# Great Neapolitan Earthquake of 1857/Part I. Ch. XVI

CHAPTER XVI.
SECOND CLASS OF DETERMINANTS, OBJECTS OVERTURNED, OR PROJECTED BY SHOCK.
I therefore pass at once to the second class of seismometric determinants—viz., those derivable from the overturning or projection of objects by the shock; i.e. by the velocity impressed upon them by the wave itself, and in the direction, of its line of transit, or contrary to it; taking in also, such work of fracture, as may occur in detaching them from their contacts.
And here, as the conditions of observation are very simple, being limited chiefly, to the accurate measurements of two ordinates and an azimuth, and to some considerations as to the forms of the bodies, and of their points of attachment, and to the mutual relations of these, we may avoid that prolix detail, which wss unavoidable in treating of the deductions, as to direction from fissures and fractures.
Cap. A.—Bodies Overthrown Only, i. e. without Fracture or Adhesion overcome.
I. By Horizontal Force (Normal Wave).
$\mathrm {F} =\mathrm {MV}$
$\mathrm {V}$
$V$
In order to upset the body, the horizontal velocity impressed by the shock (whatever be the duration of the latter) must be sufficient to make it turn upon one of the arris's or angles of its base, through an angle $\phi$ formed by the line, Fig. 93, joining the centre of gravity with that angle or arcis, and the vertical through it.
Let $\alpha$ denote the distance (in feet) of this point or edge from the centre of gravity, then the statical work done in upsetting the body, whose weight is $\mathrm {W}$
* $\mathrm {W} \alpha (1-cos{\phi })$
This must equal the dynamical work acquired, which (as is well known), is equal to the work stored up in the centre of gyration, or—
* $\mathrm {W} \alpha (1-cos{\phi })={\frac {\mathrm {W} \omega ^{2}\kappa ^{2}}{2g}}$
$\omega$
$k$
$g$
Equating these two values of the work done we find
| $\omega ^{2}k^{2}=2ga(1-\cos {\phi })$ | (I.) |
but $\omega$, the angular velocity, is equal to the statical couple applied, divided by the moment of inertia, or
* $\omega ={\frac {Va\cos {\phi }}{k^{2}}}$
* $V^{2}=2g\times {\frac {k^{2}}{a}}\times {\frac {1-\cos {\phi }}{\cos ^{2}{\phi }}}$
| $l={\frac {k^{2}}{a}}$ | (II.) |
* $V^{2}=2gl\times {\frac {1-\cos {\phi }}{\cos ^{2}{\phi }}}$
$l$
1st. In the case of a solid cube overturned (Fig. 94) whose side is $\alpha$;
* $k^{2}={\frac {2\alpha ^{2}}{3}}$
* $a={\frac {\alpha }{\sqrt {2}}}$
* $l={\frac {k^{2}}{\alpha }}={\frac {2{\sqrt {2}}\alpha }{3}}$
and
* $\cos {\phi }={\frac {1}{\sqrt {2}}}$
| $V^{2}={\frac {8}{3}}g\alpha ({\sqrt {2}}-1)$ | (III.) |
and the following geometrical construction is obtained from this expression.
Let $\delta =$ the difference between the diameter and side of the cube $=\alpha \left({\sqrt {2}}-1\right)$
| $V^{2}=2g\times \left({\frac {4}{3}}\delta \right)$ | (IV.) |
Or, the height due to the horizontal velocity of the wave of shock is equal to four thirds of the difference of the diameter and side of the cube.
2nd. In the case of a solid rectangular parallelopiped overturned (Fig. 95).
The altitude being $=\alpha$, side of the base $=\beta$, and $tan{\phi }={\frac {\beta }{\alpha }}$, we have
* $k^{2}={\frac {\alpha ^{2}+\beta ^{2}}{3}}~~~~~~~~a={\frac {1}{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
​therefore
$l={\frac {k^{2}}{a}}={\tfrac {2}{3}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
substituting in (Eq. II.) we find
| $V^{2}={\tfrac {4}{3}}g\times {\sqrt {\alpha ^{2}+\beta ^{2}}}\times {\frac {1-\cos \phi }{\cos ^{2}\phi }}$ | (V.) |
or,
| $V^{2}={\tfrac {4}{3}}g\times {\frac {\alpha ^{2}+\beta ^{2}}{\alpha ^{2}}}\times \left({\sqrt {\alpha ^{2}+\beta ^{2}}}-\alpha \right)$ | (VI.) |
Let $\delta$, as before, denote the difference between the diagonal and altitude of the parallelopiped,
$\delta ={\sqrt {\alpha ^{2}+\beta ^{2}}}-\alpha$
${\mbox{and since }}{\frac {\alpha ^{2}+\beta ^{2}}{\alpha ^{2}}}=\sec ^{2}\phi$
we have the following resulting theorem—
"The height due to the horizontal velocity of wave that will overturn a rectangular parallelopiped is two thirds of the difference of the diagonal and altitude, multiplied by the square, of the ratio of the diagonal to the altitude, or of the secant of the angle $\phi$."
3rd. In the case of a solid right cylinder overturned.
The height of the cylinder, or altitude, being $\alpha$, and the diameter, or base, $\beta$, as before, we have
$k^{2}={\frac {15\beta ^{2}+16\alpha ^{2}}{48}}$
and
$a={\tfrac {1}{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
therefore
$l={\frac {k^{2}}{a}}={\frac {15\beta ^{2}+16\alpha ^{2}}{24{\sqrt {\alpha ^{2}+\beta ^{2}}}}}$
​substituting in (Eq. II.) and since
| $\cos ^{2}\phi ={\frac {\alpha ^{2}}{\alpha ^{2}+\beta ^{2}}}$ | (VII.) |
$V^{2}={\frac {15\beta ^{2}+16\alpha ^{2}}{12\alpha ^{2}}}\times g{\sqrt {\alpha ^{2}+\beta ^{2}}}\left(1-\cos \phi \right)$
4th. In the case of a hollow rectangular parallelopiped overturned (Fig. 96).
Let the edges of the parallelopiped be $\alpha ,\beta ,\gamma$ the thickness of its walls being small in relation to their lengths, and suppose it overturned round the edge or axis $\gamma$.
It is easily demonstrable that
$k^{2}={\frac {2\beta \left(\alpha ^{2}+\beta ^{2}\right)+\gamma \ \left(2\alpha ^{2}+3\beta ^{2}\right)}{6\left(\beta +\gamma \right)}}$
and
$a={\tfrac {1}{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
and therefore
| ${\frac {k^{2}}{\alpha }}={\frac {2\beta \left(\alpha ^{2}+\beta ^{2}\right)+\gamma \ \left(2\alpha ^{2}+3\beta ^{2}\right)}{3\left(\beta +\gamma \right){\sqrt {\alpha ^{2}+\beta ^{2}}}}}$ | (VIII.) |
from which, substituting the value of $l$ in Eq. II., and remembering that $\tan \phi ={\frac {\beta }{\alpha }}$ we obtain $V$, the horizontal velocity.
​5th. In the case of a hollow cylinder overturned (Fig. 97).
Let the altitude of the right circular cylinder be $\alpha$, the exterior and interior diameters of its base $\beta ,\beta '$ so that the thickness of the wall $={\frac {\beta -\beta ^{2}}{2}}$
then
$k^{2}={\frac {15\beta ^{2}+16\alpha ^{2}+3\beta '^{2}}{48}}$
which is the case of the solid cylinder when $\beta '=0$, and when the thickness of the wall is so small in relation to the other dimensions that it may be neglected
$k^{2}={\frac {9\beta ^{2}+8\alpha ^{2}}{24}}$
In either case, the distance of the centre of gravity from the axis or point of rotation at the base is
$a={\tfrac {1}{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
Hence, in the first case,
| ${\frac {k^{2}}{\alpha }}={\frac {15\beta ^{2}+16\alpha ^{2}+3\beta '^{2}}{24{\sqrt {\alpha ^{2}+\beta ^{2}}}}}$ | (IX.) |
and in the second case,
| ${\frac {k^{2}}{\alpha }}=l={\frac {9\beta ^{2}+8\alpha ^{2}}{12{\sqrt {\alpha ^{2}+\beta ^{2}}}}}$ | (X.) |
and
$V^{2}=2g\times {\frac {9\beta ^{2}+8\alpha ^{2}}{12{\sqrt {\alpha ^{2}+\beta ^{2}}}}}\times {\frac {1-\cos \phi }{\cos ^{2}\phi }}$.
​6th. In the case of a solid parallelopiped with two adherent wedges, overturned round the free or external arris.
This form is that frequently occurring, as thrown from the ends of rectangular buildings, and described in treating of fissures, the end wall being the parallelopiped, and the wedges the adherent portions of masonry, fractured from the side walls.
The general and exact treatment of this case involves expressions too complex for practical use. The case should never be appealed to for deciding the value of $\mathrm {V}$, unless the magnitude of the parallelopiped be large, in proportion to that of the wedges, the lower angle of the latter small, and not very unequal at the two ends of the parallelopiped, and the thickness of the walls $\beta$, small in proportion to the height $\alpha$. In that case a sufficiently near approximation is readily made.
$\alpha$
$\gamma$
$\gamma$
Let the thickness or third dimension of this rectangular plate be $\tau$, and let it be supposed applied to that side of the parallelopiped, to which the wedges were adherent, and added to the thickness $t$, or width of base of the parallelopiped, for the value of $\beta =t+\tau$. Further, $h$ being the height of the end wall or parallelopiped, let its altitude be assumed increased in the proportion
* $t:h::t+{\frac {1}{4}}\tau :\alpha$
$\alpha$
$\alpha$
$\beta$
Case 7th. Angular wedge thrown over upon its apex.
This is the case referred to pp. 66-72, in treating of fissures, as one of frequent occurrence, and valuable in deciding direction of wave-path. It can, however, be very nearly applied to the determination of velocity. The problem, generally treated, leads to very complex results; and approximations are equally tedious, except in the case in which the direction of the wave-path is parallel to one of the external sides of the wedge, when the wedge vanishes and the case becomes identical with the last one.
$V$
II. OBLIQUE FORCE (Subnormal, Abnormal, and Subabnormal Waves).
Let $\mathrm {O'O}$ be the wave-path passing through any building whatever, as Fig. 99.
Let $\mathrm {OX}$ be perpendicular to the lines $\mathrm {OZ}$ and $\mathrm {OY}$, and let $\alpha$, $\beta$, and $\gamma$ be the angles made by $\mathrm {O'O}$, with $\mathrm {OX..OY..OZ}$.
If $\mathrm {V}$ = the total velocity, or that in the path of the wave, and $v_{x}v_{y}v_{z}$ the components along $\mathrm {X}$, $\mathrm {Y}$and $Z$, then
* $v_{x}=\mathrm {V} \cos {\alpha }$
| $v_{y}=\mathrm {V} \cos {\beta }$ | (XI.) |
* $v_{z}=\mathrm {V} \cos {\gamma }$
$v_{x}$
$v_{y}$
$\mathrm {Y}$
If the structure is capable of being overturned in the plane of $\mathrm {YZ}$, and also in the plane of $\mathrm {XZ}$, the components $v_{x}$ and $v_{y}$ must act together, and compel it to turn round upon one of the extreme points in the line $\mathrm {OY}$. In that case, the motion ceases to be comparable with that of a compound pendulum, and is reduced to the movement of ​a body round a fixed point under the influence of gravity; one which, even in its simplest cases, would be too difficult of reduction for practical purposes, and which virtually never occurs in seismometric observation. If, therefore, the force of the wave be confined to the plane $\mathrm {XZ}$, and be emergent at an angle $e$ with the horizon, we have
* $v_{x}=\mathrm {V} \cos e$
| $v_{z}=\mathrm {V} \sin e$ | (XII.) |
$\mathrm {V}$
The general case of the subnormal wave must be distinguished into two. First, when the wave-path through the centre of gravity falls within the base, second, when it falls without the base of the structure.
$\mathrm {O'} o'\mathrm {C}$
$\mathrm {CO'}$
In either case, overthrow by the second semiphase, although mechanically possible, seldom occurs (for the same object), in parts of buildings, as some support or obstacle generally props it against the latter.
$\mathrm {OC}$
$\mathrm {CO}$
$\mathrm {A}$
The overturning couple is
* ${\frac {v_{x}\alpha -v_{z}\beta }{2}},$
$\alpha$
$\beta$
$\mathrm {M}$
* ${\frac {\mathrm {MV} (\alpha \cos e-\beta \sin e)}{2}},$
* $\omega ={\frac {\mathrm {V} (\alpha \cos e-\beta \sin e)}{2k^{2}}},$
* $\mathrm {V} ^{2}={\frac {2g\alpha k^{2}}{({\frac {\alpha \cos e-\beta \sin e}{2}})}}\times (1-\cos \phi )$
* ${\frac {1}{2}}(\alpha \cos e-\beta \sin e)=f=a\cos(\phi +e)$
| $\mathrm {V} ^{2}=2gl\times {\frac {1-\cos \phi }{\cos ^{2}(\phi +e)}}$ | (XIII.) |
$e=0$
$V^{2}=\mathrm {V} ^{2}$
But if hte structure be overturned by the second semiphase, inertia acts in the direction of transit $\mathrm {OC}$, and tends to make it overturn round the axis $\mathrm {B}$.
The overturning couple is
* ${\frac {\mathrm {MV} (\alpha \cos e+\beta \sin e)}{2}}$
* $f'={\frac {\alpha \cos e+\beta \sin e}{2}}=a\cos(\phi -e)$
$f$
$f'$
$\mathrm {A}$
$\mathrm {B}$
Substituting in Eq. I., as before, we find
| $\mathrm {V} ^{2}=2gl\times {\frac {1-\cos \phi }{\cos ^{2}(\phi -e)}}$ | (XIV.) |
$e=0$
When $\phi +e=90^{\circ }$, wave-path passes through the centre of gravity and axis of overturning, and $\mathrm {V}$ becomes infinite, so that the structure cannot be overturned by any velocity of shock during the first semiphase of the wave. When $e=90^{\circ }$, and $\phi =0$, the wave-path is vertical, and the structure cannot be overthrown in the wave-path by any velocity, but may be conceived lifted, in the second semiphase of the wave, by its own inertia of motion first impressed.
And when $\phi -e=0$, the wave-path is perpendicular to the diagonal, or $f'=a$, and the wave in its second semiphase, produces its maximum effect, that maximum in the first semiphase, of course occurring when the wave is normal.
Proceeding to the consideration of the special problems:—
8th. In the case of a solid cube overturned by subnormal wave.
Preserving the foregoing notation, the structure shall overturn round $\mathrm {A}$ (Fig. 102), in the first semiphase, or round $\mathrm {B}$ in the second semiphase, of the wave.
$\alpha$ being the side of the cube,
* $l={\frac {\sqrt {2}}{3}}\alpha$
$\phi =45^{\circ }$
| $\mathrm {V} ^{2}={\frac {4}{3}}g\alpha \times {\frac {{\sqrt {2}}-1}{\cos ^{2}(45^{\circ }\pm e)}}$ | (XV.) |
9th. In the case of a solid parallelolpiped overturned (subnormnal wave).
Here
* $l={\frac {2}{3}}{\sqrt {\alpha ^{2}+\beta ^{2}}}$
| $\mathrm {V} ^{2}={\frac {4}{3}}g\times {\sqrt {\alpha ^{2}+\beta ^{2}}}\times {\frac {1-\cos \phi }{\cos ^{2}(\phi \pm e)}}$ | (XVI.) |
10th. In the case of a solid right cylinder overturned (subnormal wave)
In this case
* $l={\frac {15\beta ^{2}+16\alpha ^{2}}{24{\sqrt {\alpha ^{2}+\beta ^{2}}}}}$
| $\mathrm {V} ^{2}={\frac {g}{12}}\times {\frac {15\beta ^{2}+16\alpha ^{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}}\times {\frac {1-\cos \phi }{\cos ^{2}(\phi \pm e)}}$ | (XVII.) |
11th. In the case of a hollow parallelopiped overturned (subnormal wave)
Here, from Eq. VIII., XIII., and XIV., we have
| $\mathrm {V} ^{2}={\frac {2g}{3}}\times {\frac {2\beta (\alpha ^{2}+\beta ^{2})+\gamma (2\alpha ^{2}+3\beta ^{2})}{(\beta +\gamma ){\sqrt {\alpha ^{2}+\beta ^{2}}}}}\times {\frac {1-\cos \phi }{\cos ^{2}(\phi \pm e)}}$ | (XVIII.) |
12th. In the case of a hollow right cylinder overturned (subnormal wave).
Here, from what has preceded, we have
| $\mathrm {V} ^{2}={\frac {g}{12}}\times {\frac {15\beta ^{2}+16\alpha ^{2}+3\beta ^{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}}\times {\frac {1-\cos \phi }{\cos ^{2}(\phi \pm e)}}$ | (XIX.) |
| $\mathrm {V} ^{2}={\frac {g}{6}}\times {\frac {9\beta ^{2}+8\alpha ^{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}}\times {\frac {1-\cos \phi }{\cos ^{2}(\phi \pm e)}}$ | (XX.) |
Cap. B.—Bodies or Structures Fractured.
If the fracturing force, or $\mathrm {MV}$, in the direction of the wave-path, $\mathrm {M}$ being the mass broken off, act transversely to the plane of fracture, the case is one simply of cohesion destroyed by an impulsive force, in which $2\mathrm {MV}$ is equal to the statical strain that would have produced the same fracture; and if the direction of the force be such as to produce rotation in the mass fractured off, there will be a dynamic couple to be taken into account; and lastly, if the plane of fracture occur so, that it is not transverse to the line of force, the latter may be resolved into one that shall be so, which is all that need be said as to direct fractures, such as those passing down vertically or diagonally, as fissures through walls, &c.; and the rather because, precious as these become as indices of direction, they should never be adopted as measures of wave-velocity, from the uncertainty that must always attend the knowledge of the coefficient of force necessary to produce fracture through the joints across the beds of masonry, &c.
Proceeding, therefore, to the determination of fracture occurring at the base, or in horizontal planes, or in those of ​the continuous beds of the masonry, or through homogeneous bodies, such as stone shafts of columns, &c.—to none of which the same uncertainty of coefficient applies—
First. Let the wave-path be normal (the force horizontal).
$\mathrm {A}$
$\mathrm {B}$
The condition for its fracture thus, without overthrow, is that the overturning moment, shall be equal to the moment of cohesion of the fractured surface of the base.
$\mathrm {A}$
$\mathrm {B}$
​In accordance with the theory of Leibnitz, we therefore have
* $\mathrm {M} Vf=\mathrm {FA} {\frac {k^{2}}{\beta }};$
$\mathrm {M} =$
$V=$ the velocity of the wave in its path (normal), $f=$ the perpendicular height of the centre of gravity above the base of fracture; $\mathrm {F} =$ the coefficient of dynamic cohesion, or the force upon the unit of surface of the material fractured, which, when suddenly applied, is sufficient to produce fracture; $\mathrm {A} =$ the area of the base of fracture in such units; $k=$ the radius of gyration of the plane of fracture with respect to the axis $\mathrm {A}$ or $\mathrm {B}$. $\beta =$ the width of $\mathrm {AB}$.
If $\mathrm {W} =$ the weight of the mass broken off, $g=$ the velocity due to gravity in one second, then
* $V=g{\frac {\mathrm {FA} }{\mathrm {W} }}\times {\frac {k^{2}}{f\beta }};$
$a$
$f={\frac {a}{2}}$
$\mathrm {L}$
| $V=g\times {\frac {\mathrm {L} }{a}}\times {\frac {K^{2}}{f\beta }}$ | (XXI.) |
| $V=2g\times {\frac {\mathrm {L} k^{2}}{a^{2}\beta }}$ | (XXII.) |
1st. In the case of a solid cube, fractured from its horizontal base.
Here
| $k^{2}={\frac {\alpha ^{2}}{3}}={\frac {\beta ^{2}}{3}};$ | (XXIII.) |
| ${\begin{cases}V={\frac {2}{3}}g\times {\frac {\mathrm {L} }{\alpha }}\\or\\V=21.46{\frac {\mathrm {L} }{\alpha }}\end{cases}}$ | (XXIII.) |
2nd. In the case of a solid parallelopiped, fractured from its horizontal base.
Here
| $k^{2}={\frac {\beta ^{2}}{3}};$ | (XXIII.) |
| ${\begin{cases}V={\frac {2g}{3}}\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\\or\\V=21.46{\frac {\mathrm {L} \beta }{\alpha ^{2}}}\end{cases}}$ | (XXIV.) |
3rd. In the case of a right circular solid cylinder, fractured from its horizontal base.
Here $\beta =D$, the diameter of the cylinder,
* $k^{2}={\frac {5}{16}}\beta ^{2}$;
| ${\begin{cases}V={\frac {5}{8}}g\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\\or\\V=20.12{\frac {\mathrm {LD} }{\alpha ^{2}}}\end{cases}}$ | (XXV.) |
4th. In the case of a hollow parallelopiped, fractured from its horizontal base.
Assuming the thickness of the walls small, as before, and $\gamma$ being that side of the parallelopiped, which is the axis of inceptive rotation,
* $k^{2}={\frac {\beta ^{2}(\beta +2\gamma )}{4(\beta +\gamma )}}$;
| ${\begin{cases}V={\frac {1}{2}}g\times {\frac {\mathrm {L} \beta (\beta +2\gamma )}{\alpha ^{2}(\beta +\gamma )}}\\or\\V=16.10\times {\frac {\mathrm {L} \beta (\beta +2\gamma )}{\alpha ^{2}(\beta +\gamma )}}\end{cases}}$ | (XXVI.) |
5th. In the case of a hollow square prismatic tower, fractured from its base.
Then, $\beta =\gamma$, and (Eq. XXVI.) becomes
| ${\begin{cases}V={\frac {3}{4}}g\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\\or\\V=24.15\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\end{cases}}$ | (XXVII.) |
6th. In the case of hollow right cylinder, fractured from its base.
Here
* $k^{2}={\frac {5\beta ^{2}+\beta '^{2}}{16}}$,
| ${\begin{cases}V={\frac {g}{8}}g\times {\frac {(5\beta ^{2}+\beta '^{2})L}{\beta \alpha ^{2}}}\\or\\V=4.02\times {\frac {(5D^{2}+D'^{2})L}{D\alpha ^{2}}}\end{cases}}$ | (XXVIII.) |
In the 5th and 6th cases, and generally in cases of solid columns or minarets, &c., or hollow prismatic or cylindriical towers, the fracture at the base, is never perfectly, and all through, horizontal. When the breadth or diameter is small, however, in proportion to the height, the irregularity of the fracture is not great, and the slope from horizontal also small; and no serious error is introduced by considering the plane of fracture as horizontal.
Secondly. The wave-path subnormal (force emergent and oblique to the horizon).
When the wave-path is subnormal, in any prismatic structure, the first and second semiphases of the wave, act upon it as already explained; the former to produce inceptive overturn upon the axis (arris of the base) $\mathrm {A}$, and the latter upon the axis $\mathrm {B}$.
Both cases are included in Eq. XXI.
* $V=g\times {\frac {\mathrm {L} }{\alpha }}\times {\frac {k^{2}}{f\beta }}$
$f$
$f$
$f'$
* $f=a\cos {(\phi +e)}$
* $f'=a\cos {(\phi -e)}$
$a$
| $V=g\times {\frac {\mathrm {L} k^{2}}{\beta a\alpha }}\times \sec {(\phi \pm e)}$ | (XXIX.) |
7th. In the case of a solid cube, fractured from its horizontal base, by subnormal wave.
Substituting for $k^{2}$ its value, and also for $a$,
* $k^{2}={\frac {\alpha ^{2}}{3}}={\frac {\beta ^{2}}{3}}$
* $a={\frac {\alpha }{\sqrt {2}}}$
| ${\begin{cases}V={\frac {\sqrt {2}}{3}}g\times {\frac {\mathrm {L} }{\alpha }}\times \sec {(45^{\circ }\pm e)}\\or\\V=15.17\times {\frac {\mathrm {L} }{\alpha }}\times \sec {(45^{\circ }\pm e)}\end{cases}}$ | (XXX.) |
8th. In the case of a solid parallelopiped, fractured from its horizontal base, by subnormal wave.
Here
* $k^{2}={\frac {\beta ^{2}}{3}}$
* $a={\frac {1}{2}}\alpha \sec {\phi }$
| ${\begin{cases}\mathrm {V} ={\frac {2}{3}}g\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\\or\\\mathrm {V} =21.46{\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\end{cases}}$ | (XXXI.) |
9th. In the case of a right circular solid cylinder, fractured from its horizontal base, by subnormal wave.
Here
* $k^{2}={\frac {5}{16}}\beta ^{2}$
* $a={\frac {1}{2}}\alpha \sec {\phi }$
| ${\begin{cases}\mathrm {V} ={\frac {5}{8}}g\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\\or\\\mathrm {V} =20.12{\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\end{cases}}$ | (XXXII.) |
10th. In the case of a hollow parallelopiped, fractured from its horizontal base, by subnormal wave.
Substituting for $k^{2}$ and $a$ their values
* $k^{2}={\frac {\beta ^{2}(\beta +2\gamma )}{4(\beta +\gamma )}}$
* $a={\frac {1}{2}}\alpha \sec {\phi }$
​we have
| ${\begin{cases}\mathrm {V} ={\frac {1}{2}}g\times {\frac {\mathrm {L} \beta (\beta +2\gamma )}{\alpha ^{2}(\beta +\gamma )}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\\or\\\mathrm {V} =16.10\times {\frac {\mathrm {L} \beta (\beta +2\gamma )}{\alpha ^{2}(\beta +\gamma )}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\end{cases}}$ | (XXXIII.) |
11th. In the case of a hollow square prismatic tower, fractured from its horizontal base, by subnormal wave.
As before, $\beta =\gamma$, and
| ${\begin{cases}\mathrm {V} ={\frac {5}{4}}g\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\\or\\\mathrm {V} =24.15\times {\frac {\mathrm {L} \beta }{\alpha ^{2}}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\end{cases}}$ | (XXXIV.) |
12th. In the case of a hollow right cylinder, fractured from its horizontal base, by subnormal wave.
Here
* $k^{2}={\frac {5\beta ^{2}+\beta '^{2}}{16}}$
* $a={\frac {1}{2}}\alpha \sec {\phi }$;
| ${\begin{cases}\mathrm {V} ={\frac {1}{8}}g\times {\frac {\mathrm {L} (5\beta ^{2}+\beta '^{2})}{\alpha ^{2}\beta }}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\\or\\\mathrm {V} =4.02\times {\frac {\mathrm {L} (5D^{2}+D'^{2})}{\alpha ^{2}D}}\times {\frac {\cos {\phi }}{\cos {(\phi \pm e)}}}\end{cases}}$ | (XXXV.) |
Thirdly. Let the structure be fractured, from its horizontal base, and also overturned, whether by a normal or a subnormal wave.
If the structure be observed fractured only, at the base, ​but not overthrown, the velocity impressed was sensibly no more than sufficient for fracture; if it be overthrown also, it was sufficient for both. Hence, if $v_{f}$ = the velocity determined by the Eq. XXI. to XXXV. for fracture only, and $v_{t}$ = that for overturning only, by the Eq. I. to XX., the total velocity of the wave will be found
| $\mathrm {V} =v_{f}+v_{t}$ | (XXXVI.) |
$e$
* ↑ Eq. XVI. has been applied in the text of Part II. under the form $\mathrm {V} ^{2}={\frac {4}{3}}g\times {\frac {(\alpha ^{2}+\beta ^{2})^{\frac {3}{2}}}{(\alpha \cos e\pm \beta \sin e)^{2}}}\times (1-\cos \phi )$
* ↑ It has been remarked that "this involves the assumptions, (1) that the body will begin to revolve as if it were absolutely rigid, and (2) that the force of adhesion, on any element of the plane of fracture, will vary, cæteris paribus, as its distance from the axis $\scriptstyle {\mathrm {A} }$, as if the force were not impulsive, but the mass had extensibility; and it is asked, is there any experimental law which sanctions this conclusion for impulsive as well as continuous forces? If the mass has extensibility in its elements perpendicular to the plane of severance, it must, in like manner, have compressibility; and in such case the mass will not tend to turn round the axis through $\scriptstyle {\mathrm {A} }$, but round some axis parallel to it, on one side of which, there will be compression, and on the other extension. If the compressions and extensions follow the same law, and have the same coefficient, this axis, or neutral line, will divide the base into two equal parts, which would eutirely change the necessary amount of the fracturing force." I should admit the correctness of the conclusion thus expressed, if I could altogether, the premises, and their applicability to the matter in hand may be disposed of in a few words. The actual extensibility of all building materials, and still more their compressibility, are so extremely snmll, that for our present purposes beth may be regarded as in the text, without sensible error, the compressibility for the small intensity of pressures we are dealing with is insensible, and therefore the position of the axis of rotation is practically that assigned to it. If the objector will point out any good ground for adopting any oth axis of rotation, I shall be ready to employ it.