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: enter image description here
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

enter image description here

enter image description here

  • $\begingroup$ The method to derive P is incorrect, as the deflection equation is valid for the cantilever beam with a uniform cross-section only. If you bend the beam with the equal opposite moment, the potential critical section is at the necking area just under the root of the fillet. $\endgroup$
    – r13
    Jul 19, 2021 at 4:01
  • $\begingroup$ @r13 My beam has a constant cross-section. I am evaluating the situation when the force P is applied perpendicular to the beam at the bottom of the "hook" feature (i.e. at the interface between the beam and this feature). I am not sure what you mean by your second sentence. I am aware that the critical point is at the root of the fillet, since bending moment will be maximum there. $\endgroup$
    – Benjamin
    Jul 19, 2021 at 4:48
  • $\begingroup$ Assume your model is as shown in the last picture and let's ignore the thickened part at the end, you still don't have a uniform cross-section, as the fillet at the left end is much rigid than the rest of the beam. Thus, in reality, your P will increase and makes the situation worse. Also, you should indicate the yield and allowable stress, or incorporate the allowable stress as a fraction of the yield stress in the parameter study, so people can compare. Also, I sense the deflection limit is quite large that may invalidate the use of elastic analytical method and the equations. $\endgroup$
    – r13
    Jul 19, 2021 at 5:18
  • $\begingroup$ @r13 Thank you for the concerns. I am assuming that the fillet has negligible effects on the stress for now, since the fillet only changes the cross-section in the last ~8% of the beam. I wanted to attack the more pertinent problem first, then I'll take care of this. Can't you (the reader) find the allowable stress by dividing material yield strength by the SF? Maybe the deflection limit you're talking about is giving me a lot of error. Could you explain more or give me a reference? I would appreciate it! $\endgroup$
    – Benjamin
    Jul 19, 2021 at 13:28
  • $\begingroup$ I think you shall use the Timoshenko Beam Theory that takes into account shear deformation and distortion. Also, unlike Euler-Bernoulli Beam Theory, the plane needs not to be plane after deformation, it opens the door for flexible materials with large deflections. You can do a google search on "Timoshenko Beams" for more info. $\endgroup$
    – r13
    Jul 19, 2021 at 15:13

1 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.

enter image description here

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 $P = \dfrac{EI \lambda\delta_v }{L^3}$.

enter image description here

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.

  • $\begingroup$ I understand your explanation, but what is $\lambda$? I could solve for $P$ in the deflection equation if only I knew $\lambda$. Is that Table 7-3 from "Mechanics of Materials"? Could you tell me what edition that is? I want to find the book so I can read this chapter. $\endgroup$
    – Benjamin
    Jul 21, 2021 at 3:23
  • $\begingroup$ λ was defined in my answer as the division (M)/(n) using the respective numerical value from columns (m) and (n) on table 7-3. Yes, it is in the book, the book is 2ed. $\endgroup$
    – r13
    Jul 21, 2021 at 13:40
  • $\begingroup$ I understand that $\lambda$ represents this fraction. But how do I eliminate it from the equation? There are two variables to be solved for, then, unless something can be substituted in for $\lambda$. $\endgroup$
    – Benjamin
    Jul 21, 2021 at 13:43
  • $\begingroup$ Set the deflection limit (δv) first, and calculate (n) = (δv/L), then find the (m) value corresponding to (n), and solving for P. For example (n) = (δv/L) = 0.066, from the table, (m) = 0.2, λ = 0.066/0.2 = 0.33, P = [(EIδv)/L^3]*0.33. $\endgroup$
    – r13
    Jul 21, 2021 at 14:18
  • 1
    $\begingroup$ Note, due to the simplification of the model and the difficulty in capture the real plastic behavior, the solution is as good as an approximation. Ultimately, you shall use a FEM program to verify the finding or be conservative in applying the load. $\endgroup$
    – r13
    Jul 22, 2021 at 1:36

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