*(Draft – Comments Welcome)*

**Outline**

- Problem Statement
- Approach
- Factored Lateral Load on Wall
- Factored Strength
- Horizontal Reinforcement
- Conclusion (thus far)
- References

**1. Problem Statement**

Given: a popular Insulated Concrete Form (ICF) system is to be used for a rather long basement (retaining) wall. The concrete core for the wall is 6-1/4 thick and has `webs’ that facilitate placement of vertical reinforcement at a distance of 1-13/16 in. (bar centerline) from the face of the concrete. For convenience it is preferred to have one horizontal reinforcement bar each ICF course, and the courses are 16-3/4 in. high each. The wall height (above footing) is to be 8′-4″ (6 courses) and the wall is supported laterally at the top by a wood frame floor system (and by concrete basement slab at bottom). The backfill against the wall will consist of a 12 in. gravel drain `chimney’ with silty sand native fill materials beyond.

Determine (and detail): the vertical and horizontal reinforcement required. Use Gr. 60 ksi reinforcement and 3000 psi (28-day compressive strength) concrete.

**2. Approach**

a) V.S. Location – I will detail the vertical rebar to be (centered) 1-13/16 in. from the inside face and thus consider the wall as acting one-way spanning from basement floor to top of wall with the tension face of the concrete being on the inside (basement living side) of the wall (away from the soil).

b) H.S. Location – will be spaced at 16-3/4 in. o.c. and shall be used to `tie’ the V.S. in place per the ICF manufacturer recommendations. (The H.S. will thus be *not* centered in the wall.)

c) Lateral Earth Pressure – in the absence of a soils investigation I will use the pressure provided in Chapter 16 of the *International Building Code* (presumably adopted by the local building authority).

d) Unbalanced backfill Height – model codes will generally require that the finish grade of the backfill not be any higher than 8 in. below the wood framing. For convenience I will design using a triangular pressure distribution for the full height (8′-4″) of the wall. This approach will be conservative as it does not account for the thickness of the basement floor; nor does it take advantage of the 8 in. clear distance to the wood above. However, the design excludes any surcharge on the wall. I am thus assuming that by using the full wall height I will accommodate modest surcharges.

e) Axial Load on Wall – will not be taken into account. Theoretically the axial compressive force in the wall could be used to offset flexural tension. I will not take this `benefit’ into account, for several reasons: first, the benefit will be modest at best (only reduced Dead loads may be considered); the axial (Dead) load varies; and, finally, if I am going to `split hairs’ and include the axial dead load, perhaps I should *also* determine the effect of the eccentricity of (all of) the axial loads (further benefiting, or *un*-benefiting).

**3. Factored Lateral Load on Wall**

Table 1610.1 of the *International Building Code* provides Active and At-rest design lateral soil loads for various soil types. Since our retaining wall is a rigid reinforced concrete wall we will use the `At-rest’ pressure value, given in the Table to be 60 psf per foot of depth for `Silty sands’. As is common with engineering calculations we will do our `calcs’ on a per-foot-of-wall basis. Thus, our calcs will be for a vertical beam, simply supported top and bottom, that is 6-1/4 in. thick and 12 in. wide.

The pressures (σ) at the top and bottom of the wall are, respectively (wall height 8′-4″ = 8.333 ft),

σ top = 60 psf / ft * 0 ft = 0 psf

σ bot = 60 psf / ft * 8.333 ft = 500 psf

and varies linearly in between.

The tributary width of the `beam’ is (also) 12 in. (1 ft), so the `line’ load on the wall is a distribute load that varies (linearly) from 0 psf x 1 ft = 0 plf at the top of the wall to ω o = 500 psf x 1 ft = 500 plf at the bottom (using the formula ω = σ x tributary width).

The `whole’ (total) load (W) on (1 ft of wall) is (for the thus `triangular’ loading),

W = ½ ω o L,

where L is the height of the wall, so

W = ½ 500 plf (8.33 ft) = 2083 lb (per foot of wall).

The reactions, maximum bending moment, moment location, and shear values for such a load are commonly found in various textbooks and design manuals, and are as follows:

Reaction at top = R top = 1/3 W = 694 lb

Reaction at bottom = R bot = 2/3 W = 1389 lb

M max = 0.1283 W L = 0.1283 (2083 lb) (8.33 ft) = 2227 lb-ft = 26,700 lb-in.

and occurs at 0.5774 L = 0.5774 (8.33 ft) = 4.8 ft from the top

The maximum shears can be taken (conservatively) to equal the reactions, … and deflection of the wall will be assumed to be small and not of importance.

**Factored Loads**

For reinforced concrete it is customary to use the `Strength Design’ approach, where structural members as proportioned (designed) such that their factored strengths are not less than factored loads.

For lateral earth pressures the load factor is 1.6, hence, the factored maximum moment and shear values are (identified by subscript `u’),

Mu = 1.6 * 26,700 lb-in. = 42,800 lb-in.,

and

Vu, bot of wall = 1.6 * 1389 lb = 2222 lb.

**4. Factored Strength of Wall**

The strength of the wall is a function of the effective depth of the steel, steel strength, and concrete strength. The `nominal’ (perfect world) flexure and shear strengths are (denoted by `n’ for moment and `c’ for concrete in shear) …

Mn = bd2 R,

where

R = ρ fy (1 – 0.59 ρ fy / f ‘c)

b = beam width (in this case 12 in.)

d = effective depth (in this case the distance from the compression (soil) face and rebar center) …

… 6-1/4 in. minus 1-13/16 in. = 4.44 in.

ρ = reinforcement ratio = as / (s d)

fy = specified reinforcement yield stress (60,000 psi in this example), and

f’c = specified concrete strength (3000 psi).

One approach in such a design is to pick a reinforcement size and spacing, and compute the strength, and then make sure that the `factored’ strength (factored by the appropriate strength reduction factor) is not less than the factored load.

Try # 5 @ 6 in. o.c.

ρ = as / (s d) = 0.31 in.2 / (6 in. * 4.44 in.) = 0.0116

R = ρ fy (1 – 0.59 ρ fy / f ‘c) = 0.0116 (60,000 psi) (1 – 0.59 * 60,000 psi / 3000 psi) = 600 psi

Mn = bd2 R = 12 in. (4.44 in.)2 (600 psi) = 142,200 lb-in. ( … per foot of wall)

**Factored Strength**

The American Concrete Institute provides (ACI 318) `strength reduction factors’ (φ) as follows:

… for the concrete in shear in a reinforced concrete beam … 0.850.75 …

… for flexure … 0.90

(These factors reduce our calculated `perfect world’ strength values to account for uncertainties in materials, workmanship, etc.)

Hence, for the bending (flexure) part of our design, the factored (reduced) strength is,

… φ Mn = 0.90 (142,200 lb-in.) = 127,800 lb-in.

We may now ask if # 5 @ 6 in. o.c. is sufficient …

Is `factored load’ less than or equal to `factored strength’? … or, in equation form …

Is … Mu = 42,800 lb-in. ≤ φ Mn = 127,800 lb-in.?

The answer is yes, way yes!

In fact, the answer is *so much so* (yes) that we might consider increasing the spacing and/or changing the bar size.

Other things equal, flexural strength varies *approximately in proportion to the amount of steel*. Hence, in this example, since we are `over-strength’ by a factor of about 128 / 43 = 2.97, let’s try spacing the steel out by about that same proportion.

Our `new’ spacing becomes … 2.97 x 6 in. = 17.9 in. … let’s try s = 16 in.

(It turns out that the `webs’ in the ICF system being used are 8 in. o.c., so 16 in. o.c. will coincide with every other web. Sweet, *hopefully*.)

Re-doing the calcs with s = 16 in. …

ρ = 0.31 (16 * 4.44) = 0.0044

R = … 248 psi

… φ Mn = 52,800 lb-in.

Is … Mu = 42,800 lb-in. ≤ φ Mn = 52,800 lb-in.? … YES!

We could space the bars at 18 in. o.c., but the forming system is more amiable to 16 (so, let’s stay at 16). *Plus* in no case should we space the bars *more than* 18 in. o.c., if we want the wall to indeed be a `wall’ (ACI 318, Ch. 14).

It is doubtful that a smaller bar size would work (at this large of spacing).

If it’s not already obvious, at some point you will start using a spreadsheet or design software to do these calcs. Not only will it save time – it will save you with number *punching* errors -as the calculations are somewhat cumbersome.

Now we are `over-strength’ by 52,800 / 42,800 = 1.23 or 23 % … that’s more reasonable. We could alternately state that the ratio of the reinforcement required to that which is provided is 42,800 / 52,800 = 0.81 … a number that we might end up using later (in development length calculations).

Let’s check shear.

The shear strength can be taken to be …

Vc = 2 b d √f ‘c = 2 (12 in.) (4.44 in) √3000 psi = 5833 lb.

The reduced shear strength is … φ Vc = 0.850.75 (5830 lb) = 49604373 lb.

Since Vu, bot of wall= 2222 lb ≤ φVc= 4960 4373 lb … we’re good! (in shear)

**5. Horizontal Reinforcement**

For horizontal reinforcement I will use the ACI 318 Ch. 14 minimum requirement, which is …

ρ min = 0.0020 for Gr. 60 rebar not larger than # 5.

In consideration of the ICF system being used, let’s try … # 5 @ 16 in. o.c., and note that the steel ratio is based on the full thickness of the wall …

ρ = 0.31 in.2 / (16-3.4 in. * 6-1/4 in.) = 0.0030 …

Is 0.0030 ≥ ρ min = 0.0020 ? … yes!

So much so, … let’s try # 4 …

ρ = 0.20 in.2 / (16-3.4 in. * 6-1/4 in.) = 0.0019 …

Is 0.0019 ≥ ρ min = 0.0020 ? … barely not.

So, let’s use the # 5 @ 16 in. o.c.

(Note: we could argue that only temperature – shrinkage steel is necessary in the horizontal direction, for which the minimum would be 0.0018. Yeah, we could.)

So, let’s summarize where we are …

V.S.: # 5 @ 16 in. o.c. placed at the `inboard’ tabs in the webs (1-13/16 in. from inside concrete face to bar centerline.)

H.S.: # 5 @ 16-3/4 in. o.c.

We need a rest. As we continue we will investigate:

a) maximum and minimum amounts of (flextural) reinforcement

b) development of the reinforcement (`bond’)

c) embedment of V.S. into the footing

d) detailing the termination of the V.S. at the top of the wall

e) splices in the reinforcement

Normally we would also tackle (more deliberately) the issues of cover and spacing of reinforcement, but in this example (rightly or wrongly) we will assume that the ICF manufacture has already tackled these issues when providing the rebar location tabs.

**6. Conclusion (thus far)**

We have now come up with the salient features of the wall reinforcement for a one-way basement retaining wall design. The rebar is on the inboard side (near the inside face of the wall) where the wall will be in tension. So far our design is structurally sound and accommodating to the particular forms to be used. There are places where we could have `split hairs’, but we have chosen not to, giving us a design that is a bit conservative, and (or but), in the end, is probably not a whole lot different than if we had (split hairs). As the design continues, details such as splices, embedment, bar termination at the top will be hammered out. But for now we need to rest.

Continuation is … here

**7. References**

*Building Code Requirements for Structural Concrete,* ACI 318, American Concrete Institute, P.O. Box 9094, Farmington hills, Michigan, 48333.

*International Building Code*, International Code Council, 4051 West Flossmoor Road, Country Club Hills, IL 60478.