← Engineering Mechanics · DSSSB JE Civil

Chapter 5 of 6

Trusses & Virtual Work

In the DSSSB JE Civil syllabus under Engineering Mechanics · 2 parts

Analysis of Plane Trusses · Virtual Work & Simple Machines

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (19 sections)

Part 1 of 2

Analysis of Plane Trusses

Last reviewed 16 Sept 2026 · 7 min read

What a truss is

A truss is a framework of straight members connected at their ends to form triangles. Because a triangle cannot change shape without changing the length of a side, a triangulated frame carries load mainly by axial forces — tension or compression — which uses material efficiently. Roof trusses, bridge trusses, transmission towers and cranes are trusses.

Assumptions of ideal truss analysis

Members are connected at frictionless pins (hinges).
Loads and reactions act only at the joints.
Members are straight and their centroidal axes meet at a point at each joint.
Self-weight of members is neglected or applied at joints.

So each member is a two-force member carrying only axial force. In real riveted or welded trusses the joints are partly rigid and small secondary stresses (bending) arise; they are usually ignored in basic analysis.

Perfect, deficient and redundant frames

With $m$ members and $j$ joints (plane truss, reactions counted separately as determinate supports):

Condition	Name	Meaning
$m = 2 j - 3$	Perfect frame	Just enough members to be stable and determinate
$m < 2 j - 3$	Deficient (imperfect) frame	Unstable — a mechanism
$m > 2 j - 3$	Redundant frame	Statically indeterminate internally

Including reactions: $D_{s} = m + r - 2 j$ (see Determinacy, Indeterminacy & Stability).

Types of trusses

Truss	Typical use and feature
King post, queen post	Small timber roof trusses
Fink (French)	Steel roofs of moderate span; short compression members
Howe	Verticals in tension, diagonals in compression (timber bridges)
Pratt	Verticals in compression, diagonals in tension under gravity load — efficient for steel
Warren	Diagonals alternately in tension and compression; with or without verticals
K-truss	Short compression panels; deep bridge trusses
Fan, north-light (saw-tooth)	Industrial roofs; north-light admits diffuse daylight
Bowstring, Baltimore, Parker	Long-span bridges with curved or polygonal top chords

Exam TipPratt vs Howe

Long members are better in tension (no buckling). The Pratt truss puts its long diagonals in tension under gravity loads, so it suits steel. The Howe truss puts its diagonals in compression and its (steel rod) verticals in tension, historically suiting timber diagonals with iron rods.

Method of joints

Take each joint as a free body with forces concurrent at the pin. Two equations per joint, so start at a joint with no more than two unknown member forces.

Find the support reactions.
Assume every unknown member force is tensile (arrow pointing away from the joint).
Solve $\sum F_{x} = 0$ and $\sum F_{y} = 0$ .
A negative answer means compression.
Move to the next joint with at most two unknowns.

Best when forces in all members are needed.

Method of sections (Ritter's method)

Cut the truss through not more than three members whose forces are unknown (and not all concurrent or all parallel). Take one part as a free body.

Take moments about the point where two of the cut members meet — this gives the third member's force directly.
Use $\sum F_{y} = 0$ for a diagonal when both chords are horizontal.

Best when forces in a few members (say the middle panel of a long truss) are needed.

Zero-force members

Identifying these first saves time.

RememberZero-force member rules (no external load at the joint)

If two non-collinear members meet at an unloaded joint, both are zero-force members.
If three members meet at an unloaded joint and two are collinear, the third is a zero-force member.

If a load or reaction acts at the joint along one member's line, apply the rules with that force treated as a member.

Zero-force members are not useless: they brace compression chords against buckling, carry load under other load cases and support the self-weight of other members.

Tension coefficient method

Useful for trusses with many inclined members and especially for space trusses. For a member AB of length $L_{A B}$ carrying tension $T_{A B}$ , define the tension coefficient $t_{A B} = T_{A B} / L_{A B}$ . Then at joint A:

\sum t_{A B} (x_{B} - x_{A}) + X_{A} = 0, \sum t_{A B} (y_{B} - y_{A}) + Y_{A} = 0

(and a $z$ equation in 3-D), where $X_{A}$ , $Y_{A}$ are external loads at A. Solve for the $t$ values, then $T = t L$ .

Part 2 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 $n$ .

Downloaded from civilgyantests.com · Not for resale. Verify every code clause and date against the official source.

Finished reading? Test yourself.

A timed chapter test from the DSSSB JE Civil series, on exactly this chapter.

Practice this chapter →