Analysis
A large part of the CSUS ME 190 Course, and engineering in general, is to quantify and justify designs planned for implementation. Analysis can be conducted on many portions of any design, such as: energy, Structure/Stresses, Heat Transfer, Circuits, Statics, Dynamics and many others. However, due to the nature of the Transforming Staircase the more important areas that were assessed and analysis was then done on: Dynamic Motion, Deflection - in the stringer and the treads, Friction, and finally a circuit analysis
Dynamic Motion - Jake Abe
Equations:
Rigid Body Equations:
Velocity:
Øv̅B=v̅A+wABXr̅BA (Eq J.1)
Acceleration:
Øa̅B=a̅A+αABXr̅BA-wAB2r̅BA (Eq J.2)
Linear Velocity:
ØvA = 0.686 in/s positive x-direction
Angular Velocity:
ØωAB = 0.021 rad/s or ωAB = 1.203 deg/s CCW
Linear Acceleration:
ØaA = 0.031 in/s^2 positive x-direction
Angular Acceleration:
ØαAB = 0.000308 rad/s^2 or αAB = 0.0177 deg/s^2 CCW
When considering the motion of the Transforming Staircase it was first necessary to determine which movement was going to be worthy of analysis and which was not. In this design the hand railing moves in unison with the stringer as symmetry dictates. However, the linear speeds of point A and B as well as the angular speed at the center of AB (all denoted in figure J1) are all independent and worthy to determine. Once the critical points are found, a dynamic model was constructed to simplify the structure to its essential elements as show in figure J2.
With the model in place, the equations used are Eq J.1 and J.2, which are rigid body equations, commonly called "the two-point therefrom". Once calculated using vector algebra, the equations give out the results for the linear and angular velocities and acceleration of the respective points.
With the calculations complete for one point of time (angle of inclination) an extrapolation was made for the rest of the inclination of the staircases motion to determine maximums, minimums and averages.
From the table above, in red, the maximum linear and angular accelerations and velocities are quite slow - which was expected considering the speed of the linear actuators providing the vertical motion at points A and B of figure J1.
Lastly, a wire frame model was constructed to confirm the points of rotation and confirm theory that the staircase would translate and rotate as expected.
The above model shows the model at 5 degree increments of inclination during the staircases motion. The fact that the red arcs (front, center, back of tread) and the blue arcs (center of structure) are parallel shows that the rotation of the step will math the rotation of the structure, therefore keeping the step level to the ground.
The above model shows the center of each step compared to the mounding point of the tread to the hand railing at the top (which is used to keep the step stable under load). The fact that the two of each colored arcs are parallel proves the original assumption that the handrail and the stringer move in unison
Stringer Deflection - Miguel Gonzalez
The Stringer is the supporting beam that provides mounting points for the steps. The stringer supports the weight of the steps, live load, and ASME limits the amount of deflection it can have. Therefore, the stringer deflection is a necessary calculation. The red arrows below highlight the free body diagram of a distributed load across the stringer (Case 2). The load is determined from the weight applied on each tread and the reaction forces from the vertical supports and actuators.
The analysis of these components required some assumptions to be made based on the linear actuator force rating. The following assumptions provide the basis for the analysis.
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Stringer is treated as a simply supported beam
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The loading on each step is combined into a distributed load over the span of the beam
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The Stringer is acted upon by forces that include half the weight of the entire floating structure that transitions vertically
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Weight of the floating structure included hardware, outer sliding columns and step assembly (hand railing, spindle)
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The deflection is analyzed for two cases
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Case 1: Assumes that the linear actuators meet a safety factor of 5 for the driving means in accordance with the ADA and ASME
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Case 2: Takes into account the actual loading conditions provided by the chosen actuators, that is a lifting force of about 400 lbs. individually
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For Case 2, the FEA study led to a maximum deflection of 0.038 in (Figure below), which yields a %6 difference in comparison to the theoretical calculation of 0.04027 in. The small differences in the results could be attributed to the geometry of the actual stringer in conjunction with the assumed rectangular box tubing used in the theoretical calculations. Furthermore, the small difference indicates that the theoretical calculations are relatively valid.
The lifting capacity for Case 1 is calculated by including a 1050 lb. load acting at the center of each step.To calculate load W, the point loads acting on the stringer are combined and dividing by the length of the beam. The two loading conditions to consider vary in load only, where the difference in the two is the magnitude of the downwards force being distributed along the stringer. In one situation the force is higher to account for the factor of safety for the constraints in structure (Case 1), and the other being the actual loading conditions considering the limitations on the actuator (Case 2).
Tread Deflection - David Sarver
Assumptions:
1) Wood decking doesn’t translate internal shear forces therefore it can be ignored.
2) Small tubing close to reaction force has been determined to be a rigid body so it will not bend under loading therefore the step acts like a beam.
Square: I = 𝑏412b^4/12
Square Tube: Is = 𝑎4−𝑏412(a^4-b^4)/12;
Round Tube: Ir = π(𝑑𝑜4−𝑑𝑖4)64(π〖(d〗_o^4-d_i^4))/64
1.5” x 1.5” x 0.12 wall Square tubing:
Is = (𝑎^4)−(𝑏^4)12(a^4-b^4)/12 = 1.54−1.26412(〖1.5〗^4-〖1.26〗^4)/12 ; Is = 0.211 in^4
0.75”OD, 0.5” ID Round tubing:
Ir = π(𝑑𝑜4−π𝑑𝑖4)64(π〖(d〗_o^4-πd_i^4))/64 = π(0.754−0.54)64(π〖(0.75〗^4-〖0.5〗^4))/64 ; Ir = 0.0125 in4
Total I of step:
It = 2(Is) + Ir = 2(0.269 in4) + 0.0125in4 ;
It = 0.4345 in4
The tread, also known as the step needs to be designed to ASME and ADA requirements. Since each individual step will translate flat into a horizontal platform, so the design will need to accommodate loads determined from the lift platform. These loads are shown below designated by the red arrows. In a worst case loading situation all the loads will be a point force in the middle of the step.
By:
Jake Abe
By:
Miguel Gonzalez
By:
David Sarver
By:
Eric Berger
By:
Laake Scates-Gervasi
Friction - Eric Berger
For the transforming staircase, friction was of great concern due to the amount of telescoping pieces. The telescoping portions are in both horizontal and vertical directions. Below is the free body diagram of a side view of the staircase with normal forces indicated.
Equations:
F = (µk)(N) (Eq E.1)
F = (µs)(N) (Eq E.2)
ΣFx = 0 = Fx + (Ws)(Wt)(Wc/2)Sin θ° (Eq E.3)
ΣFy = 0 = Fy + (Wtot)Cos θ° (Eq E.4)
Unexpectedly the table above shows that the maximum friction is at just before the highest level of inclination. A graphical representation of the normal forces and reaction forces vs Angle of inclination is shown below showcasing the results of the table.
Friction calculation at 35.74°, max inclination:
ΣFx = 0 : Fx + 169lbs * Sin(35.74°) = 0
Fx = - 98.71lbs
ΣFy = 0 : Fy + 250lbs * Cos(35.74°) = 0
Fy = - 202.92lbs
Using the free body diagram and the equations listed below, the maximum forces were calculated for the X and Y directions for the staircase at maximum inclination:
With the friction calculations complete for one instance of inclination, the friction was extrapolated for the rest of the inclination in 2 degree increments and tabulated below:
Deflection:
𝛿y = 𝐹𝐿348𝐸𝐼(FL^3)/48EI
Deflection Angle:
Φ = 𝐹𝐿216𝐸𝐼(FL^2)/16EI
The values found are then implemented into the following equations to find the maximum vertical deflection and maximum deflection angle
𝛿y = 𝐹𝐿348𝐸𝐼𝑡(FL^3)/(48EI_t )= (1050 𝑙𝑏) (38 𝑖𝑛)348(29𝑥106𝑝𝑠𝑖)(0.4345 𝑖𝑛4)((1050 lb) (〖38 in)〗^3)/(48〖(29x10〗^6 psi)(0.4345〖 in〗^4))
= -0.0949 in
Φ = 𝐹𝐿216𝐸𝐼𝑡(FL^2)/(16EI_t ) =
(1050 𝑙𝑏𝑠)(38 𝑖𝑛)216(29 𝑥 106𝑝𝑠𝑖)(0.4345 𝑖𝑛4)((1050 lbs)〖(38 in)〗^2)/(16(29 x 〖10〗^6 psi)(0.4345〖 in〗^4))
= 0.411 degrees
The FEA performed on the step structure in Figure D4 resulted in a maximum deflection of -0.059 inches. The simulations max deflection is less than the calculated value so it can be assumed that the step will not deflect more than the largest predicted value. The FEA of step also supports the assumption that the small bars on each side act as a ridged body due the 3.9x10-32 inches that the corners of the step deflect.
Circuits - Laake Scates-Gervasi
The image shown below was the initial circuit design for the Controls System. Through manufacturing and Assembly the design slightly changed, Loop 5 was omitted.
To the right is the theoretical analysis of the systems current, voltage and power draw. The idle current was determined using the specs of the selected motor controllers during the design phase. The idle current is a function of the MOSFET included in the motor controller, and contains the MOSFET internal resistance and the trans-conductance.
