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Unformatted text preview: MechEng 382 Winter 2009 Homework #10 Solutions 1. [ A&J1 , Ex. 21.1] It is found that a force F will inject a given weight of a thermosetting polymer into an intricate mould in 30s at 177 ◦ C and in 81.5s at 157 ◦ C. If the viscosity of the polymer follows an Arrhenius law, with a rate of process proportional to e Q/ ¯ RT , calculate how long the process will take at 227 ◦ C. Solution Using the pairs ( T,t ) 1 = (450K , 30s) , ( T,t ) 2 = (430K , 81 . 5s); knowing that the process (strain) rate ˙ ∝ 1 /t (with η ∝ t , also), then ˙ 2 ˙ 1 = t 1 t 2 = C exp Q ¯ RT 2 C exp Q ¯ RT 1 = exp Q ¯ R 1 T 2 1 T 1 (21.1) so that Q = ¯ R (ln t 1 ln t 2 ) 1 T 1 1 T 2 = 8 . 314(ln30 ln81 . 5) 1 450 1 430 = 8 . 039 × 10 4 J / mol = ⇒ C = exp Q ¯ RT 1 t 1 = exp 8 . 039 × 10 4 8 . 314(450) 30 = 7 . 156 × 10 7 1 / s = ⇒ t = exp Q ¯ RT C = exp 8 . 039 × 10 4 8 . 314(500) 7 . 156 × 10 7 = 3 . 5s 1 MechEng 382 Winter 2009 Homework #10 Solutions 2. [ A&J1 , Ex. 20.1] A cylindrical tube in a chemical plant is subjected to an excess internal pressure of 6MPa , which leads to a circumferential stress in the tube wall. The tube wall is required to withstand this stress at a temperature of 510 ◦ C for 9 years. A designer has specified tubes of 40mm bore and 2mm wall thickness made from a stainless alloy of iron with 15 percent by weight of chromium. The manufacturer’s specification for this alloy gives the following information 1 . Over the present ranges of stress and temperature the alloy can be considered to creep according to the equation ˙ = Aσ 5 exp Q ¯ RT (1) where A and Q are constants, ¯ R is the universal gas constant, and T is the absolute tempera ture. Given that failure is imminent at an effective creep strain of 0.01 for the present alloy, comment on the safety of the design. Solution Assume the data was collected from uniaxial tensile data such that σ = ¯ σ and ˙ = ¯ ˙ in Eq. 1 . Then fit the logarithm of Eq. 1 to the data (versus 1 /T ): ¯ ˙ = A ¯ σ 5 exp Q ¯ RT ⇐⇒ ln ¯ ˙ {z} y = Q ¯ R {z} m 1 T {z} x +ln A + 5ln ¯ σ  {z } b ; m = 2 . 9597 × 10 4 = ⇒ Q = ¯ Rm = 8 . 314( 2 . 9597 × 10 4 ) = 2 . 461 × 10 5 , b = 16 . 99 = ⇒ A = exp( b 5ln ¯ σ ) = exp(16 . 99 5ln200) = 7 . 478 × 10 5 . 1.02 1.04 1.06 1.08 1.1 1.12 1.14 x 103 108 107 106 105 1 /T (1/K) ˙ (1/s) Exp Fit Consider two possible stress states created by the inflation, one in which there is only a circumferential stress (uncapped; case M), and another (capped; case N) with combined 1 Not shown. 2 MechEng 382 Winter 2009 Homework #10 Solutions circumferential and axial stresses: σ τ 3 = σ zz + σ θθ 2 : σ M = pR 2 t , σ N = 3 pR 4 t ; [ Dowling , Fig. A.7(a)] (6.10) τ 3 = s σ zz σ θθ 2 + τ 2 rθ : τ M = pR 2 t , τ N = pR 4 t , [ Dowling , Fig. A.7(a)] (6.9) The (von Mises) effective stresses, with...
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 Winter '08
 Thouless
 Shear Stress, Stress, strain rate, retirement life, N. E. Dowling

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