With 458 and 353 lbs., the difference 105 lbs., and

W1 ={458 − 353) × 3} + 353 = 688 lbs.

D

With 600 and 556 lbs., the difference = 44 lbs., and

W1 = {600 - 556) × 3} + 556 = 688 lbs.

WD

With 660 and 646 lbs., the difference = 14 lbs., and

646) × 3+ 646 = 688 lbs.

W1 = {660 - 646) ×

WD

Ꭰ

With 685 and 683.5 lbs., the difference

1.5 lbs., and

W2 = {685 - 683·5) × 3} + 683·5 = 688 lbs.

WD

Cases of differential Strain are very numerous in practice: thus, in a Railway bridge, the weight of the structure itself is the minimum, and that weight plus the weight of the train is the maximum. Again, with long rods to deep-well pumps the maximum strain is the pressure on the bucket due to the head of water added to the weight of the rods, &c.; the minimum strain being the latter alone, &c.

(915.) "Alternating Strains."-When a strain is alternately tensile and compressive, as, for instance, with the piston-rod of a steam-engine, or again, when the transverse strain on a beam acts in both directions, up-and-down, resolving itself eventually into alternating tensile and crushing strains, the destructive action or tendency to break is very severe. Indeed, instinct teaches us that the easiest mode of breaking anything is to bend it to-and-fro repeatedly, a very moderate strain thus exerted sufficing to effect the purpose.

From Wöhler's experiments it appears that the destructive effect of alternate strains in opposite directions is expressed by the sum of those strains:—thus 5 tons tensile, alternating with 5 tons compressive strain is, in its tendency to break the material, equivalent to 10 tons acting intermittently or off-andon in one direction only.

With wrought iron we found in (914) that the safe intermittent tensile strain was 57 tons per square inch; the com

pressive strain would be the same, or 5.7 tons also, because the tensile and crushing strengths of wrought iron are equal to one another (377). With equal strains, alternately tensile and compressive, we should have therefore 5.72 2.85 tons each way, which is th of the statical breaking weight, and is equivalent to 2.85 + 2·85 = 5·7 tons off-and-on in one direction only.

(916.) But with many materials the resistance to these two strains is very unequal, as shown by Table 79, and this fact complicates the question considerably. For instance, with cast iron, the tensile breaking strain is 7·14 tons; Factor 3 gives 7.1432.28 tons statical safe load, which for an intermittent strain is reduced to 2.28 × •76 ton, and for an alternating strain, to 76 x 38 ton, which is statical breaking weight.

=

=

of the

But the crushing strength of cast iron is 43 tons, hence the statical safe load becomes 43÷3 14.3; the safe intermittent strain 14 3 x = 4.8 tons; and the alternating or crushing and tensile strains 4.8 × = 2-4 tons per square inch. We have just seen, however, that the alternating tensile strain does not exceed 38 ton or about 4th of the same kind of compressive strain, and we find from this, that where, in any material, the power of resistance to these two strains is unequal, the case is governed, and the alternating strain limited by the strength of the weaker.

(917.) We have here supposed that the tensile and compressive loads are equal to one another, which is usually the case in practice; for instance, in a piston-rod or a double-acting pumprod the two loads, or those during the up-stroke and the downstroke, are practically equal. But in some exceptional cases they may be so unequal as to enable us to utilise the whole strength of the material, or that in both directions; for instance, with cast iron, a compressive strain of 2.4 tons per square inch might be arranged so as to alternate with a tensile strain of 38 ton. But in most cases the two loads are of necessity equal to one another, and in the case of cast iron, 38 ton would become the working tensile and compressive strain.

(918.) With the transverse and torsional strains the com

plication which arises from unequal tensile and compressive strength in the material is eliminated, for although the former strains resolve themselves eventually into the tensile and compressive, the inequalities of strength adjust themselves to one another, and the transverse strength is the mean effect of the two combined.

An alternating transverse strain is very common in practice; the beam of an ordinary steam-engine, and the lever working a double-acting pump are familiar instances. Another case is that of an over-hung axle heavily loaded at the end, as in Fig. 204 if the axle were stationary it would simply be deflected from A, its normal position, to B. If while thus strained the axle could make half a revolution, it would evidently be carried round to the position C, but the weight W continuing to act, it is retained in the position B. Thus at every revolution, the shaft is in effect strained both ways, or up and down, namely, from B to C, and, as shown by Wöhler's experiments (915), the destructive effect is proportional, not to A, B, only, but to B, C.

The strain in this case is peculiar, 1st, although intermittent and alternate, or in both directions, it is effected entirely without shock: and, 2nd, the axle is strained not only in two directions, or up and down, but in all directions equally, which would probably be more destructive than an equivalent strain in two opposite directions only.

DYNAMIC FATIGUE.

(919.) The philosophy of dynamic fatigue may be easily explained: let Fig. 206 represent an unloaded beam A, deflected 8 inches or to the position B, by a strain or weight of 8 lbs., being 1 inch per lb. Now, as shown in (775), the mean strain is (80) 24 lbs., which acting through 8 inches gives as the mechanical power producing the deflection 4 x 8 = 32 inchlbs., therefore a weight D of 1 lb. falling 32 inches from E to B would deflect the beam to B as before-neglecting inertia (781).

If the elasticity of the material were perfect, the bar would sustain any number of similar blows without injury or increase

of deflection. But it is shown in (751) that when a beam (of cast iron more particularly) is deflected by a transverse strain, it never returns to its primitive form on the relief of the strain, but takes a permanent set. Now, let us suppose that in our case, a permanent set of 1 inch occurs, so that when, unloaded, the beam returns, not to A, but to G; if then again loaded with the same weight of 8 lbs. quietly laid on, the bar would deflect to B as before, and the mean strain would also be (0+8) ÷2 = 4 lbs. as before, but as the bar now deflects with that load 8 1 = 7 inches, we should have 4 x 7 28 inch-lbs. only, which would not absorb the whole force of the second blow by 32 28 4 inch-lbs. The beam would therefore be deflected below the point B by about 0.52 inch, or to 8.52 inches below A: then the mean strain during the second blow would be (08.52) 24.26 lbs., which acting through 8.52-1= 7.52 inches will give 4.26 × 7·52 = 32 inch-lbs., as required.

We have seen in (752) that the permanent set is nearly proportional to the square of the strain; if, therefore, the deflection of 8 inches gave 1 inch set, 8.52 inches would give 1 × 8.52a

821.134 inch set. The third blow will therefore deflect the beam still lower than before, say to 8.58 inches, or the point due to 8.58 lbs. dead load, the mean strain will then be (0+8·58)÷2=4·29 lbs., which acting through 8.58 −1·134 = 7·446 inches, will give a resistance of 4.29 × 7·446 = 32 inch-lbs. nearly, as required.

The permanent set due to 8.58 inches deflection will be 1 x 8.5882: = 1.15 inch: the fourth blow will therefore deflect the bar to say 8.6 inches, the mean strain being then (0 + 8·6) ÷ 2 = 4.3 lbs., which acting through 8.6 - 1.15 = 7.45 inches, will give a resistance of 4.3 x 7.45 = 32 inch-lbs., as required.

Thus with four successive and equal blows we have deflections and strains of 8.0, 8.52, 8.58, and 8.60 inches respectively, increasing with every blow, but in a rapidly diminishing ratio, namely, 52, 06, and 02 inch respectively. Evidently, although the first blow may have strained the bar to a small fraction only of the breaking weight, a succession of similar

blows would eventually break it: in fact, it becomes a question of number of blows as much as of amount of load.

The extreme slowness with which the deflection under successive blows increases, shows that the number of blows necessary to produce fracture may be very great, extending possibly to millions and occupying years (911). This will help to explain the well-known fact that parts of machinery (such as a pump-rod) often fail with a strain which is a small fraction only of the normal breaking weight, and one, moreover, which had been borne successfully for many years. Thus, the fact that a given dynamic strain has been sustained for a lengthened period, is no guarantee that it will be borne for ever.

(920.) Where a load literally falls upon a beam, as in Fig. 186, its effect must be calculated by the laws of Impact (789), as illustrated in (805). Mr. Hodgkinson made an extensive series of experiments on the power of impact with cast and wrought bars; many of his results are given by Tables 121, 122, 140, &c. The experiments in Table 140 are liable to be misunderstood; the bars were subjected not to given fractions of the breaking weights, which is the usual and most convenient course, but to strains producing given fractions of the ultimate deflection (696), which is quite a different thing with such an imperfectly elastic material as cast iron.

From the Diagram, Fig. 219, in which the deflections up to the breaking weights are shown graphically, we find that bars 1, 2, and 3 inches square are deflected to 3rd of the ultimate deflection by 434, 456, and 488 of the respective breaking weights, whereas, of course, with perfect elasticity we should have had or 333 in all cases. We have thus obtained col. 5

in Table 140.

(921.) The Table shows that with 1-inch bars, 4000 blows, deflecting the bar to of ultimate deflection, due by col. 5 to 434 of the breaking weight, failed to break the bar; another bar was not broken with 4000 blows, deflecting it to the ultimate deflection, due to 606 of the breaking weight; another broke with 3700 blows, deflecting it to the ultimate deflection, due to 684 of the breaking weight.

With 2-inch bars, two bore 4000 blows, deflecting them to