. A text book of physics, for the use of students of science and engineering . 4 and 5 cannot be shown meanwhile,but since the force represented by the closing side of the polygon ABCDA,viz. DA (Fig. 167 (&)), would equilibrate the forces acting at joints I, 2and 3, it follows that DA represents the resultant of the forces acting at4 and 5, since this resultant would also equilibrate the forces acting at1, 2 and 3. K 14G DYNAMICS CHAP. Select any suitable pole OJFig. 167 (ft)), and join O to A, B, C and drawing the link polygon in Fig. 167 (a) by selecting a point a onthe line o
. A text book of physics, for the use of students of science and engineering . 4 and 5 cannot be shown meanwhile,but since the force represented by the closing side of the polygon ABCDA,viz. DA (Fig. 167 (&)), would equilibrate the forces acting at joints I, 2and 3, it follows that DA represents the resultant of the forces acting at4 and 5, since this resultant would also equilibrate the forces acting at1, 2 and 3. K 14G DYNAMICS CHAP. Select any suitable pole OJFig. 167 (ft)), and join O to A, B, C and drawing the link polygon in Fig. 167 (a) by selecting a point a onthe line of the force AB, and drawing ab and be parallel to OB and OCrespectively in Fig. 167 (6). From a draw of parallel to OA, and fromc draw c/parallel to OD ; these lines intersect at/, and the resultant forcerepresented by DA must pass through /. Now this force is the resultantof the forces acting at 4 and 5, and the resultant and the componentsmust intersect in the same point, therefore the lines of direction of theforces acting at 4 and 5 will be found by joining/4 •0 bar (fa is al T illsola portat t. (b) Fig. 167.—Equilibrium of a rigid frame. The force polygon in Fig. 167 (b) may be closed now by drawing DEparallel to the lino A and AE parallel to the line/5. DE and EA givecompletely the forces acting at joints 4 and 5 respectively. The forces in the bars of the frame may now be obtained by consideringoach joint separately. Taking joint 3 ; there arc three forces acting, andl he triangle of forces is constructed by using CD (Fig. 167 (b)) for one side,ami drawing DH and CH parallel respectively to bars 34 and 23. Todetermine the kind of force, go round joint 3 clockwise, and find the senseof each force from the triangle of forces. The force represented by DH(Fig. 1(>7 (6)) acts away from joint 3 in Fig. 167 (a); hence the bar 34 isunder pull. The force represented by HC in Fig. 167 {b) acts towardsjoint 3 ; hence the bar i\: is under push. hIon . XI ROOF TRUSS 14
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