TEST2_ENSC3313_SP2005_KEY

TEST2_ENSC3313_SP2005_KEY - ENSC 3313, Spring 2005 TEST 2 -...

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1 ENSC 3313, Spring 2005 TEST 2 - KEY Name_________________________ Answer all 10 questions, show all work, 2points each 1. Visualize your tensile specimen of lab 1! A test, such as you did, gave: 0.1% offset load = 2,000 lb Peak load = 6,000 lb Extension at fracture = 0.4 inches Calculate the 4 property values you would expect to find in handbooks and use in design. Answer You should recollect, diameter = 0.5", hence area 0.2 in 2 , length = 2", hence elongation/2 = strain Thus, i) Elastic Modulus = ( ) psi 10 10 001 . 0 2 . 0 2000 6 × = ii) Yield Stress = psi 000 , 10 2 . 0 2000 = iii) Tensile Strength = psi 000 , 30 2 . 0 000 , 6 = iv) % Elongation (2") = ( ) % 20 100 2 4 . 0 = 2. Shown is the plane (110) and the direction [ 11 1 ] in a BCC iron unit cell. Is this a possible slip system? If so, calculate the resolved shear stress if 10,000 psi tensile stress is applied in the [010] direction. Yes, this is a slip system since the direction lies in the plane: 0 0 1 1 1 1 1 w l v k u h = + + = + + The resolved shear stress is given by the formula: λ φ σ τ cos cos = r The angle between the plane normal [110] and the stress direction [010] is 45 o so cos φ = 2 1 The cosine of the angle between the direction [ 11 1 ] and [010] is: 3 1 cos = λ Hence, psi 080 , 4 6 000 , 10 r = = τ x y z φ λ 4R R 3 4
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TEST2_ENSC3313_SP2005_KEY - ENSC 3313, Spring 2005 TEST 2 -...

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