The composite shaft ABCD is fixed between walls at A (x=0) and D (x=3L). The shaft has a solid steel core of length 3L and round cross section of uniform radius R. In the left third of the shaft (segment AB of length L) the steel core is surrounded by a copper sleeve of inner radius R perfectly bonded to the core. The outer radius of the copper sleeve is 2R.

A uniform distributed torque per unit length of magnitude q[N⋅m/m] acts only on the central segment of the shaft, BC of length L, as indicated in the figure.

The shear modulus of the steel core is Gcore=G0, while the shear modulus of the copper sleeve is Gsleeve=(1/2)G0.

The given KNOWN quantities are L[m], R[m], q[N·m/m] and G0[Pa] (enter this as G_0).

1) Obtain a symbolic expression for the x-component of the reaction torque at D in terms of q and L:

TxD=

2) Obtain expressions for the twist rate dφdx(x), (in terms of q, x, L, G0, and R), and obtain the position x0 along the shaft where the twist rate goes to zero (dφdx(x0)=0), (in terms of L).

If you have factors of π in your answer, enter π as pi.

for0≤x<L,dφdx(x)=
unanswered
forL<x≤2L,dφdx(x)=
unanswered
for2L≤x≤3L,dφdx(x)=
unanswered
dφdx(x0)=0atx0=
unanswered

3) Obtain a symbolic expression for the maximum absolute value of the shear stress in the shaft, τmax, (in terms of q, L, and R).

If you have factors of π in your answer, enter π as pi.

τmax=

4) Obtain symbolic expressions for the maximum value of the rotation field φ(x) along the shaft φmax (in terms of q, L, G0, and R), and the position along the shaft where the maximum rotation occurs, xφmax (in terms of L).

If you have factors of π in your answer, enter π as pi.

φmax=
unanswered
xφmax=

To solve this problem, we can use the principles of torsion for composite shafts. Torsion is the twisting of an object under the influence of a torque.

1) To find the x-component of the reaction torque at D, we need to consider the torque equilibrium. This means that the sum of the torques acting on the shaft must equal zero. Since a distributed torque per unit length acts only on the central segment BC, the torque at D is equal to the integral of the torque per unit length over the length of BC.

TxD = ∫q dx, for L≤x≤2L

Here, q is the distributed torque per unit length, and dx is an infinitesimal length element along the shaft.

2) To find the twist rate dφ/dx(x) along the shaft, we need to consider the torque and the shear modulus of the materials. The twist rate is defined as the rate of change of the twist angle φ with respect to the length x.

For 0≤x<L, since there is no torque acting on this segment, dφ/dx(x) = 0.

For L<x≤2L, we can use the formula for the twist rate due to the torque:

dφ/dx(x) = (q/(G0*π*R^4)) * (x-L), for L<x≤2L

For 2L≤x≤3L, the material changes to copper and its shear modulus is (1/2)G0. So, we need to use this new modulus in the formula:

dφ/dx(x) = (q/(0.5*G0*π*R^4)) * (x-L), for 2L≤x≤3L

To find the position x0 along the shaft where the twist rate goes to zero, we can solve dφ/dx(x0) = 0 for x0:

0 = (q/(G0*π*R^4)) * (x0-L) => x0 = L

Therefore, the twist rate goes to zero at x0 = L.

3) The maximum absolute value of the shear stress in the shaft, τmax, can be found using the formula:

τmax = (q*R) / (2*G0), for 0≤x≤3L

4) To find the maximum value of the rotation field φ(x) along the shaft, we need to use the formula for φ(x) due to the torque:

For 0≤x≤3L, φ(x) = (q*x^2) / (2*G0*π*R^4)

To find the position xφmax along the shaft where the maximum rotation occurs, we can take the derivative of φ(x) with respect to x and solve for x:

dφ/dx = (q*x) / (G0*π*R^4) = 0 => xφmax = 0

Therefore, the maximum rotation occurs at xφmax = 0.

Summarizing the answers:

1) TxD = ∫q dx, for L≤x≤2L
2) dφ/dx(x) = 0, for 0≤x<L
dφ/dx(x) = (q/(G0*π*R^4)) * (x-L), for L<x≤2L
dφ/dx(x) = (q/(0.5*G0*π*R^4)) * (x-L), for 2L≤x≤3L
dφ/dx(x0) = 0 at x0 = L
3) τmax = (q*R) / (2*G0)
4) φmax = (q*x^2) / (2*G0*π*R^4), for 0≤x≤3L
xφmax = 0