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Explanation:

The most probable value with equal weightage is given by:

= \(30°30'00''+30°29'40''+30°30'20''+30°31'20'' \over 4\) = 30°30'20''

Explanation:

As per given data:

\(P: \frac{d y}{d x}=\frac{x^4+3 x^2 y^2+2 y^4}{x^3 y} \)

\(\frac{d y}{d x}=\left(\frac{x^4}{x^3 y}\right)+\left(\frac{3 x^2 y^2}{x^3 y}\right)+\left(\frac{2 y^4}{x^3 y}\right) \)

\(\frac{d y}{d x}=\left(\frac{x}{y}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3 \)

\(\frac{d y}{d x}=\left(\frac{1}{y / x}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3 \)

\(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\)

So P is a homogenous ODE

\(Q: \frac{d y}{d x}=\frac{-y^2}{x^2} \)

x²dy = -y²dx

y²dx + x²dy = 0

\(M=y^2 \quad \frac{d M}{\partial y}=2 y\)

\(N=x^2 \quad \frac{\partial N}{\partial x}=2x\)

So Q is non exact ODE 

∴ The correct answer is option (3).

Explanation:

1. Flexible footing over clayey soil: 

In flexible footing, the contact pressure at the interface between footing and soil is uniformly distributed producing dish-shape pattern in clayey soil.

2. Flexible footing over Granular soil:

In granular soil, modulus of elasticty (E_s) varies across the width being maximum at the centre and minimum at edge. As E is maximum at centre, defflection is less at centre. As E is less at edge deflection is more at edge.

3. Rigid footing on Clayey soil:

In case of flexible footing, deflection is more at centre. Hence pressure developed at centre is less. Deflection is less in flexible footing at edge, hence in rigid footing pressure developed is more at edge.

4. Rigid footing on Granular soil

E×planation:

Given Information:

• The pavement is jointed plain concrete with a slab size of 3.5m × 4.5m.
• Different stresses acting on the slab:
• Wheel load stress at different locations:
• Interior: SiwlSwli S^i_{wl}

​

• Edge: SewlSwle S^e_{wl}

​

• **Corner: SCwlSwlC S^C_{wl}

​**

• Warping stress at different locations:
• Interior: SitSti S^i_t

​

• Edge: SetSte S^e_t

​

• **Corner: SCtStC S^C_t

​**

• Frictional stress: SfSf S_f

​

Understanding Critical Stress Condition at Summer Midnight

• At midnight during summer, the pavement slab cools down, leading to negative temperature gradients:
• Warping stress at the corner (SCtStC S^C_t

) is negative (due to contraction).

• Wheel load stress at the corner (SCwlSwlC S^C_{wl}

) is always positive.

• **Frictional stress (SfSf S_f

) does not significantly affect corner stresses**.

Since the warping stress is negative, the critical stress at the corner is simply the sum of the wheel load stress and the warping stress:

\(\mathrm{S}_{\mathrm{wl} }^{\mathrm{C}}+\mathrm{S}_{\mathrm{t}}^{\mathrm{C}}\)​

Explanation:

ΣM_Q↓= 0 ⇒ R_P × 3 - 3 × 2 = 0 

RP = 2kN & R_Q = 1 kN

As beam PQ is rigid & for no rotation settlement at both P & Q should be same.

ΔP = ΔQ

\(\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{~K}_{\mathrm{P}}} =\frac{\mathrm{R}_{\mathrm{Q}}}{\mathrm{~K}_{\mathrm{Q}}} \)

\(⇒ \mathrm{~K}_{\mathrm{Q}} =\frac{\mathrm{R}_{\mathrm{Q}}}{\mathrm{R}_{\mathrm{P}}} × \mathrm{K}_{\mathrm{P}}=\frac{1}{2} × 100=50 \mathrm{kN} / \mathrm{m}\)

The value of the spring constant KQ (in kN/m) for the spring at the end Q is 50 kN/m.

Explanation:

As peer given data:

f(x) = x3 - 27x + 4

​f'(x) = 3x2 - 27

​f'(x) = 0

3x2 - 27 = 0

x = ±3

x = 3 ∈ (1, 6)

So at x = 3 function has point at local minima.

Explanation:

The Working Stress Method (WSM) is a traditional structural design approach that uses elastic theory and ensures that stresses in materials remain within their permissible limits under working loads. It does not incorporate safety factors on loads, but instead applies safety factors to material strength.

Additional Information

• Ultimate Load Method (ULM) applies a factor of safety on loads to ensure failure does not occur at the ultimate stage.
• Load Factor Method is another term for ULM, incorporating load factors for safety.
• Limit State Method (LSM) considers safety factors on both loads and material strengths, making it the most advanced and widely used approach.

Explanation:

The given partial differential equation is the Laplace equation:

\(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0\)

This equation describes steady-state heat conduction in a 2D square domain.

Given Boundary Conditions:

• The square domain is 2m × 2m.
• Three boundaries are kept at T = 50.
• The fourth boundary is maintained at T = 100.
• We need to determine the temperature at the center of the domain.

Concept of Temperature Distribution:

• The Laplace equation ensures that the temperature field follows a smooth variation.
• In a square plate with three sides at a lower temperature and one side at a higher temperature, the center temperature will be an average of the boundary conditions due to symmetry.

Calculation:

Using approximate averaging techniques for steady-state conduction in such problems, the center temperature TcenterT_{$\text{center}$}Tcenter​ is roughly estimated as:

T_center​ ≈ [3(50)+100]/4 ​= [150 + 100]/4 ​= 250/4 ​= 62.5

Explanation:

The shear center is the point through which the applied transverse shear force must pass to avoid any twisting of the section.

Key Observations:

1. The section consists of a vertical web and two unequal flanges.
2. The top flange is asymmetric, extending further to the right than the bottom flange.
3. The shear center shifts towards the more extended flange to balance the torque effects.

Why is the Shear Center at Point R?

• For thin-walled open sections like C-sections or asymmetric I-sections, the shear center is generally located towards the larger flange.
• Since the top flange extends more to the right, the shear center must shift rightward along the X-X axis.
• Point R is located towards the extended top flange, making it the correct location for the shear center.

Explanation:

Given data:

Length of pile (L) = 10 m

Diameter of pile (D) = 0.7 m

Cohesion of clay (C₄) = 50 kPa

Adhesion factor (α) = 0.5 

For clay, (P) skin friction resistance = f_s A_s

P = α C_uA_s +

Taking unit wt of concrete = 25 kN/m³

Uplift pull out capacity = skin friction resistance + wt. of pile

\(P =0.5 \times 50 \times \pi \times 0.7 \times 10+\frac{\pi(0.7)^2}{4} \times \gamma_c \)

\(P =0.5 \times 50 \times \pi \times 0.7 \times 10+\frac{\pi(0.7)^2}{4} \times 25\)

P = 550 kN