Machine Design 4

DUCTILE MATERIALS will yield locally at the stress-raiser while the lower-stressed material further from the geometric discontinuity remains below the yield point. When the material yields locally, its stress-strain curve there becomes nonlinear and of low slope (see Figure 2-2, p. 32), which prevents further significant increase in stress at that point. As the load is increased, more material is yielded, bringing more of the cross section to that stress. Only when the entire cross section has been brought to the yield point will the part continue up the σ−ε curve to fracture. Thus, it is common to ignore the effects of geometric stress concentration in ductile materials under static loading. The stress for the net cross section is calculated as if the stress concentration were not there. However, the reduction in net cross-sectional area or in area moment of inertia due to the removed material is accounted for, thus producing higher stresses than for an unnotched part of the same overall dimensions. BRITTLE MATERIALS will not yield locally, since they do not have a plastic range. Thus, stress concentrations do have an effect on their behavior even under static loads. Once the stress at the stress-raiser exceeds the fracture strength, a crack begins to form. This reduces the material available to resist the load and also increases the stress concentration in the narrow crack. The part then goes quickly to failure. So, for brittle materials under static loads, the stress-concentration factor should be applied to increase the apparent maximum stress according to equation 4.31. The one exception to this is for brittle, cast materials that tend to contain many disruptions and discontinuities within their structure due to graphite flakes in the alloy, or air bubbles, foreign matter, sand particles, etc., which found their way into the molten material in the mold. These discontinuities within the material create many stress-raisers, which are also present in the test specimens used to establish the material's basic strengths. Thus the published strength data include the effects of stress concentration. Adding geometric stress-raisers to the part design, it is argued, adds little to the overall statistical effect of those already in the material. Thus, the geometric stress-concentration factor is often ignored for cast-brittle materials or any material with known defects distributed throughout its interior. But, it should be applied to stresses in other brittle materials. Stress Concentration Under Dynamic Loading Ductile materials under dynamic loading behave and fail as if they were brittle. So, regardless of the ductility or brittleness of the material, the stress-concentration factor should be applied when dynamic loads (fatigue or impact) are present. However, there are still material-related parameters to account for. While all materials are affected by stress concentrations under dynamic loads, some materials are more sensitive than others. A parameter called notch sensitivity q is defined for various materials and used to modify the geometric factors Kt and Kts for a given material under dynamic loading. These procedures will be discussed in detail in Chapter 6. Determining Geometric Stress-Concentration Factors The theory of elasticity can be used to derive stress-concentration functions for some simple geometries. Figure 4-35 shows an elliptical hole in a semi-infinite plate subjected to axial tension. The hole is assumed to be small compared to the plate and far

For a cantilever beam with a concentrated point load as shown in Figure 4-22b (repeated here), the deflection is given by equation (j) in Example 4-5, rearranged here to define the beam's spring rate with the force applied at the end of the beam (a=l): y F EI x ax x a al y F EI x lx Fx EI x l F xl y Fl EI l l Fl EI k F y EI l = − −− [ ] = −− ( ) = − ( ) = − ( ) = − = = 6 3 6 3 0 6 3 6 3 3 3 4 3 2 3 3 2 2 2 3 3 but, when = , for at = then ( .30) : Note that the spring rate k of a beam is unique to its manner of support and its loading distribution, since k depends on the particular beam's deflection equation and point of load application. We will investigate spring design in more detail in a later chapter. 4.15 STRESS CONCENTRATION All the discussion of stress distributions within loaded members has, up to now, assumed that the members' cross sections were uniform throughout. However, most real machine parts will have varying cross sections. For example, shafts often are stepped to different diameters to accommodate bearings, gears, pulleys, etc. A shaft may have grooves for snap-rings or O-rings, or have keyways and holes for the attachment of other parts. Bolts are threaded and have heads bigger than their shank. Any one of these changes in cross-sectional geometry will cause localized stress concentrations. Figure 4-34 shows the stress concentration introduced by notches and fillets in a flat bar subjected to a bending moment. Figure 4-34b shows the stress effects as measured using photoelastic techniques. Photoelastic stress analysis involves making a physical model of the part in a particular type of transparent plastic, loading it in a fixture and photographing it under polarized light, which causes the stresses to show up as "fringes" that depict the stress distribution in the part. Figure 4-34c shows a finiteelement model (FEM) of a part with the same geometry that is constrained and loaded the same way as the photoelastic specimen. Its lines denote isobars of stress levels. Note that, at the right end of the part where the cross section is uniform, the fringe lines of Figure 4-34b are straight, of uniform width, and equispaced. The FEM isobars in Figure 4-24c show a similar pattern.* This indicates a linear stress distribution across this portion of the bar well away from the notches, but at the changes in geometry the stress distribution is very nonlinear and is larger in magnitude. At the fillets where the width of the part is reduced from D to d, the fringe lines and the FEM isobars indicate a disruption and concentration of stresses at this sudden change in geometry. The same effect is seen near the left-hand end around the two notches. Figure 4-34b provides experimental evidence, and Figure 4-34c computational evidence, of the existence of stress concentration at any change in geometry. Such geometric changes are often called "stress-raisers" and should be avoided or at least minimized as much as possible in design. Unfortunately, it is not practical to eliminate all such stress-raisers, since such geometric details are needed to connect mating parts and provide functional part shapes

RECTANGULAR BEAMS The calculation of shear stress due to transverse loading typically becomes an exercise in evaluating the integral Q for the particular beam cross section. Once that is done, the maximum value of τ is easily found. For a beam with a rectangular cross section of width b and depth h, dA = b dy, and c = h / 2. Q ydA b ydy b h y a V I h y y c y c == = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∫ ∫ 1 1 2 4 4 14 2 4 2 1 2 2 1 2 and ( . ) τ = The shear stress varies parabolically across a rectangular beam as shown in Figure 4-20a. When y1 = h / 2, τ = 0 as expected. When y1 = 0, τmax = Vh2 / 8I. For a rectangle, I = bh3 / 12, which gives τmax V A = (. ) b 3 2 4 14 This is valid only for rectangular cross-sectional beams and is shown in Figure 4-20a. ROUND BEAMS Equations 4.13g apply to any cross section. The integral Q for a circular cross section is Q ydA y r y dy r y a y c y c = = −=− ( ) ∫ ∫ 1 1 2 2 3 4 15 22 2 1 2 3 2 (. ) and the shear-stress distribution is τ π π = (. ) VQ bI Vry r y r V r y r = b ( ) − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 3 2 4 4 3 1 4 15 2 1 2 3 2 2 2 4 2 1 2 2 This is also a parabolic distribution but has a smaller peak value than the rectangular section, as shown in Figure 4-20b. The maximum shear stress in a solid, circular crosssectional beam is, at the neutral axis: τmax V A = (. ) c 4 3 4 15 If the round beam is hollow and thin walled (wall thickness < about 1/10 the outside radius), the maximum shear stress at the neutral axis will be approximately τmax V A ≅ d 2 (. ) 4 15 as shown in Figure 4-20c. I-BEAMS It can be shown mathematically that the I-beam configuration in Figure 4-21a is the optimal cross-sectional shape for a beam in terms of strength-to-weight

Statically Indeterminate Beams When a beam has redundant supports as shown in Figure 4-22d (repeated below) it is said to be statically indeterminate. This example is also called a continuous beam and is quite common. Supporting beams for buildings often have multiple columns distributed under a long beam span. The magnitudes of more than two reaction forces or moments cannot be found using only the two equations of static equilibrium, ΣF = 0 and ΣM = 0. To find more than two reactions requires additional equations, and the deflection function can be used for this purpose. The deflection can be assumed to be zero at each simple support (as a first approximation), and the beam slope is known or can be closely estimated at a moment support.* These provide an additional boundary condition for each added reaction, allowing the solution to be calculated. SOLVING INDETERMINATE BEAMS WITH SINGULARITY FUNCTIONS The singularity functions provide a convenient way to set up and evaluate the equations for the loading, shear, moment, slope, and deflection functions as was demonstrated in the previous example. This approach can also be used to solve the indeterminate beam problem and is best demonstrated by another example.

for the designer to understand how stresses are distributed within beams in order to choose the correct locations for the calculation of maximum stresses. Memorization of the beam stress formulas, while useful, is not sufficient without also gaining an understanding of how and where to properly apply them. Beams in Pure Bending While it is rare in practice to encounter a beam that is loaded strictly in "pure" bending, it is nevertheless useful to explore this simplest loading case as a means of developing the theory of stresses due to bending loads. Most real beams will also be subjected to shear loading in combination with the bending moment. That case will be addressed in the next section. STRAIGHT BEAMS As an example of a pure bending case, consider the simply supported, straight beam shown in Figure 4-14. Two identical, concentrated loads P are applied at points A and B, which are each the same distance from either end of the beam. The shear and bending moment diagrams for this loading show that the center section of the beam, between points A and B, has zero shear force and a constant bending moment of magnitude M. The absence of a shear force makes it pure bending. Figure 4-15 shows a removed and enlarged segment of the beam taken between points A and B. The assumptions for the analysis are as follows: 1 The segment analyzed is distant from applied loads or external constraints on the beam. 2 The beam is loaded in a plane of symmetry.

tial moment dM. Figure 4-18 shows that at point A, the moment M(x) is increasing as a function of beam length x, due to the presence of the nonzero shear force V at that point. The normal stresses on the vertical faces of P are found from equation 4.11a. Since the normal stress due to bending is proportional to M(x), the stress σ on the lefthand face of P is less than on its right-hand face, as shown in Figure 4-19b. For equilibrium, this stress imbalance must be counteracted by some other stress component, which is shown as the shear stress τ in Figure 4-19b. The force acting on the left-hand face of P at any distance y from the neutral axis can be found by multiplying the stress by the differential area dA at that point. σdA My I = dA a (. ) 4 13 The total force acting on the left-hand face is found by integrating F My I x dA b y c 1 1 = 4 13 ∫ (. ) and similarly for the right-hand face: F M dM y I dA c x y c 2 1 = 4 13 ( ) + ∫ (. ) The shear force on the top face at distance y1 from the neutral axis is found from F b dx d xy = τ (. ) 4 13 where the product bdx is the area of the top face of element P. For equilibrium, the forces acting on P must sum to zero,

where x represents the independent variable or r/d in this case. The values of the coefficient A and exponent b for any one value of D/d are determined by nonlinear regression on several data points taken from the experimental data. The resulting values of A and b for various magnitudes of the second independent variable D/d are given in the table within the figure. A and b for other values of D/d can be interpolated. The file name that evaluates these functions and interpolates between them is also noted in this figure and in Appendix C for each of the 14 cases shown there. The stress-concentration plots and functions provided in Appendix C along with their corresponding files will prove useful in the design of machine parts throughout this text and in your practice of engineering. For loading and geometry cases not covered in Appendix C of this text, see references 3 and 4. Designing to Avoid Stress Concentrations Complicated geometry is often necessary for the proper function of machine parts. For example, a crankshaft must have particular contours for its purpose. The designer is always faced with the problem of stress concentrations at sections having abrupt changes of shape. The best that can be done is to minimize their effects. A study of the stressconcentration curves for various geometries in Appendix C will show that, in general, the sharper the corner and the larger in magnitude the change in contour, the worse will be the stress concentration. For the stepped bar in Figure 4-36, larger D/d ratios and smaller r/d ratios give worse stress concentration. From these observations, we can state some general guidelines for designing to minimize stress concentrations. 1 Avoid abrupt and/or large-magnitude changes in cross section if possible. 2 Avoid sharp corners completely and provide the largest possible transition radii between surfaces of different contours. These guidelines are fine to state and better to observe, but practical design constraints often intervene to preclude strict adherence to them. Some examples of good and bad designs for stress concentration are shown in Figures 4-37 through 4-39 along with some common tricks used by experienced designers to improve the situation. FORCE-FLOW ANALOGY Figure 4-37a shows a shaft with an abrupt step and a sharp corner, while Figure 4-37b shows the same step in a shaft with a large transition radius. A useful way to visualize the difference in the stress states in contoured parts such as these is to use a "force-flow" analogy, which considers the forces (and thus the stresses) to flow around contours in a way similar to the flow of an ideal incompressible fluid inside a pipe or duct of changing contour. (See also Figure 4-34.) A sudden narrowing of the pipe or duct causes an increase in fluid velocity at the neckdown to maintain constant flow. The velocity profile is then "concentrated" into a smaller region. Streamlined shapes are used in pipes and ducts (and on objects that are pushed through a fluid medium, such as aircraft and boats) to reduce turbulence and resistance to flow. "Streamlining" our part contours (at least internally) can have similar beneficial effects in reducing stress concentrations. The force-flow contours at the abrupt step transition in Figure 4-37a are more concentrated than in the design of Figure 4-37b.

where σ is the principal stress magnitude and nx, ny, and nz are the direction cosines of the unit vector n, which is normal to the principal plane: ˆ ˆ ˆ ˆ ˆ ˆ n n n i jk ⋅ = =++ 1 ( . ) 4 4b nnn xyz For the solution of equation 4.4a to exist, the determinant of the coefficient matrix must be zero. Expanding this determinant and setting it to zero, we obtain σσσ σσσ τ τ τ σσ σσ σσ σσσ τ τ τ στ στ στ 3 2 2 1 0 2 1 222 0 222 0 44 2 − − −= =++ =++− − − = + −−− C CC c C C C xyz xy yz zx x y y z z x x y z xy yz zx x yz y zx z xy (. ) where Equation 4.4c is a cubic polynomial in σ. The coefficients C0, C1, and C2 are called the tensor invariants, because they have the same values regardless of the initial choice of xyz axes in which the applied stresses were measured or calculated. The units of C2 are psi (MPa), of C1 psi2 (MPa2), and of C0 psi3 (MPa3). The three principal (normal) stresses σ1, σ2, σ3 are the three roots of this cubic polynomial. The roots of this polynomial are always real[2] and are usually ordered such that σ1 > σ2 > σ3. If needed, the directions of the principal stress vectors can be found by substituting each root of equation 4.4c into 4.4a and solving for nx, ny, and nz for each of the three principal stresses. The directions of the three principal stresses are mutually orthogonal. The principal shear stresses can be found from the values of the principal normal stresses using τ σ σ τ σ σ τ σ σ 13 1 3 21 2 1 32 3 2 2 2 4 5 2 = − = − = − (.) If the principal normal stresses have been ordered as shown above, then τmax = τ13. The directions of the planes of the principal shear stresses are 45° from those of the principal normal stresses and are also mutually orthogonal. The solution of equation 4.4c for its three roots can be done trigonometrically by Viete's method* or using an iterative root-finding algorithm. The file STRESS3D provided solves equation 4.4c and finds the three principal stress roots by Viete's

3 Cross sections of the beam remain plane and perpendicular to the neutral axis during bending. 4 The material of the beam is homogeneous and obeys Hooke's law. 5 Stresses remain below the elastic limit and deflections are small. 6 The segment is subjected to pure bending with no axial or shear loads. 7 The beam is initially straight. The unloaded segment in Figure 4-15a is straight but, as the bending moment is applied in Figure 4-15b, the segment becomes curved (shown exaggerated). The line from N to N along the neutral axis does not change length, but all other lines along the x direction must either shorten or lengthen in order to keep all cross sections perpendicular to the neutral axis. The outer fibers of the beam at A-A are shortened, which puts them in compression, and the outer fibers at B-B are lengthened and put in tension. This causes the bending stress distribution shown in Figure 4-15b. The bending stress magnitude is zero at the neutral axis and is linearly proportional to the distance y from the neutral axis. This relationship is expressed by the familiar bending stress equation: σx My I = − (. ) 4 11a where M is the applied bending moment at the section in question, I is the second moment of area (area moment of inertia) of the beam cross section about the neutral plane (which passes through the centroid of the cross section of a straight beam), and y is the distance from the neutral plane to the point at which the stress is calculated. The maximum bending stress occurs at the outer fibers and is expressed as σmax Mc I = (. ) 4 11b where c is the distance from the neutral plane to the outer fiber at either top or bottom of the beam. Note that these two distances will only be the same for sections that are

4.12 TORSION When members are loaded with a moment about a longitudinal axis, they are said to be in torsion, and the applied moment is then called a torque. This situation is common in shafts that transmit power, in screw fasteners, and in any situation where the applied moment vector is parallel to the long axis of a part rather than transverse to it as in the case of bending. Many machine parts are loaded with combinations of torques and bending moments, and these situations will be dealt with in later chapters. Here we wish to consider only the simple case of pure torsional loading. Figure 4-28a shows a straight bar having uniform circular cross section with a pure torque applied in such manner that no bending moment or other forces are present. This can be accomplished with a two-handled wrench such as a tap-handle, which allows a pure couple to be applied with no net transverse force. The fixed end of the bar is embedded in a rigid wall. The bar twists about its long axis, and its free end deflects through an angle θ. The assumptions for this analysis are as follows: 1 The element analyzed is distant from applied loads or external constraints on the bar. 2 The bar is subjected to pure torsion in a plane normal to its axis, and no axial, bending, or direct shear loads are present. 3 Cross sections of the bar remain plane and perpendicular to the axis. 4 The material of the bar is homogeneous, isotropic, and obeys Hooke's law. 5 Stresses remain below the elastic limit. 6 The bar is initially straight. CIRCULAR SECTIONS A differential element taken anywhere on the outer surface will be sheared as a result of the torque loading. The stress τ is pure shear and varies from zero at the center to a maximum at the outer radius, as shown in Figure 4-28b, τ ρ = T J (.3) 4 2 a where T = applied torque, ρ = radius to any point, and J = the polar area moment of inertia of the cross section. The stress is maximum at the outer surface, at radius r,

4.14 SPRING RATES Every part made of material having an elastic range can behave as a spring. Some parts are designed to function as springs, giving a controlled and predictable deflection in response to an applied load or vice versa. The "springiness" of a part is defined by its spring rate k, which is the load per unit deflection. For rectilinear motion springs, k F y = (. ) 4 27a where F is the applied load and y is the resulting deflection. Typical units are lb/in or N/m. For angular motion springs the general expression is k T = b θ (. ) 4 27 where T is the applied torque and θ is the resulting angular deflection. Typical units are in-lb/rad or N-m/rad, or sometimes expressed as in-lb/rev or N-m/rev. The spring rate equation for any part is easily obtained from the relevant deflection equation, which provides a relationship between force (or torque) and deflection. For example, for a uniform bar in axial tension, the deflection is given by equation 4.8 (p. 152), rearranged to define its axial spring rate. k F y y Fl AE k AE l = but = so = ( .2 ) : , 48 This is a constant spring rate, dependent only on the bar's geometry and its material properties. For a uniform-section round bar in pure torsion, the deflection is given by equation 4.24 (p. 178), rearranged to define its torsional spring rate: k T Tl GJ k GJ l = but so : ( . ) θ : , θ = = 4 29 This is also a constant spring rate, dependent only on the bar's geometry and material properties.

4.16 AXIAL COMPRESSION - COLUMNS Section 4.7 discussed stress and deflection due to axial tension and developed equations for their calculation, which are repeated here for convenience. σx P A = (.) 4 7 Δs Pl AE = (.) 4 8 When the axial load direction is reversed so as to put the member in compression, equation 4.7 alone may not be sufficient to determine the safe load for the member. It is now a column and may fail by buckling rather than by compression. Buckling occurs suddenly and without warning, even in ductile materials, and as such is one of the more dangerous modes of failure. You can demonstrate buckling for yourself by taking a common rubber eraser between the palms of your two hands and gradually loading it in axial compression. It will resist the load until at some point it suddenly buckles into a bowed shape and collapses. (If you are feeling stronger, you can do the same with an aluminum beverage can.) Slenderness Ratio A short column will fail in compression as shown in Figure 2-6 (p. 35), and its compressive stress can be calculated from equation 4.7. An intermediate or a long column will fail by buckling when the applied axial load exceeds some critical value. The compressive stress can be well below the material's yield strength at the time of buckling. The factor that determines if a column is short or long is its slenderness ratio Sr, S l k r = ( . 3) 4 3 where l is the length of the column and k is its radius of gyration. Radius of gyration is defined as k I A = ( . 4) 4 3 where I is the smallest area moment of inertia (second moment of area) of the column's cross section (about any neutral axis), and A is its area at the same cross section. Short Columns A short column is usually defined as one whose slenderness ratio is less than about 10. The material's yield strength in compression is then used as the limiting factor to compare to the stress calculated in equation 4.7. Long Columns A long column requires the calculation of its critical load. Figure 4-40 shows a slender column with rounded ends acted upon by compressive forces at either end, which

4.4 PLANE STRESS AND PLANE STRAIN The general state of stress and strain is three-dimensional but there exist particular geometric configurations that can be treated differently. Plane Stress The two-dimensional, or biaxial, stress state is also called plane stress. Plane stress requires that one principal stress be zero. This condition is common in some applications. For example, a thin plate or shell may also have a state of plane stress away from its boundaries or points of attachment. These cases can be treated with the simpler approach of equations 4.6. Plane Strain There are principal strains associated with the principal stresses. If one of the principal strains (say ε3) is zero, and if the remaining strains are independent of the dimension along its principal axis, n3, it is called plane strain. This condition occurs in particular geometries. For example, if a long, solid, prismatic bar is loaded only in the transverse direction, regions within the bar that are distant from any end constraints will see essentially zero strain in the direction along the axis of the bar and be in plane strain. (However, the stress is not zero in the zero-strain direction.) A long, hydraulic dam can be considered to have a plane strain condition in regions well removed from its ends or base at which it is attached to surrounding structures. 4.5 MOHR'S CIRCLES Mohr's circles* have long provided a means to do a graphical solution of equation 4.6 and find the principal stresses for the plane stress case. Many textbooks on machine design present the Mohr's circle method as a primary solution technique for determining principal stresses. Before the advent of programmable calculators and computers, Mohr's graphical method was a reasonable and practical way to solve equation 4.6. Now, however, it is more practical to find the principal stresses numerically. Nevertheless, we present the graphical method for several reasons. It can serve as a quick check on a numerical solution, and it may be the only viable method if the power to your computer fails or your calculator's batteries go dead. It also serves the useful purpose of providing a visual presentation of the stress state at a point. Mohr's circles exist for the three-dimensional stress case as well, but a graphical construction method is not available to create them directly from the applied-stress data except for the special case where one of the principal stresses is coincident with an axis of the xyz coordinate system chosen, i.e., where one plane is a plane of principal stress. However, once the principal stresses are calculated from equation 4.4c (p. 144) by a suitable root-finding technique, the 3-D Mohr's circles can be drawn by using the calculated principal stresses. A computer program called MOHR is provided on disk for that purpose. In the special 3-D stress case where one principal stress lies along a coordinate axis, the three Mohr's circles can be graphically constructed

4.8 DIRECT SHEAR STRESS, BEARING STRESS, AND TEAROUT These types of loading occur mainly in pin-jointed, bolted, or riveted connections. Possible modes of failure are direct shear of the connector (pin, rivet, or bolt), bearing failure of connector or surrounding material, or a tearing out of the material surrounding the connector. See the Case Studies later in this chapter for examples of the calculation of these types of stresses. Direct Shear Direct shear occurs in situations where there is no bending present. A pair of scissors (also called a pair of shears) is designed to produce direct shear on the material being cut. A poor-quality or worn-out pair of scissors will not cut well (even if sharp) if it allows a gap to exist between the two blades in a direction perpendicular to the blades' motion. Figure 4-11 shows a condition of direct shear and also one in which bending occurs instead. If the gap between the two shearing "blades" or surfaces can be kept close to zero, then a state of direct shear can be assumed and the resulting average stress on the shear face can be estimated from τ xy shear P A = (.) 4 9 where P is the applied load and Ashear is the shear-area being cut, i.e., the cross-sectional area being sheared. The assumption here is that the shear stress is uniformly distributed over the cross section. This is not accurate, since higher local stresses occur at the blade. In Figure 4-11a the shear blade is tight against the jaws that hold the workpiece. Thus, the two forces P are in the same plane and do not create a couple. This provides a condition of direct shear with no bending. Figure 4-11b shows the same workpiece with a small gap (x) between the shear blade and the jaws. This creates a moment arm, turning the pair of forces P into a couple and thus bending, rather than directly shearing the part. Of course, there will still be significant shearing stresses developed in addition to the bending stresses in this case. Note that it is difficult to create situations in which pure direct shear is the only loading. Even the slight clearances necessary for function can superimpose bending stresses on the applied shear stresses. We will discuss stresses due to bending in the next section. The situation depicted in Figures 4-11a and 4-12a is also called single shear, because only one cross-sectional area of the part needs to be severed to break it. Figure 4-12b shows a pivot pin in double shear. Two areas must fail before it separates. This is called a clevis-pin joint, where the yoke-shaped link is the clevis. The area to be used in equation 4-9 is now 2A. Double shear is preferred over single shear for pivot-pin designs. Single-shear pivots should only be used where it is impossible to support both ends of the pin as in some linkage cranks, which must pass over adjacent links on one side. Bolted and riveted joints are in single shear when only two flat pieces are fastened together.

6 APPLIED VERSUS PRINCIPAL STRESSES We now want to summarize the differences between the stresses applied to an element and the principal stresses that may occur on other planes as a result of the applied stresses. The applied stresses are the nine components of the stress tensor (Eq. 4.4a, p. 144) that result from whatever loads are applied to the particular geometry of the object as defined in a coordinate system chosen for convenience. The principal stresses are the three principal normal stresses and the three principal shear stresses defined in Section 4.3. Of course, many of the applied-stress terms may be zero in a given case. For example, in the tensile-test specimen discussed in Chapter 2, the only nonzero applied stress is the σx term in equation 4.4a (p. 144), which is unidirectional and normal. There are no applied shear stresses on the surfaces normal to the force axis in pure tensile loading. However, the principal stresses are both normal and shear.

A computer program called MOHR has been written and is included with this text. Program MOHR allows the input of any set of applied stresses and computes the principal normal and shear stresses using equations 4.4 and 4.5 (p. 144). The program then plots the Mohr's circles and will also display the stress function in the vicinity of the three principal stress roots. Data files that can be read into this program are also supplied. Open the file EX04-01.moh in MOHR to see the analytical solution to that example. Additional files with the name EX04-01 can be input to various commercial programs (identified by suffix) and will also calculate the principal stresses and plot the cubic stress function for Example 4-1. See the book's CD-ROM

Also, the theoretical analysis assumes that the loading is perfectly centered on the column axis. This condition is seldom realized in practice. Any loading eccentricity will cause a moment and create larger deflections than this model predicts. For these reasons, the AISC* suggests higher values for leff than the theoretical ones, and some designers use even more conservative values as shown in the third column of Table 4-4. The problem of eccentrically loaded columns is discussed in a later section. Intermediate Columns Equations 4.7 (p. 152) and 4.38c (p. 195) are plotted in Figure 4-42 as a function of slenderness ratio. The material's compressive yield strength, Syc, is used as the value of σx in equations 4.7 and the critical unit load from equation 4.38c is plotted on the same axis as the material strength. The envelope OABCO defined by these two lines and the axes would seem to describe a safe region for column unit loads. However, experiments have demonstrated that columns loaded within this apparently safe envelope will sometimes fail. The problem occurs when the unit loads are in the region ABDA near the intersection of the two curves at point B. J. B. Johnson suggested fitting a parabolic curve between point A and a tangent point D on the Euler curve (Eq. 4.38c, p. 195), which excluded the empirical failure zone. Point D is usually taken at the intersection of the Euler curve and a horizontal line at Syc / 2. The value of (Sr)D corresponding to this point can be found from equation 4.38c. S E S S E S yc r r D yc

It is very common in machine parts to have combinations of loadings that create both normal and shear stresses on the same part. There may be locations within the part where these applied stresses must be combined to find the principal stresses and maximum shear stress. The best way to demonstrate this is with an example.

Direct Bearing A pivot pin in a hole such as is depicted in Figure 4-12 may fail in other ways than in direct shear. The surfaces of the pin and hole are subjected to a direct bearing stress, which is compressive in nature. Bearing stress occurs whenever two surfaces are pressed together. This stress tends to crush the hole or pin rather than to shear it. The bearing stress is normal, compressive, and can be calculated from equation 4.7 (p. 152). If the pin is a close fit in the hole with essentially no clearance, the area used for this calculation is typically taken as the projected area of contact of pin and hole, not the circumferential area. That is, A ld bearing = ( . 0a) 4 1 where l is the length of the bearing contact and d is the diameter of the hole or pin. If there is clearance between the pin and hole then the area of contact is reduced. Grandin[7] has shown that for this case the bearing area can be approximated by A ld bearing = π 4 ( . 0b) 4 1 Figure 4-13a shows the bearing areas for the clevis-pin joint of Figure 4-12. Each of the two links joined must be checked separately for bearing failure, as either can fail independent of the other. The length l (i.e., link thickness) as well as the pin diameter can be adjusted to create sufficient bearing area and avoid failure. Tearout Failure Another possible mode of failure for pinned joints is tearout of the material surrounding the hole. This will occur if the hole is placed too close to the edge. This is a doubleshear failure, as it requires both sides of the hole to separate from the parent material. Equation 4.9 is applicable to this case, provided that the correct shear area is used. Figure 4-13b shows the tearout areas for the clevis-pin joint from Figure 4-12. It appears that the area could be calculated as the product of the link thickness and the distance from the center of the hole to the outer edge of the part, doubled to account for both sides of the hole. However, this assumption implies that the very thin wedge of material within the hole diameter adds significant shear strength. A more common and conservative assumption is to use twice the product of the total link thickness and the dimension from the edge of the hole to the outside of the part for the tearout area. It is a simple matter to provide sufficient material around holes to prevent tearout failure. A minimum of one pin-diameter of material between edge of hole and the part outer edge is a reasonable starting point for your design calculations. 4.9 BEAMS AND BENDING STRESSES Beams are very common elements in structures and machines of all kinds. Any intermittently supported member subjected to loads transverse to its length will act as a beam. Structural floor joists, roof rafters, machinery shafts, springs, and frames are a few examples of elements that are often loaded as beams. Beams will usually have some combination of normal and shear stresses distributed over their cross sections. It is important

Eccentric Columns The above discussion of column failure assumed that the applied load was concentric with the column and passed exactly through its centroid. Even though this condition

Extend horizontal tangent lines from the top and bottom extremes of each Mohr's circle to intersect the shear (vertical) axis. This determines the value of the principal shear stress associated with each pair of principal normal stresses: i.e., τ13 = 22 071, τ12 = 14 142, and τ23 = 7 929 psi. The largest of these is τmax = 22 071, not the value 14 142 found in step 7. 11 Note that it is always the circle lying between the largest and smallest principal stresses that determines the maximum shear stress. In the previous example the zero principal stress was not the smallest of the three, because one principal stress was negative. In the present example the zero principal stress is the smallest. Thus, failing to draw all three circles would have led to a serious error in the value of τmax. 12 The files EX04-02 can be opened in program MOHR and others. See the CD-ROM.

Figure 4-44 shows plots of equation 4.46c from SECANT (over valid ranges†) superimposed on the Euler-, Johnson-, and short-column plots from Figure 4-42. These curves are normalized to the compressive yield strength of the material. The curve shapes are the same for any material modulus of elasticity E; only the horizontal scale changes. The ratio of the Sr scales to the ratio of E values of different materials is given in the figure. The secant curves are all asymptotic to the Euler curve at large Sr's. At an eccentricity ratio of zero, the secant curve becomes coincident with the Euler curve up to nearly the level of the short-column line. When the eccentricity ratio becomes smaller than about 0.1, the secant functions protrude into the concentric-column empirical failure region labeled ABDA in Figure 4-42 (p. 197), i.e., they move above the Johnson line. This indicates that for eccentric intermediate columns with small eccentricity ratios, the Johnson concentric-column formula (rather than the secant formula) may be the failure criterion and should also be computed. 4.17 STRESSES IN CYLINDERS Cylinders are often used as pressure vessels or pipelines and can be subjected to internal and/or external pressure as shown in Figure 4-45. Some common applications are air or hydraulic cylinders, fluid storage tanks and pipes, and gun barrels. Some of these devices are open-ended and some are closed-ended. If open-ended, a two-dimensional stress state will exist in the cylinder walls, with radial and tangential (hoop) stress components. If close-ended, a third-dimensional stress called longitudinal or axial will also be present. These three applied stresses are mutually orthogonal and are principal, since there is no applied shear from the uniformly distributed pressure.

Figure 4-8 shows the Mohr's circle for a tensile-test specimen. In this case, the applied stress is pure tensile and the maximum principal normal stress is equal to it in magnitude and direction. But a principal shear stress of half the magnitude of the applied tensile stress acts on a plane 45° from the plane of the principal normal stress. Thus, the principal shear stresses will typically be nonzero even in the absence of any applied shear stress. This fact is important to an understanding of why parts fail, and it will be discussed in more detail in Chapter 5. The examples in the previous section also reinforce this point. The most difficult task for the machine designer in this context is to correctly determine the locations, types, and magnitudes of all the applied stresses acting on the part. The calculation of the principal stresses is then pro forma using equations 4.4 to 4.6 (pp. 144-145). 4.7 AXIAL TENSION Axial loading in tension (Figure 4-9) is one of the simplest types of loading that can be applied to an element. It is assumed that the load is applied through the area centroid of the element and that the two opposing forces are colinear along the x axis. At some distance away from the ends where the forces are applied, the stress distribution across the cross section of the element is essentially uniform, as shown in Figure 4-10. This is one reason that this loading method is used to test material properties, as was described in Chapter 2. The applied normal stress for pure axial tension can be calculated from σx P A = (.) 4 7 where P is the applied force and A is the cross-sectional area at the point of interest. This is an applied normal stress. The principal normal stresses and the maximum shear stress can be found from equations 4-6 (p. 145). The Mohr's circle for this case was shown in Figure 4-8. The allowable load for any particular tension member can be determined by a comparison of the principal stresses with the appropriate strength of the material. For example, if the material is ductile, then the tensile yield strength, Sy, could be compared to the principal normal stress and the safety factor calculated as N = Sy / σ1. Failure criteria will be dealt with in detail in Chapter 5. The change in length Δs of a member of uniform cross section loaded in pure axial tension is given by Δs Pl AE = (.) 4 8 where P is the applied force, A is the cross-sectional area, l is the loaded length, and E is Young's modulus for the material. Tension loading is very common, occurring in cables, struts, bolts, and many other axially loaded elements. The designer needs to check carefully for the presence of other loads on the member that, if present in combination with the tensile load, will create a different stress state than the pure axial tension described here.

STRAIN ENERGY IN TRANSVERSE SHEAR LOADING For transverse shear loading in a beam, the strain energy will be a function of the cross-sectional shape as well as the load and length. For a rectangular cross-sectional beam the strain energy is U V GA dx e l = ∫ 3 5 4 22 2 0 (. ) where V is the shear loading, which may be a function of x. For cross sections other than rectangular, the fraction 3/5 will be different. The effects of transverse shear loads on deflections in beams typically will be less than 6% of the effects due to bending moments when the beam's length-to-depth ratio is >10. Thus, only short beams will have significant transverse shear effects. For nonrectangular beam cross sections, 1/2 is often used in equation 4.22e rather than 3/5 to get a quick approximation of the strain energy due to transverse shear loads. Such a rough calculation will give an indication as to whether the order of magnitude of the deflection due to transverse shear is sufficient to justify a more accurate calculation. Deflection by Castigliano's Method This method is useful for calculating deflections at particular points on a system. Equation 4.21 relates force and deflection through strain energy. For a system deflected by more than one type of load, the individual effects can be superposed using a combination of equations 4.21 and 4.22. When bending and torsional loads are present, their deflection components will often be significantly larger than those due to any axial loading present. For this reason the axial effects are sometimes ignored. The deflection at points where no actual load is applied can be found by applying a "dummy load" at that point and solving equation 4.21 with the dummy load set to zero. The computation is made easier if the partial differentiation of equation 4.21 is done before performing the integration defined in equations 4.22. To find the maximum deflection, some knowledge of its location along the beam is needed. The singularity function method, on the other hand, provides the deflection function over the entire beam, from which the maxima and minima are easily found. Finding Redundant Reactions with Castigliano's Method Castigliano's method provides a convenient way to solve statically indeterminate problems. Reaction forces at redundant supports on a beam can be found from equation 4.22d by setting the deflection at the redundant support to zero and solving for the force.

Stress and strain are linearly related by Hooke's law in the elastic region of most engineering materials as discussed in Chapter 2. Strain is also a second-order tensor and can be expressed for the 3-D case as εεε εεε εεε xx xy xz yx yy yz zx zy zz a ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ (. ) 4 3 and for the 2-D case as ε ε ε ε xx xy yx yy b ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (. ) 4 3 where ε represents either a normal or a shear strain, the two being differentiated by their subscripts. We will also simplify the repeated subscripts for normal strains to εx, εy, and εz for convenience while retaining the dual subscripts to identify shear strains. The same symmetric relationships shown for shear stress components in equation 4.2 also apply to the strain components. 4.3 PRINCIPAL STRESSES The axis systems taken in Figures 4-1 and 4-2 are arbitrary and are usually chosen for convenience in computing the applied stresses. For any particular combination of applied stresses, there will be a continuous distribution of the stress field around any point analyzed. The normal and shear stresses at the point will vary with direction in any coordinate system chosen. There will always be planes on which the shear-stress components are zero. The normal stresses acting on these planes are called the principal stresses. The planes on which these principal stresses act are called the principal planes. The directions of the surface normals to the principal planes are called the principal axes, and the normal stresses acting in those directions are the principal normal stresses. There will also be another set of mutually perpendicular axes along which the shear stresses will be maximal. The principal shear stresses act on a set of planes that are at 45° angles to the planes of the principal normal stresses. The principal planes and principal stresses for the 2-D case of Figure 4-2 are shown in Figure 4-3. Since, from an engineering standpoint, we are most concerned with designing our machine parts so that they will not fail, and since failure will occur if the stress at any point exceeds some safe value, we need to find the largest stresses (both normal and shear) that occur anywhere in the continuum of material that makes up our machine part. We may be less concerned with the directions of those stresses than with their magnitudes as long as the material can be considered to be at least macroscopically isotropic, thus having strength properties that are uniform in all directions. Most metals and many other engineering materials meet these criteria, although wood and composite materials are notable exceptions. The expression relating the applied stresses to the principal stresses is

The Mohr plane, on which Mohr's circles are drawn, is arranged with its axes drawn mutually perpendicular, but the angle between them represents 180° in real space. All angles drawn on the Mohr plane are double their value in real space. The abscissa is the axis of all normal stresses. The applied normal stresses σx, σy, and σz are plotted along this axis and the principal stresses σ1, σ2, and σ3 are also found on this axis. The ordinate is the axis of all shear stresses. It is used to plot the applied shear stresses τxy, τyx, and τxz and to find the maximum shear stress. * Mohr used a sign convention for shear stresses that makes cw shear couples positive, which is not consistent with the now-standard right-hand rule. Nevertheless, this left-handed convention is still used for his circles. The best way to demonstrate the use of Mohr's circle is with examples.

The example in Figure 4-38 is a stepped shaft to which a ball bearing is to be fitted. A step is needed to locate the bearing axially as well as radially on the shaft diameter. Commercial ball- and roller bearings have quite small radii on their corners, which forces the designer to create a fairly sharp corner at the shaft step. To reduce the stress concentration at the step (a), a larger radius is needed than the bearing will allow. Three possible design modifications to create a better force flow around the step are shown in the figure. The first design (b) removes additional material at the corner in order to increase the radius and then "returns" the contour to provide the needed axial locating surface for the bearing. The second approach (c) removes material behind the step to improve the force streamlines. The third approach (d) provides a suitably large corner radius and adds a special washer that bridges the radius to provide the bearing seat. The stress concentration is reduced in each case versus the original sharp-cornered design. A similar approach of removing material to improve the force flow is seen in Figure 4-39a, which shows a snap-ring groove in a shaft with additional relief grooves provided on each side to smooth the effective transition of the cross-sectional dimension. The effect on the force flow lines is similar to that shown in Figure 4-38c. Another common source of stress concentration is a key needed to torque-couple gears, pulleys, flywheels, etc., to a shaft. The keyway groove creates sharp corners at locations of maximum bending and torsional stresses. Different key styles are available, the most common being the square key and the circular-segment Woodruff key as shown in Figures 4-38b and 4-38c. See Section 10.10 for more information on keys and keyways. Another example of removing material to reduce stress concentration (not shown) is the reduction of the unthreaded portion of a bolt shank's diameter to a dimension less than that of the root diameter of the thread. Since the thread contours create large stress concentrations, the strategy is to keep the force-flow lines within the solid (unthreaded) portion of the bolt. These examples show the usefulness of the force-flow analogy in providing a means to qualitatively improve the design of machine parts for reduced stress concentration. The designer should attempt to minimize sharp changes in the contours of internal forceflow lines by suitable choice of part shap

The amount of stress concentration in any particular geometry is denoted by a geometric stress-concentration factor Kt for normal stresses, or by Kts for shear stresses. The maximum stress at a local stress-raiser is then defined as σ σ τ τ max t nom max ts nom K K = = ( .31) 4 where σnom and τnom are the nominal stresses calculated for the particular applied loading and net cross section, assuming a stress distribution across the section that would obtain for a uniform geometry. For example, in the beam of Figure 4-34, the nominal stress distribution is linear and at the outer fiber, σnom = Mc / I. The stress at the notches would then be σmax = Kt Mc / I. In an axial-tension case, the nominal stress distribution would be as defined in Figure 4-10 (p. 152) and for torsion as defined in Figure 4-28 (p. 177). Note that the nominal stresses are calculated using the net cross section, which is reduced by the notch geometry, i.e., using d instead of D as the width at the notches in Figure 4-34. The factors Kt and Kts only take the effects of part geometry into account and do not consider how the material behaves in the face of stress concentrations. The ductility or brittleness of the material and the type of loading, whether static or dynamic, also affects how it responds to stress concentrations. Stress Concentration Under Static Loading The ductility or brittleness of the material has a pronounced effect on its response to stress concentrations under static loads. We will discuss each of these cases in turn.

The angular deflection due to the applied torque is θ = Tl JG ( . 4) 4 2 where l is length of the bar and G is the shear modulus (modulus of rigidity) of the material as defined in equation 2.5. Note that equations 4.24 apply only to circular cross sections. Any other crosssectional shape will behave quite differently. The polar area moment of inertia of a solid circular cross section of diameter d is J d = a π 4 32 (. ) 4 25 and for a hollow circular cross section of outside dia do and inside dia di is J d d b o i = π( ) − 4 4 32 (. ) 4 25 The circular cross section is the optimum shape for any bar subjected to torsional loading and should be used in all torsion situations if possible. NONCIRCULAR SECTIONS In some cases, other shapes may be necessary for design reasons. Noncircular cross sections subjected to torsion exhibit behavior that violates some of the assumptions listed above. Sections do not remain plane, and will warp. Radial lines do not remain straight, and the shear-stress distribution is not necessarily linear across the section. A general expression for the maximum shear stress due to torsion in noncircular sections is τmax T Q = (. ) 4 26a where Q is a function of cross-section geometry. The angular deflection is θ = Tl KG (. ) 4 26b where K is a function of cross-sectional geometry. Note the similarity between this equation and equation 4.24. For a closed-circular section (only), the geometry factor K is the polar moment of inertia, J. For any closed cross-sectional shape other than circular, the factor K will be less than J for the same section dimensions, which is an indication of the value of using a closed-circular section for torsional loading. This fact will be demonstrated in the next example. Expressions for Q and K for various cross sections can be found in reference 3 as well as in other sources. Table 4-3 shows expressions for Q and K for a few common cross sections and also shows the locations of the maximum shear stres

The constants C1 and C2 can be found from boundary conditions on the shear and moment functions. For example, the moment will be zero at a simply supported beam end and either zero (or known if applied) at an unsupported free end of a beam. The shear force will be zero at an unloaded free end. Note that if the reaction forces are included in the loading function q(x), then C1 = C2 = 0. The constants C3 and C4 can be found from boundary conditions on the slope and deflection functions. For example, the deflection will be zero at any rigid support, and the beam slope will be zero at a moment joint. Substitute two known combinations of values of x and y or x and θ along with C1 and C2 in equations 4.19c and 4.19d and solve for C3 and C4. Many techniques for solution of these equations have been developed such as graphical integration, the area-moment method, energy methods, and singularity functions. We will explore the last two of these. Deflection by Singularity Functions Section 3.9 presented the use of singularity functions to represent loads on the beam. These functions make it relatively simple to perform the integration analytically and can easily be programmed for computer solution. Section 3.9 also applied this approach to obtain the shear and moment functions from the loading function. We will now extend that technique to develop the beam's slope and deflection functions. The best way to explore this method is by way of examples. Accordingly, we will calculate the beam shear, moment, slope, and deflection functions for the beams shown in Figure 4-22.

The distributions of these stresses across the wall thickness for po = 0 are shown in Figure 4-46. Under internal pressure, both are maximum at the inside surface. The tangential (hoop) stress is tensile and the radial stress is compressive. When two parts are press- or shrink-fitted together with an interference, the stresses developed in the two parts are defined by equations 4.47. Their mutual elastic deflections create internal pressure on the outer part and external pressure on the inner part. Interference fits are discussed further in Section 10.12. Thin-Walled Cylinders When the wall thickness is less than about 1/10 of the radius, the cylinder can be considered thin-walled. The stress distribution across the thin wall can be approximated as uniform, and the expressions for stress simplify to σt pr t = (. ) 4 49a σr = 0 4 49 (. ) b and if closed-ended: σa pr t = c 2 (. ) 4 49 All of these equations are valid only at locations removed from any local stress concentrations or changes in section. For true pressure-vessel design, consult the ASME Boiler Code for more complete information and guidelines to safe design. Pressure vessels can be extremely dangerous even at relatively low pressures if the stored volume is large and the pressurized medium is compressible. Large amounts of energy can be released suddenly at failure, possibly causing serious injury. 4.18 CASE STUDIES IN STATIC STRESS AND DEFLECTION ANALYSIS We will now present some case studies that continue the design of devices whose forces were analyzed in the case studies of Chapter 3. The same case-study number will be retained for a given design throughout the text, and successive installments will be designated by a series of letter suffixes. For example, Chapter 3 presented six case studies labeled 1A, 2A, 3A, 4A, 5A, and 5B. This chapter will continue case studies 1 through 4 as 1B, 2B, 3B, and 4B. Some of these will be continued in later chapters and given successive letter designators. Thus, the reader can review the earlier installments of any case study by referring to its common case number. See the list of case studies in the table of contents to locate each part. Since stresses vary continuously over a part, we must make some engineering judgment as to where they will be the largest and calculate for those locations. We do not

The equation of the parabola fitted between points A and D is P A S E S S cr yc yc r = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 4 4 2 π ( . 3) Equations 4.38c and 4.43 taken together over their appropriate regions then provide a reasonable failure model for all concentrically loaded columns. If the slenderness ratio is ≤ (Sr)D, use equation 4.43, else use equation 4.38c. Note that equation 4.43 is both valid and conservative for short columns as well. Equations 4.38c and 4.43 predict failure at the calculated critical unit loads, so an appropriate safety factor must be applied to the result to reduce the allowable load accordingly. The file COLMPLOT will compute the critical load and plot the column failure curves of Figure 4-42 for any choices of Syc, Sr, E, and end-condition factor. It can be used to check the design of any concentrically loaded column or to investigate trial designs. The reader may experiment with this program file by changing the values of the above factors and observing the effects on the plotted curves.

The fixed-pinned column (Figure 4-41d) has an leff = 0.707l and the fixed-fixed column (Figure 4-41e) has an leff = 0.5l. The more rigid end constraints make these columns behave as if they were shorter (i.e., stiffer) than a pinned-pinned version and they will thus support more load. Substitute the appropriate effective length in equation 4.33 (p. 193) to obtain the proper slenderness ratio to use in any of the critical-load formulas: S l k r eff = ( . 1) 4 4 where leff takes the values shown in Table 4-4 for various end conditions. Note that the fixed-pinned and fixed-fixed conditions have theoretical leff values of 0.5l and 0.707l, respectively, but these values are seldom used because it is very difficult to obtain a joint fixation that does not allow any change in slope at the column end. Welded joints will usually allow some angular deflection, which is dependent on the stiffness of the structure to which the column is welded

The nine stress components are shown acting on the surfaces of this infinitesimal element in Figure 4-1b and c. The components σxx, σyy, and σzz are the normal stresses, so called because they act, respectively, in directions normal to the x, y, and z surfaces of the cube. The components τxy and τxz, for example, are shear stresses that act on the x face and whose directions of action are parallel to the y and z axes, respectively. The sign of any one of these components is defined as positive if the signs of its surface normal and its stress direction are the same, and as negative if they are different. Thus the components shown in Figure 4-1b are all positive because they are acting on the positive faces of the cube and their directions are also positive. The components shown in Figure 4-1c are all negative because they are acting on the positive faces of the cube and their directions are negative. This sign convention makes tensile normal stresses positive and compressive normal stresses negative. For the 2-D case, only one face of the stress cube may be drawn. If the x and y directions are retained and z eliminated, we look normal to the xy plane of the cube of Figure 4-1 and see the stresses shown in Figure 4-2, acting on the unseen faces of the cube. The reader should confirm that the stress components shown in Figure 4-2 are all positive by the sign convention stated above. Note that the definition of the dual subscript notation given above is consistent when applied to the normal stresses. For example, the normal stress σxx acts on the x face and is also in the x direction. Since the subscripts are simply repeated for normal stresses, it is common to eliminate one of them and refer to the normal components simply as σx, σy, and σz. Both subscripts are needed to define the shear-stress components and they will be retained. It can also be shown[1] that the stress tensor is symmetric, which means that τ τ τ τ τ τ xy yx yz zy zx xz = = = (.) 4 2 This reduces the number of stress components to be calculated.

The previous example points out the advantage of using circular sections whenever torsional loads are present. Remember that the amount of material and thus the weight is identical in all four designs in this example. The closed-square section has 1.6 times the angular deflection of the closed-circular section (tube). The flat plate has 691 times the deflection of the closed-circular tube. Note that the open-circular section is no better in torsion than the flat plate; it has 708 times the angular deflection of the closed tube. This kind of result is true of any open section in torsion, whether I-beam, channel, angle, square, circle, or arbitrary shape. Any open section is generally no better in torsion than a flat plate of the same cross-sectional dimensions. Obviously, open sections should be avoided for all torsionally loaded applications. Even noncircular-closed sections should be avoided, as they are less efficient in torsion than closed-circular sections. Only closed-circular sections, either hollow or solid, are recommended for torsional loading applications.

The stress distribution across the section is no longer linear but is now hyperbolic, and it is greatest on the inner surface of a rectangular cross section as shown in Figure 4-16. The convention is to define a positive moment as one that tends to straighten the beam. This creates tension on the inside and compression on the outside surface from a positive applied moment and vice versa. For pure bending loads, the expressions for the maximum stresses at inner and outer surfaces of a curved beam now become: σ σ i i i o o o M eA c r b M eA c r c = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (. ) (. ) 4 12 4 12 where the subscript i denotes the inside surface and o the outside, M is the applied moment at the section in question, A is the cross-sectional area, and ri and ro are the radii of curvature of the inner and outer surfaces. These expressions contain the ratio c/r. If the radius of curvature r is large compared to c, then the beam looks more "straight" than "curved." When c/r becomes less than about 1:10, the stresses will be only about 10% greater than those of a straight beam of the same dimensions and loading. (Note that this is not a linear relationship, as e is also a function of c and r.) It is more common to have applied forces loading a curved beam, as shown in Figure 4-17a. An example is a clamp or a hook. The free-body diagrams in Figure 4-17b show that there is now an axial force as well as a moment on the cut section. The equations for stress at the inside and outside of the beam become σ σ i i i o o o M eA c r F A d M eA c r F A e = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + (. ) (. ) 4 12 4 12 The second terms in equations 4.12d and 4.12e represent the direct axial tensile stress on the midsection of the beam. The file CURVBEAM provided computes equations 4.12 for five common shapes of curved beam cross sections: round, elliptical, trapezoidal, rectangular, and tee. This program uses numerical integration to evaluate equation 4.12a and to find the area and centroid of the cross section. Shear Due to Transverse Loading The more common condition in beam loading is a combination of both shear force and bending moment applied to a particular section. Figure 4-18 shows a point-loaded, simply supported beam and its shear and moment diagrams. We need to now consider how the shear loading affects the stress state within the beam's cross sections. Figure 4-19a shows a segment taken from the beam around point A in Figure 4-18. An element labeled P is shown cut out of the beam at point A. This element is dx wide and is cut in from the outer fiber at c to a depth y1 from the neutral axis. Note that the magnitude of the moment M(x1) on the left-hand side of P (face b1-c1) is less than the moment M(x2) on the right-hand side (face b2-c2), and their difference is the differen

Thick-Walled Cylinders In Figure 4-45, an annular differential element is shown at radius r. The radial and tangential stresses on that element for an open-ended cylinder are given by Lame's equation: σt ii oo o i io i o o i pr p r r r rr p p rr r = a − − + ( − ) ( − ) 2 2 2 2 2 2 22 2 (. ) 4 47 σr ii oo o i io i o o i pr p r r r rr p p rr r = b − − − ( − ) ( − ) 2 2 2 2 2 2 22 2 (. ) 4 47 where ri and ro are the inside and outside radii, pi and po are the internal and external pressures, respectively, and r is the radius to the point of interest. Note that these stresses vary nonlinearly throughout the wall thickness. If the ends of the cylinder are closed, the axial stress in the walls is: σa ii oo o i pr p r r r = c − − 2 2 2 2 (. ) 4 47 Note the absence of r in this equation as the axial stress is uniform throughout the wall thickness. If the external pressure po = 0, then the equations reduce to σt i i o i o r p r r r r = a − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 2 2 2 2 1 4 48 (. ) σr i i o i o r p r r r r = b − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

This example shows that singularity functions provide a good way to solve beam problems for reactions and deflections simultaneously when there are redundant reactions present. The singularity functions allow the writing of a single expression for each function that applies across the entire beam. They also are inherently computerizable in conjunction with an equation solver that will solve simultaneous equations. The singularity function method presented here is universal and will solve any problems of the types presented. There are other techniques for the solution of deflection and redundant reaction problems. Finite element analysis (FEA) will solve these problems (see Chapter 8). The area-moment method treats the moment function as if it were a "loading" function and integrates twice to obtain the deflection function. The reader is referred to this chapter's bibliography for additional information on these topics. Castigliano's method uses strain energy equations to determine the deflection at any point. 4.11 CASTIGLIANO'S METHOD Energy methods often provide simple and rapid solutions to problems. One such method useful for the solution of beam deflections is that of Castigliano. It can also provide a

We will now change the previous example only slightly to show the need for drawing all three Mohr's circles, even in the plane stress case. The change of significance makes the applied stresses σx and σy both positive, instead of opposite in sign

You have probably had a first course in stress analysis (perhaps called Strength of Materials or Mechanics of Materials) and thus should understand the fundamentals of that subject. Nevertheless, this chapter will present a review of the basics in order to set the stage for the topic of fatigue analysis in later chapters. Stress and strain were discussed in Chapter 2 on materials properties but were incompletely defined at that juncture. In this chapter we will present a more complete definition of what is meant by the terms stress, strain, and deflection. Table 4-0 shows the variables used in this chapter and references the equations, tables, or sections in which they are used. At the end of the chapter a summary section is provided that also groups all the significant equations from this chapter for easy reference and identifies the chapter section in which their discussion can be found. 4.1 STRESS Stress was defined in Chapter 2 as force per unit area with units of psi or MPa. In a part subjected to some forces, stress is generally distributed as a continuously varying function within the continuum of material. Every infinitesimal element of the material can conceivably experience different stresses at the same time. Thus, we must look at stresses as acting on vanishingly small elements within the part. These infinitesimal elements are typically modeled as cubes, shown in Figure 4-1 (p. 142). The stress components are considered to be acting on the faces of these cubes in two different manners. Normal stresses act perpendicular (i.e., normal) to the face of the cube and tend

and for the maximum bending moment as M P e y Pe l P EI max =− + ( ) = − b ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ sec 2 (.5) 4 4 The compressive stress is σc P A Mc I P A Mc Ak =− =− a 2 (.6) 4 4 Substituting the expression for maximum moment from equation 4.45b: σc P A ec k l k P EA = + b ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 4 4 4 2 sec ( . 6 ) Failure will occur at midspan when the maximum compressive stress exceeds the yield strength of the material if ductile, or its fracture strength if brittle. Setting σc equal to the compressive yield strength for a ductile material gives an expression for the critical unit load of an eccentric column: P A S ec k l k P EA c yc eff = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 4 4 4 2 sec (.6) This is called the secant column formula. The appropriate end-condition factor from Table 4-4 (p. 196) is used to obtain an effective length leff, which accounts for the column's boundary conditions. The radius of gyration k for equation 4.46c is taken with respect to the axis about which the applied bending moment acts. If the column cross section is asymmetrical and the bending moment does not act about the weakest axis, it must be checked for concentric-column failure about the axis having the smallest k as well as for failure due to eccentric loading in the bending plane. The fraction ec / k2 in equation 4.46c is called the eccentricity ratio Er of the column. A 1933 study* concluded that assuming a value of 0.025 for the eccentricity ratio would account for typical variations in loading eccentricity of concentrically loaded Euler columns. However, if the column is in the Johnson range, Johnson's formula will apply for Er's of less than about 0.1. (See Figure 4-44 and its discussion below.) Equation 4.46c is a difficult function to evaluate. Not only does it require an iterative solution, but the secant function goes to ± ∞, causing computation problems. It also yields incorrect results when the secant function goes negative. The file SECANT computes and plots equation 4.46c (as well as the Euler and Johnson formulas) over a range of slenderness ratios for any choice of eccentricity ratio and round column crosssectional parameters. Nonround columns can also be calculated in this program by declaring the area A and moment of inertia I to be input values instead of using the column's linear dimensions. When using this program, take care to plot the resulting function and note the regions, if any, in which the results are incorrect due to the secant's behavior. It will be obvious from the plots.

are coaxial and initially act through the area centroid of the column. (A section has been removed to show a reaction force and moment within the column.) The column is shown slightly deflected in the negative y direction, which shifts its area centroid out of colinearity with the applied forces at its ends. This shift of the area centroid creates a moment arm for the force to act about and puts the member in bending as well as in compression. The bending moment tends to increase the lateral deflection, which then also increases the moment arm! Once a critical value of load Pcr is exceeded, the positive feedback of this mechanism causes a sudden, catastrophic buckling. There is no visible warning. The bending moment is given by M Py = ( . 5) 4 3 For small deflections of a beam (using equation 4.17 repeated from p. 162), M EI d y dx = 2 2 ( . 7) 4 1 Combining 4.35 and 4.17 yields a familiar differential equation: d y dx P EI y 2 2 + = 0 43 ( . 6) which has the well-known solution: y C P EI x C P EI = + 1 2 sin cos ( . 7 ) x a 4 3 where C1 and C2 are constants of integration that depend on the boundary conditions defined for the column of Figure 4-40 as y = 0 at x = 0; y = 0 at x = l. Substitution of these conditions shows that C2 = 0 and C P EI 1 sin l b = 0 43 (.7) This equation will hold if C1 = 0, but that is a null solution. So C1 must be nonzero and sin P EI l c = 0 43 (.7) which will be true for P EI ln n d = = π; ,,, 123 43 K (.7) The first critical load will occur for n = 1, which gives P EI l cr = a π2 2 (.8) 4 3 This is known as the Euler-column formula for rounded-end or pinned-end columns. Note that the critical load is a function only of the column's cross-sectional geometry I, its length l, and the material's modulus of elasticity E. The strength of the

chapter. Equation 4.17 also does not include the effects of deflection due to transverse shear loads. The transverse-shear component of deflection in long beams is small compared to that due to bending and is typically ignored unless the beam's length/depth ratio is < about 10. Equation 4.17 can be differentiated twice and integrated twice to create the set of five equations 4.18 (including Eq. 4.17 repeated as Eq. 4.18c), which define beam behavior. Section 3.9 showed the relationship between the loading function q(x), the shear function V(x), and the moment function M(x). V is the first derivative and q the second derivative of equation 4.17 with respect to x. Integrating equation 4.17 once gives the beam slope θ and integrating a second time gives the beam deflection y. These relationships form the following set of beam equations: q EI d y dx = a 4 4 (. ) 4 18 V EI d y dx = b 3 3 (.8) 4 1 M EI d y dx = c 2 2 (. ) 4 18 θ = dy dx (. ) 4 18d y fx e = ( ) (. ) 4 18 The only material parameter in these equations is Young's modulus E, which defines its stiffness. Since most alloys of a given base metal have essentially the same modulus of elasticity, equations 4.18 show why there is no advantage in using a stronger and more expensive alloy when designing to minimize deflection. Higher-strength alloys typically only provide higher yield or break strengths, and designing against a deflection criterion will usually result in relatively low stresses. This is the reason that I-beams and other structural-steel shapes are made primarily in low-strength, low-carbon steels. Determining the deflection function of a beam is an exercise in integration. The loading function q is typically known and can be integrated by any one of several methods, analytical, graphical, or numerical. The constants of integration are evaluated from the boundary conditions of the particular beam configuration. Changes in section modulus across the beam require creating the M/EI function from the moment diagram before integrating for the beam slope. If the beam's area moment of inertia I and material E is uniform across its length, the moment function can just be divided by the constant EI. If the beam's cross section changes over its length, then the integration must be done piecewise to accommodate the changes in I. The integral forms of the beam equations are V q dx C x l a = + << ∫ 1 0 41 (.9) M V dx C x C x l b = + + << ∫ 1 2 0 41 (.9) θ= + + + < < ∫ M EI dx C x C x C x l c 1 2 2 3 0 41 (.9)

have time to calculate the stresses at an infinity of locations. Since the geometries of these parts are fairly complicated, short of doing a complete finite element stress analysis, we must make some reasonable simplifications in order to model them. The aim is to quickly generate some information about the stress state of the design in order to determine its viability before investing more time in a complete analysis.

hich gives an expression for the shear stress τ as a function of the change in moment with respect to x, the distance y from the neutral axis, the second moment of area I of the cross section, and the width b of the cross section at y. Equation 3.16a (p. 111) shows that the slope of the moment function dM/dx is equal to the magnitude of the shear function V at any point, so: τ xy y c V Ib = (. ) ydA f 1 4 13 ∫ The integral in equation 4.13f represents the first moment about the neutral axis of that portion of the cross-sectional area that exists outside of the value of y1 for which the shear stress is being calculated. It is conventional to assign the variable Q to the value of this integral. Q ydA g VQ Ib y c xy = ∫ 1 and then ( . ) 4 13 τ = The integral Q will obviously vary with the shape of the beam's cross section and also with the distance y1 from the neutral axis. Thus, for any particular cross section, we should expect the shear stress to vary across the beam. It will always be zero at the outer fibers because Q vanishes when y1 becomes equal to c. This makes sense, as there is no material to shear against at the outer fiber. The shear stress due to transverse loading will be a maximum at the neutral axis. These results are very fortuitous, since the normal stress due to bending is maximum at the outer fiber and zero at the neutral axis. Thus their combination on any particular element within the cross section seldom creates a worse stress state than exists at the outer fibers. The shear stress due to transverse loading will be small compared to the bending stress Mc / I if the beam is long compared to its depth. The reason for this can be seen in equation 3.16 and in the shear and moment diagrams, one example of which is shown in Figure 4-18. Since the magnitude of the moment function is equal to the area under the shear function, for any given value of V in Figure 4-18, the area under the shear function and thus the maximum moment will increase with beam length. So, while the maximum shear-stress magnitude remains constant, the bending stress increases with beam length, eventually dwarfing the shear stress. A commonly used rule of thumb says that the shear stress due to transverse loading in a beam will be small enough to ignore if the length-to-depth ratio of the beam is 10 or more. Short beams below that ratio should be investigated for transverse shear stress as well as for bending stress

is desirable, it is seldom achieved in practice, as manufacturing tolerances will usually cause the load to be somewhat eccentric to the centroidal axis of the column. In other cases, the design may deliberately introduce an eccentricity e as shown in Figure 4-43. Whatever the cause, the eccentricity changes the loading situation significantly by superposing a bending moment Pe on the axial load P. The bending moment causes a lateral deflection y, which in turn increases the moment arm to e + y. Summing moments about point A gives ∑ =− + + =− + + M M Pe Py M P e y a A ( ) = 0 44 (.4) Substituting equation 4.17 (p. 162) yields the differential equation: d y dx P EI y Pe EI b 2 2 + =− (. ) 4 44 The boundary conditions are x = 0, y = 0, and x = l / 2, dy/dx = 0, which give the solution for the deflection at midspan a

material is not a factor. Using a stronger (higher yield strength) steel, for example, will not help matters, because all steel alloys have essentially the same modulus of elasticity and will thus fail at the same critical load regardless of yield strength. Substitute equation 4.33 and the expression I = Ak2 from equation 4.34 into 4.38a: P EAk l EA l k EA S cr b r = = ⎛ ⎝ ⎞ ⎠ = π ππ 2 2 2 2 2 2 2 (.8) 4 3 Normalizing equation 4.38b by the cross-sectional area of the column, we get an expression for the critical unit load, P A E S c cr r = π2 2 (.8) 4 3 which has the same units as stress, or strength. It is the load per unit area of a rounded- (or pinned-) end column that will cause buckling to occur. Thus, it represents the strength of the particular column rather than the strength of the material from which it is made. Substituting 4.38a in 4.37a gives the deflection curve for this column as y C x l = 1 sin 4 3 π ( . 9) which is a half-period sine wave. Note that applying different boundary or end conditions will yield a different deflection curve and a different critical load. End Conditions Several possible end conditions for columns are shown in Figure 4-41. The roundedrounded and pinned-pinned conditions of Figures 4-41a and 4-41b are essentially the same. They each allow forces but not moments to be supported at their ends. Their boundary conditions are identical, as described above. Their critical unit load is defined in equation 4.38c and their deflection in equation 4.39. The fixed-free column of Figure 4-41c supports a moment and a force at its base and thus controls both the deflection, y, and the slope, y', at that end, but it controls neither x nor y movement at its tip. Its boundary conditions are y = 0 and y' = 0 at x = 0. Substitution of these conditions in equation 4.37a gives P A E S a cr r = π2 2 4 (.0) 4 4 y C x l = 1 b 2 sin 4 4 π (.0) The deflection curve of a fixed-free Euler column is a quarter sine wave, making it effectively twice as long as a pinned-pinned column having the same cross section. This column can only support 1/4 the critical load of a pinned-pinned column. This reduction can be accounted for by using an effective length leff for a column with end conditions other than the pinned-pinned conditions used to derive the critical-load equations.

method and orders them by the above convention. STRESS3D also computes the stress function (Eq. 4.4c) for a list of user-defined values of σ and then plots that function. The root crossings can be seen on the plot. Figure 4-4 shows the stress function for an arbitrary set of applied stresses plotted over a range of values of σ that includes all three roots. Table 4-1 shows the results of the computation. For the special case of a two-dimensional stress state, the equations 4.4c for principal stress reduce to* σ σ σσ σσ τ σ a b xy xy xy c a , = (. ) = 0 + ± ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ + 2 2 4 6 2 2 The two nonzero roots calculated from equation 4.6a are temporarily labeled σa and σb, and the third root σc is always zero in the 2-D case. Depending on their resulting values, the three roots are then labeled according to the convention: algebraically largest = σ1, algebraically smallest = σ3, and other = σ2. Using equation 4.6a to solve the example shown in Figure 4-4 would yield values of σ1 = σa, σ3 = σb, and σ2 = σc = 0 as labeled in the figure.† Of course, equation 4.4c for the 3-D case can still be used to solve any two-dimensional case. One of the three principal stresses found will then be zero. The example in Figure 4-4 is of a two-dimensional case solved with equation 4.4c. Note the root at σ = 0. Once the three principal stresses are found and ordered as described above, the maximum shear stress is found from equation 4.5: τ τ σ σ max = = b − 13 1 3

of Kt and Kts for various geometric parameters and types of loading can be read. Roark and Young[4] also provide tables of stress-concentration factors for a number of cases. Figure 4-36 and Appendix C contain stress-concentration functions and their plots based on data from the technical literature for a set of cases representing commonly encountered situations in machine design. In some cases, mathematical functions have been derived to fit the various empirical curves as closely as possible. In other cases list functions (table lookups) have been created that allow interpolation and automatic retrieval of the value of Kt in the process of a stress calculation. While these stress-concentration functions (SCF) are approximations of the data in the literature, they go beyond the originals in terms of usefulness, since they can be incorporated into a mathematical model of a machine-design problem. These SCF are supplied with this text as TK Solver files, which can be merged with other models or used as stand-alone tools to calculate Kt and Kts for any supplied geometry. This is preferable to looking up data from charts for each calculation. As an example, Figure 4-36 shows the stress-concentration function for the case of a stepped, flat bar in bending. (It and other cases are also shown in Appendix C.) The reduction in width from D to d at the step creates a stress-raiser, and the size of the fillet radius r is also a factor. These two geometric parameters are expressed as the dimensionless ratios r/d and D/d. The first of these is used as the independent variable in the equation and the second determines the member of the family of curves that result. This stress-concentration function is really a three-dimensional surface with the axes, r/d, D/d, and Kt. In Figure 4-36, we are looking at lines on that 3-D surface computed at different values of D/d and projected forward to the r/d-Kt plane. The geometry of the part and its stress equation are defined in the figure, as is the function that defines each stress-concentration curve. In Figure 4-36 it is an exponential function of the form

removed from the plate boundaries. The nominal stress is calculated based on the applied force and the total area, σnom = P/A. The theoretical stress-concentration factor at the edge of the hole was developed by Inglis in 1913* and is K a c t = + a ⎛ ⎝ ⎞ ⎠ 1 2 4 32 (. ) where a is the half-width of the ellipse and c is the half-height. Clearly, as the height of the elliptical hole approaches zero, creating a sharp-edged crack, the stress concentration goes to infinity. When the hole is a circle, a = c and Kt = 3. Figure 4-35 also shows a plot of Kt as a function of c/a, the reciprocal of the ratio in equation 4.32a. The function is asymptotic to Kt = 1 at large values of c/a. STRESS CONCENTRATION MEASUREMENT The theory of elasticity can provide stress-concentration values for some cases as above. Other stress-concentration factors have come from experimental investigations of parts under controlled loading. Experimental measurements can be made with strain gages, photoelastic techniques, laser holography, or other means. Finite element analysis (FEA) and boundary element analysis (BEA) techniques are increasingly being used to develop stress-concentration factors. When a stress analysis is done with these numerical techniques, the stress concentrations "come out in the wash" as long as the mesh is made sufficiently fine around areas of geometric stress raisers. (See Figure 4-51, p. 214.) EFFECT OF LENGTH-WIDTH RATIO Note that Inglis' classic 1913 analysis of this stress concentration case described above assumed a "semi-infinite plate," meaning that the loads are applied far from the hole. Recently, FEA analysis of the circular-hole-inplate case has shown that the stress concentration factor is larger for very short plates, as defined by the ratio of the plate length in the direction of loading to its width, L/W. Troyani et al.[8] show that for L/W ratios less than about 2, and for a uniform tensile load applied over width W, the Kt values vary from around Inglis' value of 3 up to as high as 11 depending on the ratio of hole radius to plate width r/W in combination with the L/W ratio. If your design uses a very short plate with hole(s), and is loaded in tension, this reference should probably be consulted to obtain a better value of Kt. This study also showed that if the tensile loading applied to the plate is the result of a uniform displacement across width W rather than a uniform force, then the value of Kt is reduced. This is because the stiffer webs around the hole are forced to carry the load when the cross section at the displaced end is kept undeformed by the uniform displacement. This points out the effect of boundary conditions on stresses in a part. Other studies by the same authors show similar sensitivity of Kt to the L/W ratio of the part. Reference [9] analyzes flat stepped bars in tension and [10] covers round stepped bars in tension. In this last case they show that uniformly loading the large diameter end and fixing the small end of a stepped rod gives increasing Kt with reduced L/W ratio, while applying a uniform load to the small diameter end of the rod and fixing the large end gives decreasing Kt with reduced L/W. While these results are quite interesting, the changes in Kt do not occur until the L/W ratio has become quite small (< 2) in which case the stepped bar or rod begins to take on the appearance of a stack of pancakes (hotcakes). This geometric configuration may seldom be encountered in practical machine design. STRESS CONCENTRATION DATA The best-known and most-referenced set of stressconcentration factor data is in Peterson[3, 5]. This book compiles the theoretical and experimental results of many researchers into useful design charts from which the values

solution to indeterminate beam problems. When an elastic member is deflected by the application of a force, torque, or moment, strain energy is stored in the member. For small deflections of most geometries, the relationship between the applied force, moment, or torque and the resulting deflection can be assumed to be linear as shown in Figure 3-17, repeated here. This relationship is often called the spring rate k of the system. The area under the load deflection curve is the strain energy U stored in the part. For a linear relationship, this is the area of a triangle. U F = δ 2 ( . 0) 4 2 where F is the applied load and δ is the deflection. Castigliano observed that when a body is elastically deflected by any load, the deflection in the direction of that load is equal to the partial derivative of the strain energy with respect to that load. Letting Q represent a generalized load and Δ a generalized deflection, Δ = ∂ ∂ U Q (. ) 4 21 This relationship can be applied to any loading case, whether axial, bending, shear, or torsion. If more than one such loading case exists on the same part, their effects can be superposed using equation 4.21 for each case. STRAIN ENERGY IN AXIAL LOADING For axial loading, the strain energy is found by substituting the expression for axial deflection (Eq. 4.8, p. 152) into equation 4.20: U F l EA = a 1 2 4 22 2 (. ) which is valid only if neither E nor A varies over the length l. If they do vary with the distance x along the member, then an integration is necessary: U F EA dx b l = ∫ 1 2 4 22 2 0 (. ) STRAIN ENERGY IN TORSIONAL LOADING For torsional loading (see next section) the strain energy is U T GK dx c l = ∫ 1 2 4 22 2 0 (. ) where T is the applied torque, G is the modulus of rigidity, and K is a geometric property of the cross section as defined in Table 4-2 (p. 172). STRAIN ENERGY IN BENDING LOADING For bending, the strain energy is U M EI dx d l = ∫ 1 2 4 22 2 0 (. ) where M is the bending moment, which may be a function of x.

symmetrical about the neutral plane. The value of c is usually taken as positive for both top and bottom surfaces, and the proper sign is then applied to the stress based on inspection of the beam loading to determine which surface is in compression (-) and which is in tension (+). An alternate form of equation 4.11b is often used: σmax M Z = (. ) 4 11c where Z is the beam's section modulus: Z I c = (. ) 4 11d These equations, though developed for the case of pure bending, are nevertheless applicable to cases, where shear strain is negligible, in which other loads in addition to the moment are applied to the beam. In such situations the effects of the combined loadings must be properly accounted for. This will be discussed in later sections. Formulas for the geometric properties (A, I, Z) of typical beam cross sections can be found in Appendix A and are also provided as computer files on disk. CURVED BEAMS Many machine parts such as crane hooks, C-clamps, punch-press frames, etc., are loaded as beams, but are not straight. They have a radius of curvature. The first six assumptions listed above for straight beams still apply. If a beam has significant curvature, then the neutral axis will no longer be coincident with the centroidal axis and equations 4.11 do not directly apply. The neutral axis shifts toward the center of curvature by a distance e as shown in Figure 4-16. e r A dA r a = −c ∫ (. ) 4 12 where rc is the radius of curvature of the centroidal axis of the curved beam, A is the cross-sectional area, and r is the radius from the beam's center of curvature to the differential area dA. Numerical evaluation of the integral can be done for complex shapes.

tio. This explains why I-beams are commonly used as floor and roof beams in large structures. Their shape puts most of the material at the outer fibers where the bending stress is maximum. This gives a large area moment of inertia to resist the bending moment. As the shear stress is maximum at the neutral axis, the narrow web connecting the flanges (called the shear web) serves to resist the shear forces in the beam. In a long beam, the shear stresses due to bending are small compared to the bending stresses, which allows the web to be thin, reducing weight. An approximate expression for the maximum shear stress in an I-beam uses only the area of the web and ignores the flanges: τmax web V A ≅ (. ) 4 16 Figure 4-21b shows the shear-stress distribution across the I-beam section depth. Note the discontinuities at the flange-web interfaces. The shear stress in the flange is small due to its large area. The shear stress jumps to a larger value on entering the web, then rises parabolically to a maximum at the neutral axis. 4.10 DEFLECTION IN BEAMS In addition to the stresses in a beam, a designer also needs to be concerned with its deflections. Any applied bending load will cause a beam to deflect, since it is made of an elastic material. If the deflection does not create strains in excess of the material's strain at its yield point, the beam will return to its undeflected state when the load is removed. If the strain exceeds that of the material's yield point, the beam will yield and "take a set" if ductile, or possibly fracture if brittle. If the beam is sized to prevent stresses that exceed the material's yield point (or other appropriate strength criterion), then no permanent set or fracture should occur. However, elastic deflections at stresses well below the material's failure levels may still cause serious problems in a machine. Deflections can cause interferences between moving parts or misalignments that destroy the required accuracy of the device. In general, designing to minimize deflections will lead to larger beam cross sections than will designing only against stress failure. Even in static structures such as buildings, deflection can be the limiting criterion in sizing floor or roof beams. You have probably walked across a residential floor that bounced noticeably with each step. The floor was undoubtedly safe against collapse due to excessive stresses, but had not been designed stiff enough to prevent undesirable deflections under normal working loads. The bending deflection of a beam is calculated by double integration of the beam equation, M EI d y dx = 2 2 ( . 7) 4 1 which relates the applied moment M, the material's modulus of elasticity E, and the cross section's area moment of inertia I to the second derivative of the beam deflection y. The independent variable x is the position along the beam length. Equation 4.17 is only valid for small deflections, which is not a limitation in most cases of beam design for machine or structural applications. Sometimes beams are used as springs, and their deflections may then exceed the limitations of this equation. Spring design will be covered in a later

to either pull it out (tensile normal stress) or push it in (compressive normal stress). Shear stresses act parallel to the faces of the cubes, in pairs (couples) on opposite faces, which tends to distort the cube into a rhomboidal shape. This is analogous to grabbing both pieces of bread of a peanut-butter sandwich and sliding them in opposite directions. The peanut butter will be sheared as a result. These normal and shear components of stress acting on an infinitesimal element make up the terms of a tensor. * Stress is a tensor of order two† and thus requires nine values or components to describe it in three dimensions. The 3-D stress tensor can be expressed as the matrix: σττ τστ τ τσ xx xy xz yx yy yz zx zy zz a ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ (. ) 4 1 where the notation for each stress component contains three elements, a magnitude (either σ or τ), the direction of a normal to the reference surface (first subscript), and a direction of action (second subscript). We will use σ to refer to normal stresses and τ for shear stresses. Many elements in machinery are subjected to three-dimensional stress states and thus require the stress tensor of equation 4.1a. There are some special cases, however, which can be treated as two-dimensional stress states. The stress tensor for 2-D is σ τ τ σ xx xy yx yy b ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (. ) 4 1 Figure 4-1 shows an infinitesimal cube of material taken from within the material continuum of a part that is subjected to some 3-D stresses. The faces of this infinitesimal cube are made parallel to a set of xyz axes taken in some convenient orientation. The orientation of each face is defined by its surface normal vector‡ as shown in Figure 4-1a. The x face has its surface normal parallel to the x axis, etc. Note that there are thus two x faces, two y faces, and two z faces, one of each being positive and one negative as defined by the sense of its surface normal vector.

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