Vertical reaction on a shaft with 3 bearing supports

Question

I am doing an analysis on a shaft loaded by a gear at point B and supported by three bearings at points A, C and D. Is it possible to solve for reaction forces in the vertical direction for points A, C and D knowing the vertical component of B is given?

Answer 1

With the given information, no. The problem is statically indeterminate. You have three unknowns (A,C,D), but only two possible equations (sum vertical forces, sum moments).

To solve it, you will need to use a method suitable for statically indeterminate systems. You'll need to know the stiffness of the bearings and the stiffness of the shaft.

However, because C and D are close to each other, but far from A, you could get an APPROXIMATE answer by merging C and D into one support halfway in between them, and then whatever answer you get divide it equally among C & D. That would be good for at least an order of magnitude estimate.

Answer 2

It is possible with some simplification, which is however generally the case when modelling a real situation. Provided that the supports are basically pinned (unrestrained rotations) and deformations are small, you can use FEA like approach. Although there may be simpler analytical solution, FEA approach does not require much thinking, you just need to be precise with stiffness coefficients and it is practical to use computer for the system of equations :).

FEA beam element

A beam is governed by set of 4 linear equations (for example from here):

$$K\cdot U = F$$

or

$$EI\left[\begin{matrix} \frac{12}{L^3} & \frac{6}{L^2} & -\frac{12}{L^3} & \frac{6}{L^2} \\ \frac{6}{L^2} & \frac{4}{L} & -\frac{6}{L^2} & \frac{2}{L} \\ -\frac{12}{L^3} & -\frac{6}{L^2} & \frac{12}{L^3} & -\frac{6}{L^2} \\ \frac{6}{L^2} & \frac{2}{L} & -\frac{6}{L^2} & \frac{4}{L} \end{matrix}\right]\cdot \left[\begin{matrix} w_i \\ \varphi_i \\ w_j \\ \varphi_j \end{matrix}\right] = \left[\begin{matrix} F_i \\ M_i \\ F_j \\ M_j \end{matrix}\right]$$

where $w_i$, $\varphi_i$, $w_j$ and $\varphi_j$ are transverse displacements and rotations at beam endpoints $i$ and $j$ with corresponding forces and moments $F_i$, $M_i$, $F_j$ and $M_j$

System of 3 elements

For your situation, you will need 3 interconnected elements, which is also governed by a system of linear equations:

$$K_s\cdot U_s = F_s$$

With lengths of the beams denoted as $a$, $b$ and $c$ and stiffness coefficients (shown here just for $a$):

$$k_{12, a} = 12\frac{E_a I_a}{L_a^3}$$ $$k_{6, a} = 6\frac{E_a I_a}{L_a^2}$$ $$k_{2, a} = 2\frac{E_a I_a}{L_a}$$

The system of equations would look like this:

$$\left[\begin{matrix} \color{blue}{k_{12, a}} & \color{blue}{k_{6, a}} & \color{blue}{-k_{12, a}} & \color{blue}{k_{6, a}} & & & & \\ \color{blue}{k_{6, a}} & \color{blue}{2k_{2, a}} & \color{blue}{-k_{6, a}} & \color{blue}{k_{2, a}} & & & \\ \color{blue}{-k_{12, a}} & \color{blue}{-k_{6, a}} & \color{blue}{k_{12, a}}+k_{12, b} & \color{blue}{-k_{6, a}}+k_{6, b} & -k_{12, b} & k_{6, b} \\ \color{blue}{k_{6, a}} & \color{blue}{k_{2, a}} & \color{blue}{-k_{6, a}}+k_{6, b} & \color{blue}{2k_{2, a}}+2k_{2, b} & -k_{6, b} & k_{2, b} & & \\ & & -k_{12, b} & -k_{6, b} & k_{12, b}+\color{green}{k_{12, c}} & -k_{6, b}+\color{green}{k_{6, c}} & \color{green}{-k_{12, c}} & \color{green}{k_{6, c}} \\ & & k_{6, b} & k_{2, b} & -k_{6, b}+\color{green}{k_{6, c}} & 2k_{2, b}+\color{green}{2k_{2, c}} & \color{green}{-k_{6, c}} & \color{green}{k_{2, c}} \\ & & & & \color{green}{-k_{12, c}} & \color{green}{-k_{6, c}} & \color{green}{k_{12, c}} & \color{green}{-k_{6, c}} \\ & & & & \color{green}{k_{6, c}} & \color{green}{k_{2, c}} & \color{green}{-k_{6, c}} & \color{green}{2k_{2, c}} \end{matrix}\right]\cdot \left[\begin{matrix} w_a \\ \varphi_a \\ w_b \\ \varphi_b \\ w_c \\ \varphi_c \\ w_d \\ \varphi_d \end{matrix}\right] = \left[\begin{matrix} F_a \\ M_a \\ F_b \\ M_b \\ F_c \\ M_c \\ F_d \\ M_d \end{matrix}\right]$$

Boundary conditions

The system of equations needs to be modified using boundary conditions, so all the unknowns are in the left vector and the right vector is completely known:

$$K_{s,mod} \cdot U_s = F_{s, mod}$$

There is a known force $F_b$ at point B, which is reflected in the right vector. There are also known lateral displacements $w_a$, $w_c$ and $w_d$, which are all 0 and this is most easily implemented by replacing corresponding 3 equations by $1\cdot w_i = 0$.

$$\left[\begin{matrix} 1 & & & & & & & \\ \color{blue}{k_{6, a}} & \color{blue}{2k_{2, a}} & \color{blue}{-k_{6, a}} & \color{blue}{k_{2, a}} & & & \\ \color{blue}{-k_{12, a}} & \color{blue}{-k_{6, a}} & \color{blue}{k_{12, a}}+k_{12, b} & \color{blue}{-k_{6, a}}+k_{6, b} & -k_{12, b} & k_{6, b} \\ \color{blue}{k_{6, a}} & \color{blue}{k_{2, a}} & \color{blue}{-k_{6, a}}+k_{6, b} & \color{blue}{2k_{2, a}}+2k_{2, b} & -k_{6, b} & k_{2, b} & & \\ & & & & 1 & & & \\ & & k_{6, b} & k_{2, b} & -k_{6, b}+\color{green}{k_{6, c}} & 2k_{2, b}+\color{green}{2k_{2, c}} & \color{green}{-k_{6, c}} & \color{green}{k_{2, c}} \\ & & & & & & 1 & \\ & & & & \color{green}{k_{6, c}} & \color{green}{k_{2, c}} & \color{green}{-k_{6, c}} & \color{green}{2k_{2, c}} \end{matrix}\right]\cdot \left[\begin{matrix} w_a \\ \varphi_a \\ w_b \\ \varphi_b \\ w_c \\ \varphi_c \\ w_d \\ \varphi_d \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ F_b \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right]$$

When you solve this modified system of equations, you will get the whole vector $U_s$. Finally, you can simply calculate the force and moment vector $F_s$ just by multiplying system stiffness matrix $K_s$ by the displacement vector $U_s$:

$$F_s = K_s\cdot U_s$$

For your situation, the resulting forces should be: -12329.700012800344, 22682.5, -57303.35631973247, 46950.556332532804.

Although you need shaft cross section for this approach, the resulting forces should be independent of it (provided the cross section is constant for the whole shaft).