1841.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



295 



cular to E F are parallel, aud the same may be said of L, and M ; hence 

 if produced they will form a parallelogram, and the resultant diagonal 

 of K, and L, will be equal, and in the opposite direction to that of M, 

 and O ; that is, the forces will be in equilibrium ; but if K, and L, be 

 greater than M, and O, their resultant diagonal will also be greater, 

 and motion must ensue, that is such a state of things cannot exist. 

 And what is true of these single forces, will be true of the forces of 

 all the fibres collectively ; or the resistances to compression and ten- 

 sion will be equal. It may be said that this explanation, involves the 

 necessity of the centre of motion N, being midway between the con- 

 tending forces, and that that does not obtain, in the case in point; but 

 this does not prevent its application, for the resistances to tension 

 though exercised nearer to the neutral axis, are greater in amount, so 

 that the effect is practically the same. 



3. The position of the neutral axis therefore, depends upon the ra- 

 tio between the tensive and compressive resistances of the material, 

 and by the application of the above principle, with the necessary data, 

 its situation may be found. The nest step will be to develop the 

 general expression for obtaining it in simple cases. 



Let A B C D be a transverse sec- 

 ♦ •ou of a rectangular beam ; assume 

 N as the middle of the neutral axis, 

 and through it draw the central line 

 E F, and in F C take any line F L, to 

 represent the resistance of all the 

 fibres in the line B C ; join N L, and 

 the triangle X L F will denote the 

 resistance of all the fibres in tension. 

 Make N G, equal to N F, and draw 

 GH perpendicular to it, and let GH, 

 be to F L, as the resistance to com- 

 pression, is to the resistance to ten- 

 sion in the material in question; join 

 N H, and produce it to K, then the 

 triangle NEK will represent the re- 

 sistance of all the fibres in compression, and will consequentlv he equal 

 to the triangle N F L. Let .r equal X F, the distance of the neutral 

 axis from the bottom ; take d for the depth, and let/) and q stand for 

 F L and G H respectively ; that is, let them denote the tensive and 



A. 



Fig. 3. 



n 



B 



U C 



compressive resistances. 



Now by similar triangles N G : G H : : N E 



or.r : y : : d — x : EK,orEK = ?i^m^, 



X 

 q {d—x) 



EK, 



(,d — x) X 



= area of triangle NEK; and ■ 



of triangle N F L; hence we have the equation. 



px q {d — J")^ 



2x 



OT p . 



q(,d — x)' 



, hence p x^ 



d ■ — 2 dx -\- x". dividing by x'-, 



p _d^ 



:q{d — xY- or ^x'- 



1 



•2 — \-\ and extracting root 

 d 



+ 1 



(10 



4. The application of the principle to the common form of girder, 

 i» the next case that suggests itself. 



Fig. 4. 



d:..: 



/ 



>!i 



Let A B C D E F G H be the section, assume X to be the midde of 

 the neutral axis, and through it draw the central line K L, and in the 

 line E O, take a line M E, to represent the resistance to fracture of all 

 the fibres between E, and O, and for the sake of convenience, let M E 

 be equal to half E O ; join X E, X F, and let N F, and H G, produced, 

 meet in P. 



Make N Q, equal to N M, and draw Q R perpendicular to it, and let 

 Q R, be to M E, as the compressive resistance in cast iron, is to the 

 tensive resistance ; join N R, and produce it to S, and as T S ; T D, 

 so make T U : T C ; join X U, and let X U, and AB, produced, meet 

 in V. 



Then in accordance with the principles as applied in the last case, 

 the figure N L P F E, will represent the resistance to fracture, of all 

 the fibres below the neutral axis ; and the figure X S U V K, the re- 

 resistance of all those above, and these will be equal. 



Xow let .r = N L the distance of the n. a. from the bottom. 



d = the whole depth. 

 d' = depth of top flange, 

 rf" =: depth of bottom flange. 

 b = breadth of top flange. 

 J' := ditto of middle rib. 

 h" = ditto of bottom flange. 

 And as before, let/) and q represent M E and Q R respectively. 



;; h"* 



p^ 



b' p b" X 



or L P ^= 37r= rr, jHI- Therefore the area of the figure 



X — d b' (x—a) ^ 



X LP F E will be represented by, 



Then N M : M F 



XX' 



NL : LPor.r-(i" 



.r : LP 



pb" 



p b" X 



Pi^-d") , ^„ ( V +b' {x-~d")\ 

 o— ^-d \ 2 ^ = 



pjx-f^) p_drjr_ / X X 



2 ^ 2 6' ^ V Kx—d") ) 



Again, asNQ :QR::NT .TS, 



thatisar — rf" : 5 t : d — d' — x : T S, 



q (d — d' — x) 



or T .S : 



X — d" 



ButTD : TS : : TC : TUoro : 'J('^—^^'—^^ • • ^' • TU 



X — d" h' 



b p q (d — d' — x) 



h' (x~d") _ bq (d — d' — x) _ 

 p ~ b' {x-^dn 



And N T : T U : : N K : K V or 



, „ b a (d — d' — X) 



d—d'~x : -i-}~- — -—--!■ : : d—x : kv, 



b' (X — d") ' 



that is T U ; 



hence K V = 



bq (d—d' — x) . (d — x ) 



b'(x — d") _bq(d—x) 



d — d' — x b'(x — d"y 

 And the area of the figure X S U V K, will therefore be represented by 

 d-d'-xXg(d-d'-x) , , /Q(d-d'-x) bq(d- x) 

 2 x-d'' + '^ y b'(x-d") +b'(x-d" j 



* As ^ represents the resistance of all fibres between and E ; T7,will 



denote that of all those between W and F; and (— -; ) that is — , will 



e.tpress the multiple, that the number of fibres between W and F. is of the 

 number between and E, and ifju equal the resistance of fibres between O 



and E, then p x 77- will equal that of all fibres between W and F. It is ne- 

 cessary that the quantity should be put in this form, instead of the simple 

 fraction , in order that the value of L P, should involve the term (p). 

 t The last note explains this also. 



