Structville

A hub for civil engineering related designs, analyses, discussion, information, and knowledge...

Trending Posts:

Search This Blog

Thursday, January 25, 2018

Structural Design of Combined Footings: Solved Example to Eurocode 2


Introduction
Combined footings are normally employed in cases where two or more columns are closely spaced such that their individual pad footings would overlap each other. They are also used where property boundaries would not permit design of separate bases. The most common method of design of combined footings is the rigid approach, which makes the following assumptions;

(1)The footing  is infinitely rigid, and therefore, the deflection of the footing does not influence the pressure distribution.
(2) The soil pressure is distributed in a straight line or a plane surface such that the centroid of the soil pressure coincides with the line of action of the resultant force of all the loads acting on the foundation.

Solved example
Two (300 x 300)mm square columns spaced at a distance 2.45m is loaded as shown below. The foundation is founded on a soil of bearing capacity 150 kN/m2. It is desired to design the footing to satisfy all requirements using concrete grade of 30 N/mmand steel of yield strength 500 N/mm2. The concrete cover is 50mm.


Solution
At serviceability limit state;
PEd = 1.0Gk + 1.0Qk
Total service load on column P1 = 665 + 122 = 787 kN
Total service load on column P2 = 825 + 145 = 970 kN
Total load on column = 787 + 970 = 1757 KN

Assuming 10% of the service load to account for the self weight of the footing;
Sw = 0.1 × 1757 = 175.7 KN

Area of footing required = Total service load/Allowable bearing capacity = (1757 + 175.7)/200 = 9.6635 m2

Adopt a rectangular base = 5m x 2m (Area provided = 10 m2)
To locate the centroid of the footing, let us take moment about column P1;
(970 × 2.45) - 1757( x) = 0
On solving, x = 1.352m from column P1

Let the projection of the footing from the lighter column be L1
Therefore;
L1 + 1.352m = (Total length of footing)/2 = 5/2 = 2.5
Therefore, L1 = 2.5 - 1.352 = 1.148m (say 1.15m)

The final disposition of the footing is given below assuming a trial footing depth of 600mm.


We can check the preliminary thickness of 600mm by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.





At ultimate limit state; 
P1 = 1.35Gk + 1.5Qk = 1.35(665) + 1.5(122) = 1080.75 kN
P2 = 1.35Gk + 1.5Qk = 1.35(825) + 1.5(145) = 1331.25 kN
Total ultimate load = 1080.75 + 1331.25 = 2412 KNVEd < VRd,max

Where;
VEd = βVEd /u0d

VRd,max = 0.5vfcd

Considering Column B3 bearing the maximum axial load, VEd = 1850 kN
β = 1.5 (an approximate value from clause 6.4.3(6))
u= column perimeter = 2(300) + 2(300) = 1200 mm
= 0.6[1 - fck/250] (strength reduction factor for concrete cracked in shear)
= 0.6[1 - 30/250] = 0.528
fcd = αccfckcfcd = (1.0 × 30)/1.5 = 20 N/mm2
VRd,max = 0.5 × 0.528 × 20 = 5.28 N/mm2

VEd = (1.5 × 1331.25 × 1000) / (1200mm × d)

Therefore;
1996874/1200d = 5.28
On solving; dmin = 315.263mm

Hence the depth is ok.

Earth pressure intensity = 2412/10 = 241.2 KN/m2
For a width of 2m, q = 241.2 × 2 = 482.4 kN/m


Structural Design
Bottom Reinforcement
We are supposed to take our maximum moment from the column face which can be easily determined as shown below.

MEd = [482.4 × (1.4 - 0.15)2]/2 = 376.875 kNm
Effective depth (d) = 600 - 50 - 10 = 540 mm (assuming B20mm bars).

MEd = 376.875 KNm/m
b = 2000mm

k = MEd/(fckbd2 )
k = (376.875 × 106)/(30 × 2000 × 5402 ) = 0.0215

Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 - 0.882k)] = z = d[0.5+ √(0.25 - (0.882 × 0.0215)] = 0.95d

As1 = MEd/(0.87fyk z)
As1 = (376.875 × 106)/(0.87 × 500 × 0.95 × 540) = 1688 mm2
Provide 10B16mm @ 225mm c/c BOT (ASprov = 2010 mm2/m)

Read Also....

Design of Cantilever Retaining Wall Supporting Lateritic Earthfill

Design of Continuous Beam and Slab footing to BS 8110-1:1997

Top reinforcement
There is no hogging moment on the footing as shown in the bending moment diagram. However, provide minimum reinforcement at the top.

Asmin = 0.0013bd = (0.0013 × 1000 × 540) = 702 mm2

Provide B12mm @ 150mm c/c BOT (ASprov = 753 mm2/m)

Transverse Reinforcement
Cantilever arm = (2.0 - 0.3)/2 = 0.85m
MEd = [482.4 × 0.852]/2 = 174.267 kNm
Effective depth (d) = 600 - 50 - 10 = 540 mm (assuming B20mm bars).

MEd = 174.267 kNm
b = 1000mm (designing per unit strip)

k = MEd/(fckbd2 )
k = (174.267 × 106)/(30 × 1000 × 5402 ) = 0.0199

Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 - 0.882k)] = z = d[0.5+ √(0.25 - (0.882 × 0.0199)] = 0.95d

As1 = MEd/(0.87fyk z)
As1 = (174.267 × 106)/(0.87 × 500 × 0.95 × 540) = 781 mm2
Provide B16mm @ 250mm c/c BOT (ASprov = 804 mm2/m)

Check for shear
Clause 6.4.2(2) of EN 1992-1-1:2004 states that control perimeters at a distance less than 2d should be considered where the concentrated force is opposed by a high pressure (e.g. soil pressure on a base), or by the effects of a load or reaction within a distance 2d of the periphery of area of application of the force.

Let us investigate this at a distance d from the face of the column according to clause 6.4.4(2) of Eurocode 2.

Using the maximum shear force VEd = 675.36 kN
VEd,red = VEd - ∆VEd
Where ∆VEd is the net upward force within the control perimeter considered i.e. upward pressure from soil minus self weight of base.

Control area = 1.38m × 1.38m = 1.9044 m2
Soil pressure = 241.2 KN/m2
Unit weight of concrete = 25 kN/m3
Hence, ∆VEd = (241.2 × 1.9044) - (25 × 0.6 × 1.9044) = 430.775 kN

VEd,red = 1331.25 - 430.775 = 900.475 kN

vEd,red = VEd,red / ud
where u is the control perimeter = 4(2d + 300) = 4(2 × 540 + 300) = 5520mm
vEd,red = (900.475× 1000)  / (5520 ×  540) = 0.3021 N/mm2

Shear Resistance of section
VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) ]  × 2d/a ≥ (Vmin × 2d/a)

CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/540) = 1.606 > 2.0, therefore, k = 1.608

Vmin = 0.035k(3/2) fck0.5
Vmin = 0.035 × (1.608)1.5 × 300.5 = 0.391 N/mm2

ρx  (in the transverse direction) = As/bd = 804/(1000 × 540) = 0.00148
ρz  (in the longitudinal direction) = As/bd = 2010/(2000 × 540) = 0.00186

ρ1 = √(ρ× ρz) = 0.00165 < 0.02

VRd,c = [0.12 × 1.608 (100 × 0.00165 × 30 )(1/3)] × 2 = 0.6577 N/mm2

0.3021 N/mm2 < 0.6577 N/mm2

We can now verify that the section is ok for punching shear.

With this, the reinforcement detailing can be drawn.


Thank you for visiting Structville today.

Have you read...
How to Estimate The Cost of Tiling a Three Bedrooms Flat

Our facebook page is at www.facebook.com/structville

No comments:

Post a Comment