Average Stresses & Component Design

Transcription

1 2 Average Stresses & Component Design 2 1

2 Lecture 2: AVERAGE STRESSES & COMONENT DESIGN TABLE OF CONTENTS age 2.1 Introduction Average Stress And Its Uses Safety Factor Design Related Definitions Critical Safety Factor Why is Component Design Important? Simplified Strength Design Ex.1: Axially Loaded Brittle Bar Ex.2: Axially Loaded Ductile Bar Ex. 3: Bar with in Connectors, Failure by Tension Break at in Ex. 4: Bar with in Connectors, Failure by in Shear Off Ex. 5: Truss Bolt Connection With Single Shear Area Ex. 6: Truss Bolt Connection With Double Shear Area

3 2.3 SAFETY FACTOR 2.1. Introduction This lecture describes a highly simplified form of the 3D stress concept, in which the objective is to get average stresses over a finite area instead of point stresses. This is applied to some simple problems in component design. The concept of strength safety factor is discussed in some detail, first in general terms and then in conjunction with examples Average Stress And Its Uses The previous lecture emphasized that stresses vary from point to point inside a body (structure). Thus σ xx = σ xx (x, y, z), and likewise for all other components. Mathematically, stresses form a tensor field within the body. Although the field viewpoint is rigorous, it may be too elaborate for practical use if its determination requires expensive and time-consuming computations, even with the help of the computer. Often the time-pressed engineer compromises by using a simplified stress analysis process based on average stresses. The key idea is not to pass to the limit as done in Lecture 1 to define stresses at a point. After appropriately cutting the body to manifest internal force(s) of interest, divide those resultants by the finite area on which they act. This shortcut is especially useful in preliminary design of 1D structural components when internal force resultants can be quickly determined from statics by a Free Body Diagram (FBD). Several illustrative examples are given in Section 2.4 and following ones Safety Factor A safety factor characterizes the margin for bad things to happen to the structure. Some actual instances of such events are shown in Figure Design Related Definitions To quantity the foregoing statement, two definitions are introduced. Failure mode. The structure experiences something that renders it disfunctional. Mode identifies what. Failure may be catastrophic: a bridge or dam collapses, an airplane wing breaks in flight, a submarine hull buckles under pressure. Or something less drastic: a building foundation settles and cracks appear. Design Load. A load system used for designing the structure and its components. For real structures there may be many such systems, identified as design load cases. For example, the design of a commercial jet aircraft may involve hundreds of load cases. Design scenarios may include normal events while in service, as well as abnormal ones that the structure is supposed to survive without catastrophe (e.g., total collapse) when there are lives at stake. Examples: earthquakes, hurricanes, emergency landings, Critical Safety Factor Assume we have a design load case with forces 1, 2,..., and that under the given load values the structure is fine. (Load cases may include applied moments, distributed loads, temperature changes, etc., but for simplicity think for now of point forces.) Suppose that the loads are proportionally 2 3

4 Lecture 2: AVERAGE STRESSES & COMONENT DESIGN Figure 2.1. Bad things can happen. Figure 2.2. The Challenger disaster. increased to s 1, s 2..., where s > 1 is a magnification factor, and that the first failure mode triggered by this increase happens at s F. This s F is called the safety factor for that load case with respect to that failure mode. If, as generally happens, there are several load cases, the process is repeated for each one. The smallest s F encountered in this sweep is the critical safety factor, which should meet design specifications. Obviously it should not be less than a certain target. The foregoing description oversimplifies actual practice. For example, some loads, such as own weight, should be kept fixed. Furthermore, statistical and cost considerations come into the picture. Not all load cases may be equally probable (how often is an aircraft hit by a meteorite?), while the cost of achieving adequate safety against all possible events may be prohibitive. 2 4

5 2.4 EX.1: AXIALLY LOADED BRITTLE BAR A detailed safety analysis that includes statistical and cost data generally requires the use of sophisticated computer programs. For preliminary component sizing, however, a shortcut based on back-of-the-envelope stress analysis may be sufficient Why is Component Design Important? The failure of a component may trigger that of the whole structure (For want of a nail the shoe was lost...) For instance, the well known 1986 disaster of the Challenger space shuttle, which cost $6.7 billions in 1971 dollars to build, was triggered by a malfunctioning $25 O-ring. See Figure Simplified Strength Design A simplified strength design based on average stresses and given safety factor proceeds as follows. A structural component such as a bar or beam is subject to known loads that come from an analysis of the whole structure. A failure mode of the isolated component is assumed. In the ensuing examples, that mode will represent a failure of the material when a stress level is reached. But which stress? Here a distinction should be made between brittle and ductile materials. Brittle materials such as fiberglass or cast iron fail by the maximum tensile normal stress reaching a σ fail level. Ductile materials such as metals or alloys fail by the maximum absolute shear stress reaching a τ fail level. Using FBD statics compute the average normal stresses σ avg or the average shear stress τ avg over the area of an appropriate cut affected by the failure mode. The strength safety factor is the ratio s F = σ fail or s F = τ fail (2.1) σ avg τ avg as appropriate to the material type and failure mode. More complicated failure criteria do exist, but will not be covered in this course; they are studied in ASEN For strength design s F is picked by practice, design codes, or experience, and inserted into (2.1), in which either σ avg or τ avg is expressed in terms of the data. Design variables are solved for, and the component sized accordingly. Several examples of this procedure follow Ex.1: Axially Loaded Brittle Bar = 30 kips x Circular cross section of diameter d & area A ; d=? x F n cut normal to bar axis Figure 2.3. Axially loaded brittle bar. Failure by tension: normal cross section breaks.. The first design example is pictured in Figure 2.3. A bar of solid circular cross section, fabricated with brittle material, is subject to axial load. Recall that a brittle material such as cast iron or 2 5

6 Lecture 2: AVERAGE STRESSES & COMONENT DESIGN fiberglass fails (breaks apart) when the maximum normal stress (here the axial stress) reaches a limit value. Material data: cast iron, which fails by normal stress reaching σ fail = 40 ksi (kilopounds/sqin). Failure mode: cross section breaks when σ fail is reached Safety Factor: s F = 8 against failure. Load case data: axial load = 30 kips (kilopounds) Find: design variable is bar diameter d in inches. Solution. Make the normal cut shown on the right of Figure 2.3. From FBD statics, the resultant internal force is F =, aligned with +x. The area of the cut, a.k.a. resisting normal-stress area or simply resisting area, is the bar cross section A = 1 4 π d 2. The average normal stress is σ avg = F A = 1 4 π d = 4 2 π d. (2.2) 2 The design condition is σ avg σ fail /s F. Substituting for σ avg and solving for d yields d + 4s F 4 30 kips 8 = = 2.76 in πσ 2 fail π 40 kips/in (2.3) 2.5. Ex.2: Axially Loaded Ductile Bar = 30 kips x Circular section of diameter d & area A ; d=? x skew cut at o 45 from x o 45 45o n t F Figure 2.4. Axially loaded ductile bar: failure by yield due to crystal slip at ±45. The next example is pictured in Figure 2.4. The configuration is identical to the previous example: a bar of solid circular cross section under axial loads, but the material is now ductile. Failure occurs by yield when the maximum shear stress reaches a limit value τ fail. Material data: ductile Al alloy, which fails by maximum shear at τ fail = 20 ksi (kilopounds/sqin). Failure mode: crystal slip at ±45 from longitudinal bar x axis. Why 45? This will be the subject of one problem in Recitation 1. Safety Factor: s F = 5 against shear failure. Load case data: axial load = 30 kips = 30, 000 lbs as in last example. Find: design variable is again diameter d in inches. 2 6

7 2.6 EX. 3: BAR WITH IN CONNECTORS, FAILURE BY TENSION BREAK AT IN Solution. Make the skew cut shown on the right of Figure 2.4. From FBD statics, the resultant internal force is F =, aligned with +x. roject on cut plane: F t = F cos 45 = F/ 2 = / 2. The area of the skew cut, a.k.a. shear area,isa s = A/ cos 45 = A 2 = 1 4 π 2 d 2. The average shear stress is τ avg = F t = / 2 = / 2 A s A 1 s 4 π 2 d = π d. (2.4) 2 The design condition is τ avg τ fail /s F. Solving for d: d + 2s F 2 30 kips 5 = = 2.19 in πτ 2 fail π 20 kips/in (2.5) 2.6. Ex. 3: Bar with in Connectors, Failure by Tension Break at in zoom /2 /2 pin diameter is d pin t cross section h ;; =? /2 /2 cut plane Figure 2.5. Axially loaded ductile bar with pin connectors: failure by tension break at pin This example is pictured in Figure 2.5. A ductile bar member (Al alloy) of rectangular cross section h t is linked to other bar members though pin connectors. The bar transmits a tensile axial force. Two failure modes at the pin are possible: tension break at minimal cross section area weakened by hole, or pin shear-off. This example considers the first possibility. Material data: Al alloy, may fail by either maximum tensile normal stress at σ fail = 268 Ma (megaascal) or by shear at τ fail = 165 Ma. Failure mode: tension break at minimal cross section weakened by pin hole; see Figure 2.5. Safety Factor: s F = 6 against tension break at minimum cross section. Load case data: axial load = 5kN= 5, 000 N applied to pins Geometric data: cross section thickness t = 5 mm; pin diameter d pin = 12 mm. Find: design variable is cross section height h in mm. Solution. Make a transverse plane cut at the pin center as shown. The bar cross section at the cut (obviously the minimum one) is called the corrected area A corr = A A pin = ht d pin t. 2 7

8 Lecture 2: AVERAGE STRESSES & COMONENT DESIGN Average normal stress over corrected area is σ avg = /A corr = /((h d pin ) t). Design condition: σ avg σ fail /s F. Solving for h: h s F σ fail t + d pin = 5000 N N/mm 2 5mm + 12 mm = 34.4 mm (2.6) 2.7. Ex. 4: Bar with in Connectors, Failure by in Shear Off ; ; zoom d ds s =? /2 ; /2 ; t cut planes pin shears off h cross section Figure 2.6. Axially loaded ductile bar with pinhole connectors: failure by pin shearing off. The structural configuration for this example is the same as in previous one. We now consider the other possible failure mode, in which a pin shears off as depicted in Figure 2.6. Material data: same as in previous example: fails by maximum tension stress at σ fail = 268 Ma or by shear at τ fail = 165 Ma. Failure mode: shear failure with a pin shearing off, see right of Figure 2.6. Safety Factor: s F = 6 against shear failure. Load case data: axial load = 5 k-n = 5,000 N applied to pins, as before Geometric data: cross section thickness t = 5 mm, pin diameter not needed. Find: design variable is distance d s defined in the figure, expressed in mm. (Actually h and d pin are not needed in this case.) Solution. Make two cuts because the force is transmitted through the two interfaces shown in the figure. Total shear area is A s = 2d s t. Resultant shear force over shear area, from FBD: F s = =. Average shear stress τ avg = F s /A s = /A s = /(2d s t). Design condition: τ avg τ fail /s F. Solving for d s : d s s F 2τ fail t = 5,000 N N/mm 2 5mm = 18.2 mm (2.7) 2 8

9 2.8 EX. 5: TRUSS BOLT CONNECTION WITH SINGLE SHEAR AREA Figure 2.7. Bolt connection with single shear area: overall sketch. (a) (b) (c) (d) t cut plane F = s a a d bolt τ avg Figure 2.8. Bolt connection with single shear area: FBD procedural details Ex. 5: Truss Bolt Connection With Single Shear Area This example deals with the design of the bolt for connecting between two truss members as shown in Figure 2.7, using the average shear stress approach. The axial force transmitted through the bolt is. Figure 2.8(a) (d) shows a set of FBDs to help understand the concept of a shear area through which the total axial force is transmitted ( flows ) from one member to the other. The cut plane is displayed in (c). The resultant tangential force on that plane is called F s for shear force. From axial equilibrium, F s =. The shear area is colored red in (d). Remark 2.1. It should be noted that the actual stress distribution in the vicinity of the shear area may be quite complicated because of high stress gradients near contact zones as well as bending due to member centerline eccentricity (more about this later). To get a realistic picture of that distribution one would need to carry out a time-consuming 3D finite element analysis. By assuming that in the place of the bolt section a uniformly distributed shear stress develops, however, a simple solution is readily found. Approximations involved in this assumption are supposed to be covered by the safety factor. Material data: metal bolt (ductile); fails by maximum shear at τ fail. Failure mode: average shear stress over shear area, colored red in (d), reaches τ fail. Safety factor: s F would typically be 4 or more to compensate for ignoring stress concentrations as well as transfer eccentricity effects. In the derivation below s F is kept symbolic. Load case data: axial load to be transmitted by bolt. Find: design variable is bolt diameter d Solution: make the cut shown in (c) and assume that the shear stress over the shear area A s = A bolt = πd 2 /4 is uniform, as discussed above. That average is given by τ avg = F s /A s = /A bolt = 4/(πdbolt 2 ). The design condition is τ avg τ fail /s F. Solving for the bolt diameter gives 4s F d bolt +. (2.8) πτ fail 2 9

10 Lecture 2: AVERAGE STRESSES & COMONENT DESIGN It only remains to plug numbers Ex. 6: Truss Bolt Connection With Double Shear Area t 1 t 1 (a) (b) (c) (d) t 2 t 2 /2 /2 /2 /2 t 1 /2 cut planes /2 d bolt τ avg Figure 2.9. Bolt connection with double shear area. One flaw of the bolt connector studied in the foregoing example is load eccentricity caused by member centerline offset. The resulting moment may cause additional bending stresses as well as alignment problems. A better design that avoids that problem is shown in Figure 2.9, in which configurational symmetry aligns inter member load transmission whence bending is eliminated. To get the effective shear area, make two cuts as pictured in Figure 2.9(c). The effective shear area doubles to A s = 2A bolt = π dbolt 2 /2 and the average shear becomes τ avg = /(2A bolt ) = 2/(π dbolt 2 ). With the same design criteria of the previous example, the bolt diameter is now given by d bolt + 2s F πτ fail. (2.9) If, s F and τ fail stay the same, this d bolt is about 30% smaller than that given by (2.8) Furthermore s F could be cut since the lack of bending reduces stress distribution uncertainties. On the other hand, this kind of connection is likely to be costlier to fabricate. Figure 2.10 shows a well fabricated instance of a double shear area connector. This would be typical, for example, of a quality car suspection device. Figure A well fabricated connector with double shear area. Note that axial loads are nicely aligned. 2 10