BIS 



APPLIED MECHANICS. 



[HBSIKTAXCB TO TORSION. 



If the bar be round, we must increase this by one-half, 

 making it 8 ; and u 8 is the cube of 2, or2 X 2 x 2, the 

 i. .iiu.l h.ir must be 2 ins. dinn 



V \\ lint the beam is loaded not in the middle. 



Jtlf. Multiply the strength product successively by 

 tho two lengths into which the weight divides the beam, 

 and divide twice by half the total length. 



EzampU 13. An ash-beam 14 feet long, has 10, 182 Ibs. 

 suspended from it 3 feet from one end (consequently 11 

 feet from the other, half the length being 7 feet). 



10,182 X 14 



260 



11X3 



6 inches broad and 8 inches deep, for 6x8x8 384. 

 V I. When the beam is fixed at one end and loaded at 

 the other. 

 Ride. Multiply the strength product by 4. 



tmpU 14. An English oak beam projects 10 feet, 

 sud carries 10,433 Ibs. at the end. 



10,433 X 10 

 190 



X 4 = 2196. 



2197. 



13 inches square, for 10 x 13 X 13 



The following rules embody the methods of dealing 

 with the strength product when found as above. 



1. Given the depth to tin. I the breadth. 



Suit. Divide the strength product twice by the depth 

 (in inches), the quotieut is the breadth (in inches). 



Example 15. Beam 6 inches deep to give strength 

 produ 



72 

 ip g = 2 inches, the breauth 



2. Given the breadth to find the depth. 



Ride. Divide by the breadth, and take the square root 

 of the quotient. 



i:.r<impU 16. Beam 8 inches broad strength product 

 1101. 



1161 

 8 



= 145, the square root of which is nearly 12, for 



12 X 12 = 144. 



3. When the beam is square in section. 



Rule. Take the cube root of the strength product. 



Example 17. Square beam, strength product 2196. 



Cube root of 2196 = 13 nearly, for 13 X 13 X 13 = 2197. 



4. When the beam is cylindrical. 



Rule. Increase the strength' product by its half, and 

 take the cube root. 



Example 18. Cylindrical beam, strength product 5i 



Add half of 5 = 2if 



Cube root of 8 = 2. 



For beams of other forms than those which have 

 square, rectangular, or circular sections, we cannot fur- 

 nish any rule. Practical men, from long experience, and 

 from a habit of observing the proportions suitable to cer- 

 tain strains, can form a tolerably accurate conception of 

 the strength due to particular forms under various con- 

 ditions. Before any mechanical work is executed, the 

 appearance of the parts on the drawing recommends 

 itself to a practised eye, or otherwise, according as the 

 detail* are studied in consonance with the just propor- 

 tions of strength or otherwise. In providing not only 

 arainst absolute fracture, but also against deflection, 

 vibration, and all such elements of degradation and 

 weakness, the strength must be made very much in ex- 

 cel* of that which the mere calculation of breaking-strain 

 would indicate ; and the different modes of providing 

 adequate firmness and permanence without extravagant 

 use of materials, are matters of ingenious contrivance for 

 which no rule can be furnished. 



4, TORSION. As, in machinery, motion is generally 

 conveyed from one part to another by means of shafts or 

 tpindle* rotating on their axes, it becomes a matter of 



considerable importance to determine the strength of 

 materials to resist a twisting or wrenching force. If HO 

 suppose an iron shaft fitted with a wheel at each end, 

 one of which is driven by some prime mover, such as a 

 team-engine, and the other conveys the power or motion 

 to some machinery, it is clear that tho whole power so 

 conveyed has to pass through the shaft ; and the resist- 

 ance of the machinery at the one end to the prime force 

 applied at the other, subjects the shaft to a torsi ve or 

 twisting strain. 



That we may simplify the view which we should take 



of this kind of strain, 

 we shall suppose a shaft 

 A (Fig. 89) fitted with 

 awheel at each end Z 

 and C, round the cir- 

 cumference of each of 

 which, in opposite di- 

 rections, a rope passes, 

 suspending weights W 

 W. The wheels and weights being equal and balanced, 

 the shaft is not caused to turn in either direction ; but 

 both weights tend to twist the shaft in itself. If instead 

 of one of the weights W' we were to substitute a fixed 

 pin or hook to which the rope might be attached, the 

 rope would still be subjected to the same strain as if it 

 had the weight suspended from it ; and as it would react 

 on its wheel with a force precisely equal to its tension, 

 the twisting effect on the shaft would not be altered. \\ . 

 may, therefore, suppose the wheel and weight entirely 

 removed from one end of the shaft, and that end fixed 

 firmly in such a manner that it cannot turn or revolve 

 on its axis, while it is throughout its whole length sub- 

 jected to a twisting strain from the action of the weight 

 at the other end. The power with which tliis weight 

 tends to twist the shaft, depends both on its magnitude 

 and on the size of the wheel round which its rope passes. 

 The radius, or half-diameter, of the wheel is the leverage 

 of the weight ; and by increasing it wo increase the tor- 

 sive strain in like proportion. For ease of calculation, 

 we will suppose the radius, or half-diameter, of the wheel 

 to be 1 foot ; and if we can find the torsion of a certain 

 weight acting at a radius of 1 foot, we can reckon that 

 of the weight at any other radius ; or having found tho 

 weight to produce a certain torsion at 1 foot radius, we 

 can easily reckon the radius at which the same weight 

 would produce some other amount of torsion. 



Thus, by doubling or trebling the radius we double or 

 treble the torsive strain ; by doubling or trebling the 

 weight, we have to place it at one-half or one-third the 

 radial distance from the centre of the shaft, in order to 

 subject it to the same torsion ; and so generally in sim- 

 ple proportion. The torsive strain, therefore, is as the 

 weight multiplied by its radial distance. One ton at a 

 radial distance of 3 feet, is equivalent to 3 tons at a radial 

 distance of 1 foot, because 1 ton X 3 feet = 3 tons X 1 

 foot. 



We have now to inquire what effect the dimensions of 

 the shaft, or the quantity of material in it, has in resist- 

 ing a certain torsive strain ; how large the shaft must be 

 to resist a given strain, or what strain a shaft of given 

 dimensions can resist. The replies to these question* 

 must depend upon the form of section and the uatuni of 

 the material employed ; but if we can ascertain, by direct 

 experiment, the torsive strength of a shaft of certain 

 material, form, and dimensions, we must endeavour to 

 find a principle on which to reckon the strength of a 

 shaft of the same form and material, but of different 

 dimensions. We may imagine two plates of metal or 

 other material, placed face to face, having such cor- 

 responding projections and hollows in their surfaces, 

 that the projections of tho one fit into the hollows of 

 the other. Wo may suppose these plates pressed to- 

 gether, and some force applied at their edges, so as to 

 push the one along the face of the other. It is man 

 that the force for this purpose must depend upon tho 

 amount of roughened surface in contact, upon the pres- 

 sure squeezing the plates together, and the peculiarities 

 of the roughness which their surfaces present. Inv 



