194 



RESISTANCE OF MATERIALS 



Similarly, by taking moments about A the stress in BF is found to be 



aa' x Oh 



Stress in BF = 



AT 



and the stress in BC, with center of moments at F, is 



G . . cc' x Oh 

 Stress in BC = . 



By observing the signs of the moments the stresses in AB, BC, 

 and BF are found to be compressive and that in AF tensile. 



In the present case, from symmetry, the stresses in the remaining 

 members of the truss are the same as in those already found. For 

 unsymmetrical loading it would be necessary to apply the above 

 method to each individual member. 



APPLICATIONS 



277. Two equal weights of 50 Ib. each are joined by a cord which passes over 

 two pulleys in the same horizontal line, distant 12 ft. between centers. A weight 

 of 5 Ib. is attached to the string midway between the pulleys. Find the sag. 



278. The rule used by the makers of cableways for finding the stress in the 



one half the span 



cable is to calculate a factor = , and multiply the load, assumed 



twice the sag 



to be at the middle, by this factor. Show how this formula is obtained. 



279. It is usual to 

 allow a sag in a cable 

 equal to one twen- 

 tieth of the span. 

 What does the nu- 

 merical factor in the 

 preceding problem 

 become in this case, 

 and how does the 

 tension in the cable 

 compare with the 

 load? 



280. Find the 

 relation between F 

 and IF and the total 



pull on the upper support in the systems of pulleys shown in Fig. 132. 



281. In a.Weston differential pulley two sheaves, of radii a and 6, are fastened 

 together, and by means of a continuous cord passing around both and also around 

 a movable pulley, support a weight W. Find the relation between F and W, neg- 

 lecting friction (Fig. 133). 



