Mechanical Engineers Handbook Episode 3 potx

frame, i.e., they are the velocity and acceleration measured by an observer moving with the rigid body (Fig. 4.10). If A is a point of the rigid body, A P RB, v Arel  0 and a Arel  0. 4.7.1 MOTION OF A POINT RELATIVE TO A MOVING REFERENCE FRAME The velocity and acceleration of an arbitrary point A relative to a point O of a rigid body, in terms of the body ®xed reference frame, are given by Eqs. (4.20) and (4.21): v A  v O  v Arel  v  OA 4X24 a A  a O  a Arel  2v  v Arel  a  OA  v Âv ÂOAX 4X25 These results apply to any reference frame having a moving origin O and rotating with angular velocity v and angular acceleration a relative to a primary reference frame (Fig. 4.11). The terms v A and a A are the velocity and acceleration of an arbitrary point A relative to the primary reference frame. The terms v Arel and a Arel are the velocity and acceleration of A relative to the secondary moving reference frame, i.e., they are the velocity and accelera- tion measured by an observer moving with the secondary reference frame. Figure 4.10 4. Planar Kinematics of a Rigid Body 107 Dynamics 4.7.2 INERTIAL REFERENCE FRAMES A reference frame is inertial if one may use it to apply Newton's second law in the form  F  ma. Figure 4.12 shows a nonaccelerating, nonrotating reference frame with the origin at O 0 , and a secondary nonrotating, earth centered reference frame with the origin at O. The nonaccelerating, nonrotating reference frame with the origin at O 0 is assumed to be an inertial reference. The acceleration of the earth, due to the gravitational attractions of the sun, moon, etc., is g O . The earth centered reference frame has the acceleration g O as well. Figure 4.11 Figure 4.12 108 Dynamics Dynamics Newton's second law for an object A of mass m, using the hypothetical nonaccelerating, nonrotating reference frame with the origin at O 0 ,maybe written as ma A  mg A   FY 4X26 where a A is the acceleration of A relative to O 0 , g A is the resulting gravitational acceleration, and  F is the sum of all other external forces acting on A. By Eq. (4.25) the acceleration of A relative to O 0 is a A  a O  a Arel Y where a Arel is the acceleration of A relative to the earth centered reference frame and the acceleration of the origin O is equal to the gravitational acceleration of the earth, a O  g O . The earth-centered reference frame does not rotate v  0. If the object A is on or near the earth, its gravitational acceleration g A due to the attraction of the sun, etc., is nearly equal to the gravitational accelera- tion of the earth g O , and Eq. (4.26) becomes  F  ma Arel X 4X27 One may apply Newton's second law using a nonrotating, earth centered reference frame if the object is near the earth. In most applications, Newton's second law may be applied using an earth ®xed reference frame. Figure 4.13 shows a nonrotating reference frame with its origin at the center of the earth O and a secondary earth ®xed reference frame with its origin at a point B. The earth ®xed reference frame with the origin at B may be assumed to be an inertial reference and  F  ma Arel Y 4X28 where a Arel is the acceleration of A relative to the earth ®xed reference frame. The motion of an object A may be analysed using a primary inertial reference frame with its origin at the point O (Fig. 4.14). A secondary reference frame with its origin at B undergoes an arbitrary motion with angular velocity v and angular acceleration a. Newton's second law for the object A of mass m is  F  ma A Y 4X29 where a A is the acceleration of A acceleration relative to O. Equation (4.29) may be written in the form  F À ma B  2v  v Arel  a  BA  v Âv  BA  ma Arel Y 4X30 where a Arel is the acceleration of A relative to the secondary reference frame. The term a B is the acceleration of the origin B of the secondary reference frame relative to the primary inertial reference. The term 2v Âv Arel is the Coriolis acceleration, and the term À2mv  v Arel is called the Coriolis force. 4. Planar Kinematics of a Rigid Body 109 Dynamics Figure 4.13 Figure 4.14 110 Dynamics Dynamics This is Newton's second law expressed in terms of a secondary reference frame undergoing an arbitrary motion relative to an inertial primary reference frame. 5. Dynamics of a Rigid Body 5.1 Equation of Motion for the Center of Mass Newton stated that the total force on a particle is equal to the rate of change of its linear momentum, which is the product of its mass and velocity. Newton's second law is postulated for a particle, or small element of matter. One may show that the total external force on an arbitrary rigid body is equal to the product of its mass and the acceleration of its center of mass. An arbitrary rigid body with the mass m may be divided into N particles. The position vector of the i particle is r i and the mass of the i particle is m i (Fig. 5.1): m   N i1 m i X The position of the center of mass of the rigid body is r C   N i1 m i r i m X 5X1 Taking two time derivatives of Eq. (5.1), one may obtain  N i1 m i d 2 r i dt 2  m d 2 r C dt 2  ma C Y 5X2 where a C is the acceleration of the center of mass of the rigid body. Figure 5.1 5. Dynamics of a Rigid Body 111 Dynamics Let f ij be the force exerted on the j particle by the i particle. Newton's third law states that the j particle exerts a force on the i particle of equal magnitude and opposite direction (Fig. 5.1): f ji Àf ij X Newton's second law for the i particle is  j f ji  F ext i  m i d 2 r i dt 2 Y 5X3 where F ext i is the external force on the i particle. Equation (5.3) may be written for each particle of the rigid body. Summing the resulting equations from i  1toN , one may obtain  i  j f ji   i F ext i  ma C Y 5X4 The sum of the internal forces on the rigid body is zero (Newton's third law):  i  j f ji  0X The term  i F ext i is the sum of the external forces on the rigid body:  i F ext i   FX One may conclude that the sum of the external forces equals the product of the mass and the acceleration of the center of mass:  F  ma C X 5X5 If the rigid body rotates about a ®xed axis O (Fig. 5.2), the sum of the moments about the axis due to external forces and couples acting on the body is  M O  I O aY where I O is the moment of inertia of the rigid body about O and a is the angular acceleration of the rigid body. In the case of general planar motion, the sum of the moments about the center of mass of a rigid body is related to its angular acceleration by  M C  I C aY 5X6 Figure 5.2 112 Dynamics Dynamics where I C is the moment of inertia of the rigid body about its center of mass C . If the external forces and couples acting on a rigid body in planar motion are known, one may use Eqs. (5.5) and (5.6) to determine the acceleration of the center of mass of the rigid body and the angular acceleration of the rigid body. 5.2 Angular Momentum Principle for a System of Particles An arbitrary system with mass m may be divided into N particles P 1 Y P 2 Y FFFY P N . The position vector of the i particle is r i  OP i and the mass of the i particle is m i (Fig. 5.3). The position of the center of mass, C ,of the system is r C   N i1 m i r i am. The position of the particle P i of the system relative to O is r i  r C  CP i X 5X7 Multiplying Eq. (5.7) by m i , summing from 1 to N , one may ®nd that  N i1 m i CP i  0X 5X8 The total angular momentum of the system about its center of mass C is the sum of the angular momenta of the particles about C , H C   N i1 CP i  m i v i Y 5X9 where v i  dr i adt is the velocity of the particle P i . The total angular momentum of the system about O is the sum of the angular momenta of the particles, H O   N i1 r i  m i v i   N i1 r C  CP i Âm i v i  r C  mv C  H C Y 5X10 Figure 5.3 5. Dynamics of a Rigid Body 113 Dynamics or the total angular momentum about O is the sum of the angular momentum about O due to the velocity v C of the center of mass of the system and the total angular momentum about the center of mass (Fig. 5.4). Newton's second law for the i particle is  j f ji  F ext i  m i dv i dt Y and the cross product with the position vector r i , and sum from i  1toN gives  i  j r i  f ji   i r i  F ext i   i r i  d dt m i v i X 5X11 The ®rst term on the left side of Eq. (5.11) is the sum of the moments about O due to internal forces, and r i  f ji  r i  f ij  r i Âf ji  f ij 0X The term vanishes if the internal forces between each pair of particles are equal, opposite, and directed along the straight line between the two particles (Fig. 5.5). Figure 5.4 Figure 5.5 114 Dynamics Dynamics The second term on the left side of Eq. (5.11),  i r i  F ext i   M O Y represents the sum of the moments about O due to the external forces and couples. The term on the right side of Eq. (5.11) is  i r i  d dt m i v i   i d dt r i  m i v i Àv i  m i v i !  dH O dt Y 5X12 which represents the rate of change of the total angular momentum of the system about the point O. Equation (5.11) may be rewritten as  M O  dH O dt X 5X13 The sum of the moments about O due to external forces and couples equals the rate of change of the angular momentum about O. Using Eqs. (5.10) and (5.13), one may obtain  M O  d dt r C  mv C  H C r C  ma C  dH C dt Y 5X14 where a C is the acceleration of the center of mass. If the point O is coincident with the center of mass at the present instant C  O, then r C  0 and Eq. (5.14) becomes  M C  dH C dt X 5X15 The sum of the moments about the center of mass equals the rate of change of the angular momentum about the center of mass. 5.3 Equations of Motion for General Planar Motion An arbitrary rigid body with the mass m may be divided into N particles P i , i  1Y 2Y FFFY N . The position vector of the P i particle is r i  OP i and the mass of the particle is m i (Fig. 5.6). Let d O be the axis through the ®xed origin point O that is perpendicular to the plane of the motion of a rigid body x Y yY d O cxY y. Let d C be the parallel axis through the center of mass C , d C kd O . The rigid body has a general planar motion, and one may express the angular velocity vector as v  ok. The velocity of the P i particle relative to the center of mass is dR i dt  ok  R i Y where R i  CP i . The sum of the moments about O due to external forces and couples is  M O  dH O dt  d dt r C  mv C H C Y 5X16 5. Dynamics of a Rigid Body 115 Dynamics where H C   i R i  m i ok  R i  is the angular momentum about d C . The magnitude of the angular momen- tum about d C is H C  H C Á k   i R i  m i ok ÂR i  Á k   i m i R i  kÂR i  Á ko   i m i R i  kÁR i  ko   i m i jR i  kj 2 o   i m i r i oY 5X17 where the term jk  R i jr i is the perpendicular distance from d C to the P i particle. The identity a  bÁc  a Áb  c has been used. The moment of inertia of the rigid body about d C is I   i m i r 2 i Y Equation (5.17) de®nes the angular momentum of the rigid body about d C : H C  I o or H C  I ok  I vX Substituting this expression into Eq. (5.16), one may obtain  M O  d dt r C  mv C I vr C  ma C I aX 5X18 If the ®xed axis d O is coincident with d C at the present instant, r  0, and from Eq. (5.18) one may obtain  M C  I aX Figure 5.6 116 Dynamics Dynamics [...]... 10 .3 18.0 15.4 30 .0 14.5 17.2 1.6 6.7 31 .0 5 .3 6.5 48.0 26.0 18.5 30 .0 16.1 27.6 71.0 124.0 106.0 207.0 100.0 119.0 11.0 46.2 214.0 36 .5 44.8 33 1.0 179.0 127.0 207.0 111.0 190.0 3. 80 7.0 5.82 11.5 6.0 6.49 0.6 2.7 11.0 1.9 2.4 17.0 9.5 7.0 11.5 6.0 10.6 26.2 48 .3 40.1 79 .3 41.4 44.7 4.1 18.6 75.8 13. 1 16.5 117.0 65.5 48 .3 79 .3 41.4 73. 1 0 .33 4 0.285 0 .32 4 0.292 0.211 0 .32 6 0 .33 0.245 0.290 0.425 0 .35 0... 0 .32 6 0 .33 0.245 0.290 0.425 0 .35 0 0 .30 7 0 .32 0 0 .32 2 0.291 0 .34 9 0 .30 5 0.098 0.297 0 .30 9 0.282 0.260 0 .32 2 0.016 0.094 0 .30 7 0.411 0.065 0 .36 8 0 .31 9 0 .31 6 0.280 0.295 0.280 169 5 13 534 487 450 556 28 162 530 710 112 636 551 546 484 510 484 26.6 80.6 83. 8 76.5 70.6 87 .3 4 .3 25.4 83. 3 111.5 17.6 100.0 86.6 85.8 76.0 80.1 76.0 Source: J E Shigley and C R Mischke, Mechanical Engineering Design McGraw-Hill,... 152 De¯ections Analysis Using Singularity Functions 1 53 Impact Analysis 157 Strain Energy 160 Castigliano's Theorem 1 63 Compression 165 Long Columns with Central Loading 165 Intermediate-Length Columns with Central Loading 169 Columns with Eccentric Loading 170 Short Compression Members 171 3 Fatigue 3. 1 3. 2 3. 3 3. 4 3. 5 142 1 73 Endurance Limit 1 73 Fluctuating Stresses 178 Constant Life Fatigue Diagram... stresses s 1 ˆ E E1 E2 ˆ ÀnE1 s2 ˆ 0 E3 ˆ ÀnE1 s3 ˆ 0 E1 ˆ s1 ns2 À E E s1 ˆ E …E1 ‡ nE2 † 1 À n2 E2 ˆ Biaxial s2 ns1 À E E s2 ˆ E …E2 ‡ nE1 † 1 À n2 E3 ˆ À ns1 ns2 À E E s3 ˆ 0 E1 ˆ s1 ns2 ns3 À À E E E s1 ˆ E E1 …1 À n† ‡ nE …E2 ‡ E3 † 1 À n À 2n2 E2 ˆ s2 ns1 ns3 À À E E E s2 ˆ E E2 …1 À n† ‡ nE …E1 ‡ E3 † 1 À n À 2n2 E3 ˆ Triaxial s3 ns1 ns2 À À E E E s1 ˆ E E3 …1 À n† ‡ nE …E1 ‡ E2 † 1 À n À 2n2... external moment Mz ˆ M k, namely, … … E y 2 dAY …1X34† M ˆ ys dA ˆ r where the second integral in the foregoing equation is the second moment of area I about the z axis It is given by … …1X35† I ˆ y 2 dAX Manipulating Eqs (1 .34 ) and (1 .35 ), we obtain 1 M ˆ X r EI …1X36† Finally, eliminating r from Eqs (1 .32 ) and (1 .36 ) yields sˆÀ My X I …1X37† Equation (1 .37 ) states that the stress s is directly proportional... shear stresses t1a2 Y t2a3 , and t1a3 are also shown in Fig 1.5a Each of these shear stresses occurs on two planes, one of the planes being shown in Fig 1.5b The principal shear stresses can be calculated by the following equations t1a2 ˆ s 1 À s2 Y 2 t2a3 ˆ s2 À s3 Y 2 t1a3 ˆ s 1 À s3 X 2 …1X 13 If the normal principal stresses are ordered …s1 b s2 b s3 †, then tmax ˆ t1a3 Mechanics Figure 1.5 Mohr's... …1X29† dx Y ds …1X30† and the strain EˆÀ where the negative sign suggests that the beam is in compression Manipulating Eqs (1.28), (1.29), and (1 .30 ), we obtain y EˆÀ X r …1X31† Since s ˆ E E, the expression for stress is sˆÀ Ey X r …1X32† 137 1 Stress Since the x axis is the neutral axis, the preceding equation states that the moment of the area about the neutral axis is zero Thus, Eq (1 .33 ) de®nes the... …1X38† where c ˆ ymax Equation (1 .38 ) can also be written in the two forms sˆ M Y I ac sˆ M Y Z …1X39† where Z ˆ I ac is called the section modulus EXAMPLE 1.4 Determine the diameter of a solid round shaft OC in Fig 1. 13, 36 in long, such that the bending stress does not exceed 10 kpsi The transversal loads of 800 lb and 30 0 lb act on the shaft Solution The moment equation about C yields € MC ˆ 36 R1... bending moment diagrams shown in Figs 1.13b and 1.13c From the bending moment diagram, we observe that the maximum bending moment is M ˆ 600…12† ˆ 7200 lb inX The section modulus is I pd 3 ˆ 0X0982d 3 X ˆ 32 c sˆ Mechanics Then, the bending stress is M 7200 ˆ X I ac 0X0982d 3 Considering s ˆ 10Y000 psi and solving for d, we obtain s 7200 3 ˆ 1X94 inX m dˆ 0X0982…10000† 1.10... Circle 121 Triaxial Stress 125 Elastic Strain 127 Equilibrium 128 Shear and Moment 131 Singularity Functions 132 Normal Stress in Flexure 135 Beams with Asymmetrical Sections 139 Shear Stresses in Beams 140 Shear Stresses in Rectangular Section Beams Torsion 1 43 Contact Stresses 147 2 De¯ection and Stiffness 2.1 2.2 2 .3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 149 Springs 150 Spring Rates for Tension, . Members 171 3. Fatigue 1 73 3.1 Endurance Limit 1 73 3.2 Fluctuating Stresses 178 3. 3 Constant Life Fatigue Diagram 178 3. 4 Fatigue Life for Randomly Varying Loads 181 3. 5 Criteria of Failure 1 83 References. following equations t 1a2  s 1 À s 2 2 Y t 2a3  s 2 À s 3 2 Y t 1a3  s 1 À s 3 2 X 1X 13 If the normal principal stresses are ordered s 1 b s 2 b s 3 , then t max  t 1a3 . Figure 1.5 Mohr's circle. 127 1.6 Equilibrium 128 1.7 Shear and Moment 131 1.8 Singularity Functions 132 1.9 Normal Stress in Flexure 135 1.10 Beams with Asymmetrical Sections 139 1.11 Shear Stresses in Beams 140 1.12 Shear