force in a statically indeterminate cantilever beam ummu

8/11/2019 Force in a Statically Indeterminate Cantilever Beam Ummu

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FORCE IN ST TIC LLY INDETERMIN TE

C NTILEVER BE M

OBJECTIVE

To observe the effect of redundant member in method of analysing type ot this structure

LEARNING OUTCOME

Application of engineering knowledge in practical application.

To enhance technical competency in structure engineering through laboratory

application.

THEORY

In a statically indeterminate 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 “unknowns” and we would not be able to complete the calculations.

Instead we will use a method known as the flexibility method, which uses an idea

know as strain energy.The mathematical approach to the flexibility method will be found in the most

appropriate text books.

Figure 1 : Idealised Statically Indetermined Cantilever Truss



Basically the flexibility method uses the idea that energy stored in the frame would be

the same for a given load wheather or not the redundant member .

In the other words, 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 redundant

member. The value for both are combined to calculate the force in the redundant

member and remaining members.

The redundant member load in given by :

P =

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 die 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

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.

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

Figure 3 : forces in The Truss due to the load on the redundant members



PROCEDURE

1. Wind the thimbweel on the „redundant‟e thumbwheel member up to the boss and

hand tighen it. Do not use any tools to tighten thumbwheel.

2.

The preload was apply 100 N downward, re- zero the load cell and carefully zero

the digital indicator.

3.

The load of 250 N was apply carefully and the frame was check is stable and secure.

4. The load was return to zero ( leaving the 100 N preload ). Recheck and re- zero the

digital indicator. Never apply loads greater than those specified on the equipment.

5.

The was load was apply in the increment was shown in the table 1, the strain readings

and digital indicator readings was records.

6.

Substract the initial (zero) strain reading (be careful with your signs) and complete

table 2.

7. The equipment member force at 250 N was calculate and enter them into the table 3.

8. The graph of load vs deflection was plot from table 1 on the same axis as Load vs

Deflection when the redundant „removed‟.

9. The calculation for the 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 asrequired.

11. The result was enter in Table 3



RESULT

Load

(N)

Strain Reading Digital

Indicator

Reading

(mm)1 2 3 4 5 6 7 8

0 143 229 -24 -41 88 14 20 23 -0.098

50 155 225 -32 -55 92 8 32 30 -0.123

100 168 220 -41 -69 96 1 45 38 -0.143

150 181 215 -50 -83 101 -6 58 45 -0.167

200 195 210 -59 -97 104 -12 71 52 -0.192

250 207 206 -67 -110 108 -19 83 59 -0.211

Table 1: Strain Reading and Frame Deflection

Load

(N)

1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 0

50 12 -4 -8 -14 4 -6 12 7

100 25 -9 -17 -28 8 -13 25 15

150 38 -14 -26 -42 13 8 38 22

200 52 -19 -35 -56 16 -26 51 29

250 64 -23 -43 -69 20 -33 63 36

Table 2 : True Strain Reading



Member Experimental Force (N) Theoretical Force (N)

1 307.78 -375.06

2 -110.61 124.94

3 -206.79 250

4 331.82 -625.06

5 96.18 -125.06

6 -158.70 176.89

7 309.97 354

8 173.12 530.89

Table 3: Measured and Theoretical in the Redundant Cantilever Truss

Member Length F n Fnl n l Pn Pn + f

1 1 -250 -0.707 -176.75 0.5 - 125.06 - 375.06

2 1 250 -0.707 -176.75 0.5 -125.06 124.943 1 250 0 0 0 0 250

4 1 -500 -0.707 353.5 0.5 -125.06 -625.06

5 1 0 -0.707 0 0.5 -125.06 -125.06

6 1.414 0 1 0 1.414 176.89 176.89

7 1.414 354 0 0 0 0 354

8 1.414 354 1 500.56 1.414 176.89 530.89

Total 854.06 4.828

Table 4 : Table for calculating the force in the redundant truss

P =

=

= 176.89 N



Calculate For Experimental Force ( Load = 250 N )

Load

(N)

1 2 3 4 5 6 7 8

0 143 229 -24 -41 88 14 20 23

250 207 206 -67 -110 108 -19 83 59

250 64 -23 -43 -69 20 -33 63 36

Given

Area, = = 22.9 mm2

Esteel = 2.10 x 105 N/mm2

AE = 22.9 x 2.10 x 105 = 4.809 x 106

Member 1

F = AEɛ

= (4.809 x 106 )(64 x 10-6)

= 307.78 N

Member 2

F = AEɛ

= (4.809 x 106 )(-23 x 10-6)

= -110.61 N

Member 3

F = AEɛ

= (4.809 x 106 )(-43 x 10-6)

= -206.79 N



Member 4

F = AEɛ

= (4.809 x 106 )(-69 x 10-6)

= 331.82 N

Member 5

F = AEɛ

= (4.809 x 106 )(20 x 10-6)

= 96.18 N

Member 6

F = AEɛ

= (4.809 x 106 )(-33 x 10-6)

= -158.70 N

Member 7

F = AEɛ

= (4.809 x 106 )(63 x 10-6)

= 309.97 N

Member 8

F = AEɛ

= (4.809 x 106 )(36 x 10-6)

= 173.12 N



Calculation For Theoretical Force (250 N)

ƩMb = 0

-HA (0.245) + 250 (0.49) = 0

-HA (0.245) = -122.5

-HA = -500

HA = 500 N

ƩHx = 0

HA = HB

HB = 500 N

ƩHY = 0

R B = 250 N

0.245 m 0.245 m

0.245 m



Joint A

FAB

500

A FAE

Ʃ FY = 0 Ʃ FX = 0

FAB = 0 FAE + 500 = 0

FAE = -500

Joint B

250

500 FBC

FBE COS 45

FBE SIN 45 FBE

FBA

Ʃ FY = 0

FBE sin 45 + 0 = 250

FBE = 353.55 N

Ʃ FX = 0

FBC + FBE cos 45 = 500

FBC + 353.55 cos 45 = 500

FBC = 500 – 250

FBC = 250 N



Joint E

FEC

FEB (353.55)

FEA (500) FED

Ʃ FY = 0

FEC + 353.55 sin 45 = 0

FEC = -250 N

Ʃ FX = 0

FED = -500 + 353.55 cos 45

FED = -250 N

Joint D

FDC

FDE (500) D

250

Ʃ FY = 0

FDC sin 45 = 250

FDC = 353.55 N

Ʃ FX = 0

FDE +353.55 cos 45 = 0

FDE = -250 N



Calculation for n

Member A

1 sin 45 + FAB = 0 FAE = - 1 cos 45

FAB = - 0.707 FAE = - 0.707

Member B

FBE sin 45 – 0.707 = 0 FBC + 1 cos 45 = 0

FBE sin 45 = 0.707 FBC = -0.707

FBE = 1



Member c

FCD cos 45 = - 0.707 + 1 cos 45 FCE + 0 sin 45 + 1 sin 45 = 0

FCD cos 45 = - 0.707 + 0.707 FCE = - 0.707

FCD = 0

Member D = 0

Member E

FEB cos 45 – 0.707 = 0 FEC + 1 sin 45 = 0

FEB cos 45 = 0.707 FEC = - 0.707

FEB = 1



CONCLUSION

For this experiment, to evaluate the data from the trusses, we use the different loads starting

with 50N, 100N, 150N, 200N and last 250N. The most important of these criteria is the

structure will able to carry load safely. The limit load for this experiment is 350N. The result

to evaluate of structural safety can only be done mathematically and the experimental force

data was collected from digital reading of equipment. And the value of experimental is

compared with the theoretical force value that be done manually as we studied in analysis

structure module.

Mostly the data that we get from the digital reading is different with the data that we calculate

manually because of the parallax error of the equipment. The equipment is not in a good

condition while we do the experiment. So that, in real life, it will be unsafe effect for the

structural engineer to evaluate a bridge design by a full-size prototype.



APPENDIX

force in a statically indeterminate cantilever beam ummu

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