Analysis of Structures  Torque Motor
Copyright Maplesoft, a division of Waterloo Maple Inc., 2008
Introduction
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Problem Statement
Solution
Step

Result

To perform matrix computations, load the Student Linear Algebra package.
Tools > Load package > Student Linear Algebra

Loading Student:LinearAlgebra

For every conceivable position the rods can assume, the following geometric relationship holds:
This is illustrated by Figure 2 to the right.
This can also be found by the law of sines from trigonometry. Recall,
The expression above can be easier manipulated to obtain the relationship to the right.
When the positions of the rods are reflected across the vertical axis, the relationship will still hold since both angles will simply be negative.
Using this relationship, an expression can be found for the angle . Later, this will be used to eliminate angle so that can be expressed as a function of just the shaft angle, .
Rightclick the geometric relationship and select Solve > Isolate Expression for > alpha.
Obtain α and β from the Greek palette.
If an equation label is desired, click on the output to ensure that the cursor is in the line and uncheck, View > Inline Document Output.

Figure 2

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The member contains two forces, and will be either a tension or compression force. If it is a tension the rod will pull on the piston, and if it is a compression the rod will push on the piston.
There are three concurrent forces acting on the piston, the required force, , the normal force, , which the cylinder exerts on the piston, and the force, , which the member exerts on the piston.

Define the vectors and .
To define P, use the assignment operator (a colon followed by an equal sign).
To enter the vector, use the Matrix palette. Set the number of rows to two and the number of columns to one and then press the Insert Vector[column] button. You could also use the Choose button and drag the mouse to select the matrix size.
Fill in an element, then press [Tab] to move to the next placeholder.
Press [Enter] to evaluate.
Repeat for N and .
Use the underscore ( _ ) to move the cursor to the subscript position, and the right arrow (→) to move back to the baseline. For example, to enter , type [T][_][R][_][1], then press the right arrow twice to move out of the subscript.


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Assuming that the inertia of the moving parts is negligible, the equation of equilibrium of the piston at any point in time is:
Two scalar equations (one in x and one in y) can be formed from this vector equation. From these scalar equations an expression for n can be obtained.

Enter the lefthand side of the vector equilibrium equation.
Press [Enter] to evaluate.
Rightclick the output and select Conversions > To List.
Individually select both elements of the list by rightclicking and selecting Select Element > 1, then rightclick again and select Select Element > 2.
For the first entry, isolate for .
Rightclick the expression and select Solve > Isolate Expression for > t[R[1]].
Substitute this newly obtained expression into the second element of the list.
Use the template from the Expression palette to evaluate the expression at a point.
Use equation labels to reference previous equations or output in the substitution template. Press [Ctrl][L], then enter the appropriate reference equation number.
To isolate n, rightclick the expression and select Solve > Isolate Expression for > n.
This expression for n will be used in the next step to simplify the moments equation about O.


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The torque, , can be found by balancing moments about O for the complete assembly.
The force N acts at a perpendicular distance of from the point O and produces a moment opposite in direction to . Since the system is in equilibrium, the principle of moments must apply and therefore the sum of moments about O must be zero. Consequently,

Enter the righthand side of the expression for .
Press [Enter] to evaluate.
Substitute this equation into the expression obtained for n.
Reference both expressions using their equation labels, while using the template from the Expression palette.Use an equation label to reference the desired point to evaluate at.
Substitute the expression obtained for into the new equation so that it is removed through cancelation.
The torque has now been obtained for the general case where p and lengths and are unspecified.


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Consider now the special case where and are 0.6 m and 0.3 m respectively, and the magnitude of P is
Substitute the specified values of p, and into the expression for torque.
Use the template from the Expression palette and reference the expression for by its equation label. Replace the field with a list of equations specifying the values of the parameters.
Press [Enter] to evaluate.
Graph .
Rightclick the expression and select Plots > Plots Builder. In the window that opens, specify the range for beta to be from 0 to 2*Pi and then click Plot.
The vertical axis in the plot can be labeled by rightclicking the plot and selecting Axes > Labels > Add Vertical. Then again, rightclick and select Axes > Labels > Edit Vertical.
This graph shows that is zero when the piston is at the top or bottom of the stroke (i.e., when is or ), and is negative during the second half of the cycle.


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Increase the length of the second from 0.3 m to 0.4 m, and again graph . Use the same strategy for substituting into as above.
In the column to the right, the graph shows that a greater torque is needed to produce the same force, P if is increased from 0.3 m to 0.4 m.
Qualitatively, is again zero when the piston is at the top or bottom of the stroke (i.e., when is or ), and is likewise negative during the second half of the cycle.


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