This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: M a F d AB ⋅ = M a 840 lbf in ⋅ = FIGURE 4-21 3. The tensile stress at this point is found from Free Body Diagrams for Problem 4-21 Moment of inertia I π d 4 ⋅ 64 = I 0.00749 in 4 = Dist to extreme fibre c 0.5 d ⋅ = c 0.313 in = Stress σ x M a c ⋅ I = σ x 35.05 ksi = 4. There are no other stress components present at this point, so σ x is the maximum principle stress here and σ 1 σ x = σ 1 35.0 ksi = σ 2 psi ⋅ = σ 3 psi ⋅ = PROBLEM 4-21 Statement: Figure P4-9 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each case two hands a required to provide forces respectively at A and B as shown. The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. The wheel nuts require a torque of 70 ft-lb. Find the maximum principle stress and maximum deflection in each wrench design. Units: ksi 10 3 psi ⋅ = Given: Distance between A and B d AB 1 ft ⋅ = Tightening torque T 70 ft ⋅ lbf ⋅ = Wrench diameter d 0.625 in ⋅ = Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to the plane of the wheel. 2. The applied torque is perpendicular to the plane of the forces. 3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD. Solution: See Figure 4-21 and Mathcad file P0421. 1. In Problem 3-21 we found that for both cases F 70 lbf ⋅ = 2. From examination of the FBDs, we see that, in both cases, the arms are in bending and the stub that holds the socket wrench is in pure torsion. The maximum bending stress in the arm will occur near the point where the arm transitions to the stub. The stress state at this transition is very complicated, but we can find the nominal bending stress there by treating the arm as a cantilever beam, fixed at the transition point. For both cases the torque in the stub is the same. Case (a) 2. The bending moment at the transition is So, the total deflection is y a y arm y stub + = y a 0.352 in = Case (b) 10. The bending moment at the transition is M b F d AB ⋅ 2 = M b 420 lbf in ⋅ = 11. The tensile stress at this point is found from11....
View Full Document
This note was uploaded on 07/23/2011 for the course EML 4501 taught by Professor Staff during the Spring '11 term at UNF.
- Spring '11
- Machine Design