ae303 combined footing design

8/18/2019 AE303 Combined Footing Design

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Combined Footings: One footing carries two or more columns. A

common case is when building columns and crane columns are close.

Combined Footings:Four and Six Column Footings


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Combined Footing:Strip Footing with Wall and Columns

Design of Combined Footings

One factor that must be considered in the design is the distancebetween the columns. A large distance will mean the development of

negative moments.


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Example: Size a rectangular footing to support two columns, spaced 16

feet apart.The exterior (left) column carries 300k (180 k DL and 120 k LL). Theinterior (right) column carries 390k (250 k DL and 140 k LL).

Assume these loads are constant.

The allowable soil pressure is 5 k/ft2

Use fc = 4 ksi and fy = 60 ksi

The external column is located 2 ft from

the property line and we want thefooting to run to the edge.


Determine the location of the “resultant” and itsposition relative to the left exterior column.

( ) ( )

( ) ( )

ft.9 Useft.04.9

k 120k 180k 140k 250

k 120k 180ft0k 140k 250ft16

i

ii

⇒=

+++

+++==

∑

∑F

F x x

Extending the footing to the property line on the left side, half the length is 9 ft +

2 ft. = 11 ft. So in order to keep the center of the footing under the resultant ofthe loads the length of the footing must be 2 * 11 ft. = 22 ft.


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Once the footing length has been established, the width can be

determined. Calculate the required area of the footing and thenuse that to calculate the width:

1 2

2

2

Actual Col I Loads 250 k 140 k 390 k

Actual Col II Loads 180 k 120 k 300 k

Total Loads 390 k 300 k 690 k

690 kReqd Area of footing 138 ft

5 k/ft

13Reqd Width of footing

DL LL

DL LL

AL AL

= + = + =

= + = + =

= + = + =

= =

=

28 ft

6.27 ft Use 6.5 ft22 ft

= ⇒


Calculate factored soil pressure:

( ) ( )

( ) ( )

( )( )

2

u ftk /6.52ft6.5ft22

k932

pressureupwardNet

k 932

k524k408

k 1401.6k 2502.1

k 1201.6k 1802.1

6.12.1LoadsFactored

==

=

+=

++

+=

+=

q

LL DL

The calculation of minimum d based on punching shearis the same as for individual footings.

For one way shear, a shear diagram should be drawnand the highest Vu value used.


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As mentioned earlier, the goal of a combined footing design is to get thecenter of the footing at the resultant of the loads.

Using a trapezoidal shape can create a design where the overhangs

(L1 and L2) can be the same.


B1

L2L1

B2

Using the previous example, design a footing so the centroid of the footing is 9’ from

the left column (X in the picture), and also L1 = L2 = 1’

Total L = 1’+16’+1’ = 18’, and the area (B1+B2)/2 * L must still equal 138 sq ft

(calculated earlier on slide 8)Solve for the quantity (B1+B2) = 15.33’ to give the correct area, but B1 and B2 must

also be the right proportion to get the centroid where we want it.

The distance from the centroid to the right edge = 8’, call this X’

If B1=0, X’ would be L/3= 6’ (too small), if B1=B2, X’ would be at L/2 = 9’ (too big)However, when B1=5.1’ and B2=10.2’ we get X’ = 8, the correct value.


2 1 2' *3 1 2

L B B X B B

+

= +

L1 L2

B1B2


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Although the goal of a combined footing design is to get the center of the

footing at the resultant of the loads, sometimes that cannot be done.

If the loads can vary, the resultant location is not constant.In that case, we have to design for multiple load possibilities..


Combined Footing Design

W= 8 Feet (Y direction)

L = 1 0.25 Fee t ( X d ir ect io n)

Lc= 30Distance Between Columns - Inches (same direction as L)

ec1= 8Distance From Left Column to Center of Footing (inches)

DL LeftColumn 85K Case 1= Full Load Both Columns

LL Left Column 94K Case 2= DL Left Col, Full Load Right Col

DL RightColumn 20K Case 3= Full Load Left Col, DL Right Col

LL RightColumn 40K Case 4= DL Only, Both Columns

Case 1 2 3 4

Left Col = 179 Left Col = 85 Left Col = 179 Left Col = 85 Loads in Kips

Right Col = 60 Right Col = 60 Right Col = 20 Right Col = 20 Loads in Kips

LC t o Res= 7.53 LC to Res= 12.41 LC to Res= 3.02 LC to Res= 5. 71 Thi s is the di stanc e f rom lef t c ol to resultant (inc hes)

e= -0.47 e= 4.41 e= -4.98 e= -2.29 This is the distance from center of footing to resultant (inches)

M dl= -4.10 M dl= 38.62 M dl= -43.62 M dl= -20.00 Dead Load Moment (K-Ft)

M ll= -5.23 M ll= 14.71 M ll= -39.05 M ll= 0.00 Live Load Moment (K-Ft)

Iy= 717.93

Actual Loads

P/A= 2.91 1.77 2.43 1.28 Soil Pressure (Ksf)M c / I= -0.07 0.38 -0.59 -0.14 Soil Pressure (Ksf)

qs max= 2.98 2.15 3.02 1.42 Max Soil Pressure (Ksf)

qs min= 2.85 1.39 1.84 1.14 Minimum Soil Pressure (Ksf)

Factored Loads (1.2*DL + 1.6*LL)

Pu= 340.40 190.00 276.40 126.00 K

Pu/A= 4.15 2.32 3.37 1.54 Soil Pressure (Ksf)

LC to Res= 7.76 13.89 2.60 5.71 Inches

e= -0.24 5.89 -5.40 -2.29 Inches

Mu c / I= -0.05 0.67 -0.89 -0.17 Soil Pressure (Ksf)

qu max= 4.20 2.98 4.26 1.71 Max Ultimate Soil Pressure (Ksf)

qu min= 4.10 1.65 2.48 1.37 Minimum Ultimate Soil Pressure (Ksf)

If Mc/I is ( - ) Then q max is on the left side of the footing


A spreadsheet will be used to keep track of all possible load combinations.

ae303 combined footing design

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