﻿56 
  Mr. 
  W. 
  H. 
  Macaulay 
  on 
  the 
  Stresses 
  

  

  Similarly, 
  

  

  T 
  dp 
  =iQ$-2p 
  + 
  l)sW-s$. 
  

   Now 
  

  

  .»+l\ 
  /A„.-P 
  U 
  n 
  .P\ 
  u 
  m+1 
  

  

  - 
  2VV 
  ^ 
  + 
  ^ 
  M 
  w+i 
  I 
  2 
  ^ 
  (l-w) 
  2 
  J 
  (l-^) 
  2 
  ' 
  

  

  Thus 
  the 
  tensions 
  of 
  all 
  the 
  bars 
  are 
  found. 
  

  

  To 
  find 
  the 
  deflection 
  at 
  the 
  point 
  k, 
  I 
  for 
  the 
  loading 
  under 
  

   consideration 
  we 
  have 
  to 
  sum 
  the 
  expressions 
  (29) 
  and 
  (30), 
  

   the 
  former 
  for 
  values 
  of 
  m 
  from 
  I 
  to 
  N 
  — 
  1, 
  and 
  the 
  latter 
  for 
  

   values 
  of 
  m 
  from 
  1 
  to 
  l 
  — 
  l. 
  Thus 
  the 
  deflection 
  is 
  

  

  "~2 
  l 
  N 
  (^ 
  T 
  o 
  + 
  ^n)— 
  R& 
  >, 
  

  

  T 
  , 
  R;n> 
  and 
  R* 
  being 
  here 
  the 
  tensions 
  for 
  the 
  case 
  of 
  the 
  

   continuous 
  loading, 
  these 
  being 
  the 
  results 
  of 
  summing 
  the 
  

   corresponding 
  tensions 
  for 
  the 
  case 
  of 
  a 
  single 
  load. 
  Re- 
  

   membering 
  that 
  IT 
  + 
  &R;n- 
  = 
  NT 
  , 
  and 
  substituting 
  the 
  

   values 
  which 
  we 
  have 
  found 
  for 
  the 
  tensions 
  T 
  and 
  R^ 
  the 
  

   expression 
  for 
  the 
  deflection 
  becomes 
  

  

  ±W\m{W-l-ilk 
  2 
  + 
  kl 
  + 
  l 
  2 
  )\+iW/JLs 
  2 
  kl 
  

  

  1TTT 
  1 
  + 
  u 
  fN-1 
  , 
  <b-\-ylru 
  2 
  1 
  ,., 
  - 
  . 
  7v 
  

  

  or 
  

  

  ^WXfkl 
  (N 
  2 
  + 
  hi 
  - 
  2) 
  + 
  iWfis 
  2 
  kl 
  

  

  The 
  results 
  obtained 
  for 
  continuous 
  uniform 
  loading 
  may 
  

   be 
  written 
  concisely 
  as 
  follows. 
  Let 
  

  

  