BTEC HNC - Science - Analyse Static Engineering Systems

8/2/2019 BTEC HNC - Science - Analyse Static Engineering Systems

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Analyse Static Engineering

SystemsEngineering ScienceBy Brendan Burr


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Brendan Burr BTEC Higher National Certificate in ElectronicsAnalyse Static Engineering Systems

Table of Contents

TABLE OF CONTENTS ........................................................... 2

TASK 1 ................................................................................ 3

Solve Shear Force and Bending Moment problems ................................ 3Solution:- ..............................................................................................................3Shear Force Diagram:- ..........................................................................................7Bending Moment Diagram:- ..................................................................................9Check:- ...............................................................................................................10

TASK 2 .............................................................................. 12

Select standard rolled steels sections for beams ................................... 12Solution:- ............................................................................................................12Shear Force Diagram:- ........................................................................................15Bending Moment Diagram:- ................................................................................16Check:- ...............................................................................................................17

Shear Stress and the Angular Deflection due to Torsion in Circular Shafts......................................................................................................... 19

Solution:- ............................................................................................................19Check:- ...............................................................................................................21

EVALUATION ..................................................................... 22

CONCLUSION ..................................................................... 22

Books ................................................................................................ 23

Catalogues ......................................................................................... 23

Websites ............................................................................................ 23

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Task 1

Solve Shear Force and Bending Moment problems

Horizontal Beam ABCD, 13m long, is simply supported at B and D.AB = 3m, BC = 2m, and CD = 8m.Concentrated load of 4kN and 7kN act at A and C respectively, and auniformly distributed load of 1kN/m extends from C to D.Neglecting the effect of gravity on the mass of the beam, sketch theShearing Force and Bending Moment Diagrams for the beam, insertingprincipal values, and determine the position and magnitude of themaximum bending moment.

Solution:-

Not To Scale

Anticlockwise Moments = Clockwise MomentsTaking Moments about D:

( ) ( ) ( )

B

B

B

=

=++=++

10

140

10525632

104137884

kNB 14=

Upwards Forces = Downwards Forces

( )

14874

817414

++=++=+

D

D

kND 5=

3


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Bending Moment Calculations:

( )

( )

( )( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) kNBM

kNBM

kNBM

kNBM

kNBM

kNBM

kNBM

kNBM

kNBM

kNBM

kNBMkNBM

kNBM

kNBM

m

m

m

m

m

m

m

m

m

m

m

m

m

m

02

81871014134

5.42

7177914124

82

6

167814114

5.102

5157714104

122

414761494

5.122

313751484

122

212741474

5.102

111731464

82

010721454

211444

1201434824

414

004

2

13

2

12

2

11

2

10

2

9

2

8

2

7

2

6

2

5

4

3

2

1

0

=

+++

+=

+++

+=

+++

+=

+++

+=

+++

+=

+++

+=

+++

+=

+++

+=

+++

=+=+

=

==

Therefore the maximum bending moment is positioned at 8 meters and has a

magnitude of +12.5 kN/m.

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Bending Moment from 5 meters to 13 meters:-

( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) kNmx

kNmx

kNmx

kNmx

kNmx

kNmx

kNmx

kNmx

kNmx

xxBM

088385.0

5.487375.0

886365.0

5.1085355.0

1284345.0

5.1283335.0

1282325.0

5.1081315.0

880305.0

835.0

2

8

2

7

2

6

2

5

2

4

2

3

2

2

2

1

2

0

2

=++

+=++

+=++

+=++

+=++

+=++

+=++

+=+++=++

++=

Differentiate:

1=

=

n

n

naxdx

dy

axy

x

x

xxdx

BM

=

+=

+=

1

3

310

315.02 1112

3=x Meters

( ) ( ) 83335.0 2 ++=MAXBM

5.12=MAX

BM kNm

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Shear Force Diagram:-

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8


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Bending Moment Diagram:-

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Check:-

For the check I entered appropriate values into an online calculator for simplysupported beams.

Location of left support,

xS1m

Location of right

support, xS2

m

Point load #1, P1 N

Location of pointload #1,

xP1

m

Point load #2, P2 N

Location of load #2, xP2 m

Couple load #1, C1 N.m

Location of couple #1,

xC1

m

Couple load #2, C2 N.m

Location of couple #2,

xC2

m

Distributed load #1, left-

hand-side, W1l

N/m

Left hand side locationof load #1, xW1l

m

Distributed load #1,

right-hand-side, W1r

N/m

Right hand side location

of load #1, xW1r

m

Distributed load #2, left-

hand-side, W2l

N/m

Left hand side location

of load #2, xW2l

m

Distributed load #2,

right-hand-side, W1r

N/m

Right hand side location

of load #2, xW1r

m

This then returns values which correspond to the calculations I made.

Variables Values Units Maximum Shear force, Vmax 10000.0N

Location of Vmax 5.0m

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Maximum Bending moment, Mmax 12500.0N.m

Location of Mmax 8m

Reaction force, R1 14000.0N


The Shear Force and Bending Moment Diagrams are also the same as thosecreated from my calculations above.

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Task 2

Select standard rolled steels sections for beams

A beam, 5m long, simply supported at its ends, and carries a uniformlydistributed load of 15kN/m over its entire length, together with aconcentrated load of 30kN at a point 1m from the left-hand support. Ifthe maximum stress allowable in the material is 100MN/m, select astandard rolled steel section from the tables.

Not To Scale

Solution:-

Anticlockwise Moments = Clockwise Moments

Taking Moments about A:

( ) ( )

C

C

C

=

=+=+

5

5.217

55.18730

55.275130

kNC 5.43=

Upwards Forces = Downwards Forces

( )

5.437530

515305.43

+=+=+

A

A

kNA 5.61=

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Shear Force Calculations:-

kNSF

kNSF

kNSF

kNSF

kNSFkNSF

kNSF

kNSF

m

m

m

m

m

m

m

m

05.434.43

5.43155.28

5.28155.13

5.13155.1

5.1155.165.16305.46

5.46155.61

5.615.610

5

99.4

4

3

2

1

99.0

0

=+=

==+=+=

+=+=+

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Bending Moment Calculations:-

( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) kNmx

kNmx

kNmx

kNmx

kNmx

xxBM

055.755.3130

3645.745.3130

5735.735.3130

6325.725.3130

5415.715.3130

5.75.3130

2

5

2

4

2

3

2

2

2

1

2

=+

+=+

+=+

+=+

+=+

+=

Differentiate:

1=

=

n

n

naxdx

dy

axy

x

x

xxdx

BM

=

+=

+=

15

5.31

5.31150

5.315.72 1112

1.2=x Meters

( ) ( ) 301.25.311.25.7 2 ++=MAXBM

075.63=MAX

BM kNm

36

6

3

1075.630

10100

10075.63

cm

M

y

IM

=

=

Therefore from the BS5950-1:2000 Table I have selected Serial Size:

305 x 165

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Shear Force Diagram:-

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Bending Moment Diagram :-

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18/23


This then returns values which correspond to the calculations I made.

Variables Values Units

Maximum Shear force, Vmax 61500.0N

Location of Vmax 0.0m

Maximum Bending moment, Mmax 63000.0N.m

Location of Mmax 2.1m



The Shear Force and Bending Moment Diagrams are also the same as thosecreated from my calculations above.

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Task 3

Shear Stress and the Angular Deflection due to Torsion in Circular Shafts

A hollow shaft has an external diameter of 150mm and an internaldiameter of 100mm. If the maximum shear stress is not to exceed50MN/m, determine the torque, the maximum power that could betransmitted by the shaft when rotating at 180 rev/min, and the angle oftwist (in degrees) over a 5m length when transmitting this power. Themodulus of the material is 80GN/m.

Solution:-

L

G

RJ

T ==

( )( )

mmGN

mmmMN

dD

Torque5/80

75/50

32

22

44

==

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )

075.0

994175024.1

075.0

10501025.40632

075.0

10501010010150

32

1075

/105010015032

36

34343

3

2344

=

=

=

=

T

T

T

m

mNmmmm

T

58900032.26=T Nm

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J

T

L

G=

( )

( )

6

6

6

6

1080

56665.666666

6665.6666665

1080

1088350049.39

58900032.26

5

1080

=

=

=

60416666666.0= Radians

18060416666666.0

180

=

=

= 387324146.2

sRAD

revrev

/6

623sec/3min/180

=

=

TPower =

658900032.26 =Power

1908484.501=Power Watts

Therefore we see that:

Total Torque (T) = 26.58900032 NmPolar Second Moment Of Area (J) = 39.88350049 x 10-6 m4Shear Stress ( ) = 50 x 103 N/m2

Radius (R) = 75 x 10-3 mModulus Rigidity (G) = 80 x 106 N/m2

Angle Of Twist ( ) = 2.387324146 o OR 0.04166666666 RadiansRadial Line Of Length (L) = 5 mTransmitted Power (P) = 501.1908484 Watts

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Check:-

( )6666.666666

5

60416666666.01080

6667.6666661075

1050

6665.6666661088350049.39

58900032.26

6

3

3

6

=

=

=

=

=

=

L

G

R

J

T

From this we can see that all values must be correct due to the consistency ofthe answers derived from the Engineers Theory of Torsion Formulae.

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Evaluation

I can honestly say that I found this assignment difficult to grasp. Now I knowhow to calculate the Bending Moments I can easily work out the maximum

Bending Moment and the position of it.In Task 1, I could work out the Reactant Forces without any difficulty, so wasable to move straight on to working out the Bending Moments. This is where Ibegan to have trouble, with the first 5 meters I didnt have any problems, itwas when I had to work out the uniformly distributed load for the last 8 meters.I was unable to find any notes that I could put into practice, because of writingit down incorrectly. This meant I had to do research into it, which took hoursafter I had managed to decipher all of the relevant information.After attaining all of the appropriate points I created a Shear Force Diagramand a Bending Moment Diagram, by using Graphmatica. This is a very goodpiece of software once you know how to use it, and by simply typing in theBending Moment equation you can create the required parabolic curve whichis formed under a UDL.To check my answer I found an online Simply Supported Beam Calculator.By entering the values taken from the question I could compare my answer tosee if there was any deviation. Fortunately for me there wasnt whichstrengthens conclusion that Task 1 is correct.Task 2 is a similar question to Task 1, however it involved a bit of furtherinvestigation into the calculation of the Engineers Theory of Bending Formula.I had already been given the Maximum Stress value in the question, all I needto do was work out the Maximum Bending Moment and then search for the

appropriate beam using the BS5950-1:2000 Table. This question has taughtme how engineers calculate which materials are suitable for the requiredapplication.I checked this answer in the same way as Task 1.Task 3 was mainly a test on ensuring the correct values are entered into thecalculation. There were values in various forms such as Meters, Centimetres,Millimetres, Giga-Newtons, Mega-Newtons and Kilo-Newtons. This gave wayfor the potential of radical inaccuracies. My check involved entering thevalues attained into the Engineers Theory of Torsion, whilst it had beenbroken down. Each of the three fractions had to equal the same value, whichthey did.

Conclusion

I feel this assignment pushed me to work the answers out for myself. It hasbeen a difficult challenge and involved a lot of research and investigation.I am very pleased with the outcome, and a lot of time and effort has gone intothe presentation of this assignment.I will be taking thorough notes in class from now on to ensure the nextassignment flows much more smoothly. I plan to work just as hard on thenext assignment as I have done on this.

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Bibliography

Through guidance from my lecturer, the following text books, catalogues andwebsites I was able to complete this assignment:

BooksHigher Engineering Mathematics (John Bird)ISBN: 0-7506-8152-7

Higher National Engineering (Mike Tooley & Lloyd Dingle)ISBN: 978-0-7506-6177-5

Catalogues

N/A

Websites

http://www.mechengcalculations.com/jmm/beam28_process.jsp

http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-405-shear-and-moment-diagrams
http://www.mechengcalculations.com/jmm/beam28_process.jsphttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-405-shear-and-moment-diagramshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-405-shear-and-moment-diagramshttp://www.mechengcalculations.com/jmm/beam28_process.jsphttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-405-shear-and-moment-diagramshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-405-shear-and-moment-diagrams

BTEC HNC - Science - Analyse Static Engineering Systems

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