COMMON ASSIGNMENT

FACULTY OF MECHANICAL ENGINEERING
BACHELOR OF ENGINEERING (HONS) MECHANICAL

EM220: EMD3M4B

MECHANICS OF MATERIAL (MEC411)

COMMON ASSIGNMENT :
STRUCTURAL ANALYSIS OF FLOWERS/PLANTS SHELF
FOR CONGESTED SPACE OUTDOOR HOUSE DECORATION

PREPARED BY:
MUHAMAD AL- HAZIQ BIN IZHAR (2019207676)
MUHAMAD ARIF HAKIMI BIN MUHAMAD ARIFIN (2019229938)
MUHAMAD ILTIZAM EZWAN BIN SARIZAM (2019685982)
MUHAMAD AQMAL FARHAN BIN MD FARID (2019294946)
AHMAD KAMAL TAJUDIN BIN KAMARULZAMIL (2019310425)

PREPARED FOR:
DR. NIK ROZLIN NIK MOHD MASDEK

SUBMISSION DATE:
15 JANUARY 2021

TABLE OF CONTENT
PROBLEM STATEMENT ........................................................................................................ 3
OBJECTIVE............................................................................................................................... 3
INTRODUCTION...................................................................................................................... 4
LITERATURE REVIEW........................................................................................................... 9
ANALYSIS AND RESULT .................................................................................................... 11
DISCUSSION .......................................................................................................................... 15
CONCLUSION ........................................................................................................................ 17
REFERENCES......................................................................................................................... 18

2

PROBLEM STATEMENT
Many people love to plant or decorate a house with flowers so that the home

environment become cheerful. Flower shelf is one way to place their plants either indoors or
outdoors. When designing a flower shelf, we need to calculate each components because if it
is wrong in terms of balance of the shelf itself can cause the shelf to broken or even a force
that is too heavy than the shelf itself. As we know flower shelf have many designs, but shelf
that have an interesting design for example has some combination by hanging, stand and
attach to wall are hard to find. The type of material use for shelf is important in order to know
it modulus of elasticity and calculate stress or strain.The material used for this shelf is wood.
Wood is one of the most reliable construction materials. Properly installed wooden shelves fit
a large number of plants, regardless of their size.

OBJECTIVE
1. To relate the fundamental principles learned in class with real life occurences.
2. To train to do analysis and apply theories to real life problem and situations.
3. To develop teamwork skills.
The purpose of this report is to relate the external forces, internal forces, maximum shear
stress, maximum normal stress and compressive stress. that can found in the shelf body by
using the dimensions of the shelf and the formula learned in class. Other than that, bending
moment diagram( BMD) and shear force diagram( SFD) also can be use to determine the
maximum shear force and maximum bending moments. By the components of shelf, the
theories of these elements also can relate. The type of reaction force developed must in
equilibrium conditions or shelf will not stable while using it. The forces on the shelf also
needs to taken care of or the shelf will fail. Therefore, all the required elements need to
calculate properly and the use material for shelf must meet durability of the shelf.

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INTRODUCTION
Structural analysis is a studies or a prediction of the response of structures that uses

the set of mechanics theories that follows a physical laws. Structural analysis establishes the
relationship between a structural member’s expected external load and the structures
corresponding developed internal stresses and displacements that occur within the member
when in service. This is necessary to ensure that the structural members satisfy the safety.
From the overall, the analysis is to compute a structures deformation, internal and external
forces, maximum shear stress, maximum tensile, compressive stress and maximum deflection,
which can be effects the support reactions and stability. As a mechanical engineers, we often
confront a challenges in designing a mechanical systems and component that related to the
structure analysis. It is imperative that these structures contain a minimum of material to
reduce cost and increase efficiency on the mechanical system. As a hands on, structural
analysis reveals the structural performance of the engineering design and ensures the
soundness of structural integrity in design without dependence on direct testing. The geometry
of these load-bearing structural components is usually complicated because of strength and
efficiency requirements.

To ensure this analysis performed an accurate analysis, we must determine few
information such as the structural loads, the support conditions, the geometry and the
properties of the materials that can be used. The results for this analysis typically include
support reactions, stresses, displacements, external forces, internal forces, maximum shear
stress, maximum tensile, compressive stress and maximum deflection. This information will
be determined by the criteria that indicate the conditions of failure. As for this analysis, we’ve
been analyse the mechanical structure of flowers or plants shelf for congested space outdoor
house decoration. There are several types of members or components of the structures that has
been take into consideration in doing this analysis,. which make up of a structure that have
different forms or shapes depending on their functional requirements. These structural
members are such as beams, tension members, roller, and pin. The features of these forms will
be briefly discussed as below:

4

1. Beam

Beams are structural members whose longitudinal dimensions are appreciably greater than
their lateral dimensions. It resists the applied loading by a combination of internal transverse
shear force and bending moment. For example, the length of the beam, as shown in the figure,
is significantly greater than its breadth, B and depth, D. The cross section of a beam can be
rectangular, circular, or triangular, or it can be of what are referred to as standard sections,
such as channels, tees, angles, and I-sections. Beams are always loaded in the longitudinal
direction.
2. Tension Members

Tension members is a structural elements that are subjecte to direct axial tensile loads force in
a direction parallel to its longitudinal axis, which tend to elongate the members. The strength
of these members is influences by several factors such as the length of connection, size and
spacing of fasteners net area of cross section, type of fabrication, connection eccentricity and
shear at the end connection. The types of tension member such as wires, cables, bar and rods.

5

3. Roller

This is the type of support which only restrains the structure from moving in one or two
perpendicular directions. However, the structure can move in the other directions and it can
also rotate. The joint that is supported by a roller support has four or five degrees of freedom.
Therefore, roller supports have one or two support reactions.
4. Pin

Pin or hinge support is used when we need to prevent the structure from moving or restrain its
translational degrees of freedom.
5. Fixed support

A fixed support prevents all movements and rotations at the point where it is attached to the
structure. The joint with fixed support has no degrees of freedom, and therefore, there are six
support reactions applied from the support to the structure.

6

As the overall members has been defined, it is the time to explains the theories that
will be used in these structural analysis. The theories that will be implemented in these
analysis will be briefly discussed as below:

1. The Equilibrium Condition

The structures are designed to be at rest when acted upon by internal or external forces . A
structure at rest must satisfy the equilibrium conditions, which require that the resultant force
and the resultant moment acting on a structure be equal to zero.

= 0 , = 0 ;

= 0 ;

2. Flexural or Bending Stress

It is a stress that had been caused by a bending moment




Where,
M = The internal bending moment at the region of interest (N.m)
c = The perpendicular distance from the neutral axis
I = The moment of inertia about the neutral axis for the cross section

3.Transverse Shear Stress
The shear stress due to bending is often referred to as transverse shear. Like the normal stress
there is a stress profile that is based off of the neutral axis of the particular cross-sectional
area. Unlike normal stress, the highest stress value occurs at the neutral axis, while there is no
stress on the walls.


= ;

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Where,
= Shear stress in member at the point located a distance y’ from the neutral axis
V = internal resultant shear force
I = Moment of inertia of entire cross sectional area computed about the neutral axis
t = Width of the member’s cross sectional area measured at the point where is to be

determined

8

LITERATURE REVIEW

In this section , we had discussed on several idea and design which can be the main
idea in our selection for the best designed. All group member had suggested one design on
which will be contributed in the finalise design. All the design will be shown in figure below.

MUHAMAD ILTIZAM EZWAN BIN SARIZAM MUHAMAD AL-HAZIQ BIN IZHAR

MUHAMAD ARIF HAKIMI BIN MUHAMAD ARIFIN AHMAD KAMAL TAJUDDIN BIN
KAMARULZAMIL

MUHAMAD AQMAL FARHAN BIN MD FARID

9

As we made our discussion, we had choose the finalise design by combining few
selected ideas and make it one best design. Below is the finalise design that had been agreed
by all the group members.

During this discussion, we have made a conclusion regarding the selection of this
design. This design being selected due to;
1. The numbers of members used are more than 3 members.
2. The uses of beam as a mechanism for transferring vertical loads horizontally towards

the support used.
3. The uses of tension wires are from the industrial implementation to make the design

more to modern looks.

Rajah 1: CATIA Drawing of beam

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ANALYSIS AND RESULT
VASE 1 (with planks)
Mass ≈ 7 kg
W = mg

= 7 × 9.81
= 68.67 N
Distributed load

=.
.

= 343.35 N/m

VASE 2 (with planks)
Mass ≈ 6 kg
W = mg

= 6 × 9.81
= 58.86 N
Distributed load

=.

.
= 235.44 N/m

HANGING VASE (↓)
Mass ≈ 3 kg
W=T

= 3 × 9.81
= 29.43 N

ROLLER SUPPORT B
∑ = 0
- 58.67(0.2) - 29.43(0.4) - 58.86(0.775) + (0.9) = 0
- 13.73 – 11.772 – 45.62 + 0.9 = 0

71.12
= 0.9

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= 79.02
PIN SUPPORT A
∑ = 0
− 68.67 − 29.43 − 58.86 + 79.02 = 0
= 77.94
SHEAR FORCE AND BENDING MOMENT DIAGRAM

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BEAM
Modulus of Elasticity (wood) = 11GPa

∑ = 0


+ 343.35 2 − 77.94( + 0.1) = 0
= −171.68 + 77.94 + 7.794
∑ = 0
− − 343.35 + 77.94 = 0
= −343.35 + 77.94

∑ = 0


+ 235.44 2 + 29.43(0.25 + ) + 68.67(0.45 + ) − 77.94(0.65 + ) = 0
+ 117.72 + 7.36 + 29.43 + 30.9 + 68.67 − 50.67 − 77.94 = 0
= −117.72 − 20.16 + 12.41
∑ = 0
− − 235.44 − 29.43 − 68.67 + 77.94 = 0
= −235.44 − 20.16

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Maximum bending moment = 17.35 N.m
Maximum shear force = 79.47 N

Maximum Normal Stress


=

1
= 12 ℎ

1
= 12 ℎ

= 1 (0.2)(0.03)
12

= 4.5 × 10

17.35(0.015)
= 4.5 × 10

= 578333.33 Pa

Maximum Shear Stress
79.47(2.25 × 10 )

= = (4.5 × 10 )(0.2) = 19867.5

= = 0.015 (0.2 × 0.015) = 2.25 × 10
2

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DISCUSSION
In the world of engineering, every problems that they faced must solve by using formulas and
calculation to prove that their solution is valid or not. In this assignment, there are many
equations that are used in the analysis to calculate all the variables in the product shown in the
figure. Each equations have their own purpose to find the correct value for each variables.

For the assumption, the beam can hold all of the forces applied where the summation
of forces in Y-axis and Moment is equal to zero. By referring to the analysis, the first equation
that are used is the weight for vase 1 and vase 2. The formula given is = where is
the mass of the vase in kilogram while is a Gravitational acceleration constant which
simplified to 9.81 . The product for these two variables is the Gravitational definition, for

each of the vase in the unit of Newton, . Main purpose of the is to calculate the weight of
the vase. For this project, three types of vases of different masses were used which is 7kg, 6kg
and 3kg. By using the weight equation, three vase weights have been identified which is 68.67
N for vase 1, 58.86 N for vase 2 and 29.43 N for the hanging vase. It is important to find the
weight of the vase so that the load that the shelf can accommodate can be estimated. This is to
prevent the shelf from failing and collapsing. Next equation that been used is the distributed
load where the weight of the vase, is divided by the length of the vase in the unit of meter.
This equation helps to calculate the force applied on the beam every meter of the vase that
will be simplified to a concentrated load which the total length of the vase will divided to 2 to
assume that the force is actually applied at the middle of the vase rather than force is applied
at the whole length of the vase. For the calculations distributed loads, two out of three vases
were used to find its distributed load which is vase 1 and vase 2. From the calculation, 343.35
N/m for the vase 1 and 235.44 N/m for the vase 2 obtained.

The product shown that there is a hanging vase on the beam which bring to the next
equation which is the tension, . Because of the vase is hanging in the air, it can be conclude
that = . The tension in the rope is simply a concentrated force applied on the beam itself.
If the weight exceeds the cable tension, it will cause the cable to break when given a load that
is more than the capacity of the cable. Referring to the beam, it is shown that there is two
reaction force will be applied which is the pinned support and the roller support. For the
pinned support, there will be Moment, force in the vertical axis and force in the horizontal
axis which differ to the roller support will create only force in the vertical axis. The analysis
continued by using the summation of Moment, at point pinned support which denoted by

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symbol A. Equation of Moment is = which is the external force applied on the beam
while is distance from a certain point of the beam. By calculation the summation, the
reaction force applied on the beam by the roller support which denoted by B is obtained
which is = 79.02 . Then, in order to find the reaction force in the pin supported beam
that denoted by A the analysis used the summation of forces in the y-axis equal to zero. By
the calculation, the force at A have been obtained which is = 77.94 . This reaction force
is important to calculate so that the support used is able to accommodate the given load so that
the vase rack is stronger.

The Shear Force Diagram (SFD) can be drawn by consider all the forces applied to the beam
either external forces or reaction forces. To draw the Bending Moment Diagram (BMD) the
analysis first must find the equation of Moment in terms of which is the distance. The beam
is assume to be cut anywhere within the distributed force then summation of moment and
forces in y-axis is calculated in terms of . The length of distributed forces is substituted into
and plotted in the BMD. From this project, two equations of moments and two equations of
force in y-direction obtained. For the moments equation, = −171.68 + 77.94 +
7.794 and = −117.72 − 20.16 + 12.41. While for the force in y-direction, =
−343.35 + 77.94 and = −235.44 − 20.16. These diagrams can be used to easily
determine the type, size and material of the members in the structure so that a specific set of
loads can be supported without structural failure. Other than that, the application of shear
diagrams and other moments is that the deflection of the beam can be easily determined using
the moment width method or the conjugation beam method.

From the SFD and BMD that have been plotted in the analysis, it can be conclude that
the maximum bending moment is 17.35 while the maximum shear force is 79.47 .

These value are continue to be calculated to find the maximum normal stress, =

where is the maximum bending moment, is the distance from the neutral axis to the outer
part of the beam and is the moment of inertia from the x-axis. Basically, the formula is
called the Flexure formula which is allows to calculate the internal bending stress that is given
by the external moment on the beam. The beam is a rectangular shape so the = ℎ . Next

is the maximum shear stress denoted by = where is the maximum shear force, is

the first moment of area, is the moment of inertia and is the width of the cross-section of
the beam. The formula essentially is to find shear stress at the neutral axis which is at the
border of the tensile and compression force occurs.

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The durability of the beam is totally depends on the material properties of the beam
itself whether it can withstand a load, can be compress or can be stretch without fail. In order
for the beam to be stable even when force is being apply to beam is a sufficient modulus of
elasticity. It generally indicates that how much the material will yield when applied pressure.
The larger the modulus of elasticity, the more rigid the material is, the more it resist to stretch
or compress.

CONCLUSION
Plant shelf is one of the ways to decorate house with plants. By considering the types,

structure, and materials of the shelf is important as to help user to know the specifications of
the shelf. When designing a plant shelf, there are many factor that should be considered to
ensure the shelf is able to withstand the weight of plants or pot.

In this assignment, we can relate and have a better grasp of understanding on the
fundamental principles learned in the class. Condition as example stress, strain, maximum
shear stress, compressive stress, maximum shear force and bending moment are train students
to do analysis and apply theories that have been learned in real life problems and situations.

Based on this assignment, five design of plant shelf are been analyzed and combined
to be one best plant shelf design. All calculation had been taken to know the specifications of
this shelf. The maximum bending moment and shear force of the beam are obtain by using
the shear force diagram and bending moment diagram. Finally, we also obtain the maximum
normal stress and shear stress of beam by using the knowledge from learning in the class. We
assume that beam can hold all of the forces applied where the summation of forces in Y- axis
and moment is equal to zero.

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REFERENCES

1. Mechanics Of Materials 10th Edition R.c. Hibbler
2. Structural Analysis – an overview | ScienceDirect Topics

Retrieved from https://www.sciencedirect.com/topics/engineering/structural-
analysis
3. Types of supports: Knowledgebase: SAFAS
Retrieved from
https://www.setareh.arch.vt.edu/safas/007_fdmtl_08_types_of_supports.html
4. Beams Flexure Formula
Retrieved from http://www.abuildersengineer.com/2014/06/beams-flexure-
formula.html#:~:text=The%20flexure%20formula%20gives%20the,is%20valid%20for
%20any%20shape.&text=which%20gives%20the%20bending%20stress,c%20from%2
0the%20neutral%20axis
5. Mechanics of materials bending shear stress
Retrieved from
https://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/
6. Static beam material
Retrieved from http://www.geom.uiuc.edu/education/calc-init/static-
beam/material.html#:~:text=The%20modulus%20of%20elasticity%20for,unit%20are
a%20is%20called%20stress%20

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