Stresses in Engineering Components (problems 14, 15, 16) (combining elastic moduli with geometry elastic behaviour) W2L3 (Courseware pg 43-46) Free Body Analysis: common technique to develop stress equations for components basically Free Body Diagrams for static equilibrium cases da da Procedure: 1. Pass a cutting plane perpendicular to the line of action of the load (force) 2. Net internal reaction will be equal to the external force (otherwise not in static equilibrium), i.e. external force (Internal force) da cut cross section (essentially the internal force across the cut section) 1
We use this free body analysis technique mainly as a tool to enable us to derive equations for stresses in components that are subject to external forces. We are going to consider two different cases: 1. Centroidally Loaded Case: This is when the externally applied force must be symmetrical about centroidal axis. An example is a pressurized gas cylinder where the external applied force is the internal pressure. It is applied symmetrically about the centroid axis (along the length of the cylinder). We will consider use the free body analysis technique to develop equations for: - axial stresses in pressure vessels - hoop stresses in pressure vessels We will start this today (W2L3) and also will work on it in the next lecture (W3L1) 2. Beam Bending: this is not a case of centroidal loading. The free body analysis technique is a handy way to determine what the stress equations look like for beams of different shapes, sizes and materials. This takes some time to do - we will start this in W3L2 and will finish in W4L1. 2
Centroidally Loaded Case: Axial stresses in Pressure /Vacuum Vessels (Courseware pg 45) P internal pressure; assume pipe is thin walled Make the imaginary cut here perp to the axial direction (because we want to develop an expression for axial stress) so r inside r outside r wall thickness (t) P Flat,rigid end cap Internal force of the wall at the cut cut F ( axial )(area) ( axial )(2πrt) external force (due to pressure acting on end) πr 2 P If nothing is moving (static equilibrium) then the internal and external forces must be equal: ( 2πrt) πr axial axial Pr 2t WE HAVE JUST DEVELOPED AN EXPRESSION FOR CALCULATING AXIAL WALL STRESS IN A CYLINDER! 2 P 3
problem 4
Centroidally Loaded Case: Hoop Stress in Cylindrical Pressure Vessels (Courseware pg 45) As with the Axial stress case in the last slide, the first step is to make a cut perpendicular to the direction that you want to get a stress expression for. Since you want a HOOP stress you make a cut perpendicular to the hoop, which is along the length. The drawing shows the cylinder cut along its length. L hoop external force due to the internal pressure P which is (radial) as shown P hoop internal reaction force hoop force acting on cut wall wall thickness t side view can think of it as: F 1 Internal reaction force (hoop) F 2 (vertical components of pressure will cancel) F 1 External force 5
What is ΣF 1? total internal reaction force (right) area ( H ) Wall cross section What is ΣF 2? total force (left) (external) (2Lt) H sum of horizontal components of force due to pressure How do we get this? Obtain an expression for the small amount of horizontal force on each small part of the wall ds (see diagram below). Then sum up the total force over the entire semicircular section by integrating. horizontal force on ds (P horiz at ds )(area of ds) (Pcosθ)(Lds) (Pcosθ)(Lrdθ) Remember this extends into the page So ΣF 2 π θ 2 Pr π θ 2 Lcosθdθ ΣF 2 Pr Lsinθ π 2 π 2 2PrL Now for static equilibrium ΣF 1 ΣF 2 2Lt H 2PrL H Pr t This is a formula for calculating the hoop stress in the wall 6
Comparing Hoop and Axial Stresses: axial Pr 2t H Pr t Interesting note here: For a given pressure the axial stress is half of the hoop stress this is why pressure cylinders fail by splitting lengthwise rather than by the ends blowing off! So here is an interesting question normally in a system we have three main stresses in the simple case these stresses are orthogonal, so here these would be axial, hoop and thickness stresses. Below is shown a small piece of the pipe wall with these stresses labelled. T H ax So why haven t we discussed thickness stress yet? Well, in a thin-wall cylinder these are essentially zero. Why? Because in order to have a stress you need two opposing forces. However, there is nothing in the thickness direction to oppose the internal gas pressure (only air on the other side) the cylinder will just expand until it is stopped by the hoop and axial stress state. (for the record, a balloon has no stresses in the thickness direction either) --- thick wall cylinders DO have a thickness stress 7 because the cross section at the outside is greater than that on the inside creates a complex stress gradient.
FINALLY... Do we have to consider Poisson s ratio in a cylinder? Certainly! Usually (if the wall is thin) there are no stresses in the thickness (radial) direction (the 3rd orthogonal direction). But we still have strains in all 3 directions, because of Poisson. So lets write down the strain expressions for all 3 components, assuming the stress in the thickness direction is zero. Note the format for the strains will be identical to those for the 3 orthogonal xyz direction strains we developed in W2L1. There is no stress in the radial direction so r 0. Given this, the 3 strains look like: ε h h Y a σ Y a h ε a σ Y Y a h ε r σ σ Y Y Note here: these equations will NOT be on the exam equation sheets. They are derived easily from the xyz equations which ARE on the formula sheet 8
problem 9
Ex. Gas and Oil Pipelines Gas transmission lines like the TransCanada pipeline are linked by compressor stations (about 100 km apart) where gas is pressurized and pumped along. Gas pressures are typically quite high, about 6.5 MPa. (vs 2.5 MPa for BBQ test) What we try to avoid For a typical 36 diameter pipe of 9mm wall thickness: 1. What is the hoop stress in the pipe wall due to its internal pressure? 2. What is the axial stress in the pipe wall due to the internal pressure? 3. What is the axial contraction due to Poisson s ratio, assuming it is free to contract? 10
1. Hoop stress in pipe wall? Pr t 6.5x(0.5x36" x0.0254) 9x10 330 hoop 3 (note that the yield stress in this material is about 380MPa) MPa 2. Axial stress in the pipe wall? ----no end caps so there is effectively no axial stress due to the end cap effects. 0.29 for steel 3. Axial contraction? ε a a Y h σ Y ε a 220GPa for steel 330MPa l 0.29 0. 000435 3 220x10 MPa l l -(.000435)(100,000) -43.5m So if this pipe were free to contract it would do so by 43.5m!! It is constrained at the compressor station but stories are told of pipes disappearing when cut! 11
Cut again: Wall Stresses in a Sphere (under pressure) P wall internal reaction force in the wall (to the right) (2πrt)( wall ) total force (to the left) due to pressure πr 2 P (not shown, but can be obtained by integrating horizontal terms over entire hemispherical area) equating: 2πrt wall πr 2 P wall Pr 2t So this is the equation for calculating wall stresses in a sphere Note this is the same as a cylinder axial stress which is half the cylinder hoop stress Hmmm this means For a given pressure, the maximum cylinder wall stress is twice that of a sphere! So if I were designing a pressure vessel... Spheres are better pressure vessels but fabrication difficulties mean that they are only used in the most 12 critical situations!!
problem 13
Further notes: Interesting side note: bathyspheres suffered from one main problem - cable breakage meant bye bye crew! So in 1946 Swiss Physicist August Piccard designed a bathyscaphe --which could surface and sink as required. It is a short Ti-alloy cylinder that is reinforced on the inner surface. The current depth record of 10,915m (deepest ocean point - Marianas Trench) was obtained in one of these by August s son Jacques Piccard (boldly going.) The Titanic sits at 3400 m and was explored by a bathyscaphe 14