145. Frame of Two Bars-Equilibrium.-PROBLEM. Figures 70, 71, and 72 represent three cases in which a frame consisting of two

bars, jointed to each other at the point L, is loaded at that point with a given force, P, and is supported by the connection of the bars at their farther extremities, S1, S2, with fixed bodies. It is required to find the stress on each bar, and the supporting forces at S, and S.

Resolve the load P (as in Article 55) into two components, R1, R2, acting along the respective lines of resistance of the two bars. Those components are the loads borne by the two bars respectively; to which loads the supporting forces at S1, S2, are equal and directly opposed.-Q. E. I.

The symbolical expression of this solution is as follows:-let i, i, be the respective angles made by the lines of resistance of the barg with the line of action of the load; then

PR, R2 sin (i + i); sin i sin .

:

:

The inward or outward direction of the forces acting along each bar indicates that the stress is a thrust or a pull, and the bar a strut or a tie, as the case may be. Fig. 70 represents the case of two ties; fig. 71 that of two struts (such as a pair of rafters abutting against two walls); fig. 72 that of a strut, L S1, and a tie, LS, (such as the gib and the tie-rod of a crane).

-

146. Frame of Two Bars-Stability.- A frame of two bars is stable as regards deviations in the plane of its lines of resistance. With respect to lateral deviations of angular position, in a direction perpendicular to that plane, a frame of two ties is stable; so also is a frame consisting of a strut and a tie, when the direction of the load inclines from the line S, S., joining the points of support.

A frame consisting of a strut and a tie, when the direction of the load inclines towards the line S, S,, and a frame of two struts in all cases, are unstable laterally, unless provided with lateral stays.

These principles are true of any pair of adjacent bars whose farther centres of resistance are fixed; whether forming a frame by themselves, or a part of a more complex frame.

147. Treatment of Distributed Loads.-Before applying the principles of Article 145, or those of the following Articles, to frames in which the load, whether external or arising from the weight of

the bars, is distributed over their length, it is necessary to reduce that distributed load to an equivalent load, or series of loads, applied at the centres of resistance. The steps in this process are as follows:I. Find the resultant load on each single bar.

II. Resolve that load, as in Article 141, into two parallel components acting through the centres of resistance at the two ends of the bar.

III. At each centre of resistance where two bars meet, combine the component loads due to the loads on the two bars into one resultant, which is to be considered as the total load acting through that centre of resistance.

IV. When a centre of resistance is also a point of support, the component load acting through it, as found by step II. of the process, is to be left out of consideration until the supporting force required by the system of loads at the other joints has been determined; with this supporting force is to be compounded a force equal and opposite to the component load acting directly through the point of support, and the resultant will be the total supporting force.

In the following Articles of this section, all the frames will be supposed to be loaded only at those centres of resistance which are not points of support; and therefore, in those cases in which components of the load act directly through the points of support also, forces equal and opposite to such components must be combined with the supporting forces as determined in the following Articles, in order to complete the solution.

3

C

B

148. Triangular Frame.-Let fig. 73 represent a triangular frame, consisting of the three bars A, B, C, connected at the three joints 1, 2, 3, viz.: C and A at 1, A and B at 2, B and C at 3. Let a load P, be applied at the joint 1 in any given direction; let supporting forces, P2, P3, be applied at the joints 2, 3; the lines of action of those two forces must be in the same plane with that of P1, and must either be parallel to it or intersect it in one point. The latter case is taken first, because its solution comprehends that of the former.

The three external forces, in virtue of Article 131, condition I., balance each other, and are therefore proportional to the three sides of a triangle respectively parallel to their directions. In fig. 73* let A B C be such a triangle, in which CA represents P1,

Fig. 73.

Then by the conditions of equilibrium of a frame of two bars (Article 145), the external force P, applied at the joint 1, and the

resistances or stresses along the bars C and A which meet at that joint, are represented in magnitude by the sides of a triangle respectively parallel to their directions. Therefore, in fig. 73*, draw CO parallel to the bar C, and AO parallel to the bar A, meeting in the point O, and those two lines will represent the stresses on the bars C and A respectively. In the same manner it is proved, that BO represents the stress on the bar B. The three lines Co, A O, BO, meet in one point O, because the components along the line of direction of a given bar, of the external forces applied at its two extremities, are equal and directly opposed.

Hence follows the following

THEOREM. If three forces be represented by the three sides of a triangle, and if three straight lines radiating from one point be drawn to the three angles of that triangle, then a triangular frame whose lines of resistance are parallel to the three radiating lines will be in equilibrio under the three given forces, each force being applied to the joint where the two lines of resistance meet, which are parallel to the radiating lines contiguous to that side of the original triangle which represents the force in question.

Also, the lengths of the three radiating lines will represent the stresses on the bars to which they are respectively parallel.

C

B

Fig. 74.

2

149. Triangular Frame under Parallel Forces.-When the three external forces are parallel to each other, the triangle of forces A B C of fig. 73* becomes a straight line C A, as in fig. 74*, divided into two segments by the point B. Let straight lines radiate from O to A, B, C; and let fig. 74 represent a triangular frame whose sides 1 2 or A, 2 3 or B, 31 or C, are respectively parallel to O A, O B, O C; then if the load CA be applied at 1 (fig. 74), A B applied at 2, and BC applied at 3, are the supporting forces required to balance it; and the radiating lines O A, BOB, OC, represent the stresses on the bars A, B, C, respectively.

H

From O let fall OH perpendicular to CA, the comA mon direction of the external forces. Then that line Fig. 74*. will represent a component of the stress, which is of equal amount in each bar. When CA, as is usually the case, is vertical, OH is horizontal; and the force represented by it is called the "horizontal thrust" of the frame. Resistance would be a more precise term; question is a pull in some parts of the frame, and a thrust in others. In fig. 74, A and C are struts, and B a tie.

Horizontal Stress or because the force in

If the frame were

exactly inverted, all the forces would bear the same proportions to each other; but A and C would be ties, and B a strut.

The trigonometrical expression of the relations amongst the forces acting in a triangular frame, under parallel forces, is as follows :— Let a, b, c, denote the respective angles of inclination of the bars A, B, C, to the line OH (that is, in general, to a horizontal line). Load CA = OH Supporting A B = OH

Then,

Forces

The sign {+}

(tan c (tan a

tan a);

tan 6);

...(1.)

BC= OH (tan 6 ± tan c);

is to be used when the two opposite directions the same direction.

inclinations are in

E

3

2

150. Polygonal Frame-Equilibrium. - The Theorem of Article 148 is the simplest case of a general theorem respecting polygonal frames consisting of any number of bars, which is arrived at in the following manner. In fig. 75, let A, B, C, D, E, be the lines of resistance of the bars of a polygonal frame, connected together at the joints, whose centres of resistance are, 1 between A and B, 2 between B and C, 3 between C and D, 4 between D and E, and 5 between E and A. In the figure, the frame consists of five bars; but the demonstration is applicable to any number. From a point O, in fig. 75* (which may be called the Diagram of Forces), draw radiating lines UA, OB, OC, OD, O E, parallel respectively to the lines of resistance of the bars; and on those radiating lines take any lengths whatsoever, to represent the stresses on the several bars, which may have any magnitudes within the limits of strength of the material. Join the points thus found by straight lines, so as to form a closed polygon ABCDEA; then it is evident that A B is the ex

B

E

Fig. 75.

C

Fig. 75.

D

ternal force, which being applied at the joint 1 of A and B, will produce the stress OA on A and OB on B; that BC is the external force which being applied at the joint 2 of B and C, will produce the stress OB on B (already mentioned) and OC on C; and so on for all the sides of the polygon of forces ABCDEA. Hence follows this

THEOREM. If lines radiating from a point be drawn parallel to the lines of resistance of the bars of a polygonal frame, then the sides of any polygon whose angles lie in those radiating lines will represent a system of forces, which, being applied to the joints of the frame, will balance each other, each such force being applied to the joint between the bars whose lines of resistance are parallel to the pair of radiating lines that enclose the side of the polygon of forces, representing the force in question. Also, the lengths of the radiating lines will represent the stresses along the bars to whose lines of resistance they are respectively parallel.

151. Open Polygonal Frame.-When the polygonal frame, instead of being closed, as in fig. 75, is converted into an OPEN frame, by the omission of one bar, such as E, the corresponding modification is made in the diagram of forces by omitting the lines O E, DE, EA. Then the polygon of external forces becomes A BCDOA; and DO and O A represent the supporting forces respectively, equal and directly opposed to the stresses along the extreme bars of the frame, D and A, which must be exerted by the foundations (called in this case abutments), at the points 4 and 5, against the ends of those bars, in order to maintain the equilibrium.

152. Polygonal Frame-Stability. The stability or instability of a polygonal frame depends on the principles already stated in Articles 138 and 139, viz., that if a bar be free to change its angular position, then if it is a tie it is stable, and if a strut, unstable; and that a strut may be rendered stable by fixing its ends.

For example, in the frame of fig. 75, E is a tie, and stable; A, B, C, and D, are struts, free to change their angular position, and therefore unstable.

But these struts may be rendered stable in the plane of the frame by means of stays; for example, let two stay-bars connect the joints 1 with 4, and 3 with 5; then the points 1, 2, and 3, are all fixed, so that none of the struts can change their angular positions. The same effect might be produced by two stay-bars connecting the joint 2 with 5 and 4.

The frame, as a whole, is unstable, as being liable to overturn laterally, unless provided with lateral stays, connecting its joints with fixed points.