4. CE470:DesignofSteelStructuresProf.Varma
CHAPTER COMPRESSION MEMBER DESIGN 4.1 INTRODUCTORY CONCEPTS Compression Members: Structural elements that are subjected to axial compressive forces only are called columns. Columns are subjected to axial loads thru the centroid. Stress: The stress in the column cross-section can be calculated as f= P A (2.1)
where, f is assumed to be uniform over the entire cross-section. This ideal state is never reached. The stress-state will be non-uniform due to: Accidental eccentricity of loading with respect to the centroid Member out-of straightness (crookedness), or Residual stresses in the member cross-section due to fabrication processes. Accidental eccentricity and member out-of-straightness can cause bending moments in the member. However, these are secondary and are usually ignored. Bending moments cannot be neglected if they are acting on the member. Members with axial compression and bending moment are called beam-columns.
4.2 COLUMN BUCKLING Consider a long slender compression member. If an axial load P is applied and increased slowly, it will ultimately reach a value Pcr that will cause buckling of the column. Pcr is called the critical buckling load of the column.
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CE470:DesignofSteelStructuresProf.Varma
P
(a)
Pcr
(b)
What is buckling? Buckling occurs when a straight column subjected to axial compression suddenly undergoes bending as shown in the Figure 1(b). Buckling is identified as a failure limit-state for columns.
P
Pcr
Figure 1. Buckling of axially loaded compression members The critical buckling load Pcr for columns is theoretically given by Equation (4.1) Pcr = 2 E I (4.1)
( K L) 2
where, I = moment of inertia about axis of buckling K = effective length factor based on end boundary conditions Effective length factors are given on page 16.1-240 of the AISC manual.
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CE470:DesignofSteelStructuresProf.Varma
In examples, homeworks, and exams please state clearly whether you are using the theoretical value of K or the recommended design values.
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EXAMPLE 4.1 Determine the buckling strength of a W 12 x 50 column. Its length is 20 ft. For major axis buckling, it is pinned at both ends. For minor buckling, is it pinned at one end and fixed at the other end. Solution Step I. Visualize the problem
x
y
Figure 2. (a) Cross-section; (b) major-axis buckling; (c) minor-axis buckling For the W12 x 50 (or any wide flange section), x is the major axis and y is the minor axis. Major axis means axis about which it has greater moment of inertia (Ix > Iy)
Figure 3. (a) Major axis buckling; (b) minor axis buckling 4
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Step II. Determine the effective lengths According to Table C-C2.2 of the AISC Manual (see page 16.1 - 240): For pin-pin end conditions about the major axis Kx = 1.0 (theoretical value); and Kx = 1.0 (recommended design value) For pin-fix end conditions about the minor axis Ky = 0.7 (theoretical value); and Ky = 0.8 (recommended design value) According to the problem statement, the unsupported length for buckling about the major (x) axis = Lx = 20 ft. The unsupported length for buckling about the minor (y) axis = Lx = 20 ft. Effective length for major (x) axis buckling = Kx Lx = 1.0 x 20 = 20 ft. = 240 in. Effective length for minor (y) axis buckling = Ky Ly = 0.8 x 20 = 16 ft. = 192 in.
Step III. Determine the relevant section properties For W12 x 50: elastic modulus = E = 29000 ksi (constant for all steels) For W12 x 50: Ix = 391 in4. Iy = 56.3 in4 (see page 1-25 of the AISC manual)
Step IV. Calculate the buckling strength Critical load for buckling about x - axis = Pcr-x = 2 E I x =
(Kx
Lx ) 2
2 29000 391 ( 240) 2
Pcr-x = 1942.9 kips Critical load for buckling about y-axis = Pcr-y =
( K y Ly )2
2 E I y
2 29000 56.3 = (192) 2
Pcr-y = 437.12 kips Buckling strength of the column = smaller (Pcr-x, Pcr-y) = Pcr = 437.12 kips
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CE470:DesignofSteelStructuresProf.Varma
Minor (y) axis buckling governs. Notes: Minor axis buckling usually governs for all doubly symmetric cross-sections. However, for some cases, major (x) axis buckling can govern. Note that the steel yield stress was irrelevant for calculating this buckling strength.
4.3 INELASTIC COLUMN BUCKLING Let us consider the previous example. According to our calculations Pcr = 437 kips. This Pcr will cause a uniform stress f = Pcr/A in the cross-section For W12 x 50, A = 14.6 in2. Therefore, for Pcr = 437 kips; f = 30 ksi The calculated value of f is within the elastic range for a 50 ksi yield stress material. However, if the unsupported length was only 10 ft., Pcr = 1119 kips, and f = 76.6 ksi. This value of f is ridiculous because the material will yield at 50 ksi and never develop f = 76.6 ksi. The member would yield before buckling. Equation (4.1) is valid only when the material everywhere in the cross-section is in the elastic region. If the material goes inelastic then Equation (4.1) becomes useless and cannot be used. What happens in the inelastic range? Several other problems appear in the inelastic range. The member out-of-straightness has a significant influence on the buckling strength in the inelastic region. It must be accounted for.
( K y L y ) 2 would be calculated as
2 E I y
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CE470:DesignofSteelStructuresProf.Varma
-
The residual stresses in the member due to the fabrication process causes yielding in the cross-section much before the uniform stress f reaches the yield stress Fy.
-
The shape of the cross-section (W, C, etc.) also influences the buckling strength. In the inelastic range, the steel material can undergo strain hardening.
All of these are very advanced concepts and beyond the scope of CE470. You are welcome to CE579 to develop a better understanding of these issues. So, what should we do? We will directly look at the AISC Specifications for the strength of compression members, i.e., Chapter E (page 16.1-32 of the AISC manual).
4.4 AISC SPECIFICATIONS FOR COLUMN STRENGTH The AISC specifications for column design are based on several years of research. These specifications account for the elastic and inelastic buckling of columns including all issues (member crookedness, residual stresses, accidental eccentricity etc.) mentioned above. The specification presented here (AISC Spec E3) will work for all doubly symmetric crosssections and channel sections. The design strength of columns for the flexural buckling limit state is equal to cPn Where,
c = 0.9
Pn = Ag Fcr KL E 4.71 r Fy
(Resistance factor for compression members) (4.2) (or Fe 0.44 F y ) Fy
When
Fy 0.658 Fe Fcr =
(4.3)
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CE470:DesignofSteelStructuresProf.Varma
When
KL E > 4.71 r Fy
(or Fe < 0.44 F y ) Fcr = [ 0.877 Fe ] (4.4)
2E
Where, Fe = KL r
2
(4.5) K = effective length factor r = governing radius of gyration
Ag = gross member area; L = unbraced length of the member;
1.0 Fcr/F y 0.39
Fy Fcr = 0.658 Fe
F y F cr = [ 0.877 Fe ]
4.71
E Fy 2E I
KL r
Note that the original Euler buckling equation is Pcr = Fe = Pcr 2E I 2E 2E = = r2 = 2 Ag ( K L ) 2 A g ( K L ) 2 K L r
( K L) 2
Note that the AISC equation for
KL E 4.71 is Fcr = 0.877Fe r Fy
The 0.877 factor tries to account for initial crookedness. For a given column section:
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Calculate I, Ag, r Determine effective length K L based on end boundary conditions. Calculate Fe, 0.44Fy, or 4.71 E Fy
If (KL/r) greater than 4.71
E , elastic buckling occurs and use Equation (4.4) Fy E , inelastic buckling occurs and use Equation Fy
If (KL/r) is less than or equal to 4.71 (4.3)
Note that the column can develop its yield strength Fy as (KL/r) approaches zero.
4.5 COLUMN STRENGTH In order to simplify calculations, the AISC specification includes Tables. Table 4-22 on page 4-318 shows KL/r vs. cFcr for various steels. You can calculate KL/r for the column, then read the value of cFcr from this table The column strength will be equal to cFcr x Ag
EXAMPLE 4.2 Calculate the design strength of W14 x 74 with length of 20 ft. and pinned ends. A36 steel is used. Solution Step I. Calculate the effective length and slenderness ratio for the problem Kx = Ky = 1.0 Lx = Ly = 240 in.
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Major axis slenderness ratio = KxLx/rx = 240/6.04 = 39.735 Minor axis slenderness ratio = KyLy/ry = 240/2.48 = 96.77 Step II. Calculate the elastic critical buckling stress The governing slenderness ratio is the larger of (KxLx/rx, KyLy/ry) Fe =
2E
KL r
2
=
2 * 29000 ( 96.77 ) 2 = 30.56 ksi
Check the limits ( KL E 4.71 ) or ( Fe 0.44 F y ) r Fy E 29000 = 4.71 = 133.68 Fy 36
Fy F Therefore, Fcr = 0.658 e
4.71
KL E 4.71 Since ; r Fy Therefore, Fcr = 21.99 ksi
F y
Design column strength = cPn = 0.9 (Ag Fcr) = 0.85 (21.8 in2 x 21.99 ksi) = 431.4 kips Design strength of column = 431 kips Check calculated values with Table 4-22. For KL/r = 97, cFcr = 19.7 ksi
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4.6 LOCAL BUCKLING LIMIT STATE The AISC specifications for column strength assume that column buckling is the governing limit state. However, if the column section is made of thin (slender) plate elements, then failure can occur due to local buckling of the flanges or the webs.
Figure 4. Local buckling of columns If local buckling of the individual plate elements occurs, then the column may not be able to develop its buckling strength. Therefore, the local buckling limit state must be prevented from controlling the column strength. Local buckling depends on the slenderness (width-to-thickness b/t ratio) of the plate element and the yield stress (Fy) of the material. Each plate element must be stocky enough, i.e., have a b/t ratio that prevents local buckling from governing the column strength.
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CE470:DesignofSteelStructuresProf.Varma
The AISC specification B4 provides the slenderness (b/t) limits that the individual plate elements must satisfy so that local buckling does not control.
The AISC specification provides two slenderness limits ( p and r) for the local buckling of plate elements.
Fy Axia l Force, F
Compact Compact
Non -Compact Slender F t Axial shortening, b
Figure 5. Local buckling behavior and classification of plate elements If the slenderness ratio (b/t) of the plate element is greater than r then it is slender. It will locally buckle in the elastic range before reaching Fy If the slenderness ratio (b/t) of the plate element is less than r but greater than p, then it is non-compact. It will locally buckle immediately after reaching Fy If the slenderness ratio (b/t) of the plate element is less than p, then the element is compact. It will locally buckle much after reaching Fy If all the plate elements of a cross-section are compact, then the section is compact. If any one plate element is non-compact, then the cross-section is non-compact If any one plate element is slender, then the cross-section is slender.
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The slenderness limits p and r for various plate elements with different boundary conditions are given in Table B4.1 on pages from 16.1-16 to 16.1-18 of the AISC Spec.
Note that the slenderness limits ( p and r) and the definition of plate slenderness (b/t) ratio depend upon the boundary conditions for the plate. If the plate is supported along two edges parallel to the direction of compression force, then it is a stiffened element. For example, the webs of W shapes If the plate is supported along only one edge parallel to the direction of the compression force, then it is an unstiffened element. Ex., the flanges of W shapes.
The local buckling limit state can be prevented from controlling the column strength by using sections that are non-compact If all the elements of the cross-section have calculated slenderness (b/t) ratio less than r, then the local buckling limit state will not control. For the definitions of b/t, p, r for various situations see Table B4.1 and Spec B5.
EXAMPLE 4.3 Determine the local buckling slenderness limits and evaluate the W14 x 74 section used in Example 4.2. Does local buckling limit the column strength? Solution Step I. Calculate the slenderness limits See Table B4.1 on pages 16.1 16 to 16.1-18. For the flanges of I-shape sections in pure compression r = 0.56 x E = 0.56 x Fy 29000 = 15.9 36
For the webs of I-shapes section in pure compression 13
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r = 1.49 x
E = 1.49 x Fy
29000 = 42.3 36
Step II. Calculate the slenderness ratios for the flanges and webs of W14 x 74 For the flanges of I-shape member, b = bf/2 = flange width / 2 Therefore, b/t = bf/2tf. For W 14 x 74, bf/2tf = 6.41 For the webs of I shaped member, b = h h is the clear distance between flanges less the fillet / corner radius of each flange For W14 x 74, h/tw = 25.4 (See Page 1-23 in AISC) (See Page 1-23 in AISC)
Step III. Make the comparisons and comment For the flanges, b/t < r. Therefore, the flange is non-compact For the webs, h/tw < r. Therefore the web is non-compact Therefore, the section is non-compact Therefore, local buckling will not limit the column strength.
4.7 COLUMN DESIGN The AISC manual has tables for column strength. See page 4-10 onwards. For wide flange sections, the column buckling strength ( cPn) is tabulated with respect to the effective length about the minor axis KyLy in Table 4-1. The table takes the KyLy value for a section, and internally calculates the KyLy/ry, and then the tabulated column strength using either Equation E3-2 or E3-3 of the specification. If you want to use the Table 4-1 for calculating the column strength for buckling about the major axis, then do the following:
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K xLx Take the major axis KxLx value. Calculate an equivalent (KL)eq = r / r x y Use the calculated (KL)eq value to find ( cPn) the column strength for buckling about the major axis from Table (4-1) For example, consider a W14 x 74 column with KyLy = 20 ft. and KxLx = 25 ft. Material has yield stress = 50 ksi (always in Table 4-1). See Table 4-1, for KyLy = 20 ft., cPn = 494 kips (minor axis buckling strength) rx/ry for W14x74 = 2.44 from Table 4-1 (see page 4-14 of AISC). For KxLx = 25 ft., (KL)eq = 25/2.44 = 10.25 ft. For (KL)eq = 10.25 ft., cPn = 774 kips (major axis buckling strength) If calculated value of (KL)eq < KyLy then minor axis buckling will govern.
EXAMPLE 4.4 Determine the design strength of an ASTM A992 W14 x 132 that is part of a braced frame. Assume that the physical length L = 30 ft., the ends are pinned and the column is braced at the ends only for the X-X axis and braced at the ends and mid-height for the Y-Y axis. Solution Step I. Calculate the effective lengths. For W14 x 132: Kx = 1.0 and rx = 6.28 in; Ky = 1.0 Ly = 15 ft. ry = 3.76 in; Ag =38.8 in2
Lx = 30 ft. and
KxLx = 30 ft. and KyLy = 15 ft. Step II. Determine the governing slenderness ratio KxLx/rx = 30 x 12 in./6.28 in.= 57.32
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CE470:DesignofSteelStructuresProf.Varma
KyLy/ry = 15 x 12 in./3.76 in. = 47.87 The larger slenderness ratio, therefore, about buckling the major axis will govern the column strength. Step III. Calculate the column strength KxLx = 30 ft. From Table 4-1, Therefore, (KL)eq = for (KL)eq = 18.0 ft. K xLx 30 = = 17.96 ft. rx / ry 6.28 / 3.76
cPn = 1370 kips (design column strength)
Step IV. Check the local buckling limits For the flanges, bf/2tf = 7.15 For the web, h/tw = 17.7 < < r = 0.56 x r = 1.49 x E = 13.5 Fy E = 35.9 Fy
Therefore, the section is non-compact. OK.
EXAMPLE 4.5 A compression member is subjected to service loads of 165 kips dead load and 535 kips of live load. The member is 26 ft. long and pinned at each end. Use A992 (50 ksi) steel and select a W shape Solution Calculate the factored design load Pu Pu = 1.2 PD + 1.6 PL = 1.2 x 165 + 1.6 x 535 = 1054 kips Select a W shape from the AISC manual Tables For KyLy = 26 ft. and required strength = 1054 kips Select W14 x 145 from page 4-13. It has cPn = 1230 kips Select W12 x 170 from page 4-16. It has cPn = 1130 kips 16
CE470:DesignofSteelStructuresProf.Varma
No W10 will work. See Page 4-19 W14 x 145 is the lightest.
Note that column sections are usually W12 or W14. Usually sections bigger than W14 are usually not used as columns.
4.8 EFFECTIVE LENGTH OF COLUMNS IN FRAMES So far, we have looked at the buckling strength of individual columns. These columns had various boundary conditions at the ends, but they were not connected to other members with moment (fix) connections. The effective length factor K for the buckling of an individual column can be obtained for the appropriate end conditions from Table C-C2.2 of the AISC Manual . However, when these individual columns are part of a frame, their ends are connected to other members (beams etc.). Their effective length factor K will depend on the restraint offered by the other members connected at the ends. Therefore, the effective length factor K will depend on the relative rigidity (stiffness) of the members connected at the ends. The effective length factor for columns in frames must be calculated as follows: First, you have to determine whether the column is part of a braced frame or an unbraced (moment resisting) frame. If the column is part of a braced frame then its effective length factor 0.5 < K 1 If the column is part of an unbraced frame then 1 < K Then, you have to determine the relative rigidity factor G for both ends of the column
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CE470:DesignofSteelStructuresProf.Varma
G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming together at an end to the summation of the rigidity (EI/L) of all beams coming together at the same end. E Ic Lc G= EI Lb b
- It must be calculated for both ends of the column.
Then, you can determine the effective length factor K for the column using the calculated value of G at both ends, i.e., GA and GB and the appropriate alignment chart
There are two alignment charts provided by the AISC manual, One is for columns in braced (sidesway inhibited) frames. See Figure C-C2.3 on page 16.1-241 of the AISC manual. 0 < K 1 The second is for columns in unbraced (sidesway uninhibited) frames. See Figure C-C2.4 on page 16.1-242 of the AISC manual. 1 < K The procedure for calculating G is the same for both cases.
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CE470:DesignofSteelStructuresProf.Varma
EXAMPLE 4.6 Calculate the effective length factor for the W12 x 53 column AB of the frame shown below. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. Assume that the same column section is used for the stories above and below.
10 10 ft. W14 x 68
10 ft. W14 x 68
A
12 ft.
W14 x 68
B
W12 x 79 W12 x 79 W12 x 79 18 ft. 20 ft. 15 ft.
18 ft.
Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible. It is an unbraced (sidesway uninhibited) frame. Lx = Ly = 12 ft. Ky = 1.0 Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB. Need to calculate Kx using alignment charts.
Step II - Calculate Kx Ixx of W 12 x 53 = 425 in4 Ixx of W14x68 = 753
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CE470:DesignofSteelStructuresProf.Varma
Ic 425 425 + L c 10 12 12 12 6.493 GA = = = = 1.021 Ib 723 723 6.360 + L b 18 12 20 12 Ic 425 425 + L c 12 12 15 12 5.3125 GB = = = = 0.835 Ib 723 723 6.360 + L b 18 12 20 12
Using GA and GB: Kx = 1.3
- from Alignment Chart on Page 3-6
Step III Design strength of the column KyLy = 1.0 x 12 = 12 ft. Kx Lx = 1.3 x 12 = 15.6 ft. rx / ry for W12x53 = 2.11 (KL)eq = 15.6 / 2.11 = 7.4 ft. KyLy > (KL)eq Therefore, y-axis buckling governs. Therefore cPn = 547 kips
4.8.1 Inelastic Stiffness Reduction Factor Modification This concept for calculating the effective length of columns in frames was widely accepted for many years. Over the past few years, a lot of modifications have been proposed to this method due to its several assumptions and limitation. Most of these modifications have not yet been accepted in to the AISC provisions.
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CE470:DesignofSteelStructuresProf.Varma
One of the accepted modifications is the inelastic stiffness reduction factor. As presented earlier, G is a measure of the relative flexural rigidity of the columns (EIc/Lc) with respect to the beams (EIb/Lb) However, if column buckling were to occur in the inelastic range ( c < 1.5), then the flexural rigidity of the column will be reduced because Ic will be the moment of inertia of only the elastic core of the entire cross-section. See figure below
rc = 10 k si rt = 5 k si Yielded Yielded zone
rt = 5 k si
Elastic core, I c
rc = 10 k si rt = 5 k si
(a) Initial state residual stress
(b) Partially y ielded state at buckling
The beams will have greater flexural rigidity when compared with the reduced rigidity (EIc) of the inelastic columns. As a result, the beams will be able to restrain the columns better, which is good for column design. This effect is incorporated in to the AISC column design method through the use of Table 4-21 given on page 4-317 of the AISC manual. Table 4-21 gives the stiffness reduction factor ( ) as a function of the yield stress Fy and the stress Pu/Ag in the column, where Pu is factored design load (analysis)
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CE470:DesignofSteelStructuresProf.Varma
EXAMPLE 4.7 Calculate the effective length factor for a W10 x 60 column AB made from 50 ksi steel in the unbraced frame shown below. Column AB has a design factor load Pu = 450 kips. The columns are oriented such that major axis bending occurs in the plane of the frame. The columns are braced continuously along the length for out-of-plane buckling. Assume that the same column section is used for the story above
W14 x 74 12 ft. W14 x 74
A
W12 x 79 W12 x 79 W12 x 79 20 ft. 15 ft.
B
18 ft.
18 ft.
Solution Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible. It is an unbraced (sidesway uninhibited) frame. Ly = 0 ft. Ky has no meaning because out-of-plane buckling is not possible. Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB. Need to calculate Kx using alignment charts.
Step II (a) - Calculate Kx Ixx of W 14 x 74 = 796 in4 Ixx of W 10 x 60 = 341 in4
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CE470:DesignofSteelStructuresProf.Varma
Ic 341 341 + L c 12 12 15 12 4.2625 GA = = = = 0.609 Ib 796 796 7.002 + L b 18 12 20 12 G B = 10 - for pin support, see note on Page 16.1-241
Using GA and GB: Kx = 1.8
- from Alignment Chart on Page 16.1-242
Note, Kx is greater than 1.0 because it is an unbraced frame.
Step II (b) - Calculate Kx inelastic using stiffness reduction factor method Reduction in the flexural rigidity of the column due to residual stress effects First calculate, Pu / Ag = 450 / 17.6 = 25.57 ksi Then go to Table 4-21 on page 4-317 of the manual, and read the value of stiffness reduction factor for Fy = 50 ksi and Pu/Ag = 25.57 ksi. Stiffness reduction factor = = 0.878 GA-inelastic = x GA = 0.878 x 0.609 = 0.535 GB = 10 - for pin support, see note on Page 16.1-241 - alignment chart on Page 16.1-242
Using GA-inelastic and GB, Kx-inelastic = 1.75
Note: You can combine Steps II (a) and (b) to calculate the Kx-inelastic directly. You dont need to calculate elastic Kx first. It was done here for demonstration purposes.
Note that Kx-inelastic< Kx. This is in agreement with the fact that the beams offer better resistance to the inelastic column AB because it has reduced flexural rigidity.
Step III Design strength of the column KxLx = 1.75 x 15 = 26.25 ft. rx / ry for W10x60 = 1.71 - from Table 4-1, see page 4-19
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(KL)eq = 26.25/1.71 = 15.35 ft.
cPn for X-axis buckling = 545 kips
Section slightly over-designed for Pu = 450 kips.
- from Table 4-1, see page 4-19
Column design strength = cPn = 545 kips EXAMPLE 4.8: Design Column AB of the frame shown below for a design load of 500 kips. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. Assume that the same column section is used for the stories above and below.
10 10 ft. W14 x 68
10 ft. W14 x 68
A
12 ft.
W14 x 68
B
W12 x 79 W12 x 79 W12 x 79 18 ft. 20 ft. 15 ft.
18 ft.
Step I - Determine the design load and assume the steel material. Design Load = Pu = 500 kips Steel yield stress = 50 ksi (A992 material)
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CE470:DesignofSteelStructuresProf.Varma
Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible. It is an unbraced (sidesway uninhibited) frame. Lx = Ly = 12 ft. Ky = 1.0 Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB. Need to calculate Kx using alignment charts. Need to select a section to calculate Kx
Step III - Select a column section Assume minor axis buckling governs. Ky Ly = 12 ft. See Column Tables in AISC-LRFD manual Select section W12x53 cPn for y-axis buckling = 547 kips
Step IV - Calculate Kx-inelastic Ixx of W 12 x 53 =425 in4 Ixx of W14x68 = 753 in4
Account for the reduced flexural rigidity of the column due to residual stress effects Pu/Ag = 500 / 15.6 = 32.05 ksi Stiffness reduction factor = = 0.66
GA =
Ic Lc
b
L
Ib Ic Lc
b
425 425 0.66 + 10 12 12 12 = 4.285 = 0.674 = 723 723 6.360 + 18 12 20 12 425 425 0.66 + 12 12 15 12 = 3.506 = 0.551 = 723 723 6.360 + 18 12 20 12 - from Alignment Chart
GB =
L
Ib
Using GA and GB: Kx-inelastic = 1.2
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CE470:DesignofSteelStructuresProf.Varma
Step V - Check the selected section for X-axis buckling Kx Lx = 1.2 x 12 = 14.4 ft. rx / ry for W12x53 = 2.11 Calculate (KL)eq to determine strength ( cPn) for X-axis buckling (KL)eq = 14.4 / 2.11 = 6.825 ft. From the column design tables, cPn for X-axis buckling = 644 kips
Step VI. Check the local buckling limits For the flanges, bf/2tf = 8.69 For the web, h/tw = 28.1 < < r = 0.56 x r = 1.49 x E = 13.5 Fy E = 35.9 Fy
Therefore, the section is non-compact. OK, local buckling is not a problem Step VII - Summarize the solution Lx = Ly = 12 ft. Ky = 1.0 Kx = 1.2 (inelastic buckling - sway frame-alignment chart method) cPn for Y-axis buckling = 518 kips cPn for X-axis buckling = 644 kips Y-axis buckling governs the design. Selected Section is W12 x 53 made from 50 ksi steel.
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CE470:DesignofSteelStructuresProf.Varma
EXAMPLE 4.9 Design Column AB of the frame shown below for a design load of 450 kips. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced continuously along the length for out-of-plane buckling. Assume that the same column section is used for the story above.
W14 x 74 12 ft. W14 x 74
A
W12 x 79 W12 x 79 W12 x 79 20 ft. 15 ft.
B
18 ft.
18 ft.
Step I - Determine the design load and assume the steel material. Design Load = Pu = 450 kips Steel yield stress = 50 ksi
Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible. It is an unbraced (sidesway uninhibited) frame. Ly = 0 ft. Ky has no meaning because out-of-plane buckling is not possible. Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB. Need to calculate Kx using alignment charts. Need to select a section to calculate Kx
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Step III. Select a section There is no help from the minor axis to select a section Need to assume Kx to select a section. See Figure below:
W14 x 74
12 ft. W14 x 74
A
W12 x 79 W12 x 79 W12 x 79 15 ft.
B
18 ft. 20 ft.
18 ft.
Kx = 2.0 Best Case Scenario fro m Pg. 6- 184
The best case scenario for Kx is when the beams connected at joint A have infinite flexural stiffness (rigid). In that case Kx = 2.0 from Table C-C2.1 Actually, the beams don't have infinite flexural stiffness. Therefore, calculated Kx should be greater than 2.0. To select a section, assume Kx = 2.0 KxLx = 2.0 x 15.0 ft. = 30.0 ft. Need to be able to calculate (KL)eq to be able to use the column design tables to select a section. Therefore, need to assume a value of rx/ry to select a section. See the W10 column tables on page 4-26. Assume rx/ry = 1.71, which is valid for W10 x 49 to W10 x 68.
(KL)eq = 30.0/1.71 = 17.54 ft. Obviously from the Tables, for (KL)eq = 17.5 ft., W10 x 60 is the first section that will have cPn > 450 kips Select W10x60 with cPn = 457.7 kips for (KL)eq = 17.5 ft.
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CE470:DesignofSteelStructuresProf.Varma
Step IV - Calculate Kx-inelastic using selected section Ixx of W 14 x 74 = 796 in4 Ixx of W 10 x 60 = 341 in4
Account for the reduced flexural rigidity of the column due to residual stress effects Pu/Ag = 450 / 17.6 = 25.57 ksi Stiffness reduction factor = = 0.863
GA =
Ic Lc
b
L
Ib
341 341 0.863 + 12 12 15 12 = 3.678 = 0.525 = 796 796 7.002 + 18 12 20 12 - for pin support - from Alignment Chart on Page 3-6
G B = 10
Using GA and GB: Kx-inelastic = 1.75 be equal to 10 instead of
Calculate value of Kx-inelastic is less than 2.0 (the assumed value) because GB was assumed to
Step V - Check the selected section for X-axis buckling Kx Lx = 1.75 x 15 = 26.25 ft. rx / ry for W10x60 = 1.71 (KL)eq = 26.25/1.71 = 15.35 ft. ( cPn) for X-axis buckling = 545.5 kips Section slightly over-designed for Pu = 450 kips. W10 x 54 will probably be adequate, Student should check by calculating Kx inelastic and
cPn for that section.
Step VI. Check the local buckling limits For the flanges, bf/2tf = 7.41 For the web, h/tw = 18.7 < < r = 0.56 x r = 1.49 x E = 13.5 Fy E = 35.9 Fy
Therefore, the section is non-compact. OK, local buckling is not a problem
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CE470:DesignofSteelStructuresProf.Varma
Step VII - Summarize the solution Ly = 0 ft. Ky = no buckling Kx = 1.75 (inelastic buckling - sway frame - alignment chart method) cPn for X-axis buckling = 545 kips X-axis buckling governs the design. Selected section is W10 x 60 (W10 x 54 will probably be adequate).
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