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2.001 - MECHANICS AND MATERIALS I Lecture #21 11/21/2006 Prof. Carol Livermore Recall from last time: Beam Bending y = 0 on neutral axis xx = −ρy (Note: purely geometric, no material properties) σxx = xx E (All other σ are equal to 0) So: σxx = −Ey ρ Force Equilibrium: A Fx = 0 σxx dA = 0 1 If E is constant in y then A A Ey dA = 0 ρ ydA − 0. Moment Equilibrium Mz = 0 M =− σxx ydA A M= A Ey 2 dA ρ Special case: E constant: M= 1 EI ρ y 2 dA I= A New this time: Recall: σxx = −Ey ρ For constant E (special case): M= EI ρ So: E M −σxx = = ρ I y −M y I EXAMPLE: Find location of neutral axis for rectangular beam σxx = 2 E is constant across cross-section. Recall force equilibrium. Ey dA = 0 A ρ E ydA = 0 ρ A b E 2 h−a ydydz = 0 ρ −b a 2 2b 2 y biggl]ha−a dz = 0 −b 2 2 b 1 2 [(h − a)2 − a2 ]dz = 0 2 −b 2 2b 1 (h2 − 2ha)dz = 0 2 −b 2 2b 1 2 (h − 2ha)z =0 2 −b 2 1 2 b b (h − 2ha)( + ) = 0 2 2 2 b 2 (h − 2ha) = 0 2 h2 = 2ha h a= 2 So the neutral axis is in the center of the beam. What if E2 > E1 in: a is the distance to neutral axis. 3 1 ρ b 2 −b 2 −a+ h 2 −a A Ey dA = 0 ρ E(y)ydA = 0 A E1 ydy + Note: Example: Moment of Inertia One material rectangular beam 4 h−a −a+ h 2 E2 ydy dz = 0 y 2 dA I= I= I= b 2 −b 2 A b 2 −b 2 dz −h 2 y 2 dy h2 dzbiggl[f racy 3 −h 2 b 2 −b 2 I= Example: h 2 3 I= Beam Design dz bh3 12 h3 12 Find σxx . FBD: Fy = 0 Vy − P = 0 Vy = P M∗ = 0 −Mz − P (a − x) = 0 Mz = −P (a − x) = 0 Mz (x) 1 = ρ EI What about shear? Distortion of planar sections of beam. The can be ignored for slender (long and skinny beams) 6 σxx = −M y −Mz (x)y P (a − x)y = = I I I σxxm ax = P a h2 6P a = 1 3 bh2 bh 12 Solve for I a. Do areas of integration b h3 b h3 b. I = 0120 − i12i What about a composite beam? of integral during derivation. This does not work because E was take out 7 Example: Skis Get good stiﬀness (bending) but give up axial stiﬀness and lower weight. Other examples: Plants Bird bones Airplanes Recall, x-axis is all for ”pure bending” 8