GATE CE 2018 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is option 1 i.e. Brake, break.

i) Brake is a device which is used for stopping or moving a vehicle.

ii) Break refers to a pause in work or during an activity.

In this sentence, the presence of the conjunction 'however' shows that one part contradicts what is stated in the former part of the sentence.

Since the first part of the sentence is an affirmative statement about codes of behaviour, the next part has to have a contrasting tone. 'Simple', 'expected' and 'strict', all speak in the favour of codes of behaviour. So, the right option is ‘unpredictable’.

Let Hari’s age (H) = x years

As per the given condition:

Hema’s age (He) = 2x + 5, Suresh’s age (S) = 10x – 13

Suresh is 3 times as old as Hema age i.e. S = 3 × He

∴ 10x – 13 = 3(2x + 5)

⇒ 6x + 15 = 10x – 13

⇒ 15 + 13 = 10x – 6x

⇒ x = 7

∴ Hema’s age = 2x + 5 = 2 × 7 + 5 = 19 years

Explanation:

Given**:** Tower A is 90 m, Tower B is 140 m, A and B are 100 m apart

The given situation can be depicted as follows:

From similar triangle theorems:

By proportionality theorem for ΔABC and ΔCDE.

$\frac{{AB}}{{AC}} = \frac{{DE}}{{CD}} \Rightarrow \frac{{20}}{X} = \frac{{70}}{{100 - X}}$

∴ x = 22.22 m

Explanation:

Given: T = K (θP + T₀)

T - Temperature 

T₀ -  Outside temperature

p -  Number of persons in the room 

From the tabular data

32.4 = K [2θ + 25]        ……(i)

42.0 = K [5θ + 30]        ……(ii)

Dividing (i) and (ii)

$\therefore \frac{{32.4}}{{42.0}} = \frac{{2{\rm{\theta }} + 25}}{{5{\rm{\theta }} + 30}} \Rightarrow \frac{{27}}{{35}} = \frac{{2{\rm{\theta }} + 25}}{{5{\rm{\theta }} + 30}} \Rightarrow {\rm{\theta }} = 1$

${\rm{Loss;}}12.5{\rm{\% }} = \frac{{12.5}}{{100}} = \frac{1}{8}$

${\rm{Multiplying;factor; (f_1)}}=\frac{{\left( {{\rm{SP}} } \right)}}{{{\rm{CP}}}}= 1 - \frac{1}{8} = \frac{7}{8}$

${\rm{Profit;}}10{\rm{\% }} = \frac{{10}}{{100}} = \frac{1}{{10}}$

${\rm{Multiplying;factor}} =\frac{{\left( {{\rm{SP}} + 108} \right)}}{{{\rm{CP}}}}= 1 + \frac{1}{{10}} = \frac{{11}}{{10}}$

As per given condition,

Selling price (SP) and cost price (CP)

$\frac{{{\rm{SP}}}}{{{\rm{CP}}}} = {{\rm{f}}_1} = \frac{7}{8},\frac{{\left( {{\rm{SP}} + 108} \right)}}{{{\rm{CP}}}} = {{\rm{f}}_2} = \frac{{11}}{{10}},$

$\frac{{{\rm{SP}}}}{{{\rm{CP}}}} + \frac{{108}}{{{\rm{CP}}}} = \frac{{11}}{{10}} \Rightarrow \frac{{108}}{{{\rm{CP}}}} = \frac{{11}}{{10}} - \frac{{{\rm{SP}}}}{{{\rm{CP}}}}$

$\frac{{108}}{{{\rm{CP}}}} = \frac{{11}}{{10}} - \frac{7}{8}$

$\frac{{108}}{{{\rm{CP}}}} = \frac{9}{{40}} \Rightarrow {\rm{CP}} = \frac{{108 \times 40}}{9} = 480$

${\rm{Loss;in;Rs}}.{\rm{;}} = {\rm{CP}} \times {\rm{Loss\% }} = 480 \times \frac{{12.5}}{{100}} = {\rm{Rs}}.{\rm{;}}60$

The price of wire made of super alloy material is proportional to the square of its length.

Price × (length)²

Price = k × l²

Length of wire = 10 m

According to given condition,

1600 = k (10)² ⇒ 1600 = k (100) ⇒ k = 16

So that price of two wires of length 4 m and 6 m is = k [l²₁ + l²₂] = 16[4² + 6²]

Total cost = 832

Check for each equation:

(i) y = 2x + 4 is true in -3 ≤ x ≤ -1

On putting x = -3, y = -2; x = -2, y = 0; and x = -1, y = 2

(ii) y = |x - 1| is also true

x = -1, y = 2; x = 0, y = 1; and x = 1, y = 0

(iii) y = ||x| - 1| is false for -1 ≤ x ≤ 2

x = -1, y = 0 is not true; x = 0, y = 1 and x = 1, y = 0; and x = 2, y = 1

(iv) y = 1 in (2 ≤ x ≤ 3) always true

∴ Equation (i), (ii) (iv) are true.

Given: ${a_n} = \frac{1}{n} - \frac{1}{{n + 2}}$

$\therefore ;{a_1} = \frac{1}{1} - \frac{1}{{1 + 2}} = 1 - \frac{1}{3}$

Similarly,

${\rm{Let}},{\rm{;}}{a_1} + {a_2} + {a_3} + \ldots + {a_{50}} = {\rm{S}}$

$\therefore {\rm{S}} = \left[ {\left( {1 - \frac{1}{3}} \right) + \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \ldots + \left( {\frac{1}{{48}} - \frac{1}{{50}}} \right) + \left( {\frac{1}{{49}} - \frac{1}{{51}}} \right) + \left( {\frac{1}{{50}} - \frac{1}{{52}}} \right)} \right]$

All like terms will cancel out and we will be left with $1 + \frac{1}{2} - \frac{1}{{51}} - \frac{1}{{52}}$

$\therefore {a_1} + {a_2} + {a_3} + \ldots + {a_{50}} = \left( {1 + \frac{1}{2}} \right) - \left( {\frac{1}{{51}} + \frac{1}{{52}}} \right)$

Among the given options, only 5 is a prime number.

Thus, if any of A, B, C, D, E, F or G represents 5,

then (A × B × C) = (B × G × E) = (D × E × F) = multiple of 5.

This would imply at least two letters represent 5, which is not possible due to the unique representation by the letters.

Hence 5 cannot be represented by the letters A, B, C, D, E, F and G.

Concept:

Singular Matrix: It is matrix with determinant value zero and hence its inverse does not exist.

Singular matrix has at least one of the eigen values as zero and product of the two singular matrix is also singular matrix.

For 2 × 2 Matrix,

\({\rm{Let;A}} = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{b}}\ {\rm{c}}&{\rm{d}} \end{array}} \right],{\rm{;}}\left| {\rm{A}} \right| = {\rm{a}} \times {\rm{d}} - {\rm{b}} \times {\rm{c}}\)

Calculation:

\({\rm{For;A}} = \left[ {\begin{array}{*{20}{c}} 2&5\ 1&3 \end{array}} \right],\left| {\rm{A}} \right| = 6 - 5 = 1\)

\({\rm{For;B}} = \left[ {\begin{array}{*{20}{c}} 3&2\ 2&3 \end{array}} \right],\left| {\rm{B}} \right| = 9 - 4 = 5\)

\({\rm{For;C}} = \left[ {\begin{array}{*{20}{c}} 2&4\ 3&6 \end{array}} \right],\left| C \right| = 12 - 12 = 0\)

\({\rm{For;D}} = \left[ {\begin{array}{*{20}{c}} 4&3\ 6&2 \end{array}} \right],\left| {\rm{D}} \right| = 8 - 18 = - 10\)

∴ Matrix \(\left[ {\begin{array}{*{20}{c}} 2&4\ 3&6 \end{array}} \right]\) is a singular matrix.

Concept:

For orthogonal matrix: A × A^T = I ⇒ A^-1 = A^T

Calculation:

For given matrix \(\left( Q \right) = \left[ {;\begin{array}{*{20}{c}} {\frac{3}{7}}&{\frac{2}{7}}&{\frac{6}{7}}\ {\frac{{ - 6}}{7}}&{\frac{3}{7}}&{\frac{2}{7}}\ {\frac{2}{7}}&{\frac{6}{7}}&{\frac{{ - 3}}{7}} \end{array}} \right]\)

∴ Q × Q^T = 1 ⇒ Q^-1 = Q^T

^{\({{\rm{Q}}^{ - 1}} = {{\rm{Q}}^{\rm{T}}} = \left[ {{\rm{;}}\begin{array}{*{20}{c}} {\frac{3}{7}}&{\frac{{ - 6}}{7}}&{\frac{2}{7}}\ {\frac{2}{7}}&{\frac{3}{7}}&{\frac{6}{7}}\ {\frac{6}{7}}&{\frac{2}{7}}&{\frac{{ - 3}}{7}} \end{array}} \right]\)}

Alternative Method:

${{\rm{Q}}^{ - 1}} = \frac{1}{{\left| {\rm{Q}} \right|}} \times {\rm{Adj;}}\left[ {\rm{Q}} \right]$

\({\rm{Adj;Q;}} = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 21}}{{49}}}&{\frac{{42}}{{49}}}&{\frac{{ - 14}}{{49}}}\ {\frac{{ - 14}}{{49}}}&{\frac{{ - 21}}{{49}}}&{\frac{{ - 42}}{{49}}}\ {\frac{{ - 42}}{{49}}}&{\frac{{ - 14{\rm{;}}}}{{49}}}&{\frac{{21}}{{42}}} \end{array}} \right]\)

$\left| {\rm{Q}} \right| = \frac{3}{7} \times \left( {\frac{{ - 9}}{{49}} - \frac{{12}}{{49}}} \right) - \frac{2}{7} \times \left( {\frac{{18}}{{49}} - \frac{4}{{49}}} \right) + \frac{6}{7} \times \left( {\frac{{ - 36}}{{49}} - \frac{6}{{49}}} \right) = {\rm{;}} - 1$

\({{\rm{Q}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\frac{3}{7}}&{\frac{{ - 6}}{7}}&{\frac{2}{7}}\ {\frac{2}{7}}&{\frac{3}{7}}&{\frac{6}{7}}\ {\frac{6}{7}}&{\frac{2}{7}}&{\frac{{ - 3}}{7}} \end{array}} \right]\)

For the given function: f(x) = x³

From graph, it can be observed that it has neither local maximum non local minimum

Alternative:

f’(x) = 3x² ⇒ f’(x) = 0 ⇒ 3x² = 0 ⇒ x = 0

f’’(x) = 6x at x = 0, f’’(x) = 0. (Neither +ve nor -ve)

So it has neither local maximum nor minimum.

Concept:

For column, the buckling load is taken as,

\({\rm{P}} = \frac{{{{\rm{\pi }}^2}{\rm{E;}}{{\rm{I}}{{\rm{min}}}}}}{{{\rm{L}}{\rm{e}}^2}}\)

E = Modulus of Elasticity, I_min = Minimum moment of inertia, and L_e = effective length of column.

Calculation:

In first case: Height = h, and cross-sectional area = a × 2a

\({\rm{I}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{12}} = {\rm{min}}\left{ {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}},\frac{{{\rm{a}} \times {{\left( {2{\rm{a}}} \right)}^3}}}{{12}}} \right. = \frac{{2{{\rm{a}}^4}}}{{12}}\)

${\rm{P}} = \frac{{{{\rm{\pi }}^2}{\rm{E;}}\left( {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}}} \right){\rm{;}}}}{{{{\rm{h}}^2}}} = \frac{{{{\rm{\pi }}^2}{\rm{E}}{{\rm{a}}^4}}}{{6{{\rm{h}}^2}}}$

\({\rm{I}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{12}} = {\rm{min}}\left{ {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}},\frac{{{\rm{a}} \times {{\left( {2{\rm{a}}} \right)}^3}}}{{12}}} \right. = \frac{{2{{\rm{a}}^4}}}{{12}}\)

In second case: Height = 1.5 h, and cross-sectional area = 3a × 0.5a

${\rm{P'}} = \frac{{{{\rm{\pi }}^2}{\rm{E;}}\left[ {\frac{{3{\rm{a}} \times {{\left( {0.5{\rm{a}}} \right)}^3}}}{{12}}} \right]}}{{{{\left( {1.5{\rm{;h}}} \right)}^2}}} = \frac{1}{{12}} \times \frac{{{{\rm{\pi }}^2}{\rm{E}}{{\rm{a}}^4}}}{{6{{\rm{h}}^2}}} = \frac{{\rm{P}}}{{12}}$

Concept:

Bearing plates are provided below the column to transfer the load to the concrete pedestal.

Area required for the base plate = Factored load / Bearing Capacity of concrete

Bearing Capacity of concrete = 0.45 × f_ck

f_ck - Characteristic strength of concrete

Calculation:

Given:

Factored load = 2000 kN, Bearing Capacity of concrete = 0.45 × 20 = 9

∴ Area required = 2000000/9 = 222222.222 mm²

Side of plate = √222222.222 = 471.40 mm = 47.1 cm

∴ A plate of 48 cm is to be provided.

For quality control of Portland cement, the test essentially done is

1) Le Chatelier Test: This test is used to measure the soundness of OPC due to lime. Lime & Magnesia are two primary compounds responsible for the soundness of cement.

2) Blaine Air Permeability: It is used to measure the fineness of the cement.

4) The Vicat Apparatus: It is used to measure setting time and consistency of concrete.

5) Tensile (Briquette) Testing Machine: It is used to measure the tensile strength of the concrete

Creep in concrete: It is defined as deformation of structure under sustained load. It is a long-term pressure or stress on concrete due to which there is a change shape of concrete structure. This deformation usually occurs in the direction of applied force, thus causing permanent deformation in concrete structure.

Segregation: Segregation of concrete is the separation of ingredients of concrete from each other. In the case of segregation, the heavy aggregate particles settle down leaving a sand cement mix on top affecting the quality adversely.

Hydration: The exothermic reaction of cement with water which liberates a considerable quantity of heat and results is setting and hardening of cement is called hydration.

Shrinkage: It is the volumetric changes of concrete structures due to the loss of moisture by evaporation. The total shrinkage of concrete depends upon the constituents of concrete, size of the member and environmental conditions.

Concept:

Here,

τ_zy = τyz = τmax

τ_max = 16T/π d³ = 0.815 MPa

In pure torsion of a circular shaft, there are no shear stress components acting in the plane of the cross-section (the x-y plane). Shear stress only exists on planes perpendicular to the longitudinal axis (z-axis) and acts in directions within the cross-sectional plane (x or y). Thus, the component τ_xy, which represents shear stress on the x-plane in the y-direction (or vice versa), is zero.

Hence,  τxy = τyx = 0 

Additional Information

Shear stress distribution:

• The value of shear stress at any level is given by
• ${\bf{τ }} = \frac{{{\bf{FA\bar Y}}}}{{{\bf{Ib}}}}$
• where F = Applied/Resisting shear force, I = Moment of Inertia of the complete cross-section about NA, b = width of the required level where shear stress is required,
• A = Area either above or below the required level, Y = centroidal distance of considered area from NA.
• Now, as we can see from the equation width 'b' is inversely proportional to the shear stress of the section.
• So from the top of the beam as the width decreases the shear stress will go on increase up to the neutral axis and beyond the neutral axis as with the increase of width down the shear stress starts decreasing.
• Shear stress will be maximum at the neutral axis.

• Shear stress acts parallel to the cross-section of the beam.
• At extreme fibre, shear stress is zero and at NA, shear stress is non-zero but maximum in some cases like circular, square, rectangle, I-section, T-section beams, etc.
• Shear stress on a cross-section of the beam varies parabolically.

​

Both the sections (X and Y) are under reinforced, hence in both the cases the flexural strength (moment of resistance) can be calculated from the tension side and is given by,

Moment of resistance = Tension force × Lever arm

When additional compression reinforcement is placed i.e. in case of section Y, neutral axis will shift upward i.e., towards compression fiber of the section which is above than that of section X.

This shows that the section Y is more under reinforced than Section X. Hence due to the presence of steel in compression side, the rotational capacity (flexural strength or moment of resistance) of the section Y increases (which means ductility increases).

Hence, one can say that Section X has less flexural strength and is less ductile than section Y.

Explanation

Terzaghi’s bearing capacity equation is given as:

Ultimate bearing capacity (q_u) = C × N_c + γ × D × N_q + 0.5 γ × B × N_γ

For cohesionless soil c = 0, footing resting on the sand surface so, D_f = 0

Case 1: When the water level is at the base of the footing

Bearing capacity \(\left( {{{\rm{q}}{{{\rm{u}}1}}}} \right) = \frac{1}{2} \times {{\rm{\gamma }}{{\rm{sub}}}} \times {\rm{B}} \times {{\rm{N}}{\rm{\gamma }}}\)

Case 2: When the water level is at a much greater depth

Bearing capacity \(\left( {{{\rm{q}}{{{\rm{u}}2}}}} \right) = \frac{1}{2}{{\rm{\gamma }}{{\rm{sat}}}}{\rm{B}}{{\rm{N}}{\rm{\gamma }}}\)

∴ Percentage reduction in bearing capacity of soil \(= \frac{{{{\rm{q}}{{{\rm{u}}2}}} - {{\rm{q}}{{{\rm{u}}1}}}}}{{{{\rm{q}}{{{\rm{u}}1}}}}} \times 100 = \frac{{{{\rm{\gamma }}{{\rm{sat}}}} - {{\rm{\gamma }}{{\rm{sub}}}}}}{{{{\rm{\gamma }}_{{\rm{sat}}}}}} \times 100\)

We know that γ_sub ≈ 0.5 γ_sat

∴ Percentage reduction = 50%.

Concept:

For circular footing, ultimate bearing capacity

qu = 1.3 CNc + γDfNq + 0.3 γBNγ

For square footing, ultimate bearing capacity

qu = 1.3 CNc + γDfNq + 0.4 γBNγ

Calculation:

As footing is placed on the surface

D_f  = 0

For circular footing, ultimate bearing capacity

q_u = 1.3 CNc + xD_fN_q + 0.3 xBN_x

C = 0, D_f­ = 0

q_u = 0.3x BN_x  …… (i)

For square footing, ultimate bearing capacity

q_u = 1.3 CN_c + x D_fN_q + 0.4 xBN_x

C = 0, D_f = 0

q_u = 0.4 x BN_x  …… (ii)

Divide (i) by (ii)

\(\frac{{{{\left( {{q_u}} \right)}{circular}}}}{{{{\left( {{q_u}} \right)}{square}}}} = \frac{{0.3xB{N_x}}}{{0.4xB{N_x}}} = \frac{3}{4}\)

Explanation:

$\frac{P}{\rho g} + \frac{{{v^2}}}{{2g}} + Z = Constant$

Bernoulli’s equation:

• It can be derived from the principle of conservation of energy.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It represented in head form (the total energy per unit weight).

Following assumption are made in deriving the Bernoulli’s equation:

1. Fluid is inviscous i.e. zero viscosity.
2. Flow is in steady-state.
3. Flow is incompressible.
4. Flow is irrotational.
5. Flow is along the streamline.

Important Points

Even if the flow is rotational, we can still apply the Bernoulli’s equation along a streamline.

Concept:

Total Annual Emission = Average Annual daily emission per vehicle × Number of Vehicle

Average Annual daily emission per vehicle = Emission per km of travel × Distance traveled (in km)

∴ Total Annual Emission = Number of vehicles × Distance travelled × Rate of emission

Calculation:

Total Annual Emission = Number of vehicles × Distance traveled × Rate of emission

∴ Total Annual Emission = 20000 × 12000 × 2 × 10^-3 (kg) = 480000 kg

Concept:

Penetration Test: It determines the hardness and softness of the bitumen. The bitumen grade is specified in terms of penetration value. Higher penetration implies softer grade.

Ductility is the property of bitumen that permits it to undergo great deformation or elongation. Use of ductility test is to measure the adhesive property of bitumen and its ability to stretch.

Elastic recovery of any modified bitumen binder is evaluated by the percentage of recoverable strain measured after an elongation during a conventional ductility test.

According to IS 73-2013, Number 30 indicates the range of viscosity of bitumen [(100 ± 20) × 30] in terms of Poise.

Calculation:

For VG 30 Bitumen:

Maximum Viscosity: (100 + 20 = 120) × 30 = 3600 Poise

Minimum Viscosity: (100 - 20 = 80) × 30 = 2400 Poise

Concept:

We know, Traffic capacity (q) = velocity (v) × density (k) ⇒ q = kV

For Case 1: When speed is constant, and density of the flow keeps on increasing at the same speed.

q = k × constant

∴ Flow becomes the linear function of density and with increase in density, flow increases linearly.

For Case 2: With decrease in the speed further, density increases linearly.

Here flow becomes the function the speed as well as density and hence flow becomes the parabolic function of density as density is varying with speed.

∴ After that V varies linearly with k, so q will vary parabolically with k.

Concept:

Intersection is an area shared by two or more roads. Its main function is to guide vehicles to their respective directions. The essence of the intersection control is to resolve crossing conflicts at the intersection for the safe and efficient movement of both vehicular traffic and pedestrians with an objective of minimum

1. Time sharing: Time shared by each vehicle during crossing the intersection. It includes the stopping time, starting time and time required to cross the intersection. Here we have only control over the delay time as other aspects depend upon the vehicular characteristics and driver characteristics.
2. Space sharing: It is the space occupied by the vehicle. Generally it relates with the density of vehicle.

So, in a well-designed signalized intersection total delay is minimized to avoid the conflicts with proper time sharing.

\(\mathop{\int }{\text{a}}^{\text{b}}\text{f}\left( \text{x} \right)\text{ }!!!!\text{ dx }!!!!\text{ }=\mathop{\int }{\text{a}}^{\text{b}}\text{f}\left( \text{a}+\text{b}-\text{x} \right)\text{ }!!~!!\text{ dx}\)

Applying the property 

\(\text{I}=\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\text{x }!!!!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }!!!!\text{ dx}=\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\left( \text{ }!!\pi!!\text{ }-\text{x} \right)\text{ }!!!!\text{ }{{\left( \cos \left( \text{ }!!\pi!!\text{ }-\text{x} \right) \right)}^{2}}\text{ }!!!!\text{ dx}\)

\(\text{I}=\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\text{ }!!\pi!!\text{ }!!!!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }!!!!\text{ dx}-\text{ }!!~!!\text{ }\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\text{x }!!!!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }!!!!\text{ dx}\)

\(\text{I}=\mathop{\int }_{0}^{\text{ }!!\pi!!\text{ }}\text{ }!!\pi!!\text{ }!!!!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }!!!!\text{ dx}-\text{ }!!~!!\text{ I}\)

\(2\times \text{I}=\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\text{ }!!\pi!!\text{ }!!!!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }!!!!\text{ dx}=\text{ }!!\pi!!\text{ }\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\frac{1+\cos 2\text{x}}{2}\text{ }!!~!!\text{ dx}\)

\(2\times \text{I}=\text{ }!!\pi!!\text{ }\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\frac{1}{2}\text{ }!!~!!\text{ dx}+\text{ }!!\pi!!\text{ }\mathop{\int }{0}^{\text{ }!!\pi!!\text{ }}\frac{\cos 2\text{x}}{2}\text{ }!!~!!\text{ dx}\)

$2\times \text{I}=\text{ }!!~!!\text{ }\frac{\text{ }!!\pi!!\text{ }}{2}\left( \text{ }!!\pi!!\text{ }-0 \right)+\frac{1}{2}\left( \sin 2\text{ }!!\pi!!\text{ }-\sin 0 \right)=\frac{{{\text{ }!!\pi!!\text{ }}^{2}}}{2}$

\(\therefore \mathbf{I}=\mathop{\int }_{0}^{\mathbf{\pi }}\mathbf{x}{{\left( \cos \mathbf{x} \right)}^{2}}\mathbf{dx}=\frac{{{\mathbf{\pi }}^{2}}}{4}\)

Concept:

The cantilever beam having poooint load at the end is shown in the figure.

The maximum deflection of free end is given as, $\delta = \frac{PL^3}{3EI}$

And we know that, the moment equation is, $\frac{{\text{M}}}{{\text{I}}} = \frac{{σ }}{{\text{y}}} = \frac{{\text{E}}}{{\text{R}}}$

Calculation:

Given:

L = 2 m, side (a) = 0.1 m, y_max = 5 mm, E = 2.0 × 1011 N/m2.

The maximum deflection

$\delta = \frac{PL^3}{3EI}$

$5 \times 10^{-3} = \frac{P \times 2^3}{3\times 2 \times 10^{11} \times I}$

$\frac PI=375\times10^6 ~N/mm^4$

The moment equation

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

\({{{σ}}{{\text{max}}}} = \frac{{\text{M}}}{{\text{z}}} = \frac{{{\text{M}} \times {{\text{y}}{{\text{max}}}}}}{{\text{I}}}\)

$σ_{max}=\frac{375 \times 10^6\times2\times0.1}{2}=37.5\times10^6Pa$

σ_max = 37.5 MPa

Free body diagram for the following arrangement are as follows:

OY = 250 - 50 = 200 m

OB = 250 mm

${\rm{BY}} = \sqrt {{{250}^2} - {{200}^2}} = 150$

Concept:

When the roller will just start to roll up, then it will loose its contact at point X. only contact will be at point B.

At that instant, ∑ M_B = 0

Calculation:

∑ M_B = 0

P × OY - W × BY = 0

⇒ P × 200 - 10 × 150 = 0

$\Rightarrow {\rm{P}} = \frac{{10 \times 150}}{{200}} = 7.5{\rm{;kN}}.$

Concept:

The generalized hooks law is given by:

\({{\text{ }!!\varepsilon!!\text{ }}{\text{xx}}}=\text{ }!!~!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{xx}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{yy}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{zz}}}}{\text{E}}\)

\({{\text{ }!!\varepsilon!!\text{ }}{\text{yy}}}=\text{ }!!~!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{yy}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{xx}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{zz}}}}{\text{E}}\)

\({{\text{ }!!\varepsilon!!\text{ }}{\text{zz}}}=\text{ }!!~!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{zz}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{yy}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{xx}}}}{\text{E}}\)

Where

ε_xx= Strain in x direction due to stresses in x, y, z direction.

ε_yy = Strain in y direction due to stresses in x, y, z direction.

ε_zz = Strain in z direction due to stresses in x, y, z direction.

ν = Poisson’s ration = 0.3

E = Young’s modulus = 2 × 10¹¹ N/m² = = 2 × 10¹¹ × 10^-6 N/mm² = 2 × 10⁵ N/mm²

Calculation:

σ_xx = 30 MPa = 30 N/mm

σ_yy = 50 MPa = 50 N/mm

\({{\text{ }!!\varepsilon!!\text{ }}{\text{zz}}}=\text{ }!!~!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{zz}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{yy}}}}{\text{E}}-\text{ }!!\nu!!\text{ }\frac{{{\text{ }!!\sigma!!\text{ }}{\text{xx}}}}{\text{E}}\)

\({{\varepsilon }_{zz}}=\frac{0}{2\times {{10}^{5}}}-0.3\frac{50}{2\times {{10}^{5}}}-0.3\frac{30}{2\times ~{{10}^{5}}}=-1.2\times {{10}^{-4}}=-120\times {{10}^{-6}}\)

For calculation of slope and deflection in simply supported beam mainly conjugate beam method is used.

Conjugate beam method: The slope at any section of a loaded beam relative to the original axis of the beam is equal to the shear force in the conjugate beam at the corresponding section.

For Bending moment diagram first calculate reactions at P and Q support

$\sum v=0$

R_P = R_Q

Taking moment about point Q

• R_P × L + M + 2M = 0

R_P = 3M/L

R_Q = 3M/L

For bending moment diagram- Calculating bending moment from left hand side-

BM_P = 0

\(\text{B}{{\text{M}}_{\text{LEFT }!!!!\text{ OF }!!!!\text{ R}}}=-\frac{3M}{L}\times \frac{L}{3}=-\text{M }!!~!!\text{ }\)

\(\text{B}{{\text{M}}_{\text{RIGHT }!!!!\text{ OF }!!!!\text{ R}}}=-\frac{3M}{L}\times \frac{L}{3}+\text{M}=0\text{ }!!~!!\text{ }\)

\(\text{B}{{\text{M}}_{\text{LEFT }!!!!\text{ }OFS}}=-\frac{3M}{L}\times \frac{2L}{3}+\text{M}=-\text{M}\)

\(\text{B}{{\text{M}}_{\text{RIGHT }!!!!\text{ }OFS}}=-\frac{3M}{L}\times \frac{2L}{3}+\text{M}+2\text{M}=\text{M}\)

$\text{B}{{\text{M}}_{\text{Q}}}=-\frac{3M}{L}\times \text{L}+\text{M}+2\text{M}=0$

Conjugate beam with loading:

Shear at any section of the conjugate beam is equal to the slope of the real beam-

Shear force at P = Reaction at P (R_A’) in the conjugate beam

Taking moment about point Q

\(-\text{ }!!!!\text{ R}_{\text{P}}^{\text{ }!!'!!\text{ }}\times \text{ }!!!!\text{ L }!!~!!\text{ }+\frac{1}{2}\times \frac{\text{L}}{3}\times \frac{\text{M}}{\text{EI}}\times \left( \frac{2\text{L}}{3}+\frac{1}{3}\times \frac{\text{L}}{3} \right)+\frac{1}{2}\times \frac{\text{L}}{3}\times \frac{\text{M}}{\text{EI}}\times \left( \frac{\text{L}}{3}+\frac{1}{3}\times \frac{\text{L}}{3} \right)-\frac{1}{2}\times \frac{\text{L}}{3}\times \frac{\text{M}}{\text{EI}}\times \left( \frac{2}{3}\times \frac{\text{L}}{3} \right)=0\)

$\text{R}_{\text{P}}^{\text{ }!!'!!\text{ }}\times \text{ }!!~!!\text{ L}=\frac{7\text{M}{{\text{L}}^{2}}}{54\text{EI}}+\frac{4\text{M}{{\text{L}}^{2}}}{54\text{EI}}-\frac{\text{M}{{\text{L}}^{2}}}{27\text{EI}}$

$\text{R}_{\text{P}}^{\text{ }!!'!!\text{ }}=\frac{\text{ML}}{6\text{EI}}$

\(\therefore \text{Slope }!!!!\text{ at }!!!!\text{ point }!!~!!\text{ P}=\text{R}_{\text{P}}^{\text{ }!!'!!\text{ }}=\frac{\text{ML}}{6\text{EI}}\)

Concept:

Pile load capacity is given by:

Q_u = Q_b + Q_s

Q = Base resistance at pile tip and Q_s = Side friction resistance of pile

Q_b = A_b × F_b = c N_c × A­_b

Q_s = A_s × F_s = α c × A_s

A_b = Area at base of the pile, F_b = Bearing capacity of pile tip, A_s = Surface area of pile, and F_s = Average skin friction, c = having undrained shear strength or Average cohesion, a = Adhesion factor, N_c = Value depends upon the D/B ratio and it varies from 6 to 9.

Design capacity of pile or Safe Load (S­_safe) = Ultimate load capacity (Q_u)/Factor of safety

Calculation:

Given: 500 = Q_u /2 ⇒ Q_u = 1000 kN

c = 50 kN, Nc = 9.0 is generally used for the piles, a = 0.75, A_b = side² (square base) = 0.5 × 0.5 = 0.25 m²

F_b = cN_c = 50 × 9 = 450 KN

A_s = 4 × side × length (Surface area of square pile) = 4 × 0.5 × L = 2L

F_s = αC = 0.75 × 50 = 37.5 KN

1000 = 0.25 × 450 + 2L × 37.5

L = 11.833 m

Explanation:

Layer wise diagram of the closed tank:

Pressure values at the bottom most part of the tank is given by-

P_base = P_air + (ρgh) _oil + (ρgh) _water + (ρgh) _Hg

196.2 × 10³ = P_air + (600 × 3 + 1000 × 2 + 13600 × 0.5) ×9.81

P_air = 92.241 KPa (gauge)

Total water required (Q) = Potable water + Water used in backwashing/Water lost in backwashing

Water used in backwashing = 5% of Potable water

\(Q~=99+\frac{5}{100}\times 99=103.95MLD\)

Rate of filtration (ROF) = 6 m/h = 6 m³/h/m² = 6000 L/h/m²

Area of one filter = L × B

L: B = 1.35:1, B = 5.2 m, L = 5.2 × 1.35 = 7.02m, and Area of one filter = 36.54 m²

\(\text{Total }!!!!\text{ area }!!!!\text{ required}=\frac{\text{Q}}{\text{ROF}}=\frac{\frac{103.95\times {{10}^{6}}}{24}}{6000}=721.875\text{ }!!~!!\text{ }{{\text{m}}^{2}}\)

\(\text{Number }!!!!\text{ of }!!!!\text{ filter }!!!!\text{ beds }!!!!\text{ required}=\text{ }!!!!\text{ }\frac{\text{Total }!!!!\text{ area }!!!!\text{ required}}{\text{Area }!!!!\text{ of }!!!!\text{ one }!!!!\text{ filter}}=\frac{721.875}{36.54}=19.755\approx 20\text{ }!!~!!\text{ filters}\)

One additional filter is provided to take care of break-down, repair and maintenance.

Total number of filter beds = 21

Concept:

Sight triangle: At uncontrolled intersections the sight lines are obstructed by the structure or any other object at the intersection. The area of unobstructed sight formed by the lines of vision is called the sight triangle.

As per IRC: 66, 1976 at the priority intersections. The sight triangle has a side of 15m along minor road and a distance equal to 8 seconds travelled at the design speed is taken on a major road.

According to the IRC, time = 8 sec

For major road, SSD = 50 × 0.278 × 8 = 111.2 m

The minimum visibility distance along major roads for different speeds are as follows:

∑Northing = 101 + 419 = 520 m

∑Southing = 437 + 83 = 520

∑Easting = 173 + 558 = 731 m

∑Westing = 96 + 634 = 730 m

∑L = ∑Northing - ∑Southing = 520 – 520 = 0

∑D = ∑Easting - ∑Westing = 731 – 730 = 1 m

$\therefore {\rm{Length;of;closure}} = \sqrt {{{\left( {\sum {\rm{L}}} \right)}^2} + {{\left( {\sum {\rm{D}}} \right)}^2}} = \sqrt {{{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} = 1{\rm{;m}}$

$\rm{{\rm{Direction;of;closure}},{\rm{;\theta }} = {\tan ^{ - 1}}\left( {\frac{{\sum {\rm{D}}}}{{\sum {\rm{L}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right) = 90^\circ }$

Concept:

If focal length of the lens (f) and the flying height (H) above M.S.L. is known, the scale can be found from the relation:

S_h = f / (H - h)

Calculation:

Given:

1 cm² = 100² m²

S = 1 cm = 100 m

1 cm = 100 × 100 cm

RF = 1/10,000

h = 650 m (elevation on ground)

f = 150 mm

f = 0.15 m

S = f/ (H-h)

$\frac{1}{{10000}} = \frac{{0.15}}{{H - 650}}$

H = (0.15 × 10000) + 650

H = 2150 m

Concept:

It is 1D wave equation in the partial differential equation given by-

$\frac{{{\partial }^{2}}\text{y}}{\partial {{\text{t}}^{2}}}={{\text{C}}^{2}}\frac{{{\partial }^{2}}\text{y}}{\partial {{\text{x}}^{2}}}$

Where C² = T/m, T = Tension in the elastic string, and M = mass per unit length

Calculation:

On comparing the above equation, we get that C = 1/5

f(x) = 3x, g(x) = 3

f (x + ct) = f (x + t/5) = 3 (x + t/5) = 3x + 3t/5

f (x - ct) = f (x - t/5) = 3 (x- t/5) = 3x - 3t/5

\(\text{U}(\left( \text{x},\text{t} \right)=\frac{1}{2}\left{ \text{f}\left( \text{x}+\text{ct} \right)+ \text{f}\left( \text{x}-\text{ct} \right)+ \frac{1}{c } \mathop{\int }_{\text{x}-\text{ct}}^{\text{x}+\text{ct}}\text{g}\left( \text{x} \right)\text{dx} \right}\)

\(=\frac{1}{2}\left{ 6x+5\mathop{\int }_{x-t/5}^{x+t/5}3~dx \right}\)

\(=\frac{1}{2}\left{ 6\text{x}+5\left[ \left( 3\text{x}+\frac{3\text{t}}{5} \right)-\left( 3\text{x}-\frac{3\text{t}}{5} \right) \right] \right}\)

$\text{U}\left( \text{x},\text{t} \right)=\frac{1}{2}\left( 6\text{x}+6\text{t} \right)$

$\therefore \text{U}\left( 1,1 \right)=\frac{1}{2}\left( 6\times 1+6\times 1 \right)=6$