BENDING AND DIRECT STRESS 249 



For these values of p and -jt Diagram 13 gives X = 1.70 and shows that 



the problem falls under Case I. Then by formula (6) 



f WK (60,000)(1.70) 



fc = -7 = (9)(20) = 567 lb - P er square inch. 



Illustrative Problem. Change the eccentricity of the preceding problem 

 to 6 in. and solve. 



T'=2Tr- 30 



For p = 0.0087 and y = 0.30, Diagram 13 shows that ^ is too great for 



the problem to come under Case I. The method of procedure for Case 

 II must then be followed. 



Diagram 14 gives A: =0.73 for the values of p and given above. With 



k = 0.73 and p = 0.0087, Diagram 15 shows L to be 0.122. Solving equa- 

 tion (11) 



, M (60,000) (6) 



Using the formula (8) gives 



fs=Hfc (tt~ 1 } = ( 15 )( 82 ) (0.73 x 20 

 The stress /' may be found by formula 7 but is always less than nXf c > 



Illustrative Problem. An arch is 20 in. deep and is reinforced with 3 

 rods 3/4 in. in diameter to each foot of width, both above and below. If 

 the rods are embedded to a depth of 2 in. and the normal component of the 

 resultant thrust on a section is 100,000 lb. for 1-ft. width of arch with an 

 eccentricity of 3.4 in., determine the maximum intensity of compressive 

 stress on the concrete. Assume n = 15. 

 _ (6) (0.4418) 

 Po -- (12) (20) " 



Diagram 13 gives K = l.Q3 and the problem comes under Case I. Then 

 by formula (6) 



WK (100, 000) (1.63) 

 fc = ~W ~ (12) ( 20) =679 l 



PROBLEMS 



80. A rectangular beam is 8 in. wide and 20 in. deep, and contains four 1/2 

 in. round rods both at the top and bottom. At a given section the nor- 

 mal component of the resultant force is 77,000 lb. and acts at a distance 

 of 2 in. from the gravity axis. Assume n = 15. (a) Compute the maxi- 

 mum unit compressive stress in the concrete, (b) How much steel 

 reinforcement would be needed to make the stress in the concrete 600 lb. 

 per square inch? 



81. Change the eccentricity in part (a) of the preceding problem to 5 in. and 

 solve. 



