the factors is constant, this occurs when they are equal; thus

That is, the apex A, lies to the left of O at a distance s, one quarter of the distance between the two weights. To find the height of A1, put x=s, and

the maximum of maxima for first half of span.

2

(4.)

It is evident that A, will lie to the right at a distance s, and that the two parabolas will intersect at D on the vertical through the centre. It is convenient to call the first half of the span field 1, and to say that this field is governed by W,; and we observe that the maximum in field 1 occurs when W1, being in its own field, lies as far to one side of O the centre, as G lies to the other.

If it be possible for W, to occupy every point in its field without W, going off the span, we say that W, can overtake its field. In the present problem it is necessary that 48, the distance between the weights, be not greater than e the half-span, in order that each weight may be able to overtake its field. The problem divides into two cases.

Case I.-4s = or <c.

On fig. 45, 4s = c; and it is evident that one weight R may be at any point of the half-span, while the other weight is not on the span. Hence the locus of the maximum bending moment at each point when only one weight is on the span is the parabola BEC, due to a rolling load R as in fig. 42.

2

The apex E of this parabola coincides with D, the intersection of the pair of parabolas. This may be seen thus:Shift the load until W is over the centre; then, since W, is over B the extremity of the span, we may either consider it not yet on the span when OE is the bending moment, or we may consider that W, is just on the span when OD is the bending moment. Also note that OE=10A, as BEC is the parabola due to the rolling load R, while BAC is the parabola which would be due to a rolling load R.

1

Ο

The apex A, is higher than the apex E; the two parabolic arcs BA, and BE intersect at B, and every point on the arc BAD is outside of BEC. Hence the locus BA,DA,C is everywhere outside of BEC, and gives the maximum bending moment for each point of span.

2

Again, on fig. 46, 48 c; that is, A, and A, are closer together than on fig. 45, hence D is higher than E and BA,DA,C is again outside of BEC. BF and HC, parts of the parabola BEC, are shown by heavy dotted lines, and they indicate the bending moments, when one weight only is on the span.

Graphical Solution; 48 c.—Figs. 45 and 46. With a scale of feet lay off BC equal to the span, and upon each

=

side of the centre lay off OS, OS, 8; apply the rollers to BC, place any parabolic segment against the rollers with its apex on the vertical through S,, shift the rollers till the curved edge passes through B and draw BA,D; similarly draw DAC. Then BA,DA,C is the diagram of maximum bending moment at each point, the common height of A, and A, being maximum for the whole span.

2

Place the segment with its apex on the vertical through the centre, and move the rollers till the curved edge passes through B and C; mark 4, and construct a scale of ft.-lbs. for verticals such that OA,= ᎪᎡᏓ .

Case II.-4s c; fig. 47.

In this case the apexes A, and A, are farther apart than on fig. 45, hence D is below E; hEf, the central portion of the parabola BEC due to only one weight on the span, lies outside of BA,DA,C and gives the maximum at each point for that portion. It will be seen that

2

OH OF 4s - c;

1

for, suppose W, at F, then W, is at B, and Ff representing the bending moment at F is the ordinate of BEC or of DA,C, according as we consider that W, is not yet on, or is just on, the span. In this case, W, cannot overtake the portion HO, neither can W, overtake the portion OF, of their respective fields. There are two equal maxima at A, and

W

-(48>C)·

S-S

2

W2 each=

A,, and a third maximum at E; and the greatest of these is the maximum for the whole span. The point E may thus be the same height as, or higher or lower than, A ̧.

Suppose that E is of the same height as A,, then A will be twice as high as A,; and we will have

OB: SB :: √2:1; that is c (c-s): √2:1

or

Fig.47.

48:2c:: (2-2): 1; hence 48 (2-2).2c,

or the distance between the weights (22) times the

span.

Thus if the distance between the weights equals or exceeds the quantity (2-2).2c, or about ths of the span, the maximum bending moment for the whole span is at the

centre.

12

Graphical Solution; 48 c.-Fig. 47. Lay off BC equal to the span; on each side of the centre lay off OS, OS=8, and OH=0F=4s - c; draw verticals through S1, H, 0, F, S1; apply the rollers to BC, place any parabolic

segment against the rollers with its apex on the vertical through S, shift the rollers till the curved edge passes through B, and draw BA,h, stopping at the vertical through H; similarly draw CA,f. Shift the segment till the apex is on the vertical through the centre, move the rollers till the curved edge passes through B and C, and draw BAC. Bisect OA in E; and plot as many points in hEf as may be necessary, by bisecting the ordinates of h'Af', or in any manner construct the parabola BEC. Then BAhEƒAС is the diagram of maximum bending moment at each point. Construct a scale for verticals such that OA R.1. The complete interpretation of the locus BA,DA,C_is shown in fig. 48. Suppose the beam to extend beyond the

0

=

supports at B and C, and to be fixed at these supports so that P and Q may act upwards or downwards. In the figure the travelling load is standing with G over B, so that PR, and Q is zero; hence at L, the point under W, the bending moment is zero, and this is the point at which CAD meets BC. Let the load move towards the left until W, is over any point as K; Q now acts downwards; at the point K, the beam is bent upwards and the bending moment

2

1

2

is negative; the value of this negative moment is Q.KC, and it is given by the downward ordinate Kb. When W, arrives over B, the bending moment at B is negative; and its value, Q.BC W,.48, is given by the ordinate Bd.

2

As the load moves farther to the left, the bending moment at each point, as W, comes over it, is of the constant value Bd; BC is now a cantilever under the downward load Q, and the bending moment at each point of BC is now negative, increasing indefinitely as the load moves towards the left. For all positions of the load, with no restriction on the value of 48, BA,DA,C gives the maximum positive bending moment at each point; and the height of à ̧ is the greatest positive bending moment that can possibly be produced by the load system.

2

Beam under a travelling load system of two unequal weights at a fixed interval apart.-Fig. 49. Let R be the total load; W1 and W, the weights numbered from the left end; let G their centre of gravity be the origin for loads, 2h, and — 2h2 being the abscissæ of W, and W,, so that the distance between the weights is 2h, + 2h. Let W, be over any point of the span whose abscissa is a measured (positive to left) from the

1

1

centre as origin, and let it be understood all through that the whole load is on the span.

As in the previous case,

is the equation to the bending moment at any point x when

1

W1 is over it.