This document provides load calculations and structural analysis for a single span portal frame with a 15 m span, 6 m column height, and 3 m rafter rise. It includes:
1) Calculation of dead, live, wind, and crane loads. Crane loads include vertical loads up to 375 kN and transverse loads of 13.9 kN.
2) Equivalent load calculations to simplify distributed loads.
3) Analysis of the frame under key load combinations, considering beam, sway, gable, and combined mechanisms. Moments up to 239.9 kNm and horizontal loads up to 14.6 kN are calculated.
4) Identification of critical load combinations for plastic analysis
Bridge foundation includes its loading conditions and how it acts on structure. This present work also deals with the basic classifications of bridge foundation. It gives emphasis on different components and purposes of bridge foundation. While working on this specific thing one should understand the concept and follow the steps.
About Suspension Bridges:
A suspension bridge is a type of bridge in which the deck (the load-bearing portion) is hung below suspension cables on vertical suspenders. The first modern examples of this type of bridge were built in the early 19th century. Bridges without vertical suspenders have a long history in many mountainous parts of the world.
Bridge foundation includes its loading conditions and how it acts on structure. This present work also deals with the basic classifications of bridge foundation. It gives emphasis on different components and purposes of bridge foundation. While working on this specific thing one should understand the concept and follow the steps.
About Suspension Bridges:
A suspension bridge is a type of bridge in which the deck (the load-bearing portion) is hung below suspension cables on vertical suspenders. The first modern examples of this type of bridge were built in the early 19th century. Bridges without vertical suspenders have a long history in many mountainous parts of the world.
tunnel lining may be permanent or temporary based upon their use and requirement. design of lining is done in two parts one is temporary or initial lining design and other is permanent design of the lining. empirical and theoretical methods are major design methods.
Modelling Building Frame with STAAD.Pro & ETABS - Rahul LeslieRahul Leslie
A basic tutorial to learn the concepts of modelling RC building in an Analysis/Design package -- STAAD.Pro & ETABS are in focus here, but concepts are applicable for any package. Good for novice in structural designing, and also B.Tech / BE / BSc (Engg) / BS students wising to do 'design of multi-storied RC building' as their final year project.
DESIGN OF CONTINUOUS MEMBERS IN PRESTRESSEDLOGESH S
• A continuous beam is having more than one span is carried by several supports.
• It is mainly used in bridge construction.
• Simple beam cannot be used for large spans, as it requires more strength and stiffness.
• But continuous PSC beam not only provides adequate strength and stiffness, but also provides sufficient ductility.
tunnel lining may be permanent or temporary based upon their use and requirement. design of lining is done in two parts one is temporary or initial lining design and other is permanent design of the lining. empirical and theoretical methods are major design methods.
Modelling Building Frame with STAAD.Pro & ETABS - Rahul LeslieRahul Leslie
A basic tutorial to learn the concepts of modelling RC building in an Analysis/Design package -- STAAD.Pro & ETABS are in focus here, but concepts are applicable for any package. Good for novice in structural designing, and also B.Tech / BE / BSc (Engg) / BS students wising to do 'design of multi-storied RC building' as their final year project.
DESIGN OF CONTINUOUS MEMBERS IN PRESTRESSEDLOGESH S
• A continuous beam is having more than one span is carried by several supports.
• It is mainly used in bridge construction.
• Simple beam cannot be used for large spans, as it requires more strength and stiffness.
• But continuous PSC beam not only provides adequate strength and stiffness, but also provides sufficient ductility.
Analysis and Design of Residential building.pptxDP NITHIN
Complete introduction to the design and design concepts, design of structural
members like slabs, beams, columns, footing etc. along with their calculation and
Detailing through structural drawings.
Comparative Study of Pre-Engineered and Conventional Steel Frames for Differe...irjes
In this paper, the conventional steel frames having triangular Pratt truss as a roofing system of 60 m
length, span 30m and varying bay spacing 4m, 5m and 6m respectively having eaves level for all the portals is at
10m and the EOT crane is supported at the height of 8m from ground level and pre-engineered steel frames of
same dimensions are analyzed and designed for wind zones (wind zone 2, wind zone 3, wind zone 4 and wind
zone 5) by using STAAD Pro V8i. The study deals with the comparative study of both conventional and preengineered
with respect to the amount of structural steel required, reduction in dead load of the structure.
Structural Analysis of a Bungalow Reportdouglasloon
Taylor's University Lakeside Campus
School of Architecture, Building & Design
Bachelor of Science (Hons) in Architecture
Building Structures (ARC 2523 / BLD 60103)
Project 2: Structural Analysis of a Bungalow
Design of Various Types of Industrial Buildings and Their ComparisonIRJESJOURNAL
ABSTRACT :- In this paper Industrial Steel truss Building of 14m x 31.50m, 20m x 50m, 28m x 70m and bay spacing of 5.25m, 6.25m and 7m respectively having column height of 6m is compared with Pre-engineering Buildings of same dimension. Design is based on IS 800-2007 (LSM) Load considered in modeling are Dead load, Live Load, Wind load along with the combinations as specified in IS. Analysis results are observed for column base as hinge base. Results of Industrial steel truss buildings are compared with the same dimensions of Pre-Engineering Building
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
NO1 Uk best vashikaran specialist in delhi vashikaran baba near me online vas...Amil Baba Dawood bangali
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About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
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Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
1. PORTAL FRAMES
Version II
36 - 19
15 m
30 m
5 m c / c
3 m
6 m
3.25 m
15 m
0.6 m 0.6 m
A
B
C
D
E
F
G
Job No: Sheet 1 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
Made By
PU
Date
Structural Steel
Design Project
Calculation Sheet
Checked By
VK
Date
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.
Load
1.0 Load Calculation
1.1 Dead Load
Weight of asbestos sheeting = 0.17 kN/m2
Fixings = 0.025 kN/m2
Services = 0.100 kN/m2
Weight of purlin = 0.100 kN/m2
---------------
Total load /m2
= 0.395 kN/m2
---------------
2. PORTAL FRAMES
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13.8 m
1 m
( 300 + 60 ) 300
B F
6.9 m
Job No: Sheet 2 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Dead load/m run = 0.395 * 5
= 1.975 kN / m
2.0 kN/m
1.2 Live Load
Angle of rafter = tan-1
(3/7.5) = 21.80
From IS: 875 (part 2) – 1987; Table 2 (cl 4.1),
= 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.
MB = 0
5
*
10)
(21.8
0.02
0.75
run
m
/
load
Live
3. PORTAL FRAMES
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Job No: Sheet 3 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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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.
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 *375
242
9
= 13.9 kN
(By proportion)
242 kN 242 kN
3.8 m
5 m
B' B
4. PORTAL FRAMES
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Job No: Sheet 4 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Date
1.4 Wind Load
Design wind speed, Vz = k1 k2 k3 Vb
From Table 1; IS: 875 (part 3) – 1987
k1 = 1.0 (risk coefficient assuming 50 years of design life)
From Table 2;IS: 875 (part 3) – 1987
k2 = 0.8 (assuming terrain category 4)
k3 = 1.0 (topography factor)
Assuming the building is situated in Chennai, the basic wind speed is
50 m /sec
Design wind speed, Vz = k1 k2 k3 Vb
Vz = 1 * 0.8 *1 * 50
Vz = 40 m/sec
Basic design wind pressure, Pd = 0.6*Vz
2
= 0.6 * (40)2
= 0.96 kN/m2
1.4.1. Wind Load on individual surfaces
The wind load, WL acting normal to the individual surfaces is given by
WL = (Cpe – Cpi ) A*Pd
(a) Internal pressure coefficient
Assuming buildings with low degree of permeability
Cpi = 0.2
5. PORTAL FRAMES
Version II
36 - 23
h
w
w
L
plan elevation
Job No: Sheet 5 of 30 Rev
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(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
Exposed area of wall per frame @ 5 m c/c
is
A = 5 * 6 = 30 m2
For walls, A pd = 30 * 0.96 = 28.8 kN
Table 1 (a): Total wind load for wall
Cpe Cpi Cpe – Cpi Total wind(kN)
(Cpe-Cpi )Apd
Wind
Angle
Wind-
ward
Lee-
ward
Wind
ward
Lee
ward
Wind
ward
Lee
ward
0.2 0.5 -0.45 14.4 -12.9
00
0.7 -0.25
-0.2 0.9 -0.05 25.9 -1.4
0.2 -0.7 -0.7 -20.2 -20.2
900
-0.5 -0.5
-0.2 -0.3 -0.3 -8.6 -8.6
(b) For roofs
Exposed area of each sloping roof per frame @ 5 m c/c is
2
2
2
m
40.4
7.5
3.0
*
5
A
6. PORTAL FRAMES
Version II
36 - 24
15 m
Job No: Sheet 6 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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For roof, Apd = 38.7 kN
Table 1 (b): Total wind load for roof
Pressure Coefficient Cpe – Cpi Total Wind Load(kN)
(Cpe – Cpi) Apd
Cpe Cpe Wind
ward
Lee
ward
Wind
ward
Lee
ward
Wind
angle
Wind Lee
Cpi
Int. Int.
-0.328 -0.4 0.2 -0.528 -0.6 -20.4 -23.2
00
-0.328 -0.4 -0.2 -0.128 -0.2 -4.8 -7.8
-0.7 -0.7 0.2 -0.9 -0.9 -34.8 -34.8
900
-0.7 -0.7 -0.2 -0.5 -0.5 -19.4 -19.4
2.0 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,
kN
6
15
*
2.0
WD 5
2.2 Superimposed Load
Superimposed Load = 2.57 kN/m
Concentrated load , purlin
kN/
6.4
6
15
*
2.57
L
W
7. PORTAL FRAMES
Version II
36 - 25
Job No: Sheet 7 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
Made By
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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 Partial Safety Factors
3.1 Load Factors
For dead load, f = 1.5
For major live load, f = 1.5
For minor live load or defined live load, f = 1.05
3.2 Material Safety factor
m = 1.10
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) 1.5D.L + 1.05 C .L + 1.5 W.L
8. PORTAL FRAMES
Version II
36 - 26
kN
2.6
2
5.2
eaves
@
2
w
35
7
kN
2
eaves
@
2
w
94
.
1
88
.
3
Job No: Sheet 8 of 30 Rev
Job Title: Portal Frame Analysis and Design
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(ii) 1.5 D.L + 1.05 C.L+ 1.5 L.L
4.1. 1.5 D.L + 1.05 C.L+ 1.5 W.L
4.1.1Dead Load and Wind Load
(a) Vertical Load
w @ intermediate points on windward side
w = 1.5 * 5.0 – 1.5 *(4.8/3) cos21.8
= 5.27 kN.
w @ intermediate points on leeward side
w = 1.5 * 5.0 – 1.5 * 7.8/3 cos21.8
= 3.88 kN
Total vertical load @ the ridge = 2.635 + 1.94 = 4.575 kN
b) Horizontal Load
H @ intermediate points on windward side
H = 1.5 * 4.8/3 sin 21.8
= 0.89 kN
9. PORTAL FRAMES
Version II
36 - 27
Job No: Sheet 9 of 30 Rev
Job Title: Portal Frame Analysis and Design
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H/2 @ eaves points = 0.89/2 = 0.445 kN
H @ intermediate purlin points on leeward side
= 1.5 * 7.8 /3 sin 21.8
= 1.45 kN
H/2 @ eaves = 0.725 kN
Total horizontal load @ the ridge = 0.725 - 0.445 = 0.28 kN
Table 3: Loads acting on rafter points
Vertical Load (kN) Horizontal Load (kN)
Windward Leeward Windward Leeward
Intermediate
Points 5.27 3.88 0.89 1.45
Eaves 2.635 1.94 0.445 0.725
Ridge 4.575 0.28
4.1.2 Crane Loading
Moment @ B = 1.05 * 225 = 236.25 kNm
Moment @ F = 1.05 * 82 = 86.1 kNm
Horizontal load @ B & @ F = 1.05 * 13.9 = 14.6 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 =
Factored moment @ B & F = 1.05 * 5.75 * 0.6 = 3.623 kNm
kN
5.75
5
*
2
0.3
2
10. PORTAL FRAMES
Version II
36 - 28
Job No: Sheet 10 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
Made By
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Structural Steel
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Calculation Sheet
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Total moment @ B = 236.25 + 3.623 = 239.9 kNm
@ F = 86.1 + 3.623 = 89.7 kNm
4.2 1.5 D.L + 1.05 C.L + 1.5 L.L
4.2.1 Dead Load and Live Load
@ intermediate points on windward side = 1.5 * 5.0 + 1.5 * 6.4
= 17.1 kN
@ ridge = 17.1 kN
@ eaves = 17.1 / 2 8.55 kN.
15 m
3 m
6 m
3.25 m
14.6 kN
14.6 kN
239.9 89.7
38.85 kN 2.1 kN
Factored Load (1.5D.L+1.05 C.L +1.5 W.L)
1.45 kN
1.45 kN
0.725 kN
0.28 kN
4.575 kN
3.88 kN
3.88 kN
1.94 kN
0.89 kN
0.89 kN
5.27 kN
5.27 kN
2.635 kN
0.445 kN
11. PORTAL FRAMES
Version II
36 - 29
Job No: Sheet 11 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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4.2.2 Crane Load
Moment @ B = 239.9 kNm
Horizontal load @ B = 14.6 kN
Moment @ F = 89.7 kNm
Horizontal load @ F = 14.6 kN
4.3 Mechanisms
We will consider the following mechanisms, namely
(i) Beam mechanism
(ii) Sway mechanism
(iii) Gable mechanism and
(iv) Combined mechanism
Factored Load (1.5 D.L + 1.05 C.L + 1.5 L.L)
3 m
6 m
3.25 m
17.1 kN
17.1 kN
17.1 kN
17.1 kN
8.55 kN
17.1 kN
15 m
14.6 kN
14.6 kN
239.9 89.7
kNm kNm
12. PORTAL FRAMES
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36 - 30
Job No: Sheet 12 of 30 Rev
Job Title: Portal Frame Analysis and Design
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4.3.1 Beam Mechanism
(1) Member CD
Case 1: 1.5 D.L + 1.05 C.L + 1.5 W.L
Internal Work done, Wi = Mp + Mp ( /2) + Mp ( + /2)
= Mp(3 )
External Work done, We = 5.27 * 2.5 - 0.89 * 1 * +5.27 * 2.5 * /2 – 0.89
* 1 * /2
= 18.43
Equating internal work done to external work done
Wi = We
Mp (3 ) = 18.43
Mp = 6.14 kNm
Case 2: 1.5 D.L + 1.05 C.L + 1.5 L.L
Internal Work done, Wi = Mp ( + /2 + + /2)
Wi = Mp 3
4.575 kN
0.89 kN
5.27 kN
0.89 kN
0.445 kN
5.27 kN
0.28 kN
2.635 kN
/2
13. PORTAL FRAMES
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36 - 31
Job No: Sheet 13 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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External work done, We = 17.1 * 2.5 + 17.1 *2.5 /2
= 64.125
Equating Wi = We,
Mp (3 ) = 64.125
Mp = 21.375 kNm
Note: Member DE beam mechanism will not
govern.
(2) Member AC
Internal Work done,
p
M
3.69
13
11
p
M
13
11
p
M
p
M
i
W
17.1 kN
17.1 kN
17.1 kN
8.55 kN
/2
14.6 kN
38.85 kN
11 /13
239.9 kNm
14. PORTAL FRAMES
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36 - 32
Job No: Sheet 14 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
Made By
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Date
Structural Steel
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Calculation Sheet
Checked By
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Date
External Work done,
Equating Wi = We, we get
3.69 Mp = 296.6
Mp = 80.38 kNm.
(3) Member EG
Internal Work done,
External Work done,
Equating Wi = We, we get
3.69 Mp = 321.7
p
M
3.69
13
11
p
M
13
11
p
M
p
M
i
W
6
.
296
*
85
.
38
*
*
9
.
239
*
*
6
.
14
13
11
3.25
2
1
13
11
13
11
3.25
We
13
11
3.25
*
)
(
2
1
*
13
11
*
3.25
*
We
7
.
321
3
.
30
9
.
239
6
.
14
14.6 kN
239.9 kNm
30.3 kN
11 /13
15. PORTAL FRAMES
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Job No: Sheet 15 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Mp = 87.2 kNm
For members AC & EG, the 1st
load combination will govern the failure
mechanism.
4.3.2 Panel Mechanism
Case 1: 1.5 D.L + 1.05 C.L + 1.5 W.L
Internal Work done, Wi = Mp ( ) + Mp ( ) + Mp ( ) + Mp ( )
= 4Mp
External Work done, We
We = 1/2 (38.85) * 6 + 14.6 * 3.25 + 239.9 - 0.445 * 6 - 0.89 * 6 -
0.89(6 )+ 0.28 * 6 + 1.45 *6 + 1.45 * 6 + 0.725 * 6 +1/2 (2.1) * 6 +
14.6 * 3.25 - 89.7 *
= 378.03
14.6 kN
239.9 kNm
14.6 kN
89.7 kNm
38.85 kN 2.1 kN
1.45
1.45
0.725
0.28
4.575
3.88
3.88
1.94
5.27
0.89
5.27
0.89
2.635
0.445
16. PORTAL FRAMES
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36 - 34
Job No: Sheet 16 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Equating Wi = Wc, we get
4Mp = 378.03
Mp = 94.51 kNm
The second load combination will not govern.
4.3.3 Gable Mechanism
Case 1: 1.5 D.L + 1.5 W.L + 1.05 C.L
Internal Work done = Mp + Mp2 + Mp (2 ) + Mp = 6Mp
External Work done, We =
-0.89 * 1 * - 0.89 * 2 * + 0.28 * 3 * + 1.45 * 4 * + 1.45 * 5 * +
0.725 * 6 * + 5.27 * 2.5 * + 5.27 * 5 * + 4.575 * 7.5 * + 3.88 * 5 *
+ 3.88 * 2.5 * + ½ * 2.1 * 6 + 14.6 * 3.25 * - 89.7*
We = 82.56
1.45
1.45
0.725
0.28
4.575
3.88
3.88
1.94
0.89
0.89
5.27
5.27
2.63
5
0.445
14.6 kN
239.9 kNm
14.6 kN
89.7 kNm
38.85 kN 2.1 kN
17. PORTAL FRAMES
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36 - 35
Job No: Sheet 17 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Equating Wi = We, we get
6Mp = 82.56
Mp = 13.76 kNm.
Case 2: 1.5 D.L + 1.5L.L + 1.05 C.L
Internal Work done, Wi = Mp + Mp (2 ) + Mp (2 ) + Mp =6Mp
External Work done, We
= 17.1 * 2.5* + 17.1 * 5 * + 17.1 * 7.5 + 17.1 * 5 * + 17.1 * 2.5 -
89.7 * + 14.6 * 3.25
= 342.5
Equating Wi = We, we get
6Mp = 342.5
Mp = 57.1 kNm
17.1
17.1
17.1
17.1
8.55
17.1
14.6 kN
14.6 kN
239.9kNm 89.7 kNm
8.55
23. PORTAL FRAMES
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Job No: Sheet 23 of 30 Rev
Job Title: Portal Frame Analysis and Design
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Case 2: 1.5 D.L + 1.5 L.L + 1.05 C.L
(i) Assuming plastic hinge is formed @ purlin point 2 and 7 and at fixed
supports.
Internal Work done
p
4M
p
M
6
p
M
6
6
p
M
6
p
M
5 /6
/6 /6
17.1
17.1
17.1
17.1
8.55
17.1
14.6 kN
14.6 kN
239.9 kNm 89.7
36 m
8.55
5 /6
/6
24. PORTAL FRAMES
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36 - 42
Job No: Sheet 24 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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External Work done =We
Equating Wi =We, we get
4Mp = 304.1
Mp = 76.03 kNm
Plastic hinge is formed @ 3 and 7
.
3
3.25
6
2.5
6
5.0
6
7.5
6
10
6
2.5
1
6
6
3.25
1
04
7
.
89
*
*
6
.
14
*
*
1
.
17
*
*
1
.
17
*
*
1
.
17
*
*
1
.
17
*
*
1
.
17
*
9
.
239
*
*
6
.
14
/3 /3
17.1
17.1
17.1
17.1
8.55
17.1
14.6 kN
14.6 kN
239.9
kNm
89.7
18 m
8.55
2 /3
25. PORTAL FRAMES
Version II
36 - 43
.
3
3.25
3
2.5
1
3
5.0
1
3
7.5
1
3
2
5.0
1
3
2
2.5
1
3
2
3
2
3.25
3
20
7
.
89
*
6
.
14
*
*
1
.
7
*
*
1
.
7
*
*
1
.
7
*
*
1
.
7
*
*
1
.
7
*
9
.
239
*
*
6
.
14
p
p
p
p
p
i
M
M
M
M
M
W
4
3
3
3
2
3
2
Job No: Sheet 25 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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Internal work done,
External Work done, We=
Equating Wi = We, we get
4Mp = 320.3
Mp = 80.1 kNm
(ii) Plastic hinged is formed at 4 and 7
Internal Work done
p
M
4
p
M
2
p
M
2
2
p
M
2
p
M
26. PORTAL FRAMES
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36 - 44
Job No: Sheet 26 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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External Work done, We =
Equating Wi = We
4Mp = 293.8
Mp = 73.45 kNm
/2
/2 /2
17.1
17.1
17.1
17.1
8.55
17.1
14.6 kN
14.6 kN
239.9 89.7
12 m
8.55
/2
8
.
293
7
.
89
25
.
3
*
6
.
14
2
*
5
.
2
*
1
.
17
2
*
0
.
5
*
1
.
17
2
*
5
.
7
*
1
.
17
2
*
0
.
5
*
1
.
17
2
*
5
.
2
*
1
.
17
2
*
9
.
239
2
*
25
.
3
*
6
.
14
27. PORTAL FRAMES
Version II
36 - 45
3
W
*
3.25
*
*2.5
1
*
5.0
*
1
2
*
7.5
*
1
2
*
5.0
*
1
2
*
2.5
*
1
2
*
2
*
*3.25
e 3
.
56
7
.
89
1
.
14
1
.
7
1
.
7
1
.
7
1
.
7
1
.
7
9
.
239
6
.
14
Job No: Sheet 27 of 30 Rev
Job Title: Portal Frame Analysis and Design
Worked Example: 1
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(iii) Plastic hinge is formed @ 5 and 7
Internal Work Done,Wi =
External work done, We =
p
M
5
i
W
p
M
p
M
2
p
M
2
p
M
6 m
/2
17.1
17.1
17.1
17.1
8.55
14.6 kN
14.6 kN
239.9
/2
17.1
89.7
28. PORTAL FRAMES
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Job No: Sheet 28 of 30 Rev
Job Title: Portal Frame Analysis and Design
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Equating Wi = We
5Mp = 356.3 *
Mp = 71.26 kNm
Design Plastic Moment = 94.51 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= 94.51 kNm
Required section modulus, Z = Mp/ fyd
From IS: 800 (Appendix I)
ISMB 300 @ 0.46 kN/ m provides
Zp = 651 * 10-3
mm3
b = 140 mm
Ti = 12.4 mm
A = 5.626 * 10 3
mm2
tw =7.5 mm
rxx =124 mm
ryy =28.4 mm
3
3
10
*
84
.
415
51
.
94
mm
1.1
250
*106
29. PORTAL FRAMES
Version II
36 - 47
1
f
T
b
4
.
9
65
4
.
2
5.
1
70
T
b
1
f
Job No: Sheet 29 of 30 Rev
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5.2 Secondary Design Considerations
5.2.1 Check for Local buckling of flanges and webs
Flanges
bf = 140/2 = 70 mm
T1 = 12.4 mm
t = 7.5 mm
Web
O.K.
5.2.2 Effect of axial force
Maximum axial force in column, P = 51.3 kN
9
.
83
40
5
.
7
300
t
d1
30. PORTAL FRAMES
Version II
36 - 48
kN
27
*10
5.
1.1
250 3
8
1
626
Job No: Sheet 30 of 30 Rev
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Axial load causing yielding, Py = fyd * A
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
= 51.3 –8.55 kN
= 42.75 kN
Maximum shear capacity Vym, of a beam under shear and moment is given
by
Vym = 0.5 Aw* fyd / 1.1
= 0.5 * 300* 7.5* 250/1.1
=255.7 kN>> 42.75 kN
Hence O.K.
0.15
0.0
127
P
P
y
4
8
3
.
51