Posted by **Uncle Sam** on Thursday, July 25, 2013 at 1:20am.

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 -
**superman**, Friday, July 26, 2013 at 12:42pm
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 -
**access014**, Friday, July 26, 2013 at 11:05pm
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 -
**ElementarySchoolStudent**, Friday, July 26, 2013 at 11:55pm
tau max=(4*t_0*L)/(pi*R^3)

- Physics,2.01x -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 10:34am
There is a error in tau max

the correct answer is

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

- Physics,2.01x -
**11YearsOldMITStudent**, Saturday, July 27, 2013 at 2:53pm
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 -
**sick yoda**, Sunday, July 28, 2013 at 3:23am
Q2_2_4

Yes , it is -5.823 cm

- Physics,2.01x -
**Simon76**, Sunday, July 28, 2013 at 1:01pm
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 -
**MAn**, Monday, July 29, 2013 at 1:42pm
Help with the rest

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