a ee aie 
COMPLEX STRESS DISTRIBUTIONS IN ENGINEERING MATERIALS. 301 
By. PER? 
d Si Parad 
oa dn! Et 
The solution of this equation is 
z=cos nx (A cosh na +B sinh nx)+sin na (C cosh nx+D sinh nx) +e 5 (4) 
Consider a length AB (=) of the pipe between successive flanges, it being 
assumed for the present purpose that the flanges are spaced equidistantly along 
the pipe, and let the origin for x be at A (fig. 7). 
It is evident, from symmetry, that 
Nass for z=o and «=I 
dx 
and also 2z=z, for =o and z=1 
where 2 is the radial displacement at the flange, as yet undetermined. These 
four conditions enable the four arbitrary constants A, B, C, and D to be 
expressed in terms of known quantities and zo, and equation (4) becomes 
a Lal 
Et Et 
— 
a 
_ 0) (cosh na cos nx —H sinh na cos ne+H cosh na sin na 
—Lsinh na sin na ; : ; : ; (5) 
I _ cosh nl—cos xl 
where ~ sinh ni--sin nl 
__sinh nil—sin nl 
sinh nl+sin nl 
When the distance J between successive flanges is large, H and L both approxi- 
mate to a value 1 and the equation giving the radial displacement becomes 
2=2z,—(z;—29) (cosh na—sinh nx) (cos na-+ sin nz) 
or 2=2;—(2;—2%) e~ (sin nv--cos na) : - : 6 4 (3) 
p2 
= » the radial displacement in a uniform tube without flanges. 
It remains to determine z,, the radial displacement at the flange. For this 
purpose it will be assumed that the flange and the part of the tube beneath 
it form a homogeneous ring of longitudinal thickness c and external radius Ri. 
The forces acting on the ring tending to increase its diameter will be those 
due to (1) the internal pressure acting on its inner surface, and (2) the reactions 
between the ring and the adjacent tube. The latter will consist of a shearing 
force acting radially outwards on the flange over the whole circumference of 
the tube. Denoting this shearing force per unit length measured along the 
circumference of the pipe by F, the resultant outward force per unit length of 
the circumference will be 
1 
where 2;= 
Pe+2F 
and the equivalent pressure per unit area of the internal surface of the flange, 
tending to increase the diameter, will be 
Pc+2F 
It will be assumed that the longitudinal stress in the flange is confined to, and 
uniformly distributed over, a ring of thickness equal to the thickness of the 
tube. The flange may then be regarded as a compound cylinder under internal 
9 Z di F : 
pressure = the inner portion whose thickness is ¢ having a longitudinal 
stress equal to that in the tube due to the pressure P. Observing that at the 
common surface the radial stress and radial displacement have equal values in 
