Combine Loads And Moments In Foundation Design - Bearing Capacity Equations And Examples

Write about the Combine loads and Moments.

Foundation t design comprises of two discrete parts: a definitive bearing capacity of the soil under the foundation, and the middle of the road settlement that the balance can experience without influencing the superstructure. A definitive bearing capacity goes for deciding the load that the soil under the foundation can deal with before shear disappointment; while, the count of the settlement caused by the superstructure ought not surpass the breaking points of the permitted twisting for security, capacity and parts of development. Research on a definitive bearing capacity issues can be done utilizing either systematic arrangements or trial analysis. The previous could be considered through hypothesis of pliancy or limited component investigation, while the last is accomplished through directing model, model and full-scale tests. A tasteful arrangement is discovered just when hypothetical outcomes concur with those acquired tentatively. A writing overview regarding the matter demonstrates that most of the bearing capacity hypotheses include homogeneous soils under the foundation. Soil properties were accepted to stay consistent for the bearing capacity analysis, and thusly expository arrangements, similar to Hansen’s bearing capacity hypothesis, coordinated with the trial results. Nonetheless, in situations where the soil properties shift with depth, a large portion of these speculations can't be executed, and the diagnostic arrangements that think about the non-homogeneity of the soil are approximations, and thus the outcomes are off base. Layered soil profiles are frequently experienced whether normally saved or misleadingly made. Inside each layer, the soil might be considered as homogeneous. A definitive load disappointment surface in the soil relies upon the shear quality parameters of the soil layers, for example, the thickness of the upper layer; the shape, size and insertion of balance; and the proportion of the thickness of the upper layer to the width of the balance. In this manner, it is vital to decide the dirt profile and to ascertain the bearing limit appropriately.  

Objectives

1. To determine the bearing capacity of shallow foundation when the foundation has only vertical loads
2. To determine the bearing capacity of shallow when the foundation has only horizontal loads
3. To determine the bearing capacity of shallow when the foundation has both the loads
4. To determine the bearing capacity of shallow when the foundation has both the loads and moment according to x and y direction 

Equation considered in Hansen’s and Meyerhof

Hansen’s Bearing capacity equations:

Here, the bearing capacity factors are given by the following expressions which depend on ?.

Equations are available for shape factors (s_c, s_q, s_γ), depth factors (d_c, d_q, d_γ) and load inclination factors (i_c, i_q, i_γ). The effects of these factors are to reduce the bearing capacity.

For the formula of Hansens and Meyerhof, we ascertain the estimation of extreme bearing capacity (q?) which the greatest value of the soil can hold up under it (that is. the bearing stress from foundation surpasses a definitive bearing limit of the soil, shear failure in soil will be happen), so we should  plan a foundation for a bearing capacity not as much as a definitive bearing ability to avert shear failure in the soil. This bearing capacity is called allowable Bearing Capacity and design is done on its determinations

Methodology

q = q_u/FS 

Meyerhof equation

Q_u = qN_q F_qs F_qd F_qi +0.5B γN γ F γ _s F γ _d  Fγ_i 

Where

c = Cohesion of the underlying soil

q = Effective stress at the level of the bottom of the foundation.

γ = unit weight of the underlying soil

B = Width of footing (= diameter for a circular foundation).

Nc, Nq, Nγ  = Bearing capacity factors

 F_qs F γs, F_cs  = Shape factors

 F_cd F_qd Fγ_d = Depth factors

F_ci F_qi Fγ_i = Inclination factors  

Examples

A square footing 2m * 2m in a soil, c = 50 kN/m³, ? =20⁰, γ = 18 kN/m³, D = 1.5 m, concentric vertical load = 50 kN, and the load is inclined at 10⁰

q_u = q_u = cN_cs_ci_c + γ*D*N_qs_qi_q + 0.5B¹γNγsγ iγ

N_c = 14.8

N_q = 6.4

Nγ = 5.4

S_c = 1.3

S_q = 1.2

Sγ = 0.8

i_c = i_q = (1 – θ/90)²

i_c = (1 – 10⁰/90)² = 0.79

iγ = (1 – 10/20)² = 0.25

q_u = 50 * 14.8 * 1.3 * 0.79 + 18 * 1.5 *6.4 * 1.2 * 0.79 + 0.5 * 18 * 2 * 5.4 * 0.8 *0.25 = 943.23 kN/m²  

A square footing D_f = 1.0 m, cohesion , c= 0 , ? = 35⁰, γ = 18 kN/m³, is designed to carry a vertical  load per metre ramp = 242 kN/m²,  breath, B = 0.7 m,  calculate the bearing capacity

Take

N_c = 46.12

N_q = 33.30

Nγ = 33.92

Hansen formula for ultimate bearing capacity q_ult of a square footing is

q_ult = cN_cF_scF_dcF_ic +  *N_q F_sqF_dqF_iq + 0.4BγNγF_sγF_dγF_i

where  = D_f *γ

since the load is vertical, all the three inclinations factors = F_ic =F_iq =F_iγ =1

F_sq = 1 + (B/L)tan? = 1 + (0.7/1)tan35⁰ = 1.5

F_dq = 1 + 2(tan?⁰)(1-sin?⁰)( D_f/B)

= 1 + 2(tan35⁰)(1-sin35⁰)(1/0.7) = 1.85

F_s γ = 1- 0.4*(B/L) = 1-0.4*0.7 = 0.72

F_dγ = 1

q_ult = 0+ 18 *33.30*1.5*1*1.85 + 0.4*0.7*18*33.92*0.72*1*1 = 1786.42 kN/m² 

Combined loads

Meyerfoh  method

A 3m wide strip foundation is to be founded on the surface of a silt soil with the properties of, c = 40 kN/m³, ? = 20², γ = 18 kN/m³, what is the bearing capacity if the foundation is subjected to a vertical load of 220 kN/m run of wall at an eccentricity of  0.3 m, together with horizontal force of 50 kN/m run of wall.

B¹ = B – 2e

       = 3 – 2 * 0.3

  = 2.4 m

Tanθ=50/220

   = 12.8⁰

i = (1 - θ⁰/90)²

i_c = (1 -12.8⁰/90)² = 0.74

iγ = (1 -12.8⁰/90)² = 0.13

N_c = 17.7 * 0.74 = 13.1

Equation considered in Hansen’s and Meyerhof

Nγ = 5 * 0.13 = 0.7

q_u = cN_c + γ*D*N_q + 0.5B¹γNγ

        = 40 * 13.1 + 0.5 * 18 * 2.4 * 0.7

         = 539.12 kN/m² 

A 3m wide strip foundation is to be founded on the surface of a silt soil with the properties of, c = 0 kN/m³,? =47⁰, what is the bearing capacity if the foundation is subjected to a vertical load of 220 kN/m run of wall at an eccentricity  = e, together with horizontal force of 50 kN/m run of wall of height  2 m.

e = overall moment /vertical load

  = 50*2/220 = 0.45 m

B¹ = B – 2e

       = 1.4 – 2 * 0.45

  = 0.5  m

H = 50 kN

V_ult = 220 kN

L/B = 2/0.5 = 4

?= 47

N_c = 187

N_q = 174

Nγ = 300

2tan?(1-sin?)² = 2tan?(1-sin47⁰)² = 0.155

N_q/N_c = 187/174 = 1.078

All b_i = g_i = 1.0, since the base and ground are horizontal

d_q,B = 1 + 2tan?(1-sin?)²D/B = 1 + 0.155(0.5/0.5) = 1.16

d_q,L = 1 + 2tan(1-sin?)²D/L = 1 + 0.155(0.5/2.0) = 1.04

d γ,_B = dγ,_L = 1.00  

i_q,B = [1 – (0.5H)/(V+A_fc_ucot?)]^2.5 = 1.0

i γ_,B [1 – (0.5H)/(V+A_fc_ucot?)]^3.5 = 1.0

i_q,L = [1 – 0.5H/(V + 0)]^2.5 = [1-0.5(382)/1060]^2.5 = 0.608

i γ_,L = [1 – 0.7H/(V + 0)]^2.5 = [1-0.7(382)/1060]^3.5 = 0.361

s_q,B = 1 + sin(Bi_q,B/L) = 1 + sin47[0.5(1)/2] = 1.18

s_q,L = 1 + sin(Li_q,L/B) = 1 + sin47[2(0.08)/0.5] = 2.78

s γ_,B = 1 – 0.4(Bi_,γ _B/Liγ_,L)

   = 1 – 0.4[0.5*1)/(2*0.361)] = 0.723 > 0.6

s γ_,L = 1 – 0.4(Li_,γ _L/Biγ_,B)

   = 1 – 0.4[(2*0.361)/(0.5*1)] = 0.422 < 0.6

q_ult = 0 +  *N_q F_sqF_dqF_iq + 0.4BγNγF_sγF_dγF_iγ

q_ult = 0 + 0.5*9.43*187*1.18*1.16*1 + 0.5*9.43*0.5*300*0.732*1*1 = 1718.2 kN/m²  

A square footing 2.5m x 2.5m is shown below. If the maximum pressure on the foundation should not exceed the allowable bearing capacity. find the bearing capacity of the foundation can carry if the water table is 1m below the foundation, and the horizontal force = 124 KN, vertical load = 300 kN and the moment = 165 kN.m

e = Overall moment /Vertical Load = (165 + 1.5H)/ 300 = 0.55 + 0.005H

but H = 124 kN

e = overall moment /vertical load

  = 186+165/300 = 1.17 m

B¹ = B – 2e

       = 2.84 – 2 * 1.17

  = 0.5  m

H = 124 kN

V_ult = 300 kN

L/B = 2/0.5 = 4

Φ = 47

N_c = 187

N_q = 174

Nγ = 103

2tan?(1-sin?)² = 2tan?(1-sin47⁰)² = 0.155

N_q/N_c = 187/174 = 1.078

All b_i = g_i = 1.0, since the base and ground are horizontal

Hansen’s Bearing capacity equations

d_q,B = 1 + 2tan?(1-sin?)²D/B = 1 + 0.155(0.5/0.5) = 1.16

d_q,L = 1 + 2tan?(1-sin?)²D/L = 1 + 0.155(0.5/2.0) = 1.04

d γ,_B = dγ,_L = 1.00

i_q,B = [1 – (0.5H)/(V+A_fc_ucot?)]^2.5 = 1.0

i γ_,B =[1 – (0.5H)/(V+A_fc_ucot?)]^3.5  = 1.0

i_q,L = [1 – 0.5H/(V + 0)]^2.5 = [1-0.5(382)/1060]^2.5 = 0.608

i γ_,L = [1 – 0.7H/(V + 0)]^2.5 = [1-0.7(382)/1060]^3.5 = 0.361

s_q,B = 1 + sin?(Bi_q,B/L) = 1 + sin47⁰[0.5(1)/2] = 1.18

s_q,L = 1 + sin(Li_q,L/B) = 1 + sin47⁰[2(0.08)/0.5] = 2.78

s γ_,B = 1 – 0.4(Bi_,γ _B/Liγ_,L)

   = 1 – 0.4[0.5*1)/(2*0.361)] = 0.723 > 0.6

s γ_,L = 1 – 0.4(Li_,γ _L/Biγ_,B)

   = 1 – 0.4[(2*0.361)/(0.5*1)] = 0.422 < 0.6   

q_ult = 0 +  *N_q F_sqF_dqF_iq + 0.5BγNγF_sγF_dγF_iγ

q_ult = 0 + 0.5*9.43*187*1.18*1.16*1 + 0.5*9.43*0.5*300*0.732*1*1 = 1718.2 kN/m² 

A square footing 2.5m x 2.5m is shown below. If the maximum pressure on the foundation should not exceed the allowable bearing capacity. Using factor of safety (FS=3), find the bearing capacity of the foundation can carry if the water table is 1m below the foundation, and the horizontal force = 124 KN

e = Overall moment /Vertical Load = (165 + 1.5H)/ 300 = 0.55 + 0.005H

but H = 124 kN

e = 0.55 + 0.005 *124 = 1.17

q_max  q_all

q_all = q_max

FS =  = 3

c = 50 kN/m²

q(effective stress) = 17 * 1.5 = 25.5

calculation of new area that maintains q_u uniform

B¹ = B – 2e = 2.5 – 2e = 0.16

L¹ = B = 2.5

B¹_used = min(B¹, L¹) = 2.5 -2e =0.16, L¹_used = 2.5 m

Effective area A¹ = (2.5 – 2e) * 2.5 = 6.25 – 5e = 0.4 m²

d = 1m

γ = γ^- = γ¹ +(d*( γ – γ¹))/B)

γ¹ = 19.5 – 10 9.5 kN/m³

d = 1.0 m

γ = 17 kN/m³

γ^- = 9.5 +(1*( 17 – 9.5))/2.5)= 12.5 kN/m³  

Shape factor

F_cs = 1 + (B¹_used/ L¹_used)(N_q/N_c)

    = 1 + (0.16/2.5)*(22.46/37.16)

  = 1.0387

F_qs = 1 + (B¹_used/ L¹_used))tan?

    = 1+( 0.16/2.5)tan30⁰ =1.03695

F γ _s = 1 -0.4 (B¹_used/ L¹_used)

        1 – 0.4(0.16/2.5) = 0.9744

Depth factor

F_cd = F_qd - (1- F_qd)/N_c* tan

     = 1.1732 - (1-1.1732)/(37.16 * tan30) = 1.18127

F_qd = 1+ 2tan30(1-sin30)²(1.5/2.5) = 1.1732

F γ _d = 1

Bearing capacity factors

For? = 30⁰, N_c 37.16, N_q 22.46, Nγ = 19.13

Substitute for

Q_u = qN_q F_qs F_qd F_qi +0.5B γN γ F γ _s F γ _d  Fγ_i

  = 25.5*22.46*1.03695 *1.1732 + 0.5 * 2.5*17*19.13*0.9744 *1 = 1092.86 kN/m² 

A square footing 2m * 2m in a soil, c = 50 kN/m³, , γ = 18 kN/m³, D = 1.5 m, concentric horizontal load = 50 kN, and the load is inclined at 10⁰

Factor of safety [FS]

q_u = q_u = cN_cs_ci_c + γ*D*N_qs_qi_q + 0.5B¹γNγsγ iγ

N_c = 14.8

N_q = 6.4

Nγ = 5.4

S_c = 1.3

S_q = 1.2

Sγ = 0.8

i_c = i_q  = (1 –θ⁰/90⁰)²

i_c = (1 –10⁰/90)²= 0.79

iγ = (1 –10⁰/20⁰)²= 0.25

q_u = 50 * 14.8 * 1.3 * 0.79 + 18 * 1.5 *6.4 * 1.2 * 0.79 + 0.5 * 18 * 2 * 5.4 * 0.8 *0.25 = 943.23 kN/m²

A square footing D_f = 2.0 m, cohesion , c= 0 ?= 35⁰, γ = 18 kN/m³, is having a concentric horizontal  load = 240 kN/m²,  breath, B = 1.0 m,  calculate the bearing capacity

Take

N_c = 46.12

N_q = 33.30

Nγ = 33.92

Hansen formula for ultimate bearing capacity q_ult of a square footing is

q_ult = cN_cF_scF_dcF_ic +  *N_q F_sqF_dqF_iq + 0.4BγNγF_sγF_dγF_iγ

where  = D_f *γ

since the load is vertical, all the three inclinations factors = F_ic =F_iq =F_iγ =1

F_sq = 1 + (B/L)tan? = 1 + (1.0/2)tan35⁰ = 1.4

F_dq = 1 + 2tan?(1-sin?)²(D_f/B) = 1 + 2(tan35⁰)(1-sin35⁰)(2/1) = 2.194

F_s γ = 1- 0.4*(B/L) = 1-0.4*1/2 = 0.8

F_dγ = 1

q_ult = 0+ 18 *33.30*1.4*1*2.194 + 0.4*1.0*18*33.92*0.8*1*1 = 2036.50 kN/m² 

The soil over the base of the foundation are utilized just to calculate the term (q) in the second term of bearing limit conditions (Hansen’s and Meyerhof) and every single other factor are calculated for the basic soil. Continuously the estimation of (q) is the successful stress at the level of the base of the foundation.

If the foundation width (B) is required, and there exist water table beneath the foundation at separate (d), you ought to make assumption on the value of  d ≤ B, and figure B, at that point make a check for your presumption. 

Reference

Hansen,  J.B.  (1961),  A  General  Formula  for  Bearing  capacity,  Danish  Geotechnical Institute, Copenhagen, Denmark, Bulletin, 11, pp. 38-46.

Kotter, F. (1903), Die Bestimmung des Drucks angekrummten, eine Aufgabe aus der Lehre vom Erddruck, Sitzungsberichte der Akademie der Wissenschaften, Berlin (in German), pp. 229-233.

Meyerhof, G.G.  (1951), The Ultimate Bearing  Capacity of  Foundations, Geotechnique,  2, pp. 301-332.

Mohapatro, B.G. (2001), Some Studies on Bearing Capacity of Shallow Foundations, Ph.D. Thesis, IIT Bombay, India.

Prandtl,  L.  (1920),  Uber  die Harte  plastischer  Korper.  Nachrichten  von  der  Konilichen Gesellschaft der wissenschaften zu Gottingen, Mathematisch-Physikalische Klasse au dem Jahre, Berlin (in German), pp. 74-85.

Reissner H.  Zum Erddruckproblem  (1924), Proceedings of  the International  Congress for Applied Mechanics, Delft, The Netherlands (in German), pp. 295-311.

Saran, S. and Agarwal, R.K. (1991), Bearing Capacity of Eccentrically Obliquely Loaded Footing, Journal of Geotechnical Engineering, ASCE, 117(11), pp. 1669-1690.

Soubra, A.H. (1999), Upper Bound Solutions for Bearing Capacity of Foundations, Journal of Geotechnical and Geoenvironmental Engineering, ASCE, 125(1), pp. 59-68 .