2.080J Structural Mechanics: Solutions to Problem Set 2

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) ti n ij i (b) A normal vector to a plane specified by f x, y , z ax by cz d 0 is given by N where f a b c f denotes the gradient. 1 The unit vector of the surface 2 x y 3z 9 is 2 1 N N n n 2 3 2 1 ( 3) 2 2 1 1 14 3 n The magnitude of normal stress 1 2 can be calculated by n t n where t is the surface traction vector t n 20 10 10 30 10 0 10 0 50 2 1 1 14 3 1 14 1 t n 14 80 50 170 2 1 1 14 3 720 14 80 50 170 Then n n The direction of the normal stress vector n 51 Force / area n is n n n 2 51 1 14 3 2 Shear stress vector can be simply calculated by subtract the normal stress vector n from the traction vector t t 1 14 80 50 170 2 360 1 7 14 3 1 14 160 / 7 10 / 7 110 / 7 3 Problem 3-2: Three invariants of a stress tensor Suppose the state of stress at a point in a x, y, z coordinate system is given by 100 1 180 0 20 0 180 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 x y 100 20 20 z I1 I2 60 2 x y y z z x 2 yz xy 2 zx ( 100 20) ( 100 20) 202 0 0 802 I2 I3 x y z 2 10000 2 xy yz zx x yz 2 y zx 2 z xy ( 100 20 20) 2 0 0 20 ( 80) 2 0 I3 168000 b) Principal stresses The eigenvalue problem for a stress tensor 4 100 0 0 20 80 0 80 0 20 is given by det Solve for 100 0 80 0 20 0 80 0 20 0 , 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 10 10 5 is given by det Solve for 15 0 , 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 19.14 1 10 x1 5 19.14 x2 15 19.14 10 Set x1 0 1 , use the first equation x2 0.414 Normalize eigenvector 1 1 12 ( 0.414) 2 0.92 0.38 1 For 9.14 1 10 x1 5 ( 9.14) x2 15 ( 9.14) 10 Set x1 1 0.414 0 1 , use the first equation x2 2.414 Normalize eigenvector 1 2 12 (2.414)2 1 Check 1 2 1 2.414 0.38 0.92 ? 7 1 0.92 0.38 0.38 0.92 0 2 1 2 ! Transformation Matrix L 0.92 L b) 1 2 0.38 0.38 0.92 Verify L ' T L If we transform the given stress tensor, will we arrive at the principal stresses? Note: use 4 decimal places to get principal stresses L T L 0.9238 0.3827 0.3827 0.9238 15 10 0.9238 5 0.3827 0.9238 10 ' 19.174 0 0 9.14 0.3827 19.174 0 0 9.14 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 V 3. Moments M and M dV dM Equilibrium of forces in x and y direction Fx 0 N dN Fy 0 V qdx (V N 0 dN dx dV ) 0 0 dV dx q ( ) Moment equilibrium of the element at point O M 0 M qdx( dx ) (V 2 dV )dx M dM 0 From equation (*), dV q dx dV dV dx dx qdx 10 Substitute the above equation into the moment equilibrium equation, we have dM q (dx) 2 Vdx q(dx) 2 2 0 Ignore the second-order terms, we have dM dM dx Vdx V To sum up, the equations of force and moment equilibrium of a beam are dN dx dV dx dM dx 0 q V 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 V 3. Moments M and M dV dM and under moderately large deflection Equilibrium of forces in t direction Ft 0 V cos qdx cos dV dx (V dV ) cos 0 q (1) Note: All terms involve cos 12 Moment equilibrium of the element at point O M 0 M qdx( dx ) (V 2 dV )dx M dM 0 From equation (1), dV q dx dV dV dx dx qdx Substitute the above equation into the moment equilibrium equation, we have q (dx) 2 Vdx q(dx) 2 2 dM 0 Ignore the second-order terms, we have dM Vdx dM dx V The moment equilibrium equation is not affected by moderately large deflection. Force Equilibrium in x direction Fx 0 ( N dN) cos dN dx N cos 0 0 (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 V* V V N sin N dw dx (3) Force equilibrium in y direction V * dV * dV * dx V * qdx q 0 q dV dx 0 Combining equation (3), we have dV dx d dw N dx dx dN dw d 2w N 2 q 0 dx dx dx use equation(2), we have the vertical force equilibrium equation dV dx d 2w N 2 q dx 0 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 P 2T cos 70 0 15 T P 2cos 70 150 2cos 70 219.3N Free body diagram where the tension T is projected in x and y direction as Tx and Ty. Moment balance at point B MB 0 Tx 0.7 Ty 0.35 M B 0 Finally, the bending moment at the midpoint of the bow is MB Tx 0.7 Ty 0.35 T cos 70 0.7 T sin 70 219.3 cos 70 124.6N m 0.35 0.7 219.3 sin 70 0.35 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 N V P sin P cos MB 0 Moment balance at point B P r sin M 0 We have the bending moment M acting at a cross section defined by the angle M P r sin 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 W 2.5wL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b L / 9 and c L / 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 b L V r w * W r dr * b L c g g W 2g 2 b L W b L c g r2 Moment balance at any point of radius r b L M r w * r g W b L c g r * r dr * W g 1 *3 r 3 W g 1 b L 3 b L 1 *2 rr 2 3 W g r b L c r b L c b L c r 1 r b L 2 r3 2 W g r2 b L c b L c r By observation maximum shear force and bending moment occurs at r=b (closest point to the center line) Vmax M max W g 1 b L 3 W g 1 3 L 9 3 b L W 2g L 9 2 L 9 3 W b L c g b2 2 2 L L 9 Vmax 3.64 1 r b L 2 b3 3 L W 2g 1 r 2 M max 2 L 9 W L g 9 L L 10 wL2 g W g b2 2 L 3.72 L 9 b L c b L c b 2 W g L 9 L L 10 L L 10 wL3a g 19 Problem 3-10: A simple beam AB supports two connective loads P and 2P that are distance d apart (see figure). The wheels may be placed at any distance x from the left support of the beam. (a) Determine the distance x that will produce maximum shear force in the beam, and also determine the maximum shear force Vmax (b) Determine the distance that will produce the maximum bending moment, and also draw the corresponding bending moment diagram. (Assume P=10kN, d=2.4m, and L=12m) C D Problem 3-10 Solution: a) Free body diagram: D C RA RB are reaction forces. Moment balance at point A: MA 0 20 RB L P x 2P x d 3Px 2 Pd 3Px L RB 2 Pd L Force balance in y direction FY RA RA 0 3P RB P 2 2d L 3x L Shear force distribution C D There are two possible maximum shear forces, 3Px L 2 Pd 2d and P 2 L L 3x , depending on L the position of the wheels x: If x 0, the two possible maximum shear forces are Vmax P max If x L d, the two possible maximum shear forces are Vmax 3 2d 2d , L L P max 3 In the range of 0 d L , by observation, the maximum shear force occurs when x the left hand side of point B. Vmax P 3 d L 10 3 2.4 12 28kN at x L d Note : 0 x d d , L L L d , at L d 21 b) From the shear force diagram, we know the bending moment diagram could be either i orii The maximum bending moment occurs at either point C or D. Thus we need only calculate moment at these two points to determine the maximum bending moment. Knowing RA 2d L P 2 3Px L 3x and RB L MC RA x MD RA x d 2 Pd , L 2d 3x 2 3 x L L P P d P 3 2d 3dx x L L P 3 2d d L 3 3x 2 L 3 2d d L 5d 3x 2 x L L Pd Pd Calculate maximum M C and M D If x 0, M C 0, M D If x L d, MC 2Pd 1 Pd 1 d L d ,MD L 0 By observation, the maximum moment occurs at x=0, where MD 2 Pd 1 d L 2 10 2.4 M max 1 2.4 12 38.4N m 38.4 N m 22