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

​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

वर्तनी की दृष्टि से आधीन शब्द सही नहीं है
व्याख्या

1. शुद्ध शब्द-अधीन
2. अशुद्ध शब्द-आधीन
   1. ​प्रातिनिधिक: शुद्ध है इसका अर्थ: प्रतिनिधि
   2. आभ्यन्तरिक : शुद्ध है ,आभ्यन्तरिक का अर्थ: जो अंदर का हो:"वह मानव की भीतरी शारीरिक

अन्य संबंधित बिंदु

• वर्तनी:- लिखने की रीति को वर्तनी कहते है। वर्तनी शब्द का अर्थ है पीछे चलना।
• उच्चारित होने वाले शब्द लेखन में प्रयोग होने वाले चिन्हों को वर्तनी कहते है।

सही उत्तर 'प्रिय - प्रिया' हैं।

• दिए गए विकल्पों में 'प्रिय - प्रिया' तत्सम और तद्भव का युग्म गलत है।
• प्रिय शब्द का तद्भव शब्द: 'पिया' हैं।

व्याख्या

अन्य संबंधित बिंदु कुछ महत्वपूर्ण तत्सम- तद्भव शब्द

Concepts:

Concept:

Distribution factor (DF):

It is the ratio of the stiffness factor for that member and the total stiffness of all the members meeting at a joint.

DFBC =

KBC = Stiffness of member BC

KBA = Stiffness of member BA

Calculation:

As we know that

Stiffness of member CB ( Far end is Roller) KBC =

Stiffness of member BC ( Far end is fixed) KBA=

DFBC =

DFBC =

DFBC = 0.64

The distribution factor for CB is 1, as there is there is no support on the right of joint C and hence its stiffness becomes 0 and putting its stiffness in the formula for DF we get the DF for CB as 1

Anti-termite treatment

• It is a process in which soil or sub soil treatment is applied to a building in preliminary stages of its construction.
• The requirement is to provide the buildings or any type of structures with a chemical protector against the sub-terrain termites.
• Anti-termite treatment is a sophisticated task which requires effective knowledge of the chemical composition, soils characteristics, various termite to be dealt with and the suitable and non suitable environmental conditions.
• The effectiveness of chemical depends upon the choice of the chemical, the dosage adopted and the thoroughness of application.
• The chemical solutions or emulsions are required to be dispersed uniformly in the soil and to the required strength so as to form an effective chemical barrier which is lethal and repellent to termites.
• The various patented chemical Insecticide such as DDT, PHP, PCP, etc. are available.

Note:

• But the following chemicals in oil solution or preferably water emulsion have proved to be successful. They are Aldrin, Chlordane, Dieldrin, Heptachlor, etc.

∴ Aldrin is most appropriate answer.

Explanation:
The given diagram is 1(1/2) brick wall of English bond.

Important Points
Important types of bond in brick masonry:

The deflection at the free end of the cantilever due to uniformly distributed load is given as:


Given that
E = 2 × 10⁵ N/mm²
I = 8 × 10⁷ mm⁴
L = 2500 mm
Δ = 5 mm
Let the deflection is

w = 16.38 N/mm
w = = 16.38 KN/m.

Explanation:
By plotting the load effect on a specific point in a structure as a function of the position of the load:

• This is the correct description of how influence line diagrams are constructed. Influence lines show how the response (e.g., shear force, bending moment, deflection) at a specific point in a structure varies as a live load moves along the structure.
• The position of the load is plotted on the x-axis, and the corresponding load effect (e.g., shear or moment) is plotted on the y-axis, resulting in a graphical representation of the load's influence on the structure.

Other Related PointsInfluence Line Diagram (ILD):

• An influence line represents the variation of either the reaction, shear, moments, or deflection at a specified point in a member as a concentrated unit force moves over the member.
• Influence lines represent the effect of a moving load only at a specified point on a member, whereas shear and moment diagram represents the effect of fixed loads at all points along the member.
• Thus influence line helps in deciding, at a glance, where should the moving loads be placed on the structure so that it creates the greatest influence at the specified point.
• ILD can be drawn for statically determinate as well as indeterminate structures.

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.

Concept :

Parallelogram Law of forces:

If two forces acting at the point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through the point.

Resultant force F_r, of any two forces F₁ and F₂ with an angle between them, can be given by vector addition as

when θ = 90° then cos θ = cos 90 degree = 0 Cos θ

Hence,

Calculation :

Given:

F₁ = 20 kg, F₂ = 15 kg and θ = 90 degree : Cos90 = 0

F_r = 25 kg

Explanation:

Determination of Liquid Limit:

• In the laboratory, the liquid limit of soil is determined with the standard liquid limit apparatus designed by A. Casagrande.
• The apparatus consists of a vulcanized rubber compound base with a brass cup suitably mounted.
• The height of the fall of the brass cup can be adjusted with the help of an adjusting screw. Before the start of the test, the height of the fall of the cup is adjusted to 10 mm.
• A grooving tool is used to cut a groove in the pat of soil placed in the cup.

Two types of grooving tools are used to cut a groove:

1. Casagrande tool which cut a groove of 2 mm width at the bottom, 11 mm width at the top, and 8 mm deep.
2. ASTM tool which cuts a groove 2 mm wide at the bottom, 13.6 mm wide at the top, and 10 mm deep.

Important Points

• ASTM tool is preferred for low-plasticity soil.
• The liquid limit is determined by reading the water content corresponding to 25 blows on the flow curve.
• 425-μm IS sieve is used to pass the soil for liquid limit calculation.

Explanation:

Moment of inertia of Solid cylinder:

Moment of inertia of Hollow Cylinder:

where outer diameter = D and inner diameter = d

Polar moment of inertia of Solid cylinder:

Polar moment of inertia of Hollow Cylinder:

where outer diameter = D and inner diameter = d

1) Static Indeterminacy of Trusses:
a) For Plane Truss
D_s = m + r - 2j

1. m + r - 2j = 0, Statically determinate plane truss.
2. m + r - 2j > 0, Statically indeterminate plane truss.
3. m + r - 2j < 0, Unstable plane truss.

b) For Space Truss
D_s = m + r - 3j
Where, m = Number of members (every member carries force), r = Number of reactions at support, and j = No. of joints.
2) Static Indeterminacy of Frames:
a) For Plane Frame
D_s = 3m + r - 3j
a) For Space Frame
Ds = 6m + r - 6j
Hence option (3) is correct.

Deep beams are structural elements loaded as simple beams in which a significant amount of the load is carried to the supports by a compression force combining the load and the reaction. As a result, the strain distribution is no longer considered linear, and the shear deformations become significant when compared to pure flexure.

In view of ample shear strength, deep beams are primarily recommended as transfer girders. These members transfer loads through-loading face to supports in the transverse direction. The deep horizontal members predominantly fail in shear rather than flexure. These beams are characterized with small span-to-depth ratio. Pile caps, corbel, brackets, foundation walls and off-shore structures are few examples of RC deep beams.

According to IS 456-2000 a beam shall be deemed to be a deep beam when the ratio of effective span-to-overall depth, l/D is less than:

1) 2.0, for simply supported beam; and

2) 2.5, for a continuous beam.

Though different codes define deep beams in different clear span-to-depth ratios, as a general rule deep beams are recognized by their relatively small span-to-depth ratio.

Hence, Deep beams are designed for bending moment and checked for shear.

Explanation:

Brittle material:

It is defined as the materials that do not have plastic deformation, their stress-strain curve has little or no deformation. If brittle material is subjected to shear stress crack occurs at an inclination of 45^o to the side of the element.

Other Related Points

1. Brittle materials fracture rather than yield.
2. Brittle fracture in tension is due to normal stress only.
3. Brittle fracture in compression is due to a combination of normal stress and shear stress.

Concept:

The assumption for the truss analysis are as follows:

1. Truss members are designed to take the axial load.
2. The forces are assumed to act at the joint only.
3. All joints are hinged.
4. Members of a truss do not carry ant Flexural load.
5. Hence the deflection of a truss depends on axial rigidity only.

Note:

Beams and frames carry the flexural load as well as axial load.

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

Explanation:
In compaction test:

Compactive Energy Ratio (r) = (Compactive effort in modified compaction test)/(Compactive effort in standard compaction test)
r = (4.9 x 450 x 5 x 25)/(2.6 x 310 x 3 x 25) = 4.559

Concept:
Simply Supported Beam:
A simply supported beam is one that rests on two supports and is free to move horizontally. Typical practical applications of simply supported beams with point loadings include bridges, beams in buildings, and beds of machine tools.
Calculation:

Let, both reaction at A & C = R
Taking Moment about A
∑M_A = 0
4P × 2 + W × 8 - R × 6 = 0
6R = 8P + 8W ---(i)
again tarcing moment about C
R × 6 + W × 2 - 4P × 4 = 0
6R = 16P - 2W ---(ii)
Equation equation (i) and (ii) we get,
8P + 8W = 16P - 2W
8W + 2W= 16P - 8P
10W = 8P
W = 8P/10
W = 4P/5

Explanation:
Shear strength of the soil:

The shear strength of a soil is its maximum resistance to shear stresses just before the failure. The shear strength (S) of soil at a point on a particular plane was expressed by coulomb as a linear function of the normal stress on the plane as,
S = C + σ tan ϕ
Where,
σ = Normal stress on the plane
C = Cohesion of the soil
ϕ = Angle of internal friction or angle of shearing resistance

The C and ϕ values for different soils are given below

Hence the value of the angle of internal friction is least for clays, i.e ϕ = 0

Concept:
In the case of gradually applied loads, the magnitude of the load is increased gradually from zero to the value of that the load says “W”, Hence the work done by the load on the member in stretching it equals the product of the average load and the displacement δ.
δ = WL/AE
Work done = 0.5 x W x δ
Work done = 1/2 x W²L/AE
This work done is stored in the form of strain energy in the member, Therefore, Strain energy per unit volume is equal to the,
Strain Energy per unit volume  1/2 x W²L/AE x A x L
Strain Energy per unit volume  1/2 x W²L/A²E
W/A is equal to the stress developed,
Strain Energy per unit volume  1/2 x /σ²/E
In case of suddenly applied loads,
Strain Energy stored = W x δ
Hence,
Strain Energy per unit volume = W²L/AE x A x L
Strain Energy per unit volume  = σ²/E
Comparing both the equations, we get stress developed due to application of a load suddenly is two times that due to the same load being applied gradually.

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:

Graphical method for locating principal axes:

1. Dyadic circle

2. Mohr-circle

3. Circle of inertial

Other Related Points

Ellipse of inertia is not the graphical method for locating principal axes. Momental ellipse or ellipse of inertia is a graphical method used to locate the neutral axis in case of unsymmetrical bending.

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:

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:

Given that,
Length of beam = L unit
Propped at its center at c
So we can say that deflection at c will be zero.
Let take support reaction at c = p
The deflection of a simply supported beam, at mid span, carrying uniformly distributed load 'w' per unit length is given by
S_c = 5wL⁴/38EI (Down ward)
Where,
L = span of the beam
EI = flexural rigidity of the beam
Simply supported beam, at the mid point, carrying point load 'p' at the mid point is given by,
S_c = Pl³/48EI (Up ward)
The reaction developed at the mid span is such that the net deflection is zero.
Therefore,
S_c = (PL³)/(48E I) = S_c = (swL⁴) / (384EI)
p = (5 wl)/8unit
And also R_A + R_B + p = WL
∴ RA + RB = WL - 5wl/8
RA + RB = 3wl/8
The Reaction at A & B should be equal
∴ RA = RB = 3wl/16
Let take moment at x distance from the end equal to zero.

Moment at D equal to 0
Σ M_D = 0
R_A × x - w.x. x/2 = 0
3wl/16 x X wx²/2
wx/2 =3wl/16
x = (3/8)L
So from the both ends at a distance of (3/8)L, the moment equal to zero.
So the correct answer is Option 4

Concept:
Staircase:
Stairs are a set of steps that give access from floor to floor. The room or enclosure of the building, in which stair is located is known as the staircase. A staircase provides access & communication between floors in multi-story buildings.

The following are the components of the staircase

Concept:
Method of Joints:
The method of joints is a process used to solve for the unknown forces acting on members of a truss. The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve all the unknown forces in a truss structure.

• For every joint, we will consider the unknown forces going away from the joint, and the forces are considered positive.
• So after determination, if one force comes positive then, it will mean that it is tensile and if it comes negative then it would mean that it is compressive.

​Calculation:

Taking Moment about A,
∑ M_A = 0
⇒R_C x 6 = 10 x 6 sin(60)
⇒R_C = 5√3
Consider Joint C, we get
R_C + R_CB sin(60) = 0
⇒ RCB x √3/2 = -5√3
⇒ RCB = -10 KN
Negative sign indicate that the force will be towards the joint C, hence it will compressive in nature.
So the correct answer is Option 4

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 =

ic = 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.
Important Points

• For piping, the hydraulic gradient should be more than or equals to the critical hydraulic gradient.
• For most of the soils, the critical hydraulic gradient is close to unity. So when the hydraulic gradient is nearly unity piping will occur.