a) To find the reactions at A and C, we need to consider the equilibrium of the bar. The bar is in static equilibrium, which means that the sum of all forces and moments acting on the bar is equal to zero.
Let Ra be the reaction force at point A, and Rc be the reaction force at point C. Then, the sum of forces in the x-direction is:
Ra + Rb = 0
Since there are no forces acting in the y-direction, the sum of forces in the y-direction is:
Rc = 0
The sum of moments about point A is:
To + t(BC) * BC - Ra * 3d = 0
The sum of moments about point C is:
Rb * d = 0
Since Rc = 0, we know that Rb = 0. Therefore, the reactions at A and C are:
Ra = -To/(3d - dBC)
Rc = 0
b) To find the twist angle at B, we can use the torsion formula:
Ape = (To * L * dBC)/(4 * G * I)
where G is the shear modulus of the material, and I is the polar moment of inertia of the bar. Since the bar is circular, we have:
I = (pi/32) * (3d^4 - d^4)
Plugging in the given values, we get:
Ape = (To * L * dBC)/(4 * G * (pi/32) * (3d^4 - d^4))
c) To find the maximum shear stress, we need to consider the shear stress at each point along the bar and find the maximum absolute value. The shear stress at any point x along the bar can be found using the torsion formula:
t(x) = (To * (L - x) * dBC)/(2 * G * I)
The maximum absolute value of the shear stress occurs at either point B or point C, depending on the sign of t(BC). If t(BC) is positive, then the maximum shear stress occurs at point C. If t(BC) is negative, then the maximum shear stress occurs at point B.
Therefore, Tmax = max(|t(BC)|, To/(2 * G * d^3/4))