302 REPORTS ON THE STATE OF SCIENCE, ETC. 
the two portions, the radial displacement at radius R can be shown, by the 
ordinary theory of the stress in a thick cylinder, to be approximately (i.e. neg- 
lecting squares of ) 
i= | pimal, 28 | RR) RX) ERD 
2m ef Em °R,(R+t)—R(R—2) 
2n—1 , 2F 
=) Poe ce Rit 201 qhaiig old Jol Be 
{ 2m c } — % (7) 
Now the shearing force at any point of a beam is equal to 
diz 
cr 
where B is, as before, the flexural stiffness. At any point of a longitudinal 
strip (of unit width) of the tube wall under consideration, it will be given by 
m2 t® d®z 
PET aS asks a2): (gy, 
Performing the required differentiation on the expression for z given by equa- 
tion (6), and putting =0, the expression (8) becomes 
_1 Em! 
3 m—1 
which, from a consideration of the signs, will be the shearing force at the end 
of the strip acting inwards on the flange. The value of F, acting outwards, 
will therefore be 
> n(2;—Z9) 
The substitution of this value of F in equation (7) enables z, to be determined. 
2m—1,2 m? Eg, 
— Se = Ry 
A 2m 3 m?— 
Thus Zo= = 
2 1 42 m Ea £ 
J 38m—l Cc 
and this value of z, substituted in equation (6) will enable the radial displace- 
ment at any point to be determined, and from this the circumferential stress 
by equation (1). 
It will be seen from equation (6) that the expression for the radial displace- 
ment contains a periodic function of the distance from the flange, and that the 
longitudinal profile consists of a series of waves whose amplitude rapidly 
diminishes with the distance from the flange. 
Thus, if the equation be written in the form 
2=21—(21—20) P(@) 
| 
‘ : T oT OF 
¢(x) reaches a negative maximum when a=-, —, —.. 
nn 
For steel, taking 1 =(0'295, these values become 
m 
x—2'44,/tR, 7°32 VIR, 12:20VtR.... 
and (x)= —*0432, —-0019, —-00008... . 
It would appear therefore that the constraint imposed by a flange produces 
an appreciable increase in radial displacement over a region extending from 
ive. x=1'83 ViR to x=4:27 IR 
