2 ft 24 ft 24 ft 24 ft 2 ft Problem 4 B E F G 3 ft DL=190 kip DL-100 kip DL=180 kip LL = 60 kip LL= 120 kip DL=110 ki • The plan of a mat foundation with column loads is shown in Figure 2. Use the rigid method to calculate the soil pressures at point A, B, C, D, E, F, G, H, , J, K, L, M and N. The size of the mat is 76 ft x 96 ft, all columns are 24 in x 24 in in section, and qaloet = 1.5 kip/ft?. Verify that the soil pressures are less than the net allowable bearing capacity. LL = 120 kip LL = 70 ki 30 ft DL-180 kip DL=360 kip LL = 120 kip LL= 200 kip Ty LL= 250 kip LL =120 k DL-400 kip DL=200 kip 30 ft DL-190 kip DL=400 kip DL=T40 kip DL=200 kip

2 ft 24 ft 24 ft 24 ft 2 ft Problem 4 B E F G 3 ft DL=190 kip DL-100 kip DL=180 kip LL = 60 kip LL= 120 kip DL=110 ki • The plan of a mat foundation with column loads is shown in Figure 2. Use the rigid method to calculate the soil pressures at point A, B, C, D, E, F, G, H, , J, K, L, M and N. The size of the mat is 76 ft x 96 ft, all columns are 24 in x 24 in in section, and qaloet = 1.5 kip/ft?. Verify that the soil pressures are less than the net allowable bearing capacity. LL = 120 kip LL = 70 ki 30 ft DL-180 kip DL=360 kip LL = 120 kip LL= 200 kip Ty LL= 250 kip LL =120 k DL-400 kip DL=200 kip 30 ft DL-190 kip DL=400 kip DL=T40 kip DL=200 kip

Principles of Foundation Engineering (MindTap Course List)

8th Edition

ISBN:9781305081550

Author:Braja M. Das

Publisher:Braja M. Das

Chapter5: Ultimate Bearing Capacity Of Shallow Foundations: Special Cases

Section: Chapter Questions

Problem 5.5P

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