360R-06 Design of Slabs-on-Ground
—Effect of eliminating sawcut contraction
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- 360R-56 ACI COMMITTEE REPORT
| 13.14—Effect of eliminating sawcut contraction
joints with post-tensioning or shrinkage- compensating concrete The total amount of drying shrinkage of concrete is magnified when it is placed in large blocks without inter- mediate sawcut contraction joints. The construction joints surrounding 10,000 to 12,000 ft 2 (930 to 1100 m 2 ) of post- tensioned or shrinkage-compensated concrete slabs will commonly open much more than construction joints for the same areas of conventional portland-cement concrete slabs. This is because the intermediate sawcut contraction joints within the latter slabs will take up most of the shrinkage. Post-tensioned and shrinkage-compensated slabs do not have intermediate sawcut contraction joints. Where vehicle traffic will cross construction joints in post-tensioned or shrinkage-compensated slabs-on-ground, the joints should Fig. 13.4—Effect of slab length on warping and warping stress in an exposed highway slab (from Eisenmann [1971]). 360R-56 ACI COMMITTEE REPORT be doweled, and the top edges of the construction joints should be protected with back-to-back steel bars or angles, epoxy-armored edges, or by other equally durable material. 13.15—Summary and conclusions Designers of enclosed slabs-on-ground can reduce shrinkage cracking and shrinkage curling by considering the features that affect these phenomena. The following checklist indicates factors that should be addressed. SUBGRADE CONDITIONS • Before and during slab installation, check for smoothness, dryness, and permeability of the base and subgrade. Measure the base and subgrade moisture content. • Do not use a vapor retarder/barrier unless required to control moisture transmission through the slab. If it is used, decide whether an aggregate blotter should be used over the vapor retarder/barrier. Fig. 13.5—Slab length increases beyond a certain amount do not increase warping stress in the slab interior (Kelly 1939). (Note: 1 psi = 0.006895 MPa; 1 ft = 0.3048 m.) Fig. 13.6—Representative radial stresses for an effective temperature difference of 30 °F between top and bottom (Leonards and Harr 1959). (Note: 1 psi = 0.006895 MPa; 1 in. = 25.4 mm; 1 °F [ ΔT] = 0.56 °C.) |