1 Contents 1. Flight Vehicle Structures and Materials 1.1 Introduction & Design Consideration 1.2 Review of Basic Structural Elements 1.3 Structures and Load Transfer 1.4 Flight Vehicle Materials 1.5 Structural Modelling: 3-rib Wingbox 2. Structural Mechanics Issues in Flight Vehicles 2.1 Analysis of Composite Laminate 2.2 Statics and Dynamics for Flight Vehicle Structures 2.3 Fastener Analysis 2.4 Sandwich Analysis 2.5 Design Issues on Structural Members 3. Structural Failure Issues in Flight Vehicles 2. Structural Mechanics Issues in Flight Vehicles 2.3 Fastener Analysis
Transcript
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Contents
1. Flight Vehicle Structures and Materials 1.1 Introduction & Design Consideration 1.2 Review of Basic Structural Elements 1.3 Structures and Load Transfer 1.4 Flight Vehicle Materials 1.5 Structural Modelling: 3-rib Wingbox 2. Structural Mechanics Issues in Flight Vehicles 2.1 Analysis of Composite Laminate 2.2 Statics and Dynamics for Flight Vehicle Structures 2.3 Fastener Analysis 2.4 Sandwich Analysis 2.5 Design Issues on Structural Members 3. Structural Failure Issues in Flight Vehicles
2. Structural Mechanics Issues in Flight Vehicles
2.3 Fastener Analysis
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Introduction to Fasteners
• The ideal flight vehicle structure - The single complete unit - Of the same material - Involving one manufacturing operation • Real World - Many types of materials - Requirements of repair and maintenance - Several main units held to other units by main or primary fittings or connections - Many primary and secondary connections involving fittings, bolts, rivets, welding, etc. - Main or primary fittings involve more weight and cost per unit volume than any other parts - Fitting and joint design plays an important part in aerospace structural design
Economy in fitting design
• Use a minimum number of fittings, particularly those fittings connecting units which carry large loads.
• Fittings near the centerline of the airplane are far more costly than splices or fittings placed farther outboard where member sizes and loads are considerably smaller
• Poor layout of major fitting arrangement requires very expensive tools and jigs for shop fabrication and assembly
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Solid rivet
Flush(countersunk) head Universal head
Solid rivet indication
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Solid rivet
Video for solid rivet
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Blind rivet
• Blind rivets are tubular and are supplied with a mandrel through the center. The rivet assembly is inserted into a hole drilled through the parts to be joined and a specially designed tool used to draw the mandrel into the rivet. This expands the blind end of the rivet and then the mandrel snaps off. (These are also commonly called pop rivets from the sound and feel through the setting tool when the mandrel breaks.) These types of Blind rivets have non-locking mandrels and are avoided for critical structural joints because the mandrels may fall out, due to vibration or other reasons, leaving a hollow rivet that will have a significantly lower load carrying capability than solid rivets. Furthermore, because of the mandrel they are more prone to failure from corrosion and vibration. Prior to the adoption of blind rivets, installation of a solid rivet typically required two assemblers: one person with a rivet hammer on one side and a second person with a bucking bar on the other side. Seeking an alternative, inventors such as Carl Cherry and Lou Huck experimented with other techniques for expanding solid rivets. Unlike solid rivets, blind rivets can be inserted and fully installed in a joint from only one side of a part or structure, "blind" to the opposite side.
Blind rivet
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Blind Rivet
Video for blind rivet
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Hi Lok
Hi Lok
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Video for Hi Lok
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Nutplate
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Nutplate
Screws
6-point star-shaped pattern
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Bolts
• Use primarily to transfer relatively large shear or tension loads from one structural member to another
• Bolts connecting parts having relative motion or stress reversal should have close tolerances to decrease shock load
Bolts
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Bolts
Bolts
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Bolts
Nuts
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Nuts
Bushings
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Bolt failure mode in double shear
Figure 2.3.1 Bolted connection in which bolt is loaded in double shear.
Bolt failure mode in single shear
Figure 2.3.2 (a) Bolted connection in which bolt is loaded in single shear, (b) failure in the hexagon head bolt.
• Often less than double shear because of bending and prying
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Displacement field around the fastener hole
Fig. 2.3.3 Specimen preparations and mechanical setup for open-hole tensile test: measuring area covered with the grating (in mm, { }: one blanket).
Fig. 2.3.4 Displacement map around the bolt hole obtained by phase-shifting moiré interferometry.
Strain field around the hole
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Fig. 2.3.5 In-plane strain maps obtained by phase-shifting Moiré interferometry and analytic model.
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Prob 2.3.1 FEM for open hole specimens
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• Solve an open hole specimen problem using FEM package where you can choose hole and specimen dimensions.
• Compare different materials and discuss stress concentration around the holes.
Bearing stress
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• If two bodies are pressed against each other, compressive forces are developed on the area of contact. The pressure caused by these surface loads is called bearing stress. If the bearing stress is large enough, it can locally crush the material, which in turn can lead to more serious problems. In order to reduce bearing stresses, engineers sometimes employ bearing plates, the purpose of which is to distribute the contact forces over a larger area.
• As an illustration of bearing stress, consider the lap joint formed by the two plates that are riveted together as shown in Fig. (a). The bearing stress caused by the rivet is not constant; it actually varies from zero at the sides of the hole to a maximum behind the rivet as illustrated in Fig. (b). The difficulty inherent in such a complicated stress distribution is avoided by the common practice of assuming that the bearing stress is uniformly distributed over a reduced area. The reduced area A is taken to be the projected area of the rivet:
where is the thickness of the plate and represents the diameter of the rivet, as shown in the FBD of the upper plate in Fig. (c). From this FBD we see that the bearing force equals the applied load P (the bearing load will be reduced if there is friction between the plates), so that the bearing stress becomes
(a)
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Ex 2.3.1 Shear stress in rivet and bearing stress in plate
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• The lap joint shown in Fig. (a) is fastened by four rivets of 3/4-in. diameter. Find the maximum load P that can be applied if the working stresses are 14 ksi for hear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates.
Sol) • We will calculate P using each of the two design criteria. The
largest safe load will be the smaller of the two values. Figure (b) shows the FBD of the lower plate. In this FBD, the lower halves of the rivets are in the plate, having been isolated from their top halves by a cutting plane. This cut expose the shear forces V that act on the cross sections of the rivets. We see that the equilibrium condition is V = P/ 4.
• Design for Shear Stress in Rivets The value of P that would cause the shear stress in the rivets to reach its working value is found as follows:
414 10
3 4⁄
4
24700
Ex 2.3.1 (Conti)
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• Design for Bearing Stress in Plate The shear force V=P/4 that acts on the cross section of one rivet is equal to the bearing force due to the contact between the rivet and the plate. The value of P that would cause the bearing stress to equal its working value is computed from Eq. (a):
418 10 7 8⁄ 3 4⁄
47300 • Choose the Correct Answer Comparing the above solutions, we conclude that the maximum safe load P that can be applied to the lap joint is
24700 with the shear stress in the rivets being the governing design criterion.
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Prob. 2.3.2 Lap joint
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• The lap joint is connected by three 20-mm-diameter rivets. Assuming that the axial load P = 50 kN is distributed equally among the three rivets, find
(a) the shear stress in a rivet; (b) the bearing stress between a plate and a rivet;
and (c) the maximum average tensile stress in each
plate.
• Assume that the axial load P applied to the lap joint is distributed equally among the three 20-mm-diameter rivets. What is the maximum load P that can be applied if the allowable stresses are 60 MPa for shear in rivets, 110 MPa for bearing between a plate and a rivet, and 140 MPa for tension in the plates?
Prob. 2.3.3 Critical loads in the fastener and attachment lugs
• Lugs for the attachment of a vertical stabilizer to the fuselage part are bolt-connected as shown in the right figure. The main part of the lug for the stabilizer has rectangular cross section with width b1= 1.5 in and thickness t = 0.5 in. At the connection the lug is enlarged to a width b2= 3.0 in. The bolt, which transfers the load from the lug for stabilizers to the two gussets of the lugs for fuselage, has diameter d = 1.0 in.
Figure 2.3.7 Attachment lugs and bolt.
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• Determine the allowable value of the tensile load P in the lug for stabilizer based upon the following considerations:
(a) The allowable stress in the main part of the lug is 16,000 psi. (b) The allowable stress in the lug at its cross section through the bolt is 11,000 psi. (The permissible stress at this section is lower because of the stress concentrations around the hole. Net stress (n) and net-section area (the gross area minus the missing material at the hole) should be considered. (c) The allowable bearing stress between the lug and the bolt is 26,000 psi. (d) The allowable shear stress in the bolt is 6,500 psi. (Note: The factors of safety for tension, bearing, and shear were taken into account when determining the allowable stresses.)
Modeling 2.3.1 Lugs and fastener assembly modeling using CATIA
Figure 2.3.8 Attachment lugs and bolt.
Lug for fuselage
Lug for stabilizer
Washer
Bolt
• Draw out the fastener and lug configuration using CATIA by giving appropriate values for the dimensions not indicated.
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Practice of CATIA
Multimedia. Fastener and lug modeling using CATIA.
Rev 2.3 Shear flow in flight vehicle skin
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• The fact that the "Stresses are uniform through the wall thickness makes it possible to deal with the product of the shearing stress and the wall thickness instead of the stress without any ambiguity.
• For reasons which will become apparent we define this product as the shear flow
; thus
(a)
where is the wall thickness. • The resultant torque of q
about an arbitrary moment center O is
Fig. 2.3.9 Resultant of shear flow.
2 2 (b)
where /2 is the area of the shaded triangular element and the integral is the area enclosed by the curved line AB and the radial line OA and OB.
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Shear flow in a wing
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Fig. 2.3.10 Reacting shear flow in a multi-cell structure.
• Because the shear flow in any wall is constant between points where it joins other walls and the shear flow in any interior wall is equal to the difference of the shear flow in the adjoining exterior walls, we may select the unknowns in a thin-walled multicell structure as shown in Fig. above. The shear flow around the i-th cell is designated , and so the shear flows in the outer walls of this cell are equal to , and the shear flow in the web between cells i and i+1 is . A structure with n cells will therefore have n unknown shear flows.
• The shear flows must be in equilibrium with the applied torque. Using Eq. (b) in Rev. 2.3 to determine the resisting torque of each of the hear flows , we find that for equilibrium
2∑
where is the area enclosed by the midline of the i-th cell.
Shear flow transmitted through joints
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• We note from this equation that if the cross section consists of but a single cell, the problem is statically determinate, and the shear flow is
(c)
and, from Eq. (a) in Rev 2.3, the shear stress is given by
(d)
where is the wall thickness at the point where the stress is determined. We note that this eqn was derived from equilibrium considerations alone, and the results are therefore applicable to inelastic as well as elastic deformations. • The difference () in the angle of twist between the rear and front faces in Fig. 2.3.10 is
∮ ∗ (e) Bredt’s equation
where is an arbitrary reference modulus and ∗ is a modulus-weighted thickness defined by
∗ (f)
• The force on the fasteners of a joint can be determined once the shear flow that is transmitted through the joint is known. The shear flow gives the force that is transmitted across a unit length of the joint, so that
(g)
where is the shear force per fastener and n is the number of fasteners per unit length of the joint.
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Ex 2.3.2 Force per rivet in the fuselage splice
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• The right cross section is the aft section of the fuselage of a jet airplane. The lower surface, which is exposed to the jet exhaust, is made of stainless steel ( 11.5 10 ), and the rest of the skin is 7075-T6 aluminum ( 3.910 ).
• The skin splice at the junction of the steel and aluminum consists of two rows of rivets with spacing of ⁄ in.
• The section is subjected to a torque of 100,000 in.-lb.
(a) Determine the shearing stresses in the aluminum and steel (b) Determine the angle of twist which occurs over a 50-in. length of the structure (c) Determine the force per rivet in the splice.
Solution
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• The enclosed area is π 30 2830 . , so that the stresses obtained from Eq. (d) are
,100,000
2 2830 0.051346
,,
.276
• The value of is found from Eq. (e) where (since ∗ is piecewise constant) we may express the line integral as ∑Δ ∗⁄ . Noting that
34 60 141.5 .
14 60 47.2 .
and taking 3.9 10 , so that ∗ 0.051 .
∗ 0.064.
.0.1888 . from Eq. (f)
• We find from Eq. (e)
100,0004 2830 3.9 10
141.50.051
47.20.1888
2.42 10 .⁄
• The total angle of twist for the 50-in. length is 2.42 10 50 1.21 10 • From Eq. (c), 100,000 2⁄ 2830 17.66 .⁄ • The number of rivets per inch is 2 ⁄ 2.67, so that from Eq. (g) 17.66 2.67⁄
6.61 per rivet.
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Failure criteria for mechanically fastened joints
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• Predicting fastener loads and joint strength is seldom straightforward; when uncertain, you should make conservative assumptions. For design, you should multiply the predicted loads or stresses not only by the factor of safety but also by an appropriate fitting factor (Generally, 1.15). Structural failures occur primarily during test, and using a fitting factor doesn't usually make structure much heavier.
• Threaded Fasteners • Most military specifications for threaded fasteners provide minimum tensile
strengths (in terms of load, not stress). Many also specify shear strengths, which apply if the threads are out of the shear plane and bending stresses are low. These values are suitable as design allowables. When they aren't available, calculate allowables using the eqns provided below.
• Mode 1: tension • A threaded fastener with a suitable head will fail first in the threaded section,
usually at the thread nearest the head:
(a)
where is the fastener's allowable ultimate tensile strength (force), is the allowable ultimate tensile stress for the fastener material, and At is the tension area.
Tension areas for unified screw threads (Fine threads)
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• The areas in the third column are based on the minor diameters for MIL-S-8879 threads. MIL-HDBK-5 [1990] recognizes that these values are conservative: threads lie on a helix angle, and there is no planar section that has such a small area.
• The larger areas in the fourth column were derived from tests and are the basis for the minimum tensile strengths provided in most fastener specifications.
Thread (diameter-threads per in)
Shank diameter (in)
Tension Area, At (in2)
Per MIL-HDBK-5 Per Industrial Fasteners Institute [1965]
• In shear applications, we should keep the threads away from the shear plane so that fastener strength is dictated by the shank area:
(b)
where is the allowable ultimate shear strength for the fastener per shear plane (see later discussion of single and double shear), is the allowable ultimate shear stress for the fastener material, and d is the bolt shank diameter. • However, when a fastener is installed into a tapped hole or threaded insert, its
threads are in the shear plane, and its shear strength will be considerably less:
(c)
where is the tension area.
Interaction of tension and shear
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• When a fastener will be subjected to simultaneous tension and shear, we use the following interaction eqns for design:
1 (shank in shear plane)
(Astronautic Structures Manual [NASA, 1976])
1 (threads in shear plane)
where and are the design ultimate shear and tension, respectively, and and are the allowable values, as determined by Eqs. (a) through (c). • Note the difference in the above interaction eqns: a fastener with its shank in
the shear plane can develop a higher combined loading of shear and tension because the two types of loading want to cause failure at different locations.
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Interaction of tension and bending
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• In shear, a fastener will also develop bending stresses. • These are significant only if the shear is transferred across a gapped or
shimmed joint, and for lug-and-clevis joints. Tension and bending stresses combine directly.
• Eqn or interaction of tension and bending when the threaded section is in bending:
1
where and are the design ultimate and allowable values of tensile force and and
are the applied and allowable values of bending stress for a circular section whose diameter equals 4 ⁄ . • This eqn also applies if the shank is in
bending, but the applied and allowable values for tension and bending should be based on the shank area.
Tension joints
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• Mode 1 (Bolt tension failure) - as discussed in Eq. (a). • Mode 2 (Nut or insert failure) - Nuts and solid, threaded inserts have design
strengths in terms of bolt tension. It's not necessary to understand how the nut or insert fails; what's important is to make sure its strength is compatible with the selected bolt. Even if the full bolt strength is not required for a given application, we should always use a nut or insert that is at least as strong as the bolt. It's too easy to forget to check the nut or insert later if we decide to increase the installation preload or if the design load increases. Also, nut or insert failures are more brittle than bolt failures, and we always want our design to be as ductile as possible to allow loads to redistribute before an overloaded fastener fails.
Fig. 2.3.11 A tension joint (loading the fastener in tension).
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Mode 3 (End-pad shear failure)
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• In this failure mode, the end pad ruptures in shear around the washer or bolt-head circumference, leaving the bolt and a sandwiched plug of material behind. Assuming the load is evenly distributed under the washer, we calculate the end-pad shear stress, , as follows:
where is the bolt tension load, is the washer diameter, and is the end-pad thickness. As with other potential failure modes directly related to bolt tension, it's handy to determine an allowable bolt tension for this failure mode:
where is the allowable ultimate bolt tension for end-pad shear. • This failure mode is a potential concern when the bolt is tucked close to the
fitting walls. If the bolt is tucked close to just one or two walls, a different form of end-pad shear may be more critical, as shown in Fig. below.
Fig. 2.3.12 Potential end-pad shear failure for a tension joint
Shear joints
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• In many industries, joints are designed to transfer shear loads by friction between clamped plates. But for spacecraft we don't use as high a factor of safety as most industries, so the uncertainty of friction as a load path presents greater risk.
• Although we should preload shear fasteners to keep joints from shifting back and forth, we should ensure joints have adequate strength without friction. This will protect against the remote chance that our installation process fails to achieve the proper preload, or that vibration and shock make the fastener lose preload or suddenly dislodge the friction load path. This conservatism applies only to static strength analysis, however; we normally assume a reasonable coefficient of friction for fatigue, fracture-mechanics, and dynamic analyses.
• Mode 1: Fastener shear failure • The design ultimate fastener shear load, equals the total ultimate load carried
by the fastener divided by the number of shear planes (two in Fig. below). Compare this value to the allowable from Eqs. (b) and (c), as applicable.
Fig. 2.3.13 Joint in double shear.
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Mode 2 (Bearing failure in attachment fittings)
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• Bearing yield implies significant permanent deformation of the hole, whereas the bearing ultimate strength is the highest load the fitting can withstand. Figure below shows an example of a fitting that has yielded in bearing.
• We calculate bearing stress, , as
(same as Eq (a) in the bearing stress slide)
where is the bearing load (total fastener load exerted on a fitting), d is the bolt diameter, and t is the fitting thickness. For example, the bearing stress in each of the thin plates shown in Fig. 2.3.13 would be 0.5 ⁄ , and the bearing stress in the thick plate would be ⁄ .
Fig. 2.3.14 Bearing yield. A ductile material under a gradually increasing bearing load will eventually yield as shown. The result is permanent deformation at the joint.
For a spacecraft, this could cause a sensor to miss its target or prevent a mechanism from operating.
Mode 2 (Bearing allowable)
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• The bearing yield allowables provided in MIL-HDBK-5 are based on a permanent deformation of 2% of the bolt diameter. This means a joint with two fittings can permanently deform a total of 4% of the bolt diameter before it has “yielded.”
• For certain joints, such as those in precision mechanisms, may not be able to function with this much deformation. In these cases, we must first identify the permissible amount of permanent deformation, then determine allowables by test or use an uncertainty factor.
• Bearing strength depends on the ratio ⁄ , where is the edge distance, measured from the center of the hole to the edge of the fitting in the direction of applied load (shown in Fig. 2.3.13), and d is the bolt diameter. MIL-HDBK-5 provides bearing allowables for ⁄ 1.5 and ⁄ 2.0. Based on these values, we determine allowables for a joint as follows: ⁄ 2.0 - Use the allowable for ⁄ 2.0 1.5 ⁄ 2.0 - Linearly interpolate between the provided values ⁄ 1.5 - Use the lug-analysis method, or determine allowables by test
• Ratio of fitting thickness to bolt diameter, ⁄ . The allowables provided in MIL-HDBK-5 apply for 0.18 ⁄ 1.00.
• For applications outside this range, we should determine allowables by test or use an uncertainty factor.
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Mode 3 (Shear tear-out of the fitting)
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• Mode 3: Shear tear-out of the fitting • Figure below shows how a fitting can fail by shear tear-out. We calculate the
fitting's shear stress, , as follows:
where is the bearing load acting on the fitting, and is the shear area defined in Fig. below). • If the edge-distance ratio, ⁄ , is greater than or equal to 1.5, we don't need to
check the shear tear-out failure mode. Bearing allowables are determined by test, so any tendency toward shear tear-out would be reflected in the allowable.
Fig. 2.3.15 Shear tear-out. In this mode of failure, the load from the fastener tears
out a plug of material from the fitting.
Mode 4 (Net tension failure in attachment fittings)
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• The fitting shown in Fig. below must carry the load, p, in tension across its net area.
Fig. 2.3.16 Stress concentration for an aluminum plate with a round hole.
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How shims affect the strength of shear joints
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• A shim is an element used to fill a gap when necessary to produce an assembly that meets engineering drawing requirements. To a stress analyst, a shim is any part that separates the load-carrying elements of a shear joint without carrying load itself. A shim introduces a space over which the shear load must transfer. This causes a bending moment on the fastener and higher peak contact stresses on the surface of the hole. These effects interact to reduce the joint's strength.
• Even without shims, no fastener will be in pure shear; shear transfer always causes a moment. This fact becomes obvious when we look at the FBD of a shear-loaded bolt. Figures a and b show these diagrams for two types of joints.
• The addition of a shim magnifies the fastener's bending moment and causes contact stresses to concentrate more at the mating surfaces, as shown in Fig. c.
Fig. 2.3.17 Approximate linear elastic distributions of bending stress for various shear joints.
Ex 2.3.3 Single-shear test with a shim
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• Tests show that a shim affects a joint's yield strength more than strength. Figure below shows a 5/8" bolt (d= 5/8“, A-286 alloy, 160 ksi ultimate tensile strength) that was bent during a single-shear test at Martin Marietta Astronautics. The test setup included a 0.093"-thick shim and 0.75"-thick aluminum fittings (t = 0.75"). The bolt was tightened to a nominal preload of 22,000 lb. The applied shear load, aligned as shown in Fig. 2.3.17b, reached 29,000 lb on this bolt, and the joint deformed 0.23" (of this, 0.17" was permanent, as a result of yielding, and 0.06’’ was elastic). This deformation was not solely the result of bolt bending – the fitting shown in Fig. 2.3.14 was one of those used in the test. Based on the MIL-HSBK-5 criteria for joint yield (permanent deformation of 4% of bolt diameter, which in this case was 0.025"), this joint yielded at Pexy=16,700 lb. In comparison, based on the bearing yield allowable, of 94 ksi for the tested fitting, we would calculate an allowable yield load,
= 0.625(0.75)(94,000) = 44,100 lb
which is 2.64 times the yield load determined by test (Pexy)! • Not all of this reduction in yield strength was because of the shim. A single shear joint will
typically yield at lower load than a comparable double-shear joint because of the tendency of the fastener to overturn from the unsymmetrical contact forces (prying force).
• Also, the fitting thickness was greater than the bolt diameter ( ⁄ =1.2 >1), so the MIL-HDBK-5 bearing allowable doesn’t apply.
Fig. 2.3.18 High-strength bolt that was bent in a single-shear test with a shim. The yield strength of this joint (permanent
deformation of 4% of the bolt diameter) was 38% of the bearing yield value calculated based on the full thickness of the fittings.
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Conclusion for shim study
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• With ductile materials for bolts and fittings, the ultimate strength for a joint with thick fittings in single shear is not affected by shims.
• Shims greatly reduce a joint's yield strength. Recommendations:
• Avoid shims in alignment-critical shear joints. • When yielding in a shimmed joint would jeopardize the mission,
- bond the shim with adhesive to one of the fittings, - determine allowables by test, or use a yield fitting factor or conservative analysis method.
Recognizing Potential Joint Failures
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• Any given joint may have unique modes of failure. The key to identifying potential failure modes is to follow all load paths through the joint. Failures usually occur where sections change.
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Shear failure of a tension member
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Shear failure of an end cap
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Shear failure of a rod-end clevis fitting
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Rivet failure
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• Rivets are good in shear, but poor in tension. During installation, a rivet’s protruding end is hammered to form a head. This process causes the shank to expand to fill a standard-sized hole (specified in MIL-HDBK-5), thus keeping the joint from shifting during service. A head formed in this manner, however, is unable to react much applied tension. A rivet's tensile strength is typically less than 25% of its shear strength.
• Rivets have either protruding-heads or flush-heads (countersunk). Flush-head rivets are used to reduce air drag for flight structures and to avoid interfering with other parts. They aren't as strong as protruding-head rivets because shear loads tend to pry more on the head. Protruding-head rivets can fail in shear; the joined parts (often made of sheet metal) can also fail in bearing, as discussed for threaded fastener shear joints.
• The ratio ⁄ (thickness of thinner sheet to rivet diameter) influences the shear strengths of protruding-head and flush-head rivets. As the sheet gets thinner, bearing strength decreases, as expected, but so does the rivet's shear strength if ⁄ 0.33 for single-shear joints because of the magnified effect of contact
stresses. For a double-shear joint, shear strength is reduced if ⁄ 0.67, where t is the thickness of the middle sheet. We determine joint bearing allowables for protruding-head rivets the same way we do for threaded fasteners, using Eq. (a). To prevent premature joint failure, we should make the joint bearing critical.
• For flush-head rivets, there is no recognized distinction between shear and bearing failure modes. MIL-HDBK-5 simply provides joint allowables for ultimate and yield.
