Group 6
Fan Yan
Yiran Li
Runmiao Shi
1.General
The objective of our crane project is to design and build a mechanism that is powered by a standard servo motor, which will lift a cylindrical weight. Our design was to build a 6’’*6’’ rectangular base and with a hollow triangular cross-sectioned arm that held the motor and a short arm on the front of it that held the counterweight that in turn lifts the weight. Once the motor is switched, the moment from the counterweight would lift the counterweight. Our goal was to build a crane that weigh no more than 20.0 oz, and it must lift the sliding weight by at least 2” . Our final crane was 19.95 oz, and achieved a 4.1 in lift.
2.Theoretical Predictions of Performance
As shown by the calculation below, the motor applies a torque of 13.35 oz·in, 23.4% of its total torque. To allow the lever arm to lift the 1-pound weight, a counterweight of 4.16oz.
Assumptions in this section:
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1. Deformation of the crane arm is negligible. (That is, the deflection of the point C).
2. Deformation of the hook is negligible.
3. Deformation of the servo arm is negligible.
4. Rotating angle of servo arm is from -45 ° to 45 °.(maximum angle with horizontal does not exceed 45 °)
Stress Analysis 2.1:
General model for prediction of the theoretical servo torque:
Since sum of moment at C equals to zero,
Wcounter *Lcounter*cosθ+Mservo-Wbearing*Llift*cosθ = 0
Mservo = Wbearing*Llift*cosθ - Wcounter *Lcounter*cosθ=( Wbearing*Llift - Wcounter *Lcounter)*cosθ
Stress Analysis 2.2:
Theoretical prediction of the crane performance:
Now we substitute in theoretical values to analyze the unknown theoretical servo torque.
Since cosθ reaches its maximum value at θ = ±45° where -45° < θ < 45°
Let θ = ±45°
Mservo =(16*3- 4.16*7)*cos ±45°
Mservo= 13.35 oz·in
Therefore, fraction of the servo’s theoretical with its maximum torque
= 13.35/57 =0.234 < 1
Stress Analysis 2.3:
In Conclusion, theoretical torque the crane uses to lift the bearing is 13.35 oz·in, which only uses 0.234 of the maximum servo torque. This means the crane able to lift the bearing to the servo's maximum angle (all the way from -45° to 45°)
3. Theoretical Lift Distance Calculation:
Stress Analysis 3.1:
Assumptions in this section:
1 No plastic deformation occurs.
2. Deformation of the hook is negligible.
3. Deformation of the servo arm is negligible.
4. Rotating angle of servo arm is from -45 ° to 45 °.(maximum angle with horizontal does not exceed 45 °).
5.The cross-sectional shape is a hollow triangle.
Stress Analysis 3.2:
General model for the distance over which the weight should theoretically be lifted:
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1) First we temporarily ignore the deflection of the crane arm and only think of the geometry. Based on the second section, the servo is able to lift all the way from negative 45 degrees to positive 45 degrees, we can calculate the ideal theoretical distance over which the bearing can be lifted without considering the deflection of crane arm.
According to the geometry of triangle above, the ideal theoretical distance is:
Dlift=Llift*cosΘ*2
2) Then we consider the deformation of crane arm and add the deflection at C due to deformation to get a more precise value of that distance.
The deflection at C according to the table shall be:
Ddeform = -PL^3/(3EI) =-Wbearing * (Larm)^3/(3*Eal*I)
Add this deflection to the previous geometric distance and get a more precise theoretical lifting distance.
Distance lifted = Dlift + Ddeform
Theoretical distance over which the weight should be lifted:
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1) Plug in theoretical values to get an ideal lifting distance
Dlift=3*cos45°*2=4.24 in
Only considering in geometry, the ideal distance is 4.24 in.
2) Plug in theoretical values to get a more precise theoretical lifting distance by considering deformation.
The moment of inertia is calculated as cutting the I of smaller triangle out of I of the larger triangle:
Thickness of the arm = 0.2 in, b1=1.4 in, h1=2.62 in, b2 = 1 in, h2=2.22 in
Moment of inertia = (b1*h1^3-b2*h2^3)/36=0.3955 in^4
Ddeform = -16oz*(28.4in)^3/(3*10000000psi*0.3955in^4) = -0.0309 in
Distance lifted = 4.24 in + ( -0.0309 in) = 4.21 in
Stress Analysis 3.3:
In conclusion, the distance over which the weight should theoretically be lifted is 4.21 in.