Session 9 – 10 MAT FOUNDATION Course : S0484/Foundation Engineering Year : 2007 Version : 1/0 Session 9 – 10 MAT FOUNDATION

Topic: Introduction Type of Mat Foundation Bearing Capacity Settlement

INTRODUCTION A mat foundation (or raft foundation) is continuous in two directions capable of supporting multiple columns, wall or floor loads. It has dimensions from 20 to 80 ft or more for houses and hundreds of feet for large structures such as multi-story hospitals and some warehouses, or consisting of stiffening beams placed below a flat slab are useful in unstable soils such as expansive, collapsible or soft materials where differential movements can be significant (exceeding 0.5 inch).

TYPE OF MAT FOUNDATION FLAT PLATE

TYPE OF MAT FOUNDATION FLAT PLATE THICKENED UNDER COLUMN

TYPE OF MAT FOUNDATION BEAMS AND SLAB

TYPE OF MAT FOUNDATION SLAB WITH BASEMENT WALL

BEARING CAPACITY GROSS ULTIMATE BEARING CAPACITY FOR SATURATED CLAYS WITH  = 0 AND VERTICAL LOADING CONDITION (=0  Nc = 5.14 ; Nq = 1.00 ; N =0.00) WITH: SO:

BEARING CAPACITY NET ULTIMATE BEARING CAPACITY NET ALLOWABLE BEARING CAPACITY (FS = 3)

BEARING CAPACITY PLOT OF qall(net)/cu against Df/B with factor of safety = 3

BEARING CAPACITY Unit weight =  Df Q Where: Q = dead weight of the structure and the live load A = area of the raft

EXAMPLE 1 Problem: Determine the net ultimate bearing capacity of a mat foundation measuring 45 ft x 30 ft on a saturated clay with cu = 1950 lb/ft2,  = 0 and Df = 6.5 ft Solution:

FACTOR OF SAFETY For Saturated Clay

EXAMPLE 2 – PROBLEM Unit weight =  Df Q The mat dimension is 60 ft x 100 ft. The total dead and live load on the mat is 25x103 kip. The mat is placed over a saturated clay having a unit weight of 120 lb/ft3 and cu = 2800 lb/ft2. Given Df = 5 ft, determine the factor of safety against bearing capacity failure

EXAMPLE 2 - SOLUTION

EXAMPLE 3 – PROBLEM Consider a mat foundation 90 ft x 120 ft in plan, as shown in the following figure. The total dead load and live load on the raft is 45x103 kip. Estimate the consolidation settlement at the center of the foundation Q 90 ft x 120 ft 6 ft 5 ft 40 ft 18 ft sand Normally consolidated clay sat = 118 lb/ft3 Cc = 0.28 ; eo = 0.9 Water table  = 100 lb/ft3 sat = 121.5 lb/ft3

EXAMPLE 3 – SOLUTION For Q = 45x106 lb, the net load per unit area is

EXAMPLE 3 – SOLUTION In order to calculate pav, the loaded area can be divided into four areas, each measuring 45 ft x 60 ft. The average stress increase in the clay layer below the corner of each rectangular area can be calculated by using the following formula: 90 ft 45 ft x 60 ft 120 ft Where: H1 = the depth of top elevation of clay layer H2 = the depth of bottom elevation of clay layer Ia(H1) ; Ia(H2) = influence factor

EXAMPLE 3 – SOLUTION

EXAMPLE 3 – SOLUTION

EXAMPLE 3 – SOLUTION