Lecture 3 The Concept of Stress, Generalized Stresses and Equilibrium Problem 3-1: Cauchy’s Stress Theorem Cauchy’s stress theorem states that in a stress tensor field there is a traction vector t that linearly depends on the outward unit normal n: a. Express Cauchy’s stress theorem in index form. b. Suppose the state of stress at a point in x, y, z coordinate system is given by the matrix below. Determine the normal stress n and the shear stress on the surface defined by . Problem 3-1 Solution: (a) tnii ji (b) A normal vector to a plane specified by fx,,yz axbyczd0 is given by a Nf b c where f denotes the gradient. 1 The unit vector of the surface 23xy z9 is 2 N 1 n 1 N 2122(3)2 3 2 1 n 1 14 3 The magnitude of normal stress n can be calculated by n tn where t is the surface tra ction vector 20 10 10 2 80 11 tn 10 30 0 1 50 14 14 10 0 50 3170 Then 80 2 11720 tn 50 1 n 14 14 14 170 3 n 51Force / area The direction of the normal stress vector n is n 2 51 n 1 nn 14 3 2 Shear stress vector can be simply calculated by subtract the normal stress vector n from the traction vector t 80 2 160 / 7 13601 t 50 1 10 / 7 14 7 14 14 1703110 / 7 3 Problem 3-2: Three invariants of a stress tensor Supposethestateofstressatapointinax,y,zcoordinatesystemisgivenby 100 1 180 0200 180 0 20 a. Calculate the three invariants of this stress tensor. b. Determine the three principal stresses of this stress tensor. Problem 3-2 Solution: a) Invariants of stress tensor Recall: these values do not change no matter the coordinate system selected I1 xyz100 20 20 I1 60 I 22 2 2 xy yz zx xy yz zx (10020)(10020)20220080 I2 10000 I 2 22 2 3 x yz xyyzzx xyz yzx zxy (1002020)20020(80)2 0 I3 168000 b) Principal stresses The eigenvalue problem for a stress tensor 4 100 0 80 0200 80 0 20 is given by 100 0 80 det 0 20 0 0 80 0 20 Solve for , we have three eigenvalues 1 140 2 60 3 20 The principal stresses are the three eigenvalues of the stress tensor xp 140 yp 60 zp 20 5 Problem 3-4: Transformation Matrix Suppose the state of stress at a point relative to a x, y, z coordinate system is given by: 15 10 10 5 Try to find a new coordinate system (x’, y’) that corresponds to the principal directions of the stress tensor. a. Find the principal stresses. b. Determine the transformation matrix . c. Verify . Problem 3-4 Solution: a) Determine principal stresses The eigenvalue problem for a stress tensor 15 10 10 5 is given by 15 10 det 0 10 5 Solve for , we have two eigenvalues 1 19.14 2 9.14 The principal stresses are the two eigenvalues of the stress tensor xp 19.14 yp 9.14 6 b) Determine transformation matrix Calculate the eigenvectors for the stress tensor For 1 19.14 15 19.14 10 x1 0 10 5 19.14 x2 Set x1 1, use the first equation x2 0.414 Normalize eigenvector 1 1 1 122( 0.414) 0.414 0.92 1 0.38 For 1 9.14 15 ( 9.14) 10 x1 0 10 5 (9.14) x2 Set x1 1, use the first equation x2 2.414 Normalize eigenvector 1 1 2 1(22 2.414)2.414 0.38 1 0.92 Check 12 ? 7 12 0.920.38 0.380.92 0 12! Transformation Matrix L 12 0.92 0.38 L 0.38 0.92 b) Verify T ' L L If we transform the given stress tensor, will we arrive at the principal stresses? Note: use 4 decimal places to get principal stresses T 0.9238 0.3827 15 10 0.9238 0.3827 19.174 0 LL 0.3827 0.9238 10 5 0.3827 0.9238 0 9.14 19.174 0 ' 09 .14 8 Problem 3-5: Beam Equilibrium Derive the equation of force and moment equilibrium of a beam using the equilibrium of an infinitesimal beam element of the length dx. Problem 3-5 Solution: Consider an element of a beam of length dx subjected to 1. Distributed loads q 2. Shear forces V and Vd V 3. Moments M and M dM Equilibrium of forces in x and y direction dN FN00dNN 0 x dx dV FV0(qdxVdV)0 q () y dx Moment equilibrium of the element at point O dx MM0( qdx)(VdV)dxMdM0 2 From equation (*), dV q dx dV dV dx qdx dx 10 Substitute the above equation into the moment equilibrium equation, we have q dM ()dx 22Vdx q()dx 0 2 Ignore the second-order terms, we have dM Vdx dM V dx To sum up, the equations of force and moment equilibrium of a beam are dN 0 dx dV q dx dM V dx 11 Problem 3-6: Moderately large deflections in beams Explain which of the three equilibrium equations below is affected by the finite rotations in the theory of the moderately large deflections of beam. 1. Moment equilibrium 2. Vertical force equilibrium 3. Axial force equilibrium Problem 3-6 Solution: Consider an element of a beam of length dx subjected to 1. Distributed loads q 2. Shear forces V and Vd V 3. Moments M and M dM and under moderately large deflection Equilibrium of forces in t direction FVt 0cosqdxcos(VdV)cos0 dV q dx (1) Note: All terms involve cos 12 Moment equilibrium of the element at point O dx MM0( qdx)(VdV)dxMdM0 2 From equation (1), dV q dx dV dV dx qdx dx Substitute the above equation into the moment equilibrium equation, we have q dM ()dx 22Vdx q()dx 0 2 Ignore the second-order terms, we have dM Vdx dM V dx The moment equilibrium equation is not affected by moderately large deflection. Force Equilibrium in x direction FNx 0(dN)cosNcos0 dN 0 dx (2) The axial force equilibrium equation is not affected by moderately large deflection. 13 Define effective shear force V * as the sum of the cross-sectional shear V and the projection of the axial force into the vertical direction dw VV* Nsin VN dx (3) Force equilibrium in y direction Vd** VVq*dx0 dV * q 0 dx Combining equation (3), we have dV d dw dV dN dw d 2w NqNq2 0 dx dx dx dx dx dx dx use equation(2), we have the vertical force equilibrium equation dV d 2w Nq0 dx dx2 The vertical force equilibrium equation is affected by moderately large deflection. 14 Problem 3-7: At full draw, and archer applies a pull of 150N to the bowstring of the bow shown in the figure. Determine the bending moment at the midpoint of the bow. Problem 3-6 Solution: P is the pulling force at point A and T is the tension in the string. Force balance at point A Fx 0 PT 2cos70 0 15 P 150 T219.3N 2cos702cos70 Free body diagram where the tension T is projected in x and y direction as Tx and Ty. Moment balance at point B M B 0 TTxy0.7 0.35 MB0 Finally, the bending moment at the midpoint of the bow is MTBx0.7 Ty0.35 TTcos 700.7 sin 70 0.35 219.3 cos700.7 219.3sin 70 0.35 124.6Nm 16 Problem 3-8: A curved bar ABC is subjected to loads in the form of two equal and opposite force P, as shown in the figure. The axis of the bar forms a semicircle of radius r. Determine the axial force N, shear force V, and bending moment M acting at a cross section defined by the angle Problem 3-8 Solution: The free body diagram and force balance at angle : We can determine axial force N, shear force V according to the force balance diagram NP sin VP cos Moment balance at point B M 0 B Pr sin M 0 We have the bending moment M acting at a cross section defined by the angle MPrsin 17 Problem 3-9: The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) with an angular acceleration . Each of the two arms has weight w per unit length and supports a weight Ww 2.5 L at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming bL /9and cL /10 Problem 3-9 Solution: Shear force density distribution and free body diagram: Where weight at the tip of arm W is considered as lumped mass 18 Force balance at any point of radius r bL wW Vr **drbLc r gg WW 2 bLr2 bLc 2gg Moment balance at any point of radius r bL wW Mr **rrdr*bLcbLcr r gg bL WW11*3 *2 rrr bLcbLcr gg32 r W1132W bLr32rbLrbLcbLc r g32g By observation maximum shear force and bending moment occurs at r=b (closest point to the center line) WW 2 VbLb2 bLc max 2gg 22 WL LWLL LL 29gg9 9 10 wL2 V 3.64 max g W113232W Mbmax Lbrb LbbL c b L c b g32g 33 22 WL11L LLWLLL LrLLL g39929 9g9 1010 wL3a M 3.72 max g 19 Problem 3-10: A simple beam AB supports two connective loads P and 2P that are distance d apart (see figure).