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Joint Types and Behavior

Joint Types and Behavior. Rigid Pavement Design Course Jointing Patterns

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Joint Types

and Behavior

Rigid Pavement Design Course

Jointing Patterns

Rigid Pavement Design Course

Joint Functions

1) Control cracking

2) Provide space and freedom of movement

3) Facilitate construction

Slab Stiffness Components: Joint

• Thickness : shear capacity

• - Slab action : basin area

• Crack width : shear capacity

• Subbase/Slab interface friction

• Load transfer

• 1) Shear – aggregate interlock

2) Dowel

Rigid Pavement Design Course

A B

Contraction Joints• Control trans. Cracking

• Formed by weakened joint

- Saw Cutting

- Grooving

- Insert Strip

- With and without dowel

Rigid Pavement Design Course

Construction Joints• Planned interruptions

• Isolation joint

• Transverse or Longitudinal

ED

Rigid Pavement Design Course

Isolation Joint• Space for expansion

- temp/moisture increase

- Use compressible filler, LT or slab thickening

C

Rigid Pavement Design Course

Longitudinal Joints

•Between paving lanes

- can be a construction or contraction joint

• Butt or Keyway joint

• Placed with tie bar

• Wide pavement sections:

- combination or tied and doweled joints

Rigid Pavement Design Course

Longitudinal Joints

Tie Bar

Tie Bolt

Dowel Bar

F

H

G

Rigid Pavement Design Course

Load Transfer Systems

Aggregate Interlock• Simple• Small openings• Low traffic• Wears out

Mechanical load transfer• Many designs• Round dowel bars most popular• Installed in a single row

Keyed Joints• Long. Joints; 8” slabs or greater

Bent or ThreadedZJoint

Not Recommended

Tied Joints

•# 4 or #5 bars;

•24” to 48” long

•18” - 48” spacing

Rigid Pavement Design Course

0% Load Transfer

60.02δδloadedtot

0δunloaded

unloaded

δloaded

δ

tot

δ

Deflection of Loaded Slab

Deflection of unloaded Slab

Total Deflection

Where:

Applied Wheel Load (P)

0.026"Loaded δ 0 Unloadedδ

Rigid Pavement Design Course

Applied Wheel Load (P)

100% Load Transfer

30.01δ2

1δδ

totunloadedloaded

loadedunloadedtotδδδ

100%100δ

δLT

L

unlδ

0.013"Loaded δ 0.013" Unloadedδ

Rigid Pavement Design Course

Load Transfer

U 100L

LTE x

σ

Total U L fe L LLTE

LTE LTE

LTEσ

LTEδ

LTEδ

0

1.0(.55,100% LTE)

(.67,44% LTE)(1.0,0% LTE)

100u

L

LTE x

Total U L

1fe

L LTE

/

/

e w LTE

e w o LTE

Rigid Pavement Design Course

DowelBar

Subgrade

Critical Stress forMid Slab Loading

SlabThickness

Slab Length (L)

Single Axle Loading

Agg

Subbase

he

w

s

a

Traffi

c Lan

e

Should

erH

inge

Join

t

Hin

ge Jo

int

Do

3

4 2

Eh=

12 1- k

Tied Shoulder Effect

Rigid Pavement Design Course

7mm Wide

Lubricated, smooth dowel bar

d/2

(a)

(b)

d/4 except forearly entry sawcut(25 mm min.)

Load Transfer – Crack Width Requirements

Rigid Pavement Design Course

Load Transfer – Crack Width Factors

Rigid Pavement Design Course

101520253035404550

6 7 8 9 10 11 12 13 14 15 16

Slab Thickness (in)

Cra

ck W

idth

(m

ils)

Load Transfer – Crack Width Requirements:Thickness Factors

Rigid Pavement Design Course

Emergency Joints?

Rigid Pavement Design Course

Eff

ect

ive

ne

ss

, pe

rce

nt

Loading Cycles, 100 000

Influence of Joint Opening on Effectiveness, 9 Inch Concrete Slab, 6 Inch Gravel Subbase (82)

0 1 2 3 4 5 6 7 8 9 100

40

60

80

100

200.085

0.065

0.045

0.035

Joint Opening 0.025 in.

Rigid Pavement Design Course

Eff

ect

ive

ne

ss

, pe

rce

nt

Loading Cycles, 100 000

0 1 2 3 4 5 6 7 8 9 100

20

40

60

80

100

Influence of Joint Opening on Effectiveness, 7 Inch Concrete Slab, 6 Inch Gravel Subbase (82)

0.065 0.045

0.035

0.025

Joint Opening 0.015 in.

Rigid Pavement Design Course

Influence of joint opening on effectiveness (9 in. slab, cement stabilized subbase, k=542 pci)

Loading Cycles, 100 000

Eff

ect

ive

ne

ss

, pe

rce

nt

0 1 2 3 4 5 6 7 8 9 100

20

40

60

80

100

9-in. Slab

0.065-in.

0.065-in.

0.035-in. joint opening

Rigid Pavement Design Course

Functions of Subbase

1.) To Provide a stable construction platform

2.) To control the depth of frost penetration

3.) Prevent erosion of the pavement support

4.) Provide uniform slab support

5.) Facilitate drainage

6.) Provide increased slab support

Rigid Pavement Design CourseEffect of Slab Thickness on Maximum Shear Stresses at

Joint Interface

Ma

x. S

he

ar S

tre

ss

at

Join

t In

terf

ac

e, p

si

60

50

40

30

20

10

010 12 14 16 18 20 24

Agg 107

Agg=106

Agg=105

Agg=5x104

Agg=104

Agg=5x103

Agg=103

Agg=102

Rigid Pavement Design Course

Jo

int

Eff

. %

10,000

Subgrade Mod. X Mod. Of Relative Stiff., psi-1

Figure 4-18. Relation between Joint Efficiency (Eff) and

Spring Stiffness (Agg)

0.2 0.6 1.0 1.4 1.8 2.2 2.6 3.0 3.4

100

80

60

40

20

0

Agg=106psi

Agg=105psi

Agg=5x104psi

Agg=104 psi

Agg=5x103psi

Agg=103psi

Agg=102psi

Rigid Pavement Design Course

Modulus of Aggregate Stiffness, in.

Effect of Aggregate Interlock in Reducing Maximum Tensile Edge Stresses

Max

. T

ensi

le E

dg

e S

tres

s x

(Sla

b T

hic

knes

s)2 ,

k-in

.in

.

Free E

dge

Interior

Edge W/Agg

Agg 106

30 60 100

144

120

96

72

48

24

0

Agg=105

30 60 100

144

120

96

72

48

24

0

Rigid Pavement Design Course

30 60 100

144

120

96

72

48

24

0

Agg=104

30 60 100

144

120

96

72

48

24

0

Modulus of Aggregate Stiffness, in.

Max

. T

ensi

le E

dg

e S

tres

s x

(Sla

b T

hic

knes

s)2 ,

k-in

.in

.

Agg102

Rigid Pavement Design Course

Effect of Aggregate Interlock in Reducing Maximum Edge Deflection

Load

Subgrade Mod. x (Mod. Of Relative Stiff.)2, in.

Ma

x. E

dg

e D

efl

ect

ion

, in

.

0 0.08 0.16 0.24 0.32

0.12

0.10

0.08

0.06

0.04

0.02

00 0.08 0.16 0.24 0.32

0.12

0.10

0.08

0.06

0.04

0.02

0

Free

Edg

eInterio

rEdge W

/Agg

Agg 106Agg=105

Rigid Pavement Design Course

0 0.08 0.16 0.24 0.32

0.12

0.10

0.08

0.06

0.04

0.02

0

0.12

0.10

0.08

0.06

0.04

0.02

00 0.08 0.16 0.24 0.32

Agg=104 Agg102

Ma

x. E

dg

e D

efl

ect

ion

, in

.

Load

Subgrade Mod. x (Mod. Of Relative Stiff.)2, in.

Rigid Pavement Design Course A Typical Finite-Element Mesh Used for Analysis

of Keyed Joints

P=200 lb/in.

16 "

60 "

45" 45"

Rigid Pavement Design CourseTensile Stress Contours for a Standard Key on a 10 in. Cement

Stabilized Base

P=200 psi

12" 12"

Esub = 10,000 psi

200 150 100 50

0

50

100

150

200

0 0

0

200

150

100

50

400350300

300

200

100 0

Rigid Pavement Design Course

Scale

Pounds / in.

Distribution of Nodal Forces Normal to the Contact Boundaries for Different Key Designs

Standard

Key

0.2h

0.1h

483

Deep Key

0.2h

0.2h 651

Large Key

0.4h

0.1h

822

0 100 200 300

Rigid Pavement Design Course

Round Key

0.2h

0.1h

859

0.2h

0.1h

Round Smooth Key

h

1409

Scale

Pounds / in.

0 100 200 300

Rigid Pavement Design Course

Effect of Key Design on Maximum Tensile Stress in the Slab*

Key Design KeywayPsi (MPa)

Key Psi (MPa)

Standard Key 312 (2.15) 586 (4.04)Large Key 1023 (7.05) 1229 (8.47)Deep Key 259 (1.78) 589 (4.06)

Round Key 451 (3.11) 501 (3.45)Round Smooth

Key201 (344) 344 (2.37)

Z-Key 721 (4.97) 274 (1.89)

* Slab thickness was 16 in. (40.6 cm)

Rigid Pavement Design Course

(a) Doweled Joint

Figure 4-37. Possible Joint Designs for the Example Problem

(b) Tied Joint with Aggregate Interlock

(c) Joint With Aggregate Interlock On Stab. Base

12"

12" 12"

4"

Rigid Pavement Design Course

(e) Thickened Edge Joint (f) Butt Joint

(d) Butt Joint on Stab. Base

Figure 4-37. Continued

12"

6"

12" 16"