MEE 320 Chapter 4, Spring 2018

MEE 320: Strength of Materials Chapter 4 Pure Bending MEE 320: Strength of Materials Pure Bending This part is devoted to the study of pure bending and of the stresses and deformations it causes. In the picture below, the athlete holds the barbell with his hands placed at equal distances from the weights. This results in pure bending in the center portion of the bar The normal stresses and the curvature resulting from pure bending will be determined in this part of the course. We will account for stress concentrations where an abrupt change of geometry occurs in a flat bar 2 MEE 320: Strength of Materials Pure Bending Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane In the middle portion CD of the bar, the weight and reactions can be replaced by two equal and opposite moment couples 3 MEE 320: Strength of Materials Bar Clamp The bar clamp exerts forces on two pieces of lumber as they are being glued together. Equal and opposite forces are exerted by the lumber on the clamp. The straight portion of the clamp is subjected to an eccentric loading. Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple 4 Axial load + couple MEE 320: Strength of Materials Cantilever Beam The cantilever beam supports a transverse load P at its free end. Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple 5 Shear force + couple MEE 320: Strength of Materials Principle of Superposition Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. 6 MEE 320: Strength of Materials Symmetric Member in Pure Bending From statics, a couple M consists of two equal and opposite forces. The sum of the components of the forces in any direction is zero. The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment. 7 MEE 320: Strength of Materials Symmetric Member in Pure Bending The sum of the components of the forces in any direction is equal to zero since no external forces are applied 8 The sum of the moments of the elementary forces is equal to the external moments applied The stress distribution in a given cross section cannot be determined from statics alone. It is statically indeterminate and may be obtained only by analyzing the deformations produced in the member. 0 0 0          dA F dA F dA F xz z xy y x x        M dA y M dA z M dA z y M x z x y xy xz x                0 0 MEE 320: Strength of Materials Bending Deformations Beam with a plane of symmetry in pure bending: 9 0 0   xz xy   0 0   xz xy   • Member remains symmetric • Bends uniformly to form a circular arc • Cross-sectional plane passes through arc center and remains planar • All the faces keep 90o angles after deformation: MEE 320: Strength of Materials Bending Deformations The length of top surface decreases and the length of bottom face increases. Hence, a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change. In this scenario: Mz > 0 The surfaces above the neutral surface are under compression, i.e. stresses and strains are negative. The surfaces below the neutral surface are under tension, i.e. stresses and strains are positive. 10 Also σy = σz = τyz = 0 MEE 320: Strength of Materials The strain at a distance y from E: Initial length Final length Deformation Strain Max Strain Strain Due to Bending Neutral axis: Arc DE is the intersection of the neutral surface and the plane of symmetry    y L    11      y y L x        L         y y L L          c m  m x c y     MEE 320: Strength of Materials Stress Due to Bending For a linearly elastic material, For equilibrium, 0 0       dA c y dA F m x x   First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid. m x m x x c y c E y E           12 0 0     y Ay y dA MEE 320: Strength of Materials Stress Due to Bending For static equilibrium,     S M I Mc c I y dA c M dA c y y dA y M m m m m x                       2 13 m x c y     I My x    S M I Mc m    I is the moment of inertia or second moment of the cross section S is the elastic section modulus c I S y dA I    2 MEE 320: Strength of Materials Beam Section Properties The maximum normal stress due to bending, S M I Mc m    A beam section with a larger section modulus will have a lower maximum stress. Consider a rectangular beam cross section, Ah bh h bh c I S 6 1 2 6 1 3 12 1 2     Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending. Structural steel beams are designed to have a large section modulus 14 MEE 320: Strength of Materials Deformation Due to Bending The deformation of the member by the bending moment M is measured by the curvature of the neutral surface c m    1 15 I Mc E m m m      I Mc Ec Ec m 1 1     EI   M 1 MEE 320: Strength of Materials Deformations: Transverse Cross-Section Deformation due to bending moment M is quantified by the curvature of the neutral surface EI   M 1 Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,         y y x z x y       Expansion above the neutral surface and contraction below it cause an in-plane curvature. The curvature of the transverse cross section is called anticlastic curvature      1 16 MEE 320: Strength of Materials Example 1 17 4-1: Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B. MEE 320: Strength of Materials Example 1 18 MEE 320: Strength of Materials Example 2 19 4-11: Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. MEE 320: Strength of Materials Example 2 20 MEE 320: Strength of Materials Example 3 21 4-19: Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied. MEE 320: Strength of Materials Example 3 22 MEE 320: Strength of Materials Example 4 23 4-24: A 60 N.m couple is applied to the steel bar shown. (a) Assuming that the couple is applied about the z axis as shown, determine the maximum stress and the radius of curvature of the bar. (b) Solve part a, assuming that the couple is applied about the y axis. Use E = 200 GPa. MEE 320: Strength of Materials Example 4 24 MEE 320: Strength of Materials Example 5 25 4-31: A W200 × 31.3 rolled-steel beam is subjected to a couple M of moment 45 kN.m. Knowing that E = 200 GPa and v = 0.29, determine (a) the radius of curvature ρ, (b) the radius of curvature ρ’ of a transverse cross section. MEE 320: Strength of Materials Members Made of Several Materials   y x   Consider a composite beam formed from two materials with E1 and E2. The normal strain still varies linearly with the distance y from the neutral axis. The stress in material 1 will be different than that in material 2. 26 Normal Strain Normal Stress       E y E E y E x x 2 2 2 1 1 1       MEE 320: Strength of Materials Members Made of Several Materials The neutral axis does not pass through the section centroid of the composite section. 27 Strain Stress MEE 320: Strength of Materials Members Made of Several Materials Static Equilibrium: The total normal force F resulting from the stress distribution normal to the cross section should equal to zero: 28      1 1 1 1 1 A A E y dA dA F   0 2 1    F F F The total force F1 exerted on the cross sectional area A1 of material 1 is: The total force F2 exerted on the cross sectional area A2 of material 2 is:      2 2 2 2 2 A A E y dA dA F   MEE 320: Strength of Materials Members Made of Several Materials 1 2 E E n  29 This equation can be easily solved as follows: First let The location of the neutral surface is found from the static equilibrium equation: 0 2 1 2 1        A A E y dA E y dA F   0 ) ( 0 2 1 2 1 1 1 1 1               A A A A E y ndA E y dA F nE y dA E y dA F     If you now multiply the dimension of A2 parallel to the neutral surface by n, the force equation can be rewritten as: 0 ' 1 1 2 1        A A E y dA E y dA F   MEE 320: Strength of Materials Members Made of Several Materials 30 Then the equation is simplified to: The integral of F depends now on E1 only. Physically, it defines the summation of the normal forces resulting from the application of the normal stresses on A’ made of material 1. As if you converted the cross-section A2 made of material 2 to another cross-section A2’ made of material 1. 0 ' 1     A E y dA F  The new converted cross section is: ' ' 2 1 A A A   Since it is essential for the distance y of each element from the neutral axis to remain the same, the dimension parallel the neutral axis should change to n×b. MEE 320: Strength of Materials Members Made of Several Materials I My x    31 The stress σx at any point of the transformed member is obtained as  x 1 The stress σ1 at any point in material 1 of the original member is obtained as follows,   nx 2 The neutral surface of the cross section A is simply the centroid of the new converted cross section A’ The stress σ2 at any point in material 2 of the original member is calculated as: Original Area Force Transformed n×Area Force MEE 320: Strength of Materials Members Made of Several Materials The deformation of a composite member is determined using the transformed section as E I M 1 1   32 E1 is the Young’s modulus of material 1 I is the moment of inertia of the transformed section with respect to its neutral axis MEE 320: Strength of Materials Reinforced Concrete Beams Concrete beams subjected to bending moments are reinforced by steel rods. The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load only since concrete is weak in tension. To determine the location of the neutral axis,     0 0 2 2 2 1       n A d n A x x b x n A d x bx s s s In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where n = Es/Ec. The normal stress in the concrete and steel x s x n I My       33 MEE 320: Strength of Materials Stress Concentrations Stress concentrations may occur in the vicinity of : • points where the loads are applied • abrupt changes in cross section I m  K Mc  34 MEE 320: Strength of Materials Example 6 35 4-33: A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis MEE 320: Strength of Materials Example 6 36 MEE 320: Strength of Materials Example 7 37 4-40: A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa) are bonded together to form the composite beam shown. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 35 N.m, determine the maximum stress in (a) the aluminum strip, (b) the copper strip. MEE 320: Strength of Materials Example 7 38 MEE 320: Strength of Materials Example 8 39 4-52: A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 20 GPa for the concrete and 200 GPa for the steel. Using an allowable stress of 9 MPa for the concrete and 140 MPa for the steel, determine the largest allowable positive bending moment in the beam. MEE 320: Strength of Materials Example 8 40 MEE 320: Strength of Materials Example 9 41 4-59: The rectangular beam shown is made of a plastic for which the value of the modulus of elasticity in tension is one-half of its value in compression. For a bending moment M = 600 N.m, determine the maximum (a) tensile stress, (b) compressive stress. MEE 320: Strength of Materials Example 9 42 MEE 320: Strength of Materials Example 10 43 4-65: A couple of moment M = 2 kN.m is to be applied to the end of a steel bar. Determine the maximum stress in the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 10 mm, as shown in Fig. a, (b) if the bar is redesigned by removing the material to the left and right of the dashed lines as shown in Fig. b. MEE 320: Strength of Materials Example 10 44