accents to the symbols h, e, c, t, q, and 8.

the moment of inertia of the whole section

Hence we have for

1. When ct, c1 = st, the section is that of a beam with double unequal flanges; then, neglecting the rivet holes, eq. (1) and (3) become

{ (2e + k)h3 — 2e(h−t)3 + (2e ̧ + k)h ̧3 — 2 ̧ ̧(h ̧− t ̧)3 } (6)

...

where the breadth, ▲ B, of the top flange is (2e+k), and its thickness t, and so on.

2. When e1 =e=c1=c, and t1=t, in eq. (3), the section becomes the same as that of Art. 37.

3. When e,e, or the flanges are of the same length, eq. (5) and (6) also apply to a hollow beam, fig. 10, having the top and bottom plates of unequal thickness, viz., t and t1, and the sum of the thicknesses of the side plates k. And if we further take t1t, the resulting expression for I gives the moment of inertia of a simple hollow beam, as in eq. (1) Art. 33.

61. To find the moment of inertia, I, of the system of rectangular cells, described in Art. 35, about any axis N o parallel to KG the neutral axis of the cells. See fig. 108.

Here the moment of inertia, Io, of these cells about their neutral axis is given in eq. (9) Art. 35; putting x=KN=GO, we have, by eq. (1) Art. 13,

Ix = I + x2 K (1)

...

62. To find, approximately, the moment of inertia, 1', of a system of rectangular cells, A B C D, forming a portion of the section of a beam, about the neutral axis, N o, of the beam, parallel to the edge A B or C D. See fig. 108. Suppose the whole of the vertical plates to be collected in the two side plates A D and в c; and let b = A B, the exterior breadth; b1 a b, the interior breadth; e = A D = B C, the exterior depth; e, a db c, the interior depth; g = K N = Go, the distance between the two axes N O and K G; a = the section of the material; then

=

Hence it appears that the moment of inertia of the system is equal to the area of the section multiplied by the square of the distance of the centre of gravity from the neutral axis.

63. Two equal plates, ▲ B, A, B1, forming a portion of a beam, having the same inclination; to find their moment of inertia about the neutral axis, No, parallel

Fig. 120.

sistance of the rectangle n m whose base is equal to

n

very nearly, which is the same result as that given in the foregoing article.

To find the moment of inertia of a complex girder, such as an iron ship, composed of a series of parts.

..

64. Let ao, ap an represent the sectional areas of the portions composing the transverse section of the beam; a, a,, a the distance of the centres of gravity of these areas respectively from the upper edge A B of the section; Q0, Q1,... Qn the moments of inertia of these areas respectively about their respective centres of gravity; h the distance of the neutral axis of the whole section from the upper edge.

The distance of the centre of gravity of the area a from the neutral axis is (h-a), and so on to the others; hence we have from eq. (1), Art. 13,

where it will be observed that (h-a) is always positive. Hence we have the following general theorem:

General Theorem. I. If the section of a compound girder be composed of a series of sectional areas, the moment of inertia of the whole section, about its neutral axis, is equal to the sum of the moments of inertia of the different sections about their respective centres of gravity added to the sum of all the areas multiplied respectively by the square of the distance of their centres of gravity from the neutral axis.

N.B.-It may happen that the moment of inertia, q,, of some part of the section can be most readily found by referring at once to the neutral axis, N o, of the whole section; then, in this case, the value of Q, must be added to eq. (2).

65. If Q, Q, be taken as the moments of inertia of the sectional areas which meet or pass through the line of the neutral axis, or which may have a considerable depth, as for example the side plates of a tubular girder; and supposing the other sectional areas to be accumulated towards the upper and lower parts of the section of the beam; then the depths of these sectional areas being small as compared with their respective distances from the neutral axis; Q2,..., Qn may be neglected without incurring any considerable error.

n

•*• 1% = Q% + Q1 +Σ ̧a αn ( h − α ̧)2

And if Q, Q, be neglected, then

I。 = Σ" a (h—a1)2
Στη

Io

...

(4)

Hence we have the following general theorem:

General Theorem. II. When the different depths of the sectional areas of a series of plates, forming a compound girder, are small as compared with the distance of their respective centres of gravity from the neutral axis, then the moment of inertia of the whole section is very nearly equal to the sum of all the areas multiplied respectively by the square of the distance of their centres of gravity from the neutral axis.

Special investigations of this theorem are given in Art. 62 and 63, showing the magnitude of the parts neglected.

66. To determine the strength, &c., of a Tubular Beam.A B C D, composed of a series of cells, A F, with angle-iron, at the top part, and of thick solid plates, c D, at the bottom part, the cells being formed as described in fig. 121.

Here the section of the beam may be divided into three portions, viz., the cells at the top, the bottom plates, and the side plates.

For the position of the neutral axis NO

we have by eq. (1), Art. 64,

Fig. 121.

A

B

I

a

α

a

E

where a, a, a, are the areas of the material in the side plates, in the top cells, and in the bottom plates, respectively; and so on.

If DA C, the depth of the beam, then N h=D-h.

From eq. (2), Art. 64, we get

G

C D about its centre of gravity = a2 k22, where k, is the

thickness of the plate.

=

12

the moment inertia of the side plates about their centre of gravity=a d2.

Where do do

the length of the plates, with the relation,

A

a=d+, the depth of the cells ▲ E being d.

2

If the beams be loaded in the middle and supported at the ends, then м =

wl, and

which gives the load corresponding to any assumed value of Conversely s or s, may be determined for any given

s or S1.

load, w.

1. The following is a more simple and practical method of calculating (approximately) the strength of these beams:

Putting a, a, for the areas of the top and bottom parts respectively; g, g, for the distance of their respective centres of gravity from No; and G = g+g,; neglecting the side plates,— for the position of the neutral axis, we have

And from General Theorem II., Art. 65, we get

I。 a g2 + a1 g12 = a g (g + g1) or a1 J1 (g+91)
=ag G or a1 g1 G (5)

...

If K be put for the whole section, and taking =

= a constant, we get

M = c(a + α1) D = C K D ... (7)

Here the strength varies as the product of the top or bottom areas multiplied by the depth.

8gG S9 G h D h1 D

When M =

wl we get

A similar formula may be derived for beams with double

flanges, as described in fig. 119.

From the experiments on the model Conway tube, we find s1 = 16.5 tons; hence we have

which is the formula usually employed for calculating the strength of these beams. Where w is expressed in tons, a, being the area of the bottom plate in inches, D the whole depth of the beam, and the distance between the two supports expressed in the same units as D.