🇮🇳 UPSC IES/ESE (Engineering Services) · subject
UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines Syllabus
The Engineering Mechanics, Strength of Materials and Theory of Machines syllabus for UPSC IES/ESE (Engineering Services) in full: 6 chapters, 24 topics and 4 sub-topics, every one listed below, plus 60 Engineering Mechanics, Strength of Materials and Theory of Machines flashcards with answers.
Engineering Mechanics, Strength of Materials and Theory of Machines syllabus: all 6 chapters
Expand any chapter to see its topics and sub-topics.
-
Statics and Dynamics of Rigid Bodies
4 topics- Force systems, equilibrium and free body diagrams
- Trusses, frames and analysis of forces in members
- Friction, centroid and moment of inertia
- Kinematics and kinetics of particles and rigid bodies
- Work-energy and impulse-momentum principles
- D'Alembert's principle
-
Stress, Strain and Elastic Behaviour
4 topics- Simple stresses and strains, elastic constants
- Compound stresses and Mohr's circle
- Thermal stresses and strain energy
- Theories of failure
-
Bending, Shear and Torsion
4 topics- Shear force and bending moment diagrams
- Bending stresses and shear stresses in beams
- Torsion of circular shafts and power transmission
- Deflection of beams by various methods
- Double integration and Macaulay's method
- Moment-area and conjugate beam
-
Columns, Springs and Pressure Vessels
4 topics- Euler and Rankine theories of columns
- Springs in series and parallel, leaf and helical springs
- Thin and thick cylinders and spheres
- Buckling and stability considerations
-
Kinematics and Dynamics of Machines
4 topics- Mechanisms, kinematic pairs and inversions
- Velocity and acceleration analysis of mechanisms
- Cams, gears and gear trains
- Flywheels and governors
-
Vibrations and Balancing
4 topics- Free and forced vibrations of single degree of freedom systems
- Damped vibrations and resonance
- Whirling of shafts and critical speed
- Static and dynamic balancing of rotating and reciprocating masses
UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines flashcards — 60 questions with answers
What three scalar equations express the equilibrium of a coplanar (2D) force system?
$\sum F_x = 0$, $\sum F_y = 0$, and $\sum M_z = 0$ (the sum of forces along each axis and the sum of moments about any point all vanish).
What is a free body diagram (FBD) and why is it drawn?
A sketch of a single isolated body showing all external forces, reactions, and moments acting on it (with the body removed from its supports). It is drawn to systematically apply the equilibrium equations and solve for unknown forces.
State the necessary and sufficient conditions for equilibrium of a rigid body in 3D.
The resultant force and resultant couple must both be zero: $\sum \vec{F} = 0$ and $\sum \vec{M} = 0$, giving six scalar equations ($\sum F_x=\sum F_y=\sum F_z=0$ and $\sum M_x=\sum M_y=\sum M_z=0$).
For a plane truss with $m$ members, $j$ joints and $r$ reactions, what condition makes it statically determinate?
$m + r = 2j$. If $m + r > 2j$ the truss is statically indeterminate; if $m + r < 2j$ it is a mechanism (unstable).
What key assumptions underlie the analysis of an ideal pin-jointed truss?
Members are straight and connected by frictionless pins, loads and reactions act only at joints, member weights are neglected (or applied at joints), so each member carries only an axial force (pure tension or compression).
What is a zero-force member in a truss and give one rule for identifying it.
A member carrying no force under a given loading. Rule: at an unloaded joint where only two non-collinear members meet, both are zero-force members; if three members meet with two collinear and no external load, the non-collinear member is zero-force.
Compare the method of joints and the method of sections for truss analysis.
Method of joints applies $\sum F_x=0,\ \sum F_y=0$ at each joint (good for finding all member forces, max 2 unknowns per joint). Method of sections cuts the truss and uses $\sum F$ and $\sum M$ on a portion (good for finding a few specific member forces directly, up to 3 unknowns per cut).
State the laws of dry (Coulomb) friction including the limiting friction relation.
Limiting friction $F = \mu_s N$ is independent of contact area, proportional to normal reaction $N$, and acts opposite to impending motion. Once sliding, kinetic friction $F_k = \mu_k N$ with $\mu_k < \mu_s$. The angle of friction $\phi$ satisfies $\tan\phi = \mu$.
Define the angle of repose and relate it to the coefficient of friction.
The maximum inclination of a plane at which a body remains on the verge of sliding under its own weight. It equals the angle of friction $\phi$, so $\tan\alpha = \mu_s$.
Give the centroid location of a triangle, a semicircle, and a quarter circle.
Triangle: at $\frac{h}{3}$ from the base (intersection of medians). Semicircle of radius $r$: $\bar{y} = \frac{4r}{3\pi}$ from the diameter. Quarter circle of radius $r$: $\bar{x} = \bar{y} = \frac{4r}{3\pi}$ from the straight edges.
State the parallel axis theorem for moment of inertia.
$I = I_G + A d^{2}$, where $I_G$ is the second moment of area about the centroidal axis, $A$ is the area, and $d$ is the perpendicular distance between the parallel axes.
Give the area moment of inertia of a rectangle ($b\times h$) and a solid circle (diameter $d$) about their centroidal axes.
Rectangle about centroidal axis parallel to base: $I = \frac{b h^{3}}{12}$. Solid circle: $I = \frac{\pi d^{4}}{64}$.
State the perpendicular axis theorem for plane areas.
For a plane lamina, the polar moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two perpendicular in-plane axes through the same point: $I_z = I_x + I_y$ (also written $J = I_{xx} + I_{yy}$).
Distinguish kinematics from kinetics.
Kinematics studies motion (displacement, velocity, acceleration) without regard to the forces causing it. Kinetics relates the motion to the forces and masses producing it (e.g., via $\vec{F} = m\vec{a}$).
Write the equations of uniformly accelerated rectilinear motion.
$v = u + at$, $\quad s = ut + \frac{1}{2}at^{2}$, $\quad v^{2} = u^{2} + 2as$, where $u$ is initial velocity, $v$ final velocity, $a$ acceleration, and $s$ displacement.
For a rigid body rotating about a fixed axis, relate linear and angular kinematic quantities.
$v = r\omega$, tangential acceleration $a_t = r\alpha$, and normal (centripetal) acceleration $a_n = r\omega^{2} = \frac{v^{2}}{r}$, where $r$ is distance from the axis, $\omega$ angular velocity, $\alpha$ angular acceleration.
State the work-energy principle for a particle.
The net work done by all forces equals the change in kinetic energy: $W_{net} = \Delta KE = \frac{1}{2}m v_2^{2} - \frac{1}{2}m v_1^{2}$.
State the impulse-momentum principle for a particle.
The impulse of the resultant force equals the change in linear momentum: $\int F\,dt = m v_2 - m v_1$. For constant force, $F\,t = m(v_2 - v_1)$.
What is conserved in a collision and how is the coefficient of restitution defined?
Linear momentum is conserved: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$. The coefficient of restitution $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$, with $e=1$ for perfectly elastic and $e=0$ for perfectly plastic impact.
State D'Alembert's principle.
A moving body can be treated as in equilibrium by adding an inertia force $-m\vec{a}$ (and inertia couple $-I\alpha$) that is equal and opposite to the mass-times-acceleration term. Thus $\sum \vec{F} - m\vec{a} = 0$, reducing a dynamics problem to a statics-like equilibrium problem.
Show the other 40 Engineering Mechanics, Strength of Materials and Theory of Machines flashcards
Define normal stress and longitudinal (axial) strain.
Normal stress $\sigma = \frac{P}{A}$ (force per unit area). Longitudinal strain $\varepsilon = \frac{\delta L}{L}$ (change in length per original length); both are used in Hooke's law $\sigma = E\varepsilon$.
Write the formula for elongation of a prismatic bar under axial load $P$.
$\delta = \frac{PL}{AE}$, where $L$ is the length, $A$ the cross-sectional area, and $E$ Young's modulus.
Define the three elastic constants $E$, $G$, and $K$, and Poisson's ratio $\nu$.
$E$ = Young's modulus (axial stress/strain), $G$ = modulus of rigidity (shear stress/shear strain), $K$ = bulk modulus (volumetric stress/volumetric strain), $\nu$ = Poisson's ratio = $-\frac{\text{lateral strain}}{\text{longitudinal strain}}$.
State the relations among the elastic constants $E$, $G$, $K$ and $\nu$.
$E = 2G(1+\nu)$, $\quad E = 3K(1-2\nu)$, $\quad E = \frac{9KG}{3K+G}$.
Write the normal and shear stress on an oblique plane at angle $\theta$ for a 2D state $(\sigma_x,\sigma_y,\tau_{xy})$.
$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta$ and $\tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta$.
Give the principal stresses for a 2D stress state $(\sigma_x,\sigma_y,\tau_{xy})$.
$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2} + \tau_{xy}^{2}}$, and the maximum in-plane shear $\tau_{max} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2} + \tau_{xy}^{2}}$.
What does Mohr's circle represent and what are its centre and radius?
A graphical representation of the 2D stress transformation, plotting $\sigma$ vs $\tau$. Centre at $\left(\frac{\sigma_x+\sigma_y}{2},\,0\right)$; radius $R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}} = \tau_{max}$. Principal stresses are the horizontal-axis intercepts.
Write the formula for thermal stress in a fully restrained bar.
$\sigma = E\,\alpha\,\Delta T$, where $\alpha$ is the coefficient of thermal expansion and $\Delta T$ the temperature change. If the supports yield by $\delta$, then $\sigma = E\left(\alpha\,\Delta T - \frac{\delta}{L}\right)$.
Define strain energy and give the strain energy stored in an axially loaded bar.
Strain energy is the energy stored due to elastic deformation (resilience). For an axial bar: $U = \frac{P^{2}L}{2AE} = \frac{\sigma^{2}}{2E} \times \text{Volume}$.
Define proof resilience and modulus of resilience.
Proof resilience is the maximum strain energy stored up to the elastic limit. Modulus of resilience is the proof resilience per unit volume, equal to $\frac{\sigma_e^{2}}{2E}$ where $\sigma_e$ is the elastic-limit stress.
Compare strain energy under gradual, sudden, and impact axial loading.
For the same final load $P$: stress under gradual load $= \frac{P}{A}$; under sudden load the stress is doubled $= \frac{2P}{A}$; under impact (load $P$ falling height $h$) the instantaneous stress is $\sigma = \frac{P}{A}\left(1 + \sqrt{1 + \frac{2AEh}{PL}}\right)$.
Name the principal theories of elastic failure.
(1) Maximum principal stress (Rankine), (2) Maximum shear stress (Tresca/Guest), (3) Maximum principal strain (St. Venant), (4) Total strain energy (Haigh/Beltrami), (5) Maximum distortion (shear strain) energy (von Mises-Hencky).
State the maximum shear stress (Tresca) theory failure criterion.
Failure occurs when the maximum shear stress reaches the shear at yield in uniaxial tension: $\frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_y}{2}$, i.e. $\sigma_1 - \sigma_3 = \sigma_y$. It is conservative and best suited to ductile materials.
State the distortion energy (von Mises) theory failure criterion.
Yielding begins when the distortion energy equals that at uniaxial yield: $\frac{1}{\sqrt{2}}\sqrt{(\sigma_1-\sigma_2)^{2}+(\sigma_2-\sigma_3)^{2}+(\sigma_3-\sigma_1)^{2}} = \sigma_y$. It is the most accurate for ductile materials.
Which failure theory is recommended for brittle materials and why?
The maximum principal stress (Rankine) theory, $\sigma_1 = \sigma_{ut}$, because brittle materials fail by fracture (separation) rather than yielding, and fracture is governed by the largest tensile principal stress.
State the differential relationships among load $w$, shear force $V$, and bending moment $M$.
$\frac{dV}{dx} = -w$ (load intensity) and $\frac{dM}{dx} = V$. Hence $\frac{d^{2}M}{dx^{2}} = -w$. Bending moment is maximum where shear force is zero (or changes sign).
What is the shape of the SFD and BMD for a simply supported beam under a central point load $W$?
SFD: constant $+\frac{W}{2}$ from the left support, dropping to $-\frac{W}{2}$ after midspan (rectangular steps). BMD: triangular, zero at supports and maximum $M_{max} = \frac{WL}{4}$ at the centre.
Give the maximum bending moment for a simply supported beam under a uniformly distributed load $w$ over span $L$.
$M_{max} = \frac{wL^{2}}{8}$ at midspan; the SFD is linear from $+\frac{wL}{2}$ to $-\frac{wL}{2}$ and the BMD is parabolic.
What is the significance of a point of contraflexure?
It is a point on the beam where the bending moment changes sign (passes through zero), i.e. the curvature reverses from sagging to hogging. The beam tends to be straight there.
State the bending (flexure) equation and define each term.
$\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}$, where $M$ = bending moment, $I$ = second moment of area, $\sigma$ = bending stress at distance $y$ from the neutral axis, $E$ = Young's modulus, $R$ = radius of curvature.
Define section modulus and give it for a rectangular section.
Section modulus $Z = \frac{I}{y_{max}}$ relates moment to maximum bending stress via $\sigma_{max} = \frac{M}{Z}$. For a rectangle $b\times h$: $Z = \frac{b h^{2}}{6}$.
Write the formula for transverse shear stress distribution in a beam.
$\tau = \frac{V Q}{I b}$, where $V$ = shear force, $Q = A\bar{y}$ = first moment of area above the level considered, $I$ = second moment of area, $b$ = width at that level. Shear stress is maximum at the neutral axis and zero at the extreme fibres.
Compare maximum shear stress to average shear stress for rectangular and circular sections.
Rectangular: $\tau_{max} = \frac{3}{2}\tau_{avg}$ (occurs at the neutral axis). Circular: $\tau_{max} = \frac{4}{3}\tau_{avg}$. In both, average $\tau_{avg} = \frac{V}{A}$.
State the torsion equation for a circular shaft and define the terms.
$\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}$, where $T$ = torque, $J$ = polar moment of inertia, $\tau$ = shear stress at radius $r$, $G$ = modulus of rigidity, $\theta$ = angle of twist, $L$ = length.
Give the polar moment of inertia for solid and hollow circular shafts.
Solid shaft of diameter $d$: $J = \frac{\pi d^{4}}{32}$. Hollow shaft of outer/inner diameters $D,d$: $J = \frac{\pi (D^{4} - d^{4})}{32}$.
Write the relation between power, torque, and speed for a transmission shaft.
$P = \frac{2\pi N T}{60}$ (with $N$ in rpm, $T$ in N·m, $P$ in watts), or equivalently $P = T\omega$ where $\omega = \frac{2\pi N}{60}$ rad/s.
Define torsional rigidity and torsional stiffness of a shaft.
Torsional rigidity is $GJ$ (the product resisting twist). Torsional stiffness is the torque per unit angle of twist: $k_t = \frac{T}{\theta} = \frac{GJ}{L}$.
Write the governing differential equation of the elastic curve (deflection of beams).
$EI\frac{d^{2}y}{dx^{2}} = M(x)$, where $y$ is deflection, $M(x)$ the bending moment. Integrating once gives slope $\frac{dy}{dx}$ and twice gives deflection $y$, with constants found from boundary conditions.
Give the maximum deflection of a simply supported beam with central point load $W$ and with UDL $w$.
Central point load: $y_{max} = \frac{W L^{3}}{48 EI}$. Uniformly distributed load: $y_{max} = \frac{5 w L^{4}}{384 EI}$, both at midspan.
Give the maximum deflection of a cantilever with an end point load $W$ and with UDL $w$.
End point load: $y_{max} = \frac{W L^{3}}{3 EI}$. Uniformly distributed load over the whole span: $y_{max} = \frac{w L^{4}}{8 EI}$, both at the free end.
What is Macaulay's method and what is its key advantage?
A double-integration technique using bracket (step/Macaulay) functions $\langle x-a\rangle$ that are taken as zero when negative. Its advantage is that a single bending-moment expression covers the whole beam, so one pair of integration constants suffices even with several discrete loads.
State the two Mohr's moment-area theorems.
Theorem I: the change in slope between two points equals the area of the $\frac{M}{EI}$ diagram between them. Theorem II: the vertical deviation of a point from the tangent at another equals the moment of that $\frac{M}{EI}$ area about the first point.
Explain the conjugate beam method for finding beam deflections.
An imaginary 'conjugate' beam of the same length is loaded with the $\frac{M}{EI}$ diagram of the real beam. The shear in the conjugate beam equals the real beam's slope, and the bending moment in the conjugate beam equals the real beam's deflection. Supports are altered per conjugate-beam boundary rules.
State Euler's formula for buckling of a long column and define equivalent length.
Critical (crippling) load $P_{cr} = \frac{\pi^{2} E I}{L_e^{2}}$, where $L_e$ is the equivalent (effective) length depending on end conditions. It applies to long, slender columns failing by elastic buckling.
Give the effective length $L_e$ for the four standard column end conditions (actual length $L$).
Both ends hinged: $L_e = L$. Both ends fixed: $L_e = \frac{L}{2}$. One end fixed, other hinged: $L_e = \frac{L}{\sqrt{2}}$. One end fixed, other free: $L_e = 2L$.
Define slenderness ratio and the limitation of Euler's theory.
Slenderness ratio $= \frac{L_e}{k}$ where $k = \sqrt{\frac{I}{A}}$ is the least radius of gyration. Euler's theory is valid only for long columns (high slenderness); it fails for short columns because it can predict a buckling stress exceeding the crushing strength.
State Rankine's formula for columns and explain why it is preferred.
$\frac{1}{P_R} = \frac{1}{P_c} + \frac{1}{P_e}$, i.e. $P_R = \frac{\sigma_c A}{1 + a\left(\frac{L_e}{k}\right)^{2}}$, where $P_c$ is crushing load and $P_e$ Euler load. It is preferred because it applies to both short and long columns, approaching crushing for short and Euler for long columns.
Compare springs in series and in parallel for stiffness.
Series (same force, displacements add): $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots$ (softer combination). Parallel (same displacement, forces add): $k_{eq} = k_1 + k_2 + \cdots$ (stiffer combination).
Write the deflection and stiffness formulas for a close-coiled helical spring under axial load $W$.
Deflection $\delta = \frac{64 W R^{3} n}{G d^{4}}$ and stiffness $k = \frac{W}{\delta} = \frac{G d^{4}}{64 R^{3} n}$, where $R$ = mean coil radius, $d$ = wire diameter, $n$ = number of active coils, $G$ = modulus of rigidity.
Contrast helical and leaf (laminated) springs in function.
A helical spring resists axial load primarily through torsion of the wire and is used for energy storage and shock absorption (e.g., vehicle suspensions, valves). A leaf spring is a flat semi-elliptical/cantilever beam resisting load by bending, used in vehicle suspensions to carry large transverse loads while distributing stress along its length.
UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines FAQ
What is in the UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines syllabus?
UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines is split into 6 chapters: Statics and Dynamics of Rigid Bodies, Stress, Strain and Elastic Behaviour, Bending, Shear and Torsion, Columns, Springs and Pressure Vessels, Kinematics and Dynamics of Machines and Vibrations and Balancing. Together they contain 24 topics and 4 sub-topics.
How many chapters are in UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines?
6 chapters, containing 24 topics and 4 sub-topics. The largest chapter is Statics and Dynamics of Rigid Bodies, with 4 topics.
Are there flashcards for UPSC IES/ESE (Engineering Services) Engineering Mechanics, Strength of Materials and Theory of Machines?
Yes — a 60-card Engineering Mechanics, Strength of Materials and Theory of Machines deck, printed in full in the flashcards section of this page and available in the Examius app with spaced repetition. The free plan covers one exam of your choice, including its flashcards; Examius Pro, a one-time purchase, unlocks every other exam and deck.