Spacing of Nails, Bolts, and Rebar

Introduction

A common situation I run into in both the practice and teaching of engineering and building construction is having to calculate the spacingof something like nails, bolts, or steel reinforcing bars. Some parts of the Building Code (e.g., the International Building Code, or the International Residential Code) are quite specific; for example, "8d (nail) @ 16 in. o.c." ... or ... "1/2 in. diameter anchor bolts @ 6 ft o.c. (wording massaged)." We don't have to calculate anything; we just have to do it. But there are plethora situations where we may need to determine the spacing, particularly in `engineered' construction. And, what I find is that the information available to designers, and students, is generally lacking in the how to do such calculations. Or, if it is somewhere in that textbook, or manual, it is hiding. So, here goes.

Nails

A common situation involving nails is one in which we have to transfer a load from one piece of building material to another. For example, consider the attachment of a rim joist to the top plate of a wall using slant nails (toe-nails or stitch nailing) to resist shearing action between the rim joist and top of wall. The situation typically involves the amount of load needing to be transferred, say, per linear foot (also called `unit shear'), and some shear capacity per nail. Determining the shear capacity of the nail in this condition is a conversation in itself, but let's say the allowable shear capacity of this particular nail, in such and such species of wood(s), and in such and such condition(s) ... is 109 lb. We call this allowable shear capacity, with all the particulars of the nail and materials and other conditions involved ... Z'. Now let's say that the unit shear that needs to be transferred is 245 plf (lb/ft). The spacing, s, is ...

... s = Z' / v = ...

where,

... s is the spacing,

... Z' is the allowable shear capacity of the nail adjusted to the particular situation, and,

... v is the unit shear being transferred.

So,

... s = Z' / v = (109 lb / nail ) / (245 lb / ft) = 0.4449 ft / nail.

So, we could say, ... such and such nails @ 0.44 ft o.c. ("on center")

But then someone would say, "What the ...!"

And we would then further calculate ...

... s = (0.4449 ft / nail) x (12 in. per ft) = 5.34 in. per nail ... or ... 5.34 in. o.c.

But that is still kind of weird, and would be very cumbersome to actually execute, so we would `round' the spacing to, say, ... 5 in. o.c. ... or maybe 5-1/4 in. o.c.

(Note that in spacing calculations we round down.)

Someone is going to actually have to mark, `aim', and nail these nails, so while 5-1/4 is more `precise', "5 in. o.c." would be more realistic, and certainly nicer on the person having to mark where the nails will go.

Obviously, by nailing @ 5 in. o.c. we are actually providing more shear capacity than the `v' required. And we could easily calculate the capacity, by rearranging the above equation;

... v = Z' / s = ...

... v = 109 lb per nail / 3 in. per nail = 36.33 lb per in. ... which is ... 36.33 lb per in. x 12 in. per ft ... = 436 plf.

Bolts

The calculation of bolt spacing in many applications is similar to that of nails, except that in many applications we will leave the spacing in terms of `ft'. For example, if this same unit shear (245 plf) is being transferred from a mud sill plate to a concrete wall, using bolts with an allowable shear capacity of, say, 710 lb per bolt, then,

... s = Z' / v = 705 lb per bolt / 245 lb per foot = 2.878 ft per bolt ...

Probably we would round down to ...

... s = 2.5 ft o.c.

Rebar

Spacing of rebar can be a bit different. A common situation in determining rebar spacing is one in which we are given (or have to determine, but I won't go into the `determine' part here) a certain required ratio of reinforcement area to concrete area. For example, to resist the cracking in concrete due to temperature and shrinkage effects, a ratio of steel area to concrete are of 0.0018 is specified for concrete slabs (for a certain grade of reinforcement, etc., etc.). Picture this ratio as the ratio of steel cross section exposed per gross concrete area exposed with an imaginary section or `slice' through (across) the slab. In equation form ...

... ρ = As / Ag ...

where,

... ρ = the reinforcement ratio (prescribed or calculated),

As = total steel cross section area of a reinforcement in the particular cross section, and

Ag = the total (gross) cross section area (concrete and steel).

In terms of calculating spacing of rebar, generally we pick a bar size, and then look at the area of concrete `protected', or `reinforced' by a single bar. In equation form, using this same idea of a ratio of reinforcement,

... ρ = a_s / (s h ) ...

where,

... a_s is the cross section area of a single reinforcing bar,

... s is the spacing that we are after, and

... h is the thickness of the concrete (slab).

Rearranged for s, ...

... s = a_s / (ρ h ).

So, let's say we have a slab that is 6 in. thick, and want to find the required spacing of # 4 bars using this 0.0018 ratio. The cross section area of a single # 4 bar is 0.20 in.²., so

... s = 0.20 in.² / (0.0018 x 6 in.) = 18.5 in. ...

But we wouldn't specify 18 point five ... we'd round (down) to ... 18 in. o.c.

In some cases our calculations (ratios) are based on effective depth (or thickness), d ...

The spacing calc is similar,

... s = a_s / (ρ d).

Sometimes we are given a particular spacing of a bar (size), and are allowed to determine the `equivalent' spacing of another bar (size). We can go back to the `ρ' equation, holding ρ the same ...

... ρ _{new bar} = a_{s new bar} / (s _{new bar} x h) = ρ _{given bar} = a_{s given bar} / (s _{given bar} x h).

Rearranging,

... s _{new bar} = s _{given bar} ( a_{s new bar} / a_{s given bar} ).

Strictly speaking, we don't even need to know ρ or h.

For example, let's say we're given # 7 bars 35 in. o.c., and are asked to determine an equivalent spacing of # 6 bars.

... as for # 7 is 0.60 in.²;

... as for # 6 is 0.44 in.²; so

... s _{# 6} = s _{# 7} ( a_{s # 6} / a_{s # 7} ) = 37 in. ( 0.44 / 0.60) = 27 in. o.c.

This makes sense; if we use a smaller size bar we will need more of them (closer spacing) to provide equivalent strength (or protection).

Cool.

Conclusion

Hopefully this is helpful. Either memorize the equations provided, or write them down, or print this out and be able to tab to them, or remember the principles involved in obtaining them. The same can be done for the spacing of other fasteners or fastening devices. For example, instead of stitch nailing sometimes metal plates are used to transfer the shear load from a rim joist to a top plate. Let's say the allowable capacity of the plate is 450 lb. Then,

... s plate = (450 lb / plate) / (245 lb / ft) = 1.84 ft per plate ... or ... 22 in. o.c.

So,

" ... such and such plates @ 22 in. o.c."

Yeah!

References

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.

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