Scribd
Upload
68 views3 pages

RC Footing Design and Shear Check

This document summarizes the design of a reinforced concrete footing. It provides the key dimensions, loads, and calculations for: 1. Determining the minimum depth and thickness of the footing to resist bending moments. The depth is calculated as 430 mm and thickness as 500 mm. 2. Checking the shear stress, which is found to be below the allowable limit. 3. Checking the punching shear stress, which is also below the allowable limit. 4. Calculating the required steel reinforcement in two perpendicular directions. The reinforcement is specified as 6 φ12 bars/m in one direction and 6 φ10 bars/m in the other.

Uploaded by

ali
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
68 views3 pages

RC Footing Design and Shear Check

This document summarizes the design of a reinforced concrete footing. It provides the key dimensions, loads, and calculations for: 1. Determining the minimum depth and thickness of the footing to resist bending moments. The depth is calculated as 430 mm and thickness as 500 mm. 2. Checking the shear stress, which is found to be below the allowable limit. 3. Checking the punching shear stress, which is also below the allowable limit. 4. Calculating the required steel reinforcement in two perpendicular directions. The reinforcement is specified as 6 φ12 bars/m in one direction and 6 φ10 bars/m in the other.

Uploaded by

ali
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 3

Footing (6):

Pw = 620.7683 kN
PUlt = 891.5433 kN
C (55 x 25) cm
qall = 150 kN/m3
tpc = 30 cm

1 – Design the Critical Section for moment (Depth of RC footing):


- Actual Normal Stress on RC Footing (UL)
Pul 891.5433
Fact = = =2 13 .3 kN/m2
BRC x LRC 1. 9 x 2.2
BRC −a 1. 9−0. 25
Z= = =0. 8 25 m
2 2
Force = Stress x Area = Fact x Z x B = 213.3 x 0.825 x 2.2 = 387 kN
0. 83
Mact = (387) x =¿ 160 kN.m
2

Mact 1 60 x 106
d = C1 x
√ Fcu x b
=5 x
√30 x 220 0
=2 46 mm

t = 246 + 70mm (Cover) = 316 mm (take the Minimum)


then t = 500 mm, d = 430 mm Thickness in the other direction will be the same

2 – Check of Shear:
- Critical Section for shear.
d 0.43
L = Z− =0.83− =0. 6 1 m
2 2
(Qu) for 1m
Qu = Fact x L x 1 = 2 13 x 0. 6 1 x 1=130kN
- Actual Shear Stress (qu).
Qu 13 0 x 103
qu = = =0.3 N/mm2
b x d 1000 x 430

30
Allowable Shear Stress (qsu) = 0.16 x
√ 1.5
=0.715 N/mm2 (Safe Shear Stresses)

3 – Check of Punching Shear:


a + d = 0. 25+0.43=0. 68 m
b + d = 0.55 + 0.43 = 0.98 m
- Punching Force (Qp).
Qp = Pul – (Fact) x ((a + d) x (b + d)) = 891.5433 – ((2 13) x (0.68 x 0.98)) = 750 kN
- Punching Shear Area.

Ap = ( 2 x ( 25 0+ 430 ) +2 x ( 55 0+430 ) ) x 430=1427600 mm2

Qp 750 x 10 3
qpu = = =0.52 N/mm2
2 x ( a+d ) +2 x ( b+ d ) x d 1427600

- Allowable Punching Shear Stress.

0. 25 30
(
qpcu = 0.316 x 0.5+
0. 55 )
x
√
1.5
=1. 35 N/mm2 (Safe Punching Shear)

4 – Reinforcement of the Footings:


- Direction (1)
Fact = 213 kN
Brc−a 1. 9−0. 25
Z1 = = =0. 8 25 m
2 2

Mact1 = ( 213 x 0. 8 25 x 1.85 ) x ( 0.83


2 )
=1 60 kN . m

J = 0.826

Mact 1 60 x 106
As = = =1249 mm2
J x Fy x d 0.826 x 360 x 430
As 1249
As (mm2/m) = = =675 = 6 ϕ12 /m
BRC 1.85
- Direction (2)
Fact = 213 kN
LRC−b 2.2−0.55
Z1 = = =0. 825 m
2 2

Mact1 = ( 2 13 x 0. 8 25 x 1.75 ) x ( 0.725


2 )
=138 kN . m

J = 0.826

Mact 138 x 106


As = = =1079mm2
J x Fy x d 0.826 x 360 x 430
As 1079
As (mm2/m) = = =617 = 6 ϕ10 /m
BRC 1.75

You might also like