. The elements of railroad engineering . lel forces acting downwards, as shown in Fig. 222. It is requiredto find their resultant moment (algebraic sum of the moments) and themoment of their resultant, all moments to be taken about the point C. 478 STRENGTH OF MATERIALS. Solution.—Lay off O i = 30 lb. = F,, 1-2 = 20 lb. = F, and 2-3 = 20lb. = F3, and 03 is the value of the resultant. Choose some point P asa pole and draw the rays. Take any point, as d, on any force, as Fi,and complete the equilibrium polygon be deb; then, the line of actionof the resultant must pass through e. Through C draw C
. The elements of railroad engineering . lel forces acting downwards, as shown in Fig. 222. It is requiredto find their resultant moment (algebraic sum of the moments) and themoment of their resultant, all moments to be taken about the point C. 478 STRENGTH OF MATERIALS. Solution.—Lay off O i = 30 lb. = F,, 1-2 = 20 lb. = F, and 2-3 = 20lb. = F3, and 03 is the value of the resultant. Choose some point P asa pole and draw the rays. Take any point, as d, on any force, as Fi,and complete the equilibrium polygon be deb; then, the line of actionof the resultant must pass through e. Through C draw C i parallel toR and prolong d e. The moment of R about C equals the pole distance//, multiplied by the intercept // /, since // / is that part of the line drawnthrough C parallel to R, and included between the lines be and de ofthe equilibrium polygon which meet upon R. Measuring hi\.o thescale of distances (1 in. =40 ft.), it equals 23 ft. Measuring H \.o thescale of forces, it equals 40 lb. Consequently, the moment of R about \ 501-. Scale of forces 1=40 of distance 1=40. Fig. 222. E C= 40 X 23 = 920 Considering the force F3, the intercept is ^/,since F3 is parallel to R, and, consequently, to Cz; also, g tls that partof the line C/included between the sides e d and de, which meet on ^/, it is found to equal 28 ft. Hence, the moment of F3about C= 40 X 28 = 1,120 The moment of F^ about C=I/x/g= 40 X 13 = 520 The moment of F, = Hx//i = 4:0 XlS = Now F3 and F^ have positive moments, since they tend to causerotation in the direction of the hands of a watch, while Fi has a nega-tive moment, since it tends to cause rotation in the opposite , adding -the moments algebraically, the resultant moment= 1,120 + 520 - 720 - 920 , the same as the moment of theresultant. STRENGTH OF MATERIALS. 479 Having described the fundamental principles of GraphicalStatics, the subject of Strength of Materials will n
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