308 SUPPLEMENT TO CHAP. XIY. fCHAP. VI. 



CHAPTER VI. 



PARTIAL UNIFORM LOADING. 



27. Notation. In the preceding discussion of the arch we have con- 

 sidered the influence of a single concentrated load only, and this, as we 

 have repeatedly seen in the case of the simple and continuous girder, etc., 

 is sufficient for full and accurate solution. When once we are able to find 

 and tabulate the strains in every piece due to a single load in any position, 

 the thorough solution becomes simply a question of time. 



It may often happen, however, that we may wish to determine the strains 

 for a full load only, or for a uniformly distributed load extending from 

 one end to some given point. In such case it would be unnecessarily tedi- 

 ous to obtain our result by the successive determination and addition of all 

 the intermediate apex loads. We may easily deduce from the preceding 

 the general formulae for partial loading also. 



As before, we shall let a = the half span, h = the rise, I the moment of 

 inertia, and A the area of the cross-section. But we shall represent by p 

 the load per unit of length of horizontal projection, and by z the distance 

 of the end of the load extending from the left, from the crown. This dis- 

 tance 2, from the crown to the end of load, is then positive towards the 

 left. In the circular arch the angle subtended by this distance z we call #. 

 The angle is then positive to the left of the vertical. The angle sub- 

 tended by the half span is, as before, a. For /3 = a, then, or for e = a, 

 there is no load upon the span. For /3 = 0, or = 0, the load extends 

 from the left to the centre. For ft = a, or z = a, the load covers the 

 whole span. Pis. 23 and 24, Figs. 91 and 92, still hold good, therefore, for 

 our notation. We have only to conceive, instead of the concentrated load 

 P, a uniformly distributed load, per horizontal unit, extending from left end 

 as far as the position of P. This much being premised as to notation, we 

 shall treat, as before, the three cases of arch hinged at crown and ends, 

 hinged at ends only, and without hinges. 



A. ARCH HINGED AT CROWN AND ENDS. 



2. Reaction. This case is too simple to demand any extended 

 notice, in view of what has already been said. We. have from eq. (16), Art. 

 8, for the reaction at the left or loaded end, for concentrated load, 



. 



2 a 



If now we put P = p d z, and integrate, we have 



where O is the constant of integration. By taking the proper limits, we 

 can eliminate this constant, and thus obtain the reaction for load covering 



