Chapter 8. Example 4: Drilled-Shaft Footing
8.4 Design Calculations (First Load Case)
8.4.1 Step 1: Determine the Loads
The forces imposed on the column will flow through the footing to each of the four drilled shafts. Please recall that strut-and-tie models (i.e., truss models) are incapable of resisting bending moments. In order to properly model the flow of forces through the footing, the axial force and bending moment (Figure 8.2) need to be redefined in terms of an equivalent force system (refer to Figure 8.4). The equivalent set of forces will be applied to the strut-and-tie model and, by definition, should produce the same axial load and moment as those shown in
Figure 8.2. Since the applied forces flow through the footing to four drilled shafts, the equivalent set of forces should consist of four vertical loads, each corresponding to a drilled- shaft reaction (each force in Figure 8.4(b) represents two of the loads that will be applied to the STM).
(a) (b)
Figure 8.4: Developing an equivalent force system from the applied force and moment To develop the equivalent force system, the elastic stress distribution over the column cross section is determined. The location of each of the four loads comprising the equivalent force system (relative to the column cross section) is then defined. Lastly, the magnitude of each force is calculated by establishing equilibrium.
The column cross section and corresponding linear stress distribution are illustrated in Figure 8.5. The positions of the four loads that comprise the equivalent force system are also shown in the column cross section. The two loads acting on the left are compressive (pushing downward on the footing), while the two loads acting on the right are tensile (pulling upward on the footing).
C
T x
z
C
T
=
Applied Force and Moment Equivalent Force System
Figure 8.5: Linear stress distribution over the column cross section and the locations of the loads comprising the equivalent force system (first load case)
The locations of the two compressive forces are based on the linear stress diagram. The line of action for both forces coincides with the centroid of the compressive portion of the stress diagram, located 1.72 feet from the left face of the column. The compressive forces are transversely positioned at the quarter points of the column depth, Dcol, or 1.56 feet from the top and bottom of the column section in Figure 8.5.
The longitudinal column steel configuration of Figure 8.5 (detailed in Figure 8.6 below) is an assumption. This reinforcement should be specified on the basis of the final column design, which is beyond the scope of this design example. The reinforcement on the right face (or tension face) of the column will be most effective (relative to the bars elsewhere in the cross section) in resisting the tension due to the applied bending moment. The two tensile forces (which complete the equivalent force system) are therefore conservatively assumed to act at the centroid of this tension-face reinforcement, located 0.30 feet from the right face. Moreover, each of the tensile forces is transversely positioned at the centroid of either the lower or upper half of the selected column reinforcement. Each of the vertical ties (corresponding to the tensile forces) located beneath the column (Ties BI and CJ in Figure 8.8) therefore consists of six bars.
Dcol= 6.25’ 1.56’1.56’3.13’
1.72’ 3.44’ 2.35’
5.15’ 2.35’
1.58’
Wcol= 7.50’
3.44’
x y
1549 psi
Centroid of 6 – No. 11 Bars Centroid of 6 – No. 11 Bars
Column Bars Considered to Carry Forces in Ties BI and
CJ of STM
Neutral Axis
0.30’
A B
D C
C
T
= Applied Load
Figure 8.6: Assumed reinforcement layout of the column section
The magnitudes of the compressive and tensile forces are now determined so that the equivalent force system produces the same axial load, moment, and linear stress distribution as those respectively shown in Figures 8.2 and 8.5. This is accomplished by establishing equilibrium between the original and equivalent force systems. In the following equations, FComp
is the total compressive force acting on the footing, or the sum of the loads acting at points A and D in Figure 8.5, and FTens is the total tensile force, or the sum of the loads acting at points B and C.
(
) (
) Solving:
In the second equation, 7.50 ft is the value of Wcol, and 1.72 ft and 0.30 ft are taken from Figure 8.5. The four loads acting on the STM from the column are then determined as follows:
( om pression)
(Tension) These forces are shown acting on the STM of Figure 8.8.
x y
2.25” Clear
No. 5 Stirrups Wcol= 7.50’
12 – No. 11 Bars
Dcol= 6.25’ 10 –No. 11 Bars
11 Equal Spaces
11 Equal Spaces