FE Civil · Chapter 7 · 7–11 exam questions

FE Civil Mechanics of Materials

This chapter tests internal forces, stress-strain analysis, deformations, and stability of structural members.

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What the FE tests in Mechanics of Materials

Stress, Strain & Material Behavior

As a civil engineer, you compute stresses and deformations to check whether structural members are safe and serviceable — axial stress in columns, torsional shear in shafts, and material properties from stress-strain curves. These fundamentals underpin every structural calculation.

Beams

As a civil engineer, you design beams by constructing shear and moment diagrams, computing bending and shear stresses from internal forces, and checking deflections against serviceability limits. From floor joists to bridge girders, beam analysis is the most frequently tested mechanics topic.

Combined Loading & Stability

As a civil engineer, real members experience combined stresses — axial plus bending, shear plus torsion. You find principal stresses and maximum shear using Mohr's circle, and check slender columns against Euler buckling to prevent sudden stability failures.

Key Mechanics of Materials formulas

$\sigma = \frac{P}{A}$
Axial StressFE Handbook p. 130
$\delta = \frac{PL}{AE}$
Axial DeformationFE Handbook p. 130
$\tau = \frac{Tc}{J}$
Torsional ShearFE Handbook p. 133
$\sigma = \frac{Mc}{I}$
Flexural StressFE Handbook p. 135
$\tau = \frac{VQ}{Ib}$
Transverse ShearFE Handbook p. 135
$\delta_{max} = \frac{PL^3}{48EI}$
Simply Supported Beam (midpoint load)FE Handbook p. 140
$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$
Principal StressesFE Handbook p. 131
$P_{cr} = \frac{\pi^2 EI}{(KL)^2}$
Euler's Buckling LoadFE Handbook p. 136
$\sigma_1 = \frac{nMy}{I_T}, \; n = \frac{E_1}{E_2}$
Transformed (Composite) SectionFE Handbook p. 136
$M_p = F_y Z$
Plastic Moment (Z = plastic section modulus)FE Handbook p. 281

Sample Mechanics of Materials problems

Q1. A steel rod with a cross-sectional area of $500\,\text{mm}^2$ and length of $2\,\text{m}$ is subjected to a tensile force of $100\,\text{kN}$ . If $E = 200\,\text{GPa}$ , what is the elongation?

Answer: 2.0 mm

Explain it simply

Use $\delta = PL/(AE)$ . $P = 100{,}000$ N, $L = 2{,}000$ mm, $A = 500$ mm $^2$ , $E = 200{,}000$ MPa. $\delta = (100{,}000 \times 2{,}000)/(500 \times 200{,}000) = 200{,}000{,}000/100{,}000{,}000 = 2.0$ mm. The main trap is unit confusion -- make sure everything is in consistent units (N, mm, MPa).

Q2. A material is loaded beyond its yield point and then unloaded. The unloading path on the stress-strain curve:

Answer: Is parallel to the initial elastic loading line

Explain it simply

When you unload a material from beyond yield, it springs back elastically -- the unloading line has the same slope as the original elastic region (same $E$ ). But it does not return to zero strain -- there is permanent plastic deformation. The material "remembers" it was stretched. This is why the unloading line is parallel to, not on, the original curve.

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Common Mechanics of Materials mistakes on the FE

Using diameter instead of radius for c in σ = Mc/I and τ = Tc/J — c is the distance from the neutral axis.
Forgetting to use the effective length factor K in Euler's buckling formula — K depends on end conditions.
Sign errors in shear/moment diagrams — remember V = dM/dx and the relationship between loading and shear.
Using I (area moment) instead of J (polar moment) for torsion problems, or vice versa.
Not checking units — mixing MPa and kPa, or mm and m, is the #1 source of wrong answers.
For a composite section, multiply the stiffer material width by n = E₁/E₂; the stiffer material carries n times the stress.
Use the PLASTIC section modulus Z for Mp = Fy·Z — using the elastic S gives only the smaller yield moment.

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