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may be made similar to that form which has the most pleasing or convenient shape, found above as a model.

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Indeed this principle is exceeding fruitful in its practical consequences. It is easy to perceive that it contains the whole theory of the construction of arches: for each stone of an arch may be considered as one of the rafters or beams in the foregoing frames, since the whole is sustained by the mere principle of equilibration, and the method, in its application, will afford some elegant and simple solutions of the most difficult cases of this important problem.

PROBLEM 32.

Of all Hollow Cylinders, whose Lengths and the Diameters of the Inner and Outer Circles continue the same, it is required to show what will be the Position of the Inner Circle when the Cylinder is the Strongest Laterally.

Since the magnitudes of the two circles are constant, the area of the solid space included between their two circumferences, will be the same, whatever be the position of the inner circle, that is, there is the same number of fibres to be broken, and in this respect the strength will be always the same. The strength then can only vary according to the situation of the centre of gravity of the solid part, and this again will depend on the place where the cylinder must first break, or on the manner in which it is fixed.

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sequently when it opens first below, or ends above, as in the 2d figure annexed.

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PROBLLM 33.

To determine the Dimensions of the Strongest Rectangular Beam, that can be cut out of a Given Cylinder.

Let

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(D2 — b2) b

b2;

D2bb3 is a

maximum: in fluxions D2b-3b2b=0=p2 -362, or d2=362; also d2=D2-b2 =3b2 —b2=262. Conseq. b2 d2. :

Da

L

1 : 2 : 3, that is, the squares of the breadth, and of the depth, and of the cylinder's diameter, are to one another respectively as the three numbers 1, 2, 3. Corol. 1. Hence results this easy practical construction: divide the diameter Ac into three equal parts, at the points E, F ; erect the perpendiculars EB, FD; and join the points B, D to the extremities of the diameter: so shall ABCD be the rectangular end of the beam as required. For, because AE, AB, ac are in continued proportion (theor. 87 Geom.), theref. AE: AC:: AB2 in like manner AF: AC:: AD AC2; hence AE AF AC: : AB2

[graphic]

AD2: AC2:: 1:23.

: AC2; and

: :

Corol. 2. The ratios of the three b, d, D, being as the three✓ 1, 2, 3, or as 1, 1:414 1.732, are nearly as the three 5, 7, 8.6, or more nearly as 12, 17, 20·8.

Corol. 3. A square beam cut out of the same cylinder,' would have its side = DD/2. And its solidity would be to that of the strongest beam, as D2 to D2 V 2, or as 3 to 2/2, or as 3 to 2.828; while its strength would be to that of the strongest beam, as (D)3 to D / X÷D2, or as 1 √ 2 to &√3, or as 92 to 8/3, or nearly as 101 to 110.

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Corol. 4. Either of these beams will exert the greatest lateral strength, when the diagonal of its end is placed vertically, by art. 252 Statics.

Corol. 5. The strength of the whole cylinder will be to that of the square beam, when placed with its diagonal vertically, as the area of the circle to that of its inscribed square. For, the centre of the circle will be the centre of gravity of both beams, and is at the distance of the radius from the lowest point in each of them; conseq. their strengths will be as their arrears, by art. 243 Statics.

PROBLEM

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PROBLEM 34.

To determine the Difference in the Strength of a Triangular
Beam, according as it lies with the Edge or with the Flat Side
Upwards.

In the same beam, the area is the same, and therefore the
strength can only vary with the distance of the centre of
gravity from the highest or lowest point; but in a triangle, the
distance of the centre of gravity from an angle, is double of
its distance from the opposite side: therefore the strength of
the beam will be as 2 to 1 with the different sides upwards,
under different circumstances, viz. when the centre of gra-
vity is farthest from the place where fracture ends, by art. 243
Statics, that is, with the angle upwards when the beam is
supported at both ends; but with the side upwards, when it
is supported only at one end, (art. 252 Statics), because in the
former case the beam breaks first below, but the reverse in
the latter case.

PROBLEM 35.

Given the Length and Weight of a Cylinder or Prism, placed
Horizontally with one end firmly fixed, and will just support
a given weight at the other end without breaking; it is requir-
ed to find the Length of a Similar Prism or Cylinder, which,
when supported in like manner at one end, shall just bear
without breaking another given weight at the unsupported end.
Let I denote the length of the given cylinder or prism, d
the diameter or depth of its end, w its weight, and u the
weight hanging at the unsupported end; also let the like
capitals L, D. w, u denote the corresponding particulars of
the other prism or cylinder. Then, the weights of similar
solids of the same matter being as the cubes of their lengths,

L3

as 13: L3, the weight of the prism whose length,
3 :: w,

is L.

]3

Now wl will be the stress on the first beam by its own
weight w acting at its centre of gravity, or at half its length;
and lu the stress of the added weight u at its extremity, their
sum (1+u)l will therefore be the whole stress on the given
beam in like manner the whole stress on the other beam,
3
L
whose weight is w or

L 3

13w, or
w, will be (w+U)L OF ( (273w+v)L.

]8

But the lateral strength of the first beam is to that of the
second, as d3 to D3 (art. 246. Statics), or as 13 to 13
3 and the
strengths and stresses of the two beams must be in the same
ratio, to answer the conditions of the problem: therefore as

(w+u)l

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T

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L3

3

(w+u)l: W+U)L :: 13 : L3; this analogy, turned into

213

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W

7L2 + — 13σ = 0, a cubic equa

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tion from which the numeral value of L may be easily determined, when those of the other letters are known.

Corol. 1. When u vanishes, the equation gives L3 = w+2u1, whence w: w+2u:: 1: L, for the

w+2u

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W

length of the beam, which will but just support its own weight:

Corol. 2. If a beam just only support its own weight, when fixed at one end; then a beam of double its length fixed at both ends, will also just sustain itself: or if the one just break, the other will do the same.

PROBLEM 36.

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Given the Length and Weight of a Cylinder or Prism, fixed Horizontally as in the foregoing problem, and a weight which, when hung at a given point, Breaks the Prism; it is required to determine how much longer the Prism, of equal Diameter or of equal Breadth and Depth, may be extended before it Break, either by its own weight, or by the addition of any other adventitious weight.

Let I denote the length of the given prism, w its weight, and a weight attached to it at the distance d from the fixed end; also let L denote the required length of the other prism, and u the weight attached to it at the distance D. Now the strain occasioned by the weight of the first beam is wl, and that by the weight u at the distance d, is du, their sum wl + du being the whole strain. In like manner wL + Du is

LW

the strain on the second beam; but: L:: W: =w the ī

2013
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weight of this beam, theref. +DU its strain. But the

strength of the beam, which is just sufficient to resist these

WL2

strains, is the same in both cases ; therefore + DU =

21

wl + du, and hence, by reduction, the required length
wl+2du – 2nu.
i = √(1X
L

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Corol. 1. When the lengthened beam just breaks by its

own

own weight, then u O or vanishes, and the required length

becomes L = ✓ (1 × wl+2du).

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Corol. 2. Also when u vanishes, if d becomes =

l, then

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Let AB be a beam moveable about the end ▲, so as to make any angle BAC with the plane of the horizon ac: it is required to determine the position of a prop or supporter DE of a given length, which shall sustain it with the greatest ease in any given position; also to ascertain the angle BAC when the least force which can sustain AB, is greater than the least force in any other position.

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WX

W

A Dn

the force which acting at G in the direction me, is sufficient to sustain the beam; and by the nature

WX

Wx

of the lever, AE : Ag=r:: the requisite force át & : —,

AG

AE

the force capable of supporting it at E in a direction perp. to wx Wx

AB or parallel to me; and again as AF : AE :: : the

AE AF

force or pressure actually sustained by the given prop De in a direction perp. to AF. And this latter force will manifestly be the least possible when the perp. AF upon DE is the greatest possible, whatever the angle BAC may be, which is when the triangle ADE is isosceles, or has the side AD AE, by an obvious corol. from the latter part of prob. 6, Division of Surfaces, vol. 1.

Secondly, for a solution to the latter part of the problem, we have to find when is a maximum: the angles D and E

Wx

AF

being always equal to each other, while they vary in magnitude by the change in the position of AB. Let AF produced meet on in H: then, in the similar triangles ADF, Ahn, it will

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