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Chapter 4 - State of stress and strength hypothese. The following will be discussed in this chapter: State of stress at a point, plane stress, mohr’s circle, special cases of plane stress, stress – strain relations, strength hypotheses.
STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, Giai Phong Str 55, Hai Ba Trung Dist Hanoi, Vietnam CHAPTER State of Stress and Strength Hypothese 1/10/2013 Contents 4.1 State of stress at a point 4.2 Plane Stress 4.3 Mohr’s Circle 4.4 Special cases of plane stress 4.5 Stress – Strain relations 4.6 Strength Hypotheses 1/10/2013 4.1 State of stress at a point • External loads applied to the body => The body is deformed =>The stress is occurred • At a point K on the arbitrary section, there are types of stress: normal stress s and shearing stress t • The state of stress at a point K is a set of all stresses components acting on all sections, which go through this point y z K n x • The most general state of stress at a point may be represented by components, s x ,s y ,s z normal stresses t xy , t yz , t zx shearing stresses (Note: t xy t yx , t yz t zy , t zx t xz ) 1/10/2013 4.1 State of stress at a point • Principal planes: no shear stress acts on • Principal directions: the direction of the principal planes • Principal stresses: the normal stress act on the principal plane • There are three principal planes , which are perpendicular to each other and go through a point • Three principal stresses: s1, s2, s3 with: s1 ≥ s2 ≥ s3 • Types of state of stress: - Simple state of stress: of principal stresses equal to zeros - Plane state of stress: of principal stresses equal to zeros - General state of stress: all principal stresses differ from zeros 1/10/2013 4.2 Plane Stress • Plane Stress – the state of stress in which two faces of the cubic element are free of stress For the illustrated example, the state of stress is defined by s x , s y , t xy and s z t zx t zy • State of plane stress occurs in a thin plate subjected to the forces acting in the mid-plane of the plate y tyx sy txy O sx y sy tyx txy x z x sx 4.2 Plane Stress Sign Convention： • Normal Stress: positive: tension; negative: compression • Shear Stress: positive: the direction associated with its subscripts are plus-plus or minus-minus; negative: the directions are plus-minus or minus-plus 4.2.1 Complementary shear stresses： • The shear stresses with the same subscripts in two orthogonal planes (e.g txy and tyx) are equal y 1/10/2013 4.2 Plane Stress sy u Sign Convention： txy sx u O y sx v x txy Asin 1/10/2013 >0 - counterclockwise； su >0 – pull out t uv - clockwise F su 0 u tuv tyx sy A Acos 4.2.2 Stresses on Inclined Planes： s u A s x A cos2 t xy A cos sin s y A sin t yx A sin cos F v 0 τuv A - τ xy Acos α - σ x Acosαsinα +τ yx Asin2 α +σ y Asinαcosα = 4.2 Plane Stress 4.2.2 Stresses on Inclined Planes： su u s x s y s x s y su cos 2 t xy sin 2 2 txy sx sy sy 1/10/2013 tuv y tyx x s x s y t uv sin 2 t xy cos 2 v - > 0: counterclockwise from the x axis to u axis 4.2 Plane Stress 4.2.3 Principal stresses are maximum and minimum stresses ： By taking the derivative of su to and setting it equal to zero: 2t xy ds u => tg2 p =d sx sy p1, p p 90 p s max, s 1,2(3) 1/10/2013 sx s y s x s y t xy 10 4.5 Stress – Strain relations - Plane stress xy t xy G x s x s y E y s y s x E 1 s s E s s E Normal stress – unit volume change relation V a1a2a3 V1 a1 ( )a2 ( )a3 ( ) V1 V 1 V 2 2 (s1 s s ) (s x s y s z ) E E 1/10/2013 19 4.5 Stress – Strain relations Strain energy s1 1 u us ut s t 2 Principal element: t = => s2 s3 1 u s 1 s 2 s 3 2 s 12 s 22 s 32 2 s 1s s 3s s 1s 2E 1/10/2013 20 4.5 Stress – Strain relations Shape Deformation=> Change Volume s1 s -stb stb = stb s2 -stb s2 s3 s3 -stb stb b a s tb 1/10/2013 (s1 s s ) c 21 4.5 Stress – Strain relations Strain-energy density u Strain-energy density for changes of shape ushape Strain-energy density for changes of volume uvol u = ushape uvol uvol ushape 1/10/2013 2 ( s s s )2 6E 1 2 s s s s s s 6E 22 4.6 Strength Hypothesis - For a bar under tensile loading, one can conclude at which stress failure will occur from the stress – strain diagram To prevent such failure, an allowable stress sallow is introduced and it is postulated that the stresses in the bar must not exceed sallow, i.e: s ≤ sallow - In an arbitrary structural member, a spatial stress state is present and it is necessary to determine the circumstances under which the load carrying capacity is lost and the material starts to fail - There exists no experimental setup which can provide a general answer, hypothesis on the basic of specific experiments are used These so-called strength hypothesis allow us to calculate according to a specific rule, an equivalent stress se from the normal and shear stresses It is assumed that the stress se , when applied to the uniaxial case of a bar, has the same effect regarding failure through plastic flow or fracture as the given spatial stress state in the body under consideration 1/10/2013 23 4.6 Strength Hypothesis - Since the stress state in the body and in a tensile bar are then said to be equivalent, the stress se is called equivalent stress Therefore, if a structural element shall not lose its load carrying capacity, the equivalent stress must not exceed the allowable stress: s e s allow Maximum – normal – stress hypothesis: It is assumed that the material starts to fail when the largest principal stress reaches a critical value Strength condition: s e1 s s allow Maximum – normal – strain hypothesis: This hypothesis is based on the assumption that the failure occurs when the maximum normal strain reaches a critical value Strength condition : s e s s s s allow 1/10/2013 24 4.6 Strength Hypothesis Maximum – shear – stress hypothesis: This hypothesis is based on the assumption that the failure occurs when the maximum shear stress reaches a critical value Strength condition : s e3 s1 s s allow Maximum – distortion – energy hypothesis: Here it is assumed that the material state becomes critical when the energy needed for the “distortion” of a material element reaches a critical value Strength condition : s e s12 s 22 s 32 s1s s1s s 2s s allow Mohr strength hypothesis: Mohr considers that the main factor causing failure is the maximum shear stress, however, the normal stress on the section on which the maximum shear stress is has important effect on failure 1/10/2013 25 4.6 Strength Hypothese Strength condition： s M [s L ] s1 s s [s y ] tuv M K L P su N [s y] O2 s O3 O1 s [ s L] o 1/10/2013 26 Sample Problem 4.1 From the established sign convention, it is seen that 1/10/2013 27 Sample Problem 4.1 1/10/2013 28 Sample Problem 4.2 The state of plane stress state at a point is represented by the element in Fig Determine the principal stresses and the principal directions of this state of plane stress With β =60o 10KN/cm2 β Solution With coordinates xy shown in fig., We have is the angle measured from the x axis to the normal axis u of the inclined plane (counterclockwise) We have: s y 4kN / cm ; 4KN/cm2 u t xy 6kN / cm ; s u 10kN / cm2 ; 1/10/2013 6KN/cm2 150o y β 6KN/cm2 x 4KN/cm2 29 Sample Problem 4.2 From Equations (*) su s x s y s x s y cos 2 t xy sin 2 u y s x 18,928kN / cm2 β 6KN/cm2 • The Principal directions: tg 2 2t xy s x s y x 1 19,4o ; 1 90o 109,4o • The Principal stresses: s 1,2 1/10/2013 sx s y s x s y t xy 4KN/cm2 s1 21,041kN / cm2 s 1,887kN / cm2 30 Homework 1/10/2013 31 Homework 1/10/2013 32 THANK YOU FOR ATTENTION ! 1/10/2013 33 ... u K max uv O sy t t s1 13 4. 4 Special Cases of Plane Stress 4. 4.1 Uniaxial tension 4. 4.2 Pure shear 1/10/2013 14 4 .4 Special Cases of Plane Stress 4. 4.3 Special plane stress t t t s I.. .CHAPTER State of Stress and Strength Hypothese 1/10/2013 Contents 4. 1 State of stress at a point 4. 2 Plane Stress 4. 3 Mohr’s Circle 4. 4 Special cases of plane stress 4. 5 Stress –... s2 ≥ s3 • Types of state of stress: - Simple state of stress: of principal stresses equal to zeros - Plane state of stress: of principal stresses equal to zeros - General state of stress: all