TG
JC-NRLF
THEORY
lAROHBS.
Prof. W. ALLAN,
Formerly of Washington <nifi I.w University, '
NEW YORK:
D. VAN NOSTRAND COMPAN.i', •^3 MURRAY STREET AND 27 WARREN STREET.
1890,
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THEORY
ARCHES
BY
Prof. W. ALLAN,
Formerly of Washington and Lee University, Lexington, Fa.
NEW YORK: D. VAN NOSTRAND COMPANY,
23 MURRAY STREET AND 27 WARREN STREET.
1890.
COPYRIGHT. D. VAN NOSTRAND COMPANY.
1890.
PREFACE TO SECOND EDITION.
The original edition of this Mono- graph was reprinted from the pages of " Van Nostrand's Engineering Maga- zine," to which it was contributed by the late Prof. W. Allan. It had been pri- marily prepared by him from a series of notes, which notes had been litho- graphed for the use of his classes in studying "Kankine's " works.
In printing this second edition, there- fore, it has not been thought necessary to make any changes whatever, as the text is simply an amplification and ex- planation of the " Theory of Arches " as given by Prof. Eankine, and as Prof. Eankine himself has been dead some years, his treatment of the subject as de- veloped in his Manual is probably as
127692
complete as it ever will be. It would seem, therefore, that this little book will answer the purposes for which it was intended for all time to come. To the reader and student of Kankine's works it will undoubtedly be found interesting as
well as useful.
THE PUBLISHERS. May, 1890.
THEORY OF ARCHES.
The following is an amplification and explanation of Professor Rankine's chap- ters on this subject.
Perhaps the clearest way of developing the " Theory of Arches " is to begin with the consideration of the forces which act upon a suspended chain or cord. The force in the chain or cord is just the op- posite of that upon an arch— that is, it is tension instead of compression, but the relations between the "external" and " internal " forces, or, what is the same, between the loads and the resist- ances they produce, are strictly analo- gous.
Let CAB (Fig. 1) be a cord suspend- ed at C and B and loaded in any manner over its whole length. Consider the forces acting on this cord. Suppose it
6
attached to a hook at B and to another at C. A cord without stiffness cannot exert a pull except in the direction of its length; therefore the "pulls" in the rope at C and B, and exerted at these points on the suspending hooks, must be in the direction of the tangents at FIG. 1.
those points. The load is supposed to be distributed over the cord, but we may find its resultant. Let P be this result- ant and P F its direction. The three
forces, viz., the pulls at C and B, and the resultant of the load, P, are all in the same vertical plane ; they are the only forces acting on the cord ; and as they are in equilibrium, the directions of these three forces must meet in one point, and the forces themselves must be pro- portional to the three sides of a triangle drawn parallel to their directions.
G N F (Fig. 1) is such a triangle. The known directions of the pulls at B and C, and of P, give us the angles in this triangle ; and if we know also the mag- nitude of the load P, represented by the line G F, we can determine that of the pulls at B and C. For
(PullatB=GN) :GF : rsinGFN isinGNF.
(PullatC=NF) : GF : isinNGF : sinGNF.
The analysis we have made for the whole cord may be applied to any part of it. Thus, if we consider any arc B' A' (Fig. 2) of the cord, and the load on that arc, we have three forces in the same plane in equilibrium. For at A' and B' the other parts of the cord may
8
be replaced by two hooks, and the pulls on these hooks, exerted by the cord at A' and B', will be, as before, in the direction of the tangents at those points. The re- sultant P' of the load on A' B' must pass through the point of intersection of the tangents, and if the direction of that resultant be as indicated in the fig- ure, then G' N' F' will be the triangle of forces.
FIG. 2.
The principles above explained enable us to calculate the " pulls " at all points of a loaded chain or cord, and conse- quently to fix its size and strength to bear a given load ; or to determine the amount, distribution and direction of the load necessary to produce assumed " pulls " in a cord of a given shape.
Thus, suppose in the half of the loaded cord of (Fig. 1) we draw the tangents at A and B (as is done in Fig. 3), the re- sultant of the load P must pass through F, the point of intersection of the tan- gents. If the direction and amount of P be known, lay off F G to represent it.
FIG. 3.
X
Then, as above explained,
N F = pull at A and
N G = pull at B.
Suppose, on the other hand, we as- sume the pulls at A and B to be equal, we lay off on the two tangents (Fig. 4) equal lengths, F S and F N, to repre- sent these equal pulls, and upon them construct a parallelogram. Then F G
10
gives the magnitude and direction of the resultant of the load that must be put on the cord to produce the given pulls. A cord is in equilibrium when it is balanced under the load applied. Change the distribution of the load and the
FIG. 4.
cord at once changes shape and as- sumes the form necessary to equilib- rium under the new load.
Thus, if P (Fig. 5) equals the direc- tion of the resultant of the new load on the cord from the horizontal point A to the point of support B, draw the tan- gent A F, until it meets the direction of the load P, at F ; then draw F B. The
11
cord A B will have so changed its form that F B (Fig. 5) will now be the direc- tion of the tangent at B.
FORMS OF CORDS UNDER VARIOUS LOADS.
Let us now investigate the various curves which a cord will assume under different distributions of the load.
FIG. 5.
Case I. Suppose the load be alto- gether vertical, and to be distributed uniformly along the horizontal.
Let equal weights be hung, for in- stance, along a cord C B (Fig. 6) so that the horizontal distance between the
12
threads by which the weights are sus- pended shall be everywhere equal. Or, draw little elementary triangles along the curve, so that the bases of all these little triangles shall be equal, and let the threads holding the weights cut the mid- dle of these bases. Then each weight FIG. 6.
may be considered as the resultant of the load on 'the element of the curve which constitutes the hypothenuse of the little triangle to which it is attached. Such a load is vertical and is uniformly distributed along the horizontal.
To determine the curve of the cord. Obtain the resultant of the load between
13
the horizontal point A and the point B (Fig. 7). This resultant, as the little forces are all parallel, is equal to the sum of them, and it is vertical in direction.
Fia. 7.
It will also evidently bisect A T. Draw it, and from its point of intersection with A T draw the line F B, which, as has been shown, must be tangent at B. Prolong B F to I, then the subtangent I J is seen to be bisected at the vertex A of the curve. Hence the curve C B is a parabola.
14
The triangle B F T has its sides par- allel to the forces acting on the half cord A B ; so that if B T be taken to repre-
sent P,
B F = pull at B F T = pull at A. Let T = equal tension at any point along the
cord.
H = value of T at the horizontal point A, or the " horizontal pull" on the cord.
i = inclination of the tangent at any point to the horizontal.
Then as the arc A F (Fig. 7) may stand for any part of the curve counting from the horizontal point A towards one of the points of suspension, we have the following general equations from the triangle B F T :
(1.)
Tan i--.- *JL (2)
- H - H - dx
(p being = the load per unit of horizon- tal distance, A the origin of co-ordinates, A T = axis of X and A J = axis of y).
From equations (1) and (2) we can solve three problems.
1. Given the curve, and the load, to find T and H.
15
2. Given the curve, and T and H, to find P.
3. Given the load, and T and H, to find the curve.
For a full discussion of this case, see Rankine's "Civil Engineering."
Such a distribution of the load as we
FIG. 8.
have discussed in the above case is ap- proximated to in suspension bridges, and sometimes in wood, iron, or steel arches, but not usually in stone or brick ones.
16
Case II. Let the load still be vertical, bnt distributed uniformly along the curve.
That is, divide the arc C A B (Fig. 8) into elements each of a unit in length ; then the load on these elements is constant throughout. It is easily seen that such a load is not, as in the last case,uniform along the horizontal, for the bases of the little triangles of which the hypothenuses are now equal, diminish in extent as we go from A towards B or C. A chain of uni- form material and cross-section, and act- ed on by nothing but its own weight, is in the condition described, and, as is well known, the curve assumed by it is the " common catenary."
Let p = weight of a unit's length of the cord, then if p m = horizontal pull on the cord at A = H, m is called the modulus of the catenary, and represents the length of cord of the same kind as C B, the weight of which would equal the pull at A. The weight on A B = P = p s when s — length of cord A B.
The triangle of forces for any arc A D
17
(Fig. 9) can be found as before, by drawing the tangents at A and D, and the line rep- resenting the force P vertically through their intersections. The triangle D F T will represent the forces ; D T being = P = p s, and F T = H = p m, and D F = T = tension at D. Then FIG. 9.
F T pm m d x
From the differential equation d y_ s d x m
(4.)
18
we obtain the linear equation of the curve. In doing so it is most conveni- ent to take the origin at a point O, whose distance below the vertex A is = m. The line Q O X (Fig. 10) is called the directrix of the catenary.
The equations of the catenary are
IMj ) X 'X- ) i/7/2_|_j»j2
:~2~(E™ -E m-'] •~iengthofaro»(5.)
_ m ( _%_ x_ )
x = m. hy. log. j j, / #» V. (7.)
— — )— /i/ . — — - — j w r m9
Area AGE D=fyxd — m s (8.)
r> j- --
Radius of curv. = p =— - — = — ,. A .
m8 m (10.)
Since the area A O E D — m s, and m = a constant, the area varies as s. But the load on the arc A D (=p s) also
*JS = Base of Naperian Logarithms.
19
varies as s, since p is constant. Hence a convenient mode of representing the load on any arc, A D^ Suppose a sheet of
metal C Q T B A C (Fig. 10), bounded below by the "directrix," Q T, to be suspended from the curve. Let the weight of this metal corresponding to m
20
units of its surface be = p. That is, let
P . w m — p, or w — m
The weight of a strip a unit in breadth extending from A to O is then = p = the weight of a unit's length of the cord. Then the part of the sheet A ODE whose weight — w m s = p s rep- resents the weight P on the arc A D. So A O B T represents the weight on A B, and C Q T B the whole weight on C A B. In the horizontal pull at A we have
H = p m — w m2 (!!•)
and at any pointD
T= i/H2-]-P2 =p \/s*-\-m*=py=w my. (12.)
The property above explained may be illustrated in another way.
Construct on A B (Fig. 11) a series of lit- tle triangles with all their bases equal. Let the weights of the little arcs constituting the hypothenuses of these triangles be represented by balls suspended by threads from the middle of each little arc. Take the length of the thread cor- responding to the ball at A as = m ;
21
make the lengths of all the threads pro- portional to the weights of the balls hung to them ; then the lower ends of these lines will all be on the directrix O X.
FIG. 11.
That is, the intensity of the load on a catenary along the horizontal line ( — weight on a unit of horizontal distance) varies as the ordinates of the catenary, when those ordinates are measured from the directrix.
22
It makes no difference in the form of the curve A B (Fig. 11), to increase or diminish the weights, provided the pro- portion among them is preserved. Thus we may assume the cord and the sheet C Q T B (Fig. 10), to be of a different material in which a unit's length of the cord shall in weight =p', and the weight of the sheet per unit of surface shall = w', and A B will be unchanged. Note, however, that we cannot change the depth A O of the sheet (Fig. 10), nor the length of the lines (Fig. 11), without changing the curve ; for if the lines ended in O' X' for instance, instead of O X, then
A O would not be equal to A O'. D E D E'
Hence, the modulus (m = A O) fixes the catenary, or, if we assume the catena- ry, this determines the modulus. Thus, if we assume three points, B, A, C (Fig. 10), on the catenary, the distance A O is thereby determined ; and if we assume A O and the point A we cannot generally assume B and C.
This often interferes with the use of
23
the " common catenary " in the building of arches [in which case the curve is in- verted, the metal sheet A O T D is re- placed by a wall of uniform material, and the tension on its cord, C B (Fig. 10), is replaced by a thrust along CAB (Fig.
FIG. 12.
r
12)]. For we are often compelled to make the curve pass through three points, while yet the value of A O is fixed.
But this difficulty may be obviated by the use of the transformed catenary, which we will now discuss.
24
'Case III. By the principle of Parallel Projections, if any cord or arched rib is 'balanced under a system of forces which =are represented in the figure by lines, and =a parallel projection be made of the <;urve of the cord or rib and of the lines representing the forces, then the new curve will represent a cord or rib that will be balanced under the forces repre- sented by the new lines.
Imagine a cylindrical surface con- structed upon CQ T B A C (Fig. 10) as a base. To simplify matters, suppose the elements of the cylinder to be perpen- dicular to the plane of the base. Cut this cylinder by a plane inclined to the base, and we shall get a " Transformed Catenary," and the shape of the sheet of metal under which it will be balanced ; for the new curve and surface cut out by the inclined plane are the parallel projections of the curve CAB and the surface C Q T B A C (Fig. 10). Let this inclined plane be so placed that it shall intersect the plane of the base in the straight line C B (Fig. 10) or in one paral-
25
lei to it. Then all horizontal lines (or those parallel to C B or Q T) will be un- changed in length in the parallel projec- tion, while all vertical lines (those parallel to AO, etc.,) will be lengthened in a con- stant ratio whose magnitude will depend
FIG. 13 (a).
upon the inclination of the cutting plane. Make a vertical section of the cylinder on the line O Y. Then if the cutting plane passes through C B we get the triangle O U Y (Fig. 13a) cut out of the wedge to which the cylinder reduces
26
in this case. In the triangle, U Y is the ordinateof the vertex of the transformed
FIG. 13 (b).
catenary, corresponding to O A in the common catenary, and all lines parallel to U V are evidently increased over the
27
corresponding ones of which they are the parallel projections, in the same ratio that U V exceeds O A. Laid down in the same plane the two curves are CAB and C' A' B' (Fig. 13 b).
It is easy to pass from a given cate- nary to a transformed catenary whose ordinates shall be shorter instead of longer than those of the given curve, by erecting an oblique cylinder on the given catenary and surface C Q T B, and cut- ting it by a plane less oblique than the base. So, too, the horizontal dimen- sions can be changed instead of the ver- tical, by making the cutting plane meet the base in a line parallel to O Y, instead of in one parallel to Q T.
The equations of the curve C' A' B' (Fig. 136) are ttfus obtained. The ab- scissas are the same as those in C A B, but the ordinates are changed, so that (if y = general ordinate of C' A' B' and y0 — A' O, the ordinate at the vertex A')
y' : y : : A' o : A o : : y0 : m.
•'•*'= IT •yny-y'-^
28
In the equations of the common cat- enary substitute y' for y and we have the equations of C' A' B'.
From equation (6)
ra ,_ ra ( j£_ j*L I
-fc *=-*-'• (V3 +E ~ m )
• y' _ I!*! A. - ^ (.
" 2 (E m + E w f (13.)
So equation (7) becomes
* = «thy. log.
So equation (8) or area A' O E D'.
(15.)
etc., etc., etc.
The " triangle of forces "FED (Fig. 135), for any arc A D of the catenary, be- comes FED' for the arc A' D' of the transformed catenary — that is, since the horizontal lines and forces are un- changed.
Tension at vertex A' = H' = H =wm* (16.)
29
Load on A' D' is increased in ratio of A' O to A O or of D' E to D E.
A'D E~ ' m (17.)
(D' E represents this load.) Then tension at D' is
T' = |/p» H_ H* (18«)
and
~&x ~ 2m I E #i — E~ ~wj (19.)
In this curve we can assume the direc- trix Q T, the distance A' O ( = y0) and also the points B' and C'. These quan- tities assumed, we determine m (the modulus of the corresponding common catenary) from equation (14) and then by equation (13) find points of the trans- formed catenary.
From equations (18) and (19) we can solve three problems similar to those given under the head of Case I.
Case IV. So far we have discussed the forms of cords under loads par- allel and altogether vertical. Let us
30
take up the cases of loads varying in direction.
Suppose (as Case IV.) that the load be uniform and normal at every point to the cord. Such a load is represented in (Fig. 14), the load on each element d s of the curve being constant and perpen- dicular to it.
FIG. 14.
It is first to be noted that the pull or tension on a cord under any load which is everywhere normal to it, must be con- stant. That is, the pull along the cord at A and B, and at all other points, is one and the same. That the tension at
31
B in the cases previously discussed is greater than at A, is due to the fact that the elements of the load between A and B have in those cases tangential com- ponents, which go to change the value of
FIG. 15.
the pull along the cord. But in the present case, the load being everywhere normal, there are no such tangential com- ponents, and therefore the " pull " does not change.
32
Take any two adjoining elements of the cord d s (= D E, Fig. 15) and d s> (= E F, Fig. 15), each of such length as to correspond to equal elements of the load. The little loads on these lines we will represent by p d s, and p' d s'. Note that, unless the load be uniform all around the cord, d s will not be equal to d s'. The equal loads p d s and p' d s' being normal respectively to D E and E F, their resultant which lies in the direction O K (Fig. 15) bisects the angle between p d s and p d s', and also the angle D E F between d s and d s', which last is the angle between the direction of the pulls T and T' on the cord at D F. Hence the parallelogram of forces ( as shown at R") will be a rhombus, or
Again, take three elements, D E, E F, F H (Fig. 16), of the cord, each bearing the normal load pds=pds = pd s. In place of the little arcs, we use for clearness the chords of those arcs. Since the load around the whole curve
33
CAB (Fig. 14) is supposed to be uni- form, the arcs bearing the equal ele- ments (p d s) of that load must also be equal, or DE = E F = F H. We have above proved T — T". Hence the three sides D E, E F, and F H will arrange themselves symmetrically as in (Fig. 16).
FIG. 16.
Now, every other piece of the cord con- taining three elements will assume ex- actly the same slope as D H, since each such piece must equal D H in length and must be acted on by an equal and precisely similar system of forces. Con- sequently, the little chords D E, etc.r must constitute a regular polygon, and
34
the curve in which they are inscribed must be constant in curvature, in other words — a circle.
Therefore the curve of the cord CAB (Fig. 14) is the arc of a circle.
To form the triangle of forces for any point of a loaded circle as for A D (Fig.
FIG. 17.
X
x P
17), draw the tangents at the extremities A and D. From the intersection, F, of these, lay off F N = F S, to represent the equal pulls at A and D. Then the diagonal F G = the resultant of the load, and the triangle F N G or F S G represents the forces acting on A D.
It is often easier to deal with a uni- form normal load by resolving it into its vertical and horizontal components.
35
The load on an element D ~Et=ds of the quadrant A B (Fig. 18) is = p d s. The horizontal component of this load = p d s sin. 6^ where 6 = the angle made by the direction of p d s with the verti-
FIG. 18.
cal (or what is the same, the angle made by the tangent of ds with the horizontal). The vertical component = p d s cos. 6. Consider the horizontal component (p ds sin. 6) with reference to the vertical space over which it is distributed. This
36
space is E K (Fig. 18) = d s sin. 6. Hence the intensity of the horizontal component
_p d s sin. e
~~ d s sin. e
So the vertical component (p d s cos. 6) is distributed over a horizontal space = D K — d s cos. 6, and therefore its intensity is
pdscos.o
d scos.o
=P>
But p = the intensity of the normal force. Hence the original normal force at each point is equivalent to a horizon- tal and a vertical force, at that point, of equal intensity.
If we then construct little triangles on the curve A B (Fig. 19) such that their vertical sides shall be constant in length, the horizontal forces on these sides will be represented by lines of constant length. Transfer these forces in their lines of direction to A Y. A Y is the sum of all the vertical sides of the lit- tle triangles, and as the horizontal inten- sity is constant and equal to p, we have
37
(if r — radius of the circle) p (A Y) ~ p r = total horizontal force on quadrant AB.
Similarly, if we draw a set of triangles on A B with all their horizontal sides of
the same length, we may see that the total vertical force on A B is
Hence,
1. The resultant of the entire normal
38
force on the quadrant A B is equal to the resultant of a horizontal and a vertical force each of which is — p r.
2. Therefore in the parallelogram of
forces for the quadrant (Fig. 20), F S,
which represents the pull along the cord
at B, is the vertical component of P,
FIG. 20.
while N F = pull at A, is the horizontal component of P. Each of these forces = p r. Therefore the constant pull all along the cord is == p r.
If we make the pull at the vertical point (B) = V, we have
H = V = T = pr . . (20.)
39
In practice a uniform normal force ex- ists in the case of a cylinder filled with steam, or in a vertical cylinder filled with liquid. Thrust instead of tension along A B exists when the normal force pushes inwards, as in the tubes of a steam boiler or an empty vertical cylin- der immersed in water. In reference to arches, this discussion has its princi- pal value as introductory to those that follow.
Case V. In this case we obtain the curve and forces by parallel projections from the circle.
If we suppose a cylinder erected upon the circle (Fig. 21) as abase, and cut it by an inclined plane whose line of intersec- tion with the plane of the base shall be parallel to A I, we will get an ellipse whose vertical axis A' I' (Fig. 21) will = A I, and whose horizontal axis C' Br will be greater than C B. All lines parallel to A I will be unchanged in length, while all parallel to C B will be increased in the proportion of C' B' to C B. Now, by the principle of parallel projections,
40
the ellipse, which is the parallel projec- tion of the circle, will be balanced under the forces which are the parallel projec- tions of those under which the circle is balanced.
As we have seen, the circle is the curve assumed by the ring under a uniform horizontal and vertical force at each point of the same kind, and equal in in- tensity; for such a system of forces is equivalent to a constant normal force around the curve. For convenience, these forces are represented in Fig. (21) along the two diameters, each little line representing the force on a unit of dis- tance. The pull around the ring is of course tangential to it, and is everywhere the same ( = p r). This pull is repre- sented at A and B by the arrows there.
In the ellipse, the vertical lines being unchanged, the total vertical force on the elliptic ring (= the sum of all the little vertical lines) is the same as it was in the circle, and if we call the vertical force on a quadrant V (= B M) for the circle and
FIG. 21. i
i 1 1 i JrjTT* i 1 1 i
^ M
42
V (=: B' M') for the ellipse, we will have V = V . . . (21.) Notice, however, that in the ellipse the force V is distributed over the distance O' B' and not over a distance = O B. Hence the intensity of the force V, or the amount of that [ force on each unit of distance, is not the same as in the circle. In the ellipse (Fig. 21) each little verti- cal line represents, therefore, the force on a distance greater than a unit. Let O' B' = c O B. Then to obtain the intensity of V, divide it by the space over which it is distributed. Thus, let
V H
^-QB^^XO
represent the vertical and horizontal in- tensities in the circle. We have al- ready seen that in the circle
Py = Px = p.
Let py and p'x represent the vertical and horizontal intensities in the ellipse. Then
«' - V - Y =J^
py~O' B'~c.OB c (22.)
43
The lines representing the " pulls " at B and C (as B N) are also unchanged. Hence the pulls at those points in the elliptic ring are the same as in the circu- lar ; that is, they are equal to V = V.
The horizontal lines are all increased in length in the ratio 1 : c. Hence the sum of the lines representing the hori- zontal force on a quadrant of the ellipse (as I' S') is greater than the correspond- ing line (I S) in the circle in the above ratio. Therefore if H' == the horizontal force on the elliptical quadrant,
H' = c . H . . (23.)
The length over which this force H' is distributed (A' O') does not change, however, and hence the little horizontal lines in both figures represent the force on a unit of distance. Hence the inten- sity of the horizontal force in the ellipse has increased just as the length of the lines, or from the equation
H' c. H Px=&-0'=^0 = Cpx (24.)
The horizontal pull in the ring at A' or
44
I' being equal to the horizontal force on a quadrant is
H' = <j.H = e. V= c. V . . (25.)
Hence the " pull " around the ellipse is not constant as it was in the circle. The pulls at B' and A' are as
V : H' : : i : c. But
AT : C'B' : : i : c.
Therefore,
1. The pulls in an elliptical ring are as the axes to which they are parallel.
Again the intensities in the ellipse are p'y.p'x: : ~- : cpx : : -- :c: : i : c2
c c ,
And
(AT)2 : (C'B')8 : : i : c*
Therefore,
2. The intensities of the forces in an ellipse are as the squares of the axes to which they are parallel.
From this proportion we have
c=J^- . . . (26.)
45
It will be noted in the elliptic ring that the resultant of the little horizontal and vertical loads at any point is not normal to the curve except at the ex- tremities of the axes.
Let us determine the pulls and the relations between the forces at other points besides the extremities of the vertical and horizontal axes of the ellipse.
In the circle (Fig. 22) if we resolve the forces along any two rectangular axes, as A1 I, and C, Bt, we shall have evidently the same relations between them as when resolved along a vertical and horizon- tal axis. Now the three parallel lines, viz., the diameter, A.l Ip and the tangents at Cj and Bx, are projected in the ellipse into three parallel lines, viz.: A',, I',, and the tangents at C', and B\. Similarly Cj, B,, and the tangents at A1 and Ij continue parallel in the ellipse. Hence rectangular diameters of the circle be- come conjugate in the ellipse. The lines representing the forces perpendicular to C, Bt in the circle become parallel to O'
46 Fie. 22.
47
I\ in the ellipse, and are changed in length just as O' I\ is changed from O Ir So the forces which are parallel to Cl O in the circle become parallel to C\ O' in the ellipse, and vary as C\ O' does from 0,0.
Let O' !',= / and O 0',= /' and let the total force parallel to O' I', on a quadrant (such as C\ I\ or I', B',) of the ellipse be = V, and that parallel to O' B/ be=H1. Then if r = radius of the circle, we have (since the force on a quadrant of the circle as Cl II is = H = V = T)
H:Hl
Hj is equal to the pull along the ring at A't or I', and V, is that at 0^ and B',.
JLeucQ proposition 1 may be applied generally to all conjugate diameters in the ellipse ; that is,
3. The total pulls along the ring at the extremities of any two conjugate diam-
48
eters are as the diameters to which they are parallel.
Again, the intensities being equal to the total loads divided by the surfaces over which they are distributed, let
p'y = intensity of load parallel to O' I\
P'x " " " C' O'
i
Then
Vi V r' r1 "I
y — f\i~rv~ ~~ // j^y // i \j \j i r.T r
p' Hi llr" _ r"
1~~O' I/ ~ r . r" ~PX r'
• . > r . r" . r'
i' i ' ' y r" r' r'
~^~ - - r - Hence for proposition 2, we may read, 4. The intensities of a pair of conju- gate loads are to each other as the squares of the conjugate diameters to which they are respectively parallel.
To pass from one set of conjugate forces on the ellipse to another ; let p'x and p'y be the intensities parallel to one
set of conjugate diameters. HI and Vi be total pulls parallel to same set
of conjugate diameters. r" r' be the conjugate semi-diameters.
49
Also let
i i'
be the corresponding quantities for the other set. Then
r" , r
P X = PX - ,~ . ' . PX —P X Tf
i r i r
r"\ , , r'r"i
r\ l i | r\ r" \ Also,
TT »»• ~ TT Hi r '
Hi = — — ' orH= — ,- ^(29.) ^
Similarly
TT / Hr ' • TT' ±Al T 1
HI =-^T- . .H1= ~^-
Vi'=Vi -V
The ellipse (Figs. 21 and 22) is the form assumed by a cord under a load composed of horizontal and vertical com- ponents which are constant along the horizontal and vertical lines, but which differ from each other in intensity.
The diameter C' B' of the ellipse (Fig. 21) might have been made shorter
50
instead of longer than that of the circle, if required.
Cor. — If one set of the forces are ver- tical and the other not horizontal, but inclined at an angle to the horizon (Fig. 23), we still have an ellipse, the direc- tions of the forces giving the directions of two conjugate diameters (A.\ O' and
B\ O'). Then, if p'x = the intensity of
i the inclined force and ^— intensity of
the vertical force, we have by proposition
4,
p'x :p'v : : (Bi ' O')2 : (A! ' O')2
So from proposition 3, if V1 = pull along the cord at B/ or C/ and H1 = that at A/
H! : Vi : : BI ' O' : A! ' O'.
From the first of these propositions we have the ratio of the conjugate dia- meters ; and from the second we find the pulls at the extremities of those diam- eters.
Knowing two conjugate diameters and the angle (90°-^*) between them we can readily obtain the ellipse.
51
To obtain the pulls at the extremities of any diameter, such as C1 B,.
This is merely passing from one set of conjugate diameters to another and FIG. 23.
equation (29) gives the pull at B,, for
instance, as
O'K
52
(O' K being conjugate to C, O' BJ, etc., etc.
An important fact is now to be noted. Whenever the load on a cord is entirely normal to it, at that point the pull along the cord is equal to the intensity of the normal load multiplied by the radius of curvature.
For the cord at that point is similarly situated to a circular cord of the same curvature and under a load of the same intensity.
Thus, in the ellipse (Fig. 21) the action of the load at the extremities of the axes is entirely normal, for at A' and I' the horizontal component of the load vanishes and leaves only the vertical, which, at these points, is normal to the curve. So at C' and B' only the horizon- tal load has value, and its action is there normal to the curve.
Consider the elementary arc, d s, at A', for instance, which is subjected to this normal load. It is balanced under the equal pulls T = T' (Fig. 24) coming from the adjoining parts of the cord, and
53
the normal load p d s, which gives it its curvature. Imagine a circle under a constant normal force of intensity = p. Take an equal little arc d s of it, loaded with a normal load = p ds. Then, if it be acted on at its two ends by ten- sions = T = T', it is evident that it will have the same curvature as the arc of the ellipse ; or, conversely, if it has the same curvature, the pull around the circle must be — T=Tf.
FIG. 24.
Hence, having given the load on the curve at any point where it is normal, we determine easily the pull along the cord at that point. For, in the circle,
H = V = T = px r = py r = p r, and in the ellipse at A'
H' =p'yp
54
Where p — radius of curvature. If A' O' = r and O' B' = c r in the ellipse (Fig. 21) we have at A'
p = = ca r.
r
.'.H' = p' y. c2 r r= -^-c2 r — c
So in the parabola under uniform ver- tical loads (Case I.) we have seen that H = 2 p m (Bankine's C. E., p 165). But H = p p == % p m (since p = 2 m at the vertex).
If the load be everywhere normal to the cord the above equation will apply to every point, or
T =pp
be a general equation of the curve.
And further, when the load is every- where normal we have already seen that the pull along the cord must be constant, as there is no tangential force to change it. Hence,
T = p p = a constant. (31.)
"When the load p is constant, of course,
55
p must be constant too, and we have the circle already discussed. When^? varies, p must vary inversely as p.
FIG. 25.
Case VI. Ifp increases in value just in proportion to the distance of the
56
points of the cord above a horizontal line M N (Fig. 25), the cord assumes the shape of the hydrostatic arch. This curve possesses geometrically the loops shown in the figure and may be extend- ed indefinitely, but for our purpose it is evidently only necessary to discuss that part between the points C and B (Fig. 26) where the tangents are vertical.
Taking L (Fig. 26) for the origin, if the intensity of the load then be y Q (=. A. L). multiplied by a constant, or w y0, then at any other point it is = w y. Hence the equation of the curve is T=pp=wyp=wy0p0 = a constant
(y0 and pQ are the values of the ordinate and radius of curvature at A).
Let us resolve the normal load on C A B as we did in the circle, into its horizontal and vertical components. As was the case in the circle, these will be for each point equal in intensity to each other and also to the normal force, or
p = px — pv- But these quantities are no longer
58
constant (as in the circle) all along the curve, but vary from point to point.
If we form the parallelogram of forces for any arc A D (as in Fig. 27),t.he side N F = F S, since H = T = a constant, and F G must represent the resultant of the whole load on A D both in amount and direction.
The vertical component F E of F G is equal to the vertical component S X of SF, or
Vertical load on A D = T sin i = H sin i.
At B the vertical load = T = H = Y (since i = 90° there).
So the horizontal component of the total load on A D is G E, and since
NF = GS = GE + FX we have horizontal load on
A D = G E = N F-F X = H-H cos i = H (1— cos i).
At B, i = 90° .-. Horizontal load on A B = H On the arc D B
Horizontal load = H— H (I—cost) = H cos i.
59
The vertical load on A D may be thus expressed :
H sin i =
Q o0 sinj
FIG. 27.
The horizontal load thus :
60 wf ydy=w. y ~y° (33.)
And if yi = ordinate of B, the hori- zontal load on A B is
H = wy^~^°2 (34.)
For formula for radius of curvature see Bankine, C. E
The equation T = H = w>2/0 p0 = w y p, enables us to solve problems simi- lar to those under the parabola.
Case VII. If we construct a curve from the last one by using the same or- dinates and by changing all the abscissas in the ratio c : 1, so that the new co-or- dinates of a point shall be y and c x, and at the same time change the hori- zontal forces in the same proportion, leaving the vertical ones unchanged, the new curve and new system of forces so obtained will evidently be parallel projections of the former, and will be balanced. This new curve C' A B' (Fig. 28) is the " Geostatic," and bears a relation to the " Hydrostatic " strict-
61
ly analogous to that between the el- lipse and circle. Hence,
The total vertical load on A B' = V — 1
V = pull along cord at B'. l,q~ N
Total horizontal load on A B' = H' = f ^OJ c H = pull along cord at A'. J
FIG. 28.
62 The intensities are
H' £ H
For horizontal load p'x =rrr =TTT =^?o; I
DA. O A.
(V H Pa. and jt?y referring to the hy- drostatic curve.)
The load at A and B' and C' being al- together normal (it is not so at the other points), let
p'0 and p\ be the radii of curvature at A and B'.
Then
In the hydrostatic H= Pyp0. .'. cH — R' = cpyp0.
Py < •'• Po =
P'O = 0« Po • •
So
V =p'xp'i =cpxp'i = V.
But in the hydrostatic
•'•PxPi =
. •' -
(38.)
63
These radii are useful in drawing the geostatic curve.
Case VIII. So far we have dis- cussed the curves assumed by cords un- der loads distributed according to some simple law. But it is possible to dis-
FIG. 29.
cuss the more general problem : Given a load that varies and is distributed in any manner, required the curve which it will cause the cord to take ; or converse- ly, given a curve, required the character and distribution of the load to produce it. The most useful form of the prob- lem is that in which we assume the
64
shape of the cord, and the vertical com- ponents of the load, and require to be found the intensity and distribution of the horizontal components of the load necessary to produce equilibrium.
To illustrate : Assume the curve to be a circle, and the vertical load to be uni- form in intensity, we see at once that the horizontal load should be also uni- form, and of intensity equal to that of the vertical load.
But generally : Let C A B (Fig. 29) be some assumed curve, and let the ver- tical load be known in amount and dis- tribution. Making some changes in the signification of the letters heretofore used, now let
Y = vertical load on any arc A D.
Y! — vertical load on the semi-cord A B.
H = horizontal load on any arc A D.
H! = " half -cord A B.
H0 = pull along cord at A (the quantity
heretofore denoted by H). Px and py — the horizontal and vertical
intensities as heretofore. PQ = value of py at the point A. P0 and PI = radii of curvature at A and B.
65 The vertical load on an arc A D is
V = pydx . . (39.) •/ o
Again at the horizontal point A, the vertical projection of the element of the curve being = zero, the load is entirely vertical, and consequently at that point is normal to the curve. Hence the pull along the cord at A is
H0 =PoP0.
To discuss the forces upon an arc A D. Draw tangents at A and D. They meet at F (Fig. 29), through which point the resultant of the total load on A D must pass. The vertical load is also = the vertical component of the pull along the cord at D, for these two forces, be- ing the only vertical ones connected with A D, must needs balance each other, therefore,
Lay off F N = H0. Lay off F E ver-
tical and ~y pydx. Complete the rect-
angle F E S X. The pull along the cord at D = F S = F E cosec i = V cosec i . . (40.)
66
Also,
S E = F X = V cot i = horizontal compound of pull along the cord at D . . (41.)
But the horizontal pull at A is
H0 =FN = GS.
.-. G E = H0 - V cot = H = resultant of horizontal load on A D (42.)
The intensity of this horizontal load may be expressed thus :
_d H _ 6 (V cot i) _ \ PX~~ dy '- dy dy (43.)
At B the vertical load = Vt. Let this be represented by B K (Fig. 29). If the cord be itself vertical at that point, B K — . Vj will be equal to the pull along it at B. If the cord is in- clined as in the figure, draw its tangent atB, and
B L — B K cosec e\ = Vi cosec it = pull
along the cord. and
K L = B K cot ii •= Vj cot i± = horizontal
component of this pull. H0 — Vj cot il = H! = resultant of entire
horizontal load on A B.
67
It may often happen that S E = V cot i = horizontal component of the pull along the cord at D (Fig. 30) is greater than GS = FN = H0 = horizontal pull along the cord at A. In such cases G E — H0 — Y cot i is negative, which indicates that the horizontal load be- tween A and D, for at least a part of
FIG. 30.
this distance, must be contrary in direc- tion to that heretofore discussed ; that it must exert an inward pull instead of an outward one (Fig. 30). If this "inward pull " were removed or replaced by an outward one, the curve would evidently be flattened about A.
68
We may illustrate geometrically the relation between the forces in all parts of AB.
The vertical load and curve being- given, draw F Ev (Fig. 32) = the fatal vertical load on A B, and lay off on it
F E' = vertical load on the arc A D'. F E" = " « « A D", etc.
Draw a horizontal line at F and lay off F N and F K, each = H0 = pull at A. Draw through F lines parallel to the
69
tangents at D' D" D"', etc., and through E' E" E'", etc., lines parallel to the horizon. Then the oblique lines F S', F S", etc., represent the pulls along the cord at V D", etc., while E' S', E" S', etc., represent the horizontal components of these pulls. Lay off from each point S' S", etc., horizontal lines, each equal to F N, and draw through the points G' G", etc., thus obtained, a curve. It will evidently be similar to that drawn through K S' S", etc., and the line G' E' will represent the resultant of the hori- zontal load that must be distributed along the curve from A to D'; G" E", the resultant of the horizontal load be- tween A and D", and so on.
(Fig. 32) is really formed from the parallelogram of forces for the arcs A D', etc.; this parallelogram being at D' = F N G' S', in which E' S' is the horizon- tal component of the pull at D and W G' = the resultant of the horizontal load on A D'.
As the abscissas of the curve F G' G"r etc., increase to the left of F Ev from
70
the^point F to G'' (which correspond to D" on the curve), the horizontal load
FIG. 3 1 .
acts outward 011 the arc A D". The ab- scissas then diminish to G'". Hence
71
between D" and D'" on the curve, the horizontal load must act inwards as shown in (Fig. 31). From Of" the abscissas increase until we reach Gv. Hence the horizontal load acts outward throughout the remainder of the cord. The points n and n' correspond to those arcs on which the resultant of the hor- izontal load is zero. Thus on the arc A n the negative horizontal load is just equal to the positive, and hence their sum — zero. So on the arc A n'.
Note that the abscissas of the curve F G' . . . Gv are not the intensities of the horizontal loading, but that each such abscissa represents the algebraic sum of the entire horizontal load between A and the point to which the abscissa corre- sponds. The intensity in question has already been shown to be
dB.
dy
In this expression d H = the differ- ence of two neighboring abscissas of the curve F G' . . .Gv; as for instance, d H
72
— G' E' — G" E". And d ?/ = vertical projection of the arc D' D'' of the cord to which the above corresponds.
ARCHES.
FIG. 33.
ii
Let us imagine the curve of the cord to be reversed, and the cord itself to be replaced by a thin metal strip, which, like the cord, shall be practically without transverse stiffness, but, unlike the cord, shall be able to resist a compressive force in the direction of its length at every point. Let the loads be distrib- uted as heretofore, except that where there are horizontal components of the load, these should act inward, where upon the cord they acted outward, and vice versa. We then have what is called
73
a " linear arch or rib " ; and the curve assumed by it will be identical with that of the cord under equal and similarly distributed loads. If the loading is changed in distribution, the rib will change in shape just as the cord would do under similar circumstances.
In practice there are no "linear arches," but the discussion of them enables us to determine the form of equilibrium for real arches. If we know the form that a linear arch would as- sume under a given load, we can find the "line of pressures" in the real arch. This line and the value of the thrusts at all its points enable us to solve the problems that arise in arch building.
1. Suppose, for instance, we desire to construct an arch to bear a uniform ver- tical load, such as that discussed in Case I. The shape of the linear rib under such a load is a parabola. We then have,
1° Step. Assume this curve for the in- trados CAB (Fig, 34). If the arch and the load be of homogeneous material, the shape of the extrados, or outside of
74
the load, will be M Y M', the vertical distance between C A and M Y being constant.
2° Step. Is to determine the depth A L of the keystone. This depth is al- ways greater than necessary simply to prevent the crushing of the material of
FIG. 34.
the arch under the thrust at the crown. Prof. Eankine's empirical rule derived from the best examples is to make the depth of the keystone in feet
In single arches 1
" V. 12 x radius of curva'e at the crown. ! ,,/, v
In arches of a series = V.17 x radius of curva'e at the crown. J
75
3° Step. Determine whether the " line of pressures" can He in the "middle third " of the ring of voussoirs. It should be restricted to the middle third to pre- vent the voussoirs tending to open at any of the joints.
We can test this as follows :
Suppose the voussoirs to be constant in depth all around the arch as in (Fig. 34). Consider any part of the arch in- cluded between the vertical plane (A L) at the crown, and a vertical plane at any other point, as D' P'. The calculated horizontal thrust along the linear rib, which coincides in shape with the soffit C A, is indicated by the arrow with its head at A. Let the horizontal thrust of the rib at D' be indicated by the ar- row with its head at D' pointing in an opposite direction to that at A. At the crown take A K not greater than % A L. Imagine a left-handed couple ap- plied to A L in the vertical plane of the arch, whose force = H = the thrust at A, and whose lever- arm = A K. Apply an equal and opposite couple on the plane
76
D' P', with a force H', equal to the hori- zontal thrust of the rib at D'. Its lever- arm Dr P' must then
H. AK
H'
In; the parabola H = H' .-. D' P' = A K. These couples being equal and op- posite do not change the conditions of equilibrium of the section of the arch L D', but they transfer the line in which the thrust acts from A D' to K P. We can repeat the process as often as we choose by taking parts L D", etc.; and if the curve drawn through the points K P' P", etc., lies within the middle third of the arch-ring, the arch is suffi- ciently stable.
In the case before us, the horizontal thrust being constant for every point of the rib C A, the lever-arms D' P', D" P", etc., are also equal, and therefore the " line of pressures " K K' is merely the parabola raised vertically a distance = A K. If K K' does not lie in the middle third, a slight increase in the voussoirs,
77
especially towards the springing, will usually remove the difficulty.
4° Step. The joints between the vous- soirs, such as D' G (Fig. 35) are usually made normal to the soffit A C, but whether this be done or not, the direction of G D' must be such that at S, where the
line of pressures cuts it, the angle in- cluded between S N (the normal to G D') and S T (the tangent at S to K K') may be less than the angle of friction of the material of the voussoirs. The best pos- sible direction for the joints D' G, etc.,
78
would be to make them perpendicular to KK'.
The horizontal component of the thrust (H) along the curve of pressures in a parabolic arch, is, as we have seen, constant ; but the thrust along the curve (T) increases from A to C, and its value at any point may be determined by the formulae in Case I.
Parabolic stone or brick arches are noi common, because it is rare to have such a distribution of the load as that sup- posed above.
2. But if we reverse the curves dis- cussed under Cases II. and III., we have a form of arch much more frequently ap- plicable.
Thus, suppose the arch and its back- ing to be homogeneous, and that the extrados of this loading is horizontal (M Y), and suppose the action of the load to be entirely vertical. Then the arch and its backing are similar to the metal sheet and the cord discussed in the cases just referred to, and therefore the form of the linear rib under such a
79
load will be a catenary or transformed catenary— usually the latter. FIG. 36.
Assume this curve for the soffit C A
80
B (Fig. 36) ; determine the depth A L ; the line of the pressures K K'; and the
FIG. 37.
direction of the joints ; as in the last case. In this case, as in the parabola,
81
H is constant, and hence the curve of pressures is merely the curve C A raised vertically through a distance = A K.
Example. — Let the data for a required arch be (Fig. 37) span C B = 10' ; rise O A = 4'; height of extrados* M Y above springing at C = 10'. Let the arch and brickwork be of solid brick- work whose weight w per cubic foot = 112 Ibs.
The equation of the transformed cate- nary passing this C A B is
Where y0 = A Y — 6' (the origin be- ing at Y and the axis of abscissas hori- zontal).
First find m, the modulus of the cor- responding common catenary. By eq.
(14)
V
82
At the point C xr = 5 ft. and y' =
10ft.
.-. m = 4.54 ft. = YN.
Then determine points of the curve, thus ; for
x = 1, y — for x = 2, y = for # = 3, # — for x = 4, y = etc
Describe the curve through these points.
The thrust at the crown A is (for a unit of length of the arch)
H = wm* from eq. (16). .-. H = (112) (4.54)8 = 2,308.3 Ibs. From eq. (15) area A Y M C =
E™ - E~ m = 36.32 sq. ft.
Weight of load A Y M C = P = (112) (36.32)
= 4,067.84 Ibs.
From eq. (18) thrust at C = T = 4/p«^-H2 = 4,677.1 Ibs.
Inclination at C . Tan i, = *^±=^ dx, 2m
\ xl —*- ) \ EW — E m ) :
.-. »\ = 60° 33'.
83
The formula for depth of keystone will be satisfied by making the depth of the arch A L = length of one brick = 9",
oc
CO
for this gives 9" X 12" =108 square inches to bear the thrust H = 2,308.3 Ibs., or T = 4,677.1 Ibs. The latter is the greatest thrust in the arch.
84
It is easy to see that K K' will be in the middle third, for even at C the dis- tance of the point of the curve K K' vertically over C, from the nearest point of C A, is approximately
6 X cos (90° - 60° 32') = 5"+.
The extrados of the transformed cat- enary need not be the directrix M Y ; it may be another transformed catenary, provided these catenaries have the same directrix.
To illustrate: Suppose the weight of a unit of the material between CAB and M Y — w. Then the intensity of verti- cal pressure at any point G of CAB (Fig. 38) is = w y. If a heavier build- ing material were used this vertical press- ure could be brought upon G by a less height of it. Let this heavier material have a weight per unit = w and let
2 w =—w.
Then a column of the heavier material over G and of a height = f y would give
85
the same pressure as the whole column of the lighter, or
o
--w'y . . (47.)
At each point CAB (Fig. 38) lay off two-thirds of the vertical ordinate, and through these points draw C" A" B." The upper service of the load may have this form, and yet CAB still be the shape of the linear arch balanced under the applied forces. The equation of C A B being
y =•• -T 1 E™ + E
that of C'' A" B" is evidently
**>
E "I (48.)
The principle of this example is gen- eral.
When the extrados is a transformed catenary, note that, since in all the for- mulae under Case III., w = the weight corresponding to a unit of surface of
86
the space between CAB and M Y, we must make in these formulae
w = n w'
Where w' = weight of the building AA "
material and n =•
Y A
In arches of this class no provision is needed for horizontal thrust on the spandrels, as the arch is equilibrated under vertical loads alone.
In all stone or brick arches, the changes in the curve of pressures K K' due to passing loads are usually slight, because the weight of such passing loads is generally small compared with the weight of the arch itself and its backing.
3. The simplest practical case in which a uniform normal load (such as that discussed in Case IV.) can be ap- plied to an arch is when it is subjected to water pressure, the arch ring being horizontal instead of vertical. Such a pressure will exist on an empty well con- structed in a reservoir or other body of
87
water (Fig. 39). For each horizontal layer of the well wall may be considered as subjected to a uniform normal press-
FIG. 39.
ure of an intensity due to the depth of the water at that layer. This intensity will of course diminish (and so will the
. 88
pressure on the wall) from one layer to another as we come towards the top.
The soffit of such a well should be of circular form (Case IV). The thickness of the wall at any depth must be determined by the thrust, which is constant all around any given layer, and is
T = p r = w y r. . . (49.)
Where w = weight of a unit of water and y = depth of water at the layer in question.
In determining the line of pressures consider a section of the wall befween two yertical planes not parallel as here- tofore, but both normal to the soffit. Take for the lever-arm of the couple at A (Fig. 39) a distance A K = \ A L. The force is still to be = H = T.
At D apply an equal couple with force = the thrust along the soffit at that point, which is also = T = H. Then the lever-arm must be equal to A K. Hence we see that the curve of pressures is a circle parallel to the soffit, and may pass through the middle of the arch ring.
89
This kind of arch may be used for dams or the walls of reservoirs. (See Fig. 40.)
4. There is no case in ordinary prac- tice where the pressures upon an arch are strictly identical with those on an el-
90
liptical cord, for in this case the press- ure must be constant in intensity along both the horizontal and vertical projec- tions of the arch, but the intensity along the horizontal must differ from that along the vertical in a constant ratio (Fig. 41). But, as Prof. Rankine says,
FIG. 41.
H, \ 1 1 1 j
the curve of equilibrium for the arch of a tunnel through earth, when the depth below the surface is great compared with the rise of the arch itself, approximates to an ellipse.
The pressures in a mass of earth are intermediate in character between those
91
existing in a solid and those in a liquid mass. Thus a little cube of earth (Fig. 42) under the weight of the superincum- bent column of earth /?, presses down-
FIG. 42.
J
ward with a force equal to its own weight and that of the column above. It also presses out horizontally with a force less than this downward force,
92
but always bearing a constant ratio to it. If the little cube were solid it would have no horizontal push ; if liquid,
FIG. 43.
xL Y
that horizontal push would equal its pressure downward. If the upper sur- face of the earth is inclined, the outward
93
push which always remains parallel to it becomes inclined too, and is then "con- jugate " to the vertical.
If M Y (Fig. 43) is the surface of the earth, when Y A is great compared with A O, then Y A and M C differ so slightly that we may assume them to be equal. We then have on the arch a uniform ver- tical load whose intensity =
Py = (Y A) x weight of a unit of the earth=w#0,
and a horizontal load whose uniform in- tensity px is equal to the vertical inten- sity (py ) multiplied by a constant. Let
f)r
- = c9 (a constant). Py Then
px = c*wy0 and c — \ ^L . py
From the discussion of Case V. we see that c must be the ratio of the axes of the ellipse to which the pressures are re- spectively parallel. Hence if the arch be a semi-ellipse and O B be given, we have
OB OB
?— r^ = c .'. O A =— O A c
94
From these data draw the curve of the soffit.
The thrust along the soffit at A
= H = Py 9o = ^y0$V AtC or Bit is V = pxgim At other points it may be gotten from eq. (27) Case V.
We can determine the curve of press-
FIG. 44.
JL
ures by a method similar to that used in the last case. Here, however, the curve K K' will not be parallel to C A, since the thrusts along C A are not con- stant, but increase from A to C. Assume
95
A K (Fig. 44) =i f A L, then the arch must be so proportioned that K K' shall fall within the middle third.
If the arch C A B is not to be a ^mi- ellipse (as above assumed) but only a seg- nient of one, a few trials will enable us to get the ellipse from the data already given.
The strictly true curve of equilibrium required by earth pressure is the Geo- static arch.
5. An arch built with the curve dis- cussed in Case VI. is known as the Hy- drostatic arch, from the fact that the loading there described is similar to the pressure of the water upon a vertical arch.
For if M Y (Fig. 45) be the surface of the water, then its pressure on C A B is normal and proportioned at each point to the depth below M Y. This press- ure, as has been shown, may be resolved into a vertical and horizontal pressure at each point, this vertical and horizon- tal pressure being equal in intensity to each other at every point, and also to
96
the normal pressure of which they are the components.
to ^
I
The above form of arch may be applied in two cases.
97
(1) To bear the pressure of water or other liquid. Thus in the case of a river tunnel (such as those at Chicago) where the top of the tunnel is practically on a level with the bottom of the river, we might use the hydrostatic arch.
The equation of the curve is
The vertical load on the half -arch A B — I pdx=\=wy} pl=thrust along arch at B.
* Xi
The horizontal pressure against A B
y^-y<f y=iD - s — - = H = w#0/>o . (51.
y\
The 'thrust along the arch is constant, or T = H = V.
The rise A O (= a), the depth AY (= f/0), and the radii at A and B (p0 and p,) are connected by the following ap- proximate equations. The co-ordinates of B being xl and yl? let
(52)
98
Po
= V' a,f° = a +^7 = in1*-^) (53-)
The line of pressures in a hydrostatic
arch, since T is constant, is parallel to the soffit, as in circular arches.
99
Example. — Suppose the span to be 50 ft. (Fig. 46) and the depth A Y = 16 ft.
Find first the rise A O. In eq. (52) x1 = 25' y0 = 16', and a few trials show that a = rise = 20' about.
Hence
p0 = 32Jft. and^ = 14.1 ft.
With these data describe the curve of the soffit — the radius at any other point besides A and B being given by the equa- tion
The thrust at A = H = wyQ $>0. Here w = 62.4 Ibs.
. . H = 32,448 Ibs.
The rule for the depth of keystone in a single arch gives
Depth A L = V .12 x 32T5 = 1-9 ft-
This is ample. It only gives about 120 Ibs. per sq. in. as the pressure at the crown.
T being — H, the depth of the arch- ring may be uniform.
100
(2) The hydrostatic arch is also used when the loading is homogeneous ma- sonry up to the extrados M Y, provided the spandrels be suited to sustain a horizontal thrust at each point of the arch equal to the vertical load at that point.
As all stone or brick arches sink at the crown when the centers are removed, they will exert at other points an out- ward horizontal thrust. Now if we as- sume that this horizontal thrust is at every point equal in intensity to the ver- tical loading at that point, the curve of equilibrium under such a system of forces is the hydrostatic curve. This is the assumed condition of the forces act- ing in the Neuilly and other bridges of this class.
When the spandrels cannot be made firm and solid this form should not be used, but when they can be, as in the successive arches of a stone bridge, it is advantageous rather than otherwise, to have such a thrust from the arch against the spandrel ; while the hydrostatic
101
curve of given span and rise gives a greater water-way than the correspond- ing catenary would.
The catenary needs no resistance from the spandrel, being balanced under the vertical load alone.
Example. — Let the span be 100 ft. and rise 30 ft. Then the depth of load- ing at the crown (= A Y, Fig. 46) will be found from equation 52
Then OQ = 91.7ft.
Hence H = wy^Q (putting w = 160 Ibs.) = 107,600 Ibs. Depth of keystone
= A/TT2^T9r7 = 3.3 ft.
This gives a pressure of 32,280 Ibs. to the sq. ft., or about 225 Ibs. to the sq. in. 6. If the vertical forces vary as in the hydrostatic arch, and the horizontal are not equal to them, but differ at each point in a constant ratio, the curve of equilibrium (Fig. 47) becomes the Geo-
102
static curve discussed in Case VII. This curve derives its name from the fact that the system of pressures above described is similar to that exerted by a mass of loose earth against CAB. Let M T = the horizontal surface of the earth ; then. FIG. 47.
M T
at each point D of the arch there is a vertical pressure of intensity (p'y) pro- portional to the depth (y) of D below M Y, and a horizontal pressure whose in- tensity is less than p ' y in a constant ratio, or p'x = c*p'y
103
(c2 being taken to represent the ratio of the intensities).
Assume a hydrostatic arch whose verti- cal dimensions shall be identical with those of the geostatic arch, and whose span (C B) (Fig. 48) shall be connected
oo
with the span of the geostatic arch (C' B') by the equation
C'B'
(55.)
104
The intensity of the vertical pressure (the horizontal is like it) in this hydro- static arch must be
Py =
From these data deduce a hydrostatic arch and then pass by parallel projec- tions to the required geostatic arch.
Equations (35) (36) (37) (38) give the values of the quantities needed in dis- cussing the Geostatic arch.
Example 1. — Let the span of the geo- static arch (C' B'= 100 ft.) be given; also the depth of the loading (A Y = 20 ft.) ; also the ratio of the pressures (c2 — J); and the weight of a cubic ft. of the loading = w — 100 Ibs. Whence
P'y* = Wo = 2,000 Ibs. Then since
™ =
c = 58 Ibs.
" .2000 = l,154.71bs.
We. find from equations (52) (53) (54) for the hydrostatic arch
105
Else = a = O A = 57.7 ft.
P0 = 140.93 ft.
P! =363 ft.
H = V = T = pyo Po = 1,154.7 X 140.93 = 162,700 Ibs. nearly.
In the geostatic arch we have from equations (35) (36) (37) and (38)
Thrust at B = V = V = 162,700 Ibs.
A = H' = cH = 94,300 Ibs. nearly. Po' =46.97 ft. p,' = 62.65ft.
Example 2. — Suppose the span = 100 ft. depth, A Y = yQ = 20 ft. and rise, a = 30 ft. given ; to find c and thence the hydrostatic arch.
From equation (52) we find
b = 40.71,
and thence in same equations xl = 39. Hence the span of the hydrostatic arch
= 2xl = 78 ft. Andasc.CB= C'B'
Then proceed as in the last example. In
106
this example the hydrostatic arch is the smaller of the two.
The line of pressures in a geostatic arch is found as it was in the elliptic.
The geostatic is the true curve of equilibrium under earth pressure, but when A Y (Fig. 48) is great compared with A O, it approximates the ellipse described through the points C' A B' as already stated.
7. Convenience, or other reasons, will often dictate the form of the arch with- out reference to the loading, and again necessity may make the vertical load different from any and all the cases we have discussed. In such instances Case VIII. will enable us to determine the character and amount of the horizontal forces which must be applied through the resistance of the spandrel, when once the form of the arch and the vertical load are known.
When the horizontal forces thus re- quired are thrusts directed against the arch, it is generally possible so to build the spandrel that the arch may be se-
107
cure, but when they are the opposite, or outward pulls on the arch, then it is difficult to insure stability, as to do so requires tension between the arch and the spandrel. In such cases it is best to change the form of the arch.
The discussion of Case VIII. of cords enables us to determine the necessary data in the case of similar linear arches under similar loads.
Fig. (50) gives the geometrical con-
108
struction of the triangle of forces at every point of the semi-arch A B (Fig. 49).
We may discuss a given linear arch CAB under a given vertical load, by de- termining :
1. The thrust at crown ; which is
H0=jpo/>0. (560
2. Total horizontal thrust required on any arc A D', A D", etc. This, from equation (42), is
H = H0 - V cot. t. (57.)
If this be negative the spandrel must exert a pull instead of a thrust.
On the half-arch A B the above equa tion becomes
H! = H0 - VT cot. »\. (58.)
On any arc B D'v, counting from B upwards, the total spandrel thrust is
H, - H = — Vi cot. a, + V cot. i. (59.)
This last expression has at least one maximum value corresponding to some arc B D. In the Fig. (49) this value corresponds to the arc B D'".
109
I
110
Let this maximum value be denoted by Hm and let im = the inclination at D'". Then
Hw = — VA cot. t\ -f- V cot. ew = E'" G'". (Fig. 50) (60.)
~D"' is known as the " point of rupture." There the action of the spandrel ceases to be a thrust, and must, above that point, for some distance at least, become tension.
3. The intensity of the horizontal spandrel thrust or pull in any layer (as between D'" and D'v) is from equation (43)
'
d cv cot.
_
dy ~ dy dy
When H is positive (that is thrust) px is negative, as it should be, since it is equal to the increment of the abscissas of the curve F G' G", etc. (Fig. 50), and these increments are decreasing from G"' to Gv.
At the point of rupture
#» = 0. (61.)
Ill
We can determine the point of rupture in three ways : First, by constructing the Fig. (50) and finding the inclination (£m) corresponding to the maximum abscissa E'" G'". Secondly, by substi- tuting the various values of i and V in the value of
(H j - H) (eq. 59),
and getting the maximum value of the expression. The i which gives this max- imum value corresponds to the point of rupture. Thirdly, by solving equation (61) #, = 0.
4. The thrust along the rib at every point is from equation 40,
T = V cosec. », (62.)
and it is represented by the inclined lines F S' F S", etc., Fig. (50).
The horizontal component of this thrust is
Hr = V cot. i — the abscissas of K 8', etc. , which are always equal to H0, the thrust
112
at the crown minus the spandrel thrust between A and the point in question
. •. Hr = V cot. i = H0 - H. (63.)
This is evidently a maximum at the point of rupture, or, since at the point of rup- ture,
H = HJ — Hm,
we have
HB = H0 — Hx -|- Hm.
But
H0 - H! = Y! COt. i, ... HB = V, cot. i, 4- Hm. (64.)
This horizontal thrust of the rib at D'" is therefore to be balanced by the hori- zontal reaction of the abutment at B ( — Vj cot. i^) together with the resistance of the spandrel between B and D'" ( = Hm). When the arch is vertical at B, Vx cot. iv= 0.
5. In single arches it is necessary to know the point of application of the re- sultant of the forces represented by (V, cot. i^ -f Hm) in order to determine the stability of the abutments. Take mo- ments with reference to the axis of ab-
113
scissas M Y. Then if yB = ordinate of point in question, and ym and yl be the ordinates of D'" and B, we have *
= (Vt cot. i1)yl +v& H
= (Vl cot. *,) yj + / y #B dy.
«/ ym
/k ypxdy Vm
HB
(65.)
In this we neglect the spandrel forces above D'" so far as they affect the sta- bility of the abutment. This can be done with safety.
The line of pressures and depth of keystone are determined as heretofore.
Example 1. — Let the assumed form of the soffit be a semi-circle, and let the loading consist of the arch and backing of homogeneous masonry carried up to a horizontal "extrados" M Y (Fig. 51).
Place the radius of the arch — r
Depth A Y = a r Heaviness of the material — w
114
Take the origin of co-ordinates at A and
1C
6
*
express the co-ordinates in terms of the
115
inclination i of the arch as on p. 217
RanJcinds C. E.
Then
Thrust at crown = H0 — p0 p0 = (war) v = war
Vertical load on any arc = V = wr2
cos. * sin. i i
Spandrel thrust on any arc A D H = H0 - V cot. i = wr2
( . cos. 2 i . i cos. i )
1 a - (1 + a) cos. i -\ -- s -- {- =—. — . I ] 2 ' 2 sin. i j
On A B this becomes (since the arch is vertical at B and C)
H! = war9 = H0 . ' . Hm — V cot. im
Intensity of spandrel thrust d (V cot. 0
- --"
,. t — cos. i sin. t
(1 + a) - cos. . -
The point of rupture is found by put-
116
ting px = o and finding the value of im by trials. As a first approximation
1 + 30 im = arc. cos. -
a
Thrust along the rib = T = V cosec i. At B this is
Y! — tw»/a+l - -V
So
HB — Y! cot. e\ + B.m = Hm = wr8
cos. im -
and
ra /»90°
^R — HR" / • PX sin' * (! — cos- 0 47 ^m
^Example 2. — Let
r = 20' A Y = 2.5'.
Then
25 1
* = "io = 8 w = 15° lbs- (Fig-
Then
H0 = ^«^= 7,500 lbs.
( 9 . cos. i sin. £ ?,
V - 60,000 sin.*- - ----
117
At B, V = 60,000 - -- ~ = 20,376 Ibs.
s
Angle of rupture im — arc. cos. -^-
.-1. 6875 = 46° 34'
118
HR = Hm = 60,000
_
8
.Six. 947 | = 8>154 2 j" Ibs.
(Fig. 53) shows the manner in which
the forces vary. From A to D (Fig. 52) there must be a pull in the spandrel to
119
produce equilibrium. The total amount of this pull is small, being
=8,154 - 7,500 = 654 Ibs.
To rid the arch of it, so that the part D1 A D shall either be balanced under the vertical load alone or exert a thrust outwards, instead of a pull inwards, we flatten the arc D1 A D. A few trials will determine this flattening near enough for practice.
Thus if D1 A D (Fig. 54) is to be bal- anced under the vertical load alone, find the center of gravity of the section D1 A and its load. Draw a vertical line P through this point, then if we can draw a line from any point in the middle third of the joint D1 parallel to the tan- gent to the arch there, and from its in- tersection with P draw a line parallel to the arch at A which will intersect A L within the middle third, then the extreme points of the line of pressures in the section A D1 will be within the middle third, and the line of pressures will gen- erally be altogether within it.
FIG. 54.
121
The new radius required for the arc D1 A D may also be determined roughly by putting H0 = war* = HR and thence getting r1, since, if D1 A D is to be bal- anced under vertical load alone, the hori- zontal thrust at every point of it must be the same and = HB, the thrust at D1 andD.
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THE VAN NOSTRAND SCIENCE SERIES. 7
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U.C. BERKELEY
No. 10
No. 10}
Navy.
IN BRIDGE MEM- E.
ODS OF DETECT- '!s. Baker.
GAS - ENGINE. By nn.<iid edition. With additional /F. E. Idell, M.E.
CODbOf
2.— THE GAJXGE AND SANITARY PLUMB- p \V. P. Gerhard. Sixth edition. Re-
No.
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