Let the weight W (fig. 5) be sustained by a beam A B, which rests on two props at C and D.

The pressure on the prop at C is equal to W. BE: A B.
The pressure on the prop at D is equal to W. A E : A B.
The units of strain at E are equal to W. A E. BE: A B.
The units of strain at G are equal to W. A E. B G : A B.
The units of strain at Fare equal to W.B E. A F' : A B.
The greatest strain, which is produced by the weight W, is at E.
The units of strain at the middle of the beam, produced by the
W. AE.

weight Wacting at E, are equal to 2

Let A B = 18 feet, and a weight of 112 lbs. be placed at E, which is 8 feet from A.

Apply these numbers to the above formulæ and their results.

The pressure on the prop at C is equal to

The pressure on the prop at D is equal to

10 × 112
18
8 × 112
18

= 62.5 lbs.

49.8 lbs.

The units of strain at E are equal to

10 × 8 × 112

= 497.77.

= 448.

18

The units of strain on the middle are equal to

8 × 112
2

When the weight W is laid on the middle of the beam A B, the

W. A B

units of strain on the middle are equal to 4

W.AB 8

If the weight W be uniformly distributed along the beam A B, the units of strain on the middle of it will be equal to which is only one half the strain that is produced by the weight having been laid on the middle.

E

Fig. 6.

When the beam A B (fig. 6), supports a weight W, at E, it is equally strong between the points A and B, if the upper sides, AE, BE, be two parabolas whose vertex is A and B respectively. Fig. 7. W

H

Let the weight W have a bearing E F (fig. 7), equal on both sides of the centre G, and also let the weight be equally distributed on the bearing E F

The units of strain at G are equal to

W.AB
4

W.EF
8

Now, if the weight W were a sphere, and were laid on the middle of the beam at G, the units of strain at G would be equal to W.AB

4

If the same weight be formed into a cube, whose side is EF, the units of strain at the centre G will be less than in the case of the W.EF sphere by 8

Let A B be any beam suspended vertically from the point A (fig. 8): and let the sectional area be constant from A to B, where a weight W lbs. is acting to extend the beam.

Put a area of the section of the beam in square inches.

1=

e=

E

e equal to l.

length of the beam in feet before the
weight is applied to elongate it.
the elongation produced by the weight W.
weight which would be necessary to make
The quantity E is called the modulus

Fig. 8.

A

of elasticity of the material of which the beam is composed.

B

W

In the case of the beam being compressed by the weight W acting in the opposite direction,

Put c = compression produced by the weight W.

C= = force which is necessary to make c equal to half of (1).

The quantity C is called the modulus of elasticity of the material, when it is subject to compression.

Units of work done to elongate the beam e feet =

We

2

Units of work done to compress the beam c feet =

W c

2

Mean results of experiments on four different kinds of Cast-iron bars, 10 feet long and 1 square inch in section.

Hence, the breaking weight per square inch of section is 14793 lbs. 6.6 tons nearly; and the ultimate extension is .1859 inches, or 8 of the whole length, 10 feet.

845

If we deduct the set 0209 from 1859, we shall have 165 inches for the elongation produced by the weight 14793 lbs.

.. E modulus of elasticity ==

=

14793 × 10 x 12

=10758545.

.165

.. Breaking weight = 6.6 tons x area of section in square inches. If the weight 5269 be taken, the modulus of elasticity will be considerably increased. Deduct .00175 the set from .05, leaving .04825 inches for the elongation due to the weight 5269 lbs.

This difference in the modulus of elasticity arises from the circumstance of the law of elasticity not being proportional to the weight.

The bar broke with a weight of 24 tons per square inch of section. Hence the tensile force of wrought iron is nearly four times as great as the tensile force of cast iron.

TABLE

Of the Compressive Strength of Wrought Iron.
The Bar was 10 feet long and 1 square inch section.

The crushing force of wrought iron is 12 tons per square inch. It is a curious fact, that cast iron is decreased in length nearly double what wrought iron is, by the same weight; but the wrought iron bar will sink to any degree with little more than 12 tons per square inch, whilst cast iron will bear 43.56 tons to produce the same effect.

A wrought bar will bear a compression of 3 without its utility being destroyed.

Compression of Cast Iron.

of its length,

Mean results of experiments on four different kinds of Cast Iron, 10 feet long, and 1 square inch in section.

The crushing or compressive force of cast iron per square inch is 43.56 tons, which has been obtained from eleven kinds of cast iron. But the tensile force of cast iron is 66 tons; therefore the compressive force is equal to the square of the tensile force, or (6-6)2. Transverse Strength of Beams.

To find the neutral line, forces of extension, forces of compression, moments of extension, and moments of compression of a beam subject to transverse flexure.

Let the form of the section of the beam be that of the figure ABDE, where BC, HE, represent sections of the top and bottom ribs, FG that of the vertical one connecting them, and NO pass through the neutral line.

Put a, a' = N I, N K, respectively.