Faculty of Built Environment and Engineering
School of Civil and Environmental Engineering
CIVN 4015: Civil Engineering Design
Final Year Design Project: Structures
Group 4
Internal Supervisor: Dr. Ryan Bradley
External Supervisor: Mr. Daniel Surat, PREng
Compiled by:
Student Number
908593
1391559
1612449
1631963
1633852
1669086
Student’s Name and Surname
Zamanguni Ntozakhe
Musa Mdhluli
Makhosonke Mhlanga
Nonhlanhla Mhlongo
Maxwell Mohau Majoro
Jackson Mahlaule
Executive Summary
The purpose of this report is to detail the design and analysis of a warehouse. This warehouse
needs to have a bay which will allows for circulation of vehicles and have an overhead crawl
beam that can be used to remove equipment whenever necessary. The warehouse is designed
as a duo-pitched portal frame. The software which was used to determine the design loads as
well as deflections was Prokon. The loads combinations which were used for the software
analysis were determined using the Ultimate Limit State and the Serviceability Limit State.
The Ultimate Limit State was used to ensure that the sections used in the design can carry the
applied loads. The Serviceability Limit State was used to ensure that the sections used do not
experience excessive deformations when the structure is occupied.
Table of Contents
Executive Summary .......................................................................................................................... 2
List of figures .................................................................................................................................... 5
List of tables ...................................................................................................................................... 6
1.Introduction .................................................................................................................................... 7
1.1 Background ................................................................................................................................. 7
1.2 Scope of the project ..................................................................................................................... 7
1.3 Methods used to achieve the design outcomes .............................................................................. 8
1.4 Role of each group member ......................................................................................................... 9
2. Reference documents ................................................................................................................... 10
3.Preliminary Design Considerations ............................................................................................... 11
3.1 Preliminary Design 1 – Latticed Portal Frame ............................................................................ 11
3.2 Preliminary Design 2 – Trussed Portal Frame ............................................................................ 12
3.3 Preliminary Design 3 – Rolled Section Portal Frame .................................................................. 14
3.4 Preliminary Design 4 – Pre-engineered built-up Portal Frame .................................................... 15
4.Structure geometry ....................................................................................................................... 17
4.1 General arrangement .................................................................................................................. 17
4.1.1 General arrangement considerations ........................................................................................ 18
4.1.2 Design solution for the 15m bay .............................................................................................. 19
5.Analysis assumptions ................................................................................................................... 20
5.1 Wind loads ................................................................................................................................ 20
5.2 Crane loads ................................................................................................................................ 21
5.3 Imposed Loads .......................................................................................................................... 21
5.4 Dead Loads ............................................................................................................................... 21
5.5 3-D Structural steel framework .................................................................................................. 22
5.6 Members and connection conditions .......................................................................................... 22
5.7 Support conditions ..................................................................................................................... 23
5.8 End stop .................................................................................................................................... 23
6.Loads ........................................................................................................................................... 24
6.1 Dead loads ................................................................................................................................. 24
6.2 Wind loads ................................................................................................................................ 24
6.3 Imposed loads ............................................................................................................................ 26
6.4 Crane load ................................................................................................................................. 26
6.5 Geotechnical considerations ....................................................................................................... 26
7.Load combinations ....................................................................................................................... 27
8.Analysis results ............................................................................................................................ 31
8.1 Prokon Analysis ........................................................................................................................ 31
8.1.1 Member Selection ................................................................................................................... 31
8.1.2 Displacements ........................................................................................................................ 31
8.1.3 Reactions ................................................................................................................................ 32
8.1.4 Code checking ........................................................................................................................ 33
8.1.5 Connection design .................................................................................................................. 33
Rafter to Column: ........................................................................................................................ 33
Bracing beam to rafter connection: .............................................................................................. 34
Summary of connection details .................................................................................................... 35
Column to Rafter Connection:.................................................................................................. 35
8.2 Foundation ................................................................................................................................ 36
8.3 Ground slab design .................................................................................................................... 36
Conclusion ...................................................................................................................................... 37
References....................................................................................................................................... 38
Acknowledgements ......................................................................................................................... 39
Appendix 1 – Meeting Minutes ........................................................................................................ 40
Appendix 2 – Cross-disciplinary and stakeholder engagement GA 8 form........................................ 68
Appendix 3 – Preliminary Design 1 (Latticed Portal Frame) ............................................................ 69
Appendix 4 – Preliminary Design 2 (Trussed Portal Frame) ............................................................. 70
Appendix 5 – Preliminary Design 3 (Rolled Section Portal Frame) .................................................. 71
Appendix 6 – Preliminary Design 4 (Pre-engineered built-up Portal Frame) ..................................... 72
Appendix 7 – Dead Loads ............................................................................................................... 73
Appendix 8 – Wind Loads ............................................................................................................... 81
Appendix 9 - Imposed Loads ........................................................................................................... 82
Appendix 10 – Member designs....................................................................................................... 83
Appendix 11 – Foundation Design ................................................................................................... 84
Appendix 12 – Ground slab design .................................................................................................. 85
List of figures
Figure 1 Latticed portal frame ............................................................................................. 11
Figure 2: Trussed frame ...................................................................................................... 12
Figure 3: Rolled section ...................................................................................................... 14
Figure 4: Pre-engineered built-up Portal Frame ................................................................... 15
Figure 5: Front view of the structure ................................................................................... 18
Figure 6 Plan view of the structure ...................................................................................... 18
Figure 7 Elevation of the structure....................................................................................... 19
Figure 8 Plan view of the structure ...................................................................................... 20
Figure 9 Elevation of the structure....................................................................................... 20
Figure 10: summary of the connection details ...................................................................... 35
List of tables
Table 1 details the role which each group member plays in the completion of the project ...... 9
Table 2 contains a list of documents used to design the structure. ........................................ 10
Table 3 contains the dead loads which are used to analyse the structure. .............................. 24
Table 4 contains the imposed loads which act on the structure. ............................................ 26
Table 5 contains the forces that act on the crawl beam. ........................................................ 26
Table 6 contains the legend for the symbols used to determine the load combinations. ........ 27
Table 7 contains the load combinations considered for the analysis of the structure. ............ 27
Table 8 contains the load combination cases for ULS and SLS ............................................ 29
Table 9 contains the allowable deflections for the structural members ................................. 32
Table 10 contains the dominant ULS load combination ....................................................... 32
Table 11 contains the dominant SLS load combination ........................................................ 32
Table 12 contains the reaction forces used to design the gable column foundation. .............. 32
Table 13: code checking of structural members ................................................................... 33
Table 14 contains the dimensions of the foundation as well as the loads acting on it. ........... 36
Table 15 contains the loads that are used for the ground slab design .................................... 36
1.Introduction
Final year civil engineering students at the University of the Witwatersrand were tasked to
design a warehouse. The end-span of the structure needs to be 15m wide on both sides to allow
for circulation of vehicles and all other bays will be 7.5m wide. The warehouse is to be framed
using portal frames of constant width and height. All fire, acoustic, mechanical, fabrication
costs, site erection costs on optimization, and electrical requirements for the warehouse must
be ignored.
1.1 Background
Portal frames are generally low-rise structures, comprising columns and horizontal or pitched
rafters, connected by moment-resisting connections. Resistance to lateral and vertical actions
is provided by the rigidity of the connections and the bending stiffness of the members, which
is increased by a suitable haunch or deepening of the rafter sections. This form of continuous
frame structure is stable in its plane and provides a clear span that is unobstructed by bracing.
Pitched portal frames are commonly used in single storey buildings.
1.2 Scope of the project
The project aims to produce a design for a 1200m2 portal framed warehouse, with specified
parameters detailed in the general arrangement section. The design will be done in compliance
with the South African national standards and relevant building codes.
Listed below are the deliverables for the project:
Detailed design of the warehouse, member sizes specified, connection details specified
Detailed drawings of the design, including sketches of the preliminary structures and
the final structure.
Design of a crane runway.
Listed below are the items which are outside of the scope of the project:
Gutters are not accounted for.
Fabrication costs and site erection costs are not taken into consideration in the report.
Load calculations do not account for thermal and seismic loading.
Acquiring materials and erection of structure.
1.3 Methods used to achieve the design outcomes
The structure is designed to meet the Ultimate Limit State requirements, and these requirements
are for checking the strength of the members selected in designing the structure. The structure
is also designed to meet the Serviceability Limit State requirements, and these requirements
are for checking that when the structure is in use, the deflections are not too excessive.
Listed below are the steps which are taken to design the structure:
1. Selecting the type of portal frame which is to be used: This step involves conducting
research on the types of portal frames (only four types are considered) which can be
used to design the warehouse. One of the four options presented is to be selected to
design the warehouse.
2. Determine the general arrangement of the structure: A decision is taken on the type of
bracing, which is used for the structure, and where it is best (the bay) to brace the
structure. The orientation of the columns is also determined in this step. Positioning of
the gable columns is also determined in this step.
3. Determine the loads acting on the structure: Using the project brief, the loads acting on
the structure are determined. The SANS code is used to compute the wind loads.
Research is conducted to determine the best sheeting to use for the structure.
4. Determine the load combinations: The SANS code is used to determine the load factors
as well as the load cases for the structure.
5. Analyze the structure on a design software: The structure is analyzed on Prokon to
determine if the selected members (sections for rafters, columns etc.) can withstand the
loads acting on the structure, and if the deflection that occur are within the limits stated
in the SANS code. It is also used to determine if the connections and the foundation
can carry the loads.
6. Hand calculations to determine if the members selected, connections and foundation
meet the SANS code requirements: Using the checks stipulated in the code, the
members selected, the connections and foundation are verified.
7. Produce engineering drawings: AutoCAD is used to produce all the drawings necessary
for the project.
1.4 Role of each group member
Each group member is expected to contribute towards the completion of this project.
Table 1 details the role which each group member plays in the completion of the project
Group member
Zamanguni Ntozakhe
Role played by the group member
Research advantages and
disadvantages for Preliminary design
2.
Determine the load cases that would
be analyzed on the software
Design the purlins, girts and
connections
Researched advantages and
disadvantages for Preliminary design
3
Musa Mdhluli
Makhosonke Mhlanga
Nonhlanhla Mhlongo
Design gable columns
Model the structure on Prokon
Determine the dead loads acting on
the structure.
Write the introductory chapter
(excluding scope)
Design the ground slab, concrete
foundation, and tie beams
Research advantages and
disadvantages for Preliminary design
1.
Design the crawl beam, and door
frames.
Compile the report.
Determine the loads acting on crawl
beam.
Determine the wind load
Mohau Majoro
Jackson Mahlaule
Responsible for Autocad drawings
Discuss the scope of work for the
project
Responsible for the Autocad
drawings
Determine the live loads acting on
the structure.
2. Reference documents
The table below contains a list of documents which are used in the design process for the
project. These codes are prescribed in the project brief.
Table 2 contains a list of documents used to design the structure.
Scope
Code/ Handbook
Description
Loading
SANS 10160-1:2018 Ed1.2
Basis of structural design and actions for
buildings and industrial structures
Part 1: Basis of Structural Design
SANS 10160-2:2011 Ed1.1
Part 2: Self-weight and imposed loads
SANS 10160-3:2018 Ed2.0
Part 3: Wind actions
SANS 10160-5:2011 Ed1.1
Part 5: Basis for geotechnical design and
action
SANS 10162-1:2011 Ed2.1
The structural use of steel Part 1: Limitstates design of hot-rolled steelwork
The Red Book
Southern African Steel Construction
Handbook
The Blue Book
Design of Structural Steelwork to SANS
10162
The Green Book
Structural Steel Connections
Concrete
SABS 0100-1:2000 Ed2.2
The structural use of concrete Part 1:
Design
Foundation
Design
Concrete Industrial Floors on
Ground
Provides a guide for designing ground
slabs
Structural
Steel
3.Preliminary Design Considerations
There are many types of portal frames that can be used. There are four types of portal frames
that have been selected and discussed below are the advantages and disadvantages of each type
of portal frame.
3.1 Preliminary Design 1 – Latticed Portal Frame
(Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo)
Figure 1 Latticed portal frame
A Lattice Portal frame is constructed by using angles or tube steel members (Andrews, n.d.).
The structural integrity of the latticed portal frames comes from the moment resistance of the
connections and its ability to carry bending moment (Andrews, n.d.).
Written below are the advantages and disadvantages of using latticed portal frames (Andrews,
n.d.):
Advantages:
They are considered a good option for longer than spans.
They are not expensive to construct, as they can be prefabricated.
They are lightweight structures which make them easy to erect.
Disadvantages:
There are more factors that need to be considered in the design of the latticed
portal frame before it is deemed structurally sound. Each member in the
structure must be checked extensively for:
1. The effects of shear forces as well as axial forces
2. Buckling: the local buckling of the member, lateral bucking as well as column bucking
3. Deflection that occurs under service loads.
All the connections for the structure need to meet the following conditions:
economy, strength, rotational capacity, and stiffness.
Lattice portal frames have a low fire resistance, and this is problematic because
the stability of the structure is compromised quickly when a fire occurs.
The graphical representation (sketch) of a latticed portal frame can be found in Appendix 3.
3.2 Preliminary Design 2 – Trussed Portal Frame
(Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo)
Figure 2: Trussed frame
A truss is a structure formed of rod members organized in one or more triangle-shaped
configurations. The members must be triangulated because the joints are pinned (do not
transmit moments). On the other hand, a frame is a structure made up of beam members that
can be placed in any direction and are joined rigidly or with pins at joints. The members support
bending as well as axial loads (Doyle, 1991).
Written below are the advantages and disadvantages of using the Trussed Portal Frame:
Advantages:
High strength and resistance to tension and pressure
During production there is quality control of steel unlike concrete structures
Possibility of structural developments
Prefabricated parts can be used to construct
Structural development
High installation speed
Less space needed to occupy material
Able to connect numerous pieces using nuts, bolts, and welds
Can be used at high altitudes
Disadvantages:
There could be poor connections between welds, bolts, and nuts
Requires a lot of material not so economical
Angles should be precise to avoid failure
3.3 Preliminary Design 3 – Rolled Section Portal Frame
(Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo)
Figure 3: Rolled section
Generally fabricated from UB sections with a substantial eaves' haunch section, which may be
cut from a rolled section or fabricated from plate (Mohammed, 2021).
Written below are the advantages and disadvantages of using a Rolled section Portal Frame.
Advantages:
-
-
Use of uniform sections, standard universal beams
Less members used, reduces cost
Reduced complexity of fabrication, time of fabrication reduced
The use of a haunch at the eaves reduces the required depth of rafter by increasing the
moment resistance of the member where the applied moments are highest (institute,
2022).
The haunch also adds stiffness to the frame, reducing deflections, and facilitates an
efficient bolted moment connection (Institute, 2022)
(HRS) frame is the optimum system for low and short frames because its high
constructability and low unit price compensate any waste of material due to using
standard sections (Mohammed, 2021).
Disadvantages:
-
Members are too heavy
More time is needed to develop the structural designs
Of site fabrication not possible (Geometry), due to difficulty in transportation
-
More accuracy is required to build these types of structures. The portal frames are
placed at regular intervals
We cannot construct any structure above the portal frames. It is very difficult to
construct structure above the portal frame, due to its sloppy head
3.4 Preliminary Design 4 – Pre-engineered built-up Portal Frame
(Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo)
Figure 4: Pre-engineered built-up Portal Frame
Pre-engineered built- up Portal Frames are used because they are roughly 30% lighter than
other systems because steel sections are used efficiently (Ebid, et al., 2021).
Written below are the advantages and disadvantages of using the Pre-engineered built- up
Portal Frame (Ebid, et al., 2021):
Advantages:
Quicker to build
Easy to construct
Has the ability to Build Long Spans
It Is Possible to Use It for Temporary Construction
It is Reliable
Disadvantages:
High Maintenance Costs
Less Flexibility on-site
Potential for Leaks due to many joints
Design Difficulties
Transportation cost to on-site
Poor Fire Resistance
4.Structure geometry
The preliminary design which is to be designed is the Preliminary Design 3 which is the Rolled
Section Portal Frame. The reasons for selecting this preliminary design are stated below.
Uniform (same size) structural members ensure ease of designing connections and joining the
members. The design is not complex, this reduces the overall cost of the structure. The structure
is easy to construct due to minimal use of members. The overall disadvantages of the structure
are ‘better’ as compared to those of the other structures put into consideration for adoption.
4.1 General arrangement
The structure is 60m long and 20m wide. A duo-pitch portal frame is used, and the roof slope
is 10 degrees. The eaves height is 8m and the apex height is 9.76m. The end span of the
structure is 15m on both sides of the structure. There are six 7.5m wide bays on both sides of
the structure.
The 15m bay is fitted with two roller shutter doors that are 4m high and 4m wide. The two
doors have a 4m distance between them. The structure does not have gutters. The steel super
structure is supported by concrete plinths which extend 300mm above a ground slab. An
overhead crawl beam is needed, and it runs along the length of the structure.
The roof system which consists of external sheeting and purlins needs to be able to carry
imposed loads and wind loads to the structural frame of the building (Mahachi, 2004). The
purlins support the external sheeting on the roof. The sagbars are used to provide lateral support
to the bottom flange of the purlins, when it goes into compression due to reverse loading caused
by wind suction. The sagbars are also used to decrease the in-plane deflections of the roof
(Mahachi, 2004). The rafters support the purlins and sagbars and are also used to transfer the
loads from the purlins to the columns.
For the wall system, girts are used to carry external sheeting. Even on the walls, the sagbars
are used to reduce the vertical in-plane deflections due to the self-weight of sheeting and girts
(Mahachi, 2004). The columns support the girts and sagbars and are used to transfer the loads
from the rafters and girts to the concrete plinths.
The bracing system of the structure consists of tie beams, gable columns and diagonal bracing
elements. The sides of the structure are braced using the tie beams and the diagonal bracing
elements (Mahachi, 2004). This is used to carry the wind loads and horizontal cranes loads that
act on the structure. The gable columns on the gable wall which is on the front and rear end of
the structure are used to carry the wind load to the foundation (through the concrete plinths).
The gable columns are used to brace the front and last portal frame, and the wind load on the
gable wall are carried along the purlins towards the bracing systems.
The steel superstructure is supported by concrete plinths which transfers the loads from the
columns to the foundation. The foundation transfers the load from the concrete plinths to the
ground below the structure.
4.1.1 General arrangement considerations
The gable columns are placed 5m apart to provide support/ brace the front and rear end of the
structure.
The image shown below displays the front view of the structure.
Figure 5: Front view of the structure
The structure is braced at the end bay. The roof is also braced on the 15m bay.
The images shown below display the elevation and plan view of the structure.
Figure 6 Plan view of the structure
Figure 7 Elevation of the structure
4.1.2 Design solution for the 15m bay
It is not possible to use an eaves beam that is longer that is longer than 12m. A truss is used to
span the 15m bay. Two rafter beams are placed within the 15m bay and they are placed there
to provide support for the purlins that support the roof cladding. The roof within this 15m bay,
is braced to provide a medium for the transfer of loads.
The images shown below detail the side view and the top view of the structure.
Figure 8 Plan view of the structure
Figure 9 Elevation of the structure
5.Analysis assumptions
5.1 Wind loads
These are the assumptions that are made regarding the computation of wind loads.
Determining the internal pressure:
The structure that needs to be designed does not have a dominant wall, because both sides of
the structure have a 15m bay that accommodates two doors that have the same dimensions.
Note 2 of clause 8.3.9.6 of SANS 10162-3 is used to determine the internal pressure coefficient.
It states that in a situation where it is not justified or possible to estimate the opening ratio, the
internal pressure coefficient can be taken as +0.2 (Anon., 2018).
Written below are the assumptions that are made as the wind loads are inputted onto Prokon.
Wind loading effects the entire shell of the structure and the loads are applied to the purlins,
girts, columns and gable columns to model its effect. On the roof wind loading is assumed to
act as a pressure onto the sheeting which in turn is distributed to the purlins as a line load and
they are transferred to the rafters as distributed load. Therefore, the wind load is modelled as a
line load on the purlins and transferred to the rafters as distributed load along the member.
On the walls where sheeting occurs, it is assumed that wind loading acts as a pressure onto the
sheeting which is then distributed to girt as a line load and then taken up by the columns to the
foundation. Therefore, the wind load is modelled as a line load to the girts. The same principle
applies to the girts on the gable columns.
5.2 Crane loads
These are the assumptions that are made regarding the computation of crane loads.
Crawl beam loads:
For the overhead crawl beam, the loads are classified into two categories. The self-weight of
the crawl beam is considered as a dead load since the beam itself is a permanent fixture. The
effect of the loads that act on the crawl beam due to the crab (the self-weight of the crab and
the hoist capacity) is considered using the crane load factor.
Written below are the assumptions which are made to account for the loads due to the crane
for the Prokon analysis. The loading from the crane was applied to the structure in the frame
analysis as follows. Reactions from the crane beams at the supports which were applied to the
crane supports as loads in the main structure were determined in a separate analysis. This
analysis (which is described in detail under the crane beam section) used influence lines to
determine what the worst case of the reactions at the crane beam supports would be when the
crane moves along the crane runway girder, and since the crane runway girders were simply
supported between the crane beams supports, this occurred when the moving crane was
directly over the crane supports. The reactions, applied as point loads in the structure at the
crane beam supports were excluded the self-weight of the crane runway girders because the
runway girders were modelled in the main frame analysis too, where their self-weight was
accounted for.
5.3 Imposed Loads
Written below are the assumptions that are made to input the imposed loads onto Prokon for
the analysis of the structure.
Live load (imposed loads) for the portal frame occurs on the roof only. Therefore, the same
assumptions that apply to wind loading on the roof, apply to the live load modelling. Live
loading is assumed to act as a pressure onto the sheeting which in turn is distributed to the
purlins as a line load. Therefore, the live load is modelled as a line load on the purlins.
5.4 Dead Loads
Written below are the assumptions that are made to input the dead loads onto Prokon for the
analysis of the structure.
Dead loads (self-weight) acting on the structure are taken into account by the analysis by adding
the self-weight of the elements to the permanent load case. Roof sheeting, insulation, and
services not part of the self-weight of the structural elements are assumed to be a pressure on
the purlins and girts (no services) which are then distributed to the purlins and girts as live
loads. These loads included the sheeting and insulation and services which were prescribed in
the design criteria.
Written below are the assumptions that are made for the 3-D model of the structure for the
Prokon analysis.
5.5 3-D Structural steel framework
The 3-dimensional structural frame analysis of the warehouse, including the columns, gable
columns, rafters, crane beam supports, bracing beams, tie beams, purlins, girts and sag bars
was carried out using structural analysis software, Prokon.
For the structural analysis the following aspects of the structural arrangement and the loading
onto the structure were considered to best model reality and abide by the relevant codes of
practise and are discussed here in this section.
Members and connection conditions
Support conditions
Loading
Load combinations
Connection design
5.6 Members and connection conditions
The connection types chosen between members were chosen based on conditional
requirements such as stability, load paths, economy, and deflection limits. The following
connection conditions were chosen between connecting members.
Column to rafter connection is assumed to be completely rigid, a condition which is
necessary to have stability in the plane of the portal frame.
Rafter to rafter (apex) connection is assumed to be completely rigid, a connection
which is necessary to have stability in the portal frame and have continuity in the
rafters.
Tie beams to columns will be modelled as pin connections as it is only necessary to
transfer lateral (axial) loads through the tie beams for stability in the direction
perpendicular to the portal frame.
Truss to column connection is modelled as pin connections as it is only necessary to
transfer lateral (axial) loads through the truss
Bracing members will be modelled as pin connections as it is only necessary to
transfer lateral (axial) loads through the bracing members which transfers the loads
perpendicular to the portal frames into the foundation supports. Where bracing in a
bay cross over each other the one beam is modelled as continuous while the other is
modelled as two beams pin jointed to either side of the continuous member of
bracing. This ensure the effective length of the bracing is to where the beams cross
over in the centre of the bay.
Gable column to rafter connection can be simply connected (pinned) or rigidly
connected to the rafter. In our model it is assumed to be pinned to the rafter because
this is a simpler connection to design and there won’t be moment forces transferred to
the rafter form the gable column and it is not necessary to have a rigid connection at
this point.
Crawl beam support to apex connection is simply supported at this point when
designing as to create a moment of zero at this point
Purlins are modelled as continuous over the rafters to limit deflection and will be
rigidly connected to the rafter. Where spans exceed transport limits for length (12 m).
The girts are modelled as continuous over columns and gable columns where
possible. In some instances, on the gable columns, girts are simply supported between
gables and columns because they are only one span long.
5.7 Support conditions
The support conditions for the columns and gable columns were altered during the design of
the structure to achieve certain outcomes. Column to foundation connection will be designed
and built to resist rotation in the direction of the portal frames only, i.e., the connection will
be fixed in the plane of the portal frame. The following instances describe what support
conditions were chosen under certain circumstances and the reason they were selected.
The supports were modelled as pinned connections when designing the members and
checking for lateral stability in the structure. These conditions ensure that there is no
resistance to rotation at the base of the building and if the building can resist the
lateral loads under these conditions, it is effectively braced. Furthermore, under these
conditions members are designed to effectively redistribute the loads if settlement or
failure of one or more of the foundations occurs.
It must be noted that since the support conditions modelled here do not match reality,
the deflections, particularly in the plane of the portal frame, are exaggerated and can
be ignored. Deflections were checked when support conditions more accurately
matched what the as built conditions would be.
The support conditions were modelled as fixed in the direction of the portal frames
and pinned in the transverse directions to best model the actual built conditions.
Under these conditions, the deflections were checked to be within the limits under
serviceability limit state and the reactions (with the correct loading combinations)
were used to design the foundations. An additional check was done to ensure that all
the member sizes selected were sufficiently strong under the fixed support conditions
too.
5.8 End stop
The End stop is put in place to prevent the crib from sliding off the crawl beam. The End stop
is assumed to be a horizontal roller because the only force it transmits is the buffer force. It is
designed as a simple connection. The section that is used for the end stop is the same as the
section which is used for the crawl beam. Since this is a large section, the only check that
needs to be considered is for “Bolts in shear for supporting member”.
There are several assumptions that have been made to determine the loads that act on the
structure. All the assumptions that have been made are detailed below.
6.Loads
Loading has been applied according to the design criteria: SANS10160-2 (Self-weight and
imposed loads), SANS 10160-1 (Basis of structural design), SANS10160-3 (Wind loading)
and SANS10160-6 (Cranes and machinery) to all structural members.
The loads shown below are used to design the warehouse. Thermal and seismic loads do not
need to be considered for the design of the structure. Actions during execution of the project
do not need to be accounted for in the design of the structure.
6.1 Dead loads
The dead load of the structure consists of the weight of the structure itself and all the materials
or finishes on the structure which cannot be removed from the structure (if they are removed
the structural integrity of the structure would be undermined) (Parrot, 2005).
Table 3 contains the dead loads which are used to analyse the structure.
Specific Use
Sheeting
Crawl Beam
Door
Dead Load
6.28 𝑘𝑔/𝑚2
0.452 𝑘𝑔/𝑚2
9 𝑘𝑔/𝑚2
The hand-calculations are presented in Appendix 7.
6.2 Wind loads
The wind can cause an uplift of the roof of a structure, as well as push on the sides of the
structure. Quantifying the wind load is important because in most case, wind loads cause
structural failure for roofs (Parrot, 2005).
Listed below are the parameters which are considered to compute the wind loads:
Fundamental value of basic wind speed (Vb,()): 36 m/s (3s gust)
Mean return period = 50 years
Upstream terrain category = B (Low vegetation or separation of at least 20 obstacle
heights.)
Topography factor c0(z) = 1.00
Site height Above Mean Sea Level = 1600m
The hand-calculations are presented in Appendix 8.
6.3 Imposed loads
Imposed loads are induced when the structure is occupied, these loads are produced by people,
equipment, or vehicles (Parrot, 2005).
Table 4 contains the imposed loads which act on the structure.
Category
Specific use
Example
E2
F
Industrial use
Parking areas for
light vehicles of
<=25kN gross
vehicle weight
Inaccessible roofs
Ground slab/ floor
Garages, parking
areas and parking
halls
H1
Inaccessible roof
during construction
Imposed Load
(kN/m2)
5.0
2.0
0.25
The hand-calculations are presented in Appendix 9.
6.4 Crane load
The loads shown in the table are loads that act on the structure (crawl beam) due to crab.
Table 5 contains the forces that act on the crawl beam.
Description of the force (load)
Self-weight of the crab
Hoist capacity
Horizontal skewing force
Buffer force
Force (kN)
3
20
5
5
6.5 Geotechnical considerations
The allowable bearing pressure is assumed to be 150 kPa and the founding horizon is 1m below
the ground slab.
Listed below are ground soil properties:
The angle of shearing resistance is 30 degrees (unfactored)
The dry density is 18 kN/m3
The saturated density is 20 kN/m3
7.Load combinations
The table below shows all the load cases that are used to analyse the structure on Prokon. The
load cases are determined for the Ultimate Limit State as well as the Serviceability Limit State.
The table below provides an explanation of the symbols used in the table that contains the load
combinations used for the analysis of structure.
Table 6 contains the legend for the symbols used to determine the load combinations.
DEFINITIONS
Structural internal failure
STR
STR-P
Dominate Structural internal failure
GK
Permanent Load
QK,1
Imposed Load
QK,2
Imposed Load as crane
W1
Wind load at 0 degrees
W1
Wind load at 90 degrees
ULS
Ultimate limit state
SLS
Serviceability limit state
The table below contains all the load combinations which can be used to analyse the structure.
These load combinations are determined in accordance with the Ultimate Limit State as well
as the Serviceability Limit State.
Table 7 contains the load combinations considered for the analysis of the structure.
Load
Combination
Leading
Factor
Permanent
Load Gk
Imposed
Qk,1
SLS
LC1
GK
1,1
LC2
GK
1,1
1
LC3
GK
1,1
1
LC4
GK
1,1
1
LC5
GK
1,1
1
LC6
GK
1,1
1
1,0X0,6
LC7
GK
1,1
1
1,0X0,6
LC8
GK
1,1
1
LC9
GK
1,1
1
LC10
GK
1,1
LC11
GK
1,1
LC12
GK
1,1
1,0X0,6
1
LC13
GK
1,1
1,0X0,6
1
LC14
GK
1,1
LC15
GK
1,1
0,6
Impose
d Qk,2
Wind
W1
1,0X0,6
0,6X0,3
0,6X0,3
1
0,6X0,3
0,6X0,3
0,6X0,3
1
0,6X0,3
0,6X0,3
0,6X0,3
0,6
0,6
Wind W2
0,6
LC16
GK
1,1
1,0X0,6
0,6
LC17
LC18
GK
GK
1,1
1,1
1,0X0,6
1,0X0,6
0,6
LC19
GK
1,1
0,6
LC20
GK
1,1
LC21
GK
1,1
1,0X0,6
LC22
LC23
GK
GK
1
1
1
LC24
GK
1
1
LC25
GK
1
1
LC26
GK
1
1
LC27
GK
1
1
1X0,6
LC28
GK
1
1
1X0,6
LC29
GK
1
LC30
GK
1
LC31
LC32
GK
GK
1
1
LC33
GK
1
LC34
GK
1
LC35
GK
1
LC36
LC37
GK
GK
1
1
0,6
LC38
GK
1
0,6
LC39
GK
1
LC40
GK
1
LC41
LC42
GK
GK
1
1
STR-P
LC25
GK
1,35
LC26
LC27
GK
GK
1,35
1,35
LC28
GK
1,35
LC29
GK
1,35
LC30
GK
1,2
LC31
QK
1,2
1,6
LC32
QK
1,2
1,6
LC33
QK
1,2
1,6
LC34
LC35
QK,2
QK,2
1,2
1,2
0,6
0,6
1,0X0,6
1,0X0,6
0,6
0,6
0,6
1,0X0,6
0,6X0,3
0,6X0,3
0,6X0,3
0,6X0,3
1
0,6
1
1
1
0,6X0,3
0,6
1
0,6X0,3
0,6
1
0,6X0,3
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
0,6
ULS
1
1
1
1
STR
1,6X0,6
1,6X0,6
1,6X0
1,6
1,6
1,6X0
LC36
QK,2
1,2
1,6
LC37
LC38
W1
W1
1,2
1,2
LC39
W1
1,2
LC40
W1
1,2
LC41
W2
1,2
LC42
LC43
W2
W2
1,2
1,2
1,6X0,6
LC44
W2
1,2
1,6X0,6
1,6X0
1,6
1,6
1,6X0,6
1,6X0,6
106X0
1,6X0,6
1,6
1,6X0,6
1,6
1,6
1,6X0,6
1,6
1,6
1,6X0,6
1,6
The table below contains the load combinations which are used to analyse the structure. These
load combinations are determined in accordance with the Ultimate Limit State as well as the
Serviceability Limit State.
Table 8 contains the load combination cases for ULS and SLS
Load
combinations
DL
LL
ULS
1.2
1.6
SLS
1.1
1.0
LC2
DL
LL
CL
1.2
1.6
0.96
1.1
1.0
0.6
LC3
DL
CL
1.2
1.6
1
1.1
LC4
DL
LL
CL
1.2
0.96
1.6
1.1
0.6
1
LC5
DL
WL_0
1.2
1.6
1.1
0.6
LC6
DL
WL_90
1.2
1.6
1.1
0.6
LC7
DL
CL
WL_0
1.2
0.96
1.6
1.1
0.6
0.6
Load cases
LC1
LC8
DL
LL
WL_90
1.2
0.96
1.6
1.1
0.6
0.6
LC9
DL
WL_0
0.9
1.6
1
1
LC10
DL
WL_90
0.9
1.6
1
1
LC11
DL
WL_270
1.2
1.6
1.1
0.6
LC12
DL
CL
WL_270
1.2
0.96
1.6
1.1
0.6
0.6
LC13
DL
LL
WL_270
1.2
0.96
1.6
1.1
0.6
0.6
8.Analysis results
8.1 Prokon Analysis
The Prokon outputs from the models analysed are presented in appendix C. the results were
used for the design of the various members, connections and supports. The members were
designed to meet ULS and SLS criteria of SANS 10162-1:2011 Ed 2.1.
8.1.1 Member Selection
Member selection was carried out to the ability to resist the forces experienced by the members
under the ultimate limit state conditions and the ability to limit deflection under serviceability
limit state conditions. SANS 10162-1:2011Ed2.1 was the code used to determine that members
satisfied the requirements for strength and deflections. All of the members seen in Table 5
below can resist the factored loads applied to them without failing. In some cases lighter
sections could have been used, but in order to remain within the deflection limits, the member
size was increased.
The table below details the final members to be used in the structure.
Table 5: Members selected
Columns
Rafters
Purlins
Girts
Tie Beams
Sag Tie
Bracing
Gable Columns
Crane Beam
Sag Rods
Truss Member 1
Truss Member 2
356x171x67
356x171x67
125x75x20x3
125x75x20x3
139.7x4.0
80x80x6
139.7x5.0
356x171x67
305x165x40
70x70x6
70x70x6
125x75x20x2
I - Section
I - Section
Cold formed Section
Cold formed Section
Hollow Section
Cold formed Section
Hollow Section
I - Section
I - Section
Angle
Angel
Channel
The hand-calculations are presented in Appendix 10.
8.1.2 Displacements
Under serviceability conditions it is important that the structural members of the warehouse do
not deflect more than what is allowed for their purpose. Excessive deflections can lead to
improper functioning of the building and can lead the occupants of the building to feel unsafe
in its vicinity. According to SANS10162-1 Table (D.1) structural members should not exceed
deflections of more than the span of the member divided by a limit value. The maximum
vertical and horizontal displacement were assessed against the code limits. The limits are in
accordance to SANS 10162-1:2011 Ed 2.1, Limit states design of hot rolled steelwork. The full
analysis output of the displacement can be found in appendix.
Table 9 contains the allowable deflections for the structural members
Structural
Member
Deflection
type
Deflection limit
SANS10162-1
Table (D.1)
Span - L - (mm)
Limit
value (mm)
Maximum
deflection
(mm)
L/limit (mm)
Door Column
Lateral
200
4000
20
-1.84
Rafter both side
Vertical
240
10154
42
25.66
Tie beam
Vertical
300
7500
25
-18.85
Bracing
Crane runway
beam
Column
Vertical
300
8500
28
-17.91
Vertical
240
200
7500
8000
31
40
-17.91
8.1.3 Reactions
When designing for the foundations, the support conditions of the model on Prokon was
changed from pinned to a fixed support. The highest reaction forces and the relevant load
combination obtained from Prokon are tabled below. The full reaction force output is found in
appendix. The largest ULS load combination is used to design flexural steel while the largest
SLS load combination is used to verify bearing pressure and overturning.
Table 10 contains the dominant ULS load combination
Load
Dominant
combinations Reaction
Ultimate Limit State
Horizontal Vertical Moment
LC11
LC2
Horizontal -52.26
27.55
Vertical
-33.02
148.42
157.3
-1.06
LC11
Moment
157.3
-33.02
-52.26
Table 11 contains the dominant SLS load combination
Load
Dominant
combinations Reaction
Serviceability Limit State
Horizontal Vertical Moment
LC11
LC2
Horizontal -14.11
117.04
Vertical
LC11
Moment
77.5
13.01
21.17
31.32
-69.66
34.21
25.38
The following reaction forces were used to design the gable column foundations.
Table 12 contains the reaction forces used to design the gable column foundation.
Load
Dominant
combinations Reaction
Ultimate Limit State
Horizontal
Vertical
LC11
Horizontal -5.44
3.63
LC1
Vertical
0.47
24.72
Load
Dominant
combinations Reaction
SLS
Horizontal
Vertical
LC11
Horizontal -3.32
9.02
LC2
Vertical
0.31
20.84
8.1.4 Code checking
In accordance to the code, the minimum slenderness ratio for tension is 300 and 200 for
compression. The ratio indicates the tendency of members to fail in buckling in that direction.
The following table lists the worst code ratios for each member section.
Table 13: code checking of structural members
Element
Column
Gable column
Rafter
Tie beam
Vertical Bracing
Section Size
356x171x67 I
356x171x67 I
356x171x67 I
139.7x4.0 O
139.7*5 O
L/r
170
200
57
133
97
Critical axis
y-y axis
y-y axis
Fails in y-y axis
x-x axis
x-x axis
8.1.5 Connection design
The column to rafter connection at the eaves beam level. At this point, the following members
connect: the rafter connects rigidly to the column flange, tie beams connection simply to the
web of the column, the column (wall) bracing connects to the web of the column and the roof
bracing connects to the web of the rafter. Connection design for each of the elements
connecting to this point was carried out as follows.
Rafter to Column:
The design link function in the Prokon software was used to design a rigid connection for the
column to rafter connection. By inspection, the connection with the highest moment forces,
shear forces and axial forces was used to design the connection. Furthermore, the connection
was designed where support conditions of the columns were modelled as pinned and as fixed,
because this slightly changed the forces being transferred through the connection. It was
decided that a flush end plate with a haunch welded to the rafter in the shop and then bolted
onto the column on site would be used as the connection. A haunch was used to limit deflection
as much as possible in rafter, even if this means overdesigning the connection. A haunch of the
same size as the rafter beam is used and extends 400 mm from the centre of the rafter and 2000
mm away from the column. Under these conditions, the design link programme was then used
to optimize the connection for plate thicknesses, welds and the number of M20 bolts to be used.
The limit state checks for this design under fixed support conditions for the highest loaded
connection can be found in appendix.
Bracing beam to rafter connection:
The rafter bracing beams do not clash with tie beams thus this the connection is simpler to
design. The same principles of the tie beam and bracing beam to the column were followed for
this connection. A circular hollow section bracing is capped with a plate and a fin plate which
is bolted to a fin plate on the web at the base of the rafter beams. The connection consists of 10
mm plates with M20 bolts and 6 mm fillet welds. The limit state checks of this connection are
also found in the appendix.
Summary of connection details
A summary of the connection details above are as follows:
Column to Rafter Connection:
Figure 10: summary of the connection details
8.2 Foundation
The foundation was designed in accordance with SANS 10160-5: 2021 Edition 1.2, Basis of
geotechnical design and actions. It is designed as a rectangular base with a concrete plinth.
The reaction forces obtained from the structural analysis were used to design foundation for
the portal frame columns and the gable columns. Although the columns are carrying different
quantities of load, all portal frame column foundations are the same. This also applies to
gables foundations.
The table below summaries the foundation designed. Foundation and reinforcement details
are found in Appendix 11.
Table 14 contains the dimensions of the foundation as well as the loads acting on it.
Variable
Quantity
a
3m
b
2.75m
P
41.36kN
Hx
22.81kN
Mx
104.33kNm
LC 7 was found to be the worst-case scenario after our analysis
8.3 Ground slab design
The suspended floor slab of the office building is designed according to the SABS 01001:2000 Ed2.2 and SANS 10160 codes. The suspended slab was designed for thickness and
cracking.
The loads on the slab were analysed using Prokon. The self-weight of the concrete is added
by the software and the load combinations are used from table. A 50mm Screed layer is also
assumed on top of the concrete slab to be able to achieve a smooth and level surface. A finite
element shell analysis model was analysed to get the maximum positive moment, maximum
negative moment, maximum shear forces in the x and y directions as well as the maximum
deflection for the slab. The dead and live load used to calculate the ULS and SLS are
tabulated below.
The hand-calculations are presented in Appendix 12
Table 15 contains the loads that are used for the ground slab design
Type of load
Quantity of load
DL
4.65kN/m
LL
7kN/m
The ULS was calculated and found to be 16.78kNm
The SLS was calculated and found to be 12.12kNm
Conclusion
The aim of the project was to design a warehouse in accordance with South African standards.
The selection of all structural members was optimized by weight only. The project was
successfully completed with use of computer software (AutoCAD and Prokon), industry
consultations (Mr Daniel Surat and Mr Thamsanqa Shangase), and relevant documents
(indicated under references and acknowledgements). The design of the warehouse facilitated a
learning of fundamentals of structural design and significant knowledge on technical insights
in designing a portal frame.
References
Andrews,
N.,
n.d.
Home
Sweet
home.
[Online]
Available at: https://www.ehow.com/info_10075491_advantages-disadvantages-latticedportal-frames.html
[Accessed 6 October 2022].
Anon., 2018. South African National Standard Basis of structural design and actions for
buildings and industrial buildings Part 3: Wind loads. 2 ed. Groenkloof: South African Bureau
of Standards.
Doyle,
J.
F.,
1991.
Springer
link.
[Online]
Available
at:
https://link.springer.com/chapter/10.1007/978-94-011-34200_4#:~:text=A%20truss%20is%20a%20structure,or%20by%20pins%20at%20joints.
[Accessed 10 October 2022].
EL-Alghoury, M.A., Ebid, A.M. and Mahddi, 2020. Decision support system to select optimum
steel portal frame coverage system.. Ains Shams Engineering journal, Volume 12, pp. 73-82.
institute,
S.,
Available
at:
[Accessed 10 october 2022].
2022.
SteelConstruction.info.
[Online]
https://www.steelconstruction.info/Portal_frames
Mahachi, J., 2004. Design of structural steelwork to SANS 10162. Pretoria: CSIR .
Miller,
B.,
2019.
GreenGarage.
[Online]
Available at: https://greengarageblog.org/15-arch-bridges-advantages-and-disadvantages-tiedthrough-and-truss
Mohammed A, E.-. A. A. M. E. I. M. M., 2021. ScienceDirect. [Online]
Available
at:
https://www.sciencedirect.com/science/article/pii/S2090447920301829
[Accessed 10 October 2022].
Parrot, G., 2005. Structural Steel Design to SANS 10162 - 1: 2005. 2nd ed. Durban: SHADES
Technical Publications.
PR Salter, A. M. C. K., 2004. Design of a single span steel portal frames to BS 5950- 1:2000,
Silwood Park: The Steel Construction Institute.
Acknowledgements
CIVN4015A: Civil Engineering Design Project: Structures, 2022
Elvin A. (2021). CIVN3010: Structural Steel Design lecture notes. Johannesburg: University
of the Witwatersrand
Rathod G.W. (2022). CIVN4004: Geotechnical Engineering 2 – Foundations. Johannesburg:
University of the Witwatersrand
SANS Codes
Appendix 1 – Meeting Minutes
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4... of ……Structural Design…. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: ………03/10/2022….. Time: 12:00 – 13:00……
Computer Lab……………..
Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963
Name & Student No. of Secretary & signature: Makhosonke Mhlanga 1612449
Names & Student Nos. of those Present
& signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Musa Mdhluli 1391559
Names & Student Nos. of those Absent: Mohau Majoro 1633852
Venue:
…Hillman
S/N
Item
Matters from Previous meeting
1. Election of group member
Items for consideration of current meeting
1. Word document
2. WhatsApp Group
Action
Responsible
person
Nonhlanhla Mhlongo Present
was
automatically members
chosen as the group
leader.
A word document Nonhlanhla
that tracks changes Mhlongo
and contribution of
each group member.
A whatsapp group Nonhlanhla
was
created
by Mhlongo
Nonhlanhla Mhlongo
for easier and faster
communication
amongst members.
3. Google Drive for collection of materials
4. Discussion of strengths and weaknesses
5. Report writing
Nonhlanhla Mhlongo Nonhlanhla
created a google Mhlongo
drive
for
group
members to upload
any
material
necessary for the
project.
Present
group Present
members
outlined members
their strengths and
weaknesses which
gave us an idea as to
how we will tackle
the project.
Tasks to draft the Present
introduction part of members
the project were
delegated to group
members
6. Preliminary designs
A task to come up Present
with
preliminary members
designs
was
delegated to group
members
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of ……Structures Design…. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: …07/10/2022……….. Time: …11:00 – 12:00…
Teams……………………..
Name & Student No. of Chair & signature: Mohau Majoro 1633852
Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086
Names & Student Nos. of those Present & signatures:
Mohau Majoro 1633852
Jackson Mahlaule 1669086
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Musa Mdhluli 1391559
Names & Student Nos. of those Absent: Zamanguni Ntozakhe 908593
Venue:
……MS
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Report Writing
Work
on
the Present
introduction
is members
underway.
-Feedback on the
introduction
was
given by Jackson and
Makhosonke
- Possibilities of
changes as we go
along
with
the
project are to be
expected, as we get
more information
2. Preliminary Design considerations
Nonhlanhladid Present
research on a latticed members
portal frame and does
not recommend it
since it contains a lot
of truss members, as
such
many
connections
will
need
to
be
considered,
increasing
the
possibility
of
calculation errors. It
is not too safe under
fire.
Advantages:
light
weight, and easy
erection, not costly to
erect.
Musaconducted
research on the rolled
section portal frame.
Advantagescontains all sections
of
the
required
beams, no tapered
beams, uses less
members. Analysis
on this beam is
easier.
Disadvantagesdesign with the
lowest
section
possible, but you
might find that you
need
stronger
member size and
offsite fabrication is
not possible.
Musa
conducted
research on the Bus
portal frame.
Advantagesprefabricated at the
factory, take all
primary
and
secondary members
to site, it is very cost
effective and time
effective, can expand
a span about 100
meters or more, it
beats truss on the
erection time and
members
are
labelled.
Items for consideration of current meeting
1. Choosing the design
Musa’s design was Present
chosen by the group. members
2. Uploading on the master document
It is important that Present
every one of us members
uploads
on
the
master document to
prevent the loss of
sections of work.
3. Report writing
Tasks to draft the Present
introduction part of members
the project were
delegated to group
members
4. Preliminary designs
A task to come up Present group
with
preliminary members
designs
was
delegated to group
members
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of …Structures Design ……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: ……10/10/2022…….. Time: …12:00 – 13:30…
……………Hillman H205……………..
Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963
Name & Student No. of Secretary & signature: Musa Mdhluli 1391559
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Jackson Mahlaule 1669086
Musa Mdhuli 1391559
Mohau Majoro 1633852
Names & Student Nos. of those Absent: Zamanguni Ntozakhe 908593
Venue:
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Report writing
Ongoing process
Present
Makhosonkemembers
finalizing reasons for
selecting the rolled
section portal frame
as the frame that the
group would design.
2. Discussing issues about group members
Participation
in Present
group meeting- it members
was requested that
people should notify
the group if they are
not able to attend
meetings and give
reasons
3. Clarity in preliminary designs
Items for consideration of current meeting
1. Work distribution
The
need
for Makhosonke
preliminary designs
for drawings
Dead
load
calculations
Wind
load
calculations
Imposed
load
calculations
Finalizing reasons
for selecting for
rolled section portal
frame as the main
frame to be designed.
Crane
load
calculations
Load combinations
Musa
Majoro
Jackson
Makhosonke
Nonhlanhla
Zamanguni
2. Submissions
Submissions
of Present
calculations by all members
group members next
week Monday
3. Consultation
Clarity needed on the Present
design of the 15m members
end bay span and the
roller shutter doors
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group 4….. of ……Structures Design …. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: ……14/10/2022…….. Time: …11:19 – 11:58…
……………………Hillman Computer Lab……..
Name & Student No. of Chair & signature: Makhosonke Mhlanga 1612449
Name & Student No. of Secretary & signature: Nonhlanhla Mhlongo 1631963
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Musa Mdhuli 1391559
Mohau Majoro 1633852
Names & Student Nos. of those Absent: None
Venue:
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Allocation of tasks
Makhosnke
Makhosonke
reminded every one
of the tasks which
were allocated to
them in the previous
meeting.
Items for consideration of current meeting
1. Load combinations
Zamanguni informed Zamanguni
the group that she
experienced
difficulty with using
the new edition of the
code to determine the
load combinations.
She informed the
group that she would
consult with the
lecturer.
2. Imposed loads
Jackson informed the Jackson
group that he had
done his calculations
but he had not
completed
the
equipment loading
calculations.
Makhosonke offered
to help him with the
calculation.
Musa also stated that
he
would
help
Jackson
with
determining
the
nodes which would
be
used
for
calculations
3. Paragraph
explaining
preliminary design was chosen
which
Makhosonke
Makhosonke
informed the group
that he was done with
the paragraph. He
said he would add
more comments as
we continue to make
progress
project.
with
the
4. Dead loads
Musa informed the
group that he had not
started with the
calculations because
he
needed
the
group’s input on
which sheeting to
select for the project.
Musa
5. Wind loads
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4... of …Structures Design……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
H208…………….
Date: …17/10/2022………. Time: …12:00 – 15:00…
Name & Student No. of Chair & signature: Jackson Mahlaule 1669086
Name & Student No. of Secretary & signature: Mohau Majoro 1633852
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Mohau Majoro 1633852
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Musa Mdhluli 1391559
Venue:
Hillman
Names & Student Nos. of those Absent: Nonhlanhla Mhlongo 1631963
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Loads on the structure
Every
member Present
presented
their members
allocated work on
loads acting on the
structure.
2. Software to use
Musa informed the Musa
group that Prokon
will not be used for
the analysis of the
structure.
Items for consideration of current meeting
1. Wind loads
Majoro still needs to Majoro
do some work on the
wind
loads and
consult with Dr.
Bradley and others
regarding the internal
pressures when there
is no dominant wall.
2. Load combinations
Zamanguni needs to Zamanguni
consult with Dr
Bradley
regarding
loading
combinations using
the new SANS code
and watch YouTube
video
on
load
combinations
and
upload the loads by
Wednesday
midnight.
3. Design for 15m bay
Every member needs Present
to do research on members
how we are going to
design the 15m bay.
4. 2 Portal Frames
A group needs to Present
produce 2 frames of members
which we are to
compare and choose
one for the design.
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of …Structure Design……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: 28/10/2022………….. Time: …12:00 – 13:00…
…………Hillman H205………………..
Name & Student No. of Chair & signature: Makhosonke Mhlanga 1612449
Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Jackson Mahlaule 1669086
Nonhlanhla Mhlongo 1631963
Musa Mdhluli 1391559
Mohau Majoro 1633852
Zamanguni Ntozakhe 908593
Names & Student Nos. of those Absent: None
Venue:
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Progress on finalizing calculations
Outstanding
calculations:
Present
members
Wind
load Mohau
calculations- clause
8.3.9.6 gives clarity
on calculation when
there is a dominant
wall is absent. We
are only waiting for a
response on the 90
degrees part.
Tables
on
load Zamanguni
combinations are yet
to be uploaded but
the
load
combinations
are
completed.
Calculations need to Present
be complete so that members
the analysis can
begin.
Items for consideration of current meeting
1. Door frames
2. Crawl beam
2. Roller shutter doors
Door frames need to
be braced. The group
will consider asking
Dr. Bradley on the
way in which the
door bracing could
be incorporated into
the structure.
Present
members
Present
members
Crawl beam will not Present
be
extended,
a members
member will support
it just beneath the
apex.
Need to find the Present
weight
members
measurements for the
type of door we are
designing for, so far,
the websites visited
do not have the
measurements
for
this specific door
type.
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of …Structures Design……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: …2/11/2022……….. Time: … 12:00 – 13:00…
………Hillman H205…………………..
Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963
Name & Student No. of Secretary & signature: Musa Mdhluli 1391559
Names & Student Nos. of those Present & signatures:
Nonhlanhla Mhlongo 1631963
Musa Mdhluli 1391559
Makhosonke Mhlanga 1612449
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Mohau Majoro 1633852
Names & Student Nos. of those Absent: None
Venue:
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Wind load calculations
Majoro explained the Majoro
challenges he still
has
with
wind
calculations.
2. Meeting attendance
The group discussed Present
the issue of time members
keeping
and
attendance
of
meetings
Items for consideration of current meeting
1. Communication issues
We
had Present
communication
members
issues amongst group
members by not
letting us know that
you will not be
available for the
meeting and telling
us late that you had
an emergency.
2. Fixing wind calculations
Everyone must give Present
input on what they members
thinks should be
done to get better
results.
3. Time
Discussed that we are Present
running out of time to members
finish the design
project.
4. Warnings
Issued out warnings Present
that not being part of members
the meeting will
decrease your overall
contribution towards
the project. Simply
reading the meeting
minutes does not
excuse your absence.
5. Writing minutes
Everyone was told to Present
write the minutes and members
upload them no later
than a day after the
meeting.
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group ….. of ………. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: 07/11/2022………….. Time: 12:00 – 12:30……
Computer Lab…………………………..
Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963
Name & Student No. of Secretary & signature: Zamanguni Ntozakhe 908593
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Musa Mdhluli 1391559
Names & Student Nos. of those Absent: Mohau Majoro 1633852
Venue:
Hillman
S/N
Item
Matters from Previous meeting
1. Update on tasks allocated in the previous
meeting.
Action
Responsible
person
Musa had a problem Musa
with
the
wind
calculations
not
being clear and was
going to seek clarity
from Majoro.
2. Prokon results
The deflections from Musa
&
the prokon results Zamanguni
were too high and not
within the code
limits. Musa told the
group that he would
consult with Dr.
Bradley.
Items for consideration of current meeting
1. Meeting the professional in industry
2. Tabulate Loads
A profession Civil Nonhlanhla
Engineer is willing to
meet the group on Ms
Teams to discuss the
progress
of
the
group. Time TBA on
WhatsApp.
Makhosonke
will Makhosonke,
tabulate the dead Nonhlanhla &
loads,
Nonhlanhla Jackson.
will tabulate the
crane loads and
Jackson will tabulate
the imposed loads.
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of …Structures Design……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: …10/11/2022……….. Time: 12:00 – 13:00……
group chat…………………………..
Name & Student No. of Chair & signature: Mohau Majoro 1633852
Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Jackson Mahlaule 1669086
Musa Mdhluli 1391559
Mohau Majoro 1633852
Zamanguni Ntozakhe 908593
Names & Student Nos. of those Absent: None
Venue:
Whatsapp
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Prokon analysis
The analysis will be Musa
done soon then
everyone will be able
to start with the
design calculations.
Items for consideration of current meeting
1. Member design calculations
Tie
beams
–
Nonhlanhla
&
Makhosonke
Ground
slab
–
Makhosonke
Concrete foundation
– Makhosonke
Door
frameNonhlanhla
Crawl
beamNonhlanhla
Gable columns –
Musa
ConnectionsZamanguni
Purlins and girts Zamanguni
Makhosonke,
Nonhlanhla,
Musa
&
Zamanguni
School of Civil and Environmental Engineering
University of the Witwatersrand
Minutes of meeting of Group …4.. of …Structures Design……. Project
(These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working:
Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary
environments.)
Important Instructions with compiling the minutes
1. There should be minutes for each of week of the Design Project
2. The Chair and Secretary should be rotated every week among members of the group
3. The minutes must be signed by all present at the meeting
4. The minutes should be incorporated in the Appendix of the Design Report
Week: ………
Date: …14/11/2022……….. Time: …12:18 – 13:11…
Computer Lab…………………………..
Venue:
Hillman
Name & Student No. of Chair & signature: Musa Mdhluli 1391559
Name & Student No. of Secretary & signature: Nonhlanhla Mhlongo 1631963
Names & Student Nos. of those Present & signatures:
Makhosonke Mhlanga 1612449
Nonhlanhla Mhlongo 1631963
Zamanguni Ntozakhe 908593
Jackson Mahlaule 1669086
Musa Mdhluli 1391559
Mohau Majoro 1633852
Names & Student Nos. of those Absent: None
S/N
Item
Action
Responsible
person
Matters from Previous meeting
1. Prokon analysis
Musa was reminded Present
that he said he would members
present the prokon
results to the group
today.
Items for consideration of current meeting
1. Prokon analysis
2. Purlins
Upon running the Musa Mdhluli
analysis
of
the
structure on Prokon.
Musa discovered that
the deflections were
not
within
the
deflection
limits
stated in the code. It
was suggested that
we have a meeting
with the industry
advisor so solve the
issue.
Zama discussed her Zamanguni
progress on the
purlin
design
calculations
and
informed the group
that she was unable
to
finish
the
calculations because
there were some
uncertainties.
However,
she
received assistance
from the group.
Appendix 2 – Cross-disciplinary and stakeholder engagement GA 8 form
Appendix 3 – Preliminary Design 1 (Latticed Portal Frame)
Appendix 4 – Preliminary Design 2 (Trussed Portal Frame)
Appendix 5 – Preliminary Design 3 (Rolled Section Portal Frame)
Appendix 6 – Preliminary Design 4 (Pre-engineered built-up Portal Frame)
Appendix 7 – Dead Loads
Appendix 8 – Wind Loads
Appendix 9 - Imposed Loads
Appendix 10 – Member designs
Appendix 11 – Foundation Design
Appendix 12 – Ground slab design
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
As partial fulfilment of GA 8 (Individual, team and multidisciplinary working),
cross-disciplinary interactions and stakeholder engagement of students are assessed in the form
of a log-report which must be completed to the satisfaction of the Supervisor.
Date
Name
07/11/202
Thami Langa
2
Profession
or
Affiliation
Structural
Issues Addressed
Engineer
uncertain about where the peak
Wind calculations: The group was
pressure must be calculated ( the
eaves height or the apex height). Mr
Langa advised that the apex height
be used since more wind pressure is
experienced as you go higher (wind
curve). After investigating the wind
calculation, it was discovered that
the way in which the coefficients
were calculated was erroneous (the
values which we got were too low).
Mr Langa provided pointers for
rectifying the errors. Written below
are the pointers which we were
given:
● Mr Langa advised that the
group use clause 8.3.9.6
Note 2.
● Mr Langa informed the
group that when all doors
Signature
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
are open, that is considered
as an accidental load
condition (clause 8.3.9.3).
Therefore the case that
should be considered is one
where all doors are closed.
● Total Cp is determined by
subtracting the Cpi from the
Cpe. The total Cp is
multiplied by the peak
pressure to obtain the net
wind pressure on the face of
the wall or the roof.
● The wind load udl is the
resultant wind pressure.
● According to the code the,
edges are where the higher
pressures will occur so it
must be used when
designing the purlins.
● When the wind hits the
walls at 0 degrees and 180
degrees, the sway movement
will mirror each other (for
the walls).
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
● When the wind hits the
walls at 90 and 270 degrees,
the pressure decreases as
you move away from the
face where the wind acts.
Prokon analysis: The group showed
Mr Langa the 3-D model which was
used on Prokon. Mr Langa then
suggested that the group first
analyze the structure in 2-D to
obtain the correct section sizes and
strength. The loads should first be
applied on a single frame to
determine the lateral deflections as
well as the sway on the structure.
Once the sizing is correct, a 3-D
model is used to determine the
stability of the structure.
After investigating the Prokon
analysis, it was discovered that the
approach used by the group was
erroneous. Mr Langa provided
pointers for rectifying the errors.
Listed below are the pointers which
the group was given:
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
● The columns were supposed
to be pinned at the bottom in
order to determine the
highest moment that would
occur at the connection
between the beam (rafter
beam) and the column. This
is the moment that is used
for the design calculations.
● The columns were supposed
to be fixed in order to
determine the moment at the
base of the column, this is
what will be used for the
foundation calculations.
● The members used for the
columns and rafters are too
light, therefore stiffer
members.
● The deflections were too
high because there were no
haunches at the apex.
● It was recommended that the
group should use SANS
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
10162- Part 2 to determine
the deflection limits.
● It was recommended that a
I- beam or Circular hollow
section be used for the
Eaves Beam.
Bracing: The group decided on
bracing three bays. Mr Langa then
told the group to conduct research
on the limits of the number of bays
that should be braced on the
structure.
Load combinations: After
investigating the load combinations,
it was discovered that they were not
well-written.The main issue was
with the load factors (they need to
be re-calculated). Mr Langa advised
that they be revised because they
were the reason why the Prokon
analysis yielded incorrect results.
For all the load cases with wind
load, the wind loads need to be
considered in all directions. When
the wind hits the structure at 90 &
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
270 degrees, it will not behave in
the same manner because the bay
spaces are not the same.
When the load factors for the ULS
and SLS need to go hand in hand
when it is inputted into Prokon.
2022/11/22
09/11/202
2
Thami Langa
Structural
Wind calculations: The group
Engineer
showed Mr Langa the adjustments
made to the wind calculations
following the meeting we had with
him (initially, the Cpi and Cpe
calculations were wrong) and the
adjustments were done to his
satisfaction. Mr Langa
recommended that an accurate
sketch that shows how the wind
pressure decrease further away from
the face at which the wind acts)
Load combinations (and factors):
The group showed Mr Langa the
adjustments made to the load
combinations (the ULS and SLS
load combinations were not
corresponding with each other) but
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
he noted that there were still errors
which needed to be corrected, (Load
case 1,2,3 and 6 were not done
correctly) .
Bracing: The group discussed how
they intended to brace the structure
(the blue book suggests that both
end bays should be braced if the
structure is more than 50m long)
and Mr Langa was satisfied with the
group’s decision. There are two
bays which will be braced, the
second and the seventh bay.
Prokon analysis: The group showed
Mr Langa the results obtained on
Prokon (which were based on the
incorrect load combination) so it
was noted that the results would
need to be revised.
15/11/202
2
Thami Langa
Structural
Load combinations: Mr Langa
Engineer
raised the concern that of the load
combinations presented by the
group, there’s no load case that
caters to the wind uplift. The other
2022/11/22
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
problem is that for all load cases
where there are wind loads, they
need to be considered for the
angles at which the wind acts on the
structure (0 , 90, and 270).
Prokon analysis: Due to the fact that
the wind loads and the load
combinations are still incorrect, the
results obtained from Prokon were
incorrect.
Wind load: The height used to
calculate the pressure was incorrect
and it needed to be changed.
17/11/202
2
Thami Langa
Structural
Ground slab design: The group was
Engineer
under the impression that the
ground slab would be designed as a
two way slab, however Mr Langa
informed the group that two-way or
one-way slabs only apply to
suspended slabs.
He further explained that the ground
slab can be considered as a flat slab.
He recommended that the group
should use a book by Brian Perry
2022/11/22
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
called ‘Concrete industrial floors on
the ground’ to get more insight on
how to design the ground slab. He
also informed the group that the
slab dimensions are actually the
width and length of the whole
structure (20m x 60m), but there are
limits regarding how long and wide
the slab will be because shrinkage
affects the slab. The slab needs to
be broken down so as to avoid
cracks within the slab by splitting it
into panels, with saw cut joints in
between the panels which are used
to control where cracking occurs
due to shrinkage and isolation joints
on the edges to allow for movement
in the three directions (contraction
and expansion). The joints are there
to enable the slab to release internal
stresses to prevent the occurrence of
cracks. He recommended that the
panels (ratio of 1:1 - width to
length) should be square in shape.
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
The group was also under the
impression that the thickness of the
slab would be assumed in order to
carry out the design calculations,
but Mr Langa corrected the group
by stating that the thickness is
determined by carrying out the
design calculations.
Foundation design: Mr Langa
informed the group that the
founding level is determined by
taking the top of the slab as the
reference height, then subtracting
the slab thickness and 800.
Concrete plinths: The group lacked
knowledge on what concrete plinths
are and so Mr Langa informed the
group that they are used to transfer
the load from the structure to the
foundation. The plinth goes from
the foundation and it ends 300mm
above the top of the slab.
Connection design: The group
informed Mr Langa that the only
software they are familiar with for
University of the Witwatersrand
School of Civil & Environmental Engineering
CIVN4015A – Civil Engineering Design
designing connections is the
eToolkit which is no longer
available to students at the
computer lab. Mr Langa then
showed the group how to use
Prokon to design the connection.
2022/11/22
I confirm that the CIVN4015A Group 4 had consultations and received
guidance from me with regards to the above mentioned for their group
project.
T. Langa - BScEng (Civil)
Structural Engineer, WSP Africa
Email: shangase.thami@gmail.com
Cell: 071 214 5686
2022/11/22
Wind
loads
-
CIUNQOIS
Reference : SANS 10160-3
roughness and Obstructions
Terrain
*
-
(tow
category Bheights) vegetation
obstacle
Terrain
20
Height to the top of
-
Structure
=
=
Cr Gm)
i.
Cr (9,76
1.36
fÉ¥É
1.36
(§÷→)"° -0,96£
m
)
=
1.36
(390%-50)%0%0,981
=
:
Wind
*
The
i.
C
building will
prob
=
have
so
-
≥
36 mfs
year
mean
return period
1.0
where : Nb, peak = 1.0 ✗ 36
Up ( 8m )
Vigo
a
Peak Wind Speed : Vp (e)
i.
1.0
Speed
Fundamental wind speed :
→
=
"
_-
Topography Factor Co G)
→
" Cm
Topography
*
→
9,761s
roughness factor : Craig
Terrain
-
of at least
0,09s
2
0
300
B
separations
to tan lot 81-0,00156
Category Zgccn) Z.cm) Zccm ) a
-
or
=
Up (9.76m)
=
=
0,964×1 ✗ 36
=
Creed
✗
Co G)
✗
Vb peak
,
36m Is
=
0,982 ✗ 1- ✗ 36
34,700mi
=
3F35mls- ✓
Wind Pressure
*
Altitude ÷
-
:
,_
=
) (1600-1500)
[
(%::÷÷)4a• isao
" ◦◦
density Gloom =P
Air
→
1600M
i.
+
+
2000
-
P's "
-
1500
-
_Ém:
Peak Wind Pressure :
i.
qp(
9,76m )
Cfp
(som)
* External
=
{ P✗VbKJ2
£6,985)( 3%3532--0,6151511
=
,
=
¥6198s ) (34/704)
Pressure
Roof Slope
For wind
qpcas
=
0
at
=Q593KPa_
Coefficients
10°
( normal long sided
-
d- 20,2m
µ
≥
*
*
e=min{b ; 2h }
i.
-
F
-
=
Min
=
min
Wind
direction
>
{
D G
J
I
E. %
e
ee
=
{60,2
{ 60,2
;
219,76)}
; 19,52
}
19,52m
b
i. @
=
19.52m
Cd
C.Fig 8 b)
F
•
B
A
*
els
*
keys
C
*
d- I
-
20 in
Sizes
Zone
45
=
.
3,9m
; %
=
1. 95m
pressures : G-
-
Zone
A
B
C
D
E
f
cpe.io
=
; 445=15 62m ; %
0,48m
Correlation Factor
-1,2
-
-
0,8
0,5
0,73
0,85
0,36
0,85
-
-
1,3/1-0,1
§
-1,0/+0,1
It
-0,45/+91
I
-0,5/-0,3
I
-0,8/-0,3
=
4,88m
For wind
90
at
( Normal to gable End )
'
E
e=min{b ; 2h }
*
=min{20,2 ; 2×9,76}
C
e=↑9,52m_
C
I
I
s
É
It
It
g
F
g
A
F
&
D
K
b
=
20,2m
A
→
Side walls
b- 20,2m
Cfig 8 b)
B
B
A
G- 19,52ms
&
*
wind
direction
Roof Pressures : %
=
0,162
I
0,25
Zone
A
Cpe 10
,
'
Correlation Factor
-1,2
B
-
c
-
D
0,5
0,71
0
0,31
0,85
E
-
t
-
g
0,8
l
85
-
45
1,3
-
It
,
,
-0,6s
I
-
g-
0,55
_
* Internal
Pressure
Coefficients
Check for dominance with
-
Area of roller shutter
roller
Shutter doors :
doors on Windward wall
=
2
(4×4)
=32_m}
Area
of roller
shutter doors
on
Leeward wall
=
=
•
•
.
There's
no
dominant wall
.
2
14×4 )
32m£
(Area of openings
are
equal]
iii
-
µ
G-
=
=
( Normal Long Sides
?EÉ
=
as
0,48
By interpolation :
i.
-
0°
Wind at
Wind at
90
Cpi
}÷%= as
G-
=
0,162
=
0
,
Is
(From the graph )
-
=
Cpi
0,14
( Normal to gable Ends
µ
i.
=
C
0,2
(From the graph]
( Normal to long Side)
Wind
at
Zone
Gp ftp.ascpe.eocp.icpr-cpe-cpiwk-qpxcpr
A
-
0°
0,62
-1,2
0,2
-1,4
-0,87
B
0,62
-0,8
0,2
-1.0
-0,62
C
0,62
-0,5
0,2
-0,7
-0,43
☐
0,62
0,73
012
0,53
0,33
E
0,62
-0,36
0,2
-0,5-6
-0,35
F
0,62
-1,3
0,2
-1,5
-0,93
9
0,62
-1,0
0,2
-1,2
-0,74
It
0,62
-0,450,2
-0,65
-0,40
I
0,62
-0,5
0,2
-0,7
-0,43
J
0,62
-98
0,2
-1.0
-0,62
Wind
at
-
90° (Normal to gable
""" "" "
""
A
-1,2
C
•
E
,
9
It
I
-
End )
" " """
"" "" "
/ / [Y f |
0,62
B
-
'
"
"
0,2
-
1,4
-0,87
0,62
-0,8
0,2
-1.0
-0,62
0,62
-0,5
0,2
-0,7
-0,43
""
""
"
""
""
0,62
-93
-0,2
-0,5
.
%,
%
-0,31
,•
+ ,,
.
.
-0,93
962
-143
0,2
-1,5
0,62
-0,65
0,2
-0,85
0,62
-0,5-5 0,2
-0,75
-
-0,53
-
-
-0,47
*
Area leads
to line
wind at
(Normal to gable
90°
loads
-
End )
0,31 KPa
94
I
?
0,43
KPa
kPa
7
0,62
0,62
KPa
kPa
[
0,87
Kpq
0,87
KPa
0,32
Kpq
^
wind
Note :
Red
=
Blue
=
Pressure Away ( Suction )
Pressure towards
For
@ 7,5m
@ End
@
distribute
design
I
15m
-
bag ! -0,62×7 s
bag : -0,62×75/2
=
0162 ✗ 2,75
-
-
0,62 ✗ 4
=
=
0,62×7,5
Roof
-2.33 KN fun
=
-1,71 AN / in
-
2,48
/ §
0,47
kPa
""
kPa
5- B-
0,5-35
µ,
µ
,
i÷a
A
Note :
Red
_-
Pressure Away
Wind
KN fun
-4,65 KN fun
-
,
0,62hPa
-4,65 KN / in
.
0
=
0,62 ✗ 11,25=-6,98 AN / in
-
-
For
=
,
bag !
load
area
.
.
For design distribute area load
=
-0 s
,
-
s
-
kPa
@ 7 , Sm i -95-5×715=-4113 kN / in
@ End
bag : -0,5-5×742=-2,06 KNIM
@ 15m
bag
:
0,5s
-
-
Wind
0,55×6,25
-
-
0,58
-
✗ 5-
-
=
=
✗ 2,5
-3,44 KNIM
-2,75 KNIM
=
-
( Normal to long side)
at
0,87 KPa
-0,62
KPa
0143
"
"
÷:
0133
-8°
KPa
÷
§
it
¥ I i.
n
et
Ñ
-
É
I
r
0,87
kPa
0,62
Rpg
0143
kPa
0,3s
kPa
1,38 KN 1m
A
B
C
ohsuEI.IT?
-6,5319N /
on
4,65kW /
in
-
-3,2619N / -5,44kW/
m
-4,35kW / m
-2,33km/ -3,88 Kmfm -3,1 KN /
m
-3,23 ANIM -1,61 KIM
"
m
-
in
2,691%-2,15 KTN
-0,88 kN_m
-2,623141m
•
→•
9
-5,55 KIM
-2,781¥
-4,63 "Tn
-3,7
It
-3
KIM
-1,5k£
-2,5k£
-2kW / in
I
-3,23 "Fn
Ktn
-2,69 Kwan
-2,15µm
-
KIM
1,3111µm -2,19 Afn -1,75 $1m
E
J
-1,08
0,83 KIM
2,48 Fn
-
-1,5-5 KNIM
1,6s KIM
24ᵗʰ 2,06141m
D
1.
-2,18 AN /in
.
-161
.÷→.÷
4,65k£ -2,3319m -3,88kt
.÷.
-3, /
ᵗᵈm
KIM
⇐
-1.85
-
ᵗᵈm
I kN /us
-1,08 RN / us
-
I ,SSKNlm
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Member Design for Combined Stresses
MOMENTS: X-X
M max = 154.6kNm @ 8.00m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-50.0
50.0
100
Lx Eff = 6.800 m
W1x = 1.00
Ly Eff = 6.800 m
W1y = 1.00
Le Eff = 8.000 m
W2 = 2.37
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC2
150
MOMENTS: Y-Y
M max = -4.889kNm @ 6.99m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-4.00
-2.00
Ver W5.0.04 - 13 Sep 2022
Combine Ver W5.0.04
Element 42-38
Evaluate current section
Section name COLUMN
Section 356x171x67
I-sections (Web vert)
SANS 10162-1:2011 13.8.2 :
a) Cross-sectional strength (Crit. pos.= 8.000 m)
Cu
0.85Mux
0.60Muy
77.6
133
.092
-- + ------- + ------- = ---- + ---- + ---- = 0.38
Cr
Mrx
Mry
2693
381
76.5
OK
2.00
4.00
AXIAL FORCE
P max = 125.4kN @ 0.00m
120
c) Lateral torsional buckling strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.75
Cr
Mrx
Mry
483
295
76.5
100
80.0
60.0
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
40.0
20.0
b) Overall member strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.74
Cr
Mrx
Mry
359
381
76.5
d) Additional check for class 1 I sections
Mux
Muy
155
4.15
--- + --- = ---- + ---- = 0.58
Mrx
Mry
295
76.5
OK
OK
OK
13.4:Shear
Vux<Vrx
Vuy<Vry
34.5 < 687.5
4.8 < 471.1
Slenderness Ratio: Lx/rx = 45
Ly/ry = 170
------------------------------------------------------------
OK
OK
OK
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
7.00
6.00
5.00
4.00
M max = .8284kNm @ 3.75m
3.00
2.00
1.00
MOMENTS: X-X
Checked by
Combine Ver W5.0.04
Element 47-68
Section name EAVES
Date
Evaluate current section
.600
Lx Eff = 6.375 m
W1x = 1.00
Ly Eff = 6.375 m
W1y = 1.00
Le Eff = 7.500 m
W2 = 1.00
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Section class: 1
.800
Critical Load Case : LC11
.200
.400
.00500
MOMENTS: Y-Y
M max = 0.000kNm @ 0.00m
Round hollow sections
SANS 10162-1:2011 13.8.3 :
a) Cross-sectional strength (Crit. pos.= 3.750 m)
Cu
Mux
Muy
12.6
.888
0.00
-- + --- + --- = ---- + ---- + ---- = 0.06
Cr
Mrx
Mry
536
23.0
23.0
.00400
.00300
.00200
AXIAL FORCE
7.00
6.00
5.00
4.00
3.00
2.00
1.00
.00100
P max = 12.64kN @ 0.00m
12.0
b) Overall member strength
Cu
U1xMux
U1yMuy
12.6
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.15
Cr
Mrx
Mry
111
23.0
23.0
c) Lateral torsional buckling strength
Cu
U1xMux
U1yMuy
12.6
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.12
Cr
Mrx
Mry
147
23.0
23.0
10.0
8.00
6.00
OK
OK
OK
13.4:Shear
7.00
6.00
5.00
4.00
3.00
2.00
1.00
4.00
2.00
Section 139.7x4.0+
Vux<Vrx
Vuy<Vry
0.4 < 196.4
0.0 < 196.4
OK
OK
Slenderness Ratio: Lx/rx = 133
Ly/ry = 133
OK
OK
------------------------------------------------------------
7.00
6.00
5.00
4.00
M max = .8284kNm @ 3.75m
3.00
2.00
1.00
MOMENTS: X-X
Combine Ver W5.0.04
Element 26-68
Section name EAVES
Evaluate current section
.600
Lx Eff = 6.375 m
W1x = 1.00
Ly Eff = 6.375 m
W1y = 1.00
Le Eff = 7.500 m
W2 = 1.00
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Section class: 1
.800
Critical Load Case : LC12
.200
.400
.00500
MOMENTS: Y-Y
M max = 0.000kNm @ 0.00m
Round hollow sections
SANS 10162-1:2011 13.8.3 :
a) Cross-sectional strength (Crit. pos.= 3.750 m)
Cu
Mux
Muy
31.2
.993
0.00
-- + --- + --- = ---- + ---- + ---- = 0.10
Cr
Mrx
Mry
536
23.0
23.0
.00400
.00300
.00200
AXIAL FORCE
7.00
6.00
5.00
4.00
3.00
2.00
1.00
.00100
P max = 31.18kN @ 0.00m
30.0
b) Overall member strength
Cu
U1xMux
U1yMuy
31.2
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.32
Cr
Mrx
Mry
111
23.0
23.0
c) Lateral torsional buckling strength
Cu
U1xMux
U1yMuy
31.2
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.25
Cr
Mrx
Mry
147
23.0
23.0
25.0
20.0
15.0
OK
OK
OK
13.4:Shear
7.00
6.00
5.00
4.00
3.00
2.00
1.00
10.0
5.00
Section 139.7x4.0+
Vux<Vrx
Vuy<Vry
0.4 < 196.4
0.0 < 196.4
OK
OK
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Slenderness Ratio: Lx/rx = 133
Ly/ry = 133
------------------------------------------------------------
OK
OK
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 11-321
Section: 139.7x5.0+ O
Lx eff = 4587
Ly eff = 4587
Lz eff = 5397
Le eff = 5397
Section Slenderness:
DT_ratio =
=
Table 4
D.
fy
T
139.00
×350
5.00
= 9 730.000
DT_ratio < 13000 => Class 1 section
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×5 397.00
47
= 97.605
X_ratio < 200 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×5 397.00
47
= 97.605
Y_ratio < 200 => OK
10.4.2.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial compression and bending
13.8
Member strength and stability — All classes of sections except class 1 and class 2 I-shaped sections
13.8.3
Cross-sectional strength
13.8.3 a
Cr =
=
f . A . fy
1000
0.9 ×2 100.00 ×350.00
1000
= 661.500 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×89 800.00 ×350.00
1×106
= 28.287 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×89 800.00 ×350.00
1×106
= 28.287 kNm
End moment factors:
k x = -1 (no X moments)
k y = -1 (no Y moments)
w x = 0.6 - 0.4 . k x
= 0.6 - 0.4 ×0.00
= 0.6000
w y = 0.6 - 0.4 . k y
= 0.6 - 0.4 ×-1.00
= 1.0000
13.8.5
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Cex =
=
Checked by
Date
p2 . E . Ix
1000 . (Kx. Lx)2
p2 ×200 000.00 ×4 730 000.00
1000 ×(0.85 ×5 397.00 )2
= 443.658 kN
Cey =
=
p2 . E . Iy
1000 . (Ky. Ly)2
p2 ×200 000.00 ×4 730 000.00
1000 ×(0.85 ×5 397.00 )2
= 443.658 kN
U1x =
wx
1-
=
Cu
Cex
1.000
31.43
1443.74
= 1.076
U1y =
wy
1-
=
Cu
Cey
1.000
31.43
1443.74
= 1.076
F13.8.3(a) =
=
Cu U1x. Mux U1y. Muy
+
+
Cr
Mrx
Mry
31.4
1.076 ×0.53 1.076 ×0.00
+
+
661.5
28.29
28.29
= 0.0676
OK
Overall member strength
Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports
and does not refer to restraint conditions)
13.8.3 b
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
f ex =
=
p2 . E
Kx. Lx 2
rx
p2 ×200000
1 ×5397 2
47.4
= 152.259 MPa
f ey =
=
p2 . E
Ky. Ly 2
ry
p2 ×200000
1 ×5397 2
47.4
= 152.259 MPa
p2 . E . Cw
+ G. J
(Kz . Lz )2
f ez =
A . ro2
p2 ×200000 ×0
=
(1 ×5397 )2
+ 77000 ×9470×103
2100 ×67.034 2
= 77.27×103 MPa
fe = min(fex, fey, fez) = 152.3 MPa
l=
=
fy
fe
350
152.26
= 1.516
Checked by
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×2 100.0 ×350 × 1 + 1.5162 2 ×1.34
1×103
-1.34
= 232.881 kN
U1x = 1.000
U1y = 1.000
Cu U1x. Mux U1y. Muy
F13.8.3(b) =
+
+
Cr
Mrx
Mry
=
31.4
1.000 ×0.53 1.000 ×0.00
+
+
232.9
28.29
28.29
= 0.1536
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×89 800.000
1×106
= 31.430 kNm
My =
=
f y. Zex
1×106
350.000 ×68 100.000
1×106
= 23.835 kNm
Mr = f . Mp
= 0.9 ×31.430
= 28.287 kNm
Cr based on weak-axis bending
ly=
=
Ky. Ly.
fy
2.
ry
p E
0.85 ×5 397.00
350
× 2
47.40
p ×200000
= 1.289
OK
13.8.3 c
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×2 100.0 ×350 × 1 + 1.289 2 ×1.34
1×103
-1.34
= 293.242 kN
F13.8.3(c) =
=
Cu U1x. Mux U1y. Muy
+
+
Cr
Mrx
Mry
31.4
1.000 ×0.53 1.000 ×0.00
+
+
293.4
28.29
28.29
= 0.1258
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 42-38
Section: 356x171x67 I
Lx eff = 6800
Ly eff = 6800
Lz eff = 8000
Le eff = 8000
Section Slenderness:
Table 4
Flanges
Bratio =
=
B.
fy
Tf
86.60
× 350
15.70
= 103.193
Bratio < 145 => Class 1 Flange
Web
Wratio_1 =
=
1100 .
fy
1100
350
1 -
×1 -
0.39 . Cu
f . Cy
0.39 ×131.8
0.9 ×2 992.5
= 57.675
Hw/Tw(34.3) < Wratio_1 => Class 1 Web
=> Section class = 1
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×8 000.00
151
= 45.033
X_ratio < 200 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×8 000.00
40
= 170.000
Y_ratio < 200 => OK
10.4.2.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial compression and bending
13.8
Member strength and stability - class 1 and 2 I-shaped sections
13.8.2
Cross-sectional strength
13.8.2 a
Cr =
=
f . A . fy
1000
0.9 ×8 550.00 ×350.00
1000
= 2 693.250 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×1 211 000.00 ×350.00
1×106
= 381.465 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×243 000.00 ×350.00
1×106
= 76.545 kNm
End moment factors:
Mux_min
kx =
Mux_max
=
91.21
176.91
= 0.5156
ky =
Muy_min
Muy_max
=
0.00
-1.31
= 0.0000×100
w x = 0.6 - 0.4 . k x
= 0.6 - 0.4 ×0.52
= 0.3920
w x = max(w x,0.4) = 0.4
w x = 1 (loads between supports)
13.8.5
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
w y = 0.6 - 0.4 . k y
= 0.6 - 0.4 ×0.00
= 0.6000
w y = 1 (loads between supports)
Cex =
=
p2 . E . Ix
1000 . (Kx. Lx)2
p2 ×200 000.00 ×195 000 000.00
1000 ×(0.85 ×8 000.00 )2
= 8 324.277 kN
Cey =
=
p2 . E . Iy
1000 . (Ky. Ly)2
p2 ×200 000.00 ×13 600 000.00
1000 ×(0.85 ×8 000.00 )2
= 580.565 kN
U1x =
wx
1 -
=
Cu
Cex
1.000
131.77
1 8 324.28
= 1.016
U1y =
wy
1-
=
Cu
Cey
1.000
131.77
1580.56
= 1.294
F13.8.2(a) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
80.7
0.85 ×1.016 ×176.91 0.6 ×1.294 ×0.00
+
+
2 693.2
381.46
76.54
= 0.4305
OK
Overall member strength
13.8.2 b
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports
and does not refer to restraint conditions)
f ex =
=
p2 . E
Kx. Lx 2
rx
p2 ×200000
1 ×8000 2
151
= 703.240 MPa
f ey =
=
p2 . E
Ky. Ly 2
ry
p2 ×200000
1 ×8000 2
39.9
= 49.102 MPa
p2 . E . Cw
+ G. J
(Kz . Lz )2
f ez =
A . ro2
p2 ×200000 ×4120×108
(1 ×8000 )2
=
+ 77000 ×560000
8550 ×156.18 2
= 267.688 MPa
fe = min(fex, fey, fez) = 49.1 MPa
l=
=
fy
fe
350
49.102
= 2.670
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 2.6698 2 ×1.34
1×103
-1.34
= 358.757 kN
U1x = 1.000
U1y = 1.000
Ky. Ly.
fy
ly =
2
ry
p .E
=
1.00 ×8 000.0
350
× 2
39.90
p ×200000
= 2.670
b = 0.6 + 0.4 . l y
= 0.6 + 0.4 ×2.269
= 1.508
=> b = 0.85
13.8.2
F13.8.2(b) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
131.8 0.85 ×1.000 ×176.91 0.85 ×1.000 ×6.46
+
+
358.7
381.46
76.54
= 0.8334
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×1 211 000.000
1×106
= 423.850 kNm
My =
=
f y. Zex
1×106
350.000 ×1 071 000.000
1×106
= 374.850 kNm
OK
13.8.2 c
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
w 2 = 1.75 + 1.05 . k x + 0.3 . k x2
= 1.75 + 1.05 ×0.516 + 0.3 ×0.516 2
= 2.372
w 2. p.
Mcr =
Ke. L
E . Iy. G . J +
p. E 2 . .
Iy Cw
Ke. L
1000
2
p×200
2.371 ×p
× 200 ×13600000 ×77 ×560000 +
×13600000 ×412000000000
1.000 ×8000
1.000 ×8000
=
1000
= 362.826 kNm
Sclass=1 and section is doubly symmetric =>
Mcr > 0.67Mp =>
0.28 . Mp
Mr = 1.15 . f . Mp. 1 Mcr
= 1.15 ×0.9 ×423.850 × 1 -
0.28 ×423.850
362.841
= 295.200 kNm
Cr based on weak-axis bending
ly=
=
Ky. Ly.
fy
2
ry
p .E
0.85 ×8 000.00
350
× 2
39.90
p ×200000
= 2.269
1
Cr =
f . A . f y. 1 + l
2. n
-n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 2.269
1×103
2 ×1.34
-1.34
= 483.521 kN
F13.8.2(c) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
131.8 0.85 ×1.000 ×176.91 0.85 ×1.000 ×6.46
+
+
483.4
295.20
76.54
= 0.8538
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Additional check for class 1 and 2 sections
F13.8.2 =
=
Checked by
Date
OK
13.8.2
Mux Muy
+
Mrx Mry
176.9
6.46
+
295.20 76.54
= 0.6837
OK
Shear
13.4
Shear buckling coefficient
kv=5.34
13.3.4.1.1 a)
hwtw =
hw
tw
=
312
9.1
= 34.286
Shear buckling factor
Sfac =
=
13.4.1.1 a)
kv
fy
5.34
350
= 0.1235
Inelastic critical plate-buckling stress in shear
f cri =
=
13.4.1.1 b)
290 . f y. k v
hw
tw
290 × 350 ×5.34
312
9.1
= 365.670 MPa
Elastic critical plate-buckling stress in shear
13.4.1.1 d)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
f cre =
=
Checked by
Date
180000 . k v
hw 2
tw
180000 ×5.34
312 2
9.1
= 817.688 MPa
hw/tw <= 440*sqrt(kv/fy) =>
13.4.1.1 a)
f sx = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Factored shear resistance X-axis:
13.4.1.1
Avx = H . tw
= 363.4 ×9.1
= 3 306.940 mm2
Vrx =
=
0.9 . Avx. f sx
1000
0.9 ×3 306.9 ×231
1000
= 687.505 kN
Vux = 39.6kN
Factored shear resistance Y-axis:
f sy = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Avy =
=
2 . B . tf . 5
6
2 ×86.6 ×15.7 ×5
6
= 2 266.033 mm2
OK
13.4.1.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vry =
=
Checked by
Date
0.90 . Avy. f sy
1000
0.90 ×2266 ×231
1000
= 471.101 kN
Vuy = 7.8kN
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 190-191
Section: 356x171x67 I
Lx eff = 3400
Ly eff = 3400
Lz eff = 4000
Le eff = 4000
Section Slenderness:
Table 4
Flanges
Bratio =
=
B.
fy
Tf
86.60
× 350
15.70
= 103.193
Bratio < 145 => Class 1 Flange
Web
Wratio_1 =
=
1100 .
fy
1100
350
1 -
×1 -
0.39 . Cu
f . Cy
0.39 ×13.2
0.9 ×2 992.5
= 58.685
Hw/Tw(34.3) < Wratio_1 => Class 1 Web
=> Section class = 1
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×4 000.00
151
= 22.517
X_ratio < 200 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×4 000.00
40
= 85.000
Y_ratio < 200 => OK
10.4.2.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial compression and bending
13.8
Member strength and stability - class 1 and 2 I-shaped sections
13.8.2
Cross-sectional strength
13.8.2 a
Cr =
=
f . A . fy
1000
0.9 ×8 550.00 ×350.00
1000
= 2 693.250 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×1 211 000.00 ×350.00
1×106
= 381.465 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×243 000.00 ×350.00
1×106
= 76.545 kNm
End moment factors:
Mux_min
kx =
Mux_max
=
-1.60
-4.31
= 0.3712
ky =
Muy_min
Muy_max
=
0.34
2.84
= 0.1197
w x = 0.6 - 0.4 . k x
= 0.6 - 0.4 ×0.37
= 0.4520
w x = 1 (loads between supports)
13.8.5
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
w y = 0.6 - 0.4 . k y
= 0.6 - 0.4 ×0.12
= 0.5520
Cex =
=
p2 . E . Ix
1000 . (Kx. Lx)2
p2 ×200 000.00 ×195 000 000.00
1000 ×(0.85 ×4 000.00 )2
= 33.30×103 kN
Cey =
=
p2 . E . Iy
1000 . (Ky. Ly)2
p2 ×200 000.00 ×13 600 000.00
1000 ×(0.85 ×4 000.00 )2
= 2 322.260 kN
U1x =
wx
1 -
=
Cu
Cex
1.000
13.20
1 33 297.11
= 1.000
U1y =
wy
1 -
=
Cu
Cey
0.552
13.20
1 2 322.26
= 0.5552
F13.8.2(a) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
13.2
0.85 ×1.000 ×1.60 0.6 ×1.000 ×2.84
+
+
2 693.2
381.46
76.54
= 0.0307
OK
Overall member strength
Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports
13.8.2 b
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
and does not refer to restraint conditions)
f ex =
=
p2 . E
Kx. Lx 2
rx
p2 ×200000
1 ×4000 2
151
= 2 812.961 MPa
f ey =
=
p2 . E
Ky. Ly 2
ry
p2 ×200000
1 ×4000 2
39.9
= 196.406 MPa
p2 . E . Cw
f ez =
(Kz . Lz )2
+ G. J
A . ro2
p2 ×200000 ×4120×108
(1 ×4000 )2
=
+ 77000 ×560000
8550 ×156.18 2
= 450.477 MPa
fe = min(fex, fey, fez) = 196.4 MPa
l=
=
fy
fe
350
196.41
= 1.335
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 1.3349 2 ×1.34
1×103
-1.34
= 1 138.894 kN
U1x = 1.000
U1y = 1.000
Ky. Ly.
fy
ly =
2
ry
p .E
=
1.00 ×4 000.0
350
× 2
39.90
p ×200000
= 1.335
b = 0.6 + 0.4 . l y
= 0.6 + 0.4 ×1.135
= 1.054
=> b = 0.85
13.8.2
F13.8.2(b) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
13.2
0.85 ×1.000 ×4.31 0.85 ×1.000 ×2.84
+
+
1 138.9
381.46
76.54
= 0.0527
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×1 211 000.000
1×106
= 423.850 kNm
My =
=
f y. Zex
1×106
350.000 ×1 071 000.000
1×106
= 374.850 kNm
OK
13.8.2 c
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
w 2 = 1.75 + 1.05 . k x + 0.3 . k x2
= 1.75 + 1.05 ×0.371 + 0.3 ×0.371 2
= 2.181
w 2. p.
Mcr =
Ke. L
E . Iy. G . J +
p. E 2 . .
Iy Cw
Ke. L
1000
2
p×200
2.181 ×p
× 200 ×13600000 ×77 ×560000 +
×13600000 ×412000000000
1.000 ×4000
1.000 ×4000
=
1000
= 865.914 kNm
Sclass=1 and section is doubly symmetric =>
Mcr > 0.67Mp =>
0.28 . Mp
Mr = 1.15 . f . Mp. 1 Mcr
= 1.15 ×0.9 ×423.850 × 1 -
0.28 ×423.850
865.968
= 378.564 kNm
Cr based on weak-axis bending
ly=
=
Ky. Ly.
fy
2
ry
p .E
0.85 ×4 000.00
350
× 2
39.90
p ×200000
= 1.135
1
Cr =
f . A . f y. 1 + l
2. n
-n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 1.135
1×103
2 ×1.34
-1.34
= 1 399.548 kN
F13.8.2(c) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
13.2
0.85 ×1.000 ×4.31 0.85 ×1.000 ×2.84
+
+
1 400.0
378.56
76.54
= 0.0506
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Additional check for class 1 and 2 sections
F13.8.2 =
=
Checked by
Date
OK
13.8.2
Mux Muy
+
Mrx Mry
4.3
2.84
+
378.56 76.54
= 0.0485
OK
Shear
13.4
Shear buckling coefficient
kv=5.34
13.3.4.1.1 a)
hwtw =
hw
tw
=
312
9.1
= 34.286
Shear buckling factor
Sfac =
=
13.4.1.1 a)
kv
fy
5.34
350
= 0.1235
Inelastic critical plate-buckling stress in shear
f cri =
=
13.4.1.1 b)
290 . f y. k v
hw
tw
290 × 350 ×5.34
312
9.1
= 365.670 MPa
Elastic critical plate-buckling stress in shear
13.4.1.1 d)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
f cre =
=
Checked by
Date
180000 . k v
hw 2
tw
180000 ×5.34
312 2
9.1
= 817.688 MPa
hw/tw <= 440*sqrt(kv/fy) =>
13.4.1.1 a)
f sx = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Factored shear resistance X-axis:
13.4.1.1
Avx = H . tw
= 363.4 ×9.1
= 3 306.940 mm2
Vrx =
=
0.9 . Avx. f sx
1000
0.9 ×3 306.9 ×231
1000
= 687.505 kN
Vux = 3.1kN
Factored shear resistance Y-axis:
f sy = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Avy =
=
2 . B . tf . 5
6
2 ×86.6 ×15.7 ×5
6
= 2 266.033 mm2
OK
13.4.1.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vry =
=
Checked by
Date
0.90 . Avy. f sy
1000
0.90 ×2266 ×231
1000
= 471.101 kN
Vuy = 0.8kN
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Member Design for Combined Stresses
4.00
3.50
3.00
2.50
2.00
M max = 15.49kNm @ 4.00m
1.50
1.00
.500
MOMENTS: X-X
2.00
4.00
6.00
8.00
10.0
12.0
14.0
4.00
3.50
3.00
2.50
2.00
1.50
1.00
.500
M max = -1.240kNm @ 0.00m
.200
.400
-2.00
-3.00
-4.00
-5.00
-6.00
-7.00
4.00
3.50
3.00
2.50
2.00
P max = -7.060kN @ 0.00m
1.50
1.00
.500
AXIAL FORCE
-1.00
Lx Eff = 3.400 m
W1x = 0.90
Ly Eff = 3.400 m
W1y = 1.00
Le Eff = 4.000 m
W2 = 1.14
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC2
MOMENTS: Y-Y
-1.20
-1.00
-.800
-.600
-.400
-.200
Ver W5.0.04 - 13 Sep 2022
Combine Ver W5.0.04
Element 192-193
Evaluate current section
Section name DOORBE
Section 356x171x67
I-sections (Web vert)
SABS 0162 - 1993 13.9 :
a) Cross-sectional strength (Crit. pos.= 4.000 m)
Tu
Mux
Muy
7.06
15.5
.860
-- + --- + --- = ---- + ---- + ---- = 0.05
Tr
Mrx
Mry
2693
381
76.5
b) Lateral torsional buckling strength
Mux
Muy
T/A
15.5
1.24
.8257
--- + --- - ------- = ---- + ---- - ----- = 0.06
Mrx
Mry
Mrx/Zpx
323
76.5
267.1
Slenderness Ratio: Lx/rx =
Ly/ry =
23
85
------------------------------------------------------------
OK
OK
OK
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 297-11
Section: 356x171x67 I
Lx eff = 8296
Ly eff = 8296
Lz eff = 9760
Le eff = 9760
Section Slenderness:
Table 4
Flanges
Bratio =
=
B.
fy
Tf
86.60
× 350
15.70
= 103.193
Bratio < 145 => Class 1 Flange
Web
Wratio_1 =
=
1100 .
fy
1100
350
1 -
×1 -
0.39 . Cu
f . Cy
0.39 ×-14.3
0.9 ×2 992.5
= 58.919
Hw/Tw(34.3) < Wratio_1 => Class 1 Web
=> Section class = 1
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×9 760.00
151
= 54.940
X_ratio < 300 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×9 760.00
40
= 207.400
Y_ratio < 300 => OK
10.4.2.2
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial tension and bending
13.9
Cross-sectional strength
13.9 a
Cr =
=
f . A . fy
1000
0.9 ×8 550.00 ×350.00
1000
= 2 693.250 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×1 211 000.00 ×350.00
1×106
= 381.465 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×243 000.00 ×350.00
1×106
= 76.545 kNm
F13.9(a) =
=
Tu Mux Muy
+
+
Tr Mrx Mry
3.8
22.03
42.22
+
+
2 693.2 381.46 76.54
= 0.6108
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×1 211 000.000
1×106
= 423.850 kNm
My =
=
f y. Zex
1×106
350.000 ×1 071 000.000
1×106
= 374.850 kNm
13.9 c
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
w 2 = 1.75 + 1.05 . k x + 0.3 . k x2
= 1.75 + 1.05 ×0.992 + 0.3 ×0.992 2
= 3.087
w 2. p.
Mcr =
Ke. L
E . Iy. G . J +
p. E 2 . .
Iy Cw
Ke. L
1000
2
p×200
2.500 ×p
× 200 ×13600000 ×77 ×560000 +
×13600000 ×412000000000
1.000 ×9760
1.000 ×9760
=
1000
= 301.641 kNm
Sclass=1 and section is doubly symmetric =>
Mcr > 0.67Mp =>
0.28 . Mp
Mr = 1.15 . f . Mp. 1 Mcr
= 1.15 ×0.9 ×423.850 × 1 -
0.28 ×423.850
301.641
= 266.088 kNm
F13.8.9(b) =
=
Mux Muy Tu. Zpx
+
Mrx Mry
Mrx. A
22 030 000.00
42 220 000.00
14 340.0 ×1 211 000.00
+
266 088 311.94 76 545 000.00
266 088 311.94 ×8 550.00
= 0.6267
Shear
13.4
Shear buckling coefficient
kv=5.34
13.3.4.1.1 a)
hwtw =
hw
tw
=
312
9.1
= 34.286
Shear buckling factor
13.4.1.1 a)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Sfac =
=
Checked by
Date
kv
fy
5.34
350
= 0.1235
Inelastic critical plate-buckling stress in shear
f cri =
=
13.4.1.1 b)
290 . f y. k v
hw
tw
290 × 350 ×5.34
312
9.1
= 365.670 MPa
Elastic critical plate-buckling stress in shear
f cre =
=
13.4.1.1 d)
180000 . k v
hw 2
tw
180000 ×5.34
312 2
9.1
= 817.688 MPa
hw/tw <= 440*sqrt(kv/fy) =>
f sx = 0.66 . f y
13.4.1.1 a)
= 0.66 ×350
= 231.000 MPa
Factored shear resistance X-axis:
Avx = H . tw
= 363.4 ×9.1
= 3 306.940 mm2
13.4.1.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vrx =
=
Checked by
Date
0.9 . Avx. f sx
1000
0.9 ×3 306.9 ×231
1000
= 687.505 kN
Vux = 4.5kN
Factored shear resistance Y-axis:
OK
13.4.1.1
f sy = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Avy =
=
2 . B . tf . 5
6
2 ×86.6 ×15.7 ×5
6
= 2 266.033 mm2
Vry =
=
0.90 . Avy. f sy
1000
0.90 ×2266 ×231
1000
= 471.101 kN
Vuy = 25.0kN
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
MOMENTS: X-X
Checked by
M max = -20.71kNm @ 0.00m
-20.0
Date
Combine Ver W5.0.04
Element 297-11
Section name GABLEC
Evaluate curre
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-10.0
10.0
20.0
MOMENTS: Y-Y
M max = -42.88kNm @ 0.00m
-40.0
-30.0
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
-10.0
1.00
-20.0
Lx Eff = 8.296 m
W1x = 0.40
Ly Eff = 8.296 m
W1y = 1.00
Le Eff = 9.760 m
W2 = 2.50
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC11
10.0
20.0
Section 356x171x67
-10.0
-15.0
-20.0
-25.0
-30.0
I-sections (Web vert)
9.00
8.00
7.00
6.00
5.00
4.00
P max = -32.90kN @ 9.76m
3.00
2.00
-5.00
1.00
AXIAL FORCE
SANS 10162-1:2011 13.9 :
a) Cross-sectional strength (Crit. pos.= 0.000 m
Tu
Mux
Muy
25.0
20.7
42.9
-- + --- + --- = ---- + ---- + ---OK= 0.62
Tr
Mrx
Mry
2693
381
76.5
b) Lateral torsional buckling strength
Mux
Muy
T/A
20.7
42.9
3.848
--- + --- - ------- = ---- + ---- OK----- = 0
Mrx
Mry
Mrx/Zpx
266
76.5
219.7
13.4:Shear
Vux<Vrx
Vuy<Vry
4.2 < 687.5
25.3 < 471.1
Slenderness Ratio: Lx/rx = 55
Ly/ry = 208
OK
OK
OK
OK
------------------------------------------------
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Member Design for Combined Stresses
MOMENTS: X-X
M max = 114.7kNm @ 0.00m
10.0
8.00
6.00
4.00
20.0
40.0
60.0
80.0
100
2.00
-60.0
-40.0
-20.0
Lx Eff = 8.632 m
W1x = 0.51
Ly Eff = 8.632 m
W1y = 1.00
Le Eff = 10.155 m
W2 = 1.99
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC2
MOMENTS: Y-Y
M max = -5.100kNm @ 0.00m
-5.00
-4.00
-3.00
10.0
8.00
6.00
4.00
2.00
-2.00
-1.00
Ver W5.0.04 - 13 Sep 2022
Combine Ver W5.0.04
Element 32-38
Evaluate current section
Section name RAFTER
1.00
2.00
AXIAL FORCE
P max = 71.81kN @ 2.03m
70.0
Section 356x171x67
I-sections (Web vert)
SANS 10162-1:2011 13.8.2 :
a) Cross-sectional strength (Crit. pos.= 0.000 m)
Cu
0.85Mux
0.60Muy
70.6
97.5
3.82
-- + ------- + ------- = ---- + ---- + ---- = 0.33
Cr
Mrx
Mry
2693
381
76.5
b) Overall member strength
Cu
0.85U1xMux
ßU1yMuy
71.6
97.5
4.34
-- + ---------- + ---------- = ---- + ---- + ---- = 0.63
Cr
Mrx
Mry
228
381
76.5
c) Lateral torsional buckling strength
Cu
0.85U1xMux
ßU1yMuy
71.6
97.5
4.34
-- + ---------- + ---------- = ---- + ---- + ---- = 0.76
Cr
Mrx
Mry
311
206
76.5
60.0
50.0
40.0
OK
OK
OK
30.0
10.0
8.00
6.00
4.00
10.0
2.00
20.0
d) Additional check for class 1 I sections
Mux
Muy
115
5.10
--- + --- = ---- + ---- = 0.62
Mrx
Mry
206
76.5
OK
13.4:Shear
Vux<Vrx
Vuy<Vry
37.3 < 687.5
3.5 < 471.1
Slenderness Ratio: Lx/rx = 57
Ly/ry = 216
------------------------------------------------------------
OK
OK
OK
FAIL
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 190-191
Section: 356x171x67 I
Lx eff = 3400
Ly eff = 3400
Lz eff = 4000
Le eff = 4000
Section Slenderness:
Table 4
Flanges
Bratio =
=
B.
fy
Tf
86.60
× 350
15.70
= 103.193
Bratio < 145 => Class 1 Flange
Web
Wratio_1 =
=
1100 .
fy
1100
350
1 -
×1 -
0.39 . Cu
f . Cy
0.39 ×19.6
0.9 ×2 992.5
= 58.631
Hw/Tw(34.3) < Wratio_1 => Class 1 Web
=> Section class = 1
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×4 000.00
151
= 22.517
X_ratio < 200 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×4 000.00
40
= 85.000
Y_ratio < 200 => OK
10.4.2.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial compression and bending
13.8
Member strength and stability - class 1 and 2 I-shaped sections
13.8.2
Cross-sectional strength
13.8.2 a
Cr =
=
f . A . fy
1000
0.9 ×8 550.00 ×350.00
1000
= 2 693.250 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×1 211 000.00 ×350.00
1×106
= 381.465 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×243 000.00 ×350.00
1×106
= 76.545 kNm
End moment factors:
Mux_min
kx =
Mux_max
=
6.03
-6.56
= -0.9192
ky =
Muy_min
Muy_max
=
0.70
-1.46
= -0.4795
w x = 0.6 - 0.4 . k x
= 0.6 - 0.4 ×-0.92
= 0.9680
13.8.5
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
w y = 0.6 - 0.4 . k y
= 0.6 - 0.4 ×-0.48
= 0.7920
w y = 1 (loads between supports)
Cex =
=
p2 . E . Ix
1000 . (Kx. Lx)2
p2 ×200 000.00 ×195 000 000.00
1000 ×(0.85 ×4 000.00 )2
= 33.30×103 kN
Cey =
=
p2 . E . Iy
1000 . (Ky. Ly)2
p2 ×200 000.00 ×13 600 000.00
1000 ×(0.85 ×4 000.00 )2
= 2 322.260 kN
U1x =
wx
1 -
=
Cu
Cex
0.968
19.62
1 33 297.11
= 0.9686
U1y =
wy
1 -
=
Cu
Cey
1.000
19.62
1 2 322.26
= 1.009
F13.8.2(a) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
19.6
0.85 ×1.000 ×6.03 0.6 ×1.009 ×1.46
+
+
2 693.2
381.46
76.54
= 0.0323
OK
Overall member strength
13.8.2 b
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports
and does not refer to restraint conditions)
f ex =
=
p2 . E
Kx. Lx 2
rx
p2 ×200000
1 ×4000 2
151
= 2 812.961 MPa
f ey =
=
p2 . E
Ky. Ly 2
ry
p2 ×200000
1 ×4000 2
39.9
= 196.406 MPa
p2 . E . Cw
+ G. J
(Kz . Lz )2
f ez =
A . ro2
p2 ×200000 ×4120×108
(1 ×4000 )2
=
+ 77000 ×560000
8550 ×156.18 2
= 450.477 MPa
fe = min(fex, fey, fez) = 196.4 MPa
l=
=
fy
fe
350
196.41
= 1.335
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 1.3349 2 ×1.34
1×103
-1.34
= 1 138.894 kN
U1x = 1.000
U1y = 1.000
Ky. Ly.
fy
ly =
2
ry
p .E
=
1.00 ×4 000.0
350
× 2
39.90
p ×200000
= 1.335
b = 0.6 + 0.4 . l y
= 0.6 + 0.4 ×1.135
= 1.054
=> b = 0.85
13.8.2
F13.8.2(b) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
19.6
0.85 ×1.000 ×6.56 0.85 ×1.000 ×1.46
+
+
1 138.9
381.46
76.54
= 0.0480
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×1 211 000.000
1×106
= 423.850 kNm
My =
=
f y. Zex
1×106
350.000 ×1 071 000.000
1×106
= 374.850 kNm
OK
13.8.2 c
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
w 2 = 1.75 + 1.05 . k x + 0.3 . k x2
= 1.75 + 1.05 ×-0.919 + 0.3 ×-0.919 2
= 1.038
w 2. p.
Mcr =
Ke. L
E . Iy. G . J +
p. E 2 . .
Iy Cw
Ke. L
1000
2
p×200
1.038 ×p
× 200 ×13600000 ×77 ×560000 +
×13600000 ×412000000000
1.000 ×4000
1.000 ×4000
=
1000
= 412.113 kNm
Sclass=1 and section is doubly symmetric =>
Mcr > 0.67Mp =>
0.28 . Mp
Mr = 1.15 . f . Mp. 1 Mcr
= 1.15 ×0.9 ×423.850 × 1 -
0.28 ×423.850
412.238
= 312.393 kNm
Cr based on weak-axis bending
ly=
=
Ky. Ly.
fy
2
ry
p .E
0.85 ×4 000.00
350
× 2
39.90
p ×200000
= 1.135
1
Cr =
f . A . f y. 1 + l
2. n
-n
1×103
1
=
0.9 ×8 550.0 ×350 × 1 + 1.135
1×103
2 ×1.34
-1.34
= 1 399.548 kN
F13.8.2(c) =
=
Cu 0.85 . U1x. Mux b . U1y. Muy
+
+
Cr
Mrx
Mry
19.6
0.85 ×1.000 ×6.56 0.85 ×1.000 ×1.46
+
+
1 400.0
312.39
76.54
= 0.0481
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Additional check for class 1 and 2 sections
F13.8.2 =
=
Checked by
Date
OK
13.8.2
Mux Muy
+
Mrx Mry
6.6
1.46
+
312.39 76.54
= 0.0402
OK
Shear
13.4
Shear buckling coefficient
kv=5.34
13.3.4.1.1 a)
hwtw =
hw
tw
=
312
9.1
= 34.286
Shear buckling factor
Sfac =
=
13.4.1.1 a)
kv
fy
5.34
350
= 0.1235
Inelastic critical plate-buckling stress in shear
f cri =
=
13.4.1.1 b)
290 . f y. k v
hw
tw
290 × 350 ×5.34
312
9.1
= 365.670 MPa
Elastic critical plate-buckling stress in shear
13.4.1.1 d)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
f cre =
=
Checked by
Date
180000 . k v
hw 2
tw
180000 ×5.34
312 2
9.1
= 817.688 MPa
hw/tw <= 440*sqrt(kv/fy) =>
13.4.1.1 a)
f sx = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Factored shear resistance X-axis:
13.4.1.1
Avx = H . tw
= 363.4 ×9.1
= 3 306.940 mm2
Vrx =
=
0.9 . Avx. f sx
1000
0.9 ×3 306.9 ×231
1000
= 687.505 kN
Vux = 0.1kN
Factored shear resistance Y-axis:
f sy = 0.66 . f y
= 0.66 ×350
= 231.000 MPa
Avy =
=
2 . B . tf . 5
6
2 ×86.6 ×15.7 ×5
6
= 2 266.033 mm2
OK
13.4.1.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vry =
=
Checked by
Date
0.90 . Avy. f sy
1000
0.90 ×2266 ×231
1000
= 471.101 kN
Vuy = 1.8kN
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Member Design for Combined Stresses
MOMENTS: X-X
M max = 154.6kNm @ 8.00m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-50.0
50.0
100
Lx Eff = 6.800 m
W1x = 1.00
Ly Eff = 6.800 m
W1y = 1.00
Le Eff = 8.000 m
W2 = 2.37
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC2
150
MOMENTS: Y-Y
M max = -4.889kNm @ 6.99m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-4.00
-2.00
Ver W5.0.04 - 13 Sep 2022
Combine Ver W5.0.04
Element 42-38
Evaluate current section
Section name COLUMN
Section 356x171x67
I-sections (Web vert)
SANS 10162-1:2011 13.8.2 :
a) Cross-sectional strength (Crit. pos.= 8.000 m)
Cu
0.85Mux
0.60Muy
77.6
133
.092
-- + ------- + ------- = ---- + ---- + ---- = 0.38
Cr
Mrx
Mry
2693
381
76.5
OK
2.00
4.00
AXIAL FORCE
P max = 125.4kN @ 0.00m
120
c) Lateral torsional buckling strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.75
Cr
Mrx
Mry
483
295
76.5
100
80.0
60.0
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
40.0
20.0
b) Overall member strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.74
Cr
Mrx
Mry
359
381
76.5
d) Additional check for class 1 I sections
Mux
Muy
155
4.15
--- + --- = ---- + ---- = 0.58
Mrx
Mry
295
76.5
OK
OK
OK
13.4:Shear
Vux<Vrx
Vuy<Vry
34.5 < 687.5
4.8 < 471.1
Slenderness Ratio: Lx/rx = 45
Ly/ry = 170
------------------------------------------------------------
OK
OK
OK
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Member Design for Combined Stresses
MOMENTS: X-X
M max = 154.6kNm @ 8.00m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-50.0
50.0
100
Lx Eff = 6.800 m
W1x = 1.00
Ly Eff = 6.800 m
W1y = 1.00
Le Eff = 8.000 m
W2 = 2.37
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Flange class: 1
Web class
: 1
Critical Load Case : LC2
150
MOMENTS: Y-Y
M max = -4.889kNm @ 6.99m
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
-4.00
-2.00
Ver W5.0.04 - 13 Sep 2022
Combine Ver W5.0.04
Element 42-38
Evaluate current section
Section name COLUMN
Section 356x171x67
I-sections (Web vert)
SANS 10162-1:2011 13.8.2 :
a) Cross-sectional strength (Crit. pos.= 8.000 m)
Cu
0.85Mux
0.60Muy
77.6
133
.092
-- + ------- + ------- = ---- + ---- + ---- = 0.38
Cr
Mrx
Mry
2693
381
76.5
OK
2.00
4.00
AXIAL FORCE
P max = 125.4kN @ 0.00m
120
c) Lateral torsional buckling strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.75
Cr
Mrx
Mry
483
295
76.5
100
80.0
60.0
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
40.0
20.0
b) Overall member strength
Cu
0.85U1xMux
ßU1yMuy
125
131
3.53
-- + ---------- + ---------- = ---- + ---- + ---- = 0.74
Cr
Mrx
Mry
359
381
76.5
d) Additional check for class 1 I sections
Mux
Muy
155
4.15
--- + --- = ---- + ---- = 0.58
Mrx
Mry
295
76.5
OK
OK
OK
13.4:Shear
Vux<Vrx
Vuy<Vry
34.5 < 687.5
4.8 < 471.1
Slenderness Ratio: Lx/rx = 45
Ly/ry = 170
------------------------------------------------------------
OK
OK
OK
OK
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
7.00
6.00
5.00
4.00
M max = .8284kNm @ 3.75m
3.00
2.00
1.00
MOMENTS: X-X
Checked by
Combine Ver W5.0.04
Element 47-68
Section name EAVES
Date
Evaluate current section
.600
Lx Eff = 6.375 m
W1x = 1.00
Ly Eff = 6.375 m
W1y = 1.00
Le Eff = 7.500 m
W2 = 1.00
Fy = 350 MPa
Fu = 480 MPa
Tension area factor (Ane/Ag) = 1.00
Section class: 1
.800
Critical Load Case : LC11
.200
.400
.00500
MOMENTS: Y-Y
M max = 0.000kNm @ 0.00m
Round hollow sections
SANS 10162-1:2011 13.8.3 :
a) Cross-sectional strength (Crit. pos.= 3.750 m)
Cu
Mux
Muy
12.6
.888
0.00
-- + --- + --- = ---- + ---- + ---- = 0.06
Cr
Mrx
Mry
536
23.0
23.0
.00400
.00300
.00200
AXIAL FORCE
7.00
6.00
5.00
4.00
3.00
2.00
1.00
.00100
P max = 12.64kN @ 0.00m
12.0
b) Overall member strength
Cu
U1xMux
U1yMuy
12.6
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.15
Cr
Mrx
Mry
111
23.0
23.0
c) Lateral torsional buckling strength
Cu
U1xMux
U1yMuy
12.6
.828
0.00
-- + ------ + ------ = ---- + ---- + ---- = 0.12
Cr
Mrx
Mry
147
23.0
23.0
10.0
8.00
6.00
OK
OK
OK
13.4:Shear
7.00
6.00
5.00
4.00
3.00
2.00
1.00
4.00
2.00
Section 139.7x4.0+
Vux<Vrx
Vuy<Vry
0.4 < 196.4
0.0 < 196.4
Slenderness Ratio: Lx/rx = 133
Ly/ry = 133
------------------------------------------------------------
OK
OK
OK
OK
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Design Code: SANS 10162-1:2011
Element : 47-68
Section: 139.7x4.0+ O
Lx eff = 6375
Ly eff = 6375
Lz eff = 7500
Le eff = 7500
Section Slenderness:
DT_ratio =
=
Table 4
D.
fy
T
139.00
×350
4.00
= 12.16×103
DT_ratio < 13000 => Class 1 section
Maximum slenderness ratio
X_ratio =
=
Kx. Lx
rx
0.85 ×7 500.00
48
= 132.812
X_ratio < 200 => OK
Y_ratio =
=
Ky. Ly
ry
0.85 ×7 500.00
48
= 132.812
Y_ratio < 200 => OK
10.4.2.1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Axial compression and bending
13.8
Member strength and stability — All classes of sections except class 1 and class 2 I-shaped sections
13.8.3
Cross-sectional strength
13.8.3 a
Cr =
=
f . A . fy
1000
0.9 ×1 700.00 ×350.00
1000
= 535.500 kN
Mrx =
=
f . Zpx. f y
1×106
0.9 ×72 900.00 ×350.00
1×106
= 22.964 kNm
Mry =
=
f . Zpy. f y
1×106
0.9 ×72 900.00 ×350.00
1×106
= 22.964 kNm
End moment factors:
k x = -1 (no X moments)
k y = -1 (no Y moments)
w x = 0.6 - 0.4 . k x
= 0.6 - 0.4 ×-1.00
= 1.0000
w y = 0.6 - 0.4 . k y
= 0.6 - 0.4 ×-1.00
= 1.0000
13.8.5
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Cex =
=
Checked by
Date
p2 . E . Ix
1000 . (Kx. Lx)2
p2 ×200 000.00 ×3 870 000.00
1000 ×(0.85 ×7 500.00 )2
= 187.966 kN
Cey =
=
p2 . E . Iy
1000 . (Ky. Ly)2
p2 ×200 000.00 ×3 870 000.00
1000 ×(0.85 ×7 500.00 )2
= 187.966 kN
U1x =
wx
1-
=
Cu
Cex
1.000
12.29
1187.97
= 1.070
U1y =
wy
1-
=
Cu
Cey
1.000
12.29
1187.97
= 1.070
F13.8.3(a) =
=
Cu U1x. Mux U1y. Muy
+
+
Cr
Mrx
Mry
12.3
1.070 ×0.83 1.070 ×0.00
+
+
535.5
22.96
22.96
= 0.0616
OK
Overall member strength
Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports
and does not refer to restraint conditions)
13.8.3 b
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
f ex =
=
p2 . E
Kx. Lx 2
rx
p2 ×200000
1 ×7500 2
47.8
= 80.179 MPa
f ey =
=
p2 . E
Ky. Ly 2
ry
p2 ×200000
1 ×7500 2
47.8
= 80.179 MPa
p2 . E . Cw
+ G. J
(Kz . Lz )2
f ez =
A . ro2
p2 ×200000 ×0
=
(1 ×7500 )2
+ 77000 ×7740×103
1700 ×67.599 2
= 76.72×103 MPa
fe = min(fex, fey, fez) = 80.2 MPa
l=
=
fy
fe
350
80.179
= 2.089
Checked by
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×1 700.0 ×350 × 1 + 2.0893 2 ×1.34
1×103
-1.34
= 111.335 kN
U1x = 1.000
U1y = 1.000
Cu U1x. Mux U1y. Muy
F13.8.3(b) =
+
+
Cr
Mrx
Mry
=
12.3
1.000 ×0.83 1.000 ×0.00
+
+
111.3
22.96
22.96
= 0.1467
Lateral torsional buckling strength
Mp =
=
f y. Zplx
1×106
350.000 ×72 900.000
1×106
= 25.515 kNm
My =
=
f y. Zex
1×106
350.000 ×55 700.000
1×106
= 19.495 kNm
Mr = f . Mp
= 0.9 ×25.515
= 22.964 kNm
Cr based on weak-axis bending
ly=
=
Ky. Ly.
fy
2.
ry
p E
0.85 ×7 500.00
350
× 2
47.80
p ×200000
= 1.776
OK
13.8.3 c
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
1
Cr =
f . A . f y. 1 + l
2. n n
1×103
1
=
0.9 ×1 700.0 ×350 × 1 + 1.776 2 ×1.34
1×103
-1.34
= 146.853 kN
F13.8.3(c) =
=
Cu U1x. Mux U1y. Muy
+
+
Cr
Mrx
Mry
12.3
1.000 ×0.83 1.000 ×0.00
+
+
146.9
22.96
22.96
= 0.1199
OK
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
S13
Apex Connection - Ver W5.0.00 - 01 Apr 2022
Title :
Code of Practice : SANS 10162-1:2011
Created : 2022/11/21 21:39:12
Notes and Assumptions
1
2
3
All bolt holes are assumed to be normal clearance holes.
All bolts are assumed to have threads in their shear planes.
It is assumed that the connection is deep enough for the flanges
to resist the compressive and tensile forces in them.
Summary
Summary of Forces and Capacities for Design to SANS 10162-1:2011
Check
Member
Type
LC
Applied
Capacity
Units
% of Cap.
?
1
Weld
Flange
LC2
119.7
143.1 kN
83.6
O.K.
2
Weld
Web
LC10
9.6
630.9 kN
1.5
O.K.
3
Bolts
Shear
LC10
1
42.2 kN
2.3
O.K.
3b
Bolts
Slip
N/A
N/A
N/A kN
N/A
N/A
4
Bolts
Combined
LC2
0.4
1.4 kN
28.7
O.K.
5
Bolts & Plate
Tension & Bending
LC2
50.1
151.6 kN
33.1
O.K.
6
Plate
Bearing
LC10
1
144.6 kN
0.7
O.K.
Input
General Settings
Bolt Tension Analysis
Plastic
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Bolt Type
Bearing
Bolt Grade
4.8
Member Ultimate Strength
480
Member Yield Strength
350
Weld Ultimate Strength
480
Connection Type
Extended End Plate : Top
Beam
356x171x67
Beam Angle
10
Haunch Depth
(mm) 400
Haunch Length
(mm) 2030.6
I1
Ultimate Limit State Loads in Beam
Shear
(kN)
Load Case
Axial
(kN)
Moment
(kNm)
Divide Factor
(to obtain
SLS Loads)
LC1
5.55
26.51
-74.55
1
LC2
6.82
27.88
-79.32
1
LC3
4.97
13.06
-39.36
1
LC4
6.59
22.50
-65.24
1
LC5
-0.87
-19.25
61.42
1
LC6
0.02
-44.30
44.36
1
LC7
1.11
-43.05
40.45
1
LC8
0.40
-17.88
56.65
1
LC9
1.29
-42.93
39.59
1
LC10
2.39
-41.68
35.68
1
LC11
-1.58
-21.95
69.27
1
LC12
-0.69
-46.99
52.21
1
LC13
0.40
-45.75
48.31
1
End Plate
Bolts
Rows of Bolts
Bolt Offsets
Width
(mm) 173.3
Extent Above Beam Flange
(mm) 51.2
Extent Below Haunch
(mm) N/A
Thickness
(mm) 16
Diameter
(mm) 20
Above Top Flange
1
Below Top Flange
2
Above Haunch
2
Below Haunch
N/A
Row Spacing
(mm) 75
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Weld Sizes
Checked by
Web
(mm) 41.2
Flange
(mm) 23.1
Above Haunch
(mm) 23.1
Beam Flanges
(mm) 3
Beam Web
(mm) 3
Date
Check 1 : Capacity of the Beam Flange Welds
The worst load is encountered for Load Case : LC2
when Fmax =119.65 kN
The Capacity of the weld is the lesser of :
0.67 . f w. Aw. Xu
Vr =
1000
=
13.13.2.2
13.13.2.2b
0.67 ×0.67 ×672.246 ×480
1000
= 144.850 kN
0.67 . fw. Am. f u
Vr =
1000
=
13.13.2.2a
0.67 ×0.67 ×950.7 ×480
1000
= 204.849 kN
Beam Flange Weld is safe
Check 2 : Capacity of the Beam Web Welds
The worst load is encountered for Load Case : LC10
when Fmax =9.591 kN
The Capacity of the weld is the lesser of :
0.67 . fw. Aw. Xu
Vr =
1000
=
13.13.2.2
13.13.2.2
0.67 ×0.67 ×2 927.878 ×480
1000
= 630.876 kN
13.13.2.2
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vr =
=
Checked by
Date
0.67 . fw. Am. f u
1000
0.67 ×0.67 ×4 140.644 ×350
1000
= 650.557 kN
Beam Web Weld is safe
Check 3 : Shear Capacity of the Bolts
The worst load is encountered for Load Case : LC10
when Vmax =0.959 kN
13.12.1.2
The resistance of any bolt is :
Vr =
=
0.60 . f b. m . 0.7 . Ab. f u
1000
0.60 ×0.80 ×1 ×0.7 ×314.159 ×400
1000
= 42.223 kN
Bolt shear is safe
Check 4 : Shear and Tension Capacity of the Bolts
The worst load is encountered for Load Case : LC2
The factor must be less than or equal to 1.4 :
Vu Tu
Factor =
+
Vr Tr
=
13.12.1.4
0.188
25.057
+
35.362 63.146
= 0.4021
Bolt shear and tension is safe
Check 5 : Bolt tension and End Plate Bending
The worst load is encountered for Load Case : LC2
Fmax = 50.114 kN
Eurocode 1993-1
8 : 6.2.4
8 : Table 6.2
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
The resistance is the smaller of the 3 possible failure modes :
Mode 1 : Complete yielding of the End Plate
R1 =
=
4 . Mpl .
1000
m
4 ×2.83576
×1000
38.8
= 292.346 kN
Mode 2 : Bolt Failure with yielding of the End Plate
R2 =
=
2 . Mpl . 1000 + n . 2 . Bt
m+n
2 ×2.83576 ×1000 + 40.9 ×2 ×78.4
38.8 + 40.9
= 151.627 kN
Mode 3 : Bolt Failure only
R3 = 2 . Bt
= 2 ×78.4
= 156.800 kN
Therefore R = R2 = 151.626
Bolt tension and end plate bending is safe
Check 6 : Bearing of the End Plate
The Bearing Capacity of the Plate at any Bolt is the lesser of :
The worst load is encountered for Load Case : LC10
when Bmax =0.959 kN
Br =
=
fbr. t . a . f u
1000
0.67 ×16 ×28.1 ×480
1000
= 144.591 kN
End plate bearing is safe
13.10.1c
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces (Forces are given per bolt and not per row)
Bolt Forces for Load Case : LC1
0 kN
0 kN
0 kN
23.53 kN
23.53 kN
Shear force per bolt : 0.09 kN
Bolt Forces for Load Case : LC2
0 kN
0 kN
0 kN
25.06 kN
25.06 kN
Shear force per bolt : 0.19 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC3
0 kN
0 kN
0 kN
12.49 kN
12.49 kN
Shear force per bolt : 0.26 kN
Bolt Forces for Load Case : LC4
0 kN
0 kN
0 kN
20.64 kN
20.64 kN
Shear force per bolt : 0.26 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC5
15.8 kN
15.8 kN
15.8 kN
2.39 kN
2.39 kN
Shear force per bolt : 0.25 kN
Bolt Forces for Load Case : LC6
13.9 kN
13.9 kN
13.9 kN
5.45 kN
5.45 kN
Shear force per bolt : 0.77 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC7
12.88 kN
12.88 kN
12.88 kN
5.28 kN
5.28 kN
Shear force per bolt : 0.86 kN
Bolt Forces for Load Case : LC8
14.57 kN
14.57 kN
14.57 kN
2.19 kN
2.19 kN
Shear force per bolt : 0.35 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC9
12.66 kN
12.66 kN
12.66 kN
5.26 kN
5.26 kN
Shear force per bolt : 0.87 kN
Bolt Forces for Load Case : LC10
11.64 kN
11.64 kN
11.64 kN
5.08 kN
5.08 kN
Shear force per bolt : 0.96 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC11
17.85 kN
17.85 kN
17.85 kN
2.74 kN
2.74 kN
Shear force per bolt : 0.23 kN
Bolt Forces for Load Case : LC12
15.95 kN
15.95 kN
15.95 kN
5.8 kN
5.8 kN
Shear force per bolt : 0.75 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC13
14.93 kN
14.93 kN
14.93 kN
5.62 kN
5.62 kN
Shear force per bolt : 0.83 kN
3
3
400
824 x 173 x 16 End plate
2031
3
92
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Base Plate Design - SANS 10162 - 2005
Material Strength Properties
fcu :
30 MPa
Bolt Grade :
8.8
Bolt fy : 640 MPa
Bolt fu : 800 MPa
fy Baseplate : 355 MPa
fu Baseplate : 470 MPa
fy Column :
355 MPa
fu Column :
470 MPa
tu Weld :
483 MPa
Column Section
I1 356x171x67
Base Plate Design Data:
Plate Shape :
Height :
Breadth :
Thickness :
Rectangular
600 mm
600 mm
30 mm
Weld Properties
Size :10 mm Fillet Weld
Bolt Properties
Diameter :
20 mm
Anchor Length :
250 mm
Compression not allowed in bolts
Bolt End Plate Properties
End Type :
Dimension :
Thickness :
Square Plate
100 x 100 mm
30 mm
Design Loads Summary
Load
Case
P
(kN)
Torsion
(kNm)
Vx
(kN)
Mx
(kNm)
Vz
(kN)
Mz
(kNm)
LC1
LC2
LC3
LC4
LC5
LC6
LC7
LC8
LC9
LC10
LC11
LC12
LC13
98.07
100.02
61.39
85.35
-5.29
-30.45
-30.82
-3.33
-28.49
-28.87
-19.82
-44.98
-45.36
-0.03
-0.03
-0.02
-0.03
-0.02
0.01
-0.01
-0.02
0.01
-0.01
-0.01
0.01
-0.01
1.14
1.19
0.7
1.02
-3.21
-1.24
0.41
-3.16
-1.19
0.46
-3.37
-1.39
0.25
90.78
94.71
51.46
78.98
62.84
31.72
75.89
66.78
35.66
79.83
51.62
20.5
64.67
27.44
28.6
15.49
23.83
24.89
36.43
41.91
26.06
37.6
43.08
21.5
33.05
38.52
-2.18
-2.31
-1.36
-1.99
6.86
2.68
-0.95
6.72
2.54
-1.09
7.14
2.96
-0.66
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Date
Factor of Safety Summary
Load
Case
Steps
LC1
LC2
LC3
LC4
LC5
LC6
LC7
LC8
LC9
LC10
LC11
LC12
LC13
FOS
Concrete
112
112
107
111
111
106
112
111
106
112
111
109
111
FOS
Bolt
T
5.84
5.55
10.90
6.84
7.62
17.10
7.82
7.25
15.50
7.36
8.94
24.90
9.59
1.68
1.60
3.08
1.94
1.77
3.06
1.49
1.69
2.83
1.43
1.98
3.62
1.66
FOS
Bolt
C
FOS
Bolt
Shear
-
8.01
7.69
14.20
9.22
8.77
6.04
5.26
8.39
5.86
5.11
10.10
6.66
5.72
FOS
Base
Plate
6.13
5.84
11.20
7.07
6.46
11.10
5.44
6.16
10.30
5.21
7.23
13.20
6.06
FOS
Weld
2.87
2.74
5.14
3.30
3.56
6.34
2.88
3.37
5.77
2.75
4.15
8.51
3.28
FOS
Bolt
T+V
1.95
1.85
3.54
2.24
2.06
2.84
1.63
1.97
2.67
1.56
2.32
3.28
1.80
FOS
Bolt
Pull-Out
2.54
2.42
4.66
2.93
2.68
4.62
2.26
2.56
4.28
2.16
3.00
5.48
2.51
Bolt Resistance Forces
Bolt Net Cross Section
25.2.2.1
p
An = . (d - 0.938 . P )2
4
p
= ×(20 - 0.938 ×2.5 )2
4
= 244.808 mm2
Tension Resistance
Tr =
=
25.2.2.1
fb. An. f u
1000
.67 ×244.81 ×800
1000
= 131.218 kN
25.2.2.2
SANS 10100-1
4.11.6.2
Tension Resistance Concrete
Trc =
0.28 . f cu. p. d . lb + 0.6 . f cu. (AnchorArea - BoltArea)
1000
=
0.28 × 30 ×p×20 ×250 + 0.6 ×30 ×(10000 - 314.16 )
1000
= 198.435 kN
6.2.4.4.3 (b)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Shear Resistance
Vr =
=
Checked by
Date
25.2.3.3
0.6 . fb. 0.7 . An. f u
1000
0.6 ×.67 ×0.7 ×244.81 ×800
1000
= 55.112 kN
Compression Resistance
Cr =
=
13.3.1
0.9 . An. f u
1000
0.9 ×244.81 ×516.13
1000
= 113.718 kN
Find Effective Compression Area
Calculate Zpl
b . tp2
Zpl =
4
=
1 ×30 2
4
= 225.000 mm3
Moment Resistance
Mr =
=
f . Zpl . f y
1000
.9 ×225 ×355
1000
= 71.888 Nm
Moment Ultimate equation
Mu = (c*b)*(c/2)*fcu
Through substitution cMax can be calculated
Effective Distance from Edge of Section
13.5 (a)
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
fy
Zpl . 2 . 0.9 .
1.15
cMax =
f cu
b.
1.5
225 ×2 ×0.9 ×
=
355
1.15
30
1×
1.5
= 79.064 mm
Calculation Sheet for Load Case : LC8
Factored loads
P:
-3.33 kN
Mz :
6.72 kNm
Vx :
-3.16 kN
Mx :
66.78 kNm
Vz :
26.06 kN
Torsion :
-0.02 kNm
Checked by
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Find Equilibruim
The actual number of Grid Point used for calculation is 1078
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Moment balancing
Sum Of Moments around X-axis = 0.0 kNm
Sum Of Moments around Y-axis = 0.0 kNm
Axial Force balancing
Sum Of Forces in Y-direction = 0.0 kN
The Shear Resistance in the Bolts Resists the Following Forces:
Forces in X-direction
Moments around Y-axis
Forces in Z-direction
Calculating Factors of Safety in Concrete
FOS =
StrainMax
Strain
=
.0035
.000483
= 7.246
Calculating Factors of Safety in Critical Bolt
Tension in Bolts
Critical Bolt Tension
FOS =
=
Tr
Tension
131.22
77.638
= 1.690
Critical Bolt Pull-Out
FOS =
=
Trc
Tension
198.44
77.638
= 2.556
Shear in Bolts
Critical Bolt Shear
Date
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
FOS =
=
Checked by
Date
Vr
Shear
55.111
6.5722
= 8.385
Shear and Tension combined in Bolts
The factor should be less than 1.4 for bolts in shear and tension
The bolt number 4 has the critical shear and tension combination
The tension in the bolt is: 77.64 kN
The shear in the bolt is: 6.56 kN
13.12.1.4
Tension and Shear Resistance combination
combinedfactor =
=
Shear Tension
+
Vr
Tr
6.5572 77.638
+
55.111 131.22
= 0.7106
0.711 <= 1.4
OK
Converted to Factor of Safety relevant to 1
FOS =
=
1.4
f actor
1.4
.71066
= 1.970
Bolt BasePlate interaction
FOS =
Resistance
Force
=
478.4
77.638
= 6.162
25.2.4
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Welds
Since unit values are used for the length and size of the weld, the capacity
of this layout is given in kN/mm
The capacity, Vr is the lesser of Vr1 and Vr2:
Resistance of parent material
Vr1 =
=
13.13.2.2 (a)
0.67 . fw. 0.707 . Size. f u
1000
0.67 ×.67 ×0.707 ×10 ×470
1000
= 1.492 kN/mm
The angle of the axis of the weld conservatively taken as 0°
Resistance of weld material
Vr2 =
=
0.67 . fw. 0.707 . Size. xu. 1 + 0.5 . sin(q)1.5
1000
0.67 ×.67 ×0.707 ×10 ×483 × 1 + 0.5 ×sin(0 )1.5
1000
= 1.533 kN/mm
Capacity of 10mm weld is 1.492kN/mm
FOS =
Resistance
Force
=
1.4916
.44301
= 3.367
13.13.2.2 (b)
30
250
30
I1 356x171x67
100
16
500
500
I1 356x171x67
Hole Size = 18
769 x 185 x 16 End Plate
100
Haunch cut from Beam section
Height : 400 mm
Width : 2000 mm
All welds are Both Sides unless otherwise indicated
All bolts are of Grade 8.8
All bolt diameters are 20 mm
30
400
369
100
30
93
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
S12
Beam - Column Connection - Ver W5.0.00 - 01 Apr 2022
Title :
Code of Practice : SANS 10162-1:2011
Created : 2022/11/21 17:02:58
Notes and Assumptions
1
All references are formated "EC3 Part : Section" eg: 8 : 3.6.2(3)a.
for Eurocode 1993-1-8 Section 3.6.2(3)a.
All bolt holes are assumed to be normal clearance holes.
All bolts are assumed to have threads in their shear planes.
It is assumed that the connection is deep enough for the flanges
to resist the compressive and tensile forces in them.
It is assumed that compressive forces in flanges and stiffeners
are conveyed through welds and not through bearing.
Axial force in the column is not considered in the design.
2
3
4
5
6
Summary
Summary of Forces and Capacities for Design to SANS 10162-1:2011
Check
Member
Type
LC
Applied
Capacity
Units
% of Cap.
?
1
Weld
Flange
LC13
250.4
384.1 kN
65.2
O.K.
2
Weld
Web
LC12
51.5
1261.8 kN
4.1
O.K.
3
Column Web
Tension Yielding
LC13
250.4
701.3 kN
35.7
O.K.
4
Column Web
Compression Crippling
LC2
236.7
447.6 kN
52.9
O.K.
5
Column Web
Compression Buckling
LC2
236.7
812.3 kN
29.1
O.K.
6
Column Web
Shear
LC13
250.4
491.1 kN
51
O.K.
7
Bolts & Flange
Tension & Bending
LC13
138.5
234.2 kN
59.1
O.K.
8
Column Flange
Bearing
LC12
6.4
284 kN
2.3
O.K.
9
Bolts & End Plate Tension & Bending
LC13
138.5
217.9 kN
63.6
O.K.
10
End Plate
Bearing
LC12
6.4
289.4 kN
2.2
O.K.
11
Bolts
Shear
LC12
6.4
84.4 kN
7.6
O.K.
12
Bolts
Shear & Tension
LC13
0.6
1.4 kN
43.7
O.K.
13
Bolts
Slip
N/A
N/A
N/A kN
N/A
N/A
Input
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
General Settings
Bolt Tension Analysis
Plastic
Bolt Type
Bearing
Bolt Grade
8.8
Member Ultimate Strength
450
Member Yield Strength
300
Weld Ultimate Strength
480
Connection Type
Flush End Plate
Column
356x171x67
I1
356x171x67
I1
Beam
Column Extent Above
Beam Angle
(mm) Continuous
10
Haunch Depth
(mm) 400
Haunch Length
(mm) 2000
Date
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Checked by
Ultimate Limit State Loads in Beam
Load Case
Shear
(kN)
Axial
(kN)
Moment
(kNm)
SLS Factor
(Divide to
get Loads)
LC1
33.36
56.68
149.85
1
LC2
34.40
57.93
154.50
1
LC3
16.26
28.18
72.92
1
LC4
27.57
46.53
123.73
1
LC5
-19.68
-36.94
-127.93
1
LC6
-57.48
-43.30
-116.95
1
LC7
-41.30
-40.15
-152.03
1
LC8
-18.63
-35.69
-123.28
1
LC9
-56.44
-42.05
-112.31
1
LC10
-40.25
-38.90
-147.38
1
LC11
-23.31
-43.47
-144.22
1
LC12
-61.11
-49.82
-133.25
1
LC13
-44.93
-46.68
-168.32
1
Width
Extent Above Beam Flange
End Plate
Extent Below Haunch
Thickness
Stiffeners
Width
Top Stiffener Thickness
Column Stiffeners
Bottom Stiffener Thickness
Shear Stiffener Thickness
Shear Stiffener Orientation
Layout
Web Plates
Thickness
Top Backing Plate
Thickness
Bottom Backing Plate Thickness
(mm) 185.2
(mm) N/A
(mm) N/A
(mm) 16
N/A
(mm) None
(mm) None
(mm) None
(mm) None
None
None
(mm) 5
(mm) None
(mm) None
Bolts
(mm) 20
N/A
2
2
N/A
(mm) 100
Rows of Bolts
Diameter
Above Top Flange
Below Top Flange
Above Haunch
Below Haunch
Row Spacing
Date
Job Number
Sheet
Job Title
Client
Your details here
Calcs by
Bolt Offsets
Welds
Web
Flange
Above Haunch
Beam & Haunch Flanges
Beam Web
Top Stiffener
Bottom Stiffener
Shear Stiffener
Checked by
Date
(mm) 42
(mm) 30
(mm) 30
6
(mm) 6
(mm) N/A
(mm) N/A
(mm) N/A
Check 1 : Capacity of the Beam Flange Welds
The worst load is encountered for Load Case : LC13
when Fmax =250.4 kN
The Capacity of the weld is the lesser of :
0.67 . fw. Am. f u
Vr =
1000
=
13.13.2.2
13.13.2.2a
0.67 ×0.67 ×1 901.4 ×450
1000
= 384.092 kN
0.67 . f w. Aw. xu. 1.5
Vr =
1000
=
13.13.2.2b
0.67 ×0.67 ×1 300.376 ×480 ×1.5
1000
= 420.292 kN
Beam Flange Weld is safe
Check 2 : Capacity of the Beam Web Welds
The worst load is encountered for Load Case : LC12
when Fmax =51.53 kN
The Capacity of the weld is the lesser of :
0.67 . fw. Am. f u
Vr =
1000
=
13.13.2.2
13.13.2.2a
0.67 ×0.67 ×8 281.289 ×450
1000
= 1 672.862 kN
13.13.2.2b
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Vr =
=
Checked by
Date
0.67 . fw. Aw. xu
1000
0.67 ×0.67 ×5 855.755 ×480
1000
= 1 261.751 kN
Beam Web Weld is safe
Check 3 : Capacity of the Column web in tension
Opposite Top flange of the beam :
The worst load is encountered for Load Case : LC2
when Tmax =173.65 kN
The Capacity of the web is :
0.9 . tw. leff . f y
Tr =
1000
=
0.9 ×9.1 ×285.422 ×300
1000
= 701.282 kN
Opposite Bottom flange of the beam :
The worst load is encountered for Load Case : LC13
when Tmax =250.4 kN
The Capacity of the web is :
0.9 . tw. leff . f y
Tr =
1000
=
0.9 ×9.1 ×285.422 ×300
1000
= 701.282 kN
Column web is safe in tension
Check 4 : Crippling Capacity of the Column web in Compression
Opposite Top flange of the beam :
The worst load is encountered for Load Case : LC13
when Bmax =196.627 kN
The Capacity of the web is :
21.3a
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Br =
=
Checked by
Date
fbi . tw. leff . f y
1000
0.8 ×9.1 ×204.942 ×300
1000
= 447.593 kN
Opposite Bottom flange of the beam :
The worst load is encountered for Load Case : LC2
when Bmax =236.674 kN
The Capacity of the web is :
Br =
=
fbi . tw. leff . f y
21.3a
1000
0.8 ×9.1 ×204.942 ×300
1000
= 447.593 kN
Column web is safe in compression for crippling
Check 5 : Buckling Capacity of the Column web in Compression
Opposite Top flange of the beam :
The worst load is encountered for Load Case : LC13
when Bmax =196.627 kN
The Capacity of the web is :
Br =
=
fbi . 640000 . twc. leff
21.3
hwc 2 .
1000
twc
0.8 ×640000 ×9.1 ×204.942
312 2
×1000
9.1
= 812.299 kN
Opposite Bottom flange of the beam :
The worst load is encountered for Load Case : LC2
when Bmax =236.674 kN
The Capacity of the web is :
21.3
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Br =
=
Checked by
Date
fbi . 640000 . twc. leff
hwc 2 .
1000
twc
0.8 ×640000 ×9.1 ×204.942
312 2
×1000
9.1
= 812.299 kN
Column web is safe in compression for buckling
Check 6 : Shear Capacity of the Column Web
The worst load is encountered for Load Case : LC13
when Vmax =250.4 kN
0.55 . f . f y. tw. h
Vr =
1000
=
13.4.1.2
0.55 ×0.9 ×300 ×9.1 ×363.4
1000
= 491.081 kN
Column web shear is safe
Check 7 : Bolt tension and Column Flange Bending
The worst load is encountered for Load Case : LC13
Fmax = 138.483 kN
Eurocode 1993-1
8 : 6.2.4
8 : Table 6.2
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
The resistance is the smaller of the 3 possible failure modes :
Mode 1 : Complete yielding of the flange
R1 =
4 . Mpl + 2 . Mbp.
1000
m
=
4 ×2.37444 + 2 ×0
×1000
33.84
= 280.667 kN
Mode 2 : Bolt Failure with yielding of the flange
R2 =
=
2 . Mpl . 1000 + n . 2 . Bt
m+n
2 ×2.37444 ×1000 + 40.05 ×2 ×156.8
33.84 + 40.05
= 234.248 kN
Mode 3 : Bolt Failure only
R3 = 2 . Bt
= 2 ×156.8
= 313.600 kN
Therefore R = R2 = 234.248
Bolt tension and Column Flange bending is safe
Check 8 : Bearing on the Column Flange
The Bearing Capacity of the flange at any Bolt is the lesser of :
The worst load is encountered for Load Case : LC12
when Bmax =6.441 kN
3 . fbr. t . d . f u
Br =
1000
=
13.10c
3 ×0.67 ×15.7 ×20 ×450
1000
= 284.013 kN
Column flange bearing is safe
Check 9 : Bolt tension and End Plate Bending
The worst load is encountered for Load Case : LC13
Fmax = 138.483 kN
Eurocode 1993-1
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
8 : 6.2.4
8 : Table 6.2
The resistance is the smaller of the 3 possible failure modes :
Mode 1 : Complete yielding of the End Plate
R1 =
=
4 . Mpl .
1000
m
4 ×2.78953
×1000
43.2
= 258.290 kN
Mode 2 : Bolt Failure with yielding of the End Plate
R2 =
=
2 . Mpl . 1000 + n . 2 . Bt
m+n
2 ×2.78953 ×1000 + 40.05 ×2 ×156.8
43.2 + 40.05
= 217.883 kN
Mode 3 : Bolt Failure only
R3 = 2 . Bt
= 2 ×156.8
= 313.600 kN
Therefore R = R2 = 217.883
Bolt tension and End Plate bending is safe
Check 10 : Bearing on the End Plate
The Bearing Capacity of the Plate at any Bolt is :
The worst load is encountered for Load Case : LC12
when Bmax =6.441 kN
3 . fbr . t . d . f u
Br =
1000
=
3 ×0.67 ×16 ×20 ×450
1000
= 289.440 kN
End plate bearing is safe
Check 11 : Shear Capacity of the Bolts
The worst load is encountered for Load Case : LC12
when Vmax =6.441 kN
13.10c
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
13.12.1.2
The resistance of any bolt is :
Vr =
=
0.60 . fb. m . 0.7 . Ab. f u
1000
0.60 ×0.8 ×1 ×0.7 ×314.159 ×800
1000
= 84.446 kN
Bolt shear is safe
Check 12 : Shear and Tension Capacity of the Bolts
The worst load is encountered for Load Case : LC13
The factor must be less than or equal to 1.4 :
Vu Tu
Factor =
+
Vr Tr
=
4.518
69.242
+
70.724 126.292
= 0.6122
Bolt shear and tension is safe
13.12.1.4
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces
Bolt Forces for Load Case : LC1
47.96 kN
47.96 kN
0 kN
0 kN
Shear force per bolt : 2.88 kN
Bolt Forces for Load Case : LC2
49.51 kN
49.51 kN
0 kN
0 kN
Shear force per bolt : 2.98 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC3
23.26 kN
23.26 kN
0 kN
0 kN
Shear force per bolt : 1.39 kN
Bolt Forces for Load Case : LC4
39.63 kN
39.63 kN
0 kN
0 kN
Shear force per bolt : 2.38 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC5
4.97 kN
4.97 kN
52.49 kN
52.49 kN
Shear force per bolt : 1.62 kN
Bolt Forces for Load Case : LC6
6.58 kN
6.58 kN
50.02 kN
50.02 kN
Shear force per bolt : 6.14 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC7
5.84 kN
5.84 kN
62.31 kN
62.31 kN
Shear force per bolt : 4.21 kN
Bolt Forces for Load Case : LC8
4.8 kN
4.8 kN
50.59 kN
50.59 kN
Shear force per bolt : 1.52 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC9
6.4 kN
6.4 kN
48.12 kN
48.12 kN
Shear force per bolt : 6.04 kN
Bolt Forces for Load Case : LC10
5.66 kN
5.66 kN
60.4 kN
60.4 kN
Shear force per bolt : 4.11 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC11
5.86 kN
5.86 kN
59.43 kN
59.43 kN
Shear force per bolt : 1.93 kN
Bolt Forces for Load Case : LC12
7.46 kN
7.46 kN
56.95 kN
56.95 kN
Shear force per bolt : 6.44 kN
Job Number
Sheet
Job Title
Your details here
Client
Calcs by
Checked by
Date
Bolt Forces for Load Case : LC13
6.72 kN
6.72 kN
69.24 kN
69.24 kN
Shear force per bolt : 4.52 kN
PRELIMINARY DESIGN 1
INNER PORTALS
PLAN VIEW
END GABLE
SIDE VIEW
PRELIMINARY DESIGN 2
INNER PORTALS
PLAN VIEW
END GABLE
SIDE VIEW
- All steelwork is of grade: S355JR
- Bearing bolts: Grade 8.8 (only M20
in connections)
- Welds: Grade E70XX
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
- All concrete to have a cube crushing
strength of fcu=30 Mpa
- All Rebar: Grade Y
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
slope =10 degrees duo-pitch
- Doors: 4x4 Roller shutter doors
7500
7500
7500
7500
7500
7500
5000
5000
5000
15000
5000
500
600
500
2750
600
3000
300
BASE PLATE
CONCRETE PLINTH
150
1000
GROUND SLAB
400
SOIL BED
FOUNDATION
500
600
1875
2750
- All steelwork is of grade: S355JR
- Bearing bolts: Grade 8.8 (only M20
in connections)
- Welds: Grade E70XX
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
- All concrete to have a cube crushing
strength of fcu=30 Mpa
- All Rebar: Grade Y
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
slope =10 degrees duo-pitch
4 @ 2000
- Doors: 4x4 Roller shutter doors
3 @ 4000
1500
1000
1500
15 @ 1000
scale 1:2
6 @ 7500
- All steelwork is of grade: S355JR
- Bearing bolts: Grade 8.8 (only M20
in connections)
- Welds: Grade E70XX
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
- All concrete to have a cube crushing
strength of fcu=30 Mpa
- All Rebar: Grade Y
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
slope =10 degrees duo-pitch
- Doors: 4x4 Roller shutter doors
INNER PORTAL RAFTER (WITH NO CRANE)
ROLLER SHUTTER DOORS(CLOSED)
INNER PORTAL RAFTER (SHOWING CRANE)
ROOF BRACING (X BRACING)
DOOR COLUMN
GABLE END PORTAL
DOOR BEAM
WALL BRACING
INNER PORTAL RAFTER
- All steelwork is of grade: S355JR
B
- Bearing bolts: Grade 8.8 (only M20
RAFTER: 356x171x67
in connections)
(I-SECTION)
RAFTER-COLUMN
BOLTS: M20
- Welds: Grade E70XX
A
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
B
- All concrete to have a cube crushing
strength of fcu=30 Mpa
EAVES HAUNCH
CLADDING
(THICKNESS=0.58mm)
- All Rebar: Grade Y
A
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
SIDE RAILS
125x75x20x2
(CHANNEL)
COLUMN: 356x171x67
slope =10 degrees duo-pitch
(I-SECTION)
- Doors: 4x4 Roller shutter doors
RAFTER-COLUMN
CONNECTION
186
728
200
PURLINS: 125x75x20x2
(CHANNEL)
400
200
RAFTER:356x171x67
SECTION A-A
SCALE 1:2
(I-SECTION)
SECTION B-B
SCALE 1:2
- All steelwork is of grade: S355JR
- Bearing bolts: Grade 8.8 (only M20
in connections)
CLADDING
(THICKNESS=0.58mm)
RAFTER-RAFTER
CONNECTION BY M20
BOLTS
- Welds: Grade E70XX
RAFTER: 356x171x67
(I-SECTION)
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
- All concrete to have a cube crushing
strength of fcu=30 Mpa
- All Rebar: Grade Y
C
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
APEX HAUNCH-CRAWL BRAM
CONNECTION BY M20 BOLTS
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
slope =10 degrees duo-pitch
- Doors: 4x4 Roller shutter doors
C
APEX-CRAWL BEAM CONNECTION
CRANE TROLLEY
CRAWL BEAM: 305x165x40
(I-SECTION)
CRANE TROLLEY
APEX-CRAWL BEAM
CONNECTION
SCALE 1:2
TROLLEY DIRECTION
OF MOTION
SECTION C-C
- All steelwork is of grade: S355JR
- Bearing bolts: Grade 8.8 (only M20
in connections)
CLADDING
(THICKNESS=0.58mm)
RAFTER-RAFTER
CONNECTION BY M20
BOLTS
- Welds: Grade E70XX
RAFTER: 356x171x67
(I-SECTION)
- The steel sheeting to be used to
cover the warehouse is: Supa-Clad
IBR890 by Global Roofing Solutions
(Pty) Ltd. (GRS)
- All concrete to have a cube crushing
strength of fcu=30 Mpa
- All Rebar: Grade Y
C
- Concrete cover to all rebar: 50mm
for all foundations and 40 mm for
ground slab
- Warehouse dimensions: 60m long,
20m wide, 8m eaves height, Roof
slope =10 degrees duo-pitch
- Doors: 4x4 Roller shutter doors
C
APEX-CRAWL BEAM CONNECTION
GABLE END COLUMN-CRAWL BEAM:
CONNECTION BY M20 BOLTS
CRANE TROLLEY
CRAWL BEAM: 305x165x40
(I-SECTION)
GABLE END COLUMN:
356x171x67 (I-SECTION)
GABLE END COLUMN:
356x171x67 (I-SECTION)
TROLLEY DIRECTION
OF MOTION
END STOP BEAM
305x165x46
(I-SECTION)
SECTION C-C
END GABLE COLUMN-CRAWL BEAM
CONNECTION
