BEAMS AND STRUCTURES 79 



For 24 ST, allowable stress would be 0.7 X 62,000 = 43,000 lb. per sq. in. 



To calculate lower chord: 



,M_ V 

 h 2 



where M = 10,000 X 35 



/i = 20 V = 10,000 



A/r • • • 1 1. ^ lO'OOO X 35 10,000 „„ _„„ ,, 



Maximum compression in lower chord = ;^^:r -= — = —22,500 lb. 



Area of compression chord is 0.719 sq. in. Hence compressive stress developed is 

 -r = ^ ^' = —31,300 lb. per sq. in. 



Maximum allowable stress, as calculated for compression member : 



0- = 45,000, yield point of material, used in order to calculate crippling stress 

 K = 10.8 (assumed) 

 t = 0.125 in. 



p 



-r = <r tanh Kt 

 A 



= 45,000 tanh (10.8 X 0.125) 



= 39,300 lb. per sq. in. 



Hence, as this is greater than the 31,300 lb. per sq. in., stress developed, the chord is 

 safe. 



To calculate upper chord: 



, . . ^ . M V 10,000 X 35 10,000 

 Maximum tension = d= -r -^ = ^ -^ — 



= 17,500 - 5,000 = 12,500 lb. 



Tension chord area = 0.237 sq. in. 



P 12,500 ^^^r.^,^ 



A ^ 237 ^ 52,700 lb. per sq. m. 



Assuming 15 per cent reduction in area on account of rivets, and for an ultimate 

 tensile strength of 62,000 lb. per sq. in., the allowable stress will be 



0.85 X 62,000 = 52,700 lb. per sq. in. 



which is not less than (and happens to be equal to) the actual stress. Hence tension 

 chord is safe. 



Stiffeners 



Compression load by equation given above is 



a = 45° tan a = 1 



