BEAM OF RECTANGULAR CROSS-SECTION UNDER ANY SYSTEM OF LOAD. 149 
-3B^-12D^ = 0. . 
(124) is identically satisfied since 
Also (122), (118), (113) imj^ly 
A., = Co = 0 . . 
(119), (120), (112) lead to 
B. + 4a = -| j 
(124). 
. (125). 
B, = - 
Also (123), (114) imj^ly 
A, - 4Cj = 2q j 
-^3 — = 0 . 
(121), (115) are identical. Together with (111) they give 
^3 = ~ j 2/,2 (®i + 
(126). 
(127). 
(128). 
Equations (118), (124), (Dio), (126), (127), (128) contain the solution we require. 
If ve substitute the values of tlie constants in the expressions for the displacements 
and stresses, we find, after some reductions : 
U = const. + X |Ai I + + 2 x 7 / 
^ 1 1 8fM(y + f,) ^ 2/j.j ^ E ^ E 
4^3 I E ^ \e ^ i 
2^^ h~\ E 
1 
2a 
V = const. — y <! A, — ^ ^ -1- 
/ 0 ox 
X* 
+ ("f + ^‘) IW 1 E - 
P = (n' + C.) + 2B,2/-'|' + |'(| + D, 
_ 3 I 
4 ^,s -f 2/^2 > 
Q = 
s = 
1 I '"’’P _ 
2 4// 
4}r ’ 
7+D,)(i-S 
qx 
a ^ 11 _ on 
^hV U 
