1204615561 2007 Engineering Studies Assessment Task

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HSC ENGINEERING STUDIES

ASSESSMENT TASK 1

CIVIL STRUCTURES

ANDREW HARVEY





REPORT BODYTruss Design:We used the first truss design, (See Appendix A & B). We chose this one because it had thelowest maximum force in any member, and therefore would take more load than designnumber two.

Analysis of Forces in all Members:F

gb

3

F ghbh gb fg ef deaead ========

32

F dcch ==

32 F cf =

Breaking Stress in First Member to Fail:The member with the most force in it is cf . So we predicted that it would fail first.

The actual weight that caused it to fail was 35kg. So the tensile strength of 6mm square balsais 396N. However this would not be true if the first member to fail was in compression, or ifthe truss failed first at a joint.

The first member to fail was unclear. Many members broke, and we couldn’t see whichmembers failed first. The position of breakages is shown below.

F2

ad

dc

de

ae

F2

cf

bhfgghef

ch

__________________________________________________________________________HSC Engineering Studies – Assessment Task 1 3 of 14





Alternate Height:

We also considered changing the height of the truss. By use of method of section the lowermiddle member had a force of 132.8N.

35 kg

171.5N

A

M

As shown by the section line, we ignore the parts on the left.

By taking moments about point A, we have 3 external forces acting on the truss. With onlyone unknown of the cental lower member, M.

( ) (( )Tension8.132

30.3875.1713000M

←=

×+×−==+→∑ N M

M A

)

This is a less force compared with our actual design having 396.06N.

There was no height limit, so we could have made the truss higher, this would mean that itcould have held a greater load before failure.




APPENDIX ADesign 1 (Warren Truss):

Free Body Diagram:

60˚

60˚

60˚

60˚ 60

˚

60˚

60˚

60˚

60˚

60˚

60˚

60˚

60˚

60˚

60˚

RL and R R can be calculated now. By looking at it you can see that R L =2

F and R R =

2 F

.

However for the sake of accuracy, these forces can be calculated, as shown below.

( ) ( )

( ) ( )

↑=

×−×==

↑=×=

×+×−==

+→

+→

∑

∑

2R

3.0R 6.00M

26.03.0

R

3.0R 6.00M

L

LR

R

R R

R

L

F

F

F F

F

The internal stresses can also be calculated.

ADC:

Close up of Joint ADC The R L force is known in both magnitude and

direction, so it is drawn in first. The forces inmembers’ ad and dc are known in direction but notmagnitude or sense. However, the sense can be

ad

dc

F2

F2

ad30

˚

dc




obtained from the above force diagram, along withthe magnitudes.

330cos2 F F

ad =°

÷= 60˚

→=°=32

30tan2

F F dc

ADE:We know the forces in member ad, so the forces in members de and ae can be calculated.

Close up of Joint ADE The forces in member ad are known in both magnitude anddirection, so it is drawn in first. The forces in members’ deand ae are known in direction but not magnitude or sense.However, the sense can be obtained from the above forcediagram, along with the magnitudes.

de and ae are easily resolved. It is an equilateral triangle because all the internal angles are60

˚

, so ad = de = ae.

de

aeSo far the forces in the truss are as follows:

CDEF:There are 4 members at joint CDEF. two of which are unresolved.

F

F2

ad

dc

de

ae

F2

adae

de

ad de

ae

60˚

60˚




Close up of Joint CDEF. The forces in members dc and de are known in bothmagnitude and direction, so they is drawn in first. The forcesin members’ ef and cf are known in direction but notmagnitude or sense. However, the sense can be obtainedfrom the above force diagram, along with the magnitudes.

To resolve cf and ef, I have split this quadrilateral into two triangles. One of them is shownbelow.

dede

cf ×=°

= 260cos

It can be seen that as the polygon is symmetrical ef = de.

So now the forces in the truss are as follows:

Final Analysis As the truss is symmetrical, and the force is applied centrally, then we can assume the sameforces for the other side of the truss.

(All triangles are equilateral (60˚

angles), and all members are 200mm.)

Where: (as previously calculated) (magnitudes only)

F

F2

ad

dc

de

ae

F2

cf

ef

ch

gb

fggh

bh

F

F2

ad

dc

de

ae

F2

cf

ef

dc

ef

cf

de

120˚

cf

120˚

60˚

60˚

efde

dc

cf30

˚

60˚

de n




3

F ghbh gb fg ef deaead ========

32

F dcch ==

32 F cf =

So with these results, I know which members are in tension and which are in compression(from the above diagram), and I can also work out the forces in each member given differentmasses applied at force F.

Now I could see how the truss would behave. I tested it with a 50kg mass applied centrally atthe top. The forces I calculated are shown on the diagram below.

(50kg) 490N

282.9N 282.9 N

As there are 11 members, each being 0.2m long. The total amount of balsa used in thisdesign is 2.2m. 3m of balsa weights 100g, so the 2.2m that was used in our truss designweighed 75g.

(25kg) 245N

122.5 N

70.7 N 282.9 N 70.7 N

141.5 N

122.5 N

141.5 N

141.5 N

141.5 N

141.5 N

141.5 N

141.5 N141.5 N

(100kg) 980N

282.9 N 1131.6 N 282.9 N

565.8N 565.8 N

565.8 N

565.8 N

565.8 N

565.8 N

565.8 N565.8 N

490N 490 N

245N

141.5 N 565.8 N 141.5 N

282.9 N282.9 N282.9 N 282.9 N

282.9 N282.9 N

245N




APPENDIX BDesign 2 (Warren Truss):

A F B

D F HE G

JI

LK

Where all triangles are equilateral.

F2

120

C F2

The forces in members ad, dc, ae, de, ef and cf have been calculated in design 1, and theyare the same for this truss design. So the following forces are known.

F

AEFGThere are 4 members at joint AEFG. two of which are unresolved.

Free Body Diagram Force DiagramClose up of Joint AEFG.

ef fg =

3

2 F ag =

CFGHThere are 4 members at joint CFGH. two of which are unresolved.

ae ag

fgef120

˚

120˚

60˚

60˚

ag

ef fg

ae

F2

ae

dc cf

deadef

F2




Free Body Diagram Force Diagram

3

3 F ch

fg gh

=

=

As the truss is symmetrical I can fill in the rest of it.

Final Analysis

ch F

cjbiag cf F

cl dc F

bk bl kl jk ijhi gh fg ef deaead F

=

====

==

============

3

3

3

2

32

3

Case 1 (25kg)141N71N283N424N

cf

gh

ch

fg

120˚

ch

120˚

60˚

60˚

ghfg

cf

F

F2

dc cf

ae

deef

ad

F2

fg gh hiij

kkl bl

bkbiag

ch cj cl




APPENDIX C – JOINTSThe joints are shown here. When we decided upon the joints we took into account whetherthe members were in tension or compression. We gave members in tension the greater

surface area of gluing.

ADC

CDEF

CFGH

ad

dc

de ef

dc fc

fg gh

fc hc




CHB

BGH

AEFGB

hb

ch

bg

bh

gh

ae bg

effg




ADE

de

ae

ad


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1204615561 2007 Engineering Studies Assessment Task