This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ' 1323 = ' 1 2 1 3  3 2 + = Stress Element for Problem 51a 3. Using equatoion 5.7c, the von Mises stress is FIGURE 51aA 3 207 = 2 = 1 1207 = 2. From Problem 41 a , the principal stresses are 1. Draw the stress element, indicating the x and y axes. Solution: See Figure 51a and Mathcad file P0501a. zx = yz = xy 500 = z = y = x 1000 = Given: Statement: A differential stress element has a set of applied stresses on it as indicated in row a of Table P51. For row a , draw the stress element showing the applied stresses. Find the principal stresses and the von Mises stress. PROBLEM 51a CoulombMohr N ACM S ut S uc S uc 1A S ut 3A  = N ACM 13.4 = ModifiedMohr N AMM S ut 1A = N AMM 16.3 = 4. From Problem 433a, the principal stresses at Point B are 1B 16.13 MPa = 2B MPa = 3B 16.13 MPa = 5. Using the two nonzero stresses, the slope of the load line on a 1 3 graph is m 3B 1B = m 1 = This intersects the failure boundaries in the fourth quadrant. For the CoulombMohr diagram (see Figure 59 in the text) there is a single, straight line in this quadrant. For the modifiedMohr theory, the load line will intersect the boundary at the point ( S ut ,  S ut ) Figure 511 in the text. 6. Determine the factor of safety at point B CoulombMohr N BCM S ut S uc S uc 1B S ut 3B  = N BCM 16.1 = PROBLEM 535a Statement: Calculate the safety factor for the bracket in Problem 533 using the CoulombMohr and th modified Mohr effective stress theories. Comment on their appropriateness. Assume a brittle material strengt as given below. Units: MPa 10 6 Pa = Given: Tensile strength S ut 350 MPa = Compressive strength S uc 1000 MPa = Solution: See Mathcad file P0535a. 1. From Problem 433a the principal stresses at point A are 1A 21.46 MPa = 2A MPa = 3A 13.08 MPa = 2. Using the two nonzero stresses, the slope of the load line on a 1 3 graph is m 3A 1A = m 0.61 = This intersects the failure boundaries in the fourth quadrant. For the CoulombMohr diagram (see Figure 59 in the text) there is a single, straight line in this quadrant. For the modifiedMohr theory, the load line will intersect the boundary at a point similar to B' in Figure 511 in the text. 3. Determine the factor of safety at point A N 21.7 = N S ut 1B = 1 is maximum so C 3 16.1 MPa = C 3 1 2 3B 1B S uc 2 S ut + S uc 3B 1B + ( 29 + = C 2 5.6 MPa = C 2 1 2 2B 3B S uc 2 S ut + S uc 2B 3B + ( 29 + = C 1 10.5 MPa = C 1 1 2 1B 2B S uc 2 S ut + S uc 1B 2B + ( 29 + = Point B N 16.3 = N S ut 1A = 1 is maximum so C 3 18.5 MPa = C 3 1 2 3A 1A S uc 2 S ut + S uc 3A 1A + ( 29 + = C 2 4.6 MPa = C 2 1 2 2A 3A S uc 2 S ut + S uc...
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
 Staff
 Machine Design, Stress

Click to edit the document details