EXTERNAL FORCES 19 



In Fig. 10 a concentrated load is assumed to be at the middle 

 of an opening spanned by a beam- of uniform weight. The reac- 



P W P + W 

 tions are as follows: Ri - R* - -= + 77- = ~ The bending 



S m & 



M PL ^ WL 



moment is M = r + ^-. 



4 o 



In Fig. 11 several concentrated loads are shown on a span and 



the reactions are to be 



found. U ~Y L-:.i....I2 



Commencing at the left f*" ""T"k "*f 



end take moments about il c\g ft e ^ ~t| 

 Ri. Thus the bending mo. 



ment at the left support is R & P*J> 



M Pa + P a + P a * l %- 9 Uniform Load on Beam Resting 



on Two Supports 

 This moment will have a 



tendency to carry the far end of the beam, at R t , downward 

 unless a supporting force is exerted to hold it in position. 



Here the principle of moments is again applied. The moment 

 of a force is equal to the force multiplied by the distance, or arm, 

 through which it acts, so it is necessary to have at /2j an upward 

 moment equal to the downward moment. This is obtained by 

 dividing the downward moment by the span length. 



Ri = (sum of the loads) R t . 

 This is proved when we con- 

 sider that the sum of the re- 

 actions is equal to the total 

 Fig. 10 Concentrated Center Load and load. The amount of each 

 Uniform Load on Beam Resting on reaction may be checked by 



taking moments from the right 

 end instead of the left and working as before. 

 Example: Let a - 3 feet and P =200 Ibs. 

 a, - 7 " " P, -300 " 

 a, - 11 " " P H - 250 " 



750 - total load. 



Span - 15 feet. 



i? (3 x 200) + (7 x 300) + (11 x 250) 



15 

 Ri - 750 - 363.33 - 386.67 Ibs. 



__ 

 - 363.33 Ibs. 



