Calculating Beam Loads and Material Alternatives in Structural Design
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Task 1
Part 1
We have to apply the equations of equilibrium and the equations of bending and shear stress to calculation in a simply supported bean using bending moment and shear force steel beam "A" with the mentioned loads. Suppose that the beam's span is L while the point load lies at L/2, the middle of the span.
Figure 1. Shear force and Bending Moment Diagram
First, using the equation for equilibrium, we are able to calculate the reaction forces at the supports:
ΣFy = 0
RA + RB - 40 = 0
RA + RB = 40
The reaction forces are equivalent because the beam is simply supported, so we can reduce the equation as follows:
RA = RB = 20 kN(Radosavovic et al. 2020)
The equation of equilibrium is able to be employed to calculate the shear force (V) at any point in the beam:
ΣFy = 0
V - 10x - 20 = 0
V = 10x + 20
Where x is the distance from the left support.
At the left support (x = 0), the shear force is:
V = 10(0) + 20 = 20 kN
At the midpoint of the span (x = L/2), the shear force is:
V = 10(L/2) + 20 = 5L + 20 kN
At the right support (x = L), the shear force is:
V = 10(L) + 20 = 10L + 20 kN (Megson, 2019)
Figure 2. Shear force Diagram
The equation of bending stress can be applied for the calculation of the bending moment (M) at any point in the beam:
M = -EI(d^2y/dx^2) (Oganov et al. 2019)
where x is the distance from the left support, y is the deflection of the beam, I is the moment of inertia, and E is the modulus of elasticity.
The moment of inertia is givenas follows if the beam is in a rectangular cross-section:
I = (b*d^3)/12
where b is the beam's breadth and d is the beam's height.
Assuming that the beam is 300 mm in height and 150 mm in breadth. The moment of inertia is then:
I = (150300^3)/12 = 6.7510^6 mm^4
The modulus of elasticity for steel is typically 200 GPa, or 200,000 N/mm^2.
At the left support (x = 0), the bending moment is:
M = -EI(d^2y/dx^2)
M = -200000(6.75*10^6)(d^2y/dx^2) (Zhang et al. 2020)
Then the beam is simply supported, the bending moment at the left support is zero:
M = 0
At the midpoint of the span (x = L/2), the bending moment is:
M = -EI(d^2y/dx^2)
M = -200000(6.75*10^6)(d^2y/dx^2)
Assuming a positive moment at the midpoint, the bending moment is:
M = 40*(L/4) - (10*(L/2)^2)/2
M = 10L - 2.5L^2
At the right support (x = L), the bending moment is:
M = -EI(d^2y/dx^2)
M = -200000(6.75*10^6)(d^2y/dx^2)
The bending moment at the right support is 0 because the beam is simply supported:
M = 0
At this point, the steel beam "A" is supported simply, the bending moment and shear force diagrams can be shown along the grid line A.
Figure 3. Bending Moment Diagram
We have to use an equation of deflection for a beam in order to calculate the deflection in the simply supported steel beam "A" under the given loads and building structure.
For a simply supported beam and a point load at the equally distributed and middle load, of the equation of deflection is:
δ = (5wL^4)/(384EI) + (P*L^3)/(48EI) (Tang et al. 2019)
where is the deflection, w is the load applied equally all over the length of the beam per unit length, L is the beam's span, E is refer here as the elastic modulus, I is the moment of inertia, and P is the midpoint point load.
We need to determine the values of w, L, E, I, and P to be able to calculate the deflection. Assume that the beam has a 6 metre span.
The modulus of elasticity for steel is typically 200 GPa, or 200,000 N/mm^2.
Assuming a rectangular cross-section for the beam, the moment of inertia is:
I = (b*d^3)/12
where b is the width of the beam and d is the height of the beam.
Let's assume the beam has a width of 150 mm and a height of 300 mm. Then, the moment of inertia is:
I = (150300^3)/12 = 6.7510^6 mm^4
Part 2
The weight of the building materials above the beam can be utilised for calculating the evenly distributed load per unit length (Torres et al 2019) . According to the details that have been provided, the walls are constructed of 10 mm block (inner face), 100 mm cavity with 75 mm of insulation, 100 mm of reconstituted stone facing up to 4 metres high, 1500 mm broad piers up to the height of the roof, and profiled steel covering between 4 and 8 metres. The building's roof peak is 9.5 metres high.
We can calculate the load per unit of length for every section by dividing the weight of the building material into sections (Mallick, & El-Korchi, 2022) . The total of the loads for every segment can then be used to determine the overall evenly distributed load per unit length.
Section 1: inner face of a 10 mm block
The block weighs about 2.2 kN/m3 in total.
The block measures 10 mm thick, or 0.01 m.
For this section, the load is as follows:
w1 = 2.2*0.01 = 0.022 kN/m
Section 2: It has a hollow space of 100 mm and insulation of 75 mm.
The mass of the insulation is approximately 0.3 kN/m3
if the insulation has a density of 30 kg/m3.
The insulation has a thickness of 75 mm, or 0.075 m.
For this portion, the load per unit length is:
w2 = 0.3*0.075 = 0.0225 kN/m.
It is made of 100 mm of reconstituted stone that is 4 metres high.
The weight of the stone is roughly 2.2 kN/m3
assuming the reconstituted stone has a density of 22 kN/m3.
The stone measures 100 mm, or 0.1 m, thick and has a height of 4 m.
For this section, the load is as follows:
w3 = 2.2*0.1 = 0.22 kN/m
Section 3: 1500 mm-wide piers that extend to the roof height
If the piers' density is 25 kN/m3, the weight of the p.
M4
When it is about finding a design solution, by applying a material other than steel could provide some benefits as well as drawbacks. The expense of using an alternative material is one advantage. According to the specific material, it might be cheaper than steel, which may improve the total value for money of a solution for design (Zhang et al. 2020) . Also, some alternative components might be cheaper or simpler to get than steel, which might have an effect on price.
Sustainability is a possible advantage of choosing an alternate material. A major source of emissions of greenhouse gases and other environmental effects is steel production. The consequences can be minimised, and a design choice may grow less harmful to the environment by using an environmentally friendly material, such bamboo or plastic that has been recycled. Using various components has a unique set of difficulties. Durability and strength are an obstacle (Yang et al. 2019) . Other materials might not be able to offer the same amount of architectural assistance or withstand the identical amount of tear and damage because steel is known for being durable and strong. This can be a major consideration, particularly for designs which have to be strong or intended to withstand use for an extended period of time. Flexibility in design is a different problem. Steel is an extremely malleable substance that is simple to mould and form into a variety of patterns (Gregor et al. 2020) . Alternative materials could not be as flexible, which might decrease the number of possible designs. This can be especially challenging in explained or advanced designs. There can be difficulties with knowledge and accessibility. Steel is an extensively used material that is popular to many designers and builders; therefore, there is a big storage facility of information for those who work with it. Alternative supplies may not be as well known, which can make it harder to find skilled employees or builders who are able to utilise the substance effectively (Zhang et al. 2020) . When it comes to developing an architectural solution, using an element apart from steel could provide benefits as well as drawbacks. Weighing the possible benefits of cost savings and environmental sustainability against the possible drawbacks of durability and strength, design flexibility, and accessibility of information is important. Regarding the particular needs to determine if a different material is the best choice, consideration has to be given to the design solution, the readily available resources, and the understanding.
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