A blade is fixed to a rigid rotor of radius R spinning at ωrad/s around the vertical z-axis (see figure). Neglect the effects of gravity. The x-axis rotates with the blade, and the origin x=0 of the axis is located at the "root" of the rotating blade (i.e., at the outer radius of the rotor as indicated in the figure.)
Given:
Young's modulus E, mass density ρ
Constant cross sectional area, A
Rotor radius R, blade length L
Angular velocity ω
Hint: if you work in the non-inertial frame of the rotating blade, where the x-axis is defined, the d’Alembert (inertial) force per unit volume is ρω2r=ρω2(R+x) along the +x direction. You can treat this force per unit volume just as you treated ρg in HW2_1. Also, the fact that the rotor is rigid means that you can consider the blade as "fixed" at x=0.
Note that we've given you a convenient MATLAB window at the bottom of this page; it may prove helpful for doing some of the integration needed in this problem, but you will not lose points if you choose not to use it.
Find a symbolic expression for the peak stress σnmax in the blade. Express your answer in terms of R, L, ρ, and ω (enter the last two as rho and omega, respectively).
Find a symbolic expression for the blade elongation δ in terms of R, L, ρ, ω, and E:
Find a symbolic expression for the displacement of the blade mid-section, ux(L/2), in terms of R, L, ρ, ω, and E:
To find the peak stress σnmax in the blade, we can start by considering a small element of the blade at a distance x from the root (x = 0). In the non-inertial frame of the rotating blade, the d'Alembert force per unit volume acting on this element is given by ρω^2(R+x), where ρ is the mass density and ω is the angular velocity. This force acts along the positive x direction.
The axial force acting on this small element can be calculated by integrating the d'Alembert force over the length of the blade. Since the blade is fixed at x = 0, the integration limits are from 0 to L (length of the blade). The axial force Fx can be expressed as:
Fx = ∫(0 to L) ρω^2(R+x) dx
Integrating this expression, we get:
Fx = ρω^2∫(0 to L) (R+x) dx
= ρω^2[Rx + (1/2)x^2] (evaluated from 0 to L)
= ρω^2(RL + (1/2)L^2)
The peak stress σnmax in the blade is given by the ratio of the axial force Fx to the cross-sectional area A of the blade. Since the cross-sectional area A is constant throughout the blade, we can express σnmax as:
σnmax = Fx / A
= (ρω^2(RL + (1/2)L^2)) / A
Therefore, the symbolic expression for the peak stress σnmax in the blade is:
σnmax = (ρω^2(RL + (1/2)L^2)) / A
To find the blade elongation δ, we can use Hooke's law, which states that the stress is proportional to the strain. In this case, the stress is the peak stress σnmax and the strain is the ratio of the elongation δ to the original length of the blade L. Therefore, we have:
σnmax = E * (δ / L)
Solving for δ, we get:
δ = σnmax * L / E
Substituting the expression for σnmax from above, we obtain the symbolic expression for the blade elongation δ:
δ = [(ρω^2(RL + (1/2)L^2)) / A] * L / E
To find the displacement of the blade mid-section ux(L/2), we consider the small element at x = L/2. The displacement of this element in the x-direction can be calculated by integrating the equation of motion.
Since the blade is rigid and fixed at x = 0, the displacement of the mid-section can be expressed as:
ux(L/2) = ∫(0 to L/2) [∫(0 to x) (1/E) {ρω^2(R+y) dy} dx]
Integrating this expression, we get:
ux(L/2) = (1/E)∫(0 to L/2)[ρω^2(Rx + (1/2)x^2)] dx
= (1/E)∫(0 to L/2)[ρω^2Rx + (1/2)ρω^2x^2] dx
= (1/E)[(ρω^2/2)Rx^2 + (1/6)ρω^2x^3] (evaluated from 0 to L/2)
= (1/E)[(1/2)ρω^2RL^2/4 + (1/6)ρω^2(L/2)^3]
= (1/E)[(1/8)ρω^2RL^2 + (1/6)ρω^2(L^3/8)]
= (1/E)[(1/8)ρω^2RL^2 + (1/48)ρω^2L^3]
Therefore, the symbolic expression for the displacement of the blade mid-section, ux(L/2), is:
ux(L/2) = (1/E)[(1/8)ρω^2RL^2 + (1/48)ρω^2L^3]