<p>ENG5312 – Mechanics of Solids II 1</p><p>Stress Transformation – A Review</p><p>Plane Stress</p><p> The general state of plane stress at a point has six independent normal </p><p> and shear stress components (s x ,s y ,s z ,t xy ,t xz ,t yz )</p><p> A state of plane stress can be defined if the general state of stress at a point can be reduced to two normal stresses and one shear stress. ENG5312 – Mechanics of Solids II 2</p><p> The state of plane stress can be expressed by stress components along x ' y ' axes, which are not coincident with the x , y axes.</p><p> </p><p>General Equations of Plane Stress</p><p> Sign convention</p><p> o Positive normal and shear stresses act in the positive co-ordinate direction on a positive face, and in the negative co-ordinate direction on a negative face.</p><p> o Angles are measured positive counter-clockwise. ENG5312 – Mechanics of Solids II 3</p><p> Using an FBD of a sectioned element along the inclined plane.</p><p> Assuming static equilibrium and summing forces in the x' and y' directions gives:</p><p>SFx' = 0 =s x'DA - (t xy DAsinq)cosq - (t xyDAcosq)sinq</p><p>- (s xDAcosq)cosq - (s yDAsinq)sinq </p><p>SFy' = 0 = t x'y'DA - (t xyDAcosq)cosq +(t xy DAsinq)sinq +(s xDAcosq)sinq - (s yDAsinq)cosq</p><p> Or</p><p>2 2 s x' =s x cos q +s y sin q +t xy (2sinq cosq)</p><p>2 2 t x'y' = (s y - s x )sinq cosq +t x y (cos q - sin q)</p><p>ENG5312 – Mechanics of Solids II 4</p><p>Using the trigonometric identities: sin 2 q = (1- cos2q)/2; cos 2 q = (1+cos2q)/2; and sin 2q = 2sinq cosq :</p><p> s x +s y жs x - s y ц s x' = +з чcos 2q +t xy sin 2q (1) 2 2 и ш</p><p>жs - s ц t = - x y sin 2q +t cos2q x'y' з ч xy (2) и 2 ш </p><p> s +s жs - s ц s = x y - x y cos2q - t sin 2q y' з ч xy (3) 2 и 2 ш </p><p> These are the general equations of plane-stress transformation. They permit the calculation of the plane stresses acting on an element oriented at an angle q to the given stress components.</p><p>Principal Stresses</p><p> Given a state of plane stress, what are the maximum and minimum values of normal stress acting at the point, and in which direction are these stresses oriented?</p><p> Differentiate Eq. (1) for s x' wrt q, and set the result equal to zero and solve:</p><p> t tan 2q = xy p (4) (s x - s y ) 2</p><p> q q q o i.e. there are two values of p , p1 and p2 which are oriented at 90 to each other.</p><p> Substituting expressions for sin 2q p and cos2q p into Eq. (1) for s x' gives:</p><p>ENG5312 – Mechanics of Solids II 5</p><p>2 s x +s y жs x - s y ц 2 s 1,2 = ұ з ч +t xy 2 и 2 ш (5)</p><p> i.e. the maximum (s 1) and minimum (s 2) in-plane normal stress acting at a point. </p><p> s1 and s 2 are the principal stresses, which act on the principal planes q q defined by p1 and p2 measured clockwise positive from the co-ordinate system for the given stresses (i.e. x, y ).</p><p> Note: No shear stress acts on the principal planes.</p><p>Maximum In-Plane Shear Stress</p><p> Given a state of stress, what is the maximum in-plane shear stress acting at the point, and in which direction does the stress act?</p><p> Differentiate Eq. (2) for t x'y' wrt q, set the result equal to zero, and solve:</p><p>- (s x - s y ) 2 tan 2q = (6) s t xy q q  Solving for s1 and s2 shows that the planes of maximum in-plane shear stress are directed 45o to the principal planes.</p><p> The maximum in-plane shear stress is:</p><p>2 жs - s ц max x y 2 (7) t in- plane = з ч +t xy и 2 ш q q  Substitution s1 and s2 into Eq. (1) for s x' gives:</p><p> s +s s = x y (8) avg 2</p><p> i.e. a normal stress acts on the plane of maximum in-plane shear stress.</p><p> Note: The existence of s avg is quite apparent on Mohr’s Circle.</p><p>ENG5312 – Mechanics of Solids II 6</p><p>Mohr’s Circle – Plane Stress</p><p> Mohr’s circle provides a graphical solution to the general equations for plane-stress transformation.</p><p> Rewrite Eqs. (1) and (2) for s x' andt x'y':</p><p> s x +s y жs x - s y ц s x' - =з чcos 2q +t xy sin 2q 2 2 и ш</p><p>жs x - s y ц t x'y' = - з чsin 2q + t xy cos2q и 2 ш </p><p> Square and add these equations to eliminate :</p><p>2 2 й s x +s y щ 2 жs x - s y ц 2 кs x' - ъ +t x'y' =з ч +t xy л 2 ы  и 2 ш </p><p> Or</p><p>2 2 2 (s x' - s avg ) + t x'y' = R (9)</p><p>Where</p><p> s x +s y s = (9a) avg 2</p><p>2 жs - s ц x y 2 (9b) R = з ч +t xy и 2 ш </p><p> Equation (9) is the equation of a circle with radius R offset in the s direction by s avg on the s,t axes.</p><p>ENG5312 – Mechanics of Solids II 7</p><p> Point A is the known state of stress s x and t xy .</p><p> s +s  Point C is the offset at s = x y avg 2</p><p>2 жs - s ц Radius x y 2 .  R = з ч +t xy и 2 ш </p><p> o  Rotation of an element by 90 will give s x' = s y and t x'y' = - t xy . To obtain this result on Mohr’s circle requires a rotation of 180o, therefore, must rotate by angle 2q on Mohr’s circle. Maximum (principal) stress occurs at B, minimum normal stress occurs at D, i.e. point with no shear stress.</p><p> Maximum in-plane shear stress occurs at points E and F, i.e. s avg exists on the plane.</p><p> A state of stress at an orientation of q counter-clockwise from the known </p><p> state is given by the point P ( 2q on Mohr’s circle). s x' and t x'y' can be read from the circle or calculated from trigonometry.</p><p>ENG5312 – Mechanics of Solids II 8</p><p>Absolute Maximum Shear Stress (3D)</p><p> If a body is subjected to a general 3D state of stress there is a unique orientation of a plane that will give the principal stresses (  min   int   max ), and there will be no shear stress components acting on the principal planes.</p><p> Assuming:  m a x in the x ' ,  m i n in the z ' and  i n t in the y ' directions,  respectively.</p><p>     </p><p> Drawing Mohr’s circle for each plane. ENG5312 – Mechanics of Solids II 9</p><p> The absolute maximum shear stress is located on the largest circle: s - s t max = max min (10) abs 2</p><p>And it has an associated normal stress: s +s s = max min (11) avg 2</p><p> The absolute maximum shear stress occurs on a plane oriented 45o (i.e. 90o on Mohr’s circle) to a principal plane (here the x',z' plane).</p><p>Absolute Maximum Shear Stress (Plane Stress)</p><p> Consider an element exposed to plane stress ( x', y' plane). Define s max in the x', s int in the y', and s min (= 0) in the z' directions, respectively. Assuming s max and s int have the same sign:</p><p> Here</p><p> m a x  m a x  a b s   x ' z '   (12) m a x 2</p><p> ENG5312 – Mechanics of Solids II 10</p><p> max  Note: The maximum in-plane shear stress t in- plane = (s max - s int ) /2 is not equal to the maximum shear stress in the body.</p><p> Consider the case where s max and s min have opposite signs (s max in the</p><p> x', s int in the z', and s min in the y ' directions, respectively).</p><p> Here s - s t max = t max = max min (13) abs in- plane 2</p><p> max  These values for t abs are important for ductile materials as they fail in shear.</p>