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114 22 570 40 224 58 106         60 99

 

In tensional strains, the length of the beam does not affect the strength; but in the beams submitted to compression, the length is a most important element, and in the table given above, the safety strains to which beams may be subjected, without crushing or bending, has been given for lengths, varying from 6 to 60 diameters.

 

PRACTICAL RULES.

Tensional Strain.

Let T = whole tensional strain.
  "   S = strength per square inch.
  "   a = sectional area in inches.
Then we have T = Sa.

Now to find the necessary sectional area for resisting any strain, we have the following general formula:

a = T ÷ S

or, by substituting the working strenths for the various materials in the formula, we have for wood,

a = T ÷ 2000

Wrought Iron,

a = T ÷ 1500

Cast Iron,

a = T ÷ 4500

But, in practice, cast iron is seldom used except to resist compression.

 

Strains of Compression. Allowing the same letters to denote the same things as above, we have for

Wood,

a = T ÷1000

Wrought Iron,

a = T ÷ 12000

Cast Iron,

a = T ÷ 25000

As this pamphlet has to do with wooden bridges only, nothing will be said of the proper relative dimensions of cast-iron columns to sustain the strains to which they may be subjected, but a table of the strength of columns will be found further on.

 

Transverse Strains.

Let W = breaking weight in lbs.
  "   s = constant in table.
  "   b = breadth in inches.
  "   d = depth in inches.
  "   L = length in inches.

Then, for the power of a beam to resist a transverse strain, we shall have,

This formula has been derived from experiments made by the most reliable authorities.

The constant, 1250, adopted for wood in the following formula, is an average constant, derived from the table, of those woods more commonly used.

Now to reduce the formula to the most convenient shape for use, we substitute the value of s, and we have

or

But, to reduce the load to the proper working strain, we must divide this equivalent by 4, the factor of safety, and we shall have

.

Let us apply the formula—

Case I. Given a span of 14 feet,
a breadth of 8 inches,
a depth of 14 inches.

Required the safe load.

The formula becomes, by substitution,

= 11,666 lbs.

 

Case II. Given the safety load 18000 lbs.,
the breadth 9 inches,
the length 14 feet.

Required the depth.

From the above formula we have

substituting

= 16 inches nearly.

 

Case III. Given the safety load 22,400 lbs.,
the depth 18 inches,
the length 14 feet.

Required the breadth.

Deriving b from the foregoing, we have,

substituting

= 9.3 inches nearly.

For a cast iron beam or girder—Mr. Hodgkinson from numerous carefully conducted experiments that, by arranging the material in the form of an inverted T—thus creating a small top flange as well as the larger bottom one, the resistance was increased, per unit of section, over that of a rectangular beam, in the ratio of 40 to 23.

In this beam the areas of the top and bottom flanges are inversely proportional to the power of the iron to resist compression and extension. Mr. Hodgkinson's formula for the dimensions of his girder, is

The factor of safety being 6 for cast iron beams—the formula for the working load will be,

and, to find area of lower flange, we shall have

The general proportions of his girders are as follows:

Length, 16
Height, 1
Area Top Flange, 1.0
Area Bottom Flange, 6.1
In the above formula for cast iron beams,
  W = weight in tons.
  a = area in square inches of bottom flange.
  d = depth in inches.
  h = length in inches.

The web uniting the two flanges must be made solid—as any opening, by causing irregularity in cooling, would seriously affect the strength of the beam.

Example.—Required the dimensions of a Hodgkinson girder—for a span of 60 feet—with a load of 10 tons in the centre.

= 37 inches nearly.

and the area of the top flange will be,