UPPSC AE Civil Paper 1 Mock Test - 1 - Civil Engineering (CE) MCQ

दिए गए विकल्पों में सही उत्तर ‘हवा’ है।

व्याख्या

• दिए गए विकल्पों में 'हवा' शब्द 'प्रभंजन' का उचित पर्यायवाची शब्द है।
• इसके अन्य शब्द हैं - प्रवात, समीरण, मातरिश्वा, बयार, पवन, वायु, हवा, समीर, वात, मारुत, अनिल, पवमान आदि।

अन्य विकल्प:

अन्य संबंधित बिंदु

दिये गए विकल्पों में से विकल्प 3 'ख' सही उत्तर है।
व्याख्या
स्पर्श व्यंजन, कण्ठ्य ध्वनि, अघोष और महाप्राण ध्वनि है: ख
महाप्राण ध्वनि -

• महाप्राण व्यंजन वह व्यंजन होतें हैं जिन्हें मुख से वायु-प्रवाह के साथ बोला जाता है, जैसे की 'ख', 'घ', 'झ' और 'फ'। ​

अघोष-

• वह व्यंजन होतें हैं जिनका उच्चारण करते समय स्वर तंत्रियाँ झंकृत नहीं हैं। जैसे- क, ख आदि।

कंठ्य वर्ण-

• जिन वर्णों का उच्चारण में कंठ का प्रयोग किया जाता है।
• क,ख,ग,घ आदि कंठ्य वर्ण हैं।

अन्य संबंधित बिंदु

दिए गए विकल्पों में से ‘तंत्र' का एक अर्थ है- "सेना"

व्याख्या

• तंत्र के अन्य अनेकार्थी शब्द- उपासना, दवा, पद्धति, कपड़ा, सूत।
• हेम का एक अर्थ "पाला" है।
• हेम अन्य अनेकार्थी शब्द - स्वर्ण, जल, तुषार, केशर का फूल।
• तारा के अनेकार्थी शब्द - आँख की पुतली, सितारा, महाराजा हरिश्चंद्र की पत्नी।
• तम के अनेकार्थी शब्द - अँधेरा, कालिख, अज्ञान, क्रोध, राहु, पाप।

अन्य संबंधित बिंदु

• जिन शब्दों के एक से अधिक अर्थ होते हैं, उन्हें 'अनेकार्थी शब्द' कहते है।

उदाहरण -

संपूर्ण' में सम् उपसर्ग है।

व्याख्या

• सम्पूर्ण में सम् उपसर्ग और पूर्ण मूल शब्द है।
• सम्पूर्ण का अर्थ- अच्छी तरह भरा हुआ। (विशेषण)

महत्वपूर्ण बिंदु

• सम उपसर्ग से शब्द – समकालीन, समदर्शी, समकोण, समकालीन, समझाना, समवार, समर्पण, समाज, समर्पित।
• 'सम्' उपसर्ग से बना शब्द है- संस्कृत, संस्कार, संगीत, सयोग।

अन्य संबंधित बिंदु

• उपसर्ग- यह वह शब्दांश या अव्यय है, जो किसी शब्द के आरंभ में जुड़कर उसके अर्थ में विशेषता लाता है या उसका अर्थ बदल देता है।
  • जैसे- प्र (उपसर्ग) + सिद्ध (मूलशब्द) = प्रसिद्ध, अभि (उपसर्ग) + मान (मूलशब्द) = अभिमान
• उपसर्ग की विशेषता -
• उपसर्ग की तीन गतियाँ या विशेषताएँ होती हैं-
  • (1) शब्द के अर्थ में नई विशेषता लाना।
• जैसे- प्र + बल= प्रबल
• अनु + शासन= अनुशासन
  • (2) शब्द के अर्थ को उलट देना।
• जैसे- अ + सत्य= असत्य
• अप + यश= अपयश
  • (3) शब्द के अर्थ में, कोई खास परिवर्तन न करके मूलार्थ के इर्द-गिर्द अर्थ प्रदान करना।
• जैसे- वि + शुद्ध= विशुद्ध
• परि + भ्रमण= परिभ्रमण

​संतान एक तद्भव शब्द है।
संतान का तत्सम शब्द संतति होता है।
व्याख्या
तद्भव शब्द संस्कृत शब्दों में विकार से उत्पन्न होने वाला शब्द है।

अन्य संबंधित बिंदु
तत्सम शब्द:- जिन शब्दों को संस्कृत से बिना किसी परिवर्तन के ले लिया जाता है उन्हें तत्सम शब्द कहते हैं।

• उदाहरण:-आम्र, आश्चर्य,अक्षि,अमूल्य,अग्नि

सही उत्तर 'इक' है।

• 'समाज' शब्द में इक प्रत्यय जोड़कर 'सामाजिक' शब्द बना है।
• सामाजिक शब्द में समाज मूल शब्द और इक प्रत्यय है: 'समाज + इक'
• इक प्रत्यय के उदाहरण :
  • शारीरिक = शरीर + इक
  • नैतिक =नीति + इक
  • पारिवारिक=परिवार + इक
  • धार्मिक=धर्म + इक
  • दैनिक=दिन + इक

व्याख्या

• ई प्रत्यय के उदाहरण :-
  • खुशी-खुश + ई
  • दोस्ती-दोस्त +ई
  • भारी-भार + ई
  • सुखी-सुख + ई
  • बोली-बोल + ई

• इत प्रत्यय के उदाहरण :-
  • प्रमाणित -प्रमाण + इत
  • व्यथित -व्यथा + इत
  • द्रवित -द्रव + इत
  • मुखरित -मुखर + इत
  • झंकृत -झंकार + इत
  • शिक्षित -शिक्षा + इत
• ईय प्रत्यय के उदाहरण :-
  • दर्शनीय- दर्शन + ईय
  • स्वर्गीय-स्वर्ग + ईय)
  • भारतीय -भारत + ईय
  • जातीय - जाती + ईय
  • मृगीय- मृग + ईय

अन्य संबंधित बिंदु

• प्रत्यय: शब्द के उपरांत जिस शब्द का प्रयोग किया जाता है वह प्रत्यय है।
• जैसे - ता, औना, अन, अत- श्रो + ता = श्रोता
• उपसर्ग​:शब्दों तथा धातुओं के पूर्व में संयुक्त हो नये सार्थक शब्दों का निर्माण करने वाले शब्दों को उपसर्ग कहते हैं।
• जैसे - प्र, सु, अति, अधि, अनु, नि- प्र + हार = प्रहार

Explanation:

Water Absorption Test

• This test helps to determine the water absorption of coarse aggregates as per IS: 2386 (Part III) – 1963. For this test, a sample not less than 2000g should be used.
• The apparatus used for this test are:-

Wire basket – perforated, electroplated, or plastic coated with wire hangers for suspending it from the balance, Water-tight container for suspending the basket, Dry soft absorbent cloth – 75cm x 45cm (2 nos.), Shallow tray of minimum 650 sq. cm area, Air-tight container of a capacity similar to the basket and Oven.

Procedure

Procedure For Aggregate Coarser Than 6.3mm:

• About 2 kg of aggregate sample is taken, washed to remove fines, and then placed in the wire basket. The wire basket is then immersed in water, which is at a temperature of 220C to 320C.
• Immediately after immersion, the entrapped air is removed from the sample by lifting the basket 25 mm above the base of the tank and allowing it to drop, 25 times at a rate of about one drop per second
• The basket, with aggregate, is kept completely immersed in water for a period of 24 ± 0.5 hours.
• The basket and aggregate are weighed while suspended in water, which is at a temperature of 220C to 320C.
• The basket and aggregates are removed from the water and dried with a dry absorbent cloth.
• The surface dried aggregates are also weighed.
• The aggregate is placed in a shallow tray and heated to 100 to 1100C in the oven for 24 ± 0.5 hours. Later, it is cooled in an airtight container and weighed.

Formula Used =

Concepts:

The straight-line method of depreciation assumes a constant rate of depreciation. It calculates how much a specific asset depreciates in one year, and then depreciates the asset by that amount every year after that.

To calculate the straight-line depreciation rate for your asset, simply subtract the salvage value from the asset cost to get total depreciation, and then divide that by useful life to get annual depreciation:

Annual depreciation = (purchase price - salvage value) / useful life

Calculation:

Cost of Construction of building i.e. Purchase Value = Rs. 30,00, 000/-

Salvage Value = Rs. 3,00, 000/-

Design life of building = 30 Years

Annual Depreciation = (30,00, 000-3,00,000)/30

⇒ Annual Depreciation = Rs. 90000 /year

Shear stress at any point in a cross-section is given by the following formula:

Where
f = shear stress
Q = First moment of area about the neutral axis of the area above the point where we want to calculate the shear stress
Q = A × y
V is Shear force
y is the centroid of the area above the point where we want to calculate the shear stress measured from the neutral axis.
A = Cross-sectional area above the point where we want to calculate the shear stress
I = moment of inertia about the neutral axis
b= width of the section where shear stress is to be calculated.
Shear stress is zero at extreme fibres
For a symmetric ‘I’ section, the intensity of shear stress is maximum at the centroid of the section which is neutral axis as shown in the figure.

Explanation:

As per IS 456: 2000, the minimum percentage of reinforcement required will be:

Mild Steel:

HYSD Bar

Where

B = width of the slab and

D = overall depth of the slab

Now,

B = unit width given= 1 m = 1000 mm

D = 100 mm

Explanation:

Parallelogram Law

• In Mathematics, the parallelogram law is the fundamental law that belongs to elementary Geometry.
• This law is also known as parallelogram identity.

Parallelogram Law of Addition:

• Parallelogram Law of Vector Addition Formula with Proof & Example.
• The Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals. In Euclidean geometry, it is necessary that the parallelogram should have equal opposite sides.
• If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point. Therefore, the resultant vector is completely represented both in direction and magnitude by the diagonal of the parallelogram passing through the point.

Explanation:

In the figure above, Cos θ = CD/BC = 4/4√ 2 = 1/√ 2
∴ θ = 45°
At joint D:
Force 10 kN and member force BD for colinear. So, force in member CD is 0, ie F_CD = 0
And, ΣV = 0 at joint D, we get
F_BD = 10 kN (Tensile)
Now, at joint B
Σ V = 0, F_BC × Cos 45° + F_BD = 0
F_BC × 1/√ 2 + 10 = 0
F_BC = - 10 × √ 2 (Compressive)
Now, at joint B
Σ H = 0, F_BC × Sin 45° + F_AB = 0
- 10 × √ 2/ √ 2 + F_AB = 0
F_AB = 10 kN (Tensile)

Explanation:
Plasticity chart :

• It is a graph between plasticity index (IP) and liquid limit (WL) in percentage which is used for the classification of fine-grained soils as per the Indian Standard Soil Classification System(ISSCS).
• If more than 50% percent of soil passes through a 75-micron sieve, then it is classified as fine-grained soil.

• The 'A' line in this chart is expressed as IP = 0.73 (WL - 20), , where IP = plasticity index , WL = liquid limit
• Depending on the point in the chart, fine soils are divided into clays (C), silts (M), or organic soils (O).
• The organic content is expressed as a percentage of the mass of organic matter in a given mass of soil to the mass of the dry soil solids.
• The 'U' line in this chart is expressed as IP = 0.9( WL - 8 ).

Calculation:
Given
W_L = 50 %
IP = 0.73 (WL - 20) = 0.73 (50-20) = 21.9 %
W_P = W_L - I_P = 50 - 21.9 = 28.1 %

According to I.S. 456-1978, the thickness of reinforced concrete footing on piles at its edges is kept less than ​15 cm.
However, from the updated IS Module of 456:2000:
As per IS 456: 2000, Clause 34.1.2,
Thickness at the Edge of Footing
In reinforced and plain concrete footings, the thickness at the edge shall be not less than 150 mm for footings on soils.
For footings on piles, the thickness at the edge shall be not less than 300 mm (30 cm) above the tops of piles.

As per Annex-G (G-1.1 (c)) of IS: 456-2000, Limiting Moment of resistance is given by following expression;

As per clause no. 38.1, The limiting values of the depth of neutral axis for different grades of steel;

∴

Soil mass is generally a three phase system as it contains soils solids at bottom along with water in middle and air at top. The space covered by water and air in the soils mass is called voids. The water and air contents are changed by changes in conditions and location: soils can be perfectly dry (have no water content) or be fully saturated (have no air content) or be partly saturated (with both air and water present).

Vair = Volume of air voids
Vw = Volume of water
V5 = Volume of solids
Wair = mass of air which is assumed to be zero
Ww = mass of water
Ws = Mass of soil solids.

Explanation:
The construction of stones bonded together with mortar is termed stone masonry.
Types of stone masonry
Based on the arrangement of the stone in the construction and the degree of refinement in the surface finish, the stone masonry can be classified broadly in the following two categories:
1. Rubble Masonry
2. Ashlar Masonry
3. Dry rubble masonry

Dry rubble masonry:
Dry rubble masonry is a technique where stones or rocks are stacked together without the use of any mortar or bonding material. This type of construction relies solely on the interlocking and gravity between the stones to create a stable and sturdy structure.

• The stones are carefully selected and placed in a way that distributes the weight evenly to ensure the stability of the wall.
• Both dry rubble masonry and Uncoursed rubble masonry techniques are commonly used in the construction of retaining walls, boundary walls, and other similar structures.
• However, dry rubble masonry is generally more suitable for smaller structures and hence is in the poorest and cheapest form of masonry.
• While uncoursed rubble masonry is preferred for larger structures as it provides more stability and structural integrity.

Rubble Masonry
In this category, the stones used are either undressed or roughly dressed having wider joints. This can be further subdivided as uncoursed, coursed, random, dry, polygonal and flint.

• (i) Uncoursed rubble masonry:
  • This is the one of the cheapest, roughest and poorest form of stone masonry but not as cheap as dry rubble masonry.
  • The stones used in this type of masonry very much vary in their shape and size and are directly obtained from quarry.
  • Uncoursed rubble masonry can be divided into the following.
    • a) Uncoursed random rubble
    • b) Uncoursed squared rubble

• (ii) Coursed rubble masonry:
  • In this type of stone masonry the uniform height stones are used in horizontal layers not less than 13 cm in height.
  • Generally, the stone beds are hammered or chisel dressed to a depth of at least 10 cm from the face.
  • The stones are arranged in such a manner so that the vertical joints do not coincide with each other.

Ashlar Masonry

• This type of masonry is built from accurately dressed stones with uniform and fine joints of about 3 mm thickness by arranging the stone blocks in various patterns.
• The backing of Ashlar masonry walls may be built of Ashlar masonry or rubble masonry.
• The size of stones blocks should be in proportion to wall thickness.

Explanation:

For the design of laced columns:

• The thickness of the lacing flat for a single lacing system should be less than 1/40 of its effective length and for a double lacing system, it should be less than 1/60 of its effective length.
• The sections to be laced are so spaced that the radius of gyration of the section about the axis perpendicular to the plane of lacing is not less than the radius of gyration about the axis in the plane of lacing.
• The effective slenderness ratio should be taken as 1.05 times the actual maximum slenderness ratio, in order to account for shear deformation effect.
• The maximum spacing of the lacing bar should be such that the minimum slenderness ratio of the component member should not be greater than 50 or 0.7 times the slenderness ratio of the member.
• The lacing for compression members should be proportioned to resist transverse shear equal to 2.5% of the axial force in the column.
• The minimum flat width should not be less than three times the nominal diameter of the end connector.
• Angle of inclination of lacing bar should be kept between 40° - 70°. Lacing is most efficient between 35° to 45°, but angle less than 40° would result in a higher length of lacing bar which may then to buckle individually.
• The slenderness ratio of the lacing bar should be less than 145.

Concept:
Influence line diagram:

• An influence line for a given function, such as a reaction, axial force, shear force, or bending moment, is a graph that shows the variation of that function at any given point on a structure due to the application of a unit load at any point on the structure.
• The influence line diagram for S.F or B.M at a section is the variation in the value of S.F or B.M at that section as the unit load transverses the span from left to right.
• IDL can be drawn for statically determinate as well as indeterminate structures.

If the structure is given unit displacement at pinned support then there will be a vertical reaction at pinned support.
The deflected shape is a cubic parabola.
The area (m2) of the influence line diagram for the reaction at the hinged end of a uniform propped cantilever beam of span 'L' m is 3L/8

Concept:
The figure below shows a semi-circular hook, fully dimensioned with respect to K.

In case of deformed bar, the value of bond stress for the various grade of concrete is greater than 60% than the plain bar.
When hooks are formed in deformed bars, the internal radius of the bend should be at least four times the diameter of the bar.
Note:
However, in the question, both deformed bars and mild steel are written. So given answer is as per deformed bars.

Concept:

Degree of Statical Indeterminacy:

• The DEGREE OF STATIC INDETERMINACY (DSI) is the number of redundant forces in the structure.
• Redundant forces are the forces that cannot be found by writing and solving only the equations of equilibrium. They must be independent.
• Therefore, the degree of static indeterminacy (DSI) represents the difference between the number of static unknowns (reactions and internal forces) and the number of static equations (equilibrium equations).

​Mechanism:

• If the segment of a beam or any other structure between plastic hinges is able to move without increasing the load, then it is called a mechanism.
• If a mechanism is formed then it means that the structure becomes unstable.

In plastic analysis to make the statically indeterminate structure a determinate structure, the number of plastic hinges required is "j", where "j" is the degree of static indeterminacy. Further, to make a mechanism, i.e. to make a structure unstable on a statically indeterminate structure, the number of plastic hinges required = j + 1, where j is static indeterminacy.

So the correct answer is Option 3

Concept:
Kernel of a section

• The kernel of the section is the area within which the line of action of eccentric load should pass so that the tensile stress is zero
• When the load acts in the kernel of the section, there is only bending.
• The kernel of the section is also called the core of the section

For the condition of no tension in the section, the following rule has to be satisfied,
e ≤ 2 × k²/d
e - Eccentricity of loading
k - Radius of gyration
d - Depth of the section
Kernel of rectangular section:

Moment of Inertia of Rectangle I = b x d³/12
Area of rectangle A = l × b
I = A × k2

k2 = d2/12

e ≤ d/6
For no tension, the load should not be placed at a distance more than d/6 on either side of the centroidal axis.
Similarly, it can be shown e ≤ b/6
Hence the kernel of the section is a rhombus

Explanation:

Sheep foot roller:

• Sheep foot roller also named tamping roller. The front steel drum of sheep foot roller consists of many rectangular-shaped boots of equal sizes fixed in a hexagonal pattern.
• In sheep foot, roller compaction is by static weight and kneading of the respective layer. This makes tamping roller better suited for clay soils.

Other Related Points

Mostly four types of rollers used are:

1. Pneumatic tired roller
2. Tamping roller/ sheep foot roller
3. Smooth wheel rollers
4. Vibratory Roller

Pneumatic tired roller:

• Pneumatic tired roller has a number of rubber tires at the front and at the rear end. A pneumatic tired roller can be used for highways, construction of dams and for both fine-grained and non-cohesive soils.

Smooth wheel rollers:

• Smooth wheel roller and vibratory rollers are the same. Both have the same characteristics. Only the difference in both is vibratory equipment. Smooth wheel roller has no vibrator attached to the drum. This makes smooth wheel roller best suited for rolling of weaker aggregates, proof rolling of subgrades and in compacting asphalt pavements.

Vibratory Roller:

• Vibratory type rollers have two smooth wheels/drums plus the vibrators. One is fixed at the front and the other one is on the rear side of the vibratory roller. Vibration is to reduce the air voids and to cause densification of granular soils. During the vibration of the soil layer, rearrangement of particles occurs due to deformation of the granular soil because of oscillation of the roller in a cycle.

Explanation:

Indicative proportions of materials are shown below for self-compactible concrete:

1) Water/powder ratio by volume is to be 0.85 to 1.10

2) Total powder content to be 160 to 240 liters (400-600 kg) per m3

3) The sand content may be more than 38% of the mortar volume.

4) Coarse aggregate content should normally be 28 to 35% volume of the mix.

5) Water/cement ratio is selected based on the strength required. In any case, water content should not exceed 200 liters/m3.

∴ Self–compacting concrete (SCC) is characterized by High powder components and less coarse aggregate.

Types of sample:
Undisturbed sample:
If the soil structure, moisture content & mineral content remains the same while sampling, the sample is called an undisturbed sample.
Disturbed sample:
These are two types
a) Representative sample:
If only soil structure gets modified but moisture content & mineral content remains the same while sampling, the sample is called a representative sample.
b) Non-representative sample:
If all the 3 i.e. soil structure, mineral content, and moisture content get modified by sampling, samples are known as non-representative samples.
Degree of disturbance:
The degree of disturbance of the sample collected by various methods can be expressed by a term called the area ratio, which is given by

Di = inner diameter of cutting edge and Do = outer diameter of cutting edge
A soil sample generally can be considered undisturbed if the area ratio is less than or equal to 10%. The area ratio for the standard spoon sampler is nearly 10% and for Shelby, tubes are 13.5 %.
NOTE: According to the given options, the most appropriate option will be 15 %

Concept:

IS 399-1963: It is used for "the classification of commercial timbers and their zonal distribution".

1. Availability - The availability of timbers is categorized under three classes indicated below:

• X: Most common, 1415 m³ (1000 tonnes} and more per year
• Y: Common, 355 m³ (250 tonnes) to 1415 m³ (I000 tonnes) per year
• Z: Less common, below 355 m³ (250 tonnes) per year
  

2. Durability - It is based on the ‘graveyard’ tests carried out in the open, at the Forest Research Institute and Colleges, Dehra Dun, in which test specimens of size 24 x 2 X 2 in. The condition of the specimens was examined at frequent intervals and from these observations, their average useful life has been calculated.

The timbers are classified for durability according to the average life of these test specimens as follows :

• High - Timbers have an average life of 120 months and over
• Moderate - Timbers have an average life of less than 120 months but of 60 months or over
• Low - Timbers have an average life of less than 60 months
  

3. Treatability - The classification is based on experiments carried out at the Forest Research Institute and Colleges, Dehra Dun, on the pressure treatments of various timbers with creosote-crude oil mixtures and with water-soluble preservatives, under conditions of treatment which are normally used for these processes.

The treatability of timbers has been classified as follows :

• Heartwood is easily treatable.
• Heartwood is treatable, but complete penetration of preservatives is not always obtained.
• Heartwood is only partially treatable.
• Heartwood refractory to treatment.
• Heartwood is very refractory to treatment, penetration of preservative being practically nil from side or end.
  

4. Refractoriness to Air Seasoning -The timbers are classified, as stated below, under three categories, depending upon their behavior with respect to cracking and splitting during normal air-seasoning practice suitable for the species concerned:

• High refractoriness (indicated ‘High’ in the tables)
• Moderate refractoriness (indicated ‘Moderate’ in the tables)
• Low refractoriness (indicated ‘Low’ in the tables).

Explanation:

The average weight and the range of weights per cubic meter(or cubic foot) at 12 percent moisture content for all the timbers have been supplied by the Forest Research Institute and Colleges, Dehra Dun, and are based generally on a very large number of samples of each species in a particular zone or from other zones.

Concept:
FOS = (Critical Hydraulic Gradient (I_c)/(Exit Hydraulic Gradient (I_e)
Critical hydraulic gradient (I_c) = (G - 1)/(1 + e)
Calculation:
Given data:
Factor of safety (FOS) = 5
Specific gravity (G) = 2.5
Porosity (n) = 0.35
e = n/(1 - n) = 0.35 / (1 - 0.35) = 0.54
I_e = (2.5 - 1)/(1 + 0.54) = 0.97
5 = 0.97 / I_e
I_e = 0.97/5 = 0.195
The maximum permissible exit gradient for soil is 0.195

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and like then the radius of the Mohr circle will be zero.

The radius of Mohr's circle,

R =

Here,

σ_x = σ_y and τ_xy = 0

∴ R = 0 means Mohr's circle reduces to a point.

Explanation:
Compaction factor test

• The compacting factor test is designed primarily for use in the laboratory but it can also be used in the field.
• It is more precise and sensitive than the slump test
• Useful for concrete mixes of very low workability and is normally used when concrete is to be compacted by vibration.
• The compaction factor is the ratio of weights of partially compacted to fully compacted concrete.
• The compaction factor ranges from 0.78 to 0.95

Important Points Different types of a slump are as follows:

Collapse slump: In this case, fresh concrete collapses completely. The mix is too wet or high workability mix, slump test isn’t appropriate for such mix.
Shear Slump: If one-half of the cone slides down in an inclined plane, it is called a shear slump. It is an indication of the lack of cohesion of the mix. Again perform the experiment to avoid a shear slump.
True Slump: Mix has high stiff consistency. In a true slump concrete just subsides shortly and more or less maintain the mould shape. This type of slump is most desirable.
Zero Slump: If concrete maintains the actual shape of the mould, it is called zero slumps which represent stiff, consistent and almost no workability.

Concept:

A beam is expected to resist bending and shear stresses. Bending stress is zero at centre and maximum at the top and bottom layer of the beam. While shear stress is maximum at the centre and zero at top and bottom layers.

In I section, the web resists shear forces, while the flange resist more than 80% of the bending moment. Beam theory shows that the I-shaped section is a very efficient form for carrying both bending and shear loads in the plane of the web.

Even though the I-cross-section has a reduced capacity in the transverse direction, and is also inefficient in carrying torsion, for which hollow structural sections are often preferred, the latter are more complicated and expensive to manufacture.

∴ For all the practical purposes, economical and efficient I-section is used.