3.Std Portal Frame 2 - Analysis

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis

Plastic Collapse Analysis -What is Plastic Collapse? -Theorems of Plastic Collapse -Sign Convention -No of Independent Mechanism -Completeness of Mechanism -Examples

Module: Steel Structure (H23 S07) Chapter: Portal Frame – Analysis Slide: What is plastic collapse?

Plastic collapse of simple beam - Formation of plastic hinges

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Collapse Mechanisms

Theorems of Plastic Collapse

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Sign Convention

Sign Conventions

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Independent Mechanisms

B

A

C

D

E

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Completeness of Mechanisms

Completeness of collapse Mechanisms (a) Complete collapse - contains one more plastic hinge that the degree of redundancy. - statically determinate. (b) Partial collapse - contains fewer hinges than complete mechanism. - statically indeterminate. But possible to determine the unknown internal forces that must satisfied for equilibrium. - Bending moment diagram which satisfies the three criterions; equilibrium, yield, and mechanism is sufficient to satisfy the uniqueness theorem. (c) Over-complete collapse - contains more hinges than complete mechanism. - simultaneous occurrence of two or more alternative mechanisms, which should be considered saparately.

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Completeness of Mechanisms

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Example 1 λV

λH C

B A

D

L/2

L/2

h

E

λV MB

λ VL

MD

MC L/2

X ( 1a )

L/2

MD MA

0

2

Overcompleted collapse at X and Y

Y

MD

h

MC ME

λ HL

4

MP

λV

L/2

4

Safe

h ME

λH

MA L/2

( 3a ) 8

MB

λH

( 2a )

MP

Combine Mechanism (3) = (1) + (2);

Alternativ ely;

At collapse; M A = - M C = M D = - M E = M P

Eqn ( 1a ) + ( 2a ) ⇒

(+ )

Equilibrium equation M A (+θ) + M C (−2θ ) + M D (+2θ ) + M E (−θ) = λ H h (+θ) + λ V 6 MP = λ H h + λ V L

2

- -( 3a )

4 MP = λ V L 4 MP = λ H h

2

L 8 MP = λ H h + λ V L 2 (+θ) 2 (cancel hinge at B) ( − ) 2 Mp 6 MP = λ H h + λ V L

2

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Example 2 C

D

E L

3Mp

B

W

2W

Mp

Mp 2L

A

2L

2L F

C

C L

W

Combinatio n (4) = (2) + (3);

2W

E

W

- 2 Mp (Cancel hinge at C)

2L A

D

At collapse; M C = ME = MP ; MD = - 3Mp

E

Equilibrium equation M C (+θ) + MD (-2θ ) + ME (+θ) = 2 W × 2 L ( +θ ) 8 MP = 4 W L

Hinges form in column, less Mp required

2W

C

W = 2 MP / L

- -( 2a )

Equilibrium equation M A (+θ) + M C (−θ) + M E (+θ) + M F (-θ ) = W × 2 L (+θ) 4 MP = 2 W L

F

- -( 3a )

W = 2 M P / L - - ( 3b )

E

B

A

Hinges form in column, less Mp required

W = 5 Mp / 3 L

F

W

- - ( 2b )

E At collapse; M A = - M C = M E = - M F = M P

Hinges form in column, less Mp required

10 Mp = 6 W L - - ( 4a )

2W

C

(3) Sway Mechanism;

W

A

Hinge form in column, less Mp required

(2) Beam Mechanism;

2W C

+ 4 MP = 2 W L - -( 3a ) 12 Mp = 6 W L

B A

8 MP = 4 W L - -( 2a )

F

- - ( 4b )

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Example 3 λV

λH

B C

D L

Uniform MP A L/2

E L/2

Sway Mechanisms;

L

Trangulars IBD and IAE are similar. Hence, therefore, IB = L and ID = DE,

λV

I D

Mechanism 2(i);

2 (i) sway

B

L

C

λV λH

B

Rotation BD = θ (anti - clockwise), and, B moves (+ ) Lθ (→). Rotation AB = θ (clockwise), rotation of hinge at A = θ , rotation of hinge at B = (-) 2 θ. Rotation DE = θ (clockwise), rotation of hinge at E = - θ , rotation of hinge at D = (+ ) 2 θ. Vertical deflection of C = (BC) (θ) = (-) (L ) (θ ) ( ↑ ) 2

D

C L

Uniform MP

At collapse; M A = - M B = M D = - M E = M P

A

E L/2

I 2 (ii) sway

L

λV

B λH

D

C

L

Uniform MP A

E L/2

IB ID BD 1 = = = , IA IE AE 2

L/2

L

L/2

L

Equilibrium equation L M A (+θ) + M B (-2θ ) + M D (2θ ) + M E (-θ ) = λ H (+ Lθ ) + λ V (- θ) 2 λVL 6 MP = λ H L − 2

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Example 3 – con’t

I

I 2 (i) sway

L

λV B

λH

λV

D

+

C

B

A

A L/2

L/2

L

I

C

+

B C

λV

B

D D

2 (ii) sway

L

λV

λV

λH

E

L

2 (ii) sway

B

B

L

I L

D C

x2

E L/2

λV λH

=

C

L L/2

L D

=

D

λH

C

L

L A

E L/2

L/2

L

x2

A

E L/2

L/2

L

Module: Steel Structure (H23 S07) Chapter: Portal Frame - Analysis Slide: Example 4 λV h1

λH

C B

D

h2

Uniform MP A

E L/2

L/2

λV h1

λV C

B

h1

D

C

λH h2

h2 A

A

E L/2

L/2

h1

J λH

C

h1 h2

A L/2

E L/2

L/2

L MP ( 4 + 4k ) = λ H ( 2h1 ) + λ V ( ) - - - (2) 2 4MP = λ H h 2 - - - (1) +

D

B

L/2

λH

Combinatio n (3) = (1) + (2) ;

I λV

D

B

E

-

L MP ( 8 + 4k ) = λ H ( 2h1 + h 2 ) + λ V ( ) 2 2 Mp (Cancel hinge at B) L MP ( 6 + 4k ) = λ H ( 2h1 + h 2 ) + λ V ( ) - - (3) 2

Module: Steel Structure (H23 S07)

10 kN/m

1 0.5 4

?

MP

B

C

MP

1.5 MP

D 1.5 MP

A

E 12

12

(x /12) 1 0.5

I B

C x

4 A 12

C B A

A pinned based pitched roof frame is subjected to an UDL of 10kN/m. Haunches are provided at joint B, C, and D, to ensure hinges will not form in these joints. Assume that the length of the haunches at eaves is sufficient to ensure that hinges will form in the column and not rafter. Determine the value of MP, its location in the rafter (roof beam), and hence the require length of the apex haunch.

Module: Steel Structure (H23 S07)

So far… -What is Portal Frame? -Portal Frame applications -Portal Frame elements -Collapse Mechanism -Design basis

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