COMBINED FLEXURE AND DIRECT STRESS. 587 



If P = direct stress in eye-bar; M\ - bending moment due to weight in in.-lb.; c distance 

 from neutral axis to extreme fiber - h/2, where h - depth of eye-bar; / - length of bar, c. to c. 

 of pins, / - thickness of eye-bar in inches; / moment of inertia of eye-bar -fa t-tf; k ia a 

 coefficient depending upon the condition of the ends being approximately 10 for eye-bars with pin 



rmls, 24 for one pin end and one fixed end, and 32 for two fixed ends; E modulus of elasticity 



p 



of sti>el 28,000,000 Ib. per sq. in.; and ft -? unit stress due to direct loads. Then 



* n 



the stress due to combined flexure and direct stress will be 



k-E 



Now, Mi = Iw-P, where w = 0.28 t-h = the weight of the bar per lineal inch; P = ft- t-h; 

 A/z; / = iV'A 5 ; k = 10; and E = 28,000,000 Ib. per sq. in.; and substituting 



Jw-/*-JA 4,900,000% 



b-h* , h'b-h-P t . /AV 



ft + 23,000,000 r ) 



10 X 28,000,000 " \/ / 



(22) 



12 10 X 28,000,000 



then /i is the extreme fiber stress in the bar due to weight, and is tension in the lower fiber and 

 compression in the upper fiber. 



If the bar is inclined, the stress obtained by formula (22) must be multiplied by the sine 

 of the angle that the bar makes with a vertical line. 



Diagram for Stress in Bars Due to Weight. Taking the reciprocal of equation (22) 



, 23,000,000 I 7 



1 = h + \LL_ = y . + 



fi 4,9OO,oooA 4,9oo,oooA 



and 



A diagram for solving equation (23) is given in Table 134, Part II, which see. The intersections 

 of the inclined lines in Table 134 correspond to depths of eye-bar that give maximum stresses 

 due to weight. 



End-Post. Design the end-post, Fig. 12, for a 160 ft. span through highway bridge. Panel 

 length, 20' o"; depth of truss c. to c. of pins, 24' o"; length of end-post, 31' 3". The direct 

 stresses are as follows: dead load stress = 30,000 Ib.; live load stress = 60,000 Ib.; impact = 

 loo/(i6o -f- 300) X 60,000 = 13,000 Ib.; total direct stress due to dead load, live load and 

 impact = 103,000 Ib. The bridge is to be a class C bridge designed according to the "General 

 Specifications for Highway Bridges," in Chapter III. From 38 of the specifications the allow- 

 able unit stress is/ e = 16,000 70 l/r. The section will be made of two channels and one cover 

 plate. Try a section made of two 10 in. channels @ 15 Ib., and one 14 in. by 5/16 in. plate, (6), 

 Fig. 12. From Table 82, Part II, the radius of gyration about the horizontal axis A -A, is r A = 3.99 

 in., and about the vertical axis B-B is, r B = 4.67 in., and the eccentricity is, e = 1.70 in. The 



allowable stress is then f e = 16,000 = 9,400 Ib. per sq. in. The required area will 



o^yy 



be = 103,000 -f- 9,400 = 10.96 sq. in. The actual area is 13.30 sq. in. While the section ap- 

 pears to be excessive, it will be investigated for stress due to weight, eccentric loading and wind 

 before rejecting it. 



The area, radii of gyration and the eccentricity may be calculated as follows. 

 To calculate the area 



area of two 10 in. channels (Table 14) = 8.92 sq. in. 



area of one 14 in. by 5/16 in. plate (Table 2) = 4.38 sq. in. 



Total area = 13.30 sq. in. 



