Finding Safety Factor for Plastic Snap Fit Cantilever

Question

I'm trying to get a safety factor of $n\geq2$ for the root of my snap fit beam (pic at bottom of the post).
I have gone through the calculations to get the SF at the root's edge from the geometric and material properties (using a variety of plastics), and my SF seems unnaturally small.
I don't have a lot of experience with this yet, so I wanted to check here to make sure I didn't forget something.
If someone would be kind enough to check my work, I would greatly appreciate it. Is my work incorrect, or have I just overestimated the flexibility of my materials?

Find: Safety factor at the root edge, $n_e$
(I also found SF at the root center, $n_c$, from transverse shear, but this isn't critical since stress from bending moment is zero there.)

Geometric properties
Beam length: $l=6.40~\text{mm}$
Maximum deflection: $y_{d,\text{max}}=1.20~\text{mm}$
Beam depth: $h=1.60~\text{mm}$
Beam width: $b=7.31~\text{mm}$
Root fillet radius: $r_f=0.48~\text{mm}$

Material properties (ex.)
SABIC LNP STAT-KON 5E003M: $S_y=50~\text{MPa}$, $E=9060~\text{MPa}$
(this is a fairly brittle plastic, but I got very small SFs for many other plastics as well)

Equations (beam root) $$I=\frac{1}{12}bh^3~~~~~~~~~~~~~~P=\frac{6y_dEI}{x^2(x-3l)}=-\frac{3y_{d,\text{max}}EI}{l^3}~~~~~~~~~~~~~~y_{d,\text{max}}=y_{d,x=l}$$ Equation for $P$ is based on the cantilever diagram at the bottom of the post.

Variable shear force and bending moment: $$V(x)=P~~~~~~~~~M(x)=P(l-x)$$ At $x=0$: $$V_{x=0}=-\frac{3y_{d,\text{max}}EI}{l^3}~~~~~~~~~M_{x=0}=-\frac{3y_{d,\text{max}}EI}{l^2}$$ Normal stress at the root edge: $$\sigma=\frac{My}{I}~~~~~~y=\frac{1}{2}h~~~~~~\sigma_\text{max,nom}=\frac{Mh}{2I}~~~~~~\sigma_\text{max}=K_t\sigma_\text{max,nom}$$

Stress concentration factor $K_t$ can be found by:
From my design, $K_t$ turns out to be $\approx1.40$.

Shear stress at root center: $$\tau_\text{max,nom}=\frac{3V}{2A}~~~~~~~~~~~\tau_\text{max}=\tau_\text{max,nom}$$ Von Mises stresses: $\sigma_c'=\sqrt{3\tau_\text{max}^2}$ at root center, $\sigma_e'=\sigma_\text{max}$ at root edge.

Safety factor: $$n_c=\frac{S_y}{\sigma_c'}~~~~~~~~~~~~~~~~n_e=\frac{S_y}{\sigma_e'}$$

Results (beam root)
For 5E003M, $n_e=0.056$ and $n_c=0.725$.

As you can see, these safety factors are terrible! As far as changing the material properties, I could decrease the width or depth of the beam, but the length must stay the same.
Any insight would be appreciated, thanks! :D

Answer 1

This is a suggested procedure to achieve your goal. I'll simplify your model to a cantilever beam with a constant cross-section as shown below, and assume the critical section is located at the root of the fillet.

The first step is to determine the maximum deflection $\delta_v$.

The second step is to find out the equation for the deflection. For this exercise, I'll ignore the shear deformation, but apply the method developed by Timoshenko to account for "large deflection".

In his book, "Mechanics of Materials", co-authored with Gere, Timoshenko provided a table for ease of pinpoint the correct equation as shown below. In the table, column (m) is the numerical value of $\dfrac{PL^2}{EI}$, and column (n) is the numerical value of $\dfrac {\delta_v}{L}$. The equation of deflection is simply (m)/(n).

• $(m)$ = $\dfrac{PL^2}{EI}$, $(n)$ = $\dfrac {\delta_v}{L}$, $\dfrac{(m)}{(n)}$ = $\dfrac{PL^2}{EI}$/$\dfrac {\delta_v}{L}$.

• Let $\lambda = \dfrac{m}{n}$ and rearrange the terms,

• $\delta_v = \dfrac {PL^3}{EI \lambda}$, or $P = \dfrac{EI \lambda\delta_v }{L^3}$.

The third step is to check shear stress:

At this point, I'll conservatively introduce the form factor to account for the escalated shear stress due to shear deformation. For a rectangular section, the form factor $f_s$ = 6/5 = 1.2, so

$\sigma = f_s*\dfrac{3P}{2A} \leq \dfrac{f_y}{n}$, where $n$ is the desired safty factor.

The fourth step is to determine the bending stress.

1. Assume elastic behavior, $\sigma_b = \dfrac{6M}{bh^2} \leq \dfrac{F_y}{n}$, or

2. assume plastic behavior, $\sigma_b = \dfrac{4nM}{bh^2} \leq F_y$

The last step is to determine $P$:

Since $M = P*a$, plug $M$ into the two equations above, you will get the $P$ that satisfies the limit of the bending stress with the desired safety factor. However, you also need to back-check/compare the $P$ derived from steps two (deflection) and three (shear stress) before making the conclusion.

If you still couldn't get satisfactory force with the desired safety factor, you will have to increase the depth/thickness of the member or adjust the deflection limit.

Hope this helps.