File DOO. 12/122/No. 3?22 AIR CORPS TECHNICAL REPORT No. 3522
AIR CORPS INFORMATION CIRCULAR
PUBLISHED BY THE CHIEF OF THE AIR CORPS, WASHINGTON, D. C.
Vol. VII June 30, 1932 No. 673
CIRCULAR RING
WITH CONCENTRATED LOADS
(AIRPLANE BRANCH REPORT~
UNITED STATES
GOVERNMENT PRINTING OFFICE
WASHINGTON : 1932
~ -:-- ---~··- - - -- _.,, ---·---- - -------~-~--~~.,_- _____ - - . -·· -
CIRCULAR RING WITH CONCENTRATED . LOADS
(Prepared by H. W. Sibert, Materiel Division, Air Corps, Wright Field, Dayton, Ohio, September 11 , 1931)
INTRODUCTION
In a recent issue of the "Airway Age" (May 2,
1931), R . A. Miller gives solutions for a circular ring
with concentrated loads. Since Mr. Miller's paper is
rather long and hard to read, it was thought desirable
to give a shorter and simpler derivation of this problem .
The following discussion applies only to a ring with
constant cross section and constant modulus of
elasticity. In every case, the external loads will be
applied symmetrically with respect to the vertical
axis of the ring.
NOTATION
Let the y-axis pass .through the center of the ring,
and let
R = radius of the neutral axis of the ring
a = the angle from the lower .Part of the y-axis to a
variable point on the neutral axis of the ring
E =modulus of elasticity
I =moment of inertia, of the entire area of the cross
section of the ring about the neutral axis of
the cross section
U =work done by bending
dS. =an element of arc along the neutral axis
M = moment at any cross section of the ring
S =shearing force on any cross section (positive
when radially outward)
P =normal force on any cross section (positive when
tension)
Au =the cross section through the point of intersection
of the neutral axis of the ring with the upper
portion of the neutral axis
p =value of Pat Au
m =value of Mat A,.
GENERAL METHOD OF SOLUTION
The solution is based on the theorem of least work;
Since the external loads are symmetrical about the
y-axis, the work done on the left and right halves of the
ring are equal, hence, it is necessary to consider only
the work done on the left side of the ring, the value of
which is
f 'Tr M2 R f 'Tr
U= o 2EI ds= 2EI o M2da (1)
· since· dS =Ra.
·Due to the symmetry of the external loads about the
y-axis, the value of S at A,. i~ zero. Hence, the
moment at any section of the beam will be known as
soon as p and m are determined. These two unknown
quantities must have values such that U is a minimum.
A necessary requirement for this to occur is that
oU/op=O and oU/om= O. Note that oU/op= (R /2El)
J:<oM2/op)da = (R/El) f ;<MoM/op)da, since
126251-32
oM2/op = 2MoM/op. Similarly, oU/ om = (R/EI) J ;(MoM/om)da. Thus the conditions oU/op=O and
oU/om=O become, respectively,
J ;(MoM/op)aa=O
J ;(MoM/om)da= O
Case 1 (Case 4 in Miller's article).
(2)
(3)
The external loads are applied as shown in Figure 1.
Figure 2 shows the forces acting on a cross-section of a
zw
7--p ~ m -j-P
P~\ m
/11 \ 1· (\
IX I
ring when a is between 0 and </>, whereas Figure 3 shows
the forces when a is between </> and 'Tr. From Figures
2 and 3, it is evident that
M=m+p R (l+cos a)+ WR(sinct> -sin a),O ~ a ~</> (4)
M=m+p R(l+cos ~ a) , </>a ~ ,,,. (5)
Hence, for all values of a, oM/op=R(I+cos a) and
oM/om= 1. Therefore, formulas (2) and (3) become,
respectively,
Rf; M(l+cos a)da=O (6)
f ; Mda=O. (7)
If (6) is divided by R, it may be wri~ten in the following
form:
f ; Mda+ f ; M COS a da= O (6')
However, from (7) , J
0
• Mda must be zero. Hence,
conditions (6') and (7) may be written as foliows:
(7)
f : M cos a d~=O (8)
The substitution of ( 4) and (5) in (7) yields the following
results:
1
)
2
0= J ;[m+ pR(l+cosa)j da+ WR J : (sin <j>-sin a)da
=..-m+..-pR+ WR[<t> sin <t>+ cos .p-1] (9. 1)
The result of substituting ( 4) and (5) in (8) is
O ~ f; [m+pR(l·+cos a)] cos ad a+ WR f: (sin 4>
- sin a) cos ada= ..-pR/2+ (WR/2)sin2 4> (i0.1)
If (9.1) and (10.1) are solved simultaneously, the
In case 1, change w to - wand repiace 4> b y e.
between 0 and O
For a;
P= - (W/.,..) sin2 ()cos a - W sin a (13.4)
S= (W/..-) sin2 () sin a-W cos a (14.4)
M =-(WR/ .,..) (1-0 sin 0- cos ll-sin2 II cos a)
- WR (sin II-sin. a) (15.4)
For a between O and .,..
P= - (W/ ..-) sin2 0 cos a
result ing values of p and mare s·= (W/..-) sin2 11 sin a
p= - (W/.,..) sin2 .p
(ll.l) M =-;(WR/.,..) (1- 0 sinll-cos ll-sin2 11cos a )
(16.4)
(17.4)
(18.4)
m= (WR/.,..) (1 = .p sin .p - cos <t>+sin2 <t>) (12.1)
From Figure 2 and from formulas ( 4), (11.1) , and
(12.1), the values of P, S, and M for between 0 and I
.pare
S=p sin a+ W cos a= -(W/ .,..) sin2 4> sin a+ W
cos a . (13.1)
P = --p cos a+ W sin a= (W j.,..) sin2 <j> cos a+ l'V
sin a (13.1)
M = (WR/ .,..) (1 - <t> sin <j>-cos .p-sin2 .p cos a)
+ WR(sin <j>-sin a) (15.1)
From Figure 3 and from formulas (5) , (11.1) , and
(12.1), the val~es of P, S, and M for a between 4>
and.,.. are
P = p cos ( ... - a)= - p cos a= (W j.,..) sin2 cj> cos a
\ (16.1)
S= p sin ( ... -a) =p sin a= - (Wf .,..) sin2 4> sin a (17.1)
M =: (WR/.,..) (1-<t> sin <j> - cos .p - sin2 .p cos a) (18.1)
Cace 3
Case 1Z
Case -it
zw .zw
Case 2. (Case 3 in Miller's article.)
Set q,=.,../2 in Case 1. For a between 0 and ..-/2
P= (W/..-) (cos a+ ... sin a) (13.2)
S= (W/ ... ) (?r cos a -sin a) (14.2)
M= (WR/?r) (l+?r/2-cos a -?r sfn a) (15.2)
For a between ?r/2 and ?r
P= (W/?r) cos a
S= - (W f.,..) sin a
M= (WR/.,..) (1 - ?r/2-cos a)
Case 3. (Case 5 in Miller's article.)
(16.2)
(17 .2)
(18.2)
Set <j> = ?r in Case 1. For a between 0 and.,..
P = W sin a (13.3)
S = W cos a (14.3)
M =(WR/?r) (2 - .,.. sin a) (15.3)
The maximum value of M occ urs at eit her a= O or a= ?r.
Thus Mmax= 2WR/..-.
Case 4. (Miller does not consider this case.)
w
w
C<1Se o
Case 5. (Case 1 in Miller's article.)
Add cases 1 and 4. It is assumed t hat 0 < .p. For
a between 0 and II, add (13.4) , (14.4) , and (15.4) to
(13. 1), (14.1), and (15.1), respectively .
Thus
P =(W/?r) (sin2 <j>-sin2 II) cos a
S= -(W/.,..) (sin2 <j>- sin2 II) sin a
(13.5)
(13.5)
M=(WR/?r)[ - .p sin <t>+O sin II - cos <t>+cos ll-(sin2 .p
-sin2 II) cos a]+ WR (sin .p-sin II) (15.5)
For a between O and <j>, add (16.4), (17.4) , and (18.4)
to (13.1), (14.1) , and (15.1) , r espectively. Hence
P= (W /?r) (sin2 <j>-sin2 II) cos a+ W sin a (16.5)
S=-(W/?r) (sin2 .p-sin2 11) sina+ W cos a (17.5)
M =(WR/?r) [- 4> sin <t>+O sin 0-cos <t> +cos ll - (sin2
.p - sin2 II) cos a]+ WR (sin <j>-sin a) (18.5)
For a between .p and ..-, add (16.4), (17.4) , and (18.4)
to (16.1), (17.1), and (18. 1), respectively. Thus
P= (W/?r) (sin2 .p-sin2 0) cos a (19.5)
S= - (W/?r) (sin2 .p-sin2 0) sin a (20.5)
M= (WR/?r) [- 4> sin <t>+ll sin 0- cos <t>+cos 11- (sin2
<j>-sin2 O) cos a] (21.5)
Case 6. (Case 2 in Miller's article.)
Let <j>=?r/2 in Case 5. For a between 0 and 0,
P= (W/ ?r) cos2 ()cos a (13.6)
S= - (W/?r) co~2 ()~in a (14.6)
M= (WR/?r) (?r/2+o sin o+cos 0-cos2 II cos a
_ - ?r sin 0) . (15.6)
For a between () and ?r/2
P= (W/?r) cos2 ()cos a+ W sin a (16.6)
S = - (W/?r) sin2 II sin a+ W cos a (17.6)
M =(WR/?r) (?r/2 + 0 sin o+cos o-cos2 o cos a - ?r
sin a) (18.6)
For a between ?r/2 and ..- .
P= (W/?r) cos2 ()cos a
S = - (W/?r) cos2 () sin a
(19.6)
(20.6)
Af= (WR/11") (- 71"12+ 0 sin o+cos O-cos2 ()cos a) (21.6)
Case 7. (Case 6 in Miller's article.)
Conditions (2) and (3) are valid for this case also.
However, f~rmulas (4) and (5) do not apply. From
Figure 5, the value of 111 for a between 0 and O is
M =m+ pR (l+cos a) + WR (cos a -cos 0) (22)·
From Figure 6, when a is between () and .,,., M is
M = m+pR (l+cos a) (23)
Note that 8 M/8m=l and 8M/8p=R (l+cos a) for
all values of a. Since conditions (7) and (8) were obtained
fr om t hese same values of 8M/8m and 8M/8p,
these conditions apply in this case also. The result of
substituting (22) and (23) in (7) is
0= J:[m+pR (l+cos a) d a+ wRf! (cos a - cos O)
d a= .,,.m+ .,,.pR+ WR (sin(} - (} cos 0) (9.7)
Substitute · (22) and (23) in (8); the final result is
0= J :[m+pR(l+cos a) cos ada+ WR J: (cos a-cosO)
cos ada=.,,.pR/2+ (WR/2) (0-sin ()cos 0) (10.7)
If these values of p and mare substituted in equations
(22) and (23), it turns out t hat M has the following
values:
P= - (p+ W) cos a= - (W/11") (sin() cos () - 0+11") cos a
(13. 7)
S=(p+ W) sin a=(W/.,,.) (sin() cos 0- 0+11") sin a
(14. 7)
M = (WR/.,,.)[O cos 0-sin o+ (sin o cos 0- 0) cos a]
+WR (cos a - cos 0) (15".7)
For a between () and 71"
P=p cos (11" - a) = -p cos a= -(W/11") (sin() cos 0-0)
cos a (16.7)
S=p sin (11"-a)=p sin a= (W/11") (sin () cos 0-0)
sin a (17.7)
M = (WR/11") [O cos (}.-sin o+ (sin 0 cos 0- 0) cos a]
(18. 7)
0
-\