MEE 320 Chapter 2, Spring 2018

MEE 320: Strength of Materials Chapter 2 Stress and Strain Axial loading MEE 320: Strength of Materials Stress & Strain: Axial Loading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate. Determination of the stress distribution within a member also requires consideration of deformations in the member. This part is concerned with deformation of a structural member under axial loading. Later, we will deal with torsional and pure bending loads. 2 MEE 320: Strength of Materials Normal Strain A P L      stress normal strain normal 3 Normal Strain: in a rod under axial loading it is defined as the deformation per unit length of that rod If a load P is applied to end C of a rod, the rod elongates, below is a plot of P versus the elongation δ MEE 320: Strength of Materials Normal Strain dx d x x          0 lim 4 When the stress does not have a constant value through the rod, the strain is defined at a given point. For example assume the case when the area A of the member is not uniform. Then the stress at a given point is: A P  Normal Strain: in this case is defined at point Q as: MEE 320: Strength of Materials Stress-Strain Test Machine used to test tensile test specimens Loaded 5 Test specimen Unloaded MEE 320: Strength of Materials Ductile Materials 6 Ductile materials: characterized by their ability to deform plastically before rupture. Ex: steel, aluminum, copper σY : yield strength, stress at which yield is initiated, elastic plastic transition σU : ultimate strength, correspond to the maximum load applied to the specimen σB : Breaking strength, correspond to rupture MEE 320: Strength of Materials Ductile Materials 7 Yield strength σY : is defined from the stress strain curve by the offset method (є = 0.002 or 0.2%) Measure of ductility: 100 % 100 % 0 0 0 0       A A A RA L L L EL B B Percent elongation Percent area reduction L0 initial length LB length at rupture A0 initial area AB area at rupture MEE 320: Strength of Materials Brittle Materials 8 Brittle materials: characterized by the fact that rupture occurs without any noticeable prior change in the rate of elongation. Ex: iron, glass σU = σB MEE 320: Strength of Materials True stress True strain 9 Engineering Strain Engineering Stress True stress versus true strain for a typical ductile material 0 0 A P L      A P L L t t     0 ln True Strain True Stress MEE 320: Strength of Materials Hooke’s Law: Modulus of Elasticity   E  Strength is affected by alloying, heat treating, and manufacturing processes but stiffness (Modulus of Elasticity) is not. Stress-strain diagrams for iron and different grades of steel. 10 Modulus of Elasticity or Young’s Modulus is defined below the yield stress as the slop of the engineering stress strain line. E : Modulus of Elasticity MEE 320: Strength of Materials Elastic vs. Plastic If the strain disappears when the stress is removed, the material is said to behave elastically. When the strain does not return to zero after the stress is removed, the material is said to behave plastically. The largest stress for which this occurs is called the elastic limit. 11 If the member is loaded again after being loaded and unloaded, the new loading curve will closely follow the earlier unloading curve. MEE 320: Strength of Materials Fatigue When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles. A member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles. 12 Fatigue is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading in the elastic range of the material. The fatigue properties are usually shown in a Stress versus Life (number of cycles to failure) diagram or S-N diagram. MEE 320: Strength of Materials Deformations Under Axial Loading AE L P E E          Assume a member subjected to a centric axial load P First calculate the normal stress in the member Then from Hooke’s Law calculate the strain: The deformation is calculated from the strain AE  PL  With variations in loading, cross-section or material properties,   i i i i i E A PL  13 A  P MEE 320: Strength of Materials Example 1 14 2-3: An 18 m long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E = 200 GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire. MEE 320: Strength of Materials Example 2 15 2-6: A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN. Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the smallest diameter rod that should be used, (b) the corresponding maximum length of the rod. MEE 320: Strength of Materials Example 3 16 2-19: both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 KN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B. MEE 320: Strength of Materials Example 3 17 MEE 320: Strength of Materials Example 4 18 2-27: Each of the links AB and CD is made of aluminum (E = 75 GPa) and has a cross-sectional area of 258 mm2. Knowing that they support the rigid member BC, determine the deflection of point E. MEE 320: Strength of Materials Example 4 19 MEE 320: Strength of Materials Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. 0 2 1       Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. 20 AE L P 1 1 1  AE L P 2 2 2  MEE 320: Strength of Materials Example 5 21 2-37: An axial force of 200 kN is applied to the assembly shown by means of rigid end plates. Determine (a) the normal stress in the aluminum shell, (b) the corresponding deformation of the assembly. MEE 320: Strength of Materials Example 5 22 MEE 320: Strength of Materials Example 6 23 2-44: The rigid bar AD is supported by two steel wires of 1.5 mm diameter (E = 200 GPa) and a pin and bracket at A. Knowing that the wires were initially taught, determine (a) the additional tension in each wire when a 1.0 kN load P is applied at D, (b) the corresponding deflection of point D. MEE 320: Strength of Materials Example 6 24 MEE 320: Strength of Materials Thermal Strain Consider a homogeneous rod AB of uniform cross section, which rests freely on a smooth horizontal surface. TL T   25 When the temperature is raised by ΔT, the rod elongates by an amount δT which is proportional to both ΔT and length of the rod L. Let α be the coefficient of thermal expansion then T T   The thermal strain associated with the deformation is In this case no stress is associated with the thermal strain MEE 320: Strength of Materials Thermal Stresses A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports.   AE PL T L P T       Treat the additional support as redundant and apply the principle of superposition. 0 0       AE L PL T P T     The thermal deformation and the deformation from the redundant support must be compatible. T E A P T AE P          26 MEE 320: Strength of Materials Example 7 27 2-52: A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (Eb =105 GPa, αb = 20.9×10-6/oC) and portion BC is made of aluminum (Ea = 72 GPa, αa = 23.9×10-6/oC). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B. MEE 320: Strength of Materials Example 7 28 MEE 320: Strength of Materials Example 8 29 2-58: Knowing that a 0.5 mm gap exists when the temperature is 24°C, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to −75 MPa, (b) the corresponding exact length of the aluminum bar. MEE 320: Strength of Materials Example 8 30 MEE 320: Strength of Materials Poisson’s Ratio For a slender bar subjected to axial loading: 0   z y x x E     The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),  z 0 y   Poisson’s ratio is defined as x z x y         strain axial strain lateral 31 E x x z y        MEE 320: Strength of Materials Generalized Hooke’s Law For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress 2) deformations are small Effect of alone 32 x  E E E x z x y x x            Calculate the effect of , and separately and then add the results x y z MEE 320: Strength of Materials Generalized Hooke’s Law 33 E E E y z y y y x             E E E z z z y z x             Effect of y Effect of z  E E E E E E E E E z y x z z y x y z y x x                         Principle of Superposition MEE 320: Strength of Materials Shearing Strain A cubic element subjected to a shear stress τxy will deform into a rhomboid. The corresponding shear strain γxy is quantified in terms of the change in angle between the sides,   xy xy f    A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains, zx zx yz yz xy xy G G G          where G is the modulus of rigidity or shear modulus. 34 MEE 320: Strength of Materials Relation Among E, , and G Young’s modulus: E  2G 1 E Relation between E, ν and G 35 Poisson ratio: ν Modulus of rigidity: G x x E   x z y      xy Gxy   MEE 320: Strength of Materials Example 9 36 2-61: A 20 mm diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P = 6 kN. Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are observed in a 150 mm length, determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio for the material. MEE 320: Strength of Materials Example 10 37 2-68: A fabric used in air-inflated structures is subjected to a biaxial loading that results in normal stresses σx = 120 MPa and σz = 160 MPa. Knowing that the properties of the fabric can be approximated as E = 87 GPa and ν = 0.34, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC. MEE 320: Strength of Materials Example 10 38 MEE 320: Strength of Materials Example 11 39 2-79: An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 10 mm when a 22 kN lateral load is applied as shown. Knowing that the maximum allowable shearing stress is 420 kPa, determine (a) the smallest allowable dimension b, (b) the smallest required thickness a. MEE 320: Strength of Materials Saint-Venant’s Principle We assumed so far that in an axially loaded member, the normal stresses are uniformly distributed in any section perpendicular to the axis of the member. Loads transmitted through rigid plates result in uniform distribution of stress and strain. If the loads are concentrated, the elements in the immediate vicinity of the points of application of the loads are subjected to very large stresses, while other elements near the ends of the member are unaffected by the loading 40 MEE 320: Strength of Materials Saint-Venant’s Principle Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points. (Starting at a distance equal to the width of the element from either end) Stress and strain distributions become uniform at a relatively short distance from the load application points. 41 MEE 320: Strength of Materials Stress Concentration Discontinuities of cross section may result in high localized or concentrated stresses. ave max   K 42 Circular Hole Stress concentration factor A P ave MEE 320: Strength of Materials Stress Concentration Discontinuities of cross section may result in high localized or concentrated stresses. 43 Fillet A P ave  ave max   K Stress concentration factor MEE 320: Strength of Materials Example 12 44 2-98: For P = 100 kN, determine the minimum plate thickness t required if the allowable stress is 125 MPa.