Distributed_Steel_for_Concrete_Floors_C96

Distributed_Steel_for_Concrete_Floors_C96 - D esigners...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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.

Page1 / 4

Distributed_Steel_for_Concrete_Floors_C96 - D esigners...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online