Question 4.CS.2B: Crimping-Tool Stress and Deflection Analysis Problem Determi...
Crimping-Tool Stress and Deflection Analysis
Problem Determine the stresses and deflections at critical points in the crimping tool shown in Figures 3-3 (repeated here) and 4-49.
Given The geometry and loading are known from Case Study 2A on p. 84. The thickness of link 1 is 0.313 in, of links 2 and 3 is 0.125 in and of link 4, 0.187 in. All material is 1095 steel with E = 30 Mpsi.
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. The number of cycles expected over the life of the tool is low, so a static analysis is acceptable. Stress concentration can be ignored due to the material’s ductility and the static loading assumption.
See Figures 3-3, and 4-49 to 4-51.
1 Link 3 is a pinned-pinned column loaded with F_{43} = 1 548 lb as calculated in Case Study 2A (p. 84) and shown in Figure 4-49. Note that l_{eff} = l from Table 4-4. We need to first check its slenderness ratio (Eq. 4.41). This requires the radius of gyration (Eq. 4.34, p. 193) for the weakest buckling direction (the z direction in this case).*
S_r=\frac{l_{e f f}}{k} (4.41)
\frac{P_{c r}}{A}=S_{y c}-\frac{1}{E}\left\lgroup \frac{S_{y c} S_r}{2 \pi}\right\rgroup^2 (4.43)
k=\sqrt{\frac{I}{A}}=\sqrt{\frac{b h^3 / 12}{b h}}=\sqrt{\frac{h^2}{12}}=\sqrt{\frac{0.125^2}{12}}=0.036 \text { in } (a)
The slenderness ratio for z-direction buckling is then
S_r=\frac{l_{e f f}}{k}=\frac{1.228}{0.036}=34 (b)
which is >10, making it other than a short column. Calculate the slenderness ratio of the tangent point between the Johnson and Euler lines of Figure 4-42 (p. 197).
\left(S_r\right)_D=\pi \sqrt{\frac{2 E}{S_y}}=\pi \sqrt{\frac{2(30 E 6)}{83 E 3}}=84.5 (c)
The slenderness ratio of this column is less than that of the tangent point between the Johnson and Euler lines shown in Figure 4-42. It is thus an intermediate-column and the Johnson-column formula (Eq. 4.43, p. 198) should be used to find the critical load.
\frac{P_{c r}}{A}=S_{y c}-\frac{1}{E}\left\lgroup \frac{S_{y c} S_r}{2 \pi}\right\rgroup^2 (4.43)
\begin{aligned} P_{c r} &=A\left[S_y-\frac{1}{E}\left\lgroup \frac{S_y S_r}{2 \pi} \right\rgroup ^2\right] \\ &=0.125(.5)\left[83000-\frac{1}{30 E 6}\left\lgroup \frac{83000(34)}{2 \pi} \right\rgroup ^2\right]=4765 lb \end{aligned} (d)
The critical load is 3.1 times larger than the applied load. It is safe against buckling. Link 2 is a shorter, wider column than link 3 and has lower axial forces so can be assumed to be safe against buckling based on the link 3 calculations.
2 Since it does not buckle, the deflection of link 3 in axial compression is (Eq. 4.7, p. 152):
x=\frac{P l}{A E}=\frac{1548(1.23)}{0.0625(30 E 6)}=0.001 \text { in } (e)
3 Any of the links could also fail in bearing in the 0.25-in-dia holes. The largest force on any pin is 1 560 lb. This worst-case bearing stress (Eqs. 4.7 and 4.10, p. 154) is then
\sigma_x=\frac{P}{A} (4.7)
A_{\text {bearing }}=l d (4.10a)
A_{\text {bearing }}=\frac{\pi}{4} l d (4.10b)
\sigma_b=\frac{P}{A_{\text {bearing }}}=\frac{P}{\text { length }(\text { dia })}=\frac{1560}{0.125(0.25)}=49920 psi (f)
There is no danger of tearout failure in links 2 or 3, since the loading is toward the center of the part. Link 1 has ample material around the holes to prevent tearout.
4 The 0.25-in-dia pins are in single shear. The worst-case direct shear stress from equation 4.9 (p. 153) is:
\tau_{x y}=\frac{P}{A_{\text {shear }}} (4.9)
\tau=\frac{P}{A_{\text {shear }}}=\frac{1560}{\frac{\pi(0.25)^2}{4}}=31780 psi (g)
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 C. Write the equations for the load, shear, moment, slope, and deflection using singularity functions, noting that the integration constants C_{1} and C_{2} will be zero:
\begin{array}{c} q=R_1\langle x-0\rangle^{-1}-F\langle x-a\rangle^{-1}+R_2\langle x-l\rangle^{-1} \\ V=R_1\langle x-0\rangle^0-F\langle x-a\rangle^0+R_2\langle x-l\rangle^0 \\ M=R_1\langle x-0\rangle^1-F\langle x-a\rangle^1+R_2\langle x-l\rangle^1 \end{array} (h)
\begin{array}{l} \theta=\frac{1}{E I}\left\lgroup \frac{R_1}{2}\langle x-0\rangle^2-\frac{F}{2}\langle x-a\rangle^2+\frac{R_2}{2}\langle x-l\rangle^2+C_3 \right\rgroup \\ y=\frac{1}{E I}\left\lgroup \frac{R_1}{6}\langle x-0\rangle^3-\frac{F}{6}\langle x-a\rangle^3+\frac{R_2}{6}\langle x-l\rangle^3+C_3 x+C_4 \right\rgroup \end{array} (i)
6 The reaction forces can be found from ΣM = 0 and ΣF = 0. (See Appendix B.)
R_1=\frac{F(l-a)}{l}=\frac{2000(1.55-0.35)}{1.55}=1548 lb (j)
R_2=\frac{F a}{l}=\frac{2000(0.35)}{1.55}=452 lb (k)
The maximum moment is 1 548(0.35) = 541.8 lb-in at the applied load. The shear and moment diagrams for link 4 are shown in Figure 4-50.
7 The beam depth at the point of maximum moment is 0.75 in and the thickness is 0.187. The bending stress is then
\sigma=\frac{M c}{I}=\frac{541.8\left\lgroup\frac{0.75}{2}\right\rgroup }{\frac{0.187(0.75)^3}{12}}=30905 psi (l)
8 The beam slope and deflection functions require calculation of the integration constants C_{3} and C_{4}, which are found by substituting the boundary conditions x = 0, y = 0 and x = l, y = 0 in the deflection equation.
\begin{aligned} 0 &=\frac{1}{E I}\left\lgroup \frac{R_1}{6}\langle 0-0\rangle^3-\frac{F}{6}\langle 0-a\rangle^3+\frac{R_2}{6}\langle 0-l\rangle^3+C_3(0)+C_4\right\rgroup \\ C_4 &=0 \end{aligned} (m)
\begin{aligned} 0 &=\frac{1}{E I}\left\lgroup \frac{R_1}{6}\langle l-0\rangle^3-\frac{F}{6}\langle l-a\rangle^3+\frac{R_2}{6}\langle l-l\rangle^3+C_3(l) \right\rgroup \\ C_3 &=\frac{1}{6 l}\left[F(l-a)^3-R_1 l^3\right]=\frac{1}{6(1.55)}\left[2000(1.55-0.35)^3-1548(1.55)^3\right]=-248.4 \end{aligned} (n)
9 The deflection equation is found by combining equations i, j, k, m, and n:
y=\frac{F}{6 l E I}\left\{(l-a)\left(x^3+\left[(l-a)^2-l^2\right] x\right)-l\langle x-a\rangle^3+a\langle x-l\rangle^3\right\} (o)
and the maximum deflection at x = 0.68 in is
\begin{aligned} y_{\max } &=\frac{F a(l-a)}{6 l E I}\left(a^2+(l-a)^2-l^2\right) \\ &=\frac{2000(0.35)(1.55-0.35)}{6(1.55)(30 E 6)(0.0066)}\left[0.35^2+(1.55-0.35)^2-1.55^2\right]=0.0005 \text { in } \end{aligned} (p)
Only a very small deflection is allowed here to guarantee the proper crimp stroke, and this amount is acceptable. The slope and deflection diagrams are shown in Figure 4-50. Also see the file CASE2B-1.
10 Link 1 is relatively massive compared to the others and the only area of concern is the jaw, which is loaded by the 2 000-lb crimp force and has a hole in the cross section at its root. While the shape of this element is not exactly that of a curved beam with concentric inside and outside radii, this assumption will be acceptably conservative if we use an outer radius equal to the smallest section dimension as shown in Figure 4-49 (p. 210). This makes its inside radius 0.6 in and its approximate outside radius 1.6 in. The eccentricity e of the curved beam’s neutral axis versus the beam’s centroidal axis r_{c} is found from equation 4.12a (p. 157), while accounting for the section’s hole in the integration.
e=r_c-\frac{A}{\int \frac{d A}{r}} (4.12a)
e=r_c-\frac{A}{\int_0^{r_o} \frac{d A}{r}}=1.1-\frac{0.313(1-0.25)}{0.313\left\lgroup \int_{0.600}^{0.975} \frac{d r}{r}+\int_{1.225}^{1.600} \frac{d r}{r} \right\rgroup}=0.103 (q)
The radius to the neutral axis (r_{n}) and the distances (c_{i} and c_{o}) from the inner and outer fiber radii (r_{i} and r_{o}) to the neutral axis are then (see Figure 4-16, p. 157)
\begin{array}{l} r_n=r_c-e=1.10-0.103=0.997 \\ c_i=r_n-r_i=0.997-0.600=0.397 \\ c_o=r_o-r_n=1.600-0.997=0.603 \end{array} (r)
11 The applied bending moment on the curved beam section is taken as the applied load times its distance to the beam’s centroidal axis.
M=F l=2000(0.7-0.6+1.1)=2400 lb -\text { in } (s)
12 Find stresses at the inner and outer fibers from equations 4.12b and 4.12c (p. 158). Reduce the beam cross sectional area by the hole area.
\sigma_i =+\frac{M}{e A}\left\lgroup\frac{c_i}{r_i} \right\rgroup (4.12b)
\sigma_o =-\frac{M}{e A}\left\lgroup\frac{c_o}{r_o} \right\rgroup (4.12c)
\begin{array}{l} \sigma_i=+\frac{M}{e A}\left\lgroup\frac{c_i}{r_i} \right\rgroup =\frac{2400}{0.103[(1.0-0.25)(0.313)]}\left\lgroup\frac{0.397}{0.60} \right\rgroup =65 kpsi \\ \sigma_o=-\frac{M}{e A}\left\lgroup\frac{c_o}{r_o} \right\rgroup =-\frac{2400}{0.103[(1.0-0.25)(0.313)]}\left\lgroup\frac{0.603}{1.60} \right\rgroup =-37 kpsi \end{array} (t)
13 There is also a direct axial tensile stress, which adds to the bending stress in the inner fiber at point P:
\begin{aligned} \sigma_a=\frac{F}{A} &=\frac{2000}{(1.0-0.25)(0.313)}=8.5 kpsi \\ \sigma_{\max } &=\sigma_a+\sigma_i=65+8.5=74 kpsi \end{aligned} (u)
This is the principal stress for point P, since there is no applied shear or other normal stress at this edge point. The maximum shear stress at point P is half this principal stress or 37 kpsi. The bending stress at the outer fiber is compressive and thus subtracts from the axial tensile stress for a net of –37 + 8.5 = –28.5 kpsi.
14 There is significant stress concentration at the hole. The theoretical stress-concentration factor for the case of a circular hole in an infinite plate is K_{t} = 3 as defined in equation 4.32a (p. 189) and Figure 4-35 (p. 188). For a circular hole in a finite plate, K_{t} is a function of the ratio of the hole diameter to the plate width. Peterson gives a chart of stress-concentration factors for a round hole in a flat plate under tension[5] from which we find that K_{t} = 2.42 for a dia / width ratio = 1/4. The local axial tensile stress at the hole is then 2.42(8.5) = 20.5 kpsi, which is less than the tensile stress at the inner fiber.
K_t=1+2\left\lgroup\frac{a}{c} \right\rgroup (4.32a)
15 While this is far from a complete stress and deflection analysis of these parts, the calculations done address the areas judged to be most likely to fail or to have problem deflections. The stresses and deflections in link 1 were also computed using the ABAQUS finite element analysis program, which gave an estimated maximum principal stress at point P of 81 kpsi compared to our estimate of 74 kpsi. The FEA mesh and stress distribution calculated by the FEA model is shown in Figure 4-51. Case Study 2D in Chapter 8 (p. 511) presents the complete FEA analysis of this assembly.
Our analysis simplified the part geometry in order to allow the use of a known closedform model (the curved beam) whereas the FEA model included all the material in the actual part but discretized its geometry. Both analyses should be recognized as only estimates of the stress states in the parts, not exact solutions.
16 Redesign may be needed to reduce these stresses and deflections, based on a failure analysis. This case study will be revisited in the next chapter after various failure theories are presented. You may examine the models for this case study by opening the files CASE2B-1, CASE2B-2, and CASE2B-3 in the program of your choice. A stress analysis of this case study is done in Chapter 8 using Finite Element Analysis (FEA).
| Table 4-4 Column End-Condition Effective Length Factors | |||
| End Conditions | Theoretical Value | AISC* Recommends | Conservative Value |
| Rounded-Rounded | l_{eff}= l | l_{eff}= l | l_{eff}= l |
| Pinned-Pinned | l_{eff}= l | l_{eff}= l | l_{eff}= l |
| Fixed-Free | l_{eff}=2l | l_{eff}=2.1l | l_{eff}=2.4l |
| Fixed-Pinned | l_{eff}=0.707l | l_{eff}=0.80l | l_{eff}= l |
| Fixed-Fixed | l_{eff}=0.5l | l_{eff}=0.65l | l_{eff}= l |
| * The American Institute of Steel Construction, in their Manual of Steel Construction. | |||
