Đang tải... (xem toàn văn)
Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1. Động lực học công trình Tài liệu tiếng Anh 1.
Chapter Particle Kinematics Part I Rectilinear Motion A B C D Rectilinear Motion of Particles Uniformly Accelerated Rectilinear Motion Relative Motion Constrained Motion A Rectilinear Motion of Particles Position, x The vector connecting the origin and the particle For the rectilinear motion, the direction may be replaced by a plus/minus sign Displacement, x The vector connecting the position of a particle at time t to that at time t + t Note that if x0 = then x = x Velocity, v x dx = t0 t dt v = lim Chapter 1, Page Rectilinear Motion of Particles Accelration Acceleration, a v dv d x a = lim = = t0 t dt dt also dv dx dv a= =v dx dt dx Chapter 1, Page Problem 1.1 (Sample Problem 11.1, Page 609) The position of a particle is defined by x = t 6t 15t + 40 (m) Determine (a) the time at which the velocity will be zero, (b) the position and distance traveled by the particle at that time, (c) the acceleration of the particle at that time, and (d) the distance traveled by the particle from t = s to t = s Velocity: dx v= = 3t 12t 15 (m/s) dt Acceleration: a= dv = 6t 12 (m/s ) dt (a) t = s (b) x5 = 60 m d05 = 100 m (c) a5 = 18 m s (d) d46 = 18 m Chapter 1, Page Determination of Motions Key Equations v= a= Knowing a = f (t) dv = f (t) dt dv = f (t)dt dx dt dv dt dv a=v dx v v0 dv = v v0 = v= t t t 0 f (t) dt f (t) dt dx = v(t) dt dx = vdt x x0 dx = x x0 = x= t v(t) dt t v(t) dt t v(t) dt + x 0 f (t) dt + v0 Chapter 1, Page Problem 1.2 (Sample Problem 11.2, Page 610) dv = 9.81 dt dv = 9.81dt Ball tossed with 10 m/s vertical velocity from window 20 m above ground Determine (a) velocity and elevation above ground at time t, (b) highest elevation reached by ball and corresponding time, and (c) time when ball will hit the ground and corresponding velocity (a) v = 10 9.81t (m/s) , y = 20 + 10t 4.905t (m) (b) y = 25.1 m when t = 1.019 s (c) v = 22.2 m s () when t = 3.28 s v 10 dv = t 9.81dt v t v = 9.81t 10 v 10 = 9.81t v = 10 9.81t (m/s) dy = 10 9.81t dt y 20 dy = t (10 9.81t) dt y = 20 + 10t 4.905t (m) Chapter 1, Page Determination of Motions Key Equations v= Knowing a = f (x) dv = f (x) dx vdv = f (x)dx v dx dt dv a= dt a=v dv dx dx = v(x) dt v v0 v dv = v v0 = 2 x x0 x x0 x f (x) dx f (x) dx v = v02 + f (x) dx x0 dx v dt = t dt = t= x x0 x x0 dx v dx v Solve for x, x = x(t) Chapter 1, Page Problem 1.3 Simple Harmonic Motion A block of mass m is attached to a spring of constant k The spring is stretched to a length of R and then released Express the velocity and position of the block with time t a= k x , x0 = R , v0 = m v v dv k = x dx m v dv = x R ( k v = (R x ) m Let R dx = R sin dt d(Rcos ) = dt R sin d = dt k x) dx m x = Rcos , SIMULATION k = 2 m then v = R sin t dt = d = t x = Rcos t v = R sin t t= x Chapter 1, Page Determination of Motions Key Equations Knowing a = f (v) Method 1: dv dt dv dx dv f (v) dt = t a=v v dv = f (v) dt dx v= dt a= Method 2: dt = t= v dv f (v) v0 dv f (v) v v0 dv = f (v) dx dx = v x = x0 + v x vdv f (v) x0 dx = v0 v0 vdv f (v) vdv f (v) Chapter 1, Page Problem 1.4 (Sample Problem 11.3, Page 611) Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity Determine v(t), x(t), and v(x) Knowing a = kv , v0 , x0 = dv = kv dt dv = kdt v v v0 t dv = k dt v v ln v = kt v0 ln dx = v0 e kt dt dx = v0 e kt dt x dx = t ve 0 kt dt t v x = e kt k 0 v x = (1 e kt ) k v = kt v0 v = v0 e kt Chapter 1, Page Problem 1.4 By eliminating t, we obtain a velocity in terms of x v = v0 e kt e kt x= Alternative Method Knowing a = kv dv = kv dx dv = kdx v = v0 v we have v0 (Continued) kt (1 e ) k v v = (1 ) v0 k v v0 x dv = k dx v = v0 kx v = v0 kx Chapter 1, Page 10 B Uniformly Accelerated Rectilinear Motion Key Equations v= dx dt a = constant dv =a dt v v0 a= dv dt dv a=v dx dv =a dx vdv = adx v t dv = a dt v = v0 + at dx = v0 + at dt x = x0 + v0t + at 2 v v0 x v dv = a dx x0 (v v02 ) = a(x x0 ) v = v02 + 2a(x x0 ) Velocity of freefall v = 2gh Chapter 1, Page 11 Problem 1.5 (Problem 11.36, Page 624) (a) y = y1 + v1t + at = 89.6 + v1 (16) + (9.81)(16)2 v1 = 72.9 m/s (b) v = v12 + 2a( y y1 ) = (72.9)2 + 2(9.81)( ymax 89.6) ymax = 360 m Chapter 1, Page 12 C Relative Motion Retilinear Motion x B A = x B x A or x B = x A + x B A v B A = v B v A or v B = v A + v B A aB A = aB a A or aB = a A + aB A Key Concept Physical meanings of x B A , v B A , and aB A : motion of B observed from A Chapter 1, Page 13 Problem 1.6 (Sample Problem 11.4, Page 620) Ball thrown vertically upward from 12 m level in elevator shaft with initial velocity of 18 m/s At same instant, open-platform elevator passes m level moving upward at m/s Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact Ball v B = 18 9.81t y B = 12 + 18t 4.905t Elevator vE = y E = + 2t (a) y B = y E , t = 3.65 s (b) v B E = v B v E = 16 9.81t vB E t =3.65 = 19.81 m/s Chapter 1, Page 14 D Constrained Motion Key Concept Degrees of freedom = number of moving parts number of constraints x A + 2x B = constant 2x A + 2x B + xC = constant x A1 + 2x B1 = x A2 + 2x B2 2x A1 + 2x B1 + xC1 = 2x A2 + 2x B2 + xC x A + 2x B = 2x A + 2x B + xC = v A + 2v B = 2v A + 2v B + vC = a A + 2aB = 2a A + 2aB + aC = Chapter 1, Page 15 Problem 1.7 (Sample Problem 11.5, Page 621) Pulley D is pulled down with a constant velocity of 75 mm/s At t = 0, collar A starts moving down from K with a constant acceleration and no initial velocity Knowing that velocity of collar A is 300 mm/s as it passes L, determine the displacement, velocity, and acceleration of block B when block A is at L SIMULATION Collar A: 300 = a At a t 200 = A t=4 s a A = 225 mm/s ( ) moving parts - cable = DOFs x, v, a Kinematic constraints: x A + 2x D + x B = const x A + 2x D + x B = v A + 2v D + v B = a A + 2aD + aB = At time t = s a A = 225 ( ), aD = aB = 225 mm/s () v A = 300 ( ), v D = 75 ( ) v B = 450 mm/s ( ) x A = 200 ( ), v D = 75 × = 100 ( ) x B = 400 mm ( ) Chapter 1, Page 16 Problem 1.8 (Problem 11.49, Page 628) moving points - cables = DOFs Kinematic constraints: xC + 2x E = x E + xW = vC + 2v E = v E + vW = (a, b) v E = 4.5 m/s ( ) vC = 9.0 m/s ( ) vW = 4.5 m/s ( ) (c) vC (d) vW E E = vC v E = 13.5 m/s ( ) = vW v E = 9.0 m/s ( ) Chapter 1, Page 17 [...]... at dx = v0 + at dt 1 x = x0 + v0t + at 2 2 v v0 x v dv = a dx x0 1 2 (v v02 ) = a(x x0 ) 2 v 2 = v02 + 2a(x x0 ) Velocity of freefall v = 2gh Chapter 1, Page 11 Problem 1. 5 (Problem 11 .36, Page 624) (a) y = y1 + v1t + 1 2 at 2 1 0 = 89.6 + v1 (16 ) + (9. 81) (16 )2 2 v1 = 72.9 m/s (b) v 2 = v12 + 2a( y y1 ) 0 = (72.9)2 + 2(9. 81) ( ymax 89.6) ymax = 360 m Chapter 1, Page 12 C Relative Motion... from A Chapter 1, Page 13 Problem 1. 6 (Sample Problem 11 .4, Page 620) Ball thrown vertically upward from 12 m level in elevator shaft with initial velocity of 18 m/s At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact Ball v B = 18 9.81t y B = 12 + 18 t 4.905t 2 Elevator... Elevator vE = 2 y E = 5 + 2t (a) y B = y E , t = 3.65 s (b) v B E = v B v E = 16 9.81t vB E t =3.65 = 19 . 81 m/s Chapter 1, Page 14 D Constrained Motion Key Concept Degrees of freedom = number of moving parts number of constraints x A + 2x B = constant 2x A + 2x B + xC = constant x A1 + 2x B1 = x A2 + 2x B2 2x A1 + 2x B1 + xC1 = 2x A2 + 2x B2 + xC 2 x A + 2x B = 0 2x A + 2x B + xC = 0 v A +... Chapter 1, Page 15 Problem 1. 7 (Sample Problem 11 .5, Page 6 21) Pulley D is pulled down with a constant velocity of 75 mm/s At t = 0, collar A starts moving down from K with a constant acceleration and no initial velocity Knowing that velocity of collar A is 300 mm/s as it passes L, determine the displacement, velocity, and acceleration of block B when block A is at L SIMULATION Collar A: 300 = a At 1. .. ) 3 moving parts - 1 cable = 2 DOFs x, v, a Kinematic constraints: x A + 2x D + x B = const x A + 2x D + x B = 0 v A + 2v D + v B = 0 a A + 2aD + aB = 0 At time t = 4 3 s a A = 225 ( ), aD = 0 aB = 225 mm/s 2 () v A = 300 ( ), v D = 75 ( ) v B = 450 mm/s ( ) x A = 200 ( ), v D = 75 × 4 = 10 0 ( ) 3 x B = 400 mm ( ) Chapter 1, Page 16 Problem 1. 8 (Problem 11 .49, Page 628) 3...Problem 1. 4 By eliminating t, we obtain a velocity in terms of x v = v0 e kt e kt x= Alternative Method Knowing a = kv dv = kv dx dv = kdx v = v0 v we have v0 (Continued) kt (1 e ) k v v = 0 (1 ) v0 k v v0 x dv = k dx 0 v = v0 kx v = v0 kx Chapter 1, Page 10 B Uniformly Accelerated Rectilinear Motion Key Equations v= dx... 16 Problem 1. 8 (Problem 11 .49, Page 628) 3 moving points - 2 cables = 1 DOFs Kinematic constraints: xC + 2x E = 0 x E + xW = 0 vC + 2v E = 0 v E + vW = 0 (a, b) v E = 4.5 m/s ( ) vC = 9.0 m/s ( ) vW = 4.5 m/s ( ) (c) vC (d) vW E E = vC v E = 13 .5 m/s ( ) = vW v E = 9.0 m/s ( ) Chapter 1, Page 17