Assignment3_Solution

# Assignment3_Solution - Solutions to Chapter 17 Problems 209...

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Solutions to Chapter 17 Problems 209 Substituting in Eq. (iii) for q 12 from Eq. (ii) S y ξ S = 2 S y r 3 π ± π 0 ( θ + sin θ )(1 + cos θ )d θ Thus ξ S = 2 r 3 π ± π 0 ( θ + θ cos θ + sin θ + sin θ cos θ )d θ i.e. ξ S = 2 r 3 π ² θ 2 2 + θ sin θ cos 2 θ 4 ³ π 0 from which ξ S = π r 3 S.17.4 The x axis is an axis of symmetry so that I xy = 0 and the shear centre, S, lies on this axis (see Fig. S.17.4). Further S x = 0 so that Eq. (17.14) reduces to q s =− S y I xx ± s 0 ty d s (i) S 12 3 t d h x S y ξ S s 1 s 2 β d t / β Fig. S.17.4 Referring to Fig. S.17.4 I xx = th 3 12 + 2 ´ td µ h 2 2 + t β β d µ h 2 2 · = th 2 µ h 12 + d

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210 Solutions Manual From Eq. (i) q 12 =− S y I xx ± s 1 0 t ² h 2 ³ d s 1 i.e. q 12 = S y th 2 I xx s 1 (ii) Also q 32 S y I xx ± s 2 0 t β ² h 2 ³ d s 2 so that q 32 = S y th 2 β I xx s 2 (iii) Taking moments about the mid-point of the web S y ξ S = 2 ± d 0 q 12 h 2 d s 1 2 ± β d 0 q 32 h 2 d s 2 (iv) Substituting from Eqs (ii) and (iii) in Eq. (iv) for q 12 and q 32 S y ξ S = S y th 2 2 I xx ± d 0 s 1 d s 1 S y th 2 2 β I xx ± β d 0 s 2 d s 2 i.e. ξ S = th 2 2 I xx ² d 2 2 β d 2 2 ³ i.e. ξ S = th 2 d 2 (1 β ) 4 th 3 (1 + 12 d / h ) / 12 so that ξ S d = 3 ρ (1 β ) (1 + 12 ρ ) S.17.5 Referring to Fig. S.17.5 the shear centre, S, lies on the axis of symmetry, the x axis, so that I xy = 0. Therefore, apply an arbitrary shear load, S y , through the shear centre and determine the internal shear ﬂow distribution. Thus, since S x = 0, Eq. (17.14) becomes q s S y I xx ± s 0 ty d s (i)
214 Solutions Manual 1 2 3 4 5 3S y 2 h 3S b 2 h ( + a ) Fig. S.17.6(b) Substituting for p and q 12 from Eq. (iv) in (ix) gives S y ξ S = 3 S y bh ( b + a ) ± b 0 hl 2 b s 2 1 d s 1 from which ξ S = l 2(1 + a / b ) S.17.7 Initially the position of the centroid, C, must be found. From Fig. S.17.7, by inspection ¯ y = a . Also taking moments about the web 23 (2 at + 2 a 2 t + a 2 t ) ¯ x = a 2 t a 2 + 2 ata from which ¯ x = 3 a /8. To ±nd the horizontal position of the shear centre, S, apply an arbitrary shear load, S y , through S. Since S x = 0 Eq. (17.14) simpli±es to q s = S y I xy I xx I yy I 2 xy ± s 0 tx d s S y I yy I xx I yy I 2 xy ± s 0 ty d s i.e.

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## This homework help was uploaded on 04/07/2008 for the course MAE 157 taught by Professor Valdevit during the Winter '08 term at UC Irvine.

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Assignment3_Solution - Solutions to Chapter 17 Problems 209...

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