tapered thin-wall circular shaft has constant wall thickness, t, length L, and diameters linearly varying between dA at the support A(x=0) and dB at its free end B(x=L). The shaft is homogeneous with shear modulus G
HW6_1A : 20.0 POINTS
Obtain a symbolic expression for the torsional stiffness of the shaft KT=Q/Φ, in terms of t, L, G, dA (you will have factors of π in your answers: enter π as "pi" ):
KT=
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HW6_1B : 20.0 POINTS
Obtain a symbolic expression for the maximum shear strain on the generic x-section along the shaft, γmax(x), in terms of t, L, G, Q, x, dA (you will have factors of π in your answers: enter π as "pi" ):
γmax(x)=
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HW6_1CX : 0.0 POINTS
CHALLENGE QUESTION! (no points, just for fun!)
This challenge question is just for fun: it gives you no points, so you do not NEED to get the right solution. Indeed it is not even graded.
For L=0.5 m, t=2 mm, dA=4 cm, and G=70 GPa, obtain the torque Q0 that you need to apply to the shaft if you want to obtain a maximum value of 2% strain.
Then, use these values to plot γmax(x) by writing MATLAB code in the blank command window below. If you succeed, take a screenshot of your plot (NOT THE CODE) and post it in the discussion forum under the "Gamma- Challenge!" thread.
Note: be careful when you write your expression for γmax(x) in MATLAB. Remember that element-wise division needs the period, so if you need to define a vector y = 1/x where you want to obtain each element of y as the inverse of the corresponding element of x, you need to define y as: y = 1./x
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HW6_2: SOLID COMPOSITE SHAFT SUBJECTED TO DISTRIBUTED TORQUE
A composite shaft of length L is constructed from an inner core of radius R and modulus Gc=5G0, and a sleeve of outer radius 2√R and modulus Gs=G0, bonded together. One end of the shaft, B, is fixed and the other, A, is free to rotate as shown in the figure. A uniform distributed torque, tx(x)=t0 (t0 = constant with units of N⋅m/m), is applied to the shaft in the direction shown in the figure.
Obtain symbolic expressions in terms of R0, G0, L, t0, x for the following quantities.
(NOTE: you will have factors of π in your answers: enter π as "pi".)
HW6_2A : 10.0 POINTS
The axial torque resultant:
T(x)=
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HW6_2B : 10.0 POINTS
The rotation field φ(x) along the shaft:
φ(x)=
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HW6_2C : 10.0 POINTS
The angle of twist:
ΦAB=
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HW6_2D : 10.0 POINTS
The maximum magnitude of shear stress, τmax, in the shaft:
τmax=
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HW6_3: STATICALLY INDETERMINATE SHAFT UNDER DISTRIBUTED LOADING
The round shaft in the figure has length L and is fixed at both ends. The shaft is loaded by a constant distributed torque t0. The modulus of the material, G, and the polar moment of inertia of the cross section, Ip, are known.
HW6_3 : 40.0 POINTS
If we want to limit the rotation of the midsection of the shaft to a maximum value, φ(L2)=φm, what is the maximum value of the distributed load, t0,m, that can be applied to the shaft?
Provide your answer as a symbolic expression in terms of L, G, Ip, φm (write as "I_p" and "phi_m"):
t0,m=
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No one has answered this question yet.
To obtain the symbolic expressions for the given quantities, we need to apply the relevant formulas and equations. Let's go through each question step by step:
HW6_1A: Torsional Stiffness of the Shaft (KT)
The torsional stiffness (KT) is defined as the ratio of the torque (Q) applied to the shaft to the twist angle (Φ) produced. We can obtain the torsional stiffness by dividing the torque by the twist angle.
KT = Q / Φ
Final Answer: KT = Q / Φ
HW6_1B: Maximum Shear Strain on the Generic x-section along the Shaft (γmax(x))
The maximum shear strain (γmax(x)) on a generic x-section along the shaft can be obtained using the formula:
γmax(x) = (Q * x) / (G * dA * t)
Final Answer: γmax(x) = (Q * x) / (G * dA * t)
HW6_1CX: Challenge Question - Torque to Obtain Maximum Strain
In this question, the goal is to find the torque (Q0) that needs to be applied to the shaft in order to obtain a maximum shear strain of 2%. We need to use the formula for maximum shear strain and rearrange it to solve for Q:
γmax(x) = (Q * x) / (G * dA * t)
2% = (Q0 * L) / (G * dA * t)
Q0 = (G * dA * t * 0.02) / L
Final Answer: Q0 = (G * dA * t * 0.02) / L
HW6_2A: Axial Torque Resultant (T(x))
The axial torque resultant (T(x)) can be calculated as the integral of the distributed torque (tx(x)) from 0 to x. Since tx(x) is a constant t0, the integral simplifies to:
T(x) = t0 * x
Final Answer: T(x) = t0 * x
HW6_2B: Rotation Field along the Shaft (φ(x))
The rotation field (φ(x)) along the shaft can be calculated as the integral of the rotation rate (ω(x)) from 0 to x. The rotation rate is equal to the axial torque resultant (T(x)) divided by the shear modulus (Gs) and the polar moment of inertia (J):
ω(x) = T(x) / (Gs * J)
φ(x) = ∫ ω(x) dx
Final Answer: φ(x) = ∫ (T(x) / (Gs * J)) dx
HW6_2C: Angle of Twist (ΦAB)
The angle of twist (ΦAB) between points A and B can be obtained by integrating the rotation rate (ω(x)) along the shaft from 0 to L:
ΦAB = ∫ ω(x) dx (from 0 to L)
Final Answer: ΦAB = ∫ (T(x) / (Gs * J)) dx (from 0 to L)
HW6_2D: Maximum Shear Stress in the Shaft (τmax)
The maximum shear stress (τmax) in the shaft can be calculated using the formula:
τmax = (T(x) * 2 * sqrt(R)) / (pi * Rs^3 - pi * R^3)
Final Answer: τmax = (T(x) * 2 * sqrt(R)) / (pi * Rs^3 - pi * R^3)
HW6_3: Maximum Distributed Torque Load (t0,m)
To limit the rotation of the midsection of the shaft to a maximum value (φ(L2) = φm), we need to solve for the maximum distributed torque load (t0,m) that can be applied. We can rearrange the formula for the angle of twist (ΦAB) and solve for t0,m:
ΦAB = (t0, m * L^3) / (3 * G * Ip)
t0, m = (ΦAB * 3 * G * Ip) / L^3
Final Answer: t0, m = (ΦAB * 3 * G * Ip) / L^3
Note: The answers provided above are symbolic expressions in terms of the given variables. To obtain numerical results, substitute the given values into the respective formulas.