Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock): Exercises

Intuition How to use this page

These problems climb a ladder. L1 just asks you to recognize which load is which. By L5 you combine axial, bending, and dynamic reasoning in one shot — exactly what a launch-vehicle stress engineer does at max-Q. Try each one with the Solution collapsed first, then open it. Every number here is machine-verified.

Recurring symbols (all built in the parent parent topic):

$F_{axial}$ = internal push/pull along the long axis (newtons, N).
$A$ = cross-section area carrying that push (square metres, $m^{2}$ ).
$σ$ = stress = force spread over area, in pascals (Pa) or megapascals (MPa = $1 0^{6}$ Pa).
$n$ = axial load factor, a plain number: "how many g's the structure feels".
$M$ = bending moment, a sideways twist measured in newton-metres (N·m).
$I$ = second moment of area, how the material is spread away from the centre-line ( $m^{4}$ ).
$c$ = distance from the centre-line to the outermost fibre (m).
$f_{n}$ = natural frequency, the "musical note" the part likes to ring at (hertz, Hz).
$Q$ = amplification, how many times bigger the shaking gets at resonance.

Level 1 — Recognition

Recall Solution L1.1

This is an axial (thrust) load. The engine pushes at the base and all the mass above resists being sped up, so the lower structure is pushed on from both ends → compressive. What it looks like: imagine pressing a soda can down with your thumb — the walls carry a squeeze.

Recall Solution L1.2

(a) sideways gust → bending (the slender body flexes like a broomstick pushed in the middle).
(b) sound → dynamic (acoustics sub-family; pressure waves at frequencies).
(c) explosive bolt → dynamic (shock sub-family; a very short, sharp transient).
(d) steady thrust → axial (a slowly-varying, quasi-static push along the axis).

Level 2 — Application

Recall Solution L2.1

(i) Force. What: use $F_{axial} = m n g$ . Why: the term $g + a$ (gravity plus engine acceleration) is bundled into $n g$ , so one number $n$ carries both effects. $F = 1500 \times 5 \times 9.81 = 7.358 \times 1 0^{4} N .$ (ii) Stress. What: $σ = F / A$ . Why: stress is force shared over the area that actually carries it. $σ = \frac{7.358 \times 1 0 ^{4}}{0.015} = 4.905 \times 1 0^{6} Pa = 4.9 MPa (compressive) .$

Recall Solution L2.2

What: use $f_{n} = \frac{1}{2 π} k / m$ . Why this tool: set the driving force to zero and the motion is simple harmonic; $k / m$ is the angular frequency, and dividing by $2 π$ converts "radians per second" into "cycles per second" (Hz).

= \frac{632.46}{2\pi} = 100.6\ \text{Hz}.$$ See [[Modal Analysis & Natural Frequencies]] for how real parts have many such frequencies.

Level 3 — Analysis

Recall Solution L3.1

(i) What: $I = π R^{3} t$ for a thin ring. Why: nearly all the metal sits a distance $R$ from the centre-line, and $I = \int y^{2} d A$ rewards material placed far out. Look at figure s01 — the wall is a thin band at radius $R$ . $I = π (1.2)^{3} (0.004) = π \times 1.728 \times 0.004 = 0.021714 m^{4} .$ (ii) What: the outer fibre ( $c = R = 1.2$ ) feels the most stretch/squeeze. Why $c = R$ : the further from the neutral axis, the more a fibre is strained when the beam curves (figure s02). $σ_{bend} = \frac{6 \times 1 0 ^{5} \times 1.2}{0.021714} = 3.316 \times 1 0^{7} Pa = 33.2 MPa .$ More on $I$ in Beam Bending & Second Moment of Area.

Figure — Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock)

Recall Solution L3.2

(i) Area of the thin ring. What: $A = 2 π R t$ (circumference × wall). Why: unrolling the thin tube gives a rectangle of length $2 π R$ and height $t$ . $A = 2 π (1.2) (0.004) = 0.030159 m^{2} .$ (ii) Axial stress. $σ_{axial} = \frac{5 \times 1 0 ^{5}}{0.030159} = 1.658 \times 1 0^{7} Pa = 16.6 MPa .$ (iii) Total. What: add them. Why: at max-Q both loads squeeze the same fibre at the same instant, so their stresses stack (figure s03). $σ_{total} = 16.6 + 33.2 = 49.7 MPa (compressive) .$ This is the worst-case sizing point — see Max-Q and Dynamic Pressure.

Level 4 — Synthesis

Recall Solution L4.1

(i) Why this formula: at resonance the peak response is $1/ (2 ζ)$ times the static one; low damping = high amplification. $Q = \frac{1}{2 ( 0.025 )} = 20.$ (ii) What Miles' equation does: it collapses the whole shaking spectrum into one RMS number by integrating the sharp resonance peak; the $π /2$ comes from the area under that peak. $g_{RMS} = \frac{π}{2} \times 120 \times 20 \times 0.05 = 188.50 = 13.73 g .$ (iii) Why $3 σ$ : random peaks follow a bell curve; $3 σ$ covers 99.7% of them, the standard design envelope. $3 σ = 3 \times 13.73 = 41.2 g .$ Background: Random Vibration & PSD and Factor of Safety & Margins of Safety.

Recall Solution L4.2

(i) Invert $f_{n} = \frac{1}{2 π} k / m$ → $k = (2 π f_{n})^{2} m$ . $k_{1} = (2 π \times 90)^{2} \times 8 = (565.49)^{2} \times 8 = 2.558 \times 1 0^{6} N/m .$ (ii) $k_{2} = (2 π \times 140)^{2} \times 8 = (879.65)^{2} \times 8 = 6.190 \times 1 0^{6} N/m .$ (iii) Why frequency scales with $k$ : $f_{n} \propto k$ , so to multiply $f_{n}$ by $140/90$ you multiply $k$ by that ratio squared. $\frac{k _{2}}{k _{1}} = (\frac{140}{90})^{2} = 2.420.$ So the bracket needs about 2.42× stiffer to clear the requirement.

Level 5 — Mastery

Recall Solution L5.1

(i) Axial. $A = 2 π R t = 2 π (0.9) (0.006) = 0.033929 m^{2}$ .

= \frac{9.7120\times10^4}{0.033929} = 2.862\times10^6\ \text{Pa} = 2.86\ \text{MPa}.$$ **(ii) Bending.** $I = \pi R^3 t = \pi (0.9)^3 (0.006) = 0.013741\ \text{m}^4$, $c = R = 0.9$. $$\sigma_\text{bend} = \frac{Mc}{I} = \frac{4.5\times10^5 \times 0.9}{0.013741} = 2.948\times10^7\ \text{Pa} = 29.48\ \text{MPa}.$$ **(iii) Total.** Add on the compressive fibre: $$\sigma_\text{total} = 2.86 + 29.48 = 32.34\ \text{MPa}.$$ **(iv) Margin.** *Why:* FoS inflates the demand; MoS tells you the spare capacity as a fraction. $$\text{MoS} = \frac{300}{1.5 \times 32.34} - 1 = \frac{300}{48.51} - 1 = 6.184 - 1 = 5.18.$$ $\text{MoS} = +5.18 \ge 0$ → **passes comfortably** (huge margin from bending being the driver). See [[Factor of Safety & Margins of Safety]].

Recall Solution L5.2

(i) $Q = 1/ (2 ζ) = 1/ (2 \times 0.02) = 25.$ (ii) Why multiply: at resonance the response is $Q$ times the static one. $σ_{peak} = 25 \times 40 = 1000 MPa .$ (iii) $MoS = \frac{500}{1.25 \times 1000} - 1 = \frac{500}{1250} - 1 = 0.4 - 1 = - 0.6.$ $MoS = - 0.6 < 0$ → FAILS. The "static" number looked safe (40 MPa vs 500 MPa yield), but resonant amplification multiplied it 25× and blew straight past yield. This is exactly why dynamic loads are analysed by frequency, not magnitude.

Recall One-line self-test

Why do axial and bending stresses add at max-Q but you never simply take the larger? ::: Because they load the same fibre at the same moment, so their stresses superpose: $σ_{total} = F / A + M c / I$ .