + {\rm{sin}}\phi }}} \right)^2}$

Where,

q = Gross bearing capacity, γ = Unit weight of soil, and ϕ = Angle of internal friction of soil

Calculation

 Width of footing (B1)= 900 mm, Settlement of footing (S₁) = 30 mm

 Width of footing (B2) = 600 mm, Settlement of plate (S2) = ?

Under the same condition,

considering, Settlement in case of clay is given by

$\frac{{{S_f}}}{{{S_p}}} = \frac{{{B_f}}}{{{B_p}}}$

$\frac{{{S_2}}}{{{S_1}}} = \frac{{{B_2}}}{{{B_1}}}$

$S_2= {{600\times 30} \over 900}$

Sf = 20 mm

concept:

A minimum seating pressure or preload of 70 g/cm2 (0.7 t/m2), shall be applied and removed before starting the load test.

Then the load is applied with regular increment, using hydraulic jacks. Usually the increment is about 1/5^th of the estimated safe bearing capacity of the soil.

Explanation:

According to IS: 10080 - 1982, the cube size recommended for compressive strength of cement is 70.6 mm, whereas, for concrete cube, the standard size is 150 mm.

Important Points

• A Cement-sand mortar mix of 1:3 is prepared and water is added at 10% of the weight of both cement and sand. In general, 200 grams of cement is used.
• Water content is P/4 +3.5 (P= standard consistency).
• Cubes are kept at a temperature 20 ± 1 0C and at a relative humidity of not less than 90% for 24 ± 0.5 hours.
• Total 3 cubes are tested for compressive strength and the average of three corresponding values is taken as the compressive strength of cement.

So the answer is 70.6 mm.

Concepts:

Procedure for determining the compressive strength of brick as per IS 3495:

Step 1:

Place the specimen with flat faces horizontal, and mortar-filled face facing upwards between two 3-ply plywood sheets each of 3 mm thickness and carefully centered between plates of the testing machine.

Step 2:

Apply load axially at a uniform rate of 14 N/mm2 per minute till failure occurs and note the maximum load at failure. The load at failure shall be the maximum load at which the specimen fails to produce any further increase in the indicator reading on the testing machine.

Step 3:

Repeat the above steps for 5 nos. of bricks and their average value is reported.

Explanation:

Vicat’s Apparatus

• It is used to find out the consistency, initial setting time and final setting time of the cement.
• In the normal consistency test, we have to find out the amount of water to be added to the cement to form a cement paste of normal consistency.
• Vicat's apparatus consists of an arrangement to hold the plunger of 10 mm diameter and two other needles which are made to freely fall into a mold filled with the cement paste and the amount of penetration of the needles of plunder can be noted using the vertical graduations from 0 mm to 50 mm.
• The length of the needle varies from 40 mm to 50 mm.
• Standard consistency is defined as the consistency of the cement paste that permits the penetration of circular vicat’s plunger of 10 mm dia and height 50 mm up to the depth of 5 - 7 mm from the bottom into the mould or 33 to 35 mm from the top of the mould.
• The needle of diameter 1 mm is used and the center needle of the attachment of the Vicat plunger projects the circular cutting edge by 0.5 mm as can be observed in the figure below:

Explanation

As per IS 1077: 1992 code, the common burnt clay bricks are classified into various classes based on their average compressive strength as mentioned below in the tabulated form:

It is clear from above table that, the minimum crushing strength of common brunt clay brick is 3.5 MPa.

concept:

The main function of gypsum in the cement is to prevent flash set of cement.

During the grinding of clinker, gypsum or plaster of paris (CaSo4) is added to adjust the setting time. The amount of gypsum is about 3% by weight of clinker. It improves the soundness of cement.

The compound C3A is characteristically fast reacting with water and may lead to an immediate stiffening of paste and this process is termed as flash set. The role of gypsum added in the manufacture of cement is to prevent such a fast reaction.

CONCEPT:

As per IS 456: 2000,

The Short term Static Modulus of elasticity of concrete Ec is given as-

\({{\rm{E}}{\rm{c}}} = 5000{\rm{;}}\sqrt {{{\rm{f}}{{\rm{ck}}}}} \)

Where,

fck = characteristic compressive strength of concrete,

fck and Ec are measured in MPa.

As per IS 456:2000, the flexural tensile strength of concrete = $0.7\sqrt {{{\rm{f}}_{{\rm{ck}}}}}$

Where, fck = 28 days characteristic compressive strength of concrete, which is the full strength of the concrete

Flexural tensile strength of concrete is always lesser than its characteristics compressive strength and it increases with an increase in the grade of the concrete.

Explanation:

The absolute maximum bending moment will occur at the centre when the load is spread equally on either side of the centre.

$\begin{array}{l} {M_{max}}; = ;\frac{{wa}}{4}\left( {L - \frac{a}{2}} \right)\ ; = ;\frac{{40 \times 5}}{4} \times \left( {10 - \frac{5}{2}} \right) \end{array}$

= 50 × 7.5 = 375 KN-m at centre

Concept:

To determine the maximum positive shear force at a specific point on a girder, we analyze the effect of different point loads on the shear force at that point. The shear force is calculated by considering the contribution of each load based on its position relative to the point of interest.

Given Data:

• Point Loads: 10 kN, 15 kN, 15 kN, and 8 kN
• Center-to-center spacing between loads: 2 m
• Span of the girder: 30 m
• Point of interest for shear force: 8 m from left support

Calculation:

1. Shear Force with 8 kN Load:

$\text{SF} = 10 \times \frac{8}{30} + 15 \times \frac{6}{30} + 15 \times \frac{4}{30} + 8 \times \frac{2}{30}$

$\text{SF} = 8.2 , \text{kN}$

2. Maximum Positive Shear Force Calculation:

$\text{SF} = 10 \times \frac{16}{30} + 15 \times \frac{18}{30} + 15 \times \frac{20}{30} + 8 \times \frac{22}{30}$

$\text{SF} = 30.2 , \text{kN}$

3. Check for Another Load Position (15 kN Load on the Section):

$\text{SF} = 10 \times \frac{18}{30} + 15 \times \frac{20}{30} + 15 \times \frac{22}{30} - 8 \times \frac{6}{30}$

$\text{SF} = 25.4 , \text{kN}$

The maximum positive shear force is 30.2 kN.

Explanation:

Stability:

1. Determinate Structures:

• A determinate structure is one in which the reactions and internal forces can be determined solely by using the equations of static equilibrium (i.e., the sum of forces and moments equals zero).
• These structures are stable under standard load conditions and are easier to analyze because they do not involve complex compatibility conditions.
• However, determinate structures lack redundancy, meaning that if one element fails, the whole structure can become unstable.

2. Indeterminate Structures:

• An indeterminate structure has more supports or internal members than necessary for equilibrium, leading to additional unknown forces that cannot be determined by static equilibrium alone. These require additional compatibility equations.
• Indeterminate structures are generally more stable because they have redundancy. If one element fails or if unexpected loads are applied, other elements can redistribute the forces, maintaining the overall stability of the structure.
• This redundancy provides enhanced safety against unexpected loading conditions, making indeterminate structures more resilient and stable compared to determinate structures.

Based on standard structural analysis theory:

• Indeterminate structures are typically considered more stable due to their redundancy, which allows for load redistribution and provides a safeguard against failure.

Concepts:

The strain energy is the energy stored in a body due to its elastic deformation.

The most general formula for strain energy is:

$U = {\frac{1}{2}} \times F \times \delta$

Where, F is the applied force and δ is deformation

However, When stress σ is proportional to strain, ϵ, the strain energy formula is: 

   $U = {\frac{1}{2}} \times V \times \sigma \times \epsilon$; V is the volume of material.

Based on the above formals, the strain energy stored in the material due to different types of forces/moments is given as:

1. Strain energy due to bending is:

   $U = \int\frac{M^{2}}{2EI}dx$

2. Strain energy due to Torsion is :

    $U = \int\frac{T^2L}{2GJ}dx$

Concept:

Distribution factor (DF):

$DF={Stiffness, of, a, member\over Sum, of, stiffness, of , all, members, at, that , joint}$

Or

$DF={Relative, stiffness, of, a, member\over sum, of, relative, stiffness, of , all, members, at, that , joint}$

Where M1, M2, M3, and M4 are moment induced in member OA, OB, OC, and OD respectively.

The summation of DF for all members at a joint is one.

Concept:

Strain energy:

It is defined as the energy adsorption of material due to strain under an applied load. It is also equal to the work done on a specimen by an external force. It can be denoted by U.

{\bf{U}} = \mathop \smallint \limits_0^{\bf{L}} \frac{{{{\bf{M}}^2}{\bf{dx}}}}{{2{\bf{EI}}}}

Where,

M = Maximum moment, E = Elastic modulus, I = Moment of inertia.

Explanation:

Strain energy will be maximum when the maximum moment will be high.

Maximum moment for different types of beam:

For the cantilever beam, ${\bf{M}} = \frac{{{\bf{W}}{{\bf{L}}^2}}}{2}$

For Fixed-beam, ${\bf{M}} = \frac{{{\bf{W}}{{\bf{L}}^2}}}{{12}}$

For simply supported beam, ${\bf{M}} = \frac{{{\bf{W}}{{\bf{L}}^2}}}{8}$

For propped cantilever beam, ${\bf{M}} = \frac{{{\bf{W}}{{\bf{L}}^2}}}{8}$

Hence, Strain energy for the uniformly distributed load will be maximum in the cantilever beam.