The most usual method of arriving at the maximum amount of spindling or hollowing out permissible in the case of any particular spar section is by trial and error, a process which is apt to become laborious in the absence of good guessing - or luck. The following tables have been got out with the object of making it possible to arrive with certainty at a suitable section at the first attempt

Blyth, J D

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ITATIONAL ADVISORY co:IM:TTE FOR AER0NUTICB 
No. 383 
5? IiDLID AND HOLLO'[ SSPARS
By J. D. Elytla 
From F1igit," August 26,. 1926 
'iashington
October, 1926
https://ntrs.nasa.gov/search.jsp?R=19930090706 2020-06-17T03:45:56+00:00Z
NATIONAL ADVISORY OOMLiITTE FOR AEPNAUTICS. 
TEcIT::cAL	 NO. 383. 
SPINDLED AND HOLLO[ SPARS . * 
By J. D. Blyth. 
The most usual method of arriving at the maximum amount of 
spindling or hollowing out permissible in the case of any par- 
ticular spar section is by trial and error, a process which is 
apt to become laborious. in the absence of good guessing - or 
luck. The following tables have been got out with the object 
of making it cossible to arrive with certainty at a suitable 
section at the first attempt. 
The following syrbol are employed 
I = Moment of inertia. 
Z	 Section modulus. 
= Bending moment. 
T = Torque. 
S = Shear force. 
Ultimate compressive strength of material. 
U	 tensile 
=	 Il	 shear	 n	 u 
We will first consider a spar of rectangular section, later-
ally loaded only, which we wish to spindle to either I, 
or	 section. 
* From "Flight," August 26, 1926.
iI.A.O.A. Teo hriical Memorandum No. 383
	 2 
B = V!idth of section. 
= Depth it	 li 
ci. = Thickness of flange. 
=
	
web. 
It should be noted that in the case of the hollow 0 sec-
tion, t is the combined- thicimess of the walls forming the 
web.
The most rapid- method gives flange and web thicknesses 
slightly greater than are actually required, Eo the spar will 
be on the safe side. The procedure is as * follows: 
First find- t, the thickness of the web. This is given by 
..- 3	 S 
Next find the value of C, which is given by 
M 
fc z 
where	 Z - B1)
In Table I, values of 	 are tabulated- for values of C 
from 0 to 1.0.
= \D.
N.A.C.A. Technical Hemorandum No. 383
	
1 
Table I. Values of ', for all Values of C. Rectangular Sections. 
C 0 1	 2 3 4 567 8 9 
.0 0 .002 .004 .005 .007 .009 .010 .012	 .014 .016 
.1 .017 .019 .021 .023 .025 .027 .028 .00	 .02 .034 
.2 .036 .038 .040 .042 .044 .046 .048 .050	 .052 .054 
.3 .056 .058 .060 .063 .065 .067 .069 .072	 .074 .076 
.4 .079 .081 .083 .086 .088 .091 .093 .096 1 .098 .101 
.5 .103 .106 .109 .111 .114 .117 .120 .123	 .126 .129 
.6 .132 .135 .138 .141 445 .148 .151 155	 .158 .162 
.7 .166 .169 .173 .177 .181 .185 .190 .194	 .193 .203 
.8 .208 .213 .218 .223 .229 .235 .241 .247	 .254 .261 
9 .268 .276 .285 .294 .305 .316 .329 .345	 .365 .393 
1.0 .500 - - - - - - -	 - -
The method will be made clear by taking an example anL work- 
it out. 
Suppose we have a spar whose section is 2 in. wide and 4 in. 
deep,
M = 16 9 000 1'o-in. 
	
S =	 900 l. 
	
fc	 5,500 Th. per square inch. 
	
=	 800	 1I•	 U	 it 
The star is to be spindled to I section. 
	
Ye get t 	 .-= 0.42 in. 6 800 
and Z = 5.33
whence C = 0.545. 
From Table I we see that when C = .54, 	 N = .114, and 
	
when C = .55,	 N = .117. Interpolating in these values we get 
a value in the present case, 
N = .116 
d = X= .464 in. 
IT.A.0.A. Technical kemorandurn No.. 383	 4 
Remembering that the values found are on the high side, we 
will make the flanges 0.45 in. deep, and the web 0.4 in.	 thick
Checking tne secticn so oota.ried, we find that we have for 
bending a factor of safety of 1.15, and for shear a factor of 
safety of 1.14. 
The section found in this way will be suitable in most cases. 
Occasions may arise, however, when it is desired to lighten the 
spar as much as possible, i.e., to spindle it to the limit. In 
this case the PTocedure is a little longer. 
First find t as before. 
Next find the value of fc	 and subtract the value so 
found from the known value of M. Calling the remainder LI', 
find the value of 0, which in this case is given by 
C -
.	 r't i0L' 
-ihere Z 	 - t)D2 
Look u" the value of X in Thhle I; and as before, ci. = XD. 
Taking as an example the spar already described. 
As before, t	 04 in.	 - 
5 00	 X IS 
Then Li' = Li - 
= 10,150 lb.-in. 
Z I = 4.27 
C =	 .4-33 
whence	 h = .087, and ci.	 0.35 in. 
N.A.C.A. Technical Memorandum No. 383 	 5 
This section has for bending a factor of safety of 1.01. 
1 i.e will next consider a spar of rectangular section, later- 
ally loaded as before, together with an end load P. In this 
case the procedure is divided into two steps, as follows: 
First find the section neglecting end load, in the manner 
already described, and let A, square inches be the area of the 
section so found. This enables us to find the value of P/A. 
Repeat the rocess, using the value f 0 -	 instead of 
or, where applicable, ft +
	
instead of	 It should be
noted that for compressive end; loads the numerical value of P 
is taken as positive, while if the end load is tensile the numer-
ical value of P is negative. 
Taking as an example the spar already worked out, and suppos- 
ing that it is subjected to a compressive end load of 1000 lb. 
Neglecting the end load, we have , as before 
-t = 0. 4 in 
d = 0.45 in. 
We get A = 3.04 sa. in., and 	 = 329..	 -,
New value of f 0 = 5,500 - 329 = 5,171. 
Since the shear stress is unaffected by the end load, t 
remains as before.
1 6, 000 -=058 
5,171 
whence X
	
0.12S 
and d = 0.5 in
1T.A.C.A. Technical Lemorand.um No. 333
	
6 
This section has a factoT of safety of 1.13. 
If the Perry correction is to he applied the method is simi-
lar. In this case the section is found 
of f c - derive(ft, and also the value 
for the spar treated as a strut- for the 
between Doints of contraflexure. Now rn 
and with the new values of M and f(.
as before, and the value 
of P03 the Zuler load 
length of the portion 
Liltiply N. by	
p
P 
proceed as before. 
7e now turn to spars of circular section. These are most 
commonly subjected to a lateral load, combined with a torsional 
load.; and. this case we will consider, first taking the case of 
wooden spars, where the thickness t of the wall is not small 
compared. with the diameter fl (this being the external diameter). 
The procedure is as follows 
First find Te, the eüihrälent torque. This is given by 
Te/ t +T
2 
Next. find the value of C, which is given by 
rr 
iO J.c 
3 iiD I 
Turn to Table Ii, and. find the value of	 corresponding to 
the value of C. 
Then	 t = ?D.
N. A. . A. Technical Liernorandum No. 393	 7 
Table II. Values o.	 for all Values of C. 
Circular Sections.. 
C 0 1 2' 3 4 516 7 8 9 
.0 0 .001 .003 .004	 .005 .006 .008 1 .009 ..010 .012 
.1 .013 .014 .016 .017	 .019 .020 .021 .023 .024 .026 
.2 .027 .029 .030 .032	 .033 .035 .036 .038 .040 .041 
.3 .043 .045 .046 .048	 .050 .051 .063 .055 .057 .058 
.4 .060 .062 .064 .066	 .068 .070 1 .072 .074 .076 .078 
.5 .080 .082 .084 .086	 .088 .090 .093 .095 .098 .100 
.6 .102 .105 .108 j .110	 .113 .116 .118 .121 .124 .127 
.7 .130 .133 .137 .140	 .143 .]A.7. .150 .154 .158 .162 
..8 .166 .170 .175 1 .179	 .1841 .189 .194 .200 .206 .212 
.9 .219 .226 .234 .243	 :250 1 .264 .277 .292 .312 .342 
1.0 .500 - - -	
- - - - - -
In e1oying this method it must be remembered that the wall 
thickness found is not greater than is actually required; and 
further, is only that required to resist the shear stress due to 
torsion, no account having been taken of the shear stress duo to 
direôt shear. In the case of such spars. the shear stress due to 
direct shear is not usually hiF;h, and with a little practice it 
will be found easy to estimate the amount (which is very small) 
by which the computed value of t must be increased for safety. 
In the case of circular metal spars, where the thickness 
of the wall is small compared with the outside diameter, the re-
quired thic1ess t is given at once by 
Uj) 
No examples have been given in the case of, spars of circular 
section, as it is thought that the cxar p1es worked out in the 
oases of siars of rectangular section show the method sufficient-
ly clearly. 


Spindled and hollow spars

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