CONTINUOUS GIRDER. [CHAP. XIII 



what we have assumed in Art. 145. "We may also find the 

 Bame shear by finding the algebraic sum of the reactions at 

 ABC and D f roir the formulae of Art. 147. This is more 

 tedious, and, as we see, unnecessary. The moments can be 

 easily found, and then the shear obtained directly from these. 



We must bear in mind that ^ always denotes the span the 

 load is upon, whether nl,pl, or I, while l m is any span in gen- 

 eral, according to the value of m. 



152. Continuous Girder with fixed ends. It is worthy of 

 remark that if n be made zero in the formulae of Art. 149, we 

 have a girder with fastened ends and variable end spans p I. 

 If in addition^? is unity, then all the spans become equal. We 

 must, however, remember that when we thus make n = 0, the 

 number of spans is 8 2 instead of s, as before, and the end 

 spans are p I ; the end supports are also 2 and s instead of 1 

 and s + 1. 



153. Examples. As illustrations of the use of the formulae 

 of Art. 150, we give a few examples. 



Ex. 1. A beam of one span is fixed horizontally at the ends. 

 What are the end moments and reactions for a concentrated 

 weight distant k \from the left end? 



Here the two outer spans of three spans are supposed zero. 

 Therefore, s 2 = 1, and s = B. The left end is 2 instead of 

 1, and the right end 3. Hence, r = 2,^? = 0, and n = I in the 

 formulae of Art. 150. We have, then, 



^ = 0, Ca = 1, GS = 2, Ct = 4) and hence, 



rt-iur Cj' 



for m = 2, I^ = 



f o TUT C 2 



for m 3, Mo = -f 



+ A 3 

 A <% + A' 



or, inserting the values of c above, 



M 3 = 1(2A-A'), M,= -J 7 (A-2A'). 



For a concentrated load, A = P ? (2 k - 3 P + tf), 

 and A / = P?(/&-^ 8 ). Hence, M., = PZ (k - 2tf + #), 

 and Ms = Pl(tf-Z*). 



For the reaction at the left end, which is in this case the 

 same as the shear, we have 



