Foundation Example This example problem is contained in Chapter 9 of the CAC Concrete Design Handbook

Example 9.7.2 - Wall Footing Determine footing width, depth and reinforcement for the wall shown in Figure 9.7.2-1. Use factored rock reaction of qsf = 2000 kPa, fc = 35 MPa, and fy = 400 MPa. Factored wall load Pf = 4500 kN/m of wall. Calculations and Discussions: 1. Required footing width: Assuming wf is negligible bf = 4500/2000 = 2.25 m 2. Calculation of effective shear depth: ab = (2.25 - 0.5) / 2 = 0.875 m Assuming that ab < 3dv, (Clause 11.3.6.2 (b) of CSA A23.3) 21.0 = . The concrete shear

resistance is ccr fv , thus vr = 0.81 MPa. = Factored shear at the critical section: Vf = qsf (ab - dv) Minimum shear depth from the maximum shear stress condition at the critical section is: vf = Vf /(1.0dv) = ( qsf ab - qsf dv)/(1.0dv); thus the effective shear depth is dv = (qsf ab)/(vf + qsf) = 0.623 m thus the assumption of ab < 3dv was correct. Therefore d = dv /0.9 = 0.692 m use d = 0.7 m. (Clause 11.3.6.3 of CSA A23.3 would give a = 0.14 value with the above dv.) 3. Flexural design: Check for deep beam action Clause 10.7 of CSA A23.3. ab /d = 0.875/0.7 = 1.25 < 2.0 thus deep beam action needs to be considered. The deep beam action will be assessed by two methods: Park and Paulays reduced lever

arm method and the Strut-and-tie method. 3.1 Park and Paulays reduced lever arm method Lever arm: dVf / Mf = 2d/ab = 2(0.7)/0.875 = 1.6, thus z = 0.4(d + Mf /Vf) = 0.4 (d + ab /2) = 0.45 m Required reinforcement: == )zf/(MA ysfs (0.5(0.8752)(2000103)/(0.85(400)(0.45)) = 5004 mm2/m Minimum reinforcement: Based on Clauses 10.5.1.2 (a) and 7.8 of CSA A23.3, and assuming the total depth of

footing as tf = 700 + 75 (cover) + 15 (assumed half bar diameter) = 790 mm, thus use tf = 800, the required minimum reinforcement is: As,min = 1600 mm2/m, which is less than the

calculated As. Maximum rebar size that can be developed into ab 0.075 = 0.8 m from Table 9.10 is max db = 25M. Reinforcement: Use: 25M @ 100 mm c/c BLL - HH (bottom lower layer, hooked-hooked) short direction 20M @ 175 mm c/c BUL (bottom upper layer) long direction Transfer of wall forces at wall base and development of wall reinforcement into the footing

has not been checked in this example see Clause 15.9 of CSA A23.3. 3.2 Design Flexural steel based on strut-and-tie method Based on d = 0.7 m the total depth of tf = 700 + 75 + 15(half bar diameter) = ~800 mm, from

steps 1 and 2, will be used for further design. Based on the force diagram on Figure 9.7.2-2, the compressive factored strut forces and

their corresponding tension (horizontal) components are summarized below.

Strut Angle () Compression - C - in strut kN Tension -T - kN A-J; I-L A = 40 CA = 500/sinA = 778 TA = 500/tanA = 596 B-J; H-L B =

50.2 651 417

C-J; G-L 64.5 554 238 D-K; F-K 70.3 531 179 E-K 90 500 0

- Resultant tension forces in the tension reinforcement are therefore:

Nodes Factored tension force - T - kN A, I 596 B, H 596 + 417 = 1013 C, G 1013 + 238 = 1251 D, F 1251+179 = 1430 E 1430

Required tension reinforcement:

== )f/(TA ysfs 1.43106/(0.85(400)) = 4206 mm2/m This is 84% of the reinforcement by Park and Paulays reduced lever arm method. Reinforcement: Based on the reinforcement development considerations in item 3.1, use: 25M @ 110 mm c/c BLL - HH (bottom lower layer, hooked-hooked) short direction 20M @ 175 mm c/c BUL (bottom upper layer) long direction Check capacity of struts: Strut A-J: hA = 250 sin40 = 161 mm; CA = 778 kN; TA = 596 kN Strain along the tension reinforcement in point A: See Figure 9.7.2-4

== )EA/(T sAx 0.596/(200000(50010-6)/0.11) = 0.65610-3 Principle tensile strain in the strut Clause 11.4.2.3 of CSA A23.3:

=++= s2xx1 cot)002.0( 0.65610-3 + (0.65610-3 + 0.002)cot240 = 4.4310-3 Limiting compressive stress in the strut Clause 11.4.2.3 of CSA A23.3:

c1ccu f85.0)1708.0/(ff += fcu = 35/(0.8 + 170(4.4310-3)) = 22.5 MPa < 29.7 MPa therefore stress level is acceptable. Strut resistance:

cuccr fAP = = 0.65(1)(0.161)(22.5) = 2.35 MN > 0.812 MN, thus strut capacity is adequate. The other struts can be checked in a similar way; however strut A-J governs for fcu limit. Check nodal stresses:

Node K: Pf = 1500 kN; Ac = 1(0.167) = 0.167 m2 thus concrete stress is fc = 1.5/0.167 = 8.98 MPa. Limiting stress based on Clause 11.4.4.1 of CSA A23.3 is cc f. 850 = 19.3 MPa > fc. OK. Concrete area (Ac) in node K was taken at the wall and footing interface. Nodes J and L: Acting compressive force (vector sum of strut forces CA, CB, CC) with angle B is CJ = 1500/sinB = 1952 kN. Strut depth hJ = (500/3)sin B = 128 mm. Concrete stress is fc = 1.952/0.128 = 15.24 MPa < 19.3 MPa. The above calculated nodal stress levels are the function of the assumed strut widths. For strain compatibility one could have assumed different strut widths at the column-footing interface, thus lowering or increasing the nodal stress levels. The strut widths could be adjusted to achieve equal stresses at nodes J, K and L. Ultimately Clause 10.8 of CSA A23.3 may be used for checking the stresses at the column-footing interface. Transfer of wall forces at wall base and development of wall reinforcement into the footing is not checked in this example.

Table 9.10 Tension Development length, d (mm) for deformed reinforcing bars with fy = 400 MPa, for normal density concrete according to Clause 12.2.3 of CSA A23.3.

bc

y4321d d

f

fkkkk45.0 =A

cf Bar Size MPa 10M 15M 20M 25M 30M 35M 45M 55M 20 320 480 640 1010 1210 1410 1810 2210 25 290 430 580 900 1080 1260 1620 1980 30 260 390 530 820 990 1150 1480 1810 35 240 370 490 760 910 1060 1370 1670 40 230 340 460 710 850 1000 1280 1570 45 210 320 430 670 800 940 1210 1480 50 200 310 410 640 760 890 1150 1400 55 190 290 390 610 730 850 1090 1330 60 190 280 370 580 700 810 1050 1280 64 180 270 360 560 680 790 1010 1240

Notes: 1. The modification factor for bar size (k4) has been included in Table value. See Clause

12.2.4 of CSA A23.3 for other modification factors. 2. After application of all modification factors, the development length must not be less than 300 mm.