of Continuous Girders. 189 



M 2 and M 3 are the greatest positive moments in the middle span , 

 to find the maximum negative moment, we take (6), 



,, c WW 2 



m = M 2 --S 9 a? + -s-j 



S 

 this becomes a maximum for x— — : inserting this in the ex- 



w ° 



pression for m, we get 



o2 g2 g2 



w 2w 2w 



Now, from (4), 



M 2 -M s 1 7 5] _ 



s *= —r~ + 2 wl= m wL 



Therefore the greatest negative moment is 



1561 W 2 7hwl 2 . 



?n = 



21632 "" 104 



Q fr-t 



and it obtains at the point where a?= — == z-t^t I- 

 r w 104 



6. In a girder of two equal spans, ends fastened horizontally, 

 to find the reactions and inflection-points due to a single load. 



=| 2 -«_£/-». \ 4 L 



3 



Let the load be placed in the first span, then r=2; since s — 2 

 = number of spans, 5=4; then from (1), 



c l — d l — 0, 

 c 2 ^# 2 = 1, 

 c 3 =d s =-2, 

 c 4 =d 4 = 7. 

 From (2) and (3) we have, 



when rc<3, M n =c n f*^^=c n ^ (4k-7k* + Sk*) ; 



whenn>2, M n =c 6 - n f, C *~^ C \ =C6-»x (*'-**)■ 



l{c 3 + 2c 4 ) 4 



Let the load be placed at a distance from the abutment equal 

 to \l, then & = i ; making also w equal to 2, 3, and 4, we get 



, T 39P/ ., 6PZ , , T 3P/ 



M ^ W M * = 256' and ^=- 256- 



