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Unformatted text preview: σ ' 1323 = σ ' σ 1 2 σ 1 σ 3 ⋅ σ 3 2 + = Stress Element for Problem 51a 3. Using equatoion 5.7c, the von Mises stress is FIGURE 51aA σ 3 207 = σ 2 = σ 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. τ zx = τ yz = τ xy 500 = σ z = σ y = σ 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 σ 1A ⋅ S ut σ 3A ⋅ = N ACM 13.4 = ModifiedMohr N AMM S ut σ 1A = N AMM 16.3 = 4. From Problem 433a, the principal stresses at Point B are σ 1B 16.13 MPa ⋅ = σ 2B MPa ⋅ = σ 3B 16.13 MPa ⋅ = 5. Using the two nonzero stresses, the slope of the load line on a σ 1 σ 3 graph is m σ 3B σ 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 σ 1B ⋅ S ut σ 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 σ 1A 21.46 MPa ⋅ = σ 2A MPa ⋅ = σ 3A 13.08 MPa ⋅ = 2. Using the two nonzero stresses, the slope of the load line on a σ 1 σ 3 graph is m σ 3A σ 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 σ 1B = σ 1 is maximum so C 3 16.1 MPa = C 3 1 2 σ 3B σ 1B S uc 2 S ut ⋅ + S uc σ 3B σ 1B + ( 29 ⋅ + ⋅ = C 2 5.6 MPa = C 2 1 2 σ 2B σ 3B S uc 2 S ut ⋅ + S uc σ 2B σ 3B + ( 29 ⋅ + ⋅ = C 1 10.5 MPa = C 1 1 2 σ 1B σ 2B S uc 2 S ut ⋅ + S uc σ 1B σ 2B + ( 29 ⋅ + ⋅ = Point B N 16.3 = N S ut σ 1A = σ 1 is maximum so C 3 18.5 MPa = C 3 1 2 σ 3A σ 1A S uc 2 S ut ⋅ + S uc σ 3A σ 1A + ( 29 ⋅ + ⋅ = C 2 4.6 MPa = C 2 1 2 σ 2A σ 3A S uc 2 S ut ⋅ + S uc σ...
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 Spring '11
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 Machine Design, Stress, Strength of materials, Trigraph, Tensile strength, Fracture mechanics

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