Geotech PM

[geo_dirt]

[Hi Santiagj,

I'm stuck with problem 11.7 (piles chapter). The problem is asking to find the allowable load for the pile and request that use the (alpha method to find Qs). I'm using to find Qu=Qp +Qs

the problem states that the pile is fully embedded in clay with Soil friction angle =0 , so I'm using for the point load Qp=9CuAp (9x70)(4x.381)=960 KN

Qs=(alphaxCuxL)p.....using figure 11.23 Cu/avg eff. stress=.5 so Qs=(.5x70x20)x 4x.381=1070 KN

Qall= 960+1070/3=676 KN close but not good.....the answear is 575 KN

I'm missing something...could you clarify?

Use the area of the pile for qp, not the perimeter

alpha is closer to 0.75 than 05

 

[chess5329]

[Hi Santiagj,
I'm stuck with problem 11.7 (piles chapter). The problem is asking to find the allowable load for the pile and request that use the (alpha method to find Qs). I'm using to find Qu=Qp +Qs

the problem states that the pile is fully embedded in clay with Soil friction angle =0 , so I'm using for the point load Qp=9CuAp (9x70)(4x.381)=960 KN

Qs=(alphaxCuxL)p.....using figure 11.23 Cu/avg eff. stress=.5 so Qs=(.5x70x20)x 4x.381=1070 KN

Qall= 960+1070/3=676 KN close but not good.....the answear is 575 KN

I'm missing something...could you clarify?

Use the area of the pile for qp, not the perimeter

alpha is closer to 0.75 than 05

geo_dirt

How you obtain alpha 0.75 ...could you be more specific?

 

[chess5329]

[Hi Santiagj,
I'm stuck with problem 11.7 (piles chapter). The problem is asking to find the allowable load for the pile and request that use the (alpha method to find Qs). I'm using to find Qu=Qp +Qs

the problem states that the pile is fully embedded in clay with Soil friction angle =0 , so I'm using for the point load Qp=9CuAp (9x70)(4x.381)=960 KN

Qs=(alphaxCuxL)p.....using figure 11.23 Cu/avg eff. stress=.5 so Qs=(.5x70x20)x 4x.381=1070 KN

Qall= 960+1070/3=676 KN close but not good.....the answear is 575 KN

I'm missing something...could you clarify?

Use the area of the pile for qp, not the perimeter

alpha is closer to 0.75 than 05

geo_dirt

How you obtain alpha 0.75 ...could you be more specific?

Think I got it.....where I heard pay attention to details!

Here is what I got! .....Thanks geo_dirt

Qp=9CuAp (9x70)(.381 x .381)=91 KN

Qs=(alphaxCuxL)p.....using figure 11.23 Cu/avg eff. stress=.5 ( alpha is .75 instead of .5)

so Qs=(.75x70x20)x 4x.381=1600 KN

Qall= 91+1600/3=565 KN much better.....the answear is 575 KN

This is a great forum!