Shear Centre Open Section 2009

Shear Centre Open Section 2009 - Example The Shear Flow in...

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1 55 Example – The Shear Flow in an Open Channel Section Subject to a Horizontal Shear Force Sx y yy x D I S s q = ) ( The initial steps, 1 and 2, are identical to those of section = s 0 y ds x t ) s ( D Step 3 – Calculation of the Distribution y D d h 1 2 3 4 t t t S x 0 0 = = xy y I S 56 0 = s G x d s = G x d x y s x d x G = ) ( s d + + = β 2 1 1 2 + = 2 1 d x G 0 = G y 1 2 34 x G y x y
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2 57 Tutorial sheet 2 1. Determine the shear flow for the tube section shown in figure 1 96 48 16 16 t=2 All units mm Figure 1 58 Shear Centre – open section So far we have not discussed the location of the applied force, only the direction has been specified. The reason is that an open section has such a low torsional stiffness that the applied shear forces must be applied at the shear centre of the open tube. This means that the definition of the shear centre can be re-cast for the open section only to being the position where the resultant shear forces act. Let us now evaluate the resultant forces
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3 59 Resultants of the Shear Flow The shear flow in the skin will have a horizontal resultant, a vertical resultant and a torsion resultant. ds skin s α q q ds ds q cos α ds q sin α ds The Vertical Resultant The Horizontal Resultant () ds sin q S S y α = ds cos q S S x α = 60 Torsional Resultant The torsional resultant can be found by taking moments about some point 0. ds s skin q ds p 0 The contribution of this component of the shear flow to the total torque is Note that this can be interpreted geometrically as the swept area shown shaded in the Figure Care must be taken when evaluating this integral. In the Figure it has be assumed that
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This note was uploaded on 03/24/2010 for the course AE A.213 taught by Professor S.pinhoandm.aliabadi during the Winter '09 term at Imperial College.

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Shear Centre Open Section 2009 - Example The Shear Flow in...

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