following:1. the length of footing excavation2. the length of bricks work under D.P.C for each stepby using:a) Centre line methodb) Long wall-short wall methodD.P.C1.00 m0.24 m0.36 my0.48 m15 mSection.NAkW8 m5 mزف الحو不ZPlanate theL

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Answered: following: 1. the length of footing…

Construction Materials, Methods and Techniques (MindTap Course List)

4th Edition

ISBN:9781305086272

Author:William P. Spence, Eva Kultermann

Publisher:William P. Spence, Eva Kultermann

Chapter8: Cast-in-place Concrete

Section: Chapter Questions

Problem 21RQ

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Refer to Problem 16.13. Design the size of the footing using the modified general ultimate bearing capacity Eq. (16.31). 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.
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.
Redo Problem 16.4 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 16.4 Redo Problem 16.1 with the following: = 16.5 kN/m3, cu = 41 kN/m3, =0, Df = 1.5 m, and factor of safety = 5.
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.
A steel pile (H-section; HP 360 1.491; see Table 18.1) is driven into a layer of sandstone The length of the pile is 18.9 m. Following are the properties of the sandstone: Unconfined compression strength = qu(lab) = 78.7 MN/m2 Angle of friction = 36 Using a factor of safety of 3, estimate the allowable point load that can be carried by the pile. Use Eq. (18.42).
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?
Redo Problem 16.1 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
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?
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.
The water table at a site is at 5 m below the ground level, and it is required to excavate to this level. The soil profile consists of a thick bed of sand where the unit weight is m = 17.0 kN/m3 above the water table and sat = 20.0 kN/m3 below the water table. The friction angle of the sand is 37. The wall of the excavation will be supported by cantilever sheet piles. How deep would you drive the sheet piles? Use the simplified analysis (Figure 15.37) with a factor of safety of 1.5 on the passive resistance. Determine the maximum bending moment in the sheet pile and the required section modulus for the sheet pile section (given an allowable stress of 190 MN/m2).
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.
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.

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following: 1. the length of footing excavation 2. the length of bricks work under D.P.C for each step by using: a) Centre line method b) Long wall-short wall method D.P.C 1.00 m 0.24 m 0.36 m y0.48 m 15 m Section. N A k W 8 m 5 m زف الحو 不 Z Plan ate the L