14. Design - Beam_Column

8/13/2019 14. Design - Beam_Column

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By.Ir.Sugeng P Budio,MSc 1

DESIGN OF BEAM-COLUMN


Because of the many variables in the interaction formulas,the design of beam-columns is essentially a trial-and-errorprocess. A trial shape is selected and then reviewed forsatisfaction of the governing interaction formula.Obviously, the closer the trial shapes to the final selection,the better. A very efficient procedure for choosing a trial

shape, originally developed for allowable stress design, hasbeen adapted for LRFD and is given in Part 2 of the Manual,Column Design. The essence of this method is to convertthe bending moments to equivalent axial loads, and ashape that will support must then be investigated with AISCEquation. The total effective axial load is given by :


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Pueff = Pu + Muxm + MuymU

Pu = actual factored axial load (kips)Mux = factored moment about the x-axis (ft-kips)Muy = factored moment about the y-axis (ft-kips)

m = a tabulated constantU = a tabulated constant

The basis of this procedure can be examined byrewriting Equation as follows. First, multiply both sidesby cPn :


nc

nyb

uync

nxb

uxncu P

M

MP

M

MPP

+

+

or

Pu + (Mux x constant) + (Muy x constant) cPn

The right side of this inequality is the design of a memberunder consideration, and the left side can be thought of as the

applied factored load to be resisted. Each of there terms on theleft must have units of force, so the constant convert thebending moments Mux and Muy to axial load components.Average values of the constant m have been computed fordifferent groups of W- and S- shapes and are given in Table Bin Part 2 of the manual. Values of U are given in the ColumnLoad Tables for each shape listed therein. To select trial shapefor a member with axial load and bending abouth both axes,proceed as follows :


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1.Select a trial value of m based on the effective lengthKL. Let U = 2

2.Compute an effective axial compressive load :Pueff = Pu + Muxm + MuymU

Use this load to select a shape from the Column Loadtables.

3.Use the value of U given in the Column Load Tablesand an improved value of m from Table B to computean improved value of Pueff. Select another shape.

4.Repeat until there is no change in Pueff

Note that the tabular values of m are for Cm = 0.85.For other values of Cm, multiply m by Cm/0.85. aftertrial shape has been selected, it must be checkedagaints the appropriate interaction formula.


A certain structural member in a braced frame must

support a factored axial compressive load of 150 kips

and factored end moments of 75-ft-kips about thestrong axis and 30 ft-kips about the weak axis. Both ofthese moments occur at one end; the other end ispinned. The effective length with respect to each axis is15 feet. There are no transverse loads on the member.Use A36 steel and select the lightest W-shape.


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The amplification factor B1 can be estimeted as 1.0

purposes of making a trial selection. For each of thetwo axes,

Mux = B1Mntx 1.0(75) = 75 ft-kipsMuy = B1Mnty 1.0(30) = 30 ft-kips

From Table B, Part 2 of the Manual, m =2.6 for Cm = 0.85.For this member,

Cm = 0.6 0.4(M1/M2) = 0.6 0.4(0/M2) = 0.6 (for both axes)


Use m = 2.6(0.6/0.85) = 1.835 and initial value of U = 2.0

Pueff = Pu + Muxm + MuymU = 150 + 75(1.835) + 30(1.835)(2) = 398

Beginning eith the smaller shapes in the

Column Load Tables,

Try W8x67 (Try W8x67 (ccPPnn = 412 kips, U = 2.03)= 412 kips, U = 2.03)

m = 3.25(0.6/0.85) = 2.294Pueff = 150 + 75(2.294) + 30(2.294)(2.03) = 461 kips

This value is slightly larger than the design strength of412 kips, so another shape must be tried.


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Try W10x68 (Try W10x68 (ccPPnn = 475 kips, U = 2.01)= 475 kips, U = 2.01)

m = 2.95(0.6/0.85) =2.082Pueff = 150 + 75(2.082)+30(2.082)(2.01)

= 432 kips < 475 kips (OK)

The W10x68 is therefore a potential trial shape.Check the W12s and W14s for other possibilities.


Try W12x58 (Try W12x58 (ccPPnn = 397 kips, U = 1.73).= 397 kips, U = 1.73).

m = 2.4(0.6/0.85) = 1.694Pueff = 150 + 75(1.694) + 30(1.694)(1.73)

= 365 kips < 397 kips (OK)

A W12x58 is therefore a potential trial shape. Thelightest W14 with a chance of working is a W14x61, and

it is heavier than the W12x58.

Use a W12x58 as a trial shape.

EquationAISCuse

2.03778.0397

150

>== nc

u

P

P


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kipsftMB

PP

CB

kipsrKL

EAP

r

LK

ntx

eu

m

g

e

x

x

===

=


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A W12x58 is compact for any value of Pu (there is no

footnote in the Column Load Table), so the designstrength is

bMny = bMpy = bZyFy = 0.90(32.5)(36) = 1053 in.-kips= 87.75 ft-kips

But Zy/Sy = 32.5/21.4 = 1.52 > 1.5 , which means that bMnyshould be taken as

b(1.5My) = b(1.5FySy) = 0.90(1.5)(36)(21.4) = 1040 in.-kips

= 86.67 ft-kips

AISC Equation

(OK)1.00.972

67.86

30

233

75

9

83778.09

8


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Use Yuras method to select a W12 shape

for the beam-column of Example


From Equation,the equivalent axial load is

kipsxx

b

M

d

MPP

yxequiv

52512

)1230(5.7

12

)1275(2150

5.72=++=++=

where the width b is assumed to be 12 inches. Fromthe Column Load tables, try W12x72 (cPn=537 kips)

Since the flange width is 12 inches as assumed, noiteration is required. For this particular problem,Yuras method produces a more conservative trialshape than the Manual method,although this may not

always be true.

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14. Design - Beam_Column