Solution: From Example 8-7 the length of the wall is 142 feet. The number of studs needed for the standard spacing is calculated using Eq. (4-2) as follows:
All four interior walls lack corners at both ends; therefore, four ad- ditional studs are needed for these walls. There are 4 corners, 8 in- tersections, and 5 doorways that will require 2 additional studs each, for a total of 34 additional studs. There are 2 windows that will require 6 additional studs each and 1 window that will require 8 additional studs for a total of 20 additional studs. There are 2 hold-downs located at corners that will require an additional stud per hold-down for 2 additional studs. The total number of studs is
167 (107⫹ 4 ⫹ 34 ⫹ 20 ⫹ 2). 䊏
Number⫽(142 ft) (12 in/ft)
16 in ⫽ 107 each
Headers in a 2⫻ 4 wall can be made from two pieces of 2⫻ material and a 1>2-inch plywood or 7>16-inch OSB spacer. Headers in a 2⫻ 6 wall can be made from three pieces of 2⫻ material and two 1>2-inch or 7>16-inch spacers. The headers must extend over all of the jack studs, requiring the header to be 3 inches wider than the opening for openings with sin- gle jack studs and 6 inches wider than the opening for open- ings with double jack studs. Openings for doors with jambs are typically framed 2 inches wider than the door size to allow for the doorjamb and shimming. A 36-inch-wide door with single jack studs would require a 41-inch-long header (36 in⫹ 3 in ⫹ 2 in). Window openings are typically framed 1>2 to 1 inch wider than the opening to allow for placement of the window.
EXAMPLE 8-9
Prepare a materials list for the headers required to complete the walls in Figure 8-5. All headers are made from 2⫻ 12s, and the wall is made from 2⫻ 4s. The doors are 32 inches wide. Openings 6 feet and wider require double jack studs.
Solution: Each header will consist of two 2⫻ 12s with a 7>16-inch OSB spacer. The 32-inch-wide doors will require a 34-inch-wide opening and a 37-inch (34 in⫹ 3 in) header. Three 37-inch headers are needed and can be cut from two 10-foot-long 2 ⫻ 12s. Approximately 10 square feet of spacer are needed for these headers. The 4-foot windows and closet doors will require 49-inch openings and 52-inch (49 in⫹ 3 in) headers. Because the bifold closet doors do not have jambs, the opening only needs to be framed 1 inch wider than the doors to allow for two layers of half- inch drywall. Four 52-inch headers are needed and can be cut from four 10-foot-long 2⫻ 12s. Approximately 20 square feet of spacer are needed for these headers.
The 6-foot window will require a 73-inch-wide opening and a 79-inch (73 in⫹ 6 in) header. This header can be cut from two 8-foot-long 2⫻ 12s. Approximately 7 square feet of spacer are needed for this header. The total amount of spacer required is ap- proximately 37 square feet (10 ft2⫹ 20 ft2⫹ 7 ft2) and can be cut from two sheets of OSB.
Order six 10-foot-long 2⫻ 12s, two 8-foot-long 2 ⫻ 12s, and
two sheets of 7>16-inch OSB. 䊏
Hold-downs
Shear panels are used to stiffen the building against lateral forces generated by wind and earthquakes. Shear panels are created by covering the wall framing with a rigid material such as OSB or plywood. Wide shear panels may be tied to the foundation with anchor bolts. Narrow shear panels re- quire hold-downs. Hold-downs are specialty framing an- chors that tie the shear panel to the foundation. Hold-downs come in a variety of types and sizes.
Blocking
Blocking is used to strengthen load-bearing walls, prevent the spread of fire, and provide backing for wall-mounted items, such as grab bars. Blocking used to strengthen load- bearing walls is placed between the studs near the midpoint
EXCEL QUICK TIP 8-5 Studs
The number of studs needed for walls is set up in a spreadsheet by entering the data and formatting the cells as follows:
The following formula needs to be entered into Cell B12: =ROUNDUP(B1*12/B2,0)+B3+B4*2+B5*2+B6*2
+B7*6+B8*8+B9+B10*3
The data for walls is entered in Cells B1 through B10. The data shown in the foregoing figure is from Example 8-8 and is formatted using the comma style, which replaces zeros with dashes.
A B C
1 Wall Length 142 ft
2 Stud Spacing 16 in
3 Walls without Corners 4 ea
4 Corners 4 ea
5 Intersections 8 ea
6 Doorways 5 ea
7 Windows, Small 2 ea
8 Windows, Large 1 ea
9 Hold Downs at Corners 2 ea
10 Other Hold Downs - ea
11
12 Studs 165 ea
Headers
Headers are constructed of solid wood, laminated lumber, or 2⫻ materials. Because of cost, most headers are constructed of 2⫻ material and require plywood or OSB spacers such that the width of the header matches the width of the wall.
of the studs as shown in Figure 8-11. Blocking used to pre- vent the spread of fire is placed at points where walls inter- sect with mechanical drops and parallel to stair stringers. Blocking in walls is estimated using the same procedures as estimating blocking in floors.
EXAMPLE 8-10
How many lineal feet of blocking are needed for the walls in Figure 8-5 if the studs in the exterior walls are blocked at their midpoints, as shown in Figure 8-11?
Solution: Only the perimeter wall will need to be blocked. Blocking is not needed at openings for doors and windows. The length of the wall that needs to be blocked is calculated as follows:
䊏
Sheathing
Wood-framed exterior walls are commonly sheathed with 7>16-inch OSB or 1>2-inch plywood. Interior walls may also be sheeted with OSB or plywood to create a shear panel. The wood sheathing is treated as a sheet good. When calculating the number of rows, one must round the rows to a common
⫽ 76 ft
Length⫽ 2(22 ft) ⫹ 2(24 ft 4 in) ⫺ 2(4 ft) ⫺ 6 ft ⫺ 32 in
fraction of a sheet such as 1>8, 1>6, 1>5 1>4, 1>3, or 1>2. Openings for doors and windows in the wall are commonly ignored because the material cut from these openings usu- ally cannot be used as wall sheathing, but it may be used for headers built after the wall is sheathed. The quantity takeoff of wall sheathing is shown in the following example.
EXAMPLE 8-11
How many 4-foot by 8-foot sheets of 7>16-inch OSB are needed to sheet the outside of the exterior walls shown in Figure 8-5 if the wall height is 9 feet?
Solution: The number of rows is calculated using Eq. (4-13) as follows:
The number of rows is rounded up to the nearest 1>8 of a sheet. The length of the outside surface of the exterior wall is calculated as follows:
The number of columns is calculated using Eq. (4-16) as follows: NumberColumns⫽ 94 ft 4 ft ⫽ 24 columns Length⫽ 2(22 ft) ⫹ 2(25 ft) ⫽ 94 ft NumberRows⫽ 9 ft 8 ft⫽ 1.125 rows
The number of sheets is calculated using Eq. (4-18) as follows:
䊏 When sheathing gable ends, the waste from the sheets used near the peak of the gable ends can often be used for the sheets near the eaves of the gable ends as shown in the following example.
EXAMPLE 8-12
Determine the number of sheets of 7>16-inch OSB required for the end of the building shown in Figure 8-12.
Solution: The rise for the gable end is calculated as follows:
The height of the wall at the peak is 14 foot 4 inches. The number of rows of sheathing is calculated using Eq. (4-13) as follows:
The number of columns of sheathing is calculated using Eq. (4-16) as follows:
The first row of sheathing will run from the bottom of the wall to a height of 8 feet and will consist of eight sheets of sheathing. The second row of sheathing will run from a height of 8 feet to the top
NumberColumns⫽ 32 ft 4 ft ⫽ 8 columns NumberRows⫽ 14.33 ft 8 ft ⫽ 2 rows Rise⫽ a32 ft 2 ba 4 12b ⫽ 5.33 ft ⫽ 5 ft 4 in Number⫽ (1.125 rows)(24 columns) ⫽ 27 sheets
of the wall, which is 1 foot at the eaves and 6 foot 4 inches (1 ft ⫹ 5 ft 4 in) at the peak. Two pieces of sheathing can be cut from one sheet on the top row; therefore, only four sheets are required. The layout of the sheathing is shown in Figure 8-13. The number on each piece of sheathing represents the sheet from which the sheathing was cut.
Because the height at the eaves (1 ft) plus the height at the peak (6 ft 4 in) is less than 8 feet, the scrap from the short side of the gable end can be used at the peak as shown in Figure 8-14. A total of 12 sheets of sheathing are needed. 䊏