Study notes

Qwestrum Engineering360 · Civil Engineering · Fluid Mechanics

Hydrostatics and Buoyancy

Pressure grows linearly with depth (p = ρgh), the force on a submerged surface is ρg·h_c·A acting at the centre of pressure below the centroid, and buoyancy equals the weight of displaced fluid.

Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

Centre of pressure below centroid for inclined submerged surface
Metacentric height GM = I/V − BG — stability of floating bodies
Pressure head p/ρg added to elevation z in piezometric head

Topic details

Introduction

Hydrostatics deals with fluids at rest, where pressure acts equally in all directions and increases linearly with depth. This underlies the design of dams, gates, tanks and floating structures.

Scope in B.Tech and GATE syllabus

The total force on a submerged plane surface is the pressure at its centroid times its area, but the force does not act at the centroid — it acts lower, at the centre of pressure, because pressure increases with depth. Locating this point correctly is essential for moment (overturning) calculations.

Why this topic matters in practice

Buoyancy (Archimedes’ principle) states that the upward force on a submerged or floating body equals the weight of the fluid it displaces; the stability of floating bodies then depends on the metacentric height, which must be positive for stable equilibrium.

Key relations & formulas

p = \rho g h

(pressure at depth h)

F_{H} = \rho g h_{c} A

(horizontal force on plane surface)

F_{B} = \rho g V_{displaced}

(Archimedes buoyancy)

Notation and sign conventions

Relation 1 —

p = \rho g h

p = \rho g h

(pressure at depth h)

Write this relation with symbols exactly as in Fluid Mechanics & Hydraulic Machines — Modi & Seth before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Relation 2 —

F_{H} = \rho g h_{c} A

F_{H} = \rho g h_{c} A

(horizontal force on plane surface)

Relation 3 —

F_{B} = \rho g V_{displaced}

F_{B} = \rho g V_{displaced}

(Archimedes buoyancy)

Fundamentals and definitions

The hydrostatic law p = ρgh follows from vertical force balance on a fluid element; pressure depends only on depth (and fluid density), not on the container shape, which is why the piezometric head p/ρg + z is constant in a static connected fluid.

Governing relations in practice

On a submerged surface the resultant force F = ρg·h_c·A uses the depth of the centroid h_c, but because pressure varies over the surface the resultant acts at the centre of pressure, located a distance I_G/(A·h_c) below the centroid along the surface. For a vertical rectangle this puts the force at one-third the depth from the base.

Design and analysis considerations

Buoyant force F_B = ρg·V_displaced acts upward through the centroid of the displaced volume (the centre of buoyancy). A floating body is in equilibrium when its weight equals the buoyant force.

Advanced theory and extensions

Stability of a floating body is governed by the metacentre M: if M lies above the centre of gravity G (positive metacentric height GM = I/V − BG), a small tilt creates a restoring couple and the body is stable; if below, it capsizes. This is critical for ships, pontoons and caissons.

Assumptions and validity limits

State assumptions explicitly before using any relation for hydrostatics and buoyancy — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.

Wrong assumptions invalidate the entire solution even when the formula is correct. In Fluid Mechanics (Civil) viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Fluid Mechanics (Civil) papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to hydrostatics and buoyancy.
4. Use equation 1:

p = \rho g h

.
5. Use equation 2:

F_{H} = \rho g h_{c} A

.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Hydrostatics and Buoyancy appears in pipes, channels, and dams. In Indian civil curricula this topic is tested because it connects theory to hydraulics for civil works.

GATE and semester exams often combine hydrostatics and buoyancy with earlier units — revise prerequisites before attempting mixed problems.

Industry interview panels sometimes ask: "Where did you use hydrostatics and buoyancy?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Placing the resultant hydrostatic force at the centroid instead of the lower centre of pressure.
• Using gauge versus absolute pressure inconsistently.
• Forgetting buoyancy acts through the centre of buoyancy, not the body’s CG.
• Taking a negative metacentric height as stable.

Quick revision checklist

Before attempting hydrostatics and buoyancy problems, confirm you can:

1. Centre of pressure below centroid for inclined submerged surface
2. Metacentric height GM = I/V − BG — stability of floating bodies
3. Pressure head p/ρg added to elevation z in piezometric head

Revise the solved examples in Fluid Mechanics & Hydraulic Machines — Modi & Seth and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Force on a vertical rectangular gate

Problem

A vertical rectangular gate 2 m wide and 3 m high has its top edge at the water surface. Find the total hydrostatic force and the depth of the centre of pressure (ρg = 9.81 kN/m³).

Solution

Centroid depth h_c = 3/2 = 1.5 m. Area A = 2 × 3 = 6 m². Total force F = ρg·h_c·A = 9.81 × 1.5 × 6 = 88.3 kN. Centre of pressure depth h_p = h_c + I_G/(A·h_c), with I_G = bd³/12 = 2 × 3³/12 = 4.5 m⁴, so h_p = 1.5 + 4.5/(6 × 1.5) = 1.5 + 0.5 = 2.0 m below the surface (i.e. at two-thirds of the depth).

Conceptual check — Hydrostatics and Buoyancy