# Spring constant of angled beam

I'm trying to solve for the dynamics of three blocks, shown below. Two blocks are pushed into a central shuttle, and their force is transmitted through a thin angled beam/pawl. The pawl bends when pushed quickly, so I've observed shuttle velocities ($$\dot{x}$$) way lower than I originally expected from modelling the pawl as a rigid kinematic constraint.

I hypothesize that I can model the dynamics better by modelling the spring constant of this angled pawl, but I'm not sure from my intro solid mechanics courses how to do so. My first guess that I could potentially adapt the fixed-guided beam bending formula like so:

$$m_1 \ddot{y} = F_y - k_1y = F_y - \frac{12EI}{L^3}\left(\frac{y}{\sin(\alpha)}\right)$$ $$m_2 \ddot{x} = k_2y = \left(\frac{k_1}{\tan(\alpha)}\right)y =\frac{12EI}{L^3}\left(\frac{y}{\sin(\alpha)\tan(\alpha)}\right)$$

Is this the correct way to model the spring constant of an angled beam? I'd also appreciate suggestions if I'm modelling the dynamics themselves wrong, but my question is focused on the spring constant first and foremost. Thank you very much!

• Is everything pin jointed? Assuming it is then -Your FBD of the pawl is not in equilibrium. You need to split the force at each end into an axial force and a transverse load. But, absent buckling there's nothing there to bend the pawl. It probably isn't important but what forces act on the m1 blocks to keep them vertical? Commented Jul 9, 2023 at 2:28
• @GregLocock If pinned, I don't think the angled beam could work as a bending spring in this case, so it would have to be fixed at least at one end. On the other hand, it could still work as a simple tension/compression spring, which should be the easiest to incorporate into the model. Commented Jul 9, 2023 at 6:41
• @GregLocock As mentioned in the OP, I assumed a fixed-guided beam would be the best approximation (fixed on the side of m1, guided in how it moves on the shuttle). That said, my prior coursework never really covered how to handle a guided condition if the beam is fixed to move on a (relatively) angled line instead of perpendicular. Commented Jul 9, 2023 at 15:50
• @TomášLétal By tension-compression spring, did you mean it in the sense I've used it in my dynamics equations (force = kx)? If so, my confusion is coming from what my K should be for an angled fixed-guided beam. The beam's mass is trivial comes to m1 and m2, so in hindsight the FBD should be nearly at equilibrium somehow? Commented Jul 9, 2023 at 15:55
• @asyndeton256 For tension-compression, $k$ would be the "spring constant" as a ratio between the axial force $F_a$ and axial elongation $u_a$. So in this case, $k = \frac{(w\cdot T)\cdot E}{L}$. The axial direction would be in along the length $L$. Commented Jul 9, 2023 at 18:56

I think the basic approach to the spring constant should be based on simple tension-compression, i.e.: $$k = \frac{(w\cdot T)\cdot E}{L}$$
If you consider the initial state without any forces and moments and then introduce just forces $$F_x$$ and $$F_y$$ in static equilibrium, there might be little to no bending ($$\alpha$$ close to 45°). For the plate to transfer bending moments, the end angles would have to deviate from the initial $$\alpha$$, which I think would be second order effect, causing nonlinear behaviour which you cannot describe by a stiffness constant anyway.
One problem might arise in compression due to $$w \ll L$$, although in such case, you would need to consider Euler beam buckling (or plate buckling if $$T$$ is more like $$L$$ than $$w$$ in this case). This will not help you much with linear stiffness, because it only allows you to calculate a critical compression force, below which the compression constant still works, but above which the behaviour becomes nonlinear.