Semester II 1434/2013 Cl Book Time lme: For the folowingstatements answer byIfor
ID: 1843963 • Letter: S
Question
Semester II 1434/2013 Cl Book Time lme: For the folowingstatements answer byIfortrueandF for faise. Anaveren thispogea 1.inasoa mass waterfowsfrompoints of hoher total headstolowerwaterheads. 2.On principal planes there exist normalandshear stresses. 3. Normally consolidated days are soferthan overconsolidatedclays. 4.Adouble drained daylayeraonsolidates fasterthansinge drained layer. 5.Cohesionless soils have higher permeability than cohesive sols. The failure plane inatriaxialsheartestishorizontal 7. In the direct shear test the failure plane is horizontal, 8. In compression test the minor principalstressiszero. 9. Overburden stresses increase with depth. 10. Sod stresses due to extemalloads increase with depth. 11. In shear tests disturbed samples should be used. 12 In consolidation testsdisturbed samples should be used. 13, CD tests are slow because the pore water pressure mustbe kept 0. 14- For loose sands and normally consolidated clays c 0. 15. For loose sandsandnormally consolidated clays ultimate or residual strength is determined.Explanation / Answer
1
T
Water always flows from hidh head to low head.
2
F
Shear stress is zero at principal planes
3
T
Overconsolidated clays are considerably tougher at a particular stress level than normaly consolidated clays as the water content and void ratioo in normaly consolidated clay is larger than overconsolidated clay which maked it softer
4
T
Time of settlement = Tv *d^2/Cv ,d = H/2 for double drainage and d = H for single drainage
5
T
That is why cohesionless soil consolidate earlier , K is proprtional to e^3/1+e ,e is void ratio ,void ratio more in cohesionless soil
6
F
It is at an angle 45 degree + (phi)
7
T
It is parallel to the sapmle layer
8
T
Unconfine compression test is atype of tri=axial stress in which confining pressure is Zero ,so minor principal stress is zero
9
T
Overburden pressure is directly proportional to depth
10
F
Only overburden pressure increases with depth
11
F
For shear tests,consolidation and permeability undisturbed samples required
12
F
For shear tests,consolidation and permeability undisturbed samples required
13
T
In CD test drainage of sample ia allowed to make pore pressure Zero which requires lot of time
14
T
For loose sands and Normaly consolidated samples C is taken to be Zero
15
T
Because peak strength is determined only for overconslidated and dense soils as they can reach maximum strength
16
F
Dense sands and overconsolidated soils do not expand much as void ratio is very less
17
F
Unconfined soil test is mainly used for clay samples
18
F
For granular infinite slopes Fs = Tan (phi)/Tan (i) .phi is friction angle and i= slope angle so no H is required
19
T
In active state wall is assumed to move away
20
F
In active state wall is assumed to move away
1
T
Water always flows from hidh head to low head.
2
F
Shear stress is zero at principal planes
3
T
Overconsolidated clays are considerably tougher at a particular stress level than normaly consolidated clays as the water content and void ratioo in normaly consolidated clay is larger than overconsolidated clay which maked it softer
4
T
Time of settlement = Tv *d^2/Cv ,d = H/2 for double drainage and d = H for single drainage
5
T
That is why cohesionless soil consolidate earlier , K is proprtional to e^3/1+e ,e is void ratio ,void ratio more in cohesionless soil
6
F
It is at an angle 45 degree + (phi)
7
T
It is parallel to the sapmle layer
8
T
Unconfine compression test is atype of tri=axial stress in which confining pressure is Zero ,so minor principal stress is zero
9
T
Overburden pressure is directly proportional to depth
10
F
Only overburden pressure increases with depth
11
F
For shear tests,consolidation and permeability undisturbed samples required
12
F
For shear tests,consolidation and permeability undisturbed samples required
13
T
In CD test drainage of sample ia allowed to make pore pressure Zero which requires lot of time
14
T
For loose sands and Normaly consolidated samples C is taken to be Zero
15
T
Because peak strength is determined only for overconslidated and dense soils as they can reach maximum strength
16
F
Dense sands and overconsolidated soils do not expand much as void ratio is very less
17
F
Unconfined soil test is mainly used for clay samples
18
F
For granular infinite slopes Fs = Tan (phi)/Tan (i) .phi is friction angle and i= slope angle so no H is required
19
T
In active state wall is assumed to move away
20
F
In active state wall is assumed to move away
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