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Steel Columns

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easily accommodates beam framing when 10' depth. Column Slenderness and End Conditions ... ductility of steel. overall column slenderness ... – PowerPoint PPT presentation

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Title: Steel Columns


1
Steel Columns
  • Steel Chapter 4

2
Steel Columns
  • compression members
  • primary function resist compressive forces
  • complicated by buckling and bending

3
Column Shapes
  • most common
  • round pipe
  • rectangular tube
  • W-shapes
  • easily accommodates beam framing when 10 depth

4
Column Slenderness and End Conditions
  • slenderness L/r
  • end restraint effects included by use of
    modifying factor (K)
  • modified slenderness KL/r
  • should be between 50 and 100
  • do not use as compression member if greater than
    200
  • determines critical stress (Fc)

5
Column Slenderness and End Conditions
6
Safe Axial Loads for Steel Columns
  • requires determination of the maximum factored
    load (Pu) or the minimum design strength
  • given that the slenderness and end conditions are
    known
  • Fc compression resistance factor 0.85
  • more direct process is to use column load tables

7
Example 1
  • A W 12x53 of A36 steel is used as a column with
    an unbraced length of 16 feet. The top is pinned
    but prevented from lateral movement and the
    bottom is totally fixed
  • Compute the maximum factored load for the column

8
Example 1
  • end conditions case (b)
  • K 0.8 (design value)
  • Table A.3
  • rx 5.23 in.
  • ry 2.48 in.
  • A 15.6 in.2
  • we can assume that both axes are unbraced, so the
    lower r value is used
  • ry 2.48

9
Example 1
  • determine column slenderness
  • use Table 4.1 to determine the critical stress
    (Fc)
  • with KL/r 62 Fc 29.4 ksi
  • determine the safe axial load for the column

10
Safe Axial Loads for Steel Columns
  • along x-axis, the column is laterally unbraced
    for its full height
  • existence of horizontal framing provides lateral
    bracing with respect to the y-axis
  • need to investigate both axes
  • KLx/rx and KLy/ry
  • choose the largest value and compute Pu in the
    same manner

11
Design of Steel Columns
  • mostly accomplished through tabulated data (safe
    load tables)
  • if no load table is available, a trial-and-error
    approach is used
  • critical stress (Fc) cannot be determined until a
    shape is selected
  • built-up sections must use this approach because
    safe load tables are not applicable to their
    design

12
Design using Single Rolled Shapes (W-shapes)
  • most common squarish H-shaped
  • nominal depth gt 8 inches
  • Table 4.2
  • A36 steel (Fy 36 ksi)
  • size range W8x24 - W14x211
  • Table 4.3
  • Grade 50 steel (Fy 50 ksi)
  • Size range W8x24 W14x211
  • general equation for design of steel columns
    using LRFD
  • Pu Fc x Pn

13
Table 4.2
14
Table 4.3
15
Example 2
  • Using Table 4.2, select a W shape A36 steel
    column section for an axial load of 100 kips dead
    load and 150 kips live load. The column has an
    unbraced length of 24 feet and the end conditions
    are pinned at the top and bottom.
  • Pu 1.2D 1.6L 1.2(100 kips) 1.6(150
    kips)
  • Pu 360 kips

16
Example 2
  • Pu 360 kips Fc x Pn
  • using Table 4.2, select a few possible choices
    where the load is greater than 360 kips

17
Example 2
  • possible choices
  • W 10 x 88 FcPn 422 kips
  • W 12 x 79 FcPn 444 kips
  • W 14 x 82 FcPn 361 kips
  • if no other parameters are required, the choice
    on which section should be economically based
  • choose W 12 x 79 because it is the lightest
    section

18
Use of Safe Load Tables
  • both tables are set up based on
  • ry (radius of gyration about the y-axis)
  • K 1.0
  • y-axis must be the most slender
  • KLx/rx lt KLy/ry
  • always the case when
  • end conditions are the same on both axes
  • unbraced lengths are equal (Lx Ly)
  • unbraced length on y-axis gt on x-axis (Lx lt Ly)
  • must look at rx/ry if y-axis not more slender

19
Use of Safe Load Tables
  • in the event that Lx gt Ly
  • need to check the ratios of unbraced lengths
    compared to the ratios of the radius of gyrations
    for each axis
  • Lx/Ly rx/ry slenderness about y-axis controls
  • use unbraced length about y-axis Ly
  • Lx/Ly gt rx/ry slenderness about x-axis controls
  • use equivalent length,

20
Example 3
  • Using Table 4.2, select an A36 steel column
    section for an axial load of 100 kips dead load
    and 150 kips live load if the unbraced length is
    24 feet about the x-axis, the unbraced length is
    8 feet about the y-axis, and the end conditions
    are pinned at the top and bottom
  • axial load is the same as in Example 2 Pu 360
    kips

21
Example 3
  • determine ratio of unbraced lengths about the two
    axes

22
Example 3
  • we can assume the x-axis will control
  • need to find equivalent length
  • possible choices
  • W8 x 58
  • W10x54
  • W12x53
  • W14x68
  • all are acceptable because
  • actual KLy 14
  • each is gt KLy 8 feet
  • with no other parameters, the most economical is
    chosen
  • W12x53

23
Design using Steel Pipe Columns
  • most frequently used as single-story columns
  • three weight categories
  • STD standard
  • XS extra strong
  • XXS double extra strong
  • Table 4.4 safe load tables for pipe columns with
    Fy 36 ksi

24
Design using Structural Tubing Columns
  • used for buildings and members of trusses
  • for columns, range form 3-in. square tubes to the
    largest possible
  • Table 4.5 design strengths for steel tubes
    ranging from 3 in. to 12 in.

25
Columns with Bending
26
Columns with Bending
  • requires the use of interaction analysis
  • for steel columns, major issues include
  • slenderness of column flanges and webs
  • ductility of steel
  • overall column slenderness
  • effects potential buckling in both axial
    compression and bending

27
Columns with Bending
  • AISC interaction formulas

28
Columns with Bending
  • for preliminary design
  • determine equivalent axial load (Pu) which
    incorporates the bending effect
  • m bending factor found in Table 4.2 and 4.3
  • if Pu/Pu lt 0.2, recalculate Pu

29
Example 4
  • We want to use a 10-in. W shape for a column in a
    situation such as that shown in the figure. The
    factored axial load from above the column is 175
    kips, and the factored beam load is 35 kips. The
    column has a unbraced height of 16 feet and a K
    factor of 1.0. Select a trial section for the
    column

30
Example 4
  • determine equivalent axial load, Pu
  • check to see if correct equation was used
  • the correct equation was used, so now we can
    select a trial size

31
Example 4
  • using Table 4.2 and Pu 235 kips, a W10x45 is
    selected

32
Example 4
  • the section must now be checked for compliance
    with the AISC interaction formulas
  • Pu 175 kips 35 kips 210 kips
  • FcPn 252 kips (from Table 4.2)
  • FbMn 175 kip-feet (Lp lt Lb lt Lr)
  • Mux (35 kips x (5/12)) 14.6 kip-ft.

33
Example 4
  • check Pu/FcPn
  • since gt than 0.2 use first equation
  • W 10x45 is acceptable

34
Column Bases
  • in order to transfer loads from column to
    foundation, only need to analyze the direct
    bearing pressure to determine an adequate base
    plate
  • plate must be larger than column footprint in
    order to accommodate anchor bolts and welding

35
Column Bases
  • design capacity of column base plate
  • Fc 0.60
  • fc ultimate compressive strength of supporting
    concrete
  • usually 3 or 4 ksi
  • A1 area of steel bearing plate
  • A2 max. area of support which is geometrically
    similar with the plate

36
Column Bases
  • basis for determining thickness
  • find required A1
  • select dimensions B and N so that m and n are
    equal
  • choice should also take into consideration the
    location of anchor bolts and attachments
  • required thickness based on flexure

37
Example 5
  • Design a base plate of A36 steel for a W 10x49
    column with an ultimate load of 350 kips. The
    column bears on a concrete footing with fc 3
    ksi
  • use max value of 2 for v(A2/A1)

38
Example 5
  • for an approximation of the required plate
    dimensions
  • v114 10.7 in.

39
Example 5
  • since 10.7 is only slightly larger than the
    actual dimensions of the column, the dimensions
    of the base plate need to be increased to
    accommodate attachments
  • increasing each side to 12 leaves m and n both
    equal to about 2
  • now we can determine the required thickness

40
Example 5
  • minimum thickness 0.775
  • 7/8 plate is acceptable where t 0.875 inches
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