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This are eccentrically loaded bolts in which the plane of loading is parallel to the bolt plane.When deriving an equation to this particular problem almost all books start by assuming that $$\frac{F_{1}}{r_{1}}=\frac{F_{2}}{r_{2}}=\frac{F_{3}}{r_{3}}=...=c$$ So, how can one make such assumptions and have a correct answer.Meaning, wouldn't a different assumption cause different results? Is there some type physical meaning to the assumption made?

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  • $\begingroup$ I don't think this is a physical requirement. You can imagine a situation where the hole for bolt 1 was drilled too big so it isn't contributing at all requiring the three other bolts to take on all of the stress, and one could have intermediate situations. I think having the stress distributed evenly across all of the bolts to minimize the stress on any individual bolt is a design desire. $\endgroup$
    – Malcolm
    Jan 12 at 15:41
  • $\begingroup$ Similarly, the assumption that the direct sheer is perfectly distributed across the 4 bolts is not a physical requirement, but is the best case to minimize the maximum shear on any bolt. $\endgroup$
    – Malcolm
    Jan 12 at 15:52

3 Answers 3

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This answer contains a paragraph about the assumptions of a rigid motion of the beam needed to get the decomposition shown in the question, a paragraph about the linear system representing the equilibrium conditions, and a last paragraph about the detail needed to get equal "translational contribution" on all the bolts (in the picture $P/4$, in general $P/n$ with $n$ bolts): the reference point of the rigid motion must be the geometric center of the bolts.

Assumptions

The decomposition of the forces as shown in the picture is consistent with the following assumptions:

  • stiff beams if compared with the joints, so that the beams can only perform rigid motion: thus the system has 3 degrees of freedom and the displacement of a point a beam is the composition of the displacement of a reference point $P$ and the rotation around it. For small displacements, these two contributions are summed as follow \begin{equation} \mathbf{u}_i = \mathbf{u}_P + \theta \times \mathbf{r}_i \ , \end{equation} Actually, it's not strictly necessary that the whole beam is treated as rigid, but only the region with the bolts.
  • force on all the bolts is linear with the displacement of the point, with the same stiffness constant (equal bolts assumption) \begin{equation} \mathbf{F}_i = - k \, \mathbf{u}_i = -k \left( \mathbf{u}_P + \theta \times \mathbf{r}_i \right) = -k \left( \mathbf{u}_P -\mathbf{r}_i \times \theta \right) \ ; \end{equation} from this expression, it's easy to realize that:
    • the first contributions is aligned with the displacement $\mathbf{u}_P$, and constant for every bolt;
    • the second contribution is an azimutal contribution, proportional to $r_i$.

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Linear system from equilibrium equations

In order to find the values of $\mathbf{u}_P$ and $\theta$, we need to write equilibrium equations \begin{equation}\begin{aligned} \text{force:} & \quad \mathbf{0} = \mathbf{F} + \sum_i \mathbf{F}_i = \mathbf{F} - k \sum_i \left( \mathbf{u}_P - \mathbf{r}_i \times \theta \right) \\ \text{moment:} & \quad \mathbf{0} = \mathbf{r}_F \times \mathbf{F}+ \sum_i \mathbf{r}_i \times \mathbf{F}_i = \mathbf{r}_F \times \mathbf{F} - k \sum_i \mathbf{r}_i \times\left( \mathbf{u}_P - \mathbf{r}_i \times \theta \right) \\ \end{aligned}\end{equation}

This linear system can be thus written as \begin{equation} k \begin{bmatrix} n \mathbb{I} & \left[-\sum_{i=1}^{n} \mathbf{r}_i \right]_{\times} \\ \left[-\sum_{i=1}^{n} \mathbf{r}_i \right]^T_{\times} & -\sum_{i=1}^{n} \mathbf{r}_{i\times} \mathbf{r}_{i \times} \end{bmatrix} \, \begin{bmatrix} \mathbf{u}_P \\ \theta \end{bmatrix} = \begin{bmatrix} \mathbb{I} \\ \mathbf{r}_{F \times} \end{bmatrix} \mathbf{F} \end{equation} This linear system contains the identity tensor $\mathbb{I}$, a tensor of static moment $\sum_{i=1}^{n} \mathbf{r}_{i \times}$ (this contribution is identically zero if we choose $P$ to be the geometric center of the bolts), and the tensor of inertia, $-\sum_{i=1}^{n} \mathbf{r}_{i\times} \mathbf{r}_{i \times}$.

Solution w/ reference point $P \equiv$ geometric center

In this case, $\sum_{i=1}^{n} \mathbf{r}_{i \times}$ and the linear system becomes \begin{equation} k \begin{bmatrix} n \mathbb{I} & \mathbb{O} \\ \mathbb{O} & -\sum_{i=1}^{n} \mathbf{r}_{i\times} \mathbf{r}_{i \times} \end{bmatrix} \, \begin{bmatrix} \mathbf{u}_P \\ \theta \end{bmatrix} = \begin{bmatrix} \mathbb{I} \\ \mathbf{r}_{F \times} \end{bmatrix} \mathbf{F} \end{equation} so that \begin{equation} \mathbf{u}_P = \dfrac{1}{n k} \mathbf{F} \qquad , \qquad \theta = \dfrac{1}{k} \left[ -\sum_{i=1}^{n} \mathbf{r}_{i\times} \mathbf{r}_{i \times} \right]^{-1} \cdot \mathbf{r}_F \times \mathbf{F} \ . \end{equation}

For a planar problem,

  • $\theta = \theta \mathbf{\hat{z}}$
  • $-\sum_{i=1}^{n} \mathbf{r}_{i\times} \mathbf{r}_{i \times} = \sum_i r_i^2 \mathbf{\hat{z}}\mathbf{\hat{z}}$
  • $\mathbf{r}_F \times \mathbf{F}$ aligned with $\mathbf{\hat{z}}$ \begin{equation} \mathbf{u}_P = \dfrac{1}{n k} \mathbf{F} \qquad , \qquad \theta = \dfrac{1}{k \sum_i r_i^2} \mathbf{r}_F \times \mathbf{F} \ . \end{equation}

From these expressions of $\mathbf{u}_P$ and $\theta$ is then possible to retrieve the actual value of the joint displacements and forces.

It's easy to realize that the "translational contribution" is the same for all the bolts, and this contribution equilibrates the external force $\mathbf{F}$, being \begin{equation} \sum_{i=1}^{n} \mathbf{F}^{tr}_i = - k \sum_{i=1}^{n} \mathbf{u}_P = - k \, n \, \mathbf{u}_P = - k \, n \, \dfrac{1}{n \, k}\mathbf{F} = -\mathbf{F} \ . \end{equation}

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  • $\begingroup$ You need to see where $F \propto r$ and not $F \propto 1/r$, since direct proportionality comes from the condition $F/r = \text{const.}$ stated in the question. You can find bold text in my answer, just before the picture $\endgroup$
    – basics
    Jan 28 at 17:27
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    $\begingroup$ yes, you are correct. My mistake. $\endgroup$ Jan 28 at 20:18
  • $\begingroup$ don't worry. If i had to count all my mistakes, it would take forever $\endgroup$
    – basics
    Jan 28 at 21:28
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I too had the same doubt but here is what I make sense of it.

$I*\alpha=P*e$ ; α=angular acceleration

$I= m_1*l_1^2 +m_2*l_2^2+m_3*l_3^2+m_4*l_4^2$

$α=P*e/(m*(l_1^2+l_2^2+l_3^2+l_4^2))$ -1

Angular acceleration of all mass particles in the system is same. Know let us assume that there was no external Torque ($P*e$) and the bolts produced the same reaction Moment. Then each bolts contribution for the net Moment must be in such that its angular acceleration about CG is same.

So,

$F_1/(l_1*m_1) = F_2/(l_2*m_2)=F_3/(l_3*m_3)=F_4/(l_4*m_4)= α $

General assumption is all bolts are same. So, $m_1=m_2=m_3=m_4=m$

$F_1/l_1 = F_2/l_2 =F_3/l_3 =F_4/l_4 = m*α$ -2

From 1 & 2

$F_1/l_1 = F_2/l_2 =F_3/l_3 =F_4/l_4 = α = P*e/(l_1^2+l_2^2+l_3^2+l_4^2)$

The last equation is given in many textbooks. So I assume the approach must be true.

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  • $\begingroup$ Welcome to Physics.SE. Please typeset your equations using MathJax: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – NDewolf
    Mar 18, 2020 at 9:44
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    $\begingroup$ Done, Thanks for the suggestion. $\endgroup$
    – Arjun Raj
    Mar 18, 2020 at 10:23
  • $\begingroup$ The angular acceleration is from the sum of all of the applied torques. Each individual torque does not have to be the same. Consider a windlass with four sailors each pushing on a separate arm, they don't all miraculously push with exactly the same force. $\endgroup$
    – Malcolm
    Jan 12 at 15:48
  • $\begingroup$ I don't think dynamics are at play here. I think this problem needs to be examined from a structural point of view. $\endgroup$ Jan 23 at 13:38
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This is the only natural explanation that I could think of.

Assume that the beam is rotating about a common point O, as shown in the figure. The loading on the beam creates a clockwise torque, which makes the bolts move tangentially to $O$. These bolts (shown as green dots) are pushing against the matrix of the column, which in turn provides an anticlockwise restoring torque, bringing the entire system to rest.

Now imagine the beam had rotated by a small angle $d\theta$.

Each of the bolts would have deformed the matrix of the column by a distance $r d\theta$, where $r$ is the distance of the bolt from the axis of rotation. By Hooke's law, the restoring force provided by the matrix of the column due to this compression would be $$F=kx=krd\theta$$.

By this argument, for a fixed angular displacement of the beam, the Forces acting on the bolt are directly proportional to its distance from the axis of rotation.

$$F\propto r$$

Applying this to all other bolts gives you the relation.

$$\dfrac{F_i}{r_i}=constant$$

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