Campos Electromagnéticos Pulsados

Examples for determining allowable bearing capacity

Example #1: Determine allowable bearing capacity and width for a shallow strip footing on cohesionless silty sand and gravel soil. Loose soils were encountered in the upper 0.6 m (2 feet) of building subgrade. Footing must withstand a 144 kN/m² (3000 lb/ft²) building pressure.

Given

· bearing pressure from building = 144 kN/m^{2 (}3000 lbs/ft²)

· unit weight of soil, g = 21 kN/m³ (132 lbs/ft³) *from soil testing, see typical g values

· Cohesion, c = 0 *from soil testing, see typical c values

· angle of Internal Friction, f = 32 degrees *from soil testing, see typical f values

· footing depth, D = 0.6 m (2 ft) *because loose soils in upper soil strata

Solution

Try a minimal footing width, B = 0.3 m (B = 1 foot).

Use a factor of safety, F.S = 3. Three is typical for this type of application. See factor of safety for more information.

Determine bearing capacity factors N_g, N_c and N_q. See typical bearing capacity factors relating to the soils' angle of internal friction.

· N_g = 22

· Nc = 35.5

· Nq = 23.2

Solve for ultimate bearing capacity,

Qu = c Nc + g D Nq + 0.5 g B Ng *strip footing eq.

Qu = 0(35.5) + 21 kN/m³(0.6m)(23.2) + 0.5(21 kN/m³)(0.3 m)(22) metric Q_u = 362 kN/m²

Q_u = 0(35.5) + 132lbs/ft³(2ft)(23.2) + 0.5(132lbs/ft³)(1ft)(22) standard Qu = 7577 lbs/ft²

Solve for allowable bearing capacity, Qa = Qu

F.S.

Qa = 362 kN/m² = 121 kN/m² not o.k. metric 3

Q_a = 7577lbs/ft² = 2526 lbs/ft² not o.k. standard 3

Since Qa < required 144 kN/m² (3000 lbs/ft²) bearing pressure, increase footing width, B or foundation depth, D to increase bearing capacity.

Try footing width, B = 0.61 m (B = 2 ft).

Qu = 0 + 21 kN/m³(0.61 m)(23.2) + 0.5(21 kN/m³)(0.61 m)(22) metric Q_u = 438 kN/m²

Qu = 0 + 132 lbs/ft³(2 ft)(23.2) + 0.5(132 lbs/ft³)(2 ft)(22) standard Qu = 9029 lbs/ft²

Q_a = 438 kN/m² = 146 kN/m² Q_a > 144 kN/m2

o.k. metric 3

Q_a = 9029 lbs/ft² = 3010 lbs/ft² Q_a > 3000 lbs/ft2

o.k. standard 3

Conclusion

Footing shall be 0.61 meters (2 feet) wide at a depth of 0.61 meters (2 feet) below ground surface. Many engineers neglect the depth factor (i.e. D Nq = 0) for shallow foundations.

This inherently increases the factor of safety. Some site conditions that may negatively effect the depth factor are foundations established at depths equal to or less than 0.3 meters (1 feet) below the ground surface, placement of foundations on fill, and disturbed/

fill soils located above or to the sides of foundations.

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Example #2: Determine allowable bearing capacity of a shallow, 0.3 meter (12-inch) square isolated footing bearing on saturated cohesive soil. The frost penetration depth is 0.61 meter (2 feet). Structural parameters require the

foundation to withstand 4.4 kN (1000 lbs) of force on a 0.3 meter (12-inch) square column.

Given

· bearing pressure from building column = 4.4 kN/ (0.3 m x 0.3 m) = 48.9 kN/m²

· bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft²

· unit weight of saturated soil, gsat= 20.3 kN/m³ (129 lbs/ft³) *see typical g values

· unit weight of water, gw= 9.81 kN/m³ (62.4 lbs/ft³) *constant

· Cohesion, c = 21.1 kN/m² (440 lbs/ft²) *from soil testing, see typical c values

· angle of Internal Friction, f = 0 degrees *from soil testing, see typical f values

· footing width, B = 0.3 m (1 ft)

Solution

Try a footing depth, D = 0.61 meters (2 feet), because foundation should be below frost depth.

Use a factor of safety, F.S = 3. See factor of safety for more information.

Determine bearing capacity factors Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

· N_g = 0

· Nc = 5.7

· Nq = 1

Solve for ultimate bearing capacity,

Qu = 1.3c Nc + g D Nq + 0.4 g B Ng *square footing eq.

Q_u =1.3(21.1kN/m²)5.7+(20.3kN/m³-9.81kN/m³)(0.61m)1+0.4(20.3kN/m³ -9.81kN/m³)(0.3m)0

Qu = 163 kN/m² metric Q_u = 1.3(440lbs/ft²)(5.7) + (129lbs/ft³ - 62.4lbs/ft³)(2ft)(1) + 0.4(129lbs/ft³ - 62.4lbs/ft³)(1ft)(0)

Qu = 3394 lbs/ft² standard Solve for allowable bearing capacity,

Qa = Qu F.S.

Qa = 163 kN/m² = 54 kN/m² Qa > 48.9 kN/m² o.k. metric 3

Qa = 3394lbs/ft² = 1130 lbs/ft² Qa > 1000 lbs/ft² o.k. standard 3

Conclusion

The 0.3 meter (12-inch) isolated square footing shall be 0.61 meters (2 feet) below the ground surface. Other considerations may be required for foundations bearing on moisture sensitive clays, especially for lightly loaded structures such as in this example.

Sensitive clays could expand and contract, which could cause structural damage. Clay used as bearing soils may require mitigation such as heavier loads, subgrade removal and replacement below the foundation, or moisture control within the subgrade.

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Example #3: Determine allowable bearing capacity and width for a foundation using the Meyerhof Method. Soils consist of poorly graded sand. Footing must withstand a 144 kN/m² (1.5 tons/ft²) building pressure.

Given

· bearing pressure from building = 144 kN/m^{2 (}1.5 tons/ft²)

· N Value, N = 10 at 0.3 m (1 ft) depth *from SPT soil testing

· N Value, N = 36 at 0.61 m (2 ft) depth *from SPT soil testing

· N Value, N = 50 at 1.5 m (5 ft) depth *from SPT soil testing Solution

Try a minimal footing width, B = 0.3 m (B = 1 foot) at a depth, D = 0.61 meter (2 feet).

Footings for single family residences are typically 0.3m (1 ft) to 0.61m (2ft) wide. This depth was selected because soil density greatly increases (i.e. higher N-value) at a depth of 0.61 m (2 ft).

Use a factor of safety, F.S = 3. Three is typical for this type of application. See factor of safety for more information.

Solve for ultimate bearing capacity

Qu = 31.417(NB + ND) (kN/m²) (metric) Q_u = NB + ND (tons/ft²) (standard) 10 10

Qu = 31.417(36(0.3m) + 36(0.61m)) = 1029 kN/m² (metric) Q_u = 36(1 ft) + 36(2 ft) = 10.8 tons/ft² (standard) 10 10

Solve for allowable bearing capacity, Qa = Qu

F.S.

Qa = 1029 kN/m² = 343 kN/m² Qa > 144 kN/m² o.k. (metric) 3

Q_a = 10.8 tons/ft² = 3.6 tons/ft² Q_a > 1.5 tons/ft² o.k. (standard) 3

Conclusion

Footing shall be 0.3 meters (1 feet) wide at a depth of 0.61 meters (2 feet) below the ground surface. A footing width of only 0.3 m (1 ft) is most likely insufficient for the structural engineer when designing the footing with the building pressure in this problem.

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Example #4: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load.

Given

· vertical column load = 66.7 kN (15 kips or 15,000 lb)

· homogeneous soils in upper 15.2 m (50 ft); silty soil

o unit weight, g = 19.6 kN/m³ (125 lbs/ft³) *from soil testing, see typical g values

o cohesion, c = 47.9 kN/m² (1000 lb/ft²) *from soil testing, see typical c values

o angle of internal friction, f = 30 degrees *from soil testing, see typical f values

· Pile Information

o driven

o steel

o plugged end

Solution

Try a pile depth, D = 1.5 meters (5 feet) Try pile diameter, B = 0.61 m (2 ft)

Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient experience with the regional soils.

Determine ultimate end bearing of pile, Q_p = A_pq_p

Ap = p(B/2)² = p(0.61m/2)² = 0.292 m² metric Ap = p(B/2)² = p(2ft/2)² = 3.14 ft2

standard qp = gDNq

g = 19.6 kN/m³ (125 lbs/ft³); given soil unit weight f = 30 degrees; given soil angle of internal friction B = 0.61 m (2 ft); trial pile width

D = 1.5 m (5 ft); trial depth, may need to increase or decrease depending on capacity check to see if D < Dc

Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts.

If D > Dc, then use Dc

Nq = 25; Meyerhof bearing capacity factor for driven piles, based on f qp = 19.6 kN/m³(1.5 m)25 = 735 kN/m² metric q_p = 125 lb/ft³(5 ft)25 = 15,625 lb/ft² standard

Qp = Apqp = (0.292 m²)(735 kN/m²) = 214.6 kN metric Q_p = A_pq_p = (3.14 ft²)(15,625 lb/ft²) = 49,063 lb standard

Determine ultimate friction capacity of pile, Qf = Afqf

A_f = pL

p = 2p(0.61m/2) = 1.92 m metric p = 2p(2 ft/2) = 6.28 ft standard L = D = 1.5 m (5 ft); length and depth used interchangeably. check Dc as above Af = 1.92 m(1.5 m) = 2.88 m² metric A_f = 6.28 ft(5 ft) = 31.4 ft² standard qf = cA + ks tan d = cA + kgD tan d

k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel

g = 19.6 kN/m³ (125 lb/ft³); given effective soil unit weight. If water table, then g - g_w D = L = 1.5 m (5 ft); pile length. Check to see if D < Dc

Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts. If D > D_c, then use D_c d = 20 deg; external friction angle, equation chosen from Broms steel piles

B = 0.61 m (2 ft); selected pile diameter

cA = 0.5c; for clean steel. See adhesion in pile theories above.

= 24 kN/m² (500 lb/ft²)

q_f = 24 kN/m² + 0.5(19.6 kN/m³)(1.5m)tan 20 = 29.4 kN/m² metric qf = 500 lb/ft² + 0.5(125 lb/ft³)(5ft)tan 20 = 614 lb/ft² standard Qf = Afqf = 2.88 m²(29.4 kN/m²) = 84.7 kN metric Q_f = A_fq_f = 31.4 ft²(614 lb/ft²) = 19,280 lb standard

Determine ultimate pile capacity,

Q_ult = Q_p + Q_f

Qult = 214.6 kN + 84.7 kN = 299.3 kN metric Q_ult = 49,063 lb + 19,280 lb = 68,343 lb standard

Solve for allowable bearing capacity, Qa = Qult

F.S.

Qa = 299.3 kN = 99.8 kN; Qa > applied load (66.7 kN) o.k. metric 3

Q_a = 68,343 lbs = 22,781 lbs Q_a > applied load (15 kips) o.k. standard 3

Conclusion

A 0.61 m (2 ft) steel pile shall be plugged and driven 1.5 m (5 feet) below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/

thaw or effects from water. The end bearing alone (neglect skin friction) is sufficient for this case. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.

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Example #5: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load.

Given

· vertical column load = 66.7 kN (15 kips or 15,000 lb)

· upper 1.5 m (5 ft) of soil is a medium dense gravelly sand

o unit weight, g = 19.6 kN/m³ (125 lbs/ft³) *from soil testing, see typical g values

o cohesion, c = 0 *from soil testing, see typical c values

o angle of internal friction, f = 30 degrees *from soil testing, see typical f values

· soils below 1.5 m (5 ft) of soil is a stiff silty clay

o unit weight, g = 18.9 kN/m³ (120 lbs/ft³)

o cohesion, c = 47.9 kN/m² (1000 lb/ft²)

o angle of internal friction, f = 0 degrees

· Pile Information

o driven

o wood

o closed end

Solution

Try a pile depth, D = 2.4 meters (8 feet) Try pile diameter, B = 0.61 m (2 ft)

Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient experience with the regional soils.

Determine ultimate end bearing of pile, Qp = Apqp

A_p = p(B/2)² = p(0.61m/2)² = 0.292 m² metric A_p = p(B/2)² = p(2ft/2)² = 3.14 ft2

standard

qp = 9c = 9(47.9 kN/m²) = 431.1 kN/m² metric q_p = 9c = 9(1000 lb/ft²) = 9000 lb/ft² standard

Qp = Apqp = (0.292 m²)(431.1 kN/m²) = 125.9 kN metric Q_p = A_pq_p = (3.14 ft²)(9000 lb/ft²) = 28,260 lb standard

Determine ultimate friction capacity of pile, Q_f = pSqfL

p = 2p(0.61m/2) = 1.92 m metric p = 2p(2 ft/2) = 6.28 ft standard

upper 1.5 m (5 ft) of soil

qfL = [ks tan d]L = [kgD tan d]L

k = 1.5; lateral earth pressure coefficient for piles, value chosen from Broms low density timber

g = 19.6 kN/m³ (125 lb/ft³); given effective soil unit weight. If water table, then g - gw

D = L = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if D < D_c D_c = 15B = 9.2 m (30 ft); critical depth for medium dense sands. This assumption is conservative, because the soil is gravelly, and this much soil unit weight for a sand would indicate dense soils. If D > Dc, then use Dc

d = f(2/3) = 20 deg; external friction angle, equation chosen from Broms timber piles B = 0.61 m (2 ft); selected pile diameter

f = 30 deg; given soil angle of internal friction

qfL = [1.5(19.6 kN/m³)(1.5m)tan (20)]1.5 m = 24.1 kN/m metric qfL = [1.5(125 lb/ft³)(5ft)tan (20)]5 ft = 1706 lb/ft standard

soils below 1.5 m (5 ft) of subgrade qfL = aSu

Suc = 2c = 95.8 kN/m² (2000 lb/ft²); unconfined compressive strength c = 47.9 kN/m² (1000 lb/ft²); cohesion from soil testing (given)

a = 1 [0.9 + 0.3(Suc - 1)] = 0.3; because S_uc > 48 kN/m², (1 ksf) Suc

L = 0.91 m (3 ft); segment of pile within this soil strata

q_fL = [0.3(95.8 kN/m²)]0.91 m = 26.2 kN/m metric q_fL = [0.3(2000 lb/ft²)]3 ft = 1800 lb/ft standard

ultimate friction capacity of combined soil layers Qf = pSqfL

Q_f = 1.92 m(24.1 kN/m + 26.2 kN/m) = 96.6 kN metric Qf = 6.28 ft(1706 lb/ft + 1800 lb/ft) = 22,018 lb standard

Determine ultimate pile capacity, Q_ult = Q_p + Q_f

Qult = 125.9 kN + 96.6 kN = 222.5 kN metric Qult = 28,260 lb + 22,018 lb = 50,278 lb standard

Solve for allowable bearing capacity, Qa = Qult

F.S.

Q_a = 222.5 kN = 74.2 kN; Q_a > applied load (66.7 kN) o.k. metric 3

Qa = 50,275 lbs = 16,758 lbs Qa > applied load (15 kips) o.k. standard 3

Conclusion

Wood pile shall be driven 8 feet below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/ thaw or effects from water. Notice how the soil properties within the pile tip location is used in the end bearing calculations.

End bearing should also consider the soil layer(s) directly beneath this layer. Engineering judgment or a change in design is warranted if subsequent soil layers are weaker than the soils within the vicinity of the pile tip. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.

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