HW8solnREV

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 25

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online