CALCULATING REPETITIVE FRAMING MEMBERS
If you could see behind the siding or under the carpeting and sub-flooring of a house, you would probably
see wood studs behind the siding and wood joists under the sub-floor. If you climbed into the attic you
would see wood rafters supporting the roof sheathing and roofing. Studs, joists and rafters are all examples
of repetitive framing members.
When architects draw house plans they don’t always show all of the repetitive members but indicate their
presence by writing a note. For example, - 2x6 Studs 16 O.C. Typ. Or 2x12 Flr. Jst. 24 O.C. Typ.
Here is the translation:
 Size of member - 2x6, 2x12
 Use - Wall studs, floor joist
 Spacing - 16 on center, 24 on-center (more about this later)
 Where will this occur - Typ. stands for typical, meaning everywhere.
ON-CENTER (O.C.)
The term on center is used to describe the distance from the center of one framing member to the center of
the next. For example the stud wall in figure 6-1 has studs laid out 12 o.c.
Most repetitive framing members will be 12 inches, 16 inches or 24 inches on center. The reason is simple.
To efficiently install sheathing (4 x 8) over repetitive members, the sum of the on-center dimensions must
total 48 or 96 inches. For example three 16 o.c spaces equals 48 inches (the width of a piece of sheathing),
and six spaces equals 96 inches (the length of a piece of sheathing).
Top Plate
Stud
12 12 12 12 12 12
Bottom Plate
Figure 6-1 Wall Section
Example 6-A:
You are framing the wall in figure 6-1, which is six feet in length. Wall studs are to be laid out twelve
inches on center. How many studs will be used?
Answer: 7 studs.
Explanation: There are twelve inches in one foot, the wall is six feet in length, and therefore six studs are
required. One stud is added making the total 7, because the count started 12 inches from the point of
beginning.
Example 6-A is easy to solve mathematically:
The unit length is converted to feet (12  12 = 1)
The total length is divided by the unit length (6  1 = 6 studs)
One stud is added (6 + 1 = 7 studs)
Total Length  unit length
+ 1 = # of Repetitive Members
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Example 6-B: You must frame a wall, which is sixteen feet in length with studs 16 O.C. How many studs
will be used?
Use the same steps as above:
Convert the on center measurement to feet
16  12 = 1.33 feet
Divide the on center measurement into total length
16  1.33 = 12
Add one to the calculated number of studs
12 + 1 = 13 studs
Example 6-C: You are framing a wall that is 22-5 3/8” in length. The studs are to be laid out 16 O.C.
How many studs are needed? This problem is much the same as Example 5-1 except that it requires the
conversion of the total length into decimal feet in order to carry out the division.
16  12 = 1.33 (memorize this equivalency)
5.375  12= .4479 (conversion of inches to feet)
22.4479  1.33 = 16.878 (round up to seventeen)
17 + 1 = 18 studs
Try these:
1. You are to frame a shed roof. The rafters are to be laid out 24 O.C. on a wall plate 21-9 in length.
How many rafters are required?
Show work here
Answer_____________
2.
You must select materials to construct two interior walls. You have already selected the plate stock and
now must determine the number of studs needed. Both walls total 37 7. How many wall studs are
needed if laid out 16 OC?
Show work here
Answer____________
3.
How many joist, spaced 16 o.c. are required for a building 59 10 long?
Show work here
Answer_____________
4.
A building is 40 feet wide and 60 feet long. Girders running the length of the building are spaced 4 feet
o.c. The outside walls rest on a concrete foundation. Supporting each girder are piers spaced 6 feet o.c.
How many girders are needed?
Show work here
Answer______________
5.
Use the information in question four to calculate the number of posts needed to support the girders.
Show work here
Answer_____________
6.
A gable roof is 46-0 long. Trusses are to be spaced 24 o.c. How many trusses are needed?
Show work here
Answer_____________
7.
How many joists spaced 16 o.c. are required for a floor 62-6 long?
Show work here
Answer____________
8.
25 rafters spaced 16 o.c. will require a wall length of ______ _______.
Show work here
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LAYING OUT EVENLY SPACED BALUSTERS
When constructing finish stairs and balustrades or exterior decks, it is necessary to install vertical balusters
to protect people from falling through open areas. When I started out building decks the maximum space
allowed between balusters was nine inches. Later nine inches was reduced to six inches. Today the
maximum spacing is four inches. Building inspectors usually scrutinize baluster spacing very carefully. The
problem is how do you maintain equal spacing between the balusters.
If you install your balusters by provide a 4 space between them every thing will be fine until you place
your last baluster, which unless you are very lucky, will leave a space less than 4 inches. As you can see
the problem with this technique is that there is no guarantee that the last space in the run will be 4.
Depending on the length of the run the last space could be just about anything between 0 and 4. Your
customer will probably not be impressed.
Outlined below is a method of calculating baluster spacing that insures all of the spaces are uniform.
Evenly Spaced Baluster Layout:
Step 1: Determine total length of run and convert to inches.
Step 2: Deduct width of newels at each end of run and add one baluster width.
Step 3: Divide result of step 2 by desired spacing between balusters plus the
width of one baluster. i.e. desired spacing = 4 + 1 1/4”baluster width
4 + 1 1/4 = 5 1/4
Step 4: The answer obtained in step three will typically be an uneven number.
If this is the case round the number up to the next whole number and divide
again.
Step 5: Subtract one baluster width from the answer obtained in step 4 and you have
the correct spacing between balusters.
Example 6-D:
You are constructing a guardrail that will have 3 1/2” newel posts at
each end. The guardrail will be supported in the field by 1 1/2” balusters.
The code requirement for spacing between balusters is no more than 4”.
What equal spacing between balusters would be necessary to maintain
the 4” code requirement.
Fill in with balusters
Newel Post
Newel Post
12 8 3/8 total run
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Step 1: 152.375
Step 2: 152.375 total
- 7
2 newel widths
145.375
+ 1.5 one baluster width
146.875
Step 3: Code spacing = 4 + 1.5 (baluster width) = 5.5
146.875  5.5 = 26.7045
Step 4: Round 26.7045 to 27
146.875”  27 = 5.4398
Step 5:
5.4398
- 1.5
3.9398 0r 3 15/16
Try these:
9. Answer the questions below based on the following information:



Balustrade run is 8- 2 5/16
4 wide newel posts at each end
1 1/4balusters and spacing between balusters of
no more than 4
Total run in inches ________
Distance between newels _________
Distance between newels plus one baluster width _________
Maximum code spacing plus one baluster width __________
Number of balusters __________
Actual spacing between balusters __________
10. Answer the questions below based on the following information:



Balustrade run is 6- 4 9/16
4” wide newel posts at one end and rosette at other end
1 3/4 balusters and spacing between balusters of
no more than 4
Number of balusters ___________
Actual spacing of balusters __________
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11. Answer the questions below based on the following information:



Balustrade run is 14 5 1/2
3 1/2” newel posts will be placed at each end and in the center of the run.
Balusters are 1 3/8 wide and may not be spaced more than 4 apart.
Number of balusters in each run __________
Actual spacing of balusters __________
12. Use information from question 11 with one addition. Each post will have a 1 3/8 baluster attached.
The center post will have a baluster attached on both sides.
Number of balusters in each run __________
Actual spacing of balusters __________
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