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 a = 750 10 6 ; b = (300 450 3) 10 6 ; c = 150 10 6 . (i) Show that xy = 0 and determine the values of x and y . (ii) Find the force components on the face ABCD of the plate if E = 10 10 6 psi, = 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 x = x cos 2 + y sin 2 + xy sin cos gives 750 10 6 = 30 = 3 x 4 + y 4 + 3 xy 4 1 (300 450 3) 10 6 = 75 = 0 . 0670 x + 0 . 9330 y + 0 . 25 xy 150 10 6 = 120 = x 4 + 3 y 4 3 xy 4 Solving these equations for x , y , xy (using Maple), we obtain x = 1200 10 6 ; y = 600 10 6 ; xy = 0 . If the corresponding stress components are x , y , xy , we have xy = 0 and E x = x y ; E y = y x and hence 12000 = x y 3 ; 6000 = y x 3 from which x = 11250 psi ; y = 2250 psi . The only stress component that acts on the surface ABCD is x and this corresponds to an axial force F = x A ABCD = 11250 . 1 4 = 4500 lb . 2 2. The strain gauge rosette of Figure 2 records the values a = 200 10 6 ; b = 250 10 6 ; c = 400 10 6 . (i) Determine the strain components x , y , 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 x = x cos 2 + y sin 2 + xy sin cos gives x = 200 10 6 ; y = 400 10 6 and 250 10 6 = 60 = x 4 + 3 y 4 + 3 xy 4 It follows that 3 xy 4 = 250 10 6 + 50 10 6 300 10 6 or xy = 0 . 3 We conclude that x , y are principal strains and hence that y = 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 max = 1 2 = 400 10 6 + 200 10 6 = 600 10 6 . Thus, the maximum in-plane shear stress is max = G 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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