Having thus explained the construction of roofs, consisting of two principal rafters, we shall proceed to the consideration of those of a more complicated form, where the rafters are more than two in number; in which case it is generally called a kirb roof.
We have already demonstrated in our article BRIDGE, that if a string or festoon of heavy bodies connected together, is suspended from its two extremities, (See Plate LXXX. Fig. 2.) they will arrange themselves into a Catenarian curve by the force of gravity; and that if this assemblage of bodies is inverted so as to rest upon the former points of suspension, it will form an arch of equilibration. The same is obviously true of any number of bars of metal or beams of wood, connected together by moveable joints, so as to take the position of equilibrium, which the force of gravity act ing upon each of the beams must necessarily give them.
The slightest consideration is sufficient to convince us that such a position of the beams is that which they should have when formed into a roof. In this position, all the rafters are in equilibrium with each other, and are acting on each other in the direction of their lengths, and consequently resisting any external and uniformly distributed strains acting in the direction of gravity with the greatest force.
In Fig. 5, for example, let AB, BC, CD, DE, be rafters moving round flexible joints, and arranging themselves in the curve of equilibrium ABCDE; this is the position which must be given them when fixed into a kirbed roof, with this difference only, that they must be placed in an inverted position.
If they have any other position different from that of equilibrium, such as is shown by the dotted lines in Fig. 5, where Jib, bc, cd, dE, are the rafters, then the rafter cd must be held in its depressed position by some external force; and, consequently, when the whole Is inverted to form a roof, the rafter cd, must have a ten dency to assume the position of equilibrium CD, and in consequence of this unbalanced force cd and all the other beams will not act upon each other in the direc tion of their lengths, and consequently will not be in their strongest position. When they are placed, on the other hand, in a position of equilibrium, the tie beam, the king post, and the braces, &c. have to per form no other office but that of preserving the rafters in their position of equilibrium.
If the strain is uniformly diffused over the roof, as in houses covered with slate or lead, or if unequal loads are symmetrically placed upon it, then the Form of the roof, or the curve of equilibrium will be symmetrical, and its two halves will be equal and similar; but if it is loaded more in one place than another, and if that place is not on the ridge, then the form of the rafters must be unsymmetrical. Thus in Fig. 5, if the rafters AB, BC, CD, DE, are made equally heavy, the curve will be as in the figure; and is symmetrical, the angle ABC being equal to the angle CDE. If BC and CD, made equally heavy, are heavier than AB and DE, which are equally heavy, then BC and CD will fall lower, increasing the angles ABC and CDE, and di minishing the angle BCD; hut still the curve passing through the joints, will be symmetrical. If the part
of the roof CD is to be loaded with lead, while all the rest is to carry only slating, then CD being much heavier than any of the other rafters, will sink in the experi ment of suspension to cd, raising the joint B to b, and depressing C to c, and D to d. The form or the roof will therefore be that of AbcdE, which is no longer symmetrical.
Although it is now easy for the practical mechanic to determine experimentally the position of the rafters of a kirb roof mechanically, by loading the centres of gravity of his experimental beams, in the same manner as the corresponding beams in the roof are to be loaded: yet it is desirable to have a mathematical me thod of determining the best form of a kirb roof.
By referring to our article BRIDGE, and to Plate LXXX. Fig. 4, where the rafters are represented as cords, it will be found to be demonstrated, l. That the tension in any part of the cord is in versely as the sine of its inclination to the vertical, and 2. That the loads on the different joints (C, C', C") or the tension produced by the weights w, /•", are directly as the sines of the angles at these joints; and inversely as the products of the sines of the angles which the rafters make with a vertical line, that is, in Fig. 4, Plate LXXX.
Sin. r C Tension d c is as Sin. rCdx Sin d C / with these data, we arc now prepared to determine the best form of a kirb roof.
Let it be required to find the form of a kirb roof ABCDE, whose rafters AB, BC, CD, DE are equal, AE being the width, and CF the height of the roof. As the points A, C, and E, are fixed, this problem re solves itself into finding the position of the point D, in the line DHG, bisecting CE perpendicularly, when the loads at the angles C and D are equal, and consequent ly in equilibrio.
From the point G, where DH intersects AL, and with the radius GE describe the circle EKC, passing through C, because CH bisects CE at right angles. Draw HK, parallel to FE, cutting the circle EC in K, and join KC. The point D, where CK cuts GFI, pro duced is the point required, and the lines CD, ED, meeting at this point, show the position of the rafters.
Produce ED till it cut the vertical bar FC in N, and having given the rafters CB, BD, the same position as CD, DF, complete the parallelogram BCDP, and draw DB, bisecting Cl' in It - Join K, F by the line which is parallel to DP, because CDP=CKF, on ac count of the parallelism of RD, QK, and the equality of CR, RI? and CQ, QV; make CS equal and parallel to FG, and upon S with the radius Si", describe the semicircle WKF, which must pass through K, because CG=SF=GE and CQ=QF. Join WK and WS, and produce BC cutting ND in O. Now the angle WKF, at the circumference, is equal to \VSF at the centre, and is therefore equal to WSC or CGF, and double of CFE, or its alternate 'angle ECS. But ECS..-----ECD+DCS, and ECD=INDC and DCS=A DCO, or the alternate angle CDP. Hence WKF_—_- NDC+CDP=NDF, and WK parallel to ND. Con sequently CF: CW = CP: CN; and hence CF=CW we have CN=CP.
Now, in the two triangles CDN, CDP, the sides are to one another as the sines of the opposite angles, as follows: