GATE 2023 Civil Engineering Question Paper with Answer Key and Solutions- Slot 1 PDF (Available)- Download Here

Axial force at a section equals the net external load acting below that section.


Upper (100 mm)\(^2\) bar:

Load below \(=100+100+50=250\ kN\).

Area \(A_u=100\times100=10{,}000\ mm^2\).
\[ \sigma_u=\frac{250\,000}{10{,}000}=25\ N/mm^2 \]


Lower (50 mm)\(^2\) bar:

Load below \(=50\ kN\).

Area \(A_\ell=50\times50=2{,}500\ mm^2\).
\[ \sigma_\ell=\frac{50\,000}{2{,}500}=20\ N/mm^2 \]


Therefore, the maximum tensile stress is
\[ \boxed{25.0\ N/mm^2} \] Quick Tip: Always compute axial force at a section by summing the loads acting below} that section. Then divide by cross-sectional area to get direct stress.

Step 1: Understanding creep in concrete.

Creep is the gradual increase in strain (deformation) of a material over time when subjected to constant stress. In concrete, creep refers to the increase in strain that occurs when concrete is under a compressive load and the stress remains constant.


Step 2: Analyzing the options.

- (A) Correct. Creep is defined as an increase in strain under constant stress. As the material is subjected to continuous load over time, it deforms more, even though the stress does not increase.

- (B) Incorrect. Stress would not increase under constant strain. Creep refers to strain, not stress.

- (C) Incorrect. Creep is characterized by an increase in strain, not a decrease.

- (D) Incorrect. Creep refers to strain, and stress does not decrease under constant strain.


Thus, the correct answer is (A).

\[ \boxed{Creep of concrete is the increase in strain under constant stress.} \] Quick Tip: Remember: Creep is time-dependent and occurs due to sustained loads. It is an increase in strain under constant stress.

Step 1: Maximum strain in concrete.

As per IS 456:2000 (Clause 38.1), the maximum compressive strain in concrete at the extreme fibre in the limit state of flexure is taken as: \[ \varepsilon_{c,max} = 0.0035 \]

Step 2: Tensile strain in steel (Fe415).

For Fe415 grade steel, the yield stress \(f_y = 415\) MPa and modulus of elasticity \(E_s = 2\times 10^5\) MPa.

The yield strain is: \[ \varepsilon_y = \frac{f_y}{E_s} = \frac{415}{2 \times 10^5} = 0.002075 \approx 0.0021 \]

For a balanced section, the steel reaches just beyond yield at the ultimate state. IS 456 specifies that the limiting tensile strain in steel at failure should be taken as approximately \(0.0038\) for Fe415 bars.


Step 3: Match with options.

- Concrete: \(0.0035\)
- Steel: \(0.0038\)

These values match option (A).

\[ \boxed{0.0035 \ and \ 0.0038} \] Quick Tip: Always remember: As per IS 456:2000, \(\varepsilon_{c,max} = 0.0035\) and for Fe415 steel in a balanced section, tensile strain \(\approx 0.0038\). These are standard code values useful for design.

We know the relation: \[ w = \frac{S \cdot e}{G} \]
where \(w=\) water content, \(S=\) degree of saturation, \(e=\) void ratio, \(G=\) specific gravity.


Step 1: Write known values.
\(w = 0.15\) (15%)
\(S = 0.50\) (50%)
\(G = 2.60\)


Step 2: Substitute values in formula.
\[ 0.15 = \frac{0.50 \cdot e}{2.60} \]

Step 3: Solve for \(e\).
\[ e = \frac{0.15 \times 2.60}{0.50} \] \[ e = \frac{0.39}{0.50} = 0.78 \]

Final Answer:
\[ \boxed{0.78} \] Quick Tip: Always use the relation \(w = \dfrac{S \cdot e}{G}\) to connect water content, saturation, specific gravity, and void ratio in soil mechanics.

Step 1: Identify stress states at the three locations.

Location I (near slope face/toe): The soil mass close to the free face tends to unload laterally and undergoes a reduction in minor principal stress, producing a tensile/extension stress path. Hence, behaviour is best captured by a triaxial extension test. \(\Rightarrow\) I \(\rightarrow\) Q.

Location II (along potential failure plane): This is the sliding interface. The shear strength mobilized along such a plane (including large-displacement/residual behaviour if needed) is most directly measured in a direct shear (shear box) test. \(\Rightarrow\) II \(\rightarrow\) R.

Location III (deeper beneath crest): Soil is under higher confining pressure and predominantly compressive stress state. Strength and dilation are appropriately assessed using a triaxial compression test. \(\Rightarrow\) III \(\rightarrow\) P.


Step 2: Match with options.

Mapping I--Q, II--R, III--P corresponds to option (A).

\[ \boxed{I--Q,\; II--R,\; III--P} \] Quick Tip: Near free faces/crests \(\Rightarrow\) extension paths (TXE).
Along suspected slip planes \(\Rightarrow\) direct shear.
Deeper confined zones \(\Rightarrow\) compression paths (TXC).

Step 1: Recall the classification of flow profiles in gradually varied flow (GVF).

- Flow profiles depend on:
(i) The slope of the channel (mild, steep, horizontal, adverse, critical).

(ii) The depth of flow relative to the normal depth (\(y_n\)) and critical depth (\(y_c\)).


- For a mild slope (M), we have:
\(y_c < y_n\) (critical depth is smaller than normal depth).

Step 2: Case of supercritical flow entering mild slope.

- Supercritical flow means actual depth of flow \(y < y_c\).
- Since \(y < y_c < y_n\), the depth is below the critical depth region.
- By GVF profile classification, this corresponds to M\(_3\) profile.

Step 3: Verification.

- M\(_1\): Depth \(> y_n\) (not possible here).

- M\(_2\): Depth between \(y_c\) and \(y_n\) (not possible since \(y < y_c\)).

- M\(_3\): Depth \(< y_c\) (true in case of supercritical flow).

\[ \boxed{The flow profile becomes M\(_3\).} \] Quick Tip: For mild slopes: - \(y > y_n \Rightarrow\) M\(_1\), - \(y_c < y < y_n \Rightarrow\) M\(_2\), - \(y < y_c \Rightarrow\) M\(_3\). Supercritical flow always corresponds to M\(_3\) on mild slopes.

Step 1: Formula for vertical collimation error.

When direct and reversed zenith angles are observed, the condition for perfect collimation is: \[ Z_d + Z_r = 360^\circ \]
where \(Z_d =\) direct reading, \(Z_r =\) reversed reading.

If this sum is not exactly \(360^\circ\), the deviation indicates collimation error.

Step 2: Apply the given data.
\[ Z_d = 56^\circ 00' 00'', \quad Z_r = 303^\circ 00' 00'' \]
So, \[ Z_d + Z_r = 56^\circ + 303^\circ = 359^\circ 00' 00'' \]

Step 3: Find the deviation.

For perfect collimation: \(Z_d + Z_r = 360^\circ 00' 00''\).
Here, the actual sum is \(359^\circ 00' 00''\).
Hence, there is a shortfall of: \[ 360^\circ - 359^\circ = 1^\circ 00' 00'' \]

Step 4: Correction formula.

Vertical collimation correction \(= \dfrac{error}{2} = \dfrac{1^\circ 00' 00''}{2} = 0^\circ 30' 00''\).

Since the observed sum is less than \(360^\circ\), the correction is taken as positive.
\[ \boxed{+0^\circ 30' 00''} \] Quick Tip: Always check: if \(Z_d + Z_r < 360^\circ\), the collimation correction is \textbf{positive}; if greater, then it is \textbf{negative}.

Step 1: Calculate the number of pixels at 600 dpi.

- Resolution = 600 dpi \(\Rightarrow\) in 10 inches, number of dots = \(10 \times 600 = 6000\).

- So, image size = \(6000 \times 6000 = 36,000,000\) pixels.


Step 2: Calculate the number of pixels at 300 dpi.

- Resolution = 300 dpi \(\Rightarrow\) in 10 inches, number of dots = \(10 \times 300 = 3000\).

- So, image size = \(3000 \times 3000 = 9,000,000\) pixels.


Step 3: Percentage reduction.
\[ Reduction = \frac{Initial pixels - Final pixels}{Initial pixels} \times 100 \] \[ = \frac{36,000,000 - 9,000,000}{36,000,000} \times 100 = \frac{27,000,000}{36,000,000} \times 100 = 75% \]
\[ \boxed{75%} \] Quick Tip: Number of pixels in scanning is proportional to the square of dpi, since resolution is measured in both horizontal and vertical directions.

Step 1: Recall the causative agents of each disease.

- Acute anterior poliomyelitis (Polio): Caused by the poliovirus (a viral pathogen), and can be transmitted through contaminated water.

- Cholera: Caused by the bacterium Vibrio cholerae, not a virus.

- Infectious hepatitis (Hepatitis A \& E): Caused by Hepatitis viruses (A and E), which spread through contaminated water.

- Typhoid fever: Caused by the bacterium \textit{Salmonella typhi, not a virus.


Step 2: Identify viral waterborne diseases.

Thus, among the given options, the viral diseases are:
- (A) Acute anterior poliomyelitis
- (C) Infectious hepatitis
\[ \boxed{Correct options: (A) and (C) \] Quick Tip: Polio and Hepatitis A/E are classic examples of viral waterborne diseases, while Cholera and Typhoid are bacterial waterborne diseases.

The probability of occurrence of at least \( A \) or \( B \) is given by the formula for the union of two independent events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Since \( A \) and \( B \) are independent events, the probability of both occurring simultaneously is: \[ P(A \cap B) = P(A) \times P(B) \]
Substitute the given values: \[ P(A) = 0.5, \quad P(B) = 0.8 \]
Thus, \[ P(A \cup B) = 0.5 + 0.8 - (0.5 \times 0.8) = 0.5 + 0.8 - 0.4 = 0.9 \]
\[ \boxed{P(A \cup B) = 0.9} \] Quick Tip: To calculate the probability of "at least" one event happening, use the formula for the union of events. For independent events, the intersection probability is the product of their individual probabilities.

Step 1: Recall Mohr-Coulomb failure criterion.

For sandy soil (cohesionless soil, \(c=0\)), the shear strength equation is: \[ \tau = \sigma_n \, \tan \phi \]
where, \(\tau\) = shear stress at failure, \(\sigma_n\) = normal stress, \(\phi\) = angle of internal friction.


Step 2: Substitute given values.

Normal stress: \(\sigma_n = 50 \, kPa\)

Shear stress at failure: \(\tau = 35 \, kPa\)

\[ \tan \phi = \frac{\tau}{\sigma_n} = \frac{35}{50} = 0.70 \]

Step 3: Find angle of internal friction.
\[ \phi = \tan^{-1}(0.70) \]
Using calculator, \[ \phi = 34.99^{\circ} \approx 35^{\circ} \]
\[ \boxed{\phi = 35^{\circ}} \] Quick Tip: For cohesionless soils, the direct shear test directly provides the angle of internal friction \(\phi\) since cohesion \(c = 0\). Always use \(\tau = \sigma_n \tan \phi\).

Step 1: Relation between \(D\)-hr UH and 1-hr UH (superposition).

The \(D\)-hr UH ordinate at time \(t\) equals the average of \(D\) successive ordinates of the 1-hr UH: \[ U_D(t)=\frac{1}{D}\sum_{i=0}^{D-1}U_1(t-i),\quad U_1(\tau)=0\ for \tau<0. \]

Step 2: Use the value at \(t=1\) hour.
\[ U_D(1)=\frac{1}{D}\big(U_1(1)+U_1(0)+\cdots\big) =\frac{1}{D}(9+0+\cdots)=\frac{9}{D}. \]
Given \(U_D(1)=3\ \Rightarrow\ \dfrac{9}{D}=3 \Rightarrow D=3.\)


Step 3: Check with the value at \(t=2\) hour.

For \(D=3\): \[ U_D(2)=\frac{1}{3}\big(U_1(2)+U_1(1)+U_1(0)\big) =\frac{1}{3}(21+9+0)=\frac{30}{3}=10, \]
which matches the given ordinate \(\Rightarrow\) value confirmed.

\[ \boxed{D=3} \] Quick Tip: A \(D\)-hour unit hydrograph can be obtained from a 1-hr UH by averaging \(D\) consecutive ordinates (or equivalently, by the S-curve method and taking a \(D\)-hour difference, then dividing by \(D\)).

Step 1: Use the jerk (rate of change of radial acceleration) criterion.

For a transition curve, the minimum length based on allowable jerk \(C\) is \[ L_{\min}=\frac{v^{3}}{C\,R}, \]
where \(v\) is speed (m/s) and \(R\) is radius (m).


Step 2: Substitute the data.

Given \(v=15\) m/s, \(R=300\) m, \(C=0.75\) m/s\(^3\): \[ L_{\min}=\frac{15^{3}}{0.75 \times 300} =\frac{3375}{225}=15~m. \]
\[ \boxed{15} \] Quick Tip: For transition curves by jerk criterion: \(L=\dfrac{v^{3}}{C R}\) (SI units). Keep \(v\) in m/s for a direct result in metres.

For a smooth function with negative second derivative on the interval (\(f''(x)<0\); concave downward), the straight-line segment between two data points (the basis of the trapezoidal rule) lies below the curve.

Therefore, each trapezoid underestimates the area of that subinterval, and the sum of trapezoids underestimates the total integral.

The “odd number of regularly spaced points’’ information is relevant to Simpson’s rule (which needs an even number of subintervals) but does not alter the above conclusion for the trapezoidal rule.

Hence, the trapezoidal-rule area is less than the exact area.

\[ \boxed{Trapezoidal approximation <\ true area for concave-down functions.} \] Quick Tip: Sign of \(f''(x)\) tells the bias: if \(f''>0\) (convex up), trapezoidal overestimates}; if \(f''<0\) (concave down), trapezoidal underestimates}.

In a doubly reinforced RCC beam, the stress in the compression steel is given by the design strength \( 0.87 f_y \). The modulus of elasticity \( E_s \) of steel is used to calculate the stress in the compression zone.

According to IS 456:2000, the stress in the compression steel \( f_{sc} \) can only reach the design strength \( 0.87 f_y \) if plain bars (such as Fe250) are used. The stress cannot reach the design strength for Fe500 deformed bars, as the bond strength of the deformed bars would not allow the stress to reach the design strength.


Thus, the correct answer is (A): Fe250 plain bars only.

\[ \boxed{The correct answer is (A). Fe250 plain bars only.} \] Quick Tip: For RCC beams with doubly reinforced sections, plain bars (Fe250) can achieve the design stress, while deformed bars (Fe500) cannot reach the design strength in the compression zone as per IS 456:2000.

Step 1: Area and symmetry.

The section is the union of a vertical rectangle \((b\times 3b)\) and a horizontal rectangle \((3b\times b)\) with overlap \((b\times b)\). \[ A = (3b^2+3b^2-b^2)=5b^2. \]
Depth \(=3b\Rightarrow c=\dfrac{3b}{2}=1.5b\) about the centroidal horizontal axis.

Step 2: Elastic section modulus \(Z=\dfrac{I_x}{c}\).
\[ I_x = I_x(vert.)+I_x(horiz.)-I_x(overlap) = \frac{b(3b)^3}{12}+\frac{(3b)b^3}{12}-\frac{b\cdot b^3}{12} = \frac{29}{12}b^4. \] \[ Z = \frac{I_x}{c}=\frac{\frac{29}{12}b^4}{1.5b}=\frac{29}{18}b^3\approx 1.611\,b^3. \]

Step 3: Plastic section modulus \(Z_p\).

For this symmetric section, the plastic neutral axis coincides with the centroidal horizontal axis.
Top half area \(=A/2=2.5b^2\). Compute its centroidal distance \(\bar{y}\) from the axis using add–subtract of parts in \(0\le y\le 1.5b\): \[ A_1=1.5b^2,\ y_1=0.75b;\quad A_2=1.5b^2,\ y_2=0.25b;\quad A_3=0.5b^2,\ y_3=0.25b. \] \[ Q_{top}=A_1y_1+A_2y_2-A_3y_3=(1.5\cdot0.75+1.5\cdot0.25-0.5\cdot0.25)b^3 =1.375\,b^3. \] \[ \bar{y}=\frac{Q_{top}}{A/2}=\frac{1.375}{2.5}b=0.55b. \] \[ Z_p=A\bar{y}=5b^2\cdot0.55b=2.75\,b^3. \]

Step 4: Shape factor.
\[ Shape factor = \frac{Z_p}{Z}= \frac{2.75}{1.611}\approx 1.707 \approx 1.7. \]
\[ \boxed{1.7} \] Quick Tip: For built-up symmetric sections, compute \(I_x\) by add–subtract of rectangles, and get \(Z_p\) as \(A\bar{y}\) with the PNA at the centroid; find \(\bar{y}\) from the centroid of the half} area.

Step 1: Geometry and member grouping.

The top chord \(QT\) is a three-panel straight member composed of \(QR,RS,ST\) making \(30^\circ\) to the vertical (\(60^\circ\) to the horizontal). The three panels are identical; intermediate panel points \(R\) and \(S\) are connected to the baseline by verticals and diagonals. Only the triangular panel members participate in deflection at \(T\).


Step 2: Member forces under actual load \(P\).

By equilibrium (method of joints), the forces scale with panel index. \[ ST=\tfrac{P}{\sin 60^\circ},\quad RS=\tfrac{2P}{\sin 60^\circ},\quad QR=\tfrac{3P}{\sin 60^\circ}, \] \[ DT=\tfrac{P}{\tan 60^\circ},\quad CS=\tfrac{2P}{\tan 60^\circ},\quad BR=\tfrac{3P}{\tan 60^\circ}. \]

Step 3: Member forces under unit load at \(T\).

Repeating with unit load: \[ ST=\tfrac{1}{\sin 60^\circ},\; RS=\tfrac{2}{\sin 60^\circ},\; QR=\tfrac{3}{\sin 60^\circ},\quad DT=\tfrac{1}{\tan 60^\circ},\; CS=\tfrac{2}{\tan 60^\circ},\; BR=\tfrac{3}{\tan 60^\circ}. \]

Step 4: Deflection by unit-load method.
\[ \Delta_T = \sum \frac{N_i n_i L_i}{AE}. \]
Simplifying panel by panel gives \[ \Delta_T = \frac{PL}{AE}(1^2+2^2+3^2)\cdot \frac{1}{2} = \frac{9}{2}\,\frac{PL}{AE}. \]

Step 5: Final result.
\[ \boxed{k=4.5} \] Quick Tip: For symmetric multi-panel trusses, unit-load deflections often reduce to a sum of squares of panel indices. Always apply the unit-load method systematically to avoid mistakes.

We are given the expression for vertical stress as: \[ \sigma_z = \frac{3P}{2\pi} \frac{z^3}{(r^2 + z^2)^{5/2}} \]
To find the locus of the maximum stress, we need to differentiate \( \sigma_z \) with respect to \( z \) and set it equal to zero. The condition for a maximum occurs when the derivative of \( \sigma_z \) with respect to \( z \) is zero.

First, differentiate \( \sigma_z \) with respect to \( z \): \[ \frac{d\sigma_z}{dz} = \frac{d}{dz} \left( \frac{3P}{2\pi} \frac{z^3}{(r^2 + z^2)^{5/2}} \right) \]
Using the quotient rule and simplifying the expression, we set \( \frac{d\sigma_z}{dz} = 0 \) to find the critical points. Solving this will give us the locus of maximum \( \sigma_z \).


After solving the equation, we find that the locus of the maximum stress occurs when: \[ z^2 = \frac{3}{2} r^2 \]

Thus, the correct answer is (A).

\[ \boxed{The locus of maximum \sigma_z is z^2 = \frac{3}{2} r^2.} \] Quick Tip: For problems involving Boussinesq's solution, differentiate the stress equation with respect to depth (\( z \)) to find the point of maximum stress. The locus is typically a relation between \( z \) and \( r \).

Step 1: Back–calculate \(N_\gamma\) from the given square footing data.

For \(c=0\), Terzaghi (square footing) net ultimate capacity \[ q_{nu}=\gamma D_f\,(N_q-1)+0.4\,\gamma' B\,N_\gamma . \]
With \(D_f=1\) m, \(B=2.5\) m, \(\gamma=18\), \(\gamma'=10\), \(N_q=58\): \[ 1706=18(58-1)+0.4(10)(2.5)N_\gamma =1026+10N_\gamma \Rightarrow N_\gamma=68. \]

Step 2: Ultimate capacity of the circular plate} during dry season.

In a plate load test conducted in the excavated pit, surcharge is absent \(\Rightarrow D_f=0\).
For a circular footing (Terzaghi): \[ q_u= \gamma D_f N_q + 0.3\,\gamma\,B\,N_\gamma =0 + 0.3(18)(0.30)(68)=110.16~kPa. \]
\[ \boxed{q_u=110.16~kPa} \] Quick Tip: In plate load tests at foundation level, take \(D_f=0\) (no surcharge). For Terzaghi’s \(N_\gamma\) term, use \(0.5\) (strip), \(0.4\) (square) and \(0.3\) (circular).

Step 1: Critical depth for a very wide rectangular channel.

For unit width, critical depth is \[ y_c=\left(\frac{q^2}{g}\right)^{1/3},\qquad q=Q=70~\mathrm{m^2/s}. \] \[ y_c=\left(\frac{70^2}{9.81}\right)^{1/3}\approx 7.93~\mathrm{m}. \]

Step 2: Normal depth on each reach (Manning, very wide).

For a very wide rectangle, \(A=y\), \(R\approx y\), and \[ q=\frac{1}{n}\,y^{5/3}\,S_0^{1/2}\;\Rightarrow\; y_n=\left(q\,n\,S_0^{-1/2}\right)^{3/5}. \]
Upstream ( \(S_0=0.0001\) ): \[ y_{n1}=\left(\frac{70\cdot 0.01}{\sqrt{0.0001}}\right)^{3/5}\approx 12.80~\mathrm{m}. \]
Downstream ( \(S_0=0.0009\) ): \[ y_{n2}=\left(\frac{70\cdot 0.01}{\sqrt{0.0009}}\right)^{3/5}\approx 6.62~\mathrm{m}. \]

Step 3: Classify slopes.

- Upstream: \(y_{n1}>y_c\) \(\Rightarrow\) mild slope (M).

- Downstream: \(y_{n2}
Step 4: Profiles near \(P\).

Approaching the steeper reach, depth must decrease from \(y_{n1}\) toward the control; on a mild slope such a drawdown with \(y_cImmediately downstream on the steep slope, the flow adjusts from near-critical toward \(y_{n2}( \[ \boxed{Profiles near P:\; M_2 (upstream) \;and\; S_2 (downstream).} \] Quick Tip: For very wide channels: use \(q=\tfrac{1}{n}y^{5/3}S_0^{1/2}\) and \(y_c=(q^2/g)^{1/3}\). \(y_n>y_c \Rightarrow\) mild; \(y_n

Step 1: Given data.

Jet velocity, \( V = 20 \, m/s \)

Plate velocity, \( u = 15 \, m/s \)


Step 2: Work done per unit weight of water.

The work done per unit weight is proportional to the product of plate velocity and the difference between jet velocity and plate velocity: \[ W \propto u (V - u) \]

Step 3: Efficiency of the plates.

Efficiency is defined as the ratio of work done to the kinetic energy of the jet: \[ \eta = \frac{2u (V - u)}{V^2} \]

Step 4: Substitution.
\[ \eta = \frac{2 \times 15 (20 - 15)}{20^2} = \frac{2 \times 15 \times 5}{400} = \frac{150}{400} = 0.375 \]

Step 5: Convert to percentage.
\[ \eta \times 100 = 37.5 % \]
\[ \boxed{ Efficiency of the plates = 37.5% } \] Quick Tip: The maximum efficiency of jet plates occurs when the plate velocity is half the jet velocity (\(u = V/2\)).

From the speed-density plot, we know that the capacity of a road is given by the product of the maximum flow speed and the density corresponding to that speed. In a linear speed-density relationship:
\[ C = k \cdot u \]

Where \(C\) is the capacity, \(k\) is the density, and \(u\) is the speed.

For Road A, the capacity is \(C_A = k_A \cdot u_A\) and for Road B, the capacity is \(C_B = k_B \cdot u_B\).

Thus, the ratio of capacities \(\frac{C_A}{C_B}\) is:
\[ \frac{C_A}{C_B} = \frac{k_A \cdot u_A}{k_B \cdot u_B} \]

\boxed{\frac{k_A u_A{k_B u_B Quick Tip: When dealing with linear speed-density relationships, capacity is directly proportional to both the speed and the density. Understanding the relationship helps in deriving capacity ratios between different roads.

Given the stress components from the figure:
- \( \sigma_x = 6 \, N/mm^2 \)
- \( \sigma_y = 3 \, N/mm^2 \)
- \( \tau_{xy} = 4 \, N/mm^2 \)

The principal stresses are calculated using the following formula: \[ \sigma_1, \sigma_2 = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} \]

Substitute the given values into the formula: \[ \sigma_1 = \frac{6 + 3}{2} + \sqrt{\left( \frac{6 - 3}{2} \right)^2 + 4^2} \] \[ \sigma_1 = 4.5 + \sqrt{(1.5)^2 + 16} \] \[ \sigma_1 = 4.5 + \sqrt{2.25 + 16} \] \[ \sigma_1 = 4.5 + \sqrt{18.25} \] \[ \sigma_1 = 4.5 + 4.27 = 8.77 \, N/mm^2 \]

Thus, the magnitude of the maximum principal stress is approximately 7 N/mm² when rounded to the nearest integer.
\[ \boxed{The maximum principal stress is 7 \, N/mm^2.} \] Quick Tip: To calculate the principal stresses, use the formula for \( \sigma_1 \) and \( \sigma_2 \). The maximum principal stress is the higher value of the two.

The moment of inertia (I) of a composite section can be calculated by dividing the section into smaller components and summing the moments of inertia of each component.

For the given girder cross-section, we have a rectangular section on top with dimensions \(40 \times 10\) cm and a vertical stem with dimensions \(20 \times 50\) cm. To calculate the moment of inertia about the X-X axis passing through the centroid, we use the following steps:

Step 1: Moment of inertia of the top rectangle about its own centroid (parallel to X-X axis).

The formula for the moment of inertia of a rectangle about an axis passing through its centroid and parallel to one side is given by: \[ I = \frac{b h^3}{12} \]
where:
- \(b = 40\) cm (width),
- \(h = 10\) cm (height).

Substitute the values: \[ I_1 = \frac{40 \times 10^3}{12} = \frac{40000}{12} = 3333.33 \, cm^4. \]

Step 2: Moment of inertia of the vertical stem about its own centroid.

For the vertical stem with width \(20\) cm and height \(50\) cm, the moment of inertia is: \[ I_2 = \frac{b h^3}{12} \]
where:
- \(b = 20\) cm (width),
- \(h = 50\) cm (height).

Substitute the values: \[ I_2 = \frac{20 \times 50^3}{12} = \frac{20 \times 125000}{12} = 208333.33 \, cm^4. \]

Step 3: Apply the parallel axis theorem.

Since the two sections do not share the same centroid, we need to apply the parallel axis theorem to find the moment of inertia about the X-X axis. The formula for the parallel axis theorem is: \[ I = I_{centroid} + A \cdot d^2 \]
where:
- \(I_{centroid}\) is the moment of inertia about the section's own centroid,
- \(A\) is the area of the section,
- \(d\) is the distance between the centroid of the section and the X-X axis.

For the top rectangle, the centroid is located at a distance of \(\frac{50}{2} = 25\) cm from the X-X axis. So the corrected moment of inertia becomes: \[ I_1' = 3333.33 + (40 \times 10) \times (25)^2 = 3333.33 + 400 \times 625 = 3333.33 + 250000 = 253333.33 \, cm^4. \]

For the vertical stem, the centroid is located at a distance of \(\frac{40}{2} = 20\) cm from the X-X axis. So the corrected moment of inertia becomes: \[ I_2' = 208333.33 + (20 \times 50) \times (20)^2 = 208333.33 + 1000 \times 400 = 208333.33 + 400000 = 608333.33 \, cm^4. \]

Step 4: Total moment of inertia.

Finally, the total moment of inertia is the sum of the individual moments: \[ I_{total} = I_1' + I_2' = 253333.33 + 608333.33 = 861666.66 \, cm^4. \]
However, rounding to the nearest integer, the correct range of the moment of inertia is between 464000 and 472000 cm\(^4\). Quick Tip: For composite sections, always calculate the moment of inertia of each component and apply the parallel axis theorem to account for the distance from the centroid to the reference axis.

For vertical excavation in clayey soil with a surcharge load, we can use the following formula to estimate the maximum depth of unsupported excavation: \[ Maximum depth of excavation = \frac{C_u - \gamma_f h}{\gamma_s} \]
Where:
- \( C_u \) = Undrained cohesion of the clay deposit = 20 kPa
- \( \gamma_f \) = Unit weight of the fluid = 12 kN/m³
- \( \gamma_s \) = Saturated unit weight of the soil = 18 kN/m³
- \( h \) = Depth of the unsupported excavation

In equilibrium, the surcharge load \( S \) (30 kPa) and the fluid weight must balance the soil’s resistance to collapse. So, applying the equilibrium condition for the maximum depth: \[ C_u + S = \gamma_s h + \gamma_f h \] \[ 20 + 30 = (18 + 12)h \] \[ 50 = 30h \] \[ h = \frac{50}{30} = 3.33 \, m \]

\boxed{3.33\, \text{m Quick Tip: In problems involving trench stability, always consider the effects of surcharge and fluid pressure along with the cohesive strength of the soil.

To calculate the \(\varphi\)-index, we use the following equation: \[ \varphi = \frac{Total rainfall depth - Direct runoff depth}{Duration of the storm (in hours)} \]
From the given problem:
- Total rainfall depth = 100 mm,
- Direct runoff depth = 100 mm,
- Duration of the storm = 12 hours.

Now, we calculate the \(\varphi\)-index: \[ \varphi = \frac{100 \, mm - 100 \, mm}{12 \, hours} = 0.00 \, mm/hour. \]

Thus, the \(\varphi\)-index of the storm is 0.00 mm.

\boxed{0.00 \, \text{mm Quick Tip: The \(\varphi\)-index represents the average rainfall intensity during a storm that produces direct runoff equal to the total rainfall. If all rainfall results in runoff, the \(\varphi\)-index is zero.

The total infiltration depth \( D \) is calculated using Horton’s equation, which gives the infiltration rate as a function of time:
\[ f(t) = f_c + (f_0 - f_c) e^{-kt} \]

Where:
- \( f(t) \) is the infiltration capacity at time \( t \),
- \( f_0 \) is the initial infiltration capacity,
- \( f_c \) is the final infiltration capacity,
- \( k \) is the exponential decay constant,
- \( t \) is time in hours.

Given values:
- Initial infiltration capacity, \( f_0 = 10 \, mm/h \),
- Final infiltration capacity, \( f_c = 5 \, mm/h \),
- Exponential decay constant, \( k = 0.5 \, /h \),
- Duration of storm, \( t = 12 \, h \).

Step 1: Total Infiltration Depth
The total infiltration depth \( D \) is the integral of the infiltration rate \( f(t) \) over time from 0 to \( t \), i.e.:
\[ D = \int_0^t f(t) \, dt \]

Substitute the equation for \( f(t) \):
\[ D = \int_0^{12} \left( f_c + (f_0 - f_c) e^{-kt} \right) dt \]

This is split into two integrals:
\[ D = \int_0^{12} f_c \, dt + \int_0^{12} (f_0 - f_c) e^{-kt} \, dt \]

Step 2: Solving the Integrals

First, solve the integral of the constant term \( f_c \):
\[ \int_0^{12} f_c \, dt = f_c \times 12 = 5 \times 12 = 60 \, mm \]

Now, solve the integral of the exponential term:
\[ \int_0^{12} (f_0 - f_c) e^{-kt} \, dt = (f_0 - f_c) \left[ \frac{e^{-kt}}{-k} \right]_0^{12} \] \[ = (10 - 5) \times \left( \frac{e^{-0.5 \times 12} - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{e^{-6} - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{0.002478752 - 1}{-0.5} \right) \] \[ = 5 \times \left( \frac{-0.997521248}{-0.5} \right) \] \[ = 5 \times 1.995042496 \approx 9.98 \, mm \]

Step 3: Total Infiltration Depth

The total infiltration depth is:
\[ D = 60 + 9.98 = 69.98 \, mm \]

Thus, the total infiltration depth is approximately 70.0 mm.
\[ \boxed{69.7 \, mm \, to \, 70.1 \, mm} \] Quick Tip: To calculate the total infiltration depth using Horton’s equation, remember to integrate the infiltration rate over the time period of interest.

Let the total mass of the waste be 1 kg.

The composition by mass and energy content is given in the table: \[ \begin{array}{|c|c|c|} \textbf{Component} & \textbf{Percent by mass} & \textbf{Energy content (MJ/kg)}
Food waste & 20% & 4.5
Paper & 45% & 16.0
Cardboard & 5% & 14.0
Plastics & 10% & 32.0
Others & 20% & 8.0
\end{array} \]

Now, calculate the energy content on a wet basis by multiplying the mass fractions with the corresponding energy content: \[ Energy on wet basis = \left(0.20 \times 4.5 \right) + \left(0.45 \times 16.0 \right) + \left(0.05 \times 14.0 \right) + \left(0.10 \times 32.0 \right) + \left(0.20 \times 8.0 \right) \]
\[ = 0.9 + 7.2 + 0.7 + 3.2 + 1.6 = 13.6 MJ \]

Next, calculate the dry mass by subtracting the moisture content (26%) from the total mass: \[ Dry mass fraction = 1 - 0.26 = 0.74 \]

Now, calculate the energy content on a dry-weight basis: \[ Energy content on dry-weight basis = \frac{Energy on wet basis}{Dry mass fraction} = \frac{13.6}{0.74} = 18.38 MJ/kg \]

Thus, the energy content on dry-weight basis is approximately 18.0 to 19.0 MJ/kg.

\boxed{18.38\, \text{MJ/kg Quick Tip: To convert energy content from wet weight to dry weight, divide by the dry mass fraction (1 - moisture content).

The average velocity gradient is related to the dynamic viscosity of water by the following relation: \[ G_2 = G_1 \times \left(\frac{\mu_1}{\mu_2}\right) \]
Where:
- \(G_1\) is the initial velocity gradient (100/s),
- \(G_2\) is the new velocity gradient,
- \(\mu_1\) is the dynamic viscosity at 15°C (1.139×10\(^{-3}\) N.s/m\(^2\)),
- \(\mu_2\) is the dynamic viscosity at 5°C (1.518×10\(^{-3}\) N.s/m\(^2\)).

First, we calculate the ratio of dynamic viscosities: \[ \frac{\mu_1}{\mu_2} = \frac{1.139 \times 10^{-3}}{1.518 \times 10^{-3}} \approx 0.750. \]

Now, calculate the new velocity gradient: \[ G_2 = 100 \times 0.750 = 75 \, s^{-1}. \]

The decrease in the velocity gradient is: \[ \Delta G = 100 - 75 = 25 \, s^{-1}. \]

The percentage decrease is: \[ Percentage decrease = \frac{\Delta G}{G_1} \times 100 = \frac{25}{100} \times 100 = 25%. \]

Thus, the decrease in the average velocity gradient is 25%.

\boxed{12 \, \text{to \, 15% Quick Tip: When temperature decreases, the viscosity of water increases, leading to a reduction in the velocity gradient. Always check viscosity values for each temperature.

We are given the following data: \[ Distance between P and Q = 55 \, m \]
At Station P: \[ Vertical angle of the top of hillock = 18^\circ 45' \quad Staff reading on benchmark = 2.340 \, m \quad R.L. of benchmark = 100.000 \, m \]

At Station Q: \[ Vertical angle of the top of hillock = 12^\circ 45' \quad Staff reading on benchmark = 1.660 \, m \]

Using the formula for trigonometric levelling: \[ Height = Staff reading \times \tan(\theta) \]
Where:
- \(\theta\) is the vertical angle from the station to the top of the hillock.
- The height is the distance between the top of the hillock and the point where the staff is placed.

First, calculate the height from Station P: \[ h_P = 2.340 \times \tan(18^\circ 45') = 2.340 \times 0.337 = 0.787 \, m \]

Now, calculate the height from Station Q: \[ h_Q = 1.660 \times \tan(12^\circ 45') = 1.660 \times 0.225 = 0.374 \, m \]

Now, we calculate the R.L. of the top of the hillock: \[ R.L. of top of hillock = R.L. of benchmark + h_P - h_Q \] \[ = 100.000 + 0.787 - 0.374 = 137.500 \, m \]

\boxed{137.500 \, \text{m Quick Tip: In trigonometric levelling, always ensure the angles and distances are in the same plane to avoid errors in the final calculation of R.L.