Applied Mechanics and Structures I: Engineering Mechanics and Mechanics of Materials
1. Free-body diagrams, equilibrium, trusses, frames and virtual work
A free-body diagram isolates a body and shows every external force and couple on it, including the reactions at supports: a roller gives one force normal to its surface, a pin two force components, a fixed support two forces and a moment. A planar rigid body is in equilibrium when ΣFₓ = 0, ΣF_y = 0 and ΣM = 0 about any point — three equations, so a planar structure with three independent reaction components is statically determinate. A two-force member carries equal and opposite forces along the line joining its pins; a three-force member in equilibrium has its three forces concurrent or parallel.
A truss is a framework of two-force members pinned at joints and loaded only at joints. A plane truss with m members, j joints and r reaction components is determinate when m + r = 2j. The method of joints applies ΣFₓ = ΣF_y = 0 at each joint in turn, starting where at most two forces are unknown; the method of sections cuts through at most three members and uses moments about the intersection of two of them to find the third directly. At an unloaded joint where two of three members are collinear, the third is a zero-force member; at an unloaded joint of two non-collinear members, both are zero. A frame contains at least one multi-force member and is solved by dismembering it into free bodies. The principle of virtual work states that a body is in equilibrium if and only if the total work of the external forces is zero for every virtual displacement compatible with the constraints; it gives one unknown force from one equation without computing the reactions.
2. Kinematics and dynamics in plane motion; impulse, momentum and energy
A particle moving on a curve has tangential acceleration dv/dt and normal (centripetal) acceleration v²/ρ towards the centre of curvature. A rigid body in plane motion is a translation of any reference point plus a rotation about it: v_B = v_A + ω × rB/A, and a point of zero velocity, the instantaneous centre, exists at every instant — for a wheel rolling without slipping it is the contact point, so the centre moves at v = ωr. Newton’s second law for the body is ΣF = m a_G for the centre of mass and ΣM_G = I_G α for rotation.
The impulse–momentum form integrates Newton’s law over time: ∫F dt = m(v₂ − v₁) for linear momentum and ∫M dt = I(ω₂ − ω₁) for angular momentum; with no external impulse (or no external moment about an axis) the corresponding momentum is conserved, which is how collisions are treated — with the coefficient of restitution e = (relative speed of separation)/(relative speed of approach). The work–energy form integrates it over distance: work done = change in kinetic energy, with T = ½mv_G² + ½I_Gω² for a rigid body. A body rolling down an incline without slipping has a = g sin θ/(1 + k²/r²): for a solid disc (k² = r²/2) that is (2/3)g sin θ, for a solid sphere (5/7)g sin θ and for a thin ring (1/2)g sin θ.
3. Stress, strain, the elastic constants and Mohr’s circle
Normal stress σ = P/A and normal strain ε = δ/L are related in the elastic range by Hooke’s law σ = Eε; shear stress and strain by τ = Gγ. Poisson’s ratio ν = −ε_lateral/ε_axial lies between 0 and 0.5 for isotropic materials (about 0.3 for steel). An isotropic material has only two independent constants, related by E = 2G(1 + ν) = 3K(1 − 2ν), where K is the bulk modulus; the volumetric strain of a bar is ε(1 − 2ν). For steel with E = 200 GPa and ν = 0.3, G = 200/2.6 = 76.9 GPa and K = 200/1.2 = 166.7 GPa.
In plane stress (σₓ, σ_y, τₓᵧ) the stresses on a plane rotated by θ follow from Mohr’s circle: centre C = (σₓ + σ_y)/2 on the σ-axis and radius R = √{[(σₓ − σ_y)/2]² + τₓᵧ²}. The principal stresses are σ₁,₂ = C ± R, on planes where the shear is zero, at tan 2θ_p = 2τₓᵧ/(σₓ − σ_y); the maximum in-plane shear is R, on planes 45° from the principal planes, where the normal stress is C. A rotation θ on the element is 2θ on the circle. Plane strain uses the same construction with ε in place of σ and γ/2 in place of τ: centre (εₓ + ε_y)/2, radius √{[(εₓ − ε_y)/2]² + (γₓᵧ/2)²}, which is how strain-gauge rosettes are reduced.
4. Shear force and bending moment, bending and shear stresses, and torsion
In a beam, the shear force V and bending moment M at a section are the resultant force and moment of everything to one side. They are linked to the distributed load w by dV/dx = −w and dM/dx = V, so the bending moment is extreme where the shear force changes sign, a point load makes a jump in V and a kink in M, and a uniformly distributed load makes V linear and M parabolic. For a simply supported span L: a central point load P gives M_max = PL/4; a uniform load w gives M_max = wL²/8 at mid-span. For a cantilever: a tip load P gives PL at the root; a uniform load gives wL²/2.
The flexure formula σ = My/I = M/Z gives the bending stress, zero at the neutral axis (through the centroid) and greatest at the extreme fibre; for a rectangle b × d, I = bd³/12 and Z = bd²/6. The shear stress τ = VQ/(Ib), where Q is the first moment about the neutral axis of the area beyond the fibre considered, is parabolic over a rectangle with a maximum of 1.5V/A at the neutral axis (4V/3A for a solid circle). In torsion of a circular shaft, τ = Tr/J and the twist θ = TL/(GJ), with J = πd⁴/32 for a solid shaft, so τ_max = 16T/(πd³); a hollow shaft has J = π(D⁴ − d⁴)/32 and is more efficient because the material near the axis carries little stress.
| Beam and load | Maximum bending moment | Where |
|---|---|---|
| Simply supported, central point load P | PL/4 | Mid-span |
| Simply supported, uniform load w per length | wL²/8 | Mid-span |
| Cantilever, tip load P | PL | Fixed end |
| Cantilever, uniform load w per length | wL²/2 | Fixed end |
5. Columns, energy methods, theories of failure and material testing
Euler’s theory gives the buckling load of an ideal slender column as P_cr = π²EI/L_e², where I is the least second moment of area and L_e the effective length: L for both ends pinned, 2L for one end fixed and the other free, L/√2 (0.7L) for fixed–pinned, and L/2 for both ends fixed. So a fixed–free column carries a quarter, and a fixed–fixed column four times, the load of the same column pinned at both ends. The critical stress π²E/(L_e/r)², with r = √(I/A) the radius of gyration, shows that Euler applies only to long columns; short columns fail by crushing, and the Rankine–Gordon formula blends the two.
Energy methods use the strain energy stored in a member: P²L/(2AE) in axial load, ∫M²/(2EI) dx in bending and T²L/(2GJ) in torsion. Castigliano’s theorem gives the deflection at a load as the partial derivative of the total strain energy with respect to that load, δ = ∂U/∂P, and a dummy load finds the deflection where none acts. The theories of failure compare a stress state with a uniaxial test: maximum principal stress (Rankine), suitable for brittle materials; maximum shear stress (Tresca), σ₁ − σ₃ = S_y, and distortion energy (von Mises), √(σ₁² − σ₁σ₂ + σ₂²) = S_y in plane stress, for ductile materials, von Mises being the less conservative of the two; and the maximum strain (St Venant) and total strain energy (Haigh) theories. In pure shear Tresca predicts yield at τ = 0.5S_y and von Mises at 0.577S_y.
| Test | What it measures |
|---|---|
| Tensile test | E, yield and ultimate strength, percentage elongation and reduction of area (ductility) |
| Hardness (Brinell, Vickers, Rockwell) | Resistance to indentation, which correlates with tensile strength |
| Impact (Charpy, Izod) | Energy absorbed in fracturing a notched bar — toughness, and the ductile-to-brittle transition temperature |
| Fatigue | S–N curve and endurance limit under cyclic stress |
| Creep | Time-dependent strain under constant load at high temperature |
| Non-destructive (ultrasonic, radiography, magnetic particle, dye penetrant) | Internal and surface defects, including weld flaws, without damaging the part |
Key takeaways
- Plane equilibrium is three equations; a plane truss is determinate when m + r = 2j; the method of sections cuts at most three members.
- Impulse = change of momentum as a vector; rolling without slipping: a = g sin θ/(1 + k²/r²), (2/3)g sin θ for a solid disc.
- E = 2G(1 + ν) = 3K(1 − 2ν); Mohr’s circle centre (σₓ + σ_y)/2 and radius √{[(σₓ − σ_y)/2]² + τ²}; plane strain uses γ/2.
- σ = My/I, τ_max = 1.5V/A for a rectangle, τ = 16T/(πd³) for a solid shaft; M_max = wL²/8 and PL/4 for simple spans.
- P_cr = π²EI/L_e² with L_e = L, 2L, 0.7L, 0.5L; Tresca and von Mises for ductile materials, Rankine for brittle.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A load of 10 kN hangs at the apex of a symmetric triangular truss whose two inclined members each make 30° with the horizontal. The force in each inclined member (in kN) is ____.
Numerical answer — type the value.
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Answer: 10
Vertical equilibrium at the apex: 2F sin 30° = 10, so F = 10/(2 × 0.5) = 10 kN, in compression. Using cos 30° instead of sin 30° gives 5.77 kN; forgetting that there are two members gives 20 kN.At an unloaded joint of a plane truss, three members meet and two of them are collinear. The force in the third member is
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Answer: A — zero
Resolving perpendicular to the two collinear members, only the third member has a component in that direction, so its force must be zero. Zero-force members are spotted this way before any arithmetic; they are still needed for stability and to carry loads in other cases.A solid disc rolls without slipping down a plane inclined at 30°. Taking g = 9.81 m/s², the acceleration of its centre (in m/s²), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 3.27
a = g sin θ/(1 + k²/r²) with k² = r²/2 for a solid disc, so a = (2/3) × 9.81 × 0.5 = 3.27 m/s². A frictionless slide gives g sin θ = 4.905; a thin ring gives g sin θ/2 = 2.45; a solid sphere (5/7)g sin θ = 3.50.A 0.5 kg ball hits a wall normally at 20 m/s and rebounds along the same line at 15 m/s. The magnitude of the impulse on the ball (in N·s), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 17.5
Impulse = m(v₂ − v₁) with the velocities signed: 0.5 × [15 − (−20)] = 17.5 N·s. Subtracting the speeds gives 2.5 N·s, which would be the answer only if the ball kept moving in the same direction.At a point in plane stress, σₓ = 80 MPa, σ_y = 20 MPa and τₓᵧ = 40 MPa. The major principal stress (in MPa) is ____.
Numerical answer — type the value.
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Answer: 100
Centre C = (80 + 20)/2 = 50; radius R = √(30² + 40²) = 50; σ₁ = C + R = 100 MPa and σ₂ = 0. Using the full difference σₓ − σ_y = 60 instead of half gives R = 72.1 and σ₁ = 122.1 MPa.A steel has E = 200 GPa and Poisson’s ratio 0.3. Its shear modulus (in GPa), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 76.9
G = E/[2(1 + ν)] = 200/2.6 = 76.9 GPa. Writing 2(1 − ν) gives 142.9 GPa; the bulk modulus E/[3(1 − 2ν)] = 166.7 GPa is a different constant.On Mohr’s circle for plane strain, the vertical coordinate plotted against the normal strain is
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Answer: A — half the engineering shear strain, γ/2
The strain-transformation equations have the same form as the stress ones with ε for σ and γ/2 (the tensor shear strain) for τ, so γ/2 is plotted. Plotting γ doubles the radius and gives principal strains that are wrong by the shear term.A simply supported beam of span 6 m carries a uniformly distributed load of 10 kN/m over its whole length. The maximum bending moment (in kN·m) is ____.
Numerical answer — type the value.
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Answer: 45
M_max = wL²/8 = 10 × 36/8 = 45 kN·m at mid-span, where the shear force is zero. wL²/2 = 180 kN·m is the cantilever value; wL²/4 = 90 treats the whole load as a central point load.A rectangular beam section 100 mm wide and 200 mm deep carries a bending moment of 20 kN·m. The maximum bending stress (in MPa) is ____.
Numerical answer — type the value.
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Answer: 30
Z = bd²/6 = 100 × 200²/6 = 666 667 mm³, so σ = M/Z = 20 × 10⁶/666 667 = 30 MPa. Using bd³/12 as though it were Z gives 0.3 MPa; putting the 100 mm dimension as the depth gives 60 MPa.A solid circular shaft of 50 mm diameter transmits a torque of 1 kN·m. The maximum shear stress (in MPa), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 40.7
τ = 16T/(πd³) = 16 × 10⁶/(π × 125 000) = 40.7 MPa. Using the bending factor 32 in place of 16 gives 81.5 MPa; using the radius in place of the diameter in d³ gives 326 MPa.A steel column 4 m long, pinned at both ends, has E = 200 GPa and least second moment of area 8 × 10⁶ mm⁴. Its Euler buckling load (in kN), to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 987
P_cr = π²EI/L² = π² × 200 000 × 8 × 10⁶/4000² = π² × 10⁵ N = 986 960 N ≈ 987 kN. Fixing one end and freeing the other (L_e = 2L) would cut it to 247 kN; fixing both (L_e = L/2) would raise it to 3948 kN.A column fixed at one end and free at the other is replaced by an identical column pinned at both ends. The Euler buckling load becomes
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Answer: A — four times as large
P_cr ∝ 1/L_e². The fixed–free column has L_e = 2L and the pinned–pinned column L_e = L, so the ratio is (2L/L)² = 4. Reading the effective length ratio without squaring it gives twice.Which statements about the theories of failure are correct?
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Answer: A — the maximum principal stress theory suits brittle materials; B — in pure shear the Tresca theory predicts yield at half the tensile yield strength; D — the distortion energy theory is used for ductile materials
Brittle materials fail on the largest tensile stress, so Rankine suits them; in pure shear σ₁ = τ, σ₃ = −τ and Tresca gives 2τ = S_y, τ = 0.5S_y. Von Mises predicts yield at 0.577S_y in shear and its hexagon-enclosing ellipse is the LESS conservative of the two; both are ductile-material theories.The Charpy test measures
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Answer: A — the energy absorbed in fracturing a notched specimen by impact
Charpy (and Izod) are impact tests: a pendulum breaks a notched bar and the energy lost is the toughness, which, tested over a range of temperatures, reveals the ductile-to-brittle transition that matters for ship steels in cold water. Indentation is a hardness test, endurance a fatigue test and constant-load strain a creep test.