Plane Stress [PDF]

CHAPTER. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7. Transformations of. Stress and Strain. Page 2.

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Fifth SI Edition

CHAPTER

7

MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek

Transformations of Stress and Strain

Lecture Notes: J. Walt Oler Texas Tech University

© 2009 The McGraw-Hill Companies, Inc. All rights reserved.

Fifth Edition

MECHANICS OF MATERIALS Introduction • The most general state of stress at a point may be represented by 6 components,  x , y , z normalstresses  xy,  yz ,  zx shearingstresses (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. The second part of the chapter is devoted to a similar analysis of the transformation of the components of strain.

7- 2

Fifth Edition

MECHANICS OF MATERIALS 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. 7- 3

Fifth Edition

MECHANICS OF MATERIALS 7.1 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 Acos cos  xy Acos sin  y Asin sin  xy Asin cos

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

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

 x   y  x  y

 cos2  xy sin2 2 2       y  x y  x y cos2  xy sin2 2 2    xy   x y sin2  xy cos2 2

7- 4

Fifth Edition

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

 x  ave2  x2y  R2 where

 ave 

 x  y

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

2

• Principal stresses occur on the principal planes of stress with zero shearing stresses.  max,min  tan2 p 

 x  y 2

2 xy  x  y

2   x  y  2   xy    2 

Note: definestwoanglesseparatedby 90o 7- 5

Fifth Edition

MECHANICS OF MATERIALS Maximum Shearing Stress Maximum shearing stress occurs for

 x   ave

2   x  y  2   xy  max  R    2    y tan2s   x 2 xy

Note: definestwoanglesseparatedby 90o and offsetfrom p by 45o

    ave 

 x  y 2

7- 6

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.1

Fig. 7.13

For the state of plane stress shown, determine (a) the principal planes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress.

7- 7

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.1

Fig. 7.13

 x  50MPa  x  10MPa

 xy  40MPa

Fig. 7.14 7- 8

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.1

Fig. 7.13

 x  50MPa  x  10MPa

 xy  40MPa

Fig. 7.16 7- 9

Fifth Edition

MECHANICS OF MATERIALS 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 600 N 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. 7- 10

Fifth Edition

MECHANICS OF MATERIALS Sample Problem 7.1 SOLUTION:

• Determine an equivalent force-couple system at the center of the transverse section passing through H. P  600N T  600N0.45m  270Nm M x  600N0.25m  150Nm

• Evaluate the normal and shearing stresses at H.

150Nm0.015m Mc  1 4 I 4  0.015m 270Nm0.015m Tc  xy     1 4 J    0 . 015 m 2 y  

 x  0  y  56.6 MPa  y  50.9 MPa 7- 11

Fifth Edition

MECHANICS OF MATERIALS Sample Problem 7.1

7- 12

Fifth Edition

MECHANICS OF MATERIALS Problems

• 7-19, 7-22

Fifth Edition

MECHANICS OF MATERIALS 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 2 xy tan2 p   x  y The direction of rotation of Ox to Oa is the same as CX to CA. 7- 14

Fifth Edition

MECHANICS OF MATERIALS 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’.

7- 15

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.2

Fig. 7.13

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.

7- 16

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.2

7- 17

Fifth Edition

MECHANICS OF MATERIALS Concept Application 7.2

• Maximum shear stress

s   p  45

 max  R

    ave

s  71.6

 max  50 MPa

   20MPa 7- 18

Fifth Edition

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

P A

 x  ,  y   xy  0

 x   y   xy 

P 2A

• Mohr’s circle for torsional loading:

 x   y  0  xy 

Tc J

x  y 

Tc  xy  0 J 7- 19

Fifth Edition

MECHANICS OF MATERIALS 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.    100 60  ave  x y   80MPa 2 2 R

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

Fifth Edition

MECHANICS OF MATERIALS Sample Problem 7.2

• Principal planes and stresses XF 48  max  OA OC  CA tan2 p    2.4 CF 20  80  52 2 p  67.4  max  132MPa  p  33.7 clockwise

 max  OA OC  BC  80  52  min  28MPa

7- 21

Fifth Edition

MECHANICS OF MATERIALS 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  52cos52.6  y  OL  OC  CL  80  52cos52.6  xy  KX   52sin52.6

 x  48.4 MPa  y  111.6 MPa  xy  41.3MPa 7- 22

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