MECHANICS OF MATERIALS REVIEW

Transcription

1 MCHANICS OF MATRIALS RVIW Notation: - normal stress (psi or Pa) - shear stress (psi or Pa) - normal strain (in/in or m/m) - shearing strain (in/in or m/m) I - area moment of inertia (in 4 or m 4 ) J - polar area moment of inertia (in 4 or m 4 ) N - revolutions per minute - modulus of elasticit (psi or Pa) G - modulus of rigidit (psi or Pa) - Poisson s ratio α - coefficient of thermal epansion (/ F or / C) M - bending moment in beams T - torque in shafts T - temperature change ( F or C) hp - horsepower ( hp 550 ft-lb/sec) failure load F.S. - factor of safet allowed load t α T - thermal strain Section : 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 (): changes the angle between lines within the material c. Average Shear Stress in Fasteners V A V Shear force on pin A Cross sectional area of pin Single Shear V P Double Shear V P d. Bearing Stress in Fasteners P dt P force d diameter of fastener t thickness of part 07/6/3

2 Section : Aial Loading a. Aial relationships If the line of action of the load, P, passes through the centroid of the resisting cross-section: aial stress δ aial strain L If the material is also linear, then: Uniaial Hooke' s Law: Where is the modulus of elasticit for the material. The relationship between aial loading and deformation becomes: b. Staticall Determinate Members Static quilibrium ΣF 0 -F F F 3 F 4 Internal Forces P AB F (Tension) P BC F F (Tension) P CD F F F 3 (Tension) Deformation P A aial deformation : δ δ AD δ AB δ BC δ CD P ABL AB A AB P BCL BC AB A BC P CDL CD BC A CD CD Since the P s were assumed in tension, negative values will indicate compression and contraction for the deformation rather than elongation. Thermal Deformation δ thermal AB α AB L AB T P L A Thermal deformation ma be added to an mechanical deformation caused b internal forces acting on the material to obtain a total deformation. 07/6/3

3 3 c. Staticall Indeterminate Members. After the rigid bearing plate contacts the lower section: quilibrium: ΣF 0 F A F B P F A F B P displacement of bearing plate Section A: A δ force δ thermal P AL A A A α A L A T A Section B: (assumes that the bearing plate contacts B) B δ force - δ thermal gap P BL B A B B - α B L B T gap Deformation relationship P A L A A A A α A L A T P BL B A B B - α B L B T gap. After horizontal bar contacts post: quilibrium: ΣM 0 (a)p A ((b)p B (bc)p Displacement relationship: A a B b Link A: A δ force δ thermal P AL A A A α A L A T A Post B: (assumes bar contacts B) B δ force - δ thermal gap P BL B A B - α B L B T gap B Deformation relationship P L P L A A B B α A LA T α A LA T gap AA A AB B a b Link A is assumed in tension and post B in compression. 07/6/3

4 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 b: shear stress : T ρ J The maimum shear stress in the shaft is on the outer surface independent of whether the shaft is solid or hollow and is given b: ma shear stress : ma T r J o The polar area moment of inertia is: 4 π ro solid section : J π 4 4 hollow section : J ( ro ri ) The calculated stresses act on the element as shown. The deformation is measured b the angle of twist (θ) of one end relative to the other and is giver b: angle of twist : θ T L J G where G is the modulus of rigidit for the material and L is the length of shaft. The shaft also has maimum and minimum normal stresses acting on a element rotated 45 from the element for which the shear stress was calculated. The maimum tensile and compressive stresses are related to the shear stress b: ten. comp. ma T r o J 07/6/3

5 5 Section 4: Beams a. Fleural or Bending Stress If the loads on a beam act in its plane of smmetr and the beam is linear elastic, the bending stresses acting normal to the cross section var linearl with the distance from the neutral ais (N.A.) and are given b: M bending stresses : I NA In the absence of aial loads: neutral ais centrodial ais In the sketch the cross-section is shown rectangular. However, the cross section, in general, can be circular, triangular, etc. The properties of man structural sections such as T-, I-, H-sections can be found in handbooks. If the section is not standard, ou must be prepared to determine the centrodial location as well as the value of I NA. The maimum bending stress occurs at the location in the beam where (M) is maimum. The section modulus provides a single parameter for design purposes. M ma ma imum bending stress ma I S S section modulus c where c is the maimum distance of material from the neutral ais. b. Shearing Stresses in Beams The transverse and longitudinal horizontal shearing stress in a beam is given b: VQ I NAt where Q is the first moment of the shaded area about the neutral ais 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). The maimum shearing stress will occur where Q t is maimum. Q is alwas maimum at the neutral surface. However, Q t the neutral surface. Check all possibilities. ma or ma not be maimum at 07/6/3

6 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 b: 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 b: F V fs, where F V 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 an cross-section of the beam will have both shear and bending acting on it. This results in a general stress element as shown, where bending M I ; and shear VQ It. Failure is most likel to occur on a cross-section where V or M are maimum. On the cross-section failure due to pure bending ma occur at the top or bottom of the cross-section, and due to pure shear ma occur at the neutral ais. Wideflange or other non-uniform cross-sections ma have principal stresses or maimum shearing stresses at the web-flange intersection or other points of change in cross section width that eceed other stresses on the cross-section. d. Supporting Beam Topics. Maimum shear and bending moment values are found most easil and reliabl using the Shear and Bending Moment Diagrams developed in Statics. The centroidal location can be determined b first moments about an ais parallel to the bending moment ais. A A i i i where A is the entire area of the cross-section and the A i are the areas of subfigures making up the cross-section. ȳ and the i 's are the perpendicular distance from the reference ais 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 (I i ) about the centroid are known then the area moment of inertia (I NA ) for the cross-section can be determined from: I where the I i are the area moments of inertia of the NA ( I i Ai d i ) i individual areas about their own centrodial ais and d is the perpendicular distance between the area centrodial ais and the neutral ais of the cross-section. 07/6/3

7 7 e. Beam Deflections Two Integration Method The deflection of straight beams is determined from the equation: I ( ) M ( ) Here () 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, and M() is the bending moment as a function of position along the beam. Integration of this equation once ields the equation of the slope as a function of position along the beam: I ( ) M ( ) d C A second integration ields the deflection or elastic curve equation: I ( ) [ M ( ) d] d C C The two integration constants C and C are evaluated using the boundar conditions imposed on the slope and deflection b the supports. The majorit of beam loading requires that the bending moment be defined using more than one analtic function. ach 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. ach set of equations has its own pair of integration constants. The additional boundar conditions come from requiring that the slope and deflection given b the equations on both sides of a boundar between changes in bending moment functions give the same value when evaluated at the boundar. L ( at boundar) R ( at boundar) at boundar at boundar L ( ) ( ) R Superposition Method The solutions for these equations for man different tpes of supports and loads are given in man 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 etensive the available beam tables are. It must be kept in mind that the table entr must be able to eactl match the portion of the load being represented using onl a scaling factor and/or mirror imaging. Loads in the tables ma have either positive or negative values. 07/6/3

8 8 Section 5: Thin-Walled Pressure Vessels Thin-walled pressure vessels are defined as having the ratio t/r 0., where t is the wall thickness and r is the internal radius of either the sphere or clinder. The pressure, p, is the gage pressure and the analsis is onl safe for positive internal pressures. The analsis assumes that the in-plane stresses are uniform across the thickness of the wall. The radial stress is zero on the eterior surface and equal to p on the interior surface. a. Sphere pr t ; (at an point and in an direction) pr pr t t On the inside surface 3 -p MAX pr p t ( ) On the outside surface r ( ) For an - coordinate sstem in the plane of the surface Clinder a aial; h hoop directions h pr t a pr t pr pr a t t h ( ) On the inside surface 3 -p MAX pr p t On the outside surface r ( ) For an - coordinate sstem in the plane of the surface, usuall the aial and hoop directions. 07/6/3

9 9 Section 6: Column Buckling Columns are long slender members under compressive aial loading. Column buckling is a stabilit problem, which means failure can occur without the material reaching the ield 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 π I π slender columns (L'/r > 00 for steel) are Pcr cr ( L ) given b uler's Buckling quation: L L' is the effective length of the column and r depends on the tpe 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 gration for the cross section r (I/A) / The critical load for intermediate columns can be found in various code handbooks. These also give the range of slenderness ratios for which the are valid. Short columns are treated using the ordinar aial loading theor. Section 7: Plane Transformations a. Stresses pinned pinned fied fied fied free fied pinned L L L L L L L 0.7L Transformation quations It is assumed that all the stresses in one direction are zero. The coordinate aes are orientated to place the z-ais in that direction. This situation is common in engineering applications. A free surface is the classic eample. The stresses representing the state of stress at a point are different when measured with respect to two different coordinate sstems that are rotated with respect to each other. If the first sstem is labeled then the '' is rotated counter-clockwise b an angle θ. The primed stresses ma be determined from the unprimed b the equations: ( θ ) Cos( θ ) Sin( θ ) ( θ ) Sin( θ ) Cos( θ ) ' ' (θ 90 ) 07/6/3

10 0 Principal Stress and Maimum Shearing Stress There will alwas be a maimum and minimum stress value, referred to as the principal stresses, occurring at some orientation. There will also be a maimum shearing stress that occurs on two different planes. The values of the principal stresses are given b: ( ), ± The plus sign is used for the larger and the minus sign for the smaller. The value of the maimum shearing stress is given b: ( ) ma ± The orientation of the plane relative to the plane is given b: θ P Tan θ P is the counter-clockwise angle from the plane to the plane. The two principal planes at perpendicular to each other and the two maimum shearing stress planes are at 45 to either of the principal planes. Mohr's Circle for Plane Stress 07/6/3

11 b. Strains 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 centered at avg and 0. The radius of the circle is equal to the maimum in-plane shearing stress. R P 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 ais represent the two principal stress planes and the points at the top and bottom of the circle the two maimum in-plane shearing stress planes. The principal stress are then given b, 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 b considering which wa the shear stress on a given plane is tring to twist the element, clockwise twist is plotted in the upper half of the - plane and counter-clockwise in the lower half of the -ϑ plane. The sign information works both was since there is a unique one-to-one mapping. Transformation quations The analsis is based on a plane strain state in which all strains in the z- direction are zero. The analsis 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 aes - and a set '-' which have the same origin but are rotated counterclockwise from the original aes b an angle θ are given b Cos ( θ) Sin ( θ ) for the normal strains and b Sin ( θ) Cos ( θ ) for the shearing strain. Note the similarit of form between these equations and the stress transformation equations. 07/6/3

12 Principal Strains and Maimum Shearing Strain As with the stresses there are maimum and minimum (principal) values of the normal strains for particular orientations at the point and maimum shearing strains. The principal strains are given b, ± and the maimum shearing strain is given b ma ± The orientation of the larger principal strain to the positive -direction is given b θ P Tan The direction of the smaller principal strain is perpendicular to the first. The directions involved with the maimum 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 - aes at a point and a point in the strain plane can be made. The same rules appl as for the stress circle with replacing and replacing. This makes the radius of the circle equal to half the in-plane maimum shearing strain. 07/6/3

13 3 Sign convention for the shear strain is based on which wa that ais has to twist to have the right angle close for a positive shear strain and open for a negative shear strain. The circle is centered at avg and 0, with a radius R ma. As with the stresses, the principal strains are located where the circle crosses the horizontal ais. Maimum shearing strains are located at the top and bottom of the circle. Strain Rosettes Determination of the strain state on a surface which is assumed to be in a state of plane stress involves measuring three independent strain gages and solving the set of three equations for the unknowns:,, and. Cos θ Sin θ Cosθ a b c Cos θ a b Cos θ c Sin Sin θ θ a b c Cosθ b c a Sinθ Sinθ Cosθ Sinθ b c a 07/6/3

14 4 Two common rosette configurations simplif the equation set considerabl ( ) ( ) Section 8: Material Properties a. Poisson's Ratio When a material is stretched in one direction it contracts in the lateral directions. The resulting longitudinal and lateral strains occur in a fied ratio known as Poisson's ratio. The value of Poisson's ratio for a given material ma be determined from a simple tension test as - lateral longitudinal. The minus sign recognizes that the two strains alwas have opposite signs. This simple definition can be used to calculate the value of Poisson s ratio onl for a uniaial stress state with the material still in the linear region where. In multidimensional stress states both strains are effected b stress induced strains in the other direction. 07/6/3

15 5 07/6/3 b. Generalized Hooke's Law Three dimensional stress state. These relationships are valid within the linear region of the materials stress-strain response. G is the modulus of rigidit (shearing modulus of elasticit) G,, and are related b the formula: G ( ) Plane stress state: z z z 0 ( ) [ ] ( )( ) ( ) ( ) [ ] ( ) [ ] ( )( ) ( ) ( ) [ ] ( ) [ ] ( )( ) ( ) ( ) [ ] z z z z z z z z z z z z z z z z G G G G G G ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) z z G G 0

16 6 RVIW PROBLMS. An aluminum bar having a constant cross sectional area of 0.5 in carries the aial loads applied at the positions shown. Find the deformation of the bar. a in. b in. c in. d in. e. None of these.. A steel rod with a cross sectional area of 0.5 in is stretched between two rigid walls. The temperature coefficient of strain is in./in./ Ε and is psi. If the tensile load is 000 lb. at 80 F, find the tensile load at 0 F. a lb. b lb. c lb. d lb. e. 9,600 lb. 3. The composite bar shown is firml attached to unielding supports at the ends and is subjected to the aial load P shown. If the aluminum is stressed to 0,000 psi, find the stress in the steel. a. 000 psi b. 000 psi c psi d. 0,000 psi e. 0,000 psi 07/6/3

17 7 4. Find the length L necessar to make the total angle of twist between the ends of the shaft equal zero. a. 3 ft. b. 4 ft. c. 6 ft. d. 9 ft. e. ft. 5. Determine the shearing stress at points A and B which are at the inside and outside surfaces of the hollow shaft. Assume elastic behavior. 6. A hollow aluminum shaft and a solid steel shaft are rigidl connected at each end. This compound shaft is then loaded as shown. Determine the maimum shearing stress in each material and the angle of twist of the free end. Assume elastic behavior. G aluminum psi G steel 0 6 psi 07/6/3

18 8 7. Determine the maimum bending moment in the beam. a ft-lb. b ft-lb. c. 700 ft-lb. d. 800 ft-lb. e ft-lb. 8. Find the maimum transverse shearing force in the beam shown. a. 450 lb. b. 800 lb. c. 50 lb. d lb. e lb. 9. B means of strain gages, the fleural stresses are found to be,000 psi at A and 4000 psi at B. Assuming the elastic limit of the material has not been eceeded; find the fleural stress at the bottom of the beam. a psi. b psi. c psi. d. 0,000 psi. e.,000 psi. 0. For the cast iron beam shown, the maimum permissible compressive stress is,000 psi and the maimum permissible tensile stress is 5000 psi. Find the maimum safe load P that can be applied to the beam as shown. a. 0 lb. b. 333 lb. c. 50 lb. d lb. e lb. 07/6/3

19 9. A -inch, 35-lb I-beam 30 ft. long is supported at 5 ft. from each end and carries a uniform distributed load of 600 lbs per ft. (which includes its own weight). Determine the maimum fleural stress in the beam.. Find the maimum vertical shearing force which ma be applied to a bo beam having the cross section shown without eceeding a horizontal shearing stress of 500 psi. a lb. b lb. c lb. d. 630 lb. e lb. 3. Find the reaction at the right end of the beam shown. a. wl/8 b. wl/4 c. 3wL/8 4. Two beams, simpl supported at their ends, jointl support a load P 3500 lb. applied to the upper 6-ft. beam at its midpoint. The beams are identical ecept for length and cross at their midpoints. find the load carried b the lower 9-ft. beam. a. 700 lb. b. 800 lb. c. 000 lb. d. 750 lb. e. 700 lb. 07/6/3

20 0 5. Determine the deflection at the end of this beam. 6. The critical uler load for the pin-ended slender column restrained at the midpoint as shown in Fig. A is 000 lb. What is the critical uler load for the same column with the midpoint restraint removed as shown in Fig. B. a. 50 lb. b. 500 lb. c. 750 lb. d. 000 lb. e lb. 7. A rectangular bar is loaded as shown. Find the maimum tensile stress developed over section A-A. 07/6/3

21 8. For stress conditions on the element shown, find the principal stresses and the plane on which the maimum principal stress acts. 9. A circular shaft of brittle material subjected to torsion fractures along a 45 angle. Failure is due to what kind of stress? a. Shearing stress. b. Compressive stress. c. Tensile stress. d. Combined stress. e. None of these. 0. Which has the higher shear stress for a given elastic torque? a. a one-inch diameter rod, or b. a two-inch diameter rod.. Identical rods of aluminum and steel are each subjected to the same elastic torque. Which rod will have the higher shear stress? a. steel b. aluminum c. both have the same stress. If G represents the modulus of rigidit (or shear modulus of elasticit), is the modulus of elasticit, and is Poisson s ratio, which of the following statements is true for an homogeneous material? a. G is independent of. b. G is 0.4. c. G is 0.5 d. G depends upon both and. e. None of these. 07/6/3

22 . c. d 3. e 4. d 5. B 5,333.3 psi A,666 psi ANSWRS IT RVIW MCHANICS OF MATRIALS 6. S 8,58 psi A 54,35 psi θ.086 rad 7. e 8. c 9. b 0. d. ma 9,00 psi. d 3. c 4. b in. 6. a 7. 8,000 psi 8. ma 605 psi n min psi 9. c 0. a. c. d 07/6/3