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LECTURE XXXI.

CONTENTS.-Strength of Beams and Girders-Definitions of Shearing Force and Bending Moment-Beam Fixed at one end and Loaded at the other -Beam Fixed at one end and Loaded Uniformly-Beam Supported at both ends and Loaded in the middle-Example I-Beam Supported at ends and Loaded anywhere Beam Supported at both ends and Loaded Uniformly-Examples II. and III.-Floating Beams-Travelling Loads-Two Loads Moving at a Fixed Distance apart-Example IV. Distributed Travelling Load - Questions.

Strength of Beams and Girders.-The subject under this heading is one that naturally divides itself into two portions. (1) The determination of the resultant effects of the applied loads at any section of a beam or girder; and (2) the nature and amount of the resistance offered by the beam or girder to rupture at that

section.

When the section under consideration is in the same plane as the load, the only effect the load has at that section is a tendency to shear the beam; but in the more general case, where the load acts at a distance from the given section, we have, in addition, a tendency to curve or bend the beam at the section. Hence the name Bending Moment is given to this latter effect.

In the accompanying figure, let A B represent a cantilever

P

B

ILLUSTRATING SHEARING AND BENDING ACTION.

or beam fixed at one end, with a load P applied at the free end; and let C be any section in the beam. At Clet there be applied two equal and opposite forces P1, P2, of the same magnitude as P. The introduction of these forces does not affect the equilibrium of the system, as P, and P, balance each other. Hence, the effect of P at the section C is equivalent to that of a

2

couple PP, with a single force P1. A general proof of this important theorem is given in Vol. I., Lecture III., Prop. II. The couple constitutes the Bending Moment (B.M.), and the single force P1, the Shearing Force (S.F.) at the section C.

DEFINITION.—The Shearing Force at any section of a beam is the algebraic sum of all the forces acting on either side of that section.

DEFINITION.-The Bending Moment at any section is the algebraic sum of the moments of all the forces acting on either side of that section.

Or, in symbols, if P denote any one of the forces acting on one side of a section, and at a distance x, from it; consider all the forces on the same side of the section as P, paying due regard to their sign—that is, if we reckon forces acting upwards as positive, we must regard those acting downwards as negative.

Beam Fixed at one end and Loaded at the other.-Let CD be a cross-section anywhere within the length of the beam at a distance of x inches from the fixed end A. To find the S.F. and B.M. at CD, we observe that the only force acting to the right of the section is W lbs.

Hence :-

It is independent of x, and therefore the same for all such Bections as CD.

The B.M. at CD is W multiplied by its distance from the section in inches. Hence :

B.M.

=

W × BD = W (L - x) inch-lbs.

. . (III)

This equation is true whatever may be the position of W on the beam, so long as L denotes its distance in inches from the fixed end, and CD is between W and the support.

In this case, the diagram of the S.F. is a straight line parallel to the base and at a distance of W lbs. from it. Since (III) is the equation of a straight line, the B.M. is therefore a quantity increasing uniformly from zero, where x L, to W L inch-lbs., where x = 0, as shown by the accompanying figure.

=

DIAGRAM OF S.F. AND B.M. FOR BEAM FIXED AT ONE
END AND LOADED AT THE OTHER.

Beam Fixed at one end, and Loaded Uniformly.-Let the load on the beam be w lbs. per inch-run, it is required to find the shearing force and bending moment at any section CD, at x inches from the fixed end. As before, consider the part of the beam to the right of CD. The only force is the weight of that portion of the load carried by BD, so that :

S.F. = w x BD = w (L x) lbs.

CA

(IV)

BEAM FIXED AT ONE END AND LOADED UNIFORMLY.

The moment of the portion of the load on BD with respect to CD is the same as if it were all concentrated at the middle point of B D. Hence :

B.M. = wx BD BD = w × BD2= w (L-x)2 inch-lbs. (V)

Equations (IV) and (V) show us that both the S.F. and B.M. vanish when x = L; and when x = 0, we get:

The diagrams of S.F. and B.M. for this case take the forms shown in the accompanying figure.

DIAGRAM OF S. F. AND B. M. FOR BEAM FIXED AT ONE END
AND LOADED UNIFORMLY.

Beam Supported at both ends, and Loaded at the Middle.—In this case we measure from the middle point of the beam. Since W is equidistant from A and B, the reactions at those points, R and R2, are equal to each other, and since their sum is W, we have:

BEAM SUPPORTED AT BOTH ENDS AND LOADED AT MIDDLE.

The only force to the right of CD is R, and its leverage is BD.