164 W. H. WARREN. 



ance of the concrete. Hence generally no reinforcement 

 will be needed near the centre where the small shearing 

 stress may be left to the concrete. 



As an example we may find the area of the reinforce- 

 ment to resist the shearing stresses in a concrete beam 

 10 inches by 10 inches cross section, reinforced horizontally 

 by 3 rods J inch in diameter if the span is 10 feet and the 

 load uniformly distributed. 



The area of 3 rods | inches diameter = 3 x 0*6 = 1*8 

 square inches. For the first foot from the ends the area 

 required if arranged vertically is 



1 - J- | = 0*648 square inches. 



10 10 



and if inclined at 45° : — 



0*648 x 1*411 = *0916 square inches. 



For the portion included between rr. 2 =2 and oc x = 1 : — 



-I x 1*8 3 \ 



1 - —J = 0*504 square inch vertically 



= 0*504 x 1*414 = 0*713 square inch inclined at 45° 

 For the portion included between a\ 2 = 3 and ol\ = 2 : — 



4x1 



5 \ 

 1 + ~) = 0*36 square inch vertically 



10 10 



= 0*30 x 1*414 = 0*509 square inch inclined at 45° 



For the portion included between cc 2 =4 and a\ =3 : — 



4 x 1*8 7 



1 + — = 0*216 square inch vertically 



= 0*216 x 1*414 = 0*305 square inch inclined at 45° 



We may provide steel stirrups having the area calculated 

 above, or inclined bars having an area equal to the vertical 

 stirrup multiplied by 1*414. The bars arranged to take 

 the shearing stresses should be rigidly connected to the 

 horizontal reinforcement in all cases. Mr. Julius Kahn 

 accomplishes this condition by the use of bars of the form 

 shown in Fig. 15. 



