THE PLASTIC BENDING
OF CURVED BARS
by
CHARLES OLIVER SMITH
B.
S., Worcester Polytechnic Institute
(1941)
SUBMITTED IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
(1947)
Signature of Author
Dept. of Mechanical Engineering
12 September 1947
Certified by
Thlsis Supervisor
Chairman, Department Committee on Graduate Stu
~b~n6'
·--
nts
12 Septemoer l94f
Proressor Joseph S. Newell
Secretary or the Faculty
Massachusetts Institute or' Tecinology
camorldge,
Massacausetts
Dear Professor Newell:
I tereoy suomit tnis tnesis entitled, "THE
PLASTIC BENDIiNG OF CURVWED BARS,"
in partial rul-
rillment or tne requirements ror the degree of
Master of Science in Meclanlcal Engineering rrom
the Massachusetts Institute of Tecnnology.
Respectrully submitted,
Charles O. Smith
Table of Contents
CONTENTS
OF
TA 3 LE
. . . . . . . . . .
iii
List of Figures
.......
Nomenclature
. . . . . . .
vl
. . . . . . . .
ix
Suggestions for Future Investigation
. . . . . . .
12
aeneral Preliminary Investigation
. . .
... .
13
. . . . . . .
20
. . . . . . .
21
Assumptions
Introduction
. .
.
. .
.
.
.
.
.
.
.
.
. .
. .
....
Purpose
.
...
Previous Contributions to the Subject
Procedure
.
.
Summary and Discussion
.
Case I
..............
General
Elastic Condition
.
.
.
.
.
Partially Plastic Condition
Stress Distribution
.
.
.
.
.
.
.
.
. .
.
22
. . . . .
25
.
.
.
.
Equation for Neutral Axis
. . . . . . . . . .
27
Bending Moment Equation
. . .
. . .. . .
30
Differential Deflection Equation
.. .. . .
32
Calculations for the Specific Case
. . . . . .
33
Stress Distribution
. . . . . . . . . . . . .
34
. . . . . . .
35
Deflection Equations
Case II
.
.
. . . . . . . . . . . . . . . . . . . . . .
56
General
Equation for Neutral Axis
Bending ioment Equatio
Stress Distribution
. . .
.
. . . . . . .
58
. . . . . . .
59
. . . . . . .
60
ii
Table of Contents
Case II,
General
Differential Deflection Equation . . . . . . . .
. . . . . . .
Calculations for the Specific Case
Stress Distribution
. . . . . .
Deflection Equation
.
Case III
.
.
.
.
.
.
.
.
.
.
.
. .. . . . .
.
.
.
S .
.
.
.
.
.
.
.
.
.
.
.
61
62
63
64
84
General
. . . S.
Equation for Neutral Axis
Bending Moment Equation
Stress Distribution
.,
.
. .
.
.
.
. . . . .
.
........
89
..
90
Differential Deflection Equation . . . . . . . .
Calculation for the Specific Case
Stress Distribution
.
. .
. .
Deflection Equation
.
.
.
Appendix
.
.
.............
Bibliography
Notes on Integration
S. . .
.
92
. . . . . . . .
93
. . . . . . . .
96
.
.
.
.
.
.
.
.
.
.
115
A-1
.........
.
91
.
.
.
87
. . .. . . .
A-3
iii
LIST
FIGURES
OF
Title
Figure
Page
19
1
Geometry of the Curved Bar
2
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress,
Case I,
49
P . 0.000984 So
3
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case I,
P
4
-
50
0.001236 So
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case I,
P = 0.001396 So
5
51
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case I,
52
P a 0.001493 So
6
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case I,
P : 0.001524 So
53
54
Case I
7
Deflection Curves,
8
Maximum Deflection vs. Load,
9
Extent of Plastic Region, Stress Distribu-
Case I
55
tion on Plane of Maximum Stress, Case II,
P = 0.001220 So
10
77
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress,
P = 0.001364 So
Case II,
78
iv
List of Figures
Figure
11
Page
Title
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case II,
P = 0.001579 So
12
79
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case II,
80
P a 0.001939 So
13
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case II,
81
P = 0.002749 So
82
14
Deflection Curves,
Case II
15
Maximum Deflection vs Load,
16
Extent of Plastic Region, Stress Distribu-
Case II
83
tion on Plane of Maximum Stress, Case III,
108
P = 0.001102 So
17
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case III,
109
P - 0.001258 So
18
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case III,
110
P = 0.001532 So
19
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress, Case III,
P = 0.001902 So
20
111
Extent of Plastic Region, Stress Distribution on Plane of Maximum Stress,
P = 0.002812 So
Case III,
112
List of' Fiures
Fi-ure
Title
21
Deflection Curves,
Case III
22
.aximum Deflection vs Load, Case III
M
Page
113
114
vi
NOMENCLATURE
a
-
a constant, see Case I
A
-
cross-sectional area or bar, square inches
b
-
width of bar, inches
c
a constant, see Case I
Cl , 02
-
constants of integration
E
-
modulus of elasticity, pounds per square inch
h
-
depth of bar, inches
I
-
moment or inertia of cross-section,(inches) 4
k
-
a constant, used for integrating, Cases II
and II
M
bending moment at any point, inch pounds,
considered positive when decreasing the curvature or the bar
Mc -
oending moment for complete yielding, inch
pounds
Mi
-
bending moment for initial yielding, inch
pounds
Ms
-
bending moment for secondary yielding, inch
pounds
P
-
applied load, pounds
Ri
-
inner radius or curved bar, inches
Ro
-
outer radius of curved bar, inches
S
-
stress at any point, pounds per square inch
So
-
yield stress, pounds per square inch
w
-
deflection, inches
vii
Nomencia ture
y
-distance
of' a point from the neutral axis,
inches, measured positive upward
Y
distance from neutral axis to bottom or bar,
-
inches
Y.
distance from neutral axis to top or bar,
-
inches
. distance from neutral axis to edge of plastic
yl
region, measured downward, inches
y"
-
distance from neutral axis to edge of' plastic
region, measured upward, inches
7
-
vertical distance between neutral axis and
horizontal axis through the centroid, inches
S-
width of oar at any point, inches
E-
unit strain at any point, inches per inch
E,
-
a Eo -
eo
unit strain on bottom of bar, inches per inch
unit strain on top of bar, inches per inch
unit strain at yield stress, inches per inch
-
radius of curvature at any point, inches
-
initial radius of curvature, inches
-
angle of position, measured from fixed end,
degrees or radians
@1
"
angle to whicn plastic region extends on
bottom or oar
02
-
-
same to top of oar
degrees or radians
viii
Nomenclature
Numbers in parentneses refer to bibliography.
ix
ASSUMPTIONS
1.
Transverse cross-sections or the bar originally
plane and normal to the center line ol the bar
remain so aiter Dending.
2.
The material of which the bar is composea is
homogeneous and isotropic.
INTRODUCTION
The general fleld or the plastic oehavlor or
materials is or great importance since many common
raoricating processes are entirely dependent upon the
plastic properties or materials.
The bending or curved
bars, in the plastic as well as in the elastic range,
is or great importance since many such elements are
used in structural engineering and electrical machinery,
walle others serve as essential machine parts.
quently, it
is
Conse-
desired to know or the betavior or such
elements and to oe aole to predict the stress conditions
and tne derlection or such curved bars under various
loading conditions.
2
PURPOSE
It is
the purpose or tlhis tnesls to develop
analytical expressions ror stress distribution ana
ror deli'ection or a curved bar, whicn is Duiit in at
one end, of a length surriclent to rorm an are or 900,
and loaded by a single load at the Tree end, this load
being applied in the plane or0curvature.
I for the geometry or the curved bar.)
(See Figure
PREVIOUS CONTRIBUTIONS TO THE SUBJECT
Several contributions have been made to knowledge
of the general field of plastic behavior of materials by
various individuals.
A great deal of the work of formu-
lating the laws of plastic behavior has been done by
NADAI (2, 5,
6, 7).
Murphy and Timoshenko (4, 11) have written textbooks
on the advanced theory of the mechanics of materials.
Both of these texts contain expressions for stress distribution and deflection of a curved bar loaded in the plane
of curvature, but these expressions apply only within the
elastic range and for a stress-strain curve in which stress
is directly proportional to strain.
WINSLOW and EDMONDS
(12) did considerable work on comparison of experimental
results with theoretical predictions but, again, this work
was only within the elastic range.
HOGAN (3) developed
expressions for design purposes for stress distribution,
deflection angle of slope, and angle of twist in circular
cantilever beams loaded normal to the plane of curvature
under various load conditions; but these expressions also
hold only for the elastic region.
NADAI (7) has developed expressions for determining
the extent of the plastic region, the stress distribution,
and deflection of a bar subjected to plastic bending using
an idealized stress-strain curve and a curve accounting
for the effect of work hardening (somewhat idealized), but
these expressions apply only to a straight beam, not to an
4
PREVIOUS CONTRIBUTIONS TO THE SUBJECT
initially curved bar.
So far as the author could determine, there have been
no publications of work on the plastic bending of a curved
bar.
5
PR 0 0 EDURE
Three separate cases were considered.
The case of a
bar with rectangular cross-section and an idealized stressstrain curve, the case of a bar with rectangular crosssection
and an arbitrary stress-strain curve of the form
S~= ~O (~-
, and the case of the bar with triangular
cross-section and an arbitrary stress-strain curve of the
form
$ = -.
Case I,
Case II,
; these being hereafter referred to as
and Case III, respectively.
The first consideration was to determine a general
method and then apply it to each of these cases separately.
This is necessary since both the cross-section of the bar
and the stress-strain curve will affect the results
obtained.
In fact, the results and their form are entirely
dependent on these two factors.
The method developed is simple in idea but very likely
to be unwieldy in execution.
(1)
The method followed is:
The use of the equilibrium condition,
,
•
SSCA
to determine the position of the neutral axis.
(2) The use of the equilibrium condition, M~
SS VA
,
to determine the bending moment in terms of the geometry of
the cross-section and the stress-strain curve.
(3) The combination of the bending moment expression
just derived and the stress-strain expression to obtain
an
expression for stress distribution in terms of the bending
moment and the geometry of the cross-section.
PROCEDURE
(4)
The determination of the expression,
from
the bending moment expression derived in Step 2., and the
substitution of this term in the differential equation
This will give expressions for the stress distribution and deflection in general terms of M for the particular cross-section and stress-strain curve being considered.
These expressions must then be used to determine the stress
distribution equation and the deflection equation for the
geometry of the bar and the particular type of loading.
For the geometry and the loading considered in this
work, the steps to be followed are:
(5) The substitution of -P P, cos9 for M in the stress
distribution relation derived in Step 3.
(6) The substitution of -P ýcos9 for M in the differential equation derived in Step 4, and the solution of
the differential equation for the deflection in terms of
P and cose.
For each of the three cases, the extent of the plastic
region, the stress distribution on the plane of maximum
stress, and the deflection curve was determined for five
different loads, varying from that required for initial
yielding to that required for complete yielding (Case I)
or for extensive yielding (Cases I and II).
The maximum
deflection versus load relation was determined for each
of the three cases.
S UMMAR Y
DISCUSSION
AND.
Case I
For the geometry of the curved bar and the type of
loading considered (Figure 1), the stress distribution is
given by:
For the completely elastic beam,
For the partially plastic-partially elastic beam,
S = -So
S
for y" y (y 2
for y' ( y
-o
y"
(elastic portion)
S : So for y' ( y <Yl
I_
S.•ea4_,_
S:jl,
__ _ P-..,
It is possible to combine the expression for stress
distribution and bending moment to eliminate y' and obtain
a new expression for stress distribution in the elastic
portion.
This expression is:
_I-y
For the geometry of the curved bar and the type of
loading considered (Figure 1), the deflection equation is:
Summary and Discussion
-
Case I
For the completely elastic beam,
For the partially plastic-partially elastic beam,
for the partially plastic portion
\,"
E.
+ 4
c.
for the completely elastic portion
Several attempts were made to find an analytical solution to the differential equation given above.
all attempts met with no success.
Unfortunately
A solution may be made
by numerical integration by the method of successive approximations (1, 8,
9).
This was done making use of the boundary
conditions that
e-)
At the point
= 0
=o,
o=
where the partially plastic-partially
elastic and the completely elastic portions of the beam
meet, the constants of integration, C1 and C2 , may be
evaluated from the boundary conditions that
(t
a
9
Summary and Discussion
-
Case I
A question might well be raised as to the validity
of the expressions derived for this case since the position
of the neutral axis does not remain fixed but shifts from
its position at initial yielding to coincide with the horizontal axis through the centroid at complete yielding.
In this particular case, with the depth of the beam small
in comparison with the radius of curvature, this neutral
axis shift is very small and therefore its effect is
negligible.
For example, in the numerical case for which
results were calculated, this neutral axis shift was 0.002"
in an original radius of curvature of 10.248".
Case II
For the geometry of the curved bar and the type of
loading considered (Figure 1),
the stress distribution is
given by:
For the geometry of the curved bar and the type of
loading considered (Figure 1), the deflection equation is:
10
Summary and Discussion
-
Case II
1ZT+
4So
These expressions for stress distribution and deflection apply to both the completely elastic and the
partially plastic-partially elastic cases.
The reason
for this is that the stress-strain curve applies to both
the elastic and plastic regions.
Case III
For the geometry of the curved bar and the type of
loading considered (Figure 1),
is
the stress distribution
given by:
4
-
11
Summary and Discussion
-
Case III
For the geometry of the curved bar and the type of
loading considered (Figure 1),
the deflection equation is:
S3 L (
+
V
These expressions for stress distribution and deflection
apply to both the completely elastic and the partially
plastic-partially elastic cases.
The reason for this is
that the stress-strain curve applies to both the elastic
and plastic regions.
12
SUGGESTIONS FOP, FUTURE INVESTIGATION
Since this thesis covers only a very narrow range of
possibilities, future investigation could well be done in
this field by using different cross-sections (circular,
trapezoidal), various other stress-strain relations, other
boundary conditions (arcs greater or smaller than 900, bars
built in at both ends, bars simply supported), and other
conditions of loading (constant moment, uniformly distributed load, uniformly varying load, more than one concentrated
load, loading normal to the plane of curvature).
An investigation which should prove interesting would
be to determine the analytical expression for stress as
a
function of strain for a particular material, derive the
stress distribution and deflection equations as was done
in this thesis, and compare the predicted results with
those obtained experimentally for a curved bar of the same
material having the same boundary and loading conditions.
This could be done with a number of materials and for a
variety of combinations of cross-sections, boundary
conditions and loading conditions.
13
GENERAL
PRELIMINARY
INVESTIGATION
Consider the general case
of a curved bar in bending
(elastic or plastic)
14
*1~
Jt~~
C
awn parallel
From the similarity of trianSles,
JL
Z0
JL.
+1h
C
e.-
JL
*
St.C
-
-
At
· i -Le
Ib
(-L
Q0+I
15
P
~h
?
"b"
----
7
+~n,
9_
{(7~.e
de
(eo - Xx
It
should be noted that for
which is
I
EQU ATON
e
=
&o=
the case of a straight beam.
a. I
FRaom
eC0
Co
E
_.
e.
?L
eo
e.e
I
-
eo
eCo
eE
~
EQ U ATION
2
16
e.e.. (Ceo - )
jeo
E QuArrtoN
) I
C o
C~
Co-
e'-.z
(..
Where
at
-
4%
-1
&,
4
-4-'.
F
are the unit strains
respectively.
-~
ý&~F-z q vp, - 10N S'
0'U
Consider the cross-section
Z. z
(
4 A m 7-4
A N zvJ
S·-,E
A
17
For equilibrium
Ss -o
over the cross-section
of the beam.
Ss4LQo
ý(1)
A
P*G.2
I a C,ý.k4.PCtO
--VF~sw
,(C, e-)
0o-
55JA
-
S~iL~~cr~) eP,(e0-
S3 I
JA
2
FnASs,
5t
e.
b.
4tE
o 4Z
Ecuw- •~N
7
Le·-ejl4E.4
orIv~
6o•-
-~e~eez)
= fC•)
NAT
S De- ý-,I(-)o,(I-C, ) )
6,
cNTON
EQw
8
18
There are six variables:
e
)
There are five equations:
Therefore,
;= (9)
can be found.
EQtVAvr-o1V
.
Deflection Equation
It has been shown (10)
that the solution of the following differential equation will
yield the deflection equation.
a
I//
zw
_
i i
,
w
.QuATION
10
Substituting Equation 9 in Equation 10, and solving
for
\WI=
A
(e) will yield deflection equation of the bar.
FIGURE
0O.oO
O
Al*
.A
osoO"
P
GEOMETRY OF T HE CURVED
CASE
1301
CASE ISI
&
CRO3SS-
CASFE MI
ECTION OF THE BAR
BAR
Case I
Consider a curved bar of rectanxular cross-section
with depth of beam small
in comparison with the radius of curvature
and an idealized stress-strain curve such as
Q
so
I
/I
I,
--
I
21
For the Completely Elastic Condition (4)
eo
S
A
EQ~vATtN
I1
L
M eA1
Equ%-TIOM
Z.
EQVNTOQN
13
A-
o
M
Q0A
E7_
=&
If the idealized stress-strain curve is assumed,
it is obvious that beyond the yield point the neutral
axis will shift toward the centroid of the crosssection until at the point of complete yielding, the
neutral axis will coincide with the horizontal axis
passing through the centroid.
22
Partial Plastic Yielding
41
-N.A.
Considering the elastic stress equation (Equation 12)
5
assume
in the elastic portion of the bar,
in which "a" and "a" are constants for the particular
section.
For the plastic portion,
-
S=
t
soO
4· (-
S
0
It
C
so
Sr
'L-I
C-
'b
'
23
C l
C,
4
cl
-e3
C4-'
';
21
ife~
Y
^h"("'bl
~t· '^h_
~cb '~
So
so
%
tI
+·I
--
e.
)Q.
j~r5
it
eo - ~
e~-~V~8~·,
EoG
illt''
lo
S-
to11_1
(jiye. '
2a-
2-
1 2)
11
-
3
gt1
R-*
ver V 2 I
gIlk
i
L
25
s
S
S,= ,*
26
sS JA
0
s7
S.
5%~t%
-a
-~·p-]
i
I5
I
±R'L-A
jr -ýLt+
s*
~'a
· ·
dil.
C .•,•"
4L.&j
C·
a
' 0
i•
•C•- "•0
,I
I
S
£c-~
24I
-- e C"S1
a c+;
s.o
\(A\ l.
--
'
oR
eDY~I_
e.- I '
ea-2-h~r
Ab
4a
eo&-~"bI
a_
II
So -,
b
*
a
--• +b',]
M.
s·iia
c..
27
go-i•
Q0
-ZZae
/eo
C.
~b*%
~
V -2"~'
=
ea'
.A-Z6I
-^
- _c
-C •
I
-
a
--A
-
-
o
.-
).
i
.- ,•')Z
_ __.-_ '
.
(A2e,)
E
Qov T1ON 15
This expression is used to determine the position
of the neutral axis.
A trial and error solution is
recommended since the expression is not readily solvable
by algebraic means.
Care should be taken to insure
accuracy since the solution involves the small difference
of two comparatively large numbers.
28
SSSIA
M
+yAj~
f4b
1j.*
NA Isaiikg)
-I
M·%s
I
,so
- S6
,
NAI
c i14
S
a.
so)
•,,
-8
-a,
,
+ SL
?-z-c'.
(ea
Az'--
,
vo
c+~'
M . S~()s
La"-~
eo Q
t.
+sa.
r:c-,.
N\C
4a
.
Olt
o,
r,
+
(C
4tz
~xi
+L9
S&.'s
-~)~
ca-z
.
3
29
V\a
S7' (C+:")
2 So
L.(•*•")z
'
•'•"]
fS-
Q0
-
eSa(z4
Q)eoxc.1'
(Z
I''
c6
•"J
L
)
(c.
Lso 04
Se.
+ S&
o+
9
s~
-
·
(o
·
'S
so~Pr
4
m
-.-S.Iý-t
So a8~~
er
k.L (e.)
J
a
M
.,)
,o·
•t
see--•~)r
(C~)
. Z pt
+
f
91ý t, - (zfA
sa eo~x
It
- £,~
,I
tjeo
e·+'a
LX
1(~
a
E% PAN S Ow
+.1
It
,-
S,$p~
\
_
eo~2~
1-
0.*1
(r~--
S<AI
Cii;Y)
B
+1
3
SoL."
\zr
"I
/
•-,,
Ne
t••U.'•
srT
"
u
'Ia
3
3kIirSt
RE ,- t
'
. f -•
S
30
56Ge
3
so
t
t
s,
S.L
e a
I
23
aI
+2,0 %1,
M
NA
E Q•TloN
4
saa'
SoO
t
3(e"-
E
QuVnT\ON
31
M·SGng
Aeoao
_o1
\
3(i -2- ')
fg
CoN PlmETE
Mc
NA-
IE
DITN,,
IIO
Ai
S<'-'W
4t
4M
1
I
3 S(o
¾M
t
,
co
Z•)
CO
( Co$
-0 >
4eo
'
®r
Az
-
v
a
A--
too
7Z
Co
-
/M Co
I
0
32
I
-L. ' -ei77
M
S•u
ST
ew+
Ade
UTING
'r
Ns
A
TION
Q.E. I* _4 e:/3Q
tO
E QuA T• oN
33
Calculations for the Specific Case
(Figure 1)
I
M =- P
e
R= o.?-•0 u
0
O. SoO
0=
Initial Yielding
R&L
From Equation 11
*o.0o
0
lo. ooO
10. .4 8 11
From Equation 12 and using y = y'
So
-M-MM...
......-
.•SD
x
004o-ooo1
.500 XSoo L
, . --- O. o.0 1o0 8 So
PO.
o00 0
4
So
kN.8.
1B
- Yl
10.000
34
Secondary Yielding
From Equation 15
y1 a -0.248"
or
-o
a
10.248"
From Equation 16a using y" z Y2 : 0.252"
Ms : -0.01058 So inlb
P : 0.001032 So lb
Complete Yielding
From inspection of the equilibrium condition,
Qo : 10. 250"
From Equation 16a or 16b using y' - y" : 0
Mc : -0.01562 So inlb
P : 0.001524 So lb
Stress Distribution
For any value of applied load up to that required
for initial yielding (P = 0.000984 So) and for the
elastic portion (completely so) of the curved bar, the
stress distribution is given by Equation 12.
For lrger values of applied load up to that
required for complete yielding, in any section which
has partially yielded, the stress is given in the
elastic portion by Equation 14,
S = -S•o.i•
-
and in the plastic portion by So for -y and by -So
for
~y.
35
Deflection Equations
0
<' "< MA.
E atua
13
toN
APp%,es
he •
2.
AG
-M
F-t
i3ur M =
OtZiAJ
+ Wi
-
-=
ct.Ga
• cEoe
ElI
o(w La ME N AR Y So %.uTI
W\
C, 1
e
-
oN
cz co (9
PART(CUL.AR SoluTLoanl
GENS
So LVT.o
1e5 + C2 cbe
e
w O, A
Ae
9 =0
vVA I Ar I c.
C,
-
o
ZE1
a
.
=O0
CZ.
CZ,-
P
2..
3
E
QAvTiON
18
36
NiK <M <M'
o (e (e,
~.q
EA__i.
+
w
.es
Aeppi.,mrE
ON
i•
con
13o...
Strictly, Equation 17 applies only 0 <
< 82, and
Equation 13 applies only G1 < @( -- , leaving the small
range @2 < 8 < 91.
expression.
For this range there would be a new
However, since for the case under considera-
tion, ie, a thin bar with large curvature, the difference
between Mi and Ms is very small and consequently the
difference between 81 and @2 is very small.
This small
difference can be neglected for Equation 17 will extend
over this region with sufficient accuracy.
Several attempts were made to find an analytical
solution to Equation 17a.
met with.
Unfortunately, no success was
Accordingly, the equation was solved by
numerical integration using the method of successive
approximations (1, 8, 9).
37
The general solution for Equation 13a is
c
=0
w
Wz
O
0\A
I
Aw
de
The four constants of integration may be determined
from the boundary conditions as stated above.
Thus the
complete deflection curve may be determined; by numerical
integration from 0 to e 1 and by substitution in Equation
13a (after evaluation of the integration constants) from
81 to
,
38
CASE I
INITIAL YIELDING
(y'
= -0.248")
M a -0.01008 So in lbs
P =
0.000984 So lbs
fow 10.248 in
Extent of Plastic Region
There is no plastic region.
39
Case I
-
Initial Yielding
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
e
Y
inches
psi
radians
w
inches
0.252
-0.968 So
0.200
-0.772 So
0.100
0.203 .
0.150
-0.582 So
0.200
0.808 e.
0.100
-0.390 So
0.300
1.802 e.
0.050
-0.196 So
0.400
3.166 lo
0
0
0
0
0.500
4.873 Ea
-0.050
0.198 So
0.600
6.886 .o
-0.100
0.397 So
0.700
9.166 E~
-0.150
0.599 So
0.800
11.66 E.
-0.200
0.803 So
0.900
14.33 E.
-0.248
1.000 So
1.000
17.10 F.
1.100
19.93 E.
1.200
22.74 E.
1.300
25.46 E.
1.400
28.04 E•
1.500
30.41 Eo
1.5708
31.93 eo
40
CASE I
YIELDING TO ONE QUARTER OF DEPTH OF BEAM
(y' a -0.186")
M = -0.01267 So in lbs
P = 0.001236 So lbs
Eo= 10.248 in
Extent of Plastic Region
Y'
y"
in ches
in ches
degrees
-0.186
0.191
0
-0.190
0.197
9024'
-0.200
0.208
16028'
-0.220
0.230
25046'
-0.230
0.241
29045'
0.252
33023'
-0.248
8
37017'
41
Case I
-
Yielding to One-quarter of Depth of Beam
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
y
S
9
inches
psi
radians
w
inches
0.252
-1.000 So
0.200
-1.000 So
0.100
4.818 e.
0.191
-1.000 So
0.200
19.13 ~o
O.150
-0.780 So
0.300
42.50 E.
0.100
-0.523
So
0.400
74.29 E.
0.050
-0.263 So
0.500
113.7 Em
0.600
159.7 Eo
0
0
O
0
-0.050
0.265 So
0.700
210.7 Fo
-0.100
0.533 So
0.800
260.8 P.
-0.150
0.804 So
0.900
308.6 e,
-0.186
1.000 So
1.000
353.6 F.
-0.200
1.000 So
1.100
395.4 Ea
-0.248
1.000 So
1.200
433.5 E.
1.300
467.4 Eo
1.400
496.8 E.
1.500
521.3 e.
1.5708
535.6 Ea
42
CASE I
YIELDING TO ONE-HALF OF DEPTH OF BEAM
(y' = -0.124")
M
PQo:
-0.01431 So in lbs
0.001396 So lbs
10.249 in
Extent of Plastic Region
yI
yn"
9-
inches
inches
degrees
-0.124
0.127
0
-0.132
0.136
906 '
-0.150
0.154
16056 '
-0.175
0.181
24054'
-0.200
0.208
31053'
-0.225
0.235
38026
0.251
420191
-0.249
45013'
'
43
Case I
-
Yielding to One-half of Depth of Beam
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
6
inches
psi
radians
inches
0.251
-1.000 S<
0.200
-1.000 S5
0.100
7.206 C.
0.150
-1.000 Sc
0.200
28.39 4.
0.127
-1.000 S(
0.300
62.35 G.
0.100
-0.789 S(
0.400
107.6 E.
0.075
-0.593 Sc
0.500
162.4
-o
0.050
-0.396 Sc
225.0
to
0.025
-0.199 Sc
0.600
0.700
0.800
0
3
3
3
0
0
0
293.9 £o
367.6 E.
-0.025
0.200 So
0.900
-0.050
0.400 So
1.000
440.7 to
509.7 -
-0.075
0.602 So
1.100
574.0
-0.100
0.805 So
1.200
632.8 E.
-0.124
1.000 So
1.300
685.5 e.
-0.150
1.000 So
1.400
731.5 &.
-0.200
1.000 So
1.500
765.4 c.
-0.249
1.000 So
1.5708
793.2 E.
.
44
CASE I
YEILDING TO THREE-QUARTERS OF DEPTH OF BEAM
(y = -0.062")
M = -0.01530 So in lbs
P =
0.001493 So lbs
Qo = 10.250 in
Extent of Plastic Region
y"I
9
inches
inches
degrees
-0.062
0.063
0
-0.100
0.102
14059'
-0.150
0.154
26031'
-0.200
0.208
37022'
0.250
46015 '
-0.250
48047'
45
Case I
Yielding to Three-quarters of Depth of Beam
-
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
e
Y
inches
radians
psi
0
w
inches
0
0.250
-1.000 So
0.200
-1.000
So
0.100
14.26 Z.
0.150
-1.000 So
0.200
54.32 Eo
0.100
-1.000 So
0.300
114.8 ýo
0.063
-1.000 So
0.400
190.8 e.
0.050
-0.797 So
0.500
278.4 E.
0.025
-0.400 So
0.600
374.4 Eo
0.700
476.3 Eo
0
0
-0.025
0.401 so
0.800
581.8 Ec
-0.050
0.805
0.900
688.2 Ea
-0.062
1.000
1.000
788.9 Es
-0.100
1.000
30
1.100
882.0 E.
-0.150
1.000
So
1.200
966.7
E,
-0.200
1.000
30
1.300
1042
~.
-0.250
1.000 so
1.400
1107
E.
1.500
1161
E.
1.5708
1192
Et
30
46
CASE I
COMPLETE YIELDING
(y'
: y" : 0)
KE: -0.01562 So in lb
P :
0.001524 So lb
e, : 10.250 in
Extent of Plastic Region
y l
inches
inches
9
degrees
0
-0.050
0.051
90141
-0.100
0.102
18054'
-0.150
0.154
28048
-0.200
0.208
38053 '
0.250
47022
'
49049
'
-0.250
'
47
Case I
Complete Yielding
-
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
w
inches
psi
radians
0.250
-1.000 So
0.200
-1. 000 So
any other
0.150
-1.000 So
value
0.100
-1.000 So
0.050
-1.000 So
0
-1.000 So
0
1.000 So
-0. 050
1.000 So
-0.100
1.000 So
-0.150
1.000
-0.200
1.000
-0.250
1.000
30
So
0
inches
0
48
Case I
Maximum Deflection
Wmax
lbs
0
inches
0
0.000492 So
15.96
0.000984 So
31.93 E.
0.001236 So
535.6 E.
0.001396 So
793.2 Eo
0.001493 So
1192
0. 001524 So
.,
eo
49
.?.00
2
0
()
I
.100
I'U
z~
STRES3
-•.
FI GURE
.
EXTENT OF PLASTIC
oo
ui
REGION
DISTRIB'TION
ON PLANE OF MAX. STRESS
STRESS
CASE 3
p = 0.o000 84 So
0
-. 100 Z
50
PLASTIC
PLASTIt
2
.2.00
.100
1*-
w
M-6.e
STRESS
0
0
U.
FIGURE
3
EXTENT OF PLASTIC REGION
STRE3S
ON PLANE
-. too ZI
DISTRIBUTION
OF MNA.
STRES5
CASE I
P= 0.00,1 36 So
0.,
J
51
PLAST IC
.aoo.•00-
.100
J
it
2
-S.
0
0-e
FIG URE 4
IL
EXTENT OF PLASTIC
REGION
STRESS
DI3TR~IUTION
ON PLANE OF MiPAX. S"TR ESS
CASE r
P = 0.00
o 31 6 So
- .100
u
1-
52
.too
.J
.100
STRESS
+4
-050
io
*
I-.
0
I.L
w
FIGVURE
5
EXTENT OF PLASTIC
3TRESS
REGION
DOSTRI BUTION
ON PLANE OF MAK.
CASE
STI
T
P= 0.001433 So
-. 1000
D
53
.aoo _
I
UJ,
.?00
.J
.100
z
STRESS
-So
00
IL
FIQURE
6
EXTENT OF PLASTIC
REGION
STRESS DISTRIBUT ION
ON PLANE OF MAX. 5TRE SS
CASE T
P = 0.00 t S,4 So
-.
0OOt
tu
0
-. 00oo
I--0
~ 0.00(4 9 s So
1000 fo
7 SOfo
2z
S500e
6 5o
o
150 (*
P=0.000
O
o
.4
0-ANGLE
FRO \
.8
.2
FIXED END - RADIANS
84 So0
55
IS00 o
"50to
It
. 0
50l
C 5~
otn
0
0.001 So
P-APPLIED LOAD -
O.OaZS.
LB8.
56
Case II
Consider a curved bar of rectangular cross-section
with depth of beam small
in comparison with the radius of curvature
and a stress-strain curve of the form
s-s.
~
EO)1
57
IZZ2
-r-
S
j1*i
f-t·- L
005
kf- 't,-e%~o
so5(
--
,-~
-Kxs
f
It.
a.
Eoev
L aw
So,
,
C,
)
- -
N.A.
58
SS
7
tAA0
So
'0
[((e~~~~~a)I.
C~·a' =. 0o,
-\
0
C4A':
- oi
V --
4."
=C
e
-cc'AS
EQUATIoN
j.t.
4AU-[
13
Using absolute values, Equation 19 becomes
This expression is used to determine the position
of the neutral axis.
A trial and error solution is
recommended since the expression is not readily solvable
by algebraic means.
59
M=
s"
(
4
S
·,
AIF
C•.)I
0
'k
C-~·i~·~
'4
Cb
'
Ift
)t+3 (A
(et
4
" Fi IFoeq I
+
.
3
e .
L
I~r
+~~S;O·a·
M=Ar C
f
ý-I?-pt
+a
.4
oo
?-
A_11i)
IL
AL
49 L *4j.
2.1
Z
-,
3 ·(
s.
SZ.
N SoL((9
LXo-e,1c
rzI-I-,+
,-Q
44A 1
eo
· ?-~blJ~V
~de.rd~
EveorloN
.0
2
)
60
Using absolute values, Equation 20 becomes
1\A
So
Ir
eo
EQ
2-0 .
TLoN
T
Care must be taken' to use a high degree of
accuracy when evaluating the expression in Equation
20 (or 20a) contained in the brackets since it
involves the very small difference of two comparatively large numbers.
S--
M'I
6quamoN 2i.
When using Equation 21, the + signs apply to +
values of y, the - signs to - values of y.
F•oM
EQuh-T toN
"
Me.
.O,1.
ý"I
tt
Su3STTUTIrlG
A
·
(pit
IN EB)QuPAtON
10
Fit
62
Calculations for the Specific Case (Figure I)
/
M
, ~
~-.-P
b = 0.250"
A trial and error solution
h = 0.500"
of Equation 19a yields the
Ro - 10.500"
result that
Ri - 10.000"
Qo
10.249"
By the substitution of M = -PC cos@ in Equation 21
--'
Pe,,,
w~·P
Equation 21a
63
Evaluating
Initial Yielding
From Equation 21a, S z So, y - y'
= yl
Mi = -0.01250 So in-lb
P - 0.001220 So lb
Secondary Yielding
From Equation 21a, S = -So, y - y"
=
72
Ms = -0.01276 So in-lb
P = 0.001245 So
lb
Complete Yielding
From Equation 21a, S =
Mc = P
So, y
= y' a y" = 0
00 in-lb
oO
lb
Stress Distribution
The stress distribution at any angle e from 0 to •
for any load P from 0 to oO is
given by Equation 21a.
This applies to both the elastic and plastic regions
since the stress-strain curve also applies to both the
elastic and plastic regions.
64
Deflection Equation
Substitution of M a -P (c os6 and eom =
coaZ
t +'
z
in Equation 22 yields
P"z
t-w+
Ae~
4c~o oL
45"f oc~:
+l0
:Fit 0\i
3~)·
,.9;
Fi
FR
)]rR,ý
)]
IF
Complementary Solution
W= C A%
C
eC
9
Particular Solution
-
?-ý)]
rw
ý-J,
U9*
4s,,.2 Lk.foL •
i·
Flo)I
,3~flr~
9 1~ d
Lre, +
General Solution
e
W3
9Act aw 09e
- a
(C-I)S
4110,T ý-p
Boundary Conditions
= 0, nd
iN
va0u
~(yi·F.·~~32
-J1
=0
Evaluation of 0l and C2
0
3
P(bf
c
Ir
r
R
3
o
.+
3
cate+
3
FRO
-
---
-
'E
3esf R~
Equation 23
This deflection equation (Equation 23) applies
for all values of P from 0 to oO , ie, it applies over
both the elastic and plastic regions.
The sign of the
deflection must be the same as that of the bending
moment before the moment term is squared and substituted
in Equation 22.
66
CASE II
INITIAL YIELDING
(y'
= -0.249)
M = -0.01250 So in lbs
P =
0.001220 So lbs
. = 10.249 in
Extent of Plastic Region
There is no plastic region.
67
Case II
-
Initial Yielding
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
w
inches
psi
0.251
-0.980
0.200
-0.877 So
0.100
1.027 Co
0.150
-0.761 So
0.200
4.075 6.
0.100
-0.623 So
0.300
9.055
0.050
-0.442 So
0.400
15.85 e.
0.040
-0.395 So
0.500
25.84
Ea
0.030
-0.342 So
o.600oo
33.85
E.
0.020
-0.280 So
0.700
44.60
E.
0.010
-0.198
So
0.800
56.11
E.
0.900
68.00
f.
radians
So
0
0
inches
0
e.
-0.010
0.198
So
1.000
80.11
Eb
-0.020
0.280 So
1.100
91.97
Ea
-0.030
0.343 So
1.200
103.3
Ea
-0.040
0.397 So
1.300
113.9
E
-0.050
0.444
123.6
Eo
-0.100
0.629 So
1.400
1.500
132.0
Co
-0. 150
0.772 30
137.2
~.
-0.200
0.894 3So
-0.249
1.000 So
So
1. 5708
68
CASE II
YIELDING TO ONE-FIFTH OF DEPTH OF BEAM
(y'
- 0.200)
M = -0.01398 So in
P =
lb
0.001364 So lb
co = 10.249 in
Extent of Plastic Region
e
inche s
inches
degrees
-0.200
0.208
0
-0.209
0.218
120 6'
-0.215
0.224
15028
-0.230
0.241
21024'
0.251
240 9'
-0.249
26037
'
'
69
Case II
-
Yielding to One-fifth of Depth of Beam
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
inches
psi
radians
inches
0.251
-1.096 So
0.200
-0.981 So
0.100
1.285 E.
0.150
-0.851 So
0.200
5.098 4e
0.100
-0.697 So
0.300
11.33
Co
0.050
-0.494 So
0.400
19.83
E.
0.040
-0.442 So
0.500
32.33
Eo
0.030
-0.383 So
o0.60o
42.35
E,
0.020
-0.313 So
0.700
55.80 E.
0.010
-0.221 So
0.800
70.20 4e
0.900
85.07
f.
O
0
0
0
-0.010
0.222
So
1.000
100.2
E4
-0.020
0.313 So
1.100
115.1
Eo
-0.030
0.384 3o
1.200
129.3
Eo
-0.040
0.444 So
1.300
142.6
Eo
-0.050
0.496 So
1.400
154.6 C.
-0.100
0.704 So0
1.500
165.2
Eo
-0.150
0.864
1. 5708
171.7
E.
-0.200
1.000
-0.249
So
1.118 So
So
70
CASE II
YIELDING TO TWO-FIFTHS OF DEPTH OF BEAM
(y'
- -0.150")
-0.01618 So in lb
FM
0.001579 So lb
10.249 in
Extent of Plastic Region
8
Y
7"
inches
inches
degrees
-0.150
0.154
0
-0.155
0.160
10026'
-0.164
0.169
170 7'
-0.180
0.187
24017'
-0.209
0.218
32022'
-0.230
0.241
36027'
0.251
37059
'
39026
'
-0.249
71
Case II
Yielding to Two-fifths of Denth of Beam
-
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
inches
psi
radians
inches
0
0
0.251
-1.269 So
0.200
-1.135 So
0.100
1.722 4o
0.150
-0.985 So.
0.200
6.832
E.
0.100
-0.807
So
0.300
15.18
Co
0.050
-0.572 So
0.400
26.57
Ea
-0.512 So
0.500o
43.32
.4
0.030
-0.443 So
0.600
56.75
Ea
0.020
-0.362 So
0.700
74.78
Eo
0.010
-0.256 So
0.800
94.07 Co
0.900
114.0
Co
Ea
0
0
-0.010
0.256 So
1.000
134.3
-0.020
0.363 So
1.100
154.2 Ea
-0.030
0.445
So
1.200
173.2 4E
-0.040O
0.514 So
1.300
191.0 E.
-0.050O
0.574 So
1.400
207.2 4o
-0.100
0.814 So
1.500
221.3 Ea
-0.150
1.000 So
1.5708
230.0 Ea
-0.200
1.158 So
-0.249
1.295 So
72
CASE II
YIELDING TO THREE-FIFTHS OF DEPTH OF BEAM
(y'
= -0.100")
M - -0.01987 So in lb
P =
e
0.001939 So lb
= 10.249 in
Extent of Plastic Region
9
y
inches
inches
degrees
-0.100
0.102
0
-0.110
0.112
17038'
-0.125
0.128
26027'
-0.150
0.155
35028'
-0.200
0.210
45017'
0.251
500 4#
-0.249
510 1'
73
Case II
-
YieldinE to Three-fifths of Depth of Beam
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
Y
inches
psi
radians
inches
0.251
-1.558 So
0.200
-1.394 So
0.100
2.595 fo
0.150
-1.210 So
0.200oo
10.28 e,
0.100
-0.990 So
0.300
22.88 do
0.050
-0.702 So
0.400
40.06 e.
0.040
-0.628 So
0.500
65.31
0.030
-0.544 So
0.600
85.55 Eo
0.020
-0.445 So
0.700
112.7 6o
0.010
-0.314 So
0.800
141.8 Eo
0.900
171.8 Cc
O0
0
0
0
E.
-0.010
0.315 So
1.000
202.4 4.
-0.020
0.445 So
1.100
232.4 E6
-0.030
0.546 So
1.200
261.2 Eo
-0.040
0.631 So
1.300
288.0 E.
-0.050
0.705 So
1.400
312.3 Co
-0.100
1.000 So
1.500
326.0
-0.150
1.228 So
1.5708
346.7 E,
-0.200
1.421 So
-0.249
1.590 So
.
CASE II
YIELDING TO FOUR-FIFTHS OF DEPTH OF BEAM
(y m -0.050")
M
P :
-0.02817 So in
lb
0.002749 So lb
o= 10.249 in
4
Extent of Plastic Region
y'
yt"
inches
inches
-0.050
0.051
-0.053
0.054
13048'
-0.058
0.059
21056'
-0.066
0.067
29035'
-0.075
0.076
35022'
-0.100
0.102
450 8'
-0.150
0.155
54°56'
-0.200
0.210
60015'
0.251
630 4'
-0.249
degrees
0
63040'
75
Case II
Yielding to Four-fifths of Depth of Beam
-
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
y
S
9
inches
psi
radians
inches
0
0.251
-2.208 So
0.200
-1.976 So
0.150
-1.715 So
0.200
20.70 e.
0.100
-1.404 So
0.300
45.99 E.
0.050
-0.995 So
0.400
80.51 e.
0.040
-0.391 So
0.500
131.3 C.
0.030
-0.772
So
0.600
172.0 e.
0.020
-0.630 So
0.700
226.5 4o
0.010
-0.446 So
0.800
285.0 Co
0.900
345.4
0
0
0
5.216 e.
Eo
-0.010
0.446 3So
1.000
406.9 Eo
-0.020
0.632 So
1.100
467.1 E.
-0.030
0.774 So
1.200
524.9
-0.040
0.3894
1.300
578.8 Eo
-0.050
1.000 So
1.400
627.6 L.
-0.100
1.418 So
1.500
670.5 &.
-0.150
1.741 So
1.5708
696.9 E
-0.200
2.015
-0.249
2.254
30
e.
76
Case II
Maximum Deflection
P
Wmax
lbs
inches
0
0
0.0002 So
3.689 1.
0.0004 So
14.76 Co
0.0006 So
33.20 E.
0.0008 so
58.29 C.
0.0010 So
92.23 Eo
0.0012 So
132.8 C.
0.0014 So
180.4 Eo
0.0016 So
236.1 e.
0.0018 So
298.8 E.
0.0020 So
368.9 Eo
0.0022 So
446.4 E.
0.0024 So
531.2 Eo
0.0026 So
623.5 E-
0.0028 So
723.1 c.
0.0030 So
830.1 eo
77
.aoo -
.J
.100 dc
Z
6TRE 56
.mso
o
20
U.
FIGURE 3
EXTENT OF PLASTIC REGION
.100
I"
STRES3 DISTRIBUTION
ON PLANE OF MAY. STR ES
CA6E 2I
P = 0.00 a?.0 So
-. eoo
jd
78
PLASTiC
PLASTIQ
.- 00
.100
STRE S S
dc
0- z
'.5.
0
FIGURE 10
EXTENT OF PLASTIC REGION
STRESS DISTRIB UTION
ON PLANE OP MAX. STRESS
CASE 3X
P.0.001364 So
.100
Z
a
79
PLASTIC
PLA ST tC
.too
STRIESS
-6.
-.
0
.tO
Ir
FIGURE II
EXTENT OF PLASTIC
REGION
-. 100
3TRESS DISTRIBUTION
ON PLANE OF MAX. STRESS
CASE :
Pm 0.00157 9 So
1
(1
80
&
,qk
.Zoo
.100 _J
al-
w
STRE38
-.LS
- zS.
2S.
a&.
z
a0
FIGURE 12
EXTENT OF PLASTIC
REGION
STRESS
ON
DISTRIBUT ON
PLANE OF MAX. STRESS
CASE
P W0.00
-.- oU.
at
-.0
I3I
33 So
-. zoo0o
81
.1oo0
.100
-map
-S
FI GURE
L
STRESS
S.
So.
Z
2
0
13
EX'TENT OF PLASTIC
-O
Ii
REGION
-. 100 Q
z
STRESS DISTRIBU10 N
ON PLANE OF MPNX. STRESS
1-C
0)
CAp E UI
p
0.00 2z749 So
·
·
·
FI GU
.E 14
DEFLECTION
-500
0
CASE
CURVES
31
0
.40 -
.8
ANGLE
FIKOM
1..
FIX
D
END-RADI
NS
1.6
I000Eo
750&o
O
Ul
lil
LL.
wd
500E.
I1506,
a
0
0.09 10o
P-APPLIED
0o ?.So
LOAD-
LBS.
0.0030
Case III
Consider a curved bar of triangular cross-section
with depth of beam small
in comparison with the radius of curvature
and a stress-strain curve of the form
5S1E.
•---Fo
T
'1I
t4
bo .
(4 X-) 0
IT-
•o0
3-
3+
ba
itT
I
-/
VNr
3
"S rc
er-H
20
86
-o
SAPA=
N= ?
dRAR
z /,
5¾
e%.
ýA
(A-
Iv(AL-
0
-A
L
i~;",)~~&a
L
(I·-a= 0
-so, (-a , ji'
ý,Q.
vt
C
÷~sai
[
%·i~
o
A)
-t
\I
I ~
\2
l~bta/
$0
a·
·~QL(
_j3Y~
ThJ
Io~
iva
o
e
-e -
.
4FS~
1~
+i~
fl]0 r
COTar
CIO
KI
G.
z
lakx
F
-e n~~·~e
Je-a
87
--'K T %-
Co
8"b'
ýTcý0
_V1s,
+3
.
-is~
ý )]
toAA,%F-
- ý0 nAýý'
2.
Fý 0
-
3 co- (
/A
I
- Z
-Con
9, 0
(it +3
1
+
V
QAJAtION
?'+
Using absolute values, Equation 24 becomes
+
Q0
Z A'e
Ae
-
,-e A)
r
a
Eq UQAT I~ON•9..
This expression is used to determine the position
of the neutral axis.
A trial and error solution is
recommended since the expression is not readily solvable
by algebraic means.
88
N=
1
s
§t
tVV~
0
cA
AA
(·ap
3
0
^b
91
zC-
r
2
4~i
~"a
eb
eo"br
÷
,•h
A•t
!
0
S3
TA
NL4
eo
C
3
3.r55\)
+
S8
12.
'I
-
e.
v
Q*
ýcý01
J.
toE
9
e.
O
3
8
8
a
89
NA
4-__
Lj3
~
3
8
8
"
a
4-
4
-
'0
CO~S~d
C
3
3
6
8
V
_
4
C
3
C
41
e/CO
Tz7.
r
Eb
3
s
^b
·a
1
. q UA
toN Z S
90
Using absolute values, Equation 25 becomes
Care must be taken to use a high degree of
accuracy when evaluating the expression in Equation
25 (or 25a) contained in the brackets since it
involves the very small difference of two comparatively large numbers.
EQUATLON 2
When using Equation 26, the
values of y, the -
signs apply to
signs to - values of y.
91
From Equation 25a
m.
A_
a~•
FRO
e
",- f,Ty
r-ý
V
Substituting in Equation 10
Q0 "
A w 4 lW
E0.
5r--0.
asd:
3,'
· a+S,
4
Yi, If
v
_
se8
r3
+:Lrsdii·--O
-Q
3
O
c~a~-~~s~s~
-?
Ueai.s
eC.
reo 4rl r'R.4
EQ AT oN 2.7
92
Calculations for the Specific Case (Figure 1)
M -Pt
0 co
b = 0.025"
A trial and error solution
h = 0.500"
of Equation 24a yields the
Ro = 10.500"
result that
Ri = 10.000"
Q0
= 10.165"
By the substitution of M = -P tocos9 in Equation 26
+( \2z
25k
E QVNT oN Z6o4L
93
Evaluating
Initial Yielding
From Equation 26a, S : -So, y : y" : y2
Mi : - 0.01120 So in-lb
P : 0.001102 So
lb
Secondary Yielding
From Equation 26a, S : So, y : y' " Y1
Ms : -0.01557 So in-lb
P = 0.001532 So
lb
Complete Yielding
From Equation 26a, S :
So, y - y' : y" : 0
Mc = - oo in-lb
P
o
lb
Stress Distribution
The stress distribution at any angle 9 from 0 to i
for any load P from 0 to
cO
is given by Equation 26a.
This applies to both the elastic and plastic regions
since the stress-strain curve also applies to both the
elastic and plastic regions.
94
Deflection Equation
Substitution of M = -PtocosS and
oe
in Equation 27 yields
Complementary Solution
Patt c, 4eo
o
e
Particular Solution
-
P
4+
:,v
CR
+
0 ?.Eo
+
a.~
AL ~
2
(
:3
*4AV
lebl
a
6
~ofA
(2
~
\yL 2.l .s13
-~vz,
95
General Solution
Ck\N
C
+A e
3
4Cace
gar
o
Lz3
co
4co
+
7\
+
Boundary Conditions
eo),
\N = O
l-e
J ae
O
Evraluation of C,and C2
C, = 0
3 t~a
E0
+
ac
-
-
0
dow
12
V~z3:t
j~i
-i~~h4~
s+s~
I5
RO
r
96
Equation 28
This deflection equation (Equation 28) applies
for all values of P from 0 to
O , le, it applies over
both the elastic and plastic regions.
The sign of
the deflection must be the same as that of the bending moment before the moment term is squared and
substituted in Equation 27.
97
CASE III
INITIAL YIELDING
(y"
z 0.335")
M a -0.01120 So in lb
P =
0.001102 So lb
o= 10.165 in
Extent of Plastic Region
There is no plastic region.
98
Case III - Initial Yielding
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
inches
psi
radians
inche s
0
0
0.335
-1.000 So
0.300
-0.948 So
0.100
0.795 Co
0.250
-0.867 So
0.200
3.154
0.200
-0.778 So
0.300o
7.010 e.
0.150
-0.675 So
0.400
12.24 E.
0.100
-0.553 So
0.500
19.55 C.
0.050
-0.392 So
0o.600
26.20
0.040
-0.351 So
0.700
34.52 Em
0.030
-0.304 So
0.800
43.43
0.020
-0.248 So
0.900
52.63 E.
0.010
-0.176 So
1.000
62.01 Eb
1.100
71.19 Eo
O
0
0
o.
,.
i.
-0.010
0.176 So
1.200
79.99 E.
-0.020
0.249 30
1.300
88.20 ~.
-0.030
0.305
1.400
95.64 E.
-0.040
0.352 So
30
1.500
102.2 C.
1.5708
106.2 E,
-0.050
0.394
-0.100
0.558 So
-0.150
0.685 So
-0.165
0.719 So
99
CASE III
YIELDING TO ONE SIXTH OF DEPTH OF BEAM (ON TOP)
(y" = 0.255")
-0.01278 So in lb
M4
P =
0.001258 So lb
= 10.165 in
o•
Extent of Plastic ReRion
y"
e
inches
degrees
0.255
0
'
0.260
7052
0.265
110 4'
0.275
15027
'
0.300
22029
'
0.335
28051
'
100
Case III
-
Yielding to One-sixth of Depth of Beam
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
S
inches
psi
radians
inches
0.335
-1.142 Sc)
0.300
-1.082 Sc2
0.100
1.036
0.250
-0.990 Sc2
0.200
4.112
0.200
-0.888 Sc)
0.300
9.138
0.150
-0.771 Sc
0.400
15.96
0.100
-0.631 Sc)
0.500
25.43
0.050
-0.447 Sc)
0.600oo
34.16
0.040
-0.400 Sc)
0.700
45.01
0.030
-0.347 Sc
0.800
56.62
0.020
-0.283 Sc
0.900
68.62
0.010
-0.200 Sc
1.000
80.84
1.100
92.81
0
0
0
0
-0.010
0.201 So
1.200
104.3
-0.020
0.284 So
1.300
115.0
-0.030
0.348 So
1.400
124.7
-0.040
0.402 So
1 3,3
-0.050
0.449 So
1.500
1.5708
-0.100
0.637 So
-0.150
0.782 So
-0.165
0.821 So
2
138.5 i
101
CASE III
SECONDARY YIELDING
-0.165",
(y'
y" = 0.171")
M = -0.01557 So in lb
P =
0.001532 So lb
e,: 10.165 in
Extent of Plastic Region
v"
inche s
e
degrees
0.171
0.175
90 7'
0.185
160 6'
0.200
22022 '
0.250
340 0'
0.300
400 391
0.335
440 1'
102
Case III
-
Secondary Yielding
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
inche s
psi
radians
0
inches
0
0.335
-1.391 So
0.300
-1.318 So
0.100
1.537 E.
0.250
-1.206 So
0.200
6.099 E.
0.200
-1.081 So
0.300
13.55 Eo
0.150
-0.939 So
0.400
23.68 ~.
0.100
-0.786 So
0.500
37.71 E.
0.050
-0.545
So
0.600
50.67
0.040
-0.487 So
0.700
66.76 E.
0.030
-0.422 So
0.800
83.98 ~.
0.020
-0.345 So
0.900
101.8 Eo
0.010
-0.244 So
1.000
119.9 ~.
1.100
137.7
O
0
,.
e.
-0.010
0.244 So
1.200
154.7 Eo
-0.020
0.346 So
1.300
170.5 E.
-0.030
0.424 So
1.400
184.9 Eo
-0.040
0.489 So
1.500
197.65
.
-0.050
0.547 So
1.5708
205.4
e.
-0.100
0.794 So
-0.150
0.953 So
-0.165
1.000 So
103
CASE III
YIELDING TO ONE-HALF OF DEPTH OF BEA•M (ON TOP)
(Y" 1
0.110")
M = -0.01933 So in
P =
q,•
lb
0.001902 So lb
10.165 in
Extent of Plastic Region
inches
inches
degrees
-0.108
0.110
0
-0.113
0.115
11059'
-0.122
0.125
200 9'
-0.146
0.150
30054'
-0.165
36021'
0.175
37019'
0.200
41051'
0.250
480 6'
0.300
52020'
0.335
54036'
104
Case III
Yielding to One-half of Depth of Beam (On Top)
-
Stress Distribution
Deflection Curve
on Plane of
Maximum Stress
y
inches
S
e
psi
radians
0
inches
0
0.335
-1.726 So
0.300
-1.636 So
0.100
2.369 4o
0.250
-1.497 So
0.200
9.400 L
0.200
-1.343 So
0.30o0
20.89 Eq
0.150
-1.165 So
0.400
36.49 E,
0.100
-0.932 So
0.500
58.12 E.
0.050
-0.676 So
0.600
78.09
E.
0.040
-0.605 So
0.700
102.9
Eo
0.030
-0.524 So
0.800
129.4
E.
0.020
-0.428 So
0.900
156.9
,
0.010
-0.303 So
1.000
184.8 eo
1.100
212.2
0
0
Ea
-0.010
0.303 So
1.200
238.4 to
-0.020
0.429
1.300
262.9
Ea
-0.030
0.526 So
1.400
285.0
4a
-0.040
0.607 So
1.500
304.5
Ea
-0.050
0.680 3So
So
1.5708
316.5
EO
-0.100
0.963 So
-0.150
1.183 So
-0.165
1.241
So
105
CASE III
YIELDING TO SIX-TENTHS OF DEPTH OF BEAM (ON TOP)
(y" - 0.050")
M - -0.02859 So in lb
P -
0.002812 So lb
te: 10.165 in
Extent of Plastic Region
e
y
inches
inches
degrees
-0.050
0.050
0
-0.054
0.055
17030
-0.064
0.065
28038'
-0. 074
0.075
35010
-0.098
0.100
44052'
-0.146'
0.150
54032'
'
'
570 o'
-0.165
0.200
59046'
0.250
630 91
0.300
65036
0.335
66056'
'
106
Case III - Yielding to Six-tenths of Depth of Beam (On To)
Stress Distribution
on Plane of
Deflection Curve
Maximum Stress
y
S
9
inches
psi
radians
w
inches
0.335
-2.553 So
0.300
-2.420 So
0.100
5.181 Za
0.250
-2.214 So
0.200
20.56 e.
0.200
-1.986 So
0.300
45.69 Ea
0.150
-1.724 So
0.400
79.81 4.
0.100
-1.411 So
0.500
127.1 Eo
0.050
-1.000 so
0.600
170.8 E.
0.040
-0.895 so
0.700
225.0 E.
0.030
-0.775 So
0.800
283.1 E.
0.020
-0.633 So
0.900
343.1 &.
0.010
-0.448 So
1.000
404.2 Co
0
0
0
0
1.100
464.0 Cc
-0.010
0.449 So
1.200
521.4 Co
-0.020
0.635 So
1.300
574.9 Ea
-0.030
0.778 So
1.400
623.4 ea
-0.040
0.898 So
1.500
666.1 4
-0.050
1.005 So
1.5708
692.3 e.
-0.100
1.425 So
-0.150
1.749 So
-0.165
1.836 So
107
Case III
Maximum Deflection
P
Wmax
lbs
inches
0
0
.0002 So
3.501 4.
.0004 So
14.00 e.
.0006 So
31.51 E.
.0008 So
55.32 C.
.0010 So
87.53 E.
.0012 So
126.0 C.
.0014 So
171.6 EO
.0016 So
224.1 E.
.0018 So
283.6 Ea
.0020 So
350.1 C~
.0022 So
423.6 E.
.0024 so
504.2 E.
.0026 So
591.7 E.
.0028 so
686.2 C.
.0030 So
787.8 ~,
108
.0oo
I,
4)
C
IUot
b.
STRESS
-So
_U
-
Q
FIGURE
16
E.TENT OF PLASTIC
0
REGION
DISTRI UTION
STRES3
ON PLANE OF MA%. STRESS
CA SE MIm
P= 0.00110.o
So
-. 00oo
t
109
PLASTIC
100
'C
J00
I-
0
*So
STRE 58
Q
0
FI QURE
I
EXTENT OF PLASTI
3TRESS
ON PLANE
DISTR
OF AAA)
CASE I=
P=
.00.I
56 So
110
.300
I
p.s
.100 Z
O
0
cd
STAE S8
emS.
FIGVRE
O U
2
18
EXTENT OF PLASTIC
DISTRIBUTION
STRESS
ON PLANE OF NIMAX. STRESS
CASE I3
P=i0.00132
0..
REGION
So5
inIQ
'
11l
PLASTIC
I
.300 ,
w
.200 Z
1
.IOO
ST RESS
b1
~
4
-0
FIGURE 1S
EXTPENT oF PLASTic
REGION
.1001
STIESI3
3TRISUt1TION
ON PLAANE OF MNAX. STRESS
,I
CASE Mt
P a 0.00180.. So
112
I
)
z
D
QQ
.100 w
2
3TRESS
___
-so
__
I
So
z
FIGURE
0.O
I-
EXTENT OF PLASTIC REGION
STRES 0DISTRIBUTION
ON PLANE OF M~X. STRESS
II
CA3E If
p * 0.00PI•a. So
"750So
I-
o
2 250to
tso.
.4
O-ANGLE
.8
FROM
FIXED EhND-
1.6
RAOIANS
oIooEo
FIGURE Et
750~.
MAX. DEFLECTION
GASE
VS. LOAD
lt=
5OOo
aoo I So
P-APPLIED
0.002 5o
LOAD -- L56.
0.OSoQ
4H
115
APPENDIX
A-1
B I B L I OGRAP H Y
1.
P.
CROUT,
- -
D.
"An Application of Polynomial
Approximation to the Solution or Integral Equations
- -
Arising in Physical Proolems"
Journal or
Mathematics and Physics, Vol XIX, No.
..1iECKi,
- -
H.
1, January i40
"The New Theory of Plasticity,
Strain Hardening, and Creep, and the Testing of the
- -
Inelastic Behavior of Metals"
ASME,
3.
Transactions or
Vol 55, 19Y3
- -
M. B.
HOGAN,
"Circular Beams Loadea Normal to
- -
the Plane or Curvature"
Transactions of ASME,
Vol 60, 195d
4.
MURPHY,
- -
- -
G.
"Advanced Mechanics of Materials"
Book, First Edition, McGraw-Hill Book Co., Inc.,
New York and London, 1946
5.
NADAI,
A.
-
-
State of Metals"
"On the Mechanics or the Plastic
- -
Transactions or ASME, Vol 52,
1930
6.
NADAI,
A.
- -
"Theories of Strength"
- -
Trans-
actions of ASME, Vol 55, 1953
7.
NADAI, A. (and WAHL, A.M.)
Book,
-
"Plasticity"
First Edition, McGraw-Hill Book Co.,
New York and London, 1931
Inc.,
-
-
A-2
BIBLIOGRAPHY
b.
SCARBOROUGH,
Analysis"
J. B.
- -
- -
BooK,
"Numerlcal Matiematical
The Jonns Hopkins Press,
Baltimore, 1930
9. "Tables or Lagrangian Interpolation Coerricients"
- -
Prepared oy tae Mathematical Taoles Project,
Works Projects Administration or tne Federal Works
Agency, Columoia University Press, New York, 1944
10.
TIMOSHENKO,
- -
S.
- -
"Theory or Elastic Staol•ity"'
Book, First Edition, McGraw-Hill Book Co., Inc.,
New York and London, 1956
11.
TIMOSHENKO, S.
- -
"Strength or Materials"
-
-
Part II (Adv Theory and Problems), Book, Second
Edition, D. Van Nostrand, New York, 1941
12.
WINSLOW, A. M. (and EDMONDS,
R. H. G)
ana Theory of Curvec Beams"
- -
ASME,
Vol 46*, 192b
- -
"Tests
Transactions or
A-3
*NOTES ON INTEGRATIONe
Integration of Equation 17a
. ÷
•
=
- o
,
°Peaoe
This differential equation was integrated by the
method of successive approximations using an interval
in
e
of 0.02 radians, which is an interval of very
slightly over 10,.
The three point method was chosen.
The basic reason for this choice was that since some
approximations had been made at various points in
arriving at the equation which was being integrated
the accuracy was not sufficient to warrant the use of
four or five point integration.
Use was made of the
integrating ahead coefficients to predict points in
advance.
are:
The coefficients for three point integration
o
A
0.41667
-0.08333
0.41666
0.66666
0,.66666
-1.33333
-0o08333
0.41667
1.91667
At 9 = 91, the equation
W
C1
becomes effective.
e
The constants, 01 and 02, are
evaluated from the boundary conditions that when
A-4
9
1el,
w = w',
dw/de a dw'/de.
For the case where P = 0.001236 So lb, 91 = 37 17'
which is
0.65 radians.
From numerical integration
w a -184.99 4.
dw/de = -518.03E.
Evaluation of Cl and 02
C1 -
-495.44 Eo
02 : 156.88 E.
For the case where P = 0.001396 So lb, e1
which is
45 13'
0.79 radians.
From numerical integration
w : -360.05 ~o
dw/de = -752.02 o
Evaluation of C1 and C2
C1 = -747.86 eo
02 = 266.22 Eo
For the case where P = 0.001493 So lb, e 1
which is 0.85 radians.
From numerical integration
w = -635.30 C°
dw/de = -1072.5 £E
Evaluation of C1 and 02
C1 z -1143.6 EC
02 z 369.01 Eo
48 47'
A-5
Integration of Certain Integrals in Cases I and II
Integrals containing terms of the form
can be readily integrated by substitution.
(oý)
For example,
consider
This new integral is of a form which can be readily
integrated by the use of any set of integral tables.