Topics Module 3 Lecture 10 SHALLOW FOUNDATIONS: ULTIMATE BEARING CAPACITY 1.1 THE GENERAL BEARING CAPACITY EQUATION Bearing Capacity Factors General Comments 1.2 EFFECT OF SOIL COMPRESSIBILITY 1.3 ECCENTRICALLY LOADED FOUNDATIONS 1.3.1 Foundation with Two-Way Eccentricity
THE GENERAL BEARING CAPACITY EQUATION The ultimate bearing capacity equations presented in equations (3, 7 and 8) are for continuous, square, and circular foundations only. The do not address the case of rectangular foundations (0 < BB/LL < 1. Also, the equations do not take into account the shearing resistnace along the failure surface in soil above the bottom of the foundation (the portion of the failure surface marked as GI and HJ in figure 3.5). In addition, the load on the foundation may be inclined. To account for all these shortcomings, Meyerhof (1963) suggested the following form of the general bearing capacity equation: qq uu = ccnn cc FF cccc FF cccc FF eeee + qqnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbbbb γγff γγγγ FF γγγγ FF γγγγ [3.25] Where cc = cohesion qq = effective stress at the level of the bottom of the foundation γγ = unit weight of soil BB = width of foundation ( diameter for a circular foundation) FF cccc, FF qqqq, FF γγγγ = shape factors FF cccc, FF qqqq, FF γγγγ = depth factors FF cccc, FF qqqq, FF γγγγ = load inclination factors NN cc, NN qq, NN γγ = bearing capacity factors The equations for determining the various factors given in equation (25) are described briefly in the following sections. Note that the original equation for ultimate bearing capacity is derived only for the plane-strain case (that is, for continuous foundations). The shape, depth, and load inclination factors are empirical factors based on experimental data. Bearing Capacity Factors Based on laboratory and field studies of bearing capacity, the basic nature of the failure surface in soil suggested by Terzaghi now appears to be correct (Vesic, 1973). However, the angle αα as shown in figure 3.5 is closer to 45 + /2 than to. If this change is accepted, the values of NN cc, NN qq, and NN γγ for a given soil friction angle will also change from those given in table 1. With = 45 + /2, the relations for NN cc and NN qq can be derived as
NN qq = tttttt 2 45 + 2 eeππ tan [3.26] NN cc = NN qq 1 cot [3.27] The equation for NN cc given by equation (27) was originally derived by Prandtl (1921), and the relation for NN qq [equation (26)] was presented by Reissner (1924). Caquot and Kerisel (1953) and Vesic (1973) gave the relation for NN γγ as NN γγ = 2 NN qq + 1 tan [3.28] Table 4 shows the variation of the preceding bearing capacity factors with soil friction angles. In many texts and reference books, the relationship for NN γγ may be different from that in equation (28). The reason is that there is still some controversy about the variation of NN γγ with the soil friction angle, φφ. In this text, equation (28) is used. Other relationships for NN γγ generally cited are those given by Meyerhof (1963), Hansen (1970), and Lundgren and Mortensen (1953). They NN γγ values for various soil friction angles are given in appendix B (table B-1, B-2, B-3). Table 4 Bearing Capacity Factors φφ NN cc NN qq NN γγ NN qq tan φφ φφ NN cc NN qq NN γγ NN qq tan φφ /NN cc /NN cc 0 5.14 1.00 0.00 0.20 0.00 26 22.25 11.85 12.54 0.53 0.49 1 5.38 1.09 0.07 0.20 0.02 27 23.94 13.20 14.47 0.55 0.51 2 5.63 1.20 0.15 0.21 0.03 28 25.80 14.72 16.72 0.57 0.53 3 5.90 1.31 0.24 0.22 0.05 29 27.86 16.44 19.34 0.59 0.55 4 6.19 1.43 0.34 0.23 0.07 30 30.14 18.40 22.40 0.61 0.58 5 6.49 1.57 0.45 0.24 0.09 31 32.67 20.63 25.99 0.63 0.60 6 6.81 1.72 0.57 0.25 0.11 32 35.49 23.18 30.22 0.65 0.62 7 7.16 1.88 0.71 0.26 0.12 33 38.64 26.09 35.19 0.68 0.65 8 7.53 2.06 0.86 0.27 0.14 34 42.16 29.44 41.06 0.70 0.67 9 7.92 2.25 1.03 0.28 0.16 35 46.12 33.30 48.03 0.72 0.70 10 8.35 2.47 1.22 0.30 0.18 36 50.59 37.75 56.31 0.75 0.73 11 8.80 2.71 1.44 0.31 0.19 37 55.63 42.92 66.19 0.77 0.75 12 9.28 2.97 1.69 0.32 0.21 38 61.35 48.93 78.03 0.80 0.78 13 9.81 3.26 1.97 0.33 0.23 39 67.87 55.96 92.25 0.82 0.81 14 10.37 3.59 2.29 0.35 0.25 40 75.31 64.20 109.41 0.85 0.84 15 10.98 3.94 2.65 0.36 0.27 41 83.86 73.90 130.22 0.88 0.87 16 11.63 4.34 3.06 0.37 0.29 42 93.71 85.38 155.55 0.91 0.90 17 12.34 4.77 3.53 0.39 0.31 43 105.11 99.02 186.54 0.94 0.93 18 13.10 5.26 4.07 0.40 0.32 44 118.37 115.31 224.64 0.97 0.97
19 13.93 5.80 4.68 0.42 0.34 45 133.88 134.88 271.76 1.01 1.00 20 14.83 6.40 5.39 0.43 0.36 46 152.10 158.51 330.35 1.04 1.04 21 15.82 7.07 6.20 0.45 0.38 47 173.64 187.21 403.67 1.08 1.07 22 16.88 7.82 7.13 0.46 0.40 48 199.26 222.31 496.01 1.12 1.11 23 18.05 8.66 8.20 0.48 0.42 49 229.93 265.51 613.16 1.15 1.15 24 19.32 9.60 9.44 0.50 0.45 50 266.89 319.07 762.89 1.20 1.19 25 20.72 10.66 10.88 0.51 0.47 a After Vesic (1973) Shape, Depth, and Inclination Factors The relationships for the shape factors, depth factors, and inclination factors recommended for use are shown in table 5. Other relationships generally found in many texts and references are shown in table B-4 (appendix B). General Comments When the water table is present at or near the foundations, the factors qq and γγ given in the general bearing capacity equations, equation (25), will need modifications. The procedure for modifying them is the same. For undrianed loading conditions (φφ = 0 concept) in clayey soils, the general loadbearing capacity equation [equation (25)] takes the form (vertical load) qq uu = ccnn cc FF cccc FF cccc + qq [3.29] Table 5 Shape, Depth, and Inclination Factors Recommended for Use Factor Relationship Source Shape a FF cccc = 1 + BBBB qq De Beer LL NN cc (1970 FF qqqq = + BB Hansen tan φφ LL (1970) FF γγγγ = 1 0.4 BB LL Where LL = length of the foundation (LL > BB) Depth b Condition (a): DD ff /BB 1 FF cccc = 1 + 0.4 DD ff BB FF qqqq = 1 + 2 tan φφ(1 sin φφ) 2 DD ff BB FF γγγγ = 1 Condition (b): DD ff /BB > 1 Hansen (1970
Inclinatio n FF cccc = 1 + 2 tan φφ(1 sin φφ) tan 1 DD ff BB FF γγγγ = 1 FF cccc = FF qqqq = 1 ββ 90 Where ββ = inclination of the load on the foundation with respect to the vert a These shape factors are empirical relations based on extensive laboratory tests. b The factors tttttt 1 (DD ff /BB) is in radians. 2 Meyerho f (1963); Hanna and Meyerho f (1981) Hence the ultimate baring capacity (vertical load) is qq net (uu) = qq uu qq = ccnn cc FF cccc FF cccc [3.30] Skempton (1951) proposed an equation for the net ultimate baring capacity for clayey soils (φφ = condition), which is similar to equation (30) : qq net (uu) = 5cc 1 + 0.2 DD ff 1 + 0.2 BB [3.31] BB LL Example 2 A square foundation (BB BB) has to be constructed as shown in figure 3.9. Assume that γγ = 105 lb/ft 3, γγ sat = 118 lb/ft 3, DD ff = 4 ft, and DD 1 = 2 ft. The gross allowable load, QQ all, with FFFF = 3 is 150,000 lb. The field standard penetration resistance, NN FF values are as follows: Depth (ft) NN FF (blow/ft) 5 4 10 6 15 6 20 10 25 5 Determine the size of the footing. Use equation (25). Solution Using equation (7 from chapter 2) and the Liao and Whitman relationship (table 4 from chaper 2), the correct standard penetration number can be determined.
Depth (ft) NN FF σσ (ton/ft 2 ) 5 4 1 [2 105 + 3 (118 62.4)] 2000 = 0.188 10 6 0.188 + 1 (5)(118 62.4) 2000 = 0.327 15 6 0.327 + 1 (5)(118 62.4) 2000 = 0.466 20 10 0.466 + 1 (5)(118 62.4) 2000 = 0.605 25 5 0.605 + 1 (5)(118 62.4) 2000 = 0.744 NN cor = NN FF 1 σσ vv 12 11 9 13 6 Figure 3.9 The average NN cor can be taken to be about 11. From equation 11 (from chapter 2), φφ 35. Given qq all = Q all B 2 = 150,000 B 2 lb/ft 2 [a] From equation (25) (note: cc = 0), qq all = qq uu FFFF = 1 3 qqnn qqff qqqq FF qqqq + 1 2 γγ BBBB γγff γγγγ FF γγγγ For φφ = 35, from table 4, NN qq = 33.3, NN γγ = 48.03. From table 5,
FF qqqq = 1 + BB tan φφ = 1 + tan 35 = 1.7 LL FF γγγγ = 1 0.4 BB = 1 0.4 = 0.6 LL FF qqqq = 1 + 2 tan φφ(1 sin φφ) 2 = 1 + 2 tan 35(1 sin 35) 2 4 BB = 1 + 1 BB FF γγγγ = 1 qq = (2)(105) + 2(118 62.4) = 321.2 lb/ft 2 So qq all = 1 (321.2)(33.3)(1.7) 1 + 1 + 3 B 1 (118 62.4)(B)(48.03)(0.6)(10 2 = 6061.04 + [b] 6061.04 B + 267.05BB Combining equations (a) and (b) 150,000 6061.04 = 6061.04 + + 267.05BB BB 2 B By trial and error, BB 4.2 ft Example 3 Refer to example1. Use the definition of factor of safety given by equation (20) and FFFF = 5 to determine the net allowable load for the foundation. Solution From example 1, qq uu = 10,736 lb/ft 2 qq = (3)(115) = 345 lb/ft 2 qq all (net ) = 10,736 345 5 Hence 2078 lb/ft 2 qq all (net ) = (2078)(5)(5) = 51,950 lb Example 4
Refer to example 1. Use equation (7) and FFFF shear for the foundation. = 1.5 determine the net allowable load Solution For cc = 320 lb/ft 2 and φφ = 20, cc dd = cc = 320 213 lb/ft2 FFFF shear 1.5 tan φφ φφ dd = tttttt 1 = tttttt 1 FFFF shear From equation (7), tan 20 1.5 = 13.64 qq all (net ) = 1.3cc dd NN cc + qq(nn 1) + 0.4γγBBBB γγ For φφ = 13.64, the values of the bearing capacity factors from table 1 are NN γγ 1.2, NN qq 3.8, NN cc 12 Hence qq all (net ) = 1.3(213)(12) + (345)(3.8 1) + (0.4)(115)(5)(1.2) = 4565 lb/ft 2 And qq all (net ) = (4565)(5)(5) = 114,125 lb 57 ton Note: There appears to be a large discrepancy between the results of examples 3 (or 1) and 4. The use of trial and error shows that, when FFFF shear is about 1.2, the results are approximated equal. EFFECT OF SOIL COMPRESSIBILITY In section 3 equation 3, 7, and 8, which were for the case of general shear failure, were modified to equations 9, 10, and 11 to take into account the change of failure mode in soil (that is, local shear failure). The change in failure mode is due to soil compressibility. In order to account for soil compressibility, Vesic (1973) proposed the following modification to equation (25), qq uu = ccnn cc FF cccc FF cccc FF cccc + qqnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbbbb γγff γγγγ FF γγγγ FF γγγγ [3.32] Where
FF cccc, FF qqqq, and FF γγγγ = soil compressibility factors The soil compressibility factors were derived by Vesic (1973) from the analogy of the expansion of cavities. According to that theory, in order to calculate FF cccc, FF qqqq, and FF γγγγ the following steps should be taken: 1. Calculate the rigidity index, II rr, of the soil at a depth approximately BB/2 below the bottom of the foundation, or II rr = GG cc+qq tan φφ [3.33] Where GG = shear modulus of the soil qq = effective overburden pressure at a depth of DD ff + BB/2 2. The critical rigidity index, II rr(cccc ), can be expressed as II rr(cccc ) = 1 exp 3.30 0.45 BB cot 45 φφ [3.34] 2 LL 2 The variation of II rr(cccc ) for BB/LL = 0 and BB/LL = 1 are given in table 6. 3. If II rr II rr(cccc ), then FF cccc = FF qqqq = FF γγγγ = 1 However, if II rr < II rr(cccc ) FF γγγγ = FF qqqq = exp 4.4 + 0.6 BB tan φφ + LL (3.07 sin φφ) (log 2II rr ) [3.35] 1+sin φφ Table 6 Variation of II rr(cccc ) with φφ aaaaaa BB/LL 11 II rr(cccc ) φφ (deg) BB LL = 0 BB LL = 1 0 13 8 5 18 11 10 25 15 15 37 20 20 55 30 25 89 44
30 152 70 35 283 120 40 592 225 45 1442 482 50 4330 1258 1 After Vesic (1973) Figure 3.10 shows the variation of FF γγγγ = FF qqqq [equation (35)] with φφ and II rr. For φφ = 0, FF cccc = 0.32 + 0.12 BB LL + 0.60 log II rr [3.36] For φφ > 0, FF cccc = FF qqqq 1 FF qqqq NN qq tan φφ [3.37] Figure 3.10 Variation of FF γγγγ = FF qqqq with II rr and φφ Example 5 For a shallow foundation, the following are given: BB = 0.6 m, LL = 1.2 m, DD ff = o. 6 m. Soil: φφ = 25 cc = 48 kn/m 2
γγ = 18 kn/m 3 Modulus of elasticity, EE = 620 kn/m 2 Poisson s ratio, μμ = 0.3 Calculate the ultimate bearing capacity. Solution From equation (33) II rr = GG cc+qq tan φφ However, GG = So II rr = EE 2(1+μμ ) EE 2(1+μμ )[cc+qq tan φφ] qq = γγ DD ff + BB 2 II rr = 0.6 = 18 0.6 + = 162 kn/m2 2 620 2(1+0.3)[48+16.2 tan 25] = 4.29 From equation (34) II rr(cccc ) = 1 2 exp 3.3 0.45 BB LL cot 45 φφ 2 1 2 0.6 exp 3.3 0.45 cot 45 25 = 62.46 1.2 2 Since II rr(cccc ) > II rr, use equations 35 and 37. FF γγγγ = FF qqqq = exp 4.4 + 0.6 BB tan φφ + LL (3.07 sin φφ) log (2II rr ) 1+sin φφ = exp 4.4 + 0.6 0.6 sin 25) log (2 4.29) tan 25 + (3.07 = 0.347 1.2 1+sin 25 FF cccc = FF qqqq 1 FF qqqq NN qq tan φφ For φφ = 25, NN qq = 10.66 (table 4),
FF cccc = 0.347 1 0.347 10.66 tan 25 = 0.216 Now, from equation (32), qq uu = ccnn cc FF cccc FF cccc FF cccc + qqnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbbbb γγff γγγγ FF γγγγ FF γγγγ From table 4, for φφ = 25, NN cc = 20.72, NN qq = 10.66, NN γγ = 10.88. From table 5, FF cccc = 1 + NN qq BB = 1 + 10.66 NN cc LL 20.72 FF qqqq = 1 + BB LL 0.6 1.2 = 1.257 0.6 tan φφ = 1 + tan 25 = 1.233 1.2 FF γγγγ = 1 0.4 BB LL = 1 0.4 0.6 1.2 = 0.8 FF cccc = 1 + 0.4 DD ff BB = 1 + 0.4 0.6 = 1.4 0.6 FF qqqq = 1 + 2 tan φφ(1 sin φφ) 2 DD ff = 1 + 2 tan 25(1 sin BB 25)2 0.6 = 1.311 0.6 FF γγγγ = 1 Thus qq uu = (48)(20.72)(1.257)(1.4)(0.216) + (0.6 18)(10.66)(1.233)(1.311)(0.347) + 1 (18)(0.6)(10.88)(0.8)(1)(0.347) = 459 kn/m2 2 ECCENTRICALLY LOADED FOUNDATIONS In several instances, as with the base of a retaining wall, foundations are subjected to moments in addition to the vertical load, as shown in figure 3.11a. In such cases the distribution of pressure by the foundation on the soil is not uniform. The distribution of nominal pressure is qq max = QQ BBBB + 6MM BB 2 LL [3.38] And qq max = QQ BBBB 6MM BB 2 LL [3.39] Where QQ = total vertical load
MM = moment on the foundation Figure 3.11b shows force system equivalent to that shown in figure 3.11a. The distance e is he eccentricity, or ee = MM QQ [3.40] Substituting equation (40) in equations (38) and (39) gives qq max = QQ 6ee 1 + BBBB BB [3.41a] And qq max = QQ 6ee 1 BBBB BB [3.41b] Figure 3.11 Eccentrically loaded foundations Note that, in these equations, when the eccentricity, e, becomes BB/6, qq max is zero. for ee > BB/6, qq min will be negative, which means that tension will develop. Because soil cannot take any tension, there will be a separation between the foundation and the soil underlying it. The nature of the pressure distribution on the soil will be as shown in figure 3.11a. the value of qq max then is qq max = 4QQ 3LL(BB 2ee) [3.42] The exact distribution of pressure is difficult to estimate.
The factor of safety for such types of loading against baring capacity failure can be evaluated by using the procedure suggested by Meyerhof (1953), which is generally referred to as the effective area method. The following is Meyerholf a step-by-step procedure for determination of the ultimate load that the soil can support and the factor of safety against bearing capacity failure. 1. Determine the effective dimensions of the foundation as BB = effective width = BB 2ee LL = effective length = LL Note that, if the eccentricity were in the direction of the length of the foundation, the value of LL would be equal to LL 2ee. The value of BB would equal BB. The smaller of the two dimensions (that is, LL and BB ) is the effective width of the foundation. 2. Use equation (25) for the ultimate bearing capacity as qq uu = ccnn cc FF cccc FF cccc FF cccc + qqnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbb NN γγff γγγγ FF γγγγ FF γγγγ [3.43] To evaluate FF cccc, FF qqqq, and FF γγγγ, use table 5 with effective length and effective width dimensions instead of LL and BB, respectively. To determine FF cccc, FF qqqq, and FF γγγγ use table 5 (do not replace BB with BB ). 3. The total ultimate load that the foundation can sustain is AA QQ ult = qq uu (BB )(LL ) [3.44] Where AA = effective area 4. The factor of safety against bearing capacity failure is FFFF = QQ ult QQ [3.45] 5. Check the factor of safety against qq max, or, FFFF = qq uu /qq max. Note that eccentricity tends to decrease the load-bearing capacity of a foundation. In such cases, placing foundation columns off center, as shown in figure 3.12, probably is advantageous. Doing so, in procedures a centrally loaded foundation with uniformly distributed pressure.
Figure 3.12 Foundation of columns with off-center loading Foundation with Two-Way Eccentricity Consider a situation in which a foundation is subjected to a vertical ultimate load QQ ult and a moment M as shown in figure 3.13a and b. For this case, the components of the moment, M, about the x and y axes can be determined as MM xx and MM yy respectively (figure 3.13). This condition is equivalent to a load QQ ult placed eccentrically on the foundation with xx = ee BB and yy = ee LL (figure 3.13d). Note that ee BB = MM yy QQ ult [3.46] And ee LL = MM xx QQ ult [3.47] If QQ ult is needed, it can be obtained as follows [equation (44)]: QQ ult = qq uu AA Where, from equation (43) qq uu = ccnn cc FF cccc FF cccc FF cccc + qqnn qq FF qqqq FF qqqq FF qqqq + 1 γγbb NN 2 γγff γγγγ FF γγγγ FF γγγγ [3.48] And AA = effective area = BB LL
Figure 3.13 Where Figure 3.14 Effective area for the case of ee LL /LL 1 6 and e B/B 1 6 BB 1 = BB 1.5 3ee BB BB LL 1 = LL 1.5 3ee LL LL [3.49a] [3.49b] The effective length, L, is the larger of the two dimensions, that is, BB 1 or LL 1. So, the effective width is BB = AA LL [3.50]
Case II ee LL /LL < 0.5 and 0 < e B /B < 1 6 3.15a.. The effective area for this condition is shown in figure AA = 1 2 (LL 1 + LL 2 )BB [3.51] Figure 3.15 Effective area for the case of ee LL /LL < 0.5 and 0 < ee BB /BB < 1 (after Highter 6 and Anders, 1985) The magnitudes of LL 1 andll 2 can be determined from figure 3.15b. The effective width is BB = AA LL 1 or LL 2 (which is larger ) [3.52] The effective length is LL = LL 1 orll 2 (which is larger) [3.53] Case III
ee LL /LL < 1 6 and 0 < e B/B < 0.5. The effective area for this condition is shown in figure 3.16a: AA = 1 2 (BB 1 + BB 2 )LL [3.54] Figure 3.16 Effective area for the case of ee LL /LL < 1 6 and 0 < ee BB/BB < 0.5 (after Highter and Anders, 1985) The effective width is BB = AA LL [3.55] The effective length is equal to LL = LL [3.56] The magnitude of BB 1 and BB 2 can be determined from figure 3.16b.
Case IV ee LL /LL < 1 and e 6 B/B < 1. Figure 3.17a shows the effective area for this case. The ratio 6 BB 2 /BB and thus BB 2 can be determined by using the ee LL /LL curves that slope upward. Similarly, the ratio LL 2 /LL and thus LL 2 can be determined by using the ee LL /LL curves that slope downward. The effective area is then Figure 3.17 Effective area for the case of ee LL /LL < 1 and 0 < ee 6 BB/BB < 1 (after Highter and 6 Anders, 1985) AA = LL 2 BB + 1 2 (BB + BB 2)(LL LL 2 ) [3.57] The effective width is BB = AA LL [3.58] The effective length is LL = LL [3.59]
Example 6 A square foundation is shown in figure 3.18. Assume that the one-way load eccentricityee = 0.15 m. Determine the ultimate load, QQ ult. Figure 3.18 Solution With cc = 0, equation (43) becomes qq uu = ccnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbb NN γγff γγγγ FF γγγγ FF γγγγ qq = (0.7)(18) = 12.6 kn/m 2 For φφ = 30, from table 4, NN qq = 18.4 and NN γγ = 22.4 BB = 1.5 2(0.15) = 1.2 m LL = 1.5 m From table 5 FF qqqq = 1 + BB tan φφ = 1 + 1.2 tan LL 1.5 30 = 1.462 FF qqqq = 1 + 2 tan φφ(1 sin φφ) 2 DD ff = 1 + (0.289)(0.7) = 1.135 BB 1.5
FF γγγγ = 1 0.4 BB = 1 0.4 1.2 = 0.68 LL 1.5 FF γγγγ = 1 So qq uu = (12.6)(18.4)(1.462)(1.135) + 1 (18)(1.2)(22.4)(0.68)(1) = 384.7 + 164.50 = 2 549.2 kn/m 2 Hence QQ ult = BB LL qq = (1.2)(1.5)(549.2) 988kN uu Example 7 Refer to example 6. Other quantities remaining the same, assume that the load has a twoway eccentricity. Given: ee LL = 0.3 m, and ee BB = 0.15 m (figure 3.19). Determine the ultimate load, QQ ult. Solution ee LL LL = 0.3 1.5 = 0.2 ee BB BB = 0.15 1.5 = 0.1 Figure 3.19 This case is similar to that shown in figure 3.15a. From figure 3.15b, for ee LL /LL = 0.2 and e B /B = 0.1 LL 1 LL 0.85; LL 1 = (0.85)(1.5) = 1.275 m
And LL 2 LL 0.21; LL 2 = (0.21)(1.5) = 1.315 m From equation (51) AA = 1(LL 2 1 + LL 2 )BB = 1 (1.275 + 0.315)(1.5) = 1.193 m2 2 From equation (53) LL = LL 1 = 1.275 m From equation (52) BB = AA = 1.193 = 0.936 m LL 1.275 Note, from equation (43) for cc = 0 qq uu = ccnn qq FF qqqq FF qqqq FF qqqq + 1 2 γγbb NN γγff γγγγ FF γγγγ FF γγγγ qq = (0.7)(18) = 12.6 kn/m 2 For φφ = 30, from table 4, NN qq = 18.4 and NN γγ = 22.4. Thus FF qqqq = 1 + BB tan φφ = 1 + 1.936 tan LL 1.275 30 = 1.424 FF qqqq = 1 0.4 BB = 1 0.4 1.936 = 0.706 LL 1.275 FF qqqq = 1 + 2 tan φφ(1 sin φφ) 2 DD ff = 1 + (0.289)(0.7) = 1.135 BB 1.5 FF γγγγ = 1 So qq ult = AA qq uu = AA (qqnn qq FF qqqq FF qqqq + 1 2 γγbb NN γγ FF γγγγ FF γγγγ = (1.193)[(12.6)(18.4)(1.424)(1.135) + (0.5)(18)(0.936)(22.4)(0.706)(1)] = 605.95 kn