A large screen is to be mounted upon a beam; the aim of this analysis is to establish whether it should be directly placed on the beam or through a fixed pillar by comparing values obtained from the classical approach and FE method. Finite elements solve by breaking up a problem into small regions and solutions are found for each region taking into account only the regions that are right next to the one being solved. Whereas, the classical approach involves the use of free body, shear force and bending moment diagrams. Furthermore, the weight of the beam will be analysed to see if there is any significance to the stress when taking it into consideration or not. In addition, the reliability of the values obtained through the different methods will also be taken into account.

CASE 1 Point Load (without weight)

Free body diagram:

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=> R1 + R2 + Fy = 0

=> R1 + R2 + 5 = 0

=> R1 = – R2 – 5

=> ( 2 x 5 ) – (|R2|x 4 ) = 0

=> 4|R2|= 10

=> |R2 |= 2.5 kN

=> R2 = – 2.5 kN

=> R1 = – ( -2.5 ) – 5

=> R1 = – 2.5 kN

Shear force diagram:

Bending Moment Diagram:

Calculating Stress:

In order to calculate stress we need to find the second moment of area using the following equation:

Ix = D^4 = ( 0.2 )^4 = 1

12 12 7500

Then use the following formulae to calculate stress ?:

? = maximum bending moment x ( y ) = 5000 x (0.1) = 3750000 Pa

Ix

Therefore maximum stress: ? = 3.75 MPa

Calculating Safety Factor:

To calculate the safety factor we need to know the corresponding ultimate tensile strength (uts) to the density of steel of 7800kg/m^3 which is 415 MPa [William10], therefore using the following equation:

Safety Factor = uts = 415 = 110.7

? 3.75

CASE 1 Point Load (with weight)

In this case we need to calculate the weight of the beam W, taking the density of steel as 7800 kg m^-3 [Hearn97]:

=> 7800 x (0.2)^2 x 5 = 1560kg

=> 1560 x 9.81 = 15303.6N (15.3 kN)

As the weight is uniformly distributed we need to calculate the weight distribution:

=> 15303.6/ 5 = 3.06 kN

Free body diagram:

=> R1 + R2 + Fy + W = 0

=> R1 + R2 + 5 + 15.3 = 0

=> R1 = – R2 – 5 – 15.3

=> – ( 2 x 5 ) – (15.3 x 2.5) + ( 4 x |R2|) = 0

=> 4|R2|= 48.25

=> R2 = – 12.06 kN

=> R1 = – ( -12.06) – 5 – 15.3

=> R1 = – 8.24 kN

Shear force diagram:

Bending Moment Diagram:

From the bending moment diagram we can see that the maximum bending moment is 10.36 kN/m.

Calculating Stress:

? = maximum bending moment x ( y ) = 10360 x (0.1) = 7770000 Pa

Ix

Therefore maximum stress: ? = 7.77 MPa

Calculating Safety Factor:

Safety Factor = uts = 415 = 53.41

? 7.77

CASE 2 Distributed Load (without weight)

Free body diagram:

=> R1 + R2 + (P x 2 ) = 0

=> R1 + R2 + ( 2 x 2.5 ) = 0

=> R1 = – R2 – 5

=> 2( 2 x 2.5 ) + ( 4 x |R2| ) = 0

=> 4|R2| = – 10

=> R2 = – 2.5 kN

=> R1 = – (-2.5 ) – 5

=> R1 = – 2.5 kN

Shear force diagram:

Bending Moment Diagram:

From this bending moment diagram we can see that the maximum is 3.75 kN/m

Calculating Stress:

? = maximum bending moment x ( y ) = 3750 x (0.1) = 2812500 Pa

Ix

Therefore maximum stress: ? = 2.81 MPa

Calculating Safety Factor:

Safety Factor = uts = 415 = 147.7

? 2.81

CASE 2 Distributed load (with weight)

As the weight is taken into account, there is a uniform weight distribution of 3.06 kN/m across the length of the beam.

Free body diagram:

=> R1 + R2 + ( P x 2.5 ) + W = 0

=> R1 + R2 + 5 + 15.3 = 0

=> R1 = – R2 – 20.3

=> – 2( 2 x 2.5 ) – (15.3 x 2.5) + (4 x |R2| ) = 0

=> 4|R2| = 48.25

=> R2 = – 12.06 kN

=> R1 = – ( -12.06) – 15.3

=> R1 = -8.24 kN

Shear force diagram:

As the bending force diagram is drawn on the basis of calculating the area of the shear force diagram, we can see that the gradient of the line between points A and B cuts the x -axis at an unknown point where y = 0. Therefore, this x value needs to be solved by first calculating the equation of the line between the points:

Gradient (m) = Change in y = 5.94 – (-5.18) = 5.56

Change in x 3 – 1

Calculating the value of c by using the points (1, 5.18):

=> y = 5.56x + c

=> -5.18 = 5.56 + c

=> c = – 10.74

Equation of line AB:

=> y = 5.56x – 10.74

=> 0 = 5.56x – 10.74

From this bending moment diagram we can see that the maximum is 9.12 kN/m

Calculating Stress:

? = maximum bending moment x ( y ) = 9120 x (0.1) = 6840000 Pa

Ix

Therefore maximum stress: ? = 6.84 MPa

Calculating Safety Factor:

Safety Factor = uts = 415 = 60.7

? 6.84

ANSYS Data:

The following shows the readings obtained from using the ANSYS program where the minus values indicate compressive stress and positive values indicate a tensile stress:

Steel beam 1 (without weight)

Component

DOF (mm)

x

0.3595

y

-0.23998

Stress (MPa)

x

-4.082

y

-1.069

Elastic

x

-0.182E-04

y

0.616E-06

Steel Beam 1 (with weight)

Component

DOF (mm)

x

0.087303

y

-0.55238

Stress (MPa)

x

-8.102

y

-3.521

Elastic

x

-3.73E-05

y

-1.80E-04

Steel Beam 2 (without weight)

Component

DOF (mm)

x

0.32974

y

-0.21355

Stress (MPa)

x

2.815

y

-1.068

Elastic

x

1.34E-04

y

-5.46E-05

Steel Beam 2 (with weight)

Component

DOF (mm)

x

0.0843254

y

-0.52593

Stress (MPa)

x

6.844

y

-3.521

Elastic

x

3.26E-05

y

-1.80E-05

Comparing Results

The following table compares the stress results obtained through the classical approach with the results of using the ANSYS program:

Beam Weight

ANSYS result (MPa)

Analysis result (MPa)

No

-4.082

3.75

Yes

-8.102

7.77

No

2.815

2.81

Yes

6.844

6.84

Conclusion

Looking at the comparative table above we can see that the stress level values obtained through the classical approach in comparison to the ANSYS data are relatively similar in case 1 whereas, in case 2 they are practically identical. In the point load case 1, you can see that there is a slight difference between the values, with no weight considered there is a difference of 0.332MPa and with weight considered there is also difference of 0.332MPa.

We can establish in case 1 and 2 that the stress level significantly increases when the weight is considered therefore, taking this into account is a vital element and produces a more accurate and realistic value of stress. Looking at the ANSYS Data table we can see a y-component of the stress value, this is a computational error which contributes to the inaccuracy of using the FE method (ANSYS).

As we need to establish which method of mounting the screen would be safer, we need to look at the values where weight is considered only in both cases. In case 1, we can see that the stress levels are higher than in case 2. Also, in case 2 (weight considered) the safety factor is higher (60.7) than in case 1 (weight considered) therefore, the higher the value, the more safe the method of mounting is. In addition, the ANSYS data shows that the values of the DOF, strain and elastic are smaller for the distributed load where the weight is considered and not considered. Therefore, mounting the screen directly onto the beam would be the correct method.

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