Now we are going to use our free body analysis to look at Beam Bending (W3L1) Problems 17, F2002Q1, F2003Q1c

Transcription

1 Now we are going to use our free body analysis to look at Beam Bending (WL1) Problems 17, F00Q1, F00Q1c One of the most useful applications of the free body analysis method is to be able to derive equations for stresses in beams no matter what the geometry or material. The procedure is no more complicated than what we just did for pressure vessels, although it has a lot more steps. When you bend something, like the cantilevered beam shown below, one surface will be in tension, the other in compression (indicated by arrows below). As you move through the thickness the stress gets smaller until it is ero at the (centre) neutral surface. As long as the beam is fairly thin there is no vertical stress in the beam (even though the applied force is vertical) all of the stress is in the longitudinal direction. In this and the next couple of lectures we will use the free body analysis method to develop equations to calculate longitudinal stresses in beams of different shapes and material types. neutral surface (ero stress) tension Try bending your pen to visualie this compression Side view of a cantilevered beam with a force W at the end. w 1

2 Try bending your pen to visualie this neutral surface (ero stress) tension compression Points to note: Vertical Forces: (not usually of interest) w Vertical external force W (down) Assume large radius of curvature () for the beam (in other words it isn t bent too much) Beam has only longitudinal stresses if it is thin with respect to its length Top surface in tension The further from Bottom surface in compression the neutral surface the higher the neutral surface in between stress (no longitudinal stress at neutral surface) Since F (external) is vertical then there is no horiontal external force, so the net internal force across a vertical cut 0 Assume initially that the beam is light Vertical reaction force at the wall (up) The following page outlines the specific steps we will work through in order to obtain an equation for longitudinal stress in a bent beam. We will be going through each step separately in this and the next lecture.

3 Free Body Analysis Steps Bending Step 1: We want to analye the longitudinal stresses so make the cut perpendicular to the length. NS Step : eaction force is not constant over the cross section, therefore we must determine the reaction force as a function of distance from the neutral surface. Y F (on da ) da Step : Static equilibrium condition. Longitudinal forces 0 across the cut, since no longitudinal external force exists. However the torques must balance as well (all beam bending problems are based on free body analysis of the torque equilibrium). Y W ( l x) da Step 4: Solve for (the radius of curvature) in terms of the x and y co-ordinates of the beam. 1 d y dx Step 5: Sub this into the equation from Step to give a nd order differential equation. d y YI W ( l x) dx Step 6: Solve the differential equation and apply boundary conditions to give (in the simplest case): W x x (the deflection y at any y l YI 6 point x along the beam) Step 7: Determine an equation relating the position on the beam surface to stress and strain (by combining above equations). surface strain at x surface stress at x cut I nd moment of cross sectional area: it defines dimensional characteristics a( l x) y ε s l ay ( l x) y τ s l end end w general internal torque eqn for bent beams

4 So now lets go through each step individually: Step 1: Make a cut perpendicular to the direction of interest We want to analye the longitudinal stress, so Make a vertical cut NS cut w (load force) Now, unlike the case for the pressuried cylinders, the internal reaction force (represented by the arrows above) is not constant throughout the cross section. So next we have to develop an expression for how the internal force varies as a function of distance away from the neutral surface (NS). 4

5 Step determine reaction force as a function of distance from the neutral surface (NS) A A Note this is not drawn to scale the curvature is much exaggerated but really should be very small (i.e. large ) δθ neutral surface (NS) Defining terms: is co-ordinate perpendicular to the NS with 0 at NS radius of curvature for the NS ± is the radius of curvature for any fibre in the bent rod So now... A. consider the small fibre A-A, subtending δθ and a distance from the NS B. the unstretched length of A-A (if beam is unbent) is the same as it would be at the NS δθ C. stretched length of A-A ( + )δθ D. stretch of A-A ( + )δθ -δθ δθ 5

6 Step cont So strain of any fibre A-A a distance away from NS ε l l δθ δθ (this strain is positive [+] above the NS, negative [-] below the NS) longitudinal strain at any ε If the beam thickness is a then at the surface a/ 1 longitudinal stress at any τ Y strain at surface: ε a 1a stress at surface: τ Ya a Eq is τ as a function of but we need F as a function of : To get this: Look at the Beam End-On (in cross section) Total force on da will be: F da τda Y da eq da NS So F da Y da (so this tells us how the reaction force varies through the cross section of the beam at any position away from the NS) [reaction forces are in at top and out at bottom in this diagram] 6

7 Step Static Equilibrium Condition: (summing the forces and torques) need co-ordinate system: y (deflection) x cut l (overall length) The cut segment is in static equilibrium under the influence of W and the internal reaction forces at the cut. W (load) Our cantilever beam is of length, l, with a load, W. We define the cut to be a distance, x, away from the wall. l -x Summing Forces: no horiontal W component (since bending is small), so internal longitudinal reaction forces0 Summing Torques: (sum torques about point Q) external torque about Q A little rearranging... Y W ( l x) da W ( l x) W ( l x) Y Y I da Q total internal reaction torque (about Q) reaction force for each da 4 distance da is from Q (note we didn t need to sum torques for the pressure vessel since axisymmetric all cancel Cut segment W I : nd moment of cross sectional area (next page) where I da 7

8 Detour!!!! What is I? The second moment of cross sectional area about an axis which is in the neutral surface and at right angles to the external force. It defines how the area is distributed about the neutral surface (sort of analogous to the moment of inertia in rotation, which defines how mass is distributed about a rotational axis). Units are m 4!! Calculating I for a bar of rectangular cross section a looking at cross section: d b da NS I da b + d See following page for I tables for common cross sections I a a da bd integrate from -a/ to +a/ 1 1 ba (units m 4 ) Further note: for composite objects you can add or subtract I for different components provided they have the same NS (further note this is just like adding moments of inertia but there is no parallel axis theorem bummer) 8

9 I values for common beam cross sections 9

10 Ex. A light metre stick clamped to the bench with the following parameters: l 0.5m (Cross section) a 7.8mm Birch wood: Y 16x10 9 Pa mass on end 1kg (Don t actually need these yet) b 5.4mm Calculate I for each orientation: I 1 1 ba 5.4x7.8 /1 1x10 9 m 4 Note correction I 1 1 ba 7.8x5.4 /1 10.6x10 9 m 4 The second orientation makes a much stiffer beam than the first, (almost 11x) even though the material is the same!! Two critical points I is essentially the stiffness of the geometry, just as Y is the inherent stiffness of the material! since da - the further (the area) is distributed away from the NS, the geometrically stiffer it is eg. I beams are designed for maximum stiffness lots of area a long way from the NS I e.g. an I-beam 10

11 Ex. Calculate I for a c-bar of the following cross section and dimensions. 6cm cm Note there are two methods you can use to do this both are shown below neutral surface cm 4cm cm Method 1: Use the previously derived formula for rectangular beam I and subtract the middle from the outside (since both have the same NS) 6cm I 1 ba 1 4cm 8cm 4cm (6 x 8 )/1 56cm 4 (4 x 4 )/1 1. cm 4 4.7cm 4 Note that the middle doesn t have a high I value why? Because it is close to the NS. So removing it doesn t change the value of I by 11 very much (only about 9%). BUT you reduce the weight by quite a bit about % if you work out the area so you have increased the stiffness/weight ratio!!!

12 Method : Break the piece into two sections (A and B) above the NS (then multiply the answer by ). NS A B I da (Look at how we did this for the rectangle earlier) 4 (d) + (6d) 0 I Section B Section A Two halves 4.7 cm 4 1