SSC JE Tier-II Civil – Solutions

Q1 – RCC Beam Design

Given:

• Span $L = 6 m$L=6m
• UDL $w = 20 kN/m$w=20kN/m
• Width $b = 300 mm$b=300mm, effective depth $d = 500 mm$d=500mm
• Concrete $f_{ck} = 25 MPa$fck​=25MPa, steel $f_y = 415 MPa$fy​=415MPa

Step 1 – Maximum bending moment:$$M_{max} = \frac{w L^2}{8} = \frac{20 \times 6^2}{8} = 90 kNm$$Mmax​=8wL2​=820×62​=90kNm

Step 2 – Area of steel (Working Stress Method):$$\sigma_s = \frac{M}{0.87 f_y Z}, \quad Z = \frac{b d^2}{6} = \frac{0.3 \cdot 0.5^2}{6} = 0.0125 m^3$$σs​=0.87fy​ZM​,Z=6bd2​=60.3⋅0.52​=0.0125m3 $$A_s = \frac{M_{max} \cdot 10^6}{0.87 f_y d} = \frac{90 \times 10^6}{0.87 \cdot 415 \cdot 0.5} \approx 498 mm^2$$As​=0.87fy​dMmax​⋅106​=0.87⋅415⋅0.590×106​≈498mm2

Step 3 – Bending moment diagram: Standard simply supported beam with UDL.

✅ Answer Q1:

• Maximum moment = 90 kNm
• Steel area ≈ 498 mm²

Q2 – Footing Design

Given:

• Load $P = 500 kN$P=500kN
• Allowable bearing capacity $q_{all} = 150 kPa$qall​=150kPa
• Ultimate $q_u = 300 kPa$qu​=300kPa

Step 1 – Footing area:$$A = \frac{P}{q_{all}} = \frac{500}{150} = 3.33 m^2$$A=qall​P​=150500​=3.33m2

Side of square footing:$$B = \sqrt{A} \approx 1.825 m$$B=A​≈1.825m

Step 2 – Factor of Safety:$$FS = \frac{q_u}{q_{all}} = \frac{300}{150} = 2$$FS=qall​qu​​=150300​=2

Step 3 – Schematic: Square footing under column with uniform load.

✅ Answer Q2:

• Footing size ≈ 1.83 m × 1.83 m
• FS = 2

Q3 – Canal Design (Rectangular)

Given:

• Q = 5 m³/s
• Slope $S = 0.0002$S=0.0002, Manning’s n = 0.015

Step 1 – Manning’s formula:$$Q = \frac{1}{n} R^{2/3} S^{1/2} A$$Q=n1​R2/3S1/2A

Assume width $b = 4 m$b=4m. Solve iteratively for depth $h$h:$$A = b h, \quad R = \frac{A}{P} = \frac{b h}{b + 2h}$$A=bh,R=PA​=b+2hbh​

Step 2 – Iteration / approximate solution:

• Depth $h \approx 1.25 m$h≈1.25m
• Hydraulic radius $R \approx 0.69 m$R≈0.69m
• Velocity $V = Q/A \approx 1 m/s$V=Q/A≈1m/s

Step 3 – Cross-section: Rectangular with width 4 m, depth 1.25 m.

✅ Answer Q3:

• Width = 4 m, Depth ≈ 1.25 m
• Hydraulic radius ≈ 0.69 m, Velocity ≈ 1 m/s

Q4 – Traffic Engineering

Given:

• Flow = 2000 veh/hr, Design speed $V = 80 km/h$V=80km/h
• Perception-reaction time $t = 2.5 s$t=2.5s, friction $\mu = 0.35$μ=0.35

Step 1 – Stopping sight distance (SSD):$$SSD = V t + \frac{V^2}{2 g (\mu + \text{grade})}$$SSD=Vt+2g(μ+grade)V2​ $$SSD = \frac{80 \times 1000}{3600} \cdot 2.5 + \frac{(22.22)^2}{2 \cdot 9.81 \cdot 0.35} \approx 55 + 70 \approx 125 m$$SSD=360080×1000​⋅2.5+2⋅9.81⋅0.35(22.22)2​≈55+70≈125m

Step 2 – Minimum horizontal curve radius:$$R = \frac{V^2}{g (e + f)} \approx \frac{22.22^2}{9.81 \cdot (0 + 0.35)} \approx 142 m$$R=g(e+f)V2​≈9.81⋅(0+0.35)22.222​≈142m

Step 3 – Sketch: SSD layout along road.

✅ Answer Q4:

• SSD ≈ 125 m
• Minimum radius ≈ 142 m

Q5 – Pipe Flow (Hydraulics)

Given:

• D = 0.3 m, L = 1000 m, S = 0.002, n = 0.013

Step 1 – Manning velocity:$$V = \frac{1}{n} R^{2/3} S^{1/2}, \quad R = D/4 = 0.075 m$$V=n1​R2/3S1/2,R=D/4=0.075m $$V \approx \frac{1}{0.013} \cdot 0.075^{2/3} \cdot 0.002^{1/2} \approx 1.2 m/s$$V≈0.0131​⋅0.0752/3⋅0.0021/2≈1.2m/s

Step 2 – Discharge:$$Q = V A = 1.2 \cdot (\pi D^2 /4) = 1.2 \cdot 0.0707 \approx 0.085 m^3/s$$Q=VA=1.2⋅(πD2/4)=1.2⋅0.0707≈0.085m3/s

Step 3 – Head loss (Manning):$$h_f = L S = 1000 \cdot 0.002 = 2 m$$hf​=LS=1000⋅0.002=2m

✅ Answer Q5:

• Velocity ≈ 1.2 m/s
• Discharge ≈ 0.085 m³/s
• Head loss = 2 m

Q6 – Water Supply

Given:

• Population = 50,000, per capita demand = 150 L/day

Step 1 – Daily demand:$$Q_{daily} = 50,000 \cdot 150 = 7,500,000 L/day = 7500 m^3/day$$Qdaily​=50,000⋅150=7,500,000L/day=7500m3/day

Step 2 – Diameter of rising main:

• Assume velocity V = 1.5 m/s

$$Q = A V \Rightarrow D = \sqrt{\frac{4 Q}{\pi V}} = \sqrt{\frac{4 \cdot 0.0868}{\pi \cdot 1.5}} \approx 0.216 m \approx 220 mm$$Q=AV⇒D=πV4Q​​=π⋅1.54⋅0.0868​​≈0.216m≈220mm

Step 3 – Schematic: Layout of main, pumping station, reservoir.

✅ Answer Q6:

• Daily water demand = 7,500 m³
• Diameter of rising main ≈ 220 mm