CHAP. XH.] METHODS COMBINED. 171 



113. Beam fixed horizontally at both ends Supports 

 on level. Consider the span D E (Fig. 78) as fixed at the sup- 

 ports so that the tangent to the deflection curves at D and E is 

 always horizontal. Conceive the span prolonged right and left 

 beyond the supports a distance equal to the span I. It is re- 

 quired to find the position of the inflection verticals. 



From equation (1) of the preceding Art. we have, since 

 n = l, * = 0, 



i a I 



~(i 1 a)l i l a* 

 and from equation (2), since is = l, 



i% I (I a) 



Now for a beam fixed at the ends the distances of the points 

 of inflection are 



a I I (I a) 



- 



Substituting these values in the equations above, we have 



OB = o and a? = + Q- That is, the position of the inflection 

 o o 



verticals is in this case independent of ike load, and always 

 equal to from the supports.* 



This remarkable property of the beam fixed at both ends 

 enables us to find the inflection points by a construction similar 

 to that for the fixed points in the unloaded spans, as given in 

 Art. 86. 



Thus we have simply to draw from C distant I from A (PI. 

 21, Fig. 79) a line in any convenient direction, as C b intersect- 



ing the inflection vertical I, which is distant from A, ~ I, at a 



Through a and the fixed end support A draw a line to inter 

 section with the weight P. Then draw P b. The intersection 

 i of this last line with A B is the inflection point. A similar 

 construction gives i^. 



We can now find the reactions and moments. Thus H O to 



* This important result, which renders possible a complete graphical solution 

 of this case, has, so far as we are aware, never before been published. 



