Q
__éDVOCATE on INTERNAL lMPROVElVIEf"i‘S%.
TABLE l...—Abstract off’ GmAi1D’s EXPERIMENTS on the 8ti'engtl1 of Timber
loaded on the End.
DIMENSIONS or THE TIMBER. Weight in r V '
NO_ of pounds thc ‘ Weight in
experp beam bore poundsthe same Ratio R T’ _
mentS_ Length. Breadth. Thickness. ‘L1PPl1ed ‘O beam Would ' E mus’
. the extra. hear loaded ~
mity. transversely
‘FEET. I JCHES INCHES.
1 8 6.21 5.0.3 93616 8598 -092
2 8 6.39 4.17 94018 "6078 ‘O64 Broken.
' 3 8_: 6.21 3.99 ' 69165 ~ 5390 3078
4 8 . 5.23. . y.«s'9: - 50526 1 . 4325 -085 Broken.
'‘5 8.628 5.15 * 4.17 50608 2 4900 ’ ‘O97. Broken.
6 7.549 6.02 5.15 115359 9980 ‘O87 '
'7 7.549 6.21 5.05 103799 9909 ‘O95
8 7.549 6.12 4.085 73095 6396 -967 Broken.
9 7.549 (1.21 3.99 63177 6336 '100 Broken.
10 7.549 4.96 3.99 44857 4921 -109
11 6.471 6.12 5.24 87494 12366 °141
12 6.471 6.21 5.15 87481 12013 '186
13 6.471 6.21 . 9.991 87979 7392 -085
14 6'471 6.30 3.99 72823 73.13 -100 Broken.
15 6.471 5.24 4.17 103622 6525 ‘O63
16 6.471 5.05 4.25 82261 6674 ‘O81
17 7.549 6.21 4.25 87443 7022 ‘O80
18 8.628 6.21 5.32 82332 9607 'l16 Broken.
-19 8.628 6.21 5.15 103863 8993 '087 .
20 8.628 7.37 6.21 137966 15584 '1l.3
21, 8.628 7.45 .21 137866 15764. ' ’ -114
1* mean . . ;. . -996’
Tit thus a'pp’ears that the force" required ‘to:
break atimber in the direction of its length,
is about ten times that which would. break it if ‘
it applied transversely at the middle ; from
which I infer that the strain in the direction
. of the gate produced by the pressure of the
opposite one , is equal to an additional strain
of one~tenth applied transversely.
A difference exists in the comparison
’made in the preceding Table and in the case
of lockgates, which it is necessary to make
some remarks upon ; viz., that a lock gate
has atransvcrse pressure acting in adclition
‘to thatiprooiuced by the other gate, so that
the end pressure is exerted upon it after it is
alrearglydeﬂected by atransverse strain winch
‘is of" course not the case in the comparison
’rnadein the Table." How far this may effect
‘the question, or how much greuteretfect the
compressive force may have in coiisequence
of the beam being already deflected, it is very
’diﬁi‘cult to determine, butfrom an examina-
tion of the subject, I am induced to think that
the deflection is so small ‘as very sliglitly to
increase the effect-’of the end prefssure._ 1
I The 7 of" flof
depend upon the degree of deﬂeictioni" the . ’
beam has sustained from the transverse pres-
sure, and if it amounted to a quantity ex-
ceeding one-twentieth of the length, (which
would make the lever by which the end "pres-
’_sure acted“ exceed one.tcnth of that by
which the transverse strain acted,) a greater
ielfect than one_.tenth would‘ be produced;
but as the ordinary load which timber is ex.
’ pected to sustain,_do_es _ not_ produce atthe
‘utmost a deﬂection‘eiiceeding one hundredth
‘,p_art___ of the length, I cannot.conccive_ the
transverse'str‘ain above named materially to
l alter the comparison,'and l have accordingly,
in the following investigation, assumed one-
t9!1’£l1.as the amount of additional strain pro-
duced by. the end pressure of the opposite
gate. 2
It now becomes necessary to get an ex-
pression for the amount of the strains above
mentioned at any angle of salience, which is
arrived at in the following manner 2...
Let AB, AC, represent the two gates,
meeting at the point A ; draw the line AD
from the point A perpendicular to BC, and
let BD, which represents half the breadth of
the lock, = l, also
l
./
.55./.;_"' '7
/’ 1-'\.
' ,/ii
l
let the pressure of water. upon the length l
of the -gate be indicated by w and the angle
ABD = .
Then the length of the AB and
any angle Q) will be expressed
by '
and the pressure upon it by
The transverse strain produced by
- this pressure on the centre of
the beam at the same angle will‘,
be ' » . ‘
lsec Cp
to seep
-éw sec go
293
It now ‘remains to ﬁnd theamount 0'
compression in the direction of the gate:
produced by the opposite gate.
Let AF represent the force or tendency
of the gate AC to turn upon the point C,
which is of course equal to half the weight
upon the gate A C,
~ or =-_ 3 10 sec 1;:
The force may be resolved into AG, FG,
theone GF is supported by an equal and op.
posite force in the gate AB, and the other
will represent the force in the direction of
the gate, the expression for which may be
found as follows;
as sine A AGF: AF: :sinc A AFG: AG
or sine p :.;w Sbctp: : cos go‘: éw
cos qi ,:
sine (p
The whole amount of transverse strain at
any angle cpcwill tl‘ieref'o1‘e be represented by
the expression,
{.10 secq) + 2‘, 1!) cosec qa '
:2;-cosec tp
V sec+ cp + coseccp 2 min
ortan qp sectp (lip —-—+ cotcpcoscccpd¢=0
\Vl1G1lCGtt1n2q9 = 1-1,, comm (P g
and tE1n3q) :: ,-Lu ‘ ’
tan q: = 3 ,/ ,‘,, 2 30 3 ,/ 100=.4641
V rztan 4 24° 54'
The salient angle of a pair of oak gates,
when the strain is a minimum, is therefore
24° 54' , ,
In the question of the best angle for lock.
gates, it becomes necessary to consider that
the length of the gate also varies as the
rsecant of the angle up. The angle 24°. 54’
is therefore not that at which, with argiven
section of’. timber, the greatest strength will
be obtainrid ; for although the strain is the
least at this angle, yet the gates, by their
greater length, are less abie to resist it than
at some intermediate angle, when the strain
is slightly increased. The expression now
becomes
sec'*’q2 + ,-'7, see c cosec q; = min
2 secztptan