ME 352 - Machine Design I Name of Student Spring ......ME 352 - Machine Design I Name of...

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ME 352 - Machine Design I Name of Student_________________________ Spring Semester 2010 Lab. Div. Number________________________ EXAM 1. OPEN BOOK AND CLOSED NOTES Thursday, February 18th, 2010

Please use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the figures that are provided on the exam to show vectors and instantaneous centers of velocity.

Problem 1 (25 Points).

Part I (13 Points). (i) Clearly number each link and label the lower pairs and the higher pairs of the mechanism shown in Figure 1(a). Then determine the mobility of this mechanism. (ii) Define vectors that would be suitable for a kinematic analysis of the mechanism. Label and show the direction of each vector on Figure 1(a). (iii) Write the vector loop equation(s) for the mechanism and: (a) Identify suitable input(s) for the mechanism. (b) Identify the known quantities, the unknown variables, and write any constraint equations. (c) Write the rolling constraint equation in terms of the position variables.

Figure 1(a). A Planar Mechanism.

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ME 352 - Machine Design I Name of Student_________________________ Spring Semester 2010 Lab. Div. Number________________________ Problem 1 (continued).

Part II (12 Points). Consider the four-bar linkage in the position shown in Figure 1(b). The angular position of the input link 2 is given by the angle o

2 120 ,θ = measured counterclockwise from the ground link which is chosen to be coincident with the fixed X-axis. The lengths of the four links are

1 2 4R O O 15 cm,= = 2 2R O A 6 cm,= = 3R AB 10 cm,= = and 4 4R O B 10 cm.= = Use Freudenstein's equation to determine the angular position of the output link 4θ as shown in the

figure. Note that the mechanism is not drawn to scale in Figure 1(b).

Figure 1(b). A Planar Four-Bar Linkage.

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ME 352 - Machine Design I Name of Student__________________________ Spring Semester 2010 Lab. Div. Number________________________

Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, the input link 2 has a constant angular velocity 2 5 rad / sω = clockwise. The link lengths and dimensions are 2O B 5 cm,= AB BC 5 cm,= = 4CO 10 cm,= and 4O D 5 cm.= (i) Write a vector loop equation that would be suitable for a kinematic analysis of the mechanism. Clearly indicate the input, the knowns, and any constraints. Draw your vectors clearly on Figure 2. (ii) Determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and 4. Give the magnitude and direction of each vector. (iv) Determine the relative velocity of point C fixed in link 3 with respect to point B fixed in link 2. Give the magnitude and the direction of this vector.

Figure 2. A Planar Mechanism.

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Problem 3 (25 Points). For the mechanism in the position shown in Figure 3, the angular velocity of the input link 2 is a constant 2 7 rad / sω = counterclockwise. Link 5 is rolling without slipping on link 3 at the point of contact C. The mechanism is drawn full scale in Figure 3, that is, 1 cm = 1 cm. (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the Kennedy circle shown below, show the locations of all the instant centers on Figure 3. Using the locations of the instant centers, determine: (iii) The first-order kinematic coefficients of links 3, 4, and 5. (iv) The magnitudes and directions of the angular velocities of links 3 and 5. (v) The magnitude and direction of the velocity of point C fixed in link 5.

Figure 3. A Planar Mechanism. The scale is 1 cm = 1 cm.

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Problem 4 (25 Points). For the mechanism in the position shown in Figure 4, the input link 2 has a velocity 2

ˆV 2 i cm / s= and an acceleration 22

ˆA 5 i cm / s .= − The roller, link 4, is rolling without slipping on the ground link 1. The known link lengths are AB 15 cm,= 1 5 cm,ρ = and 4 2 cm.ρ = (i) Write a suitable vector loop equation for a kinematic analysis of this mechanism. Indicate the input, the knowns, and the unknown variables. Draw your vectors clearly on Figure 4. (ii) Write symbolic equations for the rolling contact equation between link 4 and the ground link 1 in terms of: (a) the position variables, and (b) the first-order kinematic coefficients of the mechanism. (iii) Determine numerical values for the first-order kinematic coefficients of the mechanism. (iv) Determine the angular velocities of links 3 and 4. Give the magnitudes and directions of each vector.

Figure 4. A Planar Mechanism.

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Solution to Problem 1.

Part I. (i) 4 points. There are 6 links in the mechanism and the joint types are as shown in Figure 1(a).

Figure 1(a). Joint Types of the Mechanism.

The Kutzbach mobility criterion for a planar mechanism can be written as

1 2M 3(n 1) 2J 1J= − − − (1)

For this mechanism, the number of links, the number of lower pairs (or 1J joints), and the number of higher pairs (or 2J joints), respectively, are

1 2n 6, J 7, and J 0= = = (2)

Substituting Equation (2) into Equation (1), the mobility of the mechanism is

M 3(6 1) 2(7) 0= − − − (3a) that is

M 15 14 0 1= − − = (3b)

This is the correct answer for this mechanism, that is, for a single input there is a unique output.

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(ii) 5 points. Suitable vectors for a kinematic analysis of the mechanism are shown in Figure 1(b).

Figure 1(b). Vectors for the Mechanism.

(iii) 4 points. There are 5 unknown variables, therefore, 2 independent vector loop equations are required and one rolling contact equation. The two independent vector loop equations can be written as

Loop 1: 2 22 3 44 4 10C C

R R R R R R√Ι √ √? √ √? √√+ + − − − = (4a)

Loop 2: 4 44 33 5 9 110C C

R R R R R R√? √ √ √? √? √√+ + − − + = (4b)

(a) The input link is chosen to be the disk, denoted as link 2, and the input variable is the angle 2θ . (b) The five unknown variables are the angles 3,θ 4 ,θ 5 ,θ 6 ,θ and 9.θ

There are three constraint equations, namely

22 2θ θ α= − (5a)

33 3θ θ β= − (5b) and

44 4 90θ θ= − ° (5c)

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(c) The rolling contact equation between the wheel (denoted as link 6) and the ground link 1 can be written as

6 91

6 1 9

θ θρρ θ θ

Δ −Δ± =

Δ −Δ (6a)

The correct sign in Equation (6a) is the negative sign because there is external rolling contact, that is

6 91

6 90θ θρ

ρ θΔ −Δ

− =−Δ

(6b)

Note that Equations (6) can also be written in terms of the first-order kinematic coefficients as

6 91

6 1 9

θ θρρ θ θ

′ ′−± =

′ ′− (7a)

that is 6 91

6 90θ θρ

ρ θ′ ′−

− =′−

(7b)

Part II (12 points). The vectors for the four-bar linkage are shown in Figure 1(b).

Figure 1(b). The vector loop for the four-bar linkage.

The vector loop equation (VLE) can be written as

√ I √ ? √ ? √ √ 2 3 4 1R R R R 0+ − − = (1)

The X and Y components of Eq. (1) are

2 2 3 3 4 4 1 1cos cos cos cos 0R R R Rθ θ θ θ+ − − = (2a) and

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2 2 3 3 4 4 1 1sin sin sin sin 0R R R Rθ θ θ θ+ − − = (2b)

Freudenstein's Equation can be written as

A cos B sin C4 4θ θ+ = (3) where

1 4 1 2 4 2A 2 R R cos 2 R R cosθ θ= − (4a)

1 4 2 4B 2 R R sin 2 R R sin1 2= −θ θ (4b) and

2 2 2 2C R R R R 2 R R cos ( )3 1 2 4 1 2 1 2= − − − + −θ θ (4c)

Substituting the known data into Equations (4) gives

2A 2 15 10 cos 0 2 6 10 cos 120 360 cm= × × − × × = + (5a)

2B 2 15 10 sin 0 2 6 10 sin 120 103.92 cm= × × × − × × × = − (5b) and

2 22 2 2C 10 15 6 10 2 x 15 x 6 cos (0 120 ) 351 cm= − − − + − = − (5c)

Substituting Equations (5) into Equation (3) gives

24 4360 cos 103.92 sin 351 cmθ θ+ − = − (6)

To determine the output angle we can write this transcendental equation as an algebraic equation, (namely, a quadratic equation). The procedure is to use the tangent of the half-angle relationship; i.e.,

4Z tan ( )2θ= (7a)

which gives

22 Zsin 4 1 Z

θ =+

and 2

21 Zcos 4 1 Z

θ −=+

(7b)

Substituting Equations (7b) into Equation (3), and rearranging, gives

2(A C) Z (2 B) Z (C A) 0+ − + − = (8)

The solution to this quadratic equation can be written as

2B B (A C)(C A)ZA C

+ ± − + −=+

(9)

Substituting Equations (5) into Equation (9) gives

2103.92 ( 103.92) (360 351)( 351 360)Z(360 351)

− ± − − − − −=−

(10a)

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The two roots to this quadratic equation are

1 2Z 3.025 and Z 26.118= + = − (10b)

Writing Equation (7a) as

I 14 12 tan Zθ −= and II 1

4 22 tan Zθ −= (11)

Then substituting Equations (10b) into Equations (11) gives

I 14 2 tan 3.025θ −= and II 1

4 2 tan 26.118θ −= − (12)

Therefore, the two possible answers for the angular position of link 4 are

143.414θ = and 175.614θ = − (13)

The correct answer for the angular position of link 4, for the given open configuration shown in Figure 1(b), is

143.414θ = (14)

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Solution to Problem 2. (i) 8 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Fig. 2(a).

Figure 2(a). Suitable vectors for the planar mechanism.


√ I ? C ? C √ √ 2 23 34 1R R R R 0+ + − = (1)

The input is the angular position of link 2, that is, 2θ . For the given input position 2 315 .θ = ° The knowns are the length of link 2, the length of the ground link 1, and the angle of the ground link. The angular position of link 3 is constrained to be perpendicular to link 2, that is

23 2 90 45θ θ= + ° = ° (2)

Also, the angular position of link 4 is constrained to be perpendicular to link 3, that is

34 23 290 180 135θ θ θ= + ° = + ° = ° (3)

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(ii) 6 points. The X and Y components of the VLE, see Eq. (1), are

2 2 23 23 34 34 1 1R cos R cos R cos R cos 0θ + θ + θ − θ = (4a) and

2 2 23 23 34 34 1 1R sin R sin R sin R sin 0θ + θ + θ − θ = (4b)

Differentiating Equations (4) with respect to the input position 2θ gives

2 2 23 23 23 23 34 34 34 34R sin R sin cos R R sin cos R 0′ ′− θ − θ + θ − θ + θ = (5a) and

2 2 23 23 23 23 34 34 34 34R cos R cos sin R R cos sin R 0′ ′θ + θ + θ + θ + θ = (5b)

Then writing Equations (5) in matrix form gives

23 34 23 2 2 23 23 34 34

23 34 34 2 2 23 23 34 34

cos cos R R sin R sin R sinsin sin R R cos R cos R sin

′θ θ θ + θ + θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′θ θ − θ − θ − θ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(6)

Substituting the known data into Equation (6) gives

23

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R0.7071 radians 0.7071 radians 7.071cmsR0.7071 radians 0.7071 radians 0′+ − +⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′+ +⎣ ⎦ ⎣ ⎦⎣ ⎦ (7a)

The determinant of the coefficient matrix in Equation (7a) can be written as

DET ( 0.7071 radians)( 0.7071 radians) ( 0.7071 radians)( 0.7071 radians)= + + − + − (7b)

Therefore, the determinant of the coefficient matrix is

2DET 1.000 radians= (7c)

Using Cramer’s rule, the first-order kinematic coefficient for link 3 with respect to link 2 is

235 cmR 5 cm / rad

1 radians+′ = = +

+ (8)

The positive sign indicates that the distance from point B to point C, that is, the vector 23R , is decreasing in length as link 2 rotates clockwise.

From Cramer’s rule, the first-order kinematic coefficient for link 3 with respect to link 4 is

345 cmR 5 cm / rad

1 radians−′ = = −

+ (9)

The negative sign indicates that the distance from point C to the ground pivot 4O , that is the vector

34R , is increasing in length as link 2 rotates clockwise. Differentiating Equation (2) with respect to the input position, the third kinematic coefficient of the

mechanism is

23 3 2 1θ θ θ′ ′ ′= = = + (10)

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(iii) 7 Points. The angular velocity of link 3 can be written as

3 3 2′ω = θ ω (11)

Substituting Equation (10) into Equation (11), the angular velocity of link 3 is

3 21 5 rad / sω = + ω = − (12)

The negative sign indicates that link 3 is rotating clockwise. The angular velocity of link 4 can be written as

4 4 2′ω = θ ω (13)

Differentiating Equation (3) with respect to the input position, and using Equation (11) gives

34 4 3 1θ θ θ′ ′ ′= = = + (14)

Then substituting Equation (14) into Equation (13) gives

4 21 5 rad / sω = + ω = − (15)

The negative sign indicates that link 4 is rotating clockwise. (iv) 7 Points. The relative velocity between point C fixed in link 3 and point B fixed in link 2 can be written as

CB 23 2V R′= ω (16)

The relative velocity between point C fixed in link 3 and point B fixed in link 2 is directed along the line BC in the direction from C to B.

Substituting Equation (8) and the input angular velocity into Equation (16) gives

CBV ( 5 cm / rad)( 5 rad / s) 25 cm / s= + − = − (17)

The positive sign indicates that point C is moving towards point B, that is, the distance BC is decreasing (the angle of the velocity vector is inclined at 135 degrees clockwise to the horizontal axis).

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Solution to Problem 3. (i) and (ii) 15 Points. This mechanism has six links, therefore, the total number of instant centers is

( 1) 6 x 5 152 2

n nN −= = = (1)

There are 7 primary instant centers; namely,

I12, I23, I34, I14, I35, I56, and I16 (2)

Therefore, there are 8 secondary instant centers; namely,

I13, I15, I24, I25, I26, I36, I45, and I46 (3)

The 8 secondary instant centers can be obtained as follows: (i) The point of intersection of the line through I12I23 and the line through I14I34 is the instant center I13. (ii) The point of intersection of the line through I13I35 and the line through I16I56 is the instant center I15. (iii) The point of intersection of the line through I12I14 and the line through I23I34 is the instant center I24. (iv) The point of intersection of the line through I12I15 and the line through I23I35 is the instant center I25. (v) The point of intersection of the line through I25I56 and the line through I12I16 is the instant center I26. (vi) The point of intersection of the line through I35I56 and the line through I13I16 is the instant center I36. (vii) The point of intersection of the line through I14I15 and the line through I34I35 is the instant center I45. (viii) The point of intersection of the line through I14I16 and the line through I45I56 is the instant center I46.

The procedure to locate these 8 secondary instant centers is indicated on the Kennedy circle below, see Figure 3(a).

Figure 3(a). The Kennedy Circle.

The locations of all 15 instant centers for this mechanism are shown on Figure 3(b).

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Figure 3(b). The locations of the fifteen instant centers.

(iii) 6 Points. The first-order kinematic coefficient of link 3 can be written as

12 233

13 23

I II I

θ ′ = (4a)

The distance 12 23I I is measured as 12 23 2.55 cmI I = and the distance 13 23I I is measured as

13 23 7.88 cm.I I = Therefore, the first-order kinematic coefficient of link 3 is

32.55 cm 0.324 rad/rad7.88 cm

θ ′ = = (4b)

Note that the correct sign the first-order kinematic coefficient of link 3 is negative because the relative instant center 23I lies between the two absolute instant centers 12I and 13I , that is

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3 0.324 rad/radθ ′ = − (4c)

The first-order kinematic coefficient of link 4 can be written as

4 12 24R I I′ = (5a)

From measuring figure, 12 24 1.70 cm,I I = therefore, the first-order kinematic coefficient of link 4 is

4 1.70 cmR′ = (5b)

Note that the correct sign is negative because link 4 will move to the left for a positive change in the input (that is, the distance from O2 to point B is decreasing for a positive change in the input). Therefore, the first-order kinematic coefficient of link 4 is

4 1.70 cmR′ = − (5c)


12 255

15 25

I II I

θ ′ = (6a)



53.38 cm 0.805 rad/rad4.20 cm

θ ′ = = (6b)

Note that the correct sign is negative because the relative instant center 25I lies between the two absolute instant centers 12I and 15I , that is, the first-order kinematic coefficient of link 5 is

5 0.805 rad/radθ ′ = − (6c)


12 266

16 26

I II I

θ ′ = (7a)



61.65 cm 0.206 rad/rad8.00 cm

θ ′ = = (7b)

Note that the correct sign is negative because the relative instant center 26I lies between the two absolute instant centers 12I and 16I , that is, the first-order kinematic coefficient of link 6 is

6 0.206 rad/radθ ′ = − (7c)

(iv) 2 Points. The angular velocity of link 3 can be written as

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3 3 2 ( 0.324 rad rad)( 7 rad sec) 2.268 rad secω θ ω′= = − + = − (8)

The negative sign implies that the direction of the angular velocity of link 3 is clockwise, see Fig. 3(c). The angular velocity of link 5 can be written as


The negative sign implies that the direction of the angular velocity of link 5 is clockwise, see Fig. 3(c). Check. The relative instant center 35I does not lie between the two absolute instant centers 13I and 15 ,I therefore, the angular velocity of link 5 is in the same direction as the angular velocity of link 3.

The angular velocity of link 6 can be written as


The negative sign implies that the direction of the angular velocity of link 6 is clockwise, see Fig. 3(c).

Figure 3(c). The angular velocities of links 3, 5 and 6 and the velocities of points B and C.

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Check. The relative instant center 26I lies between the two absolute instant centers 12I and 16.I Therefore, the angular velocity of link 6 is in the opposite direction to the angular velocity of link 2, that is, clockwise. (v) 2 Points. The velocity of point C, fixed in link 3, can be written as

3C 3 13V (I C)= ω (11)

The distance 13I C is measured as 13 5.06 cm.I C = Therefore, the velocity of point C fixed in link 3 is

3CV (2.268 rad / s)(5.06 cm) 11.48 cm / s= = (12)

The direction of the velocity of point C is perpendicular to the line 13 ,I C that is, 130 degrees to the horizontal as shown in Figure 3(c). Check: The velocity of point C, fixed in link 5, can be written as

5 3C C 5 15V V (I C)= = ω (13)

The distance 15I C is measured as 15 2.04 cm.I C = Therefore, the velocity of point C, fixed in link 5, is

CV (5.635 rad / s)(2.04 cm) 11.50 cm / s= = (14)

The direction of the velocity of point C is perpendicular to the line 15I C as shown in Figure 3(c). The velocity of point B, fixed in link 4, can be written as

4 3B B 3 13V V (I B)= = ω (15)

The distance 13I B is measured as 13 5.25 cm.I B = Therefore, the velocity of point B fixed in link 4 is

4BV (2.268 rad / s)(5.25 cm) 11.91 cm / s= = (16)

The direction of the velocity of point B is along the X-axis as shown in Figure 3(c). Check: The velocity of point B, fixed in link 4 can be written as

4B 4 2V R′= ω (17)

Substituting Equation (5b) into Equation (17), the velocity of point B fixed in link 4 is

4BV (1.70 cm)(7 rad / s) 11.90 cm / s= = (18)

This answer is in good agreement with Equation (16).

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Solution to Problem 4. (i) 5 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Fig. 4(a).

Figure 4(a). Suitable vectors for the mechanism.


I √ √ √ √ ? √ ? 2 1 9 3R R R R 0− + − = (1)

where the input is the length 2R which defines the position of link 2, and the two unknown variables 9θ and 3θ are the angular positions of the arm (link 9) and link 3. (ii) 6 points. The rolling contact constraint between link 4 and the ground link 1 of the mechanism can be written as

4 91

4 1 9

θ θρρ θ θ

Δ − Δ± =

Δ − Δ (2a)

The correct sign is negative because there is external contact between links 4 and 1, that is

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4 91

4 1 9

θ θρρ θ θ

Δ − Δ− =

Δ − Δ (2b)

Equation (2b) can be written in terms of the first-order kinematic coefficients of the mechanism as

4 91

4 1 9

θ θρρ θ θ

′ ′−− =

′ ′− (3)

(iii) 9 points. The X and Y components of the VLE, see Eq. (1), are

2 2 1 1 9 9 3 3R cos R cos R cos R cos 0θ − θ + θ − θ = (4a) and

2 2 1 1 9 9 3 3R sin R sin R sin R sin 0θ − θ + θ − θ = (4b)

Differentiating Equations (4) with respect to the input position 2R gives

2 9 9 9 3 3 31cos R sin R sin 0′ ′+ θ − θ θ + θ θ = (5a) and

2 9 9 9 3 3 31sin R cos R cos 0′ ′+ θ + θ θ − θ θ = (5b)

Then writing Equations (5) in matrix form gives

9 9 3 3 9 2

9 9 3 3 3 2

R sin R sin 1cosR cos R cos 1sin

′− θ + θ θ − θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′+ θ − θ θ − θ⎣ ⎦⎣ ⎦ ⎣ ⎦

(6)

Substituting the known data into Equation (6) gives

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3

4.95 cm 12.99 cm 1 rad4.95 cm 7.5 cm 0

′θ− − −⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥′θ+ −⎣ ⎦ ⎣ ⎦⎣ ⎦

(7a)

The determinant of the coefficient matrix in Equation (7a) is

2DET ( 4.95 cm)( 7.5 cm) ( 4.95 cm)( 12.99 cm) 101.43 cm= − − − + − = + (7b)

Using Cramer’s rule, the first-order kinematic coefficient for link 9 is

9 27.5 cm 0.074 rad / cm

101.43 cm+′θ = = +

+ (8)

The positive sign indicates that link 9 is rotating clockwise as the change in the input position 2R decreases in length.

From Cramer’s rule, the first-order kinematic coefficient for link 3 is

3 24.95 cm 0.049 rad / cm

101.43 cm+′θ = = +

+ (9)

The positive sign indicates that link 3 is rotating clockwise as the change in the input position 2R decreases in length.

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The first-order kinematic coefficient of link 4 in the mechanism can be obtained from Equation (3). The equation can be written as

4 91

4 90θ θρ

ρ θ′ ′−

− =′−

(10a)

Rearranging this equation, the first-order kinematic coefficient of link 4 can be written as

14 9

4

1 ρθ θρ

⎛ ⎞′ ′= +⎜ ⎟

⎝ ⎠ (10b)

Substituting known values into Equation (10b), the first-order kinematic coefficient for link 4 is

45 cm0.074 rad/cm 1 0.259 rad/cm2 cm

θ⎛ ⎞

′ = + + = +⎜ ⎟⎝ ⎠

(11)

(iv) 5 Points. The angular velocity of link 3 can be written as

3 3 2R′ω = θ (12)

Note that the length of the vector 2R is decreasing as the input link 2 is moving to the right (the velocity of link 2 is given to the right). Therefore, the time rate of change in the length of this vector is

2 2 2 cm/sR V= − = − (13)

Substituting Eq. (9) and Eq. (13) into Eq. (12), the angular velocity of link 3 is

3 ( 0.049 rad/cm)( 2 cm/s) 0.098 rad / sω = + − = − (14)

The negative sign indicates that link 3 is indeed rotating clockwise, that is, the angular velocity of link 3 is clockwise.

The angular velocity of link 4 can be written as

4 4 2R′ω = θ (15)


4 ( 0.259 rad/cm)( 2 cm/s) 0.518 rad / sω = + − = − (16)

The negative sign means that link 4 is rotating clockwise. Aside. The angular velocity of link 9 can be written as

9 9 2R′ω = θ (17)


9 ( 0.074 rad/cm)( 2 cm/s) 0.148 rad / sω = + − = − (18)

The negative sign means that link 9 (that is, the arm) is rotating clockwise, that is, the angular velocity of link 9 is clockwise

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