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22 Pages

### Lecture Notes 2.3

Course: 125 208, Spring 2011
School: Rutgers
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Word Count: 1097

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happens What when a beam bends? g Originally, the line segments ab, ab and cd are the same length, but after bending, the top surface (ab) gets shorter and the bottom (ab) longer, and cd stays the same length. If you draw a short segment, de that is parallel to aa, the the length of the line ae is the same as the length of the line cd, and the beam has stretched (in that location) a length 1 ef. But by how...

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happens What when a beam bends? g Originally, the line segments ab, ab and cd are the same length, but after bending, the top surface (ab) gets shorter and the bottom (ab) longer, and cd stays the same length. If you draw a short segment, de that is parallel to aa, the the length of the line ae is the same as the length of the line cd, and the beam has stretched (in that location) a length 1 ef. But by how much? O Triangle OCD is similar to triangle DEF. l ef df y = = = = l cd Od If we assume Hooke' s law : = E = E y g 2 = E = E y =M I EI M= y Bending Stress M = EI 1 Curvature Material Property Applied Moment Geometric Property 3 From calculus, curvature = 1 = d2y dx 2 (1 + ( dx ) ) 2 (Note, here y is deflection, not distance from neutral axis). dy 2 3 For small deflections, (dy dx ) << 1, so denom ~ 1. 2 d2y =2 dx 1 d2y M = 2 dx EI or d2y M = EI 2 Euler - Bernoulli Equation for Bending dx 4 What if we have a cantilevered beam with an applied point force (P) on the end From our equations of static equilibrium we can find that there is a reaction force P and a reaction moment PL at x=0. Therefore the moment along the length of the beam is M ( x) = PL + Px 5 d2y = M = PL + Px dx 2 Integrating once EI PL3 y ( L) = max deflection = 3EI PL2 slope at the end y ' ( L) = 2 EI dy Px 2 EI = PLx + + C1 2 dx The beam must have no slope at x = 0 so dy dx = 0 C1 = 0 x =0 Integrating again PLx 2 Px 3 EIy = + + C2 2 6 again applying the boundary condition that at x = 0 there is no beam deflection ( y = 0) C2 = 0 Therefore P x 3 Lx 2 y= 2 EI 6 6 What about a beam with a distributed load, w d2y w 2 EI 2 = (L x ) dx 2 dy w 3 = (L x ) + C1 EI dx 6 dy w = 0 C1 = L3 dx x =0 6 w dy w 3 = (L x ) L3 dx 6 6 w w 4 EIy = (L x ) L3 x + C2 24 6 wL4 y (0 ) = 0 C2 = 24 w w3 wL4 4 EIy = (L x ) L x + 24 6 24 w 4 (L x ) 4 L3 x + L4 y ( x) = 24 EI wL4 y ( L) = max deflection = 8 EI EI w 2 M = (L x ) 2 ( ) 7 So our spring constant is just the Force/deflection For a constant end force 3EI F= 3 y L Effective spring constant k For a distributed load 8 EI F = qL = 3 y L Effective spring constant k 8 How can biomechanics be applied? Cellular Mechanics 9 What about a beam in 3 point bending with symmetry? The bending moment for the first half of the beam is M ( x) = R1 x = C2=0 because there is no deflection at x=0 P x 2 0< x< d2y P EI 2 = x 2 dx dy P x 2 = + C1 EI dx 2 2 P x3 + C1 x + C2 EIy = 26 Px 3 + C1 x EIy = 12 L 2 10 Because the beam is symmetric, the slope at x=L/2 must be zero dy EI dx P (L / 2 ) = + C1 = 0 22 2 L/2 PL2 C1 = 16 Px 3 PL2 EIy = x 12 16 11 Shear Stress Shearing Stresses For any beam subjected to a shearing force, V, both horizontal and vertical shear stresses, , are produced 12 The shear stress at a distance yo from the neutral axis is given by c V = yda Ib yo So for a rectangular beam of height h h 2 V V bydy = = y 2 Ib yo 2I h 2 yo V h2 2 = y0 2I 4 From this we see the maximum shear stress occurs at the neutral axis (yo=0) and has magnitude Vh 2 bh 3 max = 8I I= 12 max 12V 3V = = 8bh 2bh 13 Beams/Stress Summary Tension or Compression: stress=F/A stress is same throughout Bending: stress=My/I maximum=Mc/I Shear: stress= max = 3/2 V/bh So, given some complex loading at the same time! Vc ydA situation,yall of Ib 0 F Fy Fx these stresses may be acting Produces: 1) Compression (Fx) 2) Shear (Fy) 3) Bending (Fy) These stresses combine! 14 Example F Ry Rx Fx = 0 : Fx Rx = 0; Rx = Fx Fy = 0 : Fy + R y = 0; R y = Fy M = 0 : Fy L + M = 0; M = Fy L h M w 1 wh 3 12 A = wh I= What stresses do each of these cause? 15 Fx = 0 : Fx Rx = 0; Rx = Fx Fx Rx h w Fx and Rx cause uniaxial compression.how much? 1 wh 3 12 A = wh I= Fx Fx = = A wh Is it uniform across the cross-section? YES Is it uniform along the length of the beam? YES 16 Fy Fy = 0 : Fy + R y = 0; R y = Fy M = 0 : Fy L + M = 0; M = Fy L M h Fx = 0 : Fx R x = 0; R x = Fx Fy and Ry cause shear.how much? Fy and M cause bending.how much? Ry 1 wh 3 12 A = wh I= w Fy VM Fy = 0; Fy V = 0;V = Fy M = 0; Fy x + M = 0; M = Fy x 17 Fy = 0 : Fy + R y = 0; R y = Fy Fy Fy = 0; Fy V = 0;V = Fy M = 0; Fy x + M = 0; M = Fy x V -Fy Note: Results consistent with FBD of whole beam/end conditions = CLOSURE M -(Fy)(L) 18 What stress does V produce? Shear stress, How much? VQ Ib Is it constant across the cross-section? NOMax at neutral axis, zero at top and bottom 3V 3 Fy max = = 2 bh 2 wh = Is it distributed evenly along the length of the beam? YES Why? Because from our shear diagram we know the shear force, V is a constant V=-Fy 19 What stress does M produce? Normal stress, How much? My = I Is it constant across the cross-section? NOMax at top and bottom, zero at neutral axis Also, one side of the beam will be in tension (top) and one side in compression (bottom) 1 Mh 2I Is it distributed evenly along the length of the beam? NO! Why? Because from our moment diagram we know the internal bending moment, M is a function of the distance, x M=-Fy x Where is it max? max/ min = max/ min 1 Fy Lh 6 Fy L = = 2I bh 2 1 wh 3 12 A = wh I= 20 So..combined we have (max values shown): From Fx : Fx x = bh Uniform compression across xsection and along length From Fy : 3 Fy = 2 bh At neutral axis (y = 0) and uniform along length x = 6 Fy L bh x = At x = L and y = h/2 2 6 Fy L At x = L and y = -h/2 bh 2 21 So..combined we have (max values shown): From Fx : F x = x bh x=L Uniform compression From Fy : 3 Fy = 2 bh x = + 6 Fy L bh x = Tension on top Compression on bottom Zero at y=0 2 6 Fy L = bh 2 6 Fy L Fx Tension on top : x = 2 bh bh Compression on bottom : x = 6 Fy L bh 2 Fx bh COMBINED: Less Tension on top More Compression on bottom NOT Zero at y=0 22
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