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Dissertations and Theses Dissertations and Theses
1980
The effect of residual stress distribution on the The effect of residual stress distribution on the
ultimate strength of tubular beam-columns ultimate strength of tubular beam-columns
Steven L. Barrett Portland State University
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Recommended Citation Recommended Citation Barrett, Steven L., "The effect of residual stress distribution on the ultimate strength of tubular beam-columns" (1980). Dissertations and Theses. Paper 3050. https://doi.org/10.15760/etd.3045
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I
I _J
AN ABSTRACT OF THE THESIS OF Steven L. Barrett for the
Master of Science in Applied Science presented
February 22, 1980.
Title: The Effect of Residual St~ess Distribution on
the Ultimate Strength of Tubular Beam-Columns.
APPROVED BY MEMBERS OF THE THESIS COMMITTEE:
III, Chairman
H~kErzu
son
- ~~
Using data for the longitudinal residual stress
distribution in welded steel tubes, curves describing
these distributions are selected for study. Each of these
curves are checked for static balance across the tube
cross section. The curves that exhibit an imbalance are
adjusted by a combinatior1 of a simplified model for each
and the use of a computer program that is developed to
calculate the resulting forces and moments on the cross
section. The residual stress in the area of the tube wall
opposite the longitudinal weld is found to be the most
important in the adjustment to obtain exact equilibrium.
The method of adjustment is rational and based on maintain
ing a smooth curve shape that matches the· raw data the
closest and producing a curve that is balanced within the
accuracy limits required.
The balanced longitudinal residual stress distri
butions are used to determine M-P-~ relationships,
where M equals moment, P equals axial load, and ~ equals
curvature. A computer program developed by Wagner (13)
is modified to determine these relationships. The program
was originally developed for a bending axis through the
weld and the results are found to be inaccurate for
other orientations. The problem is identified as the
creation of a moment on the cross section from an uneven
distributionof:axial stress caused during the iteration
to determine axial strain from an applied axial force.
The problem is solved by adding another iteration to the
program to redistribute the axial strains and eliminate
the moment.
2
The M-P-~. relationships obtained from this computer
program for three longitudinal residual stress distributions
with five different bending axis orientations are compared
and one residual stress distribution curve is selected
to compare with Wagner's (13) distribution. The beam
column failure load computer program (13) is used with
the M-P-~ relationships generated· and the results are
compared with test data.
The conclusion is reached that residual stresses
have a large influence on the behavior of welded steel
tube beam columns. The difference in effect of each
residual stress distribution is not large. This could be
because all of the curves have the same· general shape with
very large stresses close to the weld and rapidly de
creasing away from it. With the effect of residual
stresses included, the beam column failure load program
still overestimates actual member strength. Therefore,
other conditions effecting member strength when used as
beam columns require further study and the residual
stress distribution labeled curve number three which is
Chen and Ross's (3,4) curve forced to two maximums from
x/d = 1.0 to x/d =1)-' is recommended for use in further
research.
3
THE EFFECT OF RESIDPAL STRESS
DISTRIBUTION ON THE ULTIMATE STRENGTH
OF
TUBULAR BEAM-COLUMNS
by
STEVEN L. BARRETT
A thesis submitted in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE
in
APPLIED SCIENCE
Portland State University February 1980
; I. 1
:
TO THE OFFICE OF GRADUATE STUDIES AND RESEARCH:
The members of the Committee approve the thesis
of Steven L. Barrett presented February 22, 19'80.
Wendelin H. Mueller, III, Chairman
APPROVED:
Arnold Wagne~·
Engineering
Stanley E. Rauch, Dean of Graduate Studies and Research
~
TABLE OF CONTENTS
LIST OF FIGURES . . . . . .
CHAPTER
I
II
III
IV
v
VI
INTRODUCTION . . . . . .
REVIEW OF LITERATURE .
COMPUTER PROGRAM FOR CURVE ADJUSTMENT ..
RESIDUAL STRESS DISTRIBUTION CURVES .....
M-P-0 CURVE GENERATION . . . . . . . . .
CONCLUSION . . . . . . .
REFERENCES ..
APPENDIX I.
APPENDIX II
PAGE
iv
1
5
7
25·
39
61
69
71
83
FIGURE
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
LIST OF FIGURES
Longitudinal Residual Stress in a Welded
Steel· Tube . . . ~ . . . . . . . . .
Longitudinal Residual Stress .
Area Under Curve .
Arc Length - Lever Arm
Areas Under Curve ..
Areas Under Curve.
·Flow Chart - Moment, Force Summation
Residual Stress From Tran (11)
Approximate Curve ..
Approximate Curve.
Residual Stress Tran (11)
Simplified Curve
Residual Stress Chen & Ross (3,4)
14. Residual Stress, Baianced Chen & Ross (3,4)
Curve 1. . .
15. Residual Stress First Section Average - Chen·
& Ross (3,4) - Tran (11) Curve 2.
16.
17.
Fluctuation in Curve Shape .
Residual Stress First Portion, Smoothed Curve
· from Tran (11)
PAGE
9
9
11
11
13
14
15
18
21
22
26
27
29
31
32
33
35
FIGURE
18. Residual Stress Chen & Ross (3,4) Forced to
Two Maximums Curve 3. .
19. Flow Diagram for Calculation of M-P-0 Data
(Wagner)
20. Orientation of Axes and Residual Stress
Distribution (Wagner (12,13) . . . . . 21. M-P-.0 Curve 1 (Chen & Ross)· .. . . . .. . . . ..
22. M-P-.0 Curve 2 (First Section Ayerage) . . . . 23. M-P-.0 Curve 3 (Chen & Ross Forced To Two
Maximums). .
24. Moment Imbalance . • .
.
.
.
v
PAGE
37
40
44
46
47
48
53
25. M-P-0 Curves From Improved Generation Program. 56
26. Beam Column Results - Wagner's Distribution. . 59
27. Beam Column Results - Curve 3 (Fig. 18). . . . 60
28. Longitudinal.Residual Stress Distribution
Curves . . .
29. Moment Curvature- Relationships
30. Effect of Residual Stress on Strength Under
31.
32.
Axial Load ...
Ef feet of Residual Stress ..
Example for Testing Curve Program. .
62
63
64
65
81
CHAPTER I
INTRODUCTION
The idealized model that is used to exhibit the
pehavior of steel in most mechanics of materials texts has
well defined characteristics. The idealized material is
stress-strain free prior to application of any load,
there are no thermal strains, the material is homogenous
and it has a bilinear stress strain relationship with con-
stant slope to yield and a constant value beyond. If this
material is used in an idealized member it is possible
to theoretically predict the member's exact behavior under
load. For different cross section shapes it is possible
to determine the effect of slenderness on the member.•· s
behavior if it is used as a compression member, a beam,
or a beam-column.
For a real structural member the prediction of actual
behavior is much more difficult. Actual materials are
not entirely homogenous and do not have bilinear stress
strain curves. A real structural member is not stress
free prior to loading, may have a comple~ cross section
shape which ha~ no simple closed form solution for ·
critical buckling load, may have accidental eccentricity
of load, initial crookedness and any number of stresses
induced during erection. Each of these cond~tions that
can be accurately measured in the laboratory should be
investigated to determine its individual effect on the
strength of the member. Once the relative magnitude of
the individual effect is known, a decision can be made
whether or not to include that effect. in every. member
analysis. To that· end, there has been considerable
research done on certain structural shapes in various
materials. The largest volume of research has been done
on rolled steel wide flange shapes. due to their widespread
application for structures of all types. (l,2,5,8,10)
Because the information available for the steel wide
flange ·shape is. extensive, there has been a tendency t.o
apply it to other cross section shapes where data has not
yet been developed. For any particular case· this practice
may or may not be conservative. It is not possible to
completely discount any of the previously mentioned effects
on a cross section shape other than a wide flange based on
the argument that such an effect did not alter the behavior
of the wide flange as idealized. Nor is it possible to
make a direct comparison without some data being developed
for the particular cross secti6n.
2
The complete description of the effect of all possible
initial conditions on a welded steel tube cross section for
all types of loading is too broad in scope for this study.
Therefore the single consideration of the effect of residual
3
stress was selected to study. Residual stresses are those
stresses which are present in a member after it is fabri
cated to its finished form. Usually these residual stresses
cause a reduction in strength in fatigue, fracture and
stability· although for some distributions a strengthening
at certain load levels will be exhibited. These stresses
may result from uneven cooling after hot rolling as for
structural shapes and plates, from cold bending during
fabrication, from welding or ·from cutting, drilling and
punching during fabrication. Usually residual stresses
with the largest magpitude result from cooling and welding.
In many cases, residual stresses in the region of a weld
will be greater than the yield stress. It therefore
becomes important to determine the distribution and magni
tude of residual stress in a member that has extensive
weld~ng during fabrication. Recent research has suggested
some possible distributions of both longitudinal and
circumfere.ntial residual stresses in welded steel tubes.
(3,7,11,12,13)
The primary goal of this project was to analyze and
select one longitudinal residual stress distribution
in welded steel tubes from the literature and test data
available and then determine the effect of this distri
bution on the ultimate beam-column failure load of the
tube. The circumferential residual stress was not included
since the project was simplified excluding the effects of
local buckling or ovaling from consideration. After deter
mination of the residual stress distribution, moment
curvature-axial load curves could be generated by computer.
These describe· the performance of the cross section of the
tube. Then, using these curves, an iterative open form
computer solution calculates the beam column failure ioad~
The solution obtained is c~mpared to the ideal case of no
residual stress and with the results of previous research
using different.distributions. From this comparison it
is possible to determine the relative effect of res{dual
stresses in determining failure load and the effect of
changing the distribution.
4
CHAPTER II
REVIEW OF LITERATURE
The distribution of longitudinal residual stress
described by a linear variation between three peak values
was used by Wagner (12,13) in the absence of actual test
data. This distribution gave a starting point for the
computation of M-P-0 curves, where M equals moment, P
equals axial load and 0 equals curvature with the effect
of some residual stress added. A computer program was
developed using the assumed general shape that could adjust
the value of residual stress on each element of the tube
such that force an~ rotational equilibrium would be
satisfied.
While there is a large volume of information on the
magnitude and distribution of residual stresses in wide
flange sections (4,8), there is a scarcity of information
about the distribution and magnitude.in welded steel tubes
or its effect on the beam column failure load. Some
investigators have suggested possible theoretical distribu
tions (7,) while others (4) have attempted to use more
realistic patte~ns in correlating experimental results.
They, however, discounted the necessity of static equilibrium
across the cross section considering the imbalance of
6
internal forces to be of little consequence. Further,
no consideration of the uniformity. of the stress field
through the tube wall thickness was made.
Recently however Tran (11) developed test data that
first confirmed a uniform field of longitudinal stress
through the wall of a ·tube and showed a very close
approximation to static equilibrium across the section.
CHAPTER III
COMPUTER PROGRAM FOR CURVE ADJUSTMENT
In order to determine .~he.best curve to describe the
longitudinal residual stress distribution in the welded
tube it was necessary to generate a method of rapidly
checking the balance of forces and moments that each
distribution curve implied. If the residual stress
distribution through the th~ckness of the tube wall was
not uniform, it would be necessary to check the balance
for several layers and use that information in the genera
tion of moment-curvature axial load curves. However,
Tran (11) has shown that the distribution through the tube
wall is uniform. Therefore, it is·only necessary to
consider one layer equal to the tube wall thickness. To
check the balance of forces and moments accurately and
quickly, it was decided to write a computer program that
could handle any general curve describing the residual
stress pattern.
Consider the general longitudinal residual stress
distribution shown in Figure 1. The ordinates of the
residual stresses are measured along· the X axis with the
weld located on the Y axis. To make the curve easier to
represent, it is convenient to straighten the circumference
of the tube, in effect unrolling the tube so it is flat
(Fig. 2). Now the area under any portion of this curve
will represent a force, since the distance along the
circumference multiplied by the wall thickness is area.
Area multiplied by stress, wh~ch is force per unit area,
8
is force. This force will be located at the center of
gravity of the area under the stress curve. The distance
from any single· point on the tube circumference.to the force,
multiplied by the force will equal moment. Since the tube
is under no external loading and therefore must be in
static equilibrium for internal stresses, the summation .of
forces and the summation of moments over the entire cross
section of the tube must be equal to zero.
In general the problem is to find the area (force),
center of gravity and first moment of area (moment) about
some point for segments under a general curve that the
mathematical descrip~ion for is not known. The method used
will be a linear approximation .of segments of the curve,
in other words, short straight lines between points on
the curve. Sign convention used for this development
were: Tensile stresses positive, plotted above the X
axis (Fig. 2); compressive stresses negative, plotted
below the X axis; positive moment .about any singular
point, clockwi~e; negative moment counter clockwise.
There are several possible conditions that must be
accounted for if.the program is to handle all possible
dL
Figure I. Longitudinal residual stress in o welded steel tube
+
of tu be unrolled
Figure 2. Longitudinal residual stress
9
Refering to Figure 3:
Let Y1
= First ordinate on curve
Y2
= Second ordinate on curve
x1 = First abscissa on curve
x2
= Second abscissa on curve
C.G. 1 =-Distance from Y axis to center of gravity of area1
C.G.2
= Distance from Y axis to center of gravity of area 2
Area1 = (X2 - Xl) (Yl)
C.G. 1 _= (x2 - xl) + ~1·= x2 - xl + 2x1 = xl + x2
2
Area2 = (X2 - Xl) (Y2 - Yl)
2
2
C.G.2
= 2/3 (X 2 x1 ) + x1
= 2X2
2
2~1 + 3x1 = .2x2 + x1 3 3
10
Now since this is for a tube the X distances can be thought
of as arc lengths. (Fig. 4)
C.G.x = Arc Length
~ circumference = 21J-r = 7rr 2
~ (radians) = (C.G.x)
( -n-r ) ( 211) = ( 2)
C.G.x
r
Lever arm = L = r - (COS~)r = (1 ~ COS~)r
Moment = (Force) (Lever Arm)
Force = Area , ·Lever Arm = L
M = (Fl) (L1 ) + (F 2 .L2
)
M = {Area1 ) (r -·COS (C.G. 1 ) + (Area 2 ) (r-COS (C.G. 2 ))
r
M = [cx2 X1 ) (Yl) ][ 1 cos
(Y2 - Y1 )J [ 1-COS (2x2';
(Xl + X2)]
2r
x1 ) J (rl
r
(rl + [<2x2 + x1 J
3
. 1·
Y2
Ya
y
M x2
I
Figure 3. Area under curve
y
crown ~r,,-weld of tube
Figure 4. Arc length - Lever arm
0 lo- >-: ~ E cu i...
..::.J <Cl:
11
x
12
cases.of slope of the curve and axis interception (Fig.
5 & 6). Taking each of these one at a time, we can ex-
press each.area and center of gravity in the same terms
used previously (Fig. 6). The location of the center of
gravity for case 5 and 2 are equivalent (for the X direction)
l· as are case 6 and 1 as well as case 3 and 4. Thinking of
i this in flow chart form, we have the central ·part of the
force-moment program complete (Fig. 7). Starting with
the coordihates:of a known curve, it is possible to sum
the forces and moments for all the segments of the curve
·and arrive at a total that will indicate whether or not the
section is in static equilibrium. The method could be used
for random lengths of curve segments or for ~ constant
increment. Xhe program used for·this project used a
constant increment for the ·abscissa simplifying· the input
required since the ordinates become the only required
input. These ordinates were scaled from each curve
selected for analysis. Since the summations should be
zero about any point, the crown of tube was selected for
convenience.
Assuming that an imbalance of force is discovered,
it is possible to select the largest ordinate, divide by
10.0 and use this value to modify all ordinates until a
change of sign .of the summation of forces is detected.
Reducing the increment applied t~ the ordinates, it is
possible to continue the iteration until a force balance
i
I I I
I I I I i I
I :
y
+
13
0
A1 0
A1 x
oAd J oAI
Case I 2 3 4 5 6
y y
0A1 x 0A1 x
OA1 oA1
Cose 2 8 5 Cose I a 6
Figure ~ Areas under curve
1 · I I I
.y
x
( X2- X 1 )( Y2) A1 =
2
CG _ 2 ( X2 - XI ) _ 2 X 2 + X I - 3 + >t1 - 3
(X2-X1) --Xo= {-y,)
• . Y2 -y,
A ·_ (xoHY1l ' - 2
CG1 = (;0) + X1
X2
.y
A,·=· (x2-x1HY1) 2
14
CG = { X2- x I) +· x - X2 + 2 x I 3 1-3
A _ [x2-(X1 ·tXo~ (y2)
2- , 2
CG _ 2 ~x2-x1)- xoJ . ( ) 2-
3 -t x1+x0
= 2x2 - 2x1- 2x0 + 3x 1 + 3x 0
2x 2+ x1 + xo 3
3
Figure 6. Areas under curve
l> -.. l>
N II
0
)>
.. _ l>
N
3 CJ) c 3 l>
CJ) c 3 ~
. l>
St
(") ()
~0 N-II
0
nn P0 N-
nn (;) Si> ,\)-II
0
nn G> G1 . . N-
nn (;') ~ ~---f--...llCiCf----N -
"'CD , ~
+
r~ ~
le of zero is obtained. The result of this procedure is
a shift of the X-Axis of the curve. Re-entering the
balancing program, the balance of moment for the new ordin
ates can be found. Now if the moment is not balanced, the
program becomes how to select the location and amount of
change· required to preserve force balance, the general
curve shape and bring about moment equilibrium.
In order to maintain a smooth curve, the changes
made to balance the moment should vary from ordinate to
ordinate. With a large number of ordinates the changes
possible become large and the number of possible solutions
increases. Because the balanced curve should be the same
general shape as the curve started with, each possible
solution would have to be compared with it and measured
against some criteria to find the best solution. Rather
than design a program to do that it was decided to first
find a solution by simplifying· each curve, approximating
the areas, and doing the balancing calculations by hand.
This way possible curve shapes could be evaluated by
determining the balanced curve ordinates and comparing them
with reasonable· stress values at each element, thereby
eliminating curve-s with impossible or unlikely .distributions.
Then using the more exact summation program previously
described on each selected curve, the imbalance for moment
and force could be determined. Using this information,
changes could be made to the curve and the summation pro-
17
gram run again. This method eliminates some cases and
within four to six runs through the summation program it
is possible to find a solution that satisfies the accuracy
limit required by Wagner's (13) M-P-~ Program.
As one example, the curve suggested by Tran (11) as
shown in Figure 8 might be approximated as shown in
Figure 9. The· summation of forces and moments about the
weld can be done as follows:
d = Diameter ~=Angle from weld to center of gravity
x =Arc length·to C.G. L.A. = Lever arm to center of gravity
r = Radius to center line of element layer = 0.9635 in.
~ = 0.1 ct:.=0.1 radians L.A = (1 - COS~)r = 0.005 in.
~ = 0.3 + 0.35 = 0.65 cf::.= 0.65 radians L.A..= 0.184 in.
~ = 1.00 + 2.14/2 = 2.071 O:.= 2.071 radians L.A. = 1.337
l:F = (Yi) (0.30) (4.7) - (0.7) (4.57) + (1.61) (2.14)
= 0.70 - 3.20 + 3.45 = +0.95
.LM = +0.70 (0.005) 3.20 (0.184) + 3.45 (l.337)
= 0.335 - 0.589 + 3.343 = +4.057
The points Tran's (11) data gives generate a curve
with an imbalance of positive force and positive moment.
The imbalance is perhaps withi~ tolerable experimental
error for the data gathered .( 0. 0 2Py and 0. 0 2 7My where
Py equals axial load at yield, My equals moment at yield)
but the accuracy of balance for use of the residual stress
~ '
'< I 0 0
0 Ut 0 u. (:)
I
'Tldl c '""' (I)
CX> I . . !
::0CD N en 0 .-0.c: g_
en ... """ CD en en N
-4-(JI
~0 3 -I ... Q ~
"'Ut' b .._,
:t
St
19
distr~bution in the M-P-~ program are much more restrictive
(0.00005Py and· 0.001 My). Comparing the results of the
balancing program for Tran's (11) curve and the approximate
solution, close agreement is found confirming the computer
code. A complete check of the program is given in Appendix
I. Using the simple approximation, it can be shown that
changing the portion of the curv~ from x/d = 1.0 to x/d =7r
so that the summation of forces is zero still leaves a
positive unbalanced moment.
What then can be done to develop a curve based on
Tran's (11) data that will satisfy ~F = 0 & ~M = 0? The
first portion (to x/d =·1.0) of Tran (11) & Chen-Ross (4)
curves are of the same shape altho~gh of different amplitude
and therefore, a modification of the portion of the curve
from x/d = 1.0 to x/d =n-- seemed to be a good possibility.
This is the area that previous investigators have regarded
as being of lesser importance probably due to the rapidly
decreasing magnitude of residual stress; although it is
the most important area for accomplishing static
equilibriumL
On·viewing the simplified model, it is quite apparent
that the positive area past x/d = 1.0 is far too large to
be balanced by the negative area close to the summation
axis. Changing the negative area has little effect on the
summation as does changing the even closer positive area.
Changing the area from x/d = 1.0 to x/d ='Tr so that the
force? balance does not balance moments as was previously
shown. Further adjustment is required. Changing.the area
labeled (3) in Figure 9 require~ changing (1) to maintain
20
a force balance. Changing (3) enough to get M = 0 requires
lowering the stress level to 0.016 Fy from x/d = 1.0 to
x/d =~. Then calculating ~he area and moment for the
portion of the curve from x/d = 1.0 to x/d =7'r, the amount
of change to area (1) that is required for equlibrium can
be solved for. With a known value for area (1), the maximum
stress at the weld can be solved for. This calculation
indicates the· stress at the weld would be 2.758 times
the yield stress when the ultimate stress for the tube
as determined by Tran (11) is 1.14 times the yield stress.
A residual stress of 2.758 times yield at the weld is not
reasonable or possible. Nor does the rapid decrease of
residual stress at x/d = 1. 0 to a level of 0. 016 rr/O'y through
the rest of the curve seem logical.
A more likely solution is to change area ( 3) to be
larger positive in the section x/d = 1.0 to x/d = 2.0
and balance that with a smaller negative from x/d = 2.0
to x/d =tr (Fig. 10). This shape begins to look similar
to Chen and Ross's (4) curve (see Figure 13) but the
magnitudes are different. The only viable alternative is
to assume the r.esidual stress pattern is not symmetrical
about the weld axis. There is no available evidence to
supp.ort the assumption that the residual stress distribution
-~
5.0
4.0
3.0
2.0
1.0
Stress w -1.0
-2.0 I
•3.01 -4.0
~~.o
Q)
I
I
~ = 0.1 d
@ '2.50 % 1.5 2..0 2.5 3.0 11'
® 3.20
o< = 5.7295 L.A.= (l - cos o<) r
= (I - cos o<)(0.9035) = 0.005
® ~ = 0.3 + 0.35 = 0.65 L.A.= 0.184
@ ~ = 1.00+ 2 ·!4 = 2.071 L.A.= l. 337
Figure 9... Approximate curve
·21
L
5.0
4.0
3.0
2.0
1.0
Stress
-1.0 I
-2.0~ -3.01 -4.0
-5.0
I
22
@A
3.77
\ 1'( % 1.0 2.0
@a l.26 3.0
® 3.20
Figure I 0. A pproxi mate curve
is not symmetric. Trantookmeasurements from each half
of the tube but he did not select mirror image points.
Chen and Ross (4) also took values around the tube but did
not separate the reported values in thei-r representation
of the data.
23
If the values for the portion of the curve from x/d = 0
to x/d = 1.0 are held constant and only the portion of the
simplified curve from x/d = 1.0 to x/d =7r is changed it
i~ possible to show that for any selected point of change
from positive to negative, there is only one unique set
of areas that will satisfy summation of forces and summation
of moments equal to zero.·
Summarizing then it is possible to use a simplified
model of any residual stress curve to check for an approxi
mate summation of forces and moments. Then using a more
detailed description of the curve, e.g. 20 to 100 values
aroun~ the tube scaled from a sketch· of the curve, and using
the previously described summation program, a more exact
check for balance can be made. ·rf the result is outside
the required accuracy limit, the ordinates of the curve
can be adjusted and the balance rechecked. By choosing unit
values of change over one part of the curve, an indication
of the effect of changing those ordinates on the total
balance can be obtained. Using that information for
further refinement makes the required iteration procedure
converge on the required accuracy limit very quickly.
Using this technique, it is possible to balance a possible
curve shape in less than ten trial runs with the summation
program. No attempt is made to find a mathematical ex
pression describing these curves.
24
CHAPTER IV
RESIDUAL STRESS DISTRIBUTION CURVES
Tran (11) has shown a qurve that represents the data
points obtained from his measurements by the hole drilling
technique (Fig. 11). By dividing the curve into sixty-two
segments and scaling the ordinate of the residual stress
at each point and inputing these values into the summation
program, an indication of the degree ·of imbalance.was
obtained. The summation of moments gave an imbalance of
10.9% My and the summation of forces an imbalance of
0.7% Py, bothof which are certainly within experimental
tolerance. The problem is that the M-P-~ Program requires
a much higher degree of accuracy; 0.1% My and 0.005% Py.
Therefore, some adjustment of the curve is required.
Returning to a simplified·model of this curve
(Fig. 12) an approximate balance of forces and moments
can be attempted. The areas labeled 1, 2 and 3 can be
idealized as forces concentrated at the center of gravity
of the areas. Force 1 is positive and close to the crown,
Force 2 is negative and within x/d = 1.0 and Force 3 is
positive at approximately x/d = 2.0. Forces 2 and 3
are close to equal but of opposite sign.
9Z
,,
::u CD (/)
a.. c a
-~
.!. I
9 0 (J1 0 0
I •
t 0 9 0 ~ (>I
~ 0 0
~ ~q I
9 9 0 0 0 9 0 0 N
°' ~ (J1 0 0 0 0 0 0
~ . CJ)
3 'O
:::;; (i. 0.
g .... < CD
LZ
I .1 !
Force lx very small lever arm = very small +M
Force 2x small lever arm = small -M
Force 3x larger lever arm
Summation
= large +M
Large +M
28
When Force 2 and 3 are nearly equal in magnitude, it is not
possible to obtain static equilibrium. Tperefore the curve
Tran (11) suggests cannot.be balanced without some major
changes to the ordinates of the residual stress curve.
Chen and Ross (4) have shown a curve describing
the residual stress distribution (Fig. 13) which they ob
tained from a combination of their measurements and
theoretical considerations done by Marshall (7). The first
portion of the curve from x/d = 0.0 to x/d = 1.0 is similar
in shape to Tran's but of a different magnitude. This
curve does not match all of their data points, but was
selected based on close agreement. Dividing the curve into
62 segments as was done on the previous curve, scaling the
ordinates and using the summation program gave results
that could be compared with equilibrium and the results
from Tran's curve (Fig. 13). The summation of moments gave
an imbalance of 0.4% My and a Force imbalance of 0.06% Py.
This is again outside the limit required making further
adjustment necessary. Selecting portions of the curve
to change, it was discovered that.very slight changes
in the latter (x/d ~ 1.0) portion of the curve affected the
total moment balance appreciably. The part of the curve
.,, -·
::0 CD.
~~ 0. c 0
g>
::0 0 (It
. (It
6Z
I
0 0. 0 I
with the most difference between the actual data points
and the suggested curve were selected for change. After
five iterations through the summation program, adequate
approximation of equilibrium was obtained. Comparing the
final curve with the starting curve (Fig. 14) it can be
seen that the change required was small. Even though
equilibrium is approximated with this curve, it does not
consistently match the data presented by Chen an·d Ross (4)
and Tran (11) particularly over the portion from x/d = 1.0
to x/d =Tr • Other possibilities needed to be explored.
30
One alternative is to use an average between Chen and
Ross's curve and Tran's.over the range 0 ~ x/d ~ 1.0
with a positive and negative maximum between 1. 0 ~ x/ d s11'".
As has been previously demonstrated,. this is the minimum
number of "fluctuations that can render a possible solution.
After balancing the curve to the required limit, it can
be seen that the· required curve areas imply larger positive
and negative stresses over the latter portion than can be
justified by the available data (Fig. 15).
Examining the curve shape in general, for changes
in the first maximum of compressive stress, the behavior
of the later portions of the curve can be predicted.
Assuming that the curve crosses the axis at approximate
multiples of x/d = 1.0 and beginning with a curve like
Chen and Ross•s· (Curve 1 Fig. 16) it is possible to show
what changes will need to be made to maintain equilibrium
1.00
0.5
0 i \
J
\ ~
\ %
i
y I
'
I -0
.50
-t.0
0
.&-·-----
....
....
Ba
lan
ced
C
urv
e
./
~-~
0.5
/
1.0
1.5
~
3.0
ff
2.
0 2.
5
Init
ial
Cu
rve
Fig
ure
14.
R
esi
du
al
Str
ess
, B
ala
nce
d
[Che
n 8
Ro
ss.(
3,4
) ]-
Cu
rve
I
%
w
.......
1.00
0.5
0
%y
-0.5
0
-1.0
0
--------
Che
n a
Ros
s (3
,4)
Tro
n (
II)
Fir
st
Se
ctio
n
Ave
rao
e-C
urv
e 2
---.;...
__.....
.___ __
_
o D
ata
P
oin
ts (
Che
n 8
Ro
ss) Yc
i
Fig
ure
15
. R
·esi
dual
S
tres
s [F
irst
Sec
tion
Ave
rage
-C
hen
8 R
oss
(3,4
)-T
ran
(l I)]
Cur
ve 2
w
f\
.)
/" /
I I ,_ \ ,, \
'° \
c ' ..... ' CD '
!)) ,, \
\ c
' 0
--tN
c a I
-I N
--- 0 I ::) I
I ::> I
I
0 c ..... < 0 0 I CD c: c I ,
.., .., (J)
< < I n>
" I =r a N I "O oa"t CD \
\ \
\ \ \
'
for a change in magnitude of the first negative maximum.
Curve Number 1 is in equilibrium. If Curve 2 is generated
by adjusting Point 1 to a different magnitude and it is
assumed the curve still crosses the axis at multiples of
34
x/d = 1.0, the general shape for x/d~l.O, assuming a smooth
curve, is determined by consideration of moment and force
equilibrium. Adjusting Point 1 to a larger negative mag
nitude means Point 2 must also increase in magnitude to
balance the additional negative force from the change in
Point 1. The moments ·are not in balance now since equal
forces have been added but at different lever arms.
Therefore Point 3 must be adjusted to generate more negative
moment to balance the positive moment from adjusting
Point-2. Adding negative force from 3 means point 2 will
have to increase which again results in a moment imbalance,
although one of smaller magnitude. Continuing the iteration
between the portion of the curve at 2 and 3 will result in
a unique solution with larger final magnitudes for Points
2 and 3. Therefore the portion of the curve from x/d = 1.0
·to x/d = rr is very sensitive to Ghanges in magnitude of
Point 1.
An example of this sensitivity is the consideration
of a curve generated by smoothing out Tran's curve from
x/d = 0.0 to x/.d = 1.0 (Fig. 17). After balancing the
curve as previously described, the ordinate of the positive
maximum was three times the yield stress which is consider-
c: .... (1)
:0 (1) (/)
-· 0. c Q
(J)
-.... CD (/) (/)
.... (/)
--0 0 ..... .... o· :J .. (/)
3 0 0 -=r(1) a.
0 c .... < (1)
..... .... 0 3
-t .... Q :J
-
• 0 0 I
ably greater than.the stress at the weld. This means the
single data point Tran found at approximately x/d = 0.8
is inaccurate and should not be considerated·in generating
a curve, since the condition of equilibrium demands un
realistic stresses using the point.
In general, a smooth curve shape with a maximum
tensile stress at x/d = 0, a maximum compressive stress
at x/d ~ 1.0, and zero stress at approximately x/d = 1.0,
x/d = 2.0 and x/d = 3.0 will exhibit an easily determined
behavior for changes in the magnitude of the first com
pressive maximum. It has been shown herein that using a
base curve of this shap~, small changes in the compressive
stress maximum at x/d < 1.0 require larger changes in the
remaining tensile and compressive stress regions. It is,
therefore, possible to modify the effect of a reported
data point on the smooth curve by considering the change
required in the rest of the curve. A data point which
causes, by its inclusion i~ a smoothed curve shape, an
unreasonable stress distribution for the other parts of
the curve can be identified using this me~hod.
Several other curves were generated in an attempt
to use the combined data from Chen and Ross (3,4} and Tran
36
(11). The most consistant results were obtained from a
curve using the first part of Chen and Ross's curve from
x/d = 0.0 to x/d = 1.0 and forcing the latter portion of
the curve to one positive maximum and one negative maximum
with the curve approaching zero at x/d =7r (Fig. 18).
1.00
0.5
0
%. y
2.5
1.5
2.0
-0.5
0
-1.0
0
Fig
ure
18
. R
esi
du
al
Str
ess
[C
hen
8 R
oss
(3,4
} fo
rce
d
to
two
ma
xim
um
s]-
Cur
ve 3
%
......
......
......
....
w
....J
38
Three curves of the several curves developed were
selected to generate M-P-~ curves with: a balanced version
of Chen and Ross•s ~urve (Curve l, Fig. 14), section from
I I
x/d = 0.0 to x/d = 1.0 average between Tran and Chen and
Ross 1 s (Curve 2, Fig. 15), and Chen and Ross 1 s curve
I. forced to two maximums from x/d = 1.0 to x/d =7r (Curve 3,
Fig. 18).
CHAPTER V
M-P-~ CURVE GENERATION
Once the description of the residual stress distri
bution was complete, Wagner's (13) program was used to
generate M-P-~ curves. It is important to understand
the method that·Wagner's program uses to calculate these
curves. There are four major stages: assigning the
appropriate value of residual stress and strain to each
element, application of a percentage of the stub column
yield load to each element, assigning a curvature to the
cross section, and calculating the moment corresponding
to a state of equilibrium. For one value of axial load
at one curvature, the moment is then calculated. By
repeating these steps for several axial loads at different
curvatures, it is possible to generate a family of M-P-~
curves. Several intermmediate steps are involved in each
of the four major stages (Fig. 19).
In the first stage of the program all of the required
data is read; number of layers, number of elements, diameter
of tube, wall thickness, modulus of elasticity, yield
stress, the number of axial load values, and the number
of curvature values. Then the values of axial load and
curvature are read. The program has the·option of using
START
Assign appropriate residual stress and strain to each element.
Apply axial load (P) and calculate the strain (£ = P/AE)
a
STAGE 1
(e ) value r
STAGE 2
Calculate the total strain (ft=£+·£) for each element r a
Using£tand the stress-strain relationship find the total force on the cross section (F) .
~ No,. r Adjust fa
Yes
Assign a value of curvature
Determine the strain on each element due to curvature (£~)
STAGE 3
Calculate the total strain for each element (£t=\+~+e-)
STAGE 4
Usingft andthe stress-strain relationship; determine the total force (F) and the bending moment (M) on the cross section
No
I .. 1STOP
Adjust the location of the neutral axis.
Figure 19 Flow diagram for calculation of M-P-~ data
40
41
a tabulated stress-strain curve or a bilinear stress-strain
curve and this information is entered next. The residual
stress and strain is read in and the applicable value
assigned to each element. For each layer the average
radius, arc length of the elements and the area of the
elements is calculated. From the results of Tran's work
it is apparent that consideration of layers in the tube
wall is riot required. for the description of the residual
stress-strain distribution since these do not vary through
the thickness of the wall. The strain, curvature, bending
moment and axial load at first yield are calculated
next. Then the distance from each element to the centroid
of the cross section, the total cross sectional area,
plastic modulus, flexural stiffness, plastic hinge moment
and shape factor are determined. The description.of the
problem is now complete.
The next stage is to apply the first axial load
(expressedasa percentage of Py) to the cross section and
calculate the axial strain. An iteration loop is performed
next to determine the correct value of axial strain. This
is required since it is possible for the summation of the
residual strain and the axial strain on any particular
element to exceed the yield value. In these cases the
summation of t~e elemental stress available to resist the
axial load is less than predicted by elastic theory. The
residual stress distribution is an initial condition and
cannot be changed, therefore the additional force must
be supplied by the elements that have not yet reached
42
yield. The stress distribution and magnitude is determined
by increasing the strain on all elements by the same amount
and then calculating the resulting stress using the material
stress-strain information (tabulated input of any stress
strain curve or a bilinear curve as previously described) .
The summation of available elemental stress at each increment
of strain is compared to the total applied force .and when
they are approximately equal the interation is stopped and
the values for each element are stored for further use.
The next stage involves assigning a value of curvature
to the cross section. The neutral axis is assumed initially
to be at the centroid of the cross section and the corres
ponding strain at each element is calculated. The summation
of strain due to residual stress, axial load and the imposed
curvature is limited to the strain at first yield as pre
viously described for residual stress plus axial load for
each element. The resulting strains are used with the
material stress-strain information to determine the resulting
stress for each element. These stresses are summed and the
resulting total thrust is compared with the applied axial
load. If the axial load and thrust are not equal the loca
tion of the neut~al axis is shifted and the strains recal
culated. The iteration is continued unti~ approximate
equality is reached.
The last stage is to calculate the summation of
moments for the final stress-strain state. The result is a
value of moment at a particular curvature for one value
of axial load. This gives one point for determining one
M-P-0 curve. The process for finding the moment is
repeated for each curvature at one value of axial load.
That gives all the points of the M-P-0 curve for that
particular axial load. The program then selects the next
value of axial load and begins at stage two repeating the
process of solving for moment at each value of curvature
until all values of axial ·load input have been used. The
values of moment, axial load and curvature are then output
in tabular form. This information can then be used in the
failure load program as described by Wagner (13) .
43
In order to completely study the effect of the residual
stress distribution on failure loads it is necessary to
determine the effect orientation of the axis of bending
with respect to the axis of the weid has on the M-P-0
curve. The concern is to find the weakest axis of bending
to be sure that the controlling case has been found. The
axis of bending selected by Wagner (12,13) passed through
the weld (Fig. 20) for which results for the M-P-0 curves
were compatible with theory (l,2,5,8). Considering the
axis orientation with the weld at the top of the pipe and
the bending axis through the horizontal diameter as the
zero or reference po:tnt;, any orientation can be described
as the number of degrees of rotation of the bending axis
from the reference position. Thus the orientation Wagner
---~--~.
.. ~~--·--·-----
· Wag
ner's
A
xis
of
Ben
ding
. --
-c I
T_.
....
We
ld'
28.0
ksi
-12
.3 k
si
1 •
1 R
efe
ren
ce
Ax
is
. .
Fig
ure
20
. O
rie
nta
tio
n
of.
Axe
s an
d ·R
esid
uol
Str
ess
Dis
trib
uti
on
[W
og
ne
r.(1
2,1
3))
~ ~
45
0 (12,13) selected would be at 90 from the reference. When
any other orientation was selected, the results were not
consistant with theoretical considerations (1, 2, 5, s·) •
The moments were-not approaching the value for full plastic
moment at large values of curvature. (tig~. 21~22,23)~
Clearly there was a significant problem with either the
proposed residual stress distribution or with the program.
The first step was to verify the results obtained by
Wagner for the distribution used with the orientation of
90° from the reference. Comparing the results for this
orientation of bending axis showed exact agreement with the
work done by Wagner (12,13).
Inputing zero residual stress and checking several
axis orientations also showed exact agreement with the
theoretically predicted behavior of the M-P-~ curves.
In order to pinpoint the problem Wagner's distribution
(see Figure 20) was used. The axis of bending was assumed
at the reference position and a summation of moments
accomplished for the cross section at the end of each
iteration to determine the correct axial strain. (This is
described in the second stage of the program, see Fig. 19) .
Residual stresses were output to verify that there was no
change occurring. (They are input constants and should
not change.) The total stresses and strains were
also output to verify that they did not exceed yield during
the iteration procedures. All of the output verified the
46
1.40
Zero Residual Stress~----= ~:y ~
1.20
0.40
1.00
0.80
%y 0.60
0.40
0.20
Zero R.S.""""\_
---~----- 0.90 ----------------v
0 1.0 2.0 3.0 4:0
~y
Figure 21.. M-P-' Curve I (Chen 8 Ross)
1.40-1 ~Py 0.00
Zero R.S. ~,....~ /
/ I
~
s.20--{ I
1.00
0.80
%,y·· 0.60
0.40
0.20
0
.,,,.
tl
I I I
/,
Zero .Residual Stre~} ____ _
__ ........... --------,--- - -
1.0 2.0 . 3.0 4.0
~y
0.90
Figure 2.2. ~M-.P-~ Curve 2 (First Section Average)
47
48
%y l.40 - ------ - 0.00
R.S.
1.20
·- 0.40
1,00
0.80
M/My
0.60
0.40
Zero Residual Stress)
----------------· --- 0.90 ,,,,..,,,., 0.20
0 1.0 2.0 3.0 4.0
%y
Figure 23. M-P-~ Curve 3 {Chen 8 Ross f.orced to two maximums)
l -- ---· ~
49
procedure with the exception of the summation of moments
after finding the "correct" axial strain and corresponding
stress. The summation of moments at this point should be
equal to zero since the· residual stress distribution is in
equilibrium to begin with and only axial load has been
added. Because the program does not alter the residual
strain or the· axial strain after this point in the program
any imbalance produced here will be unchanged at the end
of the iterations. The strains due to curvature are added
after this and the total limited to the described material
properties. The strain due to curvature at this value of
axial load is calculated by subtracting strain. This strain
due to curvature is used to find the stress on each element.
Summing moments for these·values the moment for the entire
cross section is obtained. The thrust is calculated and
compared to the.applied force and they are not equal. The
neutral axis is shifted and the moment and thrust are
recalculated. This procedure in no way affects the values
of strain that were solved for due to axial load. Therefore,
if the cross section is not in static equilibrium after the
solution for strains due to axial load, it will still not
be in equilibrium after the "correct 11 location of the neutral
axis has been found and the corresponding moment calculated.
The result of the summation of moments after completing
the iteration to find the ••correct" axial strain was not
equal to zero in any case tested except for those with no
r .. ---- -· residual stress and those with 90° orientation of axis of
bending with respect to the reference. The disparity
between the theoretical plastic moment and the calculated
50
moment for the P/Py ratio was compared to the moment
imbalance created during the solution for axial strain and
found to be equal. In other words the sum of the calculated
moment at some large curvature plus the moment created in
the solution for axial strain equals the theoretically
predicted plastic moment. · It would seem the simplest
solution would be to compute the moment due to curvature
by a summation using the total stress instead·of only that
part of the stress due to curvature. That way the moment
imbalance is incorporated in the final moment. The problem
with this solution is that while the M-P-0 curve has the
correct starting point and is correct for large values of
curvature, the middle portion has its shape changed con
siderably by the moment imbalance. The solution, then is
to eliminate the moment imbalance created before entering
the iteration loop to find the moment due to curvature.
In this manner a residual stress distribution that is in
static equilibrium will have an axial stress distribution
added to it and the final result will also be in static
equilibrium.. This model is then showing the same behavior
as the actual p,hysical cross section.
To eliminate the moment imbalance, the first step
was to determine how the imbalance is created. When the
"';
r:: r
I· ·~
. \
l
\ '
• \.
i
\ .
\
"•
,-,
51
axial strain is being solved for the first step is to
apply a uniform strain to the cross section computed from
the average axial stress, P/A, and the material properties
of the cross section. The axial strain is added to the
residual strain for ·each element and the stress from this
sum is limited by the described material properties (either
tabulated or bilinear stress strain curve) . If the sum of
axial strain and residual strain is larger than the strain
at first yield the available stress will be less than the
average axial stress. That means the summation of thrust
on the cross section will be less than the applied axial
load. Therefore the strain over all the elements is in-
cremented and the summation recalcul'ated. This procedure
continues until approximate agreement is reached between
the calculated thrust and the axial load. If some of the
elements above the center of gravity·of the cross section
are very close to yield from residual strain alone, the
increase in strain for axial load will put them over yield.
The stress available· for axial load from them is less than
the average, therefore the strain is adjusted. This increase
in strain doe~ not cause much if any increase in stress
available from the elements ·that were at·or over yield
in the first iteration. The other elements now have an
increase in st~ess over the average P/A. As the iteration
process continues the distribution of axial stress may
become even more irregular as other elements yield. When
,-;
f..
•4
!
{' I \
.'
"
the iteration is complete the stress distribution is no
longer uniform and the center of gravity of the resulting
thrust is no longer located at the center of gravity of
the cross section. Therefore a moment has been created
during the iteration due to the yielding of some elements.
(Fig. 24)
To explain why there was no imbalance using this
program for the axis of .bending taken through the weld
(90° to the reference) the -residual stress distribution
used must be examined. The distribution used was assumed
to be symmetrical about the axis of the weld. Therefore
the change in stress due to axial ·load on each element is
also symmetrical about the axis of the weld. Since the
change in stress is symmetrical about this·_ axis, the
resulting thrust is positioned somewhere·along the axis.
Therefore summing moments about this axis yields zero
52
even though summation of moments about any other axis would
not be. The cross section· is not in equilibrium but the
moment is only being taken about one axis in this program so
the imbalance goes undetected for this one special case
(axis of bending through the weld, 90° with respect to
reference chosen.) The M-P-0 curves generated for this
case are then unaffected by the imbalance but for all
other cases th~y would be. The case with the largest
imbalance should be with the· axis of bending at the refer-
ence position (perpendicular to the weld axis) and this was
confirmed by a .test run of the program as written.
; ., I j I !
. ' " \. 1
0 0
I +
-+
+ ~
--Residual Stress Axial Stress
0
Summation
0 0
0
0 ..J
~· ->· I I
53
Not A\lailoble
Summation (Not in Force Equilibrium)
0
Summation (In Force Equilibrium)
0
Not in Moment Equilibrium
Figure· 24. Moment ·Imbalance
\
\ \ ' ~ t ~ ~
54
The problem was identified and several possibilities
for a solution were explored. The. method selected was
to add another iteration after determining the axial
strain·to eliminate the imbalance created. The value of
curvature that would generate ·a moment equal to the
unbalanced moment was calculated assuming the section to
be elastic. The sign of this curvature was reversed there-
\ by changing the sign of the moment. Strains were calculated
\ \
\
based on this curvature and added to the strain determined
fo~ axial load plus residual stress. This in effect would
cancel the unbalanced moment leaving the cross section in
equilibrium. The curvature used is actually the reverse
of the curvature generated by the uneven axial stress
distribution, which means the section ends up with curvature
equal to zero. The procedure actually assumes a curvature
(elastic) and iterates to find the correct curvature
(produced by uneven axial stress). The iteration also must
still limit the stress to that determined from tabulated
or bilinear stress-strain curves. The iteration also
involves checking the summation of thrust on the elements
against the applied axial load for approximate agreement.
The revisions required to the program were extensive and
are shown in Appendix II. Each axis rotation requires a
separate run of the program.
Because various orientations of the bending axis
were desired, the program was also modified to automatically
·1 '
..,
i
' .
,.
"..
,;
\ '\ ~
\
55
rotate the bending axis to any whole element increment
(for example: 0 0 0 0 360 /80 elements = 4.5 /element 45 /4.5 I
element = 10 element rotation) .. Since a large number of
M-P-~ curves were going to be generated, a graphics plot
program was written to automatically plot each set of curves
on a Cal-Com plotter. The p~ots for Wagner's residual
stress distribution and the suggested curve 3 are shown in
Figure 25. The results of these calculations agreed with
that predicted by consideration of the theoretical maximum
possible moment for each axial load ratio.
Another problem was discovered when these M-P-~
curves were used directly in the beam-column failure load
program. None of the iterative solutions supply exact
closed form solutions; they converge on a solution ~ithin
an acceptable degree of error. Since some error is inherent
in each iteration, the errors may accumulate. The degree of
error is still small but it means for some cases of axial
load with the curvature equal to zero, small positive or
negative moments appear when they should be equal to zero.
When the curves with these small negative moments were used·
in the failure load program it caused instability in the
beam-column solution. It was discovered that the curve
interpolation portions of the program could not handle
negative number.s. The solution was to assign all the moments
at zero curvature to exactly zero as they should be. The
M-P-~ curve generation program could be changed to auto-
i l
I
1.4
1.2
1.0
"'I 0.8
~
~ ....... ~
0.6
0.4
0.2
0 0
; i
i
/ /
/
/ /
/.
/ /
,,,,...-·-·-·-·
·" ,,.·"' ,,..
56
P/Pv=O
P/Pv=0.4
-------·- .---·-· - ·-· -·-·-·-·-_,.
No residual stress
Assumed residual stress (curve 3)
-·-·-·-· Assumed residua I stress [Wagner 0 2,13)- Sherman ( 9)]
P/Pv =0.9
1.0 2.0 3.0 4.0 5.0 '/J l</Jy
Figure 25. M-P-¢ Curves. from improved generation program·
j
j '
I I I
I 1
I
I i-
!
p
~ , I \ . \l
\l
1 '
I !
57
matically assign moment equal to zero for curvature equal
to zero which would eliminate the need to change the output
later. In this case, the program was not changed since the
magnitude of the moment gives an indication of the composite
accuracy of the manipulations of the program.
It should be noted that the M-P-~ curves calculated
for Wagner's (12,13) residual stress distribution exhibit
a flattening effect when compared to the curves calculated
for the suggested residual ·stress distribution (Curve 3,
Figure 18). This means when using Wagner's distribution the
section is weaker than would be predicted using· the suggested
distribution (Curve 3, Figure 18). It should also be noted
that different axis orientations control the flattest (most
critical) curve at different axial load ratios.
It should be noted that the M-P-~ curves generated
using the suggested residual stress· distribution give more
consistent results with respect to the controlling axis
orientation for different slenderness ratios than by··using
the distribution·suggested by Wagner (12,13).
Once the M-P-~ curve generation and interpolation pro-
blems had been solved, the failure load program was used
for the same sections and loading conditions that Wagner
(13) used. M-P-~ curves were generated for five axes
orientations; 00 , I
0 0 0 0 . 45 , 90 , 135 , and 180 with respect to
the reference axis (perpendicular to weld axis), for
Wagner's residual stress distribution and for the suggested
., I
t
58
residual stress distribution. Each of these M-P-0 curves
were used in the beam-column failure load program. This
was done since the M-P-0 curves for Wagner's residual
stress distribution indicated different axes orientations
controlled for different P/Py ratios. The controlling·
results ~ere then graphed as a failure load curve using the
same format as shown by Sherman (9) .
Figure 26 shows the resulting curve for Wagner 1 s
distribution at an orientation of 90° with respect to the
reference axis.
The failure load results for the suggested residual
stress distribution are shown in Figure 27 since the
curves are close enough to those of Figure 26 as to make
representation on the same figure difficult at this scale.
The orientation of the axis for the suggested residual
stress distribution (Curve 3 Figure 18) is 180° with
respect to the reference axis. These curves were compared
to available test data and reasonable agreement was
indicated.
59
1.0 I I I I I 1.0 ,, 1.0 0 0.2 0.4 0.6 0.8
0.8
0.6
a?' ' Q.
0.4
0.2
0
a ----.....:--_.,, - - -----...---- -------~
I 0
'" ..... ..... .... , ' .......... ' ..... ', -- ..... - b
' -LID= 120
' --'..... ---.......... c ----....... --- ------ -~ ............ __ ----._ ..... ,
. ' -------- .... -.....
I I I r 0.2 0.4 0.6 0.8
~MMp
.a Double curvature
b Single end moment c· Single curvature
0.8
0.6
0.4
0.2
~, 0
1.0
Figure 26. Beam column results -Wagner's distribution
i.
i
60
0 0.2 0.4 . 0.6 0.8 1.0 1 .. 0 I ' ' I I I ,I .0
0.8
0.6
~ 0..
a: 0.4
02
0.8
0.6
a ----------- -------- - -----~ - -:- .... - 0.4 , ..... , . ~- ........ _ b L/O= 120 ' -.... . .... , --...... --...._ ..
"'..... ----..... c ..... ___ ~~ ---~ ----------. - ......... __ .- ..... , ---- ..... -..... ' ----- ' --..._ ' -..... ', --..... '
0.2
' ...... \ ...... ' .......... '·
~ 0 I r 0 I I I I l.G 0 0.4 0 .. 6 o.s 0.2
M/Mp
a Double curvature
b Single end moment
c Single curvature
Figure 27. Beam column results - curve 3 {Fig.18).
CHAPTER VI
CONCLUSIONS
Comparing the balanced residual s_tress curve (Curve 3)
with the curves determined by the sectioning method, hole
drilling method and Wagner•s·assumeddistribution, the
balanced curve suggested for use (Curve 3 Figure 18) appears
to be a smooth curve more closely modeling the test data
(Fig. 28). Figure 29 shows a comparison of M-P-~ curves
for no residual stress, Wagnerlsdistribution, that dedu~ed
from member behavior and the balanced residual stress
curve (Curve No. 3). ·sherman (9) presents a curve showing
the effect of residual stress on strength under axial
load which is shown in Figure 30 with the· points determined
usipg the residu~~ stresses from Curve ~ (Figure 18) added
to it. Using the· same data as Sherman used in the beam
column failure load program but with the balanced residual
stress distribution of Curve 3 (Figure 18) instead of
Wagner's curves, new points are determined which are
plotted on Sherman's Figur~ 6 to show the effect of the
balanced distribution .(Fig: 31}.
While there are large differences:·in the shape of the
M-P-~ curves as previously described (Figure 25,29), the
effect on the interaction curve for failure loads is
0.80
~/Cf
v·
., , ..
. '" . ...
......
......
......
......
......
.... -
--.
........
........
...
..........
..........
.... _ .... _
__ ....
........
. .....
..........
. -.....
... -.
..
--
..,_
__
1
Cur
ve
of
sect
ioni
ng m
etho
d
n cu
.rve
of
hole
dri
llin
g m
eth
od
----:-
·--:-
m C
urve
of
assu
med
di
stri
butio
n {S
herm
an)
_..
..,_
_ N
C
urve
of
bala
nced
d
istr
ibu
tion
{C
urve
3)
0.40
-P~~
\.
:'> \ \ ..
~-·--<::_en
\ I
' ~b
!7,/
~
I -r
-0
.00
·t
t' f.
'\ ~/
···
Cu
rve
I2:
-0.4
0
-0.8
0
Fig
ure
28
. Lo
ngitu
dina
l re
sidu
al s
tress
dis
trib
utio
n cu
rves
--
~~·~ -~--~ _
_ ...
_ _
...
....J
°' I\.)
~
~ .......... ~
1.4 I
L2+
0
o.a
0.6
0.4
0.2
0
1.0 2.0 3.0 I I I
P/Py = 0.0
..... ....- .i:;iiiii":;-" ·"",,,
~ .,,,.,, ~ .. ~ b <," /,/ c
_..-· ,. . .,,,,. ...... ...:-~---,.,,,,.,,,,,, .....
....... -::,,.- ----_,,,.,.,,,.---·
P/Pu = 0.8
;.::. ·--- - - - = .. / . ...-· ---- ,
/ ~"' . ------------· -/" ,,.- ..--· . / ;',,,. ,/'
;-)">/ •/ / ,/ ri'// .,,/
~.
1.0 2.0 3.0
</J/~y
a. No residual stress b. Assumed residual stress of curve 3 c. Assumed residual stress, Sherman d. Deduced from member behavior
Figure 29.. Moment-curvature relationships
63
4.o· I 1.4
+ 1.2
1.0
0.8
0.6
0:2
4.0
......
......
......
......
.. ""':
....
........
........
........
. -......
. _~~
........
........
......
...-.
......
......
. .....
.. ...
.........
...~~ --
-...
......
.._..
~"--
~" -
-----
--.
0 O.
!:>
I .0
1.
5 2
.0
1.0
le:::::::::
t.'
\ '
p -
' P
y-~2
J.O
. 0.8
~ 0~
6
0:
0.4
0.2.
.
Lin
ear
:. =
, 1-0
.38
5 ~
y
LID
= 1
5 L
IO •
20
F,=
50
kst
·. Par
abol
ic (
SS
RC
)
~ =
l-0
.25
~y
LIO
=3
0
L/O
=4
0
0.8
0.6
0.4
02
LI 0
=5
0
·o 0 ~1
-'--
05
--
---~
~,J"
"~'
-_:_
~~~~
~~~J
F
" 1.
0 11
)..=
2!!
/fi
L
1.5
__
_ IL
JY
o
o a
No r
esi
du
al
stre
ss
c b
Ass
um
ed
're
sidu
al s
tre
ss (
She
rman
)
6.
c D
educ
ed
fro
m m
em
be
r b
eh
avi
or
x d
Ass
umed
re
sid
ua
l st
ress
(cu
rve
3)
Fig
ure.
30
. E
ffe
ct
of
resi
du
al
stre
ss
on
stre
ng
th
unde
r a
xia
l lo
ad
°' .e.
1.0
0.8
0.6
>o,, ·Q.
' 0. .
0.4
0
0.2 I
I
0.2
Fy:: 50 ksi
p ,. ~ l -~
65
0.4 0.6 0.8 1.0 l.O
0.8
0.6
0.4 .
w 0.2
~ I L
o~
• I I ~O 0 0.2 0.4 0.6 0.8 1.0
w L2 -
f6 Mp
---- a No residual stress
------ b Assumed residual stress distribution (Sherman) ______ : c Deduced from member behavior
-·-·--·-- d Assumed residual stress distribution (curve 3)
Figure 31. Effect of residual sfress
66
slight. The maximum difference between predicted failure
loads is in the neighborhood of five percent. The failure
load curve as described from M-P-~ curves using the suggested
residual stress distribution shows an increase in predicted
strength over the fa~lure load curve using the M-P-~ curves
as described by Wagner's residual stress distribution
(Figure 26,27,30,31). This agrees with the prediction
based on the shape of the M-P-~ curves (Figure 25,29).
The interaction curve obtai·ned using the M-P-~ curve
generated based on the suggested residual stress distri
bution (Curve 3 Figure 18) does not exactly conform to the
test data as plotted but that is to be expected for several
reasons. One is the possible error in the physical test
data, another is the small errors that are inherent in
the process of curve interpolation and iterative solution
for axial load, moment and failure loadi another is the use
of a perfect bilinear stress-strain curve instead of using
tabulated data from physically testing a coupon from the
member tested. The degree of agreement is acceptable in
light of these considerations and .the use of the suggested
residual stress distribution is recommended for future
research into the behavior of welded steel tubes in com
bined bending and axial load.
It is fur.ther recommended ·that the M-P-~ curve
generation program be changed to facilitate more rapid use
of the program. A data file should be created that would
contain the stresses and strains on each element that
result for each axial load ratio used. These stresses
67
and strains can be obtained using the first portion of the
M-P-~ curve generation program as re-written in this study.
The stresses and strains that result from the application
of residual stress plus axial stress after the moment
imbalance has been resolved are the ones that· should be
saved. By storing these stresses and strains, the program
would not be required to recalculate them· for each axis
rotation thereby reducing the computer time used.
Summarizing, the residual stresses present in a
welded steel tube can be modeled by a smooth curve that
describes stresses such that the cross section is in
equilibrium.
Curve 3 (Fig. 18) models the existing test data wel~,
is in equilibrium and exhibits consistent behavior in the
M-P-~ and beam column failure load program. This curve is
suggested for use in further research on the behavior
of steel tube columns. The effect of these residual
stresses on the predicted failure load of tubular beam
columns is significant and should not be ignored. The
difference in failure load predicted by this family of
curves was five percent at the maximum, indicating that the
exact shape of .all parts of the residual stress curve is not
required to obtain·good results. Because of the large
strength reductions indicated by use of the residual stress
distribution described by Curve 3 (Figure 18) in the beam
column failure load program, it is evident that residual
stresses can not be ignored when calculating failure loads
in steel tube beam columns.
68
REFERENCES
1. "Plastic Design in Steel, A Guide and Commentary, 11
American Society of Civil Engineers 2nd Edition, 1971.
2. Beedle, L. S., i'Plast~c ·Design of Steel Frames 11,
John Wiley and Sons, Inc., 1958.
3. Chen, W. F. and Ross; D. A., "Residual Stress Measurements in Fabricated Tubular Steel Columns," Fritz Laboratories Report, No·. 393.4, July 1975.
4. Chen, W. F. and Ross, D. A., "Tests of Fabricated Tubular Columns," Journal of the· structural Division, Proceedings of ASCE, Vol. 103, No. ST3, March 1977.
5. Disque, R. O. "Applied Plastic Design in Steel" Van Norstrand Reinhold Company, 1971.
6. Dwyer and Galambos, "Plastic Behavior of Tubular Beam-Columns 11
, Journal of the Structural Divis ion, ASCE, Volume 91, ST4, August 1975.
7. Marshall, P. W., "Stability Problems in Offshore Structures", presentation at the Annual Technical Meeting of the Column Research Council, St. Louis, March 25, 1970.
8. Salmon, C. G. and Johnson, J. E., "Steel Structures, Design and Behavior" Intext Educational Publishers 1972.
9. Sherman, D. R., Erzurmlu,. H., Mueller, w., "Behavioral Study of Tubular Beam Columns." Journal of the Structural Division, ASCE, Volume 195, No. ST6, June 1979.
10. Szuladzinski, G., "Moment-Curvature for Elastoplastic Beams," Journal of the Structural Division, Proceedings of ASCE, Vol. 101, No. ST7, July 1975 ..
11. Tran, C. M., "An Investigation of the Hole-drilling Technique for Measuring Residual Stresses in Welded Fabricated Steel Tubes" Masters Thesis, Portland State University, Dec. 1977.
70
12. Wagner, A. L., Mueller, W. H. and Erzurmlu, H., Portland State University, "Design Interaction Curves for Tubular Steel Beam-Columns 11
, .Eighth Annual Offshore Technology Conference, Houston, Texas, May 3-6, 1976.
13. Wagner, A. L., "A Numerical Solution for the Ultimate Strength of Tubular Beam-Columns, 11 Masters Thesis, Portland State University, Nov. 1976.
APPENDIX I
CURVE BALANCING COMPUTER PROGRAM
CURVE BALANCING PROGRAM
DATA INPUT
NOTE: Use of the punch output option gives the user a
deck of cards with the stress-strain information on them in
proper form for direct input to the moment-thrust curvature
program (average over element) . Numbers at left indicate
card columns.
A. Control ·Data; Curve Identity Number.
FORMAT ( 2I3)
1-3 Curve identity number
4-6 Are stress-strain values to be punched on. cards? (- or 0 no, + yes)
B. Control Data.
FORMAT (I3, 5Fl2.4)
1-3 Number of elements in 1/4 tube (Maximum number = 100)
4-15 Outside diameter of tube (in.) "·\
16-27 Tube wall thickness (in.)
28-39 Distance from crown to center of moment (in.)
40-51 Modulus of elasticity (ksi)
52-63 Yield stress (ksi)
C. Res~dual Str~ss Values from Distribution Curve
FORMAT (F9. 6)
1-9 Residual stress (ksi) (Distribution curve ordinates)
FLOW DIAGRAM
RESIDUAL STRESS DISTRIBUTION CURVE BALANCING
START
READ: NC, IP
READ: NELE, QD, T, DC E, FY
CALCULATE: X(N) = Arc Distance, Radius at 4: Wall, Py, My, Accuracy Limits.
READ: RESIDUAL STRESS CURVE
ORDINATES
·e-
y (N)
DO 240 FOR EACH
ELEMENT
+8
73
CALCULATE XO, CGl, CG2 A'l, A2
GO TO 220
CALCULATE CGl, CG2, Al, A2
GO TO 220
0
CALCULATE Al, CGl A2=0,CG2=0
GO TO 220
+
+
CALCULATE CGl, CG2 Al, A2
GO TO 220
74
Al=O, A2=0 CGl=O, CG2=0 GO TO 220
CALCULATE CGl, CG2=0 Al, A2=0
Al=Al*T*FY A2=A2*T*FY
LEVER ARM 1 = 0
+
+
CALCULATE CGl, CG2=0 Al,A2=0 GO TO 220
CALCULATE LEVER ARM 1
75.
LEVER ARM 2 = 0
CALCULATE MOMENT
+
SUM OF FORCES SUM OF MOMENTS
CALCULATE LEVER ARM ·2
76
FORCE l, LEVER ARM 1 FORCE 2, LEVER ARM 2 SUM FORCES, SUM MOMENTS
4-
FOR EACH ELEMENT CALCULATE AVERAGE STRESS & STRAIN, FORCE, SUM FORCES ~OMENT, SUM MOMENTS
PRINT RESULTS AND PUNCH .
STRESS & STRAIN
1 2 3 4 0 6 7 e. 9 1l
1 1
1
12
1
3
14
1
5
16
1
7
18
.
19
2L
t 2
1
22
2
3
24
2
5
26
2
7
c (' r ,.. \
, c r. r. c
......
...
......
... ...
_ ...
......
......
......
......
....
-... --~
~·-.
~· -·
• ~·-~
---·-
-JI
-~ .
. .
TH
E PUPFL:S~
OF
T
HIS
P
fWG
RA
M
TS
T
O
:::iu
iv:
FO
HC
fS
AN
O
i'-lM
'lEN
TS
F
OR
I
ON
GI
run
i'N
AL
H
FS
ICU
AL
S
TR
ES
S[S
S
HJ
/), C
IPC
UL
AP
Cl~OSS
SC
CT
IOf\
l.
I =
j·.1U
11 r: ~~ E
R
Of
ST
H [S
S
L 0
C /\
T I
0 f\l
S •
E:
:: Y
OU
f\I G
' S
M
G U
UL
U S
• F
Y=
YE:
I L
. C:
ST
f{ E
f\l G
T ii
G
D
::
'1 UT~ I
iJ E
D
I t\
M [
T [
R
0 F
TU
13 [
T =
TH
I CK
f"W
SS
0
F :tJ
i\ l
L
n F
SE
CT
I 0
l\J •
D
C =
D 1 ~)
T M
J C t::
FR 0
r~
C H
0 \rJ
f ~
T 0
C
EN
T E
H
0 F
!VI
0 !J
I [ j
-.JT
~ NC
=
CURV
E N
UM
RER
J p
-=
0 I\
u c
l\ rrn
s p
u ~Jc
H [
D
:1
ST
RE
SS
+
ST
RA
IN
PU
NC
Hfn
nIMENSIO~
X(l
OO
>,Y
C1
00
) Tf
<E 1
\U:::
2
I l~1
R !
T F
= :-,
REA
D CIRE~0,15)
NC
,IP
1
5
F0
hl0 l/
.\T
(2
13
) P
F t\
D
< I
H [
AU
, 1
0 )
f'J
EL
E ,
0 D
, T
, 0
C ,
E ,
FY
JO
FORM
AT
<I3
,5F
l2.4
>
n= <
un-1
>;;.:
o.50
l=
f-.I
EL
E*
2-t
l D
X=
3.t
41
L9
2G
54
/(J-
l>
X<
l>=
O.u
GO
O
no
25
f\
:::2
,1
25
X<N>=OX*fi~(N~l>
OI=
<2
.0*
R>
-T
PY
=3
.1q
15
93
/4.0
*<
0D
*O
D-O
I*n
I>*
FY
XMY=<3.141593/64.D>*<OO**q•o-DI**4·0>*<~.0/0D>*FY
ST
OP
f'.=
0.00
005*
PY
S
T 0
P r·I
= 0 •
0 0
1 * X
M Y
K
=I-
1
i·I R.
I T
C
< H
v H
! T
E ,
2 9
>
NC
2
9 F
0 R
~1 1 AT
< '
* * *
TH
IS
IS
C
UR
\/ E
i'.J
U 1v1
RE
R ' ,
I 3
>
s u flilf
i.~ =
o • n
st
mA
::: o
. n
\m I
TE
(
nw
.1 T
[ •
3 0
) 3
0
FO
PM
/\T
<'
/\
RE
A
1 L
E\/
F:H
f\
l{M
A
RE
/l,
2 l
su~·
. F
OR
CE
S
SU
M
MO
ME
NT
S
• )
H E
AD
C
I f ~
[ f\
IJ '
4 0
>
( Y
( l\J
> '
N =
1 ' I
)
t~ 0
F
0 R
r~ A
T
( F 9
• 6
}
LCV
ER
AHM
-J
~
---
---
---
---------
-----
-----
-----~
2e
no
2
40
N
=1
.K
?9
IF
<
Y<
NJ)
1
00
,90
,11
0
3 u
9 u
I F
< Y
( !~
+ 1
> )
1
8 0
, 9
5 ,
1 8
0 3
1
95
A
l=O
.n
32
1
\2=
0.0
3
3
CG
l=O
.O
34
C
G2
=0
.0
35
G
O
TO
22
0
36
1
00
IF
(Y(~+l))
12
0t1
90
,17
0
37
1
10
IF
(Y(~+l))
17
0tl
9Q
,13
0
3 j
1 2
0 I
F
< Y
< N
+ 1
> -
Y (
r ,J l
>
1 5
0 •
1 3·!
-J ,
1'+
O
3~
13U
IF
CYC~+ll-Y(NJ>
14
0.1
35
,15
0
4 0
1 3
5 A
1 =
( X ( I
', + l
> -
X (
r .1 )
) * Y
( N
) lf
l C
Gl=
CX
(t-.
J+l)
+X
<r1
> l
/?.0
4~
A2
=0
.0
46
C
G2
=0
.0
t;.4
t.to
46
4
7
48
'+9
5
u
51.
t.-r,
_1
t::
53
5
4
~5
56
5
7
5t~
5':1
60
f,
1
62
6
3
GO
TO
2
20
l~U
CG
l=<
X<
N+
l)+
X(N
))/2
.0
c G
2 =
< x <
r'.l +
1 >
+ 2
• o *
x <
r 1 >
> /
3 • o
I\ 1
= ( x
( t-.
+ 1
) -
x ( ~
·! )
) * y
(.i'J
+ 1
)
A 2
= ( x
{ N +
1 )
-x
( f'J
) } * (
y ( f ·
J ) -
y (
rJ +
1 )
) I
2 •
n
GO
TO
2
?0
1
50
C
Gl=
<X
CN
+l)
+X
(Nl)
/2.0
.
-"'
-
---
----------,--------------·
CG
2=<
2•0*
XC
N+
l>+
X<
N>
>/3
.0
t\ -1
:: <
X C
N + 1
> -
X {
r·J l
> * Y
( ~ !
> A?=<XC~+l)-XCN>>*<Y<~+ll-Y(N))/2.0
en
ro ~20
l 7
0 X
0 =
{ X (
N +
1 )
-X
( ~,
) ) *
( -1
. •
U ) * (
)' ( N )
) I
( Y
{ !'J
+ 1
) -Y
( N
) )
CG
l=C
X0
/3.u
>+
X(N
) CG2=<?·0*X(N+l)+X<N>+~0)/3.0
A l =
( X O
* '( (
1\J ) )
I 2
• 0
f\2
= (
/.. C
l\;+
l >-
( X (
1'1 >
+X
O}
) *
Y (
M+
l) /
2.
0 G
O
TO
22
0
lBL
I C
Gl=
<?-.
O*
X<
N+
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Compare with program output (next page)
Figure 32. Example for te.sting curve program
81
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APPENDIX II
REVISIONS TO MOMENT-THRUST-CURVATURE PROGRAM
MOMENT-THRUST-CURVATURE PROGRAM
DATA INPUT
NOTE: The last data card must assign the outside diameter
a value of zero to stop the program.
Numbers at left indicate card columns.
A. Control Data; Cross Section and Material Properties.
B.
FORMAT (4I5, 4El5.5)
1-5 Actual stress-strain data used? (+=Yes; -l=No)
6-10 Residual stresses used? (+l=Yes; -l=No)
11-15 Number of layers of elements. (Max. = 5)
16-20· Number of elements in 1/4 circle of one layer. (The product of the last two numbers must not exceed 50.)
21-35 Outside diameter (in.)
36-50 Wall thickness (in.)
51-65 Modulus of elasticity. (ksi)
66-80 Yield stress. (ksi)
Data and Time of Run
FORMAT (4I5)
1-5 Month
6-10 Day
11-15 Year
16-20 Time (001 - 2400)
85
C. Control Data
D.
E.
FORMAT (715)
1-5 Number of P/PY values (Max. = 12)
6-10 Number of PHI/PHIY values (Max. = 25)
11-15 KSKIP (l=l stress-strain data for one half tube) ( =2 stress-strain data for entire tube)
16-20 Rotation of bending axis (Number of elements)
21-25 Curve identification number
26-30 Are curves to be plotted? (- or· 0 = No) (+ = Yes)
31-35 Are M-P-PHI values to be punched on cards? (- or 0 = No) (+ = Yes)
Axial Load Values
FORMAT· (6Fl0. 5)
1-10 P/PY values (Always Positive)
11-20
21-30
31-40
41-50
51-60
Curvature Values .
FORMAT (5Fl0.5,/,5Fl0.5,l,5Fl0.5,l,5Fl0.5,l,5Fl0.5)
1-10 PHI-PHIY values (Always Positive)
11-20
21-30
31-40
41-5_0
NOTE: The data for one problem is now complete if the
actual stress-strain data and residual stresses are not
used. If both options are u~ed, the stress-strain curve
data is read in first.
F. Stress-Strain Curve Data
1. Control Card
FORMAT (IS)
86
1-5 Number of tabulated points on stress-strain ·.curve.
2. For each tabulat~d point
FORMAT (2El5.5)
1-15 Stress value
16-30 Strain value
G~ Residual Stress Data
1. Time of Residual Stress Calculation
FORMAT (4I5)
1-5 Month
6-10 Day
11-15 Year
16-20 Time
2. For each element
FORMA'r (2El5.'5)
1-15 Stress value
16-30 Strain value
FLOW DIAGRAM
CALCULATION OF MOMENT-THRUST-CURVATURE DATA
START
Read: NBS, IRS, NLYR, NELE OD, WT, E, FY
Yes
No
Read IDl, ID2, ID3, ID4
Read NP, NPHI, KSKIP ROTA,· IDCUR, IPLT, IPUN
Read P/Py values ..
Read 0/0y values.
STOP
87
No
used?
Yes
Read stress-strain data.
residual No
Yes
Read residual stress data.
Calculate for each layer:
Average radius Arc length of elements Area of elements.
88
~ .
Calculate the following:
Strain, curvature, bending moment and axial l.oad at first yield.
The distance.from each element to the centroid of the cross section.
The total cross sectional area, plastic modulus, flexural stiffness, pla~tic hinge moment and shape factor.
Calculate Summation ,of forces and moments for residual stress distribution
summations
No
Yes
Print Warning
89
Do 500 for each P/Py value.
Apply an (a new) axial load and calculate the corresponding axial strain (P/AE) .
Is tabular stress-strain
No
No
Axial stress
Go To 600
= P/A
90
e
Go To 300
Yes
Go To 300
Comment: This is the beginning of an iteration to determine the correct axial strain.
used?
Yes
For each element:
Calculate the total strain
No
Go To 21
· (Axial strain + residual strain)
Interpolate the corresponding stress value -
SUBROUTINE INTERP
Calculate the force on the element.
Calculate the total force on
the cross section.
Go To 75
91
For each element:
Calculate the total strain (Axial strain + residual strain)
Determine the corresponding stress value from the bilinear stress-strain relationship.
Calculate the force on the element.
Calculate the total force on the cross section.
Let "DIFF 11 = The total force on the cross section - the applied. axial load.
Yes
No
Adjust the axial strain
Go To 300
92
Calculate the unbalanced moment from residual + axial strain
Calculate curvature from unbalanced moment
Locate neutral axis at the centroid of the cross section
Have iterations balance axial stress been exceeded?
No
Have iterations find neutral axis been exceeded?
No
Is tabulated stress-strain data
used?·
Yes
Yes
Yes
No
Print Stop
Print Stop
Go To 525
93
l For each element
Calculate strain due to bending Calculate the total strain (Axial strain + residual strain + strain due to unbalanced curvature)
Interpolate the corresponding stress value -
SUBROUTINE INTERP
Calculate the force on the element
Calculate total moment for curvature stress and total moment for total stress
--'-~~~~~~~~~~~~~~--.~~~~~~~~~~~~~~~~~.--~~~~ ...
IGo To 5351 I
7 For each element
Calculate strain due to bending
Calculate the total strain
Determine the corresponding stress value from the bilinear stress -strain relationship
Calculate total moment for curvature stress and total moment for total stress
i
94
Calculate total force
Let . "force 11 be the net force on the cross section ·
Is 11 force 11
equal to
No
Yes --Go To 548
Adjust neutral axis location (sign of unbalanced curvature included)
Go To 515
stress+ axial stress·+ unbalanced curvature
less than 0.001 My?
No
Find new unbalanced curvature
95
Write axial load ratio summation of moments after balancing cross section for axial + residual strain.
- ""---------..--------------------'
Is bending axis rotated?
Yes
,,
Rotate axis of bending write; number of degree
rotated
Write; residual stress, residual strain, axial stress (after balancing)
,__ ________ ......, ______________
Do 400 for each ~/~y value.
No
Assume a (a new) curvature value.
Locate the neutral axis at the centroid of the cross section.
~
96
Go To 509
No
For each element
Calculate the strain due to bending
Calculate the total strain (Axial strain (balanced) + residual strain + strain due to bending)
Interpolate the corresponding stress value -
SUBROUTINE INTERP
Calculate the force on the element
Calculate the total force and .bending.moment on the cross section.
Go To 89
Yes
No
Print Stop
Go To 23
97
G For each element:
Calculate the strain due to bending.
Calculate the total strain.
Determine the corresponding stress value from the bilinear stress-strain relationship.
Calculate the total force and bending moment on the cross section.
Let "FORCE" be the net force on the cross section.
No
Adjust the location of the neutral axis.
Go To 450
Yes
Go To 71
98
Save the total moment calculated.
400 - Continue to next ~/~y value.
Return to original residual stress distribution
500 - Continue to next· P/Py value
Print results
Call plot subroutine if specified
Punch results if specified
Go To 999
99
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