TRANSVERSE STRENGTH.] 



APPLIED MECHANICS. 



811 



IV. When the beam ig fixed at both ends, and has a 

 load uniformly distributed over its length. 



Rule. Find the strength as in I., and triple it. 



Example 6. Required the strength of a round 

 wrought-iron bar 2 inches diameter, 10 feet long, fixed 

 at ends, the load being uniformly distributed. 



1300 X 2 X 2 X_2 =104() 



10 

 Deduct one-third = 



347 



693x3=20791bs. 



V. When the beam is loaded by a weight not in the 

 middle. 



Ride. Find the strength as before (I.), multiply twice 

 by half the length, and divide successively by the length 

 of each part into which the beam is divided by the point 

 of suspension. 



Example 7. Required the strength of an ash beam 6 

 inches broad, 8 inches deep, 14 feet long, to carry a 

 weight 3 feet from one end. Half the length is 7 feet, 

 and the two parts are 11 and 3. 



250 x 6 X 8 X 8 7x7 

 ~14 X llx3 = 



10,182 Ibs., about 4 tons. 



VI. When the beam is fixed at one end and loaded at 

 the other. 



H'de. Find the strength as in I., and divide it by 4. 

 uple 8. Required the strength of an English oak 

 bearu 13 inches square, projecting 10 feet from a wall, to 

 bear a load at the end. 



190 x 13 X 13 x 13 



: 41743 Ibs. strength by I. 



20 



Divide by 4, 10436 Ibs. 



These cases include most of the circumstances which 

 occur in practice. The converse operations for finding 

 the dimensions of a beam when we know the weight it has 

 to bear, do not furnish us with the breadth and depth 

 separately (except in the case of cylindrical and square 

 beams when the depth and breadth are equal), but give 

 us a result which is the product of the breadth by the 

 square of the depth. We must, therefore, be guided by 

 other circumstances in determining one of these dimen- 

 sions ; and having determined the one, we easily find the 

 other. For instance, if we were required to provide a 

 cast-iron girder of rectangular section, such that when 

 placed on two walls 10 feet apart it should carry a load 

 of 2 tons in the middle, we should reason thus : By the 

 table, a cast-iron rectangular beam 1 inch broad and 

 1 inch deep, and 1 foot between supports, bears 850 Ibs.; 

 one 10 feet between supports must be made ten times as 

 strong to bear 850 Ibs. ; and to bear 2 tons, or 4480 Ibs. , 

 it must be about five-and-a-quarter times as strong, be- 

 cause 4480 Ibs. is about five-and-a-quarter times 850 Ibs. 



The strength, then, for the given length and weight 

 must be 5J X 10 = 62 ; or, more correctly, 



10 X 4480 



- =527 times that of a beam 1 inch broad and 



oOU 



1 inch deep. But as the strength is as the breadth mul- 

 tiplied by the square of the depth, the number 52-7 must 

 be the product of the breadth in inches, multiplied by 

 the square of the depth, or twice by the depth ; and if we 

 determine one of these dimensions, we can easily ascer- 

 tain the other : thus 



AMume breadth, 

 lin. 



2 in. 



3 in. 



4 in. 



5 in. 



then depth, 

 7J in. 

 6 in. 

 in. 

 j in. 

 3} in. 



because 



nearly 



=527 

 X5J=527 



527 



And so on, the one dimension being assumed according 

 to circumstances of convenience, or for other reasons. 

 When it is determined that the section of the beam shall 

 be square, the product found as above is the cube of the 

 breadth or depth, or it is the breadth multiplied 3 times 

 by itself ; and therefore the breadth or depth in found 



by taking the cube root of this product. In the case 

 given, the cube root of 527 is about 3J ; and we there- 

 fore conclude that if the beam be of square section, its 

 breadth or depth must be 3| inches, because 3J X 3J X 

 3J=527 When the beam is cylindrical, the diameter is 

 also determined. Since the strength of a cylindrical 

 beam is about of a square one of the same material, we 

 must make up the other by adding to its dimensions ; 

 and as J is one-half of , we must add to the product its 

 half, and then take the cube root as for a square beam. 

 In the case given, adding to 527 its half, or 26 - 35, we 

 have 79 '05, which is nearly the cube of 4J ; and we must 

 therefore make the diameter of a cylindrical beam 4 

 inches, or more correctly 4 '3 inches. 



As a proof of the accuracy of our calculation, we have 

 only to reverse the operation, and calculate the load 

 which the beam of the dimensions we have thus deter- 

 mined will bear. If the result correspond with the 

 given weight, we know that we are right. In the example 

 we have taken, we have to find the strength of a cylin- 

 drical cast-iron beam 4 '3 inches diameter and 10 feet 

 long. By the rule for bars of circular section, case I., 



850 X 4-3 X 4-3 X 43 



JQ =67o8 Ibs. 



Deduct 4=2253 



4505 



Nearly 2 tons, or 4480 Ibs., the load which the beam 

 was intended to carry. 



The following rules give the mode of calculating the 

 strength product under various circumstances of strain, 

 and then the method of dealing with the'strength product 

 when the form, proportions, or one of the sectional 

 dimensions of the beam is given. The cases are num- 

 bered in the same order as those in the calculation of the 

 load. 



I. Beams supported at both ends and loaded in the 

 middle : given the length and load to find the stremjth 

 product. 



Rule. Multiply the load (in Ibs.) by the length (in 

 feet), and divide by the number in the table. 



Example 9. Required the strength product for a beam 

 of teak 14 feet between bearings, to carry 10 tons in the 

 middle. 



10 tons -22, 400 Ibs. 



22,400x14 . 



=.1161, strength product, 

 yfv 



8 ins. broad and 12 ins. deep would suit this case, be- 

 cause 8 X 12 X 12=1152, nearly 1161. 



II. When the beam is fixed down at the ends. 



Rule. Find the strength produced as in I., and deduct 

 from it its Jrd part. 



Example 10. A cast-iron bar fixed at both ends 9 feet 

 apart has to carry 10,200 Ibs. 



10,200 X 9 



850 

 Deduct 



108 

 36 



72 = strength product. 

 2 ins. broad and 6 ins. deep, for 2 X 6 X 6 = 72. 



III. When the beam is loosely supported, and the 

 load uniformly distributed. 



Rule. Hilve the strength product. 

 Example 11. A. deal rafter 10 feet long is loaded 

 uniformly with 9438 Ibs. 



9438 X 10 v 1 - 363 



~T30 x * 



3 ina. broad and 11 ins. deep, for 3 X 11 X 11 = 363. 



IV. When the beam is fixed at both ends, and has the 

 load uniformly distributed. 



Rule. Divide the strength product by 3. 

 Example 12. A wrought-iron bar 10 feet long, fixed at 

 its ends, is loaded with 2080 Ibs., uniformly distributed. 



stre "S th P roduct - 



