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Rectangular Combined Footing BY: Solver 1 4

This document summarizes the design of a rectangular combined footing to support two columns. Key details include: 1. The footing is 5.6m long and 3.1m wide with 28mm longitudinal bars and 14mm transverse bars at 160mm spacing. 2. The two columns have factored loads of 3530kN and 5552.5kN respectively. Punching shear, one-way shear, and moment capacities were checked. 3. Reinforcement ratios and bar sizes were designed to satisfy strength and serviceability limit states along the long and short directions of the footing. Shrinkage and temperature reinforcement is also specified.

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Joe Mark Costan Avila
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70 views10 pages

Rectangular Combined Footing BY: Solver 1 4

This document summarizes the design of a rectangular combined footing to support two columns. Key details include: 1. The footing is 5.6m long and 3.1m wide with 28mm longitudinal bars and 14mm transverse bars at 160mm spacing. 2. The two columns have factored loads of 3530kN and 5552.5kN respectively. Punching shear, one-way shear, and moment capacities were checked. 3. Reinforcement ratios and bar sizes were designed to satisfy strength and serviceability limit states along the long and short directions of the footing. Shrinkage and temperature reinforcement is also specified.

Uploaded by

Joe Mark Costan Avila
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLSX, PDF, TXT or read online on Scribd
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RECTANGULAR COMBINED FOOTING

BY:
SOLVER

1 4

0.45 0.45

0.45 0.45 3.1


B =

5.60

7 16 @ 160 mm o.c
14 28 mm bars

855 855
4 28 mm bars 4 28 mm bars
14 28 mm bars
Given:

COLUMN 1: COLUMN 2:

DL = 550 DL = 945
LL = 860 LL = 1175

column dimension: column dimension:


L = 450 0.45 L = 450 0.45
B = 450 0.45 B = 450 0.45

Df = 2.2 hc = 629 0.629


conc = 23.5 H = 1.5710
soil = 17.6 Main Bar 28
Fy = 275 Conc Cov. 75
F'c = 27.5 Temp.Bar 16
qa = 245 C-C 4
X1 = 1 0.85
X2 = 4
COMPUTATION:

P1 = 1410 Pu1 = 2232


P2 = 2120 Pu2 = 3320.5
P = 3530 Pu = 5552.5

X = 2.80 qe = 202.6
L = 5.60 Af = 17.4
B = 3.1
d = 540 0.54 qu = 318.6

check "d" for two-way shear or punching shear:


5.60
0.45 3 0.45

0.99 0.99

0.45 1 0.99 0.99 2 0.45


CLOUMN 1:

Vu = 1919.7
Vc = 3738.0
Vc = 3177.3 ok

COLUMN 2:

Vu = 3008.2
Vc = 3738.0
Vc = 3177.3 ok
check "d" for one- way shear:
0 0.235 1.765

0.54 0.54 0.54

0.45
3.11
0.45 0.45

1 3.00
5.60
Solving for Vu:
0 0
ok not ok Vu1 = 232.86534
Vu2 = -483.0327
Vu3 = 973.61433
Vu4 = -830.7837
Max Vu = 973.61433
Vc = 1467.7741
ok not ok Vc = 1247.608 ok

= 0.85
= 0.9
2.368 0.838 0 b = 0.0423366

0.54 max = 0.0317525


min = 0.0050909

b = 3110
3.11
d = 540
0.45 Abar = 616

1.60
(Long Direction)
Solving for Mu:
x = 2.252461

Mu1 = 297.5846
Mu2 = 241.2969
Mu3 = -281.746
Mu4 = 909.2262
Mu5 = 941.3631
ok not ok

SECTION -> 1 2 3 4 5
Moment, Mu 3.0E+008 2.4E+008 -3E+008 909226187 080
Ru,= 0.9 0.365 0.30 0.35 1.11 1.15
0.00134 0.00108 0.00126 0.00415 0.00430
use 0.00509 0.00509 0.00509 0.00509 0.00430
As 8549 8549 8549 8549 7226
Number of Bars, N 13.9 13.9 13.9 13.9 11.7
use N 14 14 14 14 12
Location BOTTOM BOTTOM TOP BOTTOM BOTTOM

BOTTOM TOP
Shear Diagram

973.6143263397

232.8653374621
0 0

-483.0326782103

-830.7836893327

Moment Diagram

941.3630803179
909.2261869913

297.5845602408
241.2969085724

-281.7460404356
along short direction:
b1

1.33

0.45 0.45
0
0.45

7836893327

TOP BOTTOM
0
COLUMN 1:

b1 = 855 0.855
941.3630803179
69913
Mu = 240.9365

COLUMN 2:

b2 = 855 0.855
Mu = 240.9365

d1 = d2 512 (Transverse bars above longitudin


= 0.9
b2 SECTION ->
Moment, Mu
1.33 Ru,= 0.9

use
As
3.11
Number of Bars, N
0.45 use N
Location

Shrinkage and Temperature Bars:

b = 1000
Ast = 1258
N = 6.256764
Spacing = 159.827

bars above longitudinal)


1 2
240936483 240936483 TOP BOTTOM
1.19 1.19 0
0.004460 0.004460
0.00509 0.00509
2229 2229
3.6 3.6
4 4
BOTTOM BOTTOM

nd Temperature Bars:

7
160 mm o.c

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