Physics,2.01x

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The shaft ABC is a solid circular cylinder of constant outer diameter 2R and length 3L. The shaft is fixed between walls at A and C and it is composed of two segments made of different materials. The left third of the shaft (AB) is composed of a linear isotropic elastic material of shear modulus G0, while the right two-thirds of the shaft (BC) is composed of a different linear elastic material of shear modulus 2G0. The right segment, BC, is subjected to a uniform distributed torque per unit length t0[N⋅m/m].
Obtain symbolic expressions in terms of R, G0, L, t0, and x for the quantities below. In your answers, leave rationals as fractions and enter G0, t0, and π as G_0, t_0 and pi, respectively.
The x-component of the reaction torque at C:
TCx=
Q2_1_2 : 60.0 POINTS
The twist rate dφdx(x), and the position x0 along the shaft where the twist rate goes to zero (dφdx(x0)=0):
for0≤x<L,dφdx(x)=
forL<x≤3L,dφdx(x)=
dφdx(x0)=0atx0=
Q2_1_3 : 60.0 POINTS
The maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax):
τmax=
rτmax=
xτmax=
Q2_1_4 : 100.0 POINTS
The maximum value of the rotation field φ(x) along the shaft (φmax), and the position along the shaft where the maximum rotation occurs (xφmax):
φmax=
xφmax=

  • Physics,2.01x -

    Q2_1_1
    TXC=-t_0*L

    Q2=1_2
    a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4)

    b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4)

    c) x=2*L

  • Physics,2.01x -

    TXC=-3/2*t_0*L

    Q2_1_2:
    d*phi/dx = (t_0*L)/(pi*G_0*R^4)
    d*phi/dx = (t_0*(3*L-2*x))/(2*pi*G_0*R^4)
    x0 = 3/2*L

    some one for the rest?

    Q2_1_3:
    rômax = R
    xômax =3*L

  • Physics,2.01x -

    tau max=(4*t_0*L)/(pi*R^3)

  • Physics,2.01x -

    There is a error in tau max
    the correct answer is

    tau max =(3*t_0*L)/(pi*R^3)

  • Physics,2.01x -

    This is my last chance and I need to see if my calculus is correct.

    Is it correct ?

    Q2_2_4

    vA=-5.82 cm ??

  • Physics,2.01x -

    Q2_2_4
    Yes , it is -5.823 cm

  • Physics,2.01x -

    Q2_1_4
    phi_max = 9/8*t_0*L^2/(pi*G_0*R^4)

    x_phi_max = 3/2*L

  • Physics,2.01x -

    Help with the rest

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