ice raspunsuri foundations

8/12/2019 ICE Raspunsuri Foundations

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THEORETICAL SUBJECTSfor the graduating examination - Civil Engineering specialization (ICE)

Discipline: FOUNDATIONS

1. Foundations with plain concrete block and reinforced concrete pillow. Designprescriptions. Establish the foundation base dimensions.

Answer 1:Foundations with plain concrete block and reinforced concrete pillow are composed

by a plain concrete block on which rests a reinforced concrete pillow in which columns areembedded.

Fig. 1. Foundation elements Fig. 2. Contact pressure diagramDimensioning foundations with plain concrete block and reinforced concrete pillow

consist in establishing the dimensions of the foundation block (L, B, H), respectively of the pillow ( c, bc, h c), and also the reinforcement needed in the pillow.

Dimensions in plan L and B are determined so that the maximum ground pressureshould not exceed the allowable pressure of the foundation ground: p max p all.

Verification of the ground pressure: p1 = pmax pall,, p2 = pmin 0 (1)

where 0 0 0 01,26

1 x x x

N M N e p

S W L B L

In the case when relation (1) is not fulfilled, dimensions L and B of the foundation aremodified till relation (1) is verified.


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2. Reinforced concrete block foundation. Design prescriptions. Establishing foundationbase dimensions.

Answer 2:This type of foundations is realized from a reinforced concrete block in which are

embedded cast-in place columns. Dimensioning the reinforced concrete block foundationsconsist in establishing the dimensions of the foundation block (L, B, H), and also thereinforcement quantity needed in the foundation.

Fig. 1. Foundation elementsFig. 2 Calculus scheme and foundation

reinforcementDimensions in plan L and B are determined so that the maximum ground pressure

should not exceed the allowable pressure of the foundation ground: p max p all.

Verification of the ground pressure:6

p 21,2 L B H T M

L B

G N f

p1 = pmax pall,, p2 = pmin 0 (1)In the case when relation (1) is not fulfilled, dimensions L and B of the foundation are

modified till relation (1) is verified.


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3. Plain concrete continuous foundations under masonry walls. Design prescriptions anddimensioning.

Answer 3:Plain concrete continuous foundations under masonry walls presented in fig. 1 have at

the upper part a reinforced concrete girdle (min. 20 cm height) placed on the wall width andreinforced with minim 6 14, for seismic areas with a g 0,16g.

The width of the foundation block B is established function of:a) bearing capacity of the foundation ground;

b) thickness of the wall b (or elevation) that rests on the foundation and for B will betaken into account the relation: B b+10 cm;

c) minimum dimensions necessary for excavation works.The calculus for dimensioning the continuous foundations is done for 1 m foundation

length (for most loaded area) and consists in determining the foundation base width, so thatthe ground pressure will not exceed the allowable pressure of the ground ( p tr ).

Fig. 1. Calculus scheme

The necessary width of the foundation B is

determined from the condition: all f p

BGQ1

p1,2 (1)

The foundation height H , is established fulfilling theminimum design values and the stiffness condition. Withknown values for B and H is determined the self-weight ofthe foundation, G f and is verified again the relation (1). Inthe case when the condition (1) is not fulfilled, thedimension B of the foundation must be increased.


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4. Shallow foundations under heavy loaded columns. Establishing the foundation basedimensions.

Answer 4:In the case of columns subjected to great loads, the use of plain concrete foundations

or the use of reinforced concrete block foundations leads to great contact surfaces that exceed1516 m 2. In these cases are used shallow foundations with base reinforced concrete plateand reinforced concrete buttress (fig. 1): 1 column, 2 buttresses, 3 base plate, 4 plain concretelayer.

Fig. 1. Shallow foundation

Dimensions in horizontal plan of the base plate ( B and L) are established from the bearingcapacity condition: all p maxef p . Fig. 2. Foundation types

For static calculus the base plate is considered loaded with the foundation groundreactions and it rests on buttress. The minimum thickness of the base plate is 20 cm. Buttressare disposed in plan so that to ensure the loads overtaking from the column and theirtransmission to the base plate (fig. 2). The buttress calculus is performed for the bendingmoment that acts at the middle of the buttress of pier height. On x direction buttress areverified for x stresses and those on y direction for y stresses(fig. 3). Verification relations:

2

6( )

x x c cd

c

M R f h and 2

6( )

y y c cd

c

M R f h . Minimum thickness of the buttress will be

taken to be (1520) cm.


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5. Continuous beam foundations under columns. Design prescriptions. Reinforcementprinciples.

Answer 5:From constructive point of view, continuous beam foundations under columns are

reinforced concrete beams, with or without brackets, in which columns are embedded (fig.1).Function of the columns plan disposal, the longitudinal axis of the beams can be: rectilinear,

polyhedral or circular.

Fig. 1. Continuous beam foundationsunder columns: a - rectilinear; b -

polyhedral; c circular.

Fig. 2. Reinforcement

The cross-section of the continuous beam foundations is usually an inverted T cross-section, composed by a rectangular beam and a base plate symmetrically placed towards the

beam (fig. 2). From economic point of view (steel consumption), the height H of the base plate will be taken to fulfil the condition H/B = 0,25...0,35. The height of the foundation beamis: l

61

...31

H g

. The foundation beam has a longitudinal resistance reinforcement composed

by straight bars, inclined bars, stirrups and wire stitch. The static calculus and the continuous beam foundations dimensioning consists in establishing the plan dimensions of the foundation base and longitudinal and transversal resistance reinforcement calculus for the beam and forthe foundation base.


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6. Foundations on beam grids. Structure. Design prescriptions.

Answer 6:These foundations are used for multi-storey buildings with framed structure when the

foundation ground has a low bearing capacity.The grid foundation solution ensures a reduced differential settlement due to the

differences between the values of the axial loads transmitted by the columns. So, thesefoundations are composed by beams disposed on both directions, usually orthogonaldirections, the columns discharging in the intersection points of the perpendicular beams.

Fig.1This foundation system has the purpose of stiffening the construction base on both

directions, preventing the non uniform settlement.Constructive elements, shape of the transversal cross-section, longitudinal and

transversal reinforcement are realized in the same way as those for continuous foundationsunder columns, disposed on one direction.

The static calculus and the dimensioning are realized decomposing the beam system intwo direction beams. The axial loads acting at an intersection point will be distributed partially to one direction , respectively to the other direction beam function of their stiffness.


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7. Mat foundations.

Answer 7:Mat foundations are used for underground constructions realized under water level

(buried reservoirs, cooling towers, basins and so on). In this case the mat is independent fromthe construction foundations, being separated by these with impermeable joints. Because ofthis, such kind of mat doesnt work in bending, respectively doesnt contribute to loadtransmission from construction to the foundation ground, having only the purpose of creatingan impermeable tank, together with the construction basement.

Fig. 1. Calculus scheme for a mat foundationThe thickness of the mat is determined from the condition that its weight should be

sufficient so that to equilibrate the water lifting pressure, ensuring in this way also the matstability. Calculus relation for the mat thickness, corresponding to the floating condition is:

wconcrete

w h

r h , in which: h r mat minimum necessary thickness, concrete unit weight of the

concrete, w unit weight of the water, h w maximum height of the underground watertowards the level of the horizontal hydroisolation.

Mat foundations are usually realized from plain concrete or poorly reinforcedconcrete.


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9. Cast-in place piles .

Answer 9:For assuring the stability of holes walls is used excavation with bentonite slurry.

Bentonite slurry is a suspension obtained mixing bentonite which is a clay rich inmontmorillonite with water. There are two methods for driving the pile shaft using a bentoniteslurry: direct circulation system and reverse circulation system.

The first system supposes that the bentonite slurry is pumped through the drilling rigand the excavated soil material is transported by the bentonite slurry upwards. The mixture of

bentonite slurry and soil is separated by sieving and sedimentation and the clean bentoniteslurry is reintroduced in the bored pile shaft (fig. 1). The reverse circulation system introducesthe clean bentonite slurry into the drilled shaft and it is pumped out together with theexcavated soil material.

Fig. 1. Direct circulation drilling system Fig. 2. Reverse circulation drilling systemFor realizing a pile, after finishing the drilling operation till to the necessary depth it

will be introduced in the drilled shaft a reinforcement cage using a crane and the concrete will be poured by Contractor method using a pipe which will allow to fill with concrete from the

bottom of the shaft upwards the pile.


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10. Calculus of the bearing capacity for piles at vertical loads. Calculus principles.

Answer 10:

Fig. 1. Transmission of thevertical loads to piles

Pile foundations are composed by piles and a generalmat that links their upper part. F ri ction pil es transmit loads byfriction between the lateral surface and the foundation ground.Usually friction piles are used in the case when the resistantfoundation ground is found at great depths and the pile doesntreach it. Function of the load and foundation ground naturefrom the pile base, the vertical load is transmitted to theground by friction on the lateral surface and by pressures at thecontact between the base and the foundation ground:

l l v v l v R p A p A P P

where: pv is the resistance of the soil at the pile tip, Av is the pile base cross-sectionarea, p l is the average friction resistance on the pile lateral surface, Al is the area of the pilelateral surface.

In a more general way, the calculus formula for the bearing capacity of a friction pilesubjected to compression can be written as: R = k(m ipvA + Um jf ili) where: k soil non-homogeneity coefficient; m i and m j workability coefficients, whose values depend upon themethod used to perform the pile and upon the soil nature; p v ground conventional resistanceunder pile tip, in kPa; f i conventional resistance on pile lateral surface for i layer, in kPa; l i

pile length in contact with i layer, in meters; A pile cross-section area, in square meters; U pile cross-section perimeter, in meters.


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V. CASE STUDY / PROBLEMS

Discipline: FOUNDATIONS

Problem 1

For the retaining wall in the next figure draw the lateral earth pressure diagram anddetermine the resultant active thrust (value, point of application and direction) knowing:

- The height of the retaining wall H = 4,0 m;- Behind the retaining wall is found a homogeneous soil with the following

characteristics: = 18,0 kN/m 3, = 30 0, c = 0 kN/m 2;

- The angle of friction between the wall and soil, = (1/22/3) ; - The coefficient of active earth pressure, K a = 0,299.

Solution 1:If 00 20...153/2...2/1 is choosen = 17,5 0Pressure calculus at B and A level:

0K 0 p aB

aA K H p 528,21299,0418 kN/m2

Active earth pressure resultant:

056,43299,02418

K 2H

2K HH

SP2

a

2a

ide_presiundiagramei_ a kN/m

Position of the active earth pressure resultant:z = H/3 = 4/3 = 1,33 m (from the base level above)


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Problem 2Determine the width and the height of a continuous plain concrete foundation

(presented in the next figure) placed under a bearing wall realized from brick masonry,knowing:

- load Q = 178 kN/ml;- wall width b = 37,5 cm;- freezing depth h freezing = 0,7 m;- concrete = 24,0 kN/m 3;- the supporting soil consists of a compacted sand with the following characteristics:

ID = 0,8, p all = 300 kN/m2, tg allowable = 1,30.

Solution 2:It is established the foundation depth:Df = h freezing + (0,10,2) m = 0,7 + 0,1 = 0,8 m Considering 1.00 m from the length of the continuous foundation, centrically loaded,

the condition of determining the width B is:

all f p

B

GQ p

1 (1)

where G f = concrete H Bn 1 2419,02,1 B replace G f in relationship (1) and

well obtain 3001

249,02,1175 B

B => B B 300249,02,1175 =>

B(300 - 249,02,1

) 175 => B 6384,008,274

175

m => is chosen B = 0,65 m According to figure, H = D f + 0,1 => H = 0,9 mFor H = 0,9 m is verified the stiffness condition: allowable tantan

545,61375,0

9,02/)375,0(

tan B

H

alowable tan 1,30


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Problem 3Determine the corrected conventional pressure (p conv ) for an isolated plane concrete

foundation with the plan dimensions of 2,30 x 3,00 m, with the foundation depth D f = 1,80 mand the supporting soil composed by a silty clay (e = 0,8, I C = 0,75), knowing the followingdata (STAS 3300/2-85):

The corrected conventional pressure is determined according to STAS 3300/2-85 withthe relationship:

D Bconvconv C C p p , [kN/m2]

in which: conv p - basis conventional pressure ( conv p = 235 kN/m 2);

BC - width correction;

DC - depth correction.The width correction:

- for B < 5 m is determined with the relationship: 11 B K pC conv B , [kN/m 2]

where K 1 is a coefficient with a value: 0,1 for cohesionless soils except silty sands and 0,05for silty soils and cohesive soils. - for B 5 m the width correction is :

conv B pC 4,0 for cohesionless soils, except silty sands;

conv B pC 2,0 for silty sands and cohesive soils. The depth correction is determined using the relationships:

- for D f < 2 m:

4

2 f conv D

D pC

- for D f > 2 m:

22 f D D K C where: = 18,8 kN/m 2;

K 2 = 2,0 for silty and cohesive soils.

Solution 3:The conventional pressure is determined by the relationship:

conv B Dconv p p C C For B = 2,30 m (meaning B < 5 m) the width correction is determined with therelationship: 1 1 B convC p K B where K 1 = 0,05 for cohesive soils.

1 1 B convC p K B

= 325 0,05 (2,30 - 1) = 21,125 kN/m2

For D f = 1,80 m (D f < 2 m) the depth correction is determined with the relationship:

CD = conv p 4

2D f = 3254

280,1 = - 16,25 kN/m 2

The calculus conventional pressure is:

conv B Dconv p p C C = 325 + 21,125 16,25 = 329,875 kN/m2


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Problem 4For the gravity retaining wall from the following figure is performed the stability

check to overturning, knowing:- H = 4,0 m; B = 2,0 m; b = 0,8 m; a = 1,0 m; c = 0,6 m and concrete = 24 kN/m 3.- Lateral active earth pressure behind the retaining wall (P a = 43 kN);

- Friction angle between the wall and soil , = 17,50

.

Answer 4:Position of the point of application of the

lateral active earth pressure:z = H/3 = 4/3 = 1,33 mG1 =b (H -a) 1 24 = 0,8 3 1 24 = 57,6 kN G2 = (B-b- c) (H -a) 24 / 2 = (2 -0,8- 0,60) (4 -1) 24 / 2 = 21,6 kN G3 =B a 1 24= 2 1 1 24 = 48 kN

d1=B-b/2= 2-0,8/2 =1,6md2 =B-b-(B-b-c)/3= 2-0,8-(2-0,8-0,6)/3= 1 md3 = B/2 = 2/2 = 1 mStability check to overturning: M r 0,8 M s M r = P a cos z = 43 0,9537 1,33 = 54,54 kNm M s = G 1 d1 + G 2 d 2 + G 3 d3 + P a sin B =57,6 1,6 + 21,6 1 + 48 1 + 43 0,300 2=153,06 kNm

Check : 54,54 kNm 122,45 kNm


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Problem 5Determine the plain concrete block dimensions for a plain concrete block and

reinforced concrete pillow foundation knowing: loads (N = 1150 kN; M = 50 kNm; T = 6kN); bearing capacity of the foundation ground is p all = 300 kN/m

2; column dimensions (a =40 cm, b = 35 cm); pillow dimensions (h c = 30 cm, l c = 1,0 m); med = 20 kN/m 3 and tan a =

1,3.

Solution 5:Predimensioning:

LB = 1,2N/p all where L = 1,2B =>

1,2B 2 = 1,2N/p all => B =3002,1

11502,1=

2,0 m and L = 1,2 2,0 = 2,4 m The height of the plain concrete

block is determined from the stiffness

condition: atantan

tan = 7,01

H l

H 1,3 => H = 0,95 m

Check of the foundation dimensions:

/6LB)hT(HM

LBG N

p 2cf

1,2 where G f = B L (H + h c) med =2,0 2,4 (0,95 + 0,3) 20

= 120 kN

2

2

22,1 kN/m234,64

kN/m294,52

6/4,20,225,1650

4,20,21201150

p

p1 = p max = 294,52 kN/m2 300 kN/m 2 and p2 = p min = 234,64 kN/m

2 0

ice raspunsuri foundations

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