Crimping Tool Failure Analysis Problem Determine the factors of safety at critical points in the crimping tool shown in Figures 3-3 (repeated on next page) and 5-23. Given The stresses are known from Case Study 2B on p. 209. All material is AISI 1095 steel Q & T @ 800 °F with Sy = 112 kpsi

Question

Crimping Tool Failure Analysis

Problem Determine the factors of safety at critical points in the crimping tool shown in Figures 3-3 (repeated on next page) and 5-23.

Given The stresses are known from Case Study 2B on p. 209. All material is AISI 1095 steel Q & T @ 800 °F with [latex]S_{y}[/latex] = 112 kpsi (see Table A-9 in Appendix A). It is an even material.

Assumptions The most likely failure points are link 3 as a column, the holes where the pins insert, the connecting pins in shear, and link 4 in bending.

Accepted Answer

See Figures 3-3 and 5-23, and files CASE2C-1 and CASE2C-2.

1 The previous case study found the critical column load in link 3 to be 3.1 times larger than the applied load. This is the safety factor against buckling, which is expressed in terms of load rather than stress.

2 Any link can fail in bearing in the 0.25-dia holes. The bearing stress (Eqs. 4.7, p. 152, and 4.10, p. 154) is:

[latex] \sigma_x=\frac{P}{A} [/latex] (4.7)

[latex] A_{ ext {bearing }}=l d [/latex] (4.10a)

[latex] A_{ ext {bearing }}=\frac{\pi}{4} l d [/latex] (4.10b)

[latex] \sigma_b=\frac{P}{A_{ ext {bearing }}}=\frac{1560}{0.125(0.25)}=50 kpsi [/latex] (a)

3 As the only applied stress at this element, this is the principal stress and also the von Mises stress. The safety factor for bearing stress on either hole or pin is then

[latex] N=\frac{S_y}{\sigma^{\prime}}=\frac{112}{50}=2.2 [/latex] (b)

4 The 0.25-in-dia pins are in single shear. The worst-case direct-shear stress from equation 4.9 (p. 153) is

[latex] au_{x y}=\frac{P}{A_{ ext {shear }}} [/latex] (4.9)

[latex] au=\frac{P}{A_{ ext {shear }}}=\frac{1560}{\frac{\pi(0.25)^2}{4}}=32 kpsi [/latex] (c)

As the only stress on this section, this is also the maximum shear stress. The safety factor for the pins in single shear from equation 5.9a (p. 251) is

[latex] \begin{array}{c} S_y^2=\sigma_1^2+\sigma_1 \sigma_1+\sigma_1^2=3 \sigma_1^2=3 au_{\max }^2 \ \sigma_1=\frac{S_y}{\sqrt{3}}=0.577 S_y= au_{\max } \end{array} [/latex] (5.9a)

[latex] N=\frac{0.577 S_y}{ au_{\max }}=\frac{(0.577) 112}{32}=2.0 [/latex] (d)

5 Link 4 is a 1.55-in-long beam, simply supported at the pins and loaded with the 2 000- lb crimp force at 0.35 in from point [latex]C[/latex]. The beam depth at the point of maximum moment is 0.75 in and the thickness is 0.187. The bending stress is then

[latex] \sigma=\frac{M c}{I}=\frac{541.8\left\lgroup\frac{0.75}{2} ight group }{\frac{0.187(0.75)^3}{12}}=31 kpsi [/latex] (e)

As the only applied stress on this element at the outer fiber of the beam, this is the principal stress and also the von Mises stress. The safety factor for link 4 in bending is then

[latex] N=\frac{S_y}{\sigma^{\prime}}=\frac{112}{31}=3.6 [/latex] (f)

6 Link 1 has a tensile stress due to bending in the inner fiber at point [latex]P[/latex] on the curved beam superposed on an axial tensile stress at the same point. Their sum is the maximum principal stress:

[latex] \begin{array}{c} \sigma_i=+\frac{M}{e A}\left\lgroup \frac{c_i}{r_i} ight group =\frac{2400}{0.103[(1.0-0.25)(0.313)]}\left\lgroup \frac{0.397}{0.600} ight group =65 kpsi \ \sigma_a=\frac{F}{A}=\frac{2000}{(1.0-0.25)(0.313)}=8.5 kpsi \ \sigma_1=\sigma_a+\sigma_i=65+8.5=74 kpsi \end{array} [/latex] (g)

There is no applied shear stress at point [latex]P[/latex], so this is the principal stress and also the von Mises stress. The safety factor for bending at the inner fiber of the curved beam at point [latex]P[/latex] from equation 5.8a (p. 251) is

[latex] N=\frac{S_y}{\sigma^{\prime}} [/latex] (5.8a)

[latex] N=\frac{S_y}{\sigma^{\prime}}=\frac{112}{74}=1.5 [/latex] (h)

7 At the hole in link 1, there is an axial tensile stress [latex]\sigma _{a}[/latex] from equation ([latex]g[/latex]) increased by the stress concentration factor (see Case Study 2B, step 14, p. 213). The safety factor is found from equation 5.8a:

[latex] N=\frac{S_y}{\sigma^{\prime}} [/latex] (5.8a)

[latex] N=\frac{S_y}{\sigma}=\frac{112}{2.42(8.5)}=5.4 [/latex] (i)

Note that there is also a transverse shear stress at the hole which, when combined with the axial tension, reduces the safety factor at the hole to about 3.7.

8 Some of these safety factors, such as the [latex]N[/latex] = 1.5 for bending in link 1 at point [latex]P[/latex], are somewhat low to guard against user-induced overloads. The safety factors for the pins in shear could also be increased. Either a stronger steel such as SAE 4140 could be selected or the section sizes of the parts could be increased slightly. A small change in link thickness would achieve acceptable safety factors in the existing material. Note that the geometry of this tool has been simplified for this example from that of the actual device. The stresses and safety factors calculated here are not necessarily the same as those in the actual tool, which is a well-tested and safe design.

Question 5.CS.2C: Crimping Tool Failure Analysis Problem Determine the factors...

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