3.10 Failure and Reliability

Transcription

1 3.10 Failure and Reliability Any device or object that is designed and manufactured is expected to operate as advertised over a stated length of time. If the product does not function as expected then it is considered a failure. Failure can have many reasons. Failure is usually associated with reliability - the expression of confidence that the product will deliver on its expectation. Failure also has a practical side effect which is best attributed to Taguchi : When a product fails, you must replace it or fix it. In either case, you must track it, transport it, and apologize for it. Losses will be much greater than the costs of manufacture, and none of this expense will necessarily recoup the loss to your reputation. Failure is serious business and designing for actual failure is impossible because of so many variables. Instead we try and ensure that the design meets the Failure Criteria. There is no unique criteria and the designer usually satisfies the failure criteria that is appropriate for the type of the product and its underlying design. Many failure criteria are based on principal stresses rather than the standard engineering stress and strain. Since we can calculate the principal stress form the value of the engineering stress and strain at every point, we examine some of the popular failure criteria. Prior to application of such criteria we should also consider the fact that if the design is stretched beyond the elastic domain then the residual strain on the structure changes the design forever. If the bridge does not return back to its original state it is likely to cause additional problems in many ways Mechanical Structural Failures For most designs we can investigate four types of failures. a. Failure by elastic structural deflection (δ > δ max ) b. Failure by structural yielding (σ max < σ) c. Failure by Fracture (σ = σ ult ) d. Progressive Failure (failure is built slowly during life - creep and fatigue) There are failures that are difficult to quantify and difficult to investigate. a. Operating environment (moisture, dirt, dust, corrosion) b. Aging and shelf life c. Unanticipated operating current and voltage levels d. Unintended chemical reactions e. Electromagnetic interference f. Material properties may unexpectedly vary due to production and finish The design/structural engineer must determine the possible modes of failure and establish criteria to predict these failures. In order to account for these unpredictable variations a factor of safety (FS) is usually adopted. For many structural and machine applications recommended FS are available. They vary by state and country Maximum Structural Deflection In the illustration below we start with an unloaded cantilever beam. We place an end load that causes a maximum deflection at the end of the beam as shown. As the load is removed the beam should return back to its horizontal position. As the load P is increased the deflection is increased but the beam still returns to the undeformed position after load is removed. There is a maximum elastic

2 deflection that should not be exceeded. Figure Illustration of maximum structural deflection This type of failure is prevalent in vibration when the amplitude is large that parts collide. Also beams and shell may buckle under load. The failure criteria is applied on the maximum deflection of the component that includes buckling load too. A factor of safety is assumed and the deflection must remain less than the maximum elastic deflection Maximum Yield Stress In most structural design, the maximum stress is kept below the yield stress (or proportional limit). Instead of the deflection the stress is monitored. In linear elasticity we can assume that the stress at maximum elastic deflection and maximum yield stress are pretty close by. This type of failure is used for simple structures and simple loading in beams, shells, and plates. For failure criteria, the designer must calculate the maximum stress of the component that include buckling and an appropriate factor of safety to ensure the actual maximum stress will not exceed the yield stress of the material.. Instead of the maximum stress, the failure criteria have evolved t include principal stress like Tresca and Von Mises failure criteria Failure by Fracture Failure by fracture is usually associated with brittle materials since ductile materials will have already yielded and will suffer plastic deformation prior to fracture. Fracture will also depend on existing cracks as these cause stress concentrations where local stresses will exceed any stress limit under consideration. For this type of failure the maximum principal stress must be calculated with an appropriate factor of safety to ensure the stress is still within the yield limit Progressive Failure During progressive failure, a small failure or small changes is added on to the component during routine operations. This builds up to a sudden failure at a later date even it it is not apparent at the current time. A micro crack is usually the culprit. As noted before this failure can happen due to creep, due to fatigue, and due to changes in material property leading to a change in the stress-strain behavior, due to small changes in the component.

3 During Creep failure there is usually Increase in strain without increase in stress Enhanced strain at high temperature operation During fatigue failure there is usually Change in stress level because of the change in the frequency of the load cycle Repeated changes in the direction of load causing change in local material properties Failure Criteria The mechanical failures outline above are actually implemented by verifying that the component satisfies some formal failure criteria. Now with software used for structural analysis it becomes easy to obtain a detailed picture of stress, strain, displacement everywhere through structural simulation software. The software can also report on multiple failure criteria as part of the solution. We look at four popular criteria i. Maximum Shear Stress Criterion (Tresca s Hexagon) ii. Maximum Distortion Energy Criterion (Von Mises) iii. Maximum Normal Stress Criterion (Coulomb s Criteria) iv. Maximum Normal Stran Criterion (St. Venants Criteria) A component will be safe under given loading if the stress at all critical points, including areas of stress concentration, is less that than that recommended by one or more of the failure criteria indicated above Maximum Shear Stress Criterion This criteria is based on plane stress and useful for ductile materials. shear stress It is believed that ductile materials will fail in slippage along an oblique surface due to Criteria: Consider a cantilever beam with end load. The principal stress at evey point can be evaluated as indicated below. The state of stress at every point can be translated through the principal stress as: Figure Regular and principal stresses Alternate Criteria:

4 This leads to Tresca s Hexagon with the criteria that the principal stresses at every point must lie within the area of the Hexagon shown in Figure for the component to be safe. Figure Tresca s Hexagon Maximum Distortion Energy Criterion (Von Mises) A given structural component is safe If the maximum value of the distortion energy per unit volume of the material is less than the distortion energy per unit volume required to cause yield in a tensile-test specimen f the same material The distortion energy (u d ), which is the energy associated with the change of shape, is different from the strain energy, which is the change in the volume of the material. These concepts are to early for this first course in mechanics. However we can tie the distortion energy to the principal stresses. For an isotropic material under plane stress with a modulus of rigidity G Figure Principal stress and alternate criteria for distortion energy Alternate criteria The structural component is safe if the principal stresses at every point is within the area enclosed by

5 the ellipse. This is also regarded as the Von Mises Criteria Maximum Normal Stress Criterion This failure criteria is applied to brittle materials where failure is expected through rupture or fracture. It will be sudden with no yielding before failure. It would certainly apply to products created using ceramic materials. Criteria: A given structural component is safe if the maximum normal stress in the component reaches the ultimate stress in a tensile test specimen made of the same material If the ultimate stress in tension and compression are the same the criteria is regarded as the Coulomb s criteria and can be summarized as follows: Figure Coulomb s normal stress criteria If the ultimate stress in tension and compression are different the criteria is regarded as the Mohr s criteria and can be summarized as follows: Figure Mohr s normal stress criteria Maximum Normal Strain Criterion This failure criteria is applied to brittle materials where failure is expected through rupture or fracture. It will be sudden with no yielding before failure.

6 Criteria: A given structural component is safe if the maximum normal strain in the component remains smaller than the ultimate normal strain in a tensile test specimen made of the same material For material with the same strain in tension and compression the criteria is called the Saint Venant s criteria and can be summarized as: Figure St. Venant s normal strain criteria Alternate Criteria: A structural component is safe as long as the principal stresses remain within the area of the plot shown above Example The state of stress, at a critical point on the PCB guide, due to warping is a concern. The result of tensile stress tests of the same material establishes the yield stress as 250 MPa (σ Y ). Find the factor of safety with respect to yield, using (a) the maximum-shearing-stress criterion; (b) the maximum-distortion-energy criterion Data: σ x = 80 MPa; σ y = -40 MPa; τ xy = 25 MPa; σ Y = 250 MPa Assumption: None (probably ductile material) Solution:

7 (a) the maximum-shearing-stress criterion (b) the maximum-distortion-energy criterion Example Using MATLAB % Essential Foundations in Mechanics % P. Venkataraman, July 2015 % 3.10 Failure Criteria % % Problem (2D state of stress) % sigx = 80 (MPa); sigy = -40; tauxy = 25;SIGY = 250; % Find: FOS with (i) max shera stress criterion; % (ii) max distortion energy criterion % clc, clear, format compact, close all, format shortg %% Data sigx = 80; sigy = -40; tauxy = 25; SIGY = 250; %% Calculate principal stresses sigav = 0.5*(sigx + sigy); taum = sqrt((0.5*(sigx - sigy))^2 + tauxy^2); siga = sigav + taum; sigb = sigav - taum;

8 %% Apply failure criteria TAUY = SIGY/2; FS1 = TAUY/taum; FS2 = sqrt(sigy^2/(siga^2 - siga*sigb + sigb^2)); %% Print data/solution fprintf('example Failure Criteria\n') fprintf(' \n') fprintf('state of Stress :\n') fprintf('sigma x [MPa] : '),disp(sigx) fprintf('sigma y [MPa] : '),disp(sigy) fprintf('tauxy [MPa] : '),disp(tauxy) fprintf('yield stress [MPa] : '),disp(sigy) fprintf('\n\n') fprintf('principal stress siga [MPa] : '),disp(siga) fprintf('principal stress sigb [MPa] : '),disp(sigb) fprintf('maximum shear stress [MPa] : '),disp(taum) fprintf('\n\n') fprintf('(i) FS based on Maximum shear Criteria :'),disp(fs1) fprintf('(ii) FS based on Von Mises Criteria :'),disp(fs2) Fprintf('\n\n') In the Command Window Example Failure Criteria State of Stress : sigma x [MPa] : 80 sigma y [MPa] : -40 tauxy [MPa] : 25 Yield stress [MPa] : 250 Principal stress siga [MPa] : 85 Principal stress sigb [MPa] : -45 Maximum shear stress [MPa] : 65 (i) FS based on Maximum shear Criteria : (ii) FS based on Von Mises Criteria :