158 
1 is half the length of the bolster. Let I, be the moment of inertia at A, I’ the 
moment of inertia at B and I the moment of inertia at the invariable section mn. 
Assume the boister to be uniformly loaded, to be supported at a point and 
also assume that the moment of inertia I at mn 
i a 
I 5 E 
SEAR Es: 
aoe te 
that is, the moment of inertia increases directly as x increases. 
E is the modulus of elasticity of the material. 
Equating the moment of the elastic forces to the moment of the external 
forces about the neutral axis of section mn, we find 
d? y w x? 
TE Tq a. | hee 
a yes Ww sa ae tl 
or, BT x? eee a | ee ee fae —I,)x. 
ae fe 
Dividing the enumerator of the fraction by the denominator, integrating 
twice and determining the value of the constants, we find 
25, Mes, f ER pee ey 
Ey=tn—-(2[#-C+0¢ 1 1 + log. ( +E) a 
ei = nies C n? n° 
+ C® log. e one t ay ier =} Nein aw ele aebedalsteiee (A) 
I, 
where n = > and C = (i? —I,) 
To obtain the deflection at the side bearing,.it would be necessary to substi- 
tute for n its proper numerical value along with other values, and compute the 
The deflection at the side bearing is not equal this value of 
resulting value of y. 
y; but is equal to the end deflection minus this value of y. 
<: + C? + C8 [log.e 1, — log-e I*] i 
- 
When n = 1, 
wl* ellis aay 
| a 
When n — 1 and I, — 0, 
4 
Ey= ee and when n = 1 and I, — I’ the expression becomes indeterminate 
w l4 
and evaluates to E y = 31 
The truck bolster may be treated as a cantilever, with a terminal load equal 
to one-half, the center load and a length one-half the length of the bolster as shown 
in Fig. 4. 
f= 
Te) 
wreyse| 4 
