03_DesignofRigidPavement

CE5204: Pavement Design & Rehabilitation Design of Rigid Pavemen Pavements: ts: Design Methods and Procedures Dr.. Raymond Dr Raym ond Ong Department of Civil & Environmental Engineering Email: [email protected]

Semester 2, AY2012/2013

Design of Rigid Pavement Pavement Structure •

Base (often called subbase) may be omitted in rigid pavement construction, but is usually provided for: –

–

–

–

Providing sub-surface drainage to prevent prevent accumulation of water underneath a slab Preventing Preventing pumping at joints, edges and cracks of slabs Forming a working surface for pavement construction Reducing frost action


Base (often called subbase) may be omitted in rigid pavement construction, but is usually provided for: –

–

–

–

Providing sub-surface drainage to prevent prevent accumulation of water underneath a slab Preventing Preventing pumping at joints, edges and cracks of slabs Forming a working surface for pavement construction Reducing frost action


•

Concrete mix design (Refer to standard texts) Factors affecting thickness of slab: –

Soil property

–

Inclusion or omission of base

–

Inclusion or omission of steel reinforcement

–

Intensity of traffic

–

Inclusion of pavement shoulders

UK Road Note 29 Design Method •

Developed by the TRRL.

•

Various factors involved in formulating design: –

traffic

–

rate of growth

–

design life

–

conversion to standard axles

–

•

series of relationships between subbase thickness, slab thickness and weight of reinforcement reinforcement for concrete pavements

Thickness design is based on load –

Temperature stresses are largely relieved by providing joints and/or steel reinforcements

Design for JPCP and JRCP Step 1: Determine the total 80 kN ESAL for the entire design period •

•

ESAL Computation (usually require the assumption of a structural number, say SN = 5) Initial year traffic loading W 0

W0 =

∑ ( ESAL )( N ) i

i

where: ESALi = ESAL factor for vehicle type i Ni = number of passes by vehicle type i per year

Design for JPCP and JRCP Step 1: Determine the total 80 kN ESAL for the entire design period •

Design ESAL applications over entire design period W0 (1 + r ) − 1 n

W =



r

D D d l

Lane Distribution Factors Number of

Dl

Lanes

r = annual traffic growth rate

1

1

n = design period (years) (30 to 40 years)

2

0.8 – 1

3

0.6 – 0.8

Dd = directional distribution factor (usually 0.5, usually 0.3 - 0.7)

≥4

0.5 – 0.75

D = lane distribution factor

Design for JPCP and JRCP Step 2: Determine subgrade CBR and subbase thickness •

Classification of subgrade Subgrade

Subgrade CBR

Subbase thickness

Weak

CBR ≤ 2%

Subbase thickness ≥ 150 mm

Normal

2% < CBR < 15%


Very stable

CBR ≥ 15%

Need not provide subbase

Design for JPCP and JRCP Step 3: Determine slab thickness

Design for JPCP and JRCP Step 4: Reinforcement design for JRCP •

Determine weight of reinforcement fabric and area of steel per unit width of pavement

Design for JPCP and JRCP Step 4: Reinforcement design for JRCP •

Determine spacing of joints

Note: Design is based on concrete with a minimum 28-day crushing strength of 28 MPa

Worked Example For a design ESAL of 1.89 x 106, design a jointed reinforced concrete pavement given that the subgrade CBR is 8%.

Solution: •

Step 2: Determine subbase thickness CBR = 8% ⇒ Subbase thickness = 80 mm

Subgrade

Subgrade CBR

Subbase thickness

Weak

CBR ≤ 2%


Normal

2% < CBR < 15%


Very stable

CBR ≥ 15%



Solution: •


•

Step 3: Determine slab thickness

Worked Example Step 3: Determine slab thickness


Solution: •


•

Step 3: Determine slab thickness Minimum slab thickness = 178 mm (use 180 mm)

•

Step 4: Reinforcement Design

Worked Example Step 4: Reinforcement design for JRCP •

Determine weight of reinforcement fabric and area of steel per unit width of pavement

Worked Example Step 4: Reinforcement design for JRCP •

Determine spacing of joints


Solution: •


•

Step 3: Determine slab thickness Minimum slab thickness = 178 mm (use 180 mm)

•

Step 4: Reinforcement Design Reinforcement design chart ⇒ use a 3.41 kg/m2 standard long mesh. Joint spacing design chart ⇒ use a joint spacing of 21 m

Design for CRCP •

•

Step 1: Determine the total 80 kN ESAL for the entire design period. Step 2: Determine subgrade CBR and subbase thickness Subgrade

Subgrade CBR

Subbase thickness

Weak

CBR ≤ 2%


Normal

2% < CBR < 15%


Very stable

CBR ≥ 15%


Refer to JPCP/JRCP guide for more details

Step 3: Determine slab thickness for CRCP

Design for CRCP •

Step 1: Determine the total 80 kN ESAL for the entire design period.

•

Step 2: Determine subgrade CBR.

•

Step 3: Determine slab thickness.

•

Step 4: Design for other elements. Reinforcement : Long mesh reinforcement ≥ 5.5 kg/m2 or Deformed bar reinforcement ≥ 650 mm2/m slab width Joints: - No transverse joints except for unavoidable construction joints - Longitudinal joint spacing ≤ 4.5 m Surfacing: (wearing course + basecourse) ≥ 90 mm

PCA Method for Rigid Pavement Design •

•

Portland Cement Association (PCA) thickness design procedure: –

Published in 1984.

–

Applied to JPCP, JRCP and CRCP

Based upon: –

–

–

•

Theoretical studies by Westergaard and Pickett. Model tests and in-service pavement data Road test data

Design Concept: –

Fatigue Analysis + Erosion Analysis

Miner’s Hypothesis Cumulative Damage Concept (Miner’s Hypothesis) •

Miner’s hypothesis states that every load i has a damage d i on the material and each damage d i is equal. So we can sum up the damage using:

Dr =

ni

∑ d = ∑ N i

i

i

i

where ni is the number of repetitions of the load i and Ni is the allowable number of repetitions of the load i .

Stress Ratio •

Damage on pavement due to a critical loading repetition can be represented by stress ratio SF:

Stress Ratio

1

Stress Ratio =

•

Applied Stress Allowable Stress

Since the critical stresses in the pavement slab are flexural , stress ratio can be expressed as: Stress Ratio =

Critical Flexural Stress Modulus of Rupture

0 Number of repetitions

Fatigue Analysis •

Apply Miner’s hypothesis to relate the damage ratio accumulated over the design period due to all load (i.e. total fatigue Dr, fatigue): Dr , fatigue =

ni

∑ SR = ∑ N i

i

i

i

where SRi is the stress ratio of axle load group i , ni is the expected number of repetitions of the axle load group i and and Ni is the allowable number of repetitions of the axle load group i over over the design period. 1 Dr , fa Pavement Pa vement will not fail. ⇒ fati tigu gue e ≤

>1 Dr , fatigu fatigue e

⇒

Pavement Pa vement will fail.

Critical Loading Position for Fatigue Fatigue Analysis •

Fatigue analysis is based on the edge stress midway between transverse joints with the most critical loading defined below:

Transverse joint Tandem axle load

Free edge or shoulder joint

Concrete shoulder (if used)

Traffic lane

Accounting for Wheel Path Variation •

To account for variation of wheel path (i.e. not all wheel passes will produce the most critical loading), adjustments were made to the critical stress to obtain equivalent stress factor (ESF) and the corresponding stress ratio factor (SRF): Stress Ratio Factor SRF =

Dr , fatigue =

ni

∑ N = ∑ i

i

Equivalent Equivalent Stress St ress Factor ESF Modulus of Rupture MR

f ( SRF i )

i

SRF and ESF are therefore used as the basis b asis for fatigue analysis

Rationale for Erosion Analysis •

•

Joint-related failure is an important mode of distress in addition to fatigue cracking. Repetitive traffic load causes softening and erosion of material under and beside slab, leading to joint related problems problems such as pumping, faulting and cracking.

Critical Loading Position for Erosion Analysis •

Erosion analysis is based on the most critical deflection with the axle load placed at the corner of the slab:

Transverse joint Traffic lane

Free edge or shoulder joint

Tandem axle load

Concrete shoulder if used

Erosion Analysis •

Apply Miner’s hypothesis to relate the damage ratio accumulated over the design period due to all load (i.e. total erosion Dr, erosion): Dr ,erosion =

ni

∑ N = ∑ i

i

f ( ERF i )

i

where ERF i is the erosion factor of axle load group i , ni is the expected number of repetitions of the axle load group i and Ni is the allowable number of repetitions of the axle load group i over the design period. ERF is used as the basis for erosion analysis and design charts/ tables are used to derive the total erosion

Design Considerations and Factors •

•

Decide whether doweled joints and concrete shoulders are to be used. Thickness design is governed by four design factors: –

–

Concrete modulus of rupture Subgrade and subbase support

–

Design period

–

Traffic

Concrete Modulus of Rupture •

•

Flexural strength of concrete is defined by its modulus of rupture M.R. which is determined at 28 days For ASTM C78-10: Third point loading is on a 6 in. by 6 in. by 24 in. beam with support points at 18 in. apart. ( PL 3) d 2 2 PL = f r = M .R. = 3 2 bd 12 bd where f c is the modulus of rupture, P is the failure load, L is the span length, b and d are the beam width and depth respectively.

•

ACI relationship:

•

90 day strength:

f r = 0.75 f c′

28-day compressive strength of concrete (MPa)

( M.R. at 90-day ) = 1.1× ( M.R. at 28-day )

Subgrade and Subbase Support •

•

Subgrade and subbase support is defined by the modulus of subgrade reaction k (in pci) When a base or subbase is used, revise k as follow: Subgrade k value (pci) 50 100 200 300

Untreated Subbase k values (pci) 4 in. 6 in. 8 in. 10 in. 65 75 85 110 130 140 160 190 220 230 270 320 320 330 370 430

Subgrade k value (pci) 50 100 200 300

Cement-Treated Subbase k values (pci) 4 in. 6 in. 8 in. 10 in. 170 230 310 390 280 400 520 640 470 640 830 -

Note: 1 in. = 25.4 mm; 1 pci = 271.3 kN/m3

Design Period and Traffic •

Use of axle load distribution

Steps to determine axle load distribution in design period 1. Initial year traffic loading for each axle group i , W 0 ,i 2. Design repetitions for axle group i over entire design period n  W0, i (1 + r i ) − 1  D D Wd ,i = d , i l ,i

r i

r = annual traffic growth rate n = design period (years) (typically 30 to 40 years of PCCP) Dd = directional distribution factor (usually 0.5, can vary between 0.3 and 0.7) Dl = lane distribution factor

Load Safety Factor •

•

Each axle load must be multiplied by a load safety factor (LSF) for ‘impact effect’. The recommended LSF are: –

For interstate highways and other multilane freeways: LSF = 1.2

–

For highways and arterial streets: LSF = 1.1

–

For roads, residential streets and other streets: LSF = 1.0

–

In special cases, use of LSF = 1.3 may be justified for a premium facility.

Trial and Error Thickness Design Procedure Step 1: Determine axle load repetitions over the design period. (1) Axle Load 52T 50T . . . . . . . . . 22S 20S (kip) (tonne)

(2) (1) × LSF

LSF value depends on road class (Step 2)

(kip) (tonne)

Fatigue Analysis (4) (5) Allowable Fatigue Repetitions Percent

Erosion Analysis (3) (6) (7) Expected Allowable Damage Repetitions Repetitions Percent For each value in For each value in (2), read from Table (2), read from Table 4, 5, 6 or 7 for Determine the expected repetitions 2 or 3 for SRF and Erosion Factor and use it in Figure 7 to axle load.use it inis Figure or for each This also8 known Design determine allowable 9 to determine as the axle load distribution used for Input load repetitions. (3)/(4) allowable load (3)/(6) (Step 1) repetitions. design. Note: Slab thickness is assumed. k and Note: Slab thickness M.R. are obtained is assumed. k and from (3) and (4). M.R. are obtained from (3) and (4). Σ = Σ = Total Total Fatigue Erosion

Trial and Error Thickness Design Procedure Step 2: Determine LSF for each axle load group. (1) Axle Load 52T 50T . . . . . . . . . 22S 20S (kip) (tonne)

(2) (1) × LSF


(kip) (tonne)

Fatigue Analysis (4) (5) Allowable Fatigue Repetitions Percent

Erosion Analysis (3) (6) (7) Expected Allowable Damage Repetitions Repetitions Percent For each value in For each value in (2), read from Table (2), read from Table 4, 5, 6 or 7 for Based on the road Erosion functional class, 2 or 3 for SRF and Factor and use it indetermine Figure 7 to it in Figure 8 oris the LSF use value (which Design determine allowable 9 to determine common for all axle load group). (3)/(6) Input load repetitions. (3)/(4) allowable load (Step 1) repetitions. Note: Slab thickness is assumed. k and Note: Slab thickness M.R. are obtained is assumed. k and from (3) and (4). M.R. are obtained from (3) and (4). Σ = Σ = Total Total Fatigue Erosion

Trial and Error Thickness Design Procedure

(1) Axle Load 52T 50T . . . . . . . . . 22S 20S (kip) (tonne)

(2) (1) × LSF


(kip) (tonne)

Fatigue Analysis Erosion Analysis (3) (4) (5) (6) (7) Expected Allowable Fatiguecombined Allowable Step 3: Determine modulusDamage of Repetitions Repetitions Percent Repetitions Percent subgrade reaction k if each subbase For value inis For each value in (2), read from Table used. (2), read from Table 4, 5, 6 or 7 for 2 or 3 for SRF and Erosion Factor and 4: Determine modulus of rupture of use Step it in Figure 7 to use it in Figure 8 or Design determine allowable M.R. 9 to determine concrete Input load repetitions. (3)/(4) allowable load (3)/(6) (Step 1) repetitions. Step 5: Select a trial slab thickness T0. Note: Slab thickness is assumed. k and Note: Slab thickness M.R. are obtained is assumed. k and from (3) and (4). M.R. are obtained from (3) and (4). Σ = Σ = Total Total Fatigue Erosion

Trial and Error Thickness Design Procedure Step 6: Determine the total fatigue and total erosion for trial slab thickness T0. Recall: Equivalent stress and stress ratio factors for fatigue analysis Erosion factors for erosion analysis

Fatigue and Erosion Analysis Computation Table For trial slab thickness T0: (1) Axle Load 52T 50T . . . . . . . . . 22S 20S (kip) (tonne)

(2) (1) × LSF


(kip) (tonne)

(3) Expected Repetitions

Design Input (Step 1)

Fatigue Analysis (4) (5) Allowable Fatigue Repetitions Percent For each value in (2), read from Table 2 or 3 for SRF and use it in Figure 7 to determine allowable load repetitions.

Erosion Analysis (6) (7) Allowable Damage Repetitions Percent For each value in (2), read from Table 4, 5, 6 or 7 for Determine Erosionallowable Factor and use it in due Figureto 8 or repetitions fatigue 9 to determine for each axle load (3)/(4) allowable load (3)/(6) repetitions.

Note: Slab thickness is assumed. k and M.R. are obtained from (3) and (4).

Note: Slab thickness is assumed. k and How are to obtained M.R. compute? from (3) and (4).

Σ =

Σ =

Total Fatigue

Total Erosion

Equivalent stresses for slabs w/o concrete shoulders

Stress ratio factors vs allowable load repetitions for slabs w or w/o concrete shoulders Recall: Stress Ratio Factor SRF

=

Equivalent Stress Factor ESF Modulus of Rupture MR Known From previous table

Refer to lecture notes for more tables for ESF. There is only one chart for SRF.


(2) (1) × LSF


(kip) (tonne)




(3)/(4)


Erosion Analysis (6) (7) Allowable Damage Repetitions Percent For each value in (2), read from Table 4, 5, 6 or 7 for Compute total Erosion Factor and use fatigue it in Figure 8 or 9 to determine allowable load (3)/(6) repetitions. Note: Slab thickness is assumed. k and M.R. are obtained from (3) and (4).

Σ =

Σ =

Total Fatigue

Total Erosion


(2) (1) × LSF


(kip) (tonne)




(3)/(4)


Determine allowable Σ = Total repetitions due to erosion

Fatigue

Erosion Analysis (6) (7) Allowable Damage Repetitions Percent For each value in (2), read from Table 4, 5, 6 or 7 for Erosion Factor and use it in Figure 8 or 9 to determine allowable load (3)/(6) repetitions. Note: Slab thickness is assumed. k and M.R. are obtained from (3) and (4).

Σ = Total Erosion

Erosion Factors for Slabs with Doweled Joints and no Shoulders

Erosion factors vs allowable load repetition for slabs w/o concrete shoulders

Refer to lecture notes for more tables for ERF and more charts for allowable load repetitions due to erosion (w and w/o shoulders).

Fatigue and Erosion Analysis Computation Table Compute total erosion

For trial slab thickness T0: (1) Axle Load 52T 50T . . . . . . . . . 22S 20S (kip) (tonne)

(2) (1) × LSF


(kip) (tonne)




(3)/(4)


Erosion Analysis (6) (7) Allowable Damage Repetitions Percent For each value in (2), read from Table 4, 5, 6 or 7 for Erosion Factor and use it in Figure 8 or 9 to determine allowable load (3)/(6) repetitions. Note: Slab thickness is assumed. k and M.R. are obtained from (3) and (4).

Σ =

Σ =

Total Fatigue

Total Erosion

Trial and Error Thickness Design Procedure Step 7: Check if both total fatigue and total erosion < 1. If so, the design thickness is obtained. If not, repeat Steps 5 and 6 with a new thickness until both total fatigue and total erosion < 1.

Read your lecture notes and go through the example problem.

AASHTO Rigid Pavement Design •

•

•

•

Result of AASHO road test Currently still used in USA, 1993 edition, with latest revision in 1998. Correlates engineering properties of materials to structural and functional performances of pavement Empirical approach (although there are some mechanistic components in it)

AASHTO Rigid Pavement Design Procedure Step 1: Determine Effective Modulus of Subgrade Reaction k Purpose: •

•

•

Take into consideration seasonal variation of subgrade and subbase conditions. Effective modulus of subgrade reaction k is computed for each month of the year and is used to estimate the relative damage a pavement would receive in each month. Need to account for rigid stratum and potential loss of support due to subbase erosion.

AASHTO Rigid Pavement Design Procedure Step 1: Determine Effective Modulus of Subgrade Reaction k •

•

•

Determine subgrade soil modulus (MR)i and subbase modulus (E SB)i for each month of the year. Select subbase thickness (minimum 4 in., usually 6 in.). Determine composite (k ∞)i for each month of the year using Figure 20 of lecture notes. (∞ indicates infinite depth of subgrade soil.)

Chart for estimating composite modulus of subgrade reaction k ∞


If the depth of subgrade soil layer to a rigid stratum is less than 10 in., make correction to monthly (k ∞)i values using Figure 21 of lecture notes to obtain the corrected monthly k i values.


Compute effective modulus of subgrade reaction k . –

–

–

For each monthly k i value, determine relative damage ui from Figure 22 with an assumed slab thickness. Calculate annual average u. Compute effective k corresponding to u using Figure 22 of lecture notes


Adjust effective modulus of subgrade reaction k to account for potential loss of support arising from subbase erosion, using Figure 23 and Table 7.

Correction of effective modulus of subgrade reaction for potential loss of subgrade

AASHTO Rigid Pavement Design Procedure Step 2: Traffic Analysis A: ESAL Computation B: Initial Year Traffic Loading W 0 W0 =

∑ ( ESAL )( N ) i

i

i

C: Design ESAL applications

where ESALi = ESAL factor of vehicle type i Ni = No. of passes made by vehicle type i per year

n  W0 (1 + r ) − 1  D D W = d L

r

where r = annual traffic growth factor n = design period Dd = directional distribution factor (usually 0.5, vary 0.3-0.7)

Number of Lanes

DL

1

100%

2

80-100%

3

60-80%

≥4

50-75%

56

AASHTO Rigid Pavement Design Procedure Step 3: Reliability Evaluation To account for uncertainty in traffic prediction and chance variation in pavement performance prediction. R (reliability level) and So (overall standard deviation of pavement performance) are incorporated in the AASHTO design process

AASHTO Rigid Pavement Design Procedure Step 3: Reliability Evaluation Recommended R levels: Urban

Rural

85-99.9

80-99.9

Principal Arterials

80-99

75-95

Collectors

80-95

75-95

Local

50-80

50-80

Freeways

Note: Higher R means lower risk that pavement designed would not perform to expectation. Use high R for heavy traffic, high difficulty for traffic diversion and high public expectation.

AASHTO Rigid Pavement Design Procedure Step 3: Reliability Evaluation Recommend So value: From AASHO Road Test, flexible pavements:

So = 0.35

rigid pavements:

So = 0.25

AASHTO Rigid Pavement Design Procedure Step 4: Performance Criteria Design serviceability loss: ∆ PSI = Po – Pt where

Po = initial serviceability index Pt = design terminal serviceability index

Recommended criteria: Pt = 2.5 for major highways 2.0 for minor highways Based on AASHO road test: Po = 4.2 for flexible pavements 4.5 for rigid pavements

AASHTO Rigid Pavement Design Procedure Step 5: Properties of Concrete Concrete elastic modulus Ec •

•

By ASTM Test C469 By ACI correlation for normal weight, normal density Portland Cement Concrete: Ec = 57, 000 f c′

(in psi)

= 4700 f c′ (in MPa) where f c′ = compressive strength in psi (MPa). Concrete modulus of rupture, Sc′ •

Mean Sc′ value can be determined after 28 days using third-point loading test.

AASHTO Rigid Pavement Design Procedure Step 6: Load Transfer Coefficient J •

To account for degree of load transfer across joints.

Values of J: Asphalt Shoulder With Load

PJP or JRCP CRCP

No Load

Tied Concrete Shoulder With Load

No Load

Transfer

Transfer

Transfer

Transfer

Device

Device

Device

Device

3.2

3.8-4.4

2.5-3.1

3.6-4.2

2.9-3.2

-

2.3-2.9

2

AASHTO Rigid Pavement Design Procedure Step 7: Drainage Coefficients C d Quality of Drainage

Percentage of Time Saturated

Excellent

1.40 – 1.35 1.35 – 1.30 1.30 – 1.20

1.20

Good

1.35 – 1.25 1.25 – 1.15 1.15 – 1.00

1.00

Fair

1.25 – 1.15 1.15 – 1.05 1.00 – 0.80

0.80

Poor

1.15 – 1.05 1.05 – 0.80 0.80 – 0.60

0.60

Very Poor

1.05 – 0.95 0.95 – 0.75 0.75 – 0.40

0.40

Quality of Drainage Water Removed in

Excellent 2 hr

Good

Fair

Poor

Very Poor

1 day 1 wk 1 mth No drainage

AASHTO Rigid Pavement Design Procedure Step 8: Determine Slab Thickness


AASHTO Rigid Pavement Reinforcement Design •

•

•

Reinforcement is assumed not to contribute to slab structural capacity nor to prevent cracking, but to hold cracks intact. Crack formation is caused by constraints to slab contraction due to subgrade restraint. Reinforcement must be designed to withstand tensile stresses and control slab elongation that can lead to excessive crack width

Jointed Reinforced Concrete Pavement Steel Design •

The amount of steel reinforcement required in JRCP can be computed as that needed to resist the tensile stress:

1  µr  γ c BLh  2   As = f s

where

Load transfer by grain interlock

B, L and h are the width, length and thickness of the slab

γ c the density of the concrete slab (150 pcf or 23.6 kN/m 3) µ r is the friction coefficient between the slab base & foundation f s is the allowable steel stress.

Jointed Reinforced Concrete Pavement Steel Design •

The percentage steel reinforcement ρs (either longitudinal or transverse reinforcement) can be calculated as:

ρ s =

As Bh

Note:

×100% =

γ c Lµ r 2 f s

×100%

f s = 0.75 f y = 30,000 psi (Grade 40) 45,000 psi (Grade 60) 48,750 psi (Wire fabric)

Material beneath slab

Friction factor µ r Material beneath slab Friction factor µ r The above procedure applies to: Surface treatment River gravel 1.5 Transverse steel design in2.2 jointed reinforced concrete pavement Lime stabilization 1.8 Crushed stone 1.5 Longitudinal steel design in jointed reinforced concrete pavement Asphalt stabilization 1.8 Sandstone 1.2 Transverse steel design in CRCP

Cement stabilization

1.8

Natural subgrade

0.9

Worked Example Determine the longitudinal steel reinforcement requirement for a 30 ft. (9.14 m) long JRCP constructed on crushed stone subbase.

Solution: Assume Density of the concrete slab γc = 0.09 pci (2,500 kg/m3) Allowable working stress of steel reinforcement f s = 30, 000 psi (210 MPa) µr = 1.5 for crushed stone subbase. Percent steel reinforcement:

ρ s =

γ c Lµ r 2 f s

( 0.09 )( 30 × 12 )(1.5) ×100% = × 100% = 0.081% 2 ( 30, 000 )

Longitudinal Steel Design for CRCP Limiting Criteria Crack spacing consideration: 3.5 ft ≤ x ≤ 8.0 ft Derived from spalling and punch-out considerations. Upper spacing limit is to minimize crack spalling and lower limit is to minimize developments of punch-outs.

Crack width consideration: Cw ≤ 0.04 in. (1.0 mm) Based on spalling and water penetration considerations.

Steel stress consideration: σs ≤ 0.75 f y To guard against steel fracture and to ensure that steel does not undergo any plastic deformation.

CRCP Longitudinal Steel Design Procedure Part 1: Crack Spacing Consideration – (P max)1 and (Pmin)1 •

Step 1: Select design crack spacing (3.5 ft to 8.0 ft)

•

Step 2: Determine steel and concrete thermal coefficient. Thermal coefficient of expansion for reinforcement steel:

αs = 5.0 × 10-6 in/in/oF = 9.0 × 10-6 in/in/oC


Step 2: Determine steel and concrete thermal coefficient. Thermal coefficient of expansion for concrete αc varies with water/cement ratio, age, richness of mix, relative humidity and aggregate type. Type of Coarse

-6 o c (10 in/in/ F)

Aggregate

Type of Coarse

-6 o c (10 in/in/ F)

Aggregate

Quartz

6.6

Granite

5.3

Sandstone

6.5

Basalt

4.8

Gravel

6.0

Limestone

3.8

Given thermal coefficients, compute

s /

c ratio


Step 3: Select bar diameter: ½ ″, 5/8″ or ¾″

•

Step 4: Determine concrete shrinkage at 28 days. Shrinkage can be approximately related to concrete strength: Indirect tensile

Shrinkage Z (in./in.)

strength f t (psi) ≤ 300

0.0008

400

0.0006

500

0.00045

600

0.0003

≥ 700

0.0002

Note: f t is determined by indirect tensile test at 28 days according to AASHTO T198 or ASTM C496. f t ≈ 0.86 × Modulus of Rupture

CRCP Longitudinal Steel Design Procedure Part 1: Crack Spacing Consideration – (Pmax)1 and (Pmin)1 Step 5: Tensile stress caused by wheel load due to initial construction or truck traffic. Use Figure 25 to determine

σw Input parameters: Slab thickness D, maximum wheel load, and effective k


Step 6: Determine from Figure 26 of lecture notes:

Min percent steel reinforcement ( P min)1 corresponding to x = 8.0 ft Max percent steel reinforcement ( P max)1 corresponding to x = 3.5 ft

Min % steel reinforcement for crack spacing criterion Part 1: Crack Spacing Consideration – (P max)1 and (Pmin)1 •

Step 6: Determine from Figure 26 of lecture notes:

Min percent steel reinforcement (Pmin)1 corresponding to x = 8.0 ft Max percent steel reinforcement (Pmax)1 corresponding to x = 3.5 ft

CRCP Longitudinal Steel Design Procedure Part 2: Crack Width Consideration – (Pmin)2 •

•

•

Step 1: Select design crack width (take C w = 0.04″ to determine Pmin). Step 2: Use the bar diameter: ½ ″, 5/8″ or ¾″ from Part 1 Step 3. Step 3: Tensile stress caused by wheel load due to initial construction or truck traffic. Use Figure 25 of lecture notes to determine σw. (Same as Part 1 Step 5)

CRCP Longitudinal Steel Design Procedure Part 2: Crack Width Consideration – (Pmin)2 •

Step 4: Determine concrete tensile strength f t f t can be obtained from indirect tensile test at 28 days according to AASHTO T198 or ASTM C496 f t ≈ 0.86 × Modulus of Rupture (f t was used as an input in Part 1 Step 4)

•

Step 5: Determine from Figure 27 of lecture notes, minimum percent steel reinforcement (Pmin)2.

Min % steel reinforcement for crack width criterion

CRCP Longitudinal Steel Design Procedure Part 3: Steel Stress Consideration – (Pmin)3 •

Step 1: Determine the allowable steel working stress, σs.

Value of steel working stress σs (psi): 28-day Indirect Tensile Strength (psi)

Rebar Size No. 4

No. 5

No. 6

≤ 300

65

57

54

400

67

60

55

500

67

61

56

600

67

63

58

700

67

65

59

≥ 800

67

67

60


Step 2: Determine design temperature drop DT.

∆T = T H − T L where TH = average daily high temperature during the month the pavement is constructed TL = average daily low temperature during the coldest month of the year •

Step 3: Determine concrete shrinkage at 28 days. (Same as Part 1 Step 4)


Step 4: Tensile stress caused by wheel load due to initial construction or truck traffic. (Same as Part 1 Step 5)

•

Step 5: Determine concrete tensile strength f t . (Same as Part 2 Step 4)

•

Step 6: Determine from Figure 28 of lecture notes, minimum percent steel reinforcement (Pmin)3.

Min % steel reinforcement for steel stress criterion

CRCP Longitudinal Steel Design Procedure Part 4: Finalize Design •

Step 1: The design percent longitudinal steel should fall within:

Pmin = max

{( P ) , ( P ) , ( P ) } min 1

min

2

min

3

and

Pmax = ( Pmax )1 If Pmin > Pmax, design is infeasible and the design has to be revised: change bar diameter, slab thickness or concrete materials until Pmin < Pmax.

CRCP Longitudinal Steel Design Procedure Part 4: Finalize Design •

Step 2: Determine the number of steel bars required. N min =

Pmin ( Slab thickness )( Slab width )

( 4 )φ

100 π

2

and N max =

Pmax ( Slab thickness )( Slab width )

( 4)

100 π

where φ = bar diameter.

φ 2


Mechanistic Empirical Pavement Design •

•

Mechanistic-Empirical Pavement Design (MEPD) method uses empirical relationships between cumulative damage and pavement distress to determine the adequacy of a pavement structure to carry the expected traffic load on the pavement. NCHRP 1-37: AASHTO Interim Mechanistic-Empirical Pavement Design Guide (MEPD-G) – AASHTO Interim Design Method

Overview of Pavement Design Procedure

Overview of Pavement Design Procedure

Design Procedure •

•

•

Select a trial design (using AASHTO 1993 procedure). Select appropriate performance criteria (threshold values) and design reliability level for design. Obtain all inputs for the trial pavement structure under consideration. –

–

–

Input level 1: Uses parameter directly measured from site and therefore is site-specific. Input level 2: Use of correlation and regression analyses to obtain input parameters from data at other sites that have been obtained at Level 1. Input level 3: Best estimates or default values obtained from national data.

03_DesignofRigidPavement

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