net bearing capacity = ultimate bearing capacity - surcharge (the load which was previously present at that depth due to soil) q(u) = cNc + qNq + 0.5BYNY (Y=gamma) For clays and cohesive soils, since angle of internal friction = 0 => NY = 0, Nq = 1 So, q(u) = cNc + qNq Now, since we have not filled back the soil , surcharge (q) = 0. So, qNq = 0 (which is the contribution to bearing capacity by the surcharge q) . Hence, q(u) = cNc Now, q(net) = q(u) - surcharge (q) = q(u) - 0 =cNc Remember, qNq is the increase in bearing capacity due to the surcharge (q).
Please help me with the solution: SPT test is carried out on a saturated sand layer at 6 m depth The SPT count for first 15 cm is 8 nos second 15 cm is 10 nos and for third 15 cm is 10 nos The corrected ’N value for soil at 6 m depth If Ysoil is 19 kN/m3 will be .
At 941 why did you not consider 0.5B¥N¥ Even though surcharge is zero bcz we remove it Then how can for cohesive soils plate capacity = foundation capacity
aree bhai tu kitna bhi accha ho kitna bhi accha smjha le...but jbtk tu apni writing skill nhi sudharta jbtk teri videos bkwas hai....you can explain it with simple on notebook
Excellent sir 👍👍
Excellent video sir
nicely explained ....Thankzz alot bhai....
Plz sir solve some numericals on plate load test .....
Superb
9:16 why net ultimate bearing capacity is not C Nc - yDf as ultimate bearing capacity is C Nc only
It will be CNc-yDf if there is surcharge over the plate.
net bearing capacity = ultimate bearing capacity - surcharge (the load which was previously present at that depth due to soil)
q(u) = cNc + qNq + 0.5BYNY (Y=gamma)
For clays and cohesive soils, since angle of internal friction = 0 => NY = 0, Nq = 1
So, q(u) = cNc + qNq
Now, since we have not filled back the soil , surcharge (q) = 0. So, qNq = 0 (which is the contribution to bearing capacity by the surcharge q) . Hence, q(u) = cNc
Now, q(net) = q(u) - surcharge (q) = q(u) - 0 =cNc
Remember, qNq is the increase in bearing capacity due to the surcharge (q).
Kindly upload video on this topics numerical
Thanks for videos.. but why don't you do videos on Shallow foundations- Settlement criteria? It is continuous to Bearing capacity criteria na
Excelent sir
when square & circular plate is used
plz explain things,only blindly saying is not enough plz give logic behind it.
If both pit for plate load and foundation depth is same then Df is also zero in case of bearing capacity formula for foundation???
Please help me with the solution:
SPT test is carried out on a saturated sand layer at 6 m depth The SPT count for first 15 cm is 8 nos second 15 cm is 10 nos and for third 15 cm is 10 nos The corrected ’N value for soil at 6 m depth If Ysoil is 19 kN/m3 will be .
At 6.45 min for Square footing shape factor 1.3 come in front of CNc
thks
Give the video in hindi for better understanding and with presentation..thanxxxxx..
At 941 why did you not consider 0.5B¥N¥
Even though surcharge is zero bcz we remove it
Then how can for cohesive soils plate capacity = foundation capacity
Writing bhai sahab bhut gandi hai aapki.......
aree bhai tu kitna bhi accha ho kitna bhi accha smjha le...but jbtk tu apni writing skill nhi sudharta jbtk teri videos bkwas hai....you can explain it with simple on notebook
Sir plz hindi bnawo sir smjne hm jaiso ko bahut asani hogi sir plz....n video 👍no he sir
Hindi m banao video
Please sir national languages me dale
Sir plz hindi bnawo sir smjne hm jaiso ko bahut asani hogi sir plz....n video 👍no he sir