17 - Curling Stresses in Concrete Slabs - Civil Engineering - Curling Stresses in Concrete... · Calculate the curling stresses in a concrete slab 25′ 12′ 8″ thick subject to

Rigid Pavement Mechanics

Curling Stresses

Major Distress Conditions

• Cracking– Bottom-up transverse cracks– Top-down transverse cracks– Longitudinal cracks– Corner breaks– Punchouts (CRCP)

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Major Distress Conditions

Cracks

Punchouts

Corner Break

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Major Distress Conditions

• Cracking– Bottom-up transverse cracks– Top-down transverse cracks– Longitudinal cracks– Corner breaks– Punchouts (CRCP)

• Joint Faulting• Pumping

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Major Distress Conditions

Faulting

Pumping

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Sources of Stress

• Wheel Loads• Curing Shrinkage• Thermal Contraction/Expansion• Thermal Curling• Moisture Warping

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Design Considerations

• Slab Thickness• Base Type and Thickness• Drainage• Joint Spacing• Temperature Steel• Dowel Bars• Tie Bars

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Beam Bending

qo = q × b (applied load per unit length)

qo

ko

R R

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Beam Bending

Shear and Moment Relationships

odV qdx

dM Vdx 2

2 od M qdx

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Beam Bending

Moment-Curvature Relationships

MEI

2

2

1

d wR dx 2

2 d wM EIdx

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Beam Bending

Governing Differential Equation for Beam Bending

2 2

2 2

od d wEI qdx dx

4

4 od wEI qdx

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Beams on Elastic Foundations

qo = q × b (applied load per unit length)

qo

ko

R R

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Beams on Elastic Foundations

qo

ko

qo = q × b (applied load per unit length)

p = ko w = k b w (resistance per unit length)Modulus of Subgrade Reaction

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Beams on Elastic Foundations

Shear and Moment Relationships

ookdx

wV qd

dM Vdx 2

2 o od M kd

w qx

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Beam Bending

Moment-Curvature Relationships

MEI

2

2

1

d wR dx 2

2 d wM EIdx

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Beams on Elastic Foundations

Governing Differential Equation for Beam Bending

2 2

2 2

o od d wEI qdx

k wdx

4

4 o od wE q

xk

dwI

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Beams on Elastic Foundations

Solution of the Homogeneous Differential Equation

1 2 3 4sin cos sin cos x xw e C x C x e C x C x

4

4 ok

EI

4

4 0 od wEI k wdx

(flexibility)

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Slabs on Elastic Foundations

Moment-Curvature Relationships

2

2 d wM EIdx

Beams

2 2

2 2

xw wM Dx y

Slabs

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Slabs on Elastic Foundations

Slab Stiffness

3

212 1EhD

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Slabs on Elastic Foundations

Governing Differential Equation for Slab Bending

Modulus ofSubgrade Reaction

4

4 o okdEIx

ww qd

Beams

4 4 4

4 2 2 42w w wD kw qx x y y

Slabs

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Slabs on Elastic Foundations

Solution of the Homogeneous Differential Equation

4

4 ok

EIBeam Flexibility

3

4 4 212 1

D Eh

k kSlab Stiffness

Radius ofRelativeStiffness

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Stresses Due to Curling

Temperature Curling

DOWNWARD CURLINGWARM

COOL

UPWARD CURLING

WARM

COOL

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Moisture Warping

DOWNWARD WARPINGWET

DRY

UPWARD WARPING

WET

DRY

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Reference Axes

x

y

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Infinite Slab Curling

Hooke’s Law for an Infinite Elastic Plate

y x

y E E

yx

x E E

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Slab Bent About y-axis

x

y

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Infinite Slab Curling

If slab is bent about the y axis, y = 0

y x

0

y xy E E

0 y x

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Infinite Slab Curling

Slab bent about y axis

21x

xE

21 x x x

x E E E

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Slab Bent About x-axis

x

y

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Infinite Slab Curling

If slab is bent about the x axis, x = 0

x y

0

yxx E E

0 x y

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Infinite Slab Curling

Slab bent about x axis

21

yy

E

21

y y yy E E E

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Infinite Slab Curling

Strains due to temperature difference t through slab depth

2

x yt

Coefficient ofThermal Expansion

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Coefficient of Thermal Expansion

AggregateType

Coefficient(10-6 in/in/oF)

Quartz 6.6Sandstone 6.5

Gravel 6.0Granite 5.3Basalt 4.8

Limestone 3.8Average 5.5

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Infinite Slab Curling

Stresses due to curling about y axis due to temperature difference

2 2

2

1 2 1

2 1

xx

y x

E E t

E t

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Infinite Slab Curling

Stresses due to curling about x axis due to temperature difference

2 2

2

1 2 1

2 1

yy

x y

E E t

E t

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Infinite Slab Curling

Maximum curling stresses in an infinite slab

, 2 22 1 2 1x y

E t E t

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Finite Slab Curling

Curling stresses in the Interior of a Finite Slab

, 1 22 22 1 2 1x y

E t E tC C

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Finite Slab Curling

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Finite Slab Curling

Curling stresses at the Edges of a Finite Slab

, 1 2x yE tC

0 0 y x

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Temperature Gradients

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SourceSlab

ThicknessTemperature

Gradient

AASHO Road Test 6.5" 3.2F/in 1.5F/in

Arlington Road Test 6" 3.7F/in

9" 3.4F/in

Typical Assumption 6“ – 9" 3.0F/in 1.5F/in

Example

Calculate the curling stresses in a concrete slab 25′12′8″ thick subject to a daytime temperature difference of 24°F (i.e., a temperature gradient of 3°F/in). Assume the slab is resting on a foundation with a 200-psi/in modulus of subgrade reaction.

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Standard Assumptions

• Assume =0.15(useforallPCC)• Assume =5.510–6 in/in/°F(typical)• Assumefc =5000psi(typical)• Estimate psi

• Estimate psi

• Estimate psiNotneededforthisproblem

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