Steel Structures

SE-505

Plastic Analysis and Design of Structures

M.Sc. Structural Engineering

Example

Design the following two bay single story frame with the working loads as shown in the figure.

Assume that the positive hinges are formed at the midspans and use the following load combinations:

1. 1.2D + 1.6L

2. 1.2D + 1.6L + 0.8W

3. 1.2D + 0.5L + 1.3W

30m 20m

8m

D = 5 kN/mL = 5 kN/m

25 kN/m

1. Structure And Loading

First And Second CombinationsLet w = 1.2 × 5 + 1.6 × 5 = 14 kN/mHorizontal load = 0.8 × 25 × 8 / 2 = 80 kN

= 5.72 w

Third Combination

Let w = 1.2 × 5 + 0.5 × 5 = 8.5 kN/mHorizontal load = 1.3 × 25 × 8 / 2 = 130 kN

= 15.30 w

5.72 w or2.68 (5.72 w)

Mp

Mp k1 Mp

k1 Mpk2 Mp

A

B

CD

E F

15 w 10 w7.5 w 12.5 w 5 w

2. Relative Mp Values

Let

Plastic moment capacity for beam BC = Mp

Then plastic moment capacity for column AB = Mp

The factor k1 will be calculated according to the ratio of simply supported moments for the spans BC and CD.

This ratio simplifies to the ratio of the square of spans, if the load is the same.

2

2

BC

CD

LL

2

2

3020

k1 = = = 0.44

Mp 0.44 Mp

k2 Mp

Use same Mp value of k1 Mp for the column DE.

Find k2 for the equilibrium of Joint – C, but having certain minimum value.

k2 = 1 − 0.44 = 0.56(not less than 0.5 times the larger Mp value)

3. Analysis i) Beam BC Mechanism 15w

θ θ

2θ

B CWE = 15w × 15θ = 225 w θWI = 4 Mp θw = 0.0178 Mp

10w

θ θ

2θ

C D

ii) Beam CD Mechanism

WE = 10w × 10θ = 100 w θ

WI = 0.44 Mp 4θ

w = 0.0176 Mp

iii) Panel Mechanism

WE = 5.72w × 8θ = 45.76 w θor = 15.3w × 8θ = 122.4 w θWI = 2 Mp θ + 2 × 0.56Mp θ

+ 2 × 0.44Mp θ = 4 Mp θ

w = 0.0874 Mp or 0.0326 Mp

5.72 w or15.3 w

iv) Joint C Mechanism

WE = 0

WI = 2 Mp θ θ

θθ

v) Combined Mechanism [(i) + (iii)]

WE = (225 + 45.76) w θ= 270.76 w θ

or = (225 + 122.4) w θ= 347.4 w θ

WI = 6 Mp θ

w = 0.0222 Mp or 0.0173 Mp

5.72 w or15.3 w

15 w 10 w

4. Reactions For Critical Mechanisms

A. Reactions For Mechanism (v) For 3rd Combination

Mp = 8.5 / 0.0173 = 491.3 kN-m

2 × 491.3 − 7.45 × 8 + 8.5 × 152 / 2 − VA × 15 = 0⇒ VA = 125.28 kN

VE = 8.5 × 20 / 2 = 85 kN

VF = 8.5 × 50 − 85 − 125.28 = 214.2 kN

HA = 130.3 − 68.8 − 54.05 = 7.45 kN

130.3 kN

8.5 kN/m

A

BC D

EF

491.3 275.1 216.2

216.2275.1

491.3 216.2

491.37.4568.8 54.05

54.05

214.2

68.87.45

125.28 85

B. Reactions For Mechanism (v) For 2nd Combination

Mp = 14 / 0.0222 = 630.6 kN-m

630.6 + 77.6 × 8 + 14 × 152 / 2 − VA × 15 = 0⇒ VA = 188.4 kN

VE = 14 × 20 / 2 = 140 kN

VF = 14 × 50 − 188.4 − 140 = 371.6 kN

HA = 80.08 − 88.3 − 69.4

= 77.6 kN (towards the right)

80.08 kN

14 kN/m

A

BC D

EF

630.6 353.2 277.5

277.5353.2

630.6 277.5

630.677.688.3 69.4

69.4

371.6

88.377.6

188.4 140

C. Reactions For Mechanisms (i) And (ii) For 1st Combination

14 kN/m

A

BC D

EF

397.75 222.75 175

350795.5

350

795.5

65.6

350

83.5149.2

210 140

795.5

445.5 350

* Moments at bases of columns may be taken equal to half of the top moments as carry-overs.

* Beams have equal end moments and hence the vertical reactions are equal to simply supported reactions.

5. Bending Moment Diagrams

6. Selection Of Sections

If the sections for dmin of beams are not available or are highly uneconomical, Ireq for partially fixed ends may be checked.

7. Check Moment MagnificationRevise the column and beam sections accordingly.

8. Check Compactness Of Sections9. Satisfy Interaction Equations For Columns10. Check Interaction Equations For Other

Combinations11. Check Shear12. Decide Lateral BracingConsider both the critical lengths Lp and Lpd. Also decide the unbraced lengths of columns to prevent buckling about minor axis.

13. Design Three Connections

14. Design Three Base Plates

Example

Design the given gable frame with haunched connections for the following load cases:

1.2D + 1.6L

1.2D + 0.5L + 1.3W

9m 9m

6m

D = 5 kN/m, L = 10 kN/m

10 kN/m

2.5m

1. LoadingCase I

Let w = 1.2 × 5 + 1.6 × 10 = 22 kN/m

Case II

Let w = 1.2 × 5 + 0.5 × 10 = 11 kN/m

Horizontal load = 1.3 × 10 × 8.52 / 2 = 78.27 kN= 7.12 w

2. Relative Plastic Moment ValuesMp – values of the girder at the end of the haunch (point – A) and at the maximum moment section (point – B) close to the crown should be the same, although the moments will be opposite in direction.

Mp – values of the column may be different and can be found from the moment diagram.

Let

Mp = value of Mp required at the haunch point

22 95.2 +

3341.99

×

MA and MB = value of Mp required for the girder

Length of haunch along the girder = L / 10 to L / 6

= 3 m

Length of the haunch along the column

= 2/3rd of length along the beam = 2 m

Total inclined length of one girder =

= 9.341 m

Horizontal length of haunch = = 2.89 m

3. AnalysisCase I:

x

2.89m

A

A

B

B

6H 6H8.5H

891 kN-m

289.222 2×

95.2 H

Let

Plastic moment required at the haunch point

= Mp = 6H

Due to the presence of the haunch, let the hinges be formed at section A and B.

MA = 22 × 9 × 2.89 − − 6H −

= 480.3 − 1.134 Mp

× 2.89

222 2x×

95.2 HMB = 22 × 9 × x − − 6H −

= 198 x − 11 x2 − Mp − 0.046 Mp × x

× x

Now, MB = − MA

(as both must be equal and opposite)198 x − 11 x2 − Mp − 0.046 Mp × x

= 480.3 − 1.134 Mp

Mp = x

xx046.0134.2

3.48019811 2

+++−

To get the minimum collapse load or the maximum required Mp value,

dxdM p

= 0

(−11x2 + 198 x + 480.3) (0.046) − (2.134 + 0.046 x) (−22 x + 198) = 0

0.506 x2 + 46.948 x − 400.436 = 0

x = 7.863

Mp = 543.8 kN-m

MA = − 136.3 kN-m

MB = 136.3 kN-m

Column moment = 362.53 kN-m

22 kN/m

120.65543.8

362.53

543.8

2 m

543.8 / 6 = 90.63

198

Case II:11 kN/m

2.5 m

HH − T

T = 78.27 kN

6 m

A

B

18 m

x

A

A

B

B

6H 6H8.5H

445.5 kN-m

469.62 kN-m

The negative moment on the right side (section A) is more as the simply supported positive moment here is less.

However the positive moment (section B) is greater on the left side.

289.211 2×

1889.262.469 ×

95.2 H

Let plastic moment required at the haunch point

= Mp = 6H

Due to the presence of the haunch, let the hinges be formed at section A and B.

MA = 99 × 2.89 − +

− 6H − × 2.89

= 315.57 − 1.134 Mp

211 2x×

1862.469 x×

95.2 H

MB = 99 × x − + 469.62 −

− 6H − × x

MB = 72.91 x − 5.5 x2 − Mp − 0.0463 Mp × x+ 469.62

Now, MB = − MA

(as both must be equal and opposite)

72.91 x − 5.5 x2 − Mp − 0.0463 Mp × x + 469.62

= 315.57 − 1.134 Mp

Mp = xxx

0463.0134.219.78591.725.5 2

+++−

To get the minimum collapse load or the maximum required Mp value,

dxdM p = 0

(−5.5x2 + 72.91 x + 785.19) (0.0463) − (2.134 + 0.0463 x) (−11 x + 72.91) = 0

0.25465 x2 + 23.474 x − 119.236 = 0

x = 4.827 m

Mp = 428.0 kN-m

MA = − 169.8 kN-m

MB = 169.8 kN-m

Column moment = 285.3 kN-m

11 kN/m

428.0

285.3

78.27

2 m

428 / 6 = 71.33

125.1

6.94

72.91

41.645

4. Initial Selection Of SectionsA. Girder:

Mp = 169.8 kN-m

Z = 2509.0108.169 6

××

W 360 × 44 is selected with the following properties:

A = 5710 mm2; d = 352 mm; bf = 171 mm; tf = 9.8 mm; tw = 6.9 mm

φb Mp = 174.40 kN-m; Lp = 1.88 m; Lr = 5.53 m; rx = 146mm; ry = 37.8 mm

= 755 × 103 mm3

Concluded