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HW9+Solution

# HW9+Solution - HOMEWORK PROBLEMS Week 9 Due Friday 11...

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HOMEWORK PROBLEMS Week 9: Due Friday 11 November 1. Find the distance c defining the centroid C of the symmetric beam section of Figure 1 and hence find the centroidal second moment of area (moment of inertia) I x for bending about a horizontal axis. C c 100 50 15 15 15 all dimensions in mm Figure 1 Solution The section can be regarded as the difference between a large rectangle ( 100 × 50 ) and a smaller rectangle ( 70 × 35 ) as shown schematically in Figure 1.1. 2 1 all dimensions in mm Figure 1.1 The properties of these areas are listed in the following table, where we measure y from the bottom of the section:- 1

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i 1 2 A i 5000 - 2450 ¯ y i 25 32.5 ¯ y i - ¯ y 7.2058 14.7058 I i x 1 . 042 × 10 6 - 0 . 250 × 10 6 In explanation of these values note that ¯ y 2 = 50 + 15 2 = 32 . 5 ; A = A 1 + A 2 = 5000 - 2450 = 2550 A ¯ y = A 1 ¯ y 1 + A 2 ¯ y 2 = 5000 × 25 - 2450 × 32 . 5 = 45375 c = ¯ y = 45375 2550 = 17 . 7941 mm . I 1 x = 100 × 50 3 12 = 1 . 0417 × 10 6 ; I 2 x = - 70 × 35 3 12 = - 0 . 2501 × 10 6 We then have I x = I 1 x + I 2 x + A 1 y 1 - ¯ y ) 2 + A 2 y 2 - ¯ y ) 2 = 1 . 0417 × 10 6 - 0 . 2501 × 10 6 + 5000 × 7 . 2058 2 - 2450 × 14 . 7058 2 = 521 , 400 mm 4 . 2
2. The steel beam in Figure 2(a) is loaded by two equal forces P as shown. A strain gauge located at the bottom of the section records a tensile strain ǫ x = 0 . 6 × 10 - 3 . The cross-section of the beam is shown in Figure 22.4(b). Determine the value of P . ( E steel = 210 GPa). A B P P 400 80 80 80 strain gauge 2 2 12 14 (a) (b) all dimensions in mm. Figure 2 Solution The loading on the beam is symmetrical and hence the reactions will be both equal to P . The shear force diagram therefore has the form of Figure 2.1 and we conclude (by summing the area under the curve) that the bending moment throughout the central section (and hence at the location of the strain gauge) is M = 80 P . P P 80 80 P P all dimensions in mm. Figure 2.1 To find the maximum bending stress, we must first determine the location of the centroid and the centroidal second moment of area I x for the cross section. We regard the cross section as the sum of a horizontal rectangle 14 × 2 and a vertical rectangle 2 × 12 , and measure y from the bottom of the section. We can then construct the following table:- 3

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i 1 2 A i 28 24 ¯ y i 1 8 ¯ y i - ¯ y
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