Foundation or Footing Design Theory and Example Part 2-July 8-2017

Foundation or Footing Theories and Design: Part 2
Isolated Foundation or Footing transmit column load to the underlying soil.
Figure: Tributary area for columns shown in shades/colors
Design of Square Isolated Footings
Figure: Shapes of isolated footings.
Shear
35
Moments
36
The bending moment in a square reinforced concrete footing with a square column is the same
about both axes because of symmetry. If the column is not square, the moment will be larger in
the direction of the shorter column dimension. It should be noted, however, that the effective
depth of the footing cannot be the same in the two directions because the bars in one direction
rest on top of the bars in the other direction. The effective depth used for calculations might be
the average for the two directions or, more conservatively, the value for the bars on top.
The critical section for bending is taken at the face of a reinforced concrete column or halfway
between the middle and edge of a masonry wall or at a distance halfway from the edge of the
base plate and the face of the column if structural steel columns are used.
Two way or Punching Shear
One way or Beam Shear
Example.1: Design footings for the interior column of a building. Assume, base of footing
location 5’ below ground level . Permissible soil pressure , qa = 6 ksf .Gravity loads: PDL =
351 kips , PLL = 56.4 kips, Service Moment, Mservice = 75.4 k-ft, column size = 16” x 16”
Pu
c
5ft
h d
qu
1.Design Data :
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Assume, Weight of soil plus concrete above footing base =135 pcf
(When, soil is wet packet use, weight of soil = 130 pcf; otherwise 100 pcf)
Interior column = 16” x 16”
f’c = 4000 psi (for both footing and column), fy = 60ksi
2.Load Combination :
a) Gravity loads: from column, PDL = 351 kips, PLL = 56.4 kips
Service Moment for column, Mservice = 75.4 k-ft
3. Base Area of Footing Calculation:
Weight of surcharge, soil plus concrete depth for footing = (135pcf ×5ft/1000) = 0.675 ksf
Total surcharge = 0.675+0.00, extra surcharge if any = 0.675 ksf
so, Net permissible soil pressure = ( 6 – 0.675 ) = 5.325 ksf
351+56.4
Area of footing , Af = ( 5.325 ) = 76.5 sq. ft.
Try 9ft × 9ft square footing . ( Af = 81 sq. ft.)
Now, we know, Sectional Modulus, S = I/c, where c is the distance from the neutral axis to the
most extreme fibre, I= moment of inertia, or, S = I/c = ( bh3/12 ) / (h/2)
so, S = bh2/6
9 × 92
S=( ) = 121.5 ft3
62
As, Af = 81 ft
𝑃 𝑀 407.4 75.4
Now, Stress, q = 𝐴𝑓 + 𝑆𝑓 = 81 + 121.5 =5.02+0.62 =5.64 ksf > 5.325 ksf ( Not OK )
9.5 × 9.52
Try, 9.5 ft × 9.5 ft square footing ( Af = 90.25 ft2 ),S = ( ) = 121.5 ft3= 142.9 ft3
6
351+56.4 75.4
q = 90.25 + 142.9
= 5.04 ksf < 5.325 ksf (Ok)
4. Footing Thickness :
Now, ρ for bending =As/bd [Note: Minimum steel ratio for shrinkage and temperature in slab for
60 grade steel, is ρ = 0.0018. Now assume, ρ = 0.002 in the Rn equation below, since slab
thickness, h > d]
Rn = ρ fy (1- 0.5𝜌𝑓𝑦/(0.85𝑓′𝑐))
0.5×0.002×60000
= 0.002 x 60,000 ( 1- 0.85 × 4000 )
= 117.9 psi
2 𝑀𝑢 𝑀𝑢 ×1000
d required = 𝜑 𝑅𝑛 = 0.9 × 117.9 = 9.43 𝑀𝑢 ........... (1)
𝑷𝒖 𝒄𝟐
However, wL.(L/2) Mu = ( 𝑨𝒇 )( 𝟐 ) ....... (2) [as, ( Pu / Af ) = w = qu] [Note; wL.(L/2) ]
Now, from eqn. (1) , d²req = 9.43 Mu
𝑃𝑢 𝑐2
= 9.43x (𝐴𝑓 )( 2 ) ( from eqn. 2)
𝑃𝑢 𝑐²
= 9.43 ∗ ∗
𝐴𝑓 2
𝑷𝒖
so, drequired = 2.2 c 𝑨𝒇
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Now, bearing pressure for strength design = Factored load/ Area of footing = Pu/Af
Pu = 1.2DL(351)+1.6LL(56.4) = 511.44
Bearing pressure =511.44/90.25 =5.67 ksf, or qu net =5.67 ksf
16
(9.5− )
12
Footing projection , c = 𝑓𝑡 = 4.08′
2
𝑃𝑢
h = 2.2 c + 4 (considering, more than 3” required clear cover + bar dia.)
𝐴𝑓
511.44
h = 2.2 *4.08* +4
90.25
= 25.4 in
Try, h = 27 in.
so, d = 27-4 = 23 in.
One way shear check:

Considering beam Shear or One way shear :
Beam shear , Vb = qu × Beam strip length × (c-d)
=5.67×9.5×2.167
=116.72 kips
Allowable shear, Va = 2φ√𝑓′𝑐 b d
= (2×.75×√4000× (9.5×12) ×23)/1000
= 248.71 Kips
Since, Va > Vb , therefore, OK.
Two –way or Punching

Shear
<C-d><d>
Beam or One way Shear
Considering punching Shear or Two way shear :

Punching shear , Vp = Factored load, or qu× Footing area – Punched out area× qu
= 9.5×9.5×5.67 – [(39×39)/144] ×5.67 = 451.55 Kips
Allowable shear, Va = 4φ√𝑓′𝑐 b d
= (4*.75*√4000*(39*4)*23)/1000
= 680.77 Kips
Since, Va > Vp , therefore OK.
The Bending Moment:
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Mu =(5.67*9.5*(4.08))^2)/2 = 448.3 k-ft
[Note: Mu = (qu × b × c2)/2; considering moment for the whole width of footing]
𝑀𝑢
Reinforcements : As = 𝑎
∅𝑓𝑦 (𝑑− )
2
=448.3*12/[0.9*60*(23-8/2)]; [assumed, a = 8 inches]
= 5.24 in2 ;
𝐴𝑠 𝑓𝑦
a= .85𝑓 ′ 𝑐𝑏
= (5.43*60)/*(.85*4*12) = =7.98 inches; very close to 8 inches assumed, therefore, OK;
As required= 5.24 in2.
Now, Minimum reinforcement for flexure,

As (minimum) = 200*b*d/fy
=(200*9.5*12*23)/60000 =8.74 in2
As (minimum) =8.74 in2; within 9.5 ft footing width, controls.
Since, it is square footing, use 9 #9 bar (Area provided 9 in2) in each direction.
Figure: Detail of reinforcement
40

Foundation or Footing Design Theory and Example Part 2-July 8-2017

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Foundation or Footing Theories and Design: Part 2

Isolated Foundation or Footing transmit column load to the underlying soil.

Figure: Tributary area for columns shown in shades/colors

Design of Square Isolated Footings

Figure: Shapes of isolated footings.

Two way or Punching Shear

One way or Beam Shear

One way shear check:

Two –way or Punching

Considering punching Shear or Two way shear :

The Bending Moment:

As required= 5.24 in2.

Now, Minimum reinforcement for flexure,

Figure: Detail of reinforcement

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