Question

PLEASE SOLVE İT WİTH GENERAL BEARİNG CAPACİTY EQUATİONS.Thank you ?.İt is so important for me like a midterm exam!

Question 2: For the footing given below, calculate the ultimate bearing capacity (quor qr) and net ultimate bearing capacity (anr) of the footing given below using General Bearing Capacity Equation. General shear failure occurs in the soil and the relevant information are given below. The ground water is at a depth of 3 m below ground surface. It has been observed that there is no eccentricity. However, resultant load is acting with an inclination of a = 75° from horizontal. D = 3 m бm x 6т Soil ' = 5 kPa d' = 30° Ym = 18 kN/m3 Ysar = 20 kN/m3 Center Line

Answer 1

Solution:- The ultimate bearing capacity (qu) using general bearing capacity ( i.e. using Meyerhof approach) is found by using equation given in below images. This equation is dependent on C = unit cohesion, q = effective overburden pressure at the base level of foundation = (γ × Df) , Df = depth of foundation, Shape factors, depth factors, load inclination factors, bearing capacity factors, B = width of foundation. Meyerhof's Bearing capacity factors, shape factors, depth factors and load inclination factors are first found out by using equation shown in below images. The ultimate bearing capacity (qu) is then found out by putting all the calculated and given values in the General bearing capacity equation. The net ultimate bearing capacity (qnf) is then found out by subtracting (effective overburden pressure at the base level of the foundation, q = (γ × Df)) from the ultimate bearing capacity (qu). The whole solution is shown in below images.

The obtained value of ultimate bearing capacity using General bearing capacity equation, qu = 1571.522 KN/m2 and the obtained value of net ultimate bearing capacity using General bearing capacity equation, qnf = 1517.522 KN/m2.

The general Bearing capacity equation :- qu= CNC Sedcic + q Nq Sq dq iq & I BX Ny Se deix where c-unit conesion = 5kPa = 5 kN/m² q = effective overbueden pressure at the base level of the foundation - X Dt. - effective unit weight above the base level of foundation. x = 18KN/m3 = effective unit weight of soil below the foundation base (=Y' - (20-9-81) 7 Of = depth of foundation = 3m. 1 = 10.19KN ( m3 Sc, Sq, Sx = Shape factors dc, dq , dx-depth factor ic, iq, ix = load inclination factors Be width of foundation = 6m No, No, Ny - bearing capacity factors. - Nq = e(1 tand) , tan(45+0,2) 0= 30° Nq = ( (T kan zo) tan" (45 + 36/2) Nq = 18.40 - > ☺ =) NC = (Nq-1) Lot 6 = (18.4-1) Lot 30° INC - 30-1 -> → Ne - (Nq-1) tan (1-40) - (18-4-1) tan (1.4x30') Ne = 15.7 - © > (4

Shape factors: Meyerhof's correction factors. S = 1 + 0.2 B tan (45 + 8/2) S 1 + 0.2 (6) tans (45 + 30/2) S-16 - > Sq = Sy= 1 + 01 (③) tan? (45+$12) For 4710 Sq = Sy = 1+ 0.1(E) dan? (45 + 6 30/2) Sq = Sy = 1-3 - Meyerhof's depth factors :- de = 1+ 0.2 (0) tan (45 + 9/2) de = 1+ 0.2 ( 3 ) tan (45 + 36 ) de = 1.173 --> @ dq=dx = 1 + 0.11 0/) tan (45 + 8/2) dq=dx = 1+ 0.1 (3) tan ( 45 + 30 ) [dq = dx = 1.087 - m

ot's wad inclination factors :- Lc = iq = (- ) = 75° 4:19. (-) (=iq = 0.027] x=(1-22 = ( 1 2 3 4 5 [x = 2:25] = > ☺ 5027] Putting all values from eq @ to eq 6 in eq , we get Ultimate Bearing capacity (qu) is equal to :- Yu = (5x 30.1X 1-6 X 1-173 X 0.027) +(18x3X 18.4X1-3 x 1.087 X 0.027) +(0.5X6X10-19X15.7X 1-3 X 1.687 * 2.25) qu = 7.6264 + 37.9095 + 1525.9858 Yu = 1571.522 kN/m2 It is the required ultimate bearing capacity (qu) using General Bearing capacity.