EEINFORCED CONCRETE. 







61 



We may use 4 bars each 1J in. x 1\ in. 



*- /<?"-> 



using this new value of p in the equation — 



f 



CM 



1 





T 



v E? * E a F E c 



*' 

 $ 



we find x = 0*486 



•i- 



*Pf" 



-** 



Tx M . AnAA /2*4 - 0*486 \ 1 

 Hence ^ 2 = 16000 ( g j 4Q 





ftW# 



= 255*2 





and M = 1020800 inch pounds. 









This result is nearly 20f greater than the bending moment 

 produced by the load of 1000 pounds per foot run on a span 

 of 24 feet. We may however, reduce the area of the steel 

 reinforcement thus — 



M = 16000 r 2 ' 4 - ' 486 ) P = 216 



bh 



.-.1-A =0-021 



.'. a = 4*22 square inches. 



To find the safe load applied at the centre on a beam 

 10 x 10 in. when supported at points 10 feet apart : the 

 area of the steel reinforcement is 1*8 square inches. In 



this case p = 0*018 and p 2 = 0*000324. u = 0*85 ~ = 15 



x = /225 x 0*000324 + 25*5 x 0*018 - 0*27 = 0*45 



.\-^-= 16000 X 0*018 ( /2 ' 55 " °' 46 ^ = 201*60 



bh 2 \ 3 / 



.*. M = 210600 inch pounds. 



The actual bending moment which produced failure was 

 604800 inch pounds, so that the factor of safety is — 



604800 



201600 



= 3 



