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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
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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
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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
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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 pre- load 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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APPENDIX
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