AT BROUGHTON. 



328 5 

 =— ~= 164.2 tons: dividing this quantity by 



27, the number of tons required to tear asunder 



a square inch bolt of iron, we have i^ = 6.08, 



the square inches in section w^hich one of the 

 vertical links must have to support the weight. 

 A strength which is exceeded by the links in the 

 bridge. 



From the estimates above, it appears that 

 the tensions of the bottom and top link of the 

 catenary, and of the side chain, are 333, 351, 

 and 397 respectively; and therefore their strength 

 ought to be as these numbers. But the whole 

 chain is uniform, hence the top link is weaker 

 than the middle part in the ratio of 333 to 351, 

 and the side chain will bear less than the middle 

 in the proportion of 333 to 397, or of 6 -to 6 

 nearly. Attention to this subject would cause 

 a saving in iron, which otherwise, besides its 

 expense, does harm by unnecessary weight. 



6. If the link ce were not vertical, we might 

 obtain the tensions as above, by equating 

 among themselves all the forces of the same 

 kind whether horizontal or vertical, and thus 

 arrive at two equations from which to take the 

 unknown quantities; or we might proceed as 

 below. 



