Combined footings
Contents:
• Introduction
• Service load design: determine size of footing
• Example 3.1: Determine size of a combined footing
• Structural analysis
• Example 3.2. Determine maximum sear and moment of a combined footing
• !einforced concrete design of combined footing
• Example 3.3: !einforced concrete design of a combined footing
Introduction "ombined footings and strap footings are normal used #en one of columns is sub$ected to large eccentric loadings. %en t#o columns are reasonably close& a combined footing is designed for bot columns as so#n in 'igure 3.1. %en t#o columns are far apart& a strap is designed to transfer eccentric moment bet#een t#o columns as so#n in 'igure 3.1. (e goal is to ave uniform bearing pressure and to minimize differential settlement bet#een columns.
'igure 3.1 "ombined footing and strap footing Design procedure:
Service load design: 1. Determine te size of combined footing. Structural analysis: 2. )erform structural analysis to determine moment and sear in various section of te footing. !einforced concrete design: 3. "ec* puncing sear + direct sear
,. Design longitudinal reinforcements. -. Design transverse reinforcements. . Design column do#els.
Service load design: determine size of footing (e size of te footing sall be determined to ave uniform bearing pressure under te footing so tat differential settlement is minimized. (e resultant of bearing pressures needs to coincide #it te resultant of column loads. (e procedures are as follo#s: 1. Determine te location of te resultant of column loads. 2. "alculated te re/uired lengt of footing. (e lengt of te footing is t#ice te distance from te edge footing of te exterior column to te resultant of column loads. 3. "alculate te #idt of footing. (e re/uired area of footing is te total column load divided by allo#able net soil bearing pressure. (e #idt of footing is te re/uired footing area divided by te lengt of footing. Example 3.1: Determine size of a combined footing
0iven: •
"olumn information:
•
"olumn : ive load ,4 *ips& Dead load -4 *ips
•
"olumn 5: ive load 64 *ips& Dead load 144 *ips.
•
Distance bet#een t#o columns: 1- ft.
•
'ooting information:
•
llo#able soil bearing capacity7 3444 psf
•
Distance from column to edge of footing: 1 ft.
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llo#able soil bearing capacity 3444 psf
•
%eigt of soil above footing 124 psf
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Dept of footing 2,8
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Dept of soil above footing 128
!e/uirements: Determine te size of a combined footing. Solution: (otal column load of ,49-44 *ips (otal column load of 5 649144 164 *ips (a*e moment about & ocation of resultant from 164;1-<=49164> 14 ft. (e lengt of footing 2;=1491> 22 ft
?se 22 ft
@et soil bearing capacity 3444A2;1-4A1242-64 psf !e/uired footing area =49164><2.-614,.B ft2. !e/uired #idt of footing 14,.B<22,.6C
Structural analysis:
?se - ft
Structural analysis of a combined footing is te same as analyzing an invert simply support beam supported by t#o columns #it factored soil pressure as loading. (e procedures are as follo#s: 1. "alculate factored footing pressure. 2. "alculate maximum sear at an effective dept from te face of column 3. "alculate maximum positive and negative moment in te footing. aximum positive moment occurs at face of column. aximum negative moment occurs bet#een t#o columns at zeroAsear. It is #ort to mention tat because of load factors& te centroid of factored column loads does not necessary located at te center of te footing. It means tat te factored footing pressure is no longer uniform. (e correct #ay to solve te problem to analyze te footing #it trapezoid sape of factored footing pressure. Example 3.. Determine maximum s!ear and moment of a combined footing
0iven: •
combined footing as so#n in Example 3.1
•
"olumn size: 1 ft by 1ft for bot + 5
Design code: "I 316A4!e/uirements: Determine maximum sear and moment in longitudinal direction. Solution: 'actored column loads: "olumn : ) ua 1.2;-491.;,412, *ips "olumn 5: ) ub 1.2;14491.;642,6 *ips ocation of resultant from column 2,6;1-<=12,92,6>14 ft. Since te location of resultant is at center of footing& factored footing pressure is uniform. 'actored footing pressure per linear foot of footing& u =12,92,6><221. *
t point 1: F u 1.;1.-A12, A6.B *ips t point 3: F u 1.;=1.-91,>A12,136 *ips t point ,: F u 1.;=1.-91,91>A12,A2,6 A3.2 *ips oment diagram: t point 1: u 1.;1.- 2<2A12,;4.- A,3 ftA*ips t point 2: ocation of point 2: from triangular relation bet#een point 1 and point 3 in sear diagram G 1,;6.B<=6.B9122.>.2,C
from inside face of column
u 1.;=1.-9.2,> 2<2A12,;=4.-9.2,>A32.- ftA*ips t point 3: u 1.;=1.-91,> 2<2A12,;=4.-91,>232.1 ftA*ips t point ,: u 1.;-.- 2<22--. ftA*ips
"einforced concrete design: Design procedure: 1. "ec* bot puncing sear and direct sear. (e critical section of puncing sear is at H effective dept from face of column. (e critical section of direct sear is at one effective dept of column. 'or column at te edge of footing te critical section of puncing sear only as tree sides along te column. "ritical sections of puncing sear and direct sear are so#n belo#.
2. Design longitudinal reinforcements. ongitudinal reinforcements are design based on te maximum moments from structural analysis. !einforcement for negative moment sould be placed near top face of te footing. )ositive reinforcement sould be placed near bottom face of te footing. 3. Design transverse reinforcements. (ransverse reinforcements are designed based on moment in te transverse direction at face of column. (ey sould be placed near bottom face of te footing. ,. Design column do#els. Example 3.3: "einforced concrete design of a combined footing
0iven: •
combined footing #it loading& sear& and moment as so#n in example 3.1 + 3.2
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"ompressive strengt of concrete for footing at 26 days: ,444 psi
•
ield strengt of rebar: 4 *si
Design code7 "I 316A4!e/uirement: "ec* sear stresses and design flexural reinforcements. Solution: a. "ec* puncing sear for column ssume te reinforcements are J bars& te effective dept d 2,K A 3K =cover> A 4.B-K =one bar size> 24.3 K 1.BL 'actored footing pressure =12,92,6><=22;->3.36 *ips9=128924.38>66.8
(e punc sear stress can be calculated as vu M12,A=3.36>=191.B>=4.-9191.B<2>N=1444><=24.3;66.> -B psi (e sear strengt of concrete is φ vc 4.B- x , x ,444 16.B psi
O.P.
C!ec# punc!ing s!ear for column $
(e perimeter of puncing sear is P ,;=12924.3>12.28
(e punc sear stress can be calculated as vu M2,6A=3.36>=191.B> 2N=1444><=24.3;12.2> 6-.2 psi
Q 16.B psi
O.P.
b. "ec* direct sear: (e critical section of direct sear is at one effective dept from te face of column. 'rom Example 3.2& te maximum direction sear is 136 *ips at inside face of column 5. (e distance from zero sear =point 2> to te maximum direct sear =point 3> is 1,A.2, B.BC. 'rom triangular relationsip& te direct sear at critical section is
V u 136;=B.BA1.2>
(e sear strengt of concrete for footing section& φ V c φ vc;b;d = =4.B- x 2 x ,444>;4;24.3<144411-. *ips ≈ 11.B *ips =less tan 1R difference> O.P.
c. Determine aximum positive reinforcement in longitudinal direction (e maximum positive moment at te face of te column 5 is M u 2--. *Aft.
for -C #idt of footing
?se trail metod for reinforcement design ssume dept of stress bloc*& a 4.K. ( uN M=2--.>=12>N=24.,A4.<2>N 1B4.6 *ips "alculate ne# a& a (=f cL>=b>N 1B4.6=,>=4>N 4.6, s (
≈ 4.8
.
(e reinforcement ratio is ρ s=24.,> 4.4423
inimum reinforcement ratio& ρ
4.4433 ρ min = =,<3>;4.44234.4431
min
?se ρ min 4.4431& A s = =4.4431 >=4>=24.,> 3.6 in 2.
?se AJ bars in bot directions& A s = 3. in2. d. aximum negative reinforcement in longitudinal direction (e maximum negative moment bet#een column is M u 32.- *Aft.
for -C #idt of footing
?se trail metod for reinforcement design
ssume a 1.2K. ( =32.->=12>=24.,A1.2<2>N 221.6 *ip "alculate ne# a& a 221.6=,>=4>N 1.1K s 221.6 <4 3.B in 2
≈ 1.28
.
(e reinforcement ratio is ρ s=24.,> 4.4431
inimum reinforcement ratio& ρ
4.4433 Q ρ min = =,<3>;4.44314.44,1
min
?se ρ 4.4433& A s = =4.4433 >=4>=24.,> ,.4 in2.
?se 14AJ bars in bot directions& A s = ,., in2. e. Determine reinforcement in transverse direction
(e distance from face of column to te edge of te footing is l =-T 1><2 2L
(e factored moment at te face of te column is M u =3.36>=2>2<2 .B *Aft. per foot #idt of footing
?se trail metod for reinforcement design ssume a 4.1K. ( =.B>=12>=24.,A4.1<2>N ,., *ip "alculate ne# a& a ,.,=,>=12>N 4.11 ≈ 4.18 assumed s ,.,<4 4.4B3 for one foot section.
(e reinforcement ratio is ρ s=24.,> 4.4443
inimum reinforcement ratio& ρ
4.4433 Q ρ min = =,<3>;4.44434.444,
min
?se ρ min 4.444, A s = =4.444, >=22>=12>=24.,> 2.2 in2.
?se 1 J, bars& s 4.2;1- 3 in2. aximum spacing =22;12A3A3><1- 1B.28 O.P.
Qaximum spacing& 168
e. Designing column do#els. (e bearing capacity of concrete at column base is P c =4.B>=4.6->=,>=12>=12> 3,2.B *ips
%ic is greater tan factored column loads of bot and 5. (e minimum do#el area is A s,min = =4.444->=12>=12> 4.B2 in 2
?se , A J, do#els A s = 4.6 in2 (e footing is so#n in belo#
