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Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Failure Analysis of a Fin Design for.

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Presentation on theme: "Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Failure Analysis of a Fin Design for."— Presentation transcript:

1 Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Failure Analysis of a Fin Design for the Micro-Mechanical Fish By Michael Petralia Wood MicroRobotics Laboratory Fin Design by Kyla Grigg December 12, 2006

2 Design of the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo GoalPrototype Design The final design will look like a fish fin, but to prototype the mechanical system, square elements are being used.

3 Design of the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Brass Plate Carbon Fiber Tendon Glass Fiber Laminas Silicone Rubber Shape Memory Alloy (SMA)

4 Forces acting on the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo F SMA F Tendon F Rubber F SMA

5 Forces acting on the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo θ min F SMA θ min = 6.5 o Assuming the rubber stretches at most 1 mm, The SMA’s provide a maximum pull of, F SMA = 1.47 N The horizontal force from the SMA to the rubber will be, F rubber = 2F SMA sin(θ min ) = 0.333 N Case 1: Maximum Deflection Assuming this load is equally distributed over the right edge of the fin, F rubber / A = 0.2641 N/mm 2

6 Forces acting on the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo F SMA The maximum angle is, θ max = 22 o F max = F rubber = 1.10 N Though this would mean the fin has not moved, let’s take this as our worst case scenario. The maximum horizontal force from the SMA will be, Again, assuming this load is equally distributed over the right edge of the fin, F rubber / A = 0.8740 N/mm 2 Case 2: Maximum Force θ max

7 Forces acting on the fin Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo The force in the tendon is the force necessary to keep the fin in place during swimming. It will be the horizontal force from the SMAs for a given angle. Because the deflection will be relatively small, it should be safe to assume the force in the tendon will act parallel to the surface of the fin. Although it would mean there was no deflection of the fin, let’s use the maximum horizontal force provided by the SMA for analysis. Force from the Tendon F tendon = 1.10 N

8 Properties of glass fiber Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo According to Chou * unidirectional lamina can be treated as a homogeneous, orthotropic continuum. Additionally, for circular cross-section fibers randomly distributed in the unidirectional lamina, the lamina can be considered transversely isotropic. The result is that we only have five independent material constants: E 11, E 22, v 12, G 12, v 23 These values of these constants were provided by the manufacturer: * Chou, Microstructural Design of Fiber Composites (1992) + This value was not provided by the manufacturer, but it was necessary to assume a value in order to please ABAQUS. E 11 (GPa)E 22 (GPa)v 12 G 12 (GPa)v 23 + 5070.335

9 Orientation of the glass fiber Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo + = Horizontal FibersVertical Fibers Crossed Fibers

10 List of Analyses Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo The following situations were analyzed for the cases of maximum deflection and maximum forces. To look at the worst case scenario, the tendon force was included in each analysis. One Layer:Horizontal fiber direction Vertical fiber direction Two Layers:Horizontal fiber direction Vertical fiber direction Crossed fiber direction

11 A Note on the Results Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo The maximum tensile and compressive stresses and strains in the 1 and 2 directions were compared to the maximum allowable values as provided by the manufacturer. The orientation of the fibers was considered, and the graphs are based on the stresses and strains with respect to the fiber orientation, not with respect to the fin orientation. Because of this, in the cross fiber analyses, the front and back square were considered separately.

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22 Vertical Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Deformation Scale Factor =100 σ t,max = 58.5 MPa σ c,max = 65.3 MPa

23 Horizontal Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 53.2 MPa σ c,max = 52.3 MPa

24 Vertical Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 25.5 MPa σ c,max = 32.6 MPa

25 Horizontal Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 26.7 MPa σ c,max = 26.1 MPa

26 Crossed Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Front Square: σ t,max = 27.5 MPa σ c,max = 33.1 MPa Back Square: σ t,max = 38.9 MPa σ c,max = 48.2 MPa

27 Vertical Fibers - Max Deflection Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 25.5 MPa σ c,max = 32.6 MPa

28 Vertical Fibers - Max Force Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Deformation Scale Factor =100 σ t,max = 175.2 MPa σ c,max = 62.7 MPa

29 Horizontal Fibers - Max Forces Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 60.2 MPa σ c,max = 59.3 MPa

30 Vertical Fibers - Max Force Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 45.5 MPa σ c,max = 30.7 MPa

31 Horizontal Fibers - Max Force Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 28.6 MPa σ c,max = 28.4 MPa

32 Crossed Fibers - Max Force Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Front Square: σ t,max = 31.6 MPa σ c,max = 37.7 MPa Back Square: σ t,max = 37.5 MPa σ c,max = 44.8 MPa

33 Vertical Fibers - Max Force Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo σ t,max = 45.5 MPa σ c,max = 30.7 MPa

34 Table of Maximum Principal Stresses and Strains Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo Number of LayersFiber DirectionCase σ t,max σ c,max ε t,max ε c,max 1Vertical58.565.30.00440.0059 Horizontal53.252.30.00640.0051 VerticalMax Deflection25.532.60.00220.0063 2Horizontal26.926.10.00320.0067 Crossed - Front 27.533.10.00170.0027 Crossed - Back 38.948.20.00170.0027 1Vertical175.262.70.01330.0025 Horizontal60.259.30.00630.0031 VerticalMax Force45.530.70.00340.0033 2Horizontal28.628.40.00320.0018 Crossed - Front31.637.70.00190.0016 Crossed - Back37.544.80.00190.0016

35 Conclusions Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo As was previously thought, it is necessary to employ two layers of glass-fiber. The crossed-fiber configuration provides lower strains in the the 2 direction, but the horizontal stresses will approach the limit of the material in the vertical-fiber square. Despite the higher strains, the 2 layer, horizontal-fiber configuration does not experience any stresses or strains greater than 50% of the maximum allowable value. It is recommended that the fin be constructed using two horizontal-fiber squares.

36 Questions? Harvard University Division of Engineering and Applied Sciences Eng-Sci 240: Solid Mechanics Professor Zhigang Suo


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