All Courses
All Courses
Courses by Software
Courses by Semester
Courses by Domain
Tool-focused Courses
Machine learning
POPULAR COURSES
Success Stories
ASTRACT In this project, a composite pallet is designed to replace the common steel pallet of a cement machine in order to enhance ease of translation of the blocks on the production line. The material is chosen using general performance index and the bending of the plate is analyzed using Ansys and Matlab. Theoretical…
Leslie Enos
updated on 13 Jun 2021
ASTRACT
In this project, a composite pallet is designed to replace the common steel pallet of a cement machine in order to enhance ease of translation of the blocks on the production line. The material is chosen using general performance index and the bending of the plate is analyzed using Ansys and Matlab. Theoretical calculations are made in matlab and validated for the experimental bending simulation in Ansys. The stacking angles and layer numbers is varied to develop the best performance of the composite pallet.
PROBLEM STATEMENT
A cement block manufacturing company uses a steel pallet to enhance translation of the building blocks along a line from the cement making machine. These steel pallets are durable and strong, but they began to rust over time and their heavy weight reduces cycling time on the production line. There is a need to solve this problem to enable higher production of the cement blocks and reduce maintenance cost on the pallets.
Figure 1 steel pallet for concrete manufacturing (1)
AIM
I aim to solve this problem using finite element theory. A new pallet is going to be designed with the most optimum materials used by considering elimination of rust, low weight, and high durability to reduce maintenance cost.
METHOD
First, a material which fits the above requirements should be selected. Material index equation for selecting the best performing material is used to select the best materials for this simulation.
Performance index should be minimized to achieve high strength and stiffness while keeping weight minimum. The table below shows a rough estimate of performance index of materials and their cost.
Table 1 Performance Index
From the above image, composites which has the highest performance index is selected.
ANALYSIS
METHOD
A composite laminate is two or more laminae bonded together in the direction of lamina thickness to act as an integral structural element. Laminae principal material directions are oriented to produce a structural element capable of resisting load in several directions. Each lamina can be spot by its location in the laminate, its material, and its orientation with reference axis.
Diagram of [0/90/0/90] cross-ply laminate
Classical laminate theory (CLT) has been used to calculate deformation of composite plate. For CLT, due to simplicity over energy and vibrational principles, Newtonian approach has been used in which summing up forces and moments on the plate is often used to develop the governing differential equations. The governing equations consisting the behavior of the boundary conditions. In the present analysis,
Assumptions
E= Elastic Modulus
V=Poisson Ratio
G=Shear Modulus
Transformed global stiffness matrix is expressed below
Where S =sin θ , C =cos θ
Flexural stiffness (D-Value)
h values taken from 0 at the bottom through the thickness with the middle plate being 0
Naiver Solution for Simply Supported Composite Plates (Bending)
Finite Element Modelling
The analysis of the laminate composite plate is done with ANSYS. Using ACE Pre-Design Modeler, the pallet is designed in 2D with the following dimensions.
For the analysis, there are going to be two cases,
CASE_1
CASE_2
Material Properties
A new material data is created and all values of elastic modulus and poisson ratio is specified.
Using ACP setup, the fabric and stackups is specified as seen in the picture below
A rosette coordinate is specified to indicate the direction of the angle ply.
Static Structural
Nodal displacement (free, free, 0), Nodal displacement (free, free, 0) Nodal Rotation (free, fixed,free)
Nodal Rotation (fixed, free, free) ,Nodal displacement (fixed, fixed, free), Nodal displacement (fixed, free, fixed)
PRESSURE
The pressure to be applied should correlate with the realistic pressure which would be applied on the pallet on daily basis. The average mass of one block is 40kg. With the dimension of the pallet, 24 blocks would fit on one pallet at a time.
Total mass = 40 x 24 = 960kg
Force = 960kg x 9.8= 9408 kg/
Pressure = = 4704 Pa
OUTPUT REQUEST
RESULTS
CASE_1
CASE_2 (8Ply)
Validation of Results
In order to validate the results given by ANSYS, the experiment is run in Mat lab and the results are compared.
Code and results for case_1
% Material properties
E1=43*(10^9)
E2=8.9*(10^9)
v12=0.27
v21=(E2*v12)/E1
G12=4.5*10^9
%Q values calculation
Q11=E1/(1-(v12*v21))
Q12=(v21*E1)/(1-(v12*v21))
Q21=Q12;
Q22=E2/ (1-(v12*v21))
Q66=G12
% D values calculations
% 0 degrees
Q11=E1/(1-(v12*v21))
Q12=(v21*E1)/(1-(v12*v21))
Q21=Q12;
Q22=E2/ (1-(v12*v21))
Q66=G12
m=1
n=0
Qbar110=(Q11*m^4)+(2*(Q12*(2*Q66))*(m^2)*(n^2))+(Q22*n^4)
Qbar120=((Q11+Q22-(4*Q66))*(m^2)*(n^2))+(Q12*((m^4)+(n^4)))
Qbar210=Qbar120
Qbar220=(Q11*(n^4))+(2*(m^2)*(n^2)*(Q12+(2*Q66)))+(Q22*m^4)
Qbar660=(Q11+Q22-(2*Q12))*(m^2)*(n^2)+(Q66*(m^2-n^2)^2)
%90 degrees
y=0
z=1
Qbar11=(Q11*y^4)+(2*(Q12*(2*Q66))*(y^2)*(z^2))+(Q22*z^4)
Qbar12=((Q11+Q22-(4*Q66))*(y^2)*(z^2))+(Q12*((y^4)+(z^4)))
Qbar21=Qbar12
Qbar22=(Q11*(z^4))+(2*(y^2)*(z^2)*(Q12+(2*Q66)))+(Q22*y^4)
Qbar66=(Q11+Q22-(2*Q12))*(z^2)*(y^2)+(Q66*(z^2-y^2)^2)
% Dvalue
D11=(1/3*(Qbar110*((-0.025^3)-(-0.05^3))))+(1/3*(Qbar11*((0.0^3)-(-0.025^3))))+(1/3*(Qbar11*((0.025^3)-(-0.0^3))))+(1/3*(Qbar110*((0.05^3)-(0.025^3))))
D12=(1/3*(Qbar120*((-0.025^3)-(-0.05^3))))+(1/3*(Qbar12*((0.0^3)-(-0.025^3))))+(1/3*(Qbar12*((0.025^3)-(-0.0^3))))+(1/3*(Qbar120*((0.05^3)-(0.025^3))))
D21=D12
D22=(1/3*(Qbar220*((-0.025^3)-(-0.05^3))))+(1/3*(Qbar22*((0.0^3)-(-0.025^3))))+(1/3*(Qbar22*((0.025^3)-(-0.0^3))))+(1/3*(Qbar220*((0.05^3)-(0.025^3))))
D66=(1/3*(Qbar660*((-0.025^3)-(-0.05^3)))+(1/3*(Qbar66*((0.0^3)-(-0.025^3))))+(1/3*(Qbar66*((0.025^3)-(-0.0^3))))+(1/3*(Qbar660*((0.05^3)-(0.025^3)))))
D1=D11 %Nm
D2=D22 %Nm
D3=D12+(2*D66) %Nm
% Bending Calculation
a=2 %dimension in meters
b=1
m1=1 %these are not modes
n1=1
P = 4704 %in pascal
Wstorage =0
x=1 %for the middle point
y=0.5
for m1=1:20
for n1=1:20
Bmn=(4*P/(m1*n1*pi^2))*(1-(-1)^m1)*(1-(-1)^n1)
Amn=Bmn/((D1*(m1*pi/a)^4)+(2*D3*((m1*pi/a)^2)*(n1*pi/b)^2)+(D2*(n1*pi/b)^4))
w= Amn*sin(m1*pi*x/a)*sin(n1*pi*y/b) %deflection for displacement
wstorage=wstorage+w
n1=n+1
end
end
wxy=wstorage*1000 %in mm
Final displacement for case_1 = 0.0423m
Code and results for case_2
% Material properties
E1=43*(10^9)
E2=8.9*(10^9)
v12=0.27
v21=(E2*v12)/E1
G12=4.5*10^9
thickness = 12.5*(10^-3) % in meters
% Q values calculation
Q11=E1/(1-(v12*v21))
Q12=(v21*E1)/(1-(v12*v21))
Q21=Q12;
Q22=E2/(1-(v12*v21))
Q66=G12
% D values calculations
% 0 degeres
m=1
n=0
Qbar110=(Q11*m^4)+(2*(Q12*(2*Q66))*(m^2)*(n^2))+(Q22*n^4)
Qbar120=((Q11+Q22-(4*Q66))*(m^2)*(n^2))+(Q12*((m^4)+(n^4)))
Qbar210=Qbar120
Qbar220=(Q11*(n^4))+(2*(m^2)*(n^2)*(Q12+(2*Q66)))+(Q22*m^4)
Qbar660=(Q11+Q22-(2*Q12))*(m^2)*(n^2)+(Q66*(m^2-n^2)^2)
% 90 degrees
y=0
z=1
Qbar11=(Q11*y^4)+(2*(Q12*(2*Q66))*(y^2)*(z^2))+(Q22*z^4)
Qbar12=((Q11+Q22(4*Q66))*(y^2)*(z^2))+(Q12*((y^4)+(z^4)))
Qbar21=Qbar12
Qbar22=(Q11*(z^4))+(2*(y^2)*(z^2)*(Q12+(2*Q66)))+(Q22*y^4)
Qbar66=(Q11+Q22-(2*Q12))*(z^2)*(y^2)+(Q66*(z^2-y^2)^2)
% D value
D11=(1/3*(Qbar110*((-0.0375^3)-(-0.05^3))))+(1/3*(Qbar11*((-0.025^3)-(-0.0375^3))))+(1/3*(Qbar110*((-0.0125^3)-(-0.025^3))))+(1/3*(Qbar11*((0.0^3)-(0.0125^3))))+(1/3*(Qbar11*((0.0125^3)-(0^3))))+(1/3*(Qbar110*((0.025^3)-(0.0125^3))))+(1/3*(Qbar11*((0.0375^3)-(0.025^3))))+(1/3*(Qbar110*((0.05^3)-(0.037^3))))
D12=(1/3*(Qbar120*((-0.0375^3)-(-0.05^3))))+(1/3*(Qbar12*((-0.025^3)-(-0.0375^3))))+(1/3*(Qbar120*((-0.0125^3)-(-0.025^3))))+(1/3*(Qbar12*((0.0^3)-(0.0125^3))))+(1/3*(Qbar12*((0.0125^3)-(0^3))))+(1/3*(Qbar120*((0.025^3)-(0.0125^3))))+(1/3*(Qbar12*((0.0375^3)-(0.025^3))))+(1/3*(Qbar120*((0.05^3)-(0.037^3))))
D21=D12
D22=(1/3*(Qbar220*((-0.0375^3)-(-0.05^3))))+(1/3*(Qbar22*((-0.025^3)-(-0.0375^3))))+(1/3*(Qbar220*((-0.0125^3)-(-0.025^3))))+(1/3*(Qbar22*((0.0^3)-(0.0125^3))))+(1/3*(Qbar22*((0.0125^3)-(0^3))))+(1/3*(Qbar220*((0.025^3)-(0.0125^3))))+(1/3*(Qbar22*((0.0375^3)-(0.025^3))))+(1/3*(Qbar220*((0.05^3)-(0.037^3))))
D1=D11 %Nm
D2=D22 %Nm
D3=D12+(2*D66) %Nm
% Bending Calculation
a=2 %dimension in meters
b=1
m1=1 %these are not modes
n1=1
P=4702 %in pascal
wstorage=0
x=1 %for the middle point
y=0.5
for m1=1:20
for n1=1:20
Bmn=(4*P/(m1*n1*pi^2))*(1-(-1)^m1)*(1-(-1)^n1)
Amn=Bmn/((D1*(m1*pi/a)^4)+(2*D3*((m1*pi/a)^2)*(n1*pi/b)^2)+(D2*(n1*pi/b)^4))
w= Amn*sin(m1*pi*x/a)*sin(n1*pi*y/b) %deflection for displacement
wstorage =wstorage+w
n1=n+1
end
end
wxy=wstorage*1000 %in mm
Final displacement for case_2 = 0.0323mm
DISCUSSION
Below is a summary of results collected
CASE |
PLY |
THICKNESS (mm) |
DEFORMATION(mm) ANSYS |
DEFORMATION(mm) MATLAB |
STRESS (Mpa) |
1 |
4 |
25 |
0.044 |
0.042 |
0.0165 |
2 |
8 |
12.5 |
0.033 |
0.032 |
0.0218 |
The deformation of the pallet decreased by some amount when the ply number was increased from 4 to eighth. The strength of the pallet increased as the ply thickness decreased with less stresses developed on the surface.
The greater the number of ply, the stronger the composite laminate. The pallet is designed with an 8 ply after this examination.
CONCLUSION
This experiment was conducted mainly to improve the steel pallet of a cement manufacturing company by creating a lighter product while retaining some optimum amount of strength. This was achieved using a composite laminate material with an 8 Ply layer and an angle configuration of [0 90 0 90 90 0 90 0]. Some advantages of this pallet is listed below
The concrete block pallet has the ability to withstand high impacts, way higher than that of a PVC or solid wood board. These impacts are inevitable in a concrete block factory; and are helpful in extending the working life of the pallet.
Since it is manufactured with fiberglass mat reinforced thermoplastics, pallets for concrete blocks are highly water-resistant surfaces. It does not need any additional coatings to be water or impact resistant.
The pallet for Concrete Blocks is an ecological alternative to other materials as the pallet can be further recycled after its use.
Due to its strength and high impact resistance, the pallets are incredibly difficult to damage and thus result in an increase in the number of cycles it can be used in a concrete block machine.
REFERENCES
https://www.alibaba.com/product-detail/Steel-Containment-Metal-Pallet-For-Concrete_60699898242.html
Leave a comment
Thanks for choosing to leave a comment. Please keep in mind that all the comments are moderated as per our comment policy, and your email will not be published for privacy reasons. Please leave a personal & meaningful conversation.
Other comments...
Week 3 - Solving second order ODEs
Python for Mechanical EngineersBy Enos LeslieMechanical Engineer23rd October 2024AIMThis project aims to write a code in Python to simulate the behaviour of simple pendulum using ODE and animate it. PROCEDUREInitially, the math, matplotlib, numpy, and scipy modules were imported for their respective functionalities.…
23 Oct 2024 01:09 PM IST
Week 2 Air standard Cycle
LS DYNA – Python for Mechanical EngineersBy Enos LeslieMechanical Engineer2nd October 2024AIMThis project aims to write a code in Python to solve and plot an otto cycle and solve for its thermal efficiency. PROCEDUREThe Otto cycle is an air-standard cycle, we will assume the gamma value to be 1.4. At the initial…
02 Oct 2024 03:07 PM IST
Week - 4 - Crash Box Simulation
LS DYNA – Crush SimulationBy Enos LeslieMechanical Engineer18th August 2024OBJECTIVEThe project aims to understand the crashworthiness of the crash box design and evaluate the effect of thickness on its energy absorption and structural integrity during an impact.PROCEDUREThe crush box is made up of a shell rectangular…
18 Aug 2024 06:16 PM IST
Bird Strike - Project - 2
LS DYNA – Bird StrikeBy Enos LeslieMechanical Engineer 09th August 2024Bird Strike in Aero EngineThis is a classic nonlinear transient dynamics problem similar to car crash and mobile drop. While accurate modelling of the problem requires advanced techniques such as SPH, this problem can be solved using generic…
11 Aug 2024 04:28 PM IST
Related Courses
Skill-Lync offers industry relevant advanced engineering courses for engineering students by partnering with industry experts.
© 2025 Skill-Lync Inc. All Rights Reserved.