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1. Mathcad - LCA (Load Case Algorithm)
by Julio C. Banks, PE.xmcd
page 1 of 3
Steel Beam Design
by Julio C. Banks, PE
Two beams made from 2-2x8 wood with intermediate supports are to be replaced
by steel beams support at each end. The shortest beam has one (1) intermediate
support and the longest beam has two (2) intermediate supports.
The existing loads on the 2-2x8 wood beams are not known and must be estimated
by the PE.
Solution
The method of estimating the existing beam loads is by using the flexural capacity
of the wood beams and the self-weight. An FEA software will be used to determine
the maximum live load that the 2x8 wood beam will be able without causing
structural failure of the beam.
ASCE 7-10, Section 2.3.2 - Basic Load Combinations
Load combinations 1 and 2 are the most critical when only D (Dead) and L (Live)
clods are to be considered.
1. 1.4D
2. 1.2D + 1.6L
One must select the maximum load case with the following procedure:
Determine the D/L ratio (or L/D) at which the maximum load case is determined
automatically by asking the question "When is Load Case No. 1 Greater than or
equal to Load Case No. 2?" i.e., when is LC1 LC2 ?
or when is 1.4 D 1.2 D 1.6 L ? Now, solve for a convenient ratio.
1.4 D 1.2 D 1.6 L 1( )
Divide Eq. 1 by 1.2
7
6
D D
4
3
L 2( )
Divide Eq. 2 by D 7
6
1
4
3
L
D
 3( )
L
D
7
6
1






3
4

L
D
1
8
 or
D
L
8 4( )
Julio C. Banks, PE

2. Mathcad - LCA (Load Case Algorithm)
by Julio C. Banks, PE.xmcd
page 2 of 3
When equation 4 is satisfied (yes-case) then load case 1 governs.
Equation 4 is now converted into an algorithm that will calculate which load case
governs and will state the governing load case. Load case No. 1 governs if D/L is
greater than or equal to 8, otherwise, Load case No. 2 governs.
The load U is to be used in in classical calculations by adding the self-weight to the
dead load.
LCA D L( ) "LCA (Load Case Algorithm)"
Solution
1
1.4 D
plf














D
L
8if
2
1.2 D 1.6 L( )
plf














otherwise

Solutionreturn

5( )
Example
The ASCE 7-10 minimum live load for storage is 20 psf. The maximum live load for
a given dead load that a 2-2x8 beam could resist without failure and a utilization of
0.96 or 96 % is 25 psf. When such a live load of 25 psf was applied to the known
tributary results in the distributed live load given in this example
Dead load: D 17.71 plf Live load: L 236.1 plf
D
L
0.08
The LCA (Load Case Algorithm) produces both, the governing load case and its
magnitude.
Load_Case_No LCA D L( )1 2
U LCA D L( )2 plf 399.0 plf
Julio C. Banks, PE

3. Mathcad - LCA (Load Case Algorithm)
by Julio C. Banks, PE.xmcd
page 3 of 3
The following 3-steps procedure produces the governing load case, U, producing the
same results as the algorithm presented in equation 5.
LRFD
1.4 D
1.2 D 1.6 L






24.79
399.01






plf
U max LRFD( ) 399.0 plf
Load_Case_No if
D
L
8 1 2






2
Julio C. Banks, PE