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Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10)

Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10). MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering. Introduction.

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Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10)

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  1. Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering Skew Loads & Non-Symmetric XSections

  2. Introduction • Will perform advanced stress and deflection analysis of beams with skew loads and non-symmetric cross sections. • Challenge: Need to calculatemoments of inertia – Iyy, Izz,and Iyz – and principal moments of inertia. α P Skew load P z P x y z y Non-Symmetric y Skew Loads & Non-Symmetric XSections

  3. Moments of Inertia • For any cross-section shape C z dA y Skew Loads & Non-Symmetric XSections

  4. Moments of Inertia • The moments of inertia can be transformed to y1-z1 coordinates by • Does this look familiar?? z1 C z dA θ y1 y Skew Loads & Non-Symmetric XSections

  5. Moments of Inertia • Similar to transformation of stress, principal angle (angle to the principal axes of inertia) can be found from • Where θPis the angle at which Iyz is zero. z1 C dA z θ y1 y Skew Loads & Non-Symmetric XSections

  6. Example • Find Iyy and Izz for a rectangle b C h z dA = dydz y Skew Loads & Non-Symmetric XSections

  7. Example • Find Iyy and Izz for a Z-section (non-symmetric about y-z) Let b = 7 in, t = 1 in, and h = 16 in. b t (all) h/2 z C h/2 y b Skew Loads & Non-Symmetric XSections

  8. Example • Find Iyy and Izz for an L-section (non-symmetric about y-z) Let b = 4 in, t = 0.5 in, and h = 6 in. t z h C y t b Skew Loads & Non-Symmetric XSections

  9. Skew Loads (3.10) • Skew loads for doubly symmetric cross sections • Beam will bend in two directions • Py = P cos α • Pz = P sin α C z Pz Py α x (origin of x-axis at fixed end) y P Skew Loads & Non-Symmetric XSections

  10. Skew Loads (3.10) • Find bending moments • Side view • From statics: Mz = Py(L-x) = P cos α (L-x) • Why is Mz positive? • beam is curving in direction of positive y Side x L-x z (in) Mz Mz Py C z Pz Py α y x y P Skew Loads & Non-Symmetric XSections

  11. Skew Loads (3.10) • Find bending moments • Top view • From statics: My = Pz(L-x) = P sin α (L-x) • Why is My positive or negative? Top z x L-x y (in) My My Pz C z Pz Py α x y P Skew Loads & Non-Symmetric XSections

  12. Skew Loads (3.10) • Bending stress • From side view • From top view • Combine to get  Top Side C z Pz Py α x y P Skew Loads & Non-Symmetric XSections

  13. Skew Loads • What about the neutral axis? • When there is only vertical bending, σxx=0 because y=0 at the neutral axis. no stress on this line N.A. (y=0) z C y Skew Loads & Non-Symmetric XSections

  14. Skew Loads • But with a skew load: • It turns out deflection will be perpendicular to this line. z C no stress on this line β y Skew Loads & Non-Symmetric XSections

  15. Skew Loads • Curvature due to moment • From side view • From top view • Where vy and vz are deflections in the positive y and z directions, respectively. Top Side C z Pz Py α x y P Skew Loads & Non-Symmetric XSections

  16. Skew Loads • Find deflection at free end. • Apply B.C.’s: vy(0)=0 & vy’(0)=0 • The tip deflection in the y-direction is C z Pz Py α x y P Skew Loads & Non-Symmetric XSections

  17. Skew Loads • Continued… • Apply B.C.’s: vz(0)=0 & vz’(0)=0 • The tip deflection in the z-direction is C z Pz Py α x y P Skew Loads & Non-Symmetric XSections

  18. Skew Loads • The resultant tip deflection is z C N.A. C z Pz Py α vz β x y y vy P δ Skew Loads & Non-Symmetric XSections

  19. Example • Consider a cantilever beam with the cross-section and load shown below. Find the stress at A and the tip deflection when α = 0oand α=1o. Let L = 12 ft, P = 10 kips, E = 30x106 psi and assume an S24x80 rolled steel beam is used. A (z=3.5 in, y=-12 in) z y C α P Skew Loads & Non-Symmetric XSections

  20. Non-Symmetric Cross-Sections • Bending of non-symmetriccross-sections • Iyz≠ 0 • Iyy & Izzare not principal axes • Use generalized flexure formula Mz C z y My Skew Loads & Non-Symmetric XSections

  21. Non-Symmetric Cross-Sections • Generalized moment-curvature formulas Mz C z y My Skew Loads & Non-Symmetric XSections

  22. Non-Symmetric Cross-Sections • A special case – which we discussed previously – is whenIyz = 0 and y & z are the principal axes. Skew Loads & Non-Symmetric XSections

  23. Example • Analysis choices • Work in principal coordinates – simple formulas • Work in arbitrary coordinates – more complex formulas • Calculate the stress at A and the tip deflection for the beam shown below. z Mz = 10,000 in-lbs(pure bending) Cross-section dimensions:6 x 4 x 0.5 in Iyy= 6.27 in4 Izz= 17.4 in4 Iyz = 6.07 in4 E = 30 x 106 psi A (z=-0.99 in, y=-4.01 in) L = 10 ft y C Skew Loads & Non-Symmetric XSections

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