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Jump to first page 1 Normal stress = Chapter 2 Mechanics of Materials Example: Estimate the normal stress on a shin bone ( 脛骨 ) ATensile stress (+) Compressive.

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Presentation on theme: "Jump to first page 1 Normal stress = Chapter 2 Mechanics of Materials Example: Estimate the normal stress on a shin bone ( 脛骨 ) ATensile stress (+) Compressive."— Presentation transcript:

1 Jump to first page 1 Normal stress = Chapter 2 Mechanics of Materials Example: Estimate the normal stress on a shin bone ( 脛骨 ) ATensile stress (+) Compressive stress (-) At a point:

2 Jump to first page 2 Shear stress ( 切應力 ) =  = F tangential to the area / A A At a point,

3 Jump to first page 3 Normal strain ( 正應變 )  = fractional change of length= x F F fixed Shear strain (?) = deformation under shear stress = l x

4 Jump to first page 4 Stress-strain curve   o Yield pt. Work hardening break Elastic deformation Hooke's law: In elastic region,   , or  /  = E E is a constant, named as Young’s modulus or modulus of elasticity Similarly, in elastic region,  /  = G, where G is a constant, named as shear modulus or modulus of rigidity.

5 Jump to first page 5 Exercise set 2 (Problem 3) Find the total extension of the bar. X 15mm W 5mm 1.2m 0.6m o dx Width of a cross-sectional element at x: Stress in this element : Strain of this element: The extension of this element : The total extension of the whole bar is : = 2.13 x 10 -4 m

6 Jump to first page 6 Bulk modulus 

7 Jump to first page 7 Poisson's ratio : For a homogeneous isotropic material n normal strain : lateral strain : Poisson's ratio : value of :0.2 - 0.5 FF x d

8 Jump to first page 8 Double index notation for stress and strain 1 st index: surface, 2 nd index: force For normal stress components : x  xx, y  yy, z  zz,  x   xx y z  zx  zy yzyz  xz  xy  yx

9 Jump to first page 9 y z Joint effect of three normal stress components

10 Jump to first page 10 Symmetry of shear stress components Take moment about the z axis, total torque = 0, (  xy  y  z )  x = (  yx  x  z )  y, hence,  xy =  yx. Similarly,  yz =  zy and  xz =  zx z y x  xy  yx xx yy zz

11 Jump to first page 11 Define pure rotation angle  rot and pure shear strain, such that the angular displacements of the two surfaces are:  1 =  rot +  def and  2 =  rot -  def. Hence,  rot = (  1 +  2 )/2 and  def = (  1 -  2 )/2 Original shear strain is “simple” strain = There is no real deformation during pure rotation, but “simple” strain  0. xx  2 = -  Example:  1 = 0 and  2 = - , so  def = (0+  )/2 =  /2 and  rot = (0-  )/2 = -  /2 Pure shear strain is  /2 xx dydy  rot 22  def

12 Jump to first page 12 Example: Show that Proof:  xx =  yy =  zz = , hence 3  =  xx +  yy +  zz = (1-2v)(  xx +  yy +  zz )/E  xx =  yy =  zz = -  p (compressive stress) For hydrostatic pressure l l l

13 Jump to first page 13 Point C moves further along x- and y-direction by distances of AD(  /2) and AD(  /2) respectively.  nn = [(AD   /2) 2 + (AD   /2) 2 ] 1/2 / [(AD) 2 + (AD) 2 ] 1/2 =  /2 True shear strain:  yx =  /2 Therefore, the normal component of strain is equal to the shear component of strain:  nn =  yx and  nn =  /2 Example : Show that  nn =  /2 A C’ C D D’

14 Jump to first page 14  yx (lW) sin 45 o x2 = 2 (l cos 45 o ) W  nn Example : Show that  nn =  nn /(2G) Consider equilibrium along n-direction: Therefore  yx =  nn From definition :  =  xy /G =  nn /G = 2  nn l l nn  yx  xy 2 l cos 45 o

15 Jump to first page 15  nn -  nn  xx =  xx /E -  yy /E- v  zz /E Set  xx =  nn = -  yy,  zz = 0,  xx =  nn  nn = (1+ )  nn /E =  nn /2G (previous example) Example : Show

16 Jump to first page 16 Ex. 12 kN forces are applied to the top & bottom of a cube (20 mm edges), E = 60 GPa, = 0.3. Find (i) the force exerted by the walls, (ii)  yy z y 12kN x (i)  xx = 0,  yy = 0 and  zz = -12  10 3 N/(20  10 -3 m) 2 = 3  10 7 Pa  xx = (  xx - v  yy - v  zz ) /E 0 = [  xx - 0 – 0.3  (- 3  10 7 )]/60  10 9   xx = -9  10 6 Pa (compressive) Force = A  xx = (20  10 -3 m) 2  (-9  10 6 Pa) = -3.6  10 3 N (ii)  yy = (  yy - v  zz - v  xx ) /E = [0 – 0.3  (- 3  10 7 ) – 0.3  (- 9  10 6 )]/60  10 9 = 1.95  10 -4

17 Jump to first page 17 Elastic Strain Energy F The energy stored in a small volume:   The energy stored :   Energy density in the material : e=extension x

18 Jump to first page 18 Similarly for shear strain : F


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