BACHELOR OF CIVIL ENGINEERING WITH HONOURS STRUCTURAL LABORATORY DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING FACULTY OF CIVIL ANG ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN O...
force in statically determinate truss
Structural analysis
ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES.pptx
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FORCE IN A STATICALLY INDETERMINATE CANTILEVER BEAM 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.
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 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 as required. 11. The result was enter in Table 3
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 FBE sin 45 = 0.707 FBE = 1
FBC + 1 cos 45 = 0 FBC = -0.707
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
DISCUSSION 1. From table 3, compare your answer to the experimental values. Comment on the accuracy of your result
Refer to the table 3, the value in experimental force were different with the theoretical value. It maybe because of parallax error with the equipment not going well while experiment and also the error comes from the environment in the lab where the equipment were sensitive with wind and vibration
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.
There have positive and negative force with tension and compression at all member. Some structures are built with more than this minimum number of truss members. Those structures may survive even when some of the members failure or deflection. This is because their member forces depend on the relative stiffness of the members to equilibrium condition described. These can be economy for structure. Failure occurs when the load (L) effect exceeds the ability (R) of the structure. The safety of the structure is should be R>L. 3. What problem could you for seen if you were to use a redundant frame in a “real life” application.
The structure will be failed if the loads are more than the ability. By my knowledge, the redundant frame always we can see in a bridge construction. The reason of that are, redundant frame is used to give the stability of the bridge structure also when we added redundancy in the structural system, there is an increase in the overall factor of safety and the redundant frame are useable for aesthetic value of the structure.
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
REFERENCES
1) Book Analysis Structure Module
Nota Mekanik Bahan dan Struktur ( Writer: Yusof B. Ahmad)
Force in Statically Determinate Cantilever Truss Report
2) Internet http://www.engineering.uiowa.edu/~design1/StructuralDesignII/Chapter5ForceMethod.pdf
