Sidebilder
PDF
ePub
[graphic][subsumed][subsumed][subsumed][merged small][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][merged small][subsumed][subsumed][subsumed][subsumed]
[graphic][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][merged small]

link polygon in fig. 107, add, as a correction, one third of its excess above the sum of the intercepts made by the second link polygon in fig. 108.

The proof of the correction on I。, which is absolute, is shown thus:-For any rectangle

11= }b(Y3 — y3) = }b(Y − y)(Y2 + Yy + y2)

Y Y 2 ·Y 2

= {bh {(Y+ y)2 - Yy} =S { (2 + ~ )2 + + ( 2 + " ) 2 — ! } }

2

2

= S{d2 + }(d2 — Yy)} = Sd2 + }(Sď2 — SYy);

3

where S is the area of the rectangle under consideration. The smaller h becomes, the more nearly does the value of Yy approach d2, and the smaller is the correction. The greater the number of rectangles into which the cross section is divided, the more nearly accurate will be the approximation given by fig. 107 alone.

In the example the correction on I, is about 4 per cent; in the rectangle EBB'E', fig. 102, the areas of the two halves each into the square of its leverage about 00 is less than I by one-twelfth, or about 8 per cent.

Any cross section can be blocked out into rectangles, and I and G easily calculated for it by the tabular method, if we have a table of squares and cubes. If the cross section has a very irregular outline, it may require to be blocked into a great many rectangles, and the construction, fig. 107, will probably give a sufficiently close approximation.

DEFINITION. A cross section for which the neutral axis divides the depth in the same ratio as the strengths of a given material to resist tension and thrust is called a Cross Section of Uniform Strength for that material.

The cross section, fig. 106 for instance, would be of uniform strength for a beam of a material whose resistance to tension and thrust is as 3 to 2, AA being the compressed skin and BB the stretched skin; for a cantilever of the same material it would be turned upside down.

Triangular cross section and sections which can be divided into triangles.

Triangular Section.-On making ƒ=h, the wedges,

figs. 89 and 90, become isosceles, and we have by substituting in the expression given thereat

G = vol. of wedge on triangular portion above N.A. = sibh2;

[merged small][ocr errors]

In the following way, I may be derived from the rectangle, fig. 109. The moment of inertia of the shaded triangle about the central axis CC is half that of the rectangle; for the triangle, then

[graphic][merged small][merged small][merged small][merged small][merged small][merged small][merged small][subsumed][merged small][subsumed][merged small][merged small][merged small][merged small][ocr errors][subsumed][merged small][merged small]

Hexagonal Section.-As an example of a built figure, we will take a hexagon about a diameter as axis. Fig. 110.

Taking a quadrant, we can divide it into three equal triangles; the breadth of each is 1b, and the height h; hence the area of each is bh, and the moment of inertia of each about its own neutral axis is xbx h3 = 112bh3. If I be the moment of inertia of the hexagon about 00, we have for a quadrant

朽。

= moment of each of the three triangles about its own neutral axis, together with the area of each into the

square of the distance of its centre from 00,

= 3x

bh

2

2

2

1152 + 16 { ( k )2 + (b)2 + (b) * } = sixbh3.

:. I = 5bh3.

Or, taking another quadrant divided into a rectangle and a

triangle

-3 b

bh

2

41. = '(4)(h)2 + (4) for rectangle

[ocr errors]
[merged small][merged small][ocr errors][merged small][merged small]

8

[blocks in formation]

(4)(h)2+(k) for triangle

16 6

[ocr errors]

Dividing the semi-hexagon above the axis into any set of convenient figures, and multiplying the area of each by the deviation of its centre of gravity, we have

[blocks in formation]

Hence for a hexagonal section about a diameter as axis,

m2 = 1, n' = 5805208, n=10417.

[graphic][ocr errors][merged small][merged small][merged small]

Rhomboidal Section.-As another example, we will take a section in the form of a rhombus, with diagonal as neutral axis. Fig. 111.

« ForrigeFortsett »