Design of Shallow Footing

A Seminar report On DESIGN OF SHALLOW FOOTING In partial fulfilment of requirements for the degree of Bachelor of Engineering In Civil Engineering

SUBMITTED BY: Deepak Kumar Roll no- 001110401089

Under the guidance of Prof. Dr. Sreyashi pal

DEPARTMENT OF CIVIL ENGINEERING Jadavpur university Kolkata -700032 (2014)

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DESIGN OF SHALLOW FOOTING

Introduction

According to Terzaghi, Terzaghi, a foundation is shallow if its depth is equal to or less than its width. Shallow foundations are used when the soil has sufficient strength within a short depth below the ground level. They need sufficient plan area to transfer the heavy loads to the base soil. A shallow foundation is a type t ype of foundation which transfers building loads to the earth very near the surface, rather than to a subsurface layer or a range of depths as does a deep foundation. Types of shallow footing 

isolated footing



strap footing



strip or continuous footing



combined footing



mat or raft footing

isolated footing:- It is circular, square or rectangular slab of uniform thickness to support an

individual column. Sometimes, it is stepped or haunched to spread the load over a larger area.

Strap footing:- It consists of two isolated footings connected with a structural strap or a lever.

The strap connects the footing such that they behave as one unit. The strap simply acts as a connecting beam. A strap footing is more economical than a combined footing when the allowable soil pressure is relatively high and distance between the columns is i s large.

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Strip or continuous footing:- A strip footing is another type of spread footing which is

 provided for a load bearing wall. A strip footing can also be provided for a row of columns which are so closely spaced that their spread footings overlap or nearly touch each other.

Combined footing:- When the spacing of the adjacent columns is so close that separate

isolated footings are not possible due to the overlapping areas of the footings or inadequate clear space between the two areas of the footings, combined footings are the solution combining two or more columns. Combined footing normally means a footing combining two columns. A combine footing may be rectangular or trapezoidal in plan. Trapezoidal footing is  provided when the load on one of the columns is larger than the other column.

Mat or raft footing:- It is a large slab supporting a number of columns and walls under entire

structure or a large part of the structure. A mat is required when the allowable soil pressure is low or where the columns and walls are so close that individual footings would overlap or nearly touch each other. Mat foundations are useful in reducing the differential settlements on non-homogeneous soils or where there is large variation in the loads on individual columns.

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How design of footing is different from others

It is worth mentioning that the design of foundation structures is somewhat different from the design of other elements of superstructure due to the reasons given below. Therefore, foundation structures need special attention of the designers. 1. Foundation structures undergo soil-structure interaction. Therefore, the behaviour of foundation structures depends on the properties of structural materials and soil. 2. Accurate estimations of all types of loads, moments and forces are needed for the present as well as for future expansion, if applicable. It is very important as the foundation structure, once completed, is difficult to strengthen in future. 3. Foundation structures, though remain underground involving very little architectural aesthetics, have to be housed within the property line which may cause additional forces and moments due to the eccentricity of foundation. 4. Foundation structures are in direct contact with the soil and may be affected due to harmful chemicals and minerals present in the soil and fluctuations of water table when it is very near to the foundation. Moreover, periodic inspection and maintenance are practically impossible for the foundation structures. Safe bearing capacity of soil (q c):- The safe bearing capacity q c of soil is the permissible soil

 pressure considering safety factors in the range of 2 to 6 depending on the type of soil. This is applicable under service load condition and, therefore, the partial safety factors f  for different load combinations are to be taken from those under limit state of serviceability.  Normally, the acceptable value of qc is supplied by the geotechnical consultant to the structural

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engineer after proper soil investigations. Gross and net bearing capacities are the two terms used in the design. Gross bearing capacity is the total safe bearing pressure just below the footing due to the load of the superstructure, self-weight of the footing and the weight of earth lying over the footing. On the other hand, net bearing capacity is the net pressure in excess of the existing overburden  pressure. Thus, we can write  Net bearing capacity = Gross bearing capacity - Pressure due to overburden soil The weight of the foundation and backfill may be taken as 10 to 15 per cent of the total axial load on the footing. Design considerations

(a) Minimum nominal cover (cl. 26.4.2.2 of IS 456) :The minimum nominal cover for the footings should be more than that of other structural elements of the superstructure as the footings are in direct contact with the soil. Clause 26.4.2.2 of IS 456 prescribes a minimum cover of 50 mm for footings. However, the actual cover may  be even more depending on the presence of harmful chemicals or minerals, water table etc. (b) Thickness at the edge of footings (cl. 34.1.2 and 34.1.3 of IS 456-2000):The minimum thickness at the edge of reinforced and plain concrete footings shall be at least 150 mm for footings on soils and at least 300 mm above the top of piles for footings on piles, as per the stipulation in cl.34.1.2 of IS 456-2000. For plain concrete pedestals, the angle α (fig.6) between the plane passing through the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and the horizontal plane shall be determined from the following expression (cl.34.1.3 of IS 456-2000). tanα ≤ 0.9{(100 qa/f ck ) + 1}1/2

Where qa = calculated maximum bearing pressure at the base of pedestal in N/mm 2, and f ck  = characteristic strength of concrete at 28 days in N/mm 2.

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(c) Bending moments (cl. 34.2 of IS 456-2000):The critical section of maximum bending moment for the purpose of designing an isolated concrete footing which supports a column, pedestal or wall shall be at the face of the column,  pedestal or wall for footing supporting a concrete column, pedestal or reinforced concrete wall. (d) Shear force (cl. 31.6 and 34.2.4 of IS 456-2000):Footing slabs shall be checked in one-way or two-way shears depending on the nature of  bending. If the slab bends primarily in one-way, the footing slab shall be checked in one-way vertical shear. On the other hand, when the bending is primarily two-way, the footing slab shall  be checked in two-way shear or punching shear. 1. One-way shear (cl. 34.2.4 of IS 456-2000):One-way shear has to be checked across the full width of the base slab on a vertical section located from the face of the column, pedestal or wall at a distance equal to effective depth of the footing slab . The design shear strength of concrete without shear reinforcement is given in Table 19 of cl.40.2 of IS 456-2000. 2. Two-way or punching shear (cls.31.6 and 34.2.4):Two-way or punching shear shall be checked around the column on a perimeter half the effective depth of the footing slab away from the face of the column or pedestal. The permissible shear stress, when shear reinforcement is not provided, shall not exceed k sτc , where k s = (0.5 + βc), but not greater than one, cβ being the ratio of short side to long side of the column, and τ c = 0.25(f ck ) 1/2 in limit state method of design, as stipulated in cl.31.6.3 of IS 456-2000.  Normally, the thickness of the base slab is governed by shear. Hence, the necessary thickness of the slab has to be provided to avoid shear reinforcement.

(e) Tensile reinforcement (cl.34.3 of IS 456-2000):The distribution of the total tensile reinforcement, calculated in accordance with the moment at critical sections shall be done as given below (i) In two-way reinforced square footing slabs, the reinforcement extending in each direction

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shall be distributed uniformly across the full width/length of the footing. (ii) In two-way reinforced rectangular footing slabs, the reinforcement in the long direction shall be distributed uniformly across the full width of the footing slab. In the short direction, a central band equal to the width of the footing shall be marked along the length of the footing, where the portion of the reinforcement shall be determined as given in the equation below. This  portion of the reinforcement shall be distributed across the central band.

Reinforcement in the central band = {2/( β+1)} (Total reinforcement in the short direction). Where β is the ratio of  longer dimension to shorter dimension of the footing slab (Fig.8).

Each of the two end bands shall be provided with half of the remaining reinforcement, distributed uniformly across the respective end band. (f) Transfer of load at the base of column (cl.34.4 of IS 456-2000):All forces and moments acting at the base of the column must be transferred to the pedestal, if any, and then from the base of the pedestal to the footing, (or directly from the base of the column to the footing if there is no pedestal) by compression in concrete and steel and tension in steel. Compression forces are transferred through direct bearing while tension forces are transferred through developed reinforcement. The permissible bearing stresses on full area of concrete shall be taken as given below from cl.34.4 of IS 456-2000: σ br  = 0.45f ck   (A1/A2) 1/2

with a condition that

(A1/A2)1/2 ≤ 2.0

where A1 = maximum supporting area of footing for bearing. A2 = loaded area at the base of the column. (g) Nominal reinforcement (cl. 34.5 of IS 456-2000):1. Clause 34.5.1 of IS 456 stipulates the minimum reinforcement and spacing of the bars in footing slabs as per the requirements of solid slab (cls.26.5.2.1 and 26.3.3b(2) of IS 456-2000, respectively).

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2. The nominal reinforcement for concrete sections of thickness greater than 1 m shall be 360 mm2 per metre length in each direction on each face, as stipulated in cl.34.5.2 of IS 456-2000. The clause further specifies that this provision does not supersede the requirement of minimum tensile reinforcement based on the depth of section. Distribution of base pressure

The foundation, assumed to act as a rigid body, is in equilibrium under the action of applied forces and moments from the superstructure and the reactions from the stresses in the soil. The distribution of base pressure is different for different types of soil. Typical distributions of  pressure, for actual foundations, in sandy and clayey soils are shown in Fig.9 and 10.

However, for the sake of simplicity the footing is assumed to be a perfectly rigid body, the soil is assumed to behave elastically and the distributions of stress and stain are linear in the soil  just below the base of the foundation, as shown in Fig.11. In the following, the distributions of base pressure are explained for (i) concentrically loaded footings, (ii) eccentrically loaded footings and (iii) unsymmetrical (about both the axes) footings. (i) Concentrically loaded footings:Figure 7 shows rectangular footing symmetrically loaded with service load P1 from the superstructure and P2 from the backfill including the weight of the footing. The assumed uniformly distributed soil pressure at the base of magnitude q is obtained from: q = (P1 + P2)/ A ……………. (1) Where ‘A’ is the area of the base of the footing. In the design problem, however, A is to be determined from the condition that the actual gross intensity of soil pressure does not exceed q c, the bearing capacity of the soil, a known given data. Thus from equation (1) A = (P1 + P2)/q c

………………(2)

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From the known value of A, the dimensions B and L are determined such that the maximum  bending moment in each of the two adjacent projections is equal, i.e., the ratio of the dimensions B and L of the footing shall be in the same order of the ratio of width b and depth D of the column. (ii) Eccentrically loaded footings:-

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In most of the practical situations, a column transfers axial load P and moment M to the footing, which can be represented as eccentrically loaded footing when a load P is subjected to an eccentricity e = M/P. This eccentricity may also be there, either alone or in combined mode, when • the column transfers a vertical load at a distance of e from the centroidal axis of the footing. • the column or the p edestal transfers a lateral load above the level of foundation, in addition to

vertical loads. Accordingly, the distribution of pressure may be of any one of the three types, depending on the magnitude of the eccentricity of the load, as shown in Figs. 13(b) to (d). The general expression of q a, the intensity of soil pressure at a distance of y from the origin is: qa = P/A± (Pe/Ix)y

……………. (3)

we would consider a rectangular footing symmetric to the column. Substituting the values of A = BL, Ix = BL3/12 and y = L/2, we get the values of q a at the left edge. qa at the left edge = (P/BL) {1 - (6e/L)}

……………..  (4)

It is evident from upper equation (4), that the three cases are possible: (A) when e < L/6, q a at the left edge is compression (+), (B) when e = L/6, q a at the left edge is zero and (C) when e > L/6, q a at the left edge is tension (-). The three cases are shown in Figs. 13(b) to (d), respectively. It is to be noted that similar three cases are also possible when eccentricity of the load is negative resulting the values of qa at the right edge as compression, zero or tension. Evidently, these soil reactions, in compression and tension, should be permissible and attainable. Case (A): when | e | ≤ L /6 :-

Figures 8(b) and (c) show these two cases, when |e| < L/6 or |e| = L/6, respectively. It is seen that the entire area of the footing is in compression having minimum and maximum values of q at the two edges with a linear and non-uniform variation. The values of q are obtained from equation (3). In the limiting case i.e., when |e| = L/6, the value of q a is zero at one edge and the other edge is having qa = 2P/BL (compression) with a linear variation. Similarly, when e = 0, the footing is subjected to uniform constant pressure of P/BL. Thus, when |e| = L/6, the maximum pressure under one edge of the footing is twice of the uniform pressure when e = 0. In a more general case, as in the case of footing for the corner column of a building, the load may have biaxial eccentricities. The general expression of qa at a location of (x,y) of the footing, when the load is having biaxial eccentricities of e x and ey is,

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qa = P/A ± P exy/Ix ± P eyx/Iy …………… (5) Similarly, it can be shown that the rectangular footing of width B and length L will have no tension when the two eccentricities are such that, 6ex/L + 6ey/B ≤ 1 Case (B): when | e | > L/6 :-

The eccentricity of the load more than L/6 results in development of tensile stresses in part of the soil. Stability, in such case, is ensured by either anchoring or weight of overburden  preventing uplift. However, it is to ensure that maximum compressive pressure on the other face is within the limit and sufficient factor of safety is available against over-turning. Accordingly, the maximum pressure in such a case can be determined considering the soil under compression part only. Further, assuming the line of action of the eccentric load coincides with that of resultant soil pressure (Fig. 14) we have: qmax = P/L'B + 12P(0.5 C)(1.5 C)/BL' = 2P/L'B …………. (6) Where L' = 3C

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Design of isolated footing wit h axi al l oad only 

Problem 1: Design an isolated footing for a square column, 400 mm x 400 mm with 12-20 mm diameter longitudinal bars carrying service loads of 1500 kN with M 20 and Fe 415. The 2

safe bearing capacity of soil is 250 kN/m  at a depth of 1 m below the ground level. Use M 20 and Fe 415. Solution 1:

Step 1: Size of the footing:Given P = 1500 kN, q c = 250 kN/m2 Assuming the weight of the footing and backfill as 10 per cent of the load. So, Pt = 15001.1 = 1650 KN  base area required (A) = P t/qc = 1650/250 = 6.6 m2 As it is square column so provide square footing having length(L) = breadth(B). So L = B =  = 2.569 m. So provide L = B = 2.6 m shown in fig. 15

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Step 2: Thickness of footing slab based on one-way shear (section 1-1 ):Factored soil pressure(q cf ) = 1500(1.5)/(2.6)(2.6) = 0.3328 N/mm 2, say, 0.333 N/mm2. 2

Assuming p = 0.25% in the footing slab, for M 20 concrete τ c = 0.36 N/mm  (Table 19 of IS

456-2000). That is allowable shear stress ( τc) = 0.36 N/mm 2.  Now design shear stress ( v) =

 

 Now τv  τc So, 0.36(2600)d ≥ 0.333(2600) (1100 –  d) So, d ≥ 528.57mm Provide d = 536 mm. The total depth becomes 536 + 50 + 16 + 8 (with 50 mm cover and diameter of reinforcing bars = 16 mm) = 610 mm. Step 3: Checking for two-way shear (section 2-2) :The critical section is at a distance of d/2 from the periphery of the column. The factored shear force = 0.333{(2600)2  –  (400 + d)2}(10)-3 = 1959.34 kN. Allowable punching or two way shear = k sτc = k s(0.25) (f ck )1/2 where k s = 0.5 + βc (cl. 31.6.3 of IS 456-2000 ). Here βc = 1.0, k s = 1.5 >/ 1; so k s = 1.0. This gives shear strength of concrete = 0.25 (f ck )1/2 = 1.118 N/mm 2 As punching surface = 4 (936)(536) = 2006784 mm 2. So allowable shear force = 1.118 2006784 = 2243.58 kN > 1959.34 kN. Hence, O.K Thus, the depth of the footing is governed by one-way shear. Step 4: Bending moment (section 3-3) :The critical section (Fig.15b, sec.3.3), is at the face of the column. So bending moment about the face of column (M u) = 0.333(2600)(1100)(550) = 523.809 kNm Moment of resistance of the footing = Rbd 2 where R = 2.76 Moment of resistance = 2.76(2600)(536)(536) = 2061.636 kNm > 523.809 kNm.  Now, Mu/bd2 = 523.809(106)/(2600)(536)(536) = 0.7013 N/mm 2. So from Table 2 of SP-16 p = 0.2034. Ast = 0.2034(2600)(536)/100 = 2834.58 mm 2. However, one-way shear has been checked assuming p = 0.25%. So, use p = 0.25%. Accordingly, Ast = 0.0025(2600)(536) = 3484 mm 2. Provide 18 bars of 16 mm diameter (= 3619 mm 2) both ways. The spacing of bars = {2600  –  2(50) –  16}/17 = 146.117 mm. The spacing is 140 mm c/c (Fig.15).

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The bending moment in the other direction is also the same as it is a square footing. The effective depth, however, is 16 mm more than 536 mm. But, the area of steel is not needed to  be determined with d = 552 mm as we are providing 0.25 per cent reinforcement based on oneway shear checking. Step 5: Development length: Ld = f s υ/4(τ bd) = 0.87(415)(16)/4(1.6)(1.2) = 47(16) = 752 mm (cl.26.2.1 of IS 456-2000). Length available = 1100  –  50 = 1050 mm > 752 mm. Step 6: Transfer of force at the base of the column: Pu = 1500(1.5) = 2250 kN Compressive bearing resistance = 0.45 f ck (A1/A2)1/2. For the column face A1/A2 = 1 and for the other face A1/A2 > 2 but should be taken as 2. In any case, the column face governs. Force transferred to the base through column at the interface = 0.45(20)(400)(400) = 1440 kN < 2250 kN. The balance force 2250  –  1440 = 810 KN has to be transferred by the longitudinal reinforcements, dowels or mechanical connectors. As it is convenient, we propose to continue the longitudinal bars (12-20 mm diameter) into the footing. Design of i solated footin g with axial l oad and moment 

Problem 2: Design one isolated footing for a column 300 mm x 450 mm with 20 no of 20 mm dia, having Pu = 1620 kN and Mu = 170 kNm using. Assume that the moment is reversible. 2

The safe bearing capacity of the soil is 200 kN/m  at a depth of 1 metre from ground level. Use M 25 and Fe 415 for the footing. Solution 2:

Step 1: Size of the footing: Given Pu = 1620 kN and M u = 170 kNm. The footing should be symmetric with respect to the column as the moment is reversible. Assuming the weights of footing and backfill as 15 per cent of Pu. The eccentricity of load P u at the base is e = M u/P(1.15) = 170(106)/1620(1.15)(103) = 91.25 mm. This eccentricity may be taken as < L/6 of the footing. The factored bearing pressure is 200(1.5) = 300 kN/m 2. For the footing of length L and width B, we, therefore, have: Pu/BL + 6M/BL2 300

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or, 1620(1.15)/BL + 6(170)/BL 2 ≤ 300 or, BL2 –  6.21L – 3.4 ≤ 0 For the economic proportion, let us keep equal projection beyond the face of the column in the two directions. This gives (L –  0.45)/2 = (B  –  0.3)/2 ……………… (1) or, B = L –  0.15 ……………. (2) Using Eq.(2) in Eq.(1), we have (L –  0.15) L2 –  6.212 – 3.4 ≤ 0 or L3 –  0.15 L2 –  6.21 L – 3.4 ≤ 0 We have L = 2.8 m and B = 2.65 m. Let us provide L = 2.85 m and B = 2.70 m (Fig.16b). We get the maximum and minimum pressures as, 1620(1.15)/(2.85)(2.70) ±170(6)/(2.7)(2.85)(2.85) = 242.105 ± 46.51 = 288.615 kN/m 2 and 195.595 kN/m2, respectively (Fig.16c). Both the values are less than 300 kN/m 2.

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Step 2: Thickness of footing slab based on one-way shear: The critical section for one way shear (sec.1-1 of Figs.16 a and b) is at a distance d from the face of the column. The average soil pressure at sec.1-1 is {288.615 –  (288.615 –   195.595)(1200  –  d)/2850} = 249.449 + 0.0326d. The one-way shear force at sec.1-1 = (2.7)(1.2  –   0.001d((249.449 + 0.0326d) kN. Assuming 0.15 per cent reinforcement in the footing slab, the shear strength of M 25 concrete = 0.29 N/mm2. Hence, the shear strength of the section = 2700(d)(0.29)(103) kN. From the condition that shear strength has to be ≥ shear force, we have 2700(d)(0.29)(10-3) = (2.7)(1.2  –   0.001d)(249.449 + 0.0326d) This gives, d2 + 15347.51534d  –  9182171.779 = 0 Solving, we get d = 576.6198. Let us assume d = 600 mm Step 3: Checking for two-way shear: The critical section for two way shear (sec.2222 of Figs.16a and b) is at a distance d/2 from the face of column. The shear resistance is obtained cl.31.6.31 of IS 456-2000, which gives 1/2 τc = (0.5 + 450/300)(0.25)(25)  but the multiplying factor (0.5 + 450/300) 1.0. 1/2

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So, we have τ c = 0.25(25)  = 1.25 N/mm .

Hence, the shear resistance = (1.25)(2){(300 + 600) + (450 + 600)}(600) = 2925 kN. Actual shear force is determined on the basis of average soil pressure at the centre line of the cross-section which is (195.595 + 288.615)/2 = 242.105 kN/m 2 (Fig.16c). So, the actual shear force = V u = (242.105){(2.7)(2.85)  –  (0.3 + 0.6)(0.45 + 0.6)} = 1634.209 kN < shear resistance (= 2925 kN). Hence, the depth of the footing is governed by one-way shear. With effective depth = 600 mm, the total depth of footing = 600 + 50 (cover) + 16 (bar dia) + 8 (half bar dia) = 674 mm. Step 4: Bending moment: (i) in the long direction (along the length = 2850 mm) Bending moment at the face of column (sec.33 of Figs.16a and b) is determined where the soil  pressure = 288.615 –  (288.615 –   195.595)(1200)/2850 = 249.45 kN/m 2. So, the bending moment Mu = 249.45(2.7)(1.2)(0.6) + (288.615  –   249.45)(2.7)(1.2)(2)/(2)(3) = 527.23 kNm.

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Mu/Bd2 = 527.23(106)/(2700)(616)(616) = 0.515 N/mm 2 < 3.45 N/mm 2 for M 25 concrete. Table 3 of SP-16 gives p = 0.1462 < 0.15 per cent as required for one-way shear. So take p = 0.15 percent Thus, Ast = 0.15(2700)(616)/100 = 2494.8 mm 2. Provide 13 bars of 16 mm diameter (area = 2613 mm 2), spacing = (2700 –  100 –  16)/12 = 215.33 mm, say 210 mm c/c. (ii) In the short direction (B = 2700 mm) The average pressure on soil between the edge and centre of the footing = (288.615 + 242.105)/2 = 265.36 kN/m 2. The bending moment is determined with this pressure as an approximation. Bending moment Mu = (265.36)(1.2)(0.6)(2.85) kNm = 544.519 kNm Mu/Ld2 = 544.519(106)/(2850)(600)(600) = 0.531 Table 3 of SP-16 gives p = 0.15068, which gives area of steel = 0.15068(2850)(600)/100 = 2576.628 mm2. Provide 13 bars of 16 mm diameter (area = 2613 mm 2) @ 210 mm c/c; i.e. the same arrangement in both directions. Step 6: Development length: Development length of 16 mm diameter bars (M 25 concrete) = 0.87(415)(16)/4(1.6)(1.4) = 644.73 mm. Length available = 1200  –  50 –  8 = 1142 mm > 644.73 mm. Hence, o.k. Desi gn of combin ed footin g 

Problem 3: Design a combined footing for two columns C1, 400 mm x 400 mm with 8 bars of 16 mm diameter carrying a service load of 800 kN and C2, 300 mm x 500 mm with 8 bars of 20 mm diameter carrying a service load of 1200 kN. The column C1 is flushed with the property line. The columns are at 3.0 m c/c distance. The safe bearing capacity of soil is 2

200 kN/m  at a depth of 1.5 m below the ground level. Use M 20 and Fe 415 for columns and footing. Solution 3:

Step 1: Size of the footing: Assuming the weight of combined footing and backfill as 15 per cent of the total loads of the columns, we have the required base area, considering q c = 200 kN/m2, Area of the base = (800 + 1200)(1.15)/200 = 11.5 m 2. It is necessary that the resultant of the loads of two columns and the centroid of the footing

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coincide so that a uniform distribution of soil pressure is obtained. Thus, the distance of the centroid of the footing y from column C1 (Fig.17 b) is:

 ¯ y = 800(0) + 1200(3)/2000 = 1.8 m (Fig.11.29.5b). Since  ¯ y is greater than half the c/c distance of column, a rectangular footing has to be designed. Let us provide 4 m x 3 m and the dimensions are shown in Fig.17b coinciding the centroid of the footing and the resultant line of action of the two loads, i.e. at a distance of 2 m from the left edge. Step 2: Thickness of footing slab based on one-way shear: Considering the footing as a wide beam of B = 3 m in the longitudinal direction, the uniformly

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distributed factored load = (800 + 1200)(1.5)/4 = 750 kN/m. Figures 18a, b and c present the column loads, soil pressure, shear force and bending moment diagrams.

The critical section of one-way shear is sec.11 (at point K) of Figs.17a and 18a, at a distance of d + 250 mm from G (the location of column C2). The one-way shear force is, Shear force = (1600  –  d –  250)1200/1600 = (1012.5  –  0.75d) kN Assuming p = 0.15 per cent reinforcement in the footing slab, the shear strength of M 20 concrete = 0.28 N/mm2. Hence, the shear strength of section 11 = (3000)d(0.28)(10-3) kN. From the condition that shear strength  shear force, we have (3000)d(0.28)(10-3)  1012.5 –  0.75d, which gives d  636.79 mm. So provide d = 650 mm and the total depth = 650 + 50 + 16 + 8 = 724 mm (assuming cover = 50 mm and the diameter of bars = 16 mm). Step 3: Checking for two-way shear: (i) Around column C2:

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The effective depth along 4.0 m is 650 + 16 = 666 mm. The critical section for the two-way shear around column C2 is at a distance of 666/2 = 333 mm from the face of the column and marked by 2222 line in Fig.17b. The two-way punching shear force, considering the soil pressure = 750/3 = 250 kN/m 2, is Vu = 1800 –   (1.166)(0.966)(250) = 1518.411 kN As per cl.31.6.3.1 of IS 456-2000, here k s = 0.5 + (500/300) but 1.0; so, k s = 1.0. Therefore, shear strength of concrete = 0.25(20)1/2 (2){(300 + 666) + (500 + 666)}(666) = 3174.92 kN > 1518.411 kN. Hence, o.k. (ii) Around column C1: The effective depth of footing is 666 mm. The critical section is marked by 3333 in Fig.17b. The two-way punching shear = 1200  –   (1.066)(0.733)(250) = 1004.65 kN. The resistance to two-way shear = 0.25(20)1/2 (2){(1066 + 733)}(666) = 2679.1 kN > 1004.65 kN. Hence, o.k. Thus, the depth of the footing is governed by one-way shear. Step 4: Bending moments (longitudinal direction): (i) Maximum positive moment: Figure 18c shows the maximum positive bending moment = 720 kNm at a distance of 1.4 m from the column C1 (at point J). With effective depth d = 666 mm, we have M/Bd2 = 720(106)/(3000)(666)(666) = 0.541 N/mm 2 Table 2 of SP-16 gives p = 0.1553 per cent Ast = 0.1553(3000)(666)/100 = 3102.894 mm 2 Provide 16 bars of 16 mm diameter (area = 3217 mm2), spacing = (3000  –  50 –  16)/15 = 195.6 mm c/c, say 190 mm c/c. Development length of 16 mm bars = 47.01(16) = 752.16 mm Length available = 1600  –  50 –  16 = 1534 mm > 752.16 mm Hence, o.k. (ii) Maximum negative moment: Figure 18c shows the maximum negative moment = 240 kNm at a distance of 800 mm from the right edge with the effective depth = 666 mm, we have M/Bd2 = 240(106)/(3000)(666)(666) = 0.018 N/mm 2 It is very nominal. So, provide 0.15 per cent steel, which gives Ast = 0.15(3000)(666)/100 = 2997 mm 2. Provide 27 bars of 12 mm (area = 3053 mm 2) at spacing = (3000  –  50 –  12)/26 = 113 mm c/c;

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say 110 mm c/c. Development length = 47.01(12) = 564 mm Length available = 800 –  50 –  12 = 738 mm > 564 mm Hence, o.k. Step 5: Design of column strip as transverse beam:

(i) Transverse beam under column C1: The width of the transverse beam is 0.75d from the face of column C1. The effective depth is 666  –  6 –  8 = 652 mm, as the effective depth in the longitudinal direction = 666 mm, bottom bar diameter in longitudinal direction = 12 mm and assuming the bar diameter in the transverse direction as 16 mm. We have to check the depth and reinforcement in the transverse direction considering one-way shear and bending moment. (A) One-way shear: The factored load for this transverse strip = 1200/3 = 400 kN/m. The section of the one-way shear in sec.44 (Fig.20) at a distance of d = 652 mm from the face of column C1. The width of the transverse strip = 400 + 0.75(652) = 889 mm.

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One-way shear force in sec.44 = (1500  –  652 –  200)(400)(10-3) = 259.2 kN Shear stress developed = 259.2(103)/(889)(652) = 0.447 N/mm 2 2

To have the shear strength of concrete ≥0.447 N/mm , the percentage of reinforcement is

determined by linear interpolation from Table 19 of IS 456-2000, so that the depth of footing may remain unchanged. Table 19 of IS 456-2000 gives p = 0.43125. Accordingly, A st = 0.43125(889)(652)/100 = 2499.65 mm2. Provide 13 bars of 16 mm (area = 2613 mm2), spacing = 889/12 = 74.08 mm c/c, say 70 mm c/c. However, this area of steel shall be checked for bending moment consideration also. (B) Bending moment at the face of column C1 in the transverse strip under column: Bending moment = (1.3)(1.3)(400)/2 = 338 kNm. We, therefore, have M/(width of strip)d2 = 338(106)/(889)(652)(652) = 0.89 N/mm 2 Table 2 of SP-16 gives p = 0.2608 < 0.43125. Hence, the area of steel as determine for one-way shear consideration is to be provided. Provide 13 bars of 16 mm @ 70 mm c/c in the column strip of width 889 mm under the column C1. Development length of 16 mm bars = 47.01(16) = 752.16 mm Length available = 1300  –  50 –  16 = 1234 mm > 752.16 mm Hence, o.k. (ii) Transverse beam under column C2: Figure 20 shows the strip of width = 500 + 0.75d + 0.75 d = 500 + 1.5(652) = 1478 mm, considering the effective depth of footing = 652 mm. (A) One-way shear: The factored load for this transverse strip = 1800/3 = 600 kN/m. The one-way shear section is marked by sec.5.5 in Fig.20 at a distance of d = 652 mm from the face of the column C2. One-way shear in sec.55 (of width = 1478 mm) = (1500  –  652 –  150)(600)(10-3) = 418.8 kN The shear stress developed = 418.8(103)/(1478)(652) = 0.434 N/mm 2 The corresponding percentage of area of steel, as obtained from Table 19 of IS 456 is, p = 0.404 per cent. Accordingly, Ast = 0.404(1478)(652)/100 = 3893.17 mm 2. Provide 20 bars of 16 mm (area = 4021 mm 2), spacing = 1478/19 = 77.78 mm c/c, say 75 mm c/c. However, this area of steel shall be checked for bending moment consideration also. (B) Bending moment at the face of column C2 in the transverse strip under column C2: The bending moment = (1.35)(1.35)(600)/2 = 546.75 kNm. We, therefore, have M/(width of strip)d2 = 546.75(106)/(1478)(652)(652) = 0.87 N/mm 2

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Table 2 of SP-16 gives: p = 0.2544 < 0.404 per cent as required for one-way shear. So, provide 20 bars of 16 mm diameter @ 75 mm c/c in the column strip of width 1478 under column C2. Development length as calculated in Step (B) of col strip C1 = 752.16 mm The length available = 1350  –  50 –  16 = 1284 mm > 752.16 mm Hence, o.k. Step 6: Distribution reinforcement:  Nominal distribution reinforcement shall be provided at top and bottom where the main reinforcement bars are not provided. The amount @ 0.12 per cent comes to 0.12(1000)(652)/100 = 782.4 mm 2/metre. Provide 12 mm diameter bars @ 140 mm c/c (area = 808 mm 2/m). All the reinforcing bars are shown in Figs.17 a and b. Design of sloped footin g with axi al l oad 

Problem 4: Design a sloped footing for a square column of 400 mm x 400 mm with 16 longitudinal bars of 16 mm diameter carrying a service load of 1400 kN. Use M 20 and Fe 415 both for 2

column and footing slab. The safe bearing capacity of soil is 150 kN/m . Solution 4:

Step 1: Size of the footing: Given P = 1400 kN and q c = 150 kN/m 2. Assuming the weight of the footing and the back file as 10 per cent of the load, the required  base area is: 1400(1.1)/150 = 10.27 m 2. Provide 3400 x 3400 mm giving 11.56 m 2. Step 2: Thickness of footing slab based on one-way shear: Factored bearing pressure = 1400(1.5)/(3.4)(3.4) = 181.66 kN/m 2 = 0.18166 N/mm2. Assuming 0.15 per cent reinforcement in the footing slab, Table 19 of IS 456-2000 gives τc for M 20 = 0.28 N/mm 2. From the condition that the one-way shear resistance ≥ one-way shear force, we have at a distance d from the face of the column (sec.1-1of Figs.21 a and b). 0.28(3400)d ≥ 0.18166(1500 –  d)(3400) or d ≥ 590.24 mm.

Provide total depth of footing as 670 mm, so that the effective depth = 670  –  50 –  16 –  8 = 596 mm.

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Step 3: Checking for two-way shear: At the critical section 2222 (Figs.21a and b), the shear resistance is, 4(400 + 596)(596)(0.25)(f ck )1/2 = 2654.73 kN. The shear force = {(3.4)(3.4)  –  (0.996)(0.996)}0.18166 = 1919.78 kN < 2654.73 kN. Hence, o.k. Step 4: Bending moment: We have to determine the area of steel in one direction as it is a square footing. So, we consider the lower effective depth which is 596 mm. The critical section is sec.33 (Figs.21a and b), where we have Mu = 3400(1500)(0.18166)(1500)/2 = 694.8495 kNm Mu/bd2 = 694.8495(106)/(3400)(596)(596) = 0.575 N/mm 2 Table 2 of SP-16 gives, p = 0.165%. Accordingly, area of steel = 0.165(3400)(596)/100 = 3343.56 mm 2. Provide 30 bars of 12 mm diameter (= 3393 mm 2), both ways.

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Step 5: Providing slope in the footing slab: Since the three critical sections (.i.e., of bending moment, two-way shear and one-way shear) are within a distance of 596 mm from the face of the column, the full depth of the footing slab is provided up to a distance of 700 mm from the face of the column. However, by providing slope the available section now is a truncated rectangle giving some less area for the one-way shear. Accordingly, the depth of the footing is increased from 670 mm to 750 mm. With a cover of 50 mm and bar diameter of 12 mm in both directions, the revised effective depth = 750  –  50 –  12 –  6 = 682 mm. Providing the minimum depth of 350 mm at the edge, as shown in

Figs.21 a and b, we check the one-way shear again, taking into account of the truncated rectangular cross-section at a distance of 682 mm from the face of the column. One-way shear force = 0.18166(1500  –  682)(3400) = 505232.792 N Area of truncated rectangle = 1800(682) + 1600(282) + 1600(682  –  282)/2 = 1998800 mm 2 The shear stress = 505232.792/1998800 = 0.2527 N/mm 2 < 0.28 N/mm 2. Hence, o.k. Step 6: Revised area of steel: The bending moment in step 5 is 694.8495 kNm at the face of the column. With d = 682 mm now, we have Mu/bd2 = 694.8495(106)/(3400)(682)(682) = 0.4394 N/mm 2 Table 2 of SP-16 gives, p is less than 0.15 per cent. Provide p = 0.15 per cent due to the one-way shear. So, Ast = 0.15(3400)(682)/100 = 3478.2 mm 2. Provide 31 bars of 12 mm (A st = 3506 mm 2),  both ways. Effectively, the number of bars has increased from 30 to 31 now.

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CONCLUCION & REMARKS

As shallow foundation transfers building load very near to the earth surface. Also design of footing is very important with respect to other design because finally all load from superstructure goes to the foundation and from foundation to the soil . So once design completed there will not be any expansion. We have seen different types of footing in this report, and each has its own property. These have different properties regarding economical aspect and availability of space. As sloped foundation is economical than isolated foundation because it reduces the volume of concrete. Also combined footing is provided when space available is less than the required for different isolated footing. Combined footing is provided when property lines of different footings overlap with each other or when column is provided eccentrically. So from the economic aspects we can say that combined footing is more economical than sloped and isolated footing. Also strap footing is another option for combined footing, b ecause it is better against both stability and economics. In strap footing a beam is provided rather than  providing slab throughout in combined footing.

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REFERENCES

1. Reinforced Concrete Design, 3rd Edition, by S.Unnikrishna Pillai and Devdas Menon, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2013. 2. NPTEL Lecture 3. Rcc Theory and design by M.G. SHAH, C.M. KALE 4. Indian Standard Plain and Reinforced Concrete  –  Code of Practice (4th Revision), IS 456: 2000, BIS, New Delhi 5. Design Aids for Reinforced Concrete to IS: 456  –  1978, BIS, New Delhi.

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