Calculated Used Bar size Cb + Ktr Spacing Spacing 3" + Сь Vs la & area #5 (0.31in²) dp S S 12 x 0.31 0.70 = 5.314" 5" 3.3125 2.5 0.8 15.21" = 2 x 2.5" #6 (0.44in²) #7 (0.60in²) #8 (0.79in²) (2) Repeat your selection using the ACI 318-19 Table 25.4.2.3. Use the top line in this ACI318-19 table if the conditions allow. Note: again table will help. Example of table is shown below: #4 dp la #5 #6 #7 #8 #9 #10 4. A strip footing is basically a one-way slab loaded by an (assumed) uniform foundation pressure. f=3.5 ksi, normal weight, fy epoxy coated. = 60 ksi, bars are not = Flexural analysis/design shows that the needed A₁ = 0.70 in²/ft. This could be met by #6 bar (0.44 in²) at 71½" or by #5, #7, #8 etc. bars at appropriate spacing. Now it turns out that bars must be chosen with la≤ 27-3 24" as shown 2710 -27|| →3"clear at ends 3"clear Eld (1) Find maximum bar size permitted and its spacing if ACI 318-22 Equation (25.4.2.4a) is used (Note: no stirrup in one-way slab) Note: Try to list a table will help you find your final answer. Example of table is shown below: Page 1

Calculated Used Bar size Cb + Ktr Spacing Spacing 3" + Сь Vs la & area #5 (0.31in²) dp S S 12 x 0.31 0.70 = 5.314" 5" 3.3125 2.5 0.8 15.21" = 2 x 2.5" #6 (0.44in²) #7 (0.60in²) #8 (0.79in²) (2) Repeat your selection using the ACI 318-19 Table 25.4.2.3. Use the top line in this ACI318-19 table if the conditions allow. Note: again table will help. Example of table is shown below: #4 dp la #5 #6 #7 #8 #9 #10 4. A strip footing is basically a one-way slab loaded by an (assumed) uniform foundation pressure. f=3.5 ksi, normal weight, fy epoxy coated. = 60 ksi, bars are not = Flexural analysis/design shows that the needed A₁ = 0.70 in²/ft. This could be met by #6 bar (0.44 in²) at 71½" or by #5, #7, #8 etc. bars at appropriate spacing. Now it turns out that bars must be chosen with la≤ 27-3 24" as shown 2710 -27|| →3"clear at ends 3"clear Eld (1) Find maximum bar size permitted and its spacing if ACI 318-22 Equation (25.4.2.4a) is used (Note: no stirrup in one-way slab) Note: Try to list a table will help you find your final answer. Example of table is shown below: Page 1

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter7: Ultimate Bearing Capacity Of Shallow Foundations: Special Cases

Section: Chapter Questions

Problem 7.2P: In Problem 7.1, if there was no bedrock present for at least 12.0 ft below the foundation, what...

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Repeat Problem 11.1 based on limit state design, using the factors given in Table 11.4. 11.1 A continuous foundation is required in a soil where , , and . The depth of the footing will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (6.10) and Table 6.1.
A flexible circular footing of radius R carries a uniform pressure q. Find the depth (in terms of R) at which the vertical stress below the center is 20% of q.
For the following cases, determine the allowable gross vertical load-bearing capacity of the foundation. Use Terzaghi’s equation and assume general shear failure in soil. Use FS = 4. Parameters for Problem 6.1
Redo Problem 16.13 with the following data: gross allowable load = 184,000 lb, = 121 lb/ft3, c = 0, =26, Df = 6.5 ft., and required factor of safety = 2.5. 16.13 A square footing (B B) must carry a gross allowable load of 1160 kN. The base of the footing is to be located at a depth of 2 m below the ground surface. If the required factor of safety is 4.5, determine the size of the footing. Use Terzaghis bearing capacity factors and assume general shear failure of soil. Given: = 17 kN/m3, c = 48 kN/m2, =31.
For the mat in Problem 16.15, what will be the depth, Df, of the mat for FS = 3 against bearing capacity failure? 16.15 Consider a mat foundation with dimensions of 18 m 12 m. The combined dead and live load on the mat is 44.5 MN. The mat is to be placed on a clay with cu = 40.7 kN/m2 and = 17.6 kN/m3. Find the depth, Df, of the mat for a fully compensated foundation.
Redo Problem 6.2 using the general bearing capacity equation [Eq. (6.28)]. A 5.0 ft wide square footing is placed at 3.0 ft depth within the ground where c = 200 lb/ft2, = 25, and = 115.0 lb/ft3. Determine the ultimate bearing capacity of the footing using Terzaghis bearing capacity equation and the bearing capacity factors from Table 6.1. What is the maximum column load that can be allowed with a factor of safety of 3.0?
Refer to Figure 5.2. Given: B = L = 1.75 m, Df = 1 m, H = 1.75 m, = 17 kN/m3, c = 0, and = 30. Using Eq. (5.6) and FS = 4, determine the gross allowable load the foundation can carry.
Repeat Problem 20.2 based on LRFD using the following factors. Load factor for dead load = 1.25 Load factor for live load = 1.75 Strength reduction factor on the ultimate bearing capacity = 0.50 20.2 A continuous foundation is required in a soil where c = 10 kN/m2, =26 and = 19.0 kN/m3. The depth of the foundation will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (16.3) and Table 16.1.
A continuous foundation is required in a soil where c=10kN/m2, =26, and =19.0kN/m3. The depth of the footing will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (6.10) and Table 6.1.
Redo Problem 16.2 using the modified general ultimate bearing capacity Eq. (16.31). 16.1 A continuous footing is shown in Figure 16.17. Using Terzaghis bearing capacity factors, determine the gross allowable load per unit area (qall) that the footing can carry. Assume general shear failure. Given: = 19 kN/m3, c = 31kN/m2, =28, Df = 1.5 m, B = 2 m, and factor of safety = 3.5. Figure 16.17 6.2 Refer to Problem 16.1. If a square footing with dimension 2 m 2 m is used instead of the wall footing, what would be the allowable bearing capacity?
The applied load on a shallow square foundation makes an angle of 15 with the vertical. Given: B = 1.83 m, Df = 0.91 m, = 18.08 kN/m3, =25, and c = 23.96 kN/m2. Use FS = 4 and determine the gross allowable (vertical component) load. Use Eq. (16.9).
Repeat Problem 11.1 based on LRFD using the following factors: load factor for dead load = 1.25 load factor for live load = 1.75 strength reduction factor on the ultimate bearing capacity = 0.50 11.1 A continuous foundation is required in a soil where , , and . The depth of the footing will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (6.10) and Table 6.1.