# Strength of materials - State of stress and strength hypothese

STRENGTH OF MATERIALS TRAN MINH TU - University of Civil Engineering, Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam 1/10/2013 1 4 CHAPTER 1/10/2013 State of Stress and Strength Hypothese 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 3 4.1. State of stress at a point 1/10/2013 4 K x y z n   • External loads applied to the body => 33 trang | Chia sẻ: huong20 | Ngày: 21/01/2022 | Lượt xem: 20 | Lượt tải: 0
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; The body is deformed =>The stress is occurred • At a point K on the arbitrary section, there are 2 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 • The most general state of stress at a point may be represented by 6 components, , , , , , , ) normal stresses shearing stresses (Note: x y z xy yz zx xy yx yz zy zx xz s s s t t t t t t t t t   1/10/2013 5 • Principal planes: no shear stress acts on 4.1. State of stress at a point • 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: 2 of 3 principal stresses equal to zeros - Plane state of stress: 1 of 3 principal stresses equal to zeros - General state of stress: all 3 principal stresses differ from zeros 6• 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 .0,, and xy  zyzxzyx ttstss • State of plane stress occurs in a thin plate subjected to the forces acting in the mid-plane of the plate. 4.2. Plane Stress sx txy sy x y z tyx x y sx txy sy O tyx 1/10/2013 7 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. Plane Stress y 4.2.1. Complementary shear stresses： • The shear stresses with the same subscripts in two orthogonal planes (e.g. txy and tyx) are equal 1/10/2013 8 0uF   2 2 cos cos sin sin sin cos 0 u x xy y yx A A A A A s s  t   s  t         su >0 – pull out  t uv - clockwise 2 uv xy x 2 yx y τ A- τ Acos α - σ Acosαsinα +τ Asin α+σ Asinαcosα=0 vF 0  x y sx txy sy O u sy tyx sx v u  A Acos Asin su tuv  txy 4.2. Plane Stress 4.2.2. Stresses on Inclined Planes： Sign Convention：   >0 - counterclockwise； 1/10/2013 9 4.2. Plane Stress tyx sy su tuv  sy sx txy 4.2.2. Stresses on Inclined Planes： x y v u -  > 0: counterclockwise from the x axis to u axis x y x y u xycos sin s s s s s  t      2 2 2 2 x y uv xysin2 cos 2 2 s s t  t     1/10/2013 10 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: xyu p x y 2d 0 => tg2 =- d ts   s s   2 2 1,2(3) 2 2 xy x y x y max, min s s s s s s t           01, 2 90 p p p p        1/10/2013 11 4.2. Plane Stress 4.2.4. Maximum in-plane shear stresses x y 0 s s p xy d 0 => tg2 = => = 45 d 2 s st     t    => An element subjected to maximum shear stresses will be 450 from the position of an element that is subjected to the principal stress 2 2 2 s s t t         xy x y max,min 4.2.5. The first invariant of plane stress The sum of the normal stresses has the same value in each coordinate system s s s s   x y u v const 1/10/2013 12 4.3. Mohr’s Circle - cos2 sin 2 2 2 sin 2 cos2 2 x y x y u xy x y uv xy s  s s  s s    t  s  s t    t  ( )2 ( )2 ( )2 ( )2 2 2 2 2 - 2 2 x y x y u uv xy s  s s  s    s  t   t        ,0 2 x y I s  s      2 2 2 x y xyR s  s    t    Using the transformation relations: Mohr’ Circle Center: Radius: 1/10/2013 13 O B AI s1 sx sy s2 M txy su tuv  u s x s y s y t xy t yx t yx t xy s x  u su tuv tmin tmax u1 u2 R K ` 01 02 Điểm cực Plane Stress  ,y xyM s t ,0 2 x y I s  s      2 2 2 x y xyR s  s    t    4.3. Mohr’s Circle 1/10/2013 14 4.4. Special Cases of Plane Stress 4.4.1. Uniaxial tension 4.4.2. Pure shear 1/10/2013 15 ss t t tt s t s t I smaxsmin max,min 3 2 2 1, 2 2 s s    t    s  s  2 21 ax 3 m 2 2 s  s s    t    t  4.4. Special Cases of Plane Stress 4.4.3. Special plane stress 1/10/2013 16 4.5. Stress – Strain relations 1. Uniaxial stress E x x s   xy E s    xz E s    2. Pure shear G xy xy t   0 zxyz  x y z sx x y z tx y 1/10/2013 17 4.5. Stress – Strain relations    zyx zyx x E EEE sss s  s  s    1 sy x y z sz txy sx 3. General state of stress - The Principle of superposition - Assumption: The normal strain causes only the normal stress. The Shear strain causes only the shear stress 1/10/2013 18 a. Normal stress – normal strain relation b. Shear stress – shear strain relation with E, , G are Young modulus, Poisson ratio, shear modulus, which the relation among them:   1 x x y z E  s  s s       1 y y x z E  s  s s       1 z z x y E  s  s s     xy xy G t   xz xz G t   yz yz G t    2 1 E G       1322 1 sss  E    1233 1 sss  E    3211 1 sss  E 4.5. Stress – Strain relations Generalized Hooke’s law 1/10/2013 19 1 x x y E  s s    1 y y x E  s s     1 1 2 1 E  s s   2 2 1 1 E  s s  xy xy G t   4.5. Stress – Strain relations - Plane stress 321 aaaV  )(a)(a)(aV 3322111 111   321 1     V VV x y z( ) ( ) E E    s s s s s s        1 2 3 1 2 1 2 4. Normal stress – unit volume change relation 1/10/2013 20 4.5. Stress – Strain relations 5. Strain energy s2 s3 s1 u u us t s t    1 1 2 2 Principal element: t = 0 => 332211 2 1 2 1 2 1 sss u    312321 2 3 2 2 2 1 2 2 1 sssssssss  E 1/10/2013 21 s2 s3 s 1 a c s3 -stb s 1 -stb s2 -stb stb b stb stb = Deformation=> Change Shape Volume tb ( )s s s s  1 2 3 1 3 4.5. Stress – Strain relations 1/10/2013 22 shape volu u u= volu E  s s s     21 2 3 1 2 ( ) 6      shapeu E  s s s s s s           2 2 2 1 2 2 3 3 1 1 6 Strain-energy density u Strain-energy density for changes of shape ushape Strain-energy density for changes of volume uvol 4.5. Stress – Strain relations 1/10/2013 23 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 24 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: e allows s 1. Maximum – normal – stress hypothesis: It is assumed that the material starts to fail when the largest principal stress reaches a critical value. Strength condition: 2. 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 : 1 1e allow s s s  2 3 1 2e allow   s s  s s s 1/10/2013 25 3. 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 : 4.6. Strength Hypothesis 3 1 3e allows s s s   4. 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 : 2 2 2 4 1 2 3 1 2 1 3 2 3e allows s s s s s s s s s s       4. 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 26 [s y] su uv o t [s L]O1O2 Strength condition：  ss s s ss  31 ][ ][ y L M O3 s1s3 M K L P N 4.6. Strength Hypothese 1/10/2013 27 Sample Problem 4.1 From the established sign convention, it is seen that 1/10/2013 28 Sample Problem 4.1 1/10/2013 29 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 Solution 210 / ;u kN cms  u 6KN/cm2 4KN/cm2 10KN/cm2 β 6KN/cm2 4KN/cm2 β x y 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: 24 / ;y kN cms  26 / ;xy kN cmt   150o  1/10/2013 30 Sample Problem 4.2 t ssss s 2sin2cos 22 xy yxyx u      218,928 /x kN cms  • The Principal directions: 2 2 xy x y tg t  s s    From Equations (*) u 6KN/cm2 4KN/cm2 β x y  1 2 119,4 ; 90 109,4 o o o       2 2 1,2 2 2 x y x y xy s s s s s t          • The Principal stresses: 2 1 21,041 /kN cms  2 2 1,887 /kN cms  1/10/2013 31 Homework 1/10/2013 32 Homework THANK YOU FOR ATTENTION ! 1/10/2013 33

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