Plastic Design of a Fixed-Fixed Beam-Columnpeople.duke.edu/~hpgavin/cee201/plastic-design.pdf ·...

Plastic Design of a Fixed-Fixed Beam-ColumnCEE 201L. Uncertainty, Design, and Optimization

Department of Civil and Environmental EngineeringDuke University

Henri P. GavinSpring, 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 distributionin which the stress equals the yield stress only at the outer-most fibers. In a symmetric crosssection of depth d, My = σyI/(d/2). If the cross section has a solid rectangular shape (dimen-sions d× b) then I = bd3/12 and My = σy(bd2/6). The section can carry additional momentbeyond the yield stress. The plastic moment, Mp is the bending moment corresponding toa bending stress distribution in which the stress equals the yield stress almost everywherewithin the cross section. For a solid rectangular cross section, Mp = σy(bd2/4).

1 Beam (with no axial load)For rectangular cross sections Mp = (3/2)My. For I-shaped cross sections Mp ≈ 1.12My.

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+σ y +σ y+σ y

−σy

−σy

−σy

stress

strain

ε

σ

ε= Eσ

M=MpM>My

strain

str

ess

σ

y

E

ε

M

d

d

M = EI φ

elastic

y (d/2)y

M=M = EI φ y

φ = ε /

yield point partiallyplastic

fullyplastic

y

σ

−ε

ε ∼0.002y

−ε y−ε yy

The top of a fixed-fixed beam of height H is displaced with a displacement D until failure:

��

��

��

��

M(x)

φ

x

H

F

D

M

o

o

−M

F

EI K

o

φy

y

p

y

y

p

M

M F

F

D = 1.5 yDD p

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:∑

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 = FH/2. Thisrelationship is simply a statement of equilibrium and is valid for any level of elastic or plasticdeformation.

Given the end-forces and end moments, the moments along the beam M(x) can be foundfrom a simple free-body diagram, M(x) = Mo + Fx. 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 yieldmoment, M(x) = EIφ(x), or M(x)/My = φ(x)/φy. For moments larger than My, morecomplicated 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′′(x),the slope, v′(x) and the deflection, v(x), can be found by integrating φ(x) and finding theconstants of integration from the boundary conditions.

In this way, the deflections along a fixed-fixed beam can be computed as the beam is deformedup to, and beyond its yield limit. An example of the results of this process are illustrated inFigures 1 and 2. In this example, the beam has a height H = 10 cm, and a solid rectangularcross section b = 3.0 cm, and d = 0.203 cm. The yield stress is σy = 35, 000 N/cm2. Fromthese 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 becomefully plastic. The center figure shows the discontinuity of the slope at the ends, and the topplot 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 befound from the stiffness, K, and the ultimate plastic force, Fp.

The stiffness relating the force F and collocated displacement D of the fixed-fixed beamshown above is

K = 12EIH3 = Eb

(d

H

)3

= Fy

Dy= Fp

Dp, (1)

From the yield analysis and plastic analysis of a rectangular cross section:

My = bd2

6 σy , Mp = bd2

4 σy . (2)

Applying the equilibrium equation above, F = 2Mo/H,

Fy = bd2

3H σy , Fp = bd2

2H σy . (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),

b = Fp

Dp

1E

(H

d

)3= K

E

(H

d

)3, (4)

CC BY-NC-ND H.P. Gavin

http://creativecommons.org/licenses/by-nc-nd/3.0/

Plastic Design of a Beam-Column 3

Inserting this into the expression for Fp, equation (3)

d2 = 2Fp

σy

H

b= 2Fp

σyHE

Dp

Fp

(d

H

)3

,

or,d = 1

2σy

E

H2

Dp= 1

2H2σy

E

K

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 bucklingload,

mg <π2EI

H2 , (6)

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 steelis 20× 106 N/cm2.

Find:the width and depth (b and d) of the rectangular beam cross-section.

d = 12σy

E

H2

Dp= 1

2 ·35000

20× 106 ·102

0.43 = 0.203 cm

b = 8E2FpD

2p

H3σ3y

= 8 · (20× 106)2 · (216.3) · (0.43)2

10.03 · (35000)3 = 3.0 cm




0

50

100

150

200

0 0.2 0.4 0.6 0.8 1

forc

e, N

deflection, cm

K=501.9 N/cm

Dy= 0.3 cm; Fy = 144.2 N

Fp=216.3 N

L= 10.0 cm

0

200

400

600

800

1000

1200

0 0.02 0.04 0.06 0.08 0.1

mom

ent,

N.c

m

curvature, 1/cm

φy=0.0172; My= 721 N.cm

Mp=1082 N.cm

b= 3.0 cm; d=0.203 cmEI=41827.1 N.cm

2

Figure 1. Moment-curvature (M − φ) for a rectangular section and force-deflection (F −D)for a fixed-fixed beam

0

2

4

6

8

10

0 0.2 0.4 0.6 0.8 1

deflection, cm

0

2

4

6

8

10

0 0.05 0.1

slope

0

2

4

6

8

10

-0.04 0 0.04

x, c

m

curvature, 1/cm

Figure 2. Distributed curvature, slope, and deflection along a fixed-fixed beam. Note that theslope is discontinuous and the curvature becomes infinite at the ends, the locations of plastichinges.




2 Beam-column (with axial compressive load), P

Beam-columns carry axial compressive loads and bending moments. The presence of largeaxial loads reduces the moment-carrying capacity of beam-columns.

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+σ y

−σy

−σy

−σy

stress

strain

ε

σ

ε= Eσ

M=MpM>My

strain

str

es

sσ

y

E

εd

d

M = EI φ

elastic

y (d/2)y

M=M = EI φ y

φ = ε /

yield point partiallyplastic

fullyplastic

y

σ

−ε

ε ∼0.002y

−ε y−ε

M

d−

aa

d/2

mm

P

l

yy

Considering the fully-plastic section in the figure above the tensile stresses (positive) have aresultant force T = σyba and the compressive (negative) stresses have a resultant force C =σyb(d− a). The net axial force must satisfy equilibrium along the beam-column C − T = P .

σyb(d− a)− σyba = P

(d− a)− a = P/(σyb)2a = d− P/(σyb)

a = 12(d− P/(σyb))

a

d= 1

2

(1− P

Py

)(7)

where Py is the tensile yield force in the beam-column, Py = σybd. The distance between thecentroids of the compressive and tensile stress blocks is 1

2(d− y) + 12y = 1

2d. The moment inthe cross section is the combined moments of C and T about a pivot point about which themoment of C equals the moment of T . Denoting the location of the pivot point as a distancel from the centroid of the tensile stress block, the moment of T about the pivot equals themoment of C about the pivot. That is, T and C form a couple about the pivot point.

T l = C(d/2− l)σybal = σyb(d− a)(d/2− l)

al = (d− a)(d/2− l) = d2/2 = da/2− dl + al

0 = d2/2− da/2− dl

l = 12(d− a) (8)




Now the ultimate moment can be found.

Mu = T l + C(d/2− l) = 2T l

= 2 σy ba12(d− a)

= 2 σy bd

2

(1− P

Py

)12

(d− d

2

(1− P

Py

))

= 2 σybd2

4

(1− P

Py

)(1− 1

2 + 12P

Py

)

= 2 σybd2

4

(1− P

Py

)(1 + P

Py

)12

Mu

Mp=

1−(P

Py

)2 (9)

The largest (ultimate) moment that can be applied to a beam-column is reduced by thepresence 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 thetensile yield load, the beam-column can not sustain any moment.

P / P

M / Mu

1

100

y

p P=0; M = Mu

y

P= P ; M = 0u

p

Equation 9 is an example of a beam-column interaction equation. In practice these kinds ofinteraction equations involve considerations for axial and lateral-torsional buckling. Theseconsiderations are ignored in the present analysis.

In the plastic design of a beam column as shown on page 1, the maximum load Fu sustainedby the beam-column is limited by the ultimate moment, Mu.




Example - Solid Rectangular Section.

A steel beam-column is 3 m tall, has a solid rectangular cross-section, and carries an axialgravity 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 stiffnessK = 100 kN/m and an ultimate force of Fu = 20 kN.

The two equations to use to find b and d are

K = Eb

(d

H

)3

(10)

and

Fu = 2Mu/H

= (2/H)Mp(1− (P/Py)2)

= 2Hσybd2

4

1−(

P

σybd

)2 (11)

Equations (10) and (11) can not be simplified much further. A numerical method (likeNewton’s method as implemented in Matlab’s fsolve) is required to solve this pair ofnonlinear algebraic equations for b and d.

1 function f = rectBC (bd)2 % e v a l u a t e the pa i r o f beam−column equat ions 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 n3 global E Sy P H K Fu45 b = bd (1); d = bd (2); % width and depth o f the beam−column s e c t i o n67 f = [ 1 - (E/K) * b * (d/H)ˆ3 ; % K = 12 EI / Hˆ38 1 - (1/ Fu) * (2/H) * Sy * (b*dˆ2 / 4) * ( 1 - ( P / (Sy*b*d) )ˆ2 ) ]; % Fu = 2 Mu / H

1 % rectBCsolve : use f s o l v e to s o l v e the beam column equat ions f o r a r e c t s e c t i o n2 global E Sy P H K Fu34 % u n i t s : kN , m, ton , sec56 % Young ’ s modulus | y i e l d s t r e s s | a x i a l load | h e i g h t | s t i f f n e s s | u l t . f o r c e7 E = 200 e6; Sy = 380 e3; P = 10*9.81; H = 3; K = 100; Fu = 20;89 % a very good i n i t i a l guess f o r b and d corresponds to P=0

10 d = 0.5*Hˆ2 * (Sy/E) * (K/Fu) % depth o f column , m11 b = (K/E) * (H/d)ˆ3 % width o f column , m1213 bd = fsolve ( ’rectBC ’, [ b d ] )1415 P_over_Py = P / (Sy*bd (1)* bd (2))16 Mu_over_My = (1 - ( P_over_Py )ˆ2)17 P_over_Pcr = P / ( (pi ˆ2 * E * b*d ˆ3/12) / Hˆ2 )

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 interactioneffects are insignificant, and the beam-column could be designed as a beam.




Example - Rectangular Tube Section.

In most cases a solid rectangular cross section is not very efficient. More commonly a beam-column 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 sectionarea, bending moment of inertia, plastic moment, and ultimate moment are

A = bd− (b− 2t)(d− 2t) (12)

I = bd3

12 −(b− 2t)(d− 2t)3

12 (13)

Mp = σy[(d− 2t)2t+ (d− t)bt

](14)

Mu = Mp

1−(P

Aσy

)2 (15)

So, as in the previous example, given the strength Fu and stiffness K, it is possible to findthe rectangular tube cross section, b and d by assuming a value for t, (e.g., t = d/20) andsolving the following two nonlinear equations for b and d using fsolve:

K − 12EIH3 = 0 and Fu −

2Mu

H= 0 (16)

1 function f = tubeBC (bd)2 % e v a l u a t e the pa i r o f beam−column equat ions 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 n3 global E Sy P H K Fu A Py Mp Mu I45 b = bd (1); d = bd (2); t = d/20; % width , depth , and tube w a l l t h i c k n e s s67 A = b*d - (b -2*t)*(d -2*t); % cross s e c t i o n area8 Py = Sy*A; % a x i a l y i e l d f o r c e9 Mp = Sy * (d -2*t)ˆ2*t + (d-t)*b*t; % p l a s t i c moment

10 Mu = Mp * ( 1 - (P/Py )ˆ2 ); % u l t i m a t e moment11 I = (1/12)*( b*dˆ3 - (b -2*t)*(d -2*t )ˆ3); % bending moment o f i n e r t i a1213 f = [ 1 - ( 12*E*I/Hˆ3 ) / K ;14 1 - ( 2* Mu/H ) / Fu ];

1 % tubeBCsolve : use f s o l v e to s o l v e the beam column equat ions f o r a tube s e c t i o n2 global E Sy P H K Fu A Py Mp Mu I34 % u n i t s : kN , m, ton , sec56 % Young ’ s modulus | y i e l d s t r e s s | a x i a l load | h e i g h t | s t i f f n e s s | u l t . f o r c e7 E = 200 e6; Sy = 380 e3; P = 10*9.81; H = 9.7; K = 100; Fu = 10;89 % i n i t i a l guess f o r b and d corresponds to P=0 f o r r e c t . s e c t i o n

10 d = 0.5*Hˆ2 * (Sy/E) * (K/Fu ); % depth o f column , m11 b = (K/E) * (H/d)ˆ3; % width o f column , m1213 bd = fsolve ( ’tubeBC ’, [ b d ] )1415 b = bd (1); d = bd (2); t = d/20;1617 P_over_Py = P / (Sy*A)18 Mu_over_Mp = (1 - ( P_over_Py )ˆ2)19 P_over_Pcr = P / ( (pi ˆ2*E*I) / Hˆ2 )



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