P1 DL = 285 kN P2 DL = 350 kN P1 LL = 225 kN P2 LL = 300 kN M1 DL = 85 kN-m M2 DL = 120 kN-m E = 200,000.00 MPa M1 LL = 55 kN-m M2 LL = 95 kN-m f'c = 27.50 MPa fy = 345.00 MPa Wconc = 23.50 kN per cu m Wsoil = 16.50 kN per cu m qa = 200.00kPa OR qe = 150 kPa D = 1.5 m Assume T = 600 mm 0.6 Remarks: SAFE!
Distance from the Strap Beam Properties: Section A-A
Property Line = m Column Spacing = 4 m Height = 800 mm
L1 = 1.75 m Width = 600 mm Remarks: SAFE! L2 2.2m 0.8 0.6 C1y = 500 mm C2y = 600 mm B1 0.5 0.6 B2 Bar diameter = 25 mm 0.025 C1x = 500 mm C2x = 600 mm Concrete cover = 75 mm 0.5 0.6 0.075
Temperature bar Ø = 16 mm Ratio, L1/B1 = (Optional) L2/B2 = 1 (Optional)
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
∑M at R1 = 0, CW is (+) A1 = R1/qe
0 = P2(X2 - L1/2) - P1(L1/2 - X1) - R2(X2 - L1/2) + M1 + M2 A1 = 3.328 sq m
R2 = 660.741 kN A2 = R2/qe
R1 = P1 + P2 - R2 A2 = 4.405 sq m
Design of Combined Strap Footing
<Program Created by: Engr. Jeremy E Caballes>
A
A = 7.733 sq m
Solve for Location of its Resultant: Design loads:
P1 = P1 DL + P2 LL Pu1 = 1.4(P1 DL) + 1.7(P1 LL) P1 = 510.000 kN Pu1 = 781.500 kN wp1 = 1563 kN/m, Mu1 = 212.5 kN-m P2 = P2 DL + P2 LL Pu2 = 1.4(P2 DL) + 1.7(P2 LL) P2 = 650.000 kN Pu2 = 1,000.000 kN wp2 = 1666.667 kN/m, Mu1 = 329.5 kN-m M1 = M1 DL + M2 LL Pu = Pu1 + Pu2 M1 = 140.000 kN-m Pu = 1,781.500 kN M2 = M2 DL + M2 LL Mu1 = 1.4(M1 DL) + 1.7(M1 LL) M2 = 215.000 kN-m Mu1 = 212.500 kN-m X1 = 0.250 m Mu2 = 1.4(M2 DL) + 1.7(M2 LL) X2 = 4.250 m Mu2 = 329.500 kN-m P = P1 + P2 ∑M at Ru1 = 0, CW is (+)
P = 1,160.000 kN 0 = Pu2(X2 - L1/2) - Pu1(L1/2 - X1) - Ru2(X2 - L1/2) + Mu1 + Mu2
P(Xc) = P1(X1) + P2(X2) + M1 +M2 Ru2 = 1,015.870 kN
Xc = 2.797 m X1 ≤ Xc ≤ X2, OK! Ru1 = Pu1 + Pu2 - Ru2
L = 5.300 m Ru1 = 765.630 kN
qu1 = Ru1/A1
Determine the Widths: = 229.059 kPa
qe = qa - qs - qc wua = qu1 x B1
qs = Wsoil x (D - T) wu1 = 437.503 kN/m Soil = 437.503 kN/m
qs = 14.850 kPa qu2 = Ru2/A2
qc = Wconc x T = 230.356 kPa
qc = 14.100 kPa wub = qu2 x B2
Therefore: qe = 150.000 kPa wu2 = 483.748 kN/m Soil = 483.748 kN/m
Thus: A = (P/qe)(1+6e/L)
A = 7.733 sq m
A = A1 + A2 1.000 xa = X1 - 0.5C1x xa = 0.000
P/qe = A1 + A2 -8.500 xb = X1 + 0.5C1x xb = 0.500
A2 = P/qe - A1 0.000 xc = L1 xc = 1.750
A.Xc = A1(L1/2) + A2X2 0.000 xd = L - L2 xd = 3.200
A.Xc = A1(L1/2) +(P/qe - A1)X2 L1 = 8.500 xe = X2 - 0.5C2x xe = 3.950
A1 = 3.328 sq m B1 = #DIV/0! xf = X2 + 0.5C2x xf = 4.550
L1 B1 = 3.328 sq m Stirrups for Strap (10 - mm Ø): 1 @ 50 mm, rest @ 256 mm symmetrical with respect to it's midspan
Since: L1 = 1.75 say 1.750 m f'c = 27.5 MPa Temperature bar Ø = 16 mm
Then: B1 = 1.902 say 1.910 m fy = 345 MPa L1 = 1.75 m
Check: L1/B1 = 0.916 0.75 ≤ L/B ≤ 1.0, OK! Wconc = 23.5 kN per cu m B1 = 1.91 m
A2 = 4.405 sq m Wsoil = 16.5 kN per cu m L2 = 2.1 m
L2 B2 = 4.405 qa = 200 kPa B2 = 2.1 m
L2 = 2.099 say 2.100 m T = 600 mm D = 1.5 m
B2 = 2.099 say 2.100 m Edge Dist. = 0.25 m d = 512.5 mm
Check: L1/B1 = 1.000 Col Dist. = 4 m d = 712.5 mm
0.75 ≤ L/B ≤ 1.0, OK! Edge Dist. = 1.05 m Bottom 5-16 @ 200 mm
L = 5.300 m 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 Distance (m)
Check for One-Way or Direct Shear: x-factor = 0.94339623 Origin = -1.500
d = T - 0.5Ø - cover Footing 1: Footing 2:
= 512.5 mm 1.651 -1.783 3.019 -1.217 d = 0.5125 m 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 None 0.000 1.066 3.019 1.066 a - a d - d e - e f - f Distance, x, y, (m) 0.000 5.063 0.813 0.813 0.000 1.066 3.726 1.066
Vu (kN) 000.000E+0 -114.890E+0 57.122E+0 114.890E+00.000 -1.264 2.066 -1.217 3.726 2.066
b (mm) 1910 2100 1750 21000.472 -1.264 2.066 -1.217 4.292 2.066
d (mm) 512.5 512.5 512.5 512.5 0.472 -1.736 1.255 -1.783 4.292 1.066
ØVc = Ø(1/6)√(f'c)bd, (kN) 727.212E+0 799.553E+0 666.294E+0 799.553E+00.000 -1.736 1.255 -1.783 3.726
Remarks None SAFE! SAFE! SAFE! 0.000 -1.264 2.066 -1.217 3.726
Press fac = 0.001 Load Diagram Origin = -5.500
SAFE! 0.000 -4.000 3.019 -4.000
b - b c - c SAFE! 1.651 -4.000 5.000 -4.000
Distance, y (m) 1.750 3.200SAFE! 1.651 -4.263 5.000 -4.290
Vu (kN) -15.870E+0 -15.870E+0SAFE! 0.000 -4.263 3.019 -4.290
b (mm) 600 600SAFE! 0.000 -4.000 3.019 -4.000
d (mm) 712.5 712.5SAFE! 0.000 -3.062 3.726 -3.000
ØVc = Ø(1/6)√(f'c)bd, (kN) 317.592E+0 317.592E+0SAFE! 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 3.726 -3.000
= 373.6381522 kN > Vu = -15.87 kN
Stirrups are not Required! 0.000 -1.264 0.000 -1.736 0.000 -4.000
Using: d = 10 mm 0.000 -1.217 0.000 -1.783 5.000 -4.000 S = Smax = d/2 4.292 -1.217 4.292 -1.783 0.000 -5.500 = 256.25 mm 4.292 -1.217 4.292 -1.783 5.000 -5.500 0.472 1.255 1.651 0.500 3.726 1.255 3.019 0.500 a - a b - b c - c d - d 0.000 2.000 0.000 2.000 1.651 3.019 4.776 0.000 -6.250 0.000 -6.250 1.651 3.019 4.776 dx = dy = 0.250 -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 Transverse Column 2: Property Line e - e f - f Datum: Section: Section: Column 1: Longitudinal Strap Beam Tie Beam: Pr ope rty Line a a b b c c d d e e f f She a r & M ome nt D ia gr a m Lo ad D iagra m
Check for Two-Way or Punching Shear: a - a b - b c - c d - d
0.000 2.000 0.000 -3.500 0.713 3.485 4.534
(a) Exterior Column: 0.000 -6.250 0.000 -6.250 0.713 3.485 4.534
xa = X1-C1x/2-d/2 0.000 xa = 0.000 m xb = X1+C1x/2+d/2 1.250 xb = 0.756 m x1 = 0.756 m 2 x1 = 0.756 m 0.000 -1.978 3.485 -2.025 y1 = 1.013 m 1 y1 = 1.013 m 0.713 -1.978 4.534 -2.025 Vu = Pu1 - qu1(x1)(y1) 0.713 -1.022 4.534 -0.975 Vu = 529.201 kN ßc = 1.000 0.000 -1.022 3.485 -0.975 bo = 2.525 m 4 4 0.000 -1.978 3.485 -2.025 ØVc = Ø(1/3)√(f'c)bod (1/3) 0.333333333 ØVc = 1,922.734 kN SAFE! -0.250 -1.022 4.534 -0.975 (b) Interior Column: 0.000 -1.022 5.250 -0.975 Xc = X2-C2x/2-d/2 0.750 Xc = 3.694 m -0.250 -1.978 4.534 -2.025 Xd = X2+C2x/2+d/2 0.750 Xd = 4.806 m 0.000 -1.978 5.250 -2.025 x2 = 1.113 m 2 x2 = 1.113 m -0.250 -1.022 5.250 -0.975 y2 = 1.113 m 2 y2 = 1.113 m -0.250 -1.978 5.250 -2.025 Vu = Pu2 - qu2(x2)(y2) Vu = 714.898 kN ßc = 1.000 0.000 0.000 3.485 0.000 bo = 4.450 m 4 4 0.000 -1.022 3.485 -0.975 ØVc = Ø(1/3)√(f'c)bod (1/3) 0.333333333 0.713 0.000 4.534 0.000 ØVc = 3,388.581 kN SAFE! 0.713 -1.022 4.534 -0.975 0.000 0.000 3.485 0.000 0.713 0.000 4.534 0.000
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005 Property Line
Punching Shear Failure
Exterior Interior Exterior Interior Exterior Interior Dimensions P roper ty Line a a b b c c d d S he a r & M om e nt D ia gra m Loa d D ia gr a m
Design of Reinforcement: d = T - 0.5Ø - cover = 512.500 mm Footing = 712.500 mm Strap 4√f`c 1.4 fy fy = 0.00380 = 0.00406 Use ρmin = 0.00406 Ab = πd² / 4 = 490.874 sq mm Mu = 0.000 -289.824 -312.836 136.054 a - a b - b c - c d - d a b c d 0.000 2.000 0.000 2.000 0.472 1.651 3.019
Mu = 000.000E+0 289.824E+6 312.836E+6 136.054E+6 0.000 -6.250 0.000 -6.250 0.472 1.651 3.019
b = 1,910.000 600.000 600.000 2,100.000 e - e f - f d = 512.500 712.500 712.500 512.500 3.726 4.292 Rn = 0.00000 1.05723 1.14118 0.27407 3.726 4.292 act p = 0.00000 0.00314 0.00339 0.00080 Use: p = 0.00406 0.00406 0.00406 0.00406 As = 3,972.246 1,734.783 1,734.783 4,367.391 -0.250 -0.967 2.769 -0.967 n = 9 4 4 9 -0.250 -1.217 2.769 -1.217 Soc = 219.000 149.000 149.000 243.000 1.901 -1.217 5.250 -1.217 Scl = 194.000 124.000 124.000 218.000 1.901 -0.967 5.250 -0.967
Location = None Top Top Bottom
164.756 244.897
e f
Mu = 164.756E+6 244.897E+6 ye = ½b of Strap m b = 1,750.000 2,100.000 yf = ½b of Strap m d = 512.500 512.500 yg = 0.300 m Rn = 0.39826 0.49333 yh = 0.300 m act p = 0.00116 0.00145 Use: p = 0.00406 0.00406 As = 3,639.493 4,367.391 n = 8 9 Soc = 228.000 243.000 Scl = 203.000 218.000 Location = 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 ρmin = or ρmin = Transverse Section Section Longitudinal g - g h - h Property Line Pr ope rty Line a a b b c c e e f f d d S hear & M om ent D iagra m Lo ad D iagram
Edge Dist = 0.08254717 0.083 0.583 3.101 0.583 1.651 0.583 1.568 0.583 4.917 0.583 3.019 0.583 0.071 1.172 4.222 1.172 0.000 1.255 5.000 1.066 -1.000 1.255 5.250 1.066 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 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 4.009 0.000 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 0.000 -4.000 3.019 -4.000 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 -0.500 -2.318 5.500 -2.408 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 Interior Exterior Interior Reinforcement Perimeter Interior Perimeter Dimension Exterior Interior Longitudinal Dimensions Interior Exterior
Reinforcement Perimeter Dimension Strap Exterior Exterior Exterior Interior Effective Depth Reinforcements Total Depth
P
ro
p
er
ty
L
ine
Given: Strap: 800 mm x 600 mmReinforcement Details of Combined Strap Footing
Strap Beam Properties: Section A-A
Stirrups for Strap (10 - mm Ø): 1 @ 50 mm, rest @ 256 mm symmetrical with respect to it's midspan
