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Published byAmelia Hawkins Modified over 5 years ago

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Group B Tedros Ghebretnsae Xinyan Li Zhen Yu Tang Ankit Panwar

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Design a fourbar to move from point A to B Point B is at a height, H=6R,4R,2R Single motor drives the input crank Preferable a Grashoff crank rocker or double crank The Chassis is fixed What is required from the motor Compactness Dynamically well behaved

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Links:Aluminum T6061 Area of each link: 1 cm 2 Weight of wheel:1 kg

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Two most important constraints were: the ground link had to lie inside the chassis the lower arm should travel from without colliding with the step. First choice was whether to synthesize the mechanism for two or three points. The main advantage for the two point mechanism was that there were more free choices. Whereas with the three point, there was a better control

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This was the first method employed by us. Total variables: 18, total equations: 12 The free choices were α 2, α 3, β 2, β 3, γ 2 and γ 3 The problems with this method: no constraints on the ground link the arms were colliding with the step None of the solutions satisfied our design requirements

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Next we employed the same technique but with fixed ground. The values of the betas (α 2 and α 3 for the upper blue link, β 2 and β 3 for the lower black link) were found first The known variables were the ground link and the coupler angles γ 2 and γ. We were able to get the solution for all the three points which satisfied the design requirements.

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The problem with this method was that the arm moves from point one to point two, but fails to trace the last point. The reason why this method fails is that it does not always guarantee that the fourbar will pass through all the three points The plot shows the same fourbar mechanism but in the crossed configuration Whereas the previous plots were for the uncrossed configuration

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Total variables: 14, total equations: 8 The free choices were C, S (this is the part extended beyond the joint of U and C), β 2, γ 2 and γ 3

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Steps taken: Determination of Position, Velocity and acceleration of all the center of masses and also that of the wheel Formulation of Torque versus input Theta 2 Plotting Torque versus Theta 2 using Matlab for different height H(6R, 4R and 2R) Analyze the plotted Torque and Determine and select the feasible one for the given mechanism size and orientation

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For heights 4R and 2R The Torque seems to stabilize for theta greater than 100 For those values of theta, it was managed to lower the Torque to little more than zero value which in turn means less work is needed to do the task, and hence a small Motor would work

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For height equal to 6R Unstable torque. There is a lot of fluctuation in magnitude and direction There is a big range between max and min values Theta2 values Values between 50 and 100 resulted in great fluctuation of the torque with extreme high values and lack of stability

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Describe moment of force or torque as moment = force × perpendicular distance from pivot to the line of action of force;

Describe moment of force or torque as moment = force × perpendicular distance from pivot to the line of action of force;

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