7_stress_transformations1.pdf

CHAPTER

7

MECHANICS OF SOLIDS Transformations of Stress and Strain

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MECHANICS OF SOLIDS Transformations of Stress and Strain Introduction Transformation of Plane Stress Principal Stresses Maximum Shearing Stress Example 7.01 Sample Problem 7.1 Mohr’s Circle for Plane Stress Example 7.02 Sample Problem 7.2 General State of Stress Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress Stresses in Thin-Walled Pressure Vessels

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MECHANICS OF SOLIDS Introduction • The most general state of stress at a point may be represented by 6 components,  x , y , z

normal stresses

 xy ,  yz ,  zx shearing stresses (Note :  xy   yx ,  yz   zy ,  zx   xz )

• Same state of stress is represented by a different set of components if axes are rotated. • The first part of the chapter is concerned with how the components of stress are transformed under a rotation of the coordinate axes.

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MECHANICS OF SOLIDS Introduction • Plane Stress - 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  x ,  y ,  xy and  z   zx   zy  0.

• State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate.

• State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface not subjected to an external force. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Transformation of Plane Stress • Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x, y, and x’ axes.  Fx  0   xA   x A cos  cos   xy A cos sin   y A sin sin   xy A sin  cos

 Fy  0   xyA   x A cos sin   xy A cos  cos   y A sin  cos   xy A sin sin

• The equations may be rewritten to yield  x   y 

 x  y 2  x  y

 xy  

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2

 

 x  y 2

 x  y 2  x  y 2

cos 2   xy sin 2 cos 2   xy sin 2

sin 2   xy cos 2

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MECHANICS OF SOLIDS Principal Stresses • The previous equations are combined to yield parametric equations for a circle,

 x   ave 2   x2y  R 2 where

 ave 

2

 x  y

 x  y  2    xy R   2  

2

• Principal stresses occur on the principal planes of stress with zero shearing stresses.  max, min  tan 2 p 

 x  y 2

2

 x  y  2    xy   2  

2 xy

 x  y

Note : defines two angles separated by 90o GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Maximum Shearing Stress Maximum shearing stress occurs for

 x   ave

2

 x  y  2    xy  max  R   2    x  y tan 2 s   2 xy Note : defines two angles separated by 90o and offset from  p by 45o

    ave 

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 x  y 2

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MECHANICS OF SOLIDS Example 7.01 SOLUTION: • Find the element orientation for the principal stresses from 2 xy tan 2 p   x  y • Determine the principal stresses from  max, min 

x  y

2

 x  y  2    xy   2  

2 For the state of plane stress shown, determine (a) the principal panes, • Calculate the maximum shearing stress with (b) the principal stresses, (c) the 2      x y 2 maximum shearing stress and the    xy  max   2 corresponding normal stress.  

x  y    2

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MECHANICS OF SOLIDS Example 7.01 SOLUTION: • Find the element orientation for the principal stresses from tan 2 p 

2 xy

 x  y



2 40   1.333 50   10 

2 p  53.1, 233 .1

 x  50 MPa  x  10 MPa

 p  26.6, 116 .6

 xy  40 MPa

• Determine the principal stresses from  max, min 

x  y 2

 20 

2

 x  y  2    xy   2  

30 2  40 2

 max  70 MPa  min  30 MPa GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Example 7.01 • Calculate the maximum shearing stress with 2

 x  y  2    xy  max   2   

30 2  40 2

 max  50 MPa

 x  50 MPa

 xy  40 MPa

 s   p  45

 x  10 MPa

 s  18.4, 71.6

• The corresponding normal stress is  x   y 50  10     ave   2

2

   20 MPa

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MECHANICS OF SOLIDS Sample Problem 7.1 SOLUTION: • Determine an equivalent force-couple system at the center of the transverse section passing through H.

• Evaluate the normal and shearing stresses at H. • Determine the principal planes and calculate the principal stresses. A single horizontal force P of 150 lb magnitude is applied to end D of lever ABD. Determine (a) the normal and shearing stresses on an element at point H having sides parallel to the x and y axes, (b) the principal planes and principal stresses at the point H. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Sample Problem 7.1 SOLUTION: • Determine an equivalent force-couple system at the center of the transverse section passing through H. P  150 lb T  150 lb 18 in   2.7 kip  in M x  150 lb 10 in   1.5 kip  in

• Evaluate the normal and shearing stresses at H. y 

1.5 kip  in 0.6 in  Mc  1  0.6 in 4 I 4

 xy  

2.7 kip  in 0.6 in  Tc  1  0.6 in 4 J 2

 x  0  y  8.84 ksi  y  7.96 ksi GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Sample Problem 7.1 • Determine the principal planes and calculate the principal stresses. tan 2 p 

2 xy

 x  y



27.96   1.8 0  8.84

2 p  61.0,119 

 p  30.5, 59.5

 max, min 

x  y 2

2

 x  y  2    xy   2   2

0  8.84  0  8.84  2      7.96  2  2 

 max  13.52 ksi  min  4.68 ksi

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Mohr’s Circle for Plane Stress • With the physical significance of Mohr’s circle for plane stress established, it may be applied with simple geometric considerations. Critical values are estimated graphically or calculated. • For a known state of plane stress  x , y , xy plot the points X and Y and construct the circle centered at C.  ave 

 x  y 2

2

 x  y  2    xy R   2  

• The principal stresses are obtained at A and B.  max, min   ave  R tan 2 p 

2 xy

 x  y

The direction of rotation of Ox to Oa is the same as CX to CA. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Mohr’s Circle for Plane Stress • With Mohr’s circle uniquely defined, the state of stress at other axes orientations may be depicted. • For the state of stress at an angle  with respect to the xy axes, construct a new diameter X’Y’ at an angle 2 with respect to XY. • Normal and shear stresses are obtained from the coordinates X’Y’.

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MECHANICS OF SOLIDS Mohr’s Circle for Plane Stress • Mohr’s circle for centric axial loading:

x 

P ,  y   xy  0 A

 x   y   xy 

P 2A

• Mohr’s circle for torsional loading:

 x   y  0  xy 

Tc J

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x y 

Tc  xy  0 J 7 - 18

MECHANICS OF SOLIDS Example 7.02

For the state of plane stress shown, (a) construct Mohr’s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress.

SOLUTION: • Construction of Mohr’s circle  ave 

x  y

50    10   20 MPa

2 2 CF  50  20  30 MPa FX  40 MPa R  CX 

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

30 2  40 2  50 MPa 7 - 19

MECHANICS OF SOLIDS Example 7.02 • Principal planes and stresses  max  OA  OC  CA  20  50  max  70 MPa

 max  OB  OC  BC  20  50  max  30 MPa FX 40  CP 30 2 p  53.1

tan 2 p 

 p  26.6

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MECHANICS OF SOLIDS Example 7.02

• Maximum shear stress  s   p  45

 max  R

    ave

 s  71.6

 max  50 MPa

   20 MPa

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MECHANICS OF SOLIDS Sample Problem 7.2

For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the SOLUTION: given element counterclockwise • Construct Mohr’s circle through 30 degrees.  x   y 100  60  ave    80 MPa 2

R GIK Institute of Engineering Sciences and Technology

2

CF 2  FX 2  20 2  482  52 MPa 7 - 22

MECHANICS OF SOLIDS Sample Problem 7.2

• Principal planes and stresses XF 48   2.4 CF 20 2 p  67.4

tan 2 p 

 p  33.7 clockwise

 max  OA  OC  CA  80  52

 max  132 MPa

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 max  OA  OC  BC  80  52

 min  28 MPa 7 - 23

MECHANICS OF SOLIDS Sample Problem 7.2

• Stress components after rotation by 30o Points X’ and Y’ on Mohr’s circle that correspond to stress components on the rotated element are obtained by rotating XY counterclockwise through 2  60

  180   60  67.4  52.6  x  OK  OC  KC  80  52 cos 52.6  y  OL  OC  CL  80  52 cos 52.6  xy  KX   52 sin 52.6  x  48.4 MPa  y  111 .6 MPa  xy  41.3 MPa

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MECHANICS OF SOLIDS General State of Stress • Consider the general 3D state of stress at a point and the transformation of stress from element rotation

• State of stress at Q defined by:  x , y , z , xy , yz , zx • Consider tetrahedron with face perpendicular to the line QN with direction cosines: x ,  y , z • The requirement  Fn  0 leads to,  n   x2x   y 2y   z 2z  2 xy x  y  2 yz  y z  2 zx z x

• Form of equation guarantees that an element orientation can be found such that  n   a2a   bb2   cc2

These are the principal axes and principal planes and the normal stresses are the principal stresses. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS

Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress

• Transformation of stress for an element rotated around a principal axis may be represented by Mohr’s circle.

• The three circles represent the normal and shearing stresses for rotation around each principal axis.

• Points A, B, and C represent the • Radius of the largest circle yields the principal stresses on the principal planes maximum shearing stress. 1 (shearing stress is zero)  max   max   min 2

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MECHANICS OF SOLIDS

Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress • In the case of plane stress, the axis perpendicular to the plane of stress is a principal axis (shearing stress equal zero). • If the points A and B (representing the principal planes) are on opposite sides of the origin, then a) the corresponding principal stresses are the maximum and minimum normal stresses for the element

b) the maximum shearing stress for the element is equal to the maximum “inplane” shearing stress c) planes of maximum shearing stress are at 45o to the principal planes. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS

Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress • If A and B are on the same side of the origin (i.e., have the same sign), then a) the circle defining max, min, and max for the element is not the circle corresponding to transformations within the plane of stress b) maximum shearing stress for the element is equal to half of the maximum stress c) planes of maximum shearing stress are at 45 degrees to the plane of stress

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MECHANICS OF SOLIDS Stresses in Thin-Walled Pressure Vessels • Cylindrical vessel with principal stresses 1 = hoop stress 2 = longitudinal stress

• Hoop stress:  Fz  0  12t x   p2r x  1 

pr t

• Longitudinal stress:

 

2  Fx  0   2 2 rt   p  r pr 2  2t 1  2 2

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MECHANICS OF SOLIDS Stresses in Thin-Walled Pressure Vessels • Points A and B correspond to hoop stress, 1, and longitudinal stress, 2

• Maximum in-plane shearing stress: 1 2

 max( in  plane )   2 

pr 4t

• Maximum out-of-plane shearing stress corresponds to a 45o rotation of the plane stress element around a longitudinal axis  max   2 

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pr 2t

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MECHANICS OF SOLIDS Stresses in Thin-Walled Pressure Vessels • Spherical pressure vessel: 1   2 

pr 2t

• Mohr’s circle for in-plane transformations reduces to a point   1   2  constant  max(in -plane)  0 • Maximum out-of-plane shearing stress  max  12  1 

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pr 4t

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS

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