## Hazards and Safety Precautions

The beam material used to perform the experiment is steel with dimensions of 40mm (width) by 1mm (height) by 600mm (length).

Elastic energy that is stored in the beam, E, is calculated as follows: ; Where m = mass of the load acting on the beam (kg), g = gravitational acceleration (9.8m^{2}/s) and h = the beam’s vertical deflection at the point of loading (m).

In this case, m = 0.293kg, g = 9.8m^{2}/s and h = 26cm – 23.5cm = 0.26m – 0.235m = 0.025m

Therefore

Safety is very important when conducting any experiment. There are a few hazards associated with performing this experiment. The first hazard is the unexpected rupturing of the beam when it gets loaded. When the beam ruptures, it can produce flying objects that may cause bodily injury or property damage. This risk has been eliminated by using a beam made of ductile metallic material that does not break easily, using a lightweight mass (only 0.293kg < 1kg) to perform the experiment, and ensuring that the vertical deflection of the beam when loaded is low (25mm < 100mm) so as to maintain low elastic energy stored in the beam and prevent the beam from reaching its elastic limit and subsequently rupturing. As calculated above, the maximum energy stored in the beam is 0.04J. This is a relatively small amount of energy that may not pose a great danger even if the mass used to perform the experiment hits the body of a person.

The second risk is falling of the mass from the beam during the experiment. If this happens, it can cause injury to people or property damage. This risk has been eliminated by using a relatively wide load so that it can be easily and firmly fixed on the beam without falling off, ensuring that both ends of the beam are firmly fixed on the supports to avoid lateral movement, and placing the load on the beam gently to prevent it from slipping. In case the load falls, the risk of bodily injury or property damage has been reduced by using a lightweight mass (<1kg) and ensuring that the vertical deflection of the beam when loaded is low (<100mm) so as to maintain low elastic energy stored in the beam.

Figure 1: Apparatus of the experiment

Calculating support reactions of the beam

Since the beam is symmetrical with a point load acting at midspan, the reaction at the two supports is equal i.e. R_{A} = R_{B} = (where P = point load = 0.293kg)

1kg = 9.80665N; 0.293kg = 2.87N

R_{A} = R_{B} =

The sketch of beam arrangement is as shown in Figure 2 below. R_{A} and R_{B} are the support reactions of the beam.

Figure 2: Beam arrangement

The sizes, distances and magnitude of the load used in the experiment were determined by measuring using the appropriate measuring instruments/devices. The sizes and distances were measured using a meter rule while the magnitude of the load was measuring using a digital weighing machine/balance. These quantities were defined based on the ease of availability, carrying and handling, available space to perform the experiment and ensuring that they do not cause bodily injury or property damage during the experiment. The beam supports in this experiment highly resemble the idealized simply supported arrangement because both ends of the beam are secured on the supports using hinge-like objects that allow the beam to rotate i.e. the beam is able to resist vertical and horizontal forces but does not resist a moment.

## Calculating Support Reactions of the Beam

Standard beam section 40 x 60 x 1mm

The principal second moments of area of the beam section are calculated as follows

The second moment of area when the axis is passing through the base of the section, I_{xx}

I_{xx} = (where b = width of the cross section and h = height of the cross section)

I_{xx} =

The second moment of area when the centroid axis is perpendicular to its base, I_{yy}

I_{yy} = (where b = width of the cross section and h = height of the cross section)

I_{yy} =

The accuracy of the above calculations has been verified by comparing the results with the ones obtained from online calculators and also the theoretical values from standard beam sections.

This being a simply supported beam with a point load at the mid-span, the maximum bending moment is at the mid-point of the beam.

Calculating support reactions of the beam

Since the beam is symmetrical with a point load acting at midspan, the reaction at the two supports is equal i.e. R_{A} = R_{B} = (where P = point load = 0.293kg)

1kg = 9.80665N; 0.293kg = 2.87N

R_{A} = R_{B} =

The maximum bending moment is calculated using the following formula

M_{max} = (where P = point load acting on the beam at mid-span and L = length of the beam)

M_{max} =

Alternatively, the bending moment at mid-span can be calculated by taking moments at the mid-span to the left hand side of the beam as follows:

300mm x 1.435N = 430.5Nmm

The formula for calculating maximum tensile or compressive stress in a rectangular beam section is given as follows, σ_{max} = (where m = maximum bending moment of the beam, c = distance from the neutral axis of the beam and I = second moment of area of the beam)

In this case, M_{max} = 430.5Nmm

I = I_{xx} =

Since the beam is a symmetrical rectangular section of height 1mm, C =

Maximum tensile stress, σ_{maxT} =

Maximum compressive stress, σ_{maxC} =

As aforementioned, the beam material is steel. Some of the strength and elastic properties of the steel beam include the following:

Modulus of elasticity – 210GPa: this is the measure of stiffness of steel. Steel has a high modulus of elasticity meaning that it exhibits less deformation when subjected to loading.

Bulk modulus – 140GPa: this is the measure of how resistant steel is to compression or compressive force. It describes the elastic properties of steel when it is subjected to pressure.

Yield strength – 250MPa: this is the maximum stress that steel can withstand without undergoing plastic deformation.

Ultimate tensile strength – 400MPa: this defines the maximum stress that steel can withstand before it fails.

Elongation at fracture – 15%: this is the ratio between the changed length or elongation and the original length of a steel test specimen after breakage.

Poisson ratio – 0.3: this is the ratio of the transversal elongation to the amount of axial compression. This property shows the relationship between change in cross-section of a material and the lengthwise stretching.

## Maximum Bending Moment and Stress in the Beam

Machinability – 65%: this defines the ease with which steel can be machined in terms of shear stress, horsepower or specific energy.

Shear modulus – 81GPa: this is the rigidity of steel defined by the ration between shear stress and shear strain of steel. It shows how steel responds to shear deformation.

Hardness – 126: this is a measure of steel’s resistance to localized plastic deformation that may be induced by abrasion or mechanical indentation.

- Photographs of the beam
- Beam without load

Maximum deflection, δ_{max} = 26cm – 23.5cm = 2.5cm = 25mm

The maximum slope of the simply supported beam is the same at each support. The maximum slope can be obtained from the geometry and dimensions of the deflected shape of the beam. Through measurements of the deflected beam obtained from the experiment, the slope at the end supports can be obtained using trigonometric ratios, as follows

Tan θ =

θ = tan^{-1} 0.092 = 5.26°

Theoretical results of maximum deflection and maximum slope of the beam can also be obtained through calculations as follows

Maximum deflection, δ_{max} = (where P = point load at the midspan, L = length of the beam, E = elastic modulus and I = second area of moment of the beam)

P = 2.87N, L = 600mm, E = 200,000N/mm^{2} and I = 3.3333mm^{4}

δ_{max} =

Maximum slope, θ_{max} = (where P = point load at the midspan, L = length of the beam, E = elastic modulus and I = second area of moment of the beam)

θ_{max} =

From the calculations above, the experimental and calculated/theoretical maximum deflection of the beam is 25mm and 19.4mm respectively while the experimental and calculated/theoretical maximum slope of the beam is 5.26° and 5.6° respectively. These results are generally reasonable because the percentage difference between experimental and theoretical values obtained is small.

- Design of asymmetric section
- Two identical sections

Three identical sections

Second moment of area of the two sections joined

Area of section 1, a_{1} = 40mm x 1mm = 40mm^{2}

Area of section 2, a_{2} = 40mm x 1mm = 40mm^{2}

x_{1} = 20mm, x_{2} = 20mm

y_{1} = 40.5mm, y_{2} = 20mm

Second moment of area about x-axis, I_{xx}

I_{xx1} =

I_{xx2} =

I_{xx(total)} = I_{xx1} + I_{xx2}

= 4205.833 + 9535.833 = 13,741.67mm^{4}

Second moment of area about y-axis, I_{yy}

I_{yy1} =

I_{yy2} =

I_{yy(total)} = I_{yy1} + I_{yy2}

= 3.3333 + 5333.333 = 5,336.67mm^{4}

Second moment of area of the three sections joined

Area of section 1, a_{1} = 40mm x 1mm = 40mm^{2}

Area of section 2, a_{2} = 40mm x 1mm = 40mm^{2}

Area of section 3, a_{3} = 40mm x 1mm = 40mm^{2}

x_{1} = 20mm, x_{2} = 20mm, x_{3} = 40.5mm

y_{1} = 40.5mm, y_{2} = 20mm, y_{3} = 40.5mm

Second moment of area about x-axis, I_{xx}

I_{xx1} =

I_{xx2} =

I_{xx3} =

I_{xx(total)} = I_{xx1} + I_{xx2} + I_{xx3}

= 1869.3 + 12808.09 + 7199.29 = 21,876.68mm^{4}

Second moment of area about y-axis, I_{yy}

I_{yy1} =

I_{yy2} =

I_{yy3} =

I_{yy(total)} = I_{yy1} + I_{yy2} + I_{yy3}

= 1869.3 + 7199.29 + 12808.09 = 21,876.68mm^{4}

The centroid locations and principal centroid axes (PCAs) of the three sections combined respectively

The neutral axis is calculated as follows

Area of section 1, a_{1} = 40mm x 1mm = 40mm^{2}; area of section 2, a_{2} = 40mm x 1mm = 40mm^{2}

Area of section 3, a_{3} = 40mm x 1mm = 40mm^{2}; x_{1} = 20mm, x_{2} = 20mm, x_{3} = 40.5mm

y_{1} = 40.5mm, y_{2} = 20mm, y_{3} = 40.5mm

The tensile and compresses stresses are calculated using the following formula

σ = (where M = maximum bending moment, y = distance between neutral axis and location of the action and I = second moment of area of the section)

in this case, M = 430.5Nmm and I = 21,876.68mm^{4}

Stress at section 3’s top:

Stress at section 1’s top:

Stress at section 1’s bottom:

Stress at section 2’s bottom:

Stress at section 3’s bottom:

Calculating support reactions of the beam

Since the beam is symmetrical with a point load acting at midspan, the reaction at the two supports is equal i.e. R_{A} = R_{B} = (where P = point load = 0.293kg)

1kg = 9.80665N; 0.293kg = 2.87N

R_{A} = R_{B} =

Alternatively, support reactions can also be determined by taking moments at the supports

Taking moments at support A,

(0.3m x 2.87N) – (0.6m x R_{B}) = 0

0.6R_{B} = 0.861N; R_{B} =

Sum of forces in y-direction is equal to zero, i.e.

R_{A} + R_{B} = 2.87N

R_{A} = 2.87N – 1.435N = 1.435N

Therefore support reactions, R_{A} = R_{B} = 1.435N

Shear force at A = 1.435N ↑ (support reaction at A)

Shear force at mid-span of the beam = 1.435N – 2.87N = -1.435N ↓

Shear force at B = 1.435N ↑ (support reaction at B)

The section of the beam is rectangular thus this exercise entails calculating shear stress at different locations within the rectangular section of the beam. This being a rectangular symmetrical beam, the highest shear stress will be at the neutral axis of the beam.

Where b = width of the beam (40mm), h = height of the beam (1mm, y_{1} = any point along the height of the beam from the neutral axis

Shear stress, τ, at any given location y_{1} along the height of the beam’s cross section is calculated using the following formula: (where V = shear force at the particular location of the beam cross section, I_{c} = cross section’s centroid moment of inertia, h = height of the beam’s cross section and y_{1} = any point along the height of the beam from the neutral axis).

For a beam with a rectangular section, I_{c} =

I_{c} =

Other known parameters are: h = 1mm, V = 1.435N

Substituting these values in the formula for calculating shear stress gives

Shear stress at neutral axis, y_{1} = 0

= 0.054N/mm^{2}

Shear stress at y_{1} = 0.1mm from neutral axis

= = 0.052N/mm^{2}

Shear stress at y_{1} = 0.2mm from neutral axis

= = 0.045N/mm^{2}

Shear stress at y_{1} = 0.3mm from neutral axis

= = 0.034N/mm^{2}

Shear stress at y_{1} = 0.4mm from neutral axis

= = 0.019N/mm^{2}

Shear stress at y_{1} = 0.5mm from neutral axis (at the free surface)

At the free surface, the shear stress is supposed to be zero

= = 0N/mm^{2}

From the calculations above, the shear stress will be highest at the central axis of the beam section.

Maximum elastic moment, M_{Y} = Z_{e} * f_{y} (Where Z_{e} = elastic section modulus and f_{y} = yield stress)

Elastic section modulus, Z_{e} = (where b = width of the beam (40mm) and h = height/depth of the beam (1mm))

Z_{e} =

Yield stress, f_{y} of the beam material is assumed to be 250MPa (250N/mm^{2})

M_{Y} = Z_{e} * f_{y} = 6.667mm^{3} x 250N/mm^{2} = 1,666.67Nmm

Plastic moment, M_{P} is calculated by multiplying yield stress, f_{y} with the plastic section modulus, Z i.e. plastic moment, M_{P} = Z_{p} * f_{y}

Plastic section modulus, Z_{P} = (where b = width of the beam (40mm) and h = height/depth of the beam (1mm))

Z_{P} =

Yield stress, f_{y} of the beam material is assumed to be 250MPa (250N/mm^{2})

M_{P} = Z_{P} * f_{y} = 10mm^{3} x 250N/mm^{2} = 2500Nmm

The shape factor, SF =

From the calculations above, plastic moment, M_{P} = 2500Nmm and elastic or yield moment, M_{Y} = 1,666.67Nmm.

SF =

The beam is assumed to be made of a perfectly elasto-plastic material. The distribution of the stresses is such that the stresses at the external surfaces of the beam are equal to the yield stress of beam material (Dulinskas, et al., 2010).

When the beam is subjected to partially plastic moments, i.e. part of the beam section remains elastic while the external threads yield; the elastically stressed parts of the beam are prevented by the yielded areas from returning to their original state even after the load has been removed from the beam. This produces residual stresses. The magnitude of residual stresses is calculated by assuming that the unloading process of the beam is completely elastic. Distribution of the unloading stress is linear and residual stresses can be determined by subtracting graphically from the stresses of the beam when it is partially or fully plastic. The maximum residual stress is at the neutral axis, and is equal to the yield stress of the beam material (f_{y} = 250N/mm^{2}). The residual stress at the external surfaces of the rectangular beam section is equal to 0.5f_{y}. T

Effective length of a column depends on the support or end conditions of the beam. The support conditions can be: fixed at both ends, pinned at both ends, fixed at one end and pinned at the other end, or fixed at one end and free at the other end. Assume that both ends of the column are pinned. If this is the case, effective length of the column is equal to the length of the column because the effective length factor, k is equal to 1. The length, L of the column is 600mm. Since both ends of the column are assumed to be pinned, the effective length, Le is equal to the length of the column, i.e. Le = k*L = 1 x 600mm = 600mm.

The formula for calculating Euler critical buckling load, P_{cr} of a column is given as follows:

P_{cr} =

Where E = modulus of elasticity (N/mm^{2}), I = moment of inertia (mm^{4}), k = effective length factor of the column, and L = length of column (Preetha, et al., 2019)

The above formula is applicable by making the following assumptions: the column material is isotropic and homogenous, column was initially straight, the compressive load is only acting on the column axially, the column is not affected by the initial stress, the column’s weight is negligible, the pinned ends of the column are frictionless, the column’s cross section is uniform throughout its length, the bending stress of the column is very large compared to the direct stress, the column’s length is very large compared to the column’s cross sectional dimensions, and the failure of the column is only by buckling.

Assume that the modulus of elasticity, E of the column material is 200GPa = 200,000N/mm^{2}.

Moment of inertia of the column section, I is calculated as follows

I = (where b = width of the column cross section and h = height/depth of the column cross section).

I =

Length of the column, L = 600mm

Effective length factor of the column, k = 1

P_{cr} =

= 18.3N

If buckling is prevented, the compressive load needed to generate yield in the column is calculated from the yield strength as follows

Compressive load = compressive strength x cross sectional area of the column

Compressive strength = 250 N/mm^{2}

Cross sectional area = 40mm x 1mm = 40mm^{2}

Compressive load = 250N/mm^{2} x 40mm^{2}

= 10kN

A column is considered to be long if the ratio of its effective length to its smallest horizontal dimension is > 12. In this case, length of column is 600mm and its least dimension is 1mm.

Therefore

This means that the subject column is long.

Also since the critical buckling load is greater than the compressive load, the column is considered long because it has a low load carrying capacity i.e. it is prone to buckling before it reaches the compressive load needed to generate yield.

References

Dulinskas, E., Zamblauskaitt, R. & Zabulionis, D., 2010. An analysis of elasto-plastic bar cross-section stress-strain state in a pure bending. Vilnius, Lithuania, Vilnius, Gediminas Technical University.

Preetha, V., Kalaivani, K., Navaneetha, S. & Senthil, K., 2019. Buckling Analysis of Columns. IOSR JOurnal of Engineering, 1(1), pp. 10-17.

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