The soil profile shown below is subjected to a uniformly distributed load (UDL) on ground surface of 600 kN/m². The underground soil consist of two normally consolidated clay (N.C.C.) layers, and between them one sand layer with properties as shown on the figure. If the ground water table is at the ground surface, and the (AP) assumed to be the same at both clay layers, Calculate the total consolidation settlement Sc from the two clay layer ? assume compression index, Cc-0.007(LL-10). 4.0 m 3.0 m 2.0 m Clay Sand Clay AP = 600 kN/m² W.-35%. LL-40%. G.-2.65 Ys at = 18 kN/m³ W. 40, LL=45%, G,=2.78 ▼ G.W.T

Structural Analysis
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Chapter2: Loads On Structures
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The soil profile shown below is subjected to a uniformly distributed load (UDL) on ground surface
of 600 kN/m². The underground soil consist of two normally consolidated clay (N.C.C.) layers, and
between them one sand layer with properties as shown on the figure.
If the ground water table is at the ground surface, and the (AP) assumed to be the same at both
clay layers, Calculate the total consolidation settlement Sc from the two clay layer? assume
compression index, Cc=0.007(LL-10).
4.0 m
3.0 m
2.0 m
Clay
Sand
Clay
AP=600 kN/m²
W, 35%. LL-40%. G.-2.65
Ys at = 18 kN/m³
W, 40, LL=45%, G,-2.78
▼ G.W.T
Transcribed Image Text:The soil profile shown below is subjected to a uniformly distributed load (UDL) on ground surface of 600 kN/m². The underground soil consist of two normally consolidated clay (N.C.C.) layers, and between them one sand layer with properties as shown on the figure. If the ground water table is at the ground surface, and the (AP) assumed to be the same at both clay layers, Calculate the total consolidation settlement Sc from the two clay layer? assume compression index, Cc=0.007(LL-10). 4.0 m 3.0 m 2.0 m Clay Sand Clay AP=600 kN/m² W, 35%. LL-40%. G.-2.65 Ys at = 18 kN/m³ W, 40, LL=45%, G,-2.78 ▼ G.W.T
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