CE-REF.COM
Shear:
Footing supporting a masonry wall: at ¼ thickness from the face of wall.
Footing supporting a concrete wall: at one effective section from the face of wall.
Moment: at face of wall;
Figure 2.4 critical shear section of masonry wall footing and concrete wall footing.
Service load design:
Design footing width based on service load. If wall footing is an exterior wall, check required frost depth.
Reinforced concrete design:
Determine footing depth and check direct shear stress at critical section.
Design transverse reinforcements based on factored moment.
Design longitudinal reinforcements for temperature and shrinkage.
Given:
Wall loads:
Live load: 2.5 kips/ft
Dead load:3 kips/ft
Wall type: 8” concrete masonry wall
Soil information:
Allowable soil bearing capacity: 2500 psf
Required frost depth: 18”
Unit weight of soil: 100 pcf
Materials used:
Concrete strength at 28 day = 3000 psi
Yield strength of rebars = 60 ksi
Design code: ACI 318-05
Requirement: Determine width, depth, and reinforcement.
Solution:
1. Determine width of footing:
Assume that the footing depth is 10 inch with 8” overburden soil. The bottom of footing is at 18” below ground surface that meet frost depth requirement.
The net soil bearing capacity is
p = 2500 – 15*10/120 – 100*8/12 = 2308 psf
The required footing width is
B = (2500 + 3000) / 2308 = 2.4 ft=29” Use 2’6” =30” width
2. Check direction shear
The ultimate footing pressure,
p_{u} = (1.2 x 3 + 1.6 x 2.5 ) / 2.5 = 3.04 kips/ft
The effective depth of footing,
d = 10" - 3" (cover) - 0.25" (half of bar size) = 6.75"
The distance from critical section to the edge of the wall is
l = 30/2 - 4 (half of wall)+2” = 13"
The direct shear stress at critical section is
v_{u} = (3.04)(1000)(13/12)/(6.75*12) = 40.5 psi
The shear strength of concrete is
v_{c} = (0.75)(2 Ö2500) = 75 psi > 40.5 psi O.K.
3. Design transverse reinforcement:
The distance from critical section of moment to the edge of the wall is
l = 15-4 = 11"
The ultimate moment at critical section is
M_{u} = (3.04)(11/12)2 / 2 = 1.27 k-ft/ft
Use trail method for reinforcement design
Factor R_{n} = (1.27)(1000)(12)/[(0.9)(12)(6.75^{2})]= 31 psi
Factor m = 60000/[(0.85)(3000)] = 23.5
The reinforcement ratio is r = (1/23.5){1-Ö[1-(2)(23.5)(31)/60000]} = 0.00103
Minimum reinforcement ratio, r = 0.0033 or r = (4/3)*0.00103 =0.0014
Use r_{min} =0.0014
A_{s} = (0.0014)(8.75)(12) = 0.14 in^{2}
Use #4 at 18" spacing, A_{s} = 0.15 in^{2}
4. Design longitudinal reinforcement
The temperature reinforcements at longitudinal direction is
A_{s} = (0.002)(12)(2.5)(8.75) = 0.525 in2
Use 3 - #4, spacing, As = 0.66 in2
The footing reinforcements is as shown below.