This design example shows the typical design of a reinforced concrete wall footing under concentric loads. It was originally designed and used in the following reference:

James Wight, Reinforced Concrete Mechanics and Design, 7th Edition , 2016, Pearson, Example 15-1

This is a very thorough textbook on reinforced concrete and we recommend it as a reference for concrete design in the United States.

Problem Statement

Design a reinforced concrete to support a concrete wall in a relatively large building. The wall is 12 inches thick and carries unfactored dead and live loads of 10 kip/ft and 12.5 kip/ft respectively. The allowable soil pressure is 5,000 psf and the its density is of 120 pcf. The bottom of the footing should be at 5 ft below ground level. Concrete strength is 3,000 psi and reinforcement strength is 60,000 psi.

Design Criteria

We will design our footing to resist its load and check it for:

• Soil bearing pressure
• Shear resistance
• Flexural resistance

We enter the given information directly into ClearCalcs. Since in this case we are given the depth to the bottom of the footing, we can enter "=5 ft -H", and the calculator will automatically update the depth of soil above the footing when we update the footing thickness - just like an Excel spreadsheet.

Looking at the reinforcement section, the concrete cover is already set to 3 inches (the minimum for footings) and the steel strength is already 60 ksi. All that's left here is to find the size and spacing required.

Entering our loads:

Notice that we don't use the reduced companion live load - in this case, since we only have dead and live loads, this won't affect the results, and since we don't know the source of the live load it's conservative not to reduce the live load. See ASCE 7-16, Cl 2.3.1 for more information.

Soil Bearing

The first thing to do is to determine the width of our footing, which is determined by the allowable soil bearing capacity. We need to estimate the required thickness of the footing, since the self-weight of the footing is usually quite significant.

The textbook recommends using a value of 1-1.5 times the wall thickness for the footing thickness. In the example, they first try with a 12 inch thick footing. Once we have this, we can calculate the self-weight:

S W = 12 in ⋅ 150 lb ft 3 = 150 psf SW = 12 \text{ in} \cdot 150 \frac{\text{lb}}{\text{ft}^3} = 150 \text{ psf} SW=12 in⋅150ft3lb​=150 psf

Once we know the self-weight, we immediately remove it from the allowable bearing pressure, together with the weight of the soil above the footing, and then divide the total load by this adjusted bearing pressure to find the required area.

A r e q ′ d = 10 kip/ft + 12.5 kip/ft 5000 psf − 150 psf − 4 ft × 120 pcf = 5.15 ft 2 ft A_{req'd}= \frac{10\text{ kip/ft} + 12.5 \text{ kip/ft}}{5000\text{psf} -150\text{ psf} - 4 \text{ ft}\times 120 \text{ pcf}} = 5.15 \frac{\text{ft}^2}{\text{ft}} Areq′d​=5000psf−150 psf−4 ft×120 pcf10 kip/ft+12.5 kip/ft​=5.15ftft2​

We thus select a footing width of 62 inches or 5.17 ft.

Checking in ClearCalcs, we can see that a 5.17 ft wide x 1 ft thick footing efficiently makes full use of the bearing capacity. However, we can already see a storm on the horizon! Our shear capacity may not be quite enough with only 12" of thickness, and our reinforcement can't fully develop - we'll have to do something about that...

Shear Capacity

After the little sneak peek we saw when checking soil bearing, we definitely want to take a look at shear. This is usually what will govern the footing's thickness in design. Since we are now dealing with concrete design, we use the ACI 318-14 standard, which is based on LRFD design. We thus need to factor the loads.

We can find a value for q u q_u qu​, the soil pressure at the factored load level, by dividing our total applied load by the footing area. In this case since we only have dead and live loads, it is clear that the governing load combination will be 1.2D + 1.6L.

q u = 1.2 × 10 kip/ft + 1.6 × 12.5 kip/ft 5.17 ft = 6190 psf q_u = \frac{1.2 \times 10\text{ kip/ft} + 1.6 \times 12.5 \text{ kip/ft}}{5.17 \text{ ft}} = 6 190 \text{ psf} qu​=5.17 ft1.2×10 kip/ft+1.6×12.5 kip/ft​=6190 psf

Note that we are taking the net bearing pressure, which does not include the weight of the soil above the footing and the self-weight. This is because these weights are cancelled out by their corresponding upwards soil reaction when considering the footing as a free-body.

With our new-found value of q u q_u qu​, we can find the factored shear. The ACI-318-14 code (Cl 7.4.3.2) specifies that the critical shear section should be taken at a distance d d d from the face of the wall. With our 12-inch thick footing, we need a minimum of 3 inches cover (ACI 318-14, Table 20.6.1.3.1). Assuming #8 size reinforcement (1" diameter), we can find d:

d = 12 in − 3 in − 1 2 × 1 in = 8.5 in d = 12\text{ in} - 3\text{ in} - \frac{1}{2}\times1\text{ in} = 8.5\text{ in} d=12 in−3 in−21​×1 in=8.5 in

We can now calculate the shear at the critical section:

V u = q u ( B 2 − b 2 − d ) = 6190 psf ( 62 in 2 − 12 in 2 − 8.5 in ) = 8.51 kip/ft \begin{aligned} V_u &= q_u \left(\frac{B}{2} -\frac{b}{2} -d \right) \\\ &= 6190 \text{ psf} \left( \frac{62\text{ in}}{2} -\frac{12\text{ in}}{2} - 8.5\text{ in}\right) \\\ &= 8.51 \text{ kip/ft} \end{aligned} Vu​ ​=qu​(2B​−2b​−d)=6190 psf(262 in​−212 in​−8.5 in)=8.51 kip/ft​

We must now find the shear resistance. Footings almost never have shear reinforcement - it is usually preferable to increase the footing thickness. We thus only need to calculate the factored concrete shear strength ϕ V c \phi V_c ϕVc​, which is given by ACI 318-14 Cl 22.5.5.1 :

ϕ V c = ϕ 2 λ f c ′ d \phi V_c = \phi 2\lambda \sqrt{f'_c}d ϕVc​=ϕ2λfc′​ ​d For shear, ACI 318-14 Table 21.2.1 specifies ϕ = 0.75 \phi = 0.75 ϕ=0.75 and we're using normal-weight concrete so λ = 1.0 \lambda = 1.0 λ=1.0. ϕ V c = 0.75 × 2 × 1 × 3000 psi × 8.5 in = 8.38 kip/ft \begin{aligned} \phi V_c &= 0.75 \times 2 \times 1 \times \sqrt{3000} \text{ psi} \times 8.5 \text{ in} \\\ &= 8.38 \text{ kip/ft} \end{aligned} ϕVc​ ​=0.75×2×1×3000 ​ psi×8.5 in=8.38 kip/ft​

As we had predicted with ClearCalcs in the previous section, we find that V u > ϕ V c V_u > \phi V_c Vu​>ϕVc​. At this point, we could either increase the concrete strength, increase the footing thickness or decide to add shear reinforcement. As previously discussed, shear reinforcement is usually avoided in footings and the concrete strength was already specified, so we choose to increase the thickness.

Increasing the thickness benefits shear resistance in two ways. First, it increases the capacity by providing a greater value of d d d. It also reduces the applied shear load since we are taking our critical section further away from the wall face. We pick a 13-inch thick footing and repeat the previous steps:

d = 9.5 in V u = 8.01 kip/ft ϕ V c = 9.37 kip/ft \begin{aligned} d &= 9.5 \text{ in} \\\ V_u &= 8.01 \text{ kip/ft} \\\ \phi V_c &= 9.37\text{ kip/ft} \end{aligned} d Vu​ ϕVc​​=9.5 in=8.01 kip/ft=9.37 kip/ft​

We see that the 1-inch increase both decreased V u V_u Vu​ and increase ϕ V c \phi V_c ϕVc​ as we liked.

Verifying with ClearCalcs, we can now look at the results again with a 13-inch thick footing:

We see that we went down from 102% to 85% utilization in shear, and the increase in bearing stress was negligible.

Flexural Capacity

The last failure mode which we need to check is the bending of the footing. We essentially have a cantilevered out concrete slab, with a uniformly distributed load from the soil's upward pressure. In the code, it is specified that we should take our critical section for bending at the column face (ACI 318-14, Cl 13.2.7.1). We can thus easily calculate the bending moment, using the typical equation for a cantilever beam:

M u = q u 2 ( B 2 − b 2 ) 2 = 6190 psf 2 ( 62 in 2 − 12 in 2 ) 2 = 13.5 kip-ft/ft \begin{aligned} M_u &=\frac{q_u}{2} \left(\frac{B}{2} - \frac{b}{2} \right)^2 \\\ &= \frac{6190\text{ psf}}{2} \left( \frac{62\text{ in}}{2} -\frac{12\text{ in}}{2}\right)^2\\\ &= 13.5 \text{ kip-ft/ft} \end{aligned} Mu​ ​=2qu​​(2B​−2b​)2=26190 psf​(262 in​−212 in​)2=13.5 kip-ft/ft​

Using the familiar approximation to find the required area of steel (with M u M_u Mu​ in kip-ft \text{kip-ft} kip-ft and d d d in inches):

A s ≈ M u 4 d = 13.5 kip-ft/ft 4 × 9.5 in = 0.355 in 2 /ft \begin{aligned} A_s &\approx \frac{M_u}{4d} \\\ &=\frac{13.5 \text{ kip-ft/ft}}{4 \times 9.5 \text{ in}} \\\ &= 0.355 \text{in}^2\text{/ft} \end{aligned} As​ ​≈4dMu​​=4×9.5 in13.5 kip-ft/ft​=0.355in2/ft​

Note that the Reinforced Concrete Mechanics and Design textbook makes use of a slightly less conservative approximation and finds

A s = 0.330 in 2 /ft A_s = 0.330\text{ in}^2\text{/ft} As​=0.330 in2/ft.

We must also verify that we are meeting minimum steel area requirements are met:

A s = 0.0018 h = 0.0018 × 13 in × 12 in/ft = 0.281 in 2 /ft A_s = 0.0018h= 0.0018\times 13 \text{ in} \times 12 \text{ in/ft} \\\ = 0.281 \text{in}^2\text{/ft} As​=0.0018h=0.0018×13 in×12 in/ft =0.281in2/ft

And the maximum spacing is the minimum of 3 H 3H 3H and 18 inches - the latter usually governs for footings. With these criteria in mind, we can select our reinforcement - using the textbook's approximation for required steel area, we find we can use either #5 bars at 11 inches O.C. or #4 bars at 7 inches, which both provide

A s = 0.34 in 2 /ft A_s = 0.34\text{ in}^2\text{/ft} As​=0.34 in2/ft.

Finding the actual moment resistance now:

a = A s f y 0.85 f c ′ b = 0.34 in 2 /ft × 60000 psi 0.85 × 3000 psi × 12 in/ft & = 0.667 in \begin{aligned} a &=\frac{A_sf_y}{0.85 f'_c b} \\\ &= \frac{0.34\text{ in}^2\text{/ft} \times60000 \text{ psi}}{0.85 \times 3000\text{psi} \times12 \text{ in/ft}}\\\&=0.667 \text{ in} \end{aligned} a &=0.667 in​=0.85fc′​bAs​fy​​=0.85×3000psi×12 in/ft0.34 in2/ft×60000 psi​​

With such a small value of a a a, it's clear that our footing will be tension controlled and thus ϕ = 0.90 \phi = 0.90 ϕ=0.90. We can find the moment capacity.

ϕ M n = ϕ A s f y ( d − a / 2 ) = 0.90 × 0.34 in 2 /ft × 60000 psi ( 9.5 in − 0.667 in 2 ) = 14.0 kip-ft/ft \begin{aligned} \phi M_n &= \phi A_s f_y\left(d -a/2 \right) \\\ &= 0.90 \times 0.34\text{ in}^2\text{/ft} \times 60000 \text{psi} \left(9.5\text{ in} - \frac{0.667\text{ in}}{2} \right) \\\ &= 14.0\text{ kip-ft/ft} \end{aligned} ϕMn​ ​=ϕAs​fy​(d−a/2)=0.90×0.34 in2/ft×60000psi(9.5 in−20.667 in​)=14.0 kip-ft/ft​

Note that in this example, d d d was kept at 9.5 inches even though it would be slightly larger, since we are using #4 bars with half the diameter d b d_b db​. This is conservative and simplifies calculations somewhat. Nevertheless, we see that ϕ M n > M u \phi M_n > M_u ϕMn​>Mu​ so our design is adequate.

Checking with ClearCalcs:

We can clearly see that indeed we have a higher capacity. Note that we automatically calculate the depth to reinforcement - thus the increase in d d d from using a smaller bar is automatically calculated which provides us with slightly more capacity! Opening our size selector (the filter button circled in dark blue), we see that at this spacing, #4 bars are the most optimal.

Development of Reinforcement

The last check we perform is on the development length, to ensure we have proper bonding of our reinforcement at the critical section. We go to ACI 314-18's chapter 25 to calculate the bonding length. For simplicity, we use Table 25.4.2.2, which gives a simple equation to calculate the development length. We are using a No.4 bar with large spacing, so we can use the least conservative formula as per the table. In this case neither the epoxy or casting position factors which further simplifies our calculation.

ℓ d = f y ψ t ψ e 25 λ f c ′ d b = 60000 psi × 1 × 1 25 × 1 × 3000 psi × 0.5 in = 21.9 in \begin{aligned} \ell_d &= \frac{f_y\psi_t \psi_e}{25 \lambda\sqrt{f'_c}}d_b\\\ &= \frac{60000\text{ psi}\times 1 \times 1}{25 \times 1 \times\sqrt{3000}\text{ psi}} \times 0.5 \text{ in} \\\ &= 21.9 \text{ in}\end{aligned} ℓd​ ​=25λfc′​ ​fy​ψt​ψe​​db​=25×1×3000 ​ psi60000 psi×1×1​×0.5 in=21.9 in​

We find the same value as in the textbook's example. We compare this to the distance to the critical section:

B 2 − b 2 = 5.17 ft 2 − 1 ft 2 = 2.09 ft = 25 in \frac{B}{2}-\frac{b}{2} = \frac{5.17 \text{ ft}}{2}-\frac{1 \text{ ft}}{2}=2.09 \text{ ft} = 25 \text{ in} 2B​−2b​=25.17 ft​−21 ft​=2.09 ft=25 in

Since 25 inches is larger than 21.9 inches, we know our bars are developed as required. Had this not been the case, we could have used hooks at the ends of the bar to significantly reduce the development length, or made use of the more detailed calculations which can be less conservative and more accurate. With ClearCalcs, it is just as easy to perform the more detailed calculations of development length, so this is what to do to provide safe and economical designs.

The development length is reduced by a huge margin when using the detailed equation! This mostly comes from the confinement factor, since our footing has large cover and spacing between bars this greatly benefits the development length.