m.sc. steel -9 (plastic design of structures

Steel Structures

SE-505

Plastic Analysis and Design of Structures

M.Sc. Structural Engineering

Steel StructuresMechanism Method of Analysis

The objective is to find out a mechanism (independent or composite) such that M ≤Mp every where (or at lest at the selected points – answer may only be an upper bound and may not be the right answer.

1. Determine the location of all possible hinges and thus the number of these hinges (H). Hinges may form at concentrated load points, connections, and points of zero shear in case of members subjected to UDL.

2. Determine the total degree of indeterminacy of the structure (I).

3. The number of independent mechanisms (M) is calculated as under:

M = H − I

Steel Structures

Independent mechanisms include beam mechanism, panel or sway mechanism, gable mechanism and joint mechanism.

• Sketch each mechanism one by one and give convenient virtual deformation (displacement or rotation) to any point, assuming all the hinges of this mechanism to be just formed.

• Calculate the external and internal works done. External work done will be by the load multiplied with its corresponding deformation. Internal work will be done only at the plastic hinges and calculated as the plastic moment multiplied with the rotation.

External WD = Σ load × virtual displacement

Internal WD = ΣMoment at the hinge × virtual hinge rotation

Steel Structures

6. Using the principle of virtual work, evaluate the collapse load for the mechanism under consideration.

WE = WI

7. Solve all the independent mechanisms and then form composite mechanisms by combining two or more independent and other composite mechanisms.

The composite mechanisms must be formed in such a way that plastic hinges are eliminated reducing the internal work done and hence the collapse load.

8. Find the lowest collapse load.

9. Check to see that M ≤Mp at all sections, if possible.

Steel Structures

Mechanism Method of Analysis

Example

L/2

P P

L/2 L/2 L/2

A B

CD

E

Determine the collapse Load.

Steel Structures

Mechanism Method of AnalysisSolution:

Mech-1

P P

L/2 L/2 L/2

A B

CD

E

θA θB

θC=2θ

θL/2

θ = θA = θB

Mech-2

L/2

θA is the virtual displacement given by us other displacements are as as a result of θA

I = 1

No. of Possible Hinges, H = 3

No. of Independent Mechanisms, M = H – I = 2

Steel Structures

Mechanism Method of AnalysisSolution: (contd…)

2LPθWE =

PPI MθM2θθ0W ×+×+×=

PI M3θW ×=

EI WW =Virtual Work Principle

2LPθθ3MP =

L6MP P

C =Same for other mechanism - 2

Steel Structures

Mechanism Method of AnalysisExample:

Mech-1

P P

Mech-2

I = 2

No. of Possible Hinges, H = 5

No. of Independent Mechanisms, M = H – I = 3

2MP MP

P

2m 2m 2m 5m 5m

Mech-3

Steel Structures


3PθθP2θPWE =×+×=Beam Mechanism - 1

θ θ/2

3θ/2

2θ

2θM

23θ2Mθ2MW PPPI ×+×+×=

θ5.5MW PI =

EI WW =3Pθθ5.5MP =

P1.83MP =

Steel Structures


6Pθθ4P2θPWE =×+×=

Beam Mechanism - 2θ 2θ

3θ

4θ

2θM3θ2Mθ2MW PPPI ×+×+×=

θM01W PI =

EI WW =6Pθθ10MP =

PM67.1P =

Steel Structures


5Pθθ5PWE =×=

Beam Mechanism - 3θ θ

2θ

5θ

02θMθMW PPI +×+×=

θM3W PI =

EI WW =5Pθθ3MP =

PM6.0P =

Final answer is the smallest

PC M6.0P =

Steel Structures

Mechanism Method of AnalysisExample:

Mech-1

P

θ θ

2θ

4.5θ

Mech-2

I = 2

H = 5

M = 3

4P4P4.5m4.5m 6m 6m 6m 6m

θθ

2θ

6θ

θθ

2θ

6θ

Mech-3

12Mp14Mp 16Mp

Steel Structures


18Pθ4.5θP4WE =×=Beam Mechanism - 1

θ12M2θ2M10W PPI ×+×+=

θM63W PI =

EI WW =Pθ81θ36MP =

P2MP =

Steel Structures


Pθ426θP4WE =×=Beam Mechanism - 2

θ4M1θ2M41θ2M1W PPPI +×+=

θM54W PI =

EI WW =

Pθ24θ54MP =

P2.25MP =

Steel Structures


Pθ42WE =Beam Mechanism - 3

0θ2M61θM14W PPI +×+=

θM46W PI =

EI WW =

Pθ24θ46MP =

PM92.1P =

PC M92.1P =

Steel Structures

I for Rigid Frames

I for rigid frame = no. of closed loops x 3 – Releases at the bottom from fixed end – any real hinge in the super structure

( )70I

14355I=

−−××=

Real hinge

Steel Structures

Elimination of Hinges in Mechanism MethodCombined mechanisms are obtained in such a way that number of

plastic hinges are reduced.

There are two different ways in in which hinge can be eliminated:

1. If all members of the joint rotate through same angle in same direction in mechanisms to be added.

2. When a member rotates by equal amount but in opposite direction in the mechanisms to be combined.

Steel Structures

Joint MechanismJoint mechanism shows the rotation of a joint w.r.t the structure.

Now this joint can rotate

2MP

MPHinge will be in column as its MPis smaller

2MP 4MP

MP

Hinge may not form in this column although it is the weakest, as it can’t take moment

Steel Structures

Joint Mechanism (contd…)

Location of hinge depends on loading and end conditions of other members

2MP 4MPLet, MP = 100 kN-m

MA = 100 x 4 = 400 kN-m = 4MP

Hinge will form in member AB although its strength is greater than other members.

4m

100kN

AB

MP

Joint mechanism determines location of hinges at joint of more than two members.

Steel Structures

Joint Mechanism (contd…)

θ

θ

θ

θ

θ

θ

CW rotation of joint CCW rotation of joint

Steel Structures

Mechanism Method of AnalysisExample: Calculate the collapse load

3I =6H =

336M =−=2 Beam mechanism and 1 sway mechanism

Joint mechanism is not required as no joint is having more than 2 members.

P P

P 2m 2m 2m

4m

2MP

MP MP4m

Solution:

Steel Structures


Beam2

Beam1

PCWIWEShape of MechType of

Mech

Mech#

θ θ/2

1.5θ

2θ

3PθθP2θP

=×

+×

θ4.5Mθ0.5M

1.5θ2MθM

P

P

P

P

=+

×+×

P1.5M

θ 2θ

3θ

4θ

6Pθθ2P

4θP

=×

+×

θ9M2θM

3θ2MθM

P

P

P

P

=×+×+×

P1.5M

Steel Structures


Mech 1 +

Mech 3

4

Sway3


Mech

Mech#

4θP× θ4MP × P1.0M

7Pθ4Pθ3Pθ

=+

θM5.6θM-θM

-θ4Mθ4.5M

P

PP

P

P

=

×+

P0.93M

θ θθ

θ

P4θ P P

θθ

θ

4θθ/21.5θ

Steel Structures


Mech 2 +

Mech 3

5


Mech

Mech#

10Pθ4Pθ6Pθ

=+

θM11θM-θM

θ13M

P

PP

P

=

−

P1.10M

Right hinge at the joint can’t be eliminated because column and beam are rotating in opposite directions.

PC M93.0P =

Steel Structures

Mechanism Method of AnalysisExample: Calculate the collapse load

632I =×=12H =

6621M =−=2 Beam mechanism and 2 sway mechanism2 Joint mechanism

Consider 1 hinge at the junction of 2-members and equal to no. of members at other joints.

1.5P

PMP

1.7MP 4m

Solution:

0.8MP0.8MP

2MP

1.7MP

4m

2P

4m 4m

1.8P

Steel StructuresMechanism Method of AnalysisSolution: (contd…)

Beam2

Beam1


Mech

Mech#

6Pθ4θ1.5P

=×

θ3.6Mθ0.8M

2θMθ0.8M

P

P

P

P

=+

×+

P0.6M

Pθ2.74θ1.8P

=×

θ8MθM2

2θ2Mθ2M

P

P

P

P

=×+

×+×

P1.11M

2θ

θθ

4θ

2θ

θθ4θ


Sway of

bottom story

4

Sway of top story

3


Mech

Mech#

4θP× 4θ0.8MP ×P0.8M

12Pθθ42P

4θP

=×

+×

θ6.8Mθ41.7M

P

P

=×

P0.567M

θ

θ

θ θθ

4θ

θθθ

4θ


Right Joint Mech

6

Left Joint Mech

5


Mech

Mech#

0θ4.5M

θM7.1θ2Mθ0.8M

P

P

P

P

=

++

∞

0 θ4.5MP ∞

θ

Clockwise

θθ

θ

θ

θ

These joint mechanisms are fictitious & don’t occur independently, will occur in some combination


7+48

1 + 37


Mech

Mech#

10Pθ4Pθ6Pθ

=+

θ5.2MθM8.0

-θM8.0θ3.2Mθ3.6M

P

P

P

P

P

=

−+

We need not to combine mech – 1 & 4 as no hinge will be eliminated.

θθ

P.52M0

θθ

θθ

Pθ22Pθ2110Pθ

=+

θM12θ6.8Mθ5.2M

P

P

P

=+

P.545M0


9+610

8 + 59


Mech

Mech#

Pθ22022Pθ

=+

θM5.11θM7.12

-θM8.02θ4.5Mθ12M

P

P

P

P

P

=××

−+

P.523M0

θ

θ

θ

Pθ22 P.50M0

θ

θ

θ

θ θθ

θM11θM7.12

-θM8.02θ4.5Mθ11.5M

P

P

P

P

P

=××

−+


10 + 211


Mech

Mech#

Pθ2.92Pθ22

7.2Pθ=+

θM51θM22

θ11Mθ8M

P

P

P

P

=×

−+

P.514M0θ θ 2θ

2θ

2θ

PC 0.50MP =

Steel Structures

Gable Mechanism

Gable Mechanism

• One hinge forms at crown and two at the beam joints.

• Somewhat similar to sway mechanism.

• wind ward side column does not move.

• The hinge formed at the crown sinks the frame.

• Gable mechanism is separate/independent mechanism.

θ

If an angle θ is known we cannot determine other angles by some simple method. For this we need to understand Instantaneous Center of Rotation

Steel Structures

Instantaneous Center of RotationThat center about which any part structure rotates

Gable Mechanism

θ

A

B

CD

E

For ED center of rotation is E, for BC C.O.R is B. for CD we have to find out the C.O.R.

If a point of a segment moves perpendicular to line passing through this point then center of rotation lies on this line.

Steel Structures

Instantaneous Center of Rotation (contd…)

A

B B’ C

C’

D D’

E

L/2

F

L/2 L/2

L/2

Portion ABC has C.O.R at A

Portion DE has C.O.R at E

Portion CD

Point C is common with ABC. AC is perpendicular line to movement CC’. Point D is common with DE, center of rotation will lie on extension of DE.

Point F is ICOR for portion CD

Steel Structures

Instantaneous Center of Rotation (contd…)

L/2

F

L/2 L/2

L/2

θ

θ

θ

θ

2θθ

θ2θ

θL/2

2θθ θ θ

θ

Steel Structures

Example: 320kN 320kN

160kN10m

10m

15m 15m 15m 15m

MP Constant

Calculate MP

Steel Structures

Solution: 2P 2P

PL

L

1.5L

MP Constant

Let P = 160 kNL = 10 m

I = 1, H = 5, M = H – I = 4 2 Beam mech, 1 sway mech, 1 Gable mech.

1.5L 1.5L 1.5L

Steel StructuresSolution: (contd…)

Beam2

Beam1


Mech

Mech#

3PLθ1.5Lθ2P

=×

θ4MP

LM1.33

LM

34

P

P =θ θ2θ

Multiply rotation with Hz. Distance to

get vertical component of displacement

θθ

2θ

3PLθ1.5Lθ2P

=×

θ4MP

LM1.33

LM

34

P

P =


GableMech

4

Sway or

Panel

3


Mech

Mech#

PLθ θ2MPL

M2 P

3PθP

1.5L2θ2P

1.5L2θ2P

=

×+

×

++ 1.5θθ

2θMP

LMP

θθθθ

θL

θ/2 θθ/2

θ


2LLθ

2θ=

θ/2 θ

θ/2

Lθ

4- Gable Mechanism

θ

θ23

L

L

L


1+36

1+45


Mech

Mech#

( )6PLθ

PLθ33=+ ( )

θM5θM2-34

P

P

=+

LM833.0 P

PLθ4( )

θM4θM2-24

P

P

=+

LM0.1 P

θ

1.5θ 1.5θ

θ θ


3+(2/3)57


Mech

Mech#

5PLθ

6PLθ32

PLθ

=

×+

( )

θM3

10θM-θM-

θ5M32θ2M

P

PP

PP

=

+

LM

32 P

θ θ32θ +

LM

32P P

C = 1023160MP ××=

m2400kNMP −=

Steel StructuresInfluence of Axial Force On Plastic MomentCapacity

• In columns axial force is present simultaneous with moments.

• In beams of frames, smaller axial force in addition to moment may be present.

• Stresses are consumed by axial force hence plastic moment capacity reduces.

Steel StructuresInfluence of Axial Force On Plastic Moment Capacity

+(-)

(-)

(+)

=

(-)

(+)

(-)

(+)

Fy

(-)

(+)

Fy

Fy

(-)

(+)

Fy

FyDue to axial compression, P

Due to sagging moment

P/A

N.A

ElasticPartially Inelastic

Inelastic


(-)

(+)

Fy

Fy

(-)

(+)

= +(-)

Fy

Fy

yo

yo

Contribution of moment. Upper area is equal to lower area

Contribution of Axial force

Fy

(b)(c) (d)

N.A

(-)

(+)

Fy

(e)

(-)(+)

Fy

Steel StructuresInfluence of Axial Force On Plastic MomentCapacity (contd…)

Case-I N.A Is Within The WebApplicable for smaller values of axial load, P.

From diagram “d”

yow F2ytP ××=

ywo F2t

Py =

If22t-dy f

o ≤ Then N.A. is with in Web

Steel StructuresInfluence of Axial Force On Plastic MomentCapacityCase-I N.A Is Within The Web (contd…)

Reduced moment capacity, MPC

FY x Area of diagram on top or bottom

= x Distance between centers of comp. & tension area.

Dia: (e)

MPC = FY (Z for total section – Z for central portion of 2yo height)

−×=

4)(2ytZFM

2ow

xyPC

( )2owxy ytZF −×=

2owyP ytF-M=

MP is full plastic moment capacity in the absence of Axial force


2y

2w

2

wyPPC F4tPtF-MM =

yw

2

P F4tP-M=

( )yyw

2

P

PC

FzF4tP-1

MM

×=

( ) 2

2

yyw

2

AA

FzF4tP-1 ×

×=


2

yw

2

P

PC

PP

z4tA-1

MM

×= yy FAP ×=Q

Py = Max. axial compressive load in the absence of bending moment& bucklingIf P = 0, P/Py = 0 then MPC/MP = 1, MPC = MP

(1)

Eq: 1 is applicable when 22tdy f

o−

≤

OR owy .2y.tFP ≤ ( )fwy 2t-d.tF=

Steel StructuresInfluence of Axial Force On Plastic MomentCapacity (contd…)

Case-II N.A Is Within The FlangeApplicable when P is high and M is less.

(-)

Fy

(+)

= +

(-)

Fy

yo

Fy

(d) (e)

Fy

yo

(-)

(+) N.A

Steel StructuresInfluence of Axial Force On Plastic MomentCapacityCase-II N.A Is Within The Flange (contd…)

Lever arm in the diagram 2

2yd2y oo

−+=

2dyo +=

Let “A” be the total area of section

( )[ ]ofy 2ydbAFP −−=

[ ]fofy b2ydbAF −−=

−+= Adb

FP

2b1y f

yfo

2d

2bA

F2bPy

fyfo +−=

−−=

yfo P

P12bA

2dy

If yo = d/2, 1-P/Py=0 so P/Py = 1


−=

4)(2yb

4dbFM

2of

2f

yPC

−−−=

2

yff

2f

y PP1

2bA

2db

4dbF

ZZ

PP1

2dA

PP1

4bAF

y

2

yf

2

y ×

−

×+

−−=


2

yf

2

yP

PC

PP1

Z4bA

PP1

Z2dA

MM

−+

−

××

=

This eq is valid for: fo t2dy −> ( )fwy 2td.tFP −>and

−+

−=

2

yfyP

PC

PP1

2bA

PP1d

2ZA

MM

(2)


1MM

98

PP

P

PC

y

=+yP

P

2.0PP

y

=

0.1

P

PC

MM 0.1

Actual Curve, Eq:2

2.0PP

y

≥for

Actual Curve, Eq:1

1MM

2PP

P

PC

y

=+ 2.0PP

y

<for

Steel StructuresInfluence of Axial Force On Plastic MomentCapacityActual curve is difficult to use so specifications allow to approximate the original equations by approximate equations.

For P/Py = 0.2 draw Hz. Line. Where this line cuts the original curve join that point with 1.0 value point on both axis. Then equation can be made for the straight lines.

AISC Interaction Equations

1.0MM

MM

98

PP

cy

ry

cx

rx

c

r ≤

++ For 2.0

PP

c

r ≥


1.0MM

MM

2PP

cy

uy

cx

rx

c

r ≤++ For 2.0PP

c

r <

Pr = required factored axial compressive strength (LRFD)= required allowable axial compressive strength (ASD)

Pc = design axial compressive strength (LRFD) = φcPn= allowable axial compressive strength (ASD) = Pn / Ω

Mr = required factored flexural strength (LRFD)= required allowable flexural strength (ASD)

Steel Structures

Mt = Required nominal flexural strength for no axial load

= MPC/φb

Suppose 2.0P

P

nc

u >φ

1.0M

M98

PP

Pb

PC

nc

u =+φφ

Mc = design flexural strength (LRFD) = φbMn= allowable flexural strength (ASD) = Mn / Ω

φc = resistance factor for compression = 0.9φb = resistance factor for flexure = 0.9

Ωc = safety factor for compression = 1.67Ωb = safety factor for flexure = 1.67

Steel StructuresInfluence of Axial Force On Plastic MomentCapacityAfter simplification

+= 0.889

PPMM

nc

utP φ

+= 0.889

PPZZ

nc

utx φ

For 2.0P

P

nc

u ≤φ

+= 00.1

PP5.0ZZ

nc

utx φ

Approximate interaction equations to reduce the trials

Steel StructuresDesign Procedure For Column (beam column)

1. Select section without considering the axial force.

2. Calculate φcFcr & φ cPn for the trial section. Also calculatePu/ φ cPn and decide which interaction equation is applicable.

3. Find out increased value of Z. Because due to axial force there is reduction in plastic moment capacity so we need to calculate the reduction.

4. Select section for this plastic section modulus.

5. Apply interaction equations. If L.H.S < 1.01, section is O.K. If too much less than 1.0 , the section is uneconomical.

( )y

reqPt 0.9F

MZ =

Steel StructuresExample:

Design a beam column having required plastic moment of 300 kN-m and an axial force (factored) of 500 kN. A-36 steel is to be used. Fy = 250 MPa. Effective length, KL = 2.0m

Solution:

( )2509.010300

0.9FM

Z6

y

reqPt ×

×==

33 mm101333×=From beam selection table

W 530 x 66

LP = 1.60m < Lu = 2.0 m

W 360 x 79

LP = 2.44m > Lu = 2.0 m In this way bracing is not required. In beams we can provide bracing but in case of column it may be difficult .

Trial Section W 360 x 79

A = 10,100 mm2 rx = 150mm, ry = 48.8mm

KLu/rmin = 2000/48.8 = 41 < 200 O.K.[Not better as a column]

Steel StructuresSolution:

MPa82.052Fcrc =φ

New value of “Z”

kN07921000

10,10082.052Pnc ≅×

=φ

0.224.02079500

PP

nc

u >==φ

+= 889.0

PPZZ

nc

ut φ

In this equation we can use the original value of Z.


( )889.024.0103331Z 3 +×=

Now select the section according to this new value of Z

W 360 x 91, A = 11,500mm2, Z = 1671 x 103 mm3, φbMP = 376 kN-mrx = 152 mm, ry = 62.2 mm, Lp = 3.11 m

LP > Lu. Check local stability yourself.

33 mm101506×=

33rKL

min

u = 212.38MPaFcrc =⇒φ


Final Selection: W 360 x 91

2442kN1000

11500212.38Pnc =×

=φ

0.20.2052442500

PP

nc

u >==φ

01.1914.0M

M98

PP

nxb

ux

nc

u <=+φφ

TYPES OF COLUMNS DEPENDING ON BUCKLING BEHAVIOURElastic Critical Buckling Stress

The elastic critical buckling stress is defined as under:Fe = Elastic critical buckling (Euler) stress

= 2

2

rKL

Eπ

The critical slenderness ratio dividing the expected elastic and the inelastic buckling is denoted by Rcand is given below:

COLUMN STRENGTH FORMULAS

The design compressive strength (φc Pn) and the allowable compressive strength (Pn / Ωc) of compression members, whose elements do not exhibit elastic local instability (only compact and non-compact sections), are given below:

φc = 0.90 (LRFD) : Pn = Fcr AgΩc = 0.90 (ASD) : Pn = Fcr Ag

Fcr = critical or ultimate compressive strength based on the limit state of flexural buckling determined as under:

Elastic Buckling

When KL / r > Rc or Fe < 0.44Fy

Fcr = 0.877 Fe (AISC Formula E3-2)

where Fe is the Euler’s buckling stress and 0.877 is a factor to estimate the effect of out-of-straightness of about 1/1500.

Inelastic Buckling and No BucklingWhen KL / r ≤ Rc or Fe > 0.44Fy

Fcr = Fy (AISC Formula E3-3)

e

y

FF

658.0

Element

Width-Thickness

Ratio

λr λr For A36 Steel

Unstiffened1. Flanges of I-shaped sections in pure compression, plates projecting from compression elements, outstanding legs of pairs of angles in continuous contact, and flanges of channels in pure compression.

15.9

2. Legs of single angle struts, legs of double angle struts with separators and other un-stiffened elements supported along one edge.

12.8

3. Stems of tees. 21.3

4. Flanges of built-up I-sections with projecting plates or angles.

tb

tb

td

tb

yFE56.0

yFE45.0

yFE75.0

y

c

FEk64.0

ck1.18

Element

Width-Thickness

Ratio

λr λr For A36 Steel

Stiffened1. Flanges of rectangular hollow sections of uniform thickness used for uniform compression.

39.7

2. Flexure in webs of doubly symmetric I-shaped sections and channels.

161.8

3. Uniform compression in webs of doubly symmetric I-shaped sections and uniform compression in all other stiffened elements.

42.3

4. Circular hollow sections in axial compression.D / t 0.11 (E / Fy) 88.6

bt

wth

bt

yFE40.1

yFE70.5

yFE49.1

Beams:Lp = Limiting laterally unbraced length for full plastic bending capacity (Mp = Zx×Fy) in uniform moment case (Cb = 1.0).

For I-shaped members including hybrid sections and channels:

Lp = y

y FEr76.1 ≅ 50 ry (for A-36 Steel)

A section may develop yielding only at some points in case of inelastic buckling, when the unbraced length is between the two limiting lengths Lp and Lr, that is, when,

Lp < Lb ≤ Lr

where Lr = limiting laterally unbraced length for inelastic torsional buckling, mm.

For doubly symmetric I-shaped members:

Lr = 27.0

76.6117.0

95.1

++

cJhS

EF

hScJ

FEr oxy

oxyts

≈y

ts FEr7.0

95.1 (very conservative estimate)

Mr = limiting buckling moment dividing elastic and inelastic buckling for Cb = 1.0, Cbwill be defined later

= 0.7Fy Sx/106 kN-m2

tsrx

wy

SCI

x

oy

ShI

2= for doubly symmetric I-sections=

rts ≈ radius of gyration of the compression flange plus one-sixth of the web for doubly symmetric I-sections

+

ff

w

f

tbth

b

61112

rts =

c = 1.0 for a doubly symmetric I-shape

ho = distance between the flange centroids= d – tf / 2

Cw = warping torsional constant for the section, mm6

For symmetrical sections, Cw = 42

22oyof hIhI

≈

If = moment of inertia of one flange in lateral direction, mm4

J = torsional constant for the section, mm4

≈ ∑ 3bt31

For the above expression, b is the long dimension and t is the short dimension of any rectangular element of the section and summation is for all the elements of that section.

When Lb > Lr Mn = FcrSx ≤ Mp

Fcr = compression flange critical buckling stress

Fcr =

+

ts

b

ox

ts

b

b

rL

hScJ

rL

EC078.012

2π

≈( )2

2

/ tsb

b

rLEC π

λp for compact section

1. Unstiffened flanges of I-shaped rolled beams, channels, tees and built-up doubly and single symmetric I-sections.

λp = = 10.8 for A-36 steelyFE /38.0

2. Unstiffened legs of single angles.

yFE /54.0λp = = 15.3 for A-36 steel

3. Stiffened flanges of HSS shapes

yFE /12.1λp = = 31.8 for A-36 steel

Web local buckling criterionWeb is locally stable when the following condition is satisfied:

λ ≤ λp

where λ = wth

and assumed web depth for stability (h) is defined as under:h = twice the distance from the neutral

axis to the inside face of the compression flange less the fillet or corner radius for rolled sections

h = twice the distance from the neutral axis to the nearest line of fasteners at the compression flange or the inside face of the compression flange when welds are used.

λp for compact sectionFor webs of doubly symmetric I-sections and channels:

λp = = 107 for A-36 steelyFE /76.3For webs of rectangular HSS (λ = h / t):

λp = = 68.7 for A-36 steelyFE /42.2

FLEXURAL STRENGTH OF BEAMS

For a safe beam, the applied moment (service moment Ma in ASD and factored moment Mu in LRFD) must be lesser than or equal to the design strength of the beam.

Mu ≤ φbMn φb = 0.90 (LRFD)

Ma ≤ Mn / Ωb = 1.67 (ASD)

where Mn = nominal flexural strength as determined by the limit state of yielding, lateral torsional buckling, or local buckling.

Lm = Lp +( )( )

( )rpb

prppb

MMCLLMMC

−

−−

= Lp +( )

BFCMMC

b

ppb

×

−

= Lp + ≤ Lr

−

b

bp

CC

BFM 1

BF = slope of moment capacity versus unbraced length for inelastic lateral torsional buckling.

=pr

rp

LLMM

−

−

When Cb = 1.0, Lm = Lp

Design moment capacity (Mn) is determined for various cases of unbraced lengths as follows:

Case I: Compact Sections, Cb ≥ 1.0, Lb ≤ Lm

Mn = Mp = Z Fy / 106 (kN – m)

Case II: Compact Sections, Cb ≥ 1.0,

Lm < Lb ≤ Lr

Mn = ≤ Mp (kN – m)

−

−−−

pr

pbxyppb LL

LLSFMMC )7.0(

Mn = Cb [Mp – BF(Lb – Lp)] ≤ Mp (kN – m)Case III: Compact Sections, Cb ≥1.0,

Lb > Lr

For doubly symmetric I-shaped and channel section members:

Mn = CbFcrSx ≤ Mp

where Fcr = 2

2

2

078.01

+

ts

b

ox

ts

b

b

rL

hScJ

rL

EC π

The variables rts and others are as defined earlier.

The design shear strength of webs is φν Vnwith φν = 0.90 (LRFD) and the allowable shear strength is Vn / Ωv with Ωv = 1.67 (ASD).

For webs of rolled I-shaped members:

Vn = 0.6FyAwCv

wth

ywFE /Web Yielding:

For ≤ 2.24

(= 63.4 for A36 steel) Cv = 1.0

Concluded

m.sc. steel -9 (plastic design of structures

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