Geometry, manufacturability and joining

A property advantage must survive manufacture

A material may offer attractive bulk properties yet require unsuitable forming, machining or joining for a design. Minimum wall thickness, defect control and achievable tolerances can change mass and performance. Compare material-and-process candidates.

Manufacturability means being able to produce the required geometry and material state with a specified route, not merely being able to sketch the shape. A thin wall may be too fragile during handling, a cutter may not reach an internal corner, and a joint may need room for a tool or inspection probe. These can impose minimum dimensions even when a simple stress calculation permits less material. A demonstrated process limit belongs to a particular route and specification; it is not a universal limit for every material in that family.

Treat the candidate as a material, its manufactured state, its dimensions and its connections. First specify the shared function and interfaces; then calculate within each candidate’s own geometry. Comparing two equal-sized blocks of raw material can miss both the load-transfer problem and the extra material needed to solve it. The worked comparison below keeps the external pin spacing fixed while changing the strip and connection route.

Follow the force before choosing an area

A force path is the route through which mechanical force passes from one attached part to the next. In a pinned strip it goes from a fixture pin into the wall of the hole, through the strip and out through the other pin. The contact against a hole is called bearing. A transverse section through one circular hole has less solid material than a section away from it. For strip width b, thickness t and hole diameter d, its net area is A_net = (b − d)t. The nominal net tensile stress is F/A_net. It is an average over that remaining section, not the maximum local stress around the hole.

Removing material also interrupts load-bearing paths. In a fibre composite, drilling can cut fibres, and damage between layers may accompany machining. Layup, edge distance and hole quality therefore matter along with the unnotched material response. Net-section tension, bearing damage and tearing toward an edge are different possible modes; passing a net-stress screen does not establish that the other modes pass. Nor can an isotropic hole stress factor simply be assigned to every laminate.

A bonded end fitting can introduce force through the sides of a strip instead of a hole through that strip. In the symmetric two-face connection shown here, two identical bonded faces at one end act in parallel. Under the stipulated symmetric loading each transfers F/2. If each has width b and overlap length ℓ, the average interfacial shear stress over their combined area is τ_avg = F/(2bℓ). Shear acts along the interface; peel denotes a tensile component that opens it. Symmetry removes the deliberately imposed eccentricity of a one-sided lap, but it does not prove zero local peel or uniform shear.

The two end connections act in series: each must transfer the whole strip force F. Their areas cannot be added together as though the load were split between the left and right ends. Actual bonded-joint stresses depend on stiffness, geometry, adhesive response and boundary conditions. The average below is a defined screening quantity; it is not a prediction of the peak stress or an adhesive’s universal strength.

Two complete candidates for the same attachment

Consider an invented room-temperature, slowly loaded tensile-link comparison. The two fixture-pin centres are 180 mm apart, the strip width is b = 20 mm, and the tensile force is F = 2.40 kN = 2400 N at either end. Opposite external forces balance; the section carries 2400 N, not their sum. Both assemblies occupy 200 mm overall length and fit a 30 mm wide in-plane envelope; sufficient out-of-plane room is supplied for this comparison. This is a bounded arithmetic screen, not an approved component design or a real material specification.

Candidate M is a machined metal strip 200 mm long with pin centres 10 mm from either end. It has two through-holes of nominal diameter 6.4 mm, with permitted diameter 6.4 ± 0.1 mm. Candidate C is an aligned fibre–polymer strip 160 mm long. At each end a pair of bonded metal arms extends to a fixture pin outside the composite strip. The composite occupies positions 20–180 mm within the same 200 mm overall length; the pins remain at 10 and 190 mm. Each end fitting has an outer plan width of 26 mm, within the 30 mm envelope. At each end both faces overlap the composite by nominally 12 ± 1 mm. The fitting and adhesive mass is counted separately, so the overlap does not add a second copy of the composite volume.

For this fictional screen, take b and the chosen thicknesses as exact, and the stated hole and overlap intervals as complete permissible geometric ranges. Use the largest hole and shortest overlap when testing the respective inequalities. The assigned finished-state stress criteria already refer to the specified direction and manufacturing route. They are supplied teaching inputs, not values inferred from the references, a measured statistical allowable, or a safety factor derived here. The process minima are also supplied for these particular routes.

For M, the net-section screen requires t ≥ F/[120(b − d_max)] = 2400/[120 × 13.5] = 1.4815 mm. Here 1 MPa = 1 N/mm², so the units give millimetres. The demonstrated route requires at least 2.0 mm; choose 2.0 mm. The worst-case net area is 27.0 mm² and the nominal net stress is 88.89 MPa, below the assigned 120 MPa criterion. Using the gross area would give only 60 MPa and would miss the removed section.

For C, the axial screen gives t ≥ F/(200b) = 0.600 mm, but its demonstrated route requires 1.0 mm; choose 1.0 mm. Its axial stress is then 120 MPa, below the assigned 200 MPa criterion. The shortest permitted overlap is 11 mm at either end, giving τ_avg = 2400/(2 × 20 × 11) = 5.455 MPa, below the assigned 6.0 MPa average-stress criterion. At the nominal 12 mm overlap it would be 5.00 MPa. Reducing the composite to 0.600 mm would meet this axial equation while violating the given manufacturing constraint.

Both candidates pass these stated stress, thickness and overlap screens. That is the extent of this result. Pin shear, bearing, edge failure, joint stress peaks, stiffness, fatigue, environment and process variation would require additional criteria and evidence before a real design could be called feasible for service. An average-stress pass cannot substitute for representative joint analysis and tests.

Include the connection, then test the fit

For nominal mass accounting, remove the actual two nominal circular holes from M’s strip volume. With millimetre dimensions, V_M = [bL − 2πd²/4]t = 7871.32 mm³. The density 2700 kg/m³ is 0.002700 g/mm³, giving m_M,strip = 21.2526 g. Candidate C has V_C = bLt = 3200 mm³ and density 0.001600 g/mm³, so m_C,strip = 5.1200 g. These are nominal geometric estimates; the hole interval was used for the conservative net-area screen, not silently substituted into a nominal mass estimate.

Add everything installed within the stated assembly boundary: M totals 21.2526 + 12.0 = 33.2526 g; C totals 5.1200 + 30.0 = 35.1200 g. C’s strip is much lighter, yet its complete assembly is 1.8674 g heavier. Lower constituent or strip density alone has not ranked the assemblies. This outcome is conditional on the given routes and fittings, not a claim that metal links generally beat composites.

The mass comparison would tie if C’s total extra connection mass were 33.2526 − 5.1200 = 28.1326 g. A new 27.0 g fitting package would give 32.1200 g and become lighter, provided it still satisfied the same geometry, force-transfer and evidence requirements. Removing mass can change stiffness, bearing, bond-end stresses or access. The subtraction identifies a useful target for a redesign; it does not prove the redesigned fitting works.

Fit is another constraint. Suppose M’s pins have diameter 6.00 ± 0.05 mm and the permitted diametral clearance is 0.20–0.60 mm. Diametral clearance is hole diameter minus pin diameter, not the radial gap on one side. Its possible range is [6.30 − 6.05, 6.50 − 5.95] = [0.25, 0.55] mm, which meets the stated interval. A nominal fit alone would not show that. This worst-case interval calculation assumes all combinations are permitted; it assigns no probabilities, process capability or assembly yield.

Inspection and assembly require their own geometry. For example, if an explicitly specified probe requires a 4 mm clear approach slot and the installed housing leaves only 3 mm, the design fails that access requirement even if its mass and stress arithmetic passes. An opening of 4.5 mm would satisfy the geometric access screen; it would not by itself establish that the method detects the required defect. Access, demonstrated detection performance, surface preparation records and representative joint evidence answer different questions.

The finished state is the candidate

Heat from welding, fibre cuts around holes or polymer cure can alter local response. Include production yield, inspection access and repair routes. A nominally superior material can lose its advantage when practical geometry and connections are included.

Fusion welding melts a local fusion region and heats neighbouring material that remains solid. That neighbouring heat-affected zone can have a changed microstructure and response. In a previously hardened aluminium alloy, for example, welding can reduce local strength, and the thermal cycle can distort the geometry. The size and sign of a property change depend on the alloy and prior state; “welded” is not a universal strength value. The process path must therefore be included in the material record.

Consider a separate invented comparison on equal 20.0 mm² sections. An unaffected-state stress criterion is 240 MPa; a matched post-weld local-region criterion is only 120 MPa. At 3.00 kN the nominal stress is 150 MPa. It passes the unaffected-state criterion but fails the local one. The local screen would require at least 3000/120 = 25.0 mm². These stipulated values illustrate why transferring the parent-state criterion is wrong; they are not actual aluminium test data, a weld recipe or an instruction to qualify a joint by area alone.

With a thermosetting adhesive, curing develops a molecular network; surface condition also affects the interface formed with the adherends, the parts being joined. Contamination, cure history and geometric defects can change the result even if the outer shape is unchanged. A nondestructive scan that finds no defect of its validated detectable class does not establish every required property of a bond. In particular, a closed or weakly adhering interface need not produce the same signal as an open void. Record the process and use tests appropriate to the failure mechanism rather than turning a clear image into a universal strength claim.

A useful comparison therefore keeps composition, treatment or cure, layup/direction, machining and surface preparation, dimensional records, joint configuration, loading and inspection scope together. A representative finished-joint specimen can test a proposed force path; an unnotched raw-material coupon cannot alone validate a hole, a bond edge or a changed local state. Repair and disassembly access may constrain the route too. Carry these conditions forward when service data are examined in the next lesson.

Worked examples

Try the reasoning

1. Can accessible inspection be a design constraint?

   Show the answer and reasoning

   Yes.

   Quality evidence may depend on it.

2. Does matching external shape guarantee matching manufacture?

   Show the answer and reasoning

   No.

   Different materials may require different routes and tolerances.

3. Keep both candidates’ dimensions and stipulated criteria unchanged, but raise the force to 3.00 kN. Check only M’s net stress, C’s strip stress and C’s shortest-overlap average. Which listed screen changes the result?

   Show the answer and reasoning

   M gives 111.11 MPa and C’s strip 150 MPa, both below their criteria. C’s 11 mm overlap gives 6.818 MPa, above 6.0 MPa.

   The connection-average criterion becomes the failing listed screen: 3000/27 = 111.11, 3000/20 = 150 and 3000/440 = 6.818. This does not check the other omitted modes for either design.

4. Change the hole specification to 6.40 ± 0.20 mm while retaining the pins 6.00 ± 0.05 mm and allowed diametral clearance 0.20–0.60 mm. Does every permitted pair meet the clearance requirement?

   Show the answer and reasoning

   No. The full clearance interval is 0.15–0.65 mm.

   The smallest clearance is 6.20 − 6.05 = 0.15 mm and the largest 6.60 − 5.95 = 0.65 mm. The nominal clearance remains 0.40 mm, but both interval ends violate the requirement. No yield percentage follows from a range without a distribution.

5. Suppose a separately supported redesign preserves every stated geometric, mechanical and access criterion but reduces C’s whole connection package from 30.0 to 27.0 g. What happens to the nominal mass ranking?

   Show the answer and reasoning

   C totals 32.1200 g, about 1.13 g lighter than M’s 33.2526 g.

   The body remains 5.1200 g, so add 27.0 g. This conditional ranking change comes from the whole joining solution; it does not change either material’s density. The question explicitly supplies the retained performance evidence rather than inferring it from mass reduction.

6. At one C end, suppose only one 20 mm × 11 mm face can transfer force while the other transfers none. At 2.40 kN, may you keep the two-face 5.455 MPa result?

   Show the answer and reasoning

   No. The remaining face’s average is 10.91 MPa, and the symmetric-joint model no longer applies.

   That single face must transfer 2400 N through 220 mm². Removing one path can also introduce eccentricity and peel; the recalculated average alone does not characterize the changed joint.

7. A proposal uses C thickness 0.600 mm because its axial stress is exactly 200 MPa. Has it retained the candidate tested by this lesson’s screen?

   Show the answer and reasoning

   No. It violates the supplied 1.0 mm minimum for the demonstrated manufacturing route.

   An equality in one stress equation does not remove a process constraint. A new route might make a thinner strip possible, but would need its own geometry, handling, joint and finished-state evidence.

8. A bonded fitting fits the probe and its scan finds no open gap. Is that sufficient evidence that every interface has the required mechanical strength?

   Show the answer and reasoning

   No.

   Geometric access and a negative result for a validated detectable feature do not establish sensitivity to every weak interface or establish a strength distribution. Retain method/defect scope, process controls and representative mechanical evidence.

9. In the separate post-weld example, a revised section has 30.0 mm² and carries 3.00 kN. What does the supplied 120 MPa local nominal-stress screen say, and what remains unproved?

   Show the answer and reasoning

   Its nominal stress is 100 MPa, so that one screen passes. Overall joint/service feasibility remains unproved.

   3000 N/30.0 mm² = 100 MPa. The calculation neither predicts the actual peak stress nor checks defects, distortion, other failure modes, cyclic loading or the relevance of the stipulated criterion to a real manufactured state.

What to carry forward

• Select material, process and geometry together.
• Evaluate the manufactured state.
• Trace which areas act in parallel and which connections act in series.
• Include process minima, tolerances and every installed connection mass before ranking candidates.
• A nominal-stress or access screen is useful evidence within its definition; it is not full joint qualification.