Question

1. Design a reinforced concrete wall footing for the data given below: Wall thickness - 14 in., D-21 k/A; L-20k/ft: -4000 psi; and / -60,000 psi. 4a - 4 ksf; Distance from bottom of footing to final grade - 4 ft.

Answer 1

Only spacing requirement is left dur to lack of time
Po - Wall Footing sy-0" Reiafone Design of wall Footing :- Step Thinken of hosting! the trial thiskress be equal Thet the wiaeth f .. Effective depth, d = 14 B6. - 22 where, cc. - clear & = dia af مار کر مر = d=14-3- 2 Step - weight fooling, wie wo Wonoste & Whill

on so y of Corrate, Wini -A stunning Y concrete = 150 eb/st and Isil 120 ft Lo 14x015 + (4-14) 20.12. *3)*0:12 =>/W = 0.52 kg Step 3 - Effective bearing capacity, ge :- qe = q e-W 3 Y-0.52 =/20 = 3.48 ksf - Step-y- Bearing Area required, Areg. Areg. = load/ge. load = D.L+ L.L. 21+ 200 41 u/st. Areq 41/3.42 - 11:78 ft g This bearing is for I foot of length of let's the width of fooling Well'-16 area wall allume

Pg - 3 /step Soutonlel pressure Eu aulioned load Bearing Area Fontored Load = 1,2(21) +1.6+(20) - 57.2 ufft falft qu 4.84 ks 57.2 (11-10) 1) woull are looking the rigrificant step-s] Factomed butiral shear, Vu 2 shear way This so critical position is at a distance d' from the support Vu= qu xbx (w-h exbx (w - hd) 4.84x 14 ( 11 1 2 - 14 (14 JY . 12 Nu = 21.58 k 66 - For Shear Capacity - loe know shear cap auts = p Vc od p 2 ſ jixba

- dve = 680 75x19000 x 12x10.5/1000 pvc = 5.976-714 Hip 2 I ove=245.98k Ve zove 11.96 Footing think nou should be increased Nou let hogy in = 2 Elda 24.3-1 2 and, Wo. 2x0.15.+ (4-2) x 0.12 کی ارومیه نبیه بری و مبارزه ea de W fre 3-46 4f 2 Ang = 41/346 = 11.85 h. zu with be one as yfyngs ceialth wall-por M و موثر بوش 20.5 - Vu = 4.8 4X 1X ( 11 12 = 1/ Vu = 1758 To 17.5454

ار رحمن ( وه2 27رود در 25 -230 غرامي pre = 23.34 > Vu / - Step : 8 Ple, gros) 2 where he w-ha و ( 1) 2 37 کرو .4 ( چهره ی پدر ( - Mu = 68.7 K-st. -Step 2 Art, regd. Me & fy (d-a), همینایی /2 م ده Act 68.75/2 037 - کیو مهم روح 37/Art: 0 syy en و زهر 0.55= 2 ( 0.00 Step to - Asmin = 0.1024(5 Ase)