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Distributed_Steel_for_Concrete_Floors_C96

Distributed_Steel_for_Concrete_Floors_C96 - Distributed...

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D esigners often specify the use of distributed steel ( welded wire fabric or small-diameter re i n f o rc- ing bars) in the upper half of con- crete floors on ground. It’s used to p re vent random cracks fro m widening. Keeping cracks tightly closed improves load tra n s f e r, helps to prevent faulting (differen- tial displacement of the slab at the c rack), and makes spalling less likely at the crack edges. Some de- signers re q u i re this distri b u t e d steel in all floors, regardless of joint spacing, while others specify its use only when joint spacings ex- ceed 15 to 20 feet. The amount of steel needed is usually calculated using the subgrade-drag equation. But does this equation give the right amount of steel for best floor performance? Subgrade-Drag Theory First proposed as a method for d e t e rmining joint spacing in con- c rete pave m e n t s, subgra d e - d ra g theory assumes that when a slab of c o n c rete dries and contra c t s, it draws itself over the subgrade from each end tow a rd the center. Fri c- tional resistance of the subgra d e p re vents full contraction. This re- sistance causes a tensile stress that increases as drying progresses and is highest at slab midlength. Adding re i n f o rcing steel doesn’t p re vent cracking; it simply helps to c o n t rol crack width. To calculate the amount of steel needed for this p u r p o s e, the friction force is set equal to the product of steel are a and a steel working stress that’s less than the steel yield strength. The needed area of steel is calculated as follows (Ref. 1): A s = FLw 2f s where: A s = cross-sectional area of steel in square inches per lineal foot of slab width F = coefficient of subgrade friction (assumed to be 1.5 to 2.0 for most floors) L = joint spacing in feet w = weight of the concrete in psf of floor area (usually assumed to be 12.5 psf per inch of slab thickness) f s = w o rking stress in the steel in psi(usually 0.67 to 0.75 times the steel yield strength) Using the Subgrade-Drag Theory The percentage of reinforcement is the ratio of the cro s s - s e c t i o n a l area of reinforcement to the cross- sectional area of the concre t e, ex- p ressed as a perc e n t a g e. The area of steel and percentage of re i n f o rc e- ment have been calculated for t h ree floors shown in Table 1. Fo r all calculations, F = 1.5, w = 12.5 psf per inch, and f s = 45,000 psi (0.75 60,000-psi yield strength). No t e that the reinforcement percentage d o e s n’t exceed 0.10% for any of these floors. Are these amounts of steel sufficient? Ge n e rally not, based on recommendations fro m other countries and from highway pavement practices. Other Recommendations In the Australian Standard for Concrete St r u c t u re s , the following f o rmula is recommended for calcu- lating the minimum reinforcement ratio for slabs on ground with con- t rol joints at regular intervals (Ref. 2): p min = 0.7 f y where: p min = minimum reinforcement ratio f y = steel yield strength in MPa Assuming a 400-MPa (58,000- psi) yield strength, the minimum reinforcement ratio expressed as a percent is 0.17%, double the high-
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