This is the point where the greatest bending takes place. The line of invariable fibres in this case consists of three curved lines, DC, BC, and AD, which run into one another. The equations of these curves are readily found.

We may suppose the beam to be fixed at D, and instead of the points of support, to be acted upon by the reactions P and Q. Let D be the origin of rectangular coordinates, the horizontal line through D being the axis of x. Then the moment of the exterior forces on a point (x, y) between D and C is,

M = P(b - x) - W (b − a − x) — § w (b − x)2 . . . . (1);

on a point between B and C the moment is,

M = P(b − x) — w (b − x)2.

(2);

(3).

Proceeding with these, as in the preceding propositions, we can obtain the equation of each curve.

In like manner the

method is extended to any number of straining forces.

Examples.

1. A rectangular beam of oak, 10 feet long, 10 inches broad, and 12 inches deep, rests in a horizontal position on two supports at its extremities; find how much it will be deflected from the horizontal line by suspending a weight of 35,000 lbs. from its middle, the modulus of elasticity of the oak being taken at 1,492,000.

The deflection is found from the equation (Art. 60)

in which the dimensions of the beam are to be taken in inches,

and the weight in pounds avoirdupois.

Now W = 35,000, 7 = 10 × 12 =

b = 10, d = 12; hence we get by substitution in the preceding

formula,

8 =

586 of an inch,

the required deflection.

2. In the last example find the radius of curvature at the lowest point.

If ρ be the radius of curvature at any point (x, y) of the elastic curve which contains the centres of gravity of the cross sections of the beam after deflection; then by (1) Art. 22,

But the greatest deflection in the present case takes place at the centre of the beam, and hence (Art. 33),

the radius of curvature at the lowest point of the elastic curve.

3. Determine the deflection in Ex. 1, when the weight of the beam is taken into account, a cubic foot of the oak weighing 60 lbs.

Let us first find the weight of the beam W'.

5

Now W' = 10 ×

6

× 1 × 60 = 500 lbs.

Then if & be the whole deflection, we have by (4), Art. 60,

the values of 1, W, E, and I being the same as in Ex. 1. Hence we readily get

=

. 595 of an inch,

which is the total deflection when the weight of the beam is included.

4. A bar of elm. 10 inches square, projects 10 feet from a building in which it is firmly fixed; find the greatest amount of deflection which it would sustain by suspending a weight of of a ton from the free extremity, the modulus of elasticity being taken at 700,000. Ans. 165 inches.

5. Find the deflection in the last example when the weight of the beam is considered, the weight of a cubic foot of the elm being 34 lbs. Ans. 1.746 inches.

6. In Ex. 4 find the deflection of the bar 8 feet from the fixed end. Ans. 1.16 inches.

7. A wall of brick-work which is 20 feet high, 14 inches in breadth, and weighs 112 lbs. per cubic foot, is supported by à beam of oak resting in a horizontal position on two walls 22 feet asunder. Find the deflection of the beam, its breadth being 5 inches, depth 12 inches, weight per cubic foot 62 lbs., and the modulus of elasticity 1,200,000. Ans. 16.1 inches.

8. In Ex. 7, determine the least radius of curvature of the elastic line of the bent beam. Ans. 450 inches.

9. Determine the corresponding deflection of the beam to (9), Art. 63, when the weight is uniformly distributed along the beam.

a3 w

48 EI w being the weight of a unit of length.

CHAPTER IX.

RESISTANCE TO TORSION.

65. A PRISM experiences a simple torsion when a cross section of the prism turns, relatively to a section indefinitely near to it, round the axis of the prism. The prism is consequently subjected to exterior forces which tend to twist it about its axis.

The angle of torsion is that which two lines originally parallel and passing respectively through the centres of the two cross sections of the prism indefinitely near to each other make between them after the distortion of the prism.

66. To find the resistance to torsion of a homogeneous prism.

Let us consider two cross sections of the prism near to each other, and the elementary fibres included between these sections. The exterior twisting forces will cause one of these sections to rotate, relatively to the other, about the axis of the prism. Wherefore one of these sections being regarded as fixed, a point in the other section will undergo an angular displacement by torsion, equal to the arc r'ß, r' being the distance of this point from the axis of the prism, and ẞ the circular measure of the angle of torsion. If we take the ratio of this displacement to the distance between the two sections, or the length of the fibre of which the point under consideration may be considered as one extremity, we shall have for the displacement of this fibre referred to the distance between the two sections, the expression

Now it is found by experiment (see Morin's Résistance des Matériaux, p. 455, edition of 1857), that the resistance t of a fibre to torsion is proportional to the expression (1) and the area of the fibre. Calling a then the area of a fibre at a distance 'from the axis of the prism, and E' a constant to be determined by experiment, we have as in Art. 29,

The distance between the two sections of the prism being the same for all the fibres of the section in which a is the area of one, the greatest displacement, and consequently the greatest resistance, is the value of (2) for the fibre the most removed from the axis of the prism, or for the greatest value of r', which we shall denote by r.

The moment of the resistance (2) about the axis of the prism is

and the sum of all the similar moments for the whole section is

which has been called the moment of polar inertia, is evidently the moment of inertia of a cross section of the prism about an axis through its centre of gravity and perpendicular to its plane. Denoting it by I, the sum of all the moments of the molecular resistances to torsion for the cross section of the solid is

Now in order that equilibrium may exist between the exterior forces which tend to produce torsion and the molecular re