DIAGRAM FOR FINDING STRESS IN BARS 151 



diagram, the reciprocal, which is the fibre stress f if may be read off the 

 right hand side. 



The use of the diagram will be illustrated by two problems : 

 Problem I. Required the stress in a 4" x i" eye-bar, 20 feet long, 

 which has a direct tension of 56,000 Ibs. 



In this case, h = 4" ', L = 20 ft., and f 2 = 14,000 Ibs. per sq. in. 

 The stress due to weight, f if is found as follows: On the bottom of 

 the diagram find h = 4 inches, follow up the vertical line to its inter- 

 section with inclined line marked L = 20, and then follow the horizontal 

 line passing through the point of intersection out to the left margin and 

 find y 2 = 3.3 tens of thousandths ; then follow the vertical line h = 4 

 inches, up to its intersection with inclined line marked f 2 = 14,000, and 

 then follow the horizontal line passing through the point of intersection 

 out to the left margin and find y^ = 7.2 tens of thousandths. 



Now to find the reciprocal of 3>i + y 2 7- 2 + 3-3 IO -5> find value 

 of y i -f- y 2 = 10.5 on lower edge of diagram, follow vertical line to its 

 intersection with inclined line marked "Line of Reciprocals" and find 

 stress / by following horizontal line to right hand margin to be 



f = 950 Ibs. per sq. in. 



By substituting in (76) and solving we get f t = 960 Ibs. per sq, in. 



Problem 2. Required the stress in a 5" x J4" eye-bar, 30 feet long, 

 which has a direct tension of 60,000 Ibs., and is inclined so that it makes 

 an angle of 45 with a vertical line. 



In this case, h = 5", L = 30 feet, f 2 = 16,000 Ibs., and = 45. 

 From the diagram as in Problem I, y 2 = 1.8 f ens of thousandths, and 

 v = 6.5 tens of thousandths, and 



/i = v -LV sin e = 120 X sin 8 



y\ ~t~ y^ 



= 850 Ibs. per sq. in. 



Relations between h, f^ / 2 and L. For any values of / 2 and L, f l 

 will be a maximum for that value of h which will make y + y 2 a min- 



