rigid pavement design course crack pattern development

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  • Slide 1
  • Rigid Pavement Design Course Crack Pattern Development
  • Slide 2
  • Rigid Pavement Design Course CRC Pavement Vetter, C.P. 1933 Reinforced Concrete Drying Shrinkage Temperature Drop Consider a unit Length (L) between cracks a. is restrained by the reinforcement b. Causes tension in concrete & compression in the steel. c. Bond stress between steel & concrete and the concrete & subgrade
  • Slide 3
  • Rigid Pavement Design Course (1)Bond stress in the vicinity of crack (2)Compression in steel and tension in the concrete increases until steel = concrete. In this region there is no bond slip or stress. d. Subsequent crack form in concrete when bond stress exceeds the concrete tensile strength.
  • Slide 4
  • Rigid Pavement Design Course Free Edge t L Longitudinal Joint Traverse Crack C L
  • Slide 5
  • Rigid Pavement Design Course Crack AsfsAsfs A s f sc AcftAcft Section XX Section YY Forces Acting on CRC Pavement Section Probable Strain Distribution Adjacent to a Crack
  • Slide 6
  • Rigid Pavement Design Course Extensive bond slip Crack L/2=nS min Good bond
  • Slide 7
  • Rigid Pavement Design Course Unrestrained shrinkage strain Concrete strain Crack width: equation 2.10a Steel strain S min or (L-2x)/2 Compression Tension Strain Crack width: equation 2.10b
  • Slide 8
  • Rigid Pavement Design Course Concrete Stress Steel stress Stressed, full restraint c.g. of bond h b/2 c Condition of no stress L cw Stresses and Strains in Fully Restrained, Cracked Reinforced Concrete for Decreasing Temperature
  • Slide 9
  • Rigid Pavement Design Course Assumptions of Vetter Analysis 1.Volumetric s are uniformly distributed. 2. Compatibility exists in bonded region. 3.Total bond force= Total Tensile Force= Total change in the steel stress 4.Total length of steel will remain unchanged. Total elongation = Total shortening 5.Equilibrium exists between forces at crack & the forces in the fully bonded region. In partially bonded region; compatibility of deformation does not exist. Crack width results from relative displacement between the steel and the concrete.
  • Slide 10
  • Rigid Pavement Design Course Stress Distribution Between Cracks Subject to Shrinkage L C of Crack u x L Bond Stress Tension b) Concrete Stresses a) Steel Stresses Compression Tension x1x1 f tz f sz c) Bond Stresses
  • Slide 11
  • Rigid Pavement Design Course (1) Center of crack spacing (2) Bond Force = Concrete tensile force = Change in steel force (3) Total length of steel bars remain unchanged total shortening= total elongation
  • Slide 12
  • Rigid Pavement Design Course Total Shortening = Total Elongation Note
  • Slide 13
  • Rigid Pavement Design Course For temp. drop Both
  • Slide 14
  • Rigid Pavement Design Course a) Steel Stresses b) Concrete Stresses Bond Stress y L c) Bond Stresses u Tension C of Crack L Stress Distribution Between Cracks Subject to Temperature Drop
  • Slide 15
  • Rigid Pavement Design Course (1) Center of crack spacing (2) Bond Force = concrete tensile force = change in steel force (3) Total length of steel bars remain unchanged total shortening= total elongation
  • Slide 16
  • Rigid Pavement Design Course For temp. drop Both
  • Slide 17
  • Rigid Pavement Design Course Average Crack Spacing (ft) Ratio of Steel bond Area to Concrete Volume x 10 -2 (in. 2 /in. 3 ) Relationship Between Steel Bond Area and Crack Spacing 2 3 4 5 6 7 20 18 16 12 8 4 0 Pavements Placed During Winter = Summer =
  • Slide 18
  • Rigid Pavement Design Course Development Length Allowable Bond Stress Design Strength of the Bar ACI Definition of Development Length (a) Assumed and Actual Bond Stress-Slip Relationships.
  • Slide 19
  • Rigid Pavement Design Course Actual Bond Stress Development Length Vetter Allowable Bond Stress Force in Bar Under Working Stress Condition Stress Transfer Length (b)
  • Slide 20
  • Rigid Pavement Design Course Bond Stress Relative Slip Between Concrete and Steel As Modeled in Computer Program Actual bond Stress-Slip Relationship (c)
  • Slide 21
  • Rigid Pavement Design Course concrete tension specimen steel bar P (a) (b) (c) ab x SbSb x-Displ. Slip Determination of Slip from Strain Functions ACI
  • Slide 22
  • Rigid Pavement Design Course
  • Slide 23
  • Slide 24
  • Relative to temperature drop Max. drop to cause L = 2x substitute for in equation For a greater temp. drop t 2 only if otherwise
  • Slide 25
  • Rigid Pavement Design Course
  • Slide 26
  • Slide 27
  • Structural Response Models Uniform Bond Stress Distribution Vetter: Shrinkage and Temperature Drop Zuk: Shrinkage Friberg: Temperature Drop
  • Slide 28
  • Rigid Pavement Design Course Hughes: Shrinkage and temperature Drop (concrete only)
  • Slide 29
  • Rigid Pavement Design Course CRCP-2: Computer Model for Shrinkage and Temperature Drop (force equilbrium) Regression Equations:
  • Slide 30
  • Rigid Pavement Design Course Percent Steel (p) ; prevent yielding ; p=p min f s =f y u L x Um Non-Uniform Bond Stress Distribution TTICRCP: Computer model for Shrinkage and Temperature Drop (force equilibrium and energy balance) Reis: Shrinkage and Temperature Drop
  • Slide 31
  • Rigid Pavement Design Course 2002 AASHTO Guide CRC Design K1K1 K2K2
  • Slide 32
  • Rigid Pavement Design Course Crack Width
  • Slide 33
  • Rigid Pavement Design Course Crack Spacing Distribution vc. spc %P