Assignment4_Solution

# Assignment4_Solution - S.17.13 The x axis is an axis of...

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S.17.13 The x axis is an axis of symmetry so that I xy = 0 and, since S x = 0, Eq. (17.15) simpliFes to q s =− S y I xx ± s 0 ty d s + q s ,0 (i) in which, from ±ig. S.17.13(a) I xx = th 3 12 + (2 d ) 3 t sin 2 α 12 = th 2 12 ( h + 2 d ) (ii) ‘Cut’ the section at 1. Then, from the Frst term on the right-hand side of Eq. (i) q b,12 S y I xx ± s 1 0 t ( s 1 sin α )d s 1 = S y t sin α 2 I xx s 2 1

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Solutions to Chapter 17 Problems 225 Fig. S.17.13(a) Substituting for I xx and sin α q b,12 = 3 S y hd ( h + 2 d ) s 2 1 (iii) and q b,2 = 3 S y d h ( h + 2 d ) Also q b,23 =− S y I xx ± s 2 0 t ² h 2 + s 2 ³ d s 2 + q b,2 so that q b,23 = 6 S y h ( h + 2 d ) ´ s 2 s 2 2 h + d 2 µ (iv) Now taking moments about the point 1 (see Eq. (17.18)) 0 = ± h 0 q b,23 d cos α d s 2 + 2 h 2 d cos α q s ,0 i.e. 0 = ± h 0 q b,23 d s 2 + hq s ,0 (v) Substituting in Eq. (v) for q b,23 from Eq. (iv) 0 = 6 S y h ( h + 2 d ) ± h 0 ´ s 2 s 2 2 h + d 2 µ d s 2 + hq s ,0
226 Solutions Manual which gives q s ,0 =− S y ( h + 3 d ) h ( h + 2 d ) (vi) Then, from Eqs (iii) and (i) q 12 = 3 S y hd ( h + 2 d ) s 2 1 S y ( h + 3 d ) h ( h + 2 d ) i.e. q 12 = S y h ( h + 2 d ) ± 3 s 2 1 d h 3 d ² (vii) and from Eqs (iv) and (vi) q 23 = S y h ( h + 2 d ) ± 6 s 2 6 s 2 2 h h ² (viii) The remaining distribution follows from symmetry. From Eq. (vii), q 12 is zero when s 2 1 = ( hd /3) + d 2 , i.e. when s 1 > d . Thus there is no change of sign of q 12 between 1 and 2. Further d q 12 d s 1 = 6 s 1 d = 0 when s 1 = 0 and q 1 S y ( h + 3 d ) h ( h + 2 d ) Also, when s 1 = d q 2 S y ( h + 2 d ) From Eq. (viii) q 23 is zero when 6 s 2 (6 s 2 2 / h ) h = 0, i.e. when s 2 2 s 2 h + ( h 2 / 6) = 0. Then s 2 = h 2 ± h 12 Thus q 23 is zero at points a distance h / 12 either side of the x axis. Further, from Eq. (viii), q 23 will be a maximum when s 2 = h /2 and q 23 ( max ) = S y /2( h + 2 d ). The complete distribution is shown in Fig. S.17.13(b).

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Solutions to Chapter 17 Problems 227 Fig. S.17.13(b)
S.17.15 Referring to Fig. P.17.15, the wall DB is 3 m long so that its cross-sectional area, 3 × 10 3 × 8 = 24 × 10 3 mm 2 , is equal to that of the wall EA, 2 × 10 3 × 12 = 24 × 10 3 mm 2 . If follows that the centroid of area of the section lies mid-way between DB and EA on the vertical axis of symmetry.Also since S y = 500 kN, S x = 0 and I xy = 0, Eq. (17.15) reduces to q s =− 500 × 10 3 I xx ± s 0 ty d s + q s ,0 (i)

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230 Solutions Manual If the origin for s is taken on the axis of symmetry, say at O, then q s ,0 is zero. Also I xx = 3 × 10 3 × 8 × (0 . 43 × 10 3 ) 2 + 2 × 10 3 × 12 × (0 . 43 × 10 3 ) 2 + 2 × (1 × 10 3 ) 3 × 10 × sin 2 60 / 12 i.e. I xx = 101 . 25 × 10 8 mm 4 Equation (i) then becomes q s =− 4 . 94 × 10 5 ± s 0 ty d s In the wall OA, y 0.43 × 10 3 mm. Then q OA = 4 . 94 × 10 5 ± s 0 12 × 0 . 43 × 10 3 d s = 0 . 25 s A and when s A = 1 × 10 3 mm, q = 250 N/mm. In the wall AB, y 0 . 43 × 10 3 + s B cos 30 . Then q AB 4 . 94 × 10 5 ± s 0 10( 0 . 43 × 10 3 + 0 . 866 s B )d s + 250 i.e. q AB = 0 . 21 s B 2 . 14 × 10 4 s 2 B + 250 When s B = 1 × 10 3 mm, q AB = 246 N/mm. In the wall BC, y = 0 . 43 × 10 3 mm. Then q BC 4 . 94 × 10 5 ± s 0 8 × 0 . 43 × 10 3 d s + 246 i.e.
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Assignment4_Solution - S.17.13 The x axis is an axis of...

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