350 



Mr. B. 0. Laws on the Strength 



load Q only at c and c x with no pull at the ends and free 

 to slide horizontally (fig. 4). 



Fur. 4. 



Lead Q only 



Obtain the deflexion at c (or c,) in each case and impose 

 the condition that these deflexions cancel each other — by 

 reversing the direction of Q in fig. 4. ; we then have the 

 beam as it actually is, and as indicated in fig. 2. The 

 solution is given below : — 



First. Consider w only (fig. 3) : — 



_E.lg = -Ml-P.^+^p3. . . (1) 



Put P = »i 2 E . I and let 6 be the inclination at any point 



fti/ 

 of the curve, then -. = tan 6 = #, since 6 must be small. 



d'u cW K , 2 w(a 2 -x 2 ) 

 — h nrtf— 



dx 2 " dx " E . I 



2E.I * 



And 



d 2 6 



= m 2 



dv w . x 



nvd + ^ 



to . x 



dx 2 ~"° cfe T e.i ' E.r 



The solution o£ this equation is : — 



6 — K.e + B.e ^r— { 



jft-Ji, , 1 



.and 



E.I^ = ~E.I^=-m.E.I(A.C--B.C") + 

 ax dx* ° ° 



where £ is ths base of the natural or Naperian logarithms. 

 When, = a, J = * = 0. .vA.«7 + B*--^=0. 



