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HW13+Solutions

# HW13+Solutions - HOMEWORK SOLUTIONS Week 13 Due Monday 13...

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Unformatted text preview: HOMEWORK SOLUTIONS Week 13: Due Monday 13 December 1. A 45 o strain-gauge rosette is mounted on a plate in the orientation shown in Figure 1. The readings obtained are ǫ a = 750 × 10 − 6 ; ǫ b = (300 − 450 √ 3) × 10 − 6 ; ǫ c = − 150 × 10 − 6 . (i) Show that γ xy = 0 and determine the values of ǫ x and ǫ y . (ii) Find the force components on the face ABCD of the plate if E = 10 × 10 6 psi, ν = 1 / 3 . (a) (b) (c) x y 45 o 45 o o 30 0.1 4 7 A B C D all dimensions in inches. Figure 1 Solution The inclination of the gauges to the x-axis are 30 o , 75 o , 120 o , respectively, so the strain tranfor- mation equation ǫ ′ x = ǫ x cos 2 θ + ǫ y sin 2 θ + γ xy sin θ cos θ gives 750 × 10 − 6 = ǫ 30 = 3 ǫ x 4 + ǫ y 4 + √ 3 γ xy 4 1 (300 − 450 √ 3) × 10 − 6 = ǫ 75 = 0 . 0670 ǫ x + 0 . 9330 ǫ y + 0 . 25 γ xy − 150 × 10 − 6 = ǫ 120 = ǫ x 4 + 3 ǫ y 4 − √ 3 γ xy 4 Solving these equations for ǫ x ,ǫ y ,γ xy (using Maple), we obtain ǫ x = 1200 × 10 − 6 ; ǫ y = − 600 × 10 − 6 ; γ xy = 0 . If the corresponding stress components are σ x ,σ y ,τ xy , we have τ xy = 0 and Eǫ x = σ x − νσ y ; Eǫ y = σ y − νσ x and hence 12000 = σ x − σ y 3 ; − 6000 = σ y − σ x 3 from which σ x = 11250 psi ; σ y = − 2250 psi . The only stress component that acts on the surface ABCD is σ x and this corresponds to an axial force F = σ x A ABCD = 11250 × . 1 × 4 = 4500 lb . 2 2. The strain gauge rosette of Figure 2 records the values ǫ a = − 200 × 10 − 6 ; ǫ b = 250 × 10 − 6 ; ǫ c = 400 × 10 − 6 . (i) Determine the strain components ǫ x ,ǫ y ,γ xy in the xy-coordinate system. (ii) Find the orientation of the line that experiences the greatest normal strain. (iii) Find the maximum in-plane shear stress, if the material has a shear modulus G = 57 GPa. . (a) (b) (c) x y o 60 Figure 2 Solution The inclination of the gauges to the x-axis are o , 60 o , 90 o , respectively, so the strain tranfor- mation equation ǫ ′ x = ǫ x cos 2 θ + ǫ y sin 2 θ + γ xy sin θ cos θ gives ǫ x = − 200 × 10 − 6 ; ǫ y = 400 × 10 − 6 and 250 × 10 − 6 = ǫ 60 = ǫ x 4 + 3 ǫ y 4 + √ 3 γ xy 4 It follows that √ 3 γ xy 4 = 250 × 10 − 6 + 50 × 10 − 6 − 300 × 10 − 6 or γ xy = 0 . 3 We conclude that ǫ x ,ǫ y are principal strains and hence that ǫ y = ǫ 1 is the maximum principal strain. In other words, the line that experiences the greatest normal strain is the y-axis. The maximum in-plane shear strain is γ max = ǫ 1 − ǫ 2 = 400 × 10 − 6 + 200 × 10 − 6 = 600 × 10 − 6 . Thus, the maximum in-plane shear stress is τ max = Gγ max = 57 × 10 3 × 600 × 10 − 6 = 34 . 2 MPa ....
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HW13+Solutions - HOMEWORK SOLUTIONS Week 13 Due Monday 13...

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