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AERSP 301 Bending of open and closed section beams Dr. Jose Palacios.

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1 AERSP 301 Bending of open and closed section beams Dr. Jose Palacios

2 Today’s Items HW 2 due today! HW 3: Extra Credit – 50% of 1 HW worth towards HWs grade HW 4: will be assigned on Monday. HW helper session coming up. Bending of Beams -- Megson Chapter 16, Reference: Donaldson – Chapters Lambda and Mu –Direct stress calculation –Bending deflections

3 Today HW 3 (EXTRA CREDIT, 50% OF 1 HW): DUE WEDNESDAY HW 4: ASSIGNED TODAY EXAM: OCTOBER 20 – 26 HOSLER – 8:15 – 10:15 PM REVIEW SESSION: OCTOBER 19 – 220 HAMMOND – 6 – 9 PM WEDNESDAY AFTER CLASS: VOTING REGISTRATION

4 Direct stress at a point in the c/s depends on: –Its location in the c/s –The loading –The geometry of the c/s Assumption – plane sections remain plane after deformation (No Warping), or cross-section does not deform in plane (i.e. σxx, σyy = 0) Sign Conventions! Megson pp 461 Direct stress calculation due to bending M – bending moment S – shear force P – axial load T – torque W – distributed load

5 Direct stress calculation due to bending (cont’d) Beam subject to bending moments M x and M y and bends about its neutral axis (N.A.) N.A. – stresses are zero at N.A. C – centroid of c/s (origin of axes assumed to be at C).

6 Neutral Surface Definition In the process of bending there is an axial line that do not extend or contract. The surface described by the set of lines that do not extend or contract is called the neutral surface. Lines on one side of the neutral surface extend and on the other contract since the arc length is smaller on one side and larger on the other side of the neutral surface. The figure shows the neutral surface in both the initial and the bent configuration.

7 The axial strain in a line element a distance y above the neutral surface is given by: Consider element  A at a distance ξ from the N.A. Direct Stress: Because ρ (bending radius of curvature) relates the strain to the distance to the neutral surface: Direct stress calculation due to bending (cont’d)

8 First Moment of Inertia Definition Given an area of any shape, and division of that area into very small, equal-sized, elemental areas (dA) and given an C x -C y axis, from where each elemental area is located (yi and xi) The first moment of area in the "X" and "Y" directions are respectively:

9 IF the beam is in pure bending, axial load resultant on the c/s is zero: 1 st moment of inertia of the c/s about the N.A. is zero  N.A. passes through the centroid, C Assume the inclination of the N.A. to C x is α Direct stress calculation due to bending (cont’d) Then The direct stress becomes:

10 Direct stress calculation due to bending (cont’d) Moment Resultants: Substituting for σ z in the above expressions for M x and M y, and using definitions for I xx, I yy, I xy

11 Direct stress calculation due to bending (cont’d) Using the above equation in: Gives: From Matrix Form

12 Direct stress calculation due to bending (cont’d) Or, rearranging terms: If M y = 0, M x produces a stress that varies with both x and y. Similarly for My, if M x =0. If the beam c/s has either C x or C y (or both) as an axis of symmetry, then I xy = 0. Then:

13 Further, if either M y or M x is zero, then: We saw that the N.A. passes through the centroid of the c/s. But what about its orientation α? At any point on the N.A. σ z = 0 Direct stress calculation due to bending (cont’d) or

14 Example Problem The beam shown is subjected to a 1500 Nm bending moment in the vertical plane. Calculate the magnitude and location of max σ z. 0 8 mm 40 mm80 mm 8 mm 1 st : Calculate location of Centroid x y

15 Example Problem (cont’d) Calculate I xx, I yy, I xy, with respect to C xy :

16 Example Problem (cont’d) M x = 1500 Nm, M y = 0 By inspection, MAX at y = mm and x = -8 mm (Max stress always further away From centroid) 0 x y

17 Deflections due to bending From strength of materials, recall that [Megson Ch ]: Beam bends about its N.A. under moments M x, M y. –Deflection normal to N.A. is ζ  Centroid C moves from C I (initial) to C F (final). –With R as the center of curvature and ρ as the radius of curvature MxMx MyMy wxwx w wywy Distributed Load

18 Deflections due to bending (cont’d) Further, Because

19 Deflections due to bending (cont’d) Inverse relation: Clearly Mx produces curvatures (deflections) in xz and yz planes even when My = 0 (and vice-versa) So an unsymmetrical beam will deflect vertically and horizontally even when loading is entirely in vertical (or horizontal) plane. What if I have something symmetric?? Like NACA 0012 airfoil?

20 Deflections due to bending (cont’d) If Cx or Cy (or both) are axes of symmetry then Ixy = 0. Then the expressions simplify to: Starting with the general expression: and integrating twice you can calculate the disp. u in the x-direction

21 Deflections due to bending (cont’d) Consider the case where a downward vertical force, W, is applied to the tip of a beam. What is the tip deflection of the beam? Integrating, Integrating again

22 Deflections due to bending (cont’d) Using z=0  u = 0, u’ = 0 –Gives: A = B = 0 Thus, –Tip deflection: If the c/s has an axis of symmetry, I xy = 0 You should do this on your own (z = L)

23 Simplifications for thin-walled sections Thin-walled  t << c/s dimensions. –Stresses constant through thickness –Terms in t 2, t 3, etc… neglected In that case I xx reduces to: What about I xy for this c/s What about I yy for this c/s  horizontal members  vertical members Example: You should do this on your own

24 Doubly Symmetrical Cross-Section Beam Bending Beam has a flexural rigidity: EI EI v’’’’ = distributed load EI w’’’ = Shear Force EI w’’ = Moment EI w’ = Slope y p(z) z Bc’s EI w’ = Displacement

25 Doubly Symmetrical Cross-Section Beam Bending Beam has a flexural rigidity: EI EI w’’’’ = distributed load EI w’’’ = Shear Force EI w’’ = Moment EI w’ = Slope y z Bc’s EI w’ = Displacement F

26 Concentrated and Partial Span Loads Diract delta function: Example  vertical force of magnitude F 0 locater at L/2 Heaviside step function: Example  vertical distributed force of magnitude f 0 (z) over the second part of the beam only z0z0 f

27 Concentrated and Partial Span Loads Example z y L/3 f0f0 FoFo z x f0f0 MoMo Bc’s w(0) = w’(0) = v(0) = v’(0) = w(L)= w’(L) = v(L) = v’(L) = 0

28 Integrating Diract Delta and Heaviside Function

29

30 OPTIONAL: Macauley’s Method Read Megson Chp 16.) y z aa aa W W2W A B C D F Ra Rf Determine the position and magnitude of the maximum upward and downward deflection of the beam: The bending moments around the left hand side at any section Z between D and F is:


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