Which of the following statements about the tensile stress-strain behaviour of mild steel and aluminium alloys are correct?

Mild steel shows distinct upper and lower yield points

For aluminium alloys the yield strength is usually defined by the 0.2% offset method

The elastic modulus of aluminium alloys is about one third of that of steel

Mild steel shows distinct upper and lower yield points

For aluminium alloys the yield strength is usually defined by the 0.2% offset method

The elastic modulus of aluminium alloys is about one third of that of steel

Aluminium alloys have a much higher elastic modulus per unit density than steel

For a long body in a state of plane strain in the xy-plane, which statement is correct?

ε_z = 0 but σ_z = ν(σ_x + σ_y) in general

Both σ_z and ε_z are non-zero and unrelated

Which of the following statements about energy methods for linear elastic structures are correct?

The partial derivative of strain energy with respect to a load gives the displacement at that load in its direction

For a redundant reaction R, the condition ∂U/∂R = 0 gives its value

The strain energy of a bar in axial load P is P²L/(2AE)

The partial derivative of strain energy with respect to a load gives the displacement at that load in its direction

For a redundant reaction R, the condition ∂U/∂R = 0 gives its value

The strain energy of a bar in axial load P is P²L/(2AE)

Strain energy is linear in the applied load

A thin-walled circular tube is slit along its full length, turning it from a closed into an open section of the same wall. Its torsional stiffness:

Falls enormously, from 4A²/∮(ds/t) to Σbt³/3

Is unchanged, since the material is the same

Falls enormously, from 4A²/∮(ds/t) to Σbt³/3

Increases, because warping is freed

In two-dimensional elasticity without body forces, the Airy stress function φ is defined by σ_x = ∂²φ/∂y², σ_y = ∂²φ/∂x², τ_xy = −∂²φ/∂x∂y. Which statements are correct?

Any such φ satisfies the two equilibrium equations identically

Compatibility requires φ to satisfy the biharmonic equation ∇⁴φ = 0

Any polynomial in x and y of degree three or less is an admissible stress function

Any such φ satisfies the two equilibrium equations identically

Compatibility requires φ to satisfy the biharmonic equation ∇⁴φ = 0

Any polynomial in x and y of degree three or less is an admissible stress function

The compatibility equation is needed only when the body is statically determinate

Aircraft Structures: Strength of Materials, Thin-Walled Sections and Structural Dynamics

Section 4 of the GATE Aerospace Engineering (AE) paper, Structures, in one chapter. It follows the syllabus’s three core headings in order. Strength of materials comes first: stress and strain and the stress-strain curves of steel and aluminium, three-dimensional Hooke’s law, plane stress and plane strain; stresses and deflections in determinate and indeterminate bars, trusses, beams and shafts; stress and strain transformation, Mohr’s circle, principal stresses, combined loading and the maximum-stress, Tresca and von Mises failure criteria; strain energy, Castigliano’s theorems and Euler buckling. Then flight-vehicle structures: bending, shear and torsion of open and closed thin-walled sections, symmetric and unsymmetric, and the loads an aircraft carries. Then structural dynamics: free and forced vibration of damped and undamped single-degree-of-freedom systems and free vibration of undamped two-degree-of-freedom systems. It closes with the special topic of equilibrium and compatibility in two-dimensional elasticity.

1. Stress, strain, material curves and Hooke’s law in three dimensions

Normal stress σ is force per unit area normal to a plane and shear stress τ is force per unit area along it; normal strain ε is change of length per unit length and shear strain γ is change of a right angle in radians. The stress-strain curve of mild steel in tension is linear to the proportional limit, then shows a distinct upper and lower yield point and a yield plateau, then strain hardening up to the ultimate tensile strength, then necking and fracture; its modulus is about 200 GPa. Aluminium alloys show no distinct yield point — the curve bends over gradually — so the 0.2% offset proof stress is used as the yield strength; the modulus is about 70 GPa, a third of steel’s, but the density (about 2.7 against 7.85 g/cm³) is also about a third, so the specific stiffness E/ρ is similar and the specific strength of the aerospace alloys is high.

For a linear, isotropic material the generalised (three-dimensional) Hooke’s law is ε_x = [σ_x − ν(σ_y + σ_z)]/E and its two companions, with γ_xy = τ_xy/G and so on; only two constants are independent, G = E/(2(1 + ν)) and the bulk modulus K = E/(3(1 − 2ν)). Two two-dimensional idealisations follow. In plane stress (thin plates and skins loaded in their plane) σ_z = τ_xz = τ_yz = 0, but ε_z = −ν(σ_x + σ_y)/E is not zero — the plate thins. In plane strain (long bodies restrained along their length) ε_z = γ_xz = γ_yz = 0, but σ_z = ν(σ_x + σ_y) is not zero. A free thermal strain αΔT adds to the normal strains; if it is restrained it produces stress αEΔT.

2. Bars, trusses, beams and shafts — determinate and indeterminate

A structure is statically determinate when equilibrium alone gives every reaction and internal force; for a plane pin-jointed truss that needs m + r = 2j (members, reactions, joints), and each member is a two-force member solved by the method of joints or of sections. An axially loaded bar extends by δ = PL/(AE). When there are more unknowns than equilibrium equations the structure is indeterminate, and the extra equations come from compatibility of deformation: a bar fixed between two walls and heated carries σ = −αEΔT; a composite bar of two materials shares the load in proportion to their axial stiffnesses AE/L.

A beam carries transverse load through a shear force V and a bending moment M, related by dM/dx = V and dV/dx = −w. The flexure formula σ = My/I gives the bending stress, the shear formula τ = VQ/(Ib) the transverse shear stress (1.5 times the mean for a rectangle), and the elastic curve EI d²v/dx² = M the deflection. A circular shaft in torsion has τ = Tr/J and twist θ = TL/(GJ), with J = πd⁴/32 for a solid shaft. An indeterminate beam — a propped cantilever, a fixed-fixed beam — is solved by removing a redundant, computing the deflection it would allow, and restoring compatibility: for a propped cantilever under uniform load w the prop carries 3wL/8, from wL⁴/(8EI) = RL³/(3EI).

Standard beam results worth knowing without derivation
Beam and load	Maximum moment	Maximum deflection
Cantilever, end load P	PL (at the root)	PL³/(3EI)
Cantilever, uniform load w	wL²/2	wL⁴/(8EI)
Simply supported, central load P	PL/4	PL³/(48EI)
Simply supported, uniform load w	wL²/8	5wL⁴/(384EI)

3. Stress transformation, Mohr’s circle, combined loading and failure criteria

In plane stress the stresses on a plane rotated by θ are σ_x′ = (σ_x + σ_y)/2 + ((σ_x − σ_y)/2) cos 2θ + τ_xy sin 2θ and τ_x′y′ = −((σ_x − σ_y)/2) sin 2θ + τ_xy cos 2θ. Mohr’s circle draws these as a circle of centre C = (σ_x + σ_y)/2 and radius R = √(((σ_x − σ_y)/2)² + τ_xy²): the principal stresses are σ1,2 = C ± R, on planes at tan 2θ_p = 2τ_xy/(σ_x − σ_y) with no shear on them, and the maximum in-plane shear stress is R, on planes 45° from the principal ones. Strains transform the same way with γ/2 in place of τ, which is how strain-gauge rosettes are reduced to principal strains. Combined loading — a shaft in bending and torsion, a pressurised tube with axial load — is handled by superposing the stresses at the critical point and then transforming.

Failure criteria compare a multiaxial state with the uniaxial yield strength σ_Y. The maximum principal stress (Rankine) criterion, σ1 = σ_Y, suits brittle materials. For ductile metals, Tresca (maximum shear stress) says yield occurs when max(|σ1 − σ2|, |σ2 − σ3|, |σ3 − σ1|) = σ_Y, and von Mises (distortion energy) when ½[(σ1 − σ2)² + (σ2 − σ3)² + (σ3 − σ1)²] = σ_Y²; in plane stress the von Mises stress is σ_vm = √(σ1² − σ1σ2 + σ2²) = √(σ_x² − σ_xσ_y + σ_y² + 3τ_xy²). In pure shear Tresca predicts yield at τ = σ_Y/2 and von Mises at τ = σ_Y/√3 = 0.577σ_Y, so Tresca is the more conservative; its hexagon lies inside the von Mises ellipse and touches it at six points.

⚠️ In plane stress the third principal stress is zero, and it counts

With σ1 = 70 MPa and σ2 = −30 MPa, the in-plane maximum shear is 50 MPa and Tresca uses |σ1 − σ2| = 100 MPa. But if both in-plane principal stresses had the same sign — say 70 and 30 MPa — the largest difference would be σ1 − σ3 = 70 − 0 = 70 MPa, not 40 MPa. Forgetting σ3 = 0 underestimates the Tresca stress whenever σ1 and σ2 have the same sign.

4. Strain energy, Castigliano’s theorems and Euler buckling

Strain energy is the work stored in elastic deformation, U = ∫σ²/(2E) dV for normal stress. For the common members: U = P²L/(2AE) in axial load, ∫M²/(2EI) dx in bending, and T²L/(2GJ) in torsion. Castigliano’s second theorem says that in a linear elastic structure the displacement at a load, in its direction, is δ_i = ∂U/∂P_i (and a rotation is ∂U/∂M_i); a fictitious load placed where a displacement is wanted and then set to zero gives the unit-load (dummy-load) method. The first theorem is the converse, P_i = ∂U/∂δ_i, with U written in terms of displacements. For an indeterminate structure the redundant reaction R minimises the strain energy, ∂U/∂R = 0 — the principle of least work. For a cantilever with end load, U = ∫(Px)²/(2EI) dx = P²L³/(6EI), and ∂U/∂P = PL³/(3EI) — the table result recovered.

A slender column under axial compression stays straight until the Euler critical load P_cr = π²EI/L_e², at which a bent equilibrium becomes possible; it buckles about the axis of least I. The effective length L_e depends on the end conditions: L for pinned-pinned, 2L for fixed-free, 0.5L for fixed-fixed and about 0.7L for fixed-pinned. In stress terms σ_cr = π²E/(L_e/r)², with r = √(I/A) the radius of gyration, so the slenderness ratio L_e/r decides whether a member fails by buckling or by yielding. A fixed-free column carries only a quarter of the pinned-pinned load.

5. Thin-walled sections — bending, shear flow, shear centre and torsion; loads on aircraft

Aircraft structures are thin-walled shells — skins, spars, stringers — and their analysis uses the shear flow q = τt (force per unit length of wall). In bending, a section symmetric about the axis of the moment gives σ = M_x y/I_xx. An unsymmetric section (I_xy ≠ 0) bends out of the plane of the load and its neutral axis is inclined to it; the direct stress is σ_z = [(M_y I_xx − M_x I_xy)/(I_xx I_yy − I_xy²)]x + [(M_x I_yy − M_y I_xy)/(I_xx I_yy − I_xy²)]y, which reduces to the symmetric formula when I_xy = 0. In an idealised section the booms (stringers and spar caps with their effective skin) carry the direct stress and the skin between them carries only constant shear flow.

Shear flow in an open section follows from equilibrium of a wall element: for a section with I_xy = 0 under shear S_y only, q(s) = −(S_y/I_xx)∫₀ˢ t y ds, starting from zero at a free edge. The resultant of that shear flow acts through the shear centre, the point through which a shear load must pass to bend the section without twisting it. It lies on any axis of symmetry; for a channel of flange width b, web height h and uniform thickness it is e = 3b²/(h + 6b) from the web centre-line, on the side away from the flanges — outside the section. A closed section adds an unknown constant shear flow q0 at a cut, found from moment equivalence or from the condition of zero twist.

In torsion a closed single-cell section carries a constant shear flow given by the Bredt-Batho formula T = 2Aq, where A is the area enclosed by the wall’s mid-line, so τ = T/(2At); its rate of twist is dθ/dz = (T/(4A²G))∮ds/t, i.e. J = 4A²/∮(ds/t). An open section of the same material is far weaker in torsion: J = Σ(1/3)bt³ and τ_max = Tt/J. Slitting a thin tube along its length cuts its torsional stiffness by orders of magnitude, which is why wings are built as closed boxes. The loads such structures carry are the aerodynamic loads (lift distribution, manoeuvres and gusts through the load factor), inertia loads (n times each mass), landing and ground loads, propulsion loads and cabin pressurisation, for which a thin cylinder has hoop stress pr/t and longitudinal stress pr/(2t). Along a wing they appear as shear force, bending moment and torque, largest at the root; engines and fuel in the wing give bending relief.

🎯 Why the shear centre of a channel is outside it

The shear flow in the two flanges of a channel runs in opposite directions and forms a couple about the web. For the resultant of the section’s shear flow to have no net moment beyond that of the applied load, the load must act at a distance e on the far side of the web, where its moment balances the flange couple. Load a channel through its centroid and it twists as well as bends — a real concern for spar and stringer design.

6. Structural dynamics — SDOF and 2-DOF vibration

A single-degree-of-freedom system m ẍ + c ẋ + k x = F(t) has undamped natural frequency ω_n = √(k/m) (f_n = ω_n/2π). The damping ratio is ζ = c/c_c with critical damping c_c = 2√(km) = 2mω_n. For ζ < 1 (underdamped) free vibration is a decaying oscillation at the damped frequency ω_d = ω_n√(1 − ζ²); ζ = 1 (critically damped) returns fastest without oscillating; ζ > 1 (overdamped) creeps back. The logarithmic decrement between successive peaks, δ = ln(x_n/x_n+1) = 2πζ/√(1 − ζ²) ≈ 2πζ for light damping, is how damping is measured. Under harmonic forcing F0 sin ωt the steady-state amplitude is X = (F0/k)/√((1 − r²)² + (2ζr)²) with r = ω/ω_n, lagging the force by φ = tan⁻¹(2ζr/(1 − r²)). At resonance, r = 1, the phase lag is 90° and the magnification is 1/(2ζ); the peak itself is at r = √(1 − 2ζ²), slightly below.

An undamped two-degree-of-freedom system [M]{ẍ} + [K]{x} = 0 has solutions {x} = {X} sin ωt only when det([K] − ω²[M]) = 0 — the frequency equation, a quadratic in ω² giving two natural frequencies, each with its mode shape (the ratio X1/X2). For two equal masses m joined to each other and to two walls by three equal springs k, [K] = [[2k, −k], [−k, 2k]]: the first mode (masses in phase, X1/X2 = 1) is at ω1 = √(k/m), because the middle spring is not stretched, and the second (out of phase, X1/X2 = −1) is at ω2 = √(3k/m). Any free motion is a combination of the two modes.

7. Equilibrium and compatibility in two-dimensional elasticity

A two-dimensional elasticity problem has three unknown stresses, three strains and two displacements. Equilibrium of an element gives two equations, ∂σ_x/∂x + ∂τ_xy/∂y + F_x = 0 and ∂τ_xy/∂x + ∂σ_y/∂y + F_y = 0. The strains come from two displacements (ε_x = ∂u/∂x, ε_y = ∂v/∂y, γ_xy = ∂u/∂y + ∂v/∂x), so they cannot be arbitrary: the compatibility equation ∂²ε_x/∂y² + ∂²ε_y/∂x² = ∂²γ_xy/∂x∂y ensures that a continuous displacement field exists, one without gaps or overlaps. With no body forces, the Airy stress function φ defined by σ_x = ∂²φ/∂y², σ_y = ∂²φ/∂x², τ_xy = −∂²φ/∂x∂y satisfies both equilibrium equations identically, and compatibility together with Hooke’s law then requires the biharmonic equation ∇⁴φ = ∂⁴φ/∂x⁴ + 2∂⁴φ/∂x²∂y² + ∂⁴φ/∂y⁴ = 0. Solving a problem means finding a biharmonic φ that also meets the boundary tractions; polynomials of degree three or less satisfy ∇⁴φ = 0 automatically.

Key takeaways

Steel has a distinct yield point and E ≈ 200 GPa; aluminium alloys use the 0.2% proof stress and E ≈ 70 GPa. G = E/(2(1 + ν)). Plane stress: σ_z = 0, ε_z ≠ 0; plane strain: ε_z = 0, σ_z = ν(σ_x + σ_y).
Mohr: centre (σ_x + σ_y)/2, radius √(((σ_x − σ_y)/2)² + τ_xy²). Tresca yields in shear at σ_Y/2, von Mises at σ_Y/√3; σ_vm = √(σ1² − σ1σ2 + σ2²).
Castigliano: δ = ∂U/∂P; least work ∂U/∂R = 0 for redundants. Euler: P_cr = π²EI/L_e², L_e = L, 2L, 0.5L, 0.7L for pinned, fixed-free, fixed-fixed, fixed-pinned.
Thin walls: q = τt; open-section shear flow starts at zero at a free edge; channel shear centre e = 3b²/(h + 6b) outside the web; closed-cell torsion T = 2Aq, J = 4A²/∮ds/t, far stiffer than an open section’s Σbt³/3.
SDOF: ω_n = √(k/m), ζ = c/(2√(km)), δ = 2πζ/√(1 − ζ²), magnification 1/(2ζ) at resonance. 2-DOF: det(K − ω²M) = 0; equal masses and three equal springs give √(k/m) and √(3k/m).
2-D elasticity: two equilibrium equations, one compatibility equation; the Airy stress function satisfies equilibrium identically and must be biharmonic, ∇⁴φ = 0.

Practice questions (19)

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.

Which of the following statements about the tensile stress-strain behaviour of mild steel and aluminium alloys are correct?
1. Mild steel shows distinct upper and lower yield points
2. For aluminium alloys the yield strength is usually defined by the 0.2% offset method
3. The elastic modulus of aluminium alloys is about one third of that of steel
4. Aluminium alloys have a much higher elastic modulus per unit density than steel
Show answer
Answer: A — Mild steel shows distinct upper and lower yield points; B — For aluminium alloys the yield strength is usually defined by the 0.2% offset method; C — The elastic modulus of aluminium alloys is about one third of that of steel
Mild steel has a yield-point phenomenon and a plateau; aluminium alloys yield gradually, so the 0.2% proof stress is used; E is about 70 against 200 GPa. But the densities are in nearly the same ratio (2.7 against 7.85 g/cm³), so E/ρ is about the same — the fourth statement is false; aluminium wins on specific strength, not specific stiffness.
A material has Young’s modulus 200 GPa and Poisson’s ratio 0.3. What is its shear modulus, in GPa (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 76.92
G = E/(2(1 + ν)) = 200/(2 × 1.3) = 200/2.6 = 76.92 GPa. Using (1 − ν) gives 142.9 GPa, and E/(3(1 − 2ν)) = 166.7 GPa is the bulk modulus, not the shear modulus.
For a long body in a state of plane strain in the xy-plane, which statement is correct?
1. σ_z = 0 and ε_z = 0
2. σ_z = 0 but ε_z ≠ 0
3. ε_z = 0 but σ_z = ν(σ_x + σ_y) in general
4. Both σ_z and ε_z are non-zero and unrelated
Show answer
Answer: C — ε_z = 0 but σ_z = ν(σ_x + σ_y) in general
Plane strain prevents deformation along z, so ε_z = [σ_z − ν(σ_x + σ_y)]/E = 0 requires σ_z = ν(σ_x + σ_y). The second option is plane stress, the thin-plate case where σ_z vanishes and the plate is free to thin.
A steel bar 2 m long with cross-sectional area 500 mm² (E = 200 GPa) carries an axial tensile load of 50 kN. What is its elongation, in mm?
Numerical answer — type the value.
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Answer: 1
δ = PL/(AE) = 50 000 × 2/(500 × 10⁻⁶ × 200 × 10⁹) = 100 000/10⁸ = 1 × 10⁻³ m = 1 mm. Leaving the area in mm² while E is in Pa gives an answer a million times too small.
A cantilever of length 1 m carries a load of 1 kN at its free end; EI = 200 kN·m². What is the tip deflection, in mm (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 1.67
δ = PL³/(3EI) = 1000 × 1³/(3 × 200 × 10³) = 1000/600 000 = 1.667 × 10⁻³ m = 1.67 mm. The same follows from Castigliano: U = P²L³/(6EI), ∂U/∂P = PL³/(3EI). Using 48 in the denominator is the simply supported central-load formula.
A propped cantilever of span 4 m (fixed at one end, simply supported at the other) carries a uniformly distributed load of 8 kN/m over its whole length. What is the reaction at the prop, in kN?
Numerical answer — type the value.
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Answer: 12
Remove the prop: the cantilever tip deflects wL⁴/(8EI). The prop force R must push it back by RL³/(3EI), so R = 3wL/8 = 3 × 8 × 4/8 = 12 kN, leaving 32 − 12 = 20 kN at the wall. Taking half the load (16 kN) treats the beam as simply supported at both ends.
At a point in plane stress, σ_x = 60 MPa, σ_y = −20 MPa and τ_xy = 30 MPa. What is the maximum principal stress, in MPa?
Numerical answer — type the value.
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Answer: 70
Mohr’s circle: centre C = (60 − 20)/2 = 20 MPa, radius R = √(40² + 30²) = 50 MPa, so σ1 = C + R = 70 MPa and σ2 = −30 MPa. Using the full difference (σ_x − σ_y = 80) instead of half of it in the radius gives R ≈ 85.4 and σ1 ≈ 105 MPa.
For the same state (σ_x = 60 MPa, σ_y = −20 MPa, τ_xy = 30 MPa, plane stress), what is the von Mises equivalent stress, in MPa (to one decimal place)?
Numerical answer — type the value.
Show answer
Answer: 88.9
With σ1 = 70 and σ2 = −30, σ_vm = √(σ1² − σ1σ2 + σ2²) = √(4900 + 2100 + 900) = √7900 = 88.9 MPa. Directly: √(σ_x² − σ_xσ_y + σ_y² + 3τ²) = √(3600 + 1200 + 400 + 2700) = √7900, the same. Writing +σ1σ2 instead of −σ1σ2 gives √3700 = 60.8 MPa. The Tresca equivalent here is σ1 − σ2 = 100 MPa.
A ductile material with uniaxial yield strength σ_Y is loaded in pure shear. The shear stresses at yield predicted by the Tresca and the von Mises criteria are, respectively:
1. σ_Y/√3 and σ_Y/2
2. σ_Y/2 and σ_Y/√3
3. σ_Y and σ_Y/2
4. σ_Y/2 and σ_Y/2
Show answer
Answer: B — σ_Y/2 and σ_Y/√3
Pure shear τ has principal stresses +τ, −τ, 0. Tresca: σ1 − σ3 = 2τ = σ_Y, so τ = 0.5σ_Y. Von Mises: √(τ² + τ² + τ²) = √3·τ = σ_Y, so τ = 0.577σ_Y. Tresca is the more conservative of the two, and the first option has them swapped.
A column 2 m long (E = 200 GPa, least second moment of area 8 × 10⁻⁷ m⁴) is fixed at its base and free at its top. What is its Euler buckling load, in kN (to one decimal place)?
Numerical answer — type the value.
Show answer
Answer: 98.7
Fixed-free gives L_e = 2L = 4 m, so P_cr = π²EI/L_e² = 9.8696 × 200 × 10⁹ × 8 × 10⁻⁷/16 = 1.579 × 10⁶/16 = 98.7 kN. Using L_e = L (pinned-pinned) gives 394.8 kN, four times too much.
Which of the following statements about energy methods for linear elastic structures are correct?
1. The partial derivative of strain energy with respect to a load gives the displacement at that load in its direction
2. For a redundant reaction R, the condition ∂U/∂R = 0 gives its value
3. The strain energy of a bar in axial load P is P²L/(2AE)
4. Strain energy is linear in the applied load
Show answer
Answer: A — The partial derivative of strain energy with respect to a load gives the displacement at that load in its direction; B — For a redundant reaction R, the condition ∂U/∂R = 0 gives its value; C — The strain energy of a bar in axial load P is P²L/(2AE)
Castigliano’s second theorem gives δ_i = ∂U/∂P_i; at an unyielding support the displacement is zero, so ∂U/∂R = 0 (least work); and U = ½Pδ = P²L/(2AE). Strain energy is quadratic in load — doubling the load quadruples U — which is exactly why superposition does not apply to energy.
A thin-walled closed single-cell tube encloses an area of 0.1 m² (measured to the wall mid-line) and has a uniform wall thickness of 2 mm. It carries a torque of 10 kN·m. What is the shear stress in the wall, in MPa?
Numerical answer — type the value.
Show answer
Answer: 25
Bredt-Batho: q = T/(2A) = 10 000/(2 × 0.1) = 50 000 N/m, and τ = q/t = 50 000/0.002 = 25 × 10⁶ Pa = 25 MPa. Forgetting the 2 in T = 2Aq gives 50 MPa; the enclosed area, not the wall area, is what matters.
A thin-walled channel section of uniform thickness has a web of height 100 mm and two flanges each 50 mm wide (mid-line dimensions). How far is its shear centre from the web mid-line, in mm (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 18.75
e = 3b²/(h + 6b) = 3 × 50²/(100 + 300) = 7500/400 = 18.75 mm, on the side of the web away from the flanges. Directly: e = b²h²t/(4I) with I = th³/12 + 2bt(h/2)² = t(83 333 + 250 000), giving 2.5 × 10⁷/(4 × 333 333) = 18.75 mm. The centroid lies on the other side, inside the flanges.
A thin-walled circular tube is slit along its full length, turning it from a closed into an open section of the same wall. Its torsional stiffness:
1. Is unchanged, since the material is the same
2. Falls by a factor of about two
3. Falls enormously, from 4A²/∮(ds/t) to Σbt³/3
4. Increases, because warping is freed
Show answer
Answer: C — Falls enormously, from 4A²/∮(ds/t) to Σbt³/3
Closed: J = 4A²t/(2πr) = 2πr³t for radius r. Open: J = (2πr)t³/3. The ratio is 3(r/t)², so for r/t = 20 the slit tube is 1200 times less stiff. The closed cell carries torque by a shear flow circulating round it; the open one only by shear across its thickness.
A mass of 10 kg is supported on a spring of stiffness 4000 N/m. What is the undamped natural frequency, in rad/s?
Numerical answer — type the value.
Show answer
Answer: 20
ω_n = √(k/m) = √(4000/10) = √400 = 20 rad/s, i.e. f_n = 20/(2π) = 3.18 Hz. Quoting 3.18 answers in hertz when the question asks for rad/s.
In the free vibration of a viscously damped SDOF system, each amplitude peak is half the previous one. What is the damping ratio (to three decimal places)?
Numerical answer — type the value.
Show answer
Answer: 0.11
δ = ln(x_n/x_n+1) = ln 2 = 0.6931. Solving δ = 2πζ/√(1 − ζ²): ζ = δ/√(4π² + δ²) = 0.6931/√(39.478 + 0.480) = 0.6931/6.321 = 0.1097 ≈ 0.110. The light-damping approximation ζ ≈ δ/(2π) gives 0.1103, close; using log10 2 = 0.301 gives 0.048.
An SDOF system with damping ratio 0.05 is driven by a harmonic force exactly at its undamped natural frequency. What is the dynamic magnification factor (steady amplitude divided by static deflection)?
Numerical answer — type the value.
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Answer: 10
At r = 1 the (1 − r²) term vanishes, leaving M = 1/(2ζr) = 1/(2 × 0.05) = 10, with the response lagging the force by 90°. Answering 20 uses 1/ζ; the true peak, at r = √(1 − 2ζ²), is only fractionally above 10.
Two equal masses m are connected in a line by three equal springs k: wall–spring–mass–spring–mass–spring–wall. What is the ratio of the second natural frequency to the first (to three decimal places)?
Numerical answer — type the value.
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Answer: 1.732
det([[2k − mω², −k], [−k, 2k − mω²]]) = 0 gives 2k − mω² = ±k, so ω1² = k/m (in phase, middle spring unstretched) and ω2² = 3k/m (out of phase). Ratio √3 = 1.732. Answering 3 compares ω² rather than ω.
In two-dimensional elasticity without body forces, the Airy stress function φ is defined by σ_x = ∂²φ/∂y², σ_y = ∂²φ/∂x², τ_xy = −∂²φ/∂x∂y. Which statements are correct?
1. Any such φ satisfies the two equilibrium equations identically
2. Compatibility requires φ to satisfy the biharmonic equation ∇⁴φ = 0
3. Any polynomial in x and y of degree three or less is an admissible stress function
4. The compatibility equation is needed only when the body is statically determinate
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Answer: A — Any such φ satisfies the two equilibrium equations identically; B — Compatibility requires φ to satisfy the biharmonic equation ∇⁴φ = 0; C — Any polynomial in x and y of degree three or less is an admissible stress function
Substituting the definitions makes ∂σ_x/∂x + ∂τ_xy/∂y = φ_yyx − φ_xyy = 0 identically, and likewise the second equation. Compatibility with Hooke’s law gives ∇⁴φ = 0, which fourth derivatives of a cubic satisfy trivially. Compatibility is what makes the continuum problem solvable at all, since equilibrium alone never determines the stresses in an elastic body.