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Calculation of the Plastic Section Modulus
Using the Computer
DOMINIQUE BERNARD BAUER
ABSTRACT
A simple spreadsheet is presented which calculates the plas- tic section modulus of structural members. The method con- sists in dividing the cross section into rectangles and arrang- ing all calculations conveniently into a spreadsheet program. The basic algorithm and the required spreadsheet formulas are given as well as a numerical example.
INTRODUCTION
With the increasing use of the limit states design of steel structures, engineers often have to calculate the plastic bend- ing resistance, M r, of structural members, which is a function of the plastic modulus, Z, of the cross section, that is,
M=$ZFy (1)
where
<|) = performance factor Fy = yield strength of steel.
Although the calculation of the plastic section modulus can be done easily by hand, it can also be done quickly and reliably using the computer. The following technical note presents a simple spreadsheet for the calculation of the plastic modulus. It is restricted to cross sections that can be approxi- mated by a series of rectangles, which should cover most situations that structural engineers encounter in the design office.
SPREADSHEET ALGORITHM
The proposed algorithm is described below. The cross section to be analyzed must first be divided into N rectangles (Figure la). Each rectangle must comprise the entire width of the cross section at any particular height. Hence, the arrangement shown in Figure la is valid, while the one shown in Figure lb is not valid. The width and the height of each rectangle will be entered into the spreadsheet, going consecutively from top to bottom of the cross section. These values are the only required input
Dominique Bernard Bauer, P. Eng., MASCE, MCSCE, B. Eng., M. Eng., Ph.D., structural engineering consultant in Montreal.
data. All the calculations presented below are arranged so that the equations can be expressed as spreadsheet formulas which will be evaluated automatically by the spreadsheet program. With a datum line placed at the top of the cross section, the vertical distance from the datum line to the centroid, yn, of the nth rectangle is equal to (Figure 2)
where
hn= height of the nth rectangle
Datum line
I n
...n...
.N
(a) valid arrangement
Datum line
(b) invalid arrangement
Figure 1.
yo=K = o.
The cross-sectional area, An, of the nth rectangle is equal to
K = bnhn (3)
where
bn= width of the nth rectangle.
The total area, Atot^ of the cross section is equal to
A.tntni — SA. (4)
The vertical distance, f , from the datum line to the neutral axis, which divides the cross section into two portions of equal areas, is determined by noting that if the neutral axis passes through the nth rectangle, we must have
-^ = y£Ai + bnfin (5)
or
H - l -5>
k=-
— ( 6 )
where hn positions the neutral axis as shown is Figure 2. The rectangle through which the neutral axis passes is determined from the fact that it is the only one for which
hn > 0 and hn < hn (7)
For all other rectangles, Equation 7 is not verified. Hence, the
vertical distance from the datum line to the neutral axis, Y , is equal to
m - l (8)
where the subscript m identifies the single rectangle for which Equation 7 is verified. For the other rectangles through which the neutral axis does not pass, the values
?n = ^hi + hn (9)
are meaningless and therefore discarded. The contribution to the plastic modulus, Zn, of each rectan- gle through which the neutral axis does not pass is equal to
Z„=Anabs(5j (^) (10)
where the distance from the neutral axis to the centroid of the nth rectangle, <3n, is equal to Figure 3a.
an = f-yn (11)
The contribution to the section modulus, Zm, of the rectangle through which the neutral axis does pass is equal to Figure 3b.
Datum line
Neutral axis
(a)
Datum line
'
i Kn 1
( r
r
| i... l-n-l
Centroid of ^ ectangle n
u (^) n
n
1 ' 1
Neutral axis i
L
yn
'
Figure 2.
Datum line
h ' i
j
1 1 —— , h^ 2 : m -fim 2
k
'
b„
m
1 1 Neutral axis ,
k ? '
i
(b)
Figure 3.
K
K
Mr ?
Ym
t
= height of the nth rectangle = distance from the top of the nth rectangle to the neutral axis of the cross section = plastic bending moment resistance = distance from the datum line to the neutral axis of the cross section = correct value of Y obtained with rectangle m (Ym = ?) = value of Y obtained with the nth rectangle (Yn * Y for all rectangles except rectangle m)
yn
z
zm
zn
<$>
= distance from the datum line to the centroid of the nth rectangle = plastic section modulus = contribution of rectangle m to the plastic section modulus = contribution of the nth rectangle to the plastic section modulus = performance factor
Column—>
U'Row
6
7
11
Note:
!
n
1
2
3
C
bn
D E
Atotaj/2 =
K
!iii
Atota] —
yn
4
58
112
Indicates input d
F
1100
2200
A,
1000
600
600
ata.
G
n-l V t 1=
0
8
108
H
n - l 2>i
0
1000
1600
I
Y =
K
8.
16.
-6.
J
24.
X
0
24.
0
K
Z =
20.
N. Axis
-87.
L
94733.3|
Zn
20666.7|
21666.7|
52400J
Figure 5.
