Plastic Design of a Fixed-Fixed Beam-Column
CEE 201L. Uncertainty, Design, and Optimization
Department of Civil and Environmental Engineering
Duke University
Henri P. Gavin
Spring, 2015
In elastic-plastic materials, stress is proportional to strain up to the yield stress, σy .
The yield moment, My is the bending moment corresponding to a bending stress distribution
in which the stress equals the yield stress only at the outer-most fibers. In a symmetric cross
section of depth d, My = σy I/(d/2). If the cross section has a solid rectangular shape (dimensions d × b) then I = bd3 /12 and My = σy (bd2 /6). The section can carry additional moment
beyond the yield stress. The plastic moment, Mp is the bending moment corresponding to
a bending stress distribution in which the stress equals the yield stress almost everywhere
within the cross section. For a solid rectangular cross section, Mp = σy (bd2 /4).
1 Beam (with no axial load)
stress
σ
For rectangular cross sections Mp = (3/2)My . For I-shaped cross sections Mp ≈ 1.12My .
d
σy
stress
σ
E
ε y∼0.002
d
strain ε
strain
ε
−σ y
−σ y
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+σ y
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+σ y
M
+σ y
−ε y
−ε y
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σ= E ε
M = EI φ
−σ y
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yield point
φ y = ε y / (d/2)
M=My= EI φ y
partially
plastic
fully
plastic
M>My
M=Mp
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H
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Mo
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M(x)
−Mo
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x
The top of a fixed-fixed beam of height H is displaced with a displacement D until failure:
Mp
Fp
My
Fy
K
EI
φy
φ
Dy Dp = 1.5 Dy
As the bending moment M approaches Mp , the curvature, φ approaches ∞, as in a hinge.
2
CEE 201L – Uncertainty, Design, and Optimization – Duke University – Spring 2015 – H.P. Gavin
Analyzing the equilibrium of moments about point o:
X
Mo = 0 : Mo + Mo − F × H = 0 −→ F =
2Mo
.
H
So the end moments in the fixed-fixed beam are related to the end forces, Mo = F H/2. This
relationship is simply a statement of equilibrium and is valid for any level of elastic or plastic
deformation.
Given the end-forces and end moments, the moments along the beam M (x) can be found
from a simple free-body diagram, M (x) = Mo + F x. The curvature along the beam φ(x)
can be found from M (x) and an M − φ relation like the one plotted above. Up to the yield
moment, M (x) = EIφ(x), or M (x)/My = φ(x)/φy . For moments larger than My , more
complicated expressions for φ(x) can be determined.
Since the curvature is very close to the derivative of the slope of the beam, φ(x) ≈ v 00 (x),
the slope, v 0 (x) and the deflection, v(x), can be found by integrating φ(x) and finding the
constants of integration from the boundary conditions.
In this way, the deflections along a fixed-fixed beam can be computed as the beam is deformed
up to, and beyond its yield limit. An example of the results of this process are illustrated in
Figures 1 and 2. In this example, the beam has a height H = 10 cm, and a solid rectangular
cross section b = 3.0 cm, and d = 0.203 cm. The yield stress is σy = 35, 000 N/cm2 . From
these quantities, the yield and plastic moments are My = 721 N.cm and Mp = 1082 N.cm.
Figure 2 shows the deflection, slope, and curvature of the beam when after ends have become
fully plastic. The center figure shows the discontinuity of the slope at the ends, and the top
plot shows that the curvatures become infinite, as in a hinge. This is called a plastic hinge.
For a beam with a rectangular cross section, the cross section dimensions, b and d can be
found from the stiffness, K, and the ultimate plastic force, Fp .
The stiffness relating the force F and collocated displacement D of the fixed-fixed beam
shown above is
!3
d
Fy
Fp
12EI
= Eb
=
=
,
(1)
K=
3
H
H
Dy
Dp
From the yield analysis and plastic analysis of a rectangular cross section:
My =
bd2
σy ,
6
Mp =
bd2
σy .
4
(2)
Applying the equilibrium equation above, F = 2Mo /H,
Fy =
bd2
σy ,
3H
Fp =
bd2
σy .
2H
(3)
These expressions can be used to determine the cross section dimensions from Fp , Dp , σy , E,
and H. From the stiffness of the beam, equation (1),
Fp 1
b=
Dp E
H
d
3
K
=
E
H
d
3
,
(4)
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3
Plastic Design of a Beam-Column
Inserting this into the expression for Fp , equation (3)
Fp H
Fp
Dp
d =2
= 2 HE
σy b
σy
Fp
2
d
H
!3
,
or,
d=
1 σy K
1 σy H 2
= H2
.
2 E Dp
2
E Fp
(5)
With this value for the depth of the beam, d, b can be found from equation (4) above.
Finally, note that if the weight mg supported by the column can not exceed the Euler buckling
load,
π 2 EI
mg <
,
(6)
H2
otherwise the columns will buckle under the compressive weight.
Example.
Given:
Fp = 216.3 N, Dp = 0.43 cm, H = 10.0 cm, and σy = 35000 N/cm2 , and the modulus of steel
is 20 × 106 N/cm2 .
Find:
the width and depth (b and d) of the rectangular beam cross-section.
d=
b=8
1 35000
102
1 σy H 2
= ·
·
= 0.203 cm
2 E Dp
2 20 × 106 0.43
E 2 Fp Dp2
(20 × 106 )2 · (216.3) · (0.43)2
=
8
·
= 3.0 cm
H 3 σy3
10.03 · (35000)3
CC BY-NC-ND H.P. Gavin
4
CEE 201L – Uncertainty, Design, and Optimization – Duke University – Spring 2015 – H.P. Gavin
moment, N.cm
1200
1000
Mp=1082 N.cm
800
φy=0.0172; My= 721 N.cm
600
400
200
b= 3.0 cm; d=0.203 cm
EI=41827.1 N.cm2
0
0
0.02
0.04
0.06
curvature, 1/cm
0.08
force, N
200
0.1
Fp=216.3 N
150
Dy= 0.3 cm;
Fy = 144.2 N
100
50
K=501.9 N/cm
L= 10.0 cm
0
0
0.2
0.4
0.6
0.8
1
deflection, cm
Figure 1. Moment-curvature (M − φ) for a rectangular section and force-deflection (F − D)
for a fixed-fixed beam
10
10
8
8
8
6
6
6
4
4
4
2
2
2
x, cm
10
0
0
-0.04
0
0.04
curvature, 1/cm
0
0
0.05
slope
0.1
0 0.2 0.4 0.6 0.8 1
deflection, cm
Figure 2. Distributed curvature, slope, and deflection along a fixed-fixed beam. Note that the
slope is discontinuous and the curvature becomes infinite at the ends, the locations of plastic
hinges.
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5
Plastic Design of a Beam-Column
2 Beam-column (with axial compressive load), P
σy
E
ε y∼0.002
strain ε
d
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elastic
σ= E ε
M = EI φ
−ε y
yield point
φ y = ε y / (d/2)
M=My= EI φ y
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−σ y
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y
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d/2
00
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strain 11
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00
ε 11
00
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00
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0
001
11
0
1
0
1
0
1
0
1
−σ y
M
mm
l
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d−a
d
−σ y
11
00
00
11
00
11
00
11
00
11
00
00
11
stress 11
00
11
00
σ 11
00
11
00
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001
11
0
0
1
0
1
0
1
P
a
stress
σ
Beam-columns carry axial compressive loads and bending moments. The presence of large
axial loads reduces the moment-carrying capacity of beam-columns.
partially
plastic
fully
plastic
M>My
M=Mp
Considering the fully-plastic section in the figure above the tensile stresses (positive) have a
resultant force T = σy ba and the compressive (negative) stresses have a resultant force C =
σy b(d − a). The net axial force must satisfy equilibrium along the beam-column C − T = P .
σy b(d − a) − σy ba = P
(d − a) − a = P/(σy b)
2a = d − P/(σy b)
1
a =
(d − P/(σy b))
2
!
1
P
a
=
1−
d
2
Py
(7)
where Py is the tensile yield force in the beam-column, Py = σy bd. The distance between the
centroids of the compressive and tensile stress blocks is 12 (d − y) + 12 y = 12 d. The moment in
the cross section is the combined moments of C and T about a pivot point about which the
moment of C equals the moment of T . Denoting the location of the pivot point as a distance
l from the centroid of the tensile stress block, the moment of T about the pivot equals the
moment of C about the pivot. That is, T and C form a couple about the pivot point.
C(d/2 − l)
σy b(d − a)(d/2 − l)
(d − a)(d/2 − l) = d2 /2 = da/2 − dl + al
d2 /2 − da/2 − dl
1
l =
(d − a)
2
Tl
σy bal
al
0
=
=
=
=
(8)
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6
CEE 201L – Uncertainty, Design, and Optimization – Duke University – Spring 2015 – H.P. Gavin
Now the ultimate moment can be found.
Mu = T l + C(d/2 − l) = 2T l
1
= 2 σy ba (d − a)
2
!
!!
P 1
d
P
d
1−
d−
1−
= 2 σy b
2
Py 2
2
Py
!
!
bd2
P
1 1P
= 2 σy
1−
1− +
4
Py
2 2 Py
!
!
2
P
P 1
bd
1−
1+
= 2 σy
4
Py
Py 2

Mu
P
= 1 −
Mp
Py
!2 
(9)

The largest (ultimate) moment that can be applied to a beam-column is reduced by the
presence of axial (compressive or tensile) load on the beam-column. If the axial load is zero,
the ultimate moment equals the plastic moment. However if the axial load is close to the
tensile yield load, the beam-column can not sustain any moment.
Mu / Mp
P=0; Mu= M p
1
P= Py ; Mu = 0
0
0
1
P / Py
Equation 9 is an example of a beam-column interaction equation. In practice these kinds of
interaction equations involve considerations for axial and lateral-torsional buckling. These
considerations are ignored in the present analysis.
In the plastic design of a beam column as shown on page 1, the maximum load Fu sustained
by the beam-column is limited by the ultimate moment, Mu .
CC BY-NC-ND H.P. Gavin
7
Plastic Design of a Beam-Column
Example - Solid Rectangular Section.
A steel beam-column is 3 m tall, has a solid rectangular cross-section, and carries an axial
gravity load of (10 ton)(9.81 m/s2 ). The yield stress of the material used is σy = 380, 000 kN/m2 .
Find the cross section dimension (b and d) so that the beam-column has a lateral stiffness
K = 100 kN/m and an ultimate force of Fu = 20 kN.
The two equations to use to find b and d are
d
K = Eb
H
!3
(10)
and
Fu = 2Mu /H
= (2/H)Mp (1 − (P/Py )2 )

2 bd2 
P
σy
1−
=
H
4
σy bd
!2 
(11)

Equations (10) and (11) can not be simplified much further. A numerical method (like
Newton’s method as implemented in Matlab’s fsolve) is required to solve this pair of
nonlinear algebraic equations for b and d.
1
2
3
function f = rectBC ( bd )
% e v a l u a t e t h e p a i r o f beam−column e q u a t i o n s f o r a s o l i d r e c t a n g u l a r s e c t i o n
global E Sy P H K Fu
4
5
b = bd (1); d = bd (2);
% w i d t h and d e p t h o f t h e beam−column s e c t i o n
6
7
8
1
2
f = [ 1 - ( E / K ) * b * ( d / H )ˆ3 ;
% K = 12 EI / Hˆ3
1 - (1/ Fu ) * (2/ H ) * Sy * ( b * d ˆ2 / 4) * ( 1 - ( P / ( Sy * b * d ) )ˆ2 ) ]; % Fu = 2 Mu / H
% rectBCsolve :
u s e f s o l v e t o s o l v e t h e beam column e q u a t i o n s f o r a r e c t s e c t i o n
global E Sy P H K Fu
3
4
% u n i t s : kN , m, ton , s e c
5
6
7
% Young ’ s modulus | y i e l d s t r e s s | a x i a l l o a d |
height | s t i f f n e s s | ult . force
E = 200 e6 ;
Sy = 380 e3 ;
P = 10*9.81; H = 3; K = 100;
Fu = 20;
8
9
10
11
% a v e r y good i n i t i a l g u e s s f o r b and d c o r r e s p o n d s t o P=0
d = 0.5* H ˆ2 * ( Sy / E ) * ( K / Fu )
% d e p t h o f column , m
b = ( K / E ) * ( H / d )ˆ3
% w i d t h o f column , m
12
13
bd = fsolve ( ’ rectBC ’ , [ b d ] )
14
15
16
17
P_over_Py = P / ( Sy * bd (1)* bd (2))
Mu_over_My = (1 - ( P_over_Py )ˆ2)
P_over_Pcr = P / ( ( pi ˆ2 * E * b * d ˆ3/12) / H ˆ2 )
The solution for this example is b = 0.173 m, d = 0.043 m, P/Py = 0.035, Mu /Mp = 0.999,
and P/Pcr = 0.349. Since P is very small compared to Py , the beam-column interaction
effects are insignificant, and the beam-column could be designed as a beam.
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8
CEE 201L – Uncertainty, Design, and Optimization – Duke University – Spring 2015 – H.P. Gavin
Example - Rectangular Tube Section.
In most cases a solid rectangular cross section is not very efficient. More commonly a beamcolumn has a rectangular closed section (a tube), or an I-shaped cross-section.
For a rectangular tube with depth d, width b, and tube-wall thickness t, the cross section
area, bending moment of inertia, plastic moment, and ultimate moment are
A = bd − (b − 2t)(d − 2t)
bd3 (b − 2t)(d − 2t)3
−
I =
12 h
12
i
2
Mp = σy (d − 2t) t + (d − t)bt

Mu = Mp 1 −
P
Aσy
(12)
(13)
(14)
!2 
(15)

So, as in the previous example, given the strength Fu and stiffness K, it is possible to find
the rectangular tube cross section, b and d by assuming a value for t, (e.g., t = d/20) and
solving the following two nonlinear equations for b and d using fsolve:
K−
1
2
3
12EI
=0
H3
and
Fu −
2Mu
=0
H
(16)
function f = tubeBC ( bd )
% e v a l u a t e t h e p a i r o f beam−column e q u a t i o n s f o r a t u b u l a r r e c t a n g u l a r s e c t i o n
global E Sy P H K Fu
A Py Mp Mu I
4
5
b = bd (1); d = bd (2);
t = d /20;
% width , depth , and
tube wall thickness
6
7
8
9
10
11
A
Py
Mp
Mu
I
=
=
=
=
=
b * d - (b -2* t )*( d -2* t );
Sy * A ;
Sy * (d -2* t )ˆ2* t + (d - t )* b * t ;
Mp * ( 1 - ( P / Py )ˆ2 );
(1/12)*( b * d ˆ3 - (b -2* t )*( d -2* t )ˆ3);
%
%
%
%
%
c r o s s s e c t i o n area
axial yield force
p l a s t i c moment
u l t i m a t e moment
b e n d i n g moment o f i n e r t i a
12
13
14
1
2
f = [ 1 - ( 12* E * I / H ˆ3 ) / K ;
1 - ( 2* Mu / H ) / Fu ];
% tubeBCsolve :
u s e f s o l v e t o s o l v e t h e beam column e q u a t i o n s f o r a t u b e s e c t i o n
global E Sy P H K Fu
A Py Mp Mu I
3
4
% u n i t s : kN , m, ton , s e c
5
6
7
% Young ’ s modulus | y i e l d s t r e s s | a x i a l l o a d |
height | s t i f f n e s s | ult . force
E = 200 e6 ;
Sy = 380 e3 ;
P = 10*9.81; H = 9.7; K = 100;
Fu = 10;
8
9
10
11
% i n i t i a l g u e s s f o r b and d c o r r e s p o n d s t o P=0 f o r r e c t . s e c t i o n
d = 0.5* H ˆ2 * ( Sy / E ) * ( K / Fu ); % d e p t h o f column , m
b = ( K / E ) * ( H / d )ˆ3;
% w i d t h o f column , m
12
13
bd = fsolve ( ’ tubeBC ’ , [ b d ] )
14
15
b = bd (1);
d = bd (2);
t = d /20;
16
17
18
19
P_over_Py = P / ( Sy * A )
Mu_over_Mp = (1 - ( P_over_Py )ˆ2)
P_over_Pcr = P / ( ( pi ˆ2* E * I ) / H ˆ2 )
CC BY-NC-ND H.P. Gavin