152 COMBINED AND ECCENTRIC STRESSES 



imum. This value of h will now be determined. Differentiating equa- 

 tion (76) with reference to / x and h, we have after solving for h 

 after placing the first derivation equal to zero 



in which h is the depth of bar which will have a maximum fibre 

 stress for any given values of / and f 2 . 



Now if we substitute the value of h in (78) back in equation (76), 

 we find that f^ will be a maximum when y^ = y 2 . 



Now in the diagram the values of y and y 2 for any given values 

 of / 2 and L will be equal for the depth of bar, h, corresponding to the 

 intersection of the f 2 and L lines. 



It is therefore seen that every intersection of the inclined f 2 and L 

 lines in the diagram, has for an abscissa a value of h, which will have a 

 maximum fibre stress / , for the given values of f 2 and L. 



For example, for L = 30 feet and f 2 = 12,000 Ibs. we find h = 

 8.3 inches and / = 1700 Ibs. For the given length L and direct fibre 

 stress / 2 , a bar deeper or shallower than 8.3 inches will give a smaller 

 value of / than 1700 Ibs. 



Eccentric Riveted Connections. The actual shearing stresses 

 in riveted connections are often *very much in excess of the 

 direct shearing stresses. This will be illustrated by the calculation 

 of the shearing stresses in the rivets in the standard connection shown 

 in Fig. 79 and Fig. 80. 



The eccentric force, P, may be replaced by a direct force, P, acting 

 through the center of gravity of the rivets and parallel to its original 

 direction, and a couple with a moment M = P x 3" = 60,000 inch-lbs. 

 Each rivet in the connection- will then take a direct shear equal to P 

 divided by n, where n is the total number of rivets in the connection, 

 and a shear due to bending moment M. 



The shear in any rivet due to moment will vary as the distance, 



