Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

check the attachment

1 answer below »
Answered Same Day May 11, 2021

Solution

Hemalatha answered on May 18 2021
145 Votes
Solution – pile foundation – 57474
Proposed multy storeyed building.
Problem statement- Find Ultimate axial load capacity (Q ult) and then using a suitable FoS, provide Allowable axial load (Qa) to use as the design load.
1a) alpha and beta method – TSA and ESA
The following figure represents the forces acting on the pile

Figure – Forces acting on the pile    Pile Installation
The expression for ultimate load capacity is given by the following equation.
Qu = q ba Ab + fs As
In this problem, the type of soil is CLAY.
Therefore, Qu = cub * Nc* Ab + α * cu * As
Where represents cub the undrained cohesion (Pile base)
Nc represents the bearing capacity factor for pile (or deep foundation);
whose value is considered as 9
α represents the reduction facto
cu represents the undrained cohesion (pile embedded length)
Ab and As represent the cross-section area of pile base and area of contact of pile respectively
Resistance of pile tip
Qb = q ba A
    = cub * Nc* A
Provide pile of square cross-section whose side is 1m.
cub is given as 150 kPa = 150 kN / m2
Nc is taken as 9
Ab = 1 m * 1m = 1 m2
Therefore Q b = 150 * 1 * 9 = 1350 kN
Resistance of shaft
Alpha method
The following table shows the valuse of reduction factor α
    Consistency of the soil
    N
    Bored piles (cast-in situ)
    Driven piles (precast piles)
    Soft – Very Soft
    Less than 4
    0.7
    1
    Medium Stiff
    Between 4 and 8
    0.5
    0.7
    Stiff
    Between 8 and 15
    0.4
    0.4
    Stiff to Hard
    Greater than 15
    0.3
    0.3
It is evident from the table that the reduction factors for stiff and hard soils, in the case of driven piles are greater than those of bored piles, whereas, both the types of piles have same value of α for medium stiff, soft and very soft soils.
Total depth of pile shaft = 9m
Layer 1 – depth = 4m (Firm to stiff)
cu = 80 kN / m2
As = 4 * 1 = 4 m2
α = 0.4 - bored
α = 0.4 – driven
Qs = α * cu * As
= 0.4 * 80 * 4
= 128 kN ---Layer 1 – both bored and driven piles
Layer 2 – depth = 5 – 4 = 1 m(Firm to stiff)
cu = 61 kN / m2
As = 1 * 1 = 1 m2
α = 0.3 - bored
α = 0.3 – driven
Qs = α * cu * As
= 0.3 * 61 * 1
= 18.3 kN ---Layer 2 – both bored and driven piles
Layer 3 – depth = 1m (soft)
cu = 55 kN / m2
As = 1 * 1 = 1 m2
α = 0.7 - bored
α = 1.0 – driven
Qs = α * cu * As
= 0.7 * 55 * 1
= 38.5 kN ---Layer 3 –bored piles
Qs = α * cu * As
= 1.0 * 55 * 1
= 55 kN ---Layer 3 –driven piles
Layer 4 –...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here