Portal Frame Design-worked Example

PORTAL FRAMES

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

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

A

30 m

6m

3.25 m

G 15 m

15 m

Load

1.0 Load Calculation 1.1

Dead Load

Weight of asbestos sheeting Fixings Services Weight of purlin

= = = =

Total load /m 2

=

Version II

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

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PORTAL FRAMES


Structural Steel Design Project Calculation Sheet Dead load/m run

1.2

= 0.395 * 5 = 1.975 kN / m ≈ 2.0 kN/m

Live Load

Angle of rafter = tan-1 (3/7.5) = 21.80 From IS: 875 (part 2) – 1987; Table 2 (cl 4.1), Live load / m run = {0.75 − 0.02 (21.8 − 10)} * 5 = 2.57 kN/m 1.3 Crane Loading Overhead electric crane capacity

= 300 kN

Approximate weight of crane girder

= 300 kN

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 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

MB = 0

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PORTAL FRAMES

Structural Steel Design Project Calculation Sheet


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

242 kN

3.8 m 5m

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) = 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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PORTAL FRAMES

Structural Steel Design Project Calculation Sheet 1.4


Wind Load

Design wind speed, Vz = k 1 k 2 k 3 Vb 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,

Vz = k 1 k 2 k 3 Vb Vz = 1 * 0.8 *1 * 50 Vz = 40 m/sec

Basic design wind pressure, Pd = 0.6*Vz2 = 0.6 * (40)2 = 0.96 kN/m 2 1.4.1. Wind Load on individual surfaces

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

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PORTAL FRAMES


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

Cpe

Cpi

Windward

Leeward

00

0.7

-0.25

900

-0.5

-0.5

0.2 -0.2 0.2 -0.2

Cpe – Cpi Wind ward

Lee ward

Total wind(kN) (C pe-Cpi )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


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

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

Pressure Coefficient

0

0

900

2.0

Cpe

Cpe

Cpi

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

Cpe – Cpi Wind ward

Lee ward

-0.528 -0.128 -0.9 -0.5

-0.6 -0.2 -0.9 -0.5

Total Wind Load(kN) (C pe – Cpi) Apd Wind Lee ward ward Int. Int. -20.4 -23.2 -4.8 -7.8 -34.8 -34.8 -19.4 -19.4

Equivalent Load Calculation

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 Frame Analysis and Design

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Design Project

Worked Example: 1 Made By

Date PU

Checked By

Calculation Sheet

Date VK

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

Partial Safety Factors

3.1

Load Factors

For dead load, γf = 1.35 For major live load, γf = 1.5 For minor live load, γf = 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

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Design Project


Date PU

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Calculation Sheet (ii)

Date VK

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 5.2 @ eaves = = 2.6 kN 2 2 w @ intermediate points on leeward side w = 1.35 * 5.0 – 1.05 * 7.8/3 cos21.8 = 4.2 kN w 4.2 @ eaves = = 2.1 kN 2 2 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

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Design Project


Date PU

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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  2 + 0.3  Dead load on the column =  2  * 5 = 5.75 kN   Factored moment @ B & F = 1.35 * 5.75 * 0.6 = 4.6 kNm

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Design Project

Date PU

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Calculation Sheet

Date VK

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 kN 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

Factored Load (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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PORTAL FRAMES

Design Project


Date PU

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Calculation Sheet 4.2.2

Date VK

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 kN

6m 20.8 kN 3.25 m

15 m Factored Load (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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Design Project


Date PU

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

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Design Project

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Calculation Sheet

Date VK

13.5 kN 13.5 kN 13.5 kN

θ/2

6.8 kN

θ External work done, We = 13.5 * 2.5 θ + 13.5 *2.5θ/2 = 50.6θ Equating Wi = We, Mp (3θ) = 50.6 θ Mp

= 16.9 kNm

Note: Member DE beam mechanism will not govern. (2) Member AC Internal Work done, 342 kNm 11    11  Wi = M pè + M p  è + è + M p  è 13    13 

θ

20.8 kN

= 3.69 M pè 11θ /13 27.2 kN



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Design Project


Date PU

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Calculation Sheet

Date VK

External Work done, We = 20.8 * 3.25 *

11 11 1  11  è + 342 * è + * 27.2 * 3.25  è  13 13 2  13 

= 383.9è Equating Wi = We, we get 3.69 Mp θ = 383.9 θ Mp = 104.1 kNm. (3) Member EG Internal Work done,

342 kNm

11    11  Wi = M pè + M p  è + è + M p è 13    13  = 3.69 M pè

20.8 kN

θ

11θ /13 External Work done, We = 20.8* 3.25 *

21.2 kN

11 1  11  è + 342 * è + (21.2) * 3.25  è  13 2  13 

= 428.3è Equating Wi = We, we get 3.69 Mp θ = 428.3θ

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Design Project

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Calculation Sheet

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Mp = 116.1 kNm For members AC & EG, the 1st 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

342 kNm

2.1 0.5

θ

θ

128 kNm

20.8 kN

20.8 kN θ

27.2 kN

θ

1.5 kN

Internal Work done, Wi = Mp (θ) + Mp (θ) + Mp (θ) + Mp (θ) = 4Mp θ External Work done, We We = 1/2 (27.2) * 6θ + 20.8 * 3.25θ + 342θ - 0.31 * 6θ - 0.62 * 6θ - 0.62 (6θ)+ 0.19 * 6θ + 1.0 *6θ + 1.0 * 6θ + 0.5 * 6θ+1/2 (1.5) * 6θ + 20.8 * 3.25θ - 128 * θ = 442.14θ

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Design Project

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Calculation Sheet

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Equating Wi = Wc, we get 4Mp θ = 442.14θ Mp = 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 kN

27.2 kN

20.8 kN θ

1.5 kN

Internal Work done = Mp θ + Mp 2θ + Mp (2θ) + Mp θ = 6Mp θ External Work done, We = -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 * 6θ + 20.8 * 3.25 * θ - 128*θ We = 60.56θ

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Design Project

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Calculation Sheet

Date VK

Equating Wi = We, we get 6Mp = 60.56θ Mp = 10.1 kNm. Case 2: 1.35 D.L + 1.05L.L + 1.5 C.L 13.5 13.5 13.5 6.8

θ

θ

θ

13.5 13.5

6.8

342

128

kNm

kNm

20.8 kN

20.8 kN

θ

Internal Work done, Wi = Mp θ + Mp (2θ) + Mp (2θ) + Mp θ =6Mp θ External Work done, We = 13.5 * 2.5*θ + 13.5 * 5 * θ + 13.5 * 7.5θ + 13.5 * 5 * θ + 13.5 * 2.5θ 128 * θ + 20.8 * 3.25θ = 243.4θ Equating Wi = We, we get 6Mp θ = 243.4θ Mp

= 40.6 kNm

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Design Project

Date PU

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Calculation Sheet

Date VK

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

5.2 0.62 2.6

5.2 0.62

4.7 0.19

4.2 1.0 4.2 1.0 2.1

0.31

0.5

342 kNm

128 kNm

20.8 kN

20.8 kN θ

27.2 kN

θ /2

1.5 kN

Internal Work done, Wi = Mp (θ ) + Mp (θ + θ/2) + Mp (θ/2 + θ/2) + Mp (θ/2) = Mp (θ + θ +θ/2 + θ/2 + θ/2 +θ/2 + θ/2) = 4 Mp θ External Work done, We= 1/2 * 27.2 * 6θ + 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) * 6θ/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 = 411.86θ Job No: Sheet 19 of 30 Rev Job Title: Portal Frame Analysis and Design

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Design Project

Date PU

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Calculation Sheet

Date VK

Equating Wi = We 4Mp θ = 411.86θ Mp = 102.9 kNm (ii) Internal work done, Wi = Mp θ /2 + Mp (θ /2 +θ/2) + Mp (θ /2 + θ ) +Mp θ Wi = 4Mp θ

θ /2

θ /2

12 m

5.2 0.62 2.6

5.2 0.62

4.7 0.19

0.31

4.2 1.0 4.2 1.0 2.1 0.5

342 kNm

128 kNm

20.8 kN

20.8 kN θ /2

27.2 kN


θ

1.5 kN


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Design Project

Date PU

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Calculation Sheet

Date VK

External Work done, è è 1 è è è  + 342* + * 27.2 * 6   − 0.31* 6 * − 0.62* 7 * 2 2 2 2 2 2 è è è è è è − 0.62* 8* + 0.19 * 9 * + 5.2 * 2.5* + 5.2 * 5.0 * + 4.7 * 7.5 * + 1.0 * 10 * 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 We = 20.8 * 3.25 *

Equating Wi = We, we get 4Mp θ = 251.35θ Mp = 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 5θ /6

Structural Steel

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θ

1.5 kN


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Calculation Sheet

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Internal work done, Wi  5è   è 5è  è  = M p + M p  +  + M p  + è  + M p (è ) 6   6  6 6  = 4M pè External Work done, We =  5è  5è 1 5è 5è è 20.8*3.25 *   + 342 * + * 27.2 * 6 * − 0.31* 6 * − 0.62* 35 * 6 2 6 6 6  6  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 We = 390.92θ Equating Wi = We, we get 4Mp θ = 390.92θ Mp

= 97.7 kNm

(iv) Internal Work done, è 2  2 è  Wi = M p  è  + M p  è +  + M p  + è  + M p (è ) 3 3  3  3  = 4M pè

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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 2θ /3

27.2 kN

θ

1.5 kN

External Work done, We = 2 1 2  2  2  20.8*3.25 *  è  + 342 * è + * 27.2 * 6 *  è  − 0.31* 6 *  è  3 2 3  3  3  è è 2 2 2  − 0.62* 7 *  è  − 0.62 * 16 * + 0.19*15 * + 5.2 * 2.5* è + 5.2* 5 * è 3 3 3 3 3  è è è è è è + 4.7* 7.5* + 0.19* 15 * + 1.0* 16 * + 1.0 * 17 * + 0.5 * 18* + 4.2 *5.0 * 3 3 3 3 3 3 1 è  + 4.2* 2.5   + 20.8* 3.25 *è − 128* è + (1.5)* 6è 2  3 We = 328.3è Equating Wi = We, we get 4Mp θ = 328.3θ Mp = 82.1 kNm



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Design Project

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Calculation Sheet

Date VK

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 kNθ

20.8 kN 5θ /6 /6

5θ /6

 5è   5è è  è  Internal Work done = M p   + M p  +  + M p  + è  + M pè 6  6   6 6  = 4M pè



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Date VK

External Work done =We  5è   5è  è  è  20.8 * 3.25 *   + 342 *   + 13.5 * 1 2.5 *   + 13.5 * 10 *   +  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 Equating Wi =We, we get 4Mp θ = 365.3θ Mp = 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 20.8 kN 2θ /3

128 20.8 kN θ 2θ /3



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Internal work done, θ 2  2 θ  Wi = M p  θ  + M p  θ +  + M p  +θ  + M p (θ ) 3 3  3  3  = 4 M pθ External Work done, We= 2 2 2 è 2  20.8 * 3.25 *  è  + 342 * è + 13.5 * 2.5 * è + 13.5 * 5.0 * è + 13.5 *7.5 * + 3 3 3 3 3  è è 13.5 * 5.0 * + 13.5 * 2.5 * + 20.8 * 3.25 è − 128θ 3 3 = 347.7è Equating Wi = We, we get 4Mp θ = 347.7θ Mp (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 pè



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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, We = θ θ θ θ θ + 342* + 13.5 * 2.5* + 13.5* 5.0 * + 13.5 * 7.5 * 2 2 2 2 2 θ θ + 13.5 * 5.0* + 13.5 * 2.5 * + 20.8* 3.25 θ − 128 θ 2 2 = 296.3θ 20.8* 3.25 *

Equating Wi = We 4Mp θ = 296.3θ Mp = 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,Wi = è  è  = M p   + M p  + è  + M p (è + è ) + M p (è ) 2 2  Wi = 5 M pè

θ

13.5 13.5

θ

13.5

13.5

6m

13.5

6.8

342 20.8 kN

128 θ /2

kN 20.8 kN θ /2

θ

External Work done, We= è  è  è  è  20.8 *3.25*   + 342 *   + 13.5* 2.5*   + 13.5* 5.0*   + 2 2 2 2 è  13.5* 7.5*   + 13.5 * 5.0 * è + 13.5*2.5è + 20.8 * 3.25 *è − 128è 2  We = 346.9 è



36 - {PAGE }

PORTAL FRAMES

Design Project


Date PU

Checked By

Calculation Sheet

Date VK

Equating Wi = We 5Mp θ = 346.9 * θ Mp = 69.4 kNm Design Plastic Moment = 116.1 kNm. 5.0 DESIGN 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 Selection of section Plastic Moment capacity required= 116 kNm Required section modulus, Z = Mp / f yd  116*106   = 250 1.15 = 533.6 * 10 3 mm 3 From IS: 800-1984 (Annexure F) ISMB 300 @ 0.46 kN/ m provides Zp = 683 * 10-3 mm3 b = 140 mm Ti = 13.1 mm A = 5.87 * 10 3 mm2 t w =7.7 mm rxx =124 mm ryy =28.6 mm



36 - {PAGE }

PORTAL FRAMES

Design Project


Date PU

Checked By

Calculation Sheet

Date VK

5.2 Secondary Design Considerations 5.2.1 Check for Local buckling of flanges and webs Flanges bf

=

T1

136 fy

bf = 140/2 = 70 mm T1 = 13.1 mm t = 7.7 mm bf T1

=

70 = 5.34 < 8.6 13.1

Web d1  1120 1600  P    ≤ − t f y  Py    f y  300  1120 1600 ≤ − (0.27 ) 7.7  250y 250 y  38.9 ≤ 68,Hence O. K 5.2.2 Effect of axial force Maximum axial force in column, P = 40.5 kN

Structural Steel

Version II

Job No: Sheet 30 of 30 Rev Job Title: Portal Frame Analysis and Design Worked Example: 1

36 - {PAGE }

PORTAL FRAMES

Design Project

Made By

Date PU

Checked By

Calculation Sheet

Date VK

Axial load causing yielding, Py = f yd * A 250 = = 5.87*10 3 1.15 = 1276 kN P 40.5 = = 0.03 < 0.15 Py 1276 Therefore the effect of axial force can be neglected. 5.2.3 Check for the effect of shear force Shear force at the end of the girder = P- w/2 = 40.5 -6.8 kN = 33.7 kN Maximum shear capacity Vym , of a beam under shear and moment is given by Vym = 0.55 Aw* f yd / 1.15 = 0.55 * 300* 7.7* 250/1.15 =276.2 kN>> 33.7 kN Hence O.K.

Version II

36 - {PAGE }

Portal Frame Design-worked Example

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