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A 5.125-ft square, 2.0-ft deep spread footing is subjected to a concentric vertical load of 62.0k and an overturning moment of 30.0 ft-k. The overturning moment acts parallel to one of the sides of the footing, and the top of the footing is flush with the ground surface and the groundwater table is at a depth of 0.5 ft below the ground surface. Determine whether the resultant force acts within the middle third of the footing, compute the minimum and maximum bearing pressures and show the distribution of bearing pressure in a sketch. Determine the size of the equivalent uniformly loaded footing and compute the equivalent bearing pressure. Solution: 1. Preliminary calculation: 1). Area of the footing: A = 2). Self weight of the footing: W, = 2). Middle third of the footing: B/6 = ft²: 3). Pore water pressure at the bottom of the footing due to ground water: u = 2. Check eccentricity: 1). Eccentricity of bearing pressure distribution: e = kips/ft²; lb; 3). Thus the bearing pressure distribution is 3. Calculate the minimum and maximum bearing pressure: 1). Qmin = ft; 2). Eequivalent bearing pressure: qeq = 2). Qmax = kips/ft²; 4. Equivalent uniformly loaded footing: 1). Eequivalent uniformly loaded footing width: B' = + kips/ft². ft; ft; k;

A 5.125-ft square, 2.0-ft deep spread footing is subjected to a concentric vertical load of 62.0k and an overturning moment of 30.0 ft-k. The overturning moment acts parallel to one of the sides of the footing, and the top of the footing is flush with the ground surface and the groundwater table is at a depth of 0.5 ft below the ground surface. Determine whether the resultant force acts within the middle third of the footing, compute the minimum and maximum bearing pressures and show the distribution of bearing pressure in a sketch. Determine the size of the equivalent uniformly loaded footing and compute the equivalent bearing pressure. Solution: 1. Preliminary calculation: 1). Area of the footing: A = 2). Self weight of the footing: W, = 2). Middle third of the footing: B/6 = ft²: 3). Pore water pressure at the bottom of the footing due to ground water: u = 2. Check eccentricity: 1). Eccentricity of bearing pressure distribution: e = kips/ft²; lb; 3). Thus the bearing pressure distribution is 3. Calculate the minimum and maximum bearing pressure: 1). Qmin = ft; 2). Eequivalent bearing pressure: qeq = 2). Qmax = kips/ft²; 4. Equivalent uniformly loaded footing: 1). Eequivalent uniformly loaded footing width: B' = + kips/ft². ft; ft; k;

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A 5.125-ft square, 2.0-ft deep spread footing is subjected to a concentric vertical load of 62.0k and an overturning moment of 30.0 ft-k. The overturning moment acts parallel to one of the sides of the footing, and the top of the footing is flush with the ground surface and the groundwater table is at a depth of 0.5 ft below the ground surface. Determine whether the resultant force acts within the middle third of the footing, compute the minimum and maximum bearing pressures and show the distribution of bearing pressure in a sketch. Determine the size of the equivalent uniformly loaded footing and compute the equivalent bearing pressure. Solution: 1. Preliminary calculation: 1). Area of the footing: A = 2). Self weight of the footing: Wf = 2). Middle third of the footing: B/6 = 3). Pore water pressure at the bottom of the footing due to ground water: u = 2. Check eccentricity: 1). Eccentricity of bearing pressure distribution: e = ft²; kips/ft²; = 3). Thus the bearing pressure distribution is 3. Calculate the minimum and maximum bearing pressure: 1). amin lb; 2). qmax kips/ft²; 4. Equivalent uniformly loaded footing: 1). Eequivalent uniformly loaded footing width: B' = 2). Eequivalent bearing pressure: deq ft; = → kips/ft². ft; ft; k;
Transcribed Image Text:A 5.125-ft square, 2.0-ft deep spread footing is subjected to a concentric vertical load of 62.0k and an overturning moment of 30.0 ft-k. The overturning moment acts parallel to one of the sides of the footing, and the top of the footing is flush with the ground surface and the groundwater table is at a depth of 0.5 ft below the ground surface. Determine whether the resultant force acts within the middle third of the footing, compute the minimum and maximum bearing pressures and show the distribution of bearing pressure in a sketch. Determine the size of the equivalent uniformly loaded footing and compute the equivalent bearing pressure. Solution: 1. Preliminary calculation: 1). Area of the footing: A = 2). Self weight of the footing: Wf = 2). Middle third of the footing: B/6 = 3). Pore water pressure at the bottom of the footing due to ground water: u = 2. Check eccentricity: 1). Eccentricity of bearing pressure distribution: e = ft²; kips/ft²; = 3). Thus the bearing pressure distribution is 3. Calculate the minimum and maximum bearing pressure: 1). amin lb; 2). qmax kips/ft²; 4. Equivalent uniformly loaded footing: 1). Eequivalent uniformly loaded footing width: B' = 2). Eequivalent bearing pressure: deq ft; = → kips/ft². ft; ft; k;
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