8. Determine the moment of the strain in Ex. 6 about a point 1 foot from A, the other conditions being the same as in that example. Ans. 333.

9. Find the moment of the strain in Ex. 6, when the weight of the uniform beam is 30 lbs., the other conditions the same. Ans. 677.

Obs. The actual strain in this example is not on the point from which the weight is suspended, but on the section of the beam corresponding to the point of application of the resultant of this weight and the weight of the beam, as in example 4.

10. A beam four feet in length is fixed at its extremities in two walls (the points of support being in the same horizontal line), and is loaded with a weight, 1 foot from one extremity; find the ratio of the moment of the strain about the middle of the beam, to that at the point where the weight is attached.

Ans. .

11. A beam AB rests on two supports A and B in the same horizontal line, and is loaded at two points C and D between A and B with weights W and W', such that W · AC = W'. BD; prove that the moments of the strain about any two points between C and D are equal to one another.

12. Find the depth of a square beam of elm, supported on two props 4 feet asunder in the same horizontal line, in order that it may securely sustain a weight of 600 lbs. at its centre, the coefficient of safety being taken one-third of that of resistance, which is 6,180. Ans. 2.757 inches.

13. A beam of Canadian oak, 18 inches deep and 14 inches broad, is fixed in two walls 12 feet asunder, resting on its broad face; find how much more weight it would sustain from its centre if placed on its narrow face, the coefficient of resistance being 10,560. Ans. 73,920 lbs.

14. A cast-iron beam 3 feet long and 1 inch in breadth is fixed at one end; find its depth in order that it may sustain

a weight of 1,100 lbs. at the other end, the coefficient of resistance of the iron being taken at 39,600. Ans. 24 inches.

15. A rectangular beam of wood which is 12 feet long, 8 inches broad, 5 inches deep, and which weighs 30 lbs. per cubic foot, is fixed at one end in a horizontal position. If its coefficient of resistance to fracture is 7,100, find what load it can bear,

(1.) When the load is suspended at the free end of the beam; (2.) When it is equally distributed along the beam.

Ans. 1,593-5 lbs. and 3,187 lbs.

16. A prismatic body of oak, 6 feet long, and having a rectangular section 6 inches broad, is fixed at one end in a horizontal position. The body is loaded with 5 cwt. and its coefficient of resistance to fracture is 12,000; find its depth in order that it may just sustain this load,

(1.) When the load is suspended from its extremity;
(2.) When it is equally distributed along the prism.

Ans. 183 of an inch; 13 of an inch.

17. A beam 12 feet long and having a square section, is fixed at both ends in a horizontal position. If its coefficient of resistance to flexure is 7,100, find the dimensions of the section in order that the beam may just sustain a load of 15,000 lbs. equally distributed along its length,

(1.) When the beam rests on one of its faces;
(2.) When it rests on one edge.

Ans. 611 inches, and 6·859 inches.

18. A cast-iron beam, 6 inches square, 20 feet long, is supported at the ends in a horizontal position with one of its diagonals vertical; if the coefficient of resistance to fracture is 17,000, find what weight will break the beam when applied at a point 8 feet from one extremity. Ans. 7,513 lbs.

19. A corn granary 14 feet in breadth rests upon 5 beams of

oak, each beam being 20 feet long, 10 inches broad, and 12 inches deep. Find the weight of an equally distributed cornheap which the beams can bear with safety, a cubic foot of corn being taken to weigh 50 lbs., and the coefficient of safety of the oak at of that of absolute strength, which is 12,000. (Each beam rests on two supports at its extremities.)

Ans. They can support a corn-heap 51 inches high.

20. Find what weight will break the beam in Ex. 18 when the beam's own weight (viz. 434 lbs. per cubic foot) is taken into account. Ans. 6,636 lbs.

21. Four rectangular beams of oak, the breadth of each being to its depth as 1 to 3, are fixed at one end in a horizontal position, and are to serve as supports to a balcony, which is to be 4 feet in length (measured in the direction of each beam) and 6 feet in breadth. Granite stones 5 inches in thickness and 170 lbs. per cubic foot in weight are to form the foundation of the balcony. Find the dimensions of each beam in order that it may support with safety a chance weight of 7 cwt. (independently of the weight of the balcony) at its extreme end, the co-efficient of safety the same as in Ex. 19.

Ans. Depth 8.3 inches.

=

Section 3.

BEAMS WITH CIRCULAR SECTIONS.

45. PROP. To find the moment of inertia (I) of a circle about a diameter.

The moment of inertia of a surface about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two perpendicular axes in its plane, drawn from the

point where the former axis meets the plane. (Art. 25.) It follows from this that the moment of inertia of a circle about an axis through its centre and perpendicular to its plane is equal to the sum of its moments of inertia about any two diameters at right angles to one another. But as the moment of inertia of a circle about every diameter must be the same, the moment of inertia about a diameter is half its moment of inertia about an axis passing through its centre and perpendicular to its plane.

Now if x be the distance of an elementary annulus from the centre of a circle, the area of this element is

and its moment of inertia with respect to a perpendicular axis to the plane of the circle through its centre is

Hence, by what precedes, the moment of inertia of the circle about a diameter is

Wherefore if this diameter be taken for the neutral axis in reference to a cylindrical beam, we have

BEAMS WITH HOLLOW CIRCULAR SECTIONS.

46. The value of I when the section is hollow is readily deduced from (1) of last Article. Let b be the interior radius of the hollow circular section, and a the exterior radius; then integrating (1) between the limits x = b to x = a, we get for I, the moment of inertia in this case,

Obs. When the circular section of the beam is the same throughout its whole length, the method of solution is the same as in Section 2, the modified formula of this section being employed instead of those of the preceding section. The following is the solution of an example in which the section varies throughout the length of the body.

1. Let the frustum of a cone be fixed in a horizontal position with one end in a vertical wall, and be strained at the other end by a load, to find at what part of the frustum fracture will

ensue.