Force in a Statically Indeterminate Cantilever Truss

BFC 31701

1.0

FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2

OBJECTIVE 1.1

To observe the effect of redundant member in a structure and understand the method of analysing type of this structure.

2.0

LEARNING OUTCOME 2.1

Aplication of engineering knowledge in practical aplication.

2.2

To enchance technical competency in structure engineering through laboratory aplication.

3.0

THEORY 3.1

In a statically indeterminated truss, static equilibrium alone cannot be used to calculated member force. If we were to try, we would find that there would be too many “unknows” and we would not be able to complete the calculations

3.2

Instead we will use a method know as the flexibility meethod, which uses an idea know as strain energy.

3.3

The mathematical approach to the flexibility method will be found in the most appropriate text books. 1 8 5

7

2 6 4

3

F

Figure 1 : Idealised Statically Indetermined cantilever Truss •

Basically the flexibility method usues the idea that energy stored in the frame would be the same for a given load wheather or not the redundant member whether or not.

•

In other word, the external energy = internal energy.

•

In practise, the loads in the frame are calculated in its “released” from (that is, without the redundant member) and then calculated with a unit load in place of the 1

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redundant member. The value fo both are combined to calculate the force in the redundant member and remaining members. •

The redundant member load in given by: P=

•

fnl

∑n l 2

The remaining member force are then given by: Member force = Pn + f Where, P = Redundant member load (N) L = length of members (as ratio of the shortest) n = load in each member due to unit load in place of redundant member (N) F = Force in each member when the frame is “release” (N)

•

Figure 2 shows the force in the frame due to the load of 250 N. You should be able to calculate these values from Experiment : Force in a statically determinate truss -250N

0

354N

250N

-500N

354N

250N

F=250N

Figure 2: Force in the “Released” Truss

•

Figure 3 shows the loads in the member due to the unit load being applied to the frame.

•

1

The redundant member is effectively part of the0 structure as the idealised in Figure 2 1

2 0

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Figure 3: Forces in the Truss due to the load on the Redundant members

4.0

PROCEDURE

3

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1. Wind the thumbwheel on the ‘redundant’ member up to the boss and hand – tighten it. Do not use any tools to tighten the thumbwheel.

2. Apply the pre-load of 100N downward, re-zero the load cell and carefully zero the digital indicator.

3. Carefully apply a load of 250N and check the frame is stable and secure. 4. Return the load to zero (leaving the 100N preload). Recheck and re-zero the digital indicator. Never apply loads greater than those specified on the equipment.

5. Apply loads in the increment shown in table 1, recording the strain readings and the digital indicator readings.

4

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6. Subtract the initial (zero) strain reading (be careful with your signs) and complete table 2. 7. Calculate the equipment member force at 250 N and enter them into table 3. 8. Plot a graph of Load vs. Deflection from Table 1 on the same axis as Load vs. deflection when the redundant ‘removed’. 9. The calculation for redundant truss is made much simpler and easier if the tabular method is used to sum up all of the “Fnl” and “n2l” terms. 10. Refer to table 4 and enter in the values and carefully calculated the other terms as required. 11. Enter your result in to Table 3.

5.0

RESULT Member strains (με)

5

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Load


1

2

3

4

5

6

7

8

Digital

(N)

Indicator reading (mm)

0 50 100 150 200 250

124 132 140 149 157 166

204 195 185 176 168 158

-33 -42

Table 1: Strain Reading and Frame Deflection

Member strains (με)

Load

1

2

3

4

5

6

7

8

0 13 26 41 53 65

0 5 -9 -13 -16 -21

0 -8 -17 -24 -32 -40

0 -13 -28 -38 -49 -62

0 5 10 16 21 25

0 -7 -14 -19 -24 -31

0 13 27 41 53 65

0 7 15 24 31 37

(N) 0 50 100 150 200 250

Table 2 : True Strain Reading

Member

Experimental Force (N)

Theoretical Force (N)

1 2

385.91

375.1 -124.9 -250 -374.9

3 4

-124.68 -237.481 -368.09

6

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5

148.43 -184.05 385.91 219.67

6 7 8

125.1 -176.9 354 177.1

Table 3: Measured and Theoretical in the Redundant Cantilever Truss

Member

Length

F

1 2 3 4 5 6 7 8

1 1 1 1 1 1.414 1.414 1.414

250 -250 -250 -500 0 0 354 354

n

Fnl

-0.707 -176.75 -0.707 176.75 0 0 -0.707 353.5 -0.707 0 1 0 0 0 1 500.56

Total

854.06

n2l

Pn

Pn + f

0.5 0.5 0 0.5 0.5 1.414 0 1.414

125.1 125.1 0 125.1 125.1 -176.9 0 -176.9

375.1 -124.9 -250 -374.9 125.1 -176.9 354 177.1

4.828

P = -Total Fnl Total n2l = -854.06 4.828 = -176.9 N Table 4: table for calculating the Forces in the Redundant Truss 6.0

CALCULATION 6.1

B

Calculation of real force, F

C

24cm

D 7

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A

E 24cm

m=8 j=5 r=3

24cm

250N

m = 2j – 3 8 > 2(5) – 3 8 > 10 – 3 8>7

So, the structure is statically internal indeterminate. CALCULATION FOR THEORETICAL FORCE (N) RAY RAX

A

1

E

8 5

2

7

1m

6 RBX

B

4

C

3

1m

D 250N

1m

∑MA =0 250 ( 2 ) - RBX = 0 500 - RBX = 0 - RBX = -500 RBX = 500N JOINT METHOD CALCULATION FBA

5

500 4

FBC 8

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MEMBER 4

MEMBER 5

∑Fx = 0 500 + FBC = 0 FBC = -500N

∑Fy = 0 FBA = 0 FBA = 0

FDE 7

FDC

D 3

MEMBER 3

250

MEMBER 7

∑Fx = 0 -FDE (1/1.414) – FDC = 0 - FDC - 354 (1/1.414) = 0 FDC = -250N

FCA FCA

∑Fy = 0 -250 + FDE(1/1.414) = 0 FDE = 354N

FCE

88

22

FCB

FCD 4

C

3 9

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MEMBER 2 ∑Fy = 0 250 + FCE = 0 FCE = -250N

FCA FCA Sin 45 FCA Cos 45

8

C FCB

4

MEMBER 8 ∑Fy = 0 - 250 + FCA (sin 45°) = 0 FCA = 250N Sin 45° = 354 N

500N

A

FAE FAC Cos 45 FAC Sin 45

FAB

FAC 10

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MEMBER 1 ∑Fx = 0 500 –FAC (Cos 45°) - FAE = 0 500 - 354 (Cos 45°) = FAE FAE = 250N

CALCULATION FOR n B

C 1

1 A

D

E

POINT A FAB 1 A

FAE

11

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Σfy = 0

Σfx = 0

FAB + (1 / 33.941) x 24 = 0

FAE + (1 / 33.941) x 24 –= 0

FAB = - 0.707N

FAE = - 0.707N

POINT B FBC

FBE

FAB

Σfy =0

Σfx =0

FAB – FBE (24 / 33.941) = 0 0.707 - FBE (0.707) = 0 FBE = 1

FBC + FBE (24 / 33.941) = 0 FBC + FBE (0.707) = 0 FBC = - 0.707

POINT C FBC

1

FCD FEC Σfx= 0

12

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FBC + (1 / 33.941) x 24 + FCD (24 /

Σfy = 0

33.941) = 0

-(1 / 33.941) x 24 – FCE – FCD = 0

0.707 + (- 0.707) + FCD (0.707) = 0

- 0.707 - FCE – 0 = 0

FCD = 0

FCE = - 0.707

POINT D

= Zero Bar

POINT E FEC FBE

FAE

FDE

Σfy = 0 (1 / 33.941) x 24 – FCE = 0 0.707 – 0.707 = 0 0 = 0 (CHECKING) Σfx = 0 FAE + FED – (1 / 33.941) x 24 = 0 0.707 + FED - 0.707 = 0 FED = 0

13

BFC 3051 TRUSS

6.3

FORCE IN A STATICALLY INDETERMINATE CANTILEVER GROUP 4

Calculation of Internal Forces AC P value, internal forces AC = P = = -

Σ F’nl / AE Σ n2 l / AE 854.06 4.828

= - 176.9 N (compression )

6.4

Example of calculation for member 1 (Pn + f) Given; Length, L = 1cm Force, F = 250N Load in each member due to unit load in place of redundant member, n = - 0.707N Area, A = πd²/4 = π(6)²/4 = 28.274 mm² = 0.283cm² Fnl = (250)(- 0.707)(1) = -176.75 N.cm n2l = (0.707) 2(1) = 0.5 cm Pn = (-176.9)(-0.707) = 125.1 Pn + f = 125 + 250 = 375 N = Theoretical Force (N)

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6.5

ε

250N

Calculation of Experimental Force (N)

(True strain reading for load 250N)

Strain for load 250N – 0N

0N 250N 250N

1 153 218

2 243 222

3 -27 -67

4 -48 -110

5 125 150

6 27 -4

7 22 87

8 8 45

65

-21

-40

-62

25

-31

65

37

Given; Area, A = πd²/4 = π (6²)/4 = 28.27mm² Modulus young, Esteel = 2.10 x 105 N/mm2 AE = 28.27 x 2.10 x 105 = 5.937 Member 1, F = AEε = (5.937 x 106) (65 x 10-6)

F = AEε = (5.937 x 106) (-62 x 10-6) =-368.09 N

=385.91 N Member 5, Member 2, F = AEε = (5.937 x 106) (-21 x 10-6)

F = AEε = (5.937 x 106) (25 x 10-6) = 148.43 N

= -124.68 N Member 6, Member 3, F = AEε = (5.937 x 106) (-40 x 10-6)

F = AEε = (5.937 x 106) (-31 x 10-6) = -184.05 N

= -237.481 N Member 7, Member 4,

F = AEε

BFC 3051 TRUSS

FORCE IN A STATICALLY INDETERMINATE CANTILEVER

= (5.937 x 106) (65 x 10-6) = 385.91 N

GROUP 4

F = AEε = (5.937 x 106) (37 x 10-6) = 219.67 N

Member 8,

6.0

DISCUSSION AND CONCLUSION

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1. From table 3, compare your answer to the experimental values. Comment on the accuracy of your result From the table 3, the experimental and theoretical forces are not accurate. We can see that there are huge difference value between experiment and theory. It is mean that, the accuracy of the result is not exact but for the compression and tension member, we can conclude that the following tension and compression is same with the value of the force is different. The experiment value is different compared to the theoretical value. Only member 2 for experiment value is near with theoretical value. It’s probably because of the error while setting the apparatus of the experiment

laboratory. Since the equipment is under the air conditioner, the

factor of wind and human mistakes will be taken in this experiment, so we can assume that the apparatus are not calibrated. 2. Compare all of the member forces and the deflection to those from statically determinate frame. Comment on them in terms of economy and safety of the structure. For the determinate truss it has extra member and for the indeterminate truss it has extra member. From that, for the determinate truss it safety for the structure however for the indeterminate truss it more safety for the structure. In the economy, the indeterminate truss will cost more expensive than determinate truss because the indeterminate truss has more member than the determinate truss. 3. What problem could you for seen if you were to use a redundunt frame in a “real life’ aplicatioin. (Hint: look at the zero value for the strain reading once you have included the redundant member by winding up thumnut).

BFC 3051 TRUSS


Redundant member is use to make beauty for the truss. When they are redundant in truss it will rising the cost because there are redundant member in that truss. 7.0

CONCLUSION From what we have learn and get form the experiment we can say that, the statically indeterminate structure can be classified if the equilibrium equations were not adequate to calculate the external reactions of all the internal forces. However there are pro and contra in this statically indeterminate analysis. The advantages of statically indeterminate are: •

The maximum stresses and deflections are smaller than statically determinate counterpart.

•

Can support loads with support loading on thinner members with increased stability.

•

Have a tendency to redistribute its load to redundant supports in the case of faulty design or overloading occurs.

The disadvantages of this truss are: • Redundant structures can induce problems such as differential displacement •

High cost of statically indeterminate structure compare to determinate structure.

REFERENCES 1. Yusof Ahmad(2001), “Mekanik Bahan dan Struktur”: Universiti Teknologi Malaysia,Skudai, Johor Darul Takzim

BFC 3051 TRUSS

FORCE IN A STATICALLY INDETERMINATE CANTILEVER

2. Bambang Prihartanto(2008), “Structural Analysis”:Universiti Tun Hussein Malaysia. 3. “Structural Use Of Steelwork In Building”:British Standard. 4. Plane Frame Example – http:// www.amazon.com

GROUP 4

Force in a Statically Indeterminate Cantilever Truss

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