m sc steel-04-prof-zahid_siddiq

Steel Structures

SE-505

Plastic Analysis and Design of Structures

M.Sc. Structural Engineering

Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

Steel StructuresEffect of Shear Force on Moment CapacityWe can ignore the effect of shear force on plastic moment capacity as it is very less in magnitude. AISC ALSO SUPPORTS THIS IDEA. British code considers this little effect.

Shear stresses are maximum near N.A. and flexural stresses are maximum near periphery.

Vn = Nominal shear capacity of web

wywn A0.6FV = 0.9v =φ

nvu VV φ≤


Steel StructuresEffect of Shear Force on Moment Capacity (contd…)

If Vu ≥ φvVn , to compensate the capacity following are the options:

1. Select larger section. (may be uneconomical)

2. Provide web doublers.Web doublers is extra plate attached with web. There is no need to provide intermediate stiffener unlike plate girder. In case of plate girder the web is non-compact and the intermediate stiffener provides post buckling shear strength by developing tension field action . Here we have compact section so the web has already got enough strength.

3. Provide diagonal stiffener to take some portion of shear strength. It is totally different from intermediate stiffener of plate girder.


Steel StructuresExample: Design the member assuming that significant bracing is available to prevent LTB. Rotation capacity up-to collapse is available.

Solution:

3m 1m

600 kN

AB C

θ/3 θ

4θ/3θ

θPWE ×=

θM37θθ

34MW PPI =

+=

θM37θP P=×

60073P

73MP ×=×=

m257kNMP −=


Steel StructuresSolution: (contd…)For the selection of section

reqt ZZ ≥

22L

5500LF

d ymin =

×= For A36 Steel

Should be minimum weight section

336

req mm1011422500.910257Z ×=

××

=

182mm22

4000dmin ==

From the beam selection table, Trial Section W 410 x 60


Steel StructuresSolution: (contd…)check the conditions for compact section:

web is continuously connected to flange

10.8λ6.92tb

Pf

f <=

107λ.646th

Pw

=<=

bP LL < No LTB

AB

85.75 kN

257 kN-m

514 kNShear Check

kNV maxu 514=

vC××= wyvnv td0.6FV φφ

maxnv 42310000.17.70742500.69.0V uVkN <=×××××=φ

C

Portion AB is safe but BC needs to be strengthened.


Steel StructuresSolution: (contd…)

51410004072506090 =×××× wt..

1. Provide web doubler

mm.tw 359=Extra thickness required = 9.35 – 7.70 = 1.65 mmProvide 5mm thick (based on corrosion resistance) web doubler in portion BC of length 1m and height equal to ‘d’ of section. Connect at regular intervals with weld or bolts. Spacing of weld or bolt can be calculated by usual design procedure.

2. Use of diagonal stiffener

Shear to be resisted by diagonal stiffener = 514 – 423 91 kN



( ) ( ) mm10703821000 22 =+Length of diagonal stiffener =

θ407 – 2 x 12.8 = 381.4

B C

Diagonal Stiffener

Bearing Stiffener

θ = 20.8791 kN

Fst

Force resisted by diagonal stiffener, Fst = 91/Sinθ = 254.90 kN (tensile)

211332509010009254 mm

..Ast =××

= For pair of stiffener on both sides



stst A.b 97= For b/t = 15.8

mm.bst 95113397 ≅×=

95 mm should not extend out of the section

stb

mmmm.tbb wfst 9585

277178

2<=

−=

−≤

So we use bst = 85 mm

mm.tst 76852

1133=

×= Use 7 mm Thick Plate

If we use rivets or bolts angle section is better and for weld rectangular plate is better.


Steel StructuresAdditional Conditions for Plastic Analysis

Study Appendix 1 AISC-2005

INELASTIC ANALYSIS AND DESIGN


Steel StructuresWAYS OF GETTING PLASTIC ANALYSIS RESULTS1. Mp – values of various sections are given and

the loads in terms of Pc are calculated.

2. Loads are given in terms of P and the required Mp – values are required to be found out.

3. Both loads P and Mp – values are given and the overload factor λ is to be calculated.

PPc=λ


Steel StructuresShakedown Phenomenon It occurs under variable repeated loading.

• Magnitude of load is changing

• The cycles of loads are repeated at intervals

Shakedown StateWhen repeated loads are applied on a structure with maximum magnitude greater then those corresponding to first yield moment(My) but all changes after first few cycles of loading take place elastically, the structure is said to be in shake-down state.

Shakedown state is attained only for a particular set of variable loads with well defined maximum and minimum values.


Steel StructuresThis phenomenon takes place due to development of residual stresses and moments each time the load exceeds My and reaches Mp.

This phenomenon takes place due to development of residual stresses and moments each time the load exceeds My and reaches Mp.

The final failure may either occur by alternating yield in tension and compression and fatigue reduces the strength after a large number of cycles, or, by accumulation of deformations due to yields in the same sense during different loading cycles.


Steel StructuresShakedown Phenomenon (contd…)

Shakedown LimitsIf the maximum intensity of each load or a combination of loads that can act on the structure is characterized by a single load factor λ, then the maximum value of λ at which shakedown can take place is called the Shakedown Limit.

Failure Conditions Related With Shakedown

1. Alternating Yield FailureCollapse due to fatigue

2. Incremental Collapse Mechanism,Rotations and deformations occur in same direction and accumulate


Steel StructuresShakedown Phenomenon (contd…)

+MP

+My

-My

-MP

2My

2My

A

B

C

Curvature

Applied Moment

Loading and unloading between B & C will never cause failure as structure will be elastic within this range.


Steel StructuresIf the load varies between any two points on the curve, not more than 2My ordinate apart, all changes will take place elastically.

The structure will adjust itself by modifying the residual stresses and will attain shakedown state.

The elastic capacity in a particular direction / sense is increased due to the development of residual moments at zero applied moment level in the opposite sense.


Steel StructuresExample:

L

A B

θ θ2θ

θL/2

diagram of AreawW uE ×=

C

ww is a variable load. It can act upwards or downwards and vary in magnitude

Load 0 to wu

wu = Maximum value action downwards which causes mechanism formation

4LθwL

2Lθ

21wW

2

uE ×=×

××= u

θMW PE 4=

⇒= θ4M4Lθw P

2

u 2P

u LM16w =



L

A B

C

w

-MP -MP+MP

BMD at mechanism state

PM34

12Lw 2

u =+12

Lw 2u+

24Lw 2

u−

BMD for unloading within elastic range

Resultant BMD for zero load+MP/3

1

2

3

It is identical to applying load from downward

PP ML

LM

34

1216

12Lw 2

2

2u =×=

+



L

A B

C

w

Reloading (downwards)

Elastic

PM34

−4PM

34

−PM32

+

Diagram 3 is the resultant BMD of 1 & 2. It shows the residual moments.Adding 3 & 4 no where moment crosses MP

Elastic change should not be more than 2My otherwise no shakedown will occur and again we will enter into inelastic range.



Mr = Residual moment at any section

Me = Moment at a section considering elastic behavior

M = resulting moment at a section at any particular stage of loading

yP M1.15M ≈

yyP MMM 253.134

<=

Shape function for I section ≈ 1.15

er MMM +=

Maximum variation of moment = 4/3 Mp


Steel StructuresFAILURE CONDITIONS RELATED WITH SHAKEDOWN LIMITS1. Alternating Yield FailureIf the shakedown limits (λs) are exceeded but the loads are kept lesser than the static collapse limits (λc), yield occurs in the extreme fibers at the critical sections, alternately in tension and compression, as the loads are varied.

This is obviously a particular case of fatigue problem, namely, the case when he stress limits are ± Fy.


Steel StructuresBecause the plastic deformations occur alternately in tension and compression, the fatigue life will depend on the degree of strain as well as the stress reversal, and work hardening will lead to changes in the stress-strain behavior away from the simple elastic-plastic relation.

Ultimately, failure of the structure will take place by the breakdown of the material, but usually after a large number of loading cycles. This form of failure is called Alternating Yield Failure.


Steel Structures2. Incremental Collapse Failure

If in each cycle of loading, plastic hinges are developed at certain locations producing unrecoverable deformations without the formation of full collapse mechanism and then similar hinges are developed at other locations, during reversal or alternate loading cycles, the deformations accumulate at all hinges where rotations occur in the same direction.


The net effect of a sufficient number of loading cycles is to get large deformations at all hinges and the situation becomes identical to a collapse mechanism involving hinges formed at different time instances by the varying loads. the resultant failure due to progressive built-up of permanent deflections is called Incremental Collapse Failure.

Steel Structures


Steel StructuresExample: To explain incremental collapse mechanism.

Lθ32wWE ×××=

A0 to w w to 0

L/3 L/3 L/3

C D

B

3θ

θ2θ

θL32

3θ2θθ

θL32

( ) θM63θ2θθMW PPI =++=

θM6Lθ.w32

P=

LM9w P

u =

For the condition when “w” is applied at C, the moment at A is:

( )wL

274

LL3

23

LwMM 2

2

y =×

==

For static mechanism at C or D:



wL274

15.1MP =

LMw P

y 87.5=If load equal to wy is applied yielding will take place at extreme fiber.

For load between 8.10MP/L and 9.0MP/L third/last hinge is not developed and static collapse does not occur, however two plastic hinges are developed.

Max. Load at C: Hinges are developed at A and C. Inelastic rotations occur in every cycle at both these points.

Max. Load at D: Hinges are developed at B and D. Inelastic rotations occur in every cycle at both these points.

8.10 is shakedown limit state, calculated by detailed analysis


Steel StructuresLoad lesser than 8.10MP/L will cause shakedown state hence no failure can occur for such case. Incremental Collapse Mechanism

When load w between 8.10 Mp/L acts at point C, limited plastic rotations occur at A and C, but the plastic hinge rotations are contained (meaning that these are not indefinite) due to absence of hinge at B.

Removal of load at C produces some permanent rotations and residual moments.

The subsequent application of the load w at D leads to containedplastic hinge rotations at D and B.


Steel Structures

This alters the residual stress state left after the removal of w from C and produces permanent rotations at D and B.

The reapplication of w at C is accompanied by further plastic deformations at A and C.

The net effect is a progressive build-up of permanent deflections both at C and D as though the beam were deforming in the plasticcollapse mechanism with hinges at all the four points.

The more closely the load approaches the static collapse value of 9.00 Mp/L, the greater is the increase of permanent deformation per loading cycle.


Steel StructuresShakedown TheoremIf any particular system of residual moments exists which would enable all variations of the applied loads between their prescribed limits to be supported in a purely elastic manner, then the structure will shakedown.

However, the actual system of residual moments existing in the structure when it has shaken down will not necessarily be the initial system.


Steel Structures

Pre MMM ≤+max

Conditions For Occurrence of Shakedown

First Condition

Pre MMM −≥+minThis prevents static and incremental collapse

Second Condition

Pymin

emax

e M74.1M2MM ≤≤−For I-section

This prevents Alternating yield failure

If a particular value of the residual moment can be found to satisfy the given conditions, then the structure may shakedown.

rM

To guard against the alternating yield failure, the difference of maximum and minimum moment at a point in one cycle should not be more than 2My.


Steel StructuresPrevious Example:

0M maxeA =

C

A BP

mineA M

34M −=

Pmax

eC M32M = 0M min

eC =

0M maxeB = P

mineB M

34M −=

First Condition At Point A and C

PrA MM ≤+0

PrCP MMM ≤+32

PrAP MMM −≥+−34

PrC MM −≥+0For Max. For Min.

Second Condition: Nowhere the difference b/w max. and min is more than 2My


Steel StructuresExample:For zero load level only residual moment exists, applying equilibrium conditions at this stage:

02

=+

− rBrArc

MMM

Let MrA = MrB = Mr (equal by symmetry)

0=− rrc MM rrc MM =It can be seen the following satisfy all he above inequalities and it qualifies as the residual moment distribution at shakedown.:

3MM P

r = It satisfies all four inequities of condition-1


Second Condition: Nowhere the difference b/w max. and min is more than 2My

Steel Structures

Pymin

emax

e M74.1M2MM ≤≤−

In this example,

PPee MMMM 74.134minmax ≤=−


Steel StructuresExample:Consider the fixed base portal frame of span and height 2L subjected to loads P & W as shown in figure. The two loads can vary independently b/w zero and “X”.

MP = X . L is constant throughout. Find the shakedown limit λs

MP Constant

L L

2L

W

P

A

B DC

E

θ θ

θ

θ2θ

λX

λXLθ 2Lθ


Steel StructuresSolution:W = 0 to λX and P = 0 to λX, where λ is the Load Factor. Static collapse occur when both these are maximum. From the collapse mechanism:

( ) XθλL2θL2θLXλW ccE =+=

( ) θ6XLθθ2θ2θMPWI ×=+++=

θ6XLXθλL3 c ×=

00.2λ c =


Steel StructuresElastic Moment (Me) at Point A to E

E

D

C

B

A

Due toValueDue toValueMinMaxMinMax

MeminMe

maxMoment due to λW

Moment due to λP

Section

PMλ847

PMλ847

P PMλ8448

− W0 0

0 0 W PPMλ8414

− PMλ8436

PMλ8428 0 P 0 W&P0 0

0 PMλ8436

− 0 PMλ8450

−PMλ8414

− 0

PMλ847 0 PMλ

8455 00 PMλ

8448

PMλ8436

PMλ8414

−

PMλ8428

W&P W&P

W&P W&P

PMλ8448

−



Maximum elastic moment =Up-to this moment the structure remains elastic. First hinge will form here.

PeMλ8455

PPe MMλ8455

=

53.1λe =

Between λe &λc (1.53 & 2.00) there is somewhere shake down.

MA

MB MCMD

ME

+ve Moment



Relationship Between MA, MB, MC, MD & ME for Equilibrium of Frame

Element BD ( )2PLλ

4L2PλMs =

×=

CDB

s M2

MMM =+

+

CDB M

2MM

2PLλ

=+

+

DBC MMM2PLλ −−= 1Columns

0M =∑0L2WλMMMM EDBA =×−+−+−

EDBA MMMMWLλ2 +−+= − 2

MA

MBMD

ME

λW



Suppose the residual moments are MrA, MrB, MrC, MrE, MrD,

Using Eq: 1 & 2 for unloaded stage i.e P & W = 0

rDrBrC MMM2 += 3

0MMMM rErDrBrA =+−+− 4

According to Shakedown Theorem at any section “i”

Priei MMM ≤+max

Priei MMM −≥+min5



Point A

Pmax

e Mλ847M =

PrAP MMMλ847

≤+

PrAP MMMλ8448

−≥+−

PrBP MMMλ8436

≤+

PrBP MMMλ8414

−≥+−

Point B

Point C

PrCP MMMλ8428

≤+

PrC MM0 −≥+

PrD MM0 ≤+

PrDP MMMλ8450

−≥+−

Point D

Point E

PrEP MMM ≤+λ8455

PrE MM0 −≥+

6



Eq: 3, 4, 6 are 12 equations + inequalities. Unknown are 5 residual moments + value of λ. Solution of inequalities is by methods of “Linear Solution of Inequalities”.

The value of λ that will be obtained is λs, Shakedown Limit.

cse λλλ →→Shakedown State Two type of failure can be there

• Alternating Yield Failure

• Incremental Collapse

The type of failure expected in this problem is incremental collapse because there is no reversal of load.



Assume that incremental collapse mechanism is same as static collapse mechanism

θ θ

θ

θ2θ

λX

λXLθ 2Lθ

A

B

C D

E

PA MM −=

PC MM +=

PD MM −=PE MM +=

PM+



Positive Hinges

Prmax

e MMM =+

PrAP MMMλ8448

−=+−

Negative Hinges

Prmin

e MMM −=+

A For negative hinge at A

PrCP MMMλ8428

=+C

PrDP MMMλ8450

−=+−D

PrEP MMMλ8455

=+E

8



3, 4 & 8 are total 6 equations. Six are the unknowns, 5 residualmoments and λ.

3, 4 & 8 are solved together and the solution is:

PrE

PrD

PrC

PrB

PrA

s

M274.0MM158.0MM351.0MM544.0MM112.0M

94.1λ

−=====

=



From Eq: 6

( ) PPPPmax

A MM274.0M112.0M94.1847M <=+=

Pmin

A M996.0M −=

( ) PPPPmax

B MM375.1M544.0M94.18436M >=+=

Pmin

B M221.0M =

MBmax has crossed MP but in our assumed mechanism there is no hinge at

B. So the assumed mechanism is incorrect.



Second Trial: Assume that simple sway mechanism is our incremental collapse mechanism. Now we shall get another set of equations like Eq:8. After solution

PrE

PrD

PrC

PrB

PrA

s

M164.0MM058.0MM148.0MM238.0MM016.0M

77.1λ

−=====

=

Sway mechanism



Pmin

EPmax

E

Pmin

DPmax

D

Pmin

CPmax

C

Pmin

BPmax

B

Pmin

APmax

A

M164.0M,MMMM,M058.0M

M148.0M,M741.0MM058.0M,MMMM,M164.0M

−==

−==

==

−==

−==

All are less than MPor more than –MP. It means assumed mechanism is correct.

λs = 1.77If is λ more than λc = 2.00 static collapse will occur. Below λs = 1.77 there will be shakedown. More closer to 2 lesser will be the number of cycles before collapse.


Steel StructuresMOMENT CHECK

When the final collapse load is obtained by considering all the possible failure mechanisms and by selecting the one corresponding to the minimum load, bending moment diagram is drawn to make sure that the plastic moment capacity is not exceeded at any point (M ≤ Mp).

Following possibilities may exist for plotting the bending moment diagram:


Steel StructuresSituation – I: The structure, after formation of collapse mechanism, is fully determinate. The reactions and unknown moments may be calculated by the equations of static equilibrium.

Sign Convention For Moments

i) Clockwise end moments are considered positive.

A C B+ ve− ve


Steel Structuresii) Within the beam, positive moments are considered positive when tension is produced on the bottom fiber or when the frame tries to open.

A C B

M2M1

M1 = M2 = MM1 = applied moment on segment CBM2 = applied moment on segment CA


Steel StructuresCommon Equilibrium Equations

i) For beams, a typical example is shown below:

LC

R

L / 2 L / 2+ve M for sagging / hogging convention

−ve M for sagging / hogging convention


Steel Structures

422LPMMM uRL

C +−=

or MMMM RLC +−=

22where M = simply supported positive

moment at the center.ii) For Joint Equilibrium:

Mb

Ma

Mc

Ma + Mb + Mc = 0


Steel Structuresiii) For Sway Equilibrium:

1

h

P

4

2 3

5 6

M1 + M2 + M3 + M4 + M5 + M6 + Ph = 0


Steel Structures2nd Method Of Applying Equilibrium Equations

In this method, applied loads, moments and reactions are worked out in terms of shear couples.

Couple due to shear and moment are always equal but opposite in direction.

External loads and reactions are both applied shears on the beam, which act in the same direction.

Moments are resisting couples and are acting in the opposite directions.


Steel Structuresi) Calculating moment at A, if moments are known at B and C.

A

C

B

100100

6050

55Total moment acting in portion CB = 100 + 60 = 160 clockwise

For equilibrium, the shear force in portion CB, having no load in between, must be that which produces equal and opposite couple.


Shear force = 160 / 5 = 32

AC

B

32

50

3232

At point C, external load provides the required shear. For equilibrium of joint C, the balance shear force must be 18 units acting on portion CA.

A C B

18

50

323218

Steel Structures


Steel StructuresAssume that MA is counterclockwise:

MC + MA = 18 × 5

100 + MA = 90

MA = − 10 or 10 units counterclockwise

ii) Calculating moment at C, if the end moments are known.

AC

B 3628

22

8


Steel Structures

In the absence of the central load, the shear in portion AB will be (36 − 8)/4 = 7 (counterclockwise).

Similarly, if only load is present in the absence of end moments, it produces a simply supported shear of 14 clockwise in portion AC and 14 counterclockwise in portion CB.


Steel StructuresA

CB 36

28

14

8

14

7 7

217

Shear due to load

Shear due to moments

Resultant shear

Moment at C may now be calculated from the left or the right side.

For portion AC:

MA (CCW) + MC (CCW) = 7 × 2 (CW)

MC = 7 × 2 − 8 = 6 (CCW or positive according to sagging / hogging concept)


Steel StructuresSituation – II: The structure, after formation of collapse mechanism, is partially redundant.

The objective of design is to utilize the strength of all members and all parts of the structure up to a maximum extent.

This means that a no. of plastic hinges in addition to the collapse mechanism are expected to be nearly formed, reducing the redundancy of the structure.

The condition having high degree of indeterminacy at the collapse load shows a design that is not the best.


The member strengths in the indeterminate parts may be reduced to get economy without affecting the collapse strength.

Steel Structures

There are two methods for analysis of the partially redundant structure.

a) Trial And Error MethodIn this method, unknown moments equal in number to the remaining degree of indeterminacy are assumed reasonably and the remaining moments and shears are found by the equilibrium equations.


Accurate values of moments at various sections are not generally needed.

The main objective is to make sure that the moment does not exceed Mp anywhere in the structure.

This method is suitable when the remaining redundancy is very less (like one or two).

Steel Structures

Example


Steel StructuresP 2P P

Mp Mp Mp

1 2 3 4 5 76

L L L / 2 L / 2 L / 2 3L / 2

θθ2θ Assumed Collapse Mechanism

WE = P θ L

WI = Mp (2θ) + Mp (θ) = 3Mp θ

P = 3Mp / L or Mp = PL / 3


Steel Structures

X = redundancy in the original structure = 2

M = no. of plastic hinges for the assumed mech. = 2

I = remaining redundancy = X − (M-1) = 1

Note that out of the M hinges, one hinge cause collapse but do not change the redundancy of the remaining structure.

Actually, moment M5 is still redundant.


Steel StructuresAssume M5 = Mp:

42

253

4PLMMM +

+−=

46

3×+−=

PLM p

= − Mp + 1.5 Mp = 0.5 Mp < Mp (OK)

( )( )LLLPM

M p

22/32/

43

6 +−=

89

375.0 ×+−=

PLM p = − 0.75Mp + 1.125 Mp

= 0.875 Mp < Mp (OK)


Steel StructuresP 2P P

Mp Mp Mp

1 2 3 4 5 76

L L L / 2 L / 2 L / 2 3L / 2

Assumed BMD+Mp

−Mp

+0.5Mp +0.875Mp

−Mp

The condition that M ≤ −Mp throughout the beam indicates that the assumed mechanism may be the critical one.


Steel StructuresExample: P = 3, Mp = 5 and P = 6 Mp /L

L L L

L/2

L/2

P

2P

5P 4P 5P

P

2P8P

1

2

3 45

6

7

8

9

10

11

12

13

14

16

5Mp

5Mp

3Mp

3Mp 3Mp

5Mp

5Mp

4Mp

4Mp

5Mp

12Mp

12Mp 12Mp

15


Steel Structures

2/55

LMM pp +

LM p

Portion 4-5: Shear =

= 20

Portion 3-4: Shear = LM

P p205 −

LM

LM pp 2030 −

LM p

=

= 10

Assuming M3 to be counterclockwise:

M3 + 5Mp = 10 Mp /L × L / 2 M3 = 0


Steel Structures

LM p

Portion 1-2: Shear = P = 6

M2 = 6 Mp /L × L / 2 = 3 Mp

2/4LM p

LM pPortion 15-16: Shear = = 8

LM p

LM p

LM p

LM pPortion 14-15: Shear = 2P − 8

= 12 − 8 = 4

Assuming M4 to be counterclockwise:M14 − 4Mp = −4 Mp /L × L / 2M14 = 2Mp


Steel Structures

P

2P

5P 4P

8P

1

2

3 45

6

7

8

9

5Mp

5Mp

3Mp

3Mp

5Mp

2.5Mp

8.5Mp2Mp

Zero

3Mp

5Mp

6Mp/L

6Mp/L

10Mp/L 20Mp/L

9Mp/L

15Mp/L

7Mp/L 41Mp/L

12Mp/L

20Mp/L


Steel Structures

2P

L/2

L/2

4P 5P

P

8P

58

9

10

11

12

13

14

15

3Mp

5Mp

5Mp

4Mp

4Mp

12Mp Zero

9Mp

3Mp

2Mp

2Mp2.5Mp

9Mp/L

15Mp/L

8.5Mp7Mp/L 41Mp/L

10Mp/L

42Mp/L

14Mp/L

8Mp/L

16


Steel Structures

2/25

LMM pp +

LM p


= 14

LM p

LM p

LM p

LM p

Portion 10-13: Shear = 5P − 14

= 30 − 14

= 16

Assuming M10 to be counterclockwise:M10 + 5Mp = 16 Mp /L × L / 2M10 = 3Mp


Steel Structures

2/33

LMM pp +

LM p


= 12

2/25

LMM pp +

LM pPortion 10-11: Shear = = 14

LMM pp 25 −

LM p

Portion 5-10: M5 = 2Mp for equilibrium

Shear without load =

= 3


Steel Structures

LM p

LM p

Load = 4P = 24

S.S. shear in portion 5-8 = 12

LM p

LM pTotal shear = (12-3) = 9

S.S. shear in portion 8-10 = 12 LM p

LM p

LM pTotal shear = (12+3) = 15

Assuming M8 to be counterclockwise:M8 + 2Mp = 9 Mp /L × L / 2M8 = 2.5Mp


Steel Structures

Portion 7-6-9-11-12 is still indeterminate.

For its approximate analysis, we may assume additional hinges in this portion wherever they are likely to be produced in case of failure of this part.

Let us assume a hinge at point – 11 in the beam and a hinge at section – 12.

Similarly, a hinge may also be considered at section – 7.


Steel Structures

2/912

LMM pp +

LM p

M11 for beam = 12Mp (clockwise)then M11 for column = 9Mp (counterclockwise)M12 for column = 12Mp (counterclockwise)

Portion 11-12: Shear = = 42

For Overall Lateral Equilibrium of the Frame

Leftward shear in member 7-6:

LM p

LM p

LM p

LM p+ 42 + 8 + 6 = 2P + 2P + P + P

= 36


Steel Structures

LM p

LM p

LM p

Shear in 7-6 = (36 − 42 − 8 − 6)

= − 20 or 20

Assuming M7 = 12Mp (clockwise, opposite to shear in member 7-6) and assuming that M6 is clockwise.

M6 + M7 = 20 LM p

M6 in column = 2Mp (counterclockwise)M6 in beam = 5Mp (clockwise)

× L/2


Steel Structures

LM p

Portion 6-11:Shear without load = 17

S.S. shear in portion 6-9 = 4P = 24 LM p

LM p

LM pTotal shear in portion 6-9 = (24-17)

= 7

S.S. shear in portion 9-11 = 24 LM p

LM p

LM p

Total shear in portion 9-11 = (24+17)

= 41


Steel StructuresAssuming M9 to be clockwise:

M9 + 5Mp = − 7 LM p

M9 = 8.5Mp (counterclockwise)

Hence, moment is not exceeded anywhere in the frame and the assume mechanism is the critical mechanism.

× L/2


Steel Structuresb) Moment Balancing Method

In this method, a pseudo-moment distribution process is used.

This method does not apply continuity conditions at the joints and hence is approximate and different from the actual moment distribution method.

The following steps are used in this method:

i) Sign convention for moments given earlier is used.


ii) Carryover factors are used to restore the equilibrium of the beam spans between the joints when end moment change during joint balancing. These represent the effect of a unit change in moment at one point in a span on the moments at other points.

Steel Structures

Carryover factors are only applied on beams in case of analysis of frames.

Case 1: A unit positive increase in the left end moment with no change at the right end produces the following increment at the center:


Steel StructuresAccording to the already derived equation, we have,

MC = ML / 2 − MR / 2 + Ms

∆MC = ½ ∆ML

As ∆MR = 0 and ∆Ms = 0

∆ML ∆MC ∆MR

+1 +1/2 0L C R


Steel StructuresCase 2: Keeping the centerline moment constant, a unit change in moment at one end produces a unit change in moment at the opposite end of same sign.

According to the already derived equation, we have,

MC = ML / 2 − MR / 2 + Ms

∆ML = ∆MR As ∆MC = 0 and ∆Ms = 0


Steel Structures

L CR

1.0

1.0

∆ML ∆MC ∆MR+1 0 +1

Case 3: A unit positive increase in the right end moment with no change at the left end produces the following increment at the center:


Steel StructuresAccording to the already derived equation, we have,

MC = ML / 2 − MR / 2 + Ms

∆MC = − ½ ∆MR

As ∆ML = 0 and ∆Ms = 0

∆ML ∆MC ∆MR

0 −1/2 +1 L C R

1.0− 1/2


Steel Structuresiii) Selection of Starting Moments:

Moments at ends of a given indeterminate span may be selected equal in magnitude, keeping in mind the known end conditions.

If one end is hinge and other is any other type of support, moment at both sides is assumed equal to zero.

If both ends are continuous, a negative bending moment at each end equal to the simply supported moment of that span less than Mp may be assumed.


The resulting central moment may be calculated as simply supported moment plus end moments.

Steel Structures

iv) Moment Balancing:

The joint equilibrium is restored by applying opposite of unbalanced moment.

v) Carryover Operation:

The extra moment applied during moment balancing disturbs the equilibrium of the beam portions between the ends. This operation restores the equilibrium.


Steel Structuresvi) Column Adjustment:

Moment corrections are made to bring back the sway equilibrium.

More moment is given to that section which is already weak and near its full capacity.

Zero moment may be assigned to a section where hinge can not form.

Moment after adjustment at a column end must not be greater than moment at its other end.

This gives critical condition for the moment check.


Steel Structuresvii) Summation of Moments:

All the assumed, joint-balancing and carryover moments are added together to get the total moments.

viii) Drawing Bending Moment Diagram

Bending moment diagram is completed using the total moments obtained above.

Example: Solve the previous continuous beam example by moment distribution method.


Steel Structures

42 LP×

LM

L p342

Simply supported moment at point-4

= = = 3/2 Mp

L

LLP

22

32××

LP83

LM

L p383

Simply supported moment at point-4

= = =

= 9/8 Mp = 1.125 Mp


P 2P P

Mp Mp Mp

1 2 3 4 5 76

L L L / 2 L / 2 L / 2 3L / 2

θθ2θ Assumed Collapse Mechanism

Assumed BMD+Mp

−Mp

+0.625Mp +0.563Mp

−0.75Mp

Steel Structures

1. Ms − +1.5 1.125 0

2. Assumed moments − +1.0 +1.0 −0.75 +0.75 +0.75 0 1.125 0

3. Joint balancing −− −− −− −0.25 −− 0 −0.75 −− −−

4. Carryover −− −− −− 0 −0.125 0 0 −0.563 −−

5. Total moments −− +1.0 +1.0 −1.0 +0.625 −0.75 +0.562 0+0.75


Steel Structures

Example

P

5P 2P

2Mp

2Mp

Mp

Mp

Mp

L L

L / 2

1

2 3 4 56

7

8 9

10


Steel Structures

P

5P 2P

1

2 3 4 5

6

7

8 9

10

θ θ

2θ

Assumed Collapse Mechanism


Steel Structures

LM p

2.3PL

WE = 5P × θ × L / 2 = 2.5 PθL

WI = 2Mp (θ + 2θ + θ) = 8 Mp θ

WE = WI ⇒ 2.5 PθL = 8 Mp θ

P = 3.2 or Mp =

I = 6M = 3 (number of hinges in the assumed

mechanism)X = I − (M − 1)

= 6 − (3 − 1) = 4 (resulting structure is 4th degree indeterminate)


Steel Structures

42 LP×

2

2.3L

LM p ×

Ms for portion 5-9 =

=

Assume M5 = M9-8 = 0.6 Mp

= 1.6 Mp

Assumed Distribution of Sway Moments

Clockwise moments are considered positive as before. As a first approximation, moment given to column 1-2 will be double of other columns because of heavy section.


Steel Structures

If M1 = M2 = M

then M6 = M7 = M9 = M10 = M / 2

The equation for sway becomes:

4M + PL / 2 = 0

or M = − PL / 8

= − (3.2 Mp / L) × L / 8

= − 0.4 Mp


Steel Structures

−0.6−0.6+0.6+0.8−1.0−1.0−1.0+2.0+2.0−2.0+2.0−0.46. Total / Final moments

−0.4−0.80.05. Column adjustment

−0.24. Carryover

−0.4−0.8−0.4+2.43. Joint balancing

−0.2−0.2+0.6+1.0−0.2−0.2-0.6+2.0+2.0−2.0−0.4−0.42. Assumed moments

+1.61. Ms

9-109-82-32-110

9876543

21Section


Steel StructuresJoint Balancing

Joint 2: The moment 2-3 is fixed; the balancing moment of +2.4 can only be applied to the column.

Joints 4,5,6: The unbalanced moment of −1.2Mp can only be given to joints 5 and 6.The moments for both the members become −2Mp equal to their full capacity.

Joint 9: The unbalanced moment of −0.4Mp may be given to column to get increase in moment.


Steel StructuresCarryover

Only one moment of −0.4Mp is applied as balancing moment on the beam.

Changing moment 9-8 will create imbalance at joint 9. Hence, keeping moment at 9-8 constant, the carryover to the central moment is −0.2Mp.

Column Unbalance

Joint 2-1 6 9

∆M +2.4 −0.8 −0.4 ∑M = +1.2


Steel Structures

The balancing moment of −1.2 is given to sections that are weaker to get critical conditions.

∆M1 = 0 (very strong)

∆M7 = −0.8 (hinge may be developed)

∆M10 = −0.4

∑∆M = −1.2

Note: ∆M7 and ∆M10 may be interchanged.


Steel Structures

P

5P 2P

2Mp2Mp

Mp

0.8Mp

Mp1

2 3 4 56

7

8 9

10

0.4Mp

2Mp Mp

Mp

0.6Mp

0.6Mp


Concluded

m sc steel-04-prof-zahid_siddiq

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