SITUATION 2 - Design of Rectangular Footing: The center of one of the columns of MIT Intramuros is 1.5 m from the property line. The supporting column given the following parameters: The 350 x 350 mm column carries a dead load of [650+10B] kN and a live load of [800-40A] KN. The footing is placed 1.8 m below the NGL. The allowable soil bearing capacity is [202.50 +30C] kPa. Use y,= 17.5 kN/m³, y 23.6 kN/m³, f. = 21 MPa, and fy = 280 MPa. Use 16 mm diameter for main reinforcements. Use the initial thickness of 750 mm. A. Determine the effective soil pressure B. Determine the size of the footing C. Determine the ultimate soil pressures D. Investigate if the thickness is adequate against one-way shear and two-way shear E. Determine the design moment of the footing. DISREGARD the evaluation of item (D) F. Determine the number of bars each way.
SITUATION 2 - Design of Rectangular Footing: The center of one of the columns of MIT Intramuros is 1.5 m from the property line. The supporting column given the following parameters: The 350 x 350 mm column carries a dead load of [650+10B] kN and a live load of [800-40A] KN. The footing is placed 1.8 m below the NGL. The allowable soil bearing capacity is [202.50 +30C] kPa. Use y,= 17.5 kN/m³, y 23.6 kN/m³, f. = 21 MPa, and fy = 280 MPa. Use 16 mm diameter for main reinforcements. Use the initial thickness of 750 mm. A. Determine the effective soil pressure B. Determine the size of the footing C. Determine the ultimate soil pressures D. Investigate if the thickness is adequate against one-way shear and two-way shear E. Determine the design moment of the footing. DISREGARD the evaluation of item (D) F. Determine the number of bars each way.
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Braja M. Das, Khaled Sobhan
Chapter16: Soil Bearing Capacity For Shallow Foundations
Section: Chapter Questions
Problem 16.15P: Refer to Problem 16.13. Design the size of the footing using the modified general ultimate bearing...
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Answer D, E and F
Let:
A=6
B=5
C=3.714
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