Mechanicsof Materials Formula Sheet

1 MECHANICS OF MATERIALS REVIEW Notation:  - normal stress (psi or Pa) - shear stress (psi or Pa)  - normal strain (in/in or m/m) γ - shearing strain (in/in or m/m) I - area moment of inertia (in4 or m4) J - polar area moment of inertia (in4 or m4) N - revolutions per minute E - modulus of elasticity (psi or Pa) G - modulus of rigidity (psi or Pa)  - Poisson’s ratio  - coefficient of thermal expansion (/°F or /°C) M - bending moment in beams T - torque in shafts T - temperature change (F or C) hp - horsepower (1 hp = 550 ft-lb/sec) failure load F.S. - factor of safety = allowed load t = T - thermal strain Section 1: Introduction a. Stress: force per unit area acting on a plane Normal stress (): force acts perpendicular to the plane. Shear stress (): force acts parallel to the plane b. Strain: deformation per unit length of dimension Normal strain (): stretches or compresses material Shear strain (γ): changes the angle between lines within the material c. Average Shear Stress in Fasteners V Single Shear  =A V = Shear force on pin V=P A = Cross sectional area of pin Double Shear P V=2 d. Bearing Stress in Fasteners P  = dt P = force d = diameter of fastener t = thickness of part 08/26/20 2 Section 2: Axial Loading a. Axial relationships If the line of action of the load, P, passes through the centroid of the resisting cross-section: axial stress =  = axial strain =  = P A  L If the material is also linear, then: Uniaxial Hooke' s Law :  =  E Where E is the modulus of elasticity for the material. The relationship between axial loading and deformation becomes: axial deformation :  = PL AE b. Statically Determinate Members Static Equilibrium F = 0 = -F1 + F2 – F3 + F4 Internal Forces PAB = F1 (Tension) PBC = F1 – F2 (Tension) PCD = F1 – F2 + F3 (Tension) Deformation PABLAB PBCLBC PCDLCD + + AABEAB ABCEBC ACDECD Since the P’s were assumed in tension, negative values will indicate compression and contraction for the deformation rather than elongation. δAD = δAB + δBC + δCD = Thermal Deformation =  AB LAB T  thermal AB Thermal deformation may be added to any mechanical deformation caused by internal forces acting on the material to obtain a total deformation. 08/26/20 3 c. Statically Indeterminate Members 1. After the rigid bearing plate contacts the lower section: Equilibrium: Fy = 0 = FA + FB – P F A + FB = P  = displacement of bearing plate Section A: A = δforce + δthermal P ALA = AAEA + ALAT Section B: (assumes that the bearing plate contacts B) PBLB B = δforce - δthermal + gap = ABEB - BLBT + gap Deformation relationship P AL A P B LB + ALAT = AAE A ABEB - BLBT + gap 2. After horizontal bar contacts post: Equilibrium: M = 0 = (a)PA + ((b)PB – (b+c)P A B Displacement relationship: a = b Link A: A = δforce + δthermal P ALA = AAEA + ALAT Post B: (assumes bar contacts B) B = δforce - δthermal + gap P B LB = A E - BLBT + gap B B Deformation relationship PA LA PB LB +  A LA T −  A L A T + gap AA E A AB EB = a b 08/26/20 Link A is assumed in tension and post B in compression. 4 Section 3: Torsion of Circular Sections a. Shear stress If the shaft has a circular cross section and the material remains in the linear-elastic region, the shear stress in the shaft varies as a linear function of the distance () from the center of the shaft and is given by: shear stress : = T J The maximum shear stress in the shaft is on the outer surface independent of whether the shaft is solid or hollow and is given by:  max = max shear stress : T ro J The polar area moment of inertia is: solid sec tion : J=  ro4 2 J =  ( ) ro4 − ri4 2 The calculated stresses act on the element as shown. hollow section : The deformation is measured by the angle of twist () of one end relative to the other and is giver by: angle of twist :  = TL JG where G is the modulus of rigidity for the material and L is the length of shaft. The shaft also has maximum and minimum normal stresses acting on a element rotated 45° from the element for which the shear stress was calculated. The maximum tensile and compressive stresses are related to the shear stress by:  ten. = −  comp. =  max = 08/26/20 T ro J 5 Section 4: Beams a. Flexural or Bending Stress If the loads on a beam act in its plane of symmetry and the beam is linear elastic, the bending stresses acting normal to the cross section vary linearly with the distance from the neutral axis (N.A.) and are given by: bending stresses :  = My I NA In the absence of axial loads: neutral axis = centrodial axis In the sketch the cross-section is shown rectangular. However, the cross section, in general, can be circular, triangular, etc. The properties of many structural sections such as T-, I-, H-sections can be found in handbooks. If the section is not standard, you must be prepared to determine the centrodial location as well as the value of INA. The maximum bending stress occurs at the location in the beam where (My) is maximum. The section modulus provides a single parameter for design purposes. M I max imum bending stress  max = max S = sec tion mod ulus = S c where c is the maximum distance of material from the neutral axis. b. Shearing Stresses in Beams The transverse and longitudinal horizontal shearing stress in a beam is given by: VQ I NAt where Q is the first moment of the shaded area about the neutral axis if the shearing stress is being evaluated along (c + a) the inside edge of the shaded area. For a rectangular section Q = (c – a)(t) 2 . Q The maximum shearing stress will occur where t is maximum. Q is always Q maximum at the neutral surface. However, t may or may not be maximum at the neutral surface. Check all possibilities. = 08/26/20 6 The shear flow or force per unit length of beam acting on the joint between sections making up a built-up cross-section is given by: VQ Shear flow = f = ( lb / in. or N / m) I NA where the area on either side of the joint is used to calculate Q. Shear flow and discrete fastener strength are related by: FV = fs, where FV is the net shearing strength of the joint fasteners on a single cross-section of the beam and s is the distance along the beam between cross-sections containing fasteners. c. Combined Stresses in Beams. In general any cross-section of the beam will have both shear and bending acting on it. This results in a general stress element as shown, My VQ where bending = I ; and shear = It . Failure is most likely to occur on a cross-section where V or M are maximum. On the cross-section failure due to pure bending may occur at the top or bottom of the cross-section, and due to pure shear may occur at the neutral axis. Wideflange or other non-uniform cross-sections may have principal stresses or maximum shearing stresses at the web-flange intersection or other points of change in cross section width that exceed other stresses on the cross-section. d. Supporting Beam Topics. Maximum shear and bending moment values are found most easily and reliably using the Shear and Bending Moment Diagrams developed in Statics. The centroidal location can be determined by first moments about any axis parallel to the bending moment axis. Ay =  Ai y i where A is the entire area of the cross-section and the Ai are the i areas of subfigures making up the cross-section. ȳ and the yi's are the perpendicular distance from the reference axis to the centroid of the associated area. If the cross-section can be divided into common shaped areas for which the location of the centroid and the area moment of inertia (Ii) about the centroid are known then the area moment of inertia (INA) for the cross-section can be determined from: I NA =  (I i + Ai d i2 ) where the Ii are the area moments of inertia of the individual areas about their own centrodial axis and d is i the perpendicular distance between the area centrodial axis and the neutral axis of the cross-section. 08/26/20 7 e. Beam Deflections Two Integration Method The deflection of straight beams is determined from the equation: EI y( x ) = M ( x ) Here y(x) is the lateral displacement of the beam from its original position as a function of position along the beam, the primes denote derivatives with respect to x, and M(x) is the bending moment as a function of position along the beam. Integration of this equation once yields the equation of the slope as a function of position along the beam: EIy (x ) =  M (x )dx + C1 A second integration yields the deflection or elastic curve equation: EIy( x) =   M ( x)dx dx + C1 x + C2   The two integration constants C1 and C2 are evaluated using the boundary conditions imposed on the slope and deflection by the supports. The majority of beam loading requires that the bending moment be defined using more than one analytic function. Each function is valid over its own portion of the beam length and results in its own set of slope and deflection equations that are valid in that portion of the beam. Each set of equations has its own pair of integration constants. The additional boundary conditions come from requiring that the slope and deflection given by the equations on both sides of a boundary between changes in bending moment functions give the same value when evaluated at the boundary. yL ( at boundary) = yR ( at boundary) yL ( at boundary) = yR ( at boundary) Superposition Method The solutions for these equations for many different types of supports and loads are given in many of the common engineering handbooks. The principal of superposition allows the solutions of different loads to be added together to give the solution for the combined loads. The limitations of this method depend on how extensive the available beam tables are. It must be kept in mind that the table entry must be able to exactly match the portion of the load being represented using only a scaling factor and/or mirror imaging. Loads in the tables may have either positive or negative values. 08/26/20 8 Section 5: Thin-Walled Pressure Vessels Thin-walled pressure vessels are defined as having the ratio t/r ≤ 0.1, where t is the wall thickness and r is the internal radius of either the sphere or cylinder. The pressure, p, is the gage pressure and the analysis is only safe for positive internal pressures. The analysis assumes that the in-plane stresses are uniform across the thickness of the wall. The radial stress is zero on the exterior surface and equal to –p on the interior surface. a. Sphere pr  = 2t ; (at any point and in any direction) pr 1  pr  1 = 2 = 2t  x =  y =  (1 −  ) E  2t  On the inside surface pr + p  MAX = 2t 2 3 = -p On the outside surface r = −  1 − ( x For any x-y coordinate system in the plane of the +  y ) surface Cylinder a = axial; h = hoop directions pr pr 2 = a = 1 = h = t 2t 1  pr  1 pr     h =  1 −   a =  (1 − 2 ) Et  E  2t 2   On the inside surface pr +p 3 = -p  MAX = t 2 On the outside surface r = − 08/26/20  1− (  x For any x-y coordinate system in the plane of the +  y ) surface, usually the axial and hoop directions. 9 Section 6: Column Buckling Columns are long slender members under compressive axial loading. Column buckling is a stability problem, which means failure can occur without the material reaching the yield or ultimate stress. Columns are divided into three classes; slender, intermediate, and short based on both material and slenderness ratio (L'/r). The critical buckling load or stress for  2 EI 2E Pcr =  cr = 2 2 slender columns (L'/r > 100 for steel) are (L )  L  given by Euler's Buckling Equation:   r  L' is the effective length of the column and depends on the type of supports at the ends. The four common support combinations are: L is the actual length of the column and r is the radius of gyration for the cross section r = (I/A)1/2 The critical load for intermediate columns can be found in various code handbooks. These also give the range of slenderness ratios for which they are valid. Short columns are treated using the ordinary axial loading theory. pinned − pinned fixed − free L = L L L = 2 L = 2L fixed − pinned L = 0.7 L fixed − fixed Section 7: Plane Transformations a. Stresses Transformation Equations It is assumed that all the stresses in one direction are zero. The coordinate axes are orientated to place the z-axis in that direction. This situation is common in engineering applications. A free surface is the classic example. The stresses representing the state of stress at a point are different when measured with respect to two different coordinate systems that are rotated with respect to each other. If the first system is labeled xy then the x'y' is rotated counter-clockwise by an angle . The primed stresses may be determined from the unprimed by the equations:  x  ( ) =  x +  x y ( ) = − 2 +  x − y 2 y' = x'( + 90°) 08/26/20 y  x − 2 y Cos(2 ) +  xy Sin(2 ) Sin( 2 ) +  xy Cos( 2 ) 10 Principal Stress and Maximum Shearing Stress There will always be a maximum and minimum stress value, referred to as the principal stresses, occurring at some orientation. There will also be a maximum shearing stress that occurs on two different planes. The values of the principal stresses are given by: 2  −  y  2   x  + ( xy ) 2 2   The plus sign is used for the larger 1 and the minus sign for the smaller 2. 1, 2 = x +y The value of the maximum shearing stress is given by: x − y 2  2  2  + ( xy )  The orientation of the 1 plane relative to the x plane is given by:  max =          1 xy −1  P = Tan     − 2   x y     2   P is the counter-clockwise angle from the x plane to the 1 plane. The two principal planes at perpendicular to each other and the two maximum shearing stress planes are at 45° to either of the principal planes. Mohr's Circle for Plane Stress 08/26/20 11 Mohr's Circle is a mapping of the normal and shear stress acting on a plane at a point in real space to the coordinates of a point in the - plane. All the points associated with the stresses on planes at a single point lie on a circle x + y and  = 0. The radius of the circle is equal to the centered at avg = 2 maximum in-plane shearing stress. 2  −  y  R =  P =  x  +  xy2 2   Mohr's Circle can best be used as a road map relating various planes and their stresses at the point. Rotation in real space from one plane to another results in a corresponding movement around the circle in the same direction, but twice as far. The coordinates of the new point represent the stresses acting on the new plane. The two points at which the circle crosses the horizontal axis represent the two principal stress planes and the points at the top and bottom of the circle the two maximum in-plane shearing stress planes. The principal stress are then given by 1,2 = avg ± R, where R = P. Sign convention for the normal stress is the usual positive to the right and negative to the left. Shear stresses are best treated by considering which way the shear stress on a given plane is trying to twist the element, clockwise twist is plotted in the upper half of the - plane and counter-clockwise in the lower half of the - plane. The sign information works both ways since there is a unique one-to-one mapping. b. Strains Transformation Equations The analysis is based on a plane strain state in which all strains in the zdirection are zero. The analysis can also be used for a plane stress state with one minor modification. A material can not have both plane stress and plane strain states at the same time. The relationship between the strains at a point measured relative to a set of axes x-y and a set x'-y' which have the same origin but are rotated counterclockwise from the original axes by an angle  are given by x + y x −  y  Cos(2 ) + xy Sin(2  ) + 2 2 2 for the normal strains and by x  =  xy x −  y  Sin( 2) + xy Cos(2 ) =− 2 2 2 for the shearing strain. Note the similarity of form between these equations and the stress transformation equations. 08/26/20 12 Principal Strains and Maximum Shearing Strain As with the stresses there are maximum and minimum (principal) values of the normal strains for particular orientations at the point and maximum shearing strains. The principal strains are given by 2 x + y 2   x −  y    xy     +   2  2   2  and the maximum shearing strain is given by 1, 2 = 2 2   −  y    xy   max =   x  +  2 2   2   The orientation of the larger principal strain to the positive x-direction is given by    xy       2   1 P = Tan −1     x −  y   2    2    The direction of the smaller principal strain is perpendicular to the first. The directions involved with the maximum shearing strain are the two directions at 45° to both of the principal directions. Mohr's Circle for Strain A Mohr's Circle mapping between the strains acting with respect to a set of x-y axes at a point and a point in the strain plane can be made. The same γ rules apply as for the stress circle with  replacing  and 2 replacing . This makes the radius of the circle equal to half the in-plane maximum shearing strain. 08/26/20