UPSC ESE Mechanical Engineering Strength of Materials Syllabus

For a simply supported beam of span $L$ with a central point load $W$, give the maximum shear force and maximum bending moment.

$$V_{max} = \frac{W}{2} \qquad M_{max} = \frac{WL}{4} \text{ (at midspan)}$$

For a simply supported beam of span $L$ under a uniformly distributed load $w$ per unit length, give $V_{max}$ and $M_{max}$.

$$V_{max} = \frac{wL}{2} \text{ (at supports)} \qquad M_{max} = \frac{wL^{2}}{8} \text{ (at midspan)}$$

For a cantilever of length $L$ with a point load $W$ at the free end, give $M_{max}$ and its location.

$$M_{max} = WL \text{ at the fixed end}$$ The bending moment is hogging (negative).

For a cantilever of length $L$ carrying a UDL $w$, give the maximum bending moment.

$$M_{max} = \frac{wL^{2}}{2} \text{ at the fixed support}$$

What is a point of contraflexure (inflexion) in a beam?

A section where the bending moment changes sign (passes through zero) and the beam curvature reverses from sagging to hogging or vice versa.

How does the bending moment diagram behave under (a) no load, (b) a UDL, and (c) a point load region?

(a) Under no load $M$ varies linearly (BMD straight). (b) Under UDL $M$ varies parabolically. (c) At a concentrated load the SFD jumps and the BMD has a sharp peak/kink.

State the torsion equation relating torque, shear stress, twist and section properties.

$$\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}$$ where $T$ = torque, $J$ = polar moment of inertia, $\tau$ = shear stress at radius $r$, $\theta$ = angle of twist, $L$ = length.

Give the polar moment of inertia $J$ for a solid circular shaft of diameter $d$ and for a hollow shaft of diameters $d_o$, $d_i$.

Solid: $$J = \frac{\pi d^{4}}{32}$$ Hollow: $$J = \frac{\pi (d_o^{4} - d_i^{4})}{32}$$

Where is the maximum shear stress located in a shaft under torsion, and what is its expression for a solid shaft?

At the outer surface ($r = d/2$): $$\tau_{max} = \frac{16T}{\pi d^{3}}$$ The shear stress varies linearly from zero at the axis to maximum at the surface.

Write the formula for the angle of twist of a shaft of length $L$ under torque $T$.

$$\theta = \frac{TL}{GJ}$$ where $GJ$ is the torsional rigidity.

What is the power transmitted by a shaft rotating at $N$ rpm under torque $T$ (in N·m)?

$$P = \frac{2\pi N T}{60} \text{ watts}$$ equivalently $P = T\,\omega$ with $\omega = \dfrac{2\pi N}{60}$.

Why are hollow shafts often preferred over solid shafts of equal weight?

For the same cross-sectional area (weight), a hollow shaft has a larger polar moment of inertia $J$ and section modulus, so it carries more torque and is stronger/stiffer in torsion per unit material.

Define torsional rigidity and torsional section modulus $Z_p$.

Torsional rigidity is $GJ$ (torque per unit twist per unit length). Polar section modulus $$Z_p = \frac{J}{r} = \frac{\pi d^{3}}{16}$$ for a solid shaft, giving $T = \tau_{max} Z_p$.

State the flexure (bending) formula relating bending moment, stress and curvature.

$$\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}$$ where $M$ = bending moment, $I$ = second moment of area, $\sigma$ = bending stress at distance $y$ from neutral axis, $R$ = radius of curvature.

What are the assumptions of simple (pure) bending theory?

Material is homogeneous, isotropic and obeys Hooke's law; plane sections remain plane after bending; the beam is initially straight with constant cross-section; the radius of curvature is large; and stresses are within the elastic limit.

Where does the maximum bending stress occur in a beam cross-section, and what is the formula?

At the extreme fibre farthest from the neutral axis ($y = y_{max}$): $$\sigma_{max} = \frac{M\,y_{max}}{I} = \frac{M}{Z}$$ where $Z = I/y_{max}$ is the section modulus.

Where is the neutral axis located in a beam under pure bending?

It passes through the centroid of the cross-section, where the bending stress is zero (no normal strain).

Give the section modulus $Z$ for a rectangular cross-section of width $b$ and depth $h$ bent about the horizontal centroidal axis.

$$I = \frac{bh^{3}}{12}, \qquad Z = \frac{I}{h/2} = \frac{bh^{2}}{6}$$

Give $I$ and $Z$ for a solid circular cross-section of diameter $d$ in bending.

$$I = \frac{\pi d^{4}}{64}, \qquad Z = \frac{\pi d^{3}}{32}$$

What is the transverse shear stress distribution formula in a beam cross-section?

$$\tau = \frac{V Q}{I b}$$ where $V$ = shear force, $Q$ = first moment of area above the level considered, $I$ = second moment of area, $b$ = width at that level.

For a rectangular beam section, what is the maximum transverse shear stress relative to the average?

It occurs at the neutral axis and equals $$\tau_{max} = \frac{3}{2}\,\tau_{avg} = \frac{3V}{2bh}$$ i.e. 1.5 times the mean shear stress.

What is the governing differential equation of the elastic curve (deflection $y$) for a beam?

$$EI\frac{d^{2}y}{dx^{2}} = M(x)$$ where $EI$ is the flexural rigidity and $M(x)$ the bending moment.

Describe the Double Integration Method for beam deflection.

Write $M(x)$, then integrate $EI\,y'' = M$ once to get slope $EI\,y'$ and again to get deflection $EI\,y$; the two constants of integration are found from boundary conditions (known slopes/deflections at supports).

What boundary conditions are used in the double integration method for (a) a simply supported beam and (b) a cantilever?

(a) Simply supported: deflection $y = 0$ at both supports. (b) Cantilever: at the fixed end both deflection $y = 0$ and slope $\dfrac{dy}{dx} = 0$.

For a simply supported beam of span $L$ with central point load $W$, give the maximum deflection.

$$y_{max} = \frac{WL^{3}}{48EI} \text{ at midspan}$$

For a simply supported beam of span $L$ under UDL $w$, give the maximum deflection.

$$y_{max} = \frac{5wL^{4}}{384EI} \text{ at midspan}$$

For a cantilever of length $L$ with point load $W$ at the free end, give the maximum slope and deflection.

$$\theta_{max} = \frac{WL^{2}}{2EI}, \qquad y_{max} = \frac{WL^{3}}{3EI} \text{ at the free end}$$

For a cantilever of length $L$ under UDL $w$, give the maximum (free-end) deflection.

$$y_{max} = \frac{wL^{4}}{8EI}$$

What is the key advantage of Macaulay's Method over plain double integration?

It handles beams with several discontinuous loads (point loads, partial UDLs, moments) using a single continuous moment expression with Macaulay (step) brackets, so only two constants of integration are needed for the whole beam instead of separate equations per segment.

State the Macaulay bracket rule for a term $\langle x-a\rangle^{n}$.

$$\langle x-a\rangle^{n} = \begin{cases} (x-a)^{n} & x \geq a \\ 0 & x < a \end{cases}$$ When integrating, treat the bracket as a whole: $\int \langle x-a\rangle^{n}\,dx = \dfrac{\langle x-a\rangle^{n+1}}{n+1}$, and discard (set to zero) any bracket that becomes negative when applying boundary conditions.

In Macaulay's method, how is a UDL that does not extend to the end of the beam handled?

The UDL is extended to the right end of the beam and an equal, opposite (cancelling) UDL is added over the extra length, so each can be written as a Macaulay term such as $\dfrac{w}{2}\langle x-a\rangle^{2}$.

State Mohr's two Moment-Area theorems.

Theorem 1: The change in slope between two points equals the area of the $\dfrac{M}{EI}$ diagram between them, $\theta_{AB} = \int_A^B \dfrac{M}{EI}\,dx$. Theorem 2: The deflection of one point relative to the tangent at another equals the first moment of that $\dfrac{M}{EI}$ area about the point where deflection is measured, $\delta = \int_A^B \dfrac{M}{EI}\,x\,dx$.

For which type of problems is the Moment Area Method most convenient?

For beams with simple/known bending moment diagrams—especially cantilevers and finding slope or deflection at a specific point—where the $M/EI$ diagram area and its centroid are easy to evaluate.

Using the moment-area method, what is the free-end deflection of a cantilever with end load $W$ (length $L$)?

The $M/EI$ diagram is a triangle of area $\dfrac{1}{2}\cdot L\cdot\dfrac{WL}{EI}$ with centroid at $\dfrac{2L}{3}$ from the fixed end, giving $$\delta = \frac{1}{2}\cdot\frac{WL^{2}}{EI}\cdot\frac{2L}{3} = \frac{WL^{3}}{3EI}$$

Compare the Double Integration, Macaulay and Moment-Area methods for finding deflections.

Double Integration: general, best for single continuous load/simple cases but tedious with many discontinuities. Macaulay: a streamlined double integration using step brackets, ideal for multiple discrete loads with only two constants. Moment-Area: a graphical/area approach using the $M/EI$ diagram, fastest when the BMD is simple and deflection/slope is needed at specific points (e.g. cantilevers).