Question 5.CS.3C: Automobile Scissors-Jack Failure Analysis Problem Determine ...
Automobile Scissors-Jack Failure Analysis
Problem Determine safety factors at critical points in a scissors jack.
Given The stresses are known from Case Study 3B on p. 214. The design load is 2 000 lb total or 1 000 lb per side. The width of the links is 1.032 in and their thickness is 0.15 in. The screw is a 1/2-13 UNC thread with root dia = 0.406 in. The material of all parts is ductile steel with E = 30 Mpsi and S_{y} = 60 kpsi.
Assumptions The most likely failure points are the links as columns, the holes in bearing where the pins insert, the connecting pins in shear, the gear teeth in bending, and the screw in tension. There are two sets of links, one set on each side, Assume the two sides share the load equally. The jack is typically used for very few cycles over its lifetime so a static analysis is appropriate.
See Figures 3-5 (opp. page) and 5-24, and the file CASE3C.
1 The stresses on this jack assembly for the position shown were calculated in the previous installment of this case study in Chapter 4 (p. 214). Please see that case also.
2 The jack screw is in axial tension. The tensile stress was found from equation 4.7 (p. 152).
\sigma_x=\frac{P}{A} (4.7)
\sigma_x=\frac{P}{A}=\frac{4(878)}{0.129}=27128 psi (a)
This is a uniaxial tension stress, so is also the principal and the von Mises stress. The safety factor is
N=\frac{S_y}{\sigma^{\prime}}=\frac{60000}{27128}=2.2 (b)
3 Link 2 is loaded as a beam-column. Its safety factor against buckling was calculated in the last installment of this case study and is N = 2.3.
4 The bearing stress in the most heavily loaded hole at C is
\sigma_{\text {bearing }}=\frac{P}{A_{\text {bearing }}}=\frac{1026}{0.15(0.437)}=15652 psi (c)
This is a uniaxial compression stress, so is also the principal and the von Mises stress. The safety factor is
N_{\text {bearing }}=\frac{S_y}{\sigma^{\prime}}=\frac{60000}{15652}=3.8 (d)
The pins’ shear stress is
\tau=\frac{P}{A_{\text {shear }}}=\frac{1026}{\frac{\pi(0.437)^2}{4}}=6841 psi (e)
This is a pure shear stress, so is also the maximum shear stress. The safety factor is
N=\frac{0.577 S_y}{\tau_{\max }}=\frac{0.577(60000)}{6841}=5.1 (f)
5 The bending stress at the root of the gear tooth on link 2 is
\sigma=\frac{M c}{I}=\frac{91(0.22)}{\frac{0.15(0.44)^3}{12}}=18727 psi (g)
This is a uniaxial bending stress, so is also the principal and the von Mises stress. The safety factor is
N=\frac{S_y}{\sigma^{\prime}}=\frac{60000}{18727}=3.2 (h)
6 This analysis should be continued, looking at other points in the assembly and, more importantly, at stresses and safety factors when the jack is in different positions. We have used an arbitrary position for this case study, but, as the jack moves to the lowered position, the link and pin forces will increase due to poorer transmission angles. A complete stress and safety-factor analysis should be done for multiple positions. You may examine the models for this case study by opening the files CASE3C-1 and CASE3C-2 in the program of your choice.
