Bolting a Wood Ledger to Flat Face Concrete Wall

Consider a 2 x 10 ledger connecting a wood frame floor to the flat (vertical) face of a concrete wall. The joists connect to the ledger via Joist Hangers. The ledger will be bolted to the wall with 5/8 in. diameter bolts either cast-in-place in the wall, or post-installed after the concrete cures. The focus of our calculations will be making sure the ledger adequately transfers the gravity load (weight of floor and whatever the floor is carrying ... people, furniture, etc.) from the joists to the wall. We will assume that the joist hangers are adequate; our role is to determine a bolt pattern that will resist the shear in the connection (downward action of the ledger with respect to the wall) as well as the prying action (tending to also pull the bolts out of the concrete).

Bolts Resisting the Downward Shear

The downward shearing will be resisted by bolts that bear on holes drilled in the wood ledger and are embedded in the concrete. In such a connection both wood and concrete (and steel) must be considered. Table 11E of the National Design Specification for Wood Construction (NDS) is a `perfect' resource for such, providing lateral reference design values for sawn lumber to concrete connections. The Table provides two sets of values (for various bolt sizes, wood species, member thickness, etc.), Z<sub>para</sub> and Z<sub>perp</sub>, which refer to direction of load parallel and perpendicular to the grain, respectively. In our case of a downward load and a horizontal ledger, flat against the wall, the load will be perpendicular to the grain. Our ledger thickness (`Side Member' in the Table) will be 1-1/2 in. actual (2 in. nominal). Using a Douglas fir ledger, we get Z_perp = 530 lb for a 5/8 in. diameter bolt. In wood design the reference design values from such a table need to be adjusted for each application, but in our example of carrying Occupancy floor live and dead loads, and as long as our ledger is dry and other things are `normal', the adjustments are essentially unity, and we get an adjusted value of 530 lb (per bolt).

For the load, let's consider the joists spanning simply 15 ft to some other support; and let's also use 40 psf Occupancy Live load (for Residential construction) and 10 psf Dead load (light frame floor system) for a total load, σ, of 50 psf. For simple spans half the load goes to each end; so, the line load, ω, from the floor into the wall is ...

... ω = σ x tributary width = 50 psf x ½ of 15 ft = 375 plf.

For a required bolt spacing, s, for shear, then,

... s = Z_perp / ω = 530 lb per bolt / 375 lb per foot = 1.41 ft per bolt ... or ... 16.96 in. per bolt.

... 16.96 is a rather awkward number, so we would say either 17 in. `on center' (o.c.), or, more probably, rounding down to an even integer, ... 16 in. o.c.

But wait!

Prying of the Ledger off the Wall

Since the joists bear downward some distance away from the face of the wall, there will be a tendency to pry the ledger off the wall. We will use bolts to resist this prying.

First let's get the prying load. Assuming the joists bear on 1-1/2 or so of hanger, and the hangers themselves are a tiny bit away from the face of the ledger, the resultants of the loads at the ends of the joists could be, say, 1.0 in. from the face of the ledger, plus1-1/2 in. the thickness of the ledger itself, equals 2-1/2 in. away from the concrete face.

Let's use two rows of bolts: one high and one low. To prevent splitting of the ledger we must install the bolts not closer than 4 bolt diameters from the edge (top edge in particular, and the measurement is bolt center to wood edge), ... so, 4 x 5/8 = 2.5 in.; let's say a nice round 3 in . Also, the rows must be spaced, another 4 diameters; say another 3 inches. If our ledger is 2 x 10 (which we have already said it is), then we have about 3 in. remaining (9.25 in. actual depth minus 3 in. minus 3 in.). And if we resist the prying action with the tension in the bolts on top, and compression of wood against concrete on the bottom, ... moment arm, say, 3.5 in., to balance the twist ...

T(approx 3.5 in.) = R( 2.5 in.),

where T is the tension in the upper bolt, and

R is the downward gravity load.

Being a bit conservative, and simplifying, let's round the distance of the downward loads from the face of concrete up to 3, and the moment arm down to 3, so ...

T(3) = R(3) ...

In our case, then, in terms of `along the wall' ...

T = 375 plf (same number, but prying instead of down)

And now let's resist this prying load with the top bolts.

The International Building Code (IBC) gives allowable service loads for bolts in concrete (Table 1912.2). Both shear and tension (pullout) loads are provided. For a 5/8 in. bolt in tension, embedded 4-1/2 in. minimum, far from an edge and distant from other bolts, the allowable service load (per bolt) is 2125 lb (in 2500 psi concrete). That is a fabulously large number, relatively. But, we can't really use it. Like in shearing, the wood will govern, and the wood isn't `part of the picture' in Table 1912.2. Table 1912.2 only considers the concrete and the bolt. In our situation what will govern will be the washer on the outside of the ledger as it bears against the wood (and nut). Washer sizes vary, but for a relatively modest size washer of outside diameter 1-5/16 in. the washer will bear on about 1.0 sq. in. of wood. The allowable compression perpendicular to grain for Douglas fir is 625 psi (NDS Supplement, Table 4A). So, 625 psi x 1.0 sq. in. gives ... 625 lb capacity for the bolt in tension (based on the washer). Obviously a bigger washer will bear on more wood, and so on. In a real rough sense (and say, the bolt hole got drilled a bit big), we could say, and simplifying, 500 to 600 lb tension per bolt (near the same number as the capacity in shear).

With the washers we have assumed, ...

... spacing = s = 500 or so lb per bolt / about 375 lb per foot = ... the same spacing as the bolts for shear.

So, we'll have two rows of bolts. The bottom row will carry the shear, and the upper row will resist the prying effect.

Baby!

Solution

So, here is our solution. We will have two rows of bolts: one high, and one low. We'll satisfy edge distance requirements; and well accommodate the prying effect of the downward load. We will be a bit conservative, but why split hairs when it comes to hanging a floor on a wall?

... two rows of 5/8 in. diameter bolts, staggered, spaced 16 in. o.c. each row, with 3 in. min. edge distance bottom and top and 3 in. min. spacing between rows. Bolts shall be cast-in-place with min. embedment of 6 in. or post-installed equivalent. Washers for bolts shall provide at least 1.0 sq. in. of wood bearing.

Depending on the location of the ledger, it may need to be treated to resist decay due to moisture.

The cool thing is that, perhaps as a rule of thumb, or an approximation, we can calculate the bolts necessary for shear, and then provide another row of the same bolts, high, at the same spacing, to resist the prying.

Obviously, for different loads, size bolts, washers, ledgers, etc. ... would produce different answers than the above. (And if we want fewer bolts on top, the fastest way to reduce the number would be to increase the washer size.)

Also, 6 in. of embedment into a wall might be pretty deep. Obviously, we would need to use a solution that accommodates required embedment, concrete cover, etc.

References

National Design Specification for Wood Construction, and Supplement, 2005, 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.

Spacing of Nails, Bolts, and Rebar, Jeff Filler, Associated Content.

This Thing Called Tributary Width, Jeff Filler, Associated Content.