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Example 1.1)

Example 1.12) It can be seen that the stress is proportional to the mass and acceleration of the turbine blade, and inversely proportional to the cross-sectional area. The acceleration of the blade is on the angular velocity, however it is constant thus eliminating the tangential component. It is worth noting the blade will experience both components of acceleration during startup.

Exercise 1.1) A difference to be noticed is the increase in mass of this turbine blade. The mass is proportional to the cross-sectional area and the length of the turbine blade (if the blade is of uniform cross-section). This should be noted in the design of the turbine blade as you could minimize the mass by design and minimize by selecting a less dense material.

Exercise 1.24) It can be seen that the stress is proportional to the mass and acceleration of the turbine blade, and inversely proportional to the cross-sectional area. It is also noted that angular acceleration is influenced by the square of angular velocity. This means the stress is also influenced by the square of angular velocity, and the increase in velocity greatly influences the stress on the turbine blade.

Exercise 1.25) If the turbine blade is to begin yielding due to normal stress from acceleration at 617 MPa, then the temperature would need to weaken the part from its rated yield strength of 1.5 GPa to 617MPa. It can be seen that a large temperature increase is needed to affect the yield strength of the turbine blade, however it would be below the melting temperature of the material to construct the blade. Deformation would occur prior to the material degradation when heat is introduced, starting at a temperature of 1033K or 760C.

Exercise 1.27) It can be noted that a thinner blade reduced the affect of the angular acceleration compared to a previous problem. Reducing \(r\) reduces radial acceleration proportionally. Halving \(r\) halves centripetal acceleration which halves the stress in the turbine blade. It can also be noted that the m factor increases the tolerance to heat. This can be dependent on the material selected, as one less tolerant to heat will deform further from the material's melting temperature.

Motorcycle incident: Engine & rear wheel locked up instantaneously, resulting in deadly crash.

1) Failure upstream in powertrain, learned from inspecting rear wheel.

2) Not maintenance failure, oil, clutch, other parts fully present and functional.

3) Found gear in transmission that didnt align in the engine. Damage such as shearing in half of shaft key.

4) Failure attributed to balance shaft interfering with crankshaft. Engine design is supposed to constrain these parts from ever colliding.

5) Theory is that one piece was loose, allowing other component that was not heat treated to fret through splines of shaft. Thus, balance shaft gained more degrees of freedom and interfered with crank shaft over several million engine cycles. Gradually the collisions become more severe and caused catastrophic failure.

*What is a balance shaft and crankshaft in an engine? Where are they located?

*What does it mean to fret through the splines of a shaft?

*What are the existing degrees of freedom in the balance shaft?

Check Gain at Phase Angle = -180. The Gain needs to be smaller than 1 for stability.

Ex: Gain is around -15 dB at -180 degrees. This is less than 1, therefore the system is stable! Corresponding Frequency: 30.9

Check Phase Angle at Gain = 0 dB. This needs to be above -180 degrees.

Ex: Phase is around -142 at 0 dB. This is more than -180, therefore the system is stable. Corresponding Frequency: 12.2 rad/s

PHASE MARGIN EQUATION

GAIN MARGIN EQUATION

Check the denominator of the transfer function of a unity feedback system. The system is, of course, unstable if the function is undefined.

These bumps shown here indicated an UNDERDAMPED system.

LAPLACE TIME DELAY

$$\mathcal{L}{f(t - a)u(t - a)} = e^{-as} F(s)$$

Time Delay Example:

After energizing relay, it takes time before power reaching other end.

Essentially, delay from signal sent to signal received to output motion carried out.

Transfer Function is in polynomial form, traditionally to see poles in the CE, denominator.

When it isn't given in polynomial form, it becomes hard to find poles.

Make Bode Diagram and look at -3 dB

Read Breakpoint Frequency, w_b $$s = a \pm jb$$

In transfer function G(s), substitute $$s=jw_b=\frac{j}{\tau}$$

Solve the Transfer Function

Convert to Polar Form $$z = a + jb , $$ $$z = r \angle \theta , \quad r = \sqrt{a^2 + b^2}$$ $$ \quad \theta = \tan^{-1}!\left(\frac{b}{a}\right)$$

$$r = k_dc$$

No transfer function for returning pass (unity)

Input=Output for returning pass.

ex: cruise control. r=desired speed, y=actual speed

• $$r-y = e$$

• \(e\) represents the tracking error, the difference between the desired output \(r\) and the actual output \(y\).

$$ u(t) = K_p e(t) + K_i \int e(t)\,dt + K_d \frac{de}{dt} $$

• The control signal \(u\) to the plant is equal to the proportional gain \(K_p\) times the magnitude of the error plus the integral gain \(K_i\) times the integral of the error plus the derivative gain \(K_d\) times the derivative of the error.

When you route error into the control system.

Example gravel depositing on weighted belt * Weight too heavy, system controls flow rate to reduce weight * Weight is sufficient, system does no work to correct error = neutral zone