A More Efficient Ledger to Concrete Bolt Determination

In our previous discussion (here) we came up with the bolts connecting a ledger to a flat face concrete wall. First we determined the bolt spacing required to resist the downward shear from the weight of the floor (and occupants, etc.). This was pretty straightforward; we simply used the lateral design reference values for bolts perpendicular to grain from Table 11E in the National Design Specification (NDS). But we acknowledged that the downward force from the floor, acting some distance from the face of the wall, will also tend to pry the ledger off the wall. And we came up with an easy remedy: let's add another row of bolts, high, same diameter, same spacing, staggered with the first row (our `bottom' row). Easy. Done. And the `determining factor' for the bolts resisting prying was the washer bearing on the face of the ledger. Now let's see what happens if we do some more refined calculations, and, if necessary, increase the size of the washers (for the bolts resisting prying).

Geometry of Prying

In the previous calculations we made some very generous (conservative) assumptions with regard to the geometry of prying. Now let's look at things a bit closer. If the upper bolts are centered 3.0 in. from the top of the ledger, and the bottom half of the ledger is in compression, with an assumed linear pressure distribution, the `moment arm' between the bolt in tension and the resultant compression force will be,

... d = (9.25 - 3.0 0) - (1/3^rd of ½ of 9.25) = 4.71 in. (and this is still probably a bit conservative).

And now if we assume that the joist hangers allow the joists to rest pretty close to the face of the ledger, and we calculate that the joists only need about ½ in. of actual bearing, we'll get, for the distance of the resultant downward load from the face of concrete ... call it `x' ...

... x = 1.5 (thickness of ledger) + ½ of 0.5 in. for (bearing distance) = 1.75 ... but let's round to 2.0 for not perfect fit, etc. ...

... x = 2.0 in.

That means our prying tension becomes (balancing the prying with the tension in the anchors) ...

T(4.71) = R(2) ...

T = R (2 / 4.71) = 375 plf (2 / 4.71) = 159 plf.

Upper Bolts Carrying BOTH Prying Tension and Downward Shear

Let's go back to our original bolt spacing based on shear. We obtained bolts @ 1.41 ft o.c., which we rounded (down) to 16 in. o.c. Now let's take these bolts, divide them into two rows, one high and one low, and let the high ones carry the prying tension, in addition to their share of shear. These upper bolts will now be spaced at 32 in. o.c. (2.66 ft o.c.). The prying tension demand on these bolts will be ...

... tension demand = 159 lb/ft times 2.66 ft per bolt = ... 424 lb per bolt.

We already saw that we can achieve this with a modest size washer (let alone a big one).

The wood fibers against the washer on the face of the ledger are different fibers than those resisting bearing in the bolt hole, so we'll assume that the washer and bolt bearing do NOT interact (or at least interact negatively). Thus, to examine the upper anchors acting in both shear and tension, we look at the bolt and the concrete. Equation 19-1 of the International Building Code (IBC) is just what we need.

(IBC Equation 19-1) ... (P_s / P_t)^5/3 + (V_s / V_t)^5/3 ≤ 1.00? ...

where

P_s = the design service pullout (tension) load on the bolt,

P_t = the allowable pullout load (from IBC Table 1912.2),

V_s = the design service shear load,

V_t = the allowable shear (Table 1912.2), and

... the ^5/3 is for an `elliptical' or `oval' interaction between shear and pullout.

Well, let's see ...

From Table 1912.2, using 2500 psi concrete, 5/8 in. diameter bolts, and enough embedment, edge distance, spacing, etc. ...

P_t = 2125 lb,

V_t = 2950 lb,

P_s = 424 lb, and

V_s = ... 530 lb (16 in. o.c. / 17 in. o.c.) = 499 lb ( ... we're placing a little bit less demand than capacity since we're spacing them out just a bit).

So,

Is ... (P_s / P_t)^5/3 + (V_s / V_t)^5/3 = (424 / 2125 )^5/3 + (499 / 2950)^5/3 = 0.07 + 0.05 = 0.12 ≤ 1.00? ...Yes!

Conclusion(s)

What does this mean? Well, at least in this case, it means that we can determine the number of anchor bolts by considering shear alone, and then as long as we stagger these bolts, half high, and half low, we will also adequately resist prying.

... 5/8 in. anchor bolts in two rows, staggered, 32 in. o.c. each row, upper row centered 3 in. from top of ledger, lower row centered 3 in. from bottom of ledger.

Note: this above more refined, more efficient, spacing compares better with prescriptive bolt spacing provided in the International Residential Code (Table R611.8(1)), which is comforting.

However, if we are unable to follow these calculations, or if we are unable to fit (or find our way) in the prescriptive bolt determination, the earlier determination of one row for shear, and another (equal) row, high, for prying, also works.

These calculations assume 2 x 10 nominal size ledger. A different size ledge (depth or thickness) will change the calculations. (Maybe we'd need bigger washers.)

References

Bolting a Wood Ledger to a Flat Face Concrete Wall, Jeff Filler, Associated Content.

National Design Specification for Wood Construction, and Supplement, 2006, American Forest & Paper Association / American Wood Council, 1111 Nineteenth St., NW, Suite 800, Washington, D.C., 20036, www.awc.org.

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

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