Chapter Structural Analysi S Equations
Chapter 8Structural Analysis EquationsLawrence A. SoltisContentsTorsion 8–4quations for deformation and stress, which are thebasis for tension members and beam and columndesign, are discussed in this chapter. The first twosections cover tapered members, straight members, andspecial considerations such as notches, slits, and size effect.A third section presents stability criteria for members subjectto buckling and for members subject to special conditions.The equations are based on mechanics principles and are notgiven in the design code format found in Allowable StressDesign or Load and Resistance Factor Design specifications.Axial Load 8–4Deformation EquationsDeformation Equations 8–1Axial Load 8–1Bending 8–1Combined Bending and Axial Load 8–3Stress Equations 8–4Bending 8–4Combined Bending and Axial Load 8–7Torsion 8–8Stability Equations 8–8Axial Compression 8–8Bending 8–9Interaction of Buckling Modes 8–10References 8–11Equations for deformation of wood members are presented asfunctions of applied loads, moduli of elasticity and rigidity,and member dimensions. They may be solved to determineminimum required cross-sectional dimensions to meet deformation limitations imposed in design. Average moduli ofelasticity and rigidity are given in Chapter 4. Considerationmust be given to variability in material properties and uncertainties in applied loads to control reliability of the design.Axial LoadThe deformation of an axially loaded member is not usuallyan important design consideration. More important considerations will be presented in later sections dealing withcombined loads or stability. Axial load produces a change oflength given byδ PLAE(8–1)where δ is change of length, L length, A cross-sectional area,E modulus of elasticity (EL when grain runs parallel to member axis), and P axial force parallel to grain.BendingStraight Beam DeflectionThe deflection of straight beams that are elastically stressedand have a constant cross section throughout their length isgiven by8–1
δ k bWL3 k sWL EIGA′(8–2)where δ is deflection, W total beam load acting perpendicularto beam neutral axis, L beam span, kb and ks constants dependent upon beam loading, support conditions, and location of point whose deflection is to be calculated, I beammoment of inertia, A′ modified beam area, E beam modulusof elasticity (for beams having grain direction parallel to theiraxis, E EL), and G beam shear modulus (for beams withflat-grained vertical faces, G GLT, and for beams with edgegrained vertical faces, G GLR). Elastic property values aregiven in Tables 4–1 and 4–2 (Ch. 4).Tapered Beam DeflectionFigures 8–1 and 8–2 are useful in the design of taperedbeams. The ordinates are based on design criteria such asspan, loading, difference in beam height (hc h0) as requiredby roof slope or architectural effect, and maximum allowabledeflection, together with material properties. From this, thevalue of the abscissa can be determined and the smallestbeam depth h0 can be calculated for comparison with thatgiven by the design criteria. Conversely, the deflection of abeam can be calculated if the value of the abscissa is known.Tapered beams deflect as a result of shear deflection in addition to bending deflections (Figs. 8–1 and 8–2), and thisshear deflection s can be closely approximated byThe first term on the right side of Equation (8–2) gives thebending deflection and the second term the shear deflection.Values of kb and ks for several cases of loading and support aregiven in Table 8–1. s 3PLfor midspan-concentrated load 10Gbh 0The moment of inertia I of the beams is given byI bh 3for beam of rectangular cross section12πd 4for beam of circular cross section 64(8–3)where b is beam width, h beam depth, and d beam diameter.The modified area A′ is given byA′ 5bh for beam of rectangular cross section69 πd 2 for beam of circular cross section40(8–4)If the beam has initial deformations such as bow (lateralbend) or twist, these deformations will be increased by thebending loads. It may be necessary to provide lateral ortorsional restraints to hold such members in line. (SeeInteraction of Buckling Modes section.)3WLfor uniformly distributed load20Gbh0The final beam design should consider the total deflection asthe sum of the shear and bending deflection, and it may benecessary to iterate to arrive at final beam dimensions. Equations (8–5) are applicable to either single-tapered or doubletapered beams. As with straight beams, lateral or torsionalrestraint may be necessary.Effect of Notches and HolesThe deflection of beams is increased if reductions in crosssection dimensions occur, such as by holes or notches. Thedeflection of such beams can be determined by consideringthem of variable cross section along their length and appropriately solving the general differential equations of the elastic curves, EI(d 2 y/dx 2) M, to obtain deflection expressionsor by the application of Castigliano’s theorem. (These procedures are given in most texts on strength of materials.)Table 8–1. Values of kb and ks for several beam loadingsLoadingBoth simply supportedConcentrated at midspanBoth simply supportedConcentrated at outerquarter span pointsUniformly distributedConcentrated at free end8–2Beam endsUniformly distributed(8–5)Deflection atMidspankbks5/3841/8Both clampedMidspan1/3841/8Both clampedMidspan1/1921/4Both simply supportedLoad point1/961/8Cantilever, one free, one clampedFree end1/31Both simply supportedCantilever, one free, one clampedMidspanMidspanFree end1/4811/7681/81/41/81/2
0.9Single taperh00.8hcL0.70.7L0.60.5Single taper0.4Single taperDouble taper0.4PL3 Bb (hc–h0)3 EE Elastic modulus of beamb Beam width0.5h –hγ c 0h0WL3 Bb (hc–h0)3 EW Total load on beam(uniformly distributed)0.6 B Maximum bending deflectionhcDouble taperh0L/2L/2LPL/ 2hch0hcSingle taperh00.8PL/ 20.30.3h –hγ c 0h00.20.2Double taperh0 0P Concentrated midspan load B Maximum bending deflectionE Elastic modulus of beamb Beam width0.10.112γ3450hc–h0γ12γ345Figure 8–1. Graph for determining tapered beam sizebased on deflection under uniformly distributed load.Figure 8–2. Graph for determining tapered beam sizeon deflection under concentrated midspan load.Effect of Time: Creep DeflectionsIn addition to the elastic deflections previously discussed,wood beams usually sag in time; that is, the deflectionincreases beyond what it was immediately after the load wasfirst applied. (See the discussion of creep in Time UnderLoad in Ch. 4.)Water PondingPonding of water on roofs already deflected by other loadscan cause large increases in deflection. The total deflection due to design load plus ponded water can be closely estimated byGreen timbers, in particular, will sag if allowed to dry underload, although partially dried material will also sag to someextent. In thoroughly dried beams, small changes in deflection occur with changes in moisture content but with littlepermanent increase in deflection. If deflection under longtimeload with initially green timber is to be limited, it has beencustomary to design for an initial deflection of about half thevalue permitted for longtime deflection. If deflection underlongtime load with initially dry timber is to be limited, ithas been customary to design for an initial deflection of abouttwo-thirds the value permitted for longtime deflection. 01 S Scr(8–6)where 0 is deflection due to design load alone, S beamspacing, and Scr critical beam spacing (Eq. (8–31)).Combined Bending and Axial LoadConcentric LoadAddition of a concentric axial load to a beam under loadsacting perpendicular to the beam neutral axis causes increasein bending deflection for added axial compression anddecrease in bending deflection for added axial tension.8–3
The deflection under combined loading at midspan forpin-ended members can be estimated closely by87where the plus sign is chosen if the axial load is tension andthe minus sign if the axial load is compression, is midspan deflection under combined loading, 0 beam midspandeflection without axial load, P axial load, and Pcr a constantequal to the buckling load of the beam under axial compressive load only (see Axial Compression in Stability Equations section.) based on flexural rigidity about the neutralaxis perpendicular to the direction of bending loads. Thisconstant appears regardless of whether P is tension or compression. If P is compression, it must be less than Pcr toavoid collapse. When the axial load is tension, it is conservative to ignore the P/Pcr term. (If the beam is not supportedagainst lateral deflection, its buckling load should be checkedusing Eq. (8–35).)6 01 P PcrEccentric LoadIf an axial load is eccentrically applied to a pin-ended member, it will induce bending deflections and change in lengthgiven by Equation (8–1). Equation (8–7) can be applied tofind the bending deflection by writing the equation in theformδ b ε0 ε01 P Pcr(8–8)where δ b is the induced bending deflection at midspan andε 0 the eccentricity of P from the centroid of the cross section.TorsionThe angle of twist of wood members about the longitudinalaxis can be computed byθ TLGK(8–9)where θ is angle of twist in radians, T applied torque, Lmember length, G shear modulus (use GLR GLT , or approximate G by EL/16 if measured G is not available), and Ka cross-section shape factor. For a circular cross section,K is the polar moment of inertia:K πD 432(8–10)where D is diameter. For a rectangular cross section,K hb3φ(8–11)where h is larger cross-section dimension, b is smaller crosssection dimension, and φ is given in Figure 8–3.8–4φ(8–7) 5hb4300.20.40.60.81.0b/hFigure 8–3. Coefficient φ for determining torsionalrigidity of rectangular member (Eq. (8 –11)).Stress EquationsThe equations presented here are limited by the assumptionthat stress and strain are directly proportional (Hooke’s law)and by the fact that local stresses in the vicinity of points ofsupport or points of load application are correct only to theextent of being statically equivalent to the true stress distribution (St. Venant’s principle). Local stress concentrationsmust be separately accounted for if they are to be limited indesign.Axial LoadTensile StressConcentric axial load (along the line joining the centroids ofthe cross sections) produces a uniform stress:ft PA(8–12)where ft is tensile stress, P axial load, and A cross-sectionalarea.Short-Block Compressive StressEquation (8–12) can also be used in compression if themember is short enough to fail by simple crushing withoutdeflecting laterally. Such fiber crushing produces a local“wrinkle” caused by microstructural instability. The memberas a whole remains structurally stable and able to bear load.BendingThe strength of beams is determined by flexural stressescaused by bending moment, shear stresses caused by shearload, and compression across the grain at the end bearingsand load points.
Straight Beam StressesThe stress due to bending moment for a simply supportedpin-ended beam is a maximum at the top and bottom edges.The concave edge is compressed, and the convex edge isunder tension. The maximum stress is given byfb MZ(8–13)where fb is bending stress, M bending moment, and Z beamsection modulus (for a rectangular cross section, Z bh2/6;for a circular cross section, Z πD3/32).This equation is also used beyond the limits of Hooke’s lawwith M as the ultimate moment at failure. The resultingpseudo-stress is called the “modulus of rupture,” values ofwhich are tabulated in Chapter 4. The modulus of rupturehas been found to decrease with increasing size of member.(See Size Effect section.)The shear stress due to bending is a maximum at the centroidal axis of the beam, where the bending stress happens tobe zero. (This statement is not true if the beam is tapered—see following section.) In wood beams this shear stress mayproduce a failure crack near mid-depth running along the axisof the member. Unless the beam is sufficiently short anddeep, it will fail in bending before shear failure can develop;but wood beams are relatively weak in shear, and shearstrength can sometimes govern a design. The maximumshear stress isfs kVA(8–14)where f s is shear stress, V vertical shear force on cross section, A cross-sectional area, and k 3/2 for a rectangularcross section or k 4/3 for a circular cross section.1.07/83/42/3Tapered Beam StressesFor beams of constant width that taper in depth at a slopeless than 25 , the bending stress can be obtained from Equation (8–13) with an error of less than 5%. The shear stress,however, differs markedly from that found in uniform beams.It can be determined from the basic theory presented by Makiand Kuenzi (1965). The shear stress at the tapered edge canreach a maximum value as great as that at the neutral axis ata reaction.Consider the example shown in Figure 8–4, in which concentrated loads farther to the right have produced a supportreaction V at the left end. In this case the maximum stressesoccur at the cross section that is double the depth of thebeam at the reaction. For other loadings, the location of thecross section with maximum shear stress at the tapered edgewill be different.For the beam depicted in Figure 8–4, the bending stress isalso a maximum at the same cross section where the shearstress is maximum at the tapered edge. This stress situationalso causes a stress in the direction perpendicular to theneutral axis that is maximum at the tapered edge. The effectof combined stresses at a point can be approximately accounted for by an interaction equation based on the Henky–von Mises theory of energy due to the change of shape. Thistheory applied by Norris (1950) to wood results in22f x2 f xy f y 1Fx2 Fxy2 Fy2(8–15)where fx is bending stress, fy stress perpendicular to theneutral axis, and fxy shear stress. Values of Fx, Fy, and Fxy arecorresponding stresses chosen at design values or maximumvalues in accordance with allowable or maximum valuesbeing determined for the tapered beam. Maximum stresses inValues of h0/h1/21/4xh0yαVyl y7/16α3/4α8/9ααParticular tapered beam where M Vxh h0 x tan θα 3V2bh0θx13/4αFigure 8–4. Shear stress distribution for a tapered beam.8–5
the beam depicted in Figure 8–4 are given byfx 3M2bh02f xy f x tan θ(8–16)f y f x tan 2 θSubstitution of these equations into the interaction Equation(8–15) will result in an expression for the moment capacityM of the beam. If the taper is on the beam tension edge, thevalues of fx and fy are tensile stresses.Example: Determine the moment capacity (newton-meters)of a tapered beam of width b 100 mm, depthh0 200 mm, and taper tan θ 1/10. Substituting thesedimensions into Equation (8–16) (with stresses in pascals)results inf x 375Mf xy 37.5Mf y 3.75MSubstituting these into Equation (8–15) and solving for Mresults inM [13.75 10 4 Fx2 102 Fxy2 1 Fy2]1/ 2where appropriate allowable or maximum values of the Fstresses (pascals) are chosen.Size EffectThe modulus of rupture (maximum bending stress) of woodbeams depends on beam size and method of loading, and thestrength of clear, straight-grained beams decreases as sizeincreases. These effects were found to be describable bystatistical strength theory involving “weakest link” hypotheses and can be summarized as follows: For two beams undertwo equal concentrated loads applied symmetrical to themidspan points, the ratio of the modulus of rupture of beam1 to the modulus of rupture of beam 2 is given by1/ mR 1 h 2 L 2(1 ma 2 L 2) R 2 h 1L 1(1 ma 1 L 1) (8–17)where subscripts 1 and 2 refer to beam 1 and beam 2, R ismodulus of rupture, h beam depth, L beam span, a distancebetween loads placed a/2 each side of midspan, and m aconstant. For clear, straight-grained Douglas-fir beams,m 18. If Equation (8–17) is used for beam 2 size (Ch. 4)loaded at midspan, then h2 5.08 mm (2 in.),L2 71.112 mm (28 in.), and a2 0 and Equation (8–17)becomes8–6 R1 361.29 R 2 h 1 L 1(1 ma 1 L 1 ) 1/ m R1 56 R 2 h 1 L 1(1 ma 1 L 1 ) 1/ m(metric)(8–18a)(inch–pound) (8–18b)Example: Determine modulus of rupture for a beam 10 in.deep, spanning 18 ft, and loaded at one-third span pointscompared with a beam 2 in. deep, spanning 28 in., andloaded at midspan that had a modulus of rupture of10,000 lb/in2. Assume m 18. Substituting the dimensionsinto Equation (8–18) produces1 / 18 56R 1 10, 000 2, 160(1 6) 7, 330 lb/in2Application of the statistical strength theory to beams underuniformly distributed load resulted in the following relationship between modulus of rupture of beams under uniformlydistributed load and modulus of rupture of beams underconcentrated loads:()1 / 18R u 1 18ac Lc hc Lc R c 3.876hu Lu (8–19)where subscripts u and c refer to beams under uniformlydistributed and concentrated loads, respectively, and otherterms are as previously defined.Shear strength for non-split, non-checked, solid-sawn, andglulam beams also decreases as beam size increases. A relationship between beam shear τ and ASTM shear blockstrength τASTM, including a stress concentration factor for there-entrant corner of the shear block, Cf, and the shear area A,isτ τ 1.9C f τ ASTMA1/ 51.3C f τ ASTMA1/ 5(metric)(8–20a)(inch–pound)(8–20b)where τ is beam shear (MPa, lb/in2), Cf stress concentrationfactor, τASTM ASTM shear block strength (MPa, lb/in2), andA shear area (cm2, in2).This relationship was determined by empirical fit to testdata. The shear block re-entrant corner concentration factor isapproximately 2; the shear area is defined as beam widthmultiplied by the length of beam subjected to shear force.Effect of Notches, Slits, and HolesIn beams having notches, slits, or holes with sharp interiorcorners, large stress concentrations exist at the corners. Thelocal stresses include shear parallel to grain and tension
0.0020.001A or B ((lb/in2 in. )-1)A or B (x10-4 (kPa mm )-1)a1.50Combined Bending and Axial Load0.0072.000.6Figure 8–5. Coefficients A and B for crackinitiation criterion (Eq. (8–21)).perpendicular to grain. As a result, even moderately lowloads can cause a crack to initiate at the sharp corner andpropagate along the grain. An estimate of the crack-initiationload can be obtained by the fracture mechanics analysis ofMurphy (1979) for a beam with a slit, but it is generallymore economical to avoid sharp notches entirely in woodbeams, especially large wood beams, since there is a sizeeffect: sharp notches cause greater reductions in strength forlarger beams. A conservative criterion for crack initiation fora beam with a slit is 6M 3V h A 2 B 1 2bh bh (8–21)where h is beam depth, b beam width, M bending moment,and V vertical shear force, and coefficients A and B are presented in Figure 8–5 as functions of a/h, where a is slitdepth. The value of A depends on whether the slit is on thetension edge or the compression edge. Therefore, use eitherAt or Ac as appropriate. The values of A and B are dependentupon species; however, the values given in Figure 8–5 areconservative for most softwood species.Effects of Time: Creep Rupture,Fatigue, and AgingSee Chapter 4 for a discussion of fatigue and aging. Creeprupture is accounted for by duration-of-load adjustment in thesetting of allowable stresses, as discussed in Chapters 4and 6.Water PondingPonding of water on roofs can cause increases in bendingstresses that can be computed by the same amplificationfactor (Eq. (8–6)) used with deflection. (See Water Pondingin the Deformation Equations section.)Concentric LoadEquation (8–7) gives the effect on deflection of adding an endload to a simply supported pin-ended beam already bent bytransverse loads. The bending stress in the member is modified by the same factor as the deflection:fb f b01 P Pcr(8–22)where the plus sign is chosen if the axial load is tension andthe minus sign is chosen if the axial load is compression, fbis net
The equations are based on mechanics principles and are not given in the design code format found in Allowable Stress Design or Load and Resistance Factor Design specifications. Deformation Equations Equations for deformation of wood members are presented as functions of applied loads, moduli of elasticity and rigidity, and member dimensions.
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
EQUATIONS AND INEQUALITIES Golden Rule of Equations: "What you do to one side, you do to the other side too" Linear Equations Quadratic Equations Simultaneous Linear Equations Word Problems Literal Equations Linear Inequalities 1 LINEAR EQUATIONS E.g. Solve the following equations: (a) (b) 2s 3 11 4 2 8 2 11 3 s
1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 14 Chapter 2 First Order Equations 2.1 Linear First Order Equations 27 2.2 Separable Equations 39 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 48 2.5 Exact Equations 55 2.6 Integrating Factors 63 Chapter 3 Numerical Methods 3.1 Euler’s Method 74
Chapter 1 Introduction 1 1.1 ApplicationsLeading to Differential Equations 1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 16 Chapter 2 First Order Equations 30 2.1 Linear First Order Equations 30 2.2 Separable Equations 45 2.3 Existence and Uniqueness of Solutionsof Nonlinear Equations 55
point can be determined by solving a system of equations. A system of equations is a set of two or more equations. To ÒsolveÓ a system of equations means to find values for the variables in the equations, which make all the equations true at the same time. One way to solve a system of equations is by graphing.
In Abrasive Jet Machining (AJM), abrasive particles are made to impinge on the work material at a high velocity. The jet of abrasive particles is carried by carrier gas or air. The high velocity stream of abrasive is generated by converting the pressure energy of the carrier gas or air to its kinetic energy and hence high velocity jet. The nozzle directs the abrasive jet in a controlled manner .
