Analysis of Industrial Gable Frames - Structville

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  Statically Determinate Structures  Steel Structures  Structural Analysis  Analysis of Industrial Gable Frames

Analysis of Industrial Gable Frames Frames by Ubani Obinna Ranks on March 12, 2018 in Statically Determinate Structures, Structures, Steel Structures, Structures, Structural Analysis

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1.0 Introduction

Industrialization Industrializatio n is one of the major keys to development and sustainable economy. Industrial structures are usually very easy to identify because of their unique features that are quite different from residential or commercial buildings. Engineers are often tasked with analysing and designing industrial frames, and to simplify the analysis for manual calculations, they are usually idealised as 2D plane frames.

2.0 Solved Example

In this post, a gable industrial frame structure is subjected to a load regime as shown below. The frame is hinged at point D, and it is desired to obtain the internal forces (bending moment, axial forces, and shear forces) due to the externally applied load.

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N/B: In practical construction, it not very advisable to hinge the structure at the crown due to the problem of excessive deflection.

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Geometrical Properties

Angle of inclination of the rafter θ = tan-1(2.5/6) = 22.619° cos θ = 0.923 sin θ = 0.385

Length of rafter (z) = sqrt(6 2 + 2.52) = 6.5m

Notations: Example: NCR = Axial Load at point C, just to the Right

FOLLOW BY EMAIL Let ∑MG = 0


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12Ay – [(10 × 122)/2] – (25 × 11) + (7 × 4) - (16 × 1) = 0 12Ay = 983 Ay = 81.917 kN

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Let ∑MDL = 0

6Ay – 8Ax - (25 × 5) – (7 × 4) - [(10cos22.619° × 6.52)/2] = 0

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6(81.917) – 8Ax - (25 × 5) – (7 × 4) - [(9.23 × 6.52)/2] = 0 Structural Design of Steel Connection

8Ax = 143.518 kN Ax = 17.939 kN

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Let ∑MA = 0

12Gy – [(10 × 122)/2] – (25 × 1) - (7 × 4) - (16 × 11) = 0 12Gy = 949

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Gy = 79.083 kN

Let ∑MDR = 0

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6Gy – 8Gx - (16 × 5) – [(10cos22.619° × 6.52)/2] = 0 6(79.083) – 8Gx - (16 × 5) – [(9.23 × 6.52)/2] = 0 8Gx = 199.515 kN

PRACTICAL ANALYSIS A OF STEEL ROOF TRUSSE EUROCODE 3: A SAMPLE

Gx = 24.939 kN

Equilibrium Check

What is wrong with this se structural detailing?

All downward vertical forces = (10 × 12) + 25 + 16 = 161 kN Upward reactive forces = 81.917 + 79.083 = 161 kN

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All rightward horizontal forces = Ax + 7kN = 17.939 + 7 = 24.939 kN All leftward horizontal forces = Gx = 24.939 kN

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Section A - B B (0 ≤ y ≤ 4.0m)

(i) Bending moment

Geotechnical Design of Ca

M y = -Ax. y = -17.939 y

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At y = 0; MA = 0

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At y = 4.0m; MBB = (-17.939 × 4) = -71.756 kNm How To Apply Wind Load o Buildings

(ii) Shear Force

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Qy = ∂M y /∂y = -17.939 kN Simple Proofs Why Shorter More Critical in Slab Desig

(iii) Axial Force

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Ny + Ay = 0 Ny = -Ay = -81.917 kN (compression)

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Section 1 - B R (0 ≤ x ≤ 1.0m)

(i) Bending moment M x = -25. x

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Q x = ∂M x /∂x = -25 kN

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(iii) Axial Force

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At x = 1.0m; MBR = (-25 × 1) = -25 kNm

(ii) Shear force


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N x - 7kN = 0 N x = 7kN (tension)

Section BUP - CB (4 ≤ y ≤ 5.5m)

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(i) Bending moment

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M y = -Ax. y + (25 × 1) - 7( y - 4)

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At y = 4m; MBUP = -(17.939 × 4) + 25 = -46.756 kNm At y = 5.5m; MCB = -(17.939 × 5.5) + 25 - 7(1.5) = -84.1645 kNm

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(ii) Shear force Qx + Ax + 7kN = 0

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Q x = -17.939 - 7 = -24.939 kN

BLOG ARCHIVE (iii) Axial force Ny + Ay - 25kN = 0

March (4)

N y = -81.917 + 25 = -56.917 kN (compression)

TAGS BEAMS

(24)

(1)

CABLE STRUCTURES

(1)

CONCRETE

CONSTRUCTION

Section CR - DL (0 ≤ z ≤ 6.5m) (Pay careful attention here)

(i) Bending moment

(5)

BRIDGES

(1

CONNECTION DESIGN

(12)

DYNAMIC ANALYSIS

(1)

ENGINEERING MATHEMATICS

(11)

EUROCODE 3

(10)

COLUMN

EUROCO

EUROCODE 7

(3)

The bending moment transferred transferred to the rafter at node C is -84.1645 kNm Summation of vertical force transferred ∑V = Ay - 25 = 81.917 - 25 = 56.917 kN

FORCE METHOD

Summation of horizontal force transferred ∑H = Ax + 7 = 17.939 + 7 = 24.939 kN

FRAMES GUIDES

(9)

(17)

(7)

FOUNDATIONS

GEOTECHNICAL ENGINEERI

(8)

INFORMATIVE POSTS

(11)

M z = (∑V.cos22.619-9°. z) - (∑H.sin22.619°. z) - [(10cos22.619° × z2)/2] - 84.1645 M z = 52.539 z - 9.591 z - 4.6154z2 - 84.1645 M z = -4.6154 z2 + 42.9425 z - 84.1645

LITERATURE REVIEW

(1) (1)

MATERIALS ENGINEERING

(1)

PLASTIC ANALYSIS

PROGRAMMING

At z = 0; MCR = -84.1645 kNm

PROMOTIONS

At z = 6.5m; MCB = -4.6154(6.5)2 + 42.948(6.5) - 84.1645 = -195 + 279.126 - 84.1645 = 0

REINFORCED CONCRETE DESIGN

(4)

OPINION

QUANTITY ESTIMATION

(33)

(5)

STAAD PRO

Maximum span moment

M z = -4.6154 z2 + 42.9425 z - 84.1645

STATICALLY DETERMINATE STRUCTURES

(10) (1

STATICALLY INDETERMINATE STRUCTURES

Maximum moment occurs at the point of zero shear

STEEL STRUCTURES

(15)

∂M z/∂z = -9.23z + 42.9425 = 0

z = 42.9425/9.23 = 4.652m

STRUCTURAL ANALYSIS THIN PLATES

Mmax = -4.6154(4.652)2 + 42.9425(4.652) - 84.1645 = -99.882 + 199.768 - 84.1645 = 15.7215 kNm

TRUSSES

(7)

(2)

(48)

TALL BUILDI

TIMBER STRUCTURES WIND LOAD

(1)

(ii) Shear force ∂M z/∂z = (∑V.cos22.619) - ( ∑H.sin22.619)

At z = 0; QCR = (56.917 × 0.923) - (24.939 × 0.385) = 42.933 kN At z = 6.5m; QDL = [(56.917 - (10 × 6)) × 0.923] - (24.939 × 0.385) = -2.8456 - 9.6015 = -12.447 kN

(iii) Axial force

N z = -(∑V.sin22.619) - (∑H.cos22.619)


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At z = 0; NCR = -(56.917 × 0.385) - (24.939 × 0.923) = -21.913 - 23.018 = -44.931 kN At z = 6.5m; NDL = [-(56.917 - (10 × 6)) × 0.385] - (24.939 × 0.923) = 1.187 - 23.018 = -21.831 kN

Coming from the right

Section G - F B (0 ≤ y ≤ 4.0m)

(i) Bending moment M y = -Gx. y = -24.939 y At y = 0; MG = 0 At y = 4.0m; MFB = (-24.939 × 4) = -99.756 kNm

(ii) Shear Force Qy = ∂M y /∂y = +24.939 kN (note that we are coming from right to left) (iii) Axial Force Ny + Gy = 0

Ny = -Gy = -79.083 kN (compression)

Section 2 - F R (0 ≤ x ≤ 1.0m)

(i) Bending moment M x = -16. x At x = 0; M2 = 0 At x = 1.0m; MFR = (-16 × 1) = -16 kNm

(ii) Shear force Q x = ∂M x /∂x = -16 kN (iii) Axial Force N x - 0 = 0

N x = 0 (no force)

Section F UP - E B (4 ≤ y ≤ 5.5m)

(i) Bending moment M y = -Gx. y + (16 × 1) At y = 4m; MFUP = -(24.939 × 4) + 16 = -83.756 kNm At y = 5.5m; MEB = -(24.939 × 5.5) + 16 = -121.1645 kNm

(ii) Shear force Qx - Gx = 0 Q x = 24.939 kN (iii) Axial force Ny + Ay - 16kN = 0 N y = -79.083 + 16 = -63.083 kN (compression)

Section E UP - DR (0 ≤ z ≤ 6.5m)

(i) Bending moment The bending moment transferred transferred to the rafter at node C is -121.1645 kNm Summation of vertical force transferred ∑V = Gy - 16 = 79.083 - 16 = 63.083 kN Summation of horizontal force transferred ∑H = Gx = 24.939 kN

M z = (∑V.cos22.619-9°. z) - (∑H.sin22.619°. z) - [(10cos22.619° × z2)/2] - 121.1645 M z = 58.225 z - 9.5915 z - 4.6154z2 - 121.1645


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M z = -4.6154 z2 + 48.6335 z - 121.1645

At z = 0; MER = -121.1645 kNm At z = 6.5m; MCB = -4.6154(6.5)2 + 48.6335(6.5) - 121.1645 = -195 + 316.1175 - 121.1645 = 0

Maximum span moment

M z = -4.6154 z2 + 48.6335 z - 121.1645 Maximum moment occurs at the point of zero shear ∂M z/∂z = -9.23z + 48.6335 = 0

z = 48.6335/9.23 = 5.269m

Mmax = -4.6154(5.269)2 + 48.6335(5.269) - 121.1645 = -128.134 + 256.2499 - 121.1645 = 6.9514 kNm

(ii) Shear force ∂M z/∂z = -(∑V.cos22.619) + (∑H.sin22.619)

At z = 0; QCR = (63.083 × 0.923) - (24.939 × 0.385) = -58.2256 + 9.6015 = -48.624 kN At z = 6.5m; QDL = -[(63.083 - (10 × 6)) × 0.923] + (24.939 × 0.385) = -2.8456 + 9.6015 = 6.756kN

(iii) Axial force

N z = -(∑V.sin22.619) - (∑H.cos22.619)

At z = 0; NCR = -(63.083 × 0.385) - (24.939 × 0.923) = -24.287 - 23.018 = -47.305kN At z = 6.5m; NDL = [-(63.083 - (10 × 6)) × 0.385] - (24.939 × 0.923) = -1.187 - 23.018 = -24.205 kN

(b) Internal Stresses Diagram

(a) Bending Moment Diagram

(b) Shear Force Diagram


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(c) Axial Force Diagram

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1 COMMENT:

MAHMOOD MUHAMMOD MARCH 12, 2018 AT 11:54 PM

Its a very informative article for structural engineers. You can get free structural engineering resources from my website: https://worldcentre.me Thanks Mahmood Reply

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