Question: In the drilled shaft design example 9 - 6 , the design load is now increased to 2 0 0 tons. What will be the

In the drilled shaft design example 9-6, the design load is now increased to 200 tons. What will
be the shaft length (30 points)? What is the settlement under load of 250 tons (20 points)?
In the example, beta (\beta ) is used to calculate side resistance in sand. Now, a different method is
proposed for unit side resistance in sand as follows (NHI Course No.132014,13.3.5.1
Cohesionless Soils).
\beta ~~(1-sin\phi ^('))((\sigma _(p)^('))/(\sigma _(v)^(')))^(sin\phi ^('))tan\phi ^(')
Where both \phi ^(') and \sigma _(p)^(') are evaluated, most commonly, by correlation to SPT N-values.
\sigma _(p)^(')= effective vertical pre-consolidation stress, which can be calculated as
(\sigma _(p)^('))/(p_(a))~~0.47(N_(60))^(m)
, m=0.8 for silty sands, the sand in example 9-6.
p_(a) is atmospheric pressure; \sigma _(v)^(')= effective vertical over-burden pressure.
You may choose a method to estimate \phi ^(') from measured SPT N-values.
The equation below is the recommended correlation for estimating \phi ^(') for the purpose of
evaluating unit side resistance of drilled shafts in cohesionless soils.
\phi ^(')=27.5+9.2log[(N_(1))_(60)]
Use alpha method for clay, and use both beta methods (the one in the example and this new one)
for sand. Compare the side resistance by both beta methods. The ultimate resistance of your
drilled shaft shall be no more than 50 tons higher than the required ultimate resistance.
Please clearly explain your steps, formula, design char(t)/(t)able (if any), and references.
In the drilled shaft design example 9 - 6 , the

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Civil Engineering Questions!