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Chapter 3 of 9

Simple Machines & Thermodynamics Basics

In the DSSSB JE Civil syllabus under Basic Mechanical Engineering · 2 parts

Virtual Work & Simple Machines · Thermodynamics Basics

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📑 Contents (17 sections)

Part 1 of 2

Virtual Work & Simple Machines

Last reviewed 16 Sept 2026 · 6 min read

Principle of virtual work

A virtual displacement is an imaginary, infinitesimally small displacement consistent with the constraints of a system, applied without changing the forces. The work of real forces through such displacements is virtual work.

FormulaPrinciple of virtual work

A system of rigid bodies with ideal (frictionless) constraints is in equilibrium if and only if the total virtual work done by the active forces is zero for every virtual displacement consistent with the constraints:

δ W = \sum F_{i} δ s_{i} + \sum M_{j} δ θ_{j} = 0

Why it helps: reactions at ideal supports do no work through compatible displacements, so they drop out. It is ideal for linkages, lifting mechanisms and finding a single unknown reaction (release that support, give it a virtual displacement, and write one equation). It is also the basis of the kinematic (mechanism) method of plastic analysis and of the unit load method for deflections.

Worked ExampleReaction of a beam by virtual work

A simply supported beam AB of span 6 m carries 30 kN at 2 m from A. Find $R_{B}$ .

Solution. Release B and lift it by $δ$ (beam rotates about A). The load point rises $δ \times 2/6 = δ /3$ .

$R_{B} δ - 30 \times \frac{δ}{3} = 0$ → $R_{B} = 10 kN$ ✔

Simple machines — definitions

A machine enables a small effort $P$ to overcome a larger load $W$ (or to move a load conveniently).

DefinitionKey quantities

Mechanical advantage $M A = \frac{W}{P}$
Velocity ratio $V R = \frac{distance moved by effort}{distance moved by load} = \frac{y}{x}$ (depends only on geometry)
Output = $W x$ ; Input = $P y$
Efficiency $η = \frac{output}{input} = \frac{W x}{P y} = \frac{M A}{V R}$
Ideal machine ( $η = 100%$ ): ideal effort $P_{i} = W / V R$ ; ideal load $W_{i} = P \times V R$
Effort lost in friction $= P - P_{i}$ ; load lost in friction $= W_{i} - W$

Law of a machine

Experiments show effort varies linearly with load:

P = mW + C

$m$ = slope; $C$ = effort needed to overcome friction with no load.

Maximum mechanical advantage (as $W \to \infty$ ): $M A_{ma x} = \frac{1}{m}$ .
Maximum efficiency: $η_{ma x} = \frac{1}{m V R}$ .

Reversibility and self-locking

A machine is reversible if the load, when the effort is removed, runs the machine backwards; otherwise it is irreversible (self-locking).

FormulaCondition for self-locking

A machine is self-locking when its efficiency is less than 50%. At exactly 50% it is on the border. Screw jacks and worm gears are usually designed to be self-locking so that a raised load stays up.

Proof in brief: if frictional loss when lifting is $F$ (in load-distance terms), reversing requires $W x > F$ ; with $F = P y - W x$ this becomes $2 W x > P y$ → $η > 50%$ for reversibility.

Velocity ratios of common machines

Machine	Velocity ratio
Lever, effort arm $a$ , load arm $b$	$a / b$
First system of pulleys ( $n$ movable pulleys)	$2^{n}$
Second system (block and tackle), $n$ pulleys in both blocks	$n$ (number of rope segments supporting the load)
Third system of pulleys ( $n$ pulleys)	$2^{n} - 1$
Weston differential pulley block (radii $R$ and $r$ of the upper block)	$\frac{2 R}{R - r}$
Wheel and axle (wheel diameter $D$ , axle $d$ )	$D / d$
Differential wheel and axle (axles $d_{1} > d_{2}$ )	$\frac{2 D}{d _{1} - d _{2}}$
Simple screw jack (lever arm $L$ , pitch $p$ )	$\frac{2 π L}{p}$
Differential screw jack (pitches $p_{1} > p_{2}$ )	$\frac{2 π L}{p _{1} - p _{2}}$
Worm and worm wheel (single-start worm, $T$ teeth on wheel, effort wheel radius $R$ , load drum radius $r$ )	$\frac{R T}{r}$
Single purchase winch crab	$\frac{L T _{2}}{r T _{1}}$ ( $L$ handle length, $r$ drum radius, $T_{1}$ pinion teeth, $T_{2}$ spur wheel teeth)
Double purchase winch crab	$\frac{L}{r} \cdot \frac{T _{2}}{T _{1}} \cdot \frac{T _{4}}{T _{3}}$

For a multi-start worm with $n$ starts, divide the VR by $n$ .

Part 2 of 2

Thermodynamics Basics

Last reviewed 16 Sept 2026 · 9 min read

Basic concepts

Term	Meaning
System	Quantity of matter or region selected for study
Surroundings	Everything outside the system; separated by the boundary
Closed system (control mass)	No mass transfer; energy (heat, work) can cross — piston–cylinder
Open system (control volume)	Mass and energy cross the boundary — turbines, pumps, nozzles
Isolated system	No mass or energy transfer
Property	Observable characteristic — pressure, temperature, volume, internal energy, enthalpy, entropy
Intensive property	Independent of mass — pressure, temperature, density
Extensive property	Depends on mass — volume, energy; per unit mass it becomes specific (intensive)
State	Condition described by properties
Process	Change of state; quasi-static — sequence of equilibrium states
Cycle	Series of processes returning to the initial state
Thermodynamic equilibrium	Mechanical, thermal and chemical equilibrium

Zeroth law

If two bodies are each in thermal equilibrium with a third body, they are in thermal equilibrium with each other — the basis of temperature measurement. $T (K) = t (° C) + 273.15$ .

Work and heat

Both are energy in transition across the boundary and are path functions (inexact differentials), not properties.
Displacement work $W = \int p d V$ (area under p–V curve).
Sign convention (common in engineering): heat added to system positive, work done by system positive.

First law of thermodynamics

Energy can neither be created nor destroyed.

FormulaFirst law

For a cycle: $\oint δ Q = \oint δ W$ For a process (closed system): $Q - W = Δ U$ Internal energy $U$ is a property (point function). Enthalpy $H = U + p V$

Perpetual motion machine of the first kind (producing work without energy input) is impossible.

Ideal gas and specific heats

$p V = m R T$ ; for air $R \approx 0.287$ kJ/kg·K.
$c_{p} - c_{v} = R$ ; $γ = c_{p} / c_{v}$ (air ≈ 1.4; $c_{p} \approx 1.005$ , $c_{v} \approx 0.718$ kJ/kg·K).
For ideal gases $Δ U = m c_{v} Δ T$ ; $Δ H = m c_{p} Δ T$ .

Non-flow processes (ideal gas)

Process	Law	Work $W = \int p d V$	Heat
Constant volume (isochoric)	$p / T$ = const	0	$Q = m c_{v} (T_{2} - T_{1})$
Constant pressure (isobaric)	$V / T$ = const	$p (V_{2} - V_{1})$	$Q = m c_{p} (T_{2} - T_{1})$
Isothermal	$p V$ = const	$p_{1} V_{1} ln \frac{V _{2}}{V _{1}} = m R T ln \frac{V _{2}}{V _{1}}$	$Q = W$ ( $Δ U = 0$ )
Reversible adiabatic (isentropic)	$p V^{γ}$ = const; $\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{γ - 1} = (\frac{p _{2}}{p _{1}})^{(γ - 1) / γ}$	$\frac{p _{1} V _{1} - p _{2} V _{2}}{γ - 1}$	0
Polytropic	$p V^{n}$ = const	$\frac{p _{1} V _{1} - p _{2} V _{2}}{n - 1}$	$Q = W \frac{γ - n}{γ - 1}$

Steady flow energy equation (open systems)

FormulaSFEE (per unit mass)

h_{1} + \frac{V _{1}^{2}}{2} + g z_{1} + q = h_{2} + \frac{V _{2}^{2}}{2} + g z_{2} + w

Device	Simplification
Nozzle	$q = w = 0$ → $V_{2} = 2 (h_{1} - h_{2}) + V_{1}^{2}$
Turbine (adiabatic)	$w = h_{1} - h_{2}$
Compressor/pump	$w = - (h_{2} - h_{1})$ (work input)
Throttling valve	$h_{1} = h_{2}$ (isenthalpic)
Boiler/condenser	$q = h_{2} - h_{1}$ , $w = 0$

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