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3 When a set of co-planer external forces and moments act on a body, the stress developed at any point ‘P’ inside the body can be completely defined by the two dimensional state of stress:  x = normal stress in X direction,  y = normal stress in Y direction, and  xy = shear stress which would be equal but opposite in X (cw) and Y (ccw) directions, respectively. The 2D stress at point P is described by a box drawn with its faces perpendicular to X & Y directions, and showing all normal and shear stress vectors (both magnitude and direction) on each face of the box. This is called the stress element of point P. Two dimensional state of stress, and the stress element X Y F1F1 F2F2 F3F3 F4F4 FnFn M M P Stress Element xx xx X  xy cw Y yy yy  xy ccw  xy xx yy yy P xx

4 The stress formulae that we have learnt thus far, can determine the 2D stresses developed inside a part, ONLY ALONG A RECTANGULAR AXIS SYSTEM X -Y, that is defined by the shape of the part. For example, X axis for a cantilever beam is parallel to its length, and Y axis is perpendicular to X. For a combined bending and axial loading (F 1, F 2 etc.) of this cantilever beam: the normal and shear stress at a point P, can be determined using the formulae, such as,  x = Mv/I+P/A,  xy =VQ/(Ib). Note that, these formulae can only determine stresses parallel to X and Y axis, and the stress element is aligned with X-Y axis. The question is, what would be the values of normal and shear stresses at the same point P, if the stresses are measured along another rectangular axis system U-V, rotated at an angle  with the X-Y axis system ? X Y F1F1 F2F2 P xx yy yy P xx X Y U V 

5  X Y xx yy yy xx uu v  uv u v uu vv vv X  F Knowing the 2D stresses at point P along XY coordinate system, we want to determine the 2D stresses for the same point P, when measured along a new coordinate system UV, which is rotated by an angle ww ith respect to the XY system. The Problem is: given  x,  y,  xy and , can we determine  u,  v,  uv ? v u X F1F1 Y F2F2 P X Y u  F2F2 P

6 1. We cut the stress element by an arbitrary plane at an angle TT his plane is normal to u-axis uu  xy xx yy  xx X Y xx yy yy   L Lsin  Lcos   xy (LBsin   x (LBcos   xy (LBcos   y (LBsin   u (LB)  uv (LB)  2. To maintain static equilibrium, let the internal normal and shear stresses  u &  uv, respectively are developed on the cut plane 3. Let, L be the length of the cut side. Then the other two sides are Lsin  & Lcos  4. If the thickness of the element is B, then the force acting on each face of the element will be equal to the stress multiplied by the area of the face. THIS IS HOW WE CAN ACHIEVE THAT U

7  xy (LBsin   x (LBcos   xy (LBcos   y (LBsin   u (LB)  uv (LB)   xy LBsin  cos    xy LBsin 2   xy LBsin  cos   xy LBcos 2   u (LB)  uv (LB)    y LBsin  cos   y LBsin 2    x LB cos 2   x LB cos  sin   Equating forces in u-direction:  u LB =  x LBcos 2   y LBsin 2  + 2  xy LBsin  cos  Or,  u =  x cos 2  +  y sin 2  + 2  xy sin  cos  ………..(1) Equating forces in v-direction:  uv LB =  xy LBcos 2  -  xy LBsin 2  -  x LBsin  cos  +  y LBsin  cos  Or,  uv =  xy (cos 2  - sin 2  ) – (  x -  y ) sin  cos  ……. (2) 5. Forces acting on the faces = force x area 6. Resolving each force in u & v directions CONTINUING

8 Replacing the square terms of trigonometric functions by double angle terms and rearranging : Equations 3, 4 & 5 gives us the 2D stress values, if measured along U-V axis which is at an angle ff rom X-Y axis. Since both sets of stresses refer to the stress of the same point, the two sets of stresses are also equivalent.  xy X Y xx yy yy  u uu  uv v uu vv vv X 

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10 Mohr’s circle implements these three equations by a graphical aid, which simplifies computation and visualization of the changes in stress values (  u,  v &  uv ) with the rotation angle oo f the measurement axis. Mohr circle is plotted on a rectangular coordinate system in which the positive horizontal axis represents positive (tensile) normal stress , and the positive vertical axis represents the positive (clockwise) shear stress (  ). Thus the plane of the Mohr circle is denoted as  plane. In this  plane, the stresses acting on two faces of the stress element are plotted.  xy Y xx yy xx yy X xx xx X cw  xy Y yy yy ccw     For a stress element Y faces have stress: (  y,-  xy ) x faces have stress: (  x &  xy )

11 uu U V  uv vv uu vv X  1.Start by drawing the original stress element with its sides parallel to XY axis, and show the normal and the shear stress vectors on the element. 2.Draw the  rectangular axis and label them. 3.On the  plane, plot X with normal and shear stress values of  x and  xy, and Y with values  y and –  xy. 4.Join X and Y points by a straight line, which intersects the horizontal  axis at C. C denotes the average normal stress  avg =(  x +  y )/2. 5.The line CX denotes X axis, and line CY denotes Y axis in Mohr circle. Name them. 6.Draw the Mohr circle using C as the center, and XY line as the diameter. 7.To find stress along the new UV axis system, draw a line UV rotated at an angle 2 ff rom the XY line. CU line denotes U axis, and CV denotes V axis. 8.The normal and shear stress values of the points U and V on the  plane denote the stresses in U and V directions, respectively. 9.This way we can find stresses for an element rotated at any desired angle . Y xx  xy yy xx yy X Normal stress axis (  Shear  stress axis (  U   xy 2  Y(  y,-  xy ) xx  y  avg  x +  y )/2 V  x y C  u  u v  v  u v X (  x,  x y )    X axis Y axis DRAWING MOHR CIRCLE

12 X Y xx  xy yy xx yy X uu U V  uv vv uu vv X  X (  x,  xy )   xy X axis 22 Shear  stress  axis (  Y(  y,  xy ) xx yy  Normal Stress axis (  Y axis  avg  x +  y )/2 U (  u,  uv ) V (  v,  xy ) a PROOF

13 Y(  y,-  xy ) xx yy  avg     xy X axis Y axis   X (  x,  xy ) Similarly, if the XY axis line is rotated by an angle 2  ‘ to make it vertical, then the shear stress maximizes and the element will have normal stress =  avg and Maximum shear stress =  max    In the Mohr circle, for a rotation of 2  a ngle, the XY axis line becomes horizontal. In the rotated axis    , the shear stress vanishes. The element will have only normal stresses  1 &  2, and  1 being the maximum normal stress.     aa re called the Principal normal stresses.  max Principal Normal Stresses     aa nd Max Shear Stress  max  11 11 22 22 Y X  avg  max x Y ’’  min  max 22 11 2’2’  avg,  max)  avg,-  max)

14  min  max  avg  max  avg  max x Y  Y xx  xy yy xx yy X 11 22 11 22 Y    X Formulea for Principal Normal Stresses & Max Shear Stress X (  x,  xy ) Y(  y,-  xy ) xx 22 11 yy  avg  max     2’2’  xy X axis Y axis  avg,  max)  avg,-  max)   Maximum shear stress element Principal normal stress element

15 Determining  u,  v &  uv Given  x,  y,  xy &  uu U V  uv vv uu vv X  Y xx  xy yy xx yy X U (  u,  uv )   xy X axis 22 Y(  y,-  xy ) xx yy Y axis  avg  x +  y )/2 V (  v,  xy )  xy C uu  uv vv X (  x,  xy )    22

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17 Y X 5,000 psi 20,000 psi 4,000 psi X ( 20k,5k ) Y(-4k,-5k) 20k  -5k  21k -4k 8k  max        5k X axis Y axis (8k,13k ) (8k,-13k) R=13K For a stress element with 1.Draw the stress element along XY axis. 2.Draw the aa xes for mohr circle 3.Plot point X for  x =20K,  xy =5k 4.Plot point Y for sy= -4K, txy=-5k 5.Draw line XY and show X & Y axes. 6.Draw the circle with XY as the diameter  x =20,000 psi,  y = psi, and  xy = 5000 psi. Draw the Mohr Circle and, draw two stress elements properly oriented for (i) the principal normal stresses, and (ii) max shear stresses element. This completes the Mohr circle. Next, the stress elements

18 Y X 5,000 psi 20,000 psi 4,000 psi X ( 20k,5k ) Y(-4k,-5k) 20k  -5k  21k -4k 8k  max        5k X axis Y axis (8k,13k ) (8k,-13k) R=13K 5k Y X   11.3  min  max 8k 13k 8k 13k x Y 33.7 The principal normal stress axis will be rotated CW Draw the principal stress axis 11.3 o CW from XY axis. Show the principal stresses. The  max axis will be rotated CCW Draw the  max stress axis 33.7 o CCW from XY axis. Show the the stresses.  21k PRINCIPAL NORMAL STRESS ELEMENT STRESS ELEMENT FOR  MAX That completes the drawing of the two stress elements

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