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 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 p y for bending in y direction) Consider 2-D case: Paul A. Lagace ? 2001 Unit 17 - 2 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 MIT - 16.20 Fall, 2002 Sum forces and moments: + ? ∑ F x = 0 : dF ? FF ++ dx + p x d x dx 2 ? S dw + ? ? S + dS dx ?? ?? dw + dw dx ? ? = 0 dx ? dx ? ? dx dx 2 ? This leaves: 2 dF ? dS dw S dw ? dx + p x d x + ? + 2 ? dx + H O T . = 0 (dx) 2 .. dx ? dx dx dx ? ? dF dx p d dx S dw dx x = ? ? ? ? ? ? ? ? (17-1) new term Paul A. Lagace ? 2001 Unit 17 - 4 MIT - 16.20 Fall, 2002 ? ∑ F z = 0 + : 2 ? F dw + ? ? F + dF dx ?? ?? dw + dw dx ? ? dx ? dx ? ? dx dx 2 ? ? dS ? + S ? ? S + dx ? + p z d x = 0 ? dx ? This results in: dS dx p d dx F dw dx z = ? ? ? ? + (17-2) new term ? ∑ M y = 0 + : dM dx ? M + M + dx + p z d x dx 2 dw dx ? dS ? ? pd x ? ? S + dx ? dx = 0 x dx 2 ? dx ? (using the previous equations) this results in: Paul A. Lagace ? 2001 Unit 17 - 5 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 = E I dw (17-4) dx 2 Do some manipulating - place (17-4) into (17-3): 2 S = d ?? EI dw ? dx ? dx 2 ?? (17-5) and place this into (17-2) to get: 2 d 2 ? EI dw ? dx 2 ? ? dx 2 ? ? ? d ?? F dw ? ? = p z (17-6) dx ? 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 --> To find the axial force F(x), place (17-5) into (17-1): 2 dF = ? p x ? d ? dw d ? EI dw ? ? dx dx ? ? dx dx ?? dx 2 ?? ?? For w small, this latter part is a second order term in w and is therefore negligible Thus: dF = ? p x (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 o Figure 17.3 Simply-supported column under end compression p x = 0 Paul A. Lagace ? 2001 Unit 17 - 7 MIT - 16.20 Fall, 2002 dF dx = 0 ? F = C 1 find C 1 via boundary condition @x = 0, F = -P o = C 1 ? F = -P o ? Beam under its own weight Figure 17.4 Representation of end-fixed column under its own weight p x = -mg dF =+ mg ? F = mgx + C 1 dx boundary condition: @ x = l , F = 0 So: mg l + C 1 = 0 ? C 1 = -mg l Paul A. Lagace ? 2001 Unit 17 - 8 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 beam-column) --> same possible boundary conditions as previously enumerated Notes : ? When E I --> 0, equation (17-6) reduces to: ? d ?? F dw ? ? = p z 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 (also note that a string cannot be clamped since it cannot carry a moment) ? If F = 0, get: 2 d 2 ? EI dw ? = p z dx 2 ? ? dx 2 ? ? and for E I constant: 4 EI dw = p z (basic bending equation) dx 4 ? For p z = 0, E I constant, and F constant (= -P), get: 4 2 EI dw + P dw = 0 (basic buckling equation) dx 4 dx 2 Buckling of Beam-Column Consider the overall geometry (assume beam-column initially straight) Paul A. Lagace ? 2001 Unit 17 - 1 0 MIT - 16.20 Fall, 2002 Figure 17.6 Representation of general configuration of beam-column Cut the beam-column: Figure 17.7 Representation of beam-column with cut to determine stress resultants ∑ M = 0 : Μ ? Μ primary + Pw = 0 due to transverse loading secondary moment (due to deflection) Paul A. Lagace ? 2001 Unit 17 - 1 1 MIT - 16.20 Fall, 2002 gives: 2 Μ = EI dw = Μ primary ? Pw dx 2 for transverse loading: 2 d 2 ? EI dw ? ? d ? F dw ? ? = p z dx 2 ?? dx 2 ? ? dx ? ? dx ? integrate twice with F = P = C 1 2 EI dw + Pw = M primary dx 2 same equation as by doing equilibrium Solve this by: ? getting homogenous solution for w ? getting particular solution for M primary ? applying boundary condition Paul A. Lagace ? 2001 Unit 17 - 1 2 MIT - 16.20 Fall, 2002 Figure 17.8 Representation of moment(s) versus applied load for beam-column large moment! Examples ? “Old” airplanes w/struts Paul A. Lagace ? 2001 Unit 17 - 1 3 MIT - 16.20 Fall, 2002 ? Space structure undergoing rotation inertial loading Final note : The beam-column is an important concept and the moments in a beam-column can be much worse/higher than beam theory or a perfect column alone Paul A. Lagace ? 2001 Unit 17 - 1 4