Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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Chapter 6, Problem 15P
To determine
If the driveway will be affected or not.
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Chapter 6 Solutions
Traffic and Highway Engineering
Ch. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 4PCh. 6 - Prob. 5PCh. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10P
Ch. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - Prob. 22PCh. 6 - Prob. 23PCh. 6 - Prob. 24PCh. 6 - Prob. 25PCh. 6 - The arrival times of vehicles at the ticket gate...Ch. 6 - Prob. 27P
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- subjectarrow_forwardExample 5.2: Design Example of a Square Foundation for a Colum Figure A.3a shows a square column foundation with the followin conditions: Live load=L=675 kN Dead load D= 1125 kN Allowable gross soil-bearing capacity 145 kN/m² Column size-0.5 m x 0.5 m -20.68 MN/m² 1,413.7 MN/m² Let it be required to design the column foundation. 1.25m-D L-675 AN D-1125 AN (a) Column section- 05 m x 0.5marrow_forwardExample 5.3: Design Example of a Rectangular Foundation for Column This section describes the design of a rectangular foundation to support a column having dimensions of 0.4 m x 0.4 m in cross section. Other details are as follows: Dead load D= 290 kN Live load L= 110 kN Depth from the ground surface to the top of the foundation = 1.2 m Allowable gross soil-bearing capacity = 120 kN/m² Maximum width of foundation = B = 1.5 m f, = 413.7 MN/m² f=20.68 MN/m² Unit weight of soil = y = 17.27 kN/m³ Unit weight of concrete = y₁ = 22.97 kN/m³ General Considerations For this design, let us assume a foundation thickness of 0.45 m (Figure A.4a). The weight of foundation/m² = 0.45 y = (0.45) (22.97) = 10.34 kN/m², and the weight of soil above the foundation/m² = (1.2)y = (1.2) (17.27) = 20.72 kN/m² Hence, the net allowable soil- bearing capacity Iqata] =120-10.34-20.72 = 88.94 kN/m² The required area of the foundation = (D+L)/4(all) = (290 + 110)/88.94 = 4.5 m². Hence, the length of the foundation…arrow_forward
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