'
S T 0
452
STO
inversely as rad. curv. X sine 3 inclination to vertical. Call-ing, therefore, as usual, the abscissa v* = *, c e = y,v C = z, we have cf = i,/c= y, cc = z; and since Cc :cf :: rad. : sin. inclin. at c ; therefore the load DC is in-i' 3
versely as rad. curv.
of curvature =
X 4-. But as is well known, the rad.z 3
;.3
therefore, by multiplication, d c
yx — x g’
—, that is, directly as
yx—x y
or as
is inversely as
yx — xy ' ' y 3
d ^4 -f- y,') and is equal to d Qr X 4*^ where C is a con-stant quantity, found by taking the real value of DC at thevertex, v, of the curve.
“ Now, in the present case, calling a v = a, we have
C V
d c = a + x. (a v + v e,) therefore a -f x = — X flux, of 4.
y *
a 2 x + x?
X
Take -r = u, and by integrating, we have u = yj/x x *
V 2 ax + x 3
; whence, by
and therefore y = \/ c x
integration,
y = c X Lo. (2 a x 2 a? -)- 2 >/ 2 a x + x 2 ) -(- b. Atthe vertex x = 0, therefore y = c X Lo. (2 a) And con-
x , T a + x + a/2 a x + x“.
sequently the ordinate y = «/ c X Lo.---
“ Lastly, to find the value of v/ c, we take some point ofthe extrados, where the ratio of x and y is known. Forexample, if the span, = 2 s, and height, = h, are given,
we have s = V C X
T a -f- It -{- k/ 2 ah -|- h hLo. -
hence */ C = s -i- Lo.
a h -f- \/ 2 ah -|- hh
and, finally, y = s X
Lo. i (a + x + V 2 a x + x 1 )Lo. J (a + h + V 2 a h + h 3 )
“ We subjoin a table, calculated by Dr. Hutton from thisformula, for an arch of 100 feet span and 40 feet rise, thethickness of the crown being taken at 6 feet. It is nearly ofthe same dimensions as the middle arcli of Blackfriars Bridge,and which may answer for any arch where these dimensionsare similarly related to each other.
EC.
DC or A E.
EC.
DC.
EC.
DC.
0
6.000
21
10.381
36
21.774
2
6.035
22
10.858
37
22.948
4
6.144
23
11.368
38
24.190
6
6.324
24
11.911
39
25.505
8
6.580
25
12.489
40
26.894
10
6.914
26
13.106
41
28.364
12
7.330
27
13.761
42
29.919
13
7.571
28
14.457
43
31.563
14
7.834
29
15.196
44
33.299
15
8.120
30
15 980
45
35.135
16
8.430
31
16.811
46
37 075
17
8.766
32
17.693
47
39.126
18
0.168
S 3
18.627
48
41 293
19
9.517
34
19.617
49
43.581
20
9.934
35
20.665
50
46.000
“ The curve of Figure 7 is accurately drawn jj|j«dimensions, and may give an idea of the form of ■>" ( | !t ,itlybrated arch. It is not destitute of grace, and is aburoomy for craft. . n ; for
“ Such, then, is the analytical theory of equilibiatia practical subject it does, we confess, appear abs’ruse-^.^g“ Let us turn, therefore, to another mode of consthis subject, which has been adopted by De la ^'. ie ’ oUr o"' nBelidor , and many others on the continent, and m 0country by the ingenious Mr. Atwood. wedg?’
“ The latter has, from the known properties of 1 ie n ,e-and the elementary laws of mechanics, exhibited a 0 ftrical construction for adjusting the equilibration of arc
nil
these
us p" r !:
Now, l et **■ ^ t i,at th e /’• A , so tM . itbi ' i 0 '
to ascertain the weight of B in proportion to e qiiu>
’ t0 the P'^ U 9t
when left" U free to e 3[" po,se > ma y continue
Produced by * q °"? K Si,10e M B is vV e
add to this j, ,, ’ i a ^ H ' ec£, on perpendicular to a ft jje ,-e-
fore, make 1 " V 01 13 derived from the wedge A. a ito h n, draw z produce M K t0 T - ^ the W 0 *
tending to makJs sV i' ght ‘‘" lgleS to K B; J W % equ " 1 -°r w r , draw the r. . 8 "P B K : take therefore b B d ^pejp e nflicuiar it' b meeting the verttc
of the^ e y sure s,
b ; b b will represent the necessary weight o ^ yve sS ftand the whole is so evident from the composi - g A 1 ' j,
_:_ _J_..tfo+inn. ' 5Ul/ ^ud 1 w
w° r6
as to require no farther demonstration. °“ t j on ina 6 '. „ndconstruction ; he has rendered the demons r oin et r y ’ ^prolix, by the unnecessary introduction ot trig p, <^ c 'bglit safter showing how tne weight of the sectl q” ce the se '< e fiod sbe found in the same way, he goes on to re ^^ ue3 . i land pressures to analytical and numerica
every form.
“ Figure 8.—The wedge A, ifin the direction v O, but is prevented by the reaction o' ^ lu _B', acting in the direction pq and in, perpendicu <u ^ t | l6sides ag, QDj and it is known,, from the P r0 P ertieS e( j,re A>wedge, that if p Q, or k'i, be to the weight of the'' ° o ( | ieas d o is to d g, the wedge A will remain at rest. , ^ awedge A be only at liberty to slide down G A, cons’ e e qui-fixed abutment, then the force p q alone will keep 1 ^ teI i-librio. The force pq being perpendicular to do, ’ u ^ putdency to make A slide either up or down on th’ 3
produce it towards N, making n m equal to r Q> 0 tli« rs »
force acting obliquely at n, may be reduced to £ " jj Cll |nfviz., m k perpendicular to ag, expressing thepressure on the abutment of A, and bn expressing ^or tendency it has to make A slide upwards A>
Again, take the vertical line a«, expressing the we . 1 ^ ntj tl* 1 * 1and draw a H at right angles to a g ; it is very eV1 j.j e jo"'’ 1a h expresses the tendency of A by its weight to sG a. ah is opposite, and is equal to nk. t j jen tl‘ e
“ For, draw the perpendiculars d d and A p>. ^ a lsotriangles ash, ag p, dg d, are evidently sinulai , ^ con ,-
the triangles o d d, o Q n, m n k, as they have p Q , d 111 *mon angle besides the right angle. Now, the ’ olC p 0l h)'is, MN, is to the weight of A, that is a a, as 0 Dsupposition; j
“ And, a a : A H :: A G : A p : '• D G : B ’
“ Therefore, m n : A h : : o d : d d :: M N : B;
“ Or M n has the same ratio to A h, that it has . tg s n u “
is, a h and N b are equal ; or the tendency <’ ten( ] e ncy
downwards by its weight, is balanced by the geC tio” *M N to make it slide upwards; wherefore, t ieremains at rest in equilibrio.
“ Considering the whole arch as comply—, , g ...
mutually balancing each other, the force P Q> w b)’. 1 'j
sary for sustaining the wedge A, will be su Ph lie req 11 ' 11 'reaction of the adjacent wedge B.
thfd
Dieted, 'V’th « ^