Theory of Machines, Vibrations and Machine Design

The last three sub-items of Section 2 are usually taught as three courses and are worth reading as one chain: a mechanism produces a motion, the motion excites a vibration, and the vibration is a load the design must survive. Mechanisms start with a count — n = 3(L − 1) − 2J − H — and the useful discipline is to do that count before anything else, because a linkage with two degrees of freedom cannot be analysed as though it had one. Grashof’s law, s + l ≤ p + q, then decides whether any link can make a full revolution, and which link you fix decides what the mechanism is called: the same four bars give a crank-rocker, a double-crank or a double-rocker depending on the choice. Epicyclic trains look intimidating and are pure bookkeeping — one table, one row for “arm fixed”, one for “everything rotated together”. Vibrations reduce almost entirely to two numbers, the natural frequency ω_n = √(k/m) and the damping ratio ζ = c/2√(km), and the single most useful result in the topic is that vibration isolation requires the forcing frequency to be above √2 ω_n — below that a mount makes things worse, and at exactly r = √2 the transmissibility is 1 for every value of damping. Machine design is then the business of comparing a computed stress state against a criterion, and the two criteria disagree: Tresca is always the more conservative, by up to 15 per cent, and knowing which one a question intends is worth more than remembering either formula.

Count the Freedoms, Then Ask Which Link You Fixed

Kutzbach’s criterion counts what is left after the constraints: n = 3(L − 1) − 2J − H, where L counts all links including the fixed one, J counts lower pairs each removing two freedoms, and H counts higher pairs each removing one. A four-bar has L = 4, J = 4, so n = 9 − 8 = 1 — one input, one motion, fully determined. A five-bar with five pins gives n = 15 − 10 = 2, which is not a mechanism but a two-degree-of-freedom system needing two inputs, and this is the check that stops you writing one loop-closure equation where two are needed. Replace one pin with a cam contact and L = 4, J = 3, H = 1 gives n = 2 as well, because a higher pair constrains less than a lower one — which is exactly why cams need a spring to hold the follower down.

🎯 Grashof decides whether, fixing decides what
Grashof’s law asks one question: can any link make a complete revolution relative to another? The condition is s + l ≤ p + q — the shortest plus the longest no greater than the other two. Test three sets of four links. (25, 40, 50, 60): 85 < 90, Grashof — full rotation is possible somewhere in this chain. (30, 40, 50, 70): 100 > 90, non-Grashof — no link can fully rotate, whichever one you fix, so all four arrangements are double rockers and this linkage can never be driven by a motor. (30, 40, 50, 60): 90 = 90 exactly, a change-point mechanism, where the links can become collinear and the output direction is momentarily indeterminate — a parallelogram linkage is the everyday example, and it is why railway signal linkages and parallel-motion drawing arms need a physical stop to pick a branch. Now the second half, which is where marks are lost: Grashof tells you rotation is possible, and which link you fix tells you what the mechanism is. Fix a link adjacent to the shortest and the shortest becomes the input crank — a crank-rocker, the arrangement in every reciprocating pump. Fix the shortest link itself and both its neighbours can rotate fully — a double crank, or drag-link, used to convert steady rotation into rotation at a varying rate. Fix the link opposite the shortest, so the shortest becomes the coupler, and you get a double rocker. Same four bars, three different machines, no change to any length.
Epicyclic train, sun 20 / planet 20 / annulus 60 teeth: the whole tabular method
OperationArmSun (20)Annulus (60)
1. Fix the arm, give the sun +z0+z−z/3 (external then internal mesh)
2. Rotate the whole train by +y+y+y+y
Sum — the general motionyy + zy − z/3
Annulus fixed: set y = z/3z/34z/30
Result — arm ÷ sun1/44:1 reduction, same sense = 1 + N_a/N_s—
Sun fixed instead: annulus in, arm out0.75 × annulus01.333:1 = 1 + N_s/N_a

Two things about that table are worth carrying into the exam. First, check the geometry before you compute anything: a sun–planet–annulus set only meshes if N_a = N_s + 2N_p, here 60 = 20 + 40 ✓, and a question quoting tooth counts that fail this is either testing you or misprinted. Second, notice that holding a different member turns the same hardware into a different gearbox — annulus fixed gives 4:1, sun fixed gives 1.333:1, and holding the arm gives −3 with the output reversed. That is the whole basis of an automatic transmission: one epicyclic set with brakes on different members provides several ratios and a reverse without moving a single gear. A flywheel finishes the same story from the energy side: it does not change the mean speed, only smooths the fluctuation, and I = E/(C_s ω²) sizes it. For an engine releasing E = 2000 J of excess work per cycle at 300 rpm with a permitted speed fluctuation of 2 per cent, ω = 31.416 rad/s and I = 2000/(0.02 × 986.96) = 101.3 kg·m² — and because ω² sits in the denominator, doubling the running speed cuts the required inertia to a quarter, which is why high-speed machinery needs surprisingly small flywheels.

Vibrations: Two Numbers, and the Threshold at √2

A single-degree-of-freedom system is fully described by ω_n = √(k/m) and ζ = c/c_c with c_c = 2√(km). Take m = 10 kg, k = 4000 N/m, c = 40 N·s/m: then ω_n = 20 rad/s = 3.183 Hz, c_c = 400 N·s/m, and ζ = 0.1. The damped natural frequency is ω_d = ω_n√(1 − ζ²) = 19.900 rad/s — only 0.50 per cent below the undamped value, which is the practical point of the algebra: for the light damping real machines have, ω_d ≈ ω_n and the shift is not worth computing. What damping does control is decay. The logarithmic decrement δ = 2πζ/√(1 − ζ²) = 0.6315 says successive peaks fall by a factor e^0.6315 = 1.88×, so the amplitude nearly halves every cycle. Springs combine the way their deflections do, not the way it feels: in parallel the stiffnesses add, k_eq = 2k, and ω_n rises by √2 to 28.28 rad/s; in series they add reciprocally, k_eq = k/2, and ω_n falls by √2 to 14.14 rad/s. A factor of two in frequency between the two arrangements of the same two springs.

🎯 Why a soft mount below √2 makes things worse, exactly
Transmissibility is the ratio of force transmitted to the foundation to force applied, TR = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²) with r = ω/ω_n. Tabulate it for ζ = 0.1. At r = 0.5 it is 1.328 — the mount is transmitting a third more force than it received. At resonance r = 1 it is 5.099, five times worse. At r = √2 it is exactly 1.000, and here is the striking part: put any ζ you like into the formula at r² = 2 and both numerator and denominator become √(1 + 8ζ²), so the transmissibility at r = √2 is 1 for every value of damping. Above that it finally drops — 0.356 at r = 2 and 0.145 at r = 3. So isolation is not a matter of choosing a soft enough mount in general; it is a matter of getting the forcing frequency above √2 ω_n, which for a fixed running speed means making the mount softer until ω_n is low enough. A mount that is not soft enough to clear √2 is not a partial improvement — it is actively harmful, and the machine would be better bolted straight to the floor. One more counter-intuitive consequence sits in the same table: above √2, more damping means worse isolation, because damping adds a force path that stiffness alone does not have. Damping is what you need to survive passing through resonance during run-up; it is not what you want once you are clear of it.
Magnification and transmissibility at ζ = 0.1
r = ω/ω_nMagnificationTransmissibilityVerdict
0.51.3221.328Worse than a rigid mount
1.05.000 = 1/2ζ5.099Resonance — five times worse
√2 = 1.4140.9621.000 exactly, for any ζThe break-even threshold
2.00.3300.356Genuine isolation begins
3.00.1250.14585 per cent of the force blocked

The critical or whirling speed of a shaft is the same number wearing a different name: a rotor runs unstably when its rotational speed reaches the lateral natural frequency of the shaft, so ω_c = √(g/δ) where δ is the static deflection under the rotor’s own weight. A shaft sagging 1 mm has ω_c = √(9810/1) = 99.05 rad/s = 946 rpm; stiffen it so it sags only 0.5 mm and the critical speed rises to 1338 rpm, by √2. That formula is the fastest sanity check in the topic and it needs no material property at all — the deflection has already absorbed E and I. It also explains a piece of machine-shop practice that sounds reckless: a rotor designed to run above its critical speed is passed through it quickly during run-up, and is quieter above than below, because at high r the rotor spins about its centre of mass rather than its geometric centre.

Machine Design: Two Criteria That Disagree, and Four Ways a Joint Fails

Design begins where the last chapter ended: you have a principal stress state, and you need to know whether the part yields. Two criteria are in the syllabus and they give different answers. Tresca (maximum shear stress) says yielding starts when σ_max − σ_min reaches S_y, taking the extremes over all three principal stresses. Von Mises (distortion energy) says it starts when √(½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]) reaches S_y. Take the state from the previous chapter, σ₁ = 116.02, σ₂ = −56.02, σ₃ = 0 MPa. Tresca gives an equivalent stress of 116.02 − (−56.02) = 172.05 MPa; von Mises gives 151.99 MPa. Tresca is 13.2 per cent more conservative, and it always is — the Tresca hexagon is inscribed in the von Mises ellipse, so they agree only at six points and Tresca is never the less cautious of the two. At S_y = 250 MPa the factors of safety are 1.453 and 1.645, and choosing the wrong criterion is a 13 per cent error in the same direction every time. The sharpest disagreement is in pure shear: Tresca predicts yielding at τ = 0.5 S_y, von Mises at τ = 0.577 S_y, a 15.5 per cent gap, and that is the case worth remembering because a shaft in torsion is exactly it.

🎯 A riveted joint fails four ways, and doubling the shear moves the answer
Take one pitch length of a single-riveted lap joint: pitch p = 60 mm, plate t = 10 mm, rivet d = 20 mm, with σ_t = 120, τ = 90 and σ_c = 180 MPa. Four modes, four numbers. Plate tearing between rivet holes: (p − d) t σ_t = 40 × 10 × 120 = 48 000 N. Rivet shear in single shear: (π/4)d² τ = 314.16 × 90 = 28 274 N. Crushing of plate or rivet: d t σ_c = 20 × 10 × 180 = 36 000 N. And for the efficiency, the solid plate it replaces: p t σ_t = 60 × 10 × 120 = 72 000 N. The joint is as strong as its weakest mode, so its strength is 28 274 N — shear governs — and the efficiency is 28 274/72 000 = 39.27 per cent. Now put the same rivet in double shear, as a butt joint with two cover plates does. Shear capacity doubles to 56 549 N, and it stops being the weakest: crushing takes over at 36 000 N, and the efficiency becomes exactly 36 000/72 000 = 50 per cent. Two lessons that generalise well past rivets. First, which mode governs is itself the answer to most design questions — doubling the shear area bought only a third more strength, because the bottleneck moved rather than disappeared. Second, an efficiency under 50 per cent is not a mistake in the arithmetic: a riveted joint really does throw away half the plate’s strength, and that is why aircraft skins moved to adhesive bonding and pressure vessels to welding.

The last idea in the section is fatigue, and its message is that a part which is safe under a steady load can fail under a fluctuating one at a fraction of the same stress. Decompose the cycle into a mean stress σ_m and an alternating amplitude σ_a, and the two do not do the same damage: the alternating part drives crack growth, the mean part merely opens the crack. The design lines combine them, and the two named in the syllabus differ only in which strength they use as the intercept: Goodman σ_a/S_e + σ_m/S_ut = 1/n and Soderberg σ_a/S_e + σ_m/S_y = 1/n. For S_e = 200, S_ut = 600, S_y = 400 MPa with σ_a = 100 and σ_m = 150 MPa, Goodman gives n = 1.333 and Soderberg n = 1.143 — Soderberg is the more conservative, always, because S_y < S_ut, and the reason is that it refuses to allow any yielding at all rather than merely refusing fracture. Two things are worth having internalised: a part with zero mean stress can still fail by fatigue, because σ_a alone consumes the whole allowance; and the endurance limit is not a material constant you can quote from a table without correction — surface finish, size, temperature and above all stress concentration cut it, which is why a fatigue failure almost always starts at a fillet, a keyway or a hole.

Key takeaways

  • Count the degrees of freedom before analysing anything: n = 3(L − 1) − 2J − H. Four-bar and six-bar give 1; a five-bar with five pins gives 2 and needs two inputs.
  • Grashof (s + l ≤ p + q) decides whether a link can rotate; which link you fix decides what the mechanism is. Fix a neighbour of the shortest for a crank-rocker, the shortest itself for a double crank, the opposite link for a double rocker.
  • An epicyclic set is a different gearbox depending on which member you hold. Sun 20 / annulus 60: annulus fixed gives 4:1 same-sense, sun fixed gives 1.333:1, arm fixed gives −3 reversed. Check N_a = N_s + 2N_p first.
  • Isolation needs r > √2, and at r = √2 the transmissibility is exactly 1 for every damping ratio. Below it a mount transmits more than it receives — 1.328 at r = 0.5, 5.099 at resonance — and above √2 more damping means worse isolation.
  • Light damping barely moves the frequency but dominates the decay. At ζ = 0.1, ω_d is 0.50 per cent below ω_n, while the logarithmic decrement 0.6315 halves the amplitude every cycle.
  • Tresca is always the more conservative criterion. On σ₁ = 116.02, σ₂ = −56.02, σ₃ = 0: Tresca 172.05 MPa against von Mises 151.99 MPa, 13.2 per cent apart; in pure shear the gap is 0.5 S_y against 0.577 S_y.
  • A joint is as strong as its governing mode, and the governing mode moves. The lap joint here fails in shear at 28.3 kN (39.27 per cent efficient); in double shear crushing takes over at 36 kN and efficiency becomes exactly 50 per cent.

Practice questions (8)

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.

  1. A planar linkage has 5 links connected by 5 revolute joints and no higher pairs. Its number of degrees of freedom is ______

    Numerical answer — type the value.

    Show answer

    Answer: 2

    2. Kutzbach: n = 3(L − 1) − 2J − H = 3(4) − 2(5) − 0 = 12 − 10 = 2. This is not a mechanism in the usual sense but a two-degree-of-freedom system: it needs two independent inputs before its configuration is determined, and giving it one input leaves the rest of the linkage free to move. The practical consequence is what the count is for — you cannot analyse it with the single loop-closure equation that solves a four-bar, and any attempt to do so produces one equation in two unknowns and no answer. Compare with the ones that do come out to 1: the four-bar, 3(3) − 2(4) = 1, and the six-bar with seven joints, 3(5) − 2(7) = 1, which is why Watt and Stephenson six-bars are the standard way to get a more complex single-input motion. A cautionary note the formula cannot supply: Kutzbach counts constraints without checking whether any are redundant, so a linkage with parallel links can compute as n = 0 while moving perfectly well — a parallelogram linkage is the standard counterexample, and it is why the count is a first check and not a proof.

  2. A four-bar chain has links of 30, 40, 50 and 70 mm. Which statement is correct?

    1. Fixing the 40 mm link gives a crank-rocker
    2. No link can make a complete revolution, whichever link is fixed
    3. Fixing the 30 mm link gives a double crank
    4. It is a change-point mechanism
    Show answer

    Answer: B — No link can make a complete revolution, whichever link is fixed

    No link can fully rotate. Test Grashof first: s + l = 30 + 70 = 100 and p + q = 40 + 50 = 90, so 100 > 90 and the chain is non-Grashof. That single inequality settles all four possible arrangements at once — a non-Grashof chain gives a double rocker whichever link you fix, so options A and C are ruled out not by checking them but by the classification failing before it starts. This is the useful shape of the law: Grashof is a property of the four lengths, and the fixing choice only names the mechanism if Grashof passes. Option D describes the boundary case s + l = p + q exactly, which these lengths miss by 10 mm — change the 70 to 60 and it would be a change-point mechanism, where the links can go collinear and the output direction becomes momentarily indeterminate. The practical reading of B: this linkage cannot be driven by a motor at all. Every one of its links rocks between limits, so it can only be driven by an oscillating input, which is why a designer checking s + l ≤ p + q is really checking whether the thing can be powered.

  3. In an epicyclic train the sun has 20 teeth, each planet 20 teeth and the annulus 60 teeth. With the annulus held fixed, the sun as input and the arm as output, the speed reduction ratio (sun speed ÷ arm speed) is ______

    Numerical answer — type the value.

    Show answer

    Answer: 4

    4. Check the geometry first: a sun–planet–annulus set meshes only if N_a = N_s + 2N_p, and 60 = 20 + 2(20) ✓. Now the tabular method, two rows. Row 1, arm fixed, sun +z: the sun drives the planet through an external mesh (reversing) and the planet drives the annulus through an internal mesh (not reversing), so the annulus gets −z N_s/N_a = −z/3. Row 2, rotate the whole train +y. Summing: arm = y, sun = y + z, annulus = y − z/3. Setting the annulus to zero gives y = z/3, so the arm turns z/3 while the sun turns 4z/3, and the ratio is 4:1 in the same sense. The closed form is worth carrying: with the annulus fixed and the arm as output, the reduction is 1 + N_a/N_s = 1 + 3 = 4, and the planet tooth count never enters it — the planets are idlers that set the geometry, not the ratio, which is exactly what a question quoting an odd planet size is testing. Hold a different member and the same hardware becomes a different gearbox: sun fixed gives 1 + N_s/N_a = 1.333:1, and arm fixed gives −N_a/N_s = −3 with the output reversed. One epicyclic set, three ratios and a reverse, which is the whole basis of an automatic transmission.

  4. A machine is mounted on isolators. To reduce the force transmitted to the foundation below the applied force, the ratio of forcing frequency to natural frequency must be

    1. less than 1
    2. greater than √2
    3. equal to 1
    4. greater than 1 but less than √2
    Show answer

    Answer: B — greater than √2

    Greater than √2. Transmissibility is TR = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²). The break-even point is where TR = 1, and putting r² = 2 makes the numerator and denominator both √(1 + 8ζ²) — so TR = 1 at r = √2 for every value of damping, and isolation begins only above it. Below √2 the mount is worse than no mount at all: with ζ = 0.1 the transmissibility is 1.328 at r = 0.5 and 5.099 at resonance, so option A does not merely fail to isolate, it amplifies. Option C is resonance, the worst possible choice. Option D is the trap that catches candidates who remember "above the natural frequency" without the √2: at r = 1.2 the transmissibility is still above 1, and the mount is still making things worse. Above the threshold it finally pays: 0.356 at r = 2 and 0.145 at r = 3. In practice the design lever is the mount, not the machine — the running speed is given, so you make the isolator softer to push ω_n down until r clears √2. And one more consequence sits in the same formula: above √2, increasing damping makes isolation worse, because the damper is an extra force path to the floor. Damping is what gets you safely through resonance during run-up, not what you want at running speed.

  5. A spring–mass–damper system has m = 10 kg, k = 4000 N/m and c = 40 N·s/m. Its logarithmic decrement is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 0.631

    0.631. First ζ: the critical damping coefficient is c_c = 2√(km) = 2√(4000 × 10) = 2 × 200 = 400 N·s/m, so ζ = 40/400 = 0.1. Then δ = 2πζ/√(1 − ζ²) = 0.6283/0.99499 = 0.6315. What the number means is more useful than the number: successive amplitude peaks fall by a factor of e^0.6315 = 1.88, so the amplitude drops to just over half its value every cycle and to under a fifth after three. Two related figures from the same inputs. The undamped natural frequency is ω_n = √(4000/10) = 20 rad/s = 3.183 Hz, and the damped frequency is ω_d = 20√0.99 = 19.900 rad/s — only 0.50 per cent lower. That contrast is the point of the topic: at the light damping real machinery has, damping barely touches the frequency and completely controls the decay, so a question asking for ω_d at ζ = 0.1 is usually testing whether you know the correction is negligible. The common slip is c_c = 2√(k/m), which is dimensionally wrong — the critical coefficient has to carry units of N·s/m, so it must be 2√(km), and checking the units is faster than recalling the formula.

  6. A point in a machine part is in plane stress with principal stresses 116 MPa and −56 MPa. Which of the following are correct?

    1. The Tresca equivalent stress is about 172 MPa
    2. The von Mises equivalent stress is about 152 MPa
    3. Tresca predicts a lower factor of safety than von Mises for this state
    4. For any stress state, von Mises is the more conservative of the two
    Show answer

    Answer: A — The Tresca equivalent stress is about 172 MPa; B — The von Mises equivalent stress is about 152 MPa; C — Tresca predicts a lower factor of safety than von Mises for this state

    A, B and C. Tresca takes the spread over all three principal stresses, and the third is zero here, so the equivalent stress is 116.02 − (−56.02) = 172.05 MPa ✓. Von Mises gives √(½[(116.02 + 56.02)² + 56.02² + 116.02²]) = 151.99 MPa ✓. Since Tresca reports the larger equivalent stress, it reports the smaller factor of safety — at S_y = 250 MPa the two give 1.453 and 1.645 — so C follows directly and is the useful practical statement. D is false, and it is false in general and not just here: the Tresca hexagon is inscribed in the von Mises ellipse, so Tresca can never predict a higher capacity, and the two agree only at the six vertices where the hexagon touches the ellipse. Tresca is therefore always the more conservative criterion, by up to 15.5 per cent, and that maximum gap occurs in pure shear, where Tresca yields at τ = 0.5 S_y and von Mises at τ = S_y/√3 = 0.577 S_y. The reason it matters is that a shaft in torsion is precisely the pure-shear case, so a design question about a shaft is exactly where the choice of criterion changes the answer most.

  7. In a single-riveted lap joint the pitch is 60 mm, plate thickness 10 mm and rivet diameter 20 mm, with allowable tensile, shear and crushing stresses of 120, 90 and 180 MPa. The efficiency of the joint, in per cent, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 39.27

    39.27 per cent. Compute all four numbers for one pitch length and the answer is the ratio of the smallest failure load to the solid-plate load. Tearing of the plate between holes: (60 − 20) × 10 × 120 = 48 000 N. Shear of the rivet, single shear: (π/4)(20²)(90) = 28 274 N. Crushing: 20 × 10 × 180 = 36 000 N. Solid plate: 60 × 10 × 120 = 72 000 N. The joint is only as strong as its weakest mode, so its strength is 28 274 N and shear governs, giving 28 274/72 000 = 39.27 per cent. Two things worth taking away. First, the governing mode is itself the answer to the design question: put the same rivet in double shear, as a butt joint with two cover plates does, and shear capacity doubles to 56 549 N — but crushing then becomes the weakest at 36 000 N, so the efficiency rises only to exactly 50 per cent, not to twice 39.27. Doubling the shear area moved the bottleneck rather than removing it. Second, an efficiency under half is not an arithmetic error: a riveted joint genuinely discards most of the plate’s strength, which is the engineering reason pressure vessels moved to welding and aircraft skins to adhesive bonding. The common mistake is dividing by the tearing load instead of the solid-plate load — that gives 58.9 per cent and answers a different question, since efficiency is defined against the unpunched plate.

  8. A shaft carries a fluctuating load with mean stress σ_m and alternating stress σ_a. For the same σ_m and σ_a, the Soderberg criterion compared with the Goodman criterion gives

    1. a lower factor of safety, because it uses the yield strength as the intercept
    2. a higher factor of safety, because it allows some yielding
    3. the same factor of safety, since both use the endurance limit
    4. a lower factor of safety only when σ_m is zero
    Show answer

    Answer: A — a lower factor of safety, because it uses the yield strength as the intercept

    A lower factor of safety, because of the yield-strength intercept. The two lines differ in exactly one term: Goodman is σ_a/S_e + σ_m/S_ut = 1/n and Soderberg is σ_a/S_e + σ_m/S_y = 1/n. Since S_y < S_ut for every engineering metal, the Soderberg denominator is smaller, the left-hand side is larger, and n comes out smaller. Numerically, with S_e = 200, S_ut = 600, S_y = 400 MPa, σ_a = 100 and σ_m = 150 MPa: Goodman gives n = 1/(0.5 + 0.25) = 1.333 and Soderberg n = 1/(0.5 + 0.375) = 1.143. Option B has the logic exactly inverted — Soderberg is stricter because it refuses any yielding, and refusing more possibilities cannot raise a safety factor. Option D is where the marks actually hang: the two criteria coincide only when σ_m = 0, because then the mean-stress term vanishes and both reduce to σ_a/S_e = 1/n. So D has it backwards — zero mean stress is the one case where they agree, not the only case where they differ. That coincidence is also a useful reminder that a part with zero mean stress can still fail in fatigue: σ_a alone consumes the whole allowance, and completely reversed loading is the harshest fatigue case, not the mildest.