Calculate the axial load carrying capacity of steel and reinforced concrete columns

Course: HND in Construction and the Built Environment

Introduction

Structural design refers to the procedural investigations of how stable, strong and rigid structures can and are. The main goal in structural analysis and design is to give a structure capabilities to resist any applied loads without failure during its set life time. The main purpose of structures is to support and transmit loads. When a structure is not well designed, and in the event that the actual applied load is more than the design specifications, the structure will inevitably fail to carry out its intended purpose and will lead to serious consequences.

A well designed and engineered structures will largely reduce the possibility of such failures which are costly.

Table of contents

• Introduction
• Task 1a
• Task 1b
• Task 1c
• Task 1d
• TASK 2
• Determining deflections Task 1(a)
• Deflection in beams and structural stability
• Various ways of supports for structures
• Most effective support way for a given scenario
• TASK 3
• 3a
• 3b
• 3A.
• 3B.
• 3C
• TASK4
• Develop two design solutions
• The structural design
• Evaluating the use of an alternative material, different from concrete or steel
• Assessing the use of Building Information Modelling
• Conclusion
• References

Unit 20 Principles of Structural Design

Assignment Title - Bending Moments & Shear Forces with Beam Design Considerations

Task 1 a Draw the beam arrangement and determine the reactions at the supports. Calculate and draw the shear force diagram and the bending moment diagram running along the beam.

Drawing of the Beam arrangements and the reactions at the support.

To calculate the reaction supports for the beam, 14m with 150KN at the midpoint, let's take end points A and B for the beam and calculate moments at each point using given information. We proceed as follows;

VA + VB = 150KN ....................i (total load acting at the center of the beam)
Taking Moments about point A as positive and moments about point B as negative we proceed as follows

VA x 0 + (150x7) - (VBx14) = 0
1050 = 14VB
VB = 75KN
From equation i VA + VB = 150KN therefore VA = 150 -75 = 75 KN
VA = 75KN
Therefore loads at end point A = 75KN and end point B = 75KN
Drawing the shear force diagram (SFD) and bending moment diagram (BMD)

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Drawing the beam arrangement

To calculate the reaction supports for the beam, 8m with 65KN at the midpoint, let's take end points A and B for the beam and calculate reacting supports at each point using given information. We proceed as follows from beam arrangement diagram above;

VA + VB = 65KN..........i (total reaction from both supports)
Taking momentum at VA as positive and at VB as negative
Momentum at VB = VA X 0 + (2x 65KN) - 8x VB = 0
130 -8VB = 0
VB = 16.25KN
From equation (i) VA + VB = 65
Therefore VA = 65 - 16.25
= 48.75KN
VA = 48.75KN

Determine and draw the shear force and bending moment diagrams

We draw the beam arrangement diagram as shown below

To calculate the reaction supports for the beam, 25m with 25KN/m at the midpoint, let's take end points A and B for the beam and calculate reacting supports at each point using given information. We proceed as follows from beam arrangement diagram above;

Taking moments at A to be positive and moments at B to be negative therefore,
Moments at B, MB = MAx0 + 12.5 x 25 -(25x MB) = 0

MB = 12.5KN
Since MB + MA = 25KN
MA = 25KN - 12.5 = 12.5KN

In (a) if the UDL of 20KN/m is added, means total additional load = (20x 14) = 280KN
Therefore total supported load = 280+ 150 = 430KN.
Therefore, VA + VB = 430KN ....................i
Moments about point A as positive and moments about point B as negative

VA x 0 + (430x7) - (VBx14) = 0
3010 = 14VB
VB = 215KN
From equation i VA + VB = 430KN therefore VA = 430 -215 = 215 KN
VA = 215KN

Drawing the shear force diagram (SFD) and bending moment diagram (BMD) in (a)

VA + VB = 50 + 25, = 75KN

Moments about A as positive and about B as negative therefore,
VAx0 + 10x 50KN + 12.5x25 - 25xVB = 0

= 500KN + 312.5KN- 25VB=0
= VB = 32.5KN
Since VA + VB = 75KN
Therefore VA = 75KN - 32.5KN
VA = 42.5KN

Task 1b Discuss the statutory requirements to ensure safety in structural designs

According to the Workplace Safety and Health (WSH) act, the individuals who are develops the risks must take full responsibility in line to mitigate it. This also applies to the architects as well as the engineers who must make sure that the risks which have been created due to their own design should be well reviewed using a systematic method as well as the resulting mitigated risks which should be handed to the contractor(Hunt 2013).

While specifying the design for a structure or a building, it is essential for the designer to have an understanding of the process (how) for constructing such structures and buildings. Such should put into account, how they should be maintained, demolished or decommissioned and cleaned safely. In doing so, must study the design and know the risks poised to those working on such structures, as well as other parties affected by it like the public or the people who are using such structures and buildings in future.
Also in a lot of engineering projects, there is lack of proper communications between the client, designer and the contractor. This leads to miscommunication of important information which may affect the safety and health risks within the project. This must be corrected by ensuring there is continuous communication and the client has a responsibility of appointing a qualified Project Safety and Health Coordinator. This way the coordinator is able to follow through the project from the design stage to construction and handover to the client.

To make sure that the design is safe, it is important to have a design review process. The (SH) Safety and Health committee is introduced which is made of main stakeholders like the client, contractor, Project safety and health coordinator, the architect and the design engineers. This committee is chaired by the project safety and health coordinator. The process should be systematic in which the risks of the design are discussed, reviewed and recorded. The outcome of the process must be safe design signed by all parties and keeping records of the resulting hazards and important information on safety health.

Task 1c Explore, explain and produce valid factors of safety for live loads, dead loads and imposed loads using current codes of practice and building regulations, a valid example can be utilised to illustrate this further.

From (CivilEngineering.org 2016),Live loads also known as the imposed loads are found on roofs and the floors are mostly made of all loads that are temporal and are placed on the building or structure. Examples such as the loads of machines, furniture and people. These type of loads are constantly changing every time. There are safety factors to consider for live loads such as the UDL load factors. With the building and construction regulations, the live loads of a building, structure shall be greatly applied load because of the nature of use. The live loads as given between clause 3.2 and 3.5 is;
Uniform distribution of load
Concentrated load in KN as directed
Reducing the imposed loads in accordance with clause 3.7 may be used as directed.
Allowance of partitioning in which the positions are not shown on the structure and horizontal imposed loads as directed in the clause 3.8 and 3.6 must be used as directed.
For example the bathrooms as well as the toilets, the living and bedrooms for all kinds of building and structures should have a UDL load of 2KN/m², assembly buildings should have UDL load of 3KN/m2.

Dead loads those that are due to the weight of the building or structure itself. These kind of loads are permanent are present all the time. Such loads mostly rely on weight of material. Such loads like the weight of floor beams, walls and columns. Under the building and construction regulations, dead loads shall include the self-weight of materials which are of permanent nature throughout the life of a structure, street, street works and shall be taken to include and not limited to the following;
The structure
All building materials are affixed to the structures such as claddings, windows etc.
Soil fill, drainage system for gardening etc.
The weights of receptacles and tanks shall be considered dead loads but the contents of such tanks shall be considered imposed loads
Where there are doubts about the permanency of loads, s then they shall be treated as imposed loads and the reduction of taken under clause 3.7 cannot be taken on the vertical members as well as beams.
There are recommended unit weights of various building materials and this is one of the important safety factors to consider when building. For example the recommended unit weights to consider for plain cement concrete is 24KN/m3, steel 78.5KN/m3, cement plaster 20KN/m3.

Task 1d Evaluate how maximum bending moments determine steel beam selection using current codes of practice and approved documents in terms of economics and safety.

Method for the selection of right size of the steel beam is mostly based on the basic calculations of the mechanical design. This involves
The initial input needed is the steel beam specifications for the load
The bending moment diagram is drawn for the specified loads and get the maximum bending moment's value that the steel is required to experience.
Choosing the right size of steel beam from the given standard beam table
Find the area moment for the inertia for the steel
Calculate the beam depth for the beam
Using the formulae M/I = f/(d/2)
In this case where f is the bending stress
M is the moment at axis
I is the area moment for the inertia at axis
Once this is accomplished a comparison of the calculated values for bending stress is made against the stress of the steel so as to check safety factors for the given design (Varma .A 2018).

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TASK 2

Determining deflections Task 1(a) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided

The formula for calculating the deflection of a point from one end is found by
y = (WX(L-X))⁄24EIL[L*L+X(L-X)] in this case where;
E is the young's modulus of elasticity for material
I is the area moment for inertia
L is the length of beam between centers of support
X is the distance from one side of beam
W is the load
Therefore given
W = 150KN
L = 14m x 1000 = 14000mm
X = 7m x 1000 = 7000mm
I = 45730cm4 x 10 =457300mm
E = 210KN/mm2
Therefore y= (150*7(14-7))⁄(24*210*457300*14000)[14000*14000+7000(7000)]

Deflection = y = 1.329x10-19 mm

TASK 1(b) Explain how deflection in beams affects structural stability.

W = 65KN
L = 8m x 1000 = 8000mm
X = 2m x 1000 = 2000mm
I = 45730cm4 x 10 =457300mm
E = 210KN/mm2
Therefore y= (65*2(8-2))⁄(24*210*457300*8000)[8000*8000+2000(6000)]

Deflection = y = 5.566x10-19 mm

TASK 1(c)  Identify and describe different methods of supports for structures in general and how are these are applied to both steel and reinforced concrete structures. Analyse different support methods and their effect on deflection in fixed structures.

W = 25KN
L = 25m x 1000 = 25000mm
X = 12.5m x 1000 = 12500mm
I = 45730cm4 x 10 =457300mm
E = 210KN/mm2
Therefore y= (25*7(25-12.5))⁄(24*210*457300*25000)[25000*25000+12500(12500)]
2187.5
Deflection = y = 4.859x10-20 mm

TASK 1(d) Assess the most effective support method for a given scenario, in terms of ease and speed of construction, economics, safety and environmental factors. You may apply these to examples to be

W = 430KN
L = 14m x 1000 = 14000mm
X = 7m x 1000 = 7000mm
I = 45730cm4 x 10 =457300mm
E = 210KN/mm2
Therefore y= (430*7(14-7))⁄(24*210*457300*14000)[14000*14000+7000(7000)]

Deflection = y = 2.665x10-18 mm

TASK 1(e)

W = 75KN
L = 25m x 1000 = 25000mm
X = 12.5m x 1000 = 12500mm
I = 45730cm4 x 10 =457300mm
E = 210KN/mm2
Therefore y= (75*12.5(12.5))⁄(24*210*457300*25000)[25000*25000+12500(12500)]
Deflection = y = 2.603x10-19 mm

Deflection in beams and structural stability

In engineering, deflection means the degree of which an element which forms the structure can change shape when a load is applied. This change can be distance or an angle or both which can be either invisible or visible this relies on the intensity of the load and also the shape of the element and material.

The deflection affects the structural stability due to the possible structural failure and the given building codes mostly help to determine the maximum allowable deflection on the structure to make sure it stays safe for the users and also its general stability and integrity is retained. For beams, it is usually given as a fraction of the length. For example the deflection of the beam should not be larger than 1/360th of its length (Varma2018). For example if the beam has length of 5m the deflection should not be larger than 13.9mm and it's measured at mid-point of the beam.
Various ways of supports for structures

Support is a main aspect in a lot of structures and it mostly helps the structures to resist loads. The support in a structure helps in transferring the load to the ground and helps in providing stability to the structure. The supports can be grouped mainly into the external supports and the internal supports. The external methods of support are mainly provided externally and do not disturb the structure elements. Such external supports include;

Fixed supports, hinged supports, roller supports etc. the fixed supports are able to give restrain against rotations and also they are able to resist any form of force and moment. Such supports have no room for deflections or the deflections are invisible.
The internal methods of support provide the support for the internal elements of the structure and support internal hinge and are able to withstand movements in any direction and allows rotation only. For the structures the axial elements have internal hinges and for the beams they have midpoint hinges.

Most effective support way for a given scenario
The most effecient support method for the structures is the internal methods compared to the external method. This method is effective because this method provides the ability to withstand movements in any direction and allows just for rotation. This way, using the hinges in the internal hinge support in structures gives the structure more stability and flexibility. Supports such as the internal roller support in a structure supports the center of the structure. This can be used in the tower cranes, harbor cranes and are able to shift horizontally from place to place meaning they can be constructed easily and means this method is safer for users. In terms of economic viability this method is economical because the supports needed are fewer compared to the external supports which can be cheaper.

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TASK 3

3a calculate the axial load carrying capacity of steel and reinforced concrete columns

Solution
From the provided information the column is short because the cross section measurements are smaller only 350mm compared to its length which is 3.75m
Given Pu = 0.4fck.Ac + 0.67fy.Asc
In this case Pu = axial load carrying capacity for the column
fck = comprehensive strength for concrete = 20mm2
fy = yield strength and for steel = Fe415
The cross sectional area of column (Ag)= (350 x 350) = 122500mm2
Area of steel in column (Asc) = No of steel bars 6 nos = 2950mm2
Area of concrete in column (Ac),
Ag = Ac + Asc
Ac = Ag - Asc
Ac = 122500 - 2950
= 119550mm2
Therefore Pu = 0.4x20x119550 + 0.67x415x2950
= 1776647.5N
Pu = 1776647.5/1000
= 1776.648KN

3b
Explore and analyse the load carrying capacity, size, weight and corrosion resistance properties of different materials used for beams and columns in fixed structures.

For the steel column to carry this load safely it must be a point where the load will be stable and as well as the column. If the column is 3.75m long selecting a more central point will be good for stability and even distribution of the load and create stability from both the end points. We first determine the central point along the length of the column which is 1.9m from either end points as shown in the beam arrangement diagram below. Using either end points we can calculate the momentums created from the axial load which is 1776.648KN. Then using the SFD and the BMD we can determine where the column will experience less deflection and this will be at the central point.

3A. Describe the concepts of slenderness ratio and effective length in the design of both steel and reinforced concrete columns separately, and comparing them appropriately.

The slenderness ratio for the reinforced concrete column and steel refers to the ratio between the end fixity, the lateral dimensions and length of the column. This ratio carries out an assessment on the ability of the reinforced concrete being able to withstand the buckling pressure. This ratio can be found by dividing column length and its radius of rotation. This slenderness ratio makes the difference between the short column and long. The design for long columns is managed by column measurements and strength of the material whereas the design for shorter columns is managed by column slenderness. Slender columns are slender because their cross sectional measurements are smaller compared to their length (Hamakareem2018).

3B. Determine the exact axial load carrying capacity of the selected steel column section provided above.

VAx0 + 1776.648KN x 1.9 - 3.75VB=0
= 3375.6312-3.75VB=0
VB= 900.2KN
VA = 1776.648-900.2
= 876.48KN
Axial load at 1.9 m point = 876.48KN

3C

The structural engineering relies mostly on knowing materials and their respective compositions so as to help understand their behavior when used in construction. The most common materials used are the steel, wood/Timber and concrete.

Material Load carrying capacity Size weight Corrosion resistance
Steel Has strength of about 400 -500MPa Small takes about 20% less compared to other materials Relatively light weight Can be corroded if no measures are taken to protect the material
Concrete Has strength of about 17 MPa - 70MPa Relatively large and requires larger workforce about 10% more compared to other material Recommendable heavy weight Resistance to corrosion
Timber Has strength of about 16- 65MPa Relatively large by volume compared to steel and concrete Much lighter compared to both steel and concrete Resistance to corrosion but if not well treated can rot

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TASK 4 Develop two design solutions, one in steel and the other in reinforced concrete, for this stated beam and the support column for this given scenario, comparing their corresponding costs for the beam both in steel and in reinforced concrete and also for the column in both steel and reinforced concrete as well.

Develop two design solutions

Beam
The welded steel is designed according to the Eurocode 3 for conditions of all the possible limits as well as serviceability limit conditions. When limit conditions are considered the steel beam is checked for needed resistance of cross-section to bending moment, local buckling, shear and the ultimate moment is found by elastic procedure.

Support Column

The structural design

For concrete the normal stress is limited to 0.85fcd in which fcd is design strength concrete under compression and limit strain in concrete is taken to be 0.35%.as shown in fig a below

Steel behaves differently because it behaves elastically up to maximum strain fyd is reached and corresponding maximum elastic strain Eyd shows elasticity modulus of steel as shown in fig below.

Evaluating the use of an alternative material, different from concrete or steel

Other than steel and concrete wood is another common material used in construction. Word is organic material. Its properties both the mechanical, acoustic and electrical makes it suitable for structures building.

Assessing the use of Building Information Modelling
The engineers have always looked for ways of improving and keeping up with the modern standards, which includes meeting productivity goals, problem solving and coordination. The building information modelling (BIM) concept can be used to meet these standards. This concept readily offers the ability for integrating the intelligent objects within a model and these objects consist of all the data about a certain component and the characteristics of geometry to how they relate together with other components and this makes the entire model rich with information. A lot of companies use the information modelling to support their complex engineering methods and mostly in the design and documentation. The key contributions of BIM to the structural engineering such as conceptual design, allowing for better analysis through the simulation. Also allows one to visualize the picture and be able to identify possible errors and come up with much better ways of solving the problem (Hunt2013).

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Conclusion

In the design of any structure, the first aspect deals with establishing loading as well as the other conditions of design that must be supported by the structure. After this then the computations and analysis for the internal gross forces such as bending moments, shear, thrust, deflections etc. together with other conditions for design. The last bit is then proportioning as well as selecting material for the connections that meets the set design conditions.

The method for choosing a certain proportion that leads to the desired outcome reflects the accumulated knowledge within the field, the tests on models as well as practical experiences.

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