This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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- c rete floors on ground. Its 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, re g a rdless 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 subgra d e - d rag 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 t h e o ry assumes that when a slab of c o n c rete dries and contra c t s, it d raws itself over the subgrade fro m 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 i n c reases as drying pro g resses and is highest at slab midlength. Adding re i n f o rcing steel doesnt 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 thats less than the steel yield strength. The needed area of steel is calculated as f o l l ows (Ref. 1): A s = FLw 2 f s w h e re: A s = c ross-sectional area of steel in s q u a re inches per lineal foot of slab width F = coefficient of subgrade fri c t i o n (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 t h i c k n e s s ) f s = w o rking stress in the steel in psi(usually 0.67 to 0.75 times the steel yield stre n g t h ) Using the Subgrade-Drag Theory The percentage of re i n f o rc e m e n t is the ratio of the cro s s - s e c t i o n a l a rea of re i n f o rcement to the cro s s - 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 re i n f o rcement perc e n t a g e d o e s nt 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 p a vement pra c t i c e s....
View Full Document
This note was uploaded on 03/27/2012 for the course CIVE 336 taught by Professor Desormeaux during the Spring '12 term at University of Louisiana at Lafayette.
- Spring '12