Hydraulic Pressure on a Wall Surface

Hydraulic pressure is the pressure exerted by a liquid at rest on any surface in contact with it. For a vertical or inclined wall retaining water (or any liquid), this pressure acts normal (perpendicular) to the wall surface at every point.

Key points:

1. Variation with depth: The pressure at any point at depth (h) below the free surface is given by: [ p = \rho g h ] where (\rho) = density of liquid, (g) = acceleration due to gravity. Thus pressure is zero at the free surface and increases linearly with depth.

2. Pressure diagram: For a vertical wall, the pressure distribution forms a triangle with base (\rho g H) at the bottom and apex at the free surface ((H) = total depth of liquid).

3. Total pressure (resultant force): The total hydrostatic force on the wall is the area of the pressure diagram: [ P = \frac{1}{2}\rho g H^2 \quad (\text{per unit length of wall}) ] or in general, (P = \rho g A \bar{h}), where (A) = wetted area and (\bar{h}) = depth of the centroid of the wetted area.

4. Centre of pressure: The resultant force does not act at the centroid but at the centre of pressure, which lies below the centroid. For a vertical rectangular wall: [ \bar{h}_{cp} = \frac{2H}{3} \text{ from the free surface} ]

5. Effect of wall inclination: For an inclined wall, the pressure intensity still depends only on vertical depth, but the total force acts normal to the inclined surface; its magnitude remains (P = \rho g A \bar{h}).

Significance in engineering: This pressure is the basis for designing retaining walls, dams, water tanks, and basement walls — the wall must resist both the total thrust and the overturning moment caused by it.

Considerations in Storing and Transporting Cement

Cement is a hygroscopic material — it absorbs moisture from the air and sets hard, losing its strength. Therefore, proper care during storage and transport is essential to preserve its quality.

A. Considerations During Storage

1. Dry and weather-proof godown: Cement must be stored in a dry, leak-proof godown with a damp-proof floor (raised platform or wooden planks about 150–200 mm above the floor) and walls not directly exposed to moisture.

2. Away from walls: Bags should be kept at least 150–300 mm away from the walls to avoid moisture absorption from the walls.

3. Stacking height: Bags should be stacked in stacks of not more than 10–12 bags high (about 3 m) to prevent lumping due to pressure. For long storage, the height should be reduced to 8 bags.

4. Proper stacking arrangement: Bags should be placed close together to reduce air circulation, and the stack should be arranged so that the oldest stock is used first (First In First Out principle).

5. Avoiding floor contact: Bags should never be placed directly on the floor; they should be placed on wooden planks or a raised platform.

6. Ventilation: The godown should have adequate ventilation to keep the air dry, but it should not allow damp air to enter.

7. Bulk storage: For bulk storage in silos, the silo must be airtight and moisture-proof, and cement should be used within the recommended period.

8. Storage period: Cement should not be stored for more than 3 months in ordinary conditions. After 3 months, its strength reduces by about 20%; hence it should be re-tested before use.

B. Considerations During Transport

1. Water-proof covering: Cement bags must be covered with tarpaulin or polythene sheets during transport to protect them from rain and moisture.

2. Avoiding damage to bags: Bags should be handled carefully — not thrown, dropped, or dragged — to prevent tearing and spilling.

3. Proper loading: Bags should be stacked uniformly and not overloaded; the load should be secured to prevent shifting during movement.

4. Avoiding contact with wet surfaces: Cement should not be placed directly on the wet floor of a truck or wagon; a dry platform or tarpaulin should be laid first.

5. Quick transport: Transport should be completed as quickly as possible, and cement should be unloaded immediately on arrival to avoid exposure to moisture.

6. Separation from other materials: Cement should be kept separate from wet materials like sand, aggregates, or chemicals during transport.

Conclusion

Cement loses its binding property on contact with moisture. Hence, storing it in a dry, raised, well-ventilated place and transporting it with proper water-proof covering and careful handling preserves its strength and ensures quality construction.

Cantilever Beam

A cantilever beam is a beam that is fixed (built-in) at one end and free at the other end, with no support at the free end. The fixed end is called the fixed support or support, and the free end is called the free end or tip.

Key features:

• It carries load over its span and transfers the entire load to the fixed support.
• The fixed support provides three reactions: vertical reaction, horizontal reaction (if axial load exists), and a fixing moment.
• The slope and deflection at the fixed end are zero; maximum slope and maximum deflection occur at the free end.
• The bending moment is maximum at the fixed end and zero at the free end (for a point load at the tip).
• The shear force is constant throughout the span for a point load at the tip.

Common examples: balcony slab, canopy over a door, projecting slab of a building, diving board, and bracket fixed to a wall.

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Factors to be Considered in Choosing Cantilever Span (Interval)

The cantilever span (projecting length) should be selected carefully because the bending moment and deflection increase rapidly with span. The main factors are:

1. Bending Moment and Strength of Section The maximum bending moment at the fixed end is $M = \frac{wL^2}{2} \quad \text{(UDL)} \qquad M = WL \quad \text{(point load at tip)}$ Since (M \propto L^2), a small increase in span greatly increases the moment. The section must have sufficient modulus of section (Z = M/\sigma) to keep bending stress within permissible limits. Hence the span is limited by the available depth of the section.

2. Deflection (Serviceability) Deflection at the free end is $\delta = \frac{WL^3}{3EI} \quad \text{(point load)} \qquad \delta = \frac{wL^4}{8EI} \quad \text{(UDL)}$ Deflection varies as (L^3) or (L^4), so the span must be limited so that (\delta) remains within the permissible limit (usually (L/250) to (L/350) for such members). This is often the governing factor for cantilevers.

3. Depth and Stiffness of the Section A cantilever needs greater depth than a simply supported beam of the same span, because the whole moment is concentrated at one end. If adequate depth cannot be provided, the span must be reduced. Stiffness (EI) must be sufficient to control deflection.

4. Shear Force and Shear Capacity Shear force (V = wL) or (W) increases linearly with span. The section must have enough shear capacity and, in RCC, adequate shear reinforcement near the support.

5. Stability and Overturning A cantilever produces a hogging moment at the support, which tends to overturn or lift the structure. The back span or counterweight (anchorage) must be long enough to resist this. The span is therefore limited by the stability of the supporting structure.

6. Type and Magnitude of Load Heavier live load, wind load, or snow load increases both moment and deflection, so a shorter span is chosen for heavy loading. Load position also matters — a load at the free end is more critical than the same load near the support.

7. Material and Permissible Stresses Steel, RCC, and timber have different permissible stresses and elastic moduli. A material with higher strength and stiffness permits a longer cantilever span.

8. Construction and Practical Considerations Availability of anchorage length, reinforcement detailing at the support, crack control in RCC (hogging cracks at the top), and architectural requirements also decide the practical span. As a rule of thumb, the cantilever span is kept to about 1/8 to 1/12 of the back span in continuous construction.

Conclusion: The cantilever interval should be chosen so that the section satisfies strength (bending and shear), deflection (serviceability), and stability (overturning) requirements, with deflection usually governing the design.

Difference Between One-Way Slab and Two-Way Slab

Note: The deciding factor is the aspect ratio $L_y/L_x$ (long span / short span). If $L_y/L_x \geq 2$, it is a one-way slab; if $L_y/L_x < 2$, it is a two-way slab.

1. Drawing HGL and EGL (2 marks)

Energy Grade Line (EGL) is the line joining the total energy head at every section:

$EGL = z + \frac{p}{\gamma} + \frac{V^2}{2g}$

Hydraulic Grade Line (HGL) is the line joining the piezometric head (pressure head + datum head):

$HGL = z + \frac{p}{\gamma}$

Key rules for drawing:

• EGL is always above HGL by the velocity head $\dfrac{V^2}{2g}$.
• EGL always slopes downward in the direction of flow (energy is lost due to friction).
• HGL and EGL are parallel when the pipe diameter (and hence velocity) is constant.
• At the free discharge end (open to atmosphere), $p = 0$, so HGL coincides with the pipe centreline (datum) and EGL lies $\dfrac{V^2}{2g}$ above it.
• At a sudden enlargement, EGL and HGL drop sharply (loss of energy); at a sudden contraction, they also fall.
• HGL may fall below the pipe centreline where pressure becomes negative (sub-atmospheric).

Sketch description:

        EGL  ________________________
             \  (friction slope)      \___
        HGL   \___________________________  <- V²/2g gap
   ─────────────────────────────────────────  pipe centreline
        [1]        [2]         [3]      [4]

• Section 1–2: uniform pipe → EGL and HGL parallel, both sloping down.
• Section 2–3: sudden enlargement → sharp drop in both lines.
• Section 3–4: pipe ends in free discharge → HGL meets centreline, EGL above it by $\dfrac{V^2}{2g}$.

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2. Difference between Pipe Flow and Open Channel Flow (2 marks)

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3. Conclusion (1 mark)

In pipe flow the conduit runs full and flow is governed by pressure difference, whereas in open channel flow a free surface exists and flow is governed by gravity. The EGL represents total energy and always lies above the HGL by the velocity head; both lines slope downward in the direction of flow due to friction losses.

Given:
Simply supported beam AB of span $L$ m carrying uniformly distributed load $W$ kN/m over the whole span.

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1. Support reactions

Let reactions at A and B be $R_A$ and $R_B$.

Total load $= W L$ kN, acting at midspan.

By symmetry,

$$R_A = R_B = \frac{W L}{2}$$

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2. Shear force calculation

Take a section at distance $x$ from A.

$$V_x = R_A - W x$$ $$V_x = \frac{W L}{2} - W x$$

At A, $x=0$:

$$V_A = \frac{W L}{2} \text{ kN}$$

At B, $x=L$:

$$V_B = \frac{W L}{2} - W L = -\frac{W L}{2} \text{ kN}$$

Shear force is zero at:

$$\frac{W L}{2} - W x = 0$$ $$x = \frac{L}{2}$$

So SFD is a straight line varying from $+\frac{W L}{2}$ at A to $-\frac{W L}{2}$ at B, crossing zero at midspan.

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3. Bending moment calculation

Bending moment at distance $x$ from A:

$$M_x = R_A x - W x \cdot \frac{x}{2}$$ $$M_x = \frac{W L x}{2} - \frac{W x^2}{2}$$

At A, $x=0$:

$$M_A = 0$$

At B, $x=L$:

$$M_B = \frac{W L^2}{2} - \frac{W L^2}{2} = 0$$

Maximum bending moment occurs where shear force is zero, i.e. at $x=\frac{L}{2}$.

$$M_{\max} = \frac{W L}{2}\cdot \frac{L}{2} - \frac{W}{2}\left(\frac{L}{2}\right)^2$$ $$M_{\max} = \frac{W L^2}{4} - \frac{W L^2}{8}$$ $$M_{\max} = \frac{W L^2}{8} \text{ kN·m}$$

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4. Results at both ends and midspan

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5. Shear force diagram (SFD)

Shear force (kN)
+WL/2 |\
      | \
      |  \
      |   \
     0|----\-------- x
      |     \
      |      \
-WL/2 |       \
      A   L/2   B

SFD is linear, sloping downward from $+\frac{W L}{2}$ at A to $-\frac{W L}{2}$ at B.

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6. Bending moment diagram (BMD)

Bending moment (kN·m)
WL²/8 |      ___
      |    /   \
      |   /     \
      |  /       \
     0|_/_________\___ x
      A    L/2    B

BMD is parabolic, zero at both supports A and B, and maximum $\frac{W L^2}{8}$ at midspan.

Purpose of Levelling in Survey

Levelling is the branch of surveying used to determine the relative vertical heights (elevations) of points on or below the earth's surface. Its main purposes are:

1. To find the elevation (R.L.) of points with respect to a datum, so that a vertical profile of the ground can be drawn.
2. To determine the difference in elevation between two or more points — needed for roads, canals, railways, water supply and drainage works.
3. To prepare contour maps and longitudinal/cross sections for design and earthwork calculation.
4. To set out works at a required gradient or depth, e.g. laying sewers, canal beds, foundations and benchmarks.
5. To establish permanent benchmarks (B.M.) for future reference.

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Methods of Levelling

1. Simple Levelling

Used when the two points are close together and the instrument can be set up between them in one setting. The difference in elevation is obtained from a single back sight and a single fore sight.

        Line of collimation
   ────────────────────────────────
        BS ↑              ↓ FS
             |          |
   ───A──────┼──────────┼────B───
      (known R.L.)   (R.L. to find)
             [Level]

$\text{Difference in elevation} = BS - FS$

2. Differential Levelling (Compound Levelling)

Used when the points are far apart, the difference in elevation is large, or obstacles lie between them. The work is done in several instrument settings, using change points (C.P.) and taking flying levels.

   ──────────────────────────────────────────────
    BS1 ↓FS1  BS2 ↓FS2  BS3 ↓FS3
   ──A──C.P.1──C.P.2──C.P.3──B──

$\Sigma BS - \Sigma FS = \text{R.L. of last point} - \text{R.L. of first point}$

3. Fly Levelling

A rapid form of differential levelling in which only back sights and fore sights are taken (no intermediate sights). It is used to connect a working site to a nearby benchmark or to carry a level line quickly over a long distance.

4. Profile (Longitudinal) Levelling

Levelling carried out along the centre line of a proposed road, canal or railway. Intermediate sights are taken at regular intervals to plot the longitudinal section and fix formation levels and gradients.

5. Cross-Section Levelling

Levelling done at right angles to the centre line at each chainage, to obtain the cross-section of the ground. It is essential for earthwork computation in roads, canals and embankments.

6. Reciprocal Levelling

Used to find the true difference in elevation across a wide river or ravine where the instrument cannot be set up between the points. Two sets of observations are taken — one from each bank — and the mean eliminates the errors of curvature, refraction and collimation.

   Station A (near bank)          Station B (far bank)
        BS1  ↓                        ↓ FS1
   ────────────────  River  ────────────────
        FS2  ↑                        ↑ BS2

$h = \frac{(h_1 + h_2)}{2}$

7. Precise Levelling

A very accurate levelling done with a precise level and invar staff, using the three-wire method. It is used for national levelling networks, geodetic surveys and establishing fundamental benchmarks.

8. Contouring

Levelling in which contours (lines of equal elevation) are located and plotted, either by the direct method (points of same R.L. traced on ground) or the indirect method (grid, radial or cross-section method). Used for reservoir, dam and township surveys.

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Note: The choice of method depends on the distance between points, required accuracy, terrain and purpose of the survey.

Relation between Dry Density and Moisture Content of Soil

For a given soil compacted with a given compactive effort, the dry density (γ_d) first increases as moisture content (w) increases, reaches a maximum value (Maximum Dry Density, MDD) at a particular moisture content called the Optimum Moisture Content (OMC), and then decreases with further increase in moisture content. This curve is called the compaction curve (Proctor curve).

Reason:

• At low moisture content, water acts as a lubricant; soil particles slide over each other and pack more closely → dry density increases.
• At OMC, lubrication is optimum and air voids are minimum → dry density is maximum.
• Beyond OMC, excess water occupies space and pushes particles apart; the soil becomes nearly saturated and cannot be compressed further → dry density decreases.

The relation is expressed as:

$\gamma_d = \frac{\gamma}{1 + w}$

where γ = bulk density of soil, w = moisture content (in decimal).

Laboratory Determination of Maximum Dry Density

The MDD and OMC are determined in the laboratory by the Standard Proctor Compaction Test (IS: 2720, Part 7) or the Modified Proctor Test (IS: 2720, Part 8).

Standard Proctor Test — Procedure

1. Apparatus: Cylindrical mould of 1000 ml capacity (internal diameter 100 mm, height 127.3 mm), detachable base plate and collar; rammer of 2.6 kg weight with 50 mm diameter face; oven; weighing balance; mixing tray; straight edge.

2. Preparation of sample: Take about 3 kg of air-dried soil passing the 4.75 mm IS sieve. Mix thoroughly with a known quantity of water (say 4% by weight) to obtain the first moisture content.

3. Filling the mould: Fix the mould to the base, attach the collar. Place the moist soil in the mould in three equal layers. Compact each layer with 25 blows of the 2.6 kg rammer, dropped freely from a height of 310 mm, distributing the blows uniformly over the surface.

4. Trimming: Remove the collar, strike off the excess soil level with the top of the mould using a straight edge. Clean the outside of the mould and weigh it with the soil (W₂).

5. Moisture content: Take a representative sample from the compacted soil, weigh it, oven-dry at 105–110 °C for 24 hours, and determine the moisture content w.

6. Bulk and dry density:

   • Bulk density, γ = (W₂ − W₁) / V, where W₁ = weight of empty mould, V = volume of mould.
   • Dry density, γ_d = γ / (1 + w).

7. Repetition: Repeat the test with the same soil by adding water in steps of about 2% (usually 5 to 6 trials), each time using fresh soil. Plot the compaction curve.

Compaction Curve

 γ_d
 (kN/m³)│
        │            ●  ← MDD
        │          ●   ●
        │        ●       ●
        │      ●           ●
        │    ●               ●
        │  ●                   ●
        └──┴────┴────┴────┴────┴────┴──► w (%)
              OMC

The peak of the curve gives the Maximum Dry Density (MDD) and the corresponding moisture content gives the Optimum Moisture Content (OMC).

Zero Air Void Line

The zero air void line (S = 100%) is drawn using:

$\gamma_d = \frac{G \cdot \gamma_w}{1 + wG}$

The compaction curve always lies below this line, since some air always remains trapped in the soil.

Typical Values

Note: The MDD obtained depends on the compactive effort — higher effort (Modified Proctor) gives higher MDD and lower OMC.

As per the Nepal Road Standard (2070), roads in Nepal are classified by function and administrative importance into the following classes:

National highways and feeder roads together form the Strategic Road Network (SRN), which is the responsibility of the Government of Nepal, while district, urban and village roads are the responsibility of local bodies.

I'll solve this question myself first, then compare with the model answer.

My solution:

Purposes of a septic tank:

1. To settle suspended solids (forming sludge) and float light matter (forming scum) due to adequate detention period.
2. To anaerobically digest the settled sludge, reducing its volume and making it stable.
3. To store sludge and scum for long intervals (1–3 years) before removal.
4. To partially purify sewage so effluent can be disposed of safely (soak pit/drain).
5. To protect public health by preventing pollution of soil, groundwater, and water sources.

Main components:

• Inlet pipe (with T/baffle)
• Settling/detention chamber (three zones: scum, clarified liquid, sludge)
• Scum layer
• Sludge layer
• Outlet pipe (T-shaped)
• Manhole/inspection cover
• Ventilation pipe
• (Often also: partition walls/baffles in multi-chamber tanks)

Comparison with model answer:

The model answer's purposes are correct and complete. The components list is correct. However, there are issues:

1. The schematic diagram is poorly drawn — it doesn't clearly show the T-shaped inlet and outlet pipes, the baffle walls, or the relative positions properly. A proper schematic should show inlet T-pipe, outlet T-pipe, scum layer, sludge layer, and clarified liquid zone distinctly.

2. The diagram lacks proper labeling of the T-pipes at inlet and outlet, which is a key feature of septic tank design.

3. For a 5-mark question (1+4), the answer is somewhat verbose but acceptable.

The diagram is the main weakness — it's more of a block representation than a proper schematic. Let me provide a corrected version with a proper schematic.

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Purposes of a Septic Tank

A septic tank is the first treatment unit of an on-site (household) sewage disposal system. Its main purposes are:

1. Settlement of solids — It retains sewage for a sufficient detention period (usually 24 hours) so that heavy suspended solids settle to the bottom as sludge and light matter (oil, grease, soap) floats to the top as scum.
2. Anaerobic digestion of sludge — The settled sludge is decomposed by anaerobic bacteria in the absence of oxygen, reducing its volume and making it more stable (less offensive) before disposal.
3. Storage of sludge and scum — It provides storage space for accumulated sludge and scum so they need to be removed only at long intervals (generally 1–3 years).
4. Partial purification of sewage — The clarified effluent is considerably reduced in suspended solids and organic load (about 60–70% removal of suspended solids and 30–40% of BOD), so it can be safely discharged into a soak pit, drain, or secondary treatment unit.
5. Protection of public health and environment — By preventing direct discharge of raw sewage, it controls foul odour, flies, and pollution of soil, groundwater, and nearby water sources.

Main Components of a Septic Tank (Schematic Diagram)

        House drain
             |
             v
      +------+------+
      |   Inlet     |
      |  T-pipe     |
      |    |        |
  +---+----+--------+-----------------------------------+
  |   |             |                                   |
  |   |   SCUM  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~    |<-- Manhole cover
  |   |             |                                   |    (for desludging)
  |   |             |                                   |
  |   |      CLARIFIED LIQUID (settling/detention zone) |
  |   |             |                                   |
  |   |             |                                   |
  |   |   .......................................       |
  |   |          SLUDGE (settled at bottom)             |
  +---+-------------+-----------------------------------+
                    |
                    v
              Outlet T-pipe
                    |
                    v
         (to soak pit / drain / secondary unit)

Main components:

Note: A septic tank is usually a rectangular, watertight, underground RCC or brick masonry tank with a minimum capacity of about 1 m³ for a small house, and a detention period of 24 hours.

Why a tender notice is published

A tender notice is published to invite sealed or electronic bids from eligible contractors, suppliers or firms for the execution of a specified work or supply, so that the work is awarded through open and fair competition. Its main purposes are:

1. To give wide publicity to the work so that the maximum number of capable bidders can participate.
2. To ensure transparency, impartiality and competitiveness in the selection of the contractor, avoiding favouritism and corruption.
3. To obtain the lowest reasonable (most economical) rate for the work within the estimated cost.
4. To inform all prospective bidders of the same conditions, so that every bidder competes on an equal footing.
5. To fulfil the legal requirement of the Public Procurement Act/Rules, under which procurement above a certain threshold must be done through open competitive bidding.

Information needed in a tender notice

A standard tender notice must contain the following particulars:

1. Name and address of the employer/office issuing the notice (e.g., Division Office, Rural Municipality) and the name of the project/work.
2. Nature and description of the work or supply, with its location.
3. Estimated cost/amount of the work or the approximate quantity of goods.
4. Source of fund and the fiscal year in which the work is to be carried out.
5. Bid security (earnest money) amount and the form in which it is to be deposited.
6. Cost of the bidding document (non-refundable) and the place from which it can be obtained.
7. Eligibility criteria of the bidders — valid licence, registration, VAT/PAN certificate, experience, class of contractor, etc.
8. Date, time and place of submission of the bid (deadline) and the date, time and place of opening of the bids.
9. Validity period of the bid and the amount of performance security required from the successful bidder.
10. Bid security validity period, method of submission (sealed/e-procurement portal) and the address for communication.
11. Name and designation of the contact person/officer for enquiries, along with telephone/fax/e-mail.
12. Reservation/quota provisions, if any, and the legal basis (Procurement Act/Rules) under which the notice is issued.

The notice is normally published in a national daily newspaper, the employer's website and the Public Procurement Monitoring Office (PPMO) portal so that it reaches all interested bidders.

Purpose of technical specification in a contract

A technical specification is the written description of the materials, workmanship, and methods of construction required for a work. In a contract it serves the following purposes:

1. Defines quality and standard — It lays down the grade, quality and standard of materials (e.g. OPC 43 grade cement, Fe 500 steel) and workmanship to be used, so the contractor knows exactly what is expected.

2. Forms the basis of the contract — Together with the drawings, bill of quantities and conditions of contract, it is a contract document. It binds the contractor legally to execute the work as specified.

3. Basis of tendering and rate — Tenderers price their rates on the basis of the specification, so all bidders quote on the same standard and the tenders become comparable.

4. Basis of supervision and quality control — The engineer uses it to inspect, supervise and accept or reject the work, and to order the contractor to remove or redo defective work.

5. Settlement of disputes — In case of dispute about quality or method of work, the specification is the reference document for deciding the matter.

6. Guide to the contractor — It tells the contractor the sequence of work, method of execution, testing requirements and precautions to be taken, reducing chances of faulty construction.

7. Basis of measurement and payment — It fixes what is included in the item rate (e.g. curing, scaffolding, wastage) and thus helps in correct measurement and payment.

In short, the technical specification ensures uniformity, quality, economy and legal clarity in the execution of a contract work.

Advantages of Weir and Barrage in Irrigation Structures

1. Advantages of a Weir

A weir is a low, solid masonry or concrete barrier constructed across a river to raise the water level and divert it into an irrigation canal.

• Simple and economical construction: The design is straightforward and requires less precision in construction compared to a barrage, so the initial cost is relatively low.
• Low maintenance cost: Since there are few moving parts (only the gate arrangement), the maintenance and operation costs are low.
• Stable and durable: The solid concrete/masonry structure is heavy and stable, and can withstand high flood flows without much risk of failure.
• Effective for small to medium rivers: It is well suited where the river bed is firm (rocky or hard soil) and the discharge is moderate.
• Raises water level reliably: It creates a dependable head for gravity flow into canals, ensuring continuous water supply for irrigation.
• No need for complex gate operation: Because the crest is fixed, operation is simple and does not require skilled labour.
• Less silt entry into canal: The raised water level and the crest help in controlling the entry of coarse silt to some extent.

2. Advantages of a Barrage

A barrage is a structure with a low crest and a series of large gates that can be opened or closed to control the water level and flow.

• Better control of water level: The gates allow the water level to be raised or lowered as required, giving flexible control over the head available for irrigation.
• Effective silt control: By manipulating the gates, the flow can be regulated to pass silt downstream and reduce silt deposition in the canal (silt excluder works effectively).
• Suitable for large rivers and flood flows: During floods, the gates can be fully opened to pass the flood discharge safely without raising the upstream water level excessively.
• Higher diversion capacity: A barrage can divert a larger quantity of water for irrigation than a weir of the same width.
• Reduced afflux and flooding risk: Because the gates can be opened, the afflux (rise in upstream water level) is small, reducing the risk of flooding of adjoining land.
• Economical for wide rivers with soft beds: Where the river is wide and the bed is alluvial, a barrage is more suitable and safer than a weir.
• Allows better regulation of canal supply: The canal head regulator can be operated in combination with the barrage gates to supply the exact required discharge.
• Facilitates navigation and flushing: Gate operation helps in flushing sediment and can assist in maintaining a navigable depth.

In short: A weir is cheap, simple and durable, best for firm beds and moderate flows; a barrage gives superior control over water level and silt, and is safer and more efficient for large rivers and flood conditions.

Curing is the process of keeping concrete moist and at a favourable temperature after placing, so that the cement continues to hydrate and the concrete attains its design strength. The main methods are:

1. Water-adding (wet) methods — water is kept in contact with the concrete surface:

• Ponding / immersion: Shallow ponds or bunds are formed on horizontal surfaces (slabs, floors, pavements) and filled with water. Cheap and very effective, but only for flat, horizontal work.
• Sprinkling: Water is sprayed or sprinkled continuously over the surface from a hose or perforated pipe. Used for vertical and inclined surfaces; needs constant supervision.
• Wet covering: Surfaces are covered with wet gunny bags, hessian cloth, canvas, sand, or sawdust kept continuously wet. Common for columns, beams, and walls.
• Steam curing: Steam is applied in a chamber or under a cover at about 60–80 °C. Gives rapid strength gain; used in precast concrete works.

2. Water-retaining (membrane) methods — moisture already in the concrete is prevented from escaping:

• Impermeable membrane: The surface is covered with polythene sheets, waterproof paper, or plastic film.
• Curing compound: A liquid resin, wax, or bituminous compound is sprayed on the surface to form a sealing film. Useful where water is scarce or for vertical/inaccessible surfaces.

3. Other methods:

• Formwork left in place: Timber or steel moulds are kept on and water is applied at the joints; the formwork itself retains moisture.
• Covering with earth or sand: Excavated surfaces or floors are covered with a layer of moist earth or sand.

General points: Curing should start as soon as the concrete is hard enough not to be damaged (about 12–24 hours after placing). As per IS 456, it should continue for at least 7 days for ordinary Portland cement concrete, at least 10 days for concrete with mineral admixtures or Portland pozzolana/blended cements, and at least 14 days in hot or dry weather (where the 7/10-day periods are extended). Water used for curing should be clean and free from harmful salts, acids, and organic matter.

Estimating the Quantity of Sanitary Sewage

1. Introduction

Sanitary sewage is the liquid waste originating from domestic, commercial, and industrial uses of water, carried through separate sanitary sewers. For the design of a sewerage system (sewers, pumping stations, treatment units), the quantity of sanitary sewage to be handled must be estimated accurately. The design is normally based on the dry weather flow (DWF), which represents the maximum flow during non-monsoon periods when infiltration is minimum.

2. Basic Principle

The quantity of sanitary sewage is estimated on the assumption that the entire water supplied to a community ultimately appears as sewage, with a small allowance for losses. Hence:

$Q_{\text{sewage}} = k \times Q_{\text{water supply}}$

where $k$ is the coefficient of utilization (fraction of water supply reaching the sewer). For a well-designed water supply system, $k$ is taken as 0.8 (i.e., about 80% of water supplied becomes sewage).

3. Steps in Estimating Sanitary Sewage Quantity

Step 1: Estimate the design population

The population at the end of the design period (usually 30 years for sewers) is forecast using the arithmetic increase, geometric increase, or incremental increase method, based on census data.

Step 2: Adopt the rate of water supply

The per capita water demand is taken as per the prevailing standard (e.g., 135 lpcd for a medium town in Nepal; 100–200 lpcd depending on the community).

Step 3: Compute average daily sewage flow

$Q_{\text{avg}} = k \times q \times P$

where $q$ = per capita water demand (lpcd), $P$ = design population, $k$ = 0.8.

Step 3a: Add industrial and commercial sewage

Where industries exist, their wastewater contribution is added separately, as it may be large and variable.

Step 4: Add infiltration and inflow

• Infiltration — groundwater entering through joints, cracks, and defective pipes. Taken as 2,500 to 5,000 litres/day/km of sewer length, or as a percentage of average flow.
• Inflow — storm water entering through manholes, roof connections, etc. (usually excluded in a separate system).

Step 5: Determine peak (maximum) flow

Sewers must be designed for the peak flow, not the average. The peak factor is obtained from Harmon's formula:

$M = 1 + \frac{14}{4 + \sqrt{P}}$

where $M$ = peak factor and $P$ = population in thousands. Then:

$Q_{\text{peak}} = M \times Q_{\text{avg}}$

Step 6: Check minimum flow (self-cleansing velocity)

The minimum flow must produce a self-cleansing velocity of about 0.75 m/s (or a minimum tractive force of 1 Pa) to prevent silt deposition. The sewer is designed for the peak flow but checked for the minimum flow.

4. Worked Example

Given: Design population $P = 50{,}000$; water supply rate $q = 135$ lpcd; $k = 0.8$; infiltration = 3,000 litres/day/km over 5 km of sewer.

Average sewage flow: $Q_{\text{avg}} = 0.8 \times 135 \times 50{,}000 = 5{,}400{,}000 \text{ litres/day} = 5.4 \text{ MLD}$

Infiltration: $= 3{,}000 \times 5 = 15{,}000 \text{ litres/day} = 0.015 \text{ MLD}$

Total average dry weather flow: $= 5.4 + 0.015 = 5.415 \text{ MLD} = 62.7 \text{ litres/s}$

Peak factor (Harmon's formula), $P = 50$ (thousands): $M = 1 + \frac{14}{4 + \sqrt{50}} = 1 + \frac{14}{11.07} = 2.26$

Peak flow: $Q_{\text{peak}} = 2.26 \times 62.7 = 141.7 \text{ litres/s} \approx 142 \text{ litres/s}$

The sewer is designed to carry 142 litres/s, while the minimum flow is checked for self-cleansing velocity.

5. Factors Affecting the Quantity of Sanitary Sewage

1. Rate of water supply — directly proportional to sewage flow.
2. Population and its growth pattern — larger and faster-growing populations give higher flows.
3. Type of area — residential, commercial, or industrial.
4. Standard of living — higher living standards increase water use and hence sewage.
5. Groundwater infiltration — depends on soil type, water table, and sewer condition.
6. Meteorological conditions — rainfall and temperature influence inflow and water use.
7. Design period — longer periods require larger forecast populations.

6. Conclusion

The quantity of sanitary sewage is estimated by taking the design population, adopting the per capita water demand, applying the coefficient of utilization (0.8), adding infiltration and industrial wastes, and then multiplying the average dry weather flow by the peak factor obtained from Harmon's formula. The sewer is then designed for the peak flow and checked for the minimum self-cleansing velocity, ensuring both adequate capacity and proper functioning of the sewerage system.

1. Sight Distance (2 marks)

Sight distance is the length of carriageway visible to a driver ahead, measured along the road, at any given instant. It is the clear, unobstructed distance over which the driver can see the road surface and any object, hazard or opposing vehicle, so that safe driving decisions can be made. It is a fundamental geometric design control because the horizontal and vertical curves, crests, and set-back distances are all designed to guarantee the required sight distance.

2. Factors Causing Restrictions to Sight Distance (3 marks)

1. Horizontal curves — the inner edge (cutting, hill slope, building, tree, wall) obstructs the line of sight across the curve.
2. Vertical curves (crests) — the road surface itself blocks the view beyond the summit, so the driver cannot see an object or vehicle on the far side.
3. Roadside obstructions — trees, hedges, signboards, parked vehicles, buildings, and utility poles close to the carriageway.
4. Terrain and embankment — steep hillsides, deep cuttings, and high embankments limit the view.
5. Weather and atmospheric conditions — fog, rain, snow, dust, and smoke reduce visibility.
6. Night conditions — inadequate or poorly placed street lighting restricts the visible distance.
7. Traffic conditions — heavy traffic, especially large vehicles, block the view of the road ahead.
8. Geometric features — narrow carriageway width, steep gradient, and inadequate superelevation/transition affect the sight line.

3. Significance of Stopping, Intermediate and Overtaking Sight Distance (5 marks)

(a) Stopping Sight Distance (SSD)

It is the minimum distance required for a vehicle travelling at design speed to stop safely before a stationary object on the road, after the driver perceives the hazard and applies the brakes.

$SSD = \text{Reaction distance} + \text{Braking distance} = 0.278\,V\,t + \frac{V^2}{254(f \pm g)}$

where $V$ = design speed (km/h), $t$ = reaction time (≈2.5 s), $f$ = coefficient of longitudinal friction, $g$ = gradient.

Significance:

• It is the minimum sight distance that must be available at every point of the road, including horizontal and vertical curves.
• It governs the design of crest vertical curves and the set-back distance of obstructions on the inner side of horizontal curves.
• It ensures the driver can stop before hitting a sudden obstruction, a pedestrian, or a stopped vehicle.
• It is the basis for safe sign placement, speed limits, and warning markings.

(b) Intermediate Sight Distance (ISD)

It is the sight distance provided when the availability of overtaking sight distance is not possible but SSD is insufficient for safe and comfortable operation; it is taken as twice the stopping sight distance.

$ISD = 2 \times SSD$

Significance:

• It is adopted on undivided roads with two-way traffic where full OSD cannot be economically provided.
• It gives the driver enough visibility to perceive an oncoming vehicle and return to the original lane safely after a failed overtaking attempt.
• It is a compromise design value between SSD and OSD, balancing safety with construction cost.
• It is used for the design of vertical curves on hilly and mountainous roads.

(c) Overtaking Sight Distance (OSD)

It is the minimum distance along the road required for a vehicle to overtake a slower vehicle safely, without interfering with the speed of an oncoming vehicle.

$OSD = d_1 + d_2 + d_3$

where $d_1$ = distance travelled during reaction and acceleration to overtaking position, $d_2$ = distance travelled by the overtaking vehicle alongside the overtaken vehicle, $d_3$ = distance travelled by the oncoming vehicle during the overtaking manoeuvre.

Significance:

• It is required on two-lane and multi-lane highways to permit safe overtaking.
• It governs the design of horizontal curves, crest curves, and no-overtaking zones.
• It determines the minimum length of overtaking sections and the placement of overtaking sight distance signs.
• It directly affects traffic capacity, level of service, and road safety on two-way roads.
• Where OSD cannot be provided, intermediate sight distance is adopted as a design compromise.

Permeability of Soil — Definition, Darcy's Law and Determination of Coefficient of Permeability

1. Definition

Permeability is the property of a soil which permits the flow of water through its interconnected voids (pore spaces). It is quantitatively expressed by the coefficient of permeability (k), defined as the average velocity of flow of water through a soil mass under a unit hydraulic gradient, at standard temperature (usually 20 °C).

$k = \frac{v}{i} = \frac{v}{h/L}$

where $v$ = discharge velocity, $i$ = hydraulic gradient, $h$ = head causing flow, $L$ = length of soil specimen.

Unit: cm/s (or m/s). Typical values: clean gravel $10^{1}$–$10^{2}$ cm/s; clean sand $10^{-1}$–$10^{0}$; silty sand $10^{-3}$; silt $10^{-5}$; clay $10^{-7}$ cm/s or less.

2. Darcy's Law

For laminar flow through saturated soil, the discharge velocity is directly proportional to the hydraulic gradient:

$v = k\,i \quad\Rightarrow\quad Q = k\,i\,A = k\,\frac{h}{L}\,A$

where $Q$ = discharge (cm³/s), $A$ = total cross-sectional area of soil (cm²).

Sketch — flow through soil under head:

   ┌──────────────┐  ← water level (head h1)
   │   water      │
   ├──────────────┤  h1
   │   soil       │  h = h1 − h2
   │  specimen    │  (head causing flow)
   ├──────────────┤  h2
   │   water      │
   └──────────────┘  ← outlet
        L  (length of specimen)

3. Methods of Determining the Coefficient of Permeability

(a) Laboratory Methods

(i) Constant Head Permeameter — for coarse-grained soils (sand, gravel), where $k > 10^{-3}$ cm/s.

Sketch:

   ┌───────────────┐
   │  constant     │  ← water level kept constant
   │  head h        │
   ├───────────────┤
   │  soil sample  │  A = area of specimen
   │  (sand/gravel)│  L = length
   ├───────────────┤
   │  porous disc  │
   └───────┬───────┘
           │  Q collected in measuring jar
           ▼   in time t

Formula: $k = \frac{Q\,L}{A\,h\,t}$

where $Q$ = volume of water collected in time $t$, $h$ = constant head difference.

(ii) Falling Head Permeameter — for fine-grained soils (silt, clay), where $k < 10^{-3}$ cm/s.

Sketch:

   ┌───────────────┐
   │  stand pipe   │  a = area of stand pipe
   │  (water falls)│
   │  h1 → h2      │
   ├───────────────┤
   │  soil sample  │  A = area of specimen
   │  (silt/clay)  │  L = length
   ├───────────────┤
   │  porous disc  │
   └───────┬───────┘
           ▼  outflow

Formula: $k = 2.303\,\frac{a\,L}{A\,t}\,\log_{10}\!\left(\frac{h_1}{h_2}\right)$

where $a$ = area of stand pipe, $h_1$ and $h_2$ = head at times $t_1$ and $t_2$, $t = t_2 - t_1$.

(b) Field Methods

(i) Pumping Test (Well Pumping Test) — most reliable for field conditions; a test well is pumped at constant rate and observation wells record the drawdown.

• Unconfined aquifer: $k = \frac{2.303\,Q\,\log_{10}(r_2/r_1)}{\pi\,(h_2^2 - h_1^2)}$
• Confined aquifer: $k = \frac{2.303\,Q\,\log_{10}(r_2/r_1)}{2\pi\,H\,(h_2 - h_1)}$

where $Q$ = pumping rate, $r_1, r_2$ = radial distances of observation wells, $h_1, h_2$ = depths of water in them, $H$ = thickness of confined aquifer.

(ii) Open End Test / Packer Test — used for rock strata; water is forced into a borehole section under pressure and $k$ is computed from the rate of absorption.

(c) Empirical Correlations (for sands)

• Hazen's formula: $k = C\,D_{10}^{2}$ (cm/s), where $D_{10}$ in cm and $C \approx 100$–150.
• Kozeny–Carman equation: relates $k$ to void ratio $e$ and specific surface of grains.

4. Note on Temperature Correction

Since viscosity changes with temperature, the value at test temperature $T$ is corrected to 20 °C:

$k_{20} = k_T \times \frac{\mu_T}{\mu_{20}}$

---

Summary: Permeability is the ease with which water flows through soil voids, governed by Darcy's law $v = ki$. It is determined in the laboratory by the constant head test (coarse soils) and falling head test (fine soils), and in the field by the pumping test (most reliable) or open end/packer test for rock.

Three strengths of a riveted lap joint

In a riveted lap joint the two plates overlap and the rivets pass through both plates. The joint can fail in three ways, so it has three strengths:

1. Strength in tearing (tension) of the plate The plate may tear along the line of rivet holes, because the holes reduce the net section.

$P_t = (p - d)\,t\,\sigma_t$

2. Strength in shearing of the rivets The rivets may shear off across the plane of contact of the two plates. In a lap joint the rivet is in single shear.

$P_s = n \times \frac{\pi}{4}d^2 \tau$

3. Strength in bearing (crushing) of the rivet or plate The rivet may crush the plate (or itself be crushed) against the hole surface.

$P_b = n \times d \times t \times \sigma_c$

where

• $p$ = pitch of rivets, $d$ = gross diameter of rivet, $t$ = thickness of plate,
• $n$ = number of rivets in one pitch length,
• $\sigma_t$ = permissible tensile stress in plate, $\tau$ = permissible shear stress in rivet, $\sigma_c$ = permissible bearing stress.

Sketches

Lap joint (plan and section)

   ┌──────────────────────────────────────┐
   │    ○         ○         ○         ○   │   ← rivets
   └──────────────────────────────────────┘
   |←  p  →|

   ────────────────────────────────┐
   Plate 1      ○    ○    ○    ○   │  t
   ────────────────────────────────┘
   ┌────────────────────────────────
   │  ○    ○    ○    ○              Plate 2
   └────────────────────────────────

Failure modes

 (a) Tearing of plate          (b) Shearing of rivet        (c) Bearing / crushing
   ─────○─────○─────             ─────┬─────┬─────             ─────[○]─────
   ─────○─────○─────             ─────┴─────┴─────             ─────[○]─────
   ↑ tear along (p−d)t           ↑ shear plane (π/4)d²         ↑ contact area d·t

Determination of design strength

The design strength of the joint is the least of the three strengths:

$P = \min\,(P_t,\;P_s,\;P_b)$

The mode giving the smallest value is the governing failure mode, and the joint is designed so that the applied load does not exceed this value. The other two modes are then automatically safe.

Efficiency of the joint

$\eta = \frac{\text{least strength of joint}}{\text{strength of solid plate}} \times 100\% = \frac{P}{p\,t\,\sigma_t}\times 100\%$

Example (brief): For $p = 60$ mm, $d = 20$ mm, $t = 10$ mm, $n = 1$, $\sigma_t = 150$ N/mm², $\tau = 100$ N/mm², $\sigma_c = 300$ N/mm²:

• $P_t = (60-20)\times 10 \times 150 = 60{,}000$ N
• $P_s = 1 \times \frac{\pi}{4}(20)^2 \times 100 = 31{,}416$ N
• $P_b = 1 \times 20 \times 10 \times 300 = 60{,}000$ N

Least value = $P_s = 31{,}416$ N → shearing governs, and the design strength of the joint is 31.42 kN.

I'll solve this question myself first, then compare with the model answer.

My solution:

The question asks for major principles and steps in the design of a T-beam bridge using Carbon's method. Carbon's method (also called the "effective width method") is specifically a method for computing the effective width of slab over which a concentrated wheel load is assumed to be distributed when analyzing the deck slab / T-beam.

Key points of Carbon's method:

• The wheel load is not assumed to be carried by the entire slab width; only an effective width of the slab acts with the beam.
• Effective width depends on: position of load (central vs. near unsupported edge), effective span, thickness of slab, and width of the web/rib.
• The load is assumed to spread at 45° through the slab thickness.

Standard formulas (IRC 21 / IRC 6):

• Load placed centrally (supported on both edges): $b_e = \dfrac{2}{3}L + b_w$ ... but actually the classic Carbon's method gives:
  • For a load at the center of the span, effective width $b_e = k \cdot x \left(1 - \dfrac{x}{L}\right) + b_w$ where $x$ is distance from the nearer support...

Let me be careful. The actual Carbon's method as given in IRC 21 (Clause 305) for effective width of slab:

For one-way slab (T-beam deck):

• Effective width $b_e = \alpha \cdot a \left(1 - \dfrac{a}{L}\right) + b_w$ where $a$ = distance of center of load from nearer support, $L$ = effective span, $b_w$ = width of rib, and $\alpha$ = coefficient depending on $b/L$ ratio (from IRC 21 Table).

Actually, IRC 21 gives:

$$b_e = \frac{2}{3} \cdot \frac{x}{L} \cdot L + b_w$$

... no.

Let me recall precisely. IRC 21-2000, Clause 305.1: Effective width for a concentrated load on a slab spanning in one direction: $b_e = \alpha \cdot a \left(1 - \frac{a}{L}\right) + b_w$ where $\alpha$ is a coefficient depending on $\frac{b}{L}$ (ratio of width of slab to effective span), given in Table 12 of IRC 21. For $\frac{b}{L} \ge 3$, $\alpha = 2.4$; for $\frac{b}{L} = 1$, $\alpha = 1.2$; etc. And $b_w$ is the width of the rib.

Hmm, but the model answer gives:

• Central load: $b_e = \frac{2}{3}L + b_w$
• Edge load: $b_e = \frac{1}{3}L + b_w$

These are actually the older/simplified formulas sometimes used (from IRC 21 old version or from textbooks like Punmia). Actually, these simplified formulas appear in some Indian textbooks for effective width:

• Load at center: $b_e = \frac{2}{3}L + b_w$ ... but this doesn't depend on position $a$.
• Actually the classic "Carbon's method" from older IRC 21 (1966) gives:
  • For load at center of span: $b_e = \frac{2}{3}L + b_w$ ... no, that's not right either.

Let me reconsider. The actual Carbon's method (as per IRC 21-1966, Clause 305) for effective width:

For a load placed at distance $a$ from the nearer support:

$$b_e = \frac{2}{3} \cdot \frac{a}{L} \cdot L + b_w$$

...

No. The correct formula from IRC 21 (both old and new) is:

$b_e = \alpha \cdot a \left(1 - \frac{a}{L}\right) + b_w$

where $\alpha$ depends on $b/L$. For $b/L \ge 3$, $\alpha = 2.4$. This gives maximum $b_e$ when $a = L/2$: $b_e = 2.4 \cdot \frac{L}{2} \cdot \frac{1}{2} + b_w = 0.6L + b_w$. Hmm, that's not $\frac{2}{3}L + b_w$.

Wait — actually for $b/L \ge 3$, $\alpha = 2.4$, so at center: $b_e = 2.4 \times \frac{L}{2} \times \frac{1}{2} + b_w = 0.6L + b_w$. But some texts give $\frac{2}{3}L + b_w$ for central load. The difference may be due to different versions.

Actually, I recall that in IRC 21-1966 (the older version), the effective width for a load at the center was given as:

$$b_e = \frac{2}{3}L + b_w$$

... no.

Let me just check: The model answer's formulas:

• Central: $b_e = \frac{2}{3}L + b_w$
• Edge: $b_e = \frac{1}{3}L + b_w$

These are actually the formulas given in many Nepali/Indian textbooks for effective width of slab in T-beam bridges (sometimes attributed to "Carbon's method" or "IRC method"). Actually, these are from the older IRC 21 or from textbooks like "Bridge Engineering" by S.P. Bindra or similar. In many sub-engineer level exams in Nepal/India, these simplified formulas are taught.

Actually, I now recall: In IRC 21-1966, Clause 305.1, the effective width for a concentrated load on a slab spanning in one direction was:

$$b_e = \frac{2}{3} \cdot \frac{a}{L} \cdot L + b_w$$

... no.

Let me just accept that the model answer's formulas are the commonly taught simplified versions for Carbon's method at this level. The key point is that the model answer is generally correct in its approach.

Now, checking the model answer for errors:

1. Effective width formulas: The model answer gives:

   • Central: $b_e = \frac{2}{3}L + b_w$ — This is a simplified formula. Actually, the correct Carbon's method formula is $b_e = \alpha \cdot a(1 - a/L) + b_w$ with $\alpha$ from table. The simplified versions $\frac{2}{3}L + b_w$ and $\frac{1}{3}L + b_w$ are sometimes given for central and edge loads respectively in older texts. However, these are not strictly "Carbon's method" as per current IRC 21. But for a 10-mark sub-engineer exam, this level of simplification is often accepted.

   Actually, I need to verify: The formula $b_e = \frac{2}{3}L + b_w$ for central load — is this correct? In IRC 21-1966, for a load at the center of the span, the effective width was given as $b_e = \frac{2}{3}L + b_w$? Let me think... No, I believe the correct formula from IRC 21 (old) for a load at center was actually $b_e = \frac{2}{3}L + b_w$ only when the load is at the center and the slab is supported on both edges. Hmm.

   Actually, I found it: In IRC 21-1966, Clause 305.1, the effective width for a concentrated load on a slab spanning in one direction is:

$$b_e = \frac{2}{3} \cdot \frac{a}{L} \cdot L + b_w$$

... no.

Let me just go with what I know: The standard Carbon's method formula (as per IRC 21-2000) is: $b_e = \alpha \cdot a \left(1 - \frac{a}{L}\right) + b_w$ where $\alpha$ depends on $b/L$. For $b/L \ge 3$, $\alpha = 2.4$. At center ($a = L/2$): $b_e = 2.4 \times \frac{L}{2} \times \frac{1}{2} + b_w = 0.6L + b_w$. This is not $\frac{2}{3}L + b_w$ (which is $0.667L + b_w$). Close but not exact.

However, many textbooks (especially for sub-engineer level) give the simplified formulas:

• Load at center: $b_e = \frac{2}{3}L + b_w$ ... wait, actually I think the correct simplified formula for central load is $b_e = \frac{2}{3}L + b_w$? No, I think it's actually $b_e = \frac{2}{3}L + b_w$ is wrong. Let me check: For a load at the center of the span, the effective width is often given as $b_e = \frac{2}{3}L + b_w$ in some texts. But actually, I recall that the formula for a load at the center is $b_e = \frac{2}{3}L + b_w$? Hmm.

Actually, I think the correct simplified formulas are:

• Load at center: $b_e = \frac{2}{3}L + b_w$ ... no, I think it's $b_e = \frac{2}{3}L + b_w$ is not right. Let me think about the physics: The load spreads at 45° through the slab thickness. The effective width should be related to the distance from the support. At the center, the load has maximum distance to spread, so $b_e$ should be maximum. The formula $b_e = \frac{2}{3}L + b_w$ gives a value that increases with $L$. But the actual Carbon's method gives $b_e = \alpha \cdot a(1 - a/L) + b_w$, which at center gives $0.6L + b_w$ for $\alpha = 2.4$. So $\frac{2}{3}L + b_w$ is an approximation.

Actually, I now recall that in IRC 21-1966, the effective width for a load at the center was given as $b_e = \frac{2}{3}L + b_w$? No, I think it was $b_e = \frac{2}{3}L + b_w$ is actually the formula for a load at the center in the old IRC 21? Let me just check: In IRC 21-1966, Clause 305.1, the effective width for a concentrated load on a slab spanning in one direction is:

$$b_e = \frac{2}{3} \cdot \frac{a}{L} \cdot L + b_w$$

... no.

I'll stop trying to recall the exact old formula. The key point is that the model answer's formulas are the commonly taught simplified versions for Carbon's method at this level. The model answer is generally correct in its approach.

2. Modular ratio: $m = \frac{280}{3\sigma_{cbc}}$. For M20, $\sigma_{cbc} = 7$ MPa (not 8.5 MPa). Wait — IRC 21 gives $\sigma_{cbc} = 8.5$ MPa for M20? Actually, IRC 21-2000 gives permissible stresses in concrete: for M20, $\sigma_{cbc} = 6.7$ MPa (for WSM). Hmm. Let me check: IRC 21-2000, Table 1: Permissible stresses in concrete. For M20: $\sigma_{cbc} = 6.7$ MPa (bending compression). For M25: $\sigma_{cbc} = 8.5$ MPa. For M30: $\sigma_{cbc} = 10$ MPa. So the model answer's $\sigma_{cbc} = 8.5$ MPa for M20 is wrong — it should be 6.7 MPa for M20. And $m = \frac{280}{3 \times 6.7} = 13.9 \approx 14$ for M20. The model answer says $m = 10$ for M20, which is also wrong. Actually, $m = \frac{280}{3\sigma_{cbc}}$ is the formula from IS 456 (old). For M20 with $\sigma_{cbc} = 7$ MPa (IS 456 old), $m = \frac{280}{3 \times 7} = 13.3 \approx 13$. For IRC 21 with $\sigma_{cbc} = 6.7$ MPa, $m = \frac{280}{3 \times 6.7} = 13.9 \approx 14$. So the model answer's $m = 10$ for M20 is incorrect. Actually, $m = 10$ is for M25? No, for M25 with $\sigma_{cbc} = 8.5$ MPa, $m = \frac{280}{3 \times 8.5} = 10.98 \approx 11$. Hmm. Actually, the modular ratio $m$ is often taken as 10 for M20 in some textbooks? No, that's not right. The modular ratio for M20 is typically 13.33 or 14. So the model answer's $m = 10$ for M20 is wrong.

   Wait — actually, in IRC 21-1966, the modular ratio was given as $m = \frac{280}{3\sigma_{cbc}}$ where $\sigma_{cbc}$ is the permissible compressive stress in concrete. For M20, $\sigma_{cbc} = 7$ MPa (old IRC), so $m = 13.33$. For M25, $\sigma_{cbc} = 8.5$ MPa, so $m = 10.98 \approx 11$. So the model answer's statement "$\sigma_{cbc} = 8.5$ MPa (M20), ... $m = 10$ for M20" is incorrect. It should be $\sigma_{cbc} = 6.7$ MPa (M20) as per IRC 21-2000, or 7 MPa as per old IRC, and $m \approx 13$ or 14 for M20.

   Actually, let me double-check IRC 21-2000: Table 1 gives permissible stresses in concrete. For M20: $\sigma_{cbc} = 6.7$ MPa. For M25: $\sigma_{cbc} = 8.5$ MPa. For M30: $\sigma_{cbc} = 10$ MPa. So the model answer's $\sigma_{cbc} = 8.5$ MPa for M20 is wrong. And $m = 10$ for M20 is wrong.

   However, some textbooks (especially older ones) use $\sigma_{cbc} = 7$ MPa for M20 and $m = 13.33$. So the model answer's values are off.

3. Lever arm and k: The formulas given are correct: $k = \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}$, $j = 1 - k/3$. These are standard WSM formulas. Correct.

4. Moment of resistance: $M = \sigma_{st} A_{st} j d$ — correct for balanced section.

5. Steel area: $A_{st} = \frac{M}{\sigma_{st} j d}$ — correct.

6. Permissible stresses: $\sigma_{st} = 200$ MPa for Fe415? Actually, IRC 21 gives $\sigma_{st} = 200$ MPa for Fe415 (in WSM). For Fe500, it's different. The model answer says "$\sigma_{st} = 200$ MPa (Fe415/Fe500)" — this is not quite right. For Fe500, $\sigma_{st} = 240$ MPa? Actually, IRC 21-2000 gives $\sigma_{st} = 200$ MPa for Fe415 and $\sigma_{st} = 240$ MPa for Fe500 (in WSM). So the model answer's statement is slightly off.

7. Shear stress formula: $\tau_v = \frac{V}{b_w j d}$ — correct.

8. Steps: The steps are generally correct and comprehensive. Step 6 says "if $k \cdot d \le D_s$, design as rectangular beam of width $b_e$" — this is correct. Step 7: "Compute $A_{st}$ from $M = \sigma_{st} A_{st} j d$" — correct. Step 8: "if $\tau_v > \tau_c$, provide stirrups; $A_{sv} = \frac{V_s \cdot s_v}{\sigma_{sv} \cdot d}$" — correct.

9. Preliminary proportioning: Depth of beam $d \approx \frac{\text{span}}{12}$ to $\frac{\text{span}}{15}$ — this is reasonable. Width of web $b_w \approx \frac{d}{3}$ to $\frac{d}{4}$ — reasonable. Slab thickness $D_s \approx 200$–$250$ mm — reasonable. Spacing of beams $S \approx 2$–$3$ m — reasonable.

10. Missing points: The model answer doesn't mention the check for shear stress in the slab, or the design of the slab itself (which is part of T-beam bridge design). Also, it doesn't mention the check for bond stress explicitly (though it mentions "shear and bond" in principle 8). It also doesn't mention the design of the cross-beam/diaphragm if any. But for a 10-mark question, the level of detail is adequate.

11. Major error: The model answer's effective width formulas are simplified and may not be exactly "Carbon's method" as per current IRC 21. The correct Carbon's method formula is $b_e = \alpha \cdot a(1 - a/L) + b_w$ with $\alpha$ from table. The model answer's formulas $\frac{2}{3}L + b_w$ and $\frac{1}{3}L + b_w$ are approximations that don't depend on the position of the load (except central vs. edge). This is a significant simplification. However, for a sub-engineer level exam, this may be acceptable.

12. Another error: The model answer says "For two wheels close together (tandem), the effective widths may overlap and the total effective width is taken as the sum of the individual widths minus the overlap." This is correct.

13. Modular ratio error: As noted, $m = 10$ for M20 is wrong. It should be $m \approx 13$ or 14 for M20.

14. Permissible stress error: $\sigma_{cbc} = 8.5$ MPa for M20 is wrong. It should be 6.7 MPa (IRC 21-2000) or 7 MPa (old IRC).

Given these errors, the model answer is not completely correct. The most significant errors are:

• Wrong permissible stress and modular ratio for M20.
• Simplified effective width formulas that don't fully represent Carbon's method.

However, the overall approach and steps are correct. For a 10-mark question, the model answer would likely get most of the marks, but it has factual errors.

Since the question asks me to reply with the full corrected answer if the model answer is not correct and complete, I will provide a corrected version.

Let me write the corrected answer:

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Design of T-Beam Bridge by Carbon's Method (IRC 21)

Major Principles

1. Effective width concept — Carbon's method assumes that a concentrated wheel load on the deck slab is not carried by the entire slab width but only by a certain effective width of slab acting as the top flange of the T-beam. The load is assumed to spread at 45° through the slab thickness.

2. Effective width of slab (IRC 21, Clause 305): The general formula for a concentrated load on a slab spanning in one direction is: $b_e = \alpha \cdot a \left(1 - \frac{a}{L}\right) + b_w$ where:

   • $a$ = distance of the centre of the load from the nearer support,
   • $L$ = effective span,
   • $b_w$ = width of the rib (web),
   • $\alpha$ = coefficient depending on the ratio $\frac{b}{L}$ (where $b$ is the width of the slab), obtained from IRC 21 Table 12. For $\frac{b}{L} \ge 3$, $\alpha = 2.4$; for $\frac{b}{L} = 1$, $\alpha = 1.2$; intermediate values by interpolation.

   For the common case where $\frac{b}{L} \ge 3$ (wide slab), the formula simplifies to: $b_e = 2.4 \cdot a \left(1 - \frac{a}{L}\right) + b_w$ The maximum effective width occurs when the load is at the centre of the span ($a = L/2$): $b_e = 2.4 \times \frac{L}{2} \times \frac{1}{2} + b_w = 0.6L + b_w$

   For a load near the unsupported edge of the slab (cantilever portion), the effective width is taken as:

$$b_e = \frac{1}{3}L + b_w$$

(approximately, or as per IRC 21 provisions for edge loads).

For two wheels close together (tandem), the effective widths may overlap; the total effective width is the sum of the individual widths minus the overlap.

3. Effective thickness of slab = actual thickness of the slab (for spread of load through the slab).

4. Loads considered:

   • Dead load of slab, wearing coat, kerb, railing, and self-weight of the T-beam web.
   • Live load as per IRC 6 (IRC Class AA tracked/wheeled or Class 70R) with impact factor as per IRC 6.
   • The live load is placed at the most unfavourable position (maximum bending moment and maximum shear) on the deck.

5. Analysis as a T-beam: The deck slab acts as the compression flange and the rib (web) as the tension zone. The neutral axis is checked against the flange thickness to decide whether the section behaves as a true T-beam or as a rectangular beam (when NA lies within the flange).

6. Design by Working Stress Method (WSM) as per IRC 21:

   • Modular ratio $m = \frac{280}{3\sigma_{cbc}}$
   • Lever arm $j = 1 - \frac{k}{3}$, where $k = \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}$
   • Moment of resistance $M = \sigma_{st} A_{st} j d$ (for balanced section)
   • Steel area $A_{st} = \frac{M}{\sigma_{st} \cdot j \cdot d}$

7. Permissible stresses as per IRC 21 (Table 1):

   • For M20 concrete: $\sigma_{cbc} = 6.7$ MPa (bending compression), $m = \frac{280}{3 \times 6.7} \approx 14$
   • For M25 concrete: $\sigma_{cbc} = 8.5$ MPa, $m = \frac{280}{3 \times 8.5} \approx 11$
   • For Fe415 steel: $\sigma_{st} = 200$ MPa
   • For Fe500 steel: $\sigma_{st} = 240$ MPa

8. Shear and bond: Check shear stress $\tau_v = \frac{V}{b_w \cdot j \cdot d}$ against permissible $\tau_{c,max}$; design shear reinforcement (stirrups) as per IRC 6/IRC 21. Check bond stress as per IRC 21.

9. Deflection and crack control: Check span/depth ratio and provide minimum reinforcement as per IRC 21.

Steps to be Followed

Step 1 — Data collection: Span, carriageway width, number of lanes, type of loading (IRC Class AA / 70R), grade of concrete and steel, permissible stresses.

Step 2 — Preliminary proportioning:

• Depth of beam $d \approx \frac{\text{span}}{12}$ to $\frac{\text{span}}{15}$
• Width of web $b_w \approx \frac{d}{3}$ to $\frac{d}{4}$ (min 250–300 mm)
• Slab thickness $D_s \approx 200$–$250$ mm (min 100 mm + wearing coat)
• Spacing of beams $S \approx 2$–$3$ m (c/c)

Step 3 — Computation of dead load: Slab load per metre, wearing coat, kerb, railing, self-weight of web; find total DL moment and shear.

Step 4 — Computation of live load: Place IRC wheel load at the critical position; determine effective width $b_e$ by Carbon's method (using the general formula $b_e = \alpha \cdot a(1 - a/L) + b_w$); compute live load moment and shear per beam; apply impact factor.

Step 5 — Total design moment and shear: $M_{total} = M_{DL} + M_{LL+IL}$, $V_{total} = V_{DL} + V_{LL+IL}$.

Step 6 — Check NA position: Assume NA within flange; if $k \cdot d \le D_s$, design as rectangular beam of width $b_e$; otherwise design as true T-beam.

Step 7 — Design of main steel: Compute $A_{st}$ from $M = \sigma_{st} A_{st} j d$; check minimum and maximum limits; select bar diameter and spacing.

Step 8 — Design of shear reinforcement: Compute $\tau_v$; if $\tau_v > \tau_c$, provide stirrups; $A_{sv} = \frac{V_s \cdot s_v}{\sigma_{sv} \cdot d}$.

Step 9 — Check deflection and development length: Verify span/depth ratio and $L_d$.

Step 10 — Detailing: Draw reinforcement details of T-beam, slab, and stirrups as per IRC 21 and SP:34.

---

Wait — the question asks me to reply with the full corrected answer only if the model answer is not correct and complete. The model answer has some errors (permissible stress and modular ratio for M20, and simplified effective width formulas). So I should provide the corrected answer.

But let me reconsider: The model answer's effective width formulas $\frac{2}{3}L + b_w$ and $\frac{1}{3}L + b_w$ are actually the formulas given in IRC 21-1966 (old version) for effective width?

Site investigation and in-situ testing programme for a 6-lane bridge across the Karnali River

1. Recommended method of site investigation

For a major bridge of this class, the investigation must be exploratory boring (deep boreholes) supplemented by geophysical methods, not merely trial pits. The reasons:

• The Karnali is a large, braided, gravel-bed river with deep alluvium (sand–gravel–boulder deposits) and possibly rock at depth; the foundation will be deep (pile or well foundation), so information is needed to 25–40 m depth or until refusal in rock — far beyond the reach of trial pits.
• Bearing capacity and settlement of a bridge foundation require undisturbed samples and in-situ tests at the actual founding level, which only boreholes can give.
• Six lanes mean heavy, eccentric and repeated (dynamic) loading, so both ultimate bearing capacity and settlement (immediate + consolidation) must be evaluated reliably.

Programme advised:

2. In-situ test advised in detail — Standard Penetration Test (SPT)

The SPT is advised as the primary in-situ test because it is simple, cheap, widely used in Nepal, gives a continuous profile of relative density and strength, and its N-value correlates directly with bearing capacity and settlement.

(a) Equipment

• Split-spoon sampler: 50.8 mm outer diameter, 35 mm inner diameter, 150 mm long barrel, split longitudinally.
• Drive weight: 63.5 kg (140 lb) standard hammer.
• Drop: 750 mm free fall.
• Rod: 16 mm diameter (A-rod) or 22 mm (BW rod).
• Guide tube / tripod with cathead and rope, or automatic trip hammer.

(b) Procedure

1. Clean the borehole bottom; the hole is advanced by auger or wash boring, keeping the hole stable with casing or drilling mud.
2. Lower the split spoon to the bottom of the hole on the drill rod.
3. Drive the spoon 450 mm into the soil by dropping the 63.5 kg hammer from 750 mm.
4. Count the blows for each 150 mm penetration:
   • First 150 mm — seating drive (not counted).
   • Next 300 mm — test drive; the number of blows for these 300 mm is the N-value (blows/300 mm).
5. If 50 blows are required for any 150 mm, or 100 blows for 300 mm, stop and record as refusal (N = 50/150 mm or 100/300 mm).
6. Withdraw the spoon, open it, describe the soil, and preserve the sample for laboratory testing.
7. Repeat at the next depth interval.

(c) Corrections to N-value

The field N is corrected for:

• Overburden pressure: (N_1 = N \times \dfrac{100}{\sigma_v'}) (in kPa), or (N_{60}) for 60% energy ratio.
• Hammer energy: (N_{60} = N \times \dfrac{ER}{60}), where ER is the energy ratio of the hammer system.
• Borehole diameter, rod length and sampler type as per IS 2131 / ASTM D1586.

(d) Use of SPT results for bridge foundation design

1. Bearing capacity of shallow foundation (if adopted):
   • From N, relative density (D_r) and friction angle (\phi) are estimated (e.g. (D_r \approx \sqrt{N/60}), (\phi \approx 28° + 0.15 N) for sands).
   • Allowable bearing pressure from Terzaghi/Meyerhof using (\phi), or from empirical charts: (q_{all} \approx 30 N) kPa (for 25 mm settlement) for sands.
2. Pile/well foundation:
   • Unit skin friction and end bearing in sand from N (e.g. (f_s \approx 2N) kPa, (q_p \approx 40N) kPa, capped).
   • Pile capacity checked by static formula and confirmed by pile load test.
3. Settlement:
   • Immediate settlement in sand from N using empirical relations (e.g. (S = \dfrac{q B}{N}) type charts, or Schmertmann’s method using CPT).
   • Consolidation settlement in clay from oedometer tests on UDS, not from N alone.
4. Liquefaction check for the sandy riverbed: use corrected (N_{60}) and CSR from IS 1893 / Seed–Idriss method, since Karnali is in a high seismic zone.

(e) Limitations

• N is affected by hammer energy, rod friction and operator skill; hence energy correction is essential.
• Gravel and boulders give refusal and misleading values — borehole may need to be advanced by chiselling or rock drilling.
• SPT does not give a continuous record like CPT; hence CPT and PLT are used as complementary tests.

Conclusion

For the 6-lane Karnali bridge, advise deep boreholes with SPT as the main in-situ test, supported by CPT, plate load test and vane shear test, plus geophysical profiling to bedrock. SPT N-values, after overburden and energy corrections, give the bearing capacity and settlement parameters for design, which must finally be verified by pile load test at the actual pier location.

Bernoulli's equation states that for steady, incompressible, frictionless (ideal) flow along a streamline, the total energy per unit weight of the fluid remains constant. It is the mathematical statement of the conservation of energy applied to a flowing fluid, expressed as:

$\frac{P}{\gamma} + \frac{V^2}{2g} + z = \text{constant}$

where each term represents energy per unit weight of fluid (unit: metre, i.e. head):

Case of a uniform diameter pipe flow:

For a pipe of constant diameter, the cross-sectional area $A$ is constant. By the continuity equation $Q = A_1V_1 = A_2V_2$, the velocity at every section is the same:

$V_1 = V_2 = V = \text{constant}$

Therefore the velocity head $V^2/2g$ is the same at all sections of the pipe. Bernoulli's equation between any two sections 1 and 2 becomes:

$\frac{P_1}{\gamma} + \frac{V^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{V^2}{2g} + z_2$

Cancelling the equal velocity heads:

$\frac{P_1}{\gamma} + z_1 = \frac{P_2}{\gamma} + z_2$

So in uniform-diameter flow, the piezometric head $(P/\gamma + z)$ is constant along the pipe; the pressure head varies only to balance the change in elevation. For an ideal fluid, the total energy line is horizontal and the hydraulic grade line is parallel to it, at a distance of $V^2/2g$ below.

HGL and TEL for uniform diameter pipe flow (ideal fluid):

   Head
     ^
     |  TEL  ______________________________________  (horizontal)
     |        |                                    |
     |        |  V²/2g                             |
     |  HGL  _|____________________________________|_ (horizontal)
     |        |                                    |
     |        |                                    |
     |        |                                    |
     |________|____________________________________|______> Pipe length
             1                                    2

• TEL (Total Energy Line): obtained by plotting the total head $(P/\gamma + V^2/2g + z)$ at each section. For an ideal fluid with no losses, it is a horizontal straight line.
• HGL (Hydraulic Grade Line): obtained by plotting the piezometric head $(P/\gamma + z)$ at each section. It lies exactly $V^2/2g$ below the TEL and, since $V$ is constant, it is also a horizontal straight line parallel to the TEL.

In practice (real fluid), friction causes both lines to slope downward in the direction of flow, but they remain parallel to each other for a uniform diameter pipe.

Significance of Hydrological Analysis for a Hydropower Project

Hydrological analysis is the backbone of planning, design and operation of any hydropower project. Its significance:

1. Determines project feasibility and capacity — The available flow (discharge) and head decide whether the project is technically and economically viable, and fix the installed capacity.
2. Sizing of hydraulic structures — Design of diversion weir/barrage, intake, headrace canal/tunnel, forebay, penstock, spillway and tailrace all depend on design flood and design discharge.
3. Energy estimation — Annual energy generation (firm energy and secondary energy) is computed from the flow duration curve, which requires hydrological data.
4. Safety against floods — Estimation of design flood (e.g. 1 in 100, 1 in 1000 year flood) protects the project from failure and protects downstream areas.
5. Economic and financial evaluation — Benefit–cost ratio, internal rate of return and payback period depend on the estimated energy and revenue, which come from hydrological analysis.
6. Sediment management — Sediment yield data help design desilting basins and assess turbine abrasion and reservoir siltation.
7. Operation and scheduling — Reservoir operation, flood forecasting and optimal power scheduling need continuous hydrological information.
8. Environmental flow — Minimum downstream release (environmental flow) is decided from low-flow analysis.

Parameters of Hydrological Analysis

1. Precipitation (rainfall and snowfall) — depth, intensity, duration, frequency, areal distribution.
2. Discharge (streamflow) — daily, monthly, seasonal and annual flows; low flow; high flow.
3. Catchment characteristics — area, shape, slope, drainage density, land use, soil type, elevation.
4. Design flood — peak flood of a given return period (e.g. 100-yr, 1000-yr).
5. Flow duration curve (FDP) — percentage of time a given discharge is equalled or exceeded.
6. Sediment yield — suspended and bed load concentration and annual sediment load.
7. Evaporation and transpiration — for reservoir water balance.
8. Groundwater and base flow — contribution to dry-season flow.
9. Water quality — for environmental assessment.
10. Meteorological parameters — temperature, humidity, wind, sunshine hours.

Estimation and Use of These Parameters in the Project

(a) Precipitation: Measured by rain gauges (non-recording and recording) and snow gauges. Point rainfall is converted to areal rainfall by arithmetic mean, Thiessen polygon or isohyetal method. Used for runoff estimation and design flood.

(b) Discharge: Measured by current meter, weir, flume or dilution method; stage–discharge (rating) curve is developed from gauge height. Where records are short, discharge is extended by:

• Rational method: $Q = \frac{CIA}{3.6}$ (Q in m³/s, I in mm/hr, A in km²)
• Empirical formulae: Dickens, Ryve's, Inglis.
• Unit hydrograph and SCS curve number methods.
• Regional flood frequency analysis (Gumbel, Log-Pearson III) for design flood.

(c) Flow Duration Curve: Daily flow data are arranged in descending order and plotted against percentage exceedance. From it:

• Q₄₀ or Q₆₀ → design discharge for run-of-river plant (installed capacity).
• Q₉₀/Q₉₅ → low-flow (firm flow) for firm capacity and environmental release.
• Q₁₀ → flushing flow.

(d) Design flood: Estimated by frequency analysis or by the unit hydrograph/SCS method; used for spillway capacity, freeboard and cofferdam design.

(e) Sediment: Measured by suspended sediment sampler; used for desilting basin design (settling velocity, trap efficiency) and reservoir life calculation.

(f) Catchment characteristics: Obtained from topographical maps, GIS and field survey; used in the rational method, unit hydrograph and runoff coefficient selection.

(g) Evaporation: Measured by pan evaporimeter; used in reservoir water balance to find net available water.

Conclusion: Hydrological analysis thus provides the basic flow, flood and sediment data on which the capacity, dimensions, safety and economics of the hydropower project depend; without it, the design would be either unsafe or uneconomical.

I'll solve this problem independently first, then compare with the model answer.

My solution:

Given: Q = 45 m³/s, S = 1/3000, z = 2, n = 0.015

For most economical trapezoidal section:

• b = 2y(√(1+z²) − z) = 2y(√5 − 2) = 0.4721y
• A = by + zy² = 0.4721y² + 2y² = 2.4721y²
• P = b + 2y√(1+z²) = 0.4721y + 4.4721y = 4.9443y
• R = A/P = y/2

Manning: Q = (1/n)AR^(2/3)S^(1/2) 45 = (1/0.015)(2.4721y²)(0.5y)^(2/3)(1/3000)^(1/2)

S^(1/2) = 0.018257 (0.5)^(2/3) = 0.62996

45 = 66.667 × 2.4721 × 0.62996 × 0.018257 × y^(8/3) 45 = 1.8943 y^(8/3) y^(8/3) = 23.756 y = 23.756^(3/8) = 3.28 m

Adopt y = 3.30 m, b = 1.56 m ≈ 1.60 m

Check: A = 1.60×3.30 + 2×3.30² = 5.28 + 21.78 = 27.06 m² P = 1.60 + 2×3.30×2.2361 = 1.60 + 14.758 = 16.358 m R = 27.06/16.358 = 1.654 m R^(2/3) = 1.399

Q = 66.667 × 27.06 × 1.399 × 0.018257 = 45.08 m³/s ✓

V = 45/27.06 = 1.663 m/s ✓ (within 1.5–3.0 m/s)

Freeboard = 0.30 + 0.05y = 0.465 ≈ 0.50 m Total depth = 3.80 m

The model answer's calculations are all correct. However, there is one significant issue: the "most economical section" assumption is inappropriate for a concrete-lined canal.

For lined canals, the design criterion is maximum permissible velocity (to prevent scouring of the lining), not minimum excavation. The most economical section (R = y/2) is a criterion for unlined canals where minimizing earthwork is the goal. For lined canals, the lining cost dominates, and the section is typically designed for the maximum permissible velocity (usually 1.5–2.5 m/s for concrete lining) to minimize the cross-sectional area and thus the lining cost.

The model answer's own velocity check (1.66 m/s) happens to fall within permissible limits, so the final dimensions are acceptable in practice. But the stated design philosophy is wrong for a lined canal — the "most economical section" derivation is a conceptual error that an examiner would likely penalize. The correct approach for a lined canal would be to assume a permissible velocity (say 1.8 m/s), compute A = Q/V, then use Manning's equation to find R and solve for the dimensions.

Since the question specifically asks to "design" a concrete-lined canal and the model answer uses the wrong design criterion (even though the numerical result is coincidentally acceptable), the answer is not fully correct.

Design of Concrete-Lined Irrigation Canal (Sub-Engineer, Lok Sewa Aayog)

Given Data

Assumed data:

• Lining material: cement concrete (as specified), so $n = 0.015$ is appropriate.
• Design criterion for a lined canal: Since the canal is concrete-lined, the cost of lining dominates the cost. The section is therefore designed for the maximum permissible velocity (to minimize the cross-sectional area and hence the lining cost), rather than as a "most economical" (minimum excavation) section. For cement concrete lining, the permissible velocity is taken as about $1.8\ \text{m/s}$ (IS 4745 / usual practice: $1.5$–$2.5\ \text{m/s}$).

---

Step 1: Area Required from Permissible Velocity

Adopt permissible velocity $V = 1.8\ \text{m/s}$:

$A = \frac{Q}{V} = \frac{45}{1.8} = 25.0\ \text{m}^2$

---

Step 2: Apply Manning's Equation to Find Hydraulic Radius $R$

Manning's formula:

$Q = \frac{1}{n}\,A\,R^{2/3}\,S^{1/2}$

$45 = \frac{1}{0.015}\times 25.0 \times R^{2/3} \times (3.333\times10^{-4})^{1/2}$

$S^{1/2} = \sqrt{3.333\times10^{-4}} = 0.01826$

$45 = 66.667 \times 25.0 \times 0.01826 \times R^{2/3}$

$45 = 30.43\, R^{2/3}$

$R^{2/3} = \frac{45}{30.43} = 1.4788$

$R = (1.4788)^{3/2} = 1.798\ \text{m}$

---

Step 3: Solve for Depth $y$ and Bed Width $b$

For a trapezoidal section with side slope $z = 2$:

$A = by + zy^2 = by + 2y^2 = 25.0 \quad \text{...(i)}$

$P = b + 2y\sqrt{1+z^2} = b + 2y\sqrt{5} = b + 4.4721y$

$R = \frac{A}{P} = \frac{25.0}{b + 4.4721y} = 1.798$

$b + 4.4721y = \frac{25.0}{1.798} = 13.904 \quad \text{...(ii)}$

From (ii): $b = 13.904 - 4.4721y$

Substitute into (i):

$(13.904 - 4.4721y)y + 2y^2 = 25.0$

$13.904y - 4.4721y^2 + 2y^2 = 25.0$

$13.904y - 2.4721y^2 = 25.0$

$2.4721y^2 - 13.904y + 25.0 = 0$

$y = \frac{13.904 \pm \sqrt{13.904^2 - 4(2.4721)(25.0)}}{2(2.4721)}$

$y = \frac{13.904 \pm \sqrt{193.32 - 247.21}}{4.9442}$

The discriminant is negative ($193.32 - 247.21 = -53.89$), which means no real solution exists for $V = 1.8\ \text{m/s}$ with this slope and roughness. This indicates that at $V = 1.8\ \text{m/s}$, the required hydraulic radius ($R = 1.798\ \text{m}$) cannot be achieved with a section of area $25\ \text{m}^2$ at this gentle slope — the section would need to be too shallow and wide.

Re-adopt a lower permissible velocity. Try $V = 1.6\ \text{m/s}$:

$A = \frac{45}{1.6} = 28.125\ \text{m}^2$

$45 = 66.667 \times 28.125 \times 0.01826 \times R^{2/3}$

$45 = 34.23\, R^{2/3}$

$R^{2/3} = 1.3146 \quad \Rightarrow \quad R = (1.3146)^{3/2} = 1.507\ \text{m}$

$b + 4.4721y = \frac{28.125}{1.507} = 18.663 \quad \text{...(ii)}$

$b = 18.663 - 4.4721y$

Substitute into $A = by + 2y^2 = 28.125$:

$(18.663 - 4.4721y)y + 2y^2 = 28.125$

$18.663y - 4.4721y^2 + 2y^2 = 28.125$

$18.663y - 2.4721y^2 = 28.125$

$2.4721y^2 - 18.663y + 28.125 = 0$

$y = \frac{18.663 \pm \sqrt{18.663^2 - 4(2.4721)(28.125)}}{2(2.4721)}$

$y = \frac{18.663 \pm \sqrt{348.31 - 278.11}}{4.9442}$

$y = \frac{18.663 \pm \sqrt{70.20}}{4.9442} = \frac{18.663 \pm 8.379}{4.9442}$

Two roots:

• $y_1 = \frac{18.663 + 8.379}{4.9442} = \frac{27.042}{4.9442} = 5.47\ \text{m}$
• $y_2 = \frac{18.663 - 8.379}{4.9442} = \frac{10.284}{4.9442} = 2.08\ \text{m}$

Adopt the shallower root $y = 2.08\ \text{m}$ (economical, less lining depth):

$b = 18.663 - 4.4721(2.08) = 18.663 - 9.302 = 9.361\ \text{m}$

Adopt $y = 2.10\ \text{m}$, $b = 9.35\ \text{m}$

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Step 4: Verify the Design

$A = by + 2y^2 = (9.35)(2.10) + 2(2.10)^2 = 19.635 + 8.82 = 28.455\ \text{m}^2$

$P = b + 2y\sqrt{5} = 9.35 + 2(2.10)(2.2361) = 9.35 + 9.392 = 18.742\ \text{m}$

$R = \frac{A}{P} = \frac{28.455}{18.742} = 1.518\ \text{m}$

$R^{2/3} = (1.518)^{2/3} = 1.321$

$Q = \frac{1}{n}\,A\,R^{2/3}\,S^{1/2} = 66.667 \times 28.455 \times 1.321 \times 0.01826$

$Q = 45.80\ \text{m}^3/\text{s}$

This is slightly higher than 45 cumecs (about 1.8% over). Adjust slightly:

Try $y = 2.08\ \text{m}$, $b = 9.36\ \text{m}$:

$A = 9.36(2.08) + 2(2.08)^2 = 19.469 + 8.653 = 28.122\ \text{m}^2$

$P = 9.36 + 2(2.08)(2.2361) = 9.36 + 9.302 = 18.662\ \text{m}$

$R = \frac{28.122}{18.662} = 1.507\ \text{m}$

$R^{2/3} = 1.3146$

$Q = 66.667 \times 28.122 \times 1.3146 \times 0.01826 = 45.00\ \text{m}^3/\text{s} \quad \checkmark$

Adopt $y = 2.08\ \text{m}$, $b = 9.36\ \text{m}$

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Step 5: Check Velocity

$V = \frac{Q}{A} = \frac{45}{28.122} = 1.60\ \text{m/s}$

For concrete lining, permissible velocity range is about $1.5$–$2.5\ \text{m/s}$:

$1.5 < 1.60 < 2.5 \quad \checkmark$

Safe against silting and scouring.

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Step 6: Freeboard

As per IS 4745 / practice for lined canals:

$F_b = 0.30 + 0.05\,y \quad (\text{for } Q > 10\ \text{m}^3/\text{s})$

$F_b = 0.30 + 0.05(2.08) = 0.30 + 0.104 = 0.404\ \text{m}$

Adopt freeboard $F_b = 0.45\ \text{m}$

Total depth of lining: $D = y + F_b = 2.08 + 0.45 = 2.53\ \text{m}$

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Final Designed Section

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Conclusion

A concrete-lined trapezoidal irrigation canal with bed width $9.36\ \text{m}$, flow depth $2.08\ \text{m}$, side slope $2:1$, and bed slope $1:3000$ will safely carry the design discharge of 45 cumecs. Since the canal is concrete-lined, the section is designed on the basis of the permissible velocity ($1.60\ \text{m/s}$, within the safe range of $1.5$–$2.5\ \text{m/s}$ for concrete lining), which minimizes the lining area and hence the cost. Adequate freeboard of $0.45\ \text{m}$ is provided.