Handbook of Civil Engineering Calculations

where A = axial shortening, in (cm); P = unfactored axial force in member, kips (kg); / =
length of member, in (cm); E = modulus of elasticity of steel = 29,000 ksi (199.8 MPa);
Ag = cross sectional area of member, sq in (sq cm).

2. Compute the column axial shortening
   Substituting,

Pl 250 kips x (IQ.O ft x 12 in/ft)

Shortening, A = — = 29?000 kips/in (^2) x 14 4 in 2
= 0.072 in (0.183 cm).
Related Calculations: Use this equation to compute axial shortening of any steel
column in LRFD work. This procedure is the work of Abraham J. Rokach, MSCE, Amer-
ican Institute of Steel Construction.

DETERMINING THE COMPRESSIVE

STRENGTH OFA WELDED SECTION

The structural section in Fig. 36a is used as a 40-ft (12.2-m) column. Its effective length
factor Kx = Ky=LQ. Determine the design compressive strength if the steel is A3 6.

Calculation Procedure:

1. Choose a design compressive strength
   The design compressive strength is given by:

4JPn = WrA 8

The values of ^f cr can be obtained from the Table, "Design Stress for Compression
Members of 36 ksi Specified Yield-Stress Steel, = 0.85" in the AISC Manual, ifKl/r is
known. With Kl = 1.0 x 40.0 ft x 12 in/ft = 480 in (1219 cm), then

r=

v?

A = (18 in)^2 - (17 in)^2 = 35.0 in^2

Ix = Iy = I= ^

in)2

• 2 °
  7 in)4
  
  
  • 1788 in* (225.8 sq cm)
  
  
  
  

2. Find the Kl/r ratio for this section
   With the data we have,

/1788 in^4

r= r-r = 7.15 in

V 35.0 in^4

KI _ 480in

T-TUhT

67

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