MAE Lab 7 - Internal Forces in Truss Members PDF

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Internal Forces in Truss Members...

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1 Hla Myat Kyaw Professor Yves Ngabonziza MAE 101 Lab 7 Title: Internal Forces in Truss Members INDEX

ABSTRACT

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INTRODUCTION

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OBJECTIVE

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MATERIAL

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EXPERIMENTAL PROCEDURE

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RESULTS

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Table 1: Data table of the experiment with the values of ƩF

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Calculations

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DISCUSSION

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CONCLUSIONS

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REFERENCES

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Abstract

2 A truss is a structural system formed by members that can carry loads in the axial direction of the members but cannot carry loads in the transverse direction, that is they have no resistance to bending. In this experiment, the determination of force that was developed in beams placed in axial direction was shown. A wire was used to build the truss structure, and additional load was applied to it. Two wire terminals were held by gauges, and they were attached to stands. The force on the wire was measured using the gauges. The right triangle rule was applied to calculate the equilibrium equation. The conditions for equilibrium will be discussed and verified.

Introduction When a truss structure is subjected to external loads, the members of the truss will try to transmit these loads to the supports. In doing so, internal forces will develop in each truss member. These internal forces may be a function of the loads applied, the inclination of the members, as well as the properties of the members. Provided the loading is coplanar with a cable, the requirements for equilibrium are formulated in an identical manner. Consider, for example, the cable as shown in the schematic diagram below, where the distances L1 and L2 and the load W are known. The problem here is to determine the seven unknowns consisting of the tensions P and Q in the two segments, the four components of reactions at A and B and the sag yc at the point C. Here for equilibrium, Σ F = 0. This equation requires that Σ F y = 0 and Σ Fx = 0. Therefore, the sum of forces in the vertical direction yield PcosӨ1 = QcosӨ2 = W and the sum of forces in the horizontal direction yield PsinӨ1 = QsinӨ2.

Objective

3 The objective of this experiment is to illustrate how internal forces develop in a simple truss structure.

Materials 1. Cables or wires

2. Two hanging scales 3. Two support stands

4. Test weights

or force gauges 5. Weight hangers

6. Protector

7. Rulers

Experimental Procedures To reduce experimental errors, extra cautions are exercised when measuring the angles suspended by the cable about an imaginary vertical axis separating the cable segments at loading point. A weight was attached by two wires and the two wires were attached to two force gauges. The wires kept vertical and the forces on the gauges were recorded. The two readings add up to the weight applied. The wires were hung to the two stands and a 2-lb weight was applied. The angles subtended by the cables were measured. The angle between the vertical stand on the left and the wire attached to it will be denoted as the angle Ө 1. The angle between the vertical stand on the right and the wire attached to it will be denoted as Ө 2. The force gauges measure the internal forces introduced in the inclined cables when the vertical weight W is applied. The force in the right-hand gauge named by Q and the value in the left-hand gauge was named by P. The process was repeated once more for different values of angles Ө1 and Ө2.

Results Table 1: Data table of the experiment with the values of ƩF

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Case

Θ1

Θ2

P

Q

1

32.34˚

49˚

1.09

1.02

2

25.77

47.35

1.08

0.15

Calculations: Weight W = 2 lb. = 2/2.205 kg ≈ (2/2.205) ×9.8 N ≈ 8.91 N Case I ƩFx (N) : P sin Ɵ1 – Q sin Ɵ2 = 1.09 (sin 32.34) – 1.02 (sin 49) = 0.2466 N ƩFy (N) : P cos Ɵ1 + Q cos Ɵ2 – W = 1.09 (cos 32.34) + 1.02 (cos 49) – 8.91 = -8.66 N Case II ƩFx (N) : P sin Ɵ1 – Q sin Ɵ2 = 0 =1.08 (sin 25.77) – 0.15 (sin 47.35) = 0.359 N

ƩFx (N)

ƩFy (N)

0.2466

-8.66

0.359

-7.836

5 ƩFy (N): P cos Ɵ1 + Qcos Ɵ2 – W = 0 =1.08 (cos 25.77) + 0.15 (cos 47.35) – 8.91 = -7.836 N Discussion In the equilibrium state, all the force components cancel each other at the horizontal and vertical axes. So, on that stage, ƩFx = 0 and ƩFy = 0. The values for each case are close to zero or below zero. Thus, the values for case 1 and 2 are close to prove the law of equilibrium that ƩM = 0 and ƩF = 0. But because of the values of ƩFy is negative in both cases, there were errors, prior to the recording the values on the date sheet, the experimental values and tried to obtain ƩF = 0 for each trial needed to calculate. Then the problem and solution might be predicted and calculated.

Conclusion In this experiment, the advantages and disadvantages of working with different kinds of beams, and different positions were observed. With different combinations of the different sizes of beams in truss projects could reduce the cost of the construction.

References

Benenson, Ganatos, Ghosn. Activities and Assignments for ENGR 10100: Engineering Design Freshman Manual. Third Edition. 2006.

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