TABLE 134. 

 DIAGRAM FOR STRESS IN EYE-BARS DUE TO WEK.HT. 





dye to weight 



Indirect fibre stress 



fr= depth of bor,inches 



/-length of bor inches 



= I2 



1.5 2 3 456789 



I&II. Depth of Bar in Inches 

 IID/V/2 in Tens of Thousandths 



Problem. Required stress due to weight of a 4 in. x i in. eye-bar, 20 ft. long, which has a 

 direct tension of 56,000 Ib. 



Then, h = 4 in.; L = 20 ft., and /i = 14,000 Ib. per sq. in. The stress due to weight, /i, 

 is found from the diagram as follows: On the bottom of the diagram, find h = 4 in.; follow up the 

 vertical line to its intersection with inclined line marked, L = 20 ft., then follow the horizontal 

 line passing through the point of intersection out to the left margin and find, y t = 3-3 tens of 

 thousandths; then follow vertical line, h = 4 in., up to its intersection with inclined line marked, 

 ft = 14,000, and then follow the horizontal line passing through the point of intersection to left 

 margin and find, y\ = 7.2 tens of thousandths. Now y\ + yt = 7-2 + 3.3 = 10.5. Find y\ 

 + yt = 10.5 on lower edge of diagram, follow vertical line to its intersection with line marked 

 "Line of Reciprocals" and find on right margin, /i = 950 Ib. sq. in. 



For a bar inclined at an angle with a vertical line multiply the fiber stress calculated for a 

 horizontal bar as above, of the same length, and multiply the fiber stress thus obtained by sin 9. 

 For example if the bar above is inclined at an angle of 45 degrees with the vertical; the fiber stress 

 due to weight is, /i = 950 x sin 8 = 950 x 0.707 = 672 Ib. 



Every imeisection of the inclined ft and L lines has for its abscissa a value of h, which will 

 have a maximum fiber stress, /i, for the given values of ft and L. For example for L = 30 ft.; 

 ft = 1 2, ooo Ib., we find h = 8.3 in., and fi = ijoolb. A deeper or shallower bar will give a smaller 

 value of /i. 



253 



