CRCP Reinforcement CRCP Source: WSDOT Pavement Guide Interactive - PDF document

CRCP Reinforcement CRCP Source: WSDOT Pavement Guide Interactive CD-ROM CRCP Reinforcement 1

Design Variables • Slab thickness, D (in) • Slab width, W s (in) • Modulus of subgrade reaction, k eff (psi/in) • Design temperature drop, Δ T d (°F) • Maximum wheel load magnitude, W (lb) • Rebar diameter, φ (in) Design Variables (cont.) • Concrete thermal coefficient, α c (in/in/°F) • Steel thermal coefficient, α s (in/in/°F) • Concrete tensile strength, f ′ t (psi) • Concrete shrinkage coefficient, Ζ (in/in) • Allowable steel working stress, σ s (psi) • Tensile stress due to wheel loads, σ w (psi) Concrete thermal coefficient Thermal Type of Coarse Coefficient (10 -6 in/in/°F) Aggregate Quartz 6.6 Sandstone 6.5 Gravel 6.0 Granite 5.3 Basalt 4.8 Limestone 3.8 2

Steel thermal coefficient − α = × ° 6 5 10 in/in/ F s Concrete Tensile Strength ′ = f 0.86 S t c Concrete shrinkage coefficient Indirect Shrinkage Tensile Strength Coefficient (psi) (in/in) 300 or less 0.0008 400 0.0006 500 0.00045 600 0.0003 700 or more 0.0002 3

Allowable steel working stress Reinforcing Bar Size Indirect Tensile Strength (psi) No. 4 No. 5 No. 6 300 or less 65,000 57,000 54,000 400 67,000 60,000 55,000 500 67,000 61,000 56,000 600 67,000 63,000 58,000 700 67,000 65,000 59,000 800 or more 67,000 67,000 60,000 Design Criteria Minimum crack spacing = 3½‘ Minimize potential for punchouts Maximum crack spacing = 8‘ Minimize potential for spalling Maximum crack width = 0.04“ Minimize water infiltration and potential for spalling Maximum allowable steel stress Prevent failure of reinforcing steel 4

Crack Spacing 1.15 ⎛ ⎞ ′ 6.70 α ⎛ ⎞ f ( ) + + + φ 2.19 1.32 1 t ⎜ 1 s ⎟ 1 ⎜ ⎟ α ⎝ ⎠ 1000 ⎝ 2 ⎠ = c X 5.20 σ ⎛ ⎞ ( ) ( ) 4.60 1.79 + + + ⎜ 1 w ⎟ 1 P 1 1000Z ⎝ ⎠ 1000 Minimum Crack Spacing 0.25 ′ 1.457 ⎛ α ⎞ ⎛ ⎞ f ( ) + + + φ 0.476 1.062 1 t ⎜ 1 s ⎟ 1 ⎜ ⎟ α ⎝ ⎠ 1000 ⎝ 2 ⎠ = − c P 1 max ⎛ σ ⎞ 1.13 ( ) ( ) + 0.217 + 0.389 ⎜ 1 w ⎟ 3. 5 ft 1 1000Z ⎝ ⎠ 1000 Maximum Crack Spacing 0.25 ′ 1.457 ⎛ ⎞ ⎛ ⎞ α f ( ) + + + φ 0.476 1.062 1 ⎜ t ⎟ ⎜ 1 s ⎟ 1 α ⎝ ⎠ 1000 ⎝ 2 ⎠ = − 1 c P 1 min 1.13 ⎛ σ ⎞ ( ) ( ) 0.217 0.389 + + w ⎜ 1 ⎟ 8. 0 f t 1 1000Z ⎝ ⎠ 10 0 0 5

Crack Width ′ 6.53 ⎛ ⎞ f ( ) 2.20 + + φ 0.00932 1 t 1 ⎜ ⎟ ⎝ ⎠ 1000 = CW 4.91 ⎛ σ ⎞ ( ) + + 4.55 w ⎜ 1 ⎟ 1 P ⎝ ⎠ 1000 Maximum Crack Width ′ 1.435 ⎛ ⎞ f ( ) + + φ 0.484 t 0 .358 1 ⎜ ⎟ 1 ⎝ 1000 ⎠ = − 2 P 1 min σ 1.079 ⎛ ⎞ ( ) 0.220 + ⎜ 1 w ⎟ 0.04 in ⎝ ⎠ 1000 Steel Working Stress ′ 4.09 Δ 0.425 ⎛ ⎞ ⎛ ⎞ f T + + 47,300 1 ⎜ t ⎟ ⎜ 1 d ⎟ ⎝ ⎠ ⎝ ⎠ 1000 100 σ = s 3.14 ⎛ σ ⎞ ( ) ( ) + + 2.74 + 0.494 1 w 1 P 1 1000Z ⎜ ⎟ ⎝ ⎠ 1000 6

Maximum Steel Stress ′ 1.493 Δ 0.155 ⎛ ⎞ ⎛ ⎞ f T + + 50.834 1 ⎜ t ⎟ ⎜ 1 d ⎟ ⎝ ⎠ ⎝ ⎠ 1000 100 = − 3 P 1 min 1.146 ⎛ σ ⎞ ( ) ( ) + σ 0.365 + δ 0.180 w ⎜ 1 ⎟ 1 1000 ⎝ ⎠ s c 1000 Required Steel ( ) = 1 2 3 P max P ,P ,P min min min min ( ) P 100 W D = min s N ( ) π φ min 2 4 ( ) P 100 W D = max s N ( ) π φ max 2 4 Example D = 9.5 in (from slab design) W s = 144 in (from slab design) k eff = 170 pci (from slab design) Δ T d = 60°F (from weather data) W = 20,000 lb (from loadometer data) f ′ t = 600 psi (from AASHTO T198) Limestone aggregate in concrete φ = 0.75 in (No. 6 bar assumed) 7

Step 1 δ c = 0.003 in/in (Table 12.22) α c = 3.8×10-6 in/in/°F (Table 12.23) α s = 5×10-6 in/in/°F (default value) σ s = 58,000 psi (Table 12.24) Step 2 – Steel Working Stress Step 3 – Min Crack Spacing 0.25 ′ 1.457 ⎛ α ⎞ ⎛ ⎞ f ( ) + + + φ 0.476 1.062 1 t ⎜ 1 s ⎟ 1 ⎜ ⎟ α ⎝ ⎠ b 1000 ⎝ 2 ⎠ = − c P 1 max σ 1.13 ⎛ ⎞ ( ) ( ) + 0.217 + δ 0.389 ⎜ 1 w ⎟ 3. 5 ft 1 1 000 ⎝ ⎠ c 1000 ( ) ( ) ( ) + 1.457 + 0.25 + 0.476 1.062 1 0.6 1 0.66 1 0.75 = − = P 1 0.70% ( ) ( ) ( ) max + 1.13 0.217 + 0.389 1 0.23 3.5 f t 1 0.3 8

Step 4 – Max Crack Spacing 0.25 ′ 1.457 ⎛ α ⎞ ⎛ ⎞ f ( ) + + + φ 0.476 1.062 1 ⎜ t ⎟ ⎜ 1 s ⎟ 1 α ⎝ ⎠ b 1000 ⎝ 2 ⎠ = c − 1 P 1 min ⎛ σ ⎞ 1.13 ( ) ( ) 0.217 0.389 + + δ ⎜ 1 w ⎟ 8. 0 ft 1 1 000 c ⎝ ⎠ 1000 ( ) ( ) ( ) + 1.457 + 0.25 + 0.476 1.062 1 0.6 1 0.66 1 0.75 = − = 1 P 1 0.42% ( ) ( ) ( ) min 1.13 0.217 0.389 + + 1 0.23 8.0 f t 1 0.3 Step 5 – Max Crack Width ′ 1.435 ⎛ ⎞ f ( ) + + φ 0.484 t 0.038 1 ⎜ ⎟ 1 b ⎝ 1000 ⎠ = − 2 P 1 min σ 1.079 ⎛ ⎞ ( ) + 0.220 1 w 0.04 i n ⎜ ⎟ ⎝ ⎠ 1 00 0 ( ) ( ) + 1.435 + 0.484 0.358 1 0.6 1 0.75 = − = 2 P 1 0.50% ( ) ( ) min + 1.079 0.220 1 0.2 3 0.04 in Step 6 – Max Steel Stress ⎛ ′ ⎞ 1.493 ⎛ Δ ⎞ 0.155 f T + + 50.834 1 ⎜ t ⎟ ⎜ 1 d ⎟ ⎝ ⎠ ⎝ ⎠ 1000 100 = − 3 P 1 min σ 1.146 ⎛ ⎞ ( ) ( ) + σ 0.365 + δ 0.180 ⎜ 1 w ⎟ 1 1000 ⎝ ⎠ s c 1000 ( ) ( ) + 1.493 + 0.155 50.834 1 0.6 1 0.6 = − = P 3 1 0.52% ( ) ( ) ( ) min + 1.146 0.365 + 0.180 1 0.2 3 58,0 00 1 0. 3 9

Step 7 – Required Steel ( ) = = P max 0.39% , 0.50% , 0.52% 0.52% min ( ) ( )( )( ) P 100 W D 0.52 100 144 9.5 = = = min s N 16.1 ( ) ( )( ) π φ min 2 2 4 π 4 0.75 b ( ) ( )( )( ) P 100 W D 0.70 100 144 9.5 = = = max s N 21.7 ( ) ( )( ) max π φ 2 π 2 4 4 0.75 b Solution < < 16.1 N 21.7 = Choose N 18 bars ( ) ( )( )( ) π φ π 2 2 N 4 18 4 0.75 = = = b P 0.58% ( ) ( )( )( ) 0.01 W D 0.01 144 9.5 s Crack Spacing 1.15 ′ 6.70 ⎛ ⎞ ⎛ ⎞ α f ( ) 2.19 + + + φ 1.32 1 t ⎜ 1 s ⎟ 1 ⎜ ⎟ α ⎝ ⎠ b 1000 2 ⎝ ⎠ = c X 5.20 ⎛ σ ⎞ ( ) ( ) 4.60 1.79 + + + δ ⎜ 1 w ⎟ 1 P 1 1000 ⎝ ⎠ c 1000 ( ) ( ) ( ) 6.70 1.15 2.19 + + + 1.32 1 0.6 1 0.5 1 0.75 ′ = = X 4.3 ( ) ( ) ( ) 5.20 4.60 1.79 + + + 1 0.23 1 0.58 1 0.3 10

Crack Width ′ 6.53 ⎛ ⎞ f ( ) + + φ 2.20 t 0.00932 1 ⎜ ⎟ 1 b ⎝ ⎠ 1000 = CW σ 4.91 ⎛ ⎞ ( ) + + 4.55 1 w 1 P ⎜ ⎟ ⎝ ⎠ 1000 ( ) ( ) + 6.53 + 2.20 0.00932 1 0.6 1 0.75 = = ′′ CW 0.03 ( ) ( ) + 4.91 + 4.55 1 0.23 1 0.58 Steel Working Stress ′ 4.09 Δ 0.425 ⎛ ⎞ ⎛ ⎞ f T + + t d 47,300 1 ⎜ ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ 1000 100 σ = s 3.14 ⎛ σ ⎞ ( ) ( ) 2.74 0.494 + + + δ ⎜ 1 w ⎟ 1 P 1 1000 ⎝ ⎠ c 1000 ( ) ( ) + 4.09 + 0.425 47,300 1 0.6 1 0.6 σ = = 51 ,707 psi ( ) ( ) ( ) s + 3.14 + 2.74 + 0.494 1 0.23 1 0.58 1 0.3 11