**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.

**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.

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 (P _{2}) = 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**

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**