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Assessment of the Reinforced Plate with Linear Elastic Analysis and Non-Linear Analysis

Executive Summary


In the model, there are 3 reinforcements attached to the flat plate. This model exposed to
different kind of loads to find the; force required to generate a deflection of 5mm in elastic
model, maximum total force carried without yielding occurs in elastic model, maximum total
force to generate a plastic strain between 0.1-0.5% and the permenant deflection after
unloading.


Linear analysis phase started with the application of the gravity load on the model with
basically calculating the weight of the model. Due to results, required data obtained with
proportional ratio calculations. Plastic properties hasn’t defined during this analysis so if
stress passes the steel’s yielding stress, steel still behaves elastic.


For the second part of the project, plastic properties defined and required data found by
iterating load. Since, plastic properties defined, its obvious that assumption must be greater
than the maximum total force carried without yielding found in the previous part.

  1. Problem Definition

Properties of the specific model of ‘reinforced plate’ has given: steel plate has 600mm of
width, 280 mm of length and 1.2 mm of thickness. There are three reinforcement bars with
thickness of 3.4 mm and they are located vertically (with 600mm length) on plate.
Reinforcement cross section’s descriptive image has been modeled in Figure 1.

Figure 1: Descriptive drawing of Reinforcement Bar


There are 24 attachment points with plate per reinforcement and all attachments are
located on the down plates of reinforcements which are colored red in Figure 1. Attachment
points have 50mm offset between them and 25mm points between the edges. Other than
that, they all diffused uniformly through reinforcements (12 attachments per red surface).


Analysis required for this model are ‘Linear Elastic’ and ‘Non-linear Analysis’.


In linear elastic analysis; first desired data is the total force P which generates maximum
deflection of 5mm on this specific model and the second is to find the total force P which
generates maximum stress closest to yield stress of 350Mpa.


In Non-linear analysis; first desired data is the total force P which generated maximum
plastic strain in an interval of 0.1%-0.5% and the second is to find the permanent deflection
occurred in the model with applying the total force P found in the first step and unloading
afterwards with assigning plastic properties of steel.

Linear Elastic Analysis:
1-) Calculate the total force P so that you get maximum deflection of 5.0 mm
2-) Assume that materials yield stress 350 MPa. Calculate the total force P so that you get
maximum stress below the yield stress.


Non-Linear Analysis: Use Material Plasticity and Contact Option
3-) Calculate the total force P so that you get maximum plastic strain about 0.1- 0.5 % .
Use plasticity option in material and define contact between flanges of the reinforcement
and plate. This is STEP-1 to determine the load
4-) Use same load in STEP-1), Unload (STEP-2 ); calculate permanent deflection.


Elastic Material
Young’s Modulus = 200000 MPa
Poisson’s Ration = 0.3

Plasticity
(Yield Stress, Plastic Strain Values )
350, 0.00
580, 0.10
650, 0.20

2. Analysis Description

Use deflection formulae for cantilever beam ( flat plate without reinforcement ),
use deflection formulae v = ( w L 4 ) / ( 8 E I )
• Calculate land P to get 5 mm deflection
• We already know load to get 5 mm deflection in FE analysis ( Linear Elastic )
• Compare the loads

2.1 Linear Analysis

In linear analysis of this reinforced plate, only Young’s Modulus and Poisson’s Ratio of the
material has defined. Since used material is S350 steel, Young’s Modulus is 200,000 MPa and
Poisson’s Ratio is ‘0.3’.


After the modeling process of the reinforced plate has done, 128.7 N of gravity load has
applied on the plate to observe the displacement. Maximum displacement on the plate has
occurred at the farthest point from the support and that is 0.889mm’s ‘see Fig.2 in
Appendix’.


Since this model is linear elastic, required total force to generate 5mm of displacement can
be calculated with the following calculations;

Total Force (P) = 128.7N * 5mm/0.889mm = 723 N

723 N of total force applied on the model and as expected: maximum displacement in the
model has found 4.999mm’s ‘see Fig.3 in Appendix’ and the maximum stress generated on
the plate is 113.8 MPa ‘see Fig.4 in Appendix’.


For the second task in linear analysis part, following calculation’s have been made to find the required total force to generate 350 MPa of stress.

Total Force (P2) = 723N * 350MPa/113.8MPa = 2224 N

2224 N of total force applied on the model and 350.2 MPa of stress has been generated(see
Fig.5 in Appendix).

2.2 Non-Linear Analysis

Plastic properties of the steel has been defined for this analysis. Also the methodology of
application of load has been changed so the total load will be applied incrementally and step
by step.


Main objective of this analysis is to find the total force ‘P’ that generates a maximum plastic
strain within the interval of 0.1-0.5%. To find ‘P’, some iterations has been done because it
requires massive calculations to get exact value. Assumed value has to be greater than ‘2224
N’ because plastic strain will not start before steel reaches yield strength. So, 3500 Newtons
of total force applied on the model and a ‘0.202%’ of plastic strain value and 25.32 mm’s of
maximum displacement attained (see Fig.6 and Fig.7 in Appendix).


To find the permanent deformation: elastic displacement under 3500N should be found.
Therefore;

Displacement(mm) = 3500N * 4.99mm/723N = 24.16 mm

Difference between 25.32mm and 24.16 mm is the permanent deflection and that is
1.16mm’s.

3. Results and Discussion

Table 1 Linear Analysis Outputs
Table 2 Non-Linear Analysis Outputs

Depending on this results, amount of generated deflection in the plastic region is greater
than the elastic region, the reason for that might be based on the change in the cross
section area which is directly increases the stress felt by steel because stress has reverse
ratio with the cross-section area.


There are plenty of options to increase the strength of reinforced plate such as; type of
steel, thickness of the plate, thickness of the reinforcement, etc. Most efficient way to do it
might be redesigning reinforcements with increasing the height.

4. Appendix

Figure 2 Deflection due to gravity load in Linear Elastic Analysis
Figure 3 Deflection due to 723 N of total force in Linear Elastic Analysis
Figure 4 Stress generated in the model due to 723 N of total force in Linear Elastic Analysis
Figure 5 Amount of stress generated under 2224 N of total force in Linear Elastic Analysis
Figure 6 Deflection due to iterated load of 3500 N in Non-Linear Analysis
Figure 7 Plastic strain values under 3500 N of iterated load in Non-Linear Analysis

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