continuously reinforced concrete pavement
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Continuously Reinforced Concrete Pavement. Factors that must be Considered by Pavement Engineer. Must Know. Short Slabs can be designed and load transfer is provided by Aggregate Interlock ( Contraction Joint ) - PowerPoint PPT PresentationTRANSCRIPT
Pavement Engineering M. Kharabsheh
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Continuously Reinforced Concrete Pavement
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Factors that must be Considered by Pavement Engineer. Must Know
Short Slabs can be designed and load transfer is provided by Aggregate Interlock (Contraction Joint)
The crack or contraction joint can be tied with Tie Bars of distributed steel to prevent the crack or joint from opening
Movement at joints can be permitted (Expansion Joints) and Dowel Bars are placed at the Joints to provide load transfer.
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Construction Joints with Tie Bars
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Contraction joint showing saw cut depth
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Aggregate Interlock (Contraction Joint)
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(Contraction Joint)
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(Contraction Joint)
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(Expasion Joint)
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Dowel Bars Storage
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Dowel Bars Construction
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Dowel Group Action
When traffic loads are applied at a joint, a portion of this load is transferred through the dowel bar into the next slab (this portion should be 50% for perfect load transfer)
The dowel bar immediately under the applied load bears a major portion of the load with other dowels assuming progressively lesser amount.
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Dowel Group Action
Maximum negative moment occurs at a distance 1.8L from the load.
The dowel bar immediately under the applied load carries full capacity (one full unit) and the portion of each dowel decrease to reach zero at a distance of 1.8L.
The corner dowel is the critical one for the edge load.
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Example
Load is 9000lb on each wheel
Assume 50% load transfer
Thickness= 10’ and Subgrade reaction = 50pci
Radius of relative Stiffness l = 51.1’ (from table 3.1 page 84)
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Example
X1 = 1 1/92 = x2/(92-12) x2 = 0.87 Effective Dowels group for load at A = xi =
1+.87+.74+.61+.48+.35+.22+.08 = 4.35 Total load carried by dowel
At point A = load from wheel
At A plus load from wheel at
Point B
Total load at A = 4500/(4.35*1)
+ 4500/(7.32*0.22) = 1035+135= 1170 lb
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Example on Bearing Stress by Dowel on Concrete
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