The shaft ACB, of constant outer diameter 2R and length 2L, is fixed to walls at both ends A and B. The left half of the shaft (AC) is solid, while the right half (CB) is hollow, with an inner diameter 2√R. The right half, CB, is subjected to a uniform distributed torque per unit length q0[N⋅m/m]. The material of the shaft is linear isotropic elastic with shear modulus G.

Obtain symbolic expressions in terms of R, G, L, q0, and x for the quantities below. (In your answers, leave rationals as fractions and enter q0 and π as q_0 and pi, respectively.)

EQ2_1_1

The x-component of the reaction torques at A and B:
TxA=
TxB=

The axial torque resultant in the bar T(x), and the position x0 along the shaft where this torque resultant goes to zero (T(x0)=0):
for0≤x≤L,T(x)=
forL≤x≤2L,T(x)=
T(x0)=0 at x0=

EQ2_1_3

The maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax):
τmax=
rτmax=
xτmax=

EQ2_1_4

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=

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

To obtain the symbolic expressions for the given quantities, we can use the equations for torsion in a cylindrical shaft.

EQ2_1_1:
To find the x-component of the reaction torques at A and B (TxA and TxB), we can use the equation for torsional moment. The torsional moment is equal to the product of shear stress (τ) and the polar moment of inertia (J), which is given by:

J = (π/2) * (R^4 - r^4)

where R is the outer radius of the shaft (2R) and r is the inner radius of the hollow section (2√R).

At x = 0 (end A):
TxA = τ * J = τ * (π/2) * (R^4 - r^4)

At x = L (end B):
TxB = τ * J = τ * (π/2) * (R^4 - r^4)

EQ2_1_3:
To find the maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax), we can use the equation for shear stress. The shear stress is given by:

τ = (r/R) * τmax

where τmax is the maximum shear stress and (r/R) is the radius ratio.

Since the maximum shear stress occurs at the outer radius of the shaft (R), we have:
τmax = (R/R) * τmax = τmax

The location of the maximum shear stress can be found at the outer radius of the shaft (rτmax = R) at any point along the shaft (xτmax).

EQ2_1_4:
To find 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), we can use the equation for the rotation angle. The rotation angle is given by:

φ(x) = (q0 / (2G)) * (1 - (x / L)^2)

where q0 is the distributed torque per unit length and G is the shear modulus.

The maximum rotation occurs at the point where the derivative of φ(x) with respect to x is zero (dφ/dx = 0).

To find φmax, we can substitute xφmax into the equation for φ(x) and solve for φmax.

xφmax can be found by solving the equation dφ/dx = 0 for x.

Note: The equations provided above are derived based on the given information and assumptions. Please use them as a guide and replace the symbols with the appropriate values to obtain the exact expressions.