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CEGE0101: Structural Mechanics
Structural Mechanics
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CEGE0101: Structural Mechanics
The aim of this work is to compare the theoretical calculations, which you learned
during the course and the numerical approach through an advanced software
commonly used in the industry (Oasys GSA).
You are going to look at shear force, bending moment, stress, strain and deflection of
a simply supported beam, loaded at two points, under different material and cross
sectional arrangement.
You will submit a report, which will be worth 20% of your total module mark.
Part 1: Theoretical Shear Force & Bending Moment in the Beam
You are going to consider a simply supported beam which is loaded at two points.
Each student will consider a different loading configuration. The spacing of the
supports and the location of the loads for each student are specified in Table 3
(Page 8). The total length (𝑙) of the beam is 1000 mm.
By drawing a free body diagram (FBD) of the system, you should find the value of
the reactions (𝑅). By making imaginary sections along the beam’s length you should
establish what the shear force (𝑉) and bending moment (𝑀) are along the beam’s
length.
Part 2: Theoretical Longitudinal Stress & Strain in the
Beam
The next step is to calculate stresses (σ) and strains (ε) in the beam which result
from the bending moment at centre point of the beam’s length (i.e. midspan 𝑙/2). In
any loading case there is an axis about which the beam is said to bend and along
which there will be zero longitudinal stress. This axis is known as the Neutral Axis
(𝑁.𝐴.). For a section in sagging, the stress will be compressive above the neutral
axis and tensile below this axis. No matter what the shape of the cross section, the
stress varies linearly with distance from the neutral axis. If we define 𝑦 as the
distance from the neutral axis, then the longitudinal stress (𝝈) is evaluated as
follows:
𝜎 =
𝑀 ∙ 𝑦
𝐼
where 𝑀 is the moment carried by the section and 𝐼 is the second moment of area.
The equation gives you the value of stress (𝜎) at all points along the dotted red line,
shown in Figure 1. The dashed black line is the neutral axis (𝑁.𝐴.). This passes
through the centroid of the section.
Figure 1. Cross Section through beam showing Neutral Axis (dashed black line)
The most significant stress points are nearly always the top and bottom of the cross
section where 𝑦 is maximum. Hence, to find the stress on each surface (top and
bottom) we can put in a value of 𝑦 = 𝑑/2 , where 𝑑 is the cross-sectional depth.
𝑦 𝑁. 𝐴.
CEGE0101: Structural Mechanics –Beam Behaviour CW
3
𝐼 is the second moment of inertia (a.k.a. moment of inertia). It is a property of the
cross section and an axis. The 𝐼 value which applies here is the one about the
neutral axis (𝑁.𝐴.). For a rectangular and circular cross section this is given by:
Rectangular Cross Sections with depth 𝑑 and width 𝑏 𝐼 =
𝑏 ∙ 𝑑
3
12
Circular Cross Sections with radius 𝑟 𝐼 =
𝜋 ∙ 𝑟
4
4
𝐼 governs the ability of a section to carry a bending moment (𝑀). The bigger it is the
larger the capacity of the section to carry load. This helps with the definition of 𝑏 and
𝑑. In case of rectangular cross sections, the “depth” (𝑑) is clearly the quantity which
most significantly increases bending strength (i.e. maximum stress capacity), as it
is in the power of 3 when compared with the width (𝑏). To calculate 𝐼 and thus 𝜎
you will need to have 𝑏 and 𝑑 or 𝑟 values of the beam’s cross section, which are
given in Table 1.
Table 1 - Cross Sectional Arrangements
Cross Sectional Arrangement Width
𝒃 (mm)
Depth
𝒅 (mm)
Arrangement 1 (Rectangular) 50 10
Arrangement 2 (Rectangular) 8 50
Arrangement 3 (Circular) 15 15
Arrangement 1 Arrangement 2 Arrangement 3
Having all necessary measures, now you can calculate the stress (𝜎) expected on
the top and bottom surfaces of the beam under two 60N loads, as long as the
bending moment is measures at any point along the length of the beam.
Strain (𝜺) is literally the ratio between how much longer something has got and its
original length (𝜀 = 𝛥𝑙 / 𝑙0). Values are typically very small and are often multiplied
by × 106 and hence referred to as micro-strain (µ𝜀).
The estimated stress (𝜎) can be converted into strain (𝜀), thanks to Hooke’s law:
𝜀 =
𝜎
𝐸
where 𝐸 is the Young’s Modulus (N.mm-2
) of the beam’s material. The materials you
are using are Steel (Grade: S235) and Aluminium (Grade: 6063). You should
establish a representative value for the elastic modulus (𝐸) of these material from
the literature and online sources.
50 mm
50 mm
8 mm
10 mm
15 mm
15 mm
CEGE0101: Structural Mechanics –Beam Behaviour CW
4
Part 3: Theoretical Deflection of the Beam
We can figure out the deflection (𝛥) of a beam with a known bending moment
distribution. The bending moment at each point in the beam actually causes
curvature (that’s the second derivative of deflection). The deflection (𝛥) of the beam
is the net result of all of this curvature.
For a beam with a single point load, the shear diagram, moment diagram as well as
relevant equations related to end reactions (𝑅), moments (𝑀) and deflections (𝛥)
throughout the length of the beam are shown in Figure 2.
To find the deflection (𝛥) under both loads at the same time you will need to think
about linear elastic behaviour (Hooke’s Law) and the principle of superposition,
which was discussed in the course. Calculate the value of deflection under each of
the applied 60N point loads and combine them.
Figure 2. Simply supported beam with a single concentrated load
applied at any location along the beam’s length
We now have a theoretical calculation of the stress and strain on each surface
(top and bottom) at the centre point of the beam (i.e. midspan 𝑙/2) under two 60N
loads, as well as the Deflection under each of the 60N point loads.
Part 4: Numerical Shear Force, Bending Moment & Deflection
Now you should analyse your simply supported beam model in Oasys GSA under
following loading cases for ONE of the considered materials (either Steel S235 or
Aluminium 6063) and ONE of the cross sectional arrangements, while keeping your
allocated loading distances (i.e. distance from supports according to Table 3 on
Page 8). You can choose which material and cross section arrangmemt to use.
Table 2 – Numerical Loading Conditions
Loading Cases Point Load 1 (N) Point Load 2 (N)
Case 1 30 30
Case 2 60 60
CEGE0101: Structural Mechanics –Beam Behaviour CW
5
Plot, compare and discuss the following Diagrams obtained from the software under
the given loading cases, material and cross sectional arrangement.
1) Loading + Fx, Fy and Fz Reaction Diagram
2) Myy Bending Moment Diagram
3) Fz Shear Force Diagram
4) Displacement Diagram + Deformed Image
For the numerical modelling using Oasys GSA, make sure you refer to the
“Structural Modelling and Analysis” course, available on UCLeXtend. Further
information on how to access this course and how to download and instal the
software is provided on CEGE0101 Moodle. Make sure you watch the training clip
and follow the provided notes of the following topic on UCLeXtend:
Simply Supported Beam - Linear Static Analysis
Submission: Short Professional Technical Report
You are required to write a brief technical report describing your method, results
and comparisons. As part of the report, you will need to answer the questions posed
here below. The report should be no longer than maximum 4 pages and contain
about 1000-1500 words. If necessary, you can add a single page of appendix.
Answer the following questions in your report using the results obtained from your
theoretical and numerical analysis. You can support your discussions and
reasoning by providing screen shots of your analysis results (diagrams or contours).
Compare and discuss the theoretical calculations you did for 60N loading
case and the numerical ones obtained from the software.
Do you see any difference when you change the material from Steel
(Grade: S235) to Aluminium (Grade: 6063)? What values and results
change ? Discuss why.
Do you see any difference when you change the cross-section
arrangements ? What values and results change ? Discuss why.
Do you see any difference when you change the loading values ? What
values and results change ? Discuss why.
Why do you think 60N is the maximum load we will try at each point ?
Report Writing & Formatting
The reports need to be typed and any diagrams must be presented neatly, properly
labelled and captioned. Lack of use of a ruler (if hand drawings) and bad writing, will
lead to a reduction in marks.
The aim of this work is to compare the theoretical calculations, which you learned
during the course and the numerical approach through an advanced software
commonly used in the industry (Oasys GSA).
You are going to look at shear force, bending moment, stress, strain and deflection of
a simply supported beam, loaded at two points, under different material and cross
sectional arrangement.
You will submit a report, which will be worth 20% of your total module mark.
Part 1: Theoretical Shear Force & Bending Moment in the Beam
You are going to consider a simply supported beam which is loaded at two points.
Each student will consider a different loading configuration. The spacing of the
supports and the location of the loads for each student are specified in Table 3
(Page 8). The total length (𝑙) of the beam is 1000 mm.
By drawing a free body diagram (FBD) of the system, you should find the value of
the reactions (𝑅). By making imaginary sections along the beam’s length you should
establish what the shear force (𝑉) and bending moment (𝑀) are along the beam’s
length.
Part 2: Theoretical Longitudinal Stress & Strain in the
Beam
The next step is to calculate stresses (σ) and strains (ε) in the beam which result
from the bending moment at centre point of the beam’s length (i.e. midspan 𝑙/2). In
any loading case there is an axis about which the beam is said to bend and along
which there will be zero longitudinal stress. This axis is known as the Neutral Axis
(𝑁.𝐴.). For a section in sagging, the stress will be compressive above the neutral
axis and tensile below this axis. No matter what the shape of the cross section, the
stress varies linearly with distance from the neutral axis. If we define 𝑦 as the
distance from the neutral axis, then the longitudinal stress (𝝈) is evaluated as
follows:
𝜎 =
𝑀 ∙ 𝑦
𝐼
where 𝑀 is the moment carried by the section and 𝐼 is the second moment of area.
The equation gives you the value of stress (𝜎) at all points along the dotted red line,
shown in Figure 1. The dashed black line is the neutral axis (𝑁.𝐴.). This passes
through the centroid of the section.
Figure 1. Cross Section through beam showing Neutral Axis (dashed black line)
The most significant stress points are nearly always the top and bottom of the cross
section where 𝑦 is maximum. Hence, to find the stress on each surface (top and
bottom) we can put in a value of 𝑦 = 𝑑/2 , where 𝑑 is the cross-sectional depth.
𝑦 𝑁. 𝐴.
CEGE0101: Structural Mechanics –Beam Behaviour CW
3
𝐼 is the second moment of inertia (a.k.a. moment of inertia). It is a property of the
cross section and an axis. The 𝐼 value which applies here is the one about the
neutral axis (𝑁.𝐴.). For a rectangular and circular cross section this is given by:
Rectangular Cross Sections with depth 𝑑 and width 𝑏 𝐼 =
𝑏 ∙ 𝑑
3
12
Circular Cross Sections with radius 𝑟 𝐼 =
𝜋 ∙ 𝑟
4
4
𝐼 governs the ability of a section to carry a bending moment (𝑀). The bigger it is the
larger the capacity of the section to carry load. This helps with the definition of 𝑏 and
𝑑. In case of rectangular cross sections, the “depth” (𝑑) is clearly the quantity which
most significantly increases bending strength (i.e. maximum stress capacity), as it
is in the power of 3 when compared with the width (𝑏). To calculate 𝐼 and thus 𝜎
you will need to have 𝑏 and 𝑑 or 𝑟 values of the beam’s cross section, which are
given in Table 1.
Table 1 - Cross Sectional Arrangements
Cross Sectional Arrangement Width
𝒃 (mm)
Depth
𝒅 (mm)
Arrangement 1 (Rectangular) 50 10
Arrangement 2 (Rectangular) 8 50
Arrangement 3 (Circular) 15 15
Arrangement 1 Arrangement 2 Arrangement 3
Having all necessary measures, now you can calculate the stress (𝜎) expected on
the top and bottom surfaces of the beam under two 60N loads, as long as the
bending moment is measures at any point along the length of the beam.
Strain (𝜺) is literally the ratio between how much longer something has got and its
original length (𝜀 = 𝛥𝑙 / 𝑙0). Values are typically very small and are often multiplied
by × 106 and hence referred to as micro-strain (µ𝜀).
The estimated stress (𝜎) can be converted into strain (𝜀), thanks to Hooke’s law:
𝜀 =
𝜎
𝐸
where 𝐸 is the Young’s Modulus (N.mm-2
) of the beam’s material. The materials you
are using are Steel (Grade: S235) and Aluminium (Grade: 6063). You should
establish a representative value for the elastic modulus (𝐸) of these material from
the literature and online sources.
50 mm
50 mm
8 mm
10 mm
15 mm
15 mm
CEGE0101: Structural Mechanics –Beam Behaviour CW
4
Part 3: Theoretical Deflection of the Beam
We can figure out the deflection (𝛥) of a beam with a known bending moment
distribution. The bending moment at each point in the beam actually causes
curvature (that’s the second derivative of deflection). The deflection (𝛥) of the beam
is the net result of all of this curvature.
For a beam with a single point load, the shear diagram, moment diagram as well as
relevant equations related to end reactions (𝑅), moments (𝑀) and deflections (𝛥)
throughout the length of the beam are shown in Figure 2.
To find the deflection (𝛥) under both loads at the same time you will need to think
about linear elastic behaviour (Hooke’s Law) and the principle of superposition,
which was discussed in the course. Calculate the value of deflection under each of
the applied 60N point loads and combine them.
Figure 2. Simply supported beam with a single concentrated load
applied at any location along the beam’s length
We now have a theoretical calculation of the stress and strain on each surface
(top and bottom) at the centre point of the beam (i.e. midspan 𝑙/2) under two 60N
loads, as well as the Deflection under each of the 60N point loads.
Part 4: Numerical Shear Force, Bending Moment & Deflection
Now you should analyse your simply supported beam model in Oasys GSA under
following loading cases for ONE of the considered materials (either Steel S235 or
Aluminium 6063) and ONE of the cross sectional arrangements, while keeping your
allocated loading distances (i.e. distance from supports according to Table 3 on
Page 8). You can choose which material and cross section arrangmemt to use.
Table 2 – Numerical Loading Conditions
Loading Cases Point Load 1 (N) Point Load 2 (N)
Case 1 30 30
Case 2 60 60
CEGE0101: Structural Mechanics –Beam Behaviour CW
5
Plot, compare and discuss the following Diagrams obtained from the software under
the given loading cases, material and cross sectional arrangement.
1) Loading + Fx, Fy and Fz Reaction Diagram
2) Myy Bending Moment Diagram
3) Fz Shear Force Diagram
4) Displacement Diagram + Deformed Image
For the numerical modelling using Oasys GSA, make sure you refer to the
“Structural Modelling and Analysis” course, available on UCLeXtend. Further
information on how to access this course and how to download and instal the
software is provided on CEGE0101 Moodle. Make sure you watch the training clip
and follow the provided notes of the following topic on UCLeXtend:
Simply Supported Beam - Linear Static Analysis
Submission: Short Professional Technical Report
You are required to write a brief technical report describing your method, results
and comparisons. As part of the report, you will need to answer the questions posed
here below. The report should be no longer than maximum 4 pages and contain
about 1000-1500 words. If necessary, you can add a single page of appendix.
Answer the following questions in your report using the results obtained from your
theoretical and numerical analysis. You can support your discussions and
reasoning by providing screen shots of your analysis results (diagrams or contours).
Compare and discuss the theoretical calculations you did for 60N loading
case and the numerical ones obtained from the software.
Do you see any difference when you change the material from Steel
(Grade: S235) to Aluminium (Grade: 6063)? What values and results
change ? Discuss why.
Do you see any difference when you change the cross-section
arrangements ? What values and results change ? Discuss why.
Do you see any difference when you change the loading values ? What
values and results change ? Discuss why.
Why do you think 60N is the maximum load we will try at each point ?
Report Writing & Formatting
The reports need to be typed and any diagrams must be presented neatly, properly
labelled and captioned. Lack of use of a ruler (if hand drawings) and bad writing, will
lead to a reduction in marks.
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