CONICAL SHAFT WITH DISTRIBUTED TORSIONAL LOAD
The conical shaft AB is made of steel, with shear modulus G0. The shaft has length L and is free at A (where x=0) and fixed at B (where x=L). The shaft is solid, with a linear taper so that the radius of the generic (circular) cross section is defined by the equation R(x)=R0(xL), where R0 is the radius at the wall, B. The shaft is subjected to an applied distributed torque per unit length tx(x)=t0(xL)4, where t0 is the magnitude of the applied distributed torque at B (x=L).
The KNOWN quantities in this problem are:
L=2m
G0=70GPa
R0=2cm
t0=2kN⋅m/m
In symbolic expressions, do NOT substitute ANY of the given numerical values of the known quantities, enter L, G0, R0, and t0 as L, G_0, R_0, and t_0, respectively, and enter π as pi.
1)Obtain a symbolic expression for the axial torque resultant in the bar T(x) in terms of x, t0, and L.
T(x)=
2)Compute the numerical value of the reaction TBx at support B, in units of kN·m. (Note: You will need to provide your answer to the second decimal digit.)
TBx=
3)Obtain a symbolic expression for the twist rate along the shaft, dφdx, in terms of x, G0, t0, and R0.
dφdx(x)=
4)Compute the numerical magnitude (absolute value) of the maximum shear stress in the shaft, τmax, in units of MPa:
τmax=
Enter symbolic expressions for the location (r,x) at which this maximum shear stress occurs, in terms of R0 and L:
r=
x=
5)Obtain a symbolic expression for the angle of rotation of the free end A of the shaft in terms of t0, G0, R0, and L:
φ(x=0)=
Compute the numerical value of the angle of rotation of the free end, in units of radians. (Note: You will need to provide your answer to the third decimal digit.):
φ(x=0)=
anyone please???
On the 4th one r=R_0
and x=L on the 4th one.please the rest someone!!!!
To solve this problem, we need to use the principles of solid mechanics and mechanics of materials.
Let's break down each question step-by-step:
1) Obtaining a symbolic expression for the axial torque resultant in the bar T(x) in terms of x, t0, and L:
The axial torque resultant can be found by integrating the distributed torque per unit length over the length of the shaft.
We have tx(x) = t0 * (x/L)^4
Integrating this expression over the length of the shaft (from 0 to x), we get:
T(x) = ∫[0, x] {t0 * (x/L)^4} dx
Evaluating the integral, we get:
T(x) = t0 * [(x^5)/(5L^4)]
So the symbolic expression for the axial torque resultant T(x) is:
T(x) = t0 * [(x^5)/(5L^4)]
2) Computing the numerical value of the reaction TBx at support B:
At support B, the axial torque resultant T(L) is equal to the reaction torque TBx.
Substituting x = L into the expression for T(x), we have:
T(L) = t0 * [(L^5)/(5L^4)]
Simplifying, we get:
T(L) = (t0 * L)/5
The numerical value of the reaction TBx at support B in units of kN·m is given by t0 * L / 5.
Plugging in the given values, we have TBx = (2kN⋅m/m * 2m) / 5 = 0.8 kN·m
So TBx = 0.8 kN·m
3) Obtaining a symbolic expression for the twist rate along the shaft, dφ/dx, in terms of x, G0, t0, and R0:
The twist rate, dφ/dx, is related to the torque, the shear modulus, and the radius of the shaft.
We know that the torque T(x) = (G0 * dφ/dx) * A, where A is the cross-sectional area of the shaft.
The cross-sectional area A is π * R(x)^2, where R(x) is the radius of the shaft at a specific point x.
Substituting R(x) = R0 * (x/L), we have:
A = π * (R0 * (x/L))^2 = π * R0^2 * (x/L)^2
Therefore, the equation becomes:
T(x) = (G0 * dφ/dx) * [π * R0^2 * (x/L)^2]
Simplifying, we get:
dφ/dx = T(x) / (G0 * [π * R0^2 * (x/L)^2])
So the symbolic expression for the twist rate dφ/dx is:
dφ/dx = T(x) / (G0 * [π * R0^2 * (x/L)^2])
4) Computing the numerical magnitude (absolute value) of the maximum shear stress in the shaft, τmax, in units of MPa, and obtaining the symbolic expressions for the location (r,x) at which this maximum shear stress occurs:
The maximum shear stress in a circular shaft occurs at the outer surface, where the radius is maximum. In this case, it occurs at r = R(x).
The shear stress can be computed using the equation τ = (T * r) / (J * G0), where T is the torque, r is the radius, J is the polar moment of inertia, and G0 is the shear modulus.
From the previous questions, we have the expression for T(x) as t0 * [(x^5)/(5L^4)].
We also know that the polar moment of inertia J = (π/2) * R(x)^4, where R(x) = R0 * (x/L).
Substituting the given values, we have G0 = 70 GPa = 70 * 10^9 Pa and R0 = 2 cm = 0.02 m.
The expression for τmax is given by:
τmax = (T(x) * R(x)) / ((π/2) * R(x)^4 * G0)
Substituting the expressions for T(x) and R(x), we have:
τmax = (t0 * [(x^5)/(5L^4)] * [R0 * (x/L)]) / ((π/2) * [R0 * (x/L)]^4 * G0)
Simplifying, we get:
τmax = (2 * t0 * (x^2) * L) / (5 * π * R0^2 * G0)
The symbolic expressions for the location (r,x) at which τmax occurs are:
r = R(x) = R0 * (x/L)
x = x
5) Obtaining a symbolic expression for the angle of rotation of the free end A of the shaft, φ(x=0), in terms of t0, G0, R0, and L, and computing the numerical value of the angle of rotation of the free end in units of radians:
The angle of rotation, φ(x=0), can be found by integrating the twist rate dφ/dx over the length of the shaft.
Using the expression for dφ/dx from question 3, we have:
φ(x=0) = ∫[0, L] {T(x) / (G0 * [π * R0^2 * (x/L)^2])} dx
Integrating this expression, we get:
φ(x=0) = ∫[0, L] {t0 * [(x^5)/(5L^4)] / (G0 * [π * R0^2 * (x/L)^2])} dx
Simplifying, we get:
φ(x=0) = (2 * t0 * L^4) / (5 * π * G0 * R0^2)
The numerical value of the angle of rotation of the free end in units of radians is given by the expression using the given values:
φ(x=0) = (2 * 2kN⋅m/m * (2m)^4) / (5 * π * 70GPa * (0.02m)^2)
Calculating this numerical value, we get:
φ(x=0) ≈ 0.131 radians (rounded to the third decimal digit)