Lecture 3 The Concept of Stress, Generalized Stresses and Equilibrium

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1 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) t i i j n i (b) A normal vector to a plane specified by is given by where f denotes the gradient. f x, y, z axbyczd 0 a N f b c 1

2 The unit vector of the surface x y3z 9 is N 1 n 1 N 1 (3) 3 1 n The magnitude of normal stress n can be calculated by n tn where t is the surface tra ction vector t n Then n tn n 51Force / area The direction of the normal stress vector n is n 51 n nn

3 Shear stress vector can be simply calculated by subtract the normal stress vector n from the traction vector t / t / / 7 3

4 Problem 3-: Three invariants of a stress tensor Supposethestateofstressatapointinax,y,zcoordinatesystemisgivenby a. Calculate the three invariants of this stress tensor. b. Determine the three principal stresses of this stress tensor. Problem 3- Solution: a) Invariants of stress tensor Recall: these values do not change no matter the coordinate system selected b) Principal stresses I 1 x The eigenvalue problem for a stress tensor y z I 1 60 I x y y z z x xy y z zx ( 1000)(1000) I 3 I x y z xy yz zx x yz y zx z xy ( 10000)000(80) 0 I

5 is given by det Solve for, we have three eigenvalues The principal stresses are the three eigenvalues of the stress tensor xp 140 yp 60 zp 0 5

6 Problem 3-4: Transformation Matrix Suppose the state of stress at a point relative to a x, y, z coordinate system is given by: 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 is given by det Solve for, we have two eigenvalues The principal stresses are the two eigenvalues of the stress tensor xp yp

7 b) Determine transformation matrix Calculate the eigenvectors for the stress tensor For Set x 1 1, use the first equation Normalize eigenvector For Set x 1 1, use the first equation Normalize eigenvector Check 1? x x x ( 0.414) ( 9.14) 10 x (9.14) x x (.414)

8 ! Transformation Matrix L L b) 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 T L L '

9 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. Problem 3-5 Solution: Consider an element of a beam of length subjected to 1. Distributed loads q. Shear forces V and V dv 3. Moments M and M dm Equilibrium of forces in x and y direction F F x y 0 dn N dn N 0 0 Moment equilibrium of the element at point O 0 dv V q ( V dv)0 q ( ) M 0 M q ( )(V dv)m dm 0 From equation (*), dv dv q dv q 10

10 Substitute the above equation into the moment equilibrium equation, we have q dm ( ) Ignore the second-order terms, we have V q( ) dm V dm V To sum up, the equations of force and moment equilibrium of a beam are 0 dn 0 dv q dm V 11

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. Vertical force equilibrium 3. Axial force equilibrium Problem 3-6 Solution: Consider an element of a beam of length subjected to 1. Distributed loads q. Shear forces V and V dv 3. Moments M and M dm and under moderately large deflection Equilibrium of forces in t direction F t Note: All terms involve cos 0V cos qcos (V dv)cos 0 dv q (1) 1

12 Moment equilibrium of the element at point O M 0 M q ( )(V dv)m dm 0 From equation (1), dv dv q dv q Substitute the above equation into the moment equilibrium equation, we have q dm ( ) Ignore the second-order terms, we have V q( ) dm V dm V The moment equilibrium equation is not affected by moderately large deflection. 0 Force Equilibrium in x direction F x 0 ( N dn)cos Ncos 0 dn 0 () The axial force equilibrium equation is not affected by moderately large deflection. 13

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 V V Nsin V N (3) Force equilibrium in y direction V * dv V q0 * * dv * q 0 Combining equation (3), we have dv d dw dv dn dw d w N q N q 0 use equation(), we have the vertical force equilibrium equation dv d w N q0 The vertical force equilibrium equation is affected by moderately large deflection. 14

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 F x 0 P Tcos

15 P 150 T 19.3N cos70 cos70 Free body diagram where the tension T is projected in x and y direction as T x and T y. Moment balance at point B M B 0 T 0.7 T 0.35 M B 0 Finally, the bending moment at the midpoint of the bow is x y M B T 0.7 T y 0.35 x Tcos Tsin cos sin Nm 16

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 Psin V Pcos Moment balance at point B M B 0 P rsin M 0 We have the bending moment M acting at a cross section defined by the angle M Prsin 17

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.5wL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b L/9and 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

18 Force balance at any point of radius r V b L w * * W r dr blc g g r W W bl r blc g g Moment balance at any point of radius r M b L w * * W r g g r r r dr * blc bl blcr W 1 *3 1 W r rr * blcblcr g 3 g r W 1 1 bl 3 r 3 rb L r W blc blc r g 3 g By observation maximum shear force and bending moment occurs at r=b (closest point to the center line) M max V W W b blc g g W L L W L L L L g 9 g max L b wl V max 3.64 g W 1 1 b 3 L b 3 rb L b g 3 W 1L 3 3 L 1 L L W L L L L r L L L g g wl 3 a M max 3.7 g W bl c b L c b g 19

19 Problem 3-10: A simple beam AB supports two connective loads P and P 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 V max (b) Determine the distance that will produce the maximum bending moment, and also draw the corresponding bending moment diagram. (Assume P=10kN, d=.4m, and L=1m) C D Problem 3-10 Solution: a) Free body diagram: C D R A R B are reaction forces. Moment balance at point A: M A 0 0

20 R B L PxP xd 3Px Pd 3Px Pd R B L L Force balance in y direction R A F Y 0 3P R B R A d 3x P L L Shear force distribution C D 3Px Pd d 3x There are two possible maximum shear forces, and P, depending on L L L L the position of the wheels x: d d If x 0, the two possible maximum shear forces are V max P max 3, L L d d If x Ld, the two possible maximum shear forces are V max P max 3, L L In the range of 0 d L, by observation, the maximum shear force occurs when x L d, at the left hand side of point B. d.4 V max P L 1 8kN at x L d Note :0 x L d 1

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 R A d 3x 3Px Pd P and RB, L L L L M M C D d 3x R x P A 3 x L L R xd Pd A d 3 3x d P3 x 3 d Pd L L L L d 5d 3x P3 d 3 x Pd L L L Calculate maximum M C and M D If x M d 0, M Pd 1 L 0, C D d If x Ld, M C Pd 1,M D 0 L By observation, the maximum moment occurs at x=0, where d.4 M D Pd L N m M max 38.4 N m

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