Question 4.CS.3B: Automobile Scissors-Jack Stress and Deflection Analysis Prob...
Automobile Scissors-Jack Stress and Deflection Analysis
Problem Determine the stresses and deflections at critical points in the scissors-jack assembly shown in Figures 3-5 (repeated here) and 4-52.
Given The geometry and loading are known from Case Study 3A (p. 88). 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 = 30E6 psi and S_{y} = 60 000 psi.
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 4-52 to 4-53 and the files CASE3B-1 and CASE3B-2.
1 The forces on this jack assembly for the position shown were calculated in the previous installment of this case study (3A) in Chapter 3 (p. 88). Please see that section and Table 3-4 (pp. 90–92) for additional force data.
2 The force in the jack screw is four times the 878-lb component F_{21x} at point A because that force is from the upper half of the jack in one plane only. The lower half exerts an equal force on the screw, and the back side doubles their sum. These forces put the screw in axial tension. The tensile stress is found from equation 4.7 (p. 152) using the 0.406- in root diameter of the thread to calculate the cross sectional area. This is a conservative assumption, as we shall see when analyzing threaded fasteners in Chapter 15.
\sigma_x=\frac{P}{A} (4.7)
\sigma_x=\frac{P}{A}=\frac{4(878)}{\frac{\pi(0.406)^2}{4}}=\frac{3512}{0.129}=27128 psi (a)
The axial deflection of the screw is found from equation 4.8 (p. 152).
\Delta s=\frac{P l}{A E} (4.8)
x=\frac{P l}{A E}=\frac{4(878)(12.55)}{0.129(30 E 6)}=0.011 \text { in } (b)
3 Link 2 is the most heavily loaded of the links due to the applied load P being slightly offset to the left of center, so we will calculate its stresses and deflections. This link is loaded as a beam-column with both an axial compressive force P between points C and D and a bending couple applied between D and E. Note that the force F_{12} is virtually colinear with the link axis. The axial load is equal to F_{12} cos(1°) = 1 026 lb and the bending couple created by F_{42} acting about point D is M = 412(0.9) = 371 in-lb. This couple is equivalent to the axial load being eccentric at point D by distance e = M/P = 371 / 1026 = 0.36 in.
The secant-column formula (Eq. 4.46c, p. 202) can be used with this effective eccentricity e accounting for the applied couple in the plane of bending; c is 1/2 of the 1.032- in width of the link. Since it is a pinned-pinned column, l_{eff} = l from Table 4-4. The radius of gyration k is taken in the xy plane of bending for this calculation (Eq. 4.34, p. 193):
\frac{P}{A}=\frac{S_{y c}}{1+\left\lgroup \frac{e c}{k^2} \right\rgroup \sec \left\lgroup \frac{l_{e f f}}{k} \sqrt{\frac{P}{4 E A}} \right\rgroup } (4.46c)
k=\sqrt{\frac{I}{A}} (4.34)
k=\sqrt{\frac{I}{A}}=\sqrt{\frac{b h^3}{12 b h}}=\sqrt{\frac{0.15(1.032)^3}{12(0.15)(1.032)}}=0.298 (c)
The slenderness ratio is l_{eff} / k = 20.13. The secant formula can now be applied and iterated for the value of P. (See Figure 4-53.)
\frac{P}{A}=\frac{S_{y c}}{1+\left\lgroup \frac{e c}{k^2} \right\rgroup \sec \left\lgroup \frac{l_{e f f}}{k} \sqrt{\frac{P}{4 E A}} \right\rgroup}=18 975 psi \\ P_{\text {crit }}=0.155(18 975)=2 937 lb (d)
The column also has to be checked for concentric-column buckling in the weaker (z) direction with c = 0.15 / 2. The radius of gyration in the z direction is found from
k=\sqrt{\frac{I}{A}}=\sqrt{\frac{b h^3}{12 b h}}=\sqrt{\frac{1.032(0.15)^3}{12(1.032)(0.15)}}=0.043 (e)
The slenderness ratio in the z direction is
S_r=\frac{l_{e f f}}{k}=\frac{6}{0.043}=138.6 (f)
This needs to be compared to the slenderness ratio (S_{r })D at the tangency between the Euler and Johnson lines to determine which buckling equation to use for this column:
\left(S_r\right)_D=\pi \sqrt{\frac{2 E}{S_y}}=\pi \sqrt{\frac{2(30 E 6)}{60 000}}=99.3 (g)
The S_{r} for this column is greater than \left(S_r\right)_D , making it an Euler column (see Figure 4-53). The critical Euler load is then found from equation 4.38a (p. 194).
P_{c r}=\frac{\pi^2 E I}{l^2}=\frac{\pi^2(30 E 6)(1.032)(0.15)^3}{12(6)^2}=2 387 lb (h)
Thus it is more likely to buckle in the weaker z direction than in the plane of the applied moment. Its safety factor against buckling is 2.3.
4 The pins are all 0.437-in dia. The bearing stress in the most heavily loaded hole at C is
\sigma_{\text {bearing }}=\frac{P}{A_{\text {bearing }}}=\frac{1 026}{0.15(0.437)}=15 652 psi (i)
The pins are in single shear and their worst shear stress is
\tau=\frac{P}{A_{\text {shear }}}=\frac{1 026}{\frac{\pi(0.437)^2}{4}}=6 841 psi (j)
5 The gear tooth on link 2 is subjected to a force of 412 lb applied at a point 0.22 in from the root of the cantilevered tooth. The tooth is 0.44 in deep at the root and 0.15 in thick. The bending moment is 412(0.22) = 91 in-lb and the bending stress at the root is
\sigma=\frac{M c}{I}=\frac{91(0.22)}{\frac{0.15(0.44)^3}{12}}=18 727 psi (k)
6 This analysis could be continued, looking at other points in the assembly and, more importantly, at stresses 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 analysis should be done for multiple positions.
This case study will be revisited in the next chapter for the purpose of failure analysis. You may examine the models for this case study by opening the files CASE3B-1 and CASE3B-2 in the program of your choice.
| Table 3–4 – part 1 Case Study 3A Given Data |
||
| Variable | Value | Unit |
| P_x | 0.00 | lb |
| P_y | –1 000.00 | lb |
| R_{p x} | –0.50 | in |
| R_{p y} | 0.87 | in |
| \theta | –45.00 | deg |
| R_{12 x} | –3.12 | in |
| R_{12 y} | –1.80 | in |
| R_{32 x} | 2.08 | in |
| R_{32 y} | 1.20 | in |
| R_{42 x} | 2.71 | in |
| R_{42 y} | 1.00 | in |
| R_{23 x} | –0.78 | in |
| R_{23 y} | –0.78 | in |
| R_{43 x} | 0.78 | in |
| R_{43 y} | –0.78 | in |
| R_{14 x} | 3.12 | in |
| R_{14 y} | –1.80 | in |
| R_{24 x} | –2.58 | in |
| R_{24 y} | 1.04 | in |
| R_{34 x} | –2.08 | in |
| R_{34 y} | 1.2 | in |
| Table 3–4 – part 2 Case Study 3A Calculated Data |
||
| Variable | Value | Value |
| F_{12 x} | 877.8 | lb |
| F_{12 y} | 530.4 | lb |
| F_{32 x} | –587.7 | lb |
| F_{32 y} | –820.5 | lb |
| F_{42 x} | –290.1 | lb |
| F_{42 y} | 290.1 | lb |
| F_{23 x} | 587.7 | lb |
| F_{23 y} | 820.5 | lb |
| F_{43 x} | –587.7 | lb |
| F_{43 y} | 179.5 | lb |
| F_{14 x} | –877.8 | lb |
| F_{14 y} | 469.6 | lb |
| F_{24 x} | 290.1 | lb |
| F_{24 y} | –290.1 | lb |
| F_{34 x} | 587.7 | lb |
| F_{34 y} | –179.5 | lb |
| 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. | |||
