P. P. P. P. Compression Members. Chap. (6). Columns:.
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Are compression members which are subjected to concentric axial compressive forces. These are to be found in trusses and as a lateral bracing members in frame building. Short columns are sometimes referred to as to as “struts” or “stanchions”.
Are members subjected to combined axial compressive and bending stresses; These are found in single storey of multi-storey framed structures. These are treated independently in this course (chap. 12 in your text book).
Stocky columns (short) fail by yielding of the material at the cross section, but most columns fail by “buckling” at loads for less then yielding forces.
Critical Buckling Load of Columns
For “slender” columns, Euler (1759) predicted the critical buckling load (Pcr)
– also known as Euler Buckling Load as:
where: E = Young Modulus of Elasticity.
I = Minor moment of Inertia.
L = Unbraced length of column.
Derivation of Euler Buckling Load:
Solution of this differential equation:
y = A cos (cx) + B sin (cx)
, A and B are constants.
The critical buckling load is a function of the section properties (A, L, r) and the modulus of elasticity for material, and is not a function of the strength or grade of the material.
Euler Buckling Formula
From boundary conditions:
y = 0 @ x = 0, and
y = 0 @ x = L, we get (A = 0) and (B sin cL = 0)
if B ≠ 0, then cL = n where n = 0, 1, 2, 3 ………
---- Euler Buckling Critical Load
where: r = minor radius of gyration
Find the critical buckling load for W 12 x 50, supported in a pinned-pinned condition, and has an over-all length of 20 feet?
rmin = ry = 1.96 inch (properties of section).
Pcr = Fcr A = 19.1 x 14.7 = 280.8 kips
The steel grade is not a factor affecting buckling.
For short (stocky) columns; Equation (C-2) gives high values for (Fcr), sometimes greater then proportional limit, Engessor (1889) proposed to use (Et) instead of (E) in Euler formula:
Et = Tangent Modulus of Elasticity
Et < E
When (Fcr) exceeds (FPR), this is called
“Inelastic Buckling”, constantly variable
(Et) need to be used to predict (Fcr)
in the inelastic zone.
Shanley (1947), resolved this inconsistency.
Depending on (L/r) value the column buckling
strength was presented as shown by Shanley.
Due to uneven cooling of hot-rolled sections,
residual stresses develop as seen here.
The presence of “residual stresses” in almost all hot-rolled sections further complicates the issue of elastic buckling and leads towards inelastic buckling.
The Euler buckling formula (C-1) is based on:
1 – Perfectly straight column. (no crookedness).
2 – Load is concentric (no eccentricity).
3 – Column is pinned on both ends.
K = Effective length factor.
(Kl) = Effective length.
(Kl/r) = Effective slenderness ratio.
The Previous conditions are very difficult to achieve in a realistic building condition, especially the free rotation of pinned ends. Thus an “effective slenderness factor” is introduced to account for various end conditions:
(C – C2.2) (page 16.1-240)
AISC (Chapter E) of LRFD code stipulates:
Pu (factored load) c Pn
Pu = Sum of factored loads on column.
c = Resistance factor for compression = 0.90
Pn = Nominal compressive strength = Fcr Ag
Fcr = Critical buckling Stress. (E3 of LFRD)
The above two equations of the LRFD code can be
illustrated as below:
* The code further stipulates that an upper value for
for column should not exceed (200).
* For higher slenderness ratio,
Equation (E-3.3) controls and
(Fy) has no effect on (Fcr).
Determine the design compressive strength (cPn) of W 14x74 with an untraced length of (20 ft), both ends are pinned, (A-36) steel is used?
Kl =1 x 20 x 12 = 240 in
Rmin = ry = 2.48
c Pn = 0.9 x Fcr x Ag = 0.9 x (21.99) x 21.8
= 433.44 kips (Answer)
Also from (table 4-22) LFRD Page 4-320
c Fcr = 19.75 ksi (by interpolation)
c Pn = c Fcr Ag = 430.55 kips
For must profiles used as column, the buckling of thin elements in the section may proceed the ever-all bucking of the member as a whole, this is called local bucking. To prevent local bucking from accruing prior to total buckling. AISC provides upper limits on width to thickness ratios (known as b/t ratio) as shown here.
See AISC (B4)
Part 1 on properties
of various sections.
Determine the design compressive strength
(c Pn) for W 12 x 65 column shown below,
(Fy = 50 ksi)?
A) By direct LRFD
Ag =19.1 in2
rx = 5.28 in
ry = 3.02 in
c Pn = 0.9 x Fcr Ag = 0.9 x 40.225 x 19.1 = 691.5 kips
Evaluate = = 54.55
Enter table 4.22 (page 4 – 318 LRFD)
cFc = 36.235 ksi (by interpolation)
Pn = Fc x Ag = 692.0 kips
C) From (Table 4.1 LRFD)
Enter table (4.1 ) page 4.17 LFRD with (KL)y = 13.7
Pn = 691.3 kips (by interpolation).
Design with Columns Load Table (4) LFRD:-
EXAMPLE C - 4
A compression member is subjected to service loads of 165 kips dead load and 535 kips live load. The member is 26 feet long and pinned in each end. Use (A572 – Gr 50) steel and select a W14 shape.
Calculate the factored load:
Pu = 1.2D + 1.6L = 1.2(165) + 1.6(535) = 1054 kips
Required design strength cPn = 1054 kips
From the column load table for KL = 26 ft, a W14 176
has design strength of 1150 kips.
Use a W14 145, But practically W14 132 is OK.
EXAMPLE C - 5
Select the lightest W-shape that can resists a factored compressive load Pu of 190 kips. The effective length is 24 feet. Use ASTM A572 Grade 50 steel.
The appropriate strategy here is to fined the lightest shape for each nominal size and then choose the lightest overall. The choices are as follows.
W4, W5 and W6: None of the tabulated shape will work.
W8: W8 58, cPn = 194 kips
W10: W10 49, cPn = 239 kips
W12: W12 53, cPn = 247 kips
W14: W14 61, cPn = 276 kips
Note that the load capacity is not proportional to the weight (or cross-sectional area). Although the W8 58 has the smallest design strength of the four choices, it is the second heaviest.
Use a W10 49.
B) Design for sections not from Column Load Tables:
For shapes not in the column load tables, a trial-and-error approach must be used. The general procedure is to assume a shape and then compute its design strength. If the strength is too small (unsafe) or too large (uneconomical), another trial must be made. A systematic approach to making the trial selection is as follows.
Select a W18 shape of A36 steel that can resist a factored load of 1054 kips.
The effective length KL is 26 feet.
Try Fcr = 24 ksi (two-thirds of Fy):
Try W18 x 192:
Ag = 56.4 in2 > 48.8in2