HW5 solutions - MechEng 382 Winter 2010 2009 Homework#5...

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MechEng 382 Homework #5 Solutions 1. A set of strain gauges on the free surface of a steel structure (with Young’s modulus of 196 GPa , Poisson’s ratio of 0.30, and yield strength of σ Y = 850 MPa ) record strains of ± xx =3 . 0 × 10 3 , ± yy = 1 . 5 × 10 3 and γ xy =2 . 0 × 10 3 . Calculate the safety factor against yield at the surface of this structure, using the von Mises yield criterion. Solution Hooke’s Law in 2D is ± ± xx ± ² = 1 E ³ 1 ν ν 1 ´± σ xx σ ² . (1) Inverting these equations isolates the normal stresses: ± σ xx σ ² = E 1 ν 2 ³ 1 ν ν 1 ± xx ± ² (2) = 196 × 10 3 1 0 . 3 2 ³ 10 . 3 0 . 31 3 . 0 × 10 3 1 . 5 × 10 3 ² = ± 549 129 ² MPa . The shear stress, meanwhile, is given by τ xy = xy = E 2(1 + ν ) γ xy = 151 MPa . (3) Since this is a free surface, the out-of-plane stresses are zero: σ zz = τ xz = τ yz = 0 MPa . Thus, the Mohr’s circle radius σ τ 3 and radius τ 3 are σ τ 3 = σ xx + σ 2 = 210 MPa 3 = µ σ xx σ 2 · 2 + τ 2 xy = 371 MPa; with principal stresses σ 1 = σ τ 3 + τ 3 = 581 MPa 2 = σ τ 3 τ 3 = 161 MPa 3 = 0 MPa 1 2010
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s n o i t u l o S 5 # k r o w e m o H 2010 r e t n i W 2 8 3 g n E h c e M The von Mises efective stress, Fnally, is ¯ σ H = 1 2 ± ( σ 1 σ 2 ) 2 , ( σ 2 σ 3 ) 2 , ( σ 1 σ 3 ) 2 = 1 2 ² 4 τ 2 3 +( σ τ 3 τ 3 ) 2 σ τ 3 + τ 3 ) 2 σ 3 =0 = ² σ 2 τ 3 +3 τ 2 3 (4) = ± 210 2 + 3(371 2 ) = 676 MPa; and the sa±ety ±actor X H σ Y ¯ σ H = 850 676 = 1 . 26 . (5) 2
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s n o i t u l o S 5 # k r o w e m o H 10 0 2 r e t n i W 2 8 3 g n E h c e M 2. A hollow, circular cylinder is made of an Al alloy with yield stress 180 MPa . It is subjected to a torque of 150 N · m and a compressive load of 12 kN. The inner and outer radii are 5mm and 10 mm , respectively. The Youngs modulus is 70 GPa and the Poisson’s ratio is 0 . 33 . (a) What is the safety factor against yield according to the von Mises yield criterion? Solution The normal stress σ zz due to the compressive load and shear stress τ (at the outer surface r = R 2 ) from the torque are σ = N A = N π ( R 2 2 R 2 1 ) = 50 . 9MPa , (6) τ = Tr J = TR 2 π 2 ( R 4 2 R 4 1 ) = 102 MPa; (7) The remaining stresses σ r = σ θ = τ rz = τ = 0 MPa. From here, the analysis proceeds as usual: σ τ 3 = σ 2 = N π ( R 2 2 R 2 1 ) = 25 . 5MPa , τ 3 = ± ² σ 2 ³ 2 + τ 2 = ´ µ N π ( R 2 2 R 2 1 ) 2 + µ 2 π 2 ( R 4 2 R 4 1 ) 2 = 105 MPa; σ 1 = σ τ 3 + τ 3 =79 . 2 = σ τ 3 τ 3 = 130 . 3 = 0 MPa ¯ σ H = 1 2 · ( σ 1 σ 2 ) 2 , ( σ 2 σ 3 ) 2 , ( σ 1 σ 3 ) 2 = ¸ σ 2 τ 3 +3 τ 2 3 σ 3 = 0, Eq. 4 = 1 π ´ N 2 ( R 2 2 R 2 1 ) 2 + 4 T 2 R 2 2 ( R 4 2 R 4 1 ) 2 (8) = 1 π ´ (12 × 10 3 ) 2 (0 . 01 2 0 . 005 2 ) 2 + 4(150 2 )(0 . 01) 2 (0 . 01 4 0 . 005 4 ) 2 = 184 MPa; X H σ Y ¯ σ H = 180 184 = 0 . 98 . (9) (b) If the load is changed to tension, what is the new safety factor?
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This note was uploaded on 05/12/2010 for the course MECHENG 382 taught by Professor Thouless during the Winter '08 term at University of Michigan.

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HW5 solutions - MechEng 382 Winter 2010 2009 Homework#5...

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