ms.] 



Ohservaiions on Suspension Bridges, 



331 



l.Ol tons weight of main chains. 

 Total suspended weight = 34 + 1.01 = 35.01 tons which 

 X 1.93= 671 tons. 

 The stretching strain of 7^ square inches = 7i 9 tons = 67| tons. 

 In large bridges constructed in England and elsewhere it is usual to 

 allow one and one-fourth or U times the section given by the foregoing 

 formula,but this I imagine is chiefly to provide against the great vibration 

 which there must necessarily be in those large bridges, but in this country 

 it would be superfluous. This calculation allows for a much greater 

 strain than is ever likely to be on the bridge at one time. 



Measure of horizontal draw and vertical pressure. Horizontal pull in- 

 wards = 



tension x cos. of direction of chains = 1-93 ^ .966 == 1.864. 

 Do. outwards. 



tension x cos. of direction of back stays do. do. nearly. . 

 V/hen the back stays are carried at the same angle, or nearly so, as the 

 chains, which is the best direction as then the horizontal draw equal 

 each other leaving very nearly a vertical pressure on the main piers, 

 which will be equal to tension x cos of direction = (^5.01 x 1.93) 

 X 257 = 17-34 tons, which multiplied by two (the vertical pressure 

 of the back slays being the same) == 34.68 tons. 



Mr. Drewry speaking of the strength of the abutments or retaining 

 piers, says their weight should not be much less than equal to twice the 

 utmost strain that can be brought on the chains by dead weight, and 

 their total resistance (i. e. weight and resistance of adhesion combined) 

 should be at least equal to four times the utmost strain. 



Method, of calculating the strength of the cross-beams to support 

 road- way. 



cube root of n. x length between supports in feet ^ suspended 



weight . . = but 



260 + 2 



since when the load is uniformly distributed over the length of a beam 

 it supports double the weight, that it would do, if the load were laid on 

 the middle,we have 



cube root of w = x 8 x 2538.65 = cube root of 117.1 — 4.85 depth, 

 520 



say and 4.85 = 1.6 breadth. 

 "3 



The breadth n bearing a certain proportion to the depth in this case 

 \ so that 4" deep by 2" broad would be ample for the strength of 

 these beams. 



The platform may be supported by two single chains made of wrought 

 iron bars of any length, and 2" diameter would be quite sufficient for 

 this bridge, or the best plan would be to let each chain consist of four 



