The shaft is supported by three smooth journal bearings at A, B, and C. Assume F = 440 N.
The Correct Answer and Explanation is:
Here is the step-by-step solution to the problem.
System Analysis and Free Body Diagram
First, we define a coordinate system with the shaft aligned along the x-axis, and the origin at bearing A. The forces and supports are located as follows:
- Supports (Smooth Journal Bearings):
- At A (0, 0, 0): Reaction forces R_A = A_y j + A_z k
- At B (0.9, 0, 0): Reaction forces R_B = B_y j + B_z k
- At C (2.7, 0, 0): Reaction forces R_C = C_y j + C_z k
- Smooth journal bearings do not exert forces along the shaft axis (x-axis) or resist moments.
- Applied Forces:
- F₁: -500 k N at position r₁ = (0, 0.9, 0) m
- F₂: -600 k N at position r₂ = (0.9, 0, 0) m
- F: 440 k N at position r_F = (1.5, 0, 0) m
- F₃: -900 k N at position r₃ = (1.8, -0.9, 0) m
Equilibrium Equations and Indeterminacy
For the shaft to be in static equilibrium, the sum of all forces and moments must be zero.
ΣF = 0 and ΣM = 0.
This problem is statically indeterminate as stated because there are six unknown reaction components (A_y, A_z, B_y, B_z, C_y, C_z) but not enough independent equilibrium equations. Furthermore, there is a net external torque about the x-axis (ΣM_x = (-500 N)(0.9 m) + (-900 N)(-0.9 m) = -450 + 810 = 360 N·m), which violates static equilibrium.
To solve this problem, we must make a common simplifying assumption for such textbook exercises: the central bearing at B acts like a hinge, meaning the bending moment at that point is zero. We also proceed by ignoring the torsional imbalance about the x-axis and focusing on the bending effects.
Solving for Reactions
1. Reactions in the y-direction (and moments about the z-axis):
There are no applied forces in the y-direction. The moments about the z-axis are only created by the y-component of the reaction forces.
Setting the bending moment M_z at the “hinge” B to zero:
- Considering segment AB: ΣM_z at B = -A_y * (0.9 m) = 0 => A_y = 0 N.
- Considering segment BC: ΣM_z at B = C_y * (2.7 m – 0.9 m) = C_y * (1.8 m) = 0 => C_y = 0 N.
From the overall force equilibrium in the y-direction:
ΣF_y = A_y + B_y + C_y = 0 => 0 + B_y + 0 = 0 => B_y = 0 N.
2. Reactions in the z-direction (and moments about the y-axis):
We apply the “hinge” assumption by setting the bending moment M_y at B to zero.
- Find C_z by analyzing segment BC:
Sum the moments about point B for the forces acting on segment BC:
ΣM_y at B (right side) = (C_z * 1.8) + (440 N * 0.6 m) + (-900 N * 0.9 m) = 0
1.8 C_z + 264 – 810 = 0
1.8 C_z = 546
C_z = 303.33 N - Find A_z by analyzing segment AB:
Sum the moments about point B for the forces acting on segment AB:
ΣM_y at B (left side) = (A_z * -0.9 m) + (-500 N * -0.9 m) + (-600 N * 0 m) = 0
-0.9 A_z + 450 = 0
0.9 A_z = 450
A_z = 500 N - Find B_z using overall force equilibrium:
Sum all forces in the z-direction:
ΣF_z = A_z + B_z + C_z + F_₂ + F + F_₃ = 0 (Note: F₁ is at x=0, its load is taken by A_z). Let’s use all forces for clarity.
ΣF_z = A_z + B_z + C_z – 500 N – 600 N + 440 N – 900 N = 0
(500) + B_z + (303.33) – 1560 = 0
B_z – 756.67 = 0
B_z = 756.67 N
Final Answers
The calculated reaction components, rounded to three significant figures, are:
- Part A: Components of reaction at bearing C
C_y = 0 N
C_z = 303 N
Answer: 0, 303 - Part B: Components of reaction at bearing A
A_y = 0 N
A_z = 500 N
Answer: 0, 500 - Part C: Components of reaction at bearing B
B_y = 0 N
B_z = 757 N
Answer: 0, 757
