SSC JE Civil Questions with Detailed Solutions

When concrete is placed in sub-zero temperatures, certain precautions are necessary to ensure proper setting and strength gain:
Pre-heating concrete materials:
This prevents the mixing water from freezing and helps maintain a suitable temperature for hydration.
Economical heating methods:
Materials are warmed just enough to achieve the required temperature without unnecessary expenditure.
Use of anti-freeze admixtures:
These additives reduce the freezing point of water in the mix, allowing hydration to proceed even at low temperatures.
Electrical heating of the concrete mass:
Applying controlled heat during curing prevents freezing and supports the development of adequate strength.

​​​​​Definition:
The effective length of a column is defined as the distance between two consecutive points of zero bending moment (or points of contra-flexure).
It represents the equivalent length of an idealized column that would buckle under the same load as the actual column.
End Condition                                 Effective Length
Both ends hinged                                    L
One end fixed and the other free            2L
Both ends fixed                                       L/2
One end fixed and the other hinged        L/√2

The effective length depends on the end support conditions of the column, as shown below:

​​​$$\begin{aligned}&\text{For a simply supported beam with a central load (P):} \\[6pt]&\delta_C = \frac{PL^3}{48EI} \\[6pt]&\theta_B = \theta_A = \frac{wL^2}{16EI} \\[10pt]&\text{Calculation:} \\[6pt]&\text{Slope at support A:} \\[4pt]&\theta_A = \frac{wL^2}{16EI} \\[4pt]&\text{Substituting values: } \theta_A = \frac{2 \times 5^2}{16EI} = \frac{50}{16EI} \\[10pt]&\text{Slope at support B:} \\[4pt]&\theta_B = \frac{wL^2}{16EI} \\[4pt]&\text{Substituting values: } \theta_B = \frac{1 \times 10^2}{16 \times 2EI} = \frac{50}{16EI} \\[10pt]&\text{Hence, } \boxed{\text{Slope of Beam A = Slope of Beam B}} \\[6pt]&\text{At midspan (C), the slope is zero and the beam deflects symmetrically.}\end{aligned}$$​​

​$$\begin{aligned}&\text{We know,} \quad E = 2G(1 + \mu) \\[6pt]&\text{Given,} \quad E = 3G \\[6pt]&\Rightarrow 3G = 2G(1 + \mu) \\[6pt]&\Rightarrow 1 + \mu = \frac{3}{2} \\[6pt]&\Rightarrow \mu = 0.5 \\[8pt]&\text{Volumetric strain} = \frac{\sigma}{E}(1 - 2\mu) \\[6pt]&\Rightarrow \text{Volumetric strain} = \frac{\sigma}{E}(1 - 2 \times 0.5) = 0 \\[8pt]&\text{Bulk Modulus, } K = \frac{1}{\text{Volumetric strain}} = \infty\end{aligned}$$​​

​​​​​The Moody chart is used in fluid mechanics to determine the Darcy-Weisbach friction factor for flow in a pipe. It plots the friction factor on the vertical axis and the Reynolds number on the horizontal axis, both on logarithmic scales. Various curves on the chart represent different relative roughness values of the pipe. This chart helps engineers assess head loss due to friction in pipes.

Relative roughness is the ratio of pipe roughness and the pipe diameter. 

​​​$$\displaystyle\text{Bernoulli:}\quad \frac{p}{\rho g}+\frac{v^{2}}{2g}+z=\text{constant} \\[6pt]\text{Given:}\; P=40\,\text{N/cm}^2=40\times10^{4}\,\text{N/m}^2,\; v=3\,\text{m/s},\; g=10\,\text{m/s}^2,\; \rho=1000\,\text{kg/m}^3 \\[6pt]\text{Kinetic head:}\quad \frac{v^{2}}{2g}=\frac{3^{2}}{2\times10}=\frac{9}{20}=0.45\ \text{m} \\[6pt]\text{Pressure head:}\quad \frac{p}{\rho g}=\frac{40\times10^{4}}{1000\times10}=\frac{40\times10^{4}}{10^{4}}=40\ \text{m}$$​​

​$$\begin{aligned}&\text{Given:} \\[4pt]&Q = 0.025 \, \text{m}^3/\text{s} \\ &b = 40 \, \text{cm} = 0.4 \, \text{m} \\ &H = 0.1 \, \text{m} \\ &g = 9.81 \, \text{m/s}^2 \\[6pt]&\text{Formula Used:} \\[4pt]&Q = C_d \times \frac{2}{3} \times b \times \sqrt{2g} \times H^{3/2} \\[6pt]&\text{Substituting the values:} \\[4pt]&0.025 = C_d \times \frac{2}{3} \times 0.4 \times \sqrt{2 \times 9.81} \times (0.1)^{3/2} \\[6pt]&\Rightarrow C_d = \frac{0.025}{\left( \frac{2}{3} \times 0.4 \times \sqrt{19.62} \times 0.03162 \right)} \\[6pt]&\Rightarrow C_d \approx 0.66\end{aligned}$$​​

When joining old and new concrete, the bond strength depends on proper surface preparation of the existing concrete. The following methods are commonly used:

Sandblasting: Abrasive method to roughen the surface and remove contaminants like laitance, dirt, or loose particles.

Improves mechanical bond.

Chipping: Manual or mechanical removal of the surface to expose coarse aggregates and increase roughness.

Ensures proper anchorage.

Grinding: Used to smooth out or uniformly remove a thin surface layer.

Helps in removing laitance and exposing a fresh surface.

In the construction of hollow concrete block walls, the volume of the hollow portion (cavity) within the block is not deducted during measurement. This is because:

The full block size (including the hollow portion) is considered for calculation.

Deductions are typically made only for openings like doors, windows, and chases — not for the hollows inside blocks, as they are part of the unit.

This aligns with standard CPWD and IS code measurement practices for block masonry.

​​​$$\text{Height of water column } (h) = 4\, \text{m} \\[4pt]\text{Width of tank } (w) = 3\, \text{m} \\[4pt]\text{Length of tank } (l) = 3\, \text{m} \\[4pt]\text{Density of water } (\rho) = 1000\, \text{kg/m}^3 \\[4pt]\text{Acceleration due to gravity } (g) = 9.81\, \text{m/s}^2 \\[10pt]\textbf{Step 1: Area of the bottom of the tank} \\[4pt]A = \text{length} \times \text{width} = 3 \times 3 = 9\, \text{m}^2 \\[10pt]\textbf{Step 2: Pressure at the bottom} \\[4pt]P = \rho g h = 1000 \times 9.81 \times 4 = 39240\, \text{N/m}^2 \\[10pt]\textbf{Step 3: Total force} \\[4pt]F = P \times A = 39240 \times 9 = 353160\, \text{N} = 353.16\, \text{kN}$$​​

​​​$$h_f = \frac{f L V^2}{2 g d}$$
$$= \frac{f L}{2 g d} V^2$$=$$= \frac{f L}{2 g d} \left( \frac{4Q}{\pi d^2} \right)^2$$(:v = Q/A= (Q/$$\frac{\pi d^2}{4}$$)​
$$= \frac{8 f Q^2}{\pi^2 g} \left( \frac{L}{d^5} \right)$$
$$\textbf{Series connection (} Q_1 = Q_2 \textbf{):}h_f = h_{f1} + h_{f2}$$
$$\frac{L_1}{d^5} = \frac{l}{d^5} + \frac{l}{d^5} = \frac{2l}{d^5}$$
$$\therefore\ L_1 = 2l$$
$$\textbf{Parallel connection (} Q_1 = Q/2 \textbf{):}$$
$$h_f = h_{f1}$$
$$\frac{Q^2 L}{d^5} = \frac{\left( \frac{Q}{2} \right)^2 l}{d^5}L_2 = \frac{l}{4}$$
$$\textbf{Ratio:}\frac{L_1}{L_2} = 8 : 1$$​​​​​​​​​​​

Effective size (D₁₀) is the particle size corresponding to 10% finer by weight in the grain size distribution curve.

We calculate cumulative retained weight and percent passing for each sieve:

Diameter (D) = 100 mm = 0.1 m

Darcy friction factor (f) = 0.05

Flow is laminar.

Formula Used:

For laminar flow in a circular pipe, the Darcy friction factor is given by:

f = 64 / Re

Re = 64 / 0.05 = 1280

Calculate the net area of a tension member which has:
 A drilled hole of diameter = 22 mm (for M20 bolt)
 Thickness of member = 10 mm
 Gross area = 1000 mm²
Formula Used:
Net Area = Gross Area − (Hole Diameter × Thickness)
Calculation:
Reduction due to hole = 22×10 = 220  mm²
Net Area = 1000−220 = 780  mm²

​​​Nature of Bending Stress in a Cantilever Beam
Case 1: Hogging Bending Moment (Load Acting Downwards)
When a cantilever experiences a hogging moment, the fibres above the neutral axis (NA) undergo tensile stress, while the fibres below the neutral axis experience compressive stress.

Case 2: Sagging Bending Moment (Load Acting Upwards)
Under a sagging moment, the stress pattern reverses. The fibres above the neutral axis are in compression, whereas the fibres below the neutral axis are in tension.