USEFUL DATA 



Problem II. Assume a beam having a long central span ami two shorter end 

 spans of equal length, as shown in Fig. 7, carrying a live load of 2,500 pounds per 

 linear foot and a dead load of 2,200 pounds per linear foot. To find the critical movaenta, 

 shears and reactions, assuming the ends simply supported. 



M,i o 



^-—Xl >; r<- 



\i O \Wl Af2 



1^ i,=2o'- 4^-- 



1 



^r■^ 



Af 4 o 





-Z2'40 



Ri 



Rz 



Rz 



ii-ao- 



Live load per ft. of span = 2500 lb. 

 Dead =2200 " 



Fig. 7 



The solution of this problem involves the same procedure as that carried out in 

 detail for Problem I, Case 1, and, therefore, the results only are tabulated below for 

 the various cases of critical loading. 



Casel 



4f. ;j ^ 



I— 



i 



3/2 = 3/3= -528,750 ft. lb. 

 Positive moment at point Xi= + 44,978 ft. lb. 

 Positive moment at point 0:2= +411,250 ft. lb. 

 Fi= +20,562 lbs. F'2= -73,438 lbs. ^2= +94,000 lb. 



Ri= 20,562 lbs. ^2= 167,4381b. 



Note.— This loading gives negative reaction, Ri 



Max. 3/2 =-544,375 ft. lb. 

 Max. V\=- 74,2191b. 

 Max. W = + 95,562 lb. 

 Max. iJ, =+ 169,7811b. 

 -2,0941b. 



Case 3 



Max. positive moment in 

 span/2=+442,500ft. lb. 



Note.— This loading gives negative reactions Ri and /J4= —2,875 lb. 



Case 4 



! 



JMll 



■ u ■>< 



■I2 ^-..f^-.. 



1—1 



Max. positive moment in 

 span/, = +116,294ft.lb. 

 Max. r,= +33,063 lb. 

 Max. Hi = +33,063 lb. 



43 



