Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 16, Problem 16.1CTP

The following table shows the boring log at a site where a multi-story shopping center would be constructed. Soil classification and the standard penetration number, N60, are provided in the boring log. All columns of the building are supported by square footings which must be placed at a depth of 1.5 m. Additionally, the settlement (elastic) of each footing must be restricted to 20 mm. Since the column loads at different location can vary, a design chart is helpful for quick estimation of footing size required to support a given load.

  1. a. Prepare a chart by plotting the variation of maximum allowable column loads with footing sizes, B = 1 m, 1.5 m, 2 m, and 3 m. Use a factor of safety of 3.
  2. b. If the gross column load from the structure is 250 kN, how would you use this chart to select a footing size?
  3. c. What would be the design footing size for the column in Part (b) if you use Terzaghi’s bearing capacity equation? For the well graded sand, assume that ϕ′ = 33°. Use Fs = 3.
  4. d. Compare and discuss the differences in footing sizes obtained in Parts b and c.

Chapter 16, Problem 16.1CTP, The following table shows the boring log at a site where a multi-story shopping center would be

(a)

Expert Solution
Check Mark
To determine

Plot the variation of maximum allowable column loads with size of footings to prepare a chart.

Explanation of Solution

Given information:

The location of depth of footing Df is 1.5 m.

The given size of the footing B is 1 m, 1.5 m, 2 m, and 3 m.

The settlement of each footing Se is 20 mm.

The given factor of safety Fs is 3.

Calculation:

For B value is 1 m:

Determine the depth factor using the relation.

Fd=1+0.33DfB

Substitute 1.5 m for Df and 1.0 m for B.

Fd=1+0.33×1.51.0=1.495

The Fd value should be less than 1.33. The calculated Fd(1.495)>1.33. Take Fd as 1.33.

The field standard penetration number N60 should be averaged up to a distance of 2 m (2×1) below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

depth=Df+2B

Substitute 1.5 m for Df and 1 m for B.

depth=1.5+2(1)=3.5m

Determine the averaged N60 value using the relation.

N60=(N60)SW+(N60)ML2

Here, (N60)SW is the field standard penetration number of well graded sand and (N60)ML is the field standard penetration number of sandy silts.

Substitute 12 for (N60)SW and 7 for (N60)ML.

N60=12+72=9.510

Determine the net allowable bearing capacity of the soil (qnet) using the relation.

qnet=N600.05Fd[Se25]

Substitute 10 for N60, 1.33 for Fd, and 20 mm for Se.

qnet=100.05×1.33[2025]=212.8kN/m2

Determine the maximum allowable column load Qall-net using the relation.

Qall-net=qnet×B2Fs

Substitute 212.8kN/m2 for qnet, 1 m for B, and 3 for Fs.

Qall-net=212.8×123=71kN

For B value is 1.5 m:

Determine the depth factor using the relation.

Fd=1+0.33DfB

Substitute 1.5 m for Df and 1.5 m for B.

Fd=1+0.33×1.51.5=1.33

The field standard penetration number N60 should be averaged up to a distance of 3 m (2×1.5) below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

depth=Df+2B

Substitute 1.5 m for Df and 1.5 m for B.

depth=1.5+2(1.5)=4.5m

Determine the averaged N60 value for 4.5 m depth using the relation.

N60=(N60)4+(N60)62

Here, (N60)4 is the field standard penetration number for 4 m depth and (N60)6 is the field standard penetration number for 6 m depth.

Substitute 7 for (N60)4 and 8 for (N60)6.

N60=7+82=7.58

Determine the net allowable bearing capacity of the soil (qnet) using the relation.

qnet=N600.08(B+0.3B)2Fd[Se25]

Substitute 8 for N60, 1.33 for Fd, and 20 mm for Se.

qnet=80.08×(1.5+0.31.5)2×1.33[2025]=153.2kN/m2

Determine the maximum allowable column load Qall-net using the relation.

Qall-net=qnet×B2Fs

Substitute 153.2kN/m2 for qnet, 1.5 m for B, and 3 for Fs.

Qall-net=153.2×1.523=115kN

For B value is 2 m:

Determine the depth factor using the relation.

Fd=1+0.33DfB

Substitute 1.5 m for Df and 2 m for B.

Fd=1+0.33×1.52.0=1.248

The field standard penetration number N60 should be averaged up to a distance of 4 m (2×2) below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

depth=Df+2B

Substitute 1.5 m for Df and 2.0 m for B.

depth=1.5+2(2)=5.5m

Determine the averaged N60 value for 5.5 m depth using the relation.

N60=(N60)1+(N60)4+(N60)63

Substitute 12 for (N60)1, 7 for (N60)4, and 8 for (N60)6.

N60=12+7+83=9

Determine the net allowable bearing capacity of the soil (qnet) using the relation.

qnet=N600.08(B+0.3B)2Fd[Se25]

Substitute 9 for N60, 1.248 for Fd, and 20 mm for Se.

qnet=90.08×(2+0.32)2×1.248[2025]=148.54kN/m2

Determine the maximum allowable column load Qall-net using the relation.

Qall-net=qnet×B2Fs

Substitute 148.54kN/m2 for qnet, 2 m for B, and 3 for Fs.

Qall-net=148.54×223=198kN

For B value is 3 m:

Determine the depth factor using the relation.

Fd=1+0.33DfB

Substitute 1.5 m for Df and 3 m for B.

Fd=1+0.33×1.53=1.165

The field standard penetration number N60 should be averaged up to a distance of 6 m (2×3) below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

depth=Df+2B

Substitute 1.5 m for Df and 3 m for B.

depth=1.5+2(3)=7.5m

Determine the averaged N60 value for 5.5 m depth using the relation.

N60=(N60)1+(N60)4+(N60)6+(N60)84

Here, (N60)8 is the field standard penetration number at 8 m depth.

Substitute 12 for (N60)1, 7 for (N60)4, 8 for (N60)6, and 19 for (N60)8.

N60=12+7+8+194=11.512

Determine the net allowable bearing capacity of the soil (qnet) using the relation.

qnet=N600.08(B+0.3B)2Fd[Se25]

Substitute 12 for N60, 1.165 for Fd, and 20 mm for Se.

qnet=120.08×(3+0.33)2×1.165[2025]=169.15kN/m2

Determine the maximum allowable column load Qall-net using the relation.

Qall-net=qnet×B2Fs

Substitute 169.15kN/m2 for qnet, 3 m for B, and 3 for Fs.

Qall-net=169.15×323=507.5kN

Summarize the calculated values as in Table (1).

Width B (m)Column load (kN)
171
1.5115
2198
3507.5

Plot the graph between the size of the footing and the column load as in Figure (1).

Principles of Geotechnical Engineering (MindTap Course List), Chapter 16, Problem 16.1CTP

(b)

Expert Solution
Check Mark
To determine

Find the footing size for the given gross column load of 250 kN.

Answer to Problem 16.1CTP

The footing size for the given gross column load of 250 kN is 2.25m_.

Explanation of Solution

Given information:

The location of depth of footing Df is 1.5 m.

The given size of the footing B is 1 m, 1.5 m, 2 m, and 3 m.

The settlement of each footing Se is 20 mm.

The given factor of safety Fs is 3.

Calculation:

Refer Figure (1).

The size of the footing is 2.25 m for the gross column load of 250 kN.

Therefore, the footing size for the given gross column load of 250 kN is 2.25m_.

(c)

Expert Solution
Check Mark
To determine

Find the design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation.

Answer to Problem 16.1CTP

The design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation is 2,173kN_.

Explanation of Solution

Given information:

The value of cohesion c is 0.

The soil friction angle ϕ is 33°.

The location of depth of footing base Df is 1.5 m.

The width of the footing B is 2.25 m.

The given factor of safety Fs is 3.

Calculation:

Determine the net ultimate bearing capacity of the soil (qu) using the relation.

qunet=1.3cNc+qNq+0.4γBNγq=1.3cNc+γDfNq+0.4γBNγγDf (1)

Here, Nc is the contribution of cohesion, Nq is the contribution of surcharge, γ is the unit weight of soil, and Nγ is the contribution of unit weight of soil.

Take the unit weight of the soil γ is 17kN/m3.

Refer Table 16.1, “Terzaghi’s bearing-capacity factors–Nc, Nq, and Nγ” in the textbook.

Take the Nc as 48.09, Nq as 32.23, and Nγ as 31.94 for the ϕ value of 33°.

Substitute 0 for c, 48.09 for Nc, 17kN/m3 for γ, 1.5 m for Df, 32.23 for Nq, 2.25 m for B, and 31.94 for Nγ.

qunet=1.3×0×48.09+17×1.5×32.23+0.4×17×2.25×31.9417×1.5=1,287.59kN/m2

Determine the net allowable bearing capacity (qall-net) that the footing can carry using the relation.

Qall-net=qall-netB2=qu-netB2Fs

Substitute 1,287.59kN/m2 for qu, 2.25 m for B, and 3.0 for Fs.

Qall-net=1,122.15×2.2523=2,173kN

Therefore, the design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation is 2,173kN_.

(d)

Expert Solution
Check Mark
To determine

Compare and discuss the differences in footing sizes obtained in parts (b) and (c).

Explanation of Solution

Given information:

The soil friction angle ϕ is 33°.

The location of depth of footing base Df is 1.5 m.

The width of the footing B is 2.25 m.

The given factor of safety Fs is 3.

Calculation:

The net allowable column load obtained by using Terzaghi’s bearing capacity equation (2,173 kN) is significantly higher than the method based on the N60 and limiting settlement value (250 kN). In actual foundation design, the bearing capacity based on shear strength and the settlement criteria need to be satisfied. It is possible that the allowable column load obtained in part c will decrease if the settlement limit is imposed. Considering the uncertainty in sampling and testing of granular materials for the shear strength determination, the N60 method and settlement criteria may be preferable for the given condition.

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