58 REINFORCED CONCRETE CONSTRUCTION 



the amount of steel to be employed will cause simultaneous maximum 



working stresses. 



From equation (7) t 



300,000 __ 

 (0.0084) (14,000) (0.870) 



Many different values of b and d will satisfy the last equation. 

 If 6 is taken as 10 in., then 



d = =293, or d= 17 1/4 in. 



Finally 



a s = (0.0084)(10X17.25) = 1.45 sq. in. 



If 1 3/4 in. is allowed between the tension surface of the concrete and the 

 center of the steel, the entire depth of the beam should be 19 in. 



PROBLEMS 



16. What will be the resisting moment for a beam with 6 = 12 in. and 

 d = 16 in., the area of steel section being 3.5 sq. in.? E s = 30,000,000. 

 E c = 3,000,000. f c = 600 Ib. per square inch. f s = 16,000 Ib. per square 

 inch. 



17. Suppose the beam of the preceding problem is 20 in. deep (d) and is 

 subjected to a bending moment of 550,000 in.-lb. Compute the 

 maximum unit stresses in the concrete and steel. 



18. Design a beam of 12.5 ft. span to carry a load of 1000 Ib. per linear foot 

 (including weight of beam), using a 1 : 2 : 4 Portland cement concrete 

 having an allowable working strength of 650 Ib. per square inch. A 

 mild steel, with an allowable working stress of 14,000 Ib. per square inch 

 is to be used. E a = 30,000,000. E c = 2,500,000. 



o 



19. By means of the formulas t = l/2f\/l/4f 2 + v* and tan 2K =, 



prove the following statements concerning homogeneous beams: 



(a) Wherever shearing stress exists, the maximum stress will not be on 

 a vertical section, but on an inclined one. 



(b) At all points in a beam where the shear is zero, the direction of 

 the maximum tension is horizontal, as at points of maximum bending 

 moment and along the outer fibers of the beam. 



(c) Wherever the horizontal fiber stress is zero (at the neutral plane 

 and at all sections of zero bending moment), the direction of the 

 maximum tension is inclined 45 degrees to the horizontal, and its 

 intensity is equal to the vertical shearing stress at the same place. 



34. Flexure Formulas for Ultimate Loads, Based on Parabolic 

 Variation of Stress in Concrete. The stress-deformation curve 

 for concrete in compression agrees very closely with the parabola; 

 in fact, so nearly so that the parabola is often used in theoretical 

 analysis to represent the distribution of compressive stress in a 



