TRANSVERSE STKArN. ] 



APPLIED MECHANICS. 



803 



the tearing asunder of the fibres commences, it must con- 

 tinue while the load continues to act, because the number 

 Fig. 63. 



! res resisting it is diminished, and those that remain, 

 have therefore more load to bear. 



NVliat we have to discover, however, is the strength of 

 thn beam while it U entire, or how much load W, at a 

 certain distance irom the wall, will bear without the 

 destruction of the fibres. Supposing a horizontal line 

 CD, drawn through the neutral axis, to form the long 

 ami of a bent lever of which the fulcrum is C, one short 

 arm C A, and the other C B, we have a certain power or 

 weight W acting at the extremity of the long arm, and a 

 number of resistances, viz., those of the fibres to exten- 

 sion, acting along the arm C A ; also a number of resis- 

 tances, viz., those of the fibres to compression, acting 

 along the arm C B. 



Now, when on one arm of a lever, a number of equal 

 forces, equally distributed over the arm, acts against a 

 given weight at the other end of the lever, the effect of 



Fig. 64. 

 D C B * F 



_|_X-l X-I--X S >. 



St- 



T 



f- 



s 



all these forces is the same as if they were all collected 

 into one force at the middle point of the arm on which 



they act (Fig. 64). This is true whether the lever be 

 straight as 1, or bent with one arm as 2, or bent with 

 two arms as 3. As a numerical example, let us suppose 

 that the short arm of 1 is 4 feet long, and at 

 each foot there hang 4 weights A, B, C, D, 

 each of 10 Ibs. , as marked by the arrows ; 

 while the length of the other arm is 5 feet. A 

 weight of 20 Ibs. hung to the longer arm 

 balances the four weights of 10 Ibs. distributed 

 along the shorter, because 



A = 10 Ibs. at 1 ft. leverage is balanced by 2 Ibs. at 5 ft. lev' 

 B = 10 Ibs. at 2 ft. 4 



C = 10 Ibs. at 3 ft. 6 



D = 10 Ibs. at 4 ft. ,, 8 



Total A, B, C, D distributed ... 20 



And 20 Ibs. at 5 feet leverage would be balanced 

 by 40 Ibs. at 2^ feet ; that is, the total of the 

 4 weights A, B, C, D hung at their middle 

 point, as marked by the dotted arrow. The 

 same law will be found true, whatever the 

 number of equally-distributed weights, and 

 whether the lever be straight or bent. Applying 

 this to the question of the beam, Fig. 63, let 

 E be the middle point of A C, and F the middle 

 of B C ; then the load W acting with leverage 

 C D has to resist the united tensive strength 

 of the fibres along C A acting at the lever- 



_ -'.tge C E, and the united compressive strength 



of those along C B acting with the leverage C F. 

 Those resistances will necessarily be equal to one an- 

 other because the neutral axis C is the fulcrum, on 

 each side of which they equalise themselves ; and 

 the teusive strength of C A, multiplied by its lever- 

 age C E, is therefore equal to the compressivo strength 

 of CB multiplied by the leverage C F ; or, as C A is 

 double of C E, and C B double of C F, and as doubles of 

 equal things are themselves equal, the teusive strength 

 of C A multiplied by the length C A is equal to the com- 

 pressive strength of C B multiplied by the length C B ; 

 and each of those products is equal to the weight W 

 multiplied by its leverage C D, the length of the beam, 

 because each is double of half the effect of W to break 

 the beam. 



To show how this reasoning may be applied numerically, 

 let us suppose that a beam of oak 1 inch thick and 6 

 inches deep, fixed in a wall so as to project 10 feet, is 

 broken by a load hung at the end ; and that while in the 

 act of breaking, the neutral axis is observed to be 2 

 inches from the under side, and therefore 4 inches from 

 the upper side of the beam, we may, from knowing the 

 tensive strength of oak, estimate the weight required to 

 break the beam. Taking this tensive force per square 

 inch (or force required to tear asunder a square inch) at 

 12 000 Ibs we have in the case before us 4 square inches 

 having a tensive force of 4 X 12,000 = 48, 000 Ibs., acting 

 with leverage of 4 inches against leverage of 10 feet or 

 120 inches ; and as 4 inches is the ^th of 120 inches, the 

 weight at D must bo the .fcth of 48,000, viz., 1,600 Ibs. 

 By such calculations as this we could compute the 

 transverse strength of materials, from knowing their 

 tensive strength, provided we knew the position of the 

 neutral axis of fracture. The determination of this, 

 however, is a point of very great difficulty ; for it must, 

 in the first place, depend upon the relative proportions 

 of compressive and tensive strength : and as we are 

 much in the dark as to the former, we cannot institute a 

 comparison between it and the latter. In the second 

 place, no material to which we can apply transverse 

 strain resists completely, and then instantaneously gives 

 way. ' While we add weight after weight, the extension 

 of fibres at one side and the compression of those at the 

 other goes on ; the beam bends or becomes deflected ; 

 greater and greater strain is thrown on the outer fibres ; 

 the tensive and compressive strength of each varies as it 

 is more and more extended or compressed ; the neutral 

 axis changes its position ; fibres, which were, during part 

 of the process, compressed, begin to be extended ; and 

 the condition of the material at and near the point 

 of fracture becomes generally so altered as it approaches 

 destruction, that we cannot clearly estimate its re.si.st- 



