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 find the horizontal position of the shear centre, S, apply an arbitrary shear load, S y , through S. Since S x = 0 Eq. (17.14) simplifies 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 I xy s 0 tx d s I yy s 0 ty d s (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 s 0 tx d s 53 24 s 0 ty d s (ii) from which q 12 = 9 S y 97 a 3 s 0 13 a 8 s d s 53 24 s 0 ( a )d s (iii) i.e. q 12 = 9 S y 97 a 3 7 as 12 + s 2 2 (iv) Taking moments about the corner 3 of the section S y ξ S = − 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 a 0 7 as 12 + s 2 2 d s
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) simplifies to q s = − S x I xx I yy I 2 xy I xx s 0 tx d s I xy s 0 ty d s from which, by comparison with Eq. (iii) q 12 = − 9 S x 97 a 3 t 16 3 s 0 t 13 a 8 s d s + s 0 t ( a )d s 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 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 a 0 (23 as 8 s 2 )d s which gives η S = 46 a 97 S.17.8

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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
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 ÅÅÅÅÅÅÅ Ixx Ÿ 0 s ty 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 can divide the entire section in three sub-sections (see figure 4a). The coordinates of the centroids of the three sub-sections, referred to the global centroidal system, are obviously given by: In[99]:= xc1 = 0 Out[99]= 0 In[100]:= yc1 = h ê 2 Out[100]= 0.05 In[101]:= xc2 = 0 Out[101]= 0 In[102]:= yc2 = 0 Out[102]= 0 In[103]:= xc3 = 0 Out[103]= 0 In[104]:= yc3 = - h ê 2 Out[104]= - 0.05 Problem 4 _ Solution.nb 2

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The three areas are the same, and given by: In[105]:= A1 = ht Out[105]= 0.0003 In[106]:= A2 =
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