This document summarizes the hazards, procedures, and results of a beam bending experiment. It describes the steel beam used, which was 600mm long, 40mm wide, and 1mm thick. A 0.293kg mass was placed at the beam's center to create a bending moment of 430.5 Nmm. Calculations show the elastic energy stored in the beam was 0.04J, posing little danger if released. Deflection and slope were measured, matching theoretical values closely. Safety measures prevented injury from beam rupture or falling masses.

1. MEC2402 Stress Analysis
Answer:
Question 1: Beam Bending Experiment
Hazards Of The Experiment
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 (9.8m2/s) and h = the beam’s
vertical deflection at the point of loading (m).
In this case, m = 0.293kg, g = 9.8m2/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

2. 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
Beam Arrangement
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. RA = RB = (where P = point load = 0.293kg)
1kg = 9.80665N; 0.293kg = 2.87N
RA = RB =
The sketch of beam arrangement is as shown in Figure 2 below. R-A and RB 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.
Cross-Section
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, Ixx
Ixx = (where b = width of the cross section and h = height of the cross section)
Ixx =

3. The second moment of area when the centroid axis is perpendicular to its base, Iyy
Iyy = (where b = width of the cross section and h = height of the cross section)
Iyy =
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.
Bending Moment
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. RA = RB = (where P = point load = 0.293kg)
1kg = 9.80665N; 0.293kg = 2.87N
RA = RB =
The maximum bending moment is calculated using the following formula
Mmax = (where P = point load acting on the beam at mid-span and L = length of the beam)
Mmax =
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
Maximum Tensile And Compressive Stresses
The formula for calculating maximum tensile or compressive stress in a rectangular beam
section is given as follows, σmax = (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, Mmax = 430.5Nmm

4. I = Ixx =
Since the beam is a symmetrical rectangular section of height 1mm, C =
Maximum tensile stress, σmaxT =
Maximum compressive stress, σmaxC =
Properties Of The Beam
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.
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.

5. Photographs of the beam
Beam without load
Deflected Shape
Maximum deflection, δmax = 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, δmax = (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/mm2 and I = 3.3333mm4
δmax =
Maximum slope, θmax = (where P = point load at the midspan, L = length of the beam, E =
elastic modulus and I = second area of moment of the beam)
θmax =
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.
Question 2: Asymmetric Beam Section
Design of asymmetric section
Two identical sections
Three identical sections
Second moment of area of the two sections joined

6. Area of section 1, a1 = 40mm x 1mm = 40mm2
Area of section 2, a2 = 40mm x 1mm = 40mm2
x1 = 20mm, x2 = 20mm
y1 = 40.5mm, y2 = 20mm
Second moment of area about x-axis, Ixx
Ixx1 =
Ixx2 =
Ixx(total) = Ixx1 + Ixx2
= 4205.833 + 9535.833 = 13,741.67mm4
Second moment of area about y-axis, Iyy
Iyy1 =
Iyy2 =
Iyy(total) = Iyy1 + Iyy2
= 3.3333 + 5333.333 = 5,336.67mm4
Second moment of area of the three sections joined
Area of section 1, a1 = 40mm x 1mm = 40mm2
Area of section 2, a2 = 40mm x 1mm = 40mm2
Area of section 3, a3 = 40mm x 1mm = 40mm2
x1 = 20mm, x2 = 20mm, x3 = 40.5mm
y1 = 40.5mm, y2 = 20mm, y3 = 40.5mm
Second moment of area about x-axis, Ixx
Ixx1 =

7. Ixx2 =
Ixx3 =
Ixx(total) = Ixx1 + Ixx2 + Ixx3
= 1869.3 + 12808.09 + 7199.29 = 21,876.68mm4
Second moment of area about y-axis, Iyy
Iyy1 =
Iyy2 =
Iyy3 =
Iyy(total) = Iyy1 + Iyy2 + Iyy3
= 1869.3 + 7199.29 + 12808.09 = 21,876.68mm4
The centroid locations and principal centroid axes (PCAs) of the three sections combined
respectively
Neutral Axis
The neutral axis is calculated as follows
Area of section 1, a1 = 40mm x 1mm = 40mm2; area of section 2, a2 = 40mm x 1mm =
40mm2
Area of section 3, a3 = 40mm x 1mm = 40mm2; x1 = 20mm, x2 = 20mm, x3 = 40.5mm
y1 = 40.5mm, y2 = 20mm, y3 = 40.5mm
Tensile And Compressive Stresses
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 = 21,876.68mm4

8. 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:
Question 3: Shear In Beams
Maximum Shear Force
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. RA = RB = (where P = point load = 0.293kg)
1kg = 9.80665N; 0.293kg = 2.87N
RA = RB =
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 RB) = 0
0.6RB = 0.861N; RB =
Sum of forces in y-direction is equal to zero, i.e.
RA + RB = 2.87N
RA = 2.87N – 1.435N = 1.435N
Therefore support reactions, RA = RB = 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 ↓

9. Shear force at B = 1.435N ↑ (support reaction at B)
Shear Stress At Different Locations
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, y1 = any point along
the height of the beam from the neutral axis
Shear stress, τ, at any given location y1 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, Ic = cross section’s centroid moment of inertia, h = height of the
beam’s cross section and y1 = any point along the height of the beam from the neutral axis).
For a beam with a rectangular section, Ic =
Ic =
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, y1 = 0
= 0.054N/mm2
Shear stress at y1 = 0.1mm from neutral axis
= = 0.052N/mm2
Shear stress at y1 = 0.2mm from neutral axis
= = 0.045N/mm2
Shear stress at y1 = 0.3mm from neutral axis
= = 0.034N/mm2
Shear stress at y1 = 0.4mm from neutral axis
= = 0.019N/mm2

10. Shear stress at y1 = 0.5mm from neutral axis (at the free surface)
At the free surface, the shear stress is supposed to be zero
= = 0N/mm2
From the calculations above, the shear stress will be highest at the central axis of the beam
section.
Question 4: Elasto-Plastic Analysis
Maximum Elastic Moment, MY
Maximum elastic moment, MY = Ze * fy (Where Ze = elastic section modulus and fy = yield
stress)
Elastic section modulus, Ze = (where b = width of the beam (40mm) and h = height/depth
of the beam (1mm))
Ze =
Yield stress, fy of the beam material is assumed to be 250MPa (250N/mm2)
MY = Ze * fy = 6.667mm3 x 250N/mm2 = 1,666.67Nmm
Plastic Moment, MP
Plastic moment, MP is calculated by multiplying yield stress, fy- with the plastic section
modulus, Z i.e. plastic moment, MP = Zp * fy
Plastic section modulus, ZP = (where b = width of the beam (40mm) and h = height/depth
of the beam (1mm))
ZP =
Yield stress, fy of the beam material is assumed to be 250MPa (250N/mm2)
MP = ZP * fy = 10mm3 x 250N/mm2 = 2500Nmm
Shape Factor
The shape factor, SF =

11. From the calculations above, plastic moment, MP = 2500Nmm and elastic or yield moment,
MY = 1,666.67Nmm.
SF =
Sketch Of Stresses
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).
Sketch Of Residual Stress
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 (fy = 250N/mm2). The residual stress at the external surfaces of
the rectangular beam section is equal to 0.5fy. T
Question 5: Buckling Analysis
Effective Length Of Column
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.
Euler Critical Buckling Load
The formula for calculating Euler critical buckling load, Pcr of a column is given as follows:
Pcr =
Where E = modulus of elasticity (N/mm2), I = moment of inertia (mm4), k = effective length
factor of the column, and L = length of column (Preetha, et al., 2019)

12. 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/mm2.
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
Pcr =
= 18.3N
Compressive Load
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/mm2
Cross sectional area = 40mm x 1mm = 40mm2
Compressive load = 250N/mm2 x 40mm2
= 10kN
A column is considered to be long if the ratio of its effective length to its smallest horizontal

13. 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.