APPLIED MECHANICS. 



[TRANSVERSE snutn. 



nee, eren it we were possessed of the most accurate 

 knowledge of its strength to resist compression or exten- 

 sion. Owing to these circumstances, it becomes necessary 

 to make distinct experiments on the transverse strength 

 of bodies, and, by moans of them, to establish certain 

 data which wo may apply in calculation. But while 

 experiment and observation furnish the data or facts on 

 which to calculate, we must still trust to reasoning for 

 our ini-thods of applying these facts ; and fortunately the 

 reasoning is of a simple character. 



As to the neutral axis of fracture, without requiring 

 to determine its position in any beam, we assume that, 

 for beams of the same material, its distance from the 

 upper or lower sides is proportional to the depth of the 

 beam. Thus, if in a beam of ash 6 inches deep, it is 

 2 inches from the lower side and 4 inches from the 

 upper ; in a beam of ash 12 inches deep it will be 

 4 inches from the lower and 8 inches from the upper 

 .-i'lc- ; in one of 1 inch deep, it will be Ird of an inch 

 from the lower, and jrds of an inch from the upper ; and 

 so on in regular proportion. We can see no reason why 

 this should not be the case ; and, so far as we can trace 

 the circumstances which determine the position of the 

 neutral axis, we see every reason for believing that it is 

 so. This assumption greatly simplifies the rest of our 

 reasoning ; fur it enables us to get rid of all calculation 

 as to the actual position of the neutral axis, and to pro- 

 ceed with the comparison of beams as to transverse 

 strength independently of it. We will suppose that we 

 have three beams of the same material, all of equal 

 lengths and breadths, but of the following depths : 

 A depth 2 inches, B depth 4 inches, C depth 6 inches, 

 and that we desire to compare their transverse strengths. 

 Let us assume, for the sake of simplicity, that the neu- 

 tral axis is midway in the depth ; then in A we have, 

 above the neutral axis, a set of fibres extending over 1 

 inch in depth, and acting with a leverage of 1 inch to 

 resist the breaking load. In B we have fibres extending 



- inches, and therefore double the number of those 

 acting in A, acting with 2 inches leverage, or double the 

 leverage of those in A. We see, then, that the strength 

 of B must be twice 2, that is, 4 times the strength of A. 

 Again, in C we have 3 inches depth of fibres and a lever- 

 age of 3 inches ; therefore C has a strength of 3 times 3, 

 or 9 times that of A. And so with any other depths, the 

 strength being always as the depth multiplied by itself, 

 or as the square of the depth. If we had assumed the 

 neutral axis to be in any other position, such as jrd of 

 the depth from the lower side, we should still have found 

 the same law to obtain ; for so long as the beams we ore 

 considering are of the some material, and of similar form, 

 the neutral axis must proportionally divide their depth, 

 and leave proportional numbers of fibres to aet with 

 leverage proportional also to the depth. 



We, therefore, have established the simple law, that 

 iu beams of the same form, length, breadth, and material, 

 the transverse strengths are as the squares of the depths ; 

 and knowing the strength of a beam 1 inch deep, we can 

 estimate the strength due to any other depth by multi- 

 plying that of the 1-inch beam twice, by the depth of the 

 in inches ; that is, by the square of the depth. 



Example. An iron beam 1 inch deep breaks with a 

 load of 2 tons : required the load that will break a simi- 

 lar beam 4 inches deep. The square of 4 is 16, and 

 2 x 1C - 32 tons, or 2 tons X 4 X 4 - 32 tons. 



Now, let us ascertain what relation the transverse 

 s'r.-ngths of beams bear to their breadths. Suppose that 

 a beam 1 inch broad carried a load of I ton, then another 

 exactly like it will also carry 1 ton ; a third will also carry 

 1 ton ; and the three placed side by side, or united into 

 one beam 3 inches broad, will of course carry 3 tons. 

 Beams, then, of a certain material of equal lengths and 

 depths, have transverse strengths which are proportional 

 to their breadths ; and if we know the strength of a beam 

 1 inch broad, we compute that of a beam having any 

 other breadth, by multiplying the strength due to 1 inch 

 of breadth, by the number of inches in the given breadth. 



mplt. A fir beam 1 inch broad carries 10 cwt. : 

 required the load carried by a fir beam 6 inches broad, of 



like length and depth. 10 cwt. x C ins. CO cwt, or 3 

 tous. 



We may now combine the laws as to bremlth an 

 into one. ami thus compute the transverse strength of a 

 beam as depending on both dimensions. This law is, 

 that the transverse strengths of beams of the same 

 material and similar in form, are as the breadths m 

 plied by the squares of their depths; and if w. know 

 the strength of a beam 1 inch brood and 1 inch < i 

 wo compute the strength of a beam having ai:y other 



'th and depth, by multiplying the stren 

 1-inch beam by the breadth in inches, and twice by the 

 depth in inches of the other. 



I '.ample. Suppose a beam 1 inch broad and 1 inch 

 deep bears 5 cwt. : required the strength of a beam 4 

 inches brood and 6 inches deep. 

 cwt. X 4 ins. x C ins. x 6 ins. =720 cwt. or 36 tons. 



The strength of a beam is irrespective of its length ; 

 but the actual weight which it can carry depends upon 

 the length, inasmuch as the leverage with which the 

 weight acts, so as to fracture the beam, is greater, the 

 greater the length. A weight of 12 tons banging at the 

 end of a beam 1 foot long, is exactly equivalent to a 

 weight of 6 tons at 2 feet distance, of 4 tons at 3 ; 

 of 3 tons at 4 feet, of 2 tons at 6 feet, of 1 ton at 1 2 

 feet, of J ton at 24 feet, and so on the weight multi- 

 plied by its distance being always o constant quantity. 

 If, then, we know the load which a beam 1 foot long will 

 carry at the end, we estimate the weight carried at any 

 other distance by dividing that carried at 1 foot by the 

 distance in feet ; and combining this element with those 

 of breadth and depth, we have the general rule em- 

 bracing all the dimensions of a beam ; i.e., multiply the 

 weight carried by a beam 1 inch broad, 1 inch deep, and 

 1 foot long, by the breadth in inches, and twice by the 

 depth in inches, and divide the product by the length in 

 feet. 



Example. A beam 1 inch square and 1 foot long, 

 carries 5 cwt. at its end : required the load carried by a 

 beam 4 inches broad, 6 inches deep, ond 18 feet long. 



5 cwt. X 4 ins. X Cins. x Gins. 



~18ft = cw '-> or 2 tons. 



So for we have discussed the strength of beams of 

 square or rectangular sections ; but similar reasoning 

 will apply to those of other forms. When the section 

 is circular, the breadth being equal to the depth, we 

 have to multiply the strengtii of a cylindrical beam 1 

 inch in diameter and 1 foot long, three times by tho 

 diameter in inches, and divide by the length in feet. 



Example. A cylindrical beam 1 inch diameter and 1 

 foot long, bears 3 cwt. ; required the load supported by 

 a cylindrical beam 4 inches in diameter and 8 feet long. 



8 cwt. X 4ins. >< 4ns. X 4 ins. 



Fig. 65. 



= 24 cwt 



When the section is of any form, such as the T shaped 

 iron girder (Fig. 65), from our ignorance as to the 

 position of the neutral axis of fracture, we are at a 

 loss how to reckon the eflect of the 

 tibres at different parts of the section, 

 and cannot therefore estimate tho 

 strength from any data obtained by 

 experiments upon beams of square 

 or rectangular section. Bnt if we 

 know the strength of a beam of 

 certain dimensions, and of the form 

 in question, we may pretty nearly 

 estimate thot of a beam of the same 

 kind whose dimensions, as to breadth 

 and depth respectively, are propor- 

 tionally greater or less. For instance, 

 to compare two T~ shaped beams of 

 which the dimensions are as foDows: 

 B=l in., C=4 ins., D-l in., 



<--B"> 



No. 1. A 3 ins., ., 



length 10 feet, is found to bear 2 tons : required the 

 strength of No. 2, where A-6 ins., B-2 ins., 0-6 ins., 

 D-1J in., length 16 feet 



