CONCENTRIC COLU Unit Units: s: 1þÿS SI
SI English
Design of Square/Rectangular Square/Rectangular Isolated
P DL = 8 P LL = 8
For SI Units put E = 200000 For English Units put E = 29000000
S T I N U ) I S ( M E T S Y S
E= f'c = fy = Wconc = Wsoil = qa =
200000 20.7 275 23.5 18 144
kN kN M DL = M LL =
MPa MPa MPa kN per cu m kN per cu m kPa
H DL = H LL =
OR
kPa
(OPTIONAL) qe =
h=
SAFE !
Assume T = 200 Shear: SAFE!!! Bending: OK!
mm
Design of Square/Rec Square/Rectangular tangular Isolated
P DL = 185 P LL = 150
For SI Units put E = 200000
. . . E L B A L I A V A T O N S I N O I
For English Units put E = 29000000
E= f'c = fy = Wconc = Wsoil = qa =
29000000 3000 60000 150 100 4000
kips kips M DL = M LL =
ps i psi psi lbs per cu ft lbs per cu ft ps f
H DL = H LL =
OR
(OPTIONAL) qe =
psf h=
SAFE !
Assume T = 24 Shear: SAFE!!! Bending: OK!
in
N Units: 1þÿSI
Footing >
For SI Units For English 3 3
kN-m kN-m
kN kN D = 0.5 m
m
S T I N U ) I S ( M E T S Y S
E= f'c = fy = Wconc = Wsoil = qa = (OPTIONAL) qe =
Footing >
For SI Units
kips-ft kips-ft
kips kips D= 5
ft
ft
Width, S = 0.5
ft
. . . E L B A L I A V A T O N S I N O I
For English
E= f'c = fy = Wconc = Wsoil = qa = (OPTIONAL) qe =
(Optional) Clear Ed
ECCENTRIC COLUMN þÿSI
SI English
Design of Square/Rectangular Isolated Footing
ut E = 200000 nits put E = 29000000 200000 20.7 275 23.5 18 144
MPa MPa MPa kN per cu m kN per cu m kPa
P DL = 38 P LL = 200
kN kN M DL = 1 M LL = 1
kN-m kN-m
H DL = 25 H LL = 222
kN kN
OR
kPa SAFE !
D= 3 h=
m
Assume T = 300 Shear: SAFE!!! Bending: OK!!!
Computed offset: 0 32 m to make the pressure unifo
Design of Square/Rectangular Isolated Footing
P DL = 185 P LL = 150
ut E = 200000 nits put E = 29000000 29000000 3000 60000 150 100 4000
MPa MPa MPa kN per cu m kN per cu m kPa
kN kN M DL = M LL =
kN-m kN-m
H DL = H LL =
kN kN
OR
kPa
D= 5 h=
SAFE !
Assume T = 24 Shear: SAFE!!! Bending: OK! e Distance =
m
mm
Computed offset: 0.32 m to make the pressure unifo
m
offset =
m from the center
m
m
m
m
Design of Square/Rectangular Isolated Footing
P DL = 8 P LL = 8
For SI Units put E = 200000 For English Units put E = 29000000
E= f'c = fy = Wconc = Wsoil = qa =
200000 20.7 275 23.5 18 144 OR (OPTIONAL) qe = 0
kN kN
MPa MPa MPa kN per cu m kN per cu m kPa
M DL = 3 M LL = 3
kN-m kN-m
H DL = 0 H LL = 0
kN kN
kPa
D = 0.5 h= 0
SAFE !
Assume T = 200
m
m
mm
Shear: SAFE!!! Bending: OK!
Width, S = 0.5 C2 = 250
mm 0.25
bar Ø = 16 Concrete cover = 75
m OR
L/S Ratio = 1.2 (OPTIONAL)
mm mm
Remarks: Standard Hook is Required... Allow Clear Spacing = 25 Remarks: OK!!!
mm
C1 = 250
mm 0.25 Computed Length, L = 0.96 m (OPTIONAL) L = 0.96 Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Effective bearing capacity:
qe = qconc = qsoil = qe =
qa - qsoil - qconc 4.700 kPa 5.400 kPa 133.900 kPa
133.9
kPa
Critical Section for One-way Shear X Origin = Scale = 0
Determine the footing dimension:
P= P= M= M=
P DL + P LL 16.000 kN M DL + M LL + (H DL+H LL)(T+h) 6.000 kN-m
e = M/P e = 0.375 q = qe =
m
qmin is positive qmin is negative r is given S is given r is given S is given 133.900 133.900 1.000 0.478 0.000 -32.000 -0.750 -19.200 -72.000 -0.191 -43.200 0.755 0.862 0.951 1.069 L = 0.755 L = 0.951
P(1+6e/L) LS
L = 0.755 S = 0.629 qmax = P(1+6e/L)/(LS) qmax = 133.900
sq m m m kPa
SAFE !
qmin = P(1-6e/L)/(LS) qmin = -66.594
Use: L = 0.951 S = 0.793 Design Loads:
Pu = Pu = Mu = Mu = Assume: eu = eu = qu max =
kPa SAFE ! Tension/Upliftment! m say m say
0.960 0.800
960 800 L/2 = 0.48 m 0.5C1 + d = 0.242 m S/2 = 0.4 m
1.4DL + 1.7LL 24.800 kN 1.4MDL + 1.7MLL + (1.4HDL+1.7HLL)(T+h) 9.300 kN-m e 0.375 m Pu(1+6eu/L)/(LS) qu max = 107.975 kPa qu max = 107.975 kPa qu min = Pu(1-6eu/L)/(LS) qu min = -43.392 kPa qu min = -43.392 kPa
0
Y Origin = -1.5 Scale = 0
-0.13 -0.13 0.13 0.13 -0.13
0 -0.29 -0.29 0 0
-0.5 -0.5 0.5 0.5 -0.5
-0.6 -0.54 -0.75 -0.6 -0.6
-0.5 -0.5 0.5 0.5 -0.5
-0.29 -0.5 -0.5 -0.29 -0.29
-0.68 1
-1.26 -1.26
0.5 1
-1.08 -1.08
-0.5 -0.5 0.5 0.5 -0.5
-1.08 -1.92 -1.92 -1.08 -1.08
-1 1
-1.5 -1.5
-1 -0.5
-1.08 -1.08
-0.13 -0.13 0.13 0.13 -0.13
-1.37 -1.63 -1.63 -1.37 -1.37
-1 -1
-1.08 -1.5
1 1 1
-1.08 -1.26 -1.5
0.25 0.25
-0.12 -2.05
0 0
-0.12 -2.25
0 0.5
-2.25 -2.25
0.5 0.5
-1.92 -2.25
0 0.25 0.5
-2.05 -2.05 -2.05
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
-0.18
Check for One-Way or Direct Shear: (a) Transverse
Eff. d = T - 0.5Ø - cover d = 117.000 b= S b = 800.000 Thus use: a = 0.685 qu max = 107.975 qu min = 0.000
mm
0.12 0.5C1 + d = 242 mm
mm
S/2 = 400 mm
m (compression zone) kPa kPa
(Neglect Upliftment Pressure)
x = L/2-C1/2-d x = 0.238 qs = 37.526 Vu = Vu = Vu = ØVc = ØVc = Check for d =
m kPa
0.5(qs+qumax) S xs 6.926 kN 6,926 N Ø(1/6)√(f'c)bd 60,329 N 36.005 mm
x = 0.238 m qs = 37.526 kPa
SAFE! SAFE!
(b) Longitudinal
Eff. d = d= b= b= y= y= Vu = Vu = Vu = ØVc = ØVc = Check for d =
T - 1.5Ø - cover 101.000 L 960.000
mm
0.1 0.5C2 + d = 0.226 m
mm
S/2-C2/2-d 0.174 m 0.5(qumin + qumax) a y 6.433 kN 6,432.907 N Ø(1/6)√(f'c)bd 62,495 N 21.254 mm
y = 0.174 m
SAFE! SAFE!
Check for Two-Way or Punching Shear:
No failure for this condition! Eff. d 101.000
0.18 0.18 -0.18
-1.68 -1.32 -1.32 -1.68 -1.68
-0.18 -0.18
-1.68 -2.25
0.18 0.18
-1.68 -2.25
-0.18 0.18
-2.25 -2.25
0.18 1
-1.68 -1.68
0.18 1
-1.32 -1.32
1 1
-1.32 -1.68
-1 0.25
-1.5 -1.5
Critical Section for Tw o-way Shear -0.18
mm
C2 + d = 351 mm C1 + d = 351 mm
ßc = 1.000 44 (1/3) 0.33
5.26
1343.08 -76176.55 Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Design of Reinforcement: (a) Long Direction.
Eff. d = T - 1.5Ø - cover d = 101.000 b= S b = 800.000 Thus use: a = 0.685 qu min = 0.000 qu max = 107.975 x= x= qm = Mu = Mu = Rn = Rn = Act p = Min p = Use p = As = As = Ab = n= n= Soc =
c = side cover = c = 0.5Soc = Ktr = (c + Ktr)/db =
mm
Soc-Ø 301.000 1.00 1.00 1.00 0.800 1.00
√f`c 4fy
= Use: ρmin = 0.00509
ρmin =
or 0
=
1.4 fy 0.01
0 0
-0.12 -1.75
-0.5 -0.5
-0.83 -1.08
-0.5 0
-0.95 -0.95
0.5 1
-1.08 -1.08
-0.68 1
-1.37 -1.37
1 1 1
-1.37 -1.5 -1.08
0.13 0.13
-0.12 -2.05
0.13 0.5
-2.05 -2.05
mm
Critical Section for Bending
m (compression zone) kPa kPa
L/2-C1/2 0.355 m 52.001 kPa 4.502 kN-m 4,502,478 N-m Mu Øbd² 0.613 MPa 0.00227 0.00509 0.00509 pbd 411.345 sq mm 201.062 sq mm As/Ab 2.046 say (S - Ø - 2cover)/(n - 1)
Soc = 317.000
Scl = Scl = α= β= αβ = γ= λ=
ρmin =
x = 0.355 m
3
mm
mm
OK!
≤ 1.70 Use: 1.00 for ≤ 19 mm Normal Weight Concrete
83.000 mm 158.500 mm Use: c = 83.000 0.000 5.188 ≤ 2.5 Use: 2.500 9fyαβγλ ld/db = 10√f'c(c+Ktr)/db
mm
Allow Ld = Ld x Mod Factor
ld/db = Ld = Mod. Factor = prov As = Mod. Factor =
17.408 diameters 278.522 mm (req As)/(prov As) 603.186 sq mm 0.682
Allow Ld = 189.940 Actual Ld = 280.000
T - Ø/2 - cover 117.000 L 960.000 S/2-C2/2 0.275
Scl = Scl = Now = Now = Outer width = Outer width = Soc =
mm
Scl = Soc-Ø Scl =
ld/db =
mm m
y = 0.275 m 0.5C1 = 0.25 m
ld/db = Ld = Mod. Factor = prov As = Mod. Factor = Allow Ld =
≤ 2.5 Use: 9fyαβγλ 10√f'c(c+Ktr)/db 17.408 diameters 278.522 mm (req As)/(prov As) 603.186 sq mm 0.948 Ld x Mod Factor
Allow Ld = 264.035
Actual Ld = S/2 - cover Actual Ld = 200.000
2.500
mm
mm
Standard hook is required!
Check for Bearing Stress:
Act Pb = 1.4PDL+1.7PLL Act Pb = 24,800
3
A1= A1= A2= A2= √(A2/A1) = All Pb =
3
0
mm
mm 0
N
C1/C2 = 1.000 L/S = 0.833 S' = 0.800 L' = 0.800
C1 x C2 62,500 sq mm S' x L' 640,000 sq mm 3.200 ≤ 2.0 Use: 2.000 Ø 0.85 f'c A1 √(A2/A1)
All Pb = 1,539,563
mm
Soc-Ø 384.000 mm (n - Nbw)/2 0.000 say (L - Bandwidth - Ø - 2cover)/2 -3.000 mm (Outer Width-cover)/(Now)
Soc =
OK !
Ktr = 0.000 (c + Ktr)/db = 4.688
Mu = 1.398 kN-m Mu = 1,397,956 N-m Mu Rn = Øbd² Rn = 0.118 MPa Act p = 0.00043 Min p = 0.00509 Use p = 0.00509 As = pbd As = 571.811 sq mm Ab = 201.062 sq mm n = As/Ab n = 2.844 say 2N Nbw = ß+1 ß = L/S ß = 1.200 Nbw = 2.585 say Band width = S Band width = 800.000 mm Soc = (Band Width)/(n-1) Soc = 400.000
mm
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
(b) Short Direction.
Eff. d = d= b= b= y= y=
mm
Actual Ld = x - cover
N
Dowels are not Required!
m m
α= β= αβ = γ= λ= c = side cover = c = 0.5Soc =
1.00 1.00 1.00 0.80 1.00 75.000 200.000
≤ 1.70 Use: 1.00 for ≤ 20 mm Normal Weight Concrete mm mm Use: c = 75.000
mm
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Details of Reinforcement: 0.96 x 0.8 (Using 16 mm Ø)
(250 mm x 250 mm) RC-Column/Pedestal Dowels are not Req'd!!! T = 200 mm D = 0.5 m
-0.75 1
-0.44 -0.44
-0.75 -0.75
-0.44 -0.53
-0.42 -0.42
-1.17 -1.83
0.5 0.5
-1.08 -0.75
d = 117 mm Edge dist. = 83 mm S = 0.8 m
-0.75 -0.5
-0.53 -0.53
1 1
-0.44 -0.65
0.42 0.42
-1.17 -1.83
-0.5 0.5
-0.75 -0.75
0 @ 0 mm L = 0.96 m 3 @ 400 mm 3 @ 317 mm f'c = 20.7 MPa fy = 275 MPa Wconc = 23.5 kN per cu m
-0.41 0.41
-0.53 -0.53
-0.41 0.41
-0.56 -0.56
-0.42 -0.42
-1.83 -2.25
-0.41 0.41
-2.3 -2.3
-0.13 -0.13 0.13 0.13 -0.13
0.25 -0.44 -0.44 0.25 0.25
-1 1
-0.13 -0.13
0.42 0.42
-1.83 -2.25
-0.5 -0.5
-1.08 -0.75
-1 1
-0.65 -0.65
0.5 1
-1.17 -1.17
0 0
-0.83 -2.25
-0.5 -0.5 0.5 0.5 -0.5
-0.44 -0.65 -0.65 -0.44 -0.44
-1 -1
-0.13 -0.65
0.5 1
-1.83 -1.83
-1 -0.5
-1.92 -1.92
0.41 0.41
-1.83 -2.25
-1.08 -1.92
-1.17 -1.83 -1.83 -1.17 -1.16
-0.41 -0.41
-1.83 -2.25
-2.25 -2.25 -2.25 -2.25
-1 -1
-0.41 -0.41 0.41 0.41 -0.42
-0.41 -0.42 0.42 0.41
-0.75 0.75
-1.5 -1.5
1 1
-1.17 -1.83
-1 -0.5
-1.08 -1.08
Wsoil = 18 kN per cu m qa = 144 kPa bar Ø = 16 mm Concrete cover = 75 mm Dowels, not required!
3 400
Footing Reinforcement Detail Programmed by:-0.25 Engr. Jeremy E. Caballes Increase =
Design of Square/Rectangular Isolated Footing
P DL = 38 P LL = 200
For SI Units put E = 200000 For English Units put E = 29000000
E= f'c = fy = Wconc = Wsoil = qa =
200000 20.7 275 23.5 18 144 OR
kN kN
MPa MPa MPa kN per cu m kN per cu m kPa
M DL = 1 M LL = 1
kN-m kN-m
H DL = 25 H LL = 222
kN kN
h= 0
m
kPa
(OPTIONAL) qe =
D= 3
SAFE !
Assume T = 300
m
mm 0.3
Shear: SAFE!!! Bending: OK!!! (Optional) Clear Edge Distance =
Computed offset: 0.32 m to make the pressure uniform offset = 0.32 m from the center
m
Width, S = 3
m OR
C2 = 300
mm
L/S Ratio = 0
0.3 bar Ø = 16 Concrete cover = 75 Allow Clear Spacing = 25 Remarks: OK!!!
mm mm mm C1 = 300
mm 0.3 Computed Length, L = 3.99 m
(OPTIONAL) L = 0
m 3.99
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
0.32 Effective bearing capacity:
qe = qconc = qsoil = qe =
qa - qsoil - qconc 7.050 kPa 48.600 kPa 88.350 kPa
Critical Section for One-way Shear 88.35
Determine the footing dimension:
P= P= M= M= M/P = offset = e= e=
P DL + P LL 238.000 kN M DL + M LL + (H DL+H LL)(T+h) 76.100 kN-m 0.320 m 0.320 m M/P - offset 0.000 m
q2 is positive q2 is negative S is given r is given S is given 88.350 88.350 1.000 1.796 0.000 -238.000 0.000 0.000 -456.600 0.000 0.000 0.000 3.989 0.000 1.197 L = 3.989 L = 1.197
X Origin = Scale = 0
P qe
A= A= L= S= q=
r is given
2.694 3.989 3.000 P/A
q = 19.886
sq m m m kPa
SAFE !
kPa
SAFE !
m say m say
3.990 3.000
q2 = P(1-6e/L)/(LS) q2 = 19.886
Use: L = 3.989 S = 3.000 Design Loads:
Pu = Pu = Mu = Mu = Assume: eu = eu = qu =
offset = 0.32 m L/2 + offset = 2.315 m L/2 - offset = 1.675 m 3990 mm 3000 mm 0.5C1 + d = 0.367 m S/2 = 1.5 m
1.4DL + 1.7LL 393.200 kN 1.4MDL + 1.7MLL + (1.4HDL+1.7HLL)(T+h) 126.820 kN-m e 0.000 m Pu/A qu = 32.849 kPa qu = 32.849 kPa qu min = Pu(1-6eu/L)/(LS) qu2 = 32.849 kPa qu = 32.849 kPa
0
X Origin = Scale = 0.01
-2
-0.12 -0.12 -0.04 -0.04 -0.12
0 -0.42 -0.42 0 0
-0.5 -0.5 0.5 0.5 -0.5
-0.75 -1 -1 -0.75 -0.75
-0.5 -0.5 0.5 0.5 -0.5
-0.42 -0.5 -0.5 -0.42 -0.42
-0.68 1
-1.91 -1.91
0.5 1
-1.62 -1.62
-0.5 -0.5 0.5 0.5 -0.5
-1.62 -2.38 -2.38 -1.62 -1.62
-1 1
-2 -2
-1 -0.5
-1.62 -1.62
-0.12 -0.12 -0.04 -0.04 -0.12
-1.96 -2.04 -2.04 -1.96 -1.96
-1 -1
-1.62 -2
1 1 1
-1.62 -1.91 -2
0.01 0.01
-0.25 -2.75
-0.08 -0.08
-0.25 -2.75
-0.08 0.5
-2.75 -2.75
0.5 0.5
-0.75 -2.75
-0.08 0.01 0.5
-2.75 -2.75 -2.75
Pressure Diagram Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Check for One-Way or Direct Shear: (a) Transverse
Eff. d = T - 0.5Ø - cover 0.22 d = 217.000 mm b= S b = 3,000.000 mm Thus use: a = 3.990 m (compression zone) a = 3,990.000 mm qu1 = 32.849 kPa qu2 = 32.849 kPa xs = a - (L/2 - offset + 0.5C1 + d) xs = 1.948 m xs = 1.948 m qs = 32.849 kPa qs = 32.849 kPa Vu = qu xs S Vu = 191.968 Vu = 191.968E+3
kN N
ØVc = Ø(1/6)√(f'c)bd ØVc = 419.598E+3
N
Critical Section for Two-way Shear
SAFE!
-0.14 -0.14 -0.02 -0.02 -0.14
-2.06 -1.94 -1.94 -2.06 -2.06
-0.14 -0.14
-2.06 -2.75
-0.02 -0.02
-2.06 -2.75
-0.14 -0.02
-2.75 -2.75
-0.02 1
-2.06 -2.06
-0.02 1
-1.94 -1.94
1 1
-1.94 -2.06
-1 0.17
-2 -2
(b) Longitudinal
Eff. d = d= b= b= L/2 - offset = y= y= Vu = Vu = Vu =
T - 1.5Ø - cover 201.000 mm L 3,990.000 mm 1,675.000 m
0.2
S/2 - C2/2 - d 1.149 qu y L 150.596 150.596E+3
0.5C2 + d = 0.351 m y = 1.149 m
ØVc = Ø(1/6)√(f'c)bd ØVc = 516.918E+3
m kN N
N
SAFE!
Check for Two-Way or Punching Shear:
Eff. d = 201.000 mm bo = 2(c1+d)+2(c2+d) bo = 2,004.000 mm Vu = qu[LS-(c1+d)(c2+d)] Vu = 384.955 kN Vu = 384.955E+3 N ØVc = Ø((1/3))√(f'c)bod ØVc = 610.883E+3 N Check for d = 162.082 mm
C2 + d = 501 mm C1 + d = 501 mm
ßc = 1.000 44 (1/3) 0.33 SAFE! SAFE!
5.19 1566.62 -390243.61
d/2 = C1/2 = (C1 + d)/2 = Pressure Diagram L/2 - offset = Difference =
100.5 150 250.5 1675 -1,424.500
mm mm mm mm mm
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Design of Reinforcement:
Eff. d = T - 1.5Ø - cover d = 201.000 b= S b = 3,000.000 Thus use: a = 3.990 qu2 = 32.849 qu1 = 32.849 xa = xa = qa = xb = xb = qb =
0.2 mm
Rn = Act p = Min p = Use p = As = As = Ab = n= n=
0.706 MPa 0.00262 0.00509 0.00509 pbd 4,407.862 sq mm 201.062 sq mm As/Ab 21.923 say 2n ß+1 L/S 1.330 18.818 say 3.000 say 320.000 mm 3,840.000 mm L/2 - offset - S/2 - 0.5Ø - cover
ß= ß= Nbw = Now = Offset = L - 2cover = Left OW =
Left OW = 92.000
n = 0.447 Soc = 0.000 Scl = 0.000
0
=
22
c-c (OW)
c-c (OW)
119.435 119.435E+6 0.706 0.00262 0.00509 0.00509 624
16
19
4
4
S 3,000.000
BW 3,000.000
OW 990
OW 990.000
Soc = 118.000
188.000
166.000
206.000
Scl = 102.000
172.000
150.000
190.000
OK!
OK!
OK!
OK!
1.0 1.0 1.0 1.0 0.80 1.0 75.0 94.000 75.000 0.000 4.688 2.500
1.0 1.0 1.0 1.0 0.80 1.0 75.0 83.000 75.000 0.000 4.688 2.500
1.0 1.0 1.0 1.0 0.8 1.0 75.0 103.000 75.000 0.000 4.688 2.500
1.00 1.00 1.00 1.00 0.80 1.00 75.000 103.000 75.000 0.000 4.688 2.500
α = 1.00 β = 1.00 αβ = 1.00 αβ ≤ 1.70 Use: 1.00 γ = 0.80 λ = 1.00 c = side cover = 75.000 c = 0.5Soc = 59.000 Use: c = 59.000 Ktr = 0.000 (c + Ktr)/db = 3.688 (c + Ktr)/db ≤ 2.5 Use: 2.500 9fyαβγλ Ld/db = 17.408 10√f'c(c+Ktr)/db Ld = 278.522 prov As = 5,027 Mod. Factor = (req As)/(prov As) = 0.987 Allow Ld = Ld x Mod Factor 274.922
19
c-c (BW)
119.435 119.435E+6 0.706 0.00262 0.00509 0.00509 3,784
Remarks: OK!
mm
b-b
0.01
114.591 114.591E+6 1.050 0.00394 0.00509 0.00509 3,070
Width = S Width = 3,000.000
mm
sq mm
mm mm
Normal Wt mm mm mm
kN-m N-mm MPa
sq mm
206.000
mm
190.000
mm
Normal Wt mm mm mm
17.408
17.408
diameters
17.408
278.522 3,820 0.990
278.522 804 0.776
mm sq mm
278.522 804.248 0.776
mm sq mm
mm
216.200
mm
275.854
216.200
xb - cover
y - cover
y - cover
Actual Ld = 2,090.000
1,450.000
1,275.000
1275.000
No Hooks!
Right OW = 732.000
mm
n = 3.553 Soc = 244.000 Scl = 228.000
say mm mm
diameters
y - cover mm
No Hooks! No Hooks!
Right OW = L - BW - Left OW - Ø - 2cover 0
119.435 119.435E+6 0.706 0.003 0.005 0.005 624.289
17.408
265.780
No Hooks!
kN-m N-mm MPa
278.522 3,217 0.954
Actual Ld = xa - cover Remarks:
mm
say mm mm
230.954 230.954E+6 2.117 0.00823 0.00509 0.00823 4,962
n = 25
m
T - 0.5Ø - cover 217.000 L 3,990.000
Mu = Mu = Rn = Act p = ρmin = Use p = As =
1.4 fy
ρmin =
or
a-a
m (compression zone) kPa kPa
Eff. d = d= b= b=
Nbw =
= Use: ρmin = 0.00509
mm
a - (L/2 - offset + 0.5C1) 2.165 m 32.849 kPa L/2 - offset - C1/2 1.525 m 32.849 kPa
yc = S/2-C2/2 yc = 1.350
√f`c 4fy
ρmin =
3
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
1,275.000 No Hooks!
mm
Check for Bearing Stress:
Act Pb = 1.4PDL+1.7PLL N C1/C2 = C1 x C2 L/S = 90,000 sq mm S' = S' x L' L' = 9,000,000 sq mm 10.000 ≤ 2.0 Use: 2.000 Ø 0.85 f'c A1 √(A2/A1)
Act Pb = 393,200
A1= A1= A2= A2= √(A2/A1) = All Pb =
1.000 0.752 3.000 3.000
xa = 2.165 m qa =Critical 32.849 kPaSection 230.95 xb = 1.525 m qb = 32.849 kPa 114.59 0.5C2 = 0.15 m yc = 1.35 m 119.43
for Bending
All Pb = 2,216,970 N Dowels are not Required!
-0.08 -0.08
-0.25 -2.25
0 0
-1.25 -2.25
-0.5 -0.5
-1.25 -1.62
-0.5 -0.08 0
-1.25 -1.25 -1.25
-0.68 1
-1.96 -1.96
1 1 1
-1.62 -1.96 -2
-0.04 -0.04
-0.25 -2.75
-0.04 0.5
-2.75 -2.75
-0.12 -0.12
-0.25 -2.75
-0.5 -0.5
-1 -2.75
-0.5 -0.12
-2.75 -2.75
0.5 1
-1.62 -1.62
Pressure Diagram Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Details of Reinforcement: 3.99 m x 3 m (Using 16 mm Ø)
(300 mm x 300 mm) RC-Column/Pedestal Dowels, not required! T = 300 mm D=3m d = 217 mm Edge dist. = 83 mm S=3m
b Center Right
a
16 @ 188 mm 19 @ 166 mm 3 @ 244 mm L = 3.99 m offset = 0.32 m L/2 - offset = 1.675 m L/2 = 1.995 m 25 @ 118 mm f'c = 20.7 MPa fy = 275 MPa Wconc = 23.5 kN per cu m Wsoil = 18 kN per cu m qa = 144 kPa bar Ø = 16 mm Concrete cover = 75 mm
Left
Footing Reinforcement Detail Increase = -0.2 Delta y = By: -0.75 Engr. Jeremy E. Caballes -0.27 -0.75 -0.27
0.75 0.5 0.9
-1.61 -1.61
-0.5 -0.5 0.5 0.5 -0.5
-0.87 -1.63 -1.63 -0.87 -0.87
-0.48 -0.46 0.3 0.48
-2 -2 -2 -2
-0.12 -0.12 -0.04 -0.04 -0.12
-1.21 -1.29 -1.29 -1.21 -1.21
0.9
-0.27
-0.75
-0.32
-0.75 -0.5
-0.32 -0.32
0.9 0.9
-0.27 -0.35
-0.48 0.48
-0.32 -0.32
-0.48 0.48
-0.33 -0.33
-0.12 -0.12 -0.04 -0.04 -0.12
0.25 -0.27 -0.27 0.25 0.25
-1 1
0.4 0.4
-1 0.9
-0.35 -0.35
-1 -1
-0.87 -1.63
-1 -0.5
-1.63 -1.63
-0.5 -0.5 0.5 0.5 -0.5
-0.27 -0.35 -0.35 -0.27 -0.27
-1 -1
0.4 -0.35
-0.75 0.75
-1.25 -1.25
-1 -0.5
-0.87 -0.87
0.48 0.48
-1.63 -2
0 0
-0.55 -1.78
-0.48 -0.48 0.48 0.48 -0.48
-0.89 -1.61 -1.61 -0.89 -0.89
-0.48 -0.48
-1.63 -2
0.3 0.3
-0.87 -2
0.9 0.9
-0.89 -1.61
-0.46 -0.46
-0.87 -2
0.5 0.5
-0.87 -0.4
-0.5 -0.08 0 0.5
-0.55 -0.55 -0.55 -0.55
0.3 0.3
-0.87 -2
-0.5 0.5
-0.4 -0.4
-0.8 -0.5
-0.89 -0.89
-0.46 -0.46
-1.63 -2
-0.08 -0.08
-0.55 -1.78
-0.8 -0.5
-1.61 -1.61
0.3 0.3
-1.63 -2
-0.5 -0.5
-0.87 -0.4
-0.8 -0.8
-0.89 -1.61
0.5 0.9
-0.89 -0.89
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
