Handbook of Civil Engineering Calculations

9. Establish the allowable spacing of the bridging
   To do this, compare the moments of inertia with respect to the principal axes. Thus, Iy -
   2(1/12)(5.5)(4.875)
   3
   
   
   • 2(0.5)(40)(2.875)
     2
     = 433 in
     4
     (18,022.8 cm
     4
     ); then IxIIy.=
     16,080/433 = 37.1.
     For this ratio, the Wood Handbook specifies that "the beam should be restrained by
     bridging or other bracing at intervals of not more than 8 ft (2.4 m)."
   
   
   
   

DETERMINING THE CAPACITY OF

A SOLID COLUMN

An 8 x 10 in (203.2 x 254 mm) column has an unbraced length of 10 ft 6 in (3.20 m). the
allowable compressive stress is 1500 lb/in
2
(10,342.5 kPa), and E = 1,760,000 lb/in
2
(12,135.2 MPa). Calculate the allowable load on this column (a) by applying the recom-
mendations of the Wood Handbook; (b) by applying the provisions of the National De-
sign Specification.

Calculation Procedure:

1. Record the properties of the member; evaluate k; classify
   the column
   Let L = unbraced length of column, in (mm); d = smaller side of rectangular section, in
   (mm);/, = allowable compressive stress parallel to the grain in short column of the same
   species, lb/in
   2
   (kPa);/= allowable compressive stress parallel to grain in column under
   investigation, lb/in
   2
   (kPa).
   The Wood Handbook divides columns into three categories: short, intermediate, and
   long. Let K denote a parameter defined by the equation K = Q.64(E/fc)°-
   5
   .
   The range of the slenderness ratio and the allowable stress for each category of col-
   umn are as follows: short column, LId ^ 11 and/=/; intermediate column, 11 < LId ^ K
   and/=/[l -^1 A(LMK)^4 ]; long column, LId > K and/= 0.274E/(L/d)^2.
   For this column, the area A - 71.3 in^2 (460.03 cm^2 ), using the dressed dimensions.
   Then/,/</= 126/7.5 = 16.8. Also, K= 0.64(1,760,000/150O)^05 = 21.9. Therefore, this is an
   intermediate column because LId lies between K and 11.


2. Compute the capacity of the member
   Use the relation capacity, Ib(N) = P = Af= 71.3(150O)[I-^1 Xs(16.8/21.9)^4 ] = 94,600 Ib
   (420,780.8 N). This constitutes the solution to part a, using data from the Wood Hand-
   book. For part Z?, data from the National Design Specification are used.


3. Compute the capacity of the column
   Determine the stress from/= 0.3QE/(L/d)^2 = 0.30(1,760,000)/16.8^2 = 1870 lb/in^2
   (12,893.6 kPa). Setting/= 1500 lb/in^2 (10,342.5 kPa) gives P = Af= 71.3(1500) =
   107,000 Ib (475,936 N). Note that the smaller stress value is used when the column ca-
   pacity is computed.

DESIGN OFA SOLID WOODEN COLUMN

A 12-ft (3.7-m) long wooden column supports a load of 98 kips (435.9 kN). Design a sol-

id section in the manner recommended in the Wood Handbook, using/ = 1400 lb/in^2

(9653 kPa) and E= 1,760,000 lb/in^2 (12,135.2 MPa).