=

a'b'o', it follows that a"b" × o'h' = a'b' x MN, and therefore a"b" × o'h' x oh = a'b' × oh x MN. But a'b' × oh= ABX MN. Therefore a"b" x o'h' × oh AB × MN2 = moment of inertia of AB about M. a'b'n' and a"b"n" are link polygons, of which the first determines the moment AB × MN, and the second determines the moment of this moment, namely, (AB x MN) x MN. The lengths a'b' and a"b" must be measured with the force scale, and the lengths oh and o'h' with the linear scale.

Fig. 206 shows the application of the method to the determination of the moment of inertia of the shaded figure about an axis a'a" in the plane of the figure. The area is divided into parallel strips, and parallel forces AB, BC, CD, DE, and EF are supposed to act at the centres of gravity of these strips, the magnitudes of the forces being proportional to the areas of the strips. The sum of the moments of these forces about the given axis is equal to a'f' × oh, and the sum of their moments of inertia is equal to a"f" × o'h' x oh. The lengths a'f" and a"f" must be measured with the area scale and the lengths oh and o'h' with the linear scale.

Exercises VIIIb

1. Three coplanar forces P, Q, and R (Fig. 207) act on a rigid body which is capable of rotation about an axis through O at right angles to the plane of the forces. P 8 lb., Q = 10 lb., and R = 6 lb. Find the resultant moment of these forces about O. Determine the magnitude and sense of the force S which will prevent the rotation of the body.

2. AB (Fig. 208) is a lever whose fulcrum is at A. Forces P and Q act on the lever as shown. P = 10 lb., and Q = 15 lb. Rotation of the lever about its fulcrum is prevented by the force R. Find the magnitude of the force R and determine the reaction of the fulcrum on the lever.

Reproduce the above diagrams three times this size and then consider that the reproduced diagrams are drawn to the scale of full size.

20 lb.,

3. A bar AB (Fig. 209) is acted on by forces P and Q as shown. P and Q 25 lb. A couple whose moment is 30 inch-pounds also acts on the bar in the same plane as P and Q and tends to give clockwise rotation to the bar. Find the additional force acting on the bar which will produce equilibrium.

4. A bent lever AOB (Fig. 210) is capable of rotation about a pin at O. The lever is at rest under the action of the forces P, Q, and R, and the reaction S of the pin on the lever. P = 15 lb., and Q = 20 lb. Find the magnitude and sense

of R and determine S.

5. Draw a square ABCD of 2 inches side, the lettering being in the clockwise direction. A force P acts through A in the plane of the square. P has a clockwise moment of 11 inch-pounds about B and an anti-clockwise moment of 19 inch-pounds about D. Determine the force P.

Another force Q also acts through A in the plane of the square. Q has a

clockwise moment of 24 inch-pounds about B and a clockwise moment of 8 inchpounds about C. Determine the force Q.

6. ABC is a triangle. AB 8 inches, BC 9 inches, and CA = 10 inches. The lettering is clockwise. A force R acting in the plane of the triangle has a clockwise moment of 40 inch-pounds about A, an anti-clockwise moment of 30 inch-pounds about B, and a clockwise moment of 12 inch-pounds about C. Determine the force R, using a linear scale of .

7. ABCD is a square of 2 inches side, lettered clockwise. A force P acts along the side AB and another force Q acts along the diagonal AC. The resultant moment of the forces P and Q about D is 10-4 inch-pounds, anti-clockwise. The resultant moment of P and Q about the middle point of BC is 25.5 inch-pounds, clockwise. Determine the forces P and Q.

8. The cantilever (Fig. 211) carries five loads each of 100 lb. as shown. Draw the bending moment and shear force diagrams.

9. The cantilever (Fig. 212) is loaded as shown, the loads being in tons. Draw the bending moment and shear force diagrams.

10. The beam (Fig. 213) rests on supports 16 feet apart and is loaded as shown, the loads being in tons. Draw the bending moment and shear force diagrams.

11. The beam (Fig. 214) rests on supports 10 feet apart and is loaded as shown, the loads being in tons. Draw the bending moment and shear force diagrams.

12. A beam is built firmly into a wall at one end and projects 24 feet from the face of that wall. The other end rests freely on a support. The beam carries a uniformly distributed load of 1 ton per foot run. It may be proved that the reaction at the support is three-eighths of the total load on the beam. Draw the bending moment and shear force diagrams.

13. A beam (Fig. 215) rests on supports 16 feet apart. The load at A is 2 tons and each of the other loads is 1 ton. Draw the bending moment and shear force diagrams.

14. A log of timber 24 feet long and of uniform cross section floats in still water. On the top of the log at points 6 feet from its ends are placed loads of 200 lb. each. Draw the bending moment and shear force diagrams.

[Note that the bending moments and shear forces on the log are those due to the given loads and a uniform upward water pressure whose total is 400 lb.] 15. The beam A (Fig. 216) rests freely on two cantilevers B and C as shown.

The given loads are in tons. Draw the diagrams of bending moment and shear force for this structure.

16. A beam rests on two supports 12 feet apart. This beam carries a distributed load which varies uniformly in intensity from nothing over one support to a maximum over the other support. The total load is 15 tons. Draw the bending moment and shear force diagrams.

17. Determine the moment of inertia of each of the figures given in Fig. 217 about a horizontal axis through its centroid.

FIG. 217.

In reproducing the above figures take the small squares as of half inch side.

105. Stress Diagrams for Framed Structures. It will be assumed that the framed structures considered are made up of bars which are connected by frictionless pin joints at their ends. It will also be assumed that the loads on the structure are concentrated at

the joints. If a bar carries a load uniformly distributed over its length this load is divided into two equal parts, and one part is placed at each end of the bar. If a bar carries a load concentrated at an intermediate point, this load is divided into two parts, which are to one another as the distances of the load from the ends of the bar; these parts are then placed one at each end of the bar, the greater part being at that end of the bar which is nearest to the original load.

In studying the equilibrium of a structure, two kinds of forces have to be considered, (1) the external forces, which for the whole structure must balance one another, and (2) the internal forces. As a consequence of the two assumptions mentioned at the beginning of this article, the bars forming the structure are subjected either to direct compression or to direct tension under the action of the external forces. It follows, therefore, that the lines of action of the internal forces are the lines which represent the bars on the diagram of the structure (called the frame diagram). At any joint, therefore, the forces acting are the internal forces acting along the bars meeting at that joint, and the external forces, if there are any, acting at that joint.

If a sufficient number of the forces acting at any joint are known, the polygon of forces for that joint can be drawn and the unknown forces determined.

The general method of drawing the complete stress diagram for a framed structure will be understood by reference to the example worked out in Fig. 218. A simple roof truss is shown carrying a load AB at its apex. The other external forces are the reactions BC and CA at the supports. The internal forces are the forces acting along the bars AD, BD, and CD. The lines of action of all the forces are known, but AB is the only force whose magnitude is known as yet.

At each joint there are three forces acting, and the polygon of forces

for each joint is therefore a triangle. The triangle of forces for the joint 2 or for the joint 3 cannot yet be drawn, because the magnitudes of all the forces at these joints are as yet unknown, but the triangle of forces for the joint 1 may be drawn, and this is shown at (m). This triangle determines the magnitudes bd and da of the internal forces in the bars BD and DA respectively. The sense of these forces is also determined, and it will be observed that the internal forces in the bars BD and DA both act towards the joint 1, therefore these bars are in compression. In drawing the triangle (m) the forces have been taken in the order in which they occur in going round the joint 1 in the watch-hand direction, beginning with the known force AB. Beginning with BA, and going round the joint in the opposite direc

tion, the triangle (8), which is similar to (m) but differently situated, is obtained.

Passing next to the joint 2, the three forces acting there are known in direction, and the magnitude of one of them, BD, has been determined by the drawing of the triangle (m) or the triangle (8). Beginning with DB, and taking the forces in the order in which they occur in going round the joint in the watch-hand direction, the triangle of forces (n) is drawn. If the forces be taken in the order in which they occur in going round the joint in the opposite direction, beginning with BD, the triangle (u) is obtained. Proceeding next to the joint 3, the triangle (o) is obtained when the forces are taken in the watch-hand order, and the triangle (v) is obtained when the forces are taken in the opposite order.

The construction of the three triangles (m), (n), and (o), or the three triangles (8), (u), and (v), determines the magnitude and sense of each of the three internal forces, and also the magnitudes and sense of the external forces BC and CA.

It is obvious that the triangles (n) and (o) may be applied to the triangle (m) so as to form the figure (r), and this figure gives all the results which were found from the separate triangles (m), (n), and (o), and this figure (r) is the complete stress diagram for the given framed structure. The figure (r) may of course be drawn at once without drawing the triangles (m), (n), and (o). It should, however, be noticed that in order that the force polygons for the different joints may be combined into one diagram, these polygons must be drawn by taking the forces in the order in which they occur in going round each joint in the same direction. (r) is the form of the stress diagram when the forces are taken in the order in which they occur when going round each joint in the watch-hand direction, and (w) is the form of the diagram when the order is reversed.

It is important to observe that in going from the joint at one end of a bar to the joint at the other end the sense of the force in that bar is reversed, because if a bar is in tension it must exert a pull at each of the joints at its ends and if it is in compression it must exert a thrust at each of the joints at its ends.

The sense of the force in any bar may be determined from the polygon of forces for the joint at either end and from the sense of one of the forces represented by one side of that polygon because the polygon of forces represents forces which are in equilibrium, and if arrow heads be placed on the sides of the polygon to show the senses of the forces these arrow heads must follow one another round the polygon.

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106. Examples of Stress Diagrams.—(1) The frame diagram a saw-tooth roof truss is shown at (u), Fig. 219. The truss is loaded at the joints as shown, the loads being in pounds. The first step is to draw the line of loads ab. g. To determine the reactions GH and HA at the supports, which are obviously vertical, choose a pole o and draw the link polygon shown at (v). The closing line of this polygon is shown dotted. From the pole o draw oh parallel to the closing line of the link polygon to meet the line of loads at h, then gh and ha are the magnitudes of the reactions GH and HA respectively.

Commencing at the joint ABKHA at the left-hand support the polygon of forces abkha is completed. Proceeding to the joint BCLKB the polygon of forces belkb is completed. Proceeding from joint to joint round the frame diagram the various polygons of forces are completed, the whole of the polygons of forces making up the complete force diagram for the truss as shown at (w).

The magnitudes of the forces in the various bars may now be scaled off from the diagram (w). The senses of the forces at any joint are determined by inspecting the polygon of forces for that joint and the sense of the force in a bar at a joint determines whether that bar is in tension or compression. Lines on the frame diagram which represent members in compression have been made bold.

(2) The lower part of Fig. 220 shows a roof truss under two systems of loads. One system is vertical and is due to the weight of