Case Studies in Yield-Limited Design

section epub:type=”chapter”>



Case Studies in Yield-Limited Design



13.1 Introduction


We now examine three applications of plasticity. The first (material selection for a spring) requires no plasticity whatever. The second (material selection for a pressure vessel) typifies plastic design of a large structure. It is unrealistic to expect no plasticity: there will always be some, at bolt holes, loading points, or changes of section. The important thing is that yielding should not spread entirely through any section of the structure—plasticity must not become general. Finally, we examine an instance (the rolling of metal strip) in which yielding is deliberately induced, to give large-strain plasticity.


13.2 Case Study 1: Elastic Design—Materials for Springs


Springs come in many shapes and have many purposes. One thinks of axial springs (a rubber band, for example), leaf springs, helical springs, spiral springs, torsion bars. Regardless of their shape or use, the best material for a spring of minimum volume is that with the greatest value of σy2/Esi2_e. Here E is Young’s modulus and σy the failure strength of the material of the spring: its yield strength if ductile, its fracture strength, or modulus of rupture if brittle. Some materials with high values of this quantity are listed in Table 13.1.



The argument, at its simplest, is as follows. The primary function of a spring is that of storing elastic energy and—when required—releasing it again. The elastic energy stored per unit volume in a block of material stressed uniformly to a stress σ (see Figure 9.1) is


Uel=σ22E



si3_e


It is this that we wish to maximize. The spring will be damaged if the stress σ exceeds the yield stress or failure stress σy; the constraint is σ ≤ σy. So the maximum energy density is


Uel=σy22E



Torsion bars and leaf springs are less efficient than axial springs because some of the material is not fully loaded: the material at the neutral axis, for instance, is not loaded at all.


The leaf spring


Even leaf springs can take many different forms, but all of them are basically elastic beams loaded in bending. For the loading shown in Figure 13.1, the beam bending results in Chapter 7 give


δ=FL34Ebt3



si5_e  (13.1)


f13-01-9780081020517
Figure 13.1 A leaf spring under load.

The elastic energy stored in the spring, per unit volume, is


Uel=12FδbtL=F2L28Eb2t4



Figure 13.2 shows that the stress in the beam is zero along the neutral axis at its center, and is a maximum at the surface, at the midpoint of the beam (because the bending moment is biggest there). The beam bending results in Chapter 7 show that the maximum surface stress is given by


σ=McI=F2×L2×t2×12bt3=3FL2bt2



si7_e  (13.3)


f13-02-9780081020517
Figure 13.2 Stresses inside a leaf spring.

Now to be successful, a spring must not undergo a permanent set during use: it must always “spring” back. The condition for this is that the maximum stress must always be less than the yield stress:


3FL2bt2<σy



Eliminating t between this and Equation (13.2) gives


Ue1=118σy2E



si9_e


So if in service a spring has to undergo a given deflection δ under a force F, the ratio of σy2/Esi2_e must be high enough to avoid a permanent set. This is why we have listed values of σy2/Esi2_e in Table 13.1: the best springs are made of materials with high values of this quantity. For this reason spring materials are heavily strengthened (see Chapter 11): by solid solution strengthening plus work-hardening (cold-rolled, single-phase brass, and bronze), solid solution and precipitate strengthening (spring steel), and so on. Annealing any spring material removes the work-hardening, and may cause the precipitate to coarsen (increasing the particle spacing), reducing σy and making the material useless as a spring.


Worked Example


Springs for a centrifugal clutch


Suppose you are asked to select a material for a spring with the following application. A spring-controlled clutch like that shown in Figure 13.3 is designed to transmit 20 horsepower at 800 rpm; the clutch is to begin to pick up load at 600 rpm. The blocks are lined with Ferodo or some other friction material. When properly adjusted, the maximum deflection of the springs is to be 6.35 mm (but the friction pads may wear, and larger deflections may occur; this is a standard problem with springs—they must often withstand extra deflections without losing their sets).


f13-03-9780081020517
Figure 13.3 Leaf springs in a centrifugal clutch.

Mechanics


The force on the spring is


F=mrω2



where m is the mass of the block, r the distance of the center of gravity of the block from the center of rotation, and ω the angular velocity. The net force each block exerts on the clutch rim at full speed is


mr(ω22ω12)



where ω2 and ω1 correspond to the angular velocities at 800 and 600 rpm (the net force must be zero for ω2 = ω1, at 600 rpm). The full power transmitted by all four blocks is given by 4μsmr (ω22ω12si14_e) × distance moved per second by inner rim of clutch at full speed, that is


power=4μsmr(ω22ω12)×ω2r



μs is the coefficient of static friction and r is specified by the design (the clutch cannot be too big). Power, ω2, and ω1 are specified in Equation (13.7), so m is also specified. The maximum force on the spring is determined by the design from F = mrω12.si16_e The requirement that this force deflect the beam by only 6.35 mm with the linings just in contact is what determines the thickness, t, of the spring via Equation (13.1) (L and b are fixed by the design).


Metallic materials for the clutch springs


Given the spring dimensions (t = 2 mm, b = 50 mm, L = 127 mm) and given δ ≤ 6.35 mm, all of which are specified by design, which material should we use? Eliminating F between equations (13.1) and (13.4) gives


σyE>6δtL2=6×6.35×2127×127=4.7×103



si17_e  (13.8)


As well as seeking materials with high values of σy2/Esi18_e, we must also ensure that the material we choose meets the criterion of Equation (13.8).


Table 13.1 shows that spring steel, the cheapest material listed, is adequate for this purpose, but has a worryingly small safety factor to allow for wear of the linings. Only the expensive beryllium–copper alloy, of all the metals shown, would give a significant safety factor (σy/E = 11.5 × 10-3).


In many designs, the mechanical requirements are such that single springs of the type considered so far would yield even if made from beryllium copper. This commonly arises in the case of suspension springs for vehicles, and so on, where both large δ (“soft” suspensions) and large F (good load-bearing capacity) are required. The solution can then be to use multi-leaf springs (Figure 13.4). t can be made small to give large δ without yield according to


σyE>6δtL2



si19_e  (13.9)


while the lost load-carrying capacity resulting from small t can be made up by having several leaves combining to support the load.


f13-04-9780081020517
Figure 13.4 Multi-leaved springs (schematic).

Nonmetallic materials


Finally, materials other than the metals originally listed earlier in Table 13.1 can make good springs. Glass, or fused silica, with σy/E as large as 58 × 10-3 is excellent, provided it operates under protected conditions where it cannot be scratched or suffer impact loading. Nylon is good—provided the forces are low—having σy/E ≈ 22 × 10-3, and it is widely used in household appliances and children’s toys (you probably brushed your teeth with little nylon springs this morning). Leaf springs for heavy trucks are now being made of CFRP: the value of σy/E (6 × 10-3) is similar to that of spring steel, and the weight saving compensates for the higher cost. CFRP is always worth examining where an innovative use of materials might offer advantages.


13.3 Case Study 2: Plastic Design—Materials for Pressure Vessels


We now look at material selection for a pressure vessel able to contain a gas at pressure p, minimizing the weight. We seek a design that will not fail by plastic collapse (i.e., general yield). But we must be cautious: structures can also fail by fast fracture, by fatigue, and by corrosion. We shall discuss these later, but for now we assume that plastic collapse is our only problem.


The body of an aircraft, the casing of a solid rocket booster: these are examples of pressure vessels that must be as light as possible.


The stress in the vessel wall (Figure 13.5) is:


σ=pr2t



si20_e  (13.10)


r, the radius of the pressure vessel, is fixed by the design. For safety, σ ≤ σy/S, where S is the safety factor. The vessel mass is


m=4πr2tρ



giving


t=mr2ρ



f13-05-9780081020517
Figure 13.5 Thin-walled spherical pressure vessel.

Substituting for t in Equation (13.10) we find that


σyS2πpr3ρm



si23_e  (13.13)


From Equation (13.13) we have


m=Spr3ρσy



si24_e  (13.14)


so for the lightest vessel we require the smallest value of (ρy). Table 13.2 gives values of (ρ/σy) for candidate materials.



The lightest pressure vessels are made of CFRP, aluminum alloy, and pressure-vessel steel. Reinforced concrete or mild steel results in a very heavy vessel. Of course, CFRP is very expensive, which is why aluminum or high-strength steel is used for things like a aircraft fuselages and SRB casings.


13.4 Case Study 3: Large-Strain Plasticity—Metal Rolling


Forging, sheet drawing, and rolling are metal-forming processes in which the section of a billet or slab is reduced by compressive plastic deformation. When a slab is rolled (Figure 13.6) the section is reduced from t1 to t2 over a length L as it passes through the rolls.


f13-06-9780081020517
Figure 13.6 The rolling of metal sheet.

At first sight, it might appear that there would be no sliding (and thus no friction) between the slab and the rolls, since these move with the slab. But the metal is elongated in the rolling direction, so it speeds up as it passes through the rolls, and some slipping is inevitable. If the rolls are polished and lubricated (as they are for precision and cold-rolling) the frictional losses are small. We shall ignore them here (though detailed treatments of rolling include them) and calculate the rolling torque for perfectly lubricated rolls.


From the geometry of Figure 13.6


L2+(rx)2=r2



si25_e


or, if x=12(t1t2)si26_e is small (as it almost always is),


L=r(t1t2)



The rolling force F must cause the metal to yield over the length L and width w (normal to Figure 13.6). Thus


F=σywL



si28_e


If the reaction on the rolls appears halfway along L then the torque on each roll is


T=FL2=σywL22



giving


T=σywr(t1t2)2



The torque required to drive the rolls increases with yield strength σy, so hot-rolling (when σy is low—see Chapter 21) takes less power than cold-rolling. It obviously increases with the reduction in sectiont1t2si31_e. And it increases with roll diameter 2r; this is one of the reasons why small-diameter rolls are used, often backed by two or more rolls of larger diameter (to stop them bending).


Rolling can be analyzed in much more detail to include important aspects, which we have ignored: friction, the elastic deformation of the rolls, and the constraint of two-dimensional strain imposed by the rolling geometry. But this case study gives an idea of why an understanding of plasticity, and the yield strength, is important in forming operations, both for metals and polymers.


Examples




  1. 13.1 Referring to Example 8.5, the sphere can fail by yield or compressive fracture at a pressure pf given by

    pf=2σftr



    si32_e


    where σf is the yield stress or the compressive fracture stress as appropriate.
    The basic design requirement is that the pressure hull has the minimum possible mass compatible with surviving the design pressure.
    By eliminating t from the equations, show that the minimum mass of the hull is given by the expression

    mf=2πr3pfρσf



    Hence obtain a merit index to meet the design requirement for the failure mechanism. [You may assume that the surface area of the sphere is 4πr2.]
  2. 13.2 Consider the pressure hull of Examples 8.5 and 13.1. For each material listed in the following table, calculate the minimum mass and wall thickness of the pressure hull for both failure mechanisms (i.e., external-pressure buckling and yield/compressive fracture) at the design pressure.

    Hence determine the limiting failure mechanism for each material. [Hint: This is the failure mechanism which gives the larger of the two values of t.]
    What is the optimum material for the pressure hull? What are the mass, wall thickness, and limiting failure mechanism of the optimum pressure hull?
  3. 13.3 The drawing on the next page shows a bolted flanged connection in a tubular supporting pier for an old bridge. You are required to determine the bending moment M needed to make the connection fail by yielding of the wrought-iron bolts. Because the data for the bridge are old, the units for the dimensions in the drawing are inches. The units for force and stress are tons and tons/square inch (tsi). Assume that the yield strength of wrought iron is 11 tsi. [Hint: Find the yield load for one bolt, then sum the moments generated by all the bolts about the axis XX.]
    u13-01-9780081020517

  4. 13.4 The following diagram gives the dimensions of a steel bicycle chain. The chain is driven by a chain wheel which has pitch diameter of 190 mm. The chain wheel is connected to the pedals by a pair of cranks set at 180° in the usual way.
    u13-02-9780081020517

    The center of each pedal is 170 mm away from the center of the chain wheel. If the cyclist weighs 90 kg, estimate the factor of safety of the chain. You may assume that a link would fail in simple tension at the position of minimum cross-sectional area and that a pin would fail in double shear. The yield strength of the steel is 1500 Nm m−2.
    u13-03-9780081020517

  5. 13.5 The diagram shows a coupling between two rotating shafts designed to transmit power from a low-speed hydraulic motor to a gearbox. The coupling sleeve was a sliding fit on the shafts and the torque was taken by the two Bissell pins as shown in the diagram. Owing to a malfunction in the gearbox, one of the pins sheared, disconnecting the drive. Assuming that the shear yield stress k is 750 MN m−2, estimate the failure torque.
  6. 13.6 The tubular casing of a solid rocket booster (SRB) for the space shuttle has a mean radius of 71.76″ and a wall thickness of 0.479″. The maximum working pressure it has to contain is 6.303 MN m−2. What is the hoop stress at this pressure? The casing is made from D6AC steel with a yield strength of 1244 MN m−2. What is the safety factor S for the casing?
  7. 13.7 The length and radius of the SRB casing are fixed by the design. Obtain an expression analogous to Equation (13.14) for the mass of the SRB casing.
  8. 13.8 When considering composite materials for a spherical pressure vessel, why is it incorrect to use the strength of the unidirectional composite? Under what pressure loading conditions would it be appropriate to use the unidirectional strength for a cylindrical pressure vessel? Referring to Table 13.2, and assuming that the unidirectional strength is three times the strength of the quasi-isotropic laminate, what effect would this have on the choice of material to give the lightest pressure vessel?
  9. 13.9 When a beam is subjected to a gradually increasing bending moment, the maximum bending stress—which is located at the surfaces furthest from the neutral axis—increases until it reaches the yield stress (when the moment is My). If the moment is increased further, a zone of plastic deformation spreads into the beam from these surfaces, until the whole of the cross section has yielded (when the moment is Mp). Calculate the value of Mp/My for the following beam cross- sections: (a) rectangle, (b) solid round bar, (c) thin-walled tube.
  10. 13.10 A metal beam is loaded in axial compression to produce a stress that is just a little less than the yield stress. At this load, there is a factor of safety of 5 on the critical load for elastic buckling (see the results for buckling of beams at the end of Chapter 7). The load is then increased until the metal yields in compression. The beam then buckles. Explain why.

Answers




  1. 13.1 Also see Equations (13.10) to (13.14). Merit index = σf/ρ.
  2. 13.2 The table below shows values calculated for the design pressure pb , pf = 200 MN m2 and the design radius r = 1 m.

    The optimum material is alumina, with a mass of 2.02 tonne, a wall thickness of 41 mm, and a limiting failure mechanism of external-pressure buckling.
  3. 13.3 When the bolts yield, the connection can be approximated as a mechanism, which hinges at X. The yield load of one bolt = 13.5 tons. The value of M for full yield = 126 ton foot (1 foot = 12 inches).
  4. 13.4 Safety factor ≈ 8.5.
  5. 13.5 12 kgf m.
  6. 13.6 From Example 7.5, σ=prt=944MNm2si34_e. S=σyσ=1.32si35_e.
  7. 13.7 m=2πrtLρ,som=1.32×2πpr2Lρσy.si36_e
  8. 13.8 Because the applied stress is not unidirectional—it’s equibiaxial (see Equation (12.10), and Figures 3.4 and 12.5). When the applied stress is unidirectional—i.e., when the only stress acting is the hoop stress, and there is no axial stress. This stress state is satisfied when the cylinder has no end caps—the opposite of the conditions shown in Example 7.5. A good example is a long pressurized water main, with the axial load removed by fixing the pipe to reinforced concrete thrust blocks. In such an application, CFRP is clearly the winner, resulting in a pressure vessel approximately three times lighter than one made from alloy steel or aluminum alloy. However, cost is still a major issue, and may well rule out CFRP for other than special applications (e.g., aerospace).
  9. 13.9 Refer to the results at the end of Chapter 7 for the elastic bending of beams and second moments of area. Refer to the results at the end of Chapter 12 for plastic moments. Also compare Figure 13.2 (elastic bending stress field) with the plastic bending stress field shown at the end of Chapter 12 (bending of beams).

    1. (a) σy=MycI=6Mybd2,Mp=σybd24,MpMy=1.5si37_e
    2. (b) σy=4Myπr3,Mp=4σyr33,MpMy=1.70si38_e
    3. (c) σy=Myπr2t,Mp=4σyr2t,MpMy=1.27si39_e

  10. 13.10 See the section on plastic buckling at the end of Chapter 12. Once the metal yields, the slope of the stress/strain curve decreases markedly. Because the buckling load is determined by the slope of the stress/strain curve (Young’s modulus below yield, and tangent modulus above), it decreases markedly as well.

Aug 9, 2021 | Posted by in General Engineer | Comments Off on Case Studies in Yield-Limited Design
Premium Wordpress Themes by UFO Themes