The loads of the building are transferred to the soil through foundations. The sizes

type of the foundation should be chosen so as,

• to ensure structural safety of the foundation (no damage at the foundation),

• to ensure that there is no settlement (no excess of the bearing capacity of soil),

• to provide economy.

The first step for the design of foundation is correct determination of soil properties

(i.e. by soil survey at the site).

The stress distribution under the foundation depends on,• type of loading (loads transferred through the foundation),• type of soil,• rigidity of the foundation.It is important to ensure that the max. stress under the foundation is smaller than the bearing capacity (allowable stress) of the soil (σz,max.≤ σz,allow.).

The types of foundations,1. Shallow foundations

a. Wall Footings: a continuous strip along the length of wall having a width larger than the wall thickness.

The types of foundations,

b. Independent Isolated Column (Single) Footings: isolated rectangular or square slabsunder the column. The length of all sides from the column face to the edge of foundationmay be different (due to presence of the bending moments). They are reinforced alongboth directions and economical solution for relatively small loads.

The types of foundations,

c. Combined Column Footings: support two or more columns. They are used to provide a uniform stress distribution over the soil. If the geometric center of the footing slab and location of the resultant forces transferred by the columns coincide, this uniform stress distribution can be achieved. They may also be used if the columns are too close to each other.

The types of foundations,

d. Strap Footings: transfers the loads of two or more columns/shear walls to the groundsafely. The straps may be either one-way or two-way. They are much more effectivecompared to isolated columns, especially when subjected to earthquake loads (i.e. Whenconsiderable bending moments are transferred by the column). Besides, thesefoundations prevent relative settlement, which may be a problem in case of isolatedfootings.

One-way strap footings Two-way strap footings

The types of foundations,

e. Raft (Mat) Foundations: When the allowable bearing capacity of the soil is low and/orthe loads that are transferred from the building are high, the use of raft foundations maybe necessary. The loads from the columns/shear walls are distributed over a large area ofsoil and relative support settlements are prevented. Also, if the ground water level ishigh, then mat foundation with water proofing may be ideal. The mat foundation may beassumed as an inverted floor system.

L (1/4~1/5)L

The analysis and design of mat foundations are performed by assuming these systems:a. Rigidb. Flexible

Rigid Foundation Assumption: It is assumed that the rigidity of the foundation isvery high compared to the rigidity of soil. The stress distribution under thefoundation will then depend on the magnitude and type of loads transferred by thestructure and weight of the foundation itself.

(Case I) (Case II)

Rigid Foundation Assumption:

(Case I)

(Case II)

Case I: If the resultant of vertical loads from the structure passes close to geometric center of themat and if the mat plate is thick, then the effect of bending moments may be ignored and stressdistribution may be assumed as uniform.

𝜎𝜎𝑧𝑧,𝑚𝑚𝑚𝑚𝑚𝑚 =∑𝑁𝑁𝑑𝑑𝐴𝐴

Case II: If the eccentricity (distance between the resultant of vertical loads and geometric centerof the mat) may not be ignored, then the stress distribution changes linearly along both axes. Thenthe maximum stress may be calculated as:

𝜎𝜎𝑧𝑧,𝑚𝑚𝑚𝑚𝑚𝑚 =∑𝑁𝑁𝑑𝑑𝐴𝐴

+∑𝑀𝑀𝑑𝑑𝑚𝑚 × 𝑦𝑦𝑚𝑚𝑚𝑚𝑚𝑚

𝐼𝐼𝑚𝑚+∑𝑀𝑀𝑑𝑑𝑚𝑚 × 𝑦𝑦𝑚𝑚𝑚𝑚𝑚𝑚

𝐼𝐼𝑚𝑚

Nd: Total vertical loads from the structureA: Area of the mat foundation Mdx: Moment along x direction (Nd x eccentricity along y-direction)Mdy: Moment along y direction (Nd x eccentricity along x-direction)Ix: Moment of inertia of mat area about x-axisIy: Moment of inertia of mat area about y-axis

ymax: max. distance from the geometric center, along y-directionxmax: max. distance from the geometric center, along x-direction

Flexible Foundation Assumption: If the the mat foundation is not rigid (lowthickness and no beams), then uniform or linear stress distribution may not bevalid.

In this case, finite element approach should be used. In this approach, elasticsprings are defined under the mat foundation, in order to represent the response ofelastic soil (spring coefficient is defined for each different type of soil).

In finite element analysis, the mat foundation should be meshed into parts whichhave approximately 1 m2 area. The multiplication of spring coefficient (ko; unit:kN/m3) and max. vertical displacement (δz,max., found by analysis) will then givethe max. stress under the foundation (σz,max.= ko × δz,max.).ko→ unit: kN/m3

δz,max. → unit: mσz,max.→ unit: kN/m2

Check:σz,max.≤ σz,allow.

The following mat foundation (for a 4-story building) will be modeled and analyzed by usingSAP2000. The thickness of the mat is selected as 0.80 m. The local site class is defined as Z3.The soil type is hard clay and corresponding modulus of subgrade reaction (ko) is assumed as20000 kN/m3. The effective depth may be taken as 0.75 m. Materials: C25/S420. Note thatcantilever portion of the mat is assumed to have a length of 0.60 m. (away from the column faceat all sides). The allowable stress of the soil will be taken as 200 kN/m2.

The analysis was performed consideringcolumn loads corresponding to«1.4G+1.6Q» combination for the analysisof building, which yields highest axialcolumn loads. The axial column loads wastaken from the analysis of 4-story building(SAP2000 modeling was demonstratedpreviously).

Examples utilized: Examples on Page 143and Page 642; Betonarme Yapıların Hesapve Tasarımı; Doğangün,A.

The column axial loads at the base which were taken from «1.4G+1.6Q» loadcombination of the analysis for 4-story building are shown below. The matfoundation will be analyzed for these point loads at the column joints (Load Case:Dead). Note that each portion between the columns will be

divided into 4×4 parts (meshing). This will provideapproximately 1 m. dimension for each meshed part.

• Change units to «kN, m, C»• File > New Model > Quick Grid Lines

• Right Click > Edit Grid Data > Modify/Show System

Change distances between grid lines according to your plan dimensions! After clicking OK two times, you will see the modified dimensions (as shown below).

• Define > Materials > Add New Material > (Region: User; Material Type: Other) >

OK > «Material Property Data» will be changed as below:For C25 concrete, modulus of elasticity is 3×107 kN/m2. The poisson’s ratioand coefficient of thermal expansion may be taken as 0.2 and 0, respectively.The weight per unit volume is taken as 25 kN/m2 in order to include the selfweight of the mat foundation in the analysis.

• Define > Section Properties > Area Sections > Select «Shell» > Add New Section

> (Type: Plate Thick; Material Name: C25; Thickness: 0.8) > OK (two times)

Thickness for both membrane and bending is taken identical.

• Click on xy view once again (although you see X-Y Plane view on the screen).

Then Draw > Quick Draw Area > «draw the area members starting from the leftbottom corner moving to the right each time» (note that «Section» should be theone that you have defined for the mat on the «Properties of Object» window).

• Set Display Options > Select «Labels» under «Areas» > you will see labels ofeach area.

• Areas 2, 3, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19 will be meshed. The number ofmeshing along the x and y directions are shown below for each area label.

Area Label

2 3 9 10 11 12 13 14 15 16 18 19

Along x-dir. 4 4 1 4 4 1 1 4 4 1 4 4

Along y-dir. 1 1 4 4 4 4 4 4 4 4 1 1

• For each area that will be meshed, select corresponding area first (you may select the areas that

will be meshed identically at the same time). For example, the areas 2, 3, 18 and 19 will all bedivided into 4 parts along x-direction and 1 part along y-direction. Select these areas at the sametime.

• Edit > Edit Areas > Divide Areas > on the «Divide Selected Areas» select «Divide Area IntoThis Number of Objects»

Along Edge from Point 1 to 2 = Along x-directionAlong Edge from Point 1 to 3 = Along y-directionDo the same meshing (divide areas) operation for the other areas and youwill see the following meshed area.

But the labeling order of the areas is notproper as you see here. Therefore, this will bere-ordered in the next step.

• Select all model• Edit > Change Labels > in «Interactive Name Change» menu > Item Type:

Element Labels – Area; First Relabel Order: Y, Second Relabel Order: X > on thesame menu > Edit > Auto Relabel > All in List > Click OK (twice).

• Set Display Options > Un-select «Labels» under «Areas» > Select «Labels» under

«Joints» > you will see labels of each joint.• Note that the labels of joints are also not in order. «Select all» so that the same

ordering will be done.• Edit > Change Labels > in «Interactive Name Change» menu > Item Type:

Element Labels – Joint; First Relabel Order: Y, Second Relabel Order: X > on thesame menu > Edit > Auto Relabel > All in List > Click OK (twice).

At the end, the labeling of each joint will be changed as you see in the next figure.

• Select all.• Assign > Joint > Restrains > on «Joints Restrains» window select only

Translation 1, Translation 2 and Rotation about 3 > OK (two times)This will restrain movement along x and y direction,and bending about z axis (torsional movement).Actually rigid supports are placed to restrain thesemovements.

• Select all.• Assign > Area > Area Springs > on «Assign Spring To Area Object Face» window,

Spring Type: Simple, Spring Stiffness per Unit Area: 20000 (kN/m2), Simple Spring Resists:Tension and Compr., Area Object Face: Bottom, Spring Tension Direction: Parallel to AreaObject Local Axes (3, meaning along positive z-direction), Options: Replace Existing Springs(may be selected).

This will add a spring at the bottomface of each area (directed towardspositive z-direction). These springsrepresent resistance of soil which wasdefined as a spring constant equal toko=20000 kN/m2.

• The following point loads are the axial loads of columns for «1.4G+1.6Q» load

combination at the base level. These axial loads are taken from the analysis of thebuilding which was demonstrated previously).

Joint Vertical Point Load (kN)

13 403.60

17 695.20

21 405.00

57 681.60

61 1085.30

65 672.70

101 423.70

105 704.10

109 430.40

After selecting the related joint,Assign > Joint Loads > Forces > Load Pattern Name: DEAD, Foce Global Z: -403.60(in the negative z-direction, along gravitational direction) > OK.Repeat this operation to define all loads defined in the next table.

• Save Project and Run Analysis (you may not run for Modal Load Case; just Dead Load Case is

enough).• In order to see the vertical displacements (settlement), choose «Show Deformed Shape». On

«Deformed Shape» window, choose «Draw displacement contours on area objects». «Areacontour component» will be Uz (vertical displacement along z-axis). You will see the change ofsettlement on the mat foundation area in x-y view (following figure).

The maximum settlement is at the bottom left corner, which is (δz,max.)0.00851 m. (negative, in the downward direction). The maximum stress:𝜎𝜎𝑧𝑧,𝑚𝑚𝑚𝑚𝑚𝑚. = 𝑘𝑘𝑜𝑜 × 𝛿𝛿𝑧𝑧,𝑚𝑚𝑚𝑚𝑚𝑚. = 20000 𝑘𝑘𝑘𝑘

𝑚𝑚3 × 0.00851

= 170.2 𝑘𝑘𝑘𝑘𝑚𝑚2 < 200 𝑘𝑘𝑘𝑘

𝑚𝑚2 SAFE!

• In order to see the distribution of bending moment about x-axis, choose «Display», «Show

Forces/Stresses», «Shells». On «Member Force Diagram» window, choose «M11» as«Component». The reinforcement along y-direction will be determined based on this bendingmoment. Due to positive moment, there will be tension at the lower side of the foundation. Dueto negative bending moment, there will be tension at the upper side of the foundation.

Therefore you may determine the bottom and topreinforcement (along the y-direction) by consideringhighest negative and positive bending moments (aboutthe x-axis), respectively.

𝐴𝐴𝑠𝑠,𝑏𝑏𝑜𝑜𝑏𝑏𝑏𝑏𝑜𝑜𝑚𝑚 = 𝑀𝑀𝑑𝑑+

𝑓𝑓𝑦𝑦𝑑𝑑×𝑗𝑗𝑙𝑙×𝑑𝑑

𝐴𝐴𝑠𝑠,𝑏𝑏𝑜𝑜𝑡𝑡 = 𝑀𝑀𝑑𝑑−

𝑓𝑓𝑦𝑦𝑑𝑑×𝑗𝑗𝑙𝑙×𝑑𝑑

As an example, the demonstrated analysis has thehighest negative bending moment about x-axis with avalue of -231.9 kN.m (as shown in the next figure).

Bending moment about x-axis: M11=Mxx

Positive sign convention according to SAP2000 (for shell members)

• In order to see the distribution of bending moment about y-axis, choose «Display», «Show

Forces/Stresses», «Shells». On «Member Force Diagram» window, choose «M22» as«Component». The reinforcement along x-direction will be determined based on this bendingmoment.

You may determine the bottom and top reinforcement(along the x-direction) by considering highest negativeand positive bending moments (about the y-axis),respectively.

𝐴𝐴𝑠𝑠,𝑏𝑏𝑜𝑜𝑏𝑏𝑏𝑏𝑜𝑜𝑚𝑚 = 𝑀𝑀𝑑𝑑+

𝑓𝑓𝑦𝑦𝑑𝑑×𝑗𝑗𝑙𝑙×𝑑𝑑

𝐴𝐴𝑠𝑠,𝑏𝑏𝑜𝑜𝑡𝑡 = 𝑀𝑀𝑑𝑑−

𝑓𝑓𝑦𝑦𝑑𝑑×𝑗𝑗𝑙𝑙×𝑑𝑑

As an example, the demonstrated analysis has thehighest negative bending moment about y-axis with avalue of -261.6 kN.m (as shown in the next figure).

Bend

ing

mom

ent a

bout

y-a

xis:

M22

=Myy

Positive sign convention according to SAP2000 (for shell members)

The mat foundation is reinforced by longitudinal bars at the upper and lower portions of the foundation, along bothx- and y-axis (mesh type reinforcement). These longitudinal reinforcements are calculated regarding the bendingmoments, as explained in the previous slides. The free edges of the foundation is reinforced by hairpin bars (nocalculation). The upper reinforcements are held in place by means of rebar chairs (no calculation). The additionalpunching reinforcement against punching (if required) is as follows:

Reference: Website «http://debug.pi.gr/Default.aspx?ch=77»