THEORETICAL SUBJECTS for the graduating examination - Civil Engineering specialization (ICE) Discipline: FOUNDATIONS 1. Foundations with plain concrete block and reinforced concrete pillow. Design prescriptions. 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 are embedded. Fig. 1. Foundation elements Fig. 2. Contact pressure diagram Dimensioning 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 ( l c, b c, h c ), and also the reinforcement needed in the pillow. Dimensions 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: p 1 = p max p all,, p 2 = p min 0 (1) N where 0 M 0x N0 6e0 x p1,2 1 S Wx L B L In the case when relation (1) is not fulfilled, dimensions L and B of the foundation are modified till relation (1) is verified.
2. Reinforced concrete block foundation. Design prescriptions. Establishing foundation base 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 foundations consist in establishing the dimensions of the foundation block (L, B, H), and also the reinforcement quantity needed in the foundation. Fig. 2 Calculus scheme and foundation Fig. 1. Foundation elements reinforcement Dimensions 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. N G f M T H Verification of the ground pressure: p1,2 2 B L B L 6 p 1 = p max p all,, p 2 = p min 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.
3. Plain concrete continuous foundations under masonry walls. Design prescriptions and dimensioning. 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 and reinforced with minim 614, 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 be taken 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 that the ground pressure will not exceed the allowable pressure of the ground (p tr ). The necessary width of the foundation B is Q G f determined from the condition: p 1,2 pall (1) B 1 The foundation height H, is established fulfilling the minimum design values and the stiffness condition. With known values for B and H is determined the self-weight of the foundation, G f and is verified again the relation (1). In the case when the condition (1) is not fulfilled, the dimension B of the foundation must be increased. Fig. 1. Calculus scheme
4. Shallow foundations under heavy loaded columns. Establishing the foundation base dimensions. 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 exceed 15 16 m 2. In these cases are used shallow foundations with base reinforced concrete plate and reinforced concrete buttress (fig. 1): 1 column, 2 buttresses, 3 base plate, 4 plain concrete layer. Fig. 1. Shallow foundation Dimensions in horizontal plan of the base plate (B and L) are established from the bearing capacity condition: p p ef max all. Fig. 2. Foundation types For static calculus the base plate is considered loaded with the foundation ground reactions and it rests on buttress. The minimum thickness of the base plate is 20 cm. Buttress are disposed in plan so that to ensure the loads overtaking from the column and their transmission to the base plate (fig. 2). The buttress calculus is performed for the bending moment that acts at the middle of the buttress of pier height. On x direction buttress are verified for σ x stresses and those on y direction for σ y stresses(fig. 3). Verification relations: 6 M 6 M x x ( ) 2 R c f y and cd y ( ) 2 R c f. Minimum thickness of the buttress will be cd hc hc taken to be δ (15 20) cm.
5. Continuous beam foundations under columns. Design prescriptions. Reinforcement principles. 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. 2. Reinforcement Fig. 1. Continuous beam foundations under columns: a - rectilinear; b - polyhedral; c circular. The cross-section of the continuous beam foundations is usually an inverted T crosssection, 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 beam is: 1 1. The foundation beam has a longitudinal resistance reinforcement composed H g... l 3 6 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 for the foundation base.
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, these foundations are composed by beams disposed on both directions, usually orthogonal directions, the columns discharging in the intersection points of the perpendicular beams. Fig.1 This 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 foundations under columns, disposed on one direction. The static calculus and the dimensioning are realized decomposing the beam system in two 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.
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 from the construction foundations, being separated by these with impermeable joints. Because of this, such kind of mat doesn t work in bending, respectively doesn t contribute to load transmission from construction to the foundation ground, having only the purpose of creating an impermeable tank, together with the construction basement. Fig. 1. Calculus scheme for a mat foundation The 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 mat stability. Calculus relation for the mat thickness, corresponding to the floating condition is: w h, in which: h r mat minimum necessary thickness, concrete unit weight of the h r concrete w concrete, w unit weight of the water, h w maximum height of the underground water towards the level of the horizontal hydroisolation. Mat foundations are usually realized from plain concrete or poorly reinforced concrete.
8. Precast reinforced concrete piles. Structure. Reinforcement principles. Answer 8: Precast reinforced concrete piles are realized from concrete with the minimum class C 18/22,5 in the case of reinforced concrete piles and minimum C25/30 in the case of prestressed concrete piles. Usually reinforced concrete piles have a rectangular cross-section, with the side of 20...45 cm, and the length between 6 and 25 m. Fig. 2. Reinforced concrete pile (1 - ring; 2 steel bar; 3 - welding; 4 steel tip) The precast piles reinforcement is necessary for overtaking the load appearing during transportation and during the driving processes. Longitudinal reinforcement is composed by 4 or 8 bars with the diameter of 14...22 mm. Transversal reinforcement is composed by steel stirrups with diameter of 6 8 mm. Distances between stirrups are variable on the pile length. For overtaking the great loads during driving, the upper part of the pile is reinforced with 3 wire meshes with diameter of 6 mm placed horizontally at a distance of 5 cm one from another. The pile tip is protected by a steel bar to which are welded longitudinal bars or by a steel tip. For manipulation, in the pile body are placed hooks at distances established by calculus so that the bending moments in supports (hooks) to be equal with those in the field. Fig. 3 Piles static scheme under self-weight
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 in montmorillonite with water. There are two methods for driving the pile shaft using a bentonite slurry: direct circulation system and reverse circulation system. The first system supposes that the bentonite slurry is pumped through the drilling rig and 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 bentonite slurry is reintroduced in the bored pile shaft (fig. 1). The reverse circulation system introduces the clean bentonite slurry into the drilled shaft and it is pumped out together with the excavated soil material. Fig. 1. Direct circulation drilling system Fig. 2. Reverse circulation drilling system For 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.
10. Calculus of the bearing capacity for piles at vertical loads. Calculus principles. Answer 10: Pile foundations are composed by piles and a general mat that links their upper part. Friction piles transmit loads by friction between the lateral surface and the foundation ground. Usually friction piles are used in the case when the resistant foundation ground is found at great depths and the pile doesn t reach it. Function of the load and foundation ground nature from the pile base, the vertical load is transmitted to the ground by friction on the lateral surface and by pressures at the contact between the base and the foundation ground: Fig. 1. Transmission of the R pl Al pv Av Pl Pv vertical loads to piles where: p v is the resistance of the soil at the pile tip, A v is the pile base cross-section area, p l is the average friction resistance on the pile lateral surface, A l is the area of the pile lateral surface. In a more general way, the calculus formula for the bearing capacity of a friction pile subjected to compression can be written as: R = k(m i p v A + U Σm j f i l i ) where: k soil nonhomogeneity coefficient; m i and m j workability coefficients, whose values depend upon the method used to perform the pile and upon the soil nature; p v ground conventional resistance under 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.
V. CASE STUDY / PROBLEMS Discipline: FOUNDATIONS Problem 1 For the retaining wall in the next figure draw the lateral earth pressure diagram and determine 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: γ = 18,0 kn/m 3, Φ = 30 0, c = 0 kn/m 2 ; - The angle of friction between the wall and soil, δ = (1/2 2/3) Φ; - The coefficient of active earth pressure, K a = 0,299. Solution 1: 0 If 1 / 2...2/3 15...20 0 is choosen = 17,5 0 Pressure calculus at B and A level: p γ 0 K 0 B a p A γ H Ka 18 40,299 21, 528 kn/m 2 Active earth pressure resultant: 2 2 H γ H K a γ H 18 4 Pa Sdiagramei_de_presiun i K a 0,299 43,056 kn/m 2 2 2 Position of the active earth pressure resultant: z = H/3 = 4/3 = 1,33 m (from the base level above)
Problem 2 Determine 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; - γ concrete = 24,0 kn/m 3 ; - the supporting soil consists of a compacted sand with the following characteristics: I D = 0,8, p all = 300 kn/m 2, tgα allowable = 1,30. Solution 2: It is established the foundation depth: D f = h freezing + (0,1 0,2) 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: Q G f p pall (1) B 1 where G f = n B H 1 concrete 1,2 B 0,9 1 24 replace G f in relationship (1) and 175 1,2 B 0,9 24 we ll obtain 300 => 175 1,2 B 0,9 24 300 B => B 1 B (300-1,2 0,9 24 ) 175 => B 175 0, 6384 m => is chosen B = 0,65 m 274,08 According to figure, H = D f + 0,1 => H = 0,9 m For H = 0,9 m is verified the stiffness condition: tan H ( B 0,375) / 2 tan alowable 1,30 0,9 6,545 0,1375 tan tan allowable
Problem 3 Determine 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 m and the supporting soil composed by a silty clay (e = 0,8, I C = 0,75), knowing the following data (STAS 3300/2-85): The corrected conventional pressure is determined according to STAS 3300/2-85 with the relationship: p p C C, [kn/m 2 ] conv conv B D in which: p conv - basis conventional pressure ( pconv = 235 kn/m 2 ); C B - width correction; C D - depth correction. The width correction: - for B < 5 m is determined with the relationship: CB pconv K1 B 1, [kn/m 2 ] where K 1 is a coefficient with a value: 0,1 for cohesionless soils except silty sands and 0,05 for silty soils and cohesive soils. - for B 5 m the width correction is: C 0, 4 for cohesionless soils, except silty sands; B p conv CB 0, 2 p conv for silty sands and cohesive soils. The depth correction is determined using the relationships: - for D f < 2 m: D f 2 C D pconv 4 - for D f > 2 m: CD K2 D f 2 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: pconv pconv CB CD For B = 2,30 m (meaning B < 5 m) the width correction is determined with the C p K B where K 1 = 0,05 for cohesive soils. relationship: B conv 1 1 C p K B B = 325 0,05 (2,30-1) = 21,125 kn/m 2 conv 1 1 For D f = 1,80 m (D f < 2 m) the depth correction is determined with the relationship: C D = p D 2 conv f 1,80 2 = 325 = - 16,25 kn/m 2 4 4 The calculus conventional pressure is: p p C C = 325 + 21,125 16,25 = 329,875 kn/m 2 conv conv B D
Problem 4 For 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 γ concrete = 24 kn/m 3. - Lateral active earth pressure behind the retaining wall (P a = 43 kn); - Friction angle between the wall and soil, δ = 17,5 0. Answer 4: Position of the point of application of the lateral active earth pressure: z = H/3 = 4/3 = 1,33 m G 1 =b (H-a) 1 24 = 0,8 3 1 24 = 57,6 kn G 2 = (B-b-c) (H-a) 24 / 2 = (2-0,8-0,60) (4-1) 24 / 2 = 21,6 kn G 3 =B a 1 24= 2 1 1 24 = 48 kn d 1 =B-b/2= 2-0,8/2 =1,6m d 2 =B-b-(B-b-c)/3= 2-0,8-(2-0,8-0,6)/3= 1 m d 3 = B/2 = 2/2 = 1 m Stability check to overturning: M r 0,8 M s M r = P a cos δ z = 43 0,9537 1,33 = 54,54 knm M s = G 1 d 1 + G 2 d 2 + G 3 d 3 + P a sin δ B = 57,6 1,6 + 21,6 1 + 48 1 + 43 0,300 2 =153,06 knm Check: 54,54 knm 122,45 knm
Problem 5 Determine the plain concrete block dimensions for a plain concrete block and reinforced concrete pillow foundation knowing: loads (N = 1150 kn; M = 50 knm; T = 6 kn); 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); γ med = 20 kn/m 3 and tanα a = 1,3. Solution 5: Predimensioning: L B = 1,2 N/p all where L = 1,2 B => 1,2 B 2 1,2 1150 = 1,2 N/p all => B = = 1,2 300 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: tanα tanα a tanα = H l 1 H 0, 7 1,3 => H = 0,95 m Check of the foundation dimensions: N G f M T(H h c ) p1,2 where G 2 f = B L (H + h c ) γ med =2,0 2,4 (0,95 + 0,3) 20 B L B L /6 = 120 kn 2 1150 120 50 6 1,25 294,52 kn/m p 1,2 2 2 2,0 2,4 2,0 2,4 / 6 234,64 kn/m p 1 = p max = 294,52 kn/m 2 300 kn/m 2 and p 2 = p min = 234,64 kn/m 2 0