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Structural Engineering Calculations

This document presents an example for calculating effective length factors (K-factors) for columns in a stepped crane column frame using the Lui approach. It provides the dimensions and properties of the stepped crane column frame. It then shows the step-by-step calculations to determine the K-factors for each column according to the Lui model equations, which account for the interaction between adjacent columns. The calculated K-factors are found to be in good agreement with previously reported theoretical values.

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290 views3 pages

Structural Engineering Calculations

This document presents an example for calculating effective length factors (K-factors) for columns in a stepped crane column frame using the Lui approach. It provides the dimensions and properties of the stepped crane column frame. It then shows the step-by-step calculations to determine the K-factors for each column according to the Lui model equations, which account for the interaction between adjacent columns. The calculated K-factors are found to be in good agreement with previously reported theoretical values.

Uploaded by

buildcon
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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FIGURE 17.27: Crane column model for effective length factor computation. (From Lui, E.M.

and
Sun, M.Q., AISC Eng. J., 32(2), 98, 1995. With permission.)


KU = KL

LL
LU

 

PL + PU
PU



IU
IL


(17.100)

where P is the applied load and subscripts U and L represent upper and lower shafts, respectively.

EXAMPLE 17.13:

Given: A stepped crane column is shown in Figure 17.28a. The example is the same frame as used
by Fraser [30] and Lui and Sun [59]. Determine the effective length factors for all columns using the
Lui approach. E = 29,000 ksi (200 GPa).
IAB
1999 by CRC Press LLC

c


IF G = IL = 30,000 in.4 (1.25 1010 mm4 )

FIGURE 17.28: A pin-based stepped crane column. (From Lui, E.M. and Sun, M.Q., AISC Eng. J.,
32(2), 98, 1995. With permission.)

AAB
IBC
ABC

=
=
=

AF G = AL = 75 in.2 (48,387 mm2 )


IEF = ICE = IU = 5,420 in.4 (2.26 109 mm4 )
AEF = ACE = AU = 34.14 in.2 (22,026 mm2 )

Solution

1. Apply a set of ctitious lateral forces with = 0.001 as shown in Figure 17.28b.
1999 by CRC Press LLC

c


2. Perform a rst-order analysis and nd


(1 )B = 0.1086 in.(2.76 mm) and (1 )F = 0.1077 in. (2.74 mm)
so,
(0.1086 + 0.1077)/2
1
 =
= 0.198 in./kips (1.131 mm/kN)
0.053 + 0.3 + 0.053 + 0.14
H
3. Calculate factors from Equation 17.81.
Since the bottom of column AB and F G is pin-based, m = 0,

AB

=
=

(3 + 4.8m + 4.2m2 )EI


3EI
= 3
3
L
L
(3)(29,000)(30,000)
= 42.03 kips/in. (7.36 mm/kN)
(396)3

F G =

42.03 + 42.03 = 84.06 kips/in. (14.72 mm/kN)

4. Calculate the K-factors for columns AB and F G using Equation 17.80.




KAB

KF G



 
1
2 (29,000)(30,000)
353 + 193
+
0.198
396
5(84.06)
(353)(396)2
= 6.55



 
1
2 (29,000)(30,000)
353 + 193
+
0.198
=
396
5(84.06)
(193)(396)2
= 8.85
=

5. Calculate the K-factors for columns BC and EF using Equation 17.100.


 


LAB
PAB + PBC
IBC
= KAB
LBC
PBC
IAB

 



5420
353
396
= 18.2
= 6.55
156
53
30,000

 


LF G
PF G + PEF
IEF
= KF G
LEF
PEF
IF G
 



5420
193
396
= 18.2
= 8.85
156
53
30,000


KBC

KEF

The K-factors calculated above are in good agreement with the theoretical values reported by Lui
and Sun [59].
1999 by CRC Press LLC

c


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