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Foundation Engineering
Foundation EngineeringFoundation Engineering
Prof. Mesut PervizpourProf. Mesut Pervizpour
Structural Design of Shallow Foundationsg
1
Spread Footing, Structural Design• Introduction (Types of foundation, Design steps, Footing size, Materials, Loads Reduction factors Cover requirements )Loads, Reduction factors, Cover requirements )
• Structural Design (Two-way shear, One-way shear, Flexure, Development lengths, Dowels, Load transfer)
• Continuous Footing• Rectangular Footing• Examples (square wall rectangular combined eccentricity)Examples (square, wall, rectangular, combined, eccentricity)
2
Loads acting on foundationLoads acting on foundation
TT
N l l d
Torsion loadsTT
PP Normal loads(compression-tension, é)
PP
MM Moment loads(in å or ç directions)
MM
VV Shear loads(in å or ç directions)
VV
3Figure 2.1 Coduto
Design LoadsDesign LoadsDead Loads (D)
Allowable Stress Design (ASD)Allowable Stress Design (ASD)Estimates of most critical actual service loads
geotechnical( )
Live Loads (L)
Snow/Rain Loads (S/R)
geotechnicalEx: ANSI/ASCE (Use the largest of design load:)
DD + L + F + H + T + (Lr or S or R)D + L + (L or S or R) + (W or E)Snow/Rain Loads (S/R)
Earth Pressure Loads (H)
Fluid Loads (F) Allowable = ultimate / FS > design load OK
D + L + (Lr or S or R) + (W or E)D + (W or E)
Alternative method for W & E (pg19 Coduto)
( )
Earthquake Loads (E)
Wind Loads (W)
Load & Resistance Factor Design (LRFD)Load & Resistance Factor Design (LRFD)Ultimate strength design, use of factored load (U):
Structural( )
Self-straining Loads (T)
Impact Loads (I)
Load factors: Pu = 1 PD + 2 PL + ….Resistance factor: Pu Pn
nominal load capacity
Ex, ACI 318-05 (Ch9): U = 1.4(D + L)
Stream Flow/Ice Loads (SF/ICE)
Centrifugal/Braking Loads (CF/BF)
U = 1.2(D +F+T )+ 1.6(L+H)+0.5(LR or S or R)U = 1.2D + 1.6(Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6W + L + 0.5(LR or S or R)Or Use ACI 318-05 Appendix C Load factor and
4
strength reduction factors BUT …..ANSI/ASCE: U = 1.4D (eqns on pg 23 Coduto)
AASHTO: U = 1.25D + 1.75L (bridges)
Load & Resistance Factor Design (LRFD)Load & Resistance Factor Design (LRFD)Ultimate strength design, use of factored load (U): Structural
Load factors: Pu = 1 PD + 2 PL + ….Resistance factor: Pu Pn nominal load capacity
ACI 318-05 (Ch9): ACI 318-05 (Appendix C 1999):
U = 1.4(D + L)
U = 1.2(D +F+T )+ 1.6(L+H)+0.5(LR or S or R)
U = 1.2D + 1.6(Lr or S or R) + (L or 0.8W)
U 1 2D 1 6W L 0 5(L S R)
U = 1.4D + 1.7L
U = 0.5(1.4D + 1.4F + 1.7T )
U = 0.9D + 1.4F
U 1 4D 1 7L 1 4FU = 1.2D + 1.6W + L + 0.5(LR or S or R) U = 1.4D + 1.7L + 1.4F
U = 1.4D + 1.7L + 1.7H ….. other
factors:
Fl 0 90
Pu = U = 1.2 D + 1.6 L (ACI)
factors:
Fl 0 90
Flexure = 0.90
Shear = 0.85
Bearing on Concrete = 0.70
Unreinforced footings = 0 65
Note: AASHTO Pu =1.25D + 1.75L
Flexure = 0.90
Shear = 0.75
Bearing on Concrete = 0.65
Unreinforced footings =
If you use ACI 318-05 Appendix C Load factor then use the same strength reduction factors together and do not mix with
Unreinforced footings = 0.65
5
Unreinforced footings = factors together and do not mix with Chapter 9 numbers.
Types of Foundations Isolated Spread footingfooting
Combined footing
Property line
C til
Wall (continuous) footingCantilever
(Strap)footing
footing
6
Types of Foundations A BCombined footings
Isolated spread footing
Wall footing
Rectangular, PA = PB
Property line
A B
Rectangular, PA < PB
A B
A B
Property line
Mat Footing Pile cap
Rectangular, PA < PB
A B
Property linePiles
Weak soil
7Strap or Cantilever
Bearing stratum
Design Considerations
Footings must be designed to carry the column loads and transmitFootings must be designed to carry the column loads and transmit
them to the soil safely while satisfying code limitations.
1. Area of the footing is based on the soil allowable bearing capacity
2. Footing thickness is based on:
• Two-way shear or punch out (diagonal) shear
• One-way (wide beam) shear
3. Steel reinforcement is based on flexure design at critical section
4. Bearing capacity of columns at their base (load transfer to footing)
5. Dowel requirements
6. Development length of bars
7. Differential settlement
8
The area of footing can be determined from the actual external loads such
Footing Size
The area of footing can be determined from the actual external loads such that the allowable soil pressure is not exceeded.
weightselfincludingloadTotalf ifA
pressure soil allowable
ggfootingof Area
WLLDL
Strength design requirements (structural design):
all
f
q
WLLDL Area
f tif
uu
Pq
Strength design requirements (structural design):
the bearing pressure at the base of footing
footing ofareaLLDL Pu 7141 .. ACI 05 - Appendix C
9
LLDL 6.12.1 Pu ACI 05 - Chapter 9
Structural DesignMaterial Selection:
Concrete compressive strength (f’c): 2000 – 3000 psi (15 – 20 Mpa)Concrete, compressive strength (f c): 2000 3000 psi (15 20 Mpa)Steel, Grade 40 (fy): 40,000 psi (300 Mpa)
Grade 60 (fy): 60,000 psi (420 Mpa)
Modes of failure: shear and flexureModes of failure: shear and flexure
Design Loads:
Factors:
Factored normal load (ACI) (appendix C), Pu = 1.4 D + 1.7 LFactored normal load (ACI-05) (chapter 9), Pu = 1.2 D + 1.6 L
Factors:Flexure = 0.90 Bearing on concrete = 0.65 (0.70)Shear = 0.75 (0.85) Unreinforced footings = (0.65)
Values in parenthesis are to be used with Appendix C equations.
Minimum cover requirements and standard dimensions:
dEffective depth, d:
d T 3i d 6 i
3 in
db
dT d = T – 3in – db 6 in
Minimum thickness, T:
T 12 in (300 mm)ACI 15.7
10
3 in(70 mm)Flexural Steel
db: nominal diameter of reinforcing steel
T 12 in (300 mm)
Round T to a multiple of 3 in or 100 mm(12, 15, 18 .. in , or 300, 400, 500 mm)
Reinforced Footing Design:Initial design for shear: find the initial estimate for effective depth, d
No shear reinforcement in footings
Shear resistance provided only by the depth of concrete above steel
One way (beam shear) or Two way (diagonal tension) shear:One-way (beam shear) or Two-way (diagonal tension) shear:
Two-way shear governs when subjected to vertical loads alone for
square footings, Check both two-way and one-way shear for applied
shear/moment loads
Designing for Shear: dOne-way Sheard/2
d/2Two-way Shear
Vuc Vnc
T
Criteria to find “d”
Vuc: factored shear force on critical surface
: resistance factor for shear = 0.75 (App. C=0.85)
Vnc = Vc + Vs dT
dVnc: nominal shear capacity on the critical surface
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Vc: nominal shear load capacity of concreteVs: nominal shear load capacity of reinforcing steel (neglected)
Reinforced Footing Design:Two-way (punching) shear design d/2
d/2Two-way Shear
P
T
d/2Pu
d
T
c1 + d
d/2
c2c2 + d
b0
12c1
b0
Reinforced Footing Design:Two-way (punching) shear design
Vuc Vnc
Assume d.
Determine b0.
b 4( +d) for square columns
1.
2.
b0 = 4(c+d)
b0 = 2(c1+d) +2(c2+d)
for square columns where one side = c
for rectangular columns of gsides c1 and c2.
Shear force Vuc acts at a section that has a length b 4( +d) 2( +d) +2( +d) d d th d
3.
bo
b0 = 4(c+d) or 2(c1+d) +2(c2+d) and a depth d;
the section is subjected to a vertical downward load Pu and vertical upward pressure qu. u p p qu
2
u u uV P q c d
V P q c d c d
for square columns
for rectang lar col mns
c
13
u u u 1 2V P q c d c d for rectangular columnsc
Reinforced Footing Design; Vuc Vnc Criteria to find “d”
Two-way (punching) shear design
For “square footings supporting square or circular columns on the interior”, the
nominal two-way shear capacity [ACI 11 12 2 1]:
4. Calculate nominal shear capacity on critical surface
nominal two way shear capacity [ACI 11.12.2.1]:
'4 0 ccnc fdbVV 1
(English Units)Nominal two-way
:b (length of critical shear surface)
'3
10 ccnc fdbVV (SI Units)
shear capacity
:0b (length of critical shear surface)
fc’: 28 – day compressive strength of concrete (psi, MPa)
Vx: in lb or NFor columns with other shapes and located
c, d, b0: in inches or mmFor columns with other shapes and located out of center ACI 11.12.2.1 (next page)
Vuc = Vnc Solve for “d”5. Set
14
uc nc
If d is not close to the assumed d, revise your assumptions
G l th i l t h it [ACI 11 12 2 1]
Vuc Vnc Criteria to find “d”Reinforced Footing Design; Two-way (punching) shear designGeneral, the nominal two-way shear capacity [ACI 11.12.2.1]:
Nominal two-way shear capacity (Smallest Value for Vnc calculated below)
(English Units) (SI Units)
'4 0 ccnc fdbVV (English Units)
4
'3
10 ccnc fdbVV
(SI Units)
21 '
42 0 c
ccnc fdbVV
'2
16
10 c
ccnc fdbVV
'c
scnc fdb
b
dVV 0
0
2
'cs
cnc fdbb
dVV 0
0
212
1
Vnc: nominal shear capacity (one critical surface lb/N)
b0: length of critical surface (in/mm)
d: effective depth (in/mm) : long side / short side (c / c ) of column
15
p ( )
c: column width (in/mm)c: long side / short side (cl / cs) of column
s: interior = 40, edge = 30, corner = 20 (columns)
Reinforced Footing Design; One-way (wide beam) shear design
Vuc Vnc
F f ti ith b di ti i di ti th iti l ti i l t d
dOne-way Shear
For footings with bending action in one direction the critical section is located at a distance d from face of column.
1 A d1. Assume a d.
2. bo = b.
dT
The ultimate shearing force at section m-m can be calculated
cL b =
3.
d
cLbqV o 22 uu
If no shear reinforcement is to be sed then d
bo
16
If no shear reinforcement is to be used, then d can be checked
Vuc Vnc Criteria to find “d”Reinforced Footing Design; One-way (wide beam) shear design
Nominal one-way shear load capacity [ACI 11.3.1.1]:
'2 fdbVV (E li h U it )Nominal
4. Calculate nominal shear capacity on critical surface
2 cocnc fdbVV
'1
fdbVV
(English Units)
(SI Units)
one-way
shear load
capacity6 cocnc fdbVV (SI Units) capacity
Vnc: nominal one-way shear capacity on the critical section (lb/N)
b : length of one way critical shear surface (in/mm)bo: length of one-way critical shear surface (in/mm)
d: effective depth (in/mm)
fc’: 28-day compressive strength of concrete (psi/MPa)
The larger of effective depth d obtained from two way and one way
Vuc = Vnc Solve for “d”5.
If d is not close to the assumed d, revise your assumptions
Set
17
The larger of effective depth, d, obtained from two-way and one-way capacity calculations is used for design purposes.
Procedure in determination of effective depth, d
Reinforced Footing Design;
• Assume a trial d (~ column width, note: T multiple of 3 in)
• Compute Vuc and Vnc and check
• Repeat first two steps to find smallest d, satisfying the shear criteria
Vuc Vnc
p p , y g
• Compute footing thickness T (using a db = 1 in)
Actual value of d will be determined from the flexural analysis
18
Actual value of db will be determined from the flexural analysis
Critical Sections
Design for Flexure (As)
Reinforced Footing Design; Muc MncFlexural design criteria:
= 0.90 for flexure in RCObtain size of member and amount and location of reinforcing steel.
b 0.85fc’
a/2
dNeutral Axis
c a = 1 c C
a/2jd
= dAs(min)
Grade 40 Neutral Axis(Axis of Zero Strain)
Asf T
d –
a/2
Grade 40As 0.002 AgGrade 60As 0.0018 Ag
Asfy T
2
adfAM ysn
As: steel area: steel ratio
0F
fAabf '850nu MM
'850y
f
fda
b: width ysc fAabf '.850
db
As '59.019.0 cyysuc ffdfAM
Calculate Muc and solve equation for As
19
'85.0 cfdb
bf
Mdd
f
bfA
c
uc
y
cs '
353.2
176.1
' 2
Critical Sections
Design for Flexure (As)
Reinforced Footing Design; Muc Mnc
Calculate Muc at the critical section for moment
Location of critical section for bending (l cantilever distance):
c c cc
FlexuralSteel c/4 cp
lB
lB
l = (B c/2) / 2
lB
l = (2B (c + c )) / 4
Concrete Columnl = (B – c) / 2
Masonry Columnl = (B – c/2) / 2
Steel Columnl = (2B – (c + cp )) / 4
for circular/octagonal column use an equivalent area square column
lc Where Muc at the critical section is:
2
2
1lqM uuc
Care to be taken that this is moment per width, for instance for a square footing with width B, the moment is:
20qu
2 uuc2
2
1lBqM uuc
Limiting As:
Reinforced Footing Design;
Minimum steel reinforcement in footing:ACI [10.5.4 and 7.12.2]
Grade 40 steel As 0.0020 Ag
Grade 60 steel As 0.0018 Ag
The minimum steel percentage () required in flexural members is 200/fy with minimum area and maximum spacing of steel bars in the direction of bending shall be as required for shrinkage temperature reinforcement (10.5.4).
Rebar sizing criteria:
• Clear space between bars, the larger of: “db”, or “4/3 times the nominal maximum
aggregate size”
• Center to center spacing of reinforcement less than 3T or 18 in (whichever less)
Rebar sizing criteria:
• Center-to-center spacing of reinforcement less than 3T or 18 in (whichever less)
21
Design Data for Steel Reinforcing Bars:
Reinforced Footing Design;
22
Reinforced Footing Design;
Development Length of Flexural Steel: ld ld-suppliedCriteria: d d suppliedAvailable (Supplied) development length (ld-supplied):
3 inld-supplied
ld-supplied = l – 3 in (70 mm)
If not satisfied, use smaller db bars
Where l, is the length of critical section for bending (cantilever segment)
lB
Critical section for bending
for bending (cantilever segment)
Transverse Reinforcement Steel Area:
A 2A
23
Atr = 2Ab Atr = 4Ab
ACI [12 2 3]
ld ld-suppliedCriteria:
*At l t 2d d t i t l t d
Required Development Length ld for Bars in Tension:
Reinforced Footing Design;
ACI [12.2.3]
fl 3
(English units)
dfl 9
(SI units)
*At least 2db, and concrete spacing at least db
b
trc
y
b
d
dKcf
f
d
l '40
3
b
trc
y
b
d
dKcf
f
d
l '10
9
2 5 b
ns
fAK yttr
tr 1500 ns
fAK yttr
tr 10
2.5
ld: min. required development length (in, mm), at least 300mm (12 in)fyt: transverse reinforcing steel yield strength (psi, MPa)
For spread footings, use of Ktr = 0 is conservative
c: spacing or cover (in, mm), smaller of bar center-to-nearest concrete, 0.5 center-center bar spacing)Atr: total x-sec. area of all transverse reinforcement within spacing s, crossing potential
l f litti th h th d l d i f t (i 2 2) b t k
24
plane of splitting through the developed reinforcement (in2, mm2), maybe taken zeros: max. center-to-center spacing of transverse reinforcement within ld (in, mm)n: number of bars or wires being developed along the plane of splitting
ld ld-suppliedCriteria:
Required Development Length ld for Bars in Tension:
Reinforced Footing Design;
List of Modifiers for Development Length Calculations:
Modifier Value
1 3 horizontal (top) reinforcement(reinforcement location)
1.3 horizontal (top) reinforcement within 12 in of concrete
1.0 all other 1.7
(coating factor)
1.5 epoxy coated with spacing < 6db
or cover < 3db
1.2 epoxy coated bars or wires1 0 uncoated bars or wires
(reinforcement factor)
1.0 uncoated bars or wires
0.8 #6 bar and smaller, and deformed wire
1 0 #7 d l
(lightweight concrete factor)
1.0 #7 and larger
1.3 lightweight aggregate concrete1.0 Normal weight concrete
25
(lightweight concrete factor) g
Design for Compressive Loads:
Load Transfer from Columns to Footings;
Nominal column bearing capacity
AfP '850 sAfP cnb 185.0Design criterion
nbu PP Pu = factored column load ( 1.2D + 1.6L ACI-05 )
A1 = cross-sectional area of the column = c2
s = (A2/A1)0.5 < 2 if c + 4d < B; otherwise, s = 1s (A2/A1) 2 if c 4d B; otherwise, s 1
A2 = (c + 4d)2
= 0.65
26
Note: on column use column fc’, and on footing used footing fc’
Column compressive forces transmitted to footing directly by bearingDesign for Compressive Loads:
Load Transfer from Columns to Footings;
Permitted bearing capacity at base of column:
Column compressive forces transmitted to footing directly by bearingUplift (tension) forces are transferred by developed reinforcing bars
Check the strength of the lower part of the column (only concrete):Permitted bearing capacity at base of column:
where A1 = loaded area (column area)1(0.85 )nb cP f A
A = 0.65A1
450 May be multiplied by for bearing on footing:1
2
A
A
2A2 measuredon this plane where A2 = area of lower base of the largest
21 1
1
(0.85 ) 2 (0.85 )nb c c
AP f A f A
A
1on this plane 2
pyramid cone contained within footing havingside slope 1 vertical to 2 horizontal (or area of portion of supporting footing that is geometrically
27
similar and concentric with column)
If bearing force (Pu) > the smallest of the allowable values carry excess by dowels
• Design for compressive loads
A minimum steel ratio = 0.005 of the column section (Ag) as compared to
Use of Dowels for Connection:
Connections with Superstructure
Wall steel
Design for compressive loads• Design for moment loads• Design for shear loads
( g) p = 0.01 as minimum reinforcement for the column itself. The number of dowel
Wall steel
Lap joint
Concrete ormasonrycolumn
bars needed is four, these may be placed at the four corners of the column. The d l b ll
Doweldowel bars are usually extended into the footing, bent at the ends, and tied to the main footingthe main footing reinforcement. The dowel diameter shall not exceed the diameter of the longitudinal
Use at least 4 dowels with As at least equal to column steel or 0.005 Ag (col area)
gbars in the column by more than 0.15 in.
28
g
Should not be greater than #11 bars (should have 90o hooks at bottom)ld of dowel (ACI 12.3)
Reinforced Footing Design; Required Development Length ld for Bars in Compression:
l lC
absd ll yb
ybab fd
f
fdl 00030020 .
'.
ld ld-suppliedCriteria:
ld: min. required development length (in)lab: basic development length (in)d : nominal bar diameter (in)
cf
db: nominal bar diameter (in) s: Modifier
As-required / As-provided for excess reinforcement0 75 for spirally enclosed reinforcement0.75 for spirally enclosed reinforcement
fc’: 28 day compressive strength of concrete (psi)
SI units:
In column & footing: 0.0750.0043b y
d b y
c
d fl d f
f
SI units:
29
c
Design for Moment Loads:
Load Transfer from Columns to Footings;
Moment in column causes some dowels to act in tension.
Dowels must be embedded at least one development length in to footing.
'c
bdh
f
dl
1200 English
'c
bdh
f
dl
100 SI
lldh: development length for 90o hooks T ldh
70mm
12db
The development length might dictate the footing thickness.
The number and size of dowels should at least be as the steel in the column.
12db
30
Try using a smaller diameter dowel to reduce the development length
Design for Shear Loads:
Load Transfer from Columns to Footings;
Shear (Vu) in column must be transferred to footing by dowels.
Footing and column poured separately (weak shear plane – cold joint)
Minimum required dowel steel area:
uVA
Vu: applied shear load
: 0.75 (0.85) for shear y
s fA ( )
: 0.6 (cold joint), 1.0 (cold joint but roughened)
Ultimate shear load Vu, can not exceed 0.2 f’c Ac.
Ac and f’c are column area and concrete compressive strength.
31
Foundation Depth:
Embedment depth D:Dmin
(i )Load P(k)
Dmin
(i )Load
P/b (k/ft)
Square & Rectangular Continuous Footings
Embedment depth, D:
Accommodate footing thickness, T
D should be at least > 300mm (12”)
(in)P(k) (in)P/b (k/ft)0-65
65-140
12
18
0-10
10-20
12
18
Depth of frost penetration
Presence of expansive soil
Potential of scour
140-260
260-420
420-650
24
30
36
20-28
28-36
36-44
24
30
36Dmin
(mm)Load P(kN)
Dmin
(mm)Load
P/b (kN/m)
0-300 300 0-170 300
300-500
500-800
800-1100
400
500
600
170-250
250-330
330-410
400
500
600800 00
1100-1500
1500-2000
2000 2700
600
700
800
900
410-490
490-570
570 650
600
700
800
900
32
2000-2700
2700-3500
900
1000
570-650
650-740
900
1000
Example 1 Square footing:21in x 21in square reinforced concrete column carries a vertical load of 380 k and a vertical live load of 270 k It is to be supported on a square spread footing that will be
ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution
vertical live load of 270 k. It is to be supported on a square spread footing that will be founded on a soil with an allowable bearing pressure of 6500 psf. The groundwater table is well below the bottom of the footing. Determine the required width, B, thickness, T, and effective depth, d.
Solution: Unfactored load: P = k
Depth of embedment for P = 650k D = in
W B2 D B2 3 * 150 450 B2Wf = B2 D c = B2 3 * 150 = 450 B2
Df
a uB
WPq
2 06500
450650000 2
B
uq
WPB f
B 06500 uq Da
B = ft (use in)
Factored load: P = 1 2 (380) + 1 6 (270) = kFactored load: Pu = 1.2 (380) + 1.6 (270) = k
Material properties: Use fc’ = 4000 psi, and fy = 60 k/in2
O l li d l d i ti l ( h li d t) l h k t h
33
Only applied load is vertical (no shear or applied moment) only check two-way shear.
Solution 1 (cont.): Two-way shear: bo = 4(c+d) =
bo =
ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution
bo = 4(c+d)Calculate qu = Pu / A = lbs / 1262
qu = psi
c=21inCalculate shear force acting on bo for a square footing:Vuc = Pu – qu (c + d)2
Vuc =
B = 10.5 ft = 126 in
uc
Vuc =
Two-way Nominal shear capacity for square footing:
'4 0 ccnc fdbVV 00044844 ,ddVnc
2012125221 ddVnc ,, 2759939,15 ddVnc
Set V = V =0.75V and solve for d:
34
Set Vuc Vnc 0.75Vnc and solve for d:
Solution 1 (cont.):
Two-way shear:
815 d2 + 18 291 d 863 304 0 d2 + 22 44 d 1 059 27 0O
863,304 – 2,352d – 56 d2 = 15,939d + 759 d2
ACIACI--05 05 ChCh9 9 Factors SolutionFactors Solution
815 d2 + 18,291 d – 863,304= 0
ina
acbbd
2
27.059,11444.2244.22
2
4 22
d2 + 22.44 d – 1,059.27 = 0Or:
Use d = in T = d + db + cover = + 1 + 3 in
Flexural design: For a concrete column: incB
l 5522
21126
2.
2
2
1lBqM uuc
22
inlbMuc
050,724,92
5.5256126 2
'59.019.0 cyysuc ffdfAM
000,60
5901240006090 sAAM
000,4
,
2412659.0124000,609.0 s
suc AM
3,792.86 As2 - 1,296,000 As + 9,724,050 = 0
Or: 2 785632146934169341
35
Or:As
2 – 341.69 As + 2,563.78 = 02
2
68.72
78.563,21469.34169.341inAs
Solution 1 (cont.): Flexural design:Check limiting As: 0.0018 Ag = 0.0018 * B * T
ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution
g s g
= 0.0018 * * = 6.35 in2 As = 7.68 in2 OK
Use 12 # 8 bars: db = 1in, Ab = 0.79 in2 As = 12 * 0.79 = 9.42 in2 > 7.68in2 OK
61262 Use same reinforcement in
Development length: ld-supplied = l – cover = 52.5 – 3 = 49.5 in
Bar spacing: inn
erBspacing 910
112
6126
1
2.
cov
Use same reinforcement in other direction
trc
y
b
d
Kcf
f
d
l '40
3
525253053.
useKc tr
c = cover + 0.5db = 3 + 0.5 = 3.5 in
bd 2.5
5252531
... usedb
inld
ld
d 462814628462852
1111
0004
00060
40
3...
,min
d d
b 52000440 .,min
ld-supplied = 49.5 in > ld-min = 28.6 in OK Lay steel to 3in to cover.
36
Lay steel to 3in to cover.
Solution 1 (cont.): Column Bearing (Load transfer):
sAfP b 1'850 s = (A /A )0 5 < 2 for c + 4d < B;
ACIACI--05 05 ChCh9 9 Factors SolutionFactors Solution
sAfP cnb 185.0 s = (A2/A1)0.5 < 2 for c + 4d < B;
21 + 4*24 = 117 < 126in
A2 = (c+4d)2 = 13,689 in2
s = (A2/A1)0.5 = (13,689 / 212)1/2 = 5.57 > 2Therefore use s = 2
lbsPnb 80099822210004850 2 ,,,. lbP 22094918009982650
nbu PP 000,888lbsPnb 22094918009982650 ,,,,.
Compression bearing OK
Dowels (column footing):Dowels (column-footing): As-min = 0.005 Ag = 0.005 * 212 = 2.205in2
Use 4 # 7 bars db = 0.875in and Ab = 0.6 in2 As = 0.6*4 = 2.4 in2 > As-min = 2.205 in2
OKDowel development length (in compression):Dowel development length (in compression):
yb
c
ybab fd
f
fdl 00030020 .
'.
000608750
Required ld = 16.60 in
37
infdinl ybab 7515000608750000300003060160004
000608750020 .,....
,
,..
Solution 1 (cont.): Dowel development length (in compression): Required ld = 16.60 in
Available length in footing:
ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution
Available length in footing:Footing thickness – cover – 2 (footing bar diameter) – dowel diameterld-available = 28 in – 3 in – 2*1 in – 0.875 in = 22.125 in
l = 22 125” > l = 16 60” OKld-available = 22.125 > ld-required= 16.60 OK
4 # 721 in
Critical section for moment
T = 28 in d = 24 in
B =126 in12 # 8
38
UsingACI – 05 Appendix 5 FactorsFor Solution of Same Examplep
39
Example 1 Square footing:21in2 square reinforced concrete column carries a vertical load of 380 k and a vertical live load of 270 k It is to be supported on a square spread footing that will be founded
ACIACI--05 Appendix C Solution05 Appendix C Solution
live load of 270 k. It is to be supported on a square spread footing that will be founded on a soil with an allowable bearing pressure of 6500 psf. The groundwater table is well below the bottom of the footing. Determine the required width, B, thickness, T, and effective depth, d.
Solution: Unfactored load: P = 380 + 270 = 650 k
Depth of embedment for P = 650k D = 36 in
W B2 D B2 3 * 150 450 B2Wf = B2 D c = B2 3 * 150 = 450 B2
Df
a uB
WPq
2 06500
450650000 2
B
uq
WPB f
B 06500 uq Da
B = 10.36 ft (use 10 ft 6in 126 in)
Factored load: P = 1 4 (380) + 1 7 (270) = 991 kFactored load: Pu = 1.4 (380) + 1.7 (270) = 991 k
Material properties: Use fc’ = 4000 psi, and fy = 60 k/in2
O l li d l d i ti l ( h li d t) l h k t h
40
Only applied load is vertical (no shear or applied moment) only check two-way shear.
Solution 1 (cont.): Two-way shear: bo = 4(c+d) = 4 (21 + d)
bo = 84 + 4d
ACIACI--05 Appendix C Solution05 Appendix C Solution
bo = 4(c+d)Calculate qu = Pu / A = 991,000 lbs / 1262
qu = 62.421 psi
c=21inCalculate shear force acting on bo for a square footing:Vuc = Pu – qu (c + d)2
Vuc = 991,000 – 62.421 (21 + d)2
B = 10.5 ft = 126 in
uc
Vuc = 963,472.3 – 2,621.7d – 62.421 d2
Two-way Nominal shear capacity for square footing:
'4 0 ccnc fdbVV 00044844 ,ddVnc
2012125221 ddVnc ,, 22860206418 ddVnc ..,
Set V = V and solve for d:
41
Set Vuc Vnc and solve for d:
963,472.3 – 2,621.7d – 62.421 d2 = 18,064.2d + 860.2 d2
Solution 1 (cont.):
Two-way shear: 963,472.3 – 2,621.7d – 62.421 d2 = 18,064.2d + 860.2 d2
922 621 d2 + 20 685 9 d 963 472 3 0 d2 + 22 42 d 1 044 28 0O
ACIACI--05 05 Appendix C SolutionAppendix C Solution
922.621 d2 + 20,685.9 d - 963,472.3 = 0
ina
acbbd 9922
2
2804411442224222
2
4 22
..,..
d2 + 22.42 d - 1,044.28 = 0Or:
Use d = 23in T = d + db + cover = 23 + 1 + 3 27 in
Flexural design: For a concrete column: incB
l 5522
21126
2.
2
2
1lBqM uuc
22
inlbM uc
52016839102
55242162126 2
.,,..
'59.019.0 cyysuc ffdfAM
00060
5901230006090,sA
AM
000423126
5901230006090,
.,. ssuc AM
3,792.86 As2 - 1,242,000 As + 10,839,016.52 = 0
Or: 2 748572144632746327
42
Or:As
2 – 327.46 As + 2,857.74 = 02
2
9782
748572144632746327inAs .
.,..
Solution 1 (cont.): Flexural design:Check limiting As: 0.0018 Ag = 0.0018 * B * T
ACIACI--05 Appendix C Solution05 Appendix C Solution
g s g
= 0.0018 * 126 * 27 = 6.124 in2 As = 8.97 in2 OK
Use 12 # 8 bars: db = 1in, Ab = 0.79 in2 As = 12 * 0.79 = 9.42 in2 > 8.97in2 OK
61262 Use same reinforcement in
Development length: ld-supplied = l – cover = 52.5 – 3 = 49.5 in
Bar spacing: inn
erBspacing 910
112
6126
1
2.
cov
Use same reinforcement in other direction
trc
y
b
d
Kcf
f
d
l '40
3
525253053.
useKc tr
c = cover + 0.5db = 3 + 0.5 = 3.5 in
bd 2.5
5252531
... usedb
inld
ld
d 462814628462852
1111
0004
00060
40
3...
,min
d d
b 52000440 .,min
ld-supplied = 49.5 in > ld-min = 28.6 in OK Lay steel to 3in to cover.
43
Lay steel to 3in to cover.
Solution 1 (cont.): Column Bearing (Load transfer):
sAfP b 1'850 s = (A /A )0 5 < 2 for c + 4d < B;
ACIACI--05 05 Appendix C SolutionAppendix C Solution
sAfP cnb 185.0 s = (A2/A1)0.5 < 2 for c + 4d < B;
21 + 4*23 = 113 < 126in
A2 = (c+4d)2 = 12,769 in2
s = (A2/A1)0.5 = (12,769 / 212)1/2 = 5.38 > 2Therefore use s = 2
lbsPnb 80099822210004850 2 ,,,. lbP 22094918009982650
nbu PP 000,991lbsPnb 22094918009982650 ,,,,.
Compression bearing OK
Dowels (column footing):Dowels (column-footing): As-min = 0.005 Ag = 0.005 * 212 = 2.205in2
Use 4 # 7 bars db = 0.875in and Ab = 0.6 in2 As = 0.6*4 = 2.4 in2 > As-min = 2.205 in2
OKDowel development length (in compression):Dowel development length (in compression):
yb
c
ybab fd
f
fdl 00030020 .
'.
000608750
Required ld = 16.60 in
44
infdinl ybab 7515000608750000300003060160004
000608750020 .,....
,
,..
Solution 1 (cont.): Dowel development length (in compression): Required ld = 16.60 in
Available length in footing:
ACIACI--05 Appendix C Solution05 Appendix C Solution
Available length in footing:Footing thickness – cover – 2 (footing bar diameter) – dowel diameterld-available = 27 in – 3 in – 2*1 in – 0.875 in = 21.125 in
l = 21 125” > l = 16 60” OKld-available = 21.125 > ld-required= 16.60 OK
4 # 721 in
Critical section for moment
T = 27 in d = 23 in
B =126 in12 # 8
45