602 ON THE CALCULATION OF THE EQUILIBRIUM AND STIFFNESS OF FRAMES. 



If there are two additional connexions R and S, with elasticities p and or, 

 x=-F 



_ 

 ) - (2S(er)) 



/ 2 (cpr) 2 (ers) S (?s) + 2 (eps) S (egr) 2 (ers) + 2 (ep?) 2e (r 3 + p) 2e (s 1 + o-^l 

 1-2 (epr) 2 (e(jr) 2e(s > + <r) - 2 (<>p*) 2 (eg*) 2e (r 1 + /) - 2 (ep?) (2 (ers))' J ' 



The expressions for the extensibility, when there are many additional pieces, 

 are of course very complicated. 



It will be observed, however, that p and q always enter into the equations 

 in the same way, so that we may establish the following general 



THEOREM. The extension in BC, due to unity of tension along DE, is 

 always equal to the tension in DE due to unity of tension in BC. Hence we 

 have the following method of determining the displacement produced at any 

 joint of a frame due to forces applied at other joints. 



1st. Select as many pieces of the frame as are sufficient to render all its 

 points stiff. Call the remaining pieces R, S, T, &c. 



2nd. Find the tension on each piece due to unit of tension in the 

 direction of the force proposed to be applied. Call this the value of p for each 

 piece. 



3rd. Find the tension on each piece due to unit of tension in the 

 direction of the displacement to be determined. Call this the value of q for 

 each piece. 



4th. Find the tension on each piece due to unit of tension along R, S, T, 

 Ac., the additional pieces of the frame. Call these the values of r, s, t, &c. 

 for each piece. 



5th. Find the extensibility of each piece and call it e, those of the 

 additional pieces being p, a, T, <fec. 



6th. R, S, T, Ac. are to be determined from the equations 

 Rp + R2 (er 1 ) + S(ers) + Tt (ert) + F2 (epr) = 0, 

 So- + RZ(ers) + S(e,f) + T2(est) +F2(eps) = 0, 



as many equations as there are quantities to be found. 



7th. x, the extension required, is then found from the equation 

 x= -FS,(epq)-RZ(erq)-S*(eqs)- Tt(eqt). 



