THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[Janiaby, 



mass (if tlie siisponsion rods. Tlic lengtli of these increases from 

 tlie centre to t}ie almtmeiit, and is equal to that of the vertical 

 co-ordinates of the catenary. 



The iiitCffi-ation of equation (2) g-ives the relation of y to .r, 

 which M ill he found sudi tliat ///'' jinipotiion hetween any two vertical 

 (•(i-ordinute.s is equiil to the sijitare of the projiortiiin Ijetireeii tlie corre- 

 x/ii»u/iiiy horirotitd/ eo-orditidte/s — a very siin|de rule for deterniininjr 

 the form uliicli the chain will nearly assume wlien subject to its 

 hrcakiufT weii^ht. 



It folhjws from this rule, that when 



.r is .j'j th of the half-span, y is ,J|, th of the deflection 

 ■r -ibths „ y ^ths „ 



.r :,i,tlis „ y 105 ths „ 



■r ,1-,ths „ y ^Sths „ 



&c. „ &c. ,, 



Now, if we sup])ose the half-platform attached to the half-diain 

 at every 20 tli of tlie lentftli of the former, we shall, by adding up 

 all tlie/z's in the above scheme, get the total length of all the ver- 

 tical roils together, which, therefore, is equal to 



^ (1 -I- 2' -f T- + 1' + + 19-) X deflection; 



and this by actual calculation is equal to the deflection X 6'175. 

 Therefore, since tlie deflection is 50 feet, the total length of all 

 \ ertical rods is :i08-75 feet. This supposes that the platform meets 

 tlie chain at the centre, the value of y at that point being zero. 



Each rod sustains one-twentietli of the weight of the half-span, 

 and therefore has a tension - 4«'. Therefore, if, as before, < be 

 the tension per square inch of the sectional area, that area, by 



w 

 principles already laid down = ^'^ -. The mass of all the rods equals 



this quantity X the length 308-75 feet, just obtained = ^ 

 Adding this to the mass of the half-chain, we have finally 



15-4. 



the mass of the half-chain and its rods ; 



X 1204-7 (A). 



Next, let it be ascertained what is the quantity of material re- 

 quired when the load is supported 

 by straight rods exclusively. It 

 has been sliown that if 7,(,u' be sus- 

 tained at A by a straight rod A B, the 



, , , 1 «) A B = 



mass of that rod = —--- .,„ • 



This expression (since ABC is a 

 right angled triangle, and therefore 

 AB' = A C- -l" BC-',) is equivalent 



Fig. 4. 



20 



A(2<i)= 



A C is i'j th of the half-span, A C = 4o (2 a)' 

 AC iliths „ AC* 



AC .l;ths „ AC' 



&c. &e. 



Adding up all the values of A C=, we get for the total mass of 

 the rods 



i" {(1+2^ + .3^+ 19^)ilSc 



+ 19 BC 



:} = 



w f 



7 I 



6-175 ^i~}r + 19 Bc}: 



761-3 (B). 



20 



-i 6-IYi '„ „ , , . 



20 < (. BC J t 



Comparing the expression (B) with (A), we see that the material 

 re(|uirod in one case is about j'^ ths of what is required in tlie other 

 case. In otlier words, if <i .•.■i(.\/ieii>iifm ImiUje, of the dimensions of the 

 Hungerford bridge, u-ere sustained hy independent oblique rectilineal 

 rods, instead of n main chain and vertical rods, a saving of nearly half 

 the material would lie effected. 



It will be hereafter shown, that great as is this advantage with 

 respect to tlie )iower of the bridge to sustain a statical load uni- 

 formly distributed, still greater advantages belong to the method 

 of suspension here advocated, when the eff'ect of moveable loads is 

 taken into consideration. 



TIIK OBLIQUE ARCH. 



On the Focus to which the Joints on the Face converge. 



Suppose a right-angled plane triangle formed of any flexible 

 material, having its two sides respectively equal to L, the axial 

 length, and ir r, the semicircle obtained by taking a section of 

 the bridge jierpendicular to the axis. If the side L be placed along 

 the top of the abutment of an obliipie bridge, and the triangle 

 wrapped over the laggings, tlie hypo'dienuse will form a spiral line 

 which is the intersection of tlie coursing joints and soffit. If a 

 straight line move along the axis ot the cylinder, so as always to 

 intersect it at right angles, and pass tiirougli the hypothenuse of 

 the above triangle, it will generate the twisted surface jiroper for 

 the beds of the stones. Mr. Buck was the first to show that the 

 joints that ajipear on tlie face of the arch, pass tlirougli a point O, 

 below the centre C, when the " section on the square," or section 

 perpendicular to the axis, is a circle. A similar expression for the 

 length of the line C O may be obtained when the sections of the 

 intrados and extrados, made by a plane perpendicular to the axis, 

 are similar ellipses. 



a, A 



Let the figure represent the elevation of an oblique bridge, cir- 

 cular or elliptic " on tlie square" — if it be segmental, the ellipses 

 must be supposed completed. Take the axis of the cylinder for 

 the axis of y, and let the plane .»■)/ be horizontal. A, B ; a, b, are 

 the semi-axes, major and minor, of the extrados and intrados re- 

 spectively. A X E, a X E, the lengths of these semi-ellipses. 

 S = arc A'P, where CN = A cos©; and « = arc up, where C j- = 

 arc a']). L = axial length, n = acute angle between the directions 



AE 



of the roads. <]. — angle of extrados ; and, therefore, tan * = 



The equations to the extradosal and intradosal spiral are, respec- 

 tively, 



X = 

 Y = 



A cos ^ 



a'e« FO) 



3 J 



Z = B sin 



X = a cos 9 

 L 



tTE 

 6 sin e 



y = „-f/- • (2) 



"i 



z — 



and the equation to the face of the arch is i/ = .r cot 0. -\- d. (3) 



Let X' Y'Z', x' y' z' , be the co-ordinates of the points in which 

 the corresponding extradosal and inti-adosal spirals meet the face 

 of the arch. Then X' Y' Z', x' y' z' , are co-ordinates of a point in 

 the face of the arch, and must satisfy equation (3). 



Y' 



X 



X' cot D. -\- d ; and y' — x' cot n -j- rf ; 



- X = (Y' - 2/ ) tan n = j; i^j^ - - j t^n n = 

 A /S s\ 

 \k~ a) 



, . , tan Cl. 

 tan 4 



In order to determine the length of CO, it will be sufficient to 

 confine our attention to the projection on the plane xz, of a straight 

 line passing through X'Y'Z', x'y'z'. The equation to this projec- 

 tion is 



y __Z' -Z' 



z — L — 



and if u- = 0, 



X' - .f' 



= CO, .-. co = 



(,r - X') ; 



' x' — z' X' 

 X' - x'~ 



