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3μx2 _ (n + 1) (d ̧2+nd„2)+12ny2

SA,

=

(2y+d1) + (n+2)d2

. . . (653).

If the flanges be exceedingly thin, d, and d, are exceedingly small and may be neglected. The equation will then become that to a parabola whose vertex is at A and its axis vertical. This may therefore be assumed as a near approximation to the true form of the curve AQC.

Where the material is cast iron, it appears by Mr. Hodgkinson's experiments (Art. 411.) that n is to be taken =6.

420. A BEAM OF UNIFORM SECTION IS supported at its exTREMITIES AND LOADED AT ANY POINT BETWEEN THEM: IT IS REQUIRED TO DETERMINE THE CONDITIONS of RUPTURE.

The point of rupture in the case of a uniform section is

WO

evidently (Art. 412.) the point C, from which the load is suspended; representing AB, AC, BC, by a, a,, and a,; and observ

ing that the pressure P1 upon the point B of the beam Wa

= so that the moment of P1, in respect to the section of

a

rupture C=Wa12, we have, by equation (637),

Wa,a,_ SI

a

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If the beam be rectangular, I=1bc3, c1 = 1c,

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where W represents the breaking weight, S the modulus of rupture, a the length, 6 the breadth, c the depth, and a,, a, the distances of the point c from the two extremities, all these dimensions being in inches.

If the load be suspended in the middle, a,a,a,

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If the beam be a solid cylinder, whose radius = c, then I= c, c,c; therefore, equation (654),

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If the beam be a hollow cylinder, whose mean radius is r, and its thickness c, I=cr(r2+1c2), c1=r+}c; therefore, equation (654),

W=z sacr(r2+{c2)
a1a2(r+ c)

(658).

If the section of the beam be that represented in Art. 411., being everywhere of the same dimensions, then, observing, that Ac1=d,A ̧+d ̧Â1, nearly, we have (equations 503 and 654)

W=

Sa A(A,d2+Ad2+A ̧d ̧2)+3(4A, A2+A, A+A2A ̧)d ̧2

6

6

2

(2A,+A3)a,ad,

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where A,, A, represent the areas of the sections of the upper and lower flanges, and A, that of the connecting rib or plate, and d1, da, d, their respective depths.

421. A BEAM IS SUPPORTED

AT ITS EXTREMITIES,

AND

LOADED AT ANY GIVEN POINT BETWEEN THEM; ITS SECTION IS OF A GIVEN GEOMETRICAL FORM, BUT OF VARIABLE DIMENSIONS: IT IS REQUIRED TO DETERMINE THE

LAW OF THIS VARIATION, SO THAT THE STRENGTH OF THE BEAM MAY BE A MAXIMUM.

W representing the breaking load upon the beam, and

B

M

a1, a, the distances of its point of suspension C, from A and B, the pressure P1 upon A is Wa. If, there

represented by a

any

fore (Art. 388.), x represent the horizontal distance of section MQ from the point of support A, and I its moment of inertia, and c, the distance from its centre of gravity to the point where rupture is about to take place (in this case its lowest point); then by equation (637)

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1st. Let the section be rectangular; let its breadth b be constant; and let its depth at the distance x from A be represented by y; therefore I=by3, c=y. Substituting in the above equation and reducing,

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The curve AC is therefore a parabola, whose vertex is at A, and its axis horizontal. In like manner the curve BC is a parabola, whose equation is identical with the above, except that a, is to be substituted in it for a,.

2d. Let the section of the beam be a circle. Representing the radius of a section at distance

from A by y, we have I=y', c=y; therefore by equation (660)

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3d. Let the section of the beam be circular; but let it be hollow, the thickness of its material being every where the same, and represented by c. If y = mean radius of cylinder at distance x from A, then I=лcу(y2+c2), ¢1= (y+c);

Stacy (4y2+c2) . ... (663).

..x= 2 Wa, 2y+c

422. THE BEAM OF GREATEST ABSOLUTE STRENGTH WHEN

LOADED AT A GIVEN

EXTREMITIES.

POINT AND SUPPORTED AT THE

Let the section of the beam be that of greatest strength

(Art. 411.). Substituting in equation (660) the value of

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as before in equation (652), and reducing,

6Wax (A,d,+A„d ̧2+cy3)(A,+A2+cy)+12A‚Â ̧y2+3(A,+A,)cy3

Sa

(y+2d)cy+2(y+d2+‡d1)Å‚+‚d,

(664).

If the section cy of the rib be every where exceedingly small as compared with the sections of the flanges, and if A2=nA1

12Wa2 (n + 1) (d ̧2+nd ̧2)+12ny2 2y+d1+ (n+2)d2

SA1a

x=

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(665).

There is a value of x in this equation for which y becomes impossible. For values less than this, the condition of uniform strength cannot therefore obtain. It is only in respect to those parts of the beam which lie between the values of x (measured from the two points of support) for which y thus becomes impossible, that the condition of greatest strength (Art. 388.) is possible. If its proper value be assigned to n (Art. 411.), this may be assumed as an approximation to the true form of beam of THE GREATEST ABSOLUTE STRENGTH. When the material is cast iron, it appears by the experiments of Mr. Hodgkinson (Art. 411.) that n=6. A, represents in all the above cases the section of the extended flange; in this case, therefore, it represents the section of the lower flange.

The depth CD at the point of suspension may be determined by substituting a, for x in equation (665); its value is thus found to be represented by the formula

Waa2
CD=SA,na

(666).

423. If instead of the depth of the beam being made to vary so as to adapt itself to the condition (Art. 388.) of uniform strength, its breadth b be made thus to vary, the depth c remaining the same; then, assuming the breadth of the upper flange at the distance x from the point of support A to be represented by y, and the section of the lower flange to be n times greater than that of the upper; observing, moreover, that in equation (503) A1=yd1, A‚=nA ̧ =nyd; neglecting also A, as exceedingly small when compared with A, and A,, and writing c for d, we have by reduction,

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Also c, being the distance of the lower surface of the beam from the common centre of gravity of the sections of the two flanges, we have c1(n+1)=c. Eliminating, therefore, the values of I and c, from equation (660),

x=

Sa

Wa, {T's(n+1) (d,2 +nd,;)a1 +ned, } y . . . . (666),

с

the equation to a straight line. Each flange is therefore in this case a quadrilateral figure, whose dimensions are determined from the greatest breadth; this last being known, for the upper flange, by substituting a, for x in the above equation, and solving in respect of y, and for the lower flange from the equation nb,d,=b,d, in which b1, b, represent the greatest breadths of the two flanges, and d1, d, their depths.

424. A BEAM IS LOADED UNIFORMLY THROUGHOUT ITS WHOLE LENGTH, AND SUPPORTED AT ITS EXTREMITIES: IT IS REQUIRED TO DETERMINE, 1. THE CONDITIONS OF ITS RUPTURE WHEN ITS CROSS SECTION IS UNIFORM THROUGHOUT; 2. THE STRONGEST FORM OF BEAM HAVING EVERY WHERE A RECTANGULAR CROSS SECTION; 3. the BEAM OF GREATEST STRENGTH IN REFERENCE BOTH TO THE FORM AND THE VARIATION OF ITS CROSS SECTION.

1. If the section of the beam be uniform, its point of rupture is determined by formula (639) to be its middle point. Representing therefore, in this case, the length of the beam by 2a, the weight on each inch of its length by μ, and its breadth by b; and observing that in this case ΡιΡι= μαμα=ιμα, we have by

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