Stress and Strain(3.8-3.12, 3.14) MAE 316 – Strength of Mechanical Components Y. Zhu Stress and Strain 1 Introduction Stress and Strain 2 MAE 316 is a continuation of MAE 314 (solid mechanics) Review topics Beam theory Columns Pressure vessels Principle stresses New topics Contact Stress Press and shrink fits Fracture mechanics Fatigue Normal Stress (3.9) Stress and Strain 3 Normal Stress (axial loading) Sign Convention σ 0 Tensile (member is in tension) σ 0 Compressive (member is in compression) A F = o Shear Stress (3.9) Stress and Strain 4 Shear stress (transverse loading) “Single” shear “Double” shear A F A P ave = = t A F A F A P ave 2 2 = = = t average shear stress Strain (3.8) Stress and Strain 5 Normal strain (axial loading) Hooke’s Law L o c = c o E = Where E = Modulus of Elasticity (Young’s modulus) Torsion (3.12) Stress and Strain 6 Torsion (circular shaft) Shear strain Shear stress Angle of twist L µu ¸ = T J µ t = TL GJ u = Where G = Shear modulus (Modulus of rigidity) and J = Polar moment of inertia of shaft cross-section T θ θ Stress and Strain Review (3.6) Stress and Strain 7 Beams in pure bending Normal stress only “Plane sections remain plane” Sign convention Positive bending moment: beam bends towards +y direction Negative bending moment: beam bends towards -y direction Right angle Beams in Bending (3.10) Stress and Strain 8 Beams in pure bending Strain Stress µ c y x ÷ = y z y v c c µ = = Where ν = Poisson’s Ratio and ρ = radius of curvature x My I o = ÷ Where I = 2 nd moment of inertia of the cross-section Beam Shear and Bending (3.10-3.11) Stress and Strain 9 Beams (non-uniform bending) Shear and bending moment Shear stress Design of beams for bending w dx dV ÷ = V dx dM = S M I c M max max max = = o allowable Factor of Safety applied = It VQ avg = t Where Q = 1 st moment of the cross-section Combined Stress in Beams Stress and Strain 10 In MAE 314, we calculated stress and strain for each type of load separately (axial, centric, transverse, etc.). When more than one type of load acts on a beam, the combined stress can be found by the superposition of several stress states. Problem 1 Stress and Strain 11 Determine the normal and shearing stresses at points K and H. The radius of the bar is 20 mm. Thin-Walled Pressure Vessels (3.14) Stress and Strain 12 Thin-walled pressure vessels Cylindrical Spherical 1 i t pr t o o = = 2 2 i l pr t o o = = Circumferential “hoop” stress Longitudinal stress 1 2 2 i pr t o o = = Column Buckling • Replace the length L with an “equivalent” or “effective” length L e . • L is the actual length of the beam & L e is the length for use in P CR . EI L P e CR 2 2 t = 2D and 3D Stress (3.6-3.7) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical & Aerospace Engineering Stress and Strain 14 Plane (2D) Stress (3.6) Stress and Strain 15 Consider a state of plane stress: σ z = τ xz = τ yz = 0. Slice cube at an angle φ to the x-axis (new coordinates x’, y’) φ φ φ φ φ φ φ φ φ Plane (2D) Stress (3.6) Stress and Strain 16 Sum forces in x’ direction and y’ direction and use trig identities to formulate equations for transformed stress. ( ) ( ) ' cos 2 sin 2 2 2 x y x y x xy o o o o o ¢ t ¢ + ÷ = + + ( ) ( ) ' cos 2 sin 2 2 2 x y x y y xy o o o o o ¢ t ¢ + ÷ = ÷ ÷ ( ) ( ) ' ' sin 2 cos 2 2 x y x y xy o o t ¢ t ¢ ÷ = ÷ + φ φ φ φ φ Plane (2D) Stress (3.6) Stress and Strain 17 Plotting a Mohr’s Circle, we can also develop equations for principle stress, maximum shearing stress, and the orientations at which they occur. 2 2 min max, 2 2 min max, 2 2 2 xy y x xy y x y x t o o t t o o o o o + | | . | \ | ÷ ± = + | | . | \ | ÷ ± + = ( ) 2 tan 2 xy P x y t ¢ o o = ÷ R R R R ave ave ÷ = = ÷ = + = min max min max t t o o o o ( ) tan 2 2 x y S xy o o ¢ t ÷ = ÷ Problem 2 Stress and Strain 18 For the given state of stress, determine the angles at which the principle stresses occur (principle planes). Also, determine the orientation of maximum shearing stress. Plane (2D) Strain Stress and Strain 19 Mathematically, the transformation of strain is the same as stress transformation with the following substitutions. 2 ¸ t c o ÷ ÷ and 2 2 min max, 2 2 2 | | . | \ | + | | . | \ | ÷ ± + = xy y x y x ¸ c c c c c ( ) tan 2 xy P x y ¸ ¢ c c = ÷ ( ) ( ) ' cos 2 sin 2 2 2 2 x y x y xy x c c c c ¸ c ¢ ¢ + ÷ = + + ( ) ( ) ' cos 2 sin 2 2 2 2 x y x y xy y c c c c ¸ c ¢ ¢ + ÷ = ÷ ÷ ( ) ( ) ( ) ' ' sin 2 cos 2 x y x y xy ¸ c c ¢ ¸ ¢ = ÷ ÷ + 3D Stress (3.7) Stress and Strain 20 Now, there are three possible principal stresses. Also, recall the stress tensor can be expressed in matrix form. ( ( ( ¸ ( ¸ z zy zx yz y yx xz xy x o t t t o t t t o 3D Stress (3.7) Stress and Strain 21 We can solve for the principle stresses (σ 1 , σ 2 , σ 3 using a stress cubic equation. Where i = 1,2,3 and the three constant I 1 , I 2 , and I 3 are expressed as follows. 0 3 2 2 1 3 = ÷ + ÷ I I I i i i o o o 2 2 2 3 2 2 2 2 1 2 xy z xz y yz x xz yz xy z y x xz yz xy z y z x y x z y x I I I t o t o t o t t t o o o t t t o o o o o o o o o ÷ ÷ ÷ + = ÷ ÷ ÷ + + = + + = 3D Stress (3.7) Stress and Strain 22 How do we find the maximum shearing stress? The most visual method is to observe a 3D Mohr's Circle. Rank principle stresses largest to smallest: σ 1 > σ 2 > σ 3 σ 3 σ 1 σ 2 1 3 max 2 o o t ÷ = Problem 3 (3.7) Stress and Strain 23 For the stress state shown below, find the principle stresses and maximum shear stress. ( ( ( ¸ ( ¸ ÷ 0 0 0 0 6 4 0 4 9 Stress tensor Problem 4 (3.7) Stress and Strain 24 Draw Mohr’s Circle for the stress state shown below. ( ( ( ¸ ( ¸ ÷ 25 0 0 0 35 30 0 30 60 Stress tensor Curved Beams (3.18) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical & Aerospace Engineering Stress and Strain 25 Curved Beams (3.18) Stress and Strain 26 Thus far, we have only analyzed stress in straight beams. However, there many situations where curved beams are used. Curved structural beams Hooks Chain links Curved Beams (3.18) Stress and Strain 27 Assumptions Pure bending (no shear and axial forces present – will add these later) Bending occurs in a single plane The cross-section has at least one axis of symmetry What does this mean? σ = -My/I no longer applies Neutral axis and axis of symmetry (centroid) are no longer the same Stress distribution is not linear Curved Beams (3.18) Stress and Strain 28 Where M = bending moment about centroidal axis (positive M puts inner surface in tension) y = distance from neutral axis to point of interest A = cross-section area e = distance from centroidal axis to neutral axis r n = radius of neutral axis Flexure formula for tangential stress: ( ) n My Ae r y o = ÷ Curved Beams (3.18) Stress and Strain 29 If there is also an axial force present, the flexure formula can be written as follows. Table 3-4 in the textbook shows r n formulas for several common cross-section shapes. ( ) n P My A Ae r y o = + ÷ Calculate the tangential stress at A and B on the curved hook shown below if the load P = 90 kN. Problem 5 (3.18) Stress and Strain 30