USEFUL DATA 



From page 38. Wi+23f,(/i+Zj) 4-^3/2= -"^ -^ 



Substituting numerical values in these two equations: 



^70A/,+10M. = - <'-^; <^^>' - <'-«««^) <"»' = -7.481.250 (I) 



10Af,+60J/. = - ('-^i <'")' - <'-™' ^^'' ^ -4.050.000 (2) 



4 4 



Multiplying equation (1) by 6 and subtracting equation (2) 

 410Ms= -40,837,500 



ifj= -99,604 ft. lb. and we find M3= -50,899 ft. lb. 



Substituting in Eq. (2) 3/3= -50,899 ft. lb. 



From page 38. F.= ^^^ + ^ 



^ (-99,604) _^ (1,800) (25) ^^3 3^^^^ 



F',= ri-Wi = 18,516-(1,800)(25) =-26.484 lb. 



_ Mz—Mj w^2 



L 2 



_ (-50,899)-(-99.604) ^ (1,800) (10) ^ 1 3,870 lb. 

 10 2 



• rs = F2-W2=13.870-(l,800) (10) = -4.1301b. 



1/ - Mi— Ml . wds 

 I 2 



-(-50,899) (1,800) (20) ^ ,^^^ 



~ 20 ^ 2 



F'4 = F3-W3=20,M5-(1.800)(20) =-15.455 lb. 



/{, = Fi= 18,516 lbs. 



R, =rj+n=13,870+26,484 = 40,3541b. 

 R, =F3+F'8 = 20,545+4,130 = 24.675 lb. 

 ^4 =F'4= 15.455 lbs. 



Distance from left support to point of zero shear, 



^i=-=r§r= 10. 29 ft. for span Zx 

 • ^'=-=TW= 7.71 ft. for span/, 



^3=^ = r^=ll 41ft.forspan/3 

 W3 1,800 



41 



