DESIGN OF END-POST. 589 



Stress Due to Weight of Member. The total weight of the member will be 

 Two 10 in. channels @ 15 lb., 31' 6" long - 945 Ib. 



One 14 in. X 5/16 in. plate @ 14.88 lb., 30' o" long - 447 Ib. 

 Details and lacing about 25 per cent 308 lb. 



Total Weight, W - 1700 lb. 



The bending moment due to weight of member is M = \W'l-t>\n B. 

 Stress due to weight 



M-c IW-l-smO.* 



Jw P-P P-t* 



A loE A loE 

 The stress due to weight in the upper fiber will be 



- = j X 1.700 X 375 X 0.645 X 3-6125 

 2II8 103.000 X375* 

 10 X 30,000,000 

 = 940 lb. per sq. in. 



The stress due to weight in the lower fiber is 



/'. = - 6.70 X 94 -* 3- 6 i 2 5 

 = 1745 lb. per sq. in. 



Stress Due to Eccentric Loading. The pins were placed i inch above the center of the channels, 

 and the stress due to eccentric loading will be 



_ M,-c _ P X (1.70 - Q-5) X c , 



T - ?-* ~ P '* 



~ loE ~ loE 



The eccentric stress in the upper fiber will be 



, _ 103,000 X 1.20 X 3.6125 



~ 211 8 - l0 ^ 000 X 375> 

 10 X 30,000,000 

 = 2,280 lb. per sq. in. 



The eccentric stress in the lower fiber is 



fe = + 6.70 X 2,280 -r- 3.6125 



= + 4,230 lb. per sq. in. 



The resultant stress due to weight and eccentric loading is/z =/+/= + 940 2,280 = 

 1,340 lb. in the upper fiber, and 1,745 + 4i 2 3 = 2485 lb. per sq. in. in the lower fiber. 



The allowable stress due to weight and eccentric loading is greater than 10 per cent of the 

 allowable stress and must be considered, with the allowable unit stress increased by 10 per cent 

 ( 48, p. 142). 



The total unit stress in the member will be, / = 103,000 -f- 13.30 + 2,485 = 7,752 + 2,485 

 = 10,237 lb. per sq. in. The allowable unit stress when weight and eccentric loading are con- 

 sidered is 9,400 X i.io = 10,340 lb. per sq. in., which is sufficient. 



Stress Due to Wind Moment. The stresses in the portal and the direct wind stresses in the 

 end-post when the end-post is assumed as pin-connected at the base are shown in (d) and (e) Fig. 

 12. The end-posts may both be assumed as fixed if the windward end-post is fixed. To fix the 

 windward end-post the bending moment must not be greater than the resisting moment which 

 will be 



M, = H-y = (90,000 - V - D')a/2 



where V = 5,060 lb. and D' = 7,000 lb. the direct stress due to wind, and a = distance center 

 to center of metal in the sides of the end-post = 8.87 in., (/), Fig. 12. (The impact stress is 

 omitted.) If yis taken equal to \d = 10' o" = 120 in., we will have 



2,000 X 120 < (90,000 5,060 7,000) 8.87/2 

 which makes 240,000 < 345,600, and the end-post may be assumed as fixed at the base. 



