Calculation Procedure:
- Assume that d = 7.5 in (19O.5 mm), and classify the column
Thus, LId= 144/7.5 = 19.2 and K= 0.64(1,760,000/140O)^05 = 22.7. This is an intermedi-
ate column if the assumed dimension is correct. - Compute the required area and select a section
For an intermediate column, the stress/= 1400(1 -^1 X 3 (19.2/22.7)4] = 1160 lb/in^2 (7998.2
kPa). Then^ = PIf= 98,000/1160 = 84.5 in^2 (545.19 cm^2 ).
Study of the required area shows that an 8 12 in (203.2 304.8 mm) column having
an area of 86.3 in^2 (556.81 cm^2 ) should be used.
INVESTIGATION OFA SPACED COLUMN
The wooden column in Fig. 2 is composed of three 3 8 in (76.2 203.2 mm) sections.
Determine the capacity of the member if/c = 1400 lb/in
2
(9653 kPa) and E = 1,760,000
lb/in
2
(12,135.2 MPa).
Calculation Procedure:
- Record the properties
of the elemental section
In analyzing a spaced column, it is necessary to as-
sess both the aggregate strength of the elements and
the strength of the built-up section. The end spacer
blocks exert a restraining effect on the elements and
thereby enhance their capacity. This effect is taken
into account by multiplying the modulus of elastici-
ty by a. fixity factor F.
The area of the column A = 19.7 in
2
(127.10
cm
2
) when the dressed sizes are used. Also, LId =
114/2.625 = 43.4; F = 2.5; K = 0.64(2.5 x
1,760,000/140O)
05
= 35.9. Therefore, this is a long
column. - Calculate the aggregate strength
of the elements
Thus,/= Q214EI(LId)
2
for a long column, or/ =
0.274(2.5)(l,760,000)/(43.4)^2 = 640 lb/in^2 (4412.8
kPa). P = 3(19.7)(640) = 37,800 Ib (168,134.4 N). - Repeat the foregoing steps FIGURE 2 Spaced col-
for the built-up member umn.
Thus, LId= 114/7.5 x 15.2; K= 22.7; therefore, this
is an intermediate column. Then / = 1400(1 -
'/3(15.2/22J)^4 ] = 1306 lb/in^2 (9004.9 kPa) > 640
lb/in^2 (4412.8 kPa).
The column capacity is therefore limited by the elements and P = 37,800 Ib (168,134.4
N).