86454170 Portal Frame Design Worked Example

PORTAL FRAMES

Job No: Sheet 1 of 30 Rev Job Title: Portal F r ame An alysis alysis and Design Design Worked Example: 1 Made By Date PU Checked By Date V K

Structural Steel Design Project Calculation Sheet Problem

Analyse and Design a single span portal frame with gabled roof. The frame has a span of 15 m, the column height 6m and the rafter rise 3m. Purlins are provided @ 2.5 m c/c.

D 5mc/c

3m C

E

B

F 0.6 m

0.6 m

30 m

6m

3.25 m

G 15 m

15 m

Load

1.0 L oad Calcul ation 1.1

Dead L oad

Weight of asbestos sheeting Fixings Services Weight of purlin

= = = =

Total load /m 2

=

Version II

0.17 kN/m2 0.025 kN/m 2 0.100 kN/m2 0.100 kN/m2 --------------0.395 kN/m 2 ---------------

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PORTAL FRAMES


Structural Steel Design Project Calculation Sheet Dead load/m run run

1.2 1.2

= 0.395 * 5 = 1.975 kN / m ≈ 2.0 kN/m

L ive L oad

Angle of rafter = tan ta n-1 (3/7.5) = 21.80 From IS: 875 (part 2) – 1987; 1987; Table 2 (cl 4.1),

} * 5 Live load / m r un = {0.75 − 0.02 (21.8 − 10) = 2.57 kN/m

1.3 1.3 Crane L oadin oadin g Overhead electric crane capacity

= 300 kN

Approximate weight w eight of crane girder

= 300 kN

Weight Weight of crab

= 60 kN

The extreme position of crane hook is assumed as 1 m from the centre line of rail. The span of crane is approximately taken as 13.8 m. And the wheel base has been taken as 3.8 m 1.3.1

Vertical load

The weight of the crane is shared equally by four four wheels on both sides. The reaction on wheel due to the lifted weight and the crab can be obtained by taking moments about the centreline of wheels. ( 300 + 60 ) B

300 6.9 m

1m 13.8 m

F

M B = 0

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PORTAL FRAMES

Structural Steel Design Project Calculation Sheet


2 R F (13.8) = (300 + 60) * 1 + 300 * (6.90) R F = 88 kN M F = 0 2 R B (13.8) = (300 + 60) * (13.8-1) (13.8-1) + 300 * (6.9) R B = 242 kN To get maximum wheel load on a frame from gantry girder BB', taking the gantry girder as simply supported. supported. 242 kN B'

3.8 m 5m

242 k B

Centre to centre distance between frames is 5 m c/c. Assuming impact @ 25% Maximum wheel Load @ B = 1.25 (242 (1 + (5-3.8)/5) (5-3.8)/5) = 375 kN. Minimum wheel Load @ B = (88 /242)*375 =136.4 kN 1.3.2

Transverse Load:

Lateral load per wheel = 5% (300 + 60)/2 = 9 kN (i.e. Lateral load is assumed as 5% of the lifted load and the weight of the trolley acting on each rail). Lateral load on each column { EMBED Equation.3 } = 13.9 kN (By proportion)

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Structural Steel Design Project Calculation Sheet 1.4

Job No: Sheet 4 of 30 Rev Job Title: Portal F r ame An alysis and Design Worked Example: 1 Made By Date PU Checked By Date VK

Win d L oad

Design wind speed, V z = k 1 k 2 k 3 V b From Table 1; IS: 875 (part 3) – 1987 k 1 = 1.0 (risk coefficient assuming 50 years of design life) From Table 2; IS: 875 (part 3) – 1987 k 2 = 0.8 (assuming terrain category 4) k 3 = 1.0 (topography factor) Assuming the building is situated in Chennai, the basic wind speed is 50 m /sec Design wind speed,

V z = k 1 k 2 k 3 V b V z = 1 * 0.8 *1 * 50 V z = 40 m/sec

Basic design wind pressure, P d = 0.6*V z 2 = 0.6 * (40)2 = 0.96 kN/m 2 1.4.1. Wind Load on individual surfaces

The wind load, W L acting normal to the individual surfaces is given by W L = (C pe – C pi ) A*P d (a) Internal pressure coefficient Assuming buildings with low degree of permeability C pi = ± 0.2

Version II

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PORTAL FRAMES

Job No: Sheet 5 of 30 Rev Job Title: Portal F r ame An alysis and Design Worked Example: 1 Made By Date PU Checked By Date VK

Structural Steel Design Project Calculation Sheet (b) External pressure coefficient

External pressure coefficient for walls and roofs are tabulated in Table 1 (a) and Table 1(b) 1.4.2

Calculation of total wind load

(a) For walls h/w = 6/15 = 0.4 L/w = 30/15 = 2.0

L

Exposed area of wall per frame @ 5 m c/c is A = 5 * 6 = 30 m2

h w

w elevation

plan For walls, A pd = 30 * 0.96 = 28.8 kN Table 1 (a): Total wind load for wall Wind Angle θ

C pe

C pi

Windward

Leeward

00

0.7

-0.25

900

-0.5

-0.5

0.2 -0.2 0.2 -0.2

C pe – C pi Wind ward

Lee ward

Total wind(kN) (C pe-C pi )Apd Wind Lee ward ward

0.5 0.9 -0.7 -0.3

-0.45 -0.05 -0.7 -0.3

14.4 25.9 -20.2 -8.6

-12.9 -1.4 -20.2 -8.6

(b) For roofs Exposed area of each sloping roof per frame @ 5 m c/c is A = 5 *

Version II

(3.0 ) 2 + (7.5) 2 = 40.4 m 2

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PORTAL FRAMES

Job No: Sheet 6 of 30 Rev Job Title: Portal F r ame Analysis and Design Worked Example: 1 Made By Date PU Checked By Date VK

Structural Steel Design Project Calculation Sheet For roof, Apd = 38.7 kN

Table 1 (b): Total wind load for roof Wind angle

0

Pressure Coefficient

0

900

2.0

C pe

C pe

C pi

Wind -0.328 -0.328 -0.7 -0.7

Lee -0.4 -0.4 -0.7 -0.7

0.2 -0.2 0.2 -0.2

C pe – C pi

Total Wind Load(kN) (C pe – C pi ) Apd Lee Wind Lee ward ward ward Int. Int. -0.6 -20.4 -23.2 -0.2 -4.8 -7.8 -0.9 -34.8 -34.8 -0.5 -19.4 -19.4

Wind ward -0.528 -0.128 -0.9 -0.5

Equ ivalent L oad Calcul ation

2.1 Dead Load Dead Load = 2.0 kN/m Replacing the distributed dead load on rafter by equivalent concentrated loads at two intermediate points on each rafter,

{ EMBED Equation.3 } 2.2 Superimposed Load 15 m

Superimposed Load = 2.57 kN/m Concentrated load , { EMBED Equation.3 }

Structural Steel

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Job No: Sheet 7 of 30 Rev Job Title: Portal F r ame An alysis and Design

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Design Project

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2.3 Crane Load Maximum Vertical Load on columns = 375 kN (acting at an eccentricity of 600 mm from column centreline) Moment on column = 375 *0.6 = 225 kNm. Minimum Vertical Load on Column = 136.4 kN (acting at an eccentricity of 600 mm) Maximum moment = 136.4 * 0.6 = 82 kNm

3.0

Parti al Safety F actor s

3.1

Load Factors

For dead load, γ f = 1.35 For major live load, γ f = 1.5 For minor live load, γ f = 1.05 3.2

Material Safety factor

γm = 1.15

4.0

Analysis

In this example, the following load combinations are considered, as they are found to be critical. Similar steps can be followed for plastic analysis under other load combinations. (i)

1.35D.L + 1.5 C .L + 1.05 W.L

Structural Steel

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Calculation Sheet (ii)

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1.35 D.L + 1.5 C.L+ 1.05 L.L

4.1. 1.35 D.L + 1.5 C.L+ 1.05 W.L 4.1.1Dead Load and Wind Load (a) Vertical Load w @ intermediate points on windward side w = 1.35 * 5.0 – 1.05 *(4.8/3) cos21.8 = 5.2 kN. w 2

@ eaves

=

5.2 2

=

2.6 kN

w @ intermediate points on leeward side w = 1.35 * 5.0 – 1.05 * 7.8/3 cos21.8 = 4.2 kN w 2

@ eaves

=

4.2 2

=

2.1 kN

Total vertical load @ the ridge = 2.6 + 2.1 = 4.7 kN b) Horizontal Load H @ intermediate points on windward side H = 1.05 * 4.8/3 sin 21.8 = 0.62 kN

Structural Steel Version II


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Calculation Sheet H/2 @ eaves points

Date

VK = 0.62/2 = 0.31 kN

H @ intermediate purlin points on leeward side = 1.05 * 7.8 /3 sin 21.8 = 1 kN H/2 @ eaves = 0.5 kN Total horizontal load @ the ridge = 0.5 - 0.31 = 0.19 kN Table 3: Loads acting on rafter points

Intermediate Points Eaves Ridge 4.1.2

Vertical Load (kN) Windward Leeward 5.2 4.2 2.6 2.1 4.7

Horizontal Load (kN) Windward Leeward 0.62 1.0 0.31 0.5 0.19

Crane Loading

Moment @ B = 1.5 * 225 = 337.5 kNm Moment @ F = 1.5 * 82 = 123 kNm Horizontal load @ B & @ F = 1.5 * 13.9 = 20.8 kN

Note: To find the total moment @ B and F we have to consider the moment due to the dead load from the weight of the rail and the gantry girder. Let us assume the weight of rail as 0.3 kN/m and weight of gantry girder as 2.0 kN/m

Dead load on the column =

 2 + 0.3    * 5 = 2  

5.75 kN

Factored moment @ B & F = 1.35 * 5.75 * 0.6 = 4.6 kNm

Structural Steel

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Total moment @ B = 337.5 + 4.6 = 342 kNm @ F = 123 + 4.6 = 128 kNm 4.7 kN 0.19 kN 4.2 kN 1.0 kN 4.2 kN 1.0 kN 2.1 kN 0.5 kN

5.2 kN 0.62 k N 5.2 kN 0.62 kN 2.6 kN 0.31 kN

3m

128

342

6m

20.8 kN 20.8 kN 3.25 m 27.2 kN

1.5 kN

15 m

F actored L oad (1.35D.L +1.5 C.L +1.05 W.L ) 4.2

1.35 D.L + 1.5 C.L + 1.05 L.L

4.2.1

Dead Load and Live Load

@ intermediate points on windward side = 1.35 * 5.0 + 1.05 * 6.4 = 13.5 kN @ ridge = 13.5 kN @ eaves = 13.5 / 2 ≈ 6.8 kN.



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Calculation Sheet 4.2.2

Date

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Crane Load

Moment @ B = 342 kNm Horizontal load @ B = 20.8 kN Moment @ F

= 128 kNm

Horizontal load @ F = 20.8 kN

13.5 kN 13.5 kN 13.5 kN 13.5 kN

13.5 kN

3m

6.8 kN

128

342 20.8 k

6m 20.8 k 3.25 m

15 m

F actored L oad (1.35 D.L + 1.5 C.L + 1.05 L.L ) 4.3

Mechanisms

We will consider the following mechanisms, namely (i) (ii) (iii) (iv)

Beam mechanism Sway mechanism Gable mechanism and Combined mechanism



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Calculation Sheet 4.3.1

Date

VK

Beam Mechanism

(1) Member CD Case 1: 1.35 D.L + 1.5 C.L + 1.05 W.L 4.7 kN 0.19 kN

5.2 kN 0.62 kN 5.2 kN

θ /2

0.62 kN 2.6 kN 0.31 kN

θ

Internal Work done, Wi = M /2) + M /2) p θ + M p ( θ p ( θ + θ = M p(3θ ) External Work done, W e = 5.2 * 2.5θ- 0.62 * 1 * θ+5.2 * 2.5 * θ /2 – 0.62 * 1 * θ /2 = 18.6 θ Equating internal work done to external work done W i = W e M p (3θ ) = 18.6 θ M p = 6.2 kNm Case 2: 1.35 D.L + 1.5 C.L + 1.05 W.L Internal Work done, W i = M p ( θ +θ /2 + θ+ θ /2) W i = M p 3θ

Structural Steel

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VK 13.5 kN 13.5 kN

13.5 kN

θ /2 6.8 kN

θ

External work done, W e = 13.5 * 2.5 θ+ 13.5 *2.5θ /2 = 50.6 θ Equating W i = W e , M p (3θ ) = 50.6 θ M p

= 16.9 kNm

Note: Member DE beam mechanism will not govern. (2) Member AC Internal Work done, 342 kNm

 

W i = M p è + M p  è +

11   11  è  + M p  è  13   13 

= 3.69 M p è

θ

20.8 kN

11θ/13 27.2 kN



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External Work done, W e

= 20.8 * 3.25 *

11 è 13

+ 342 *

11 è 13

+

1  11 * 27.2 * 3.25  è 2  13

  

= 383.9è Equating W i = W e , we get 3.69 M pθ = 383.9 θ M p = 104.1 kNm. (3) Member EG 342 kNm

Internal Work done,

 

W i = M p è + M p  è +

11   11  è  + M p  è  13   13 

θ

20.8 kN

= 3.69 M p è 11θ/13 External Work done,

21.2 kN

11 1  11  W e = 20.8 * 3.25 * è + 342 * è + (21.2) * 3.25  è  13 2  13 

= 428.3è Equating W i = W e , we get 3.69 M p θ = 428.3θ

Structural Steel

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M p = 116.1 kNm For members AC & EG, the 1 st load combination will govern the failure mechanism. 4.3.2

Panel Mechanism

Case 1: 1.35 D.L + 1.5 C.L + 1.05 W.L 4.7

5.2 0.62 0.31

0.19

5.2 0.62

4.2 1.0 4.2 1.0

2.6

2.1 0.5

θ

342 kNm

θ

128 kNm

20.8 kN

20.8 kN θ

27.2 kN

θ

1.5 kN

Internal Work done, W i = M p ( θ ) + M p ( θ ) + M p ( θ ) + M p ( θ ) = 4M pθ External Work done, W e W e = 1/2 (27.2) * 6 θ + 20.8 * 3.25θ+ 342θ- 0.31 * 6 θ - 0.62 * 6 θ - 0.62 (6 θ )+ 0.19 * 6 θ + 1.0 *6 θ + 1.0 * 6 θ + 0.5 * 6 θ+1/2 (1.5) * 6 θ + 20.8 * 3.25θ- 128 * θ = 442.14θ

Structural Steel

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Equating W i = W c , we get 4M pθ= 442.14θ M p = 110.5 kNm The second load combination will not govern. 4.3.3 Gable Mechanism Case 1: 1.35 D.L + 1.05 W.L + 1.5 C.L

5.2 0.62 2.6 0.31

5.2 0.62

θ

4.7 θ 0.19 4.2 1.0

θ 4.2 1.0 2.1 0.5

342 kNm

128 kNm

20.8 k

27.2 k

20.8 k θ

1.5 kN

Internal Work done = M ) + M p θ+ M p 2θ + M p (2θ p θ = 6M p θ External Work done, W e = -0.62 * 1 * θ - 0.62 * 2 *θ + 0.19 * 3 * θ + 1.0 * 4 * θ + 1.0 * 5 * θ+ 0.5 * 6 * θ + 5.2 * 2.5 * θ + 5.2 * 5 * θ+ 4.7 * 7.5 * θ + 4.2 * 5 * θ+ 4.2 * 2.5 * θ+ ½* 1.5 * 6 θ + 20.8 * 3.25 * θ- 128*θ W e = 60.56 θ

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Equating W i = W e , we get 6M p = 60.56 θ M p = 10.1 kNm.

Case 2: 1.35 D.L + 1.05L.L + 1.5 C.L 13.5 13.5

θ

θ

13.5 13.5

13.5

θ

6.8

6.8

342

128

20.8 kN 20.8 kN

θ

Internal Work done, W i = M pθ+ M p (2θ ) + M p (2θ ) + M pθ=6M pθ External Work done, W e = 13.5 * 2.5*θ+ 13.5 * 5 * θ+ 13.5 * 7.5θ+ 13.5 * 5 * θ+ 13.5 * 2.5θ128 * θ + 20.8 * 3.25θ = 243.4θ Equating W i = W e , we get 6M pθ= 243.4θ M p

= 40.6 kNm

Structural Steel

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Job No: Sheet 18 of 30 Rev Job Title: Por tal F r ame An alysis and Design

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4.3.4 Combined Mechanism Case1: 1.35 D.L + 1.05 W.L + 1.5 C.L

(i )

θ/2

θ /2

12 m

4.7

5.2 0.62 2.6

5.2 0.62

0.19

4.2 1.0 4.2 1.0 2.1

0.31

0.5

342 kNm

128 kNm

20.8 k

20.8 k θ

27.2 k

θ/2 1.5 kN

Internal Work done, W i = M p ( θ ) + M p ( θ + θ /2) + M /2 + θ /2) + M /2) p ( θ p ( θ = M p ( θ + θ+θ /2 + θ /2 + θ /2 +θ /2 + θ /2) = 4 M pθ External Work done, W e= 1/2 * 27.2 * 6 θ + 20.8 * 3.25* θ+ 342θ- 0.31 * 12 * θ /2 - 0.62 * 11 * θ /2 - 0.62 * 10 *θ /2 + 0.19 * 9 * θ /2 + 1.0 * 8 * θ /2 + 1.0 * 7 * θ/2 + 0.5 * 6* θ /2 + 1/2 (1.5) * 6 θ /2 + 20.8 * 3.25 * θ /2 - 128 * θ /2 – 5.2 * 2.5 * θ /2 – 5.2 * 5.0 * θ /2 – 4.7 * 7.5 * θ /2 – 4.2 * 5 * θ /2 – 4.2 * 2.5 * θ /2 = 411.86 θ Job No: Sheet 19 of 30 Rev Job Title: Portal F r ame An alysis and Design

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Equating W i = W e 4M pθ= 411.86 θ M p = 102.9 kNm

(ii) Internal work done, W i = M p θ/2 + M p ( θ /2 +θ /2) + M p ( θ /2 + θ) +M p θ W i = 4M p θ

θ /2

θ /2

12 m

4.7

5.2 0.62 2.6

5.2 0.62

0.31

0.19

4.2 1.0 4.2 1.0 2.1 0.5

342 kNm

128 kNm

20.8 k

20.8 k θ/2

27.2 k

Structural Steel

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θ 1.5 kN


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External Work done,

è è 1 è è  è  W e = 20.8 * 3.25 * + 342 * + * 27.2 * 6   − 0.31 * 6 * − 0.62 * 7 * 2 2 2 2 2  2  è è è è è è 2 2 2 2 2 2 è è è è + 1.0 * 11 * + 0.5 * 12 * + 4.2 * 5.0 * + 4.2 * 2.5 * + 20.8 * 3.25 è − 128 * è 2 2 2 2 1 + * 1.5 * 6è 2 = 251.35θ

− 0.62 * 8 * + 0.19 * 9 * + 5.2 * 2.5 * + 5.2 * 5.0 * + 4.7 * 7.5 * + 1.0 * 10 *

Equating W i = W e , we get 4M pθ= 251.35θ M p = 62.84 kNm

(iii)

θ /6

θ /6

36 m

0.31

5.2 0.62 2.6

5.2 0.62

4.7 0.19

4.2

1.0 4.2 1.0 2.1 0.5

342 kNm

128 kNm

20.8 kN

27.2 kN

20.8 kN 5 θ /6

Structural Steel

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θ

1.5 kN


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Internal work done, W i

 5è   è 5è   è  = M p   + M p  +  + M p  + è  + M p (è )  6   6 6   6  = 4M p è External Work done, W e =

 5è  5è 1 5è 5è è − 0.31 * 6 * − 0.62 * 35 *  + 342 * + * 27.2 * 6 * 6 2 6 6 6  6 

20.8 * 3.25 * 

è

è

è

è

è

− 0.62* 34 * + 0.19 *33* + 1.0 * 34 * + 1.0 * 35 * + 0.5 * 36 *

6 6 6 6 6 è è è è è + 5.2 * 12.5 * + 5.2 * 10 * + 4.7 * 7.5 * + 4.2 * 5.0 * + 4.2 * 2.5 * 6 6 6 6 6 1 + 20.8 * 3.25 * è − 128 * è + (1.5) (6 * è) 2 W e = 390.92θ Equating W i = W e , we get 4M pθ= 390.92θ M p

= 97.7 kNm

(iv) Internal Work done,

 2   2 è   è  ) W i = M p  è +  + M p  + è  + M p  è  + M p (è 3 3 3 3       = 4M p è

Structural Steel

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VK θ

/3

θ

/3

18 m

0.31

5.2 0.62 2.6

5.2 0.62

4.7 0.19

4.2

1.0 4.2 1.0 2.1 0.5

342 kNm

128 kNm

20.8 kN

20.8 kN 2 θ /3

27.2 kN

θ

1.5 kN

External Work done, W e =

2 1  2   2   2  20.8 *3.25 *  è + * 27.2 * 6 *  è  + 342 * è  − 0.31 * 6 *  è  3 3 2 3 3       è è 2 2  2  − 0.62 * 7 *  è + 5.2 * 5 * è  − 0.62 * 16 * + 0.19 * 15 * + 5.2 * 2.5 * è 3 3 3 3  3  è

è

è

è

è

è

3

3

+ 4.7* 7.5* + 0.19 * 15 * + 1.0 * 16 * + 1.0 * 17 * + 0.5 * 18 * + 4.2 *5.0 * 3 3 3 3 1  è  + 4.2 * 2.5   + 20.8 * 3.25 * è − 128 * è + (1.5) * 6è 2  3  W e = 328.3è Equating W i = W e , we get 4M pθ= 328.3θ M p = 82.1 kNm



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Case 2: 1.35 D.L + 1.05 L.L + 1.5 C.L

(i)

Assuming plastic hinge is formed @ purlin point 2 and 7 and at fixed supports.

θ /6

θ /6

36 m

13.5

13.5

13.5

13.5 13.5 6.8

6.8

342

128 20.8 kN θ

20.8 kN 5θ /6

5 θ /6

 5è   5è è   è  +  + M p  + è  + M p   + M p è 6 6 6 6      

Internal Work done = M p 

= 4M p è



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VK

External Work done =W e

 5è  + 342 *  5è  + 13.5 * 1 2.5 *  è  + 13.5 * 10 *  è  +         6   6   6   6  è è  è  13.5 *7.5 *   + 13.5 * 5.0 * + 13.5 * 2.5 * + 20.8 * 3.25 *è − 128è 6 6  6  = 365.3è 20.8 * 3.25 * 

Equating W i =W e , we get 4M pθ= 365.3θ M p = 91.3 kNm

(ii)

Plastic hinge is formed @ 3 and 7

θ /3

θ /3

18 m

13.5 13.5

13.5

13.5 13.5 6.8

6.8

342

128

20.8 kN 2 θ /3

20.8 k 2 θ /3


θ


36 - {PAGE }

PORTAL FRAMES

Design Project


Date

PU Checked By

Calculation Sheet

Date

VK

Internal work done, W i

2   2 θ   θ  = M p   θ  + M p  θ +  + M p  +θ  + M p (θ)  3   3 3   3  = 4 M pθ

External Work done, W e=

 2 20.8 * 3.25 *  è  3 13.5 * 5.0 *

è 3

 + 342 * 2 è + 13.5 * 2.5 * 2 è + 13.5 * 5.0 * 2 è + 13.5 *7.5 * è +  3 3 3 3  è

+ 13.5 * 2.5 * + 20.8 * 3.25 è − 128θ 3

= 347.7è Equating W i = W e , we get 4M pθ= 347.7 θ M p (iii)

= 86.9 kNm Plastic hinged is formed at 4 and 7

 è   è è   è  Internal Work done = M p   + M p  +  + M p  + è  + M p è  2   2 2   2  = 4 M p è



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PORTAL FRAMES


Design Project

Date

PU Checked By

Calculation Sheet

Date

VK θ /2

θ /2

12 m

13.5 13.5

13.5

13.5

13.5

6.8

6.8

128

342

20.8 kN θ

20.8 kN θ /2

θ /2

External Work done, W e = θ

θ

θ

θ

θ

20 .8 * 3 .25 * + 342 * + 13 .5 * 2 .5 * + 13 .5 * 5 .0 * + 13 .5 * 7 .5 * 2 2 2 2 2 θ

θ

2

2

+ 13 .5 * 5 .0 * + 13 .5 * 2 .5 * + 20 .8 * 3 .25 θ − 128 θ = 296 .3 θ

Equating W i = W e 4M pθ= 296.3θ M p = 74 kNm



36 - {PAGE }

PORTAL FRAMES


Design Project

Date

PU Checked By

Calculation Sheet (iv)

Date

VK

Plastic hinge is formed @ 5 and 7

Internal Work Done,W i =

 è   è  ) + M p (è ) = M p   + M p  + è  + M p (è + è  2   2  W i = 5 M p è

13.5 13.5

θ 13.5

13.5

θ

6m

13.5

6.8

342 20.8 kN

128 20.8 k

θ /2 θ /2

θ

External Work done, W e=

 è  + 342 *  è  + 13.5* 2.5*  è  + 13.5* 5.0*  è  +         2   2   2   2   è  + 13.5 * 5.0 * è + 13.5*2.5è + 20.8 * 3.25 *è − 128è    2 

20.8 *3.25*  13.5* 7.5* W e

= 346.9 è



36 - {PAGE }

PORTAL FRAMES

Design Project


Date

PU Checked By

Calculation Sheet

Date

VK

Equating W i = W e 5M pθ= 346.9 * θ M p = 69.4 kNm Design Plastic Moment = 116.1 kNm.

5.0 DESI GN For the design it is assumed that the frame is adequately laterally braced so that it fails by forming mechanism. Both the column and rafter are analysed assuming equal plastic moment capacity. Other ratios may be adopted to arrive at an optimum design solution.

5.1 Selecti on of section Plastic Moment capacity required= 116 kNm Required section modulus, Z = M p / f yd

6   116*10   = 

250 1.15 = 533 .6 * 10 3 mm 3 From IS: 800-1984 (Annexure F) ISMB 300 @ 0.46 kN/ m provides Z p = 683 * 10-3 mm 3 b = 140 mm T i = 13.1 mm A = 5.87 * 10 3 mm 2 t w =7.7 mm r xx =124 mm r yy =28.6 mm



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PORTAL FRAMES

Design Project


Date

PU Checked By

Calculation Sheet

Date

VK

5.2 Secondary D esign Consider ations 5.2.1 Check for L ocal buckli ng of f langes and webs Flanges b f T 1

=

136 f y

b f = 140/2 = 70 mm T 1 = 13.1 mm t = 7.7 mm b f T 1

=

70 13.1

= 5.34 < 8.6

Web

 1120 1600  P      ≤ −  t  f y f y  P y      300  1120 1600 ≤ − (0.27 ) 7.7  250 y 250  y  38.9 ≤ 68, Hence O. K d 1

5.2.2 Ef fect of axial f orce Maximum axial force in column, P = 40.5 kN

Structural Steel

Version II

Job No: Sheet 30 of 30 Rev Job Title: Portal F r ame An alysis and Design Worked Example: 1

36 - {PAGE }

86454170 Portal Frame Design Worked Example

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