HW4_Solution

HW4_Solution - S.17.7 Initially the position of the...

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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. q s = S y I xx I yy I 2 xy ² I xy ± s 0 tx d s I yy ± s 0 ty d s ³ (i)
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Solutions to Chapter 17 Problems 215 Fig. S.17.7 in which, referring to Fig. S.17.7 I xx = a 2 t ( a ) 2 + 2 at ( a ) 2 + t (2 a ) 3 / 12 = 16 a 3 t / 3 I yy = 2 ta 3 / 12 + 2 ta ( a / 8) 2 + t (2 a ) 3 / 12 + 2 at (5 a / 8) 2 + 4 at (3 a / 8) 2 = 53 a 3 t / 24 I xy = a 2 t ( a / 8)( a ) + 2 at (5 a / 8)( a ) =− a 3 t Substituting for I xx , I yy and I xy in Eq. (i) gives q s = 9 S y 97 a 3 t ± ² s 0 tx d s 53 24 ² s 0 ty d s ³ (ii) from which q 12 = 9 S y 97 a 3 ´ ² s 0 ± 13 a 8 s ³ d s 53 24 ² s 0 ( a )d s µ (iii) i.e. q 12 = 9 S y 97 a 3 ± 7 as 12 + s 2 2 ³ (iv) Taking moments about the corner 3 of the section S y ξ S =− ² 2 a 0 q 12 (2 a )d s (v) Substituting for q 12 from Eq. (iv) in (v) S y ξ S =− 18 S y 97 a 2 ² 2 a 0 ± 7 as 12 + s 2 2 ³ d s
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216 Solutions Manual from which ξ S =− 45 a 97 Now apply an arbitrary shear load S x through the shear centre, S. Since S y = 0 Eq. (17.14) simpliFes to q s =− S x I xx I yy I 2 xy ± I xx ² s 0 tx d s I xy ² s 0 ty d s ³ from which, by comparison with Eq. (iii) q 12 =− 9 S x 97 a 3 t ´ 16 3 ² s 0 t ± 13 a 8 s ³ d s + ² s 0 t ( a )d s µ i.e. q 12 =− 3 S x 97 a 3 (23 as 8 s 2 ) (vi) Taking moments about the corner 3 S x (2 a η S ) =− ² 2 a 0 q 12 (2 a )d s Substituting for q 12 from Eq. (vi) S x (2 a η S ) = 6 S x 97 a 2 ² 2 a 0 (23 as 8 s 2 )d s which gives η S = 46 a 97
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Problem 4 By the symmetry of the cross-section, we immediately realize that the shear center coincides with the centroid. Hence the applied shear foce passes through the shear center and will induce no twisting of the cross-section. We can then calculate the shear flow. Let's define the geometry and the loads (see Figure 4a). ü Shear force In[96]:= Sy = 1000 Out[96]= 1000 ü Wall thickness (which is the same for the web and the flanges) In[97]:= t = 0.003 Out[97]= 0.003 ü Length of flanges and web In[98]:= h = 0.1 Out[98]= 0.1 Problem 4 _ Solution.nb 1
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Given that the thickness of the web and flanges is much smaller than h, we can adopt the usual thin-wall approximation, and neglect terms that are of order t/h when compared to h. The global section is symmetric; consequently, Ixy=0. Since Sx=0, the general expression for the shear flow reduces to: q = qs0 - Sy ÅÅÅÅÅÅ Ixx Ÿ 0 s t y s We need to calculate the moment of inertia, Ixx with respect to the global centroidal system. We already know the position of the centroid by symmetry considerations. We
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This note was uploaded on 03/10/2009 for the course MAE 157 taught by Professor Valdevit during the Winter '08 term at UC Irvine.

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HW4_Solution - S.17.7 Initially the position of the...

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