INTERNAL FORCES 73 



We have in reinforced concrete a material (steel) stressed in 

 tension which is from 20 to 30 times stronger than the compressive 

 strength of the concrete. Furthermore, the weaker material has 

 a deformation under load 15 times greater than the deformation 

 of the stronger material. 



Let E, = modulus of elasticity of the steel. 

 E e = modulus of elasticity of the concrete. 



E 



n = ratiofc of deformation = -=r (usually 15). 



&e 



f, = allowable fiber stress (working stress) in the steel. 

 f e = allowable fiber stress (working stress) in the concrete. 



TO = stress ratio = --' 



Jc 



d = depth from top of concrete beam to center of gravity 



of the steel reinforcement. 

 k = depth from the top of the beam to the neutral axis 



(expressed as a percentage of d). 



j = moment arm expressed as a per cent of d. When the 

 value of d is given in inches then the moment arm is jd. 



k , . , , dk 

 j = I - - and jd = d - 



Referring to Fig. 57 ; on a piece of squared paper set off ten units 

 vertically. Measuring to the right along the top lay off n. Measur- 

 ing to the right along the bottom lay off m. Connect the ends as 

 shown to form two triangles which cross at the neutral axis. The 

 depth, k, to the neutral axis may then be measured on the paper. 

 The two triangles show an obvious geome- 

 trical relation which enables us to find k 

 by computation, thus: 

 n 



n + m 



Fig. 58 illustrates the triangle of com- 

 pressive force. The area of the triangle 



- ^ x k. The tensile force is equal to the pj g 53 



area of the steel times the fiber stress, so the area of the steel will 

 be obtained by dividing the compressive force by the steel fiber 

 stress. Units are used throughout, so the steel area will be the 

 ratio between the concrete and the steel. Then, 



