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34. PROP. A beam A B (fig. 8) is supported and loaded as in the preceding proposition; to find the moment of the strain about any point D of the beam.

Α

Р

C

Let Q and P be the pressures produced by W at the extremities B and A of the beam; then the moment of A the strain on the beam at D is either the moment of Q taken with respect to D, or the difference of the moments of P and W taken in reference to the same point D. Taking moments about the point 4, we have

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W

Fig. 8.

AC

AB

B

D

Hence the moment of the straining force on D is the moment of Q taken with respect to D, viz.:

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The same resulting expression is obtained for the straining force when P is used instead of Q. For taking moments about B, we have

P. AB W. BC, or P= W.

=

BC

AB

Wherefore the moment of the strain on D in this case is

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AD. BC-DC. AB (AC+ CD) BC - (AC+ CB) DC

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Obs. We see from this that the moment of the strain on the point D is more readily found from the pressure at B than from that at A, because in the latter case the moments both of P and Ware involved in the moment of the strain.

In a similar way the moment of the strain on any point of the beam might be obtained for any number of straining forces.

A

D

E

Fig. 9.

In this case the

B

35. PROP. A beam AB (fig. 9) is supported as in the preceding proposition; to find the moment of the strain at any point C when the straining weight is uniformly distributed along the beam.

beam is strained on each side of the point. C. Let w be the weight of the straining load distributed along A C, and w' that along BC; then if D and E be the middle points of AC and BC, w may be supposed to act at D, and w' at E. Wherefore the moment of the strain on C produced by w is by the preceding proposition

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and the moment of the strain on the same point produced by w' is

w' AC. BC

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Whence the moment of the strain produced on C by the whole straining load is

M =

AC. BC
2. AB

(w + w')

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W being the whole weight distributed uniformly along the beam.

=

Cor. 1. If C be the middle of A B, then because AC BC, AC. BC AC2 = A B2, and therefore in this case

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When a uniform beam, therefore, is strained by its own weight, the moment of the strain is

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W being the weight, and 7 the length of the beam.

Cor. 2. The strain on any point of a beam by a weight uniformly distributed is the same as that of half the weight suspended from that point.

Obs. In some treatises on the strength of the materials, in finding the moment of the strain on C from a weight uniformly distributed as in this proposition, we are told that the moment of the strain is the same as if the weight were collected at its centre of gravity, the middle of AB. But it is not difficult to see the difference between this strain and the actual strain on C. It is clear that the pressure of gravity downwards on any point between A and C has no tendency to diminish the strain at C arising from the upward reaction at B; whereas the pressure of gravity between C and B is almost in direct opposition to this strain and must diminish it.

36. PROP. A beam AB of length l is supported as in the preceding proposition; to find the moment of the strain on the beam produced by any number of straining loads.

The effective strain on the beam is evidently at the point of application G of the resultant of all the straining forces.

Let P1, P2, P3, &c. be the weights, and a1, a2, as, &c. their distances from A; then if R be the resultant of these forces, and h its distance from A, or the distance of G from A, we have by parallel forces and the theory of moments,

and

h =

R = P1 + P2 + P3 + . . . .

P11 + P2α2+ P3 α3 +....
P+ P+ P+....

Hence (Art. 34), the moment of the strain on G is

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If the beam is also strained by a weight W (including its own weight) uniformly distributed along the beam; then R' being the resultant of R and W, h' the distance of R' from A, we have

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A

37. BEAMS INCLINED TO THE HORIZON.

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P

B

1

Let P (fig. 10) be a weight acting at the extremity B of a beam AB fixed at the other extremity A, and inclined at an angle a to the horizon. The weight P may be decomposed into two components, the one P1 = P cos a, perpendicular to the direction of the beam, and which will tend to bend it, the other, P2 = P sin a, directed towards the point of fracture, and which will tend to compress it in the direction of its length

Fig. 10.

The bending moment will be found by writing P cos a for W in Article 32.

In a similar way would the bending moment be found for any other mode of fixture of the beam.

The total compression produced along the beam by P1 and P, is obtained in the following manner :

The tension per unit of surface produced on a cross section of the beam by P2 is, by Art. 6, equal to

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A being the area of the section of the beam.

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Also the tension per unit of surface due to the force of flexure P is (Art. 22)

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M being the moment of the strain about the neutral axis of the section at A, and x and I the same as in (2), Art. 22.

Hence the whole tension per unit of surface is

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CHAPTER VI.

APPLICATIONS.

Section 1.

FORMULE OF SOLUTION AND SECTION OF FRACTURE.

38. LET W be the total load and 7 the unsupported length of a beam, I the moment of inertia of one of its cross sections about the neutral axis of that section, x the distance of the most extended or the most compressed fibre from the neutral axis, and S the tension per unit of area of the section.

Then by Chap. V. the moment of the exterior forces (M) may be represented by the expression

m Wl,

m being a numerical coefficient depending upon the disposition of the load.

And by Art. 23, the moment of resistance of the beam is

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which is the general formula of solution.

The following are some of the values of m; others are readily deduced from the propositions in Chap. V.

Values of m

(1.) Beam fixed at one end and loaded at the

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(2.) Beam fixed at one end and uniformly loaded

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H

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