MT-1620 al.2002 Unit 17 The beam-Column Readings Theory of Elastic Stability Timoshenko(and Gere), McGraw-Hill, 1961(2nd edition), Ch. 1 Paul A Lagace, Ph. D Professor of aeronautics Astronautics and Engineering Systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 17 The Beam-Column Readings: Theory of Elastic Stability, Timoshenko (and Gere), McGraw-Hill, 1961 (2nd edition), Ch. 1 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 al.2002 Thus far have considered separately beam - takes bending loads column -takes axial loads Now combine the two and look at the beam-column (Note: same geometrical restrictions as on others l > cross-sectional dimensions) Consider a beam with an axial load (general case) Figure 17.1 Representation of beam-column could also have p for bending in y direction) X Consider 2-D case: Paul A Lagace @2001 Unit 17-2
MIT - 16.20 Thus far have considered separately: • beam -- takes bending loads • column -- takes axial loads Now combine the two and look at the “beam-column” (Note: same geometrical restrictions as on others: l >> cross- sectional dimensions) Consider a beam with an axial load (general case): Figure 17.1 Representation of beam-column Fall, 2002 (could also have py for bending in y direction) Consider 2-D case: Paul A. Lagace © 2001 Unit 17 - 2
MT-1620 al.2002 Cut out a deformed element d Figure 17.2 Loads and moment acting on deformed infinitesimal element of beam-column M+rdx dx 召 dx dx 积 2 X Assume small angles such that sIn ax dw cOS d Paul A Lagace @2001 Unit 17-3
MIT - 16.20 Fall, 2002 Cut out a deformed element dx: Figure 17.2 Loads and moment acting on deformed infinitesimal element of beam-column Assume small angles such that: sin dw ≈ dw dx dx dw cos ≈ 1 dx Paul A. Lagace © 2001 Unit 17 - 3
MT-1620 al.2002 Sum forces and moments dF F+F+dx+p, dx d1 ds, dw d S-+s+adx dx|=0 ax This leaves (dx) dF dsdw dw dx+p, dx S- ldx+HOT=0 dx dF → p S 17-1) ax new term Paul A Lagace @2001 Unit 17-4
MIT - 16.20 Fall, 2002 Sum forces and moments: + • ∑Fx = 0 : dF − F F + + dx + px dx dx 2 − S dw + S + dS dx dw + d w dx = 0 dx dx dx d x2 This leaves: 2 dF d S dw S dw dx + px dx + + 2 dx + H O T. = 0 (dx)2 .. dx d x dx dx ⇒ dF dx p d dx S dw dx = − x − (17-1) new term Paul A. Lagace © 2001 Unit 17 - 4
MT-16.20 al.2002 F=01+ dF, d Fa+F+=dx dx ax S-S+dx +pdx=0 This results in d/dw d x dx、dx (17-2) new term ∑M=0 M+m dM x dx+p dx dw dx dS x S+dx dx=0 dx 2 (using the previous equations)this results in Paul A Lagace @2001 Unit 17-5
MIT - 16.20 Fall, 2002 • ∑Fz = 0 + : 2 − F dw + F + dF dx dw + d w dx dx dx dx d x2 dS + S − S + dx + pz dx = 0 dx This results in: dS dx p d dx F dw dx = z + (17-2) new term • ∑ My = 0 + : dM dx − M + M + dx + pz dx dx 2 dw dx dS − p dx − S + dx dx = 0 x dx 2 dx (using the previous equations) this results in: Paul A. Lagace © 2001 Unit 17 - 5
MT-1620 al.2002 (173) dx Note: same as before( for Simple Beam Theory Recall from beam bending theory M= El W 7-4) x Do some manipulating -place(17-4)into(17-3) S El d x dx (175) and place this into (17-2)to get El P(17-6) d x dx af as dx Basic differential equation for beam-Column Bending equation -fourth order differential equation) Paul A Lagace @2001 Unit 17-6
MIT - 16.20 Fall, 2002 dM = S (17-3) dx Note: same as before (for Simple Beam Theory) Recall from beam bending theory: 2 M = EI dw (17-4) dx2 Do some manipulating - place (17-4) into (17-3): 2 S = d EI dw dx dx2 (17-5) and place this into (17-2) to get: 2 d 2 EI dw dx2 dx2 − d F dw = pz (17-6) dx dx Basic differential equation for Beam-Column -- (Bending equation -- fourth order differential equation) Paul A. Lagace © 2001 Unit 17 - 6
MT-1620 al.2002 To find the axial force F(x), place(17-5)into(17-1) d dw d E dx Px dx d: x ax d x For w small, this latter part is a second order term in w and is therefore negligible Thus dF (17-7) dx Note: Solve this equation first to find F(x distribution and use that in equation (17-6) Examples of solution to Equation (17-7) End compression P Figure 17. 3 Simply-supported column under end compression Paul A Lagace @2001 Unit 17-7
MIT - 16.20 Fall, 2002 --> To find the axial force F(x), place (17-5) into (17-1): 2 dF = − px − d dw d EI dw dx dx dx dx dx2 For w small, this latter part is a second order term in w and is therefore negligible Thus: dF = − px (17-7) dx Note: Solve this equation first to find F(x) distribution and use that in equation (17-6) Examples of solution to Equation (17-7) • End compression Po Figure 17.3 Simply-supported column under end compression px = 0 Paul A. Lagace © 2001 Unit 17 - 7
MT-1620 al.2002 dF 0→F=C d find C, via boundary condition @x=0, F=-P0=C1 Beam under its own weight Figure 17. 4 Representation of end-fixed column under its own weight Px =-mg dF dr+mg→F=mgx+C boundary condition: @X=, F=0 So:mg+C=0→C1=mg Paul A Lagace @2001 Unit 17-8
MIT - 16.20 Fall, 2002 dF dx = 0 ⇒ F = C1 find C1 via boundary condition @x = 0, F = -Po = C1 ⇒ F = -Po • Beam under its own weight Figure 17.4 Representation of end-fixed column under its own weight px = -mg dF = + mg ⇒ F = mgx + C1 dx boundary condition: @ x = l, F = 0 So: mgl + C1 = 0 ⇒ C1 = -mgl Paul A. Lagace © 2001 Unit 17 - 8
MT-16.20 al.2002 →F=-mg(-x) Helicopter blade Figure 17.5 Representation of helicopter blade (radial force due to rotation) similar to previous case Once have F(x), proceed to solve equation (17-6). Since it is fourth order, need four boundary conditions (two at each end of the beam column) same possible boundary conditions as previously enumerated Notes When El-->0, equation(17-6)reduces to d/d dx dx this is a string(second order only need two boundary conditions one at each end Paul A Lagace @2001 Unit 17-9
MIT - 16.20 Fall, 2002 ⇒ F = -mg (l - x) • Helicopter blade Figure 17.5 Representation of helicopter blade (radial force due to rotation) similar to previous case Once have F(x), proceed to solve equation (17-6). Since it is fourth order, need four boundary conditions (two at each end of the beamcolumn) --> same possible boundary conditions as previously enumerated Notes: • When EI --> 0, equation (17-6) reduces to: − d F dw = pz dx dx this is a string (second order ⇒ only need two boundary conditions -- one at each end) Paul A. Lagace © 2001 Unit 17 - 9
MT-1620 al.2002 (also note that a string cannot be clamped since it cannot carry a moment) If F=0, get d x2 El and for ei constant El dr+"P:(basic bending equation) For p,=0, EI constant, and F constant(=-P), get E/H (basic buckling equation d x d x2 Buckling of Beam-Column Consider the overall geometry(assume beam-column initially straight) Paul A Lagace @2001 Unit 17-10
MIT - 16.20 Fall, 2002 (also note that a string cannot be clamped since it cannot carry a moment) • If F = 0, get: 2 d 2 EI dw = pz dx2 dx2 and for EI constant: 4 EI dw = pz (basic bending equation) dx 4 • For pz = 0, EI constant, and F constant (= -P), get: 4 2 EI dw + P dw = 0 (basic buckling equation) dx 4 dx2 Buckling of Beam-Column Consider the overall geometry (assume beam-column initially straight) Paul A. Lagace © 2001 Unit 17 - 10
