Strength of Materials and Structural Analysis — ESE Civil

Weightage: Strength of Materials and Structural Analysis together form one of the largest blocks in the Civil Engineering papers. They appear in Prelims Paper II and in both Mains papers, usually as numerical questions with a derivation step. Every later chapter (design, foundations, bridges) depends on them.

1. Stress, strain and elastic constants

Stress is force per unit area, $\sigma =P/A$. Strain is relative deformation, $\epsilon =\delta /L$. Within the elastic limit Hooke's law gives $\sigma =E\epsilon$, so a bar of length $L$ under axial load $P$ elongates by:

$\delta =\frac{PL}{AE}$

Four elastic constants are linked for an isotropic material:

$E=2G(1+\nu )=3K(1-2\nu )$

with $\nu$ the Poisson ratio, $G$ the shear modulus and $K$ the bulk modulus. Poisson's ratio for an isotropic material lies between $-1$ and $0.5$, and for steel is about 0.3.

Temperature stress in a restrained bar is $\sigma =E\alpha \Delta T$. If the bar is free to expand there is no stress. For bars in parallel the load splits in proportion to $AE$, and strain is common.

2. Bending of beams

For a beam in pure bending the flexure formula applies:

$\frac{M}{I}=\frac{\sigma }{y}=\frac{E}{R}$

The stress is zero on the neutral axis and maximum at the extreme fibre. The section modulus is $Z=I/y_{max}$, so ${\sigma }_{max}=M/Z$.

Moments of inertia to remember: rectangle about its centroid, $b{h}^3/12$; circle, $\pi d^4/64$; and for a section moved from the centroid by $d$, the parallel axis theorem, $I=I_G+A{d}^2$.

Shear stress in a beam is $\tau =\frac{VQ}{Ib}$. For a rectangle it is parabolic, zero at the top and bottom and maximum at the neutral axis, with ${\tau }_{max}=1.5{\tau }_{avg}$. For a circle, ${\tau }_{max}=\frac{4}{3}{\tau }_{avg}$.

Bending moment diagram rules: the slope of the shear force diagram is the load intensity, the slope of the bending moment diagram is the shear force, and the moment is maximum where shear is zero.

3. Torsion

For a circular shaft:

$\frac{T}{J}=\frac{\tau }{r}=\frac{G\theta }{L}$

The polar moment is $J=\pi d^4/32$ for a solid shaft. Power transmitted is $P=2\pi NT/60$ with $N$ in rpm. A hollow shaft is stronger per unit weight than a solid one of the same material, because material near the axis is lightly stressed.

4. Deflection of beams

The governing equation is $EI\frac{d^{2}y}{d{x}^2}=M$. Standard results save time:

The moment-area and conjugate beam methods give slope and deflection from the bending moment diagram. Deflection varies with the cube of the span, so a small increase in span has a large effect.

5. Columns and buckling

A slender column fails by buckling before it crushes. Euler's critical load is:

$P_{cr}=\frac{{\pi }^{2}EI}{L_e^2}$

with effective length $L_e$:

The slenderness ratio is $L_e/r$, with $r=\sqrt{I/A}$. Euler's formula holds only for long columns, and short columns are governed by crushing strength. Because load goes with $I$, buckling occurs about the axis of least moment of inertia.

Worked example. A steel column $E=200\text{GPa}$, $I=4\times 10^{-6}{\text{m}}^4$, length 3 m, hinged at both ends. Then $P_{cr}={\pi }^2(200\times 10^9)(4\times 10^{-6})/9=877\text{kN}$, roughly.

6. Combined stresses and Mohr's circle

For a plane stress state with ${\sigma }_x$, ${\sigma }_y$ and ${\tau }_{xy}$:

${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

The maximum shear stress equals the radius of Mohr's circle, half the difference of the principal stresses. Principal planes carry no shear stress, and planes of maximum shear lie 45 degrees from them. Failure theories include maximum principal stress (brittle), maximum shear (Tresca) and distortion energy (von Mises), the last two for ductile metals.

7. Determinacy and trusses

A plane frame is statically determinate if the reactions and member forces follow from equilibrium alone.

• For a plane truss, $m+r=2j$ is determinate. More than that is indeterminate, fewer is unstable.
• For beams, count the reaction components against three equilibrium equations, less one for each internal hinge.

Methods of joints and sections find truss forces. In a loaded joint, a zero-force member arises when two non-collinear members meet with no load and no support at the joint.

8. Influence lines

An influence line shows the value of a force or moment at one section as a unit load moves along the structure. It is used for moving loads on bridges.

The Muller-Breslau principle says the influence line for a force is the deflected shape obtained by removing that restraint and giving a unit displacement.

For a simply supported beam, the maximum bending moment from a moving train of loads occurs under a load. The section sits where the centre of the span bisects the distance between that load and the resultant.

9. Indeterminate structures

Indeterminate analysis needs compatibility as well as equilibrium.

• Slope-deflection method: $M_{AB}=\frac{2EI}{L}(2{\theta }_A+{\theta }_B-3\psi )+{\text{FEM}}_{AB}$.
• Moment distribution (Hardy Cross): lock joints, balance, carry over half, repeat.
• Fixed-end moments: $w{L}^2/12$ for a UDL and $PL/8$ for a central point load.
• Stiffness of a far end fixed member is $4EI/L$, and of a far end hinged member is $3EI/L$.
• Castigliano's theorem gives deflection as the partial derivative of strain energy with respect to the load.

A three-hinged arch is determinate, with a horizontal thrust of $H=M_{beam}/y$ at the crown, while a two-hinged arch is once indeterminate.

Common traps

• Using Euler's formula for a short column. Check the slenderness ratio.
• Taking the wrong axis for buckling. Use the smaller $I$.
• Reading ${\tau }_{max}$ for a rectangle as the average. It is 1.5 times larger.
• Mixing stiffness factors. $4EI/L$ is for a fixed far end and $3EI/L$ for a hinged one.
• Forgetting the $L^3$ and $L^4$ powers in deflection formulas.

Memory aids

• "4 fixed, 3 hinged": member stiffness.
• "1 L, 2 L, half, 0.7": effective lengths.
• "Slope of shear is load; slope of moment is shear": diagram rules.

Summary

Axial, bending, shear and torsion results all derive from the flexure and torsion formulas. Deflection and buckling add the powers of span and the effective length factors, and Mohr's circle handles combined stress.

Structural analysis moves from determinate trusses and influence lines to indeterminate frames solved by slope-deflection, moment distribution or energy methods.

Exam protocol

• Write the governing formula, then substitute in consistent units.
• Draw the shear and moment diagram before using them.
• Check determinacy before choosing a method.
• Keep the standard deflection table on a one-line card.