Combined-strap-footing.xlsx 2hj4i
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Design of Combined Strap Footing
E= f'c = fy = Wconc = Wsoil = qa =
P1 DL = P1 LL = M1 DL = M1 LL =
200,000.00 MPa 27.50 MPa 345.00 MPa 23.50 kN per cu m 16.50 kN per cu m 200.00 kPa OR 150 kPa
qe =
285 kN 225 kN 85 kN-m 55 kN-m
P2 DL = P2 LL = M2 DL = M2 LL =
350 kN 300 kN 120 kN-m 95 kN-m
A Assume T =
600 mm 0.6
Remarks: SAFE! Distance from the Property Line =
D=
A
m L1 =
Column Spacing = 1.75 m
4
m
L2
C1y = B1
1.5 m
2.2 m
Strap Beam Properties: Section A-A Height = 800 mm Width = 600 mm Remarks: SAFE! 0.8 0.6
500 mm 0.5
C2y =
600 mm 0.6 B2
Property Line
Bar diameter = Concrete cover = Temperature bar Ø =
25 mm 0.025 75 mm 0.075 16 mm
C1x =
Ratio, L1/B1 =
500 mm 0.5
C2x =
(Optional)
L2/B2 =
600 mm 0.6 1 (Optional)
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005 ∑M at R1 = 0, CW is (+) 0 = P2(X2 - L1/2) - P1(L1/2 - X1) - R2(X2 - L1/2) + M1 + M2 R2 = 660.741 kN
A1 = R1/qe A1 = 3.328 A2 = R2/qe
sq m
R1 = P1 + P2 - R2 R1 = 499.259
kN
A2 = 4.405 A = A1 + A2
sq m
Solve for Location of its Resultant: P1 = P1 DL + P2 LL P1 = 510.000 kN P2 = P2 DL + P2 LL P2 = 650.000 kN M1 = M1 DL + M2 LL M1 = 140.000 kN-m M2 = M2 DL + M2 LL M2 = 215.000 kN-m X1 = 0.250 m X2 = 4.250 m P = P1 + P2 P = 1,160.000 kN P(Xc) = P1(X1) + P2(X2) + M1 +M2 Xc = 2.797 m L = 5.300 m
A = 7.733 Design loads:
X1 ≤ Xc ≤ X2, OK!
Determine the Widths: qe = qa - qs - qc qs = Wsoil x (D - T) qs = 14.850 kPa qc = Wconc x T qc = 14.100 kPa Therefore: qe = 150.000 kPa Thus: A = (P/qe)(1+6e/L) A = 7.733 sq m A = A1 + A2 P/qe = A1 + A2 A2 = P/qe - A1 A.Xc = A1(L1/2) + A2X2 A.Xc = A1(L1/2) +(P/qe - A1)X2 A1 = 3.328 sq m L1 B1 = 3.328 sq m Since: L1 = 1.75 say 1.750 Then: B1 = 1.902 say 1.910 Check: L1/B1 = 0.916 0.75 ≤ L/B ≤ 1.0, OK! A2 = 4.405 sq m L2 B2 = 4.405 L2 = 2.099 say 2.100 B2 = 2.099 say 2.100 Check: L1/B1 = 1.000 0.75 ≤ L/B ≤ 1.0, OK! L = 5.300 m
1.000 -8.500 0.000 0.000 L1 = 8.500 B1 = #DIV/0! m m
m m
sq m
Pu1 = 1.4(P1 DL) + 1.7(P1 LL) Pu1 = 781.500 kN wp1 = 1563 kN/m, Mu1 = 212.5 kN-m Pu2 = 1.4(P2 DL) + 1.7(P2 LL) Pu2 = 1,000.000 kN wp2 = 1666.667 kN/m, Mu1 = 329.5 kN-m Pu = Pu1 + Pu2 Pu = 1,781.500 kN Mu1 = 1.4(M1 DL) + 1.7(M1 LL) Mu1 = 212.500 kN-m Mu2 = 1.4(M2 DL) + 1.7(M2 LL) Mu2 = 329.500 kN-m ∑M at Ru1 = 0, CW is (+) 0 = Pu2(X2 - L1/2) - Pu1(L1/2 - X1) - Ru2(X2 - L1/2) + Mu1 + Mu2 Ru2 = 1,015.870 kN Ru1 = Pu1 + Pu2 - Ru2 Ru1 = 765.630 kN qu1 = Ru1/A1 = 229.059 kPa wua = qu1 x B1 wu1 = 437.503 kN/m Soil = 437.503 kN/m qu2 = Ru2/A2 = 230.356 kPa wub = qu2 x B2 wu2 = 483.748 kN/m Soil = 483.748 kN/m Distance (m) xa = X1 - 0.5C1x xa = 0.000 xb = X1 + 0.5C1x xb = 0.500 xc = L1 xc = 1.750 xd = L - L2 xd = 3.200 xe = X2 - 0.5C2x xe = 3.950 xf = X2 + 0.5C2x xf = 4.550 Stirrups for Strap (10 - mm Ø): 1 @ 50 mm, rest @ 256 mm symmetrical with respect to it's midspan f'c = 27.5 MPa Temperature bar Ø = 16 mm fy = 345 MPa L1 = 1.75 m Wconc = 23.5 kN per cu m B1 = 1.91 m Wsoil = 16.5 kN per cu m L2 = 2.1 m qa = 200 kPa B2 = 2.1 m T = 600 mm D = 1.5 m Edge Dist. = 0.25 m d = 512.5 mm Col Dist. = 4 m d = 712.5 mm Edge Dist. = 1.05 m Bottom 5-16 @ 200 mm Strap: 800 mm x 600 mm Bottom 9-25 @ 243 mm Ext. Col: 500 mm x 500 mm Bottom 8-25 @ 228 mm Int. Col: 600 mm x 600 mm Bottom 9-25 @ 243 mm Bar diameter = 25 mm Top 4-25, Top Concrete cover = 75 mm Top 4-25, Top Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Check for One-Way or Direct Shear: d = T - 0.5Ø - cover = 512.5 mm d = 0.5125 m
None
x-factor = 0.943396226 Origin = -1.500 Footing 1: Footing 2: 1.651 -1.783 3.019 -1.217 1.651 -2.401 3.019 -0.509 0.000 -2.401 5.000 -0.509 xa = X1 - C1x/2 - d Ve = qu1 (B1/2 - ye) (L1) 0.000 -0.599 5.000 -2.491 xb = L1 Ve = 57.122 kN 1.651 -0.599 3.019 -2.491 xc = L - L2 Vf = qu2 (B2/2 - yf) (L2) 1.651 -1.217 3.019 -1.783 xd = X2 + C2x/2 + d Vf = 114.890 kN 0.000 1.066 3.019 1.066 ye = B1/2 - C1y/2 - d 1.651 1.066 5.000 1.066 yf = B2/2 - C2y/2 - d 1.651 0.500 5.000 0.500 0.000 0.500 3.019 0.500 Longitudinal Transverse 0.000 1.066 3.019 1.066 Section: a-a d-d e-e f-f Column 1: Column 2: Distance, x, y, (m) 0.000 5.063 0.813 0.813 0.000 1.066 Vu (kN) 000.000E-2 -114.890E+0 571.216E-1 114.890E+0 0.000 -1.264 2.066 -1.217 b (mm) 1910 2100 1750 2100 0.472 -1.264 2.066 -1.217 d (mm) 512.5 512.5 512.5 512.5 0.472 -1.736 1.255 -1.783 ØVc = Ø(1/6)√(f'c)bd, (kN) 727.212E+0 799.553E+0 666.294E+0 799.553E+0 0.000 -1.736 1.255 -1.783 Remarks None SAFE! SAFE! SAFE! 0.000 -1.264 2.066 -1.217 Press fac = 0.001 Load Diagram Origin = Strap Beam SAFE! 0.000 -4.000 3.019 -4.000 Section: b-b c-c SAFE! 1.651 -4.000 5.000 -4.000 Distance, y (m) 1.750 3.200 SAFE! 1.651 -4.263 5.000 -4.290 Vu (kN) -158.704E-1 -158.704E-1 SAFE! 0.000 -4.263 3.019 -4.290 b (mm) 600 600 SAFE! 0.000 -4.000 3.019 -4.000 d (mm) 712.5 712.5 SAFE! 0.000 -3.062 3.726 -3.000 ØVc = Ø(1/6)√(f'c)bd, (kN) 317.592E+0 317.592E+0 SAFE! 0.472 -3.062 4.292 -3.000 Remarks SAFE! SAFE! 0.472 -4.000 4.292 -4.000 0.000 -4.000 3.726 -4.000 Vc = (1/6)√(f'c)bd 0.000 -3.062 -3.000 Load Diagram 3.726 = 373.638152161 kN > Vu = -15,87 kN Tie Beam: Stirrups are not Required! 0.000 -1.264 0.000 -1.736 Using: d = 10 mm 0.000 -1.217 0.000 -1.783 S = Smax = d/2 4.292 -1.217 4.292 -1.783 = 256.25 mm 4.292 -1.217 4.292 -1.783 0.472 1.255 1.651 She ar & M om e nt Diagram 0.500 3.726 1.255 3.019 0.500 Property Line a-a 0.000 2.000 0.000 2.000 0.000 -6.250 0.000 -6.250 dx = dy = 0.250 e-e f-f -0.250 -0.733 2.769 -0.733 Use: S = 256.250 mm 0.000 -0.733 3.019 -0.733 1.651 -0.733 5.000 -0.733 1.901 -0.733 5.250 -0.733
3.726 3.726 4.292 4.292 3.726 3.726 -5.500
0.000 5.000 0.000 5.000
b-b 1.651 1.651
1.066 2.066 2.066 1.066
Datum: -4.000 -4.000 -5.500 -5.500
c-c 3.019 3.019
d-d 4.776 4.776
Critical Section for One-Way or Direct Shear
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Check for Two-Way or Punching Shear: (a) Exterior Column: xa = X1-C1x/2-d/2 xb = X1+C1x/2+d/2 x1 = 0.756 m y1 = 1.013 m Vu = Pu1 - qu1(x1)(y1) Vu = 529.201 kN bo = 2.525 m ØVc = Ø(1/3)√(f'c)bod ØVc = 1,922.734 kN (b) Interior Column: Xc = X2-C2x/2-d/2 Xd = X2+C2x/2+d/2 x2 = 1.113 m y2 = 1.113 m Vu = Pu2 - qu2(x2)(y2) Vu = 714.898 kN bo = 4.450 m ØVc = Ø(1/3)√(f'c)bod ØVc = 3,388.581 kN
0.000 1.250
xa = 0.000 xb = 0.756 2 x1 = 0.756 m 1 y1 = 1.013 m
m m
ßc = 1.000 44 (1/3) 0.3333333333 SAFE!
0.750 0.750
Xc = 3.694 Xd = 4.806 2 x2 = 1.113 m 2 y2 = 1.113 m ßc = 1.000 44 (1/3) 0.3333333333 SAFE!
m m
Property Line a-a 0.000 2.000 0.000 -3.500 0.000 -6.250 0.000 -6.250 Punching Shear Failure Exterior Interior 0.000 -1.978 3.485 -2.025 0.713 -1.978 4.534 -2.025 0.713 -1.022 4.534 -0.975 0.000 -1.022 3.485 -0.975 0.000 -1.978 3.485 -2.025 Dimensions Exterior Interior -0.250 -1.022 4.534 -0.975 0.000 -1.022 5.250 -0.975 -0.250 -1.978 4.534 -2.025 0.000 -1.978 5.250 -2.025 -0.250 -1.022 5.250 -0.975 -0.250 -1.978 5.250 -2.025 Exterior Interior 0.000 0.000 3.485 0.000 0.000 -1.022 3.485 -0.975 0.713 0.000 4.534 0.000 0.713 -1.022 4.534 -0.975 0.000 0.000 3.485 0.000 0.713 0.000 4.534 0.000
b-b 0.713 0.713
c-c 3.485 3.485
d-d 4.534 4.534
Load Diagram
Shear & M om e nt Diagram
Critical Section for Two-Way or Punching Shear Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Design of Reinforcement: d = T - 0.5Ø - cover = 512.500 mm = 712.500 mm 4√f`c ρmin = fy = 0.00380 Use ρmin = 0.00406 Ab = πd² / 4 = 490.874 Mu = 0.000 Section Mu = b= d= Rn = act p = Use: p = As = n= Soc = Scl = Location =
Section Mu = b= d= Rn = act p = Use: p = As = n= Soc = Scl = Location =
Footing Strap or
1.4 fy = 0.00406
ρmin =
sq mm -289.824
-312.836 Longitudinal a b c 000.000E-2 289.824E+6 312.836E+6 1,910.000 600.000 600.000 512.500 712.500 712.500 0.00000 1.05723 1.14118 0.00000 0.00314 0.00339 0.00406 0.00406 0.00406 3,972.246 1,734.783 1,734.783 9 4 4 219.000 149.000 149.000 194.000 124.000 124.000 None Top Top 164.756 244.897 Transverse e f 164.756E+6 244.897E+6 ye = 1,750.000 2,100.000 yf = 512.500 512.500 yg = 0.39826 0.49333 yh = 0.00116 0.00145 0.00406 0.00406 3,639.493 4,367.391 8 9 228.000 243.000 203.000 218.000 Bottom Bottom
Temperature/Shrinkage Reinforcement: Ø = 16 mm Ast = 0.002bh consider: b = 1000 mm Ast = 1,200.000 sq mm n = 5.968 say 6 Soc = 200.000 mm Scl = 184.000 mm > 25 mm OK! Ext Footing:Le = 1.135 m 5 Int Footing:Le = 1.325 m 6
136.054 Property Line 0.000 2.000 0.000 -6.250 e-e f-f 3.726 4.292 3.726 4.292
d 136.054E+6 2,100.000 512.500 0.27407 0.00080 0.00406 4,367.391 9 243.000 218.000 Bottom
½b of Strap ½b of Strap 0.300 0.300
a-a 0.000 0.000
g-g -0.250 -0.250 1.901 1.901
m m m m
2.000 -6.250
b-b 0.472 0.472
h-h -0.967 -1.217 -1.217 -0.967
2.769 2.769 5.250 5.250
-0.967 -1.217 -1.217 -0.967
Load Diagram
She ar & M om e nt Diagram
Critical Section for Bending
c-c 1.651 1.651
d-d 3.019 3.019
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Reinforcements
Edge Dist = 0.08254717 Strap 1.651 0.583 3.019 0.583 0.071 1.172 4.222 1.172
Reinforcement Details of3.101 Combined Strap Footing 0.083 0.583 0.583 1.568 0.583 4.917 0.583 By: Engr. Jeremy E. Caballes Exterior
Given:
Interior
Effective Depth Total Depth 0.000 1.255 5.000 1.066 -1.000 1.255 5.250 1.066 Strap: 800 m m x 600 m m 0.000 0.583 5.000 0.500 -1.000 0.583 5.250 0.500 -1.000 1.255 5.250 1.066 -1.000 0.583 5.250 0.500 -0.500 1.066 0.000 1.066 -0.500 0.583 0.000 0.583 -0.500 1.066 -0.500 0.583 Longitudinal Dimensions 0.000 -0.509 0.236 4.009 5.000 0.000 0.000 0.000 0.500 0.236 4.009 5.000 0.236 0.000 Perimeter Dimension 4.009 0.000 Exterior Interior 5.000 0.000 0.000 -0.599 5.000 -0.509 -1.000 -0.599 6.000 -0.509 0.000 -2.401 5.000 -2.491 -1.000 -2.401 6.000 -2.491 -1.000 -0.599 6.000 -0.509 -1.000 -2.401 6.000 -2.491 Exterior Interior Reinforcement Perimeter Dimension 0.000 -4.000 3.019 -4.000 Exterior Interior 0.000 -2.401 3.019 -2.491 -0.500 -0.682 4.917 -0.592 1.651 -4.000 5.000 -4.000 0.083 -0.682 5.500 -0.592 1.651 -2.401 5.000 -2.491 -0.500 -2.318 4.917 -2.408 0.000 -4.000 3.019 -4.000 0.083 -2.318 5.500 -2.408 1.651 -4.000 5.000 -4.000 -0.500 -0.682 5.500 -0.592 Reinforcement Perimeter -0.500 -2.318 5.500 -2.408 Exterior Interior Exterior Interior 1.568 -1.783 3.101 -1.217 0.083 -3.500 3.101 -3.500 1.568 -2.318 3.101 -0.592 0.083 -2.318 3.101 -2.408 0.083 -2.318 4.917 -0.592 1.568 -3.500 4.917 -3.500 0.083 -0.682 4.917 -2.408 1.568 -2.318 4.917 -2.408 1.568 -0.682 3.101 -2.408 0.083 -3.500 3.101 -3.500 1.568 -1.217 3.101 -1.783 1.568 -3.500 4.917 -3.500
E= f'c = fy = Wconc = Wsoil = qa =
P1 DL = P1 LL = M1 DL = M1 LL =
200,000.00 MPa 27.50 MPa 345.00 MPa 23.50 kN per cu m 16.50 kN per cu m 200.00 kPa OR 150 kPa
qe =
285 kN 225 kN 85 kN-m 55 kN-m
P2 DL = P2 LL = M2 DL = M2 LL =
350 kN 300 kN 120 kN-m 95 kN-m
A Assume T =
600 mm 0.6
Remarks: SAFE! Distance from the Property Line =
D=
A
m L1 =
Column Spacing = 1.75 m
4
m
L2
C1y = B1
1.5 m
2.2 m
Strap Beam Properties: Section A-A Height = 800 mm Width = 600 mm Remarks: SAFE! 0.8 0.6
500 mm 0.5
C2y =
600 mm 0.6 B2
Property Line
Bar diameter = Concrete cover = Temperature bar Ø =
25 mm 0.025 75 mm 0.075 16 mm
C1x =
Ratio, L1/B1 =
500 mm 0.5
C2x =
(Optional)
L2/B2 =
600 mm 0.6 1 (Optional)
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005 ∑M at R1 = 0, CW is (+) 0 = P2(X2 - L1/2) - P1(L1/2 - X1) - R2(X2 - L1/2) + M1 + M2 R2 = 660.741 kN
A1 = R1/qe A1 = 3.328 A2 = R2/qe
sq m
R1 = P1 + P2 - R2 R1 = 499.259
kN
A2 = 4.405 A = A1 + A2
sq m
Solve for Location of its Resultant: P1 = P1 DL + P2 LL P1 = 510.000 kN P2 = P2 DL + P2 LL P2 = 650.000 kN M1 = M1 DL + M2 LL M1 = 140.000 kN-m M2 = M2 DL + M2 LL M2 = 215.000 kN-m X1 = 0.250 m X2 = 4.250 m P = P1 + P2 P = 1,160.000 kN P(Xc) = P1(X1) + P2(X2) + M1 +M2 Xc = 2.797 m L = 5.300 m
A = 7.733 Design loads:
X1 ≤ Xc ≤ X2, OK!
Determine the Widths: qe = qa - qs - qc qs = Wsoil x (D - T) qs = 14.850 kPa qc = Wconc x T qc = 14.100 kPa Therefore: qe = 150.000 kPa Thus: A = (P/qe)(1+6e/L) A = 7.733 sq m A = A1 + A2 P/qe = A1 + A2 A2 = P/qe - A1 A.Xc = A1(L1/2) + A2X2 A.Xc = A1(L1/2) +(P/qe - A1)X2 A1 = 3.328 sq m L1 B1 = 3.328 sq m Since: L1 = 1.75 say 1.750 Then: B1 = 1.902 say 1.910 Check: L1/B1 = 0.916 0.75 ≤ L/B ≤ 1.0, OK! A2 = 4.405 sq m L2 B2 = 4.405 L2 = 2.099 say 2.100 B2 = 2.099 say 2.100 Check: L1/B1 = 1.000 0.75 ≤ L/B ≤ 1.0, OK! L = 5.300 m
1.000 -8.500 0.000 0.000 L1 = 8.500 B1 = #DIV/0! m m
m m
sq m
Pu1 = 1.4(P1 DL) + 1.7(P1 LL) Pu1 = 781.500 kN wp1 = 1563 kN/m, Mu1 = 212.5 kN-m Pu2 = 1.4(P2 DL) + 1.7(P2 LL) Pu2 = 1,000.000 kN wp2 = 1666.667 kN/m, Mu1 = 329.5 kN-m Pu = Pu1 + Pu2 Pu = 1,781.500 kN Mu1 = 1.4(M1 DL) + 1.7(M1 LL) Mu1 = 212.500 kN-m Mu2 = 1.4(M2 DL) + 1.7(M2 LL) Mu2 = 329.500 kN-m ∑M at Ru1 = 0, CW is (+) 0 = Pu2(X2 - L1/2) - Pu1(L1/2 - X1) - Ru2(X2 - L1/2) + Mu1 + Mu2 Ru2 = 1,015.870 kN Ru1 = Pu1 + Pu2 - Ru2 Ru1 = 765.630 kN qu1 = Ru1/A1 = 229.059 kPa wua = qu1 x B1 wu1 = 437.503 kN/m Soil = 437.503 kN/m qu2 = Ru2/A2 = 230.356 kPa wub = qu2 x B2 wu2 = 483.748 kN/m Soil = 483.748 kN/m Distance (m) xa = X1 - 0.5C1x xa = 0.000 xb = X1 + 0.5C1x xb = 0.500 xc = L1 xc = 1.750 xd = L - L2 xd = 3.200 xe = X2 - 0.5C2x xe = 3.950 xf = X2 + 0.5C2x xf = 4.550 Stirrups for Strap (10 - mm Ø): 1 @ 50 mm, rest @ 256 mm symmetrical with respect to it's midspan f'c = 27.5 MPa Temperature bar Ø = 16 mm fy = 345 MPa L1 = 1.75 m Wconc = 23.5 kN per cu m B1 = 1.91 m Wsoil = 16.5 kN per cu m L2 = 2.1 m qa = 200 kPa B2 = 2.1 m T = 600 mm D = 1.5 m Edge Dist. = 0.25 m d = 512.5 mm Col Dist. = 4 m d = 712.5 mm Edge Dist. = 1.05 m Bottom 5-16 @ 200 mm Strap: 800 mm x 600 mm Bottom 9-25 @ 243 mm Ext. Col: 500 mm x 500 mm Bottom 8-25 @ 228 mm Int. Col: 600 mm x 600 mm Bottom 9-25 @ 243 mm Bar diameter = 25 mm Top 4-25, Top Concrete cover = 75 mm Top 4-25, Top Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Check for One-Way or Direct Shear: d = T - 0.5Ø - cover = 512.5 mm d = 0.5125 m
None
x-factor = 0.943396226 Origin = -1.500 Footing 1: Footing 2: 1.651 -1.783 3.019 -1.217 1.651 -2.401 3.019 -0.509 0.000 -2.401 5.000 -0.509 xa = X1 - C1x/2 - d Ve = qu1 (B1/2 - ye) (L1) 0.000 -0.599 5.000 -2.491 xb = L1 Ve = 57.122 kN 1.651 -0.599 3.019 -2.491 xc = L - L2 Vf = qu2 (B2/2 - yf) (L2) 1.651 -1.217 3.019 -1.783 xd = X2 + C2x/2 + d Vf = 114.890 kN 0.000 1.066 3.019 1.066 ye = B1/2 - C1y/2 - d 1.651 1.066 5.000 1.066 yf = B2/2 - C2y/2 - d 1.651 0.500 5.000 0.500 0.000 0.500 3.019 0.500 Longitudinal Transverse 0.000 1.066 3.019 1.066 Section: a-a d-d e-e f-f Column 1: Column 2: Distance, x, y, (m) 0.000 5.063 0.813 0.813 0.000 1.066 Vu (kN) 000.000E-2 -114.890E+0 571.216E-1 114.890E+0 0.000 -1.264 2.066 -1.217 b (mm) 1910 2100 1750 2100 0.472 -1.264 2.066 -1.217 d (mm) 512.5 512.5 512.5 512.5 0.472 -1.736 1.255 -1.783 ØVc = Ø(1/6)√(f'c)bd, (kN) 727.212E+0 799.553E+0 666.294E+0 799.553E+0 0.000 -1.736 1.255 -1.783 Remarks None SAFE! SAFE! SAFE! 0.000 -1.264 2.066 -1.217 Press fac = 0.001 Load Diagram Origin = Strap Beam SAFE! 0.000 -4.000 3.019 -4.000 Section: b-b c-c SAFE! 1.651 -4.000 5.000 -4.000 Distance, y (m) 1.750 3.200 SAFE! 1.651 -4.263 5.000 -4.290 Vu (kN) -158.704E-1 -158.704E-1 SAFE! 0.000 -4.263 3.019 -4.290 b (mm) 600 600 SAFE! 0.000 -4.000 3.019 -4.000 d (mm) 712.5 712.5 SAFE! 0.000 -3.062 3.726 -3.000 ØVc = Ø(1/6)√(f'c)bd, (kN) 317.592E+0 317.592E+0 SAFE! 0.472 -3.062 4.292 -3.000 Remarks SAFE! SAFE! 0.472 -4.000 4.292 -4.000 0.000 -4.000 3.726 -4.000 Vc = (1/6)√(f'c)bd 0.000 -3.062 -3.000 Load Diagram 3.726 = 373.638152161 kN > Vu = -15,87 kN Tie Beam: Stirrups are not Required! 0.000 -1.264 0.000 -1.736 Using: d = 10 mm 0.000 -1.217 0.000 -1.783 S = Smax = d/2 4.292 -1.217 4.292 -1.783 = 256.25 mm 4.292 -1.217 4.292 -1.783 0.472 1.255 1.651 She ar & M om e nt Diagram 0.500 3.726 1.255 3.019 0.500 Property Line a-a 0.000 2.000 0.000 2.000 0.000 -6.250 0.000 -6.250 dx = dy = 0.250 e-e f-f -0.250 -0.733 2.769 -0.733 Use: S = 256.250 mm 0.000 -0.733 3.019 -0.733 1.651 -0.733 5.000 -0.733 1.901 -0.733 5.250 -0.733
3.726 3.726 4.292 4.292 3.726 3.726 -5.500
0.000 5.000 0.000 5.000
b-b 1.651 1.651
1.066 2.066 2.066 1.066
Datum: -4.000 -4.000 -5.500 -5.500
c-c 3.019 3.019
d-d 4.776 4.776
Critical Section for One-Way or Direct Shear
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Check for Two-Way or Punching Shear: (a) Exterior Column: xa = X1-C1x/2-d/2 xb = X1+C1x/2+d/2 x1 = 0.756 m y1 = 1.013 m Vu = Pu1 - qu1(x1)(y1) Vu = 529.201 kN bo = 2.525 m ØVc = Ø(1/3)√(f'c)bod ØVc = 1,922.734 kN (b) Interior Column: Xc = X2-C2x/2-d/2 Xd = X2+C2x/2+d/2 x2 = 1.113 m y2 = 1.113 m Vu = Pu2 - qu2(x2)(y2) Vu = 714.898 kN bo = 4.450 m ØVc = Ø(1/3)√(f'c)bod ØVc = 3,388.581 kN
0.000 1.250
xa = 0.000 xb = 0.756 2 x1 = 0.756 m 1 y1 = 1.013 m
m m
ßc = 1.000 44 (1/3) 0.3333333333 SAFE!
0.750 0.750
Xc = 3.694 Xd = 4.806 2 x2 = 1.113 m 2 y2 = 1.113 m ßc = 1.000 44 (1/3) 0.3333333333 SAFE!
m m
Property Line a-a 0.000 2.000 0.000 -3.500 0.000 -6.250 0.000 -6.250 Punching Shear Failure Exterior Interior 0.000 -1.978 3.485 -2.025 0.713 -1.978 4.534 -2.025 0.713 -1.022 4.534 -0.975 0.000 -1.022 3.485 -0.975 0.000 -1.978 3.485 -2.025 Dimensions Exterior Interior -0.250 -1.022 4.534 -0.975 0.000 -1.022 5.250 -0.975 -0.250 -1.978 4.534 -2.025 0.000 -1.978 5.250 -2.025 -0.250 -1.022 5.250 -0.975 -0.250 -1.978 5.250 -2.025 Exterior Interior 0.000 0.000 3.485 0.000 0.000 -1.022 3.485 -0.975 0.713 0.000 4.534 0.000 0.713 -1.022 4.534 -0.975 0.000 0.000 3.485 0.000 0.713 0.000 4.534 0.000
b-b 0.713 0.713
c-c 3.485 3.485
d-d 4.534 4.534
Load Diagram
Shear & M om e nt Diagram
Critical Section for Two-Way or Punching Shear Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Design of Reinforcement: d = T - 0.5Ø - cover = 512.500 mm = 712.500 mm 4√f`c ρmin = fy = 0.00380 Use ρmin = 0.00406 Ab = πd² / 4 = 490.874 Mu = 0.000 Section Mu = b= d= Rn = act p = Use: p = As = n= Soc = Scl = Location =
Section Mu = b= d= Rn = act p = Use: p = As = n= Soc = Scl = Location =
Footing Strap or
1.4 fy = 0.00406
ρmin =
sq mm -289.824
-312.836 Longitudinal a b c 000.000E-2 289.824E+6 312.836E+6 1,910.000 600.000 600.000 512.500 712.500 712.500 0.00000 1.05723 1.14118 0.00000 0.00314 0.00339 0.00406 0.00406 0.00406 3,972.246 1,734.783 1,734.783 9 4 4 219.000 149.000 149.000 194.000 124.000 124.000 None Top Top 164.756 244.897 Transverse e f 164.756E+6 244.897E+6 ye = 1,750.000 2,100.000 yf = 512.500 512.500 yg = 0.39826 0.49333 yh = 0.00116 0.00145 0.00406 0.00406 3,639.493 4,367.391 8 9 228.000 243.000 203.000 218.000 Bottom Bottom
Temperature/Shrinkage Reinforcement: Ø = 16 mm Ast = 0.002bh consider: b = 1000 mm Ast = 1,200.000 sq mm n = 5.968 say 6 Soc = 200.000 mm Scl = 184.000 mm > 25 mm OK! Ext Footing:Le = 1.135 m 5 Int Footing:Le = 1.325 m 6
136.054 Property Line 0.000 2.000 0.000 -6.250 e-e f-f 3.726 4.292 3.726 4.292
d 136.054E+6 2,100.000 512.500 0.27407 0.00080 0.00406 4,367.391 9 243.000 218.000 Bottom
½b of Strap ½b of Strap 0.300 0.300
a-a 0.000 0.000
g-g -0.250 -0.250 1.901 1.901
m m m m
2.000 -6.250
b-b 0.472 0.472
h-h -0.967 -1.217 -1.217 -0.967
2.769 2.769 5.250 5.250
-0.967 -1.217 -1.217 -0.967
Load Diagram
She ar & M om e nt Diagram
Critical Section for Bending
c-c 1.651 1.651
d-d 3.019 3.019
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Reinforcements
Edge Dist = 0.08254717 Strap 1.651 0.583 3.019 0.583 0.071 1.172 4.222 1.172
Reinforcement Details of3.101 Combined Strap Footing 0.083 0.583 0.583 1.568 0.583 4.917 0.583 By: Engr. Jeremy E. Caballes Exterior
Given:
Interior
Effective Depth Total Depth 0.000 1.255 5.000 1.066 -1.000 1.255 5.250 1.066 Strap: 800 m m x 600 m m 0.000 0.583 5.000 0.500 -1.000 0.583 5.250 0.500 -1.000 1.255 5.250 1.066 -1.000 0.583 5.250 0.500 -0.500 1.066 0.000 1.066 -0.500 0.583 0.000 0.583 -0.500 1.066 -0.500 0.583 Longitudinal Dimensions 0.000 -0.509 0.236 4.009 5.000 0.000 0.000 0.000 0.500 0.236 4.009 5.000 0.236 0.000 Perimeter Dimension 4.009 0.000 Exterior Interior 5.000 0.000 0.000 -0.599 5.000 -0.509 -1.000 -0.599 6.000 -0.509 0.000 -2.401 5.000 -2.491 -1.000 -2.401 6.000 -2.491 -1.000 -0.599 6.000 -0.509 -1.000 -2.401 6.000 -2.491 Exterior Interior Reinforcement Perimeter Dimension 0.000 -4.000 3.019 -4.000 Exterior Interior 0.000 -2.401 3.019 -2.491 -0.500 -0.682 4.917 -0.592 1.651 -4.000 5.000 -4.000 0.083 -0.682 5.500 -0.592 1.651 -2.401 5.000 -2.491 -0.500 -2.318 4.917 -2.408 0.000 -4.000 3.019 -4.000 0.083 -2.318 5.500 -2.408 1.651 -4.000 5.000 -4.000 -0.500 -0.682 5.500 -0.592 Reinforcement Perimeter -0.500 -2.318 5.500 -2.408 Exterior Interior Exterior Interior 1.568 -1.783 3.101 -1.217 0.083 -3.500 3.101 -3.500 1.568 -2.318 3.101 -0.592 0.083 -2.318 3.101 -2.408 0.083 -2.318 4.917 -0.592 1.568 -3.500 4.917 -3.500 0.083 -0.682 4.917 -2.408 1.568 -2.318 4.917 -2.408 1.568 -0.682 3.101 -2.408 0.083 -3.500 3.101 -3.500 1.568 -1.217 3.101 -1.783 1.568 -3.500 4.917 -3.500
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