17 - curling stresses in concrete slabs - civil engineering - curling stresses in concrete... ·...
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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
• Cracking– Bottom-up transverse cracks– Top-down transverse cracks– Longitudinal cracks– Corner breaks– Punchouts (CRCP)
• Joint 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
Governing Differential Equation for Beam Bending
2 2
2 2
od d wEI qdx dx
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4 od wEI qdx
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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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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
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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
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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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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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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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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