EXAMPLES TO THE SECOND SERIES. 
xxxi 
Scantlings of nearly equal strength, sufficient to carry the weights in 
Figs. 2 or 1, the breaking weight being 8 times the respective weights. 
Weight uniformly 
loaded in Fig. 2. 
Scantlings by the Second Series of Tables. 
Length 1 6 feet clear bearing. 
AG = 
8 cwts. 
4X8; 5X5i; 6X3i; 7X2f; 8X2. 
DH = 
64 
II 
(9X13; lOXlOi; 11X8^; 12X7^; 13X6^; 14X5J; 
1 15X41; 16X4. 
EI = 
56 
II 
(8X14; 9X111; 10X9; 11X7J; 12X6^; 13X5^; 
1 14X4f ; 15X4; 16X3^. 
LI = 
96 
II 
11X13^ 12X101; 13X9i; 14X7|; 15X7; 16X6. 
BM = 
32 
II 
(6X14i; 7X11; 8X8; 9X6i; 10X5^; 11X4J; 
1 12X3i; 13X3i; 14X2f; 15X2i; 16X2. 
JVotes to the preceding Example, JVb. 16. 
1. If the weight DH be placed in the middle between A and C, then AD + AE = AC; 
and the stress at C from the weight DH in Fiff. 1, = stress at C from the weight DH uni- 
formly loaded ; as in the Example. 
2. If the middle of the beam or the point C be between E and L, the weight uniformly 
loaded to have tlie same stress at C as the weight EM may be found thus : — 
Let the weight EM = W' = 96 cwts., then the weight uniformly loaded to have the same 
stress at C as the weight EM, is 
. AE X EC + BL X LC 12 + 16 
W + — X W = 96 + — X 96 = 152 cwts. ; 
AC X EL 48 
and the Scantling for that weight by Table XVI., is 16 x 9i : or 15x11; 14| x llj ; 
14xl2i; 13^x131; &c. 
3. If CE = CL or AE = BL, then the required weight uniformly loaded is 
, AE 
W + — X W = 96 + f X 96 = 156 cwts. 
4. In Fiff. 1, join LK; let the weight of the triangular prism of which the section is 
LBK = W'=32 cwts.; then the weight uniformly loaded to have the same stress at C as 
LBK is, 
BL 
f X — X TV' = f X I X 32 = lOf = 10-66 cwts. ; 
and the length being 16 feet, the scantlings for 10'66 cwts. or the next greater weight are, 
4xlOh 5x7; 6x4i; 6ix4i; lx3k; lixH; 8x2f; 8^x2^; 9x2i; 
9ix 2. 
5. The stress at C from the weight LBK, is i of the stress at C from the weight BM. 
