HW8solnREV - Problem 8.31 Shear flow in circular section i3...

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Problem 8.31 Shear flow in circular section R t s θ i 2 i 3 Remove @ "Global` " DH remove all symbols L The centroidal axes are located at the center of the circle as shown above. (1) Find bending stiffness: From symmetry, H 22 = H 33 and H 23 = 0. H 22 = 2 Ε −π ê 2 π ê 2 H R Sin @ θ DL 2 tR Åθ π R 3 t Ε (2) Find shear flow distribution due to V 3 : The loading H V 3 ) is also symmetric and therefore the shear flow must vanish on the symmetry axis i 3 . As a result, f q H q=≤p ê 2 L = 0 an the problem can be analyzed as two back-to-back semicircular sections (see problem 8.18). Rather than make use of this important simplification, we will go ahead and construct a "naive" solution, making a cut at q =0 instead. Shear flow f o in open (cut) system: Q 2 θ 0 θ R Sin @ α D tR Åα R 2 t Ε H 1 + Cos @ θ DL f o θ =− V 3 Q 2 θ H 22 H 1 + Cos @ θ DL V 3 π R
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Closing shear flow f c : warp = 0 2 π f o θ + f c Gt R Åθ 2 π Rf c 2V 3 Gt 8 f c < = 8 f c . Solve @ warp m 0, f c [email protected]@ 1 DD : V 3 π R > Shear flow f o + f c : f θ = f o θ + f c êê Simplify Cos @ θ D V 3 π R NOTE : This is identical to the expression for shear flow in a semicircular section, - p /2 §q§p /2. As a result, the shear flow is zero on symmetry axis i 3 as pointed out at the beginning. (3) Find location and magnitude of max shear flow: It is clear from the expression for f q that it is maximum at q =0, p (where section crosses axis i 2 ), and the value is: f θ max = f θ ê . θ→ 0 V 3 π R 2 chapter8Esolutions.nb
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i 2 i 3 R a a t t t cut s 1 s 2 s 3 s 4 θ A B Remove @ "Global` " DH remove all symbols L Shear V 3 is applied to section. a R; H simplifies the algebra L (1) Find centroid and bending stiffnesses: Since both axes are symmetry axes, the centroid is at the center of the circle as shown in the figure above. H 22 0 2 π H R Sin @ θ DL 2 tR Åθ π R 3 t Ε H 33 0 2 π H R Cos @ θ DL 2 tR Åθ+ 2 1 12 a 3 t + at R + a 2 2 êê Simplify 1 3 R 3 t I 3 π+ 2 α I 3 + 3 α+α 2 MM Ε H 23 = 0; H due to symmetry L (2) FInd shear flow on section: The presence of the flanges will have no effect because they both lie on axis i 2 and therefore their stiffness static moment, Q, is zero. As a result, this problem reduces to s simple circular tube. Nonetheless, we will illustrate the full "naive" solution that assumes we didn't realize this. Shear flow f oi in open (cut) section: Cut tube as indicated in figure to create an open section described by s i where s 2 = R q 2 and s 4 = R q 4 . Use notation where Q ij refers to axis i i and perimetric coordinate s j . 14
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HW8solnREV - Problem 8.31 Shear flow in circular section i3...

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