Chapter 15 of Kinematics, Dynamics, and Design of Machinery 2nd Ed. by Waldron and Kinzel
February 27, 2017 | Author: jla101126 | Category: N/A
Short Description
This file contains the answers to chapter 15 (shaking forces and balances) of Waldorn and Kinzel. I hope this material h...
Description
Solutions to Chapter 15 Exercise Problems Problem 15.1 The figure below shows a system with two weights, WA and WB, which have been found to balance a system of weights (not shown) on the shaft. The weights for W A and W B are 4 and 8 lb, respectively, and the radii, rA and rB , are both 6 inches. Later, it is decided to replace W A and W B by two weights, WC and W D, where the planes for the two weights are as shown. What are the magnitudes and angular locations of WC and WD if the radius of the center of gravity for both links is 5 in? Plane C
Plane D
WB
WA
WB
rB 30˚
rA 30˚
WA
5" 9" 15"
Solution
To be dynamiclly equivalent to W A and W B, W C and W D must create the same dynamic force and dynamic moment system. The weights do not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces in planes A and B must be the same as the summation for planes C and D. Then, and
WArA cos A + WBrB cos B = WC rC cosC + WDrD cos D
(1)
WArA sin A + WBrB sin B = WC rC sinC + WDrDsin D
(2)
For an equivalent moment system, both systems will create the same net moment about the point where the shaft intersects plane B. Let the z axis point along the shaft from A to B. Then, zCk WC rC(cosC i + sinC j) + zDk WDrD(cos Di + sin D j) = zAk WArA(cos Ai + sin A j) Equating components and simplifying and
zCWC rC sinC + zDWDrD sinD = zAWArA sin A
(3)
zCWC rC cosC + zDWDrD cosD = zAWArA cos A
(4)
The angles are shown in Fig. 15.1.1. In Eqs (1)-(4), there are four unknowns: WC, WD, C, and D. The known values are: WA = 4 lb WB = 8 lb rC = 5 in rD = 5 in
rA = 6 in rB = 6 in zC = 6 in zD = 10 in
A = 30˚ B = 150˚
zA = 15 in zB = 0 in
- 580 -
WB
y
WC
WC Plane D WD
Plane C
rC
WD
rD
rA
WB
WA
rB z
6"
θA
θB θC
10" 15"
θD
Fig. 15.1.1: Equivalent counterbalance forces
To simplify Eqs. (1) - (4), define WC cosC = CC WD cosD = CD WC sinC = SC WD sinD = SD Equations (1) - (4) then become rC(CC ) + rD(CD) = WArA cos A + WBrB cos B rC(SC ) + rD(SD) = WArA sin A + WBrB sin B zCrCSC + zDrDSD = zAWArA sin A zCrCCC + zDrDCD = zAWArA cos A In matrix form, rC
0
0
zCrC
rD 0 0 zDrD
0 rC zC rC 0
0 CC WArA cos A + WBrB cos B
rD CD WArA sin A + WBrB sin B = zDrD SC zAWArA sin A 0 SD) zAWArA cos A
Substitution for the known values gives 5 0 0 CC 4(6)cos30˚+8(6)cos150˚ 5 0 0 5 5 CD 4(6)sin30˚+8(6)sin150˚
= 0 0 6(5) 10(5)
SC 15(4)(6)sin30˚ 6(5) 10(5) 0 0 SD) 15(4)(6)cos30˚ or 5
0
0
30
5 0 0 CC 20.785
0 5 5 CD 36 = 0 30 50 SC 180 50 0 0 SD) 311.77
Solving for the unknowns using MATLAB gives - 581 -
WA x
CC = WC cosC = 25.98
(5)
CD = WD cos D = 21.82
(6)
SC = WC sinC = 9
(7)
SD = WD sin D = 1.8
(8)
To solve for C , divide Eq. (7) by Eq. (5). Then, W sin C = tan1 WC cosC = tan1 9 = 160.89˚ 25.98 C C
(
)
(
)
From Eq. (5), WC = 25.98 / cos C = 25.98 / cos(160.89˚) = 27.49 lbs To solve for D , divide Eq. (8) by Eq. (6). Then, W sin D = tan1 WDcosD = tan1 1.8 = 4.71˚ 21.82 D D
(
)
(
)
From Eq. (6), WD = 21.82 / cos D = 21.82 / cos(4.71˚) = 21.89 lbs
Problem 15.2 The figure below shows a system with two weights, WA and WB, which have been found to balance a system of weights (not shown) on the shaft. The weights for W A and W B are 6 and 8 lb, respectively, and the radii, rA and rB , are both 5 inches. It is decided to replace WA and WB by two weights, WC and WD, where the planes for the two weights are as shown. What are the magnitudes and angular locations of WC and WD if the radius of the center of gravity for both links is 6 in? Plane C
WB
Plane D
WA
5"
WB
rB 30˚
rA 30˚
WA
4"
15"
Solution To be dynamiclly equivalent to W A and W B, W C and W D must create the same dynamic force and dynamic moment system. The weights do not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces in planes A and B must be the same as the summation for planes C and D. Then,
- 582 -
and
WArA cos A + WBrB cos B = WC rC cosC + WDrD cos D
(1)
WArA sin A + WBrB sin B = WC rC sinC + WDrDsin D
(2)
For an equivalent moment system, both systems will create the same net moment about the point where the shaft intersects plane B. Let the z axis point along the shaft from A to B. Then, zCk WC rC(cosC i + sinC j) + zDk WDrD(cos Di + sin D j) = zAk WArA(cos Ai + sin A j) Equating components and simplifying and
zCWC rC sinC + zDWDrD sinD = zAWArA sin A
(3)
zCWC rC cosC + zDWDrD cosD = zAWArA cos A
(4)
The angles are shown in Fig. 15.2.1. In Eqs (1)-(4), there are four unknowns: WC, WD, C, and D. The known values are: Plane C
WB
y
WC
Plane D
WC
WD
rC
rD
rA
WB
WA
WD
rB z 5"
4"
θA
θB θC
15"
Fig. 15.2.1: Equivalent counterbalance forces
WA = 6 lb WB = 8 lb rC = 6 in rD = 6 in
rA = 5 in rB = 5 in zC = 10 in zD = 19 in
A = 30˚ B = 150˚
zA = 15 in zB = 0 in
To simplify Eqs. (1) - (4), define WC cosC = CC WD cosD = CD WC sinC = SC WD sinD = SD Equations (1) - (4) then become rC(CC ) + rD(CD) = WArA cos A + WBrB cos B rC(SC ) + rD(SD) = WArA sin A + WBrB sin B zCrCSC + zDrDSD = zAWArA sin A zCrCCC + zDrDCD = zAWArA cos A In matrix form,
- 583 -
WA
θD
x
rC
0
0
zCrC
rD 0 0 zDrD
0 rC zC rC 0
0 CC WArA cos A + WBrB cos B
rD CD WArA sin A + WBrB sin B = zDrD SC zAWArA sin A 0 SD) zAWArA cos A
Substitution for the known values gives 6 0 0 CC 6(5)cos30˚+8(5)cos150˚ 6 0 0 6 6 CD 6(5)sin30˚+8(5)sin150˚
= 0 0 10(6) 19(6)
SC 15(6)(5)sin30˚ 10(6) 19(6) 0 0 SD) 15(6)(5)cos30˚ or 6 0 0 CC 8.66
6
0 0 6 6 CD 35 =
0 0 60 114 SC 225
60 114 0 0 SD) 389.71 Solving for the unknowns using MATLAB gives CC = WC cosC = 10.26
(5)
CD = WD cos D = 8.82
(6)
SC = WC sinC = 8.15
(7)
SD = WD sin D = 2.31
(8)
To solve for C , divide Eq. (7) by Eq. (5). Then, W sin C = tan1 WC cosC = tan1 8.15 = 141.55˚ 10.26 C C
(
)
(
)
From Eq. (5), WC = 10.26 / cosC = 10.26 / cos(141.55˚) = 13.11 lbs To solve for D , divide Eq. (8) by Eq. (6). Then, W sin D = tan1 WDcosD = tan1 2.31 = 14.70˚ 8.82 D D
(
)
(
)
From Eq. (6), WD = 8.82 / cos D = 8.82 / cos(14.70˚) = 9.12 lbs
- 584 -
Problem 15.3 Three rotating weights W1 , W2 , W3 are to be balanced by two weights WA and W B in planes A and B. Determine the magnitude and angular locations of the counterbalance weights necessary to balance the rotating weights. W2
r2
A
W1
θ2 r1
B
W2
W1 15 in
θ1 θ3
zB zA
r3
9 in
W3 W1 = 12 lb W2 = 9 lb W3 = 8 lb WA = ? lb WB = ? lb
W3 r1 = 2.5 in r2 = 2.5 in r3 = 2.5 in rA = 2.5 in rB = 2.5 in
1 = 30˚ 2 = 150˚ 3 = 270˚ A = ? B = ?
zA = 4 in zB = 14 in
Solution To be dynamiclly equivalent to W1 , W2 , and W3 , W A and W B must create the same dynamic force and dynamic moment system. The weights do not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces in planes 1, 2 and 3 must be the same as the summation for planes A and B. Then, and
W1r1 cos1 + W2r2 cos 2 + W3r3 cos3 = WArA cos A + WBrB cos B
(1)
W1r1 sin1 + W2r2 sin 2 + W3r3 sin 3 = WArA sin A + WBrB sinB
(2)
For an equivalent moment system, both systems will create the same net moment about the point where the shaft intersects plane 1. Let the z axis point along the shaft from A to B. Then, zAk WArA(cos Ai + sin A j) + zBk WBrB(cos Bi + sin B j) = z2k W2r2(cos2i + sin 2 j)+ z3k W3r3(cos3i + sin3 j) Equating components and simplifying and
zAWArA sin A + zBWBrB sinB = z2W2r2 sin2 + z3W3r3 sin3
(3)
zAWArA cos A + zBWBrB cosB = z2W2r2 cos2 + z3W3r3 cos3
(4)
The angles are shown in Fig. 15.3.1. In Eqs (1)-(4), there are four unknowns: WA, WB, A, and B. The known values are summarized in the problem statement. - 585 -
WA
r2
W2
rA
θ2
A
θΑ
WA
W1
r1
W1 15"
θ1 θ3
zB
rB
θΒ
W2
B
zA
WB r3
WB 9"
W3
W3
Fig. 15.3.1: Equivalent counterbalance forces
To simplify Eqs. (1) - (4), define WA cos A = CA WB cosB = CB WA sin A = SA WB sin B = SB Equations (1) - (4) then become rA(CA)+ rB(CB) = W1r1 cos1 + W2r2 cos2 + W3r3 cos 3 rA(SA) + rB(SB) = W1r1sin 1 + W2r2 sin2 + W3r3 sin3 zArASA + zBrBSB = z2W2r2 sin 2 + z3W3r3sin 3 zArACA + zBrBCB = z2W2r2 cos2 + z3W3r3 cos3 In matrix form, rA
0
0
zArA
rB 0 0 zBrB
0 rA zArA 0
0 CA W1r1 cos1 + W2r2 cos 2 + W3r3 cos3
rB CB W1r1 sin1 + W2r2 sin 2 + W3r3 sin3 = zBrB SA z2W2r2 sin 2 + z3W3r3sin 3 0 SB) z2W2r2 cos2 + z3W3r3 cos 3
Substitution for the known values gives 2.5 0 0 CC 12(2.5)cos30˚+9(2.5)cos150˚+8(2.5)cos270˚ 2.5 0 0 2.5 2.5 CD 12(2.5)sin30˚+9(2.5)sin150˚+8(2.5)sin270˚
= 0 0 4(2.5) 14(2.5)
SC 15(9)(2.5)sin150˚+9(8)(2.5)sin270˚ 4(2.5) 14(2.5) 0 0 SD) 15(9)(2.5)cos150˚+9(8)(2.5)cos270˚ or 0 CA 6.49
2.5 2.5 0
0 0 2.5 2.5 CB 6.25 =
0 0 10 35 SA 11.25
10 35 0 0 SB) 292.28 - 586 -
Solving for the unknowns using MATLAB gives CA = WA cos A = 15.32
(5)
CB = WB cos B = 12.73
(6)
SA = WA sin A = 3.95
(7)
SB = WB sin B = 1.45
(8)
To solve for A , divide Eq. (7) by Eq. (5). Then, W sin A = tan1 WA cosA = tan1 3.95 = 14.45˚ 15.32 A A
(
)
(
)
From Eq. (5), WA = 15.32 / cos A = 15.32 / cos(14.45˚) = 15.83 lbs To solve for B , divide Eq. (8) by Eq. (6). Then, W sin B = tan1 WBcosB = tan1 1.45 = 173.5˚ 12.73 B B
(
)
(
)
From Eq. (6), WB = 12.73 / cos B = 12.73 / cos(173.5˚) = 12.81 lbs
Problem 15.4 Resolve Problem 15.3 if zA and zB are 4.5 in and 12 in, respectively. Solution To be dynamiclly equivalent to W1 , W2 , and W3 , W A and W B must create the same dynamic force and dynamic moment system. The weights do not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces in planes 1, 2 and 3 must be the same as the summation in planes A and B. Then, W1r1 cos1 + W2r2 cos 2 + W3r3 cos3 = WArA cos A + WBrB cos B
(1)
W1r1 sin1 + W2r2 sin 2 + W3r3 sin 3 = WArA sin A + WBrB sinB
(2)
and
For an equivalent moment system, both systems will create the same net moment about the point where the shaft intersects plane 1. Let the z axis point along the shaft from A to B. Then, zAk WArA(cos Ai + sin A j) + zBk WBrB(cos Bi + sin B j) = z2k W2r2(cos2i + sin 2 j)+ z3k W3r3(cos3i + sin3 j) - 587 -
Equating components and simplifying and
zAWArA sin A + zBWBrB sinB = z2W2r2 sin2 + z3W3r3 sin3
(3)
zAWArA cos A + zBWBrB cosB = z2W2r2 cos2 + z3W3r3 cos3
(4)
The angles are shown in Fig. 15.4.1. In Eqs (1)-(4), there are four unknowns: WA, WB, A, and B. The known values are: W1 = 12 lb W2 = 9 lb W3 = 8 lb rA = 2.5 in rB = 2.5 in
W2
r1 = 2.5 in 1 = 30˚ r2 = 2.5 in 2 = 150˚ r3 = 2.5 in 3 = 270˚
WA
r2
rA
θ2
zA = 4.5 in zB = 12 in
A
θΑ
WA
W1
r1
W1 15"
θ1 θ3
zB
rB
θΒ
W2
B
zA
WB r3
WB 9"
W3
W3
Fig. 15.4.1: Equivalent counterbalance forces
To simplify Eqs. (1) - (4), define WA cos A = CA WB cosB = CB WA sin A = SA WB sin B = SB Equations (1) - (4) then become rA(CA)+ rB(CB) = W1r1 cos1 + W2r2 cos2 + W3r3 cos 3 rA(SA) + rB(SB) = W1r1sin 1 + W2r2 sin2 + W3r3 sin3 zArASA + zBrBSB = z2W2r2 sin 2 + z3W3r3sin 3 zArACA + zBrBCB = z2W2r2 cos2 + z3W3r3 cos3 In matrix form,
- 588 -
rA
0
0
zArA
rB 0 0 zBrB
0 rA zArA 0
0 CA W1r1 cos1 + W2r2 cos 2 + W3r3 cos3
rB CB W1r1 sin1 + W2r2 sin 2 + W3r3 sin3 = zBrB SA z2W2r2 sin 2 + z3W3r3sin 3 0 SB) z2W2r2 cos2 + z3W3r3 cos 3
Substitution for the known values gives 2.5 0 0 CC 12(2.5)cos30˚+9(2.5)cos150˚+8(2.5)cos270˚ 2.5 0 0 2.5 2.5 CD 12(2.5)sin30˚+9(2.5)sin150˚+8(2.5)sin270˚
= 0 0 4.5(2.5) 12(2.5)
SC 15(9)(2.5)sin150˚+9(8)(2.5)sin270˚ 4.5(2.5) 12(2.5) 0 0 SD) 15(9)(2.5)cos150˚+9(8)(2.5)cos270˚ or 0 0 CA 6.49
2.5 2.5
0 0 2.5 2.5 CB 6.25 =
0 0 11.25 30 SA 11.25
11.25 30 0 0 SB) 292.28 Solving for the unknowns using MATLAB gives CA = WA cos A = 19.74
(5)
CB = WB cos B = 17.15
(6)
SA = WA sin A = 4.60
(7)
SB = WB sin B = 2.10
(8)
To solve for A , divide Eq. (7) by Eq. (5). Then, W sin A = tan1 WA cosA = tan1 4.60 = 13.11˚ 19.74 A A
(
)
(
)
From Eq. (5), WA = 19.74 / cos A = 19.74 / cos(13.11˚) = 20.27 lbs To solve for B , divide Eq. (8) by Eq. (6). Then, W sin B = tan1 WBcosB = tan1 2.10 = 173.02˚ 17.15 B B
(
)
(
)
From Eq. (6), WB = 17.15 / cos B = 17.15 / cos(173.02˚) = 17.27 lbs
- 589 -
Problem 15.5 Four weights, W1 , W2 , W3 and W4 are all rotating in a single plane. Determine the magnitude and angular location of the single weight necessary to balance the four rotating weights. Assume that the radius to the center of gravity of the balancing weight is 9 in. The shaft is rotating at 1800 rpm. r2
W2
θ2
r1
W1 θ1
W1 = 12 lb W2 = 9 lb W3 = 8 lb W4 = 5 lb
θ3 θ4 r4 W4
r3
r1 = 9 in r2 = 12 in r3 = 10 in r4 = 8 in
θ1 = 30˚ θ2 = 135˚ θ3 = 270˚ θ4 = 315˚
W3 Solution To be dynamically equivalent to W1 , W2 , W3 , and W4 , the weight WA must create the same dynamic force system. The weight does not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces for weights 1, 2, 3, and 4 must be the same as for WA. Then, W1r1 cos1 + W2r2 cos 2 + W3r3 cos3 + W4r4 cos4 = WArA cos A
(1)
W1r1 sin1 + W2r2 sin 2 + W3r3 sin 3 + W4r4 sin4 = WArA sin A
(2)
and
The equivalent system is shown in Fig. 15.5.1. In Eqs (1)-(2), there are two unknowns: W A, and A. The known values are given in the problem statement. To solve for A , divide Eq. (2) by Eq. (2). Then, W r sin Wr sin + W r sin + W r sin + W r sin A = tan1 WArAcosA = tan1 W 1r 1cos 1+ W2r2cos2 + W3r3 cos3 + W4 r4 cos4 AA A 11 1 22 2 33 3 44 4
[
]
[
or
- 590 -
]
θΑ
WA W2
r2
rA
θ2
r1
W1 θ1
θ3 θ4 r4 r3
W4
W3 Fig. 15.5.1: Equivalent counterbalance forces
A = tan1 12(9)sin30˚+9(12)sin135˚+8(10)sin270˚+5(8)sin315˚ 12(9)cos30˚+9(12)cos135˚+8(10)cos270˚+5(8)cos315˚ = tan1 22.08 = 25.91˚ 45.45
[
]
From Eq. (1), WA = 45.45 / (rA cosA) = 45.45 / (9cos25.91˚) = 5.61lbs Problem 15.6 Resolve Problem 15.5 for the following set of data. W1 = 20 lb W2 = 10 lb W3 = 8 lb W4 = 6 lb
r1 = 4 in r2 = 12 in r3 = 12 in rA = 10 in
1 = 45˚ 1 = 135˚ 1 = 180˚ 1 = 270˚
Solution To be dynamically equivalent to W1 , W2 , W3 , and W4 , the weight WA must create the same dynamic force system. The weight does not depend on the rotational speed of the rotor. For an equivalent force system, the summation of the horizontal and vertical inertial forces for weights 1, 2, 3, and 4 must be the same as for WA. Then, and
W1r1 cos1 + W2r2 cos 2 + W3r3 cos3 + W4r4 cos4 = WArA cos A
(1)
W1r1 sin1 + W2r2 sin 2 + W3r3 sin 3 + W4r4 sin4 = WArA sin A
(2)
The equivalent system is shown in Fig. 15.5.1. In Eqs (1)-(2), there are two unknowns: W A, and A. The known values are given in the problem statement.
- 591 -
θΑ
WA W2
r2
rA
θ2
r1
W1 θ1
θ3 θ4 r4 W4
r3 W3
Fig. 15.5.1: Equivalent counterbalance forces
To solve for A , divide Eq. (2) by Eq. (2). Then,
[
]
[
WA rA sin A = tan 1 W1r1 sin1 + W2 r2 sin2 + W3 r3 sin 3 + W4 r4 sin 4 A = tan1 W W1 r1 cos1 + W2 r2 cos2 + W3 r3 cos3 + W4 r4 cos 4 A rA cos A
]
or
A = tan1 20(4)sin45˚+10(12)sin135˚ +8(12)sin180˚ +6(10)sin270˚ 20(4)cos45˚+10(12)cos135˚ +8(12)cos180˚ +6(10)cos270˚ = tan 1 81.42 =146.77˚ 1.242
[
]
From Eq. (1), WA = 1.242 /(rA cos A )= 1.242/ (9cos146.77˚) =16.59 lbs
Problem 15.7 For the mechanism shown, determine the magnitude and location of the shaking force acting on the frame. Determine the location with respect to point A. Also find the magnitude of the reaction force at point A and at point C. Assume that W3 >> W2 and W4 . W 3 = 2.0 lbI3 = 0.1 lb-s2 -in
2 = 6.28 rad/s CCW (constant)
- 592 -
ω2 A T12 75˚ 2 AB = 7.0 in AC = 12.0 in w = 2 in
4
w 3
B
C Solution To solve for the shaking forces, we must first conduct a velocity and acceleration analysis of the mechanism in the position shown and determine the acceleration of B3 and the 3 = 4 . Position Analysis Locate the relative position of points A, B and C and draw link 4. Only link 3 has significant mass so we need to determine only the acceleration of point B3 , which is the center of mass of link 3. Velocity Analysis: v B3 = vB2 = v B2 / A2 = vB 4 + vB2 / B4
(1)
v B4 = v B4 /C4 Now, v B2 / A2 = 2 rB2 / A2 v B2 / A2 = 2 rB2 / A2 = 6.28(7)= 43.96 in / sec( to rB2 / A2 ) v B4 /C4 = 4 rB4 /C4 vB4 /C4 = 4 rB4 /C4 ( to rB4 /C4 ) v B2 / B4 in the direction of rB4 /C4
- 593 -
b2
b3 Velocity Polygon
o b'2
20 in/s
b'3 b4
t
A
aB4 /C 4
r
a B2/ A2 2
o'
Acceleration Polygon 100 in/s 2
B
4
r
a B4 / C4 4a B / B 2 4
C 4 in
acB2 /B 4
Fig. 15.7.1: Position, velocity, and acceleration polygons.
Solve Eq. (1) graphically with the velocity polygon shown in Fig. 15.7.1. From the polygon, v B2 / B4 = 24.78 in / s Also, - 594 -
or
v B4 /C4 = 36.18 in / s v 36.18 4 = rB4 /C4 = 5.534 = 6.54 rad / s B 4 /C4
By inspection, the angular velocity is CW. Also,
3 = 4 Acceleration Analysis: a B3 = a B2 = aB2 / A2 =a B4 +a B2 / B4 a B4 = aB4 /C4 a rB2 / A2 + atB2 / A2 =a rB4 /C4 + aBt 4 /C4 + 4 aB2 / B4 + 2 4 v B2 / B4 Now, a rB2 / A2 = 2 ( 2 rB2 / A2 ) a rB2 / A2 = 2 2 rB2 / A2 = 6.282 (7)= 276.07 in / s2 in the direction opposite to rB2 / A2 a tB2 / A2 = 2 rB2 / A2 a tB2 / A2 = 2 rB2 / A2 = 0 a rB4 /C4 = 4 ( 4 rB4 /C4 ) a rB4 /E4 = 4 2 rB4 /C4 = 6.542 5.534 = 236.70 in / s2 in the direction opposite to rB 4 /C4
- 595 -
(2)
a tB4 /E 4 = 4 rB4 /E4 a tB4 /E4 = 4 rB4 /E4 ( to rB4 /E 4 ) 4 aB /B 2 4
in the direction along to rB 4 /C4
a cB2 / B4 = 2 4 vB2 / B4 = 2(6.54)(24.78)= 324.12 in / s2 ( to rB4 /C4 ) Solve Eq. (2) graphically with an acceleration polygon. From the polygon, Then,
a tB4 /C4 = 480.0 in / s2 in the direction shown in the polygon. at 2 4 = r B4 /C4 = 480.0 5.534 = 86.74 rad / s CW B 4 /C4 Next compute the inertial force and inertial torque. The inertia force is given by 2 F1 3 = m3 aB3 = 386 (276.07) = 1.43 lb in the direction opposite to a B3 . The inertia moment is given by M1 3 = I3 3 = 0.1(86.74) = 8.674 in-lbs in the direction opposite to 3 (CCW). The force and moment are shown in Fig. 15.7.2. To identify where the inertia force is located, replace the force-moment system with a force offset to create the same moment about the point B. The offset distance is given by h = M13 = 8.674 = 6.065 in 1.43 F13 The offset is in the direction which will create the moment M1 3. This is shown in Fig. 15.7.2 by the dashed line. This is the shaking force acting on the frame. To determine the reaction forces at A and C, draw a free-body diagram of each link. These are shown in Fig. 15.7.3.
- 596 -
1ω
A
2
T12 75˚ 2 4
h = 6.065 in
M13
B
3
F13
F13 = shaking force C 4 in
Fig. 15.7.2: Location of shaking force 1 F43
F12 T12
w
A
3 M13
B F23
75˚
4
2 F43
F13
1 F34
B 2 F34
F32
5.534" C F14
4 in Fig. 15.7.3: Free body diagrams.
Summing momnets about point B in link 3 gives - 597 -
or
MB = 0 = M13 + F432(w / 2) F431(w / 2) F431(w / 2) F432(w / 2) = M13 F431 F432 = M13(2 / w) = 8.674(2 / 2) = 8.674 (3)
Summation of moments about point C on link 4 gives
MC = 0 = F342(5.534 + w / 2) F341(5.534 w / 2) F341 =
F342(5.534 + w / 2) = F342(5.534 + 2 / 2) = 1.441 F2 34 (5.534 w / 2) (5.534 2 / 2)
Then from Eq. (3), 1.441F432 F432 = 8.674 F432 = 8.674 = 19.67 lb = F342 0.441 and F341 = 1.441 F342 = 1.441(19.67) = 28.34 lb Summing forces in the direction normal to link 4 gives
Fn = F341 + F342 + F14 = 0 F14 = F341 F342 = 28.34 19.67= 8.67 lb in the direction shown. Now return to link 3. Summing forces vectorially on that link gives
F = F431 + F432 + F23 + F13 = 0
(4)
Equation (4) can be solved graphically for F2 3. The result is shown in Fig. 15.7.4.
- 598 -
F12 T12
A 75˚
2
F32
5.322 in
9.55 lb
1.43 lb
F32 8.67 lb
Fig. 15.7.3: Force polygon for link 3 and free-body diagram for link 2.
From the polygon, F23 = 9.55 lb Link 2 has only two forces on it so F12 = F23 = 9.55 lb in the direction shown. To find the torque T1 2, we can sum moments about point A. This gives, T12 = F12(5.322) = 9.55(5.322) = 50.825 The torque is in the CCW direction to oppose the couple produced by F1 2 and F3 2.
- 599 -
Problem 15.8 For the mechanism and data given, determine the shaking force and its location relative to point A. Draw the shaking force vector on the figure. The force FB is 10 lb in the direction shown. For the moments of inertia of link 3, use g = 386 in/s2 and IG = m l 2 /12.
2 = 160 rad/s CCW
2 = 0 rad/s2
IG2 = 0.00369 lb-s2 -in
W 3 = 3.5 lb
W 2 = 0.95 lb W 4 = 2.5 lb 4
A 2
G2
FB
C
3
30˚
AB = 3 in BC = 12 in BG 3 = 3.6 in BG 2 = 1.2 in
G3 B
G4
Solution To determine the shaking force, we must first perform a velocity and acceleration analysis and determine the linear accelerations G2 , G3 , and G4 and the angular acceleration of link 3.. Note that the applied force FB does not influence the shaking force and can be ignored. Velocity Analysis: v B3 = vB2 = v B2 / A2 = vC3 + v B3 /C3 Now,
(1)
vC3 = vC4 v B2 / A2 = 2 rB2 / A2 vB2 / A2 = 2 rB2 / A2 = 160(3) = 480 in / sec( to rB2 / A2 ) v B4 /C4 is along the slide. v B3 /C3 = 3 rB3 /C3 vB3 /C3 = 3 rB3 /C3 ( to rB3 /C3 ) From the velocity polygon in Fig. 15.8.1, v B3 /C3 = 419.3 in/s vC3 = 290.7 in /s Then,
3 = v B3 /C3 / rB3 /C3 = 419.3/12 = 34.94 rad/s (CW)
- 600 -
4
A 2
G2
C G4
30˚ 3
G3
B b2 b3
b'2 b'3
1ar B 3 /C 3
g'3
g'2 1ar B2/A 2
1a t B 3 /C 3
c'3 o
c4
1a
C3
o'
c'4 Acceleration Polygon
c3
30,000 in/s 2
Velocity Polygon 200 in/s Fig. 15.8.1: Polygons for problem 15.8.
Acceleration Analysis:
Or,
a B3 = a B2 = aB2 / A2 = aC3 + aB3 /C3 aC3 = aC4 a rB2 / A2 + atB2 / A2 = aC3 + a rB3 /C3 + atB3 /C3
(2)
Now, a rB2 / A2 = 2 ( 2 rB / A ) a rB2 / A2 = 2 2 rB / A = 160 2 3 = 76,800 in / s2 in the direction opposite to that of rB/ A a tB2 / A2 = 2 rB/A a tB2 / A2 = 2 rB/A = 03 = 0 in / s2 a rB3 /C3 = 3 ( 3 rB/C ) a rB3 /C3 = 3 2 rB/C = 34.94 2 12 = 14,650 in / s2 in the direction opposite to that of rB/ C . a tB3 /C3 = 3 rB/C Solve Eq. (2) graphically with the acceleration polygon shown in Fig. 15.8.1. From the polygon,, - 601 -
aC4 = 76,600 in / s2 aG 2 = 46,000 in / s2 aG3 = 74,500 in / s2 Then
a tB3 /C3 = 36,800 in / s2 a tB3 /C3 36,800 3 = = = 3,070 rad / s2 CW rB/C 12 We can now conduct the inertia force analysis. There will be an inertia force associated with each center of gravity. The forces are 2 F1 2 = m2 aG2 = Wg2 aG2 = 0.95 386 46,000 = 113.2 in / s (opposite aG2 ) 3.5 74,500 = 675.5 in / s2 (oppositea ) F1 3 = m3 aG3 = Wg3 aG3 = 386 G3 W 2.5 76,600 = 496.1in / s2 (opposite a ) F1 4 = m4 aG4 = g4 aG4 = 386 G4
Only link 3 has an angular acceleration so it is the only link that has an inertia moment. The inertia moment is given by M1 3 = I3 3 =
m2 (BC)2 3.5(12)2 3 = 3,070 = 334.0 in lb (opposite 3 or CCW) 12g 386(12)
The inertia forces are shown in Fig. 15.8.2. The shaking force is the resultant (Fs ) of the inertia forces as shown in Fig. 15.8.2. The magnitude of the shaking force is given by Fs = F12 + F13 + F14 = 1,256 lbs The location of the shaking force is obtained by replacing the resultant inertia force and inertia moment by the resultant inertia force offset by the distance h where h is given by h=
334.0 M13 = = 0.265 in . F12 + F13 + F14 1,256
The shaking force is located such that it creates the moment relative to its original position. The proper location is shown in Fig. 15.8.2d.
- 602 -
A
4 2
G2 F12
G4 F14
30˚ B
M 13
G3
3
F13
Force Polygon 300 lbs
F14 A
2
G2 F12
F12 A
2
G2
4
(a)
C G4 F14
30˚ G3
3
B
F13
F 14 + F 30˚
C
F 14 + F
(b)
13
13
4
C G4
B
M 13
G3
3
(c)
F 12 + F 14+ F 13 1.667 in h = 0.265 in
A
2
G2
C G4
30˚ B
4
3
M 13
G3
(d)
F s = F 12 + F 14+ F 13
Fig. 15.8.2: Determination of shaking force. a) inertial forces displayed. b) Resolution o f F 1 3 and F1 4 into a single force. c) Resolution of F1 2, F1 3, and F 14 into the shaking force. d) Offsetting the shaking force by h to locate it relative to the frame.
Problem 15.9 For the mechanism shown, determine the magnitude and location of the shaking force acting on the frame. Determine the location with respect to point A. Draw the shaking force vector on the figure.
2 = 12 rad/s CCW
2 =0
W 2 = 0.5 NW 3 = 2.5 N
W 4 = 1.5 N
For the moments of inertia for the links, use g = 9.81 m/s2 and IG = m l 2 /12.
- 603 -
C
3
B ω2 T12
A
AB BC CD AD
4
= 10 cm = 50 cm = 30 cm = 40.0 cm D
Velocity Analysis: Start with the points at B. v B2 = 2 rB / A v B2 = 2 rB / A = 12(10) = 120 cm / s The direction is normal to AB and in the direction obtained by rotating rB/ A 90˚ in the direction of 2 . and
v B3 = vB2 vC3 = v B3 + v C3 / B3 = vC4 = vC4 /D4
(1)
Now vC3 / B3 = 3 rC/ B vC3 / B3 = 3 rC / B ( to rC / B ) vC4 /D4 = 4 rC / D v C4 / D4 = 4 rC /D ( to rC / D ) The solution to Eq. (1) is shown in the velocity polygon in Fig. 15.9.1. From the polygon, vC3 / B3 = 51.82 cm / s and vC4 /D4 = 75.50 cm / s Then,
- 604 -
C
AB BC CD AD
= 10 cm = 50 cm = 30 cm = 40.0 cm
g3 3
4 g4
B ω2 T12
A
g2
D o c4 c3
c'3
Velocity Polygon 50 cm/s
c'4 g'4 b2 b 3 o' g'3
g'2 Acceleration Polygon 500 cm/s 2 b'2
b'3
Fig. 15.9.1: Velocity and acceleration polygon for Problem 15.9
3 =
vC3 / B3 51.82 = = 1.036 rad / s CCW rC / B 50 - 605 -
and
4 =
vC4 /D4 75.50 = = 2.517rad / s CCW rC /D 30
Acceleration Analysis: Start with the points at B. a B2 = aB2 / A2 = arB2 / A2 + atB2 / A2 where a rB2 / A2 = 2 (2 rB/A ) arB2 / A2 = 2 2 rB / A = 12 210 = 1, 440 cm / s2 (opposite rB/ A ) a rB2 / A2 = 2 rB/A = 0 rB / A = 0 a B3 = a B2 and aC3 = aB3 + aC3 / B3 = aC4 = aC4 / D4 Or a B3 + arC3 / B3 + aCt 3 / B3 = aCr 4 /D 4 + aCt 4 /D4
(2)
Now, aCr 3 / B3 = 3 (3 rC / B ) arC3 / B3 = 3 2 rC / B = 1.0362 50 = 56.66 cm / s2 (opposite rC/ B ) aCt 3 / B3 = 3 rC / B aCt 3 / B3 = 3 rC / B aCr 4 /D 4 = 4 ( 4 rC /D ) aCr 4 /D4 = 4 2 rC /D = 2.517230 = 190.05 cm / s2 (opposite rC /D ) aCt 4 /D 4 = 4 rC /D aCt 4 /D4 = 4 rC /D The solution to Eq. (2) is shown in the acceleration polygon in Fig. 15.9.1. After the polygon is drawn, the acceleration of the center of gravity for each link can be determined by image. From the polygon, atC3 / B3 = 1,835 cm / s2 atC4 / D4 = 2,025 cm / s2 aG2 = 720 cm / s2 aG3 = 1,503 cm / s2 - 606 -
aG4 = 1,017 cm / s2 Then
3 =
aCt 3 / B3 1835 = = 36.7 rad / s2 CCW rC / B 50
4 =
aCt 4 /D4 1,017 = = 33.9 rad / s2 CCW rC /D 30
Inertia-Force Analysis: For the inertia force analysis, we must compute the inertia force and inertia torque for each link. For link 2, 0.5 720 F1 2 = m 2 aG2 = = 0.367 N 9.81 100 opposite to 1aG2 . m2 l 22 1 W2 l22 1 0.5 (10 /100)2 M1 2 = I2 2 = 2 = 2 = 0=0 12 12g 9.81 12 For link 3, 2.5 1,503 = 3.83 N 9.81 100
F1 3 = m3 aG3 = opposite to 1aG3 .
m3l 23 W3l 23 2.5 (50 /100)2 M 1 3 = I3 3 = 3 = 3 = 36.7 = 0.195 N m 12 12g 9.81 12 CW For link 4, F1 4 = m 4 aG4 =
1.5 1,017 = 1.55 N 9.81 100
opposite to 1aG4 . M1 4 = I4 4 =
m4l 24 W4 l24 1.5 (30 /100)2 4 = 4 = 33.9 = 0.0338 N m CW 12 12g 9.81 12
The inertia forces and moments are shown in Fig. 15.9.2a. The shaking force is the resultant of the inertial forces and moments. Before determine the resultant force system, represent each inertial force and moment by an equivalent force system. To do this, we must offset each inertial force by the distance h where, h= M F For links 2 and 3, h3 = M13 = 0.195 = 0.0509 m = 5.09 cm 3.83 F13 - 607 -
C
M 13 F13
g3 3
4 M 14
g4
F14
B A
Inertial Force Polygon 1N
F12
D
(a)
C
F13 h3
g3 F14
3 g4 4
B A
h4
Position Polygon 10 cm
F12
D
(b )
C F13
F13 + F14
g3 F14
3 g4 4
B A
F12
(c)
D
Fig. 15.9.2: a) Inertial forces. b) Replacement of forces and moments by equivalent forces. c) Resolution of F1 3 and F1 4 into a single force.
- 608 -
and h4 = M14 = 0.0338 = 0.0218 m = 2.18 cm 1.55 F14 The equivalent force systems are shown in Fig. 15.9.2b. The resultant shaking force is 4.995 N and is shown in Fig. 15.9.2c. C Fs = F12 + F13 + F14 F13 + F14 g3 3
F12 + F13 + F14 g4 4
B A
F12
Scales 10 cm 1N
D
Fig. 15.9.3: Shaking force located relative to frame
Problem 15.10 For the mechanism given, assume that 2 is 200 rad/s CCW (constant), and link 2 is balanced so that its center of mass is located at the pivot at point A. Also assume that IG2 is small enough to be neglected. For the data given, determine the shaking force and its location relative to point A. Draw the shaking force vector on the figure. IG3 = 0.0106 lb-s2 -in IG4 = 0.0531 lb-s2 -in
W 3 = 2.65 lb W 4 = 6.72 lb
- 609 -
C
G4 G3 ω2 G2
3
A D
2 T12
4
B
C AD D A BB BC 3 G 4 C G
= 8.0 in = 13.5 in = 2.2 in = 17.6 in = 8.8 in = 4.0 in
45˚
Velocity Analysis: Start with the points at B. v B2 = 2 rB / A v B2 = 2 rB / A = 200(10) = 2,000 in / s The direction is normal to AB and in the direction obtained by rotating rB/ A 90˚ in the direction of 12 . and
v B3 = vB2 vC3 = v B3 + v C3 / B3 = vC4 = vC4 /D4
(1)
Now vC3 / B3 = 3 rC/ B vC3 / B3 = 3 rC / B ( to rC / B ) vC4 /D4 = 4 rC / D v C4 / D4 = 4 rC /D ( to rC / D ) The solution to Eq. (1) is shown in the velocity polygon in Fig. 15.10.1. From the polygon, vC3 / B3 = 3,060 in / s and vC4 /D4 = 3,230 in / s Then,
- 610 -
C
G4 G3 ω2 A
G2
Position Scale 4 in
2 B
aCr 3 /B 3
4
3
D
b2 b3
45˚
b'3 b'2 Velocity Polygon
o'
1000 in/s
o
aCt 3 /B 3 g'3
g'4
c4
aCr 4 /B 4 Acceleration Polygon 500,000 in/s 2
aCt 4 /B 4 c'4
c'3
Fig. 15.10.1: Velocity and acceleration polygon for Problem 15.10
3 =
vC3 / B3 3,060 = = 173.9 rad / s CW rC / B 17.6
4 =
vC4 /D4 3,230 = = 403.8rad / s CCW rC /D 8
and
- 611 -
c3
Acceleration Analysis: Start with the points at B. a B2 = aB2 / A2 = arB2 / A2 + atB2 / A2 where arB2 / A2 = 2 ( 2 rB/ A ) arB2 / A2 = 2 2 rB/A = 200 2( 2.2) = 88,000 in / s2 (opposite rB/A ) a rB2 / A2 = 2 rB/ A = 0 rB/A = 0 a B3 = a B2 and aC3 = aB3 + aC3/ B3 = aC4 = aC4 /D4 Or aB3 + aCr 3 / B3 + atC3 / B3 = arC4 /D4 + aCt 4 /D4
(2)
Now, arC3 / B3 = 3 (3 rC/B) aCr 3 / B3 = 3 2 rC/B = 173.92 (17.6)= 532,000 in / s2 (opposite rC/B) atC3 / B3 = 3 rC / B aCt 3 / B3 = 3 rC / B arC4 /D4 = 4 (4 rC/D ) aCr 4 /D4 = 4 2 rC/D = 403.82 (8) =1,300,000 in / s2 (opposite rC /D ) atC4 /D4 = 4 rC /D aCt 4 /D4 = 4 rC/D The solution to Eq. (2) is shown in the acceleration polygon in Fig. 15.10.1. After the polygon is drawn, the acceleration of the center of gravity for each link can be determined by image. From the polygon, aCt 3 / B3 =1,512,000 in / s2 aCt 4 /D4 = 805,400 in / s2 aG3 = 733,800 in / s2 aG4 = 765,400 in / s2 Then at 2 3 = Cr 3 / B3 = 1,512,000 17.6 = 85,910 rad / s CW C/B at 2 4 = Cr 4 /D4 = 805,400 8 = 100,675 rad / s CW C/D
- 612 -
Inertia-Force Analysis: For the inertia force analysis, we must compute the inertia force and inertia torque for each link. For link 3, F1 3 = m3 aG3 = 2.65 386 733,800 = 5038 lbs opposite to aG3 . M1 3 = I3 3 = 0.0106(85,910) = 910.6 in lbs CCW For link 4, F1 4 = m 4 aG4 = 6.72 765,400 =13,300 lbs opposite to aG4 . 386 M1 4 = I4 4 = 0.0531(100,675) = 5,335.8 in lbs CCW. The inertia forces and moments are shown in Fig. 15.10.2a. The shaking force is the resultant of the inertial forces and moments. Before determine the resultant force system, represent each inertial force and moment by an equivalent force system. To do this, we must offset each inertial force by the distance h where, h= M F For links 2 and 3, h3 = M13 = 910.6 = 0.181 in F13 5038 and h4 = M14 = 5,335.8 = 0.401 in F14 13,300 The equivalent force systems are shown in Fig. 15.10.2b. The resultant shaking force is 18,3580 lbs and is shown in Fig. 15.10.3. Note that F13 and F1 4 are almost parallel. Therefore, the intersection of these two forces and the resulting point through which Fs lies is off the page.
- 613 -
C Position Scale 4 in G4
M 13 G3 ω2 G2
3
M 14
4 F14
F13
A D
2 B
45˚
Inertial Force Polygon 10,000 lbs
Position Scale 4 in
(a )
C h4
h3
G4
M 13 G3 ω2 G2
3
M 14
4 F14
F13
A D
2 B
45˚
Inertial Force Polygon 10,000 lbs (a
Fig. 15.9.2: a) Inertial forces.
b) Replacement of forces and moments by equivalent forces.
- 614 -
The shaking force passes through the intersection of these two lines. F12 + F13 + F14
Fs
=
C
F13 + F14
Position Scale 4 in G4 G3 ω2 G2
A
3
F14
F13 D
2 B
4
45˚
Inertial Force Polygon 10,000 lbs
Fig. 15.9.3: Shaking force located relative to frame
Problem 15.11 A single cylinder engine is mounted so that the crankshaft is horizontal as shown in Fig. 15.12. The engine is characterized by the following data. Rotational speed Stroke Length of connecting rod Distance from crank pin to CG of connecting rod Equivalent unbalanced weight of crank at a 3 in radius Weight of piston Weight of connecting rod
1200 rpm 6 in 12 in 4 in 6 lb 7 lb 15 lb
Determine the magnitude of the shaking force when the crank angle is 120˚ if there is no counterbalance weight. Then determine the shaking force at the crank location if a counterbalancing weight is added that is equal to mcb = mA + 2mB / 3. Solution The shaking force is given by Eq. (15.29) as
- 615 -
()
[
]
fS = R 2 (mA + mB mcb )cos + mB R cos2 i + R 2(mA mcb )sinj L For this problem when the counterbalance weight is zero, R = stroke / 2 = 6 / 2 = 3 in L = 12 in
= 1200(2 / 60) = 125.66 rad / s 15(8) m A = R2 m2 + m3b = 1 33 6 + (12) = 0.0414 lb s2 / in L 386 R
[
]
m B = m4 + m3a = 1 7 + 15(4) = 0.0311 lb s2 / in L 386 12 m cb = Rc mcb = 0 R
= 120˚ Then,
( )
[
]
fS = 3(125.66)2 (0.0414 + 0.0311 0)cos120˚+0.0311 3 cos2(120˚) i 12 + 3(125.66)2(0.0414 0)sin120˚j or fS = 1901i + 1698j lb Then, fS = (1901)2 + (1698)2 = 2550 lb For this problem when there is a counterbalance weight, R = stroke / 2 = 6 / 2 = 3 in L = 12 in
= 1200(2 / 60) = 125.66 rad / s 15(8) m A = R2 m2 + m3b = 1 33 6 + (12) = 0.0414 lb s2 / in L 386 R
[
]
m B = m4 + m3a = 1 7 + 15(4) = 0.0311 lb s2 / in L 386 12 m cb = mA + 2m B / 3 = 0.0414+ 2(0.0311) / 3 = 0.0621 lb s2 / in
= 120˚ - 616 -
Then,
( )
[
]
fS = 3(125.66)2 (0.0414 + 0.0311 0.0621)cos120˚+0.0311 3 cos2(120˚) i 12 + 3(125.66)2(0.0414 0.0621)sin120˚j or fS = 430.5i 849 j lb Then, fS = (430.5)2 + (849)2 = 952.6 lb Problem 15.12 Resolve Problem 15.11 if the stroke is 4 in, the engine speed is 1800 rpm, and the equivalent unbalanced weight of the crank at a 2 in radius is 5 lb. Solution The shaking force is given by Eq. (15.29) as
()
[
]
fS = R 2 (mA + mB mcb )cos + mB R cos2 i + R 2(mA mcb )sinj L For this problem when the counterbalance weight is zero, R = stroke / 2 = 4 / 2 = 2 in L = 12 in
= 1800(2 / 60) = 188.5 rad / s 15(8) m A = R2 m2 + m3b = 1 22 6 + (12) = 0.0414 lb s2 / in L 386 R
[
]
m B = m4 + m3a = 1 7 + 15(4) = 0.0311 lb s2 / in L 386 12 m cb = Rc mcb = 0 R
= 120˚ Then,
( )
[
]
fS = 3(188.5 )2 (0.0414 + 0.0311 0)cos120˚+0.0311 3 cos2(120˚) i 12 + 3(125.66)2(0.0414 0)sin120˚j or fS = 2761i + 2551j lb Then, - 617 -
fS = (2761)2 + (2551)2 = 3759 lb For this problem when there is a counterbalance weight, R = stroke / 2 = 4 / 2 = 2 in L = 12 in
= 1800(2 / 60) = 188.5 rad / s mA =
R2 m + m3b = 1 2 6 + 15(8) = 0.0414 lb s2 / in (12) L 386 2 R 2
[
]
m B = m4 + m3a = 1 7 + 15(4) = 0.0311 lb s2 / in L 386 12 m cb = 5/ 386 = 0.0129 lb s2 / in
= 120˚ Then,
( )
[
]
fS = 3(188.5 )2 (0.0414 + 0.0311 0.0129)cos120˚+0.0311 3 cos2(120˚) i 12 + 3(125.66)2(0.0414 0.0129)sin120˚ j or fS = 2301.2i + 1753.7j lb Then, fS = (2301.2)2 + (1753.7)2 = 2893.2 lb
- 618 -
Problem 15.13 For the mechanism and data given, determine the shaking force and its location relative to point A. Draw the shaking force vector on the figure.
2 = 210 rad/s CCW
2 = rad/s2
W 2 = 3.40 lb
IG3 = 0.1085 lb-s2 -in
W 3 = 3.40 lb
W 4 = 2.85 lb
B 45˚ A G2
3 G3 AB = 3 in BC = 12 in BG 3 = 6 in
C 4 G4
Solution To determine the shaking force, we must first perform a velocity and acceleration analysis and determine the linear accelerations for G3 , and G4 and the angular acceleration of link 3. Velocity Analysis: v B3 = vB2 = v B2 / A2 = vC3 + v B3 /C3 Now,
(1)
vC3 = vC4 v B2 / A2 = 2 rB2 / A2 v B2 / A2 = 2 rB2 / A2 = 210(3)= 630 in /sec( to rB2 / A2 ) vC4 is along the slide. v B3/C3 = 3 rB3 /C3 v B3 /C3 = 3 rB3 /C3 ( to rB3 /C3 ) From the velocity polygon in Fig. 15.13.1, v B3/C3 =458.1 in / s vC3 =519.3 in / s Then,
3 = v B3 /C3 / rB3 /C3 = 458.1/12 = 38.17 rad / s (CCW)
- 619 -
Fig. 15.13.1: Polygons for problem 15.13.
Acceleration Analysis:
Or,
a B3 = a B2 = aB2 / A2 = aC3 + aB3 /C3 aC3 = aC4 a rB2 / A2 + atB2 / A2 = aC3 + a rB3 /C3 + atB3 /C3
(2)
Now, a rB2 / A2 = 2 ( 2 rB/A ) a rB2 / A2 = 2 2 rB/A = 2102 3 =132,300 in / s2 in the direction opposite to that of rB/ A a tB2 / A2 = 2 rB/A a tB2 / A2 = 2 rB/A = 03 = 0 in / s2 a rB3/C3 = 3 ( 3 rB/C ) a rB3 /C3 = 3 2 rB /C = 38.172 12 = 17,480 in / s2 in the direction opposite to that of rB/ C . a tB3 /C3 = 3 rB/C Solve Eq. (2) graphically with the acceleration polygon shown in Fig. 15.13.1. From the polygon, - 620 -
aC4 = 94,620 in / s2 aG2 = 0 in / s2 aG3 =105,100in / s2 a tB3/C3 = 91,620 in / s2 Then atB3 /C3 91,620 3 = r = 12 = 7,635 rad/ s2 CCW B /C We can now conduct the inertia force analysis. There will be an inertia force associated with each center of gravity. The forces are 2 F1 2 = m2 aG2 = Wg2 aG2 = 0.95 386 0 = 0 in /s (opposite aG2 ) 3.5 105,10 = 952.6 in / s2 (oppositea ) F1 3 = m3 aG3 = Wg3 aG3 = 386 G3 W 2.5 94,620 = 612.8 in /s2 (opposite a ) F1 4 = m4 aG4 = g4 aG4 = 386 G4
Only link 3 has an angular acceleration so it is the only link that has an inertia moment. The inertia moment is given by M1 3 = I3 3 =
m2 (BC)2 3.5(12)2 3= 12g 386(12) 7,635= 830.6 in lb (opposite 3 or CW)
The inertia forces are shown in Fig. 15.8.2. The shaking force is the resultant (Fs ) of the inertia forces as shown in Fig. 15.8.2. The magnitude of the shaking force is given by Fs = F1 2 + F1 3 + F1 4 =1524 lbs The location of the shaking force is obtained by replacing the resultant inertia force and inertia moment by the resultant inertia force offset by the distance h where h is given by h=
M1 3 830.6 F1 2 + F1 3 + F1 4 = 1524 = 0.545 in .
The shaking force is located such that it creates the moment relative to its original position. The proper location is shown in Fig. 15.8.2d.
- 621 -
Fig. 15.8.2: Determination of shaking force. a) inertial forces displayed. b) Resolution o f F 1 3 and F1 4 into a single force. c) Offsetting the shaking force by h to locate it relative t o the frame.
- 622 -
Problem 15.14 For the engine given in Problem 15.11, lump the weight of the connecting rod at the crank pin and piston pin and draw the polar shaking force diagram for the following three cases. 1. No counterbalancing weights 2. A counterbalancing weight equal to the sum of the crank weight at the crank radius, the part of the connecting rod weight assumed to be concentrated at the crank pin, the weight of the piston, and the part of the connecting rod weight concentrated at the piston pin. 3. A counterbalancing weight equal to the sum of the crank weight at the crank radius, the part of the connecting rod weight assumed to be concentrated at the crank pin, and half of the weight concentrated at the piston pin (weight of the piston and part of the connecting rod weight concentrated at the piston pin). Solution This problem can be solved by plotting Eq. (15.29) for 0˚ 360˚. The plot can be generated manually or a MATLAB program can be written to make the plot. Here, we will use MATLAB. For the first case (no counterbalance force), the plot is shown in Fig. 15.13.1.
Fig. 15.13.1: Shaking force diagrams for no counterbalance force
For the second case, the counterbalance weight is given by m cb = mA + mB = 0.0414 + 0.0311 = 0.0725 lb s2 / in The plot of the shaking force is shown in Fig. 15.13.2.
- 623 -
Fig. 15.13.2: Shaking force diagrams for
m cb = mA + mB
For the second case, the counterbalance weight is given by m cb = mA + 0.5m B = 0.0414 + 0.0311/ 2 = 0.570 lb s2 / in The plot of the shaking force is shown in Fig. 15.13.3.
Fig. 15.13.3: Shaking force diagrams for
- 624 -
m cb = mA + 0.5m B
Problem 15.15 For the engine given in Problem 15.11, lump the weight of the connecting rod at the crank pin and piston pin and locate the counterbalancing weight at the crank radius. Determine the optimum counter balancing weight, which will give 1. The smallest horizontal shaking force 2. The smallest vertical shaking force Solution The information from Problem 15.11 is: A single cylinder engine is mounted so that the crankshaft is horizontal as shown in Fig. 15.12. The engine is characterized by the following data. Rotational speed Stroke Length of connecting rod Distance from crank pin to CG of connecting rod Equivalent unbalanced weight of crank at a 3 in radius Weight of piston Weight of connecting rod
1200 rpm 6 in 12 in 4 in 6 lb 7 lb 15 lb
The shaking force is given by Eq. (15.29) as
()
[
]
fS = R 2 (mA + mB mcb )cos + mB R cos2 i + R 2(mA mcb )sinj L For this problem when the counterbalance weight is zero, R = stroke / 2 = 6 / 2 = 3 in L = 12 in
= 1200(2 / 60) = 125.66 rad / s 15(8) m A = R2 m2 + m3b = 1 33 6 + (12) = 0.0414 lb s2 / in L 386 R
[
]
m B = m4 + m3a = 1 7 + 15(4) = 0.0311 lb s2 / in L 386 12 m cb = Rc mcb = R mcb = mcb R R Then,
- 625 -
(1)
[
]
fS = 3(125.66)2 (0.0414+ 0.0311mcb )cos + 0.0311 3 cos2 i 12
( )
+ 3(125.66)2 (0.0414 mcb )sinj or
[
]
fS x = R 2 (mA + mB mcb )cos + mB R cos2 L = 47,370[(0.0725 mcb)cos + 0.007775cos2 ]
()
(1)
Part 1 To determine the optimum value for the counterbalance force that minimizes the maximum value of the horizontal component of the shaking force, we could vary mcb and in a MATLAB program and compute
[
]
R fS x = R 2 (mA + mB mcb )cos + mB L cos2 = 47,370[(0.0725 mcb )cos + 0.007775cos2 ]
()
If this is done, it will be apparent that the maximum value for the horizontal component of the shaking force will be minimized when the first term in the expression is zero. This occurs when mA + mB mcb = 0 or mcb = mA + mB = 0.0725 lb s2 / in Then the counterbalance weight is Wcb = gmcb = 386(0.0725) = 28 lb The results from the program ShakeAnalysis is shown in Fig. 15.15.1 for this case. Note that while the horizontal component is optimal, the vertical component is not. Part 2 For the optimum value for the vertical component of the shaking force, we do not need to write a computer program to see how the vertical component of the shaking force varies with the counterbalance weight. From Eq. (1), it is clear that the vertical component of the shaking force is zero when mA mcb = 0 or mA = mcb This occurs when mcb = mA = 0.0414 lb s2 / in or - 626 -
Wcb = mcbg = 386(0.0414) = 16 lb The results from the program ShakeAnalysis is shown in Fig. 15.15.2 for this case. A very small vertical component is shown (6x10-16), but this is due to numerical errors. Note again that the value is optimum from the standpoint of the vertical component but not the horizontal component.
Fig. 15.15.1 Results from Shake for case when the counterbalance weight it 28 lb.
- 627 -
Fig. 15.15.2 Results from Shake for case when the counterbalance weight it 16 lb. The vertical component of the shaking force is exaggerated in the plot because of the plotting scale, but it is essentially zero as the vertical scale indicates.
- 628 -
Problem 15.16 The two-cylinder engine shown below has identical cranks, connecting rods, and pistons. The rotary masses are perfectly balanced. Derive an expression for the shaking forces and shaking moments using the symbols indicated if 2 = 90˚. Are the primary or secondary shaking forces balanced? What about the primary and secondary shaking moments?
A1
φ2 R A2
L
θ
B2
Β1 Β1
Plane 1 a
Plane 2
B2 Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, 1 = 2 = 0 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
For this problem,
1 = 0 and 2 = 90˚. Therefore, the shaking force components can be written as fx = mBR 2[cos sin ]
(3)
and fy = 0
(4)
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When 1 = 2 = 0 , these equations become: Mx = 0
(5) - 629 -
and n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
(6)
For 1 = 0 , a1 = 0, and 2 = 90˚, and
Mx = 0
(5)
My = mBR 2 a2 sin a2 R cos2 L
{
}
(6)
Therefore, the primary shaking forces, the primary shaking moment, and the secondary shaking moment are unbalanced. Only the secondary shaking force is balanced directly. Problem 15.17 Resolve Problem 15.16 when 2 = 180˚. Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, 1 = 2 = 0 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
For this problem,
1 = 0 and 2 = 180˚. Therefore, the shaking force components can be written as fx = mBR 2[cos (1 1)+ R cos2 (1 +1) = mBR 2 2 R cos2 L L
]
[
]
(3)
and fy = 0
(4)
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When 1 = 2 = 0 , these equations become: and
Mx = 0
(5)
n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
(6)
For 1 = 0 , a1 = 0, and 2 = 180˚, - 630 -
and
Mx = 0
(5)
My = mBR 2[a2 cos +a2 R cos2 L
]
(6)
Therefore, the secondary shaking forces, the primary shaking moment, and the secondary shaking moment are unbalanced. Only the primary shaking force is balanced directly. Problem 15.18 The four-cylinder engine shown below has identical cranks, connecting rods, and pistons. The rotary masses are perfectly balanced. Derive an expression for the shaking forces and shaking moments for the angles and offset values indicated. Are the primary or secondary shaking forces balanced? What about the primary and secondary shaking moments?
B1
Β1
B4
B2
Β4 Β2
B3
R
A4
φ4 φ
2
Plane 3
θ
A1
A2
Plane 2
Plane 1
L
Plane 4
Β3
φ 1 = 0˚ φ2 = 180˚ φ 3 = 180˚ φ 4 = 0˚ a1 = 0 a2 = a a3 = 2 a a4 = 3 a
a2
φ3 A3
a3 a4
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, i = 0, i = 1, 2, 3, 4 , and these equations become: fx =
mBR 2
and
n n n n R R + cos cos sin sin cos2 cos2 sin2 sin2i i i i L L
i=1 i=1 i=1 i=1
fy = 0
(1) (2)
- 631 -
For this problem,
1 = 3 = 0 and 2 = 3 = 180˚ . Therefore, the shaking force components can be written as cos0 sin0 cos0 sin0 + cos180 + sin180 R + cos360 R +sin360 f x = mB R 2 cos cos2 sin2 + cos180 sin + sin180 + L + cos360 L +sin360 + cos0 + sin0 + cos0 +sin0
or fx = mBR 2[cos (1 1 1+1)+ R cos2(1+ 1+1+ 1) = mBR 2[4 R cos2 L L
]
]
(3)
and fy = 0
(4)
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4 , these equations become: and
Mx = 0
(5)
n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
(6)
For 1 = 3 = 0 , 2 = 3 = 180˚ , and a1 = 0; a2 = a; a3 = 2a; a4 = 3a and
Mx = 0
(5)
0cos0 0sin0 0cos0 0sin0 +acos180 +asin180 R +acos360 R +asin360 My = mB R 2 cos cos2 sin2 +2acos180 sin +2asin180 + L +2acos360 L +2asin360 +3acos0 +3asin0 +3acos0 +3asin0
or
[
My = mBR 2 cos (a 2a + 3a) +
R cos2(a + 2a + 3a) R = mB R 2 6a cos2 L L
]
[
]
(6)
Therefore, the secondary shaking forces and the secondary shaking moment are unbalanced. The primary shaking forces and the primary shaking moments are balanced directly.
Problem 15.19 Resolve Problem 15.18 for the following values for the phase angles and offset distances.
- 632 -
1 = 0 a1 = 0
2 = 90˚ a2 = a
3 = 270˚ a3 = 2a
4 = 180˚ a4 = 3a
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, i = 0, i = 1, 2, 3, 4 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
For this problem,
1 = 0, 2 = 90˚, 3 = 270˚, 4 = 180˚ Therefore, the shaking force components can be written as cos0 sin0 cos0 sin0 +cos90 +sin90 R +cos180 R +sin180 f x = mB R 2 cos cos2 L sin2 +sin540 +cos270 sin +sin270 + L +cos540 +cos180 +sin180 +cos360 +sin360
or
[
f x = mB R 2 cos (1+ 0 + 0 1)sin (0+ 11+ 0) + R cos2 (111+ 1) R sin2 (0 + 0 0 + 0) L L =0
]
and fy = 0 The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4 , these equations become: and
Mx = 0 n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
For
1 = 0, 2 = 90˚, 3 = 270˚, 4 = 180˚ and a1 = 0; a2 = a; a3 = 2a; a4 = 3a Mx = 0 - 633 -
and 0cos0 0sin0 0cos0 0sin0 +a cos90 +a sin90 R +acos180 R +a sin190 My = mB R 2 cos sin + cos2 sin2 +2acos270 +2asin270 L +2acos540 L +2asin540 +3acos180 +3asin180 +3acos180 +3asin180
or
My = mB R 2 cos (0+ 0 + 0 3a) sin (0+ a 2a+ 0)+ R cos2 (0 a 2a + 3a) R sin2 (0 + 0+ 0+ 0) L L = mB R 2[-3a cos + a sin ]
[
]
Therefore, the shaking forces are balanced, and the secondary shaking moments are balanced. The primary shaking moments are unbalanced.
Problem 15.20 Resolve Problem 15.18 for the following values for the phase angles and offset distances.
1 = 0 a1 = 0
2 = 180˚ a2 = a
3 = 90˚ a3 = 2a
4 = 270˚ a4 = 3a
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, i = 0, i = 1, 2, 3, 4 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
For this problem,
1 = 0, 2 = 180˚, 3 = 90˚, 4 = 270˚ Therefore, the shaking force components can be written as cos0 sin0 cos0 sin0 +cos180 +sin180 R +cos360 R +sin360 f x = mB R 2 cos cos2 sin2 +cos90 sin +sin90 + L +cos180 L +sin180 +cos270 +sin270 +cos540 +sin540
or f x = mB R 2 cos[11+ 0 + 0] sin [0+ 0+11] + R cos2[1+1 11] R sin2[0 + 0 + 0+ 0] L L =0
{
}
and - 634 -
fy = 0 The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4 , these equations become: and
Mx = 0 n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
For
1 = 0, 2 = 180˚, 3 = 90˚, 4 = 270˚ and a1 = 0; a2 = a; a3 = 2a; a4 = 3a Mx = 0 and 0cos0 0sin0 0cos0 0sin0 +acos180 +asin180 R +acos360 R +asin360 My = mB R 2 cos cos2 sin2 +2acos90 sin +2asin90 + L +2acos180 L +2asin180 +3acos270 +3asin270 +3acos540 +3asin540 or [ ] R [ ] My = mB R 2 cos[0 a + 0 + 0] sin [0 + 0+ 2a 3a] + R L cos2 0+ a 2a 3a L sin2 0+ 0+ 0 + 0 R 2 = mB R 2 acos + asin 4a R L cos2 = mB R a cos + sin 4 L cos2
{ (
}
)
(
)
Therefore, the shaking forces are balanced, but both the primary and secondary shaking moments are unbalanced.
- 635 -
Problem 15.21 The six-cylinder engine shown below has identical cranks, connecting rods, and pistons. The rotary masses are perfectly balanced. Derive an expression for the shaking forces and shaking moments for the angles and offset values indicated. Are the primary or secondary shaking forces balanced? What about the primary and secondary shaking moments? Β6 B6 B1
Β1
Β5
B5 B4
B2
Β4
Β2
B3
φ6
Plane 6
Plane 5
Plane 3
A6
R
φ2
Plane 2
θ
A1
A2
Plane 1
L
Plane 4
Β3
φ 1 = 0˚ φ 2 = 240˚ φ 3 = 120˚ φ 4 = 120˚ φ 5 = 240˚ φ 6 = 0˚ a1 = 0 a2 = a a3 = 2 a a4 = 3 a a5 = 4 a a6 = 5 a
A5
φ5 φ3 A3
φ4
A4
a2 a3 a4 a5 a6
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, i = 0, i = 1, 2, 3, 4 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
- 636 -
For this problem,
1 = 0˚; 2 = 240˚; 3 = 120˚; 4 = 120˚; 5 = 240˚; 6 = 0˚. Therefore, the shaking force components can be written as
and
fx = mBR 2[cos (1 0.5 0.5 0.5 0.5+ 1) sin(0 0.866 + 0.866 + 0.866 0.866 + 0)] + R cos2 (1 0.5 0.5 0.5 0.5 +1) R sin2(0 + 0.866 0.866 0.866 + 0.866)] = 0 L L fy = 0
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4, 5, 6, these equations become: and
Mx = 0 n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
For 1 = 0˚; 2 = 240˚; 3 = 120˚; 4 = 120˚; 5 = 240˚; 6 = 0˚, and a1 = 0; a2 = a; a3 = 2a; a4 = 3a; a5 = 4a; a6 = 5a and
Mx = 0
My = mBR 2[ cos (0.5a 1.0a 1.5a 2a + 5a) sin{0.866a + 0.866(2a)+ 0.866(3a) 0.866(4a)} + R cos2 (0.5a 1.0a 1.5a 2a + 5a) R sin2{0.866(a) 0.866(2a) 0.866(3a) + 0.866(4a)} L L R R = mBR 2 cos (0) sin(0) + cos2 (0) sin2 (0) = 0 L L
[
]
]
Therefore, the primary and secondary shaking forces and shaking momenet are balanced directly. t are balanced directly.
Problem 15.22 Resolve Problem 15.21 for the following values for the phase angles and offset distances.
1 = 0˚ a1 = 0
2 = 120˚ 3 = 240˚ 4 = 60˚ a2 = a a3 = 2a a4 = 3a
- 637 -
5 = 300˚ 6 = 180˚ a5 = 4a a6 = 5a
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). For an inline engine, i = 0, i = 1, 2, 3, 4 , and these equations become:
and
n n n n fx = mBR 2 cos cosi sin sini + R cos2 cos2i R sin2 sin2i L L i=1 i=1 i=1 i=1
(1)
fy = 0
(2)
For this problem,
1 = 0˚; 2 = 120˚; 3 = 240˚; 4 = 60˚; 5 = 300˚; 6 = 180˚. Therefore, the shaking force components can be written as cos (1 0.5 0.5+ 0.5 + 0.5 1) sin (0 + 0.866 0.866 + 0.866 0.866 + 0) =0 fx = mBR 2 + R cos2(1 0.5 0.5 0.5 0.5 + 1) L R L sin2(0 0.866 + 0.866 + 0.866 0.866 + 0) and fy = 0 The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4, 5, 6, these equations become: and
Mx = 0 n n n n My = mBR 2 cos ai cosi sin ai sin i+ R cos2 ai cos2i R sin2 ai sin2i L L i=1 i=1 i=1 i=1
For 1 = 0˚; 2 = 120˚; 3 = 240˚; 4 = 60˚; 5 = 300˚; 6 = 180˚, and a1 = 0; a2 = a; a3 = 2a; a4 = 3a; a5 = 4a; a6 = 5a Mx = 0 and cos (0 0.5(a) 0.5(2a)+ 0.5(3a)+ 0.5(4a) 5a) sin (0 + 0.866(a) 0.866(2a) + 0.866(3a) 0.866(4a) + 0) My = mBR 2 + R cos2(0 0.5(a) 0.5(2a) 0.5(3a) 0.5(4a)+ 5a) L R sin2(0 0.866(a)+ 0.866(2a)+ 0.866(3a) 0.866(4a) + 0) L R R = mBR 2 cos (3a) sin (1.732a) + cos2(0) sin2 (0) L L = mBR 2[cos (3a) sin(1.732a)]
[
]
- 638 -
Therefore, the primary shaking moment is unbalanced, but the primary and secondary shaking forces and secondary shaking moment
Problem 15.23 The two-cylinder V engine shown below has identical cranks, connecting rods, and pistons. The rotary masses are perfectly balanced. Derive an expression for the shaking forces and shaking moments for the angles and offset values indicated. The V angle is = 60˚, and the phase angle is 2 = 90˚. Are the primary or secondary shaking forces balanced? What about the primary and secondary shaking moments? X 2
L
ψ A2
1
Cylinder 1
B
Cylinder 2
B
θ
φ2
A1
R a
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). n n fx = mBR 2 cos cos(i i )cos i sin sin(i i)cos i i=1 i=1 n n + R cos2 cos2(i i )cos i R sin2 sin2(i i)cos i L L
i=1 i=1
(1)
and n n fy = mB R 2 cos cos(i i )sin i sin sin(i i)sin i i=1 i=1 n n + R cos2 cos2(i i )sin i R sin2 sin2(i i)sin i L L
i=1 i=1
For this problem,
1 = 0˚; 2 = 90˚; 1 = 0˚; 2 = 60˚. - 639 -
(1)
Therefore, the shaking force components can be written as fx = mBR 2 cos (1+ 0.433) sin (0 + 0.25) + R cos2 (1+ 0.25) R sin2(0 + 0.433) L L R R = mBR 2 cos (1.433) sin (0.25) + cos2 (1.25) sin2(0.433) L L
[ [
and
]
]
R R fy = mB R 2 cos (0 + 0.750) sin (0 + 0.433) + cos2 (0 + 0.433) sin2 (0 + 0.750) L L R R = mBR 2 cos (0.750) sin(0.433) + cos2 (0.433) sin2 (0.750) L L
[ [
]
]
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4, 5, 6, these equations become: n n Mx = mBR 2 cos ai cos(i i)sin i sin ai sin(i i )sin i i=1 i=1
(15.47)
n n + R cos2 ai cos2(i i )sin i R sin2 ai sin2(i i)sin i L L
i=1 i=1
and the y component is given by n n My = mBR 2 cos ai cos(i i )cos i sin ai sin(i i)cos i i=1 i=1
(15.48)
n n + R cos2 ai cos2(i i )cos i R sin2 ai sin2(i i)cos i L L
i=1 i=1
For 1 = 0˚; 2 = 90˚; 1 = 0˚; 2 = 60˚ and a1 = 0; a2 = a Mx = mBR 2 cos(0 + 0.750a) sin (0 + 0.433a) + R cos2 (0 + 0.433a) R sin2 (0 + 0.750a) L L R R = mBR 2 cos (0.750a) sin (0.433a) + cos2(0.433a) sin2(0.750a) L L
[
[
]
]
and R R My = mBR 2 cos (0 + 0.433a) sin (0 + 0.25a)+ cos2 (0 + 0.25a) sin2 (0 + 0.433a) L L R R = mBR 2 cos (0.433a) sin (0.25a)+ cos2 (0.25a) sin2 (0.433a) L L
[
[
]
]
Therefore, the primary and secondary shaking forces and moments are unbalanced. None of the forces and moments are balanced directly.
Problem 15.24 Resolve Problem 15.23 if = 180˚ and 2 = 180˚. - 640 -
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). n n fx = mBR 2 cos cos(i i )cos i sin sin(i i)cos i i=1 i=1 n n R R cos2 cos2( )cos sin2 sin2(i i)cos i + i i i L L
i=1 i=1
and n n fy = mB R 2 cos cos(i i )sin i sin sin(i i)sin i i=1 i=1 n n + R cos2 cos2(i i )sin i R sin2 sin2(i i)sin i L L
i=1 i=1
For this problem, . Therefore, the shaking force components can be written as
and
fx = mBR 2 cos (11) sin (0 + 0) + R cos2(1 1) R sin2 (0 + 0) = 0 L L
[ ] R R f = m R [cos (0 0) sin (0 + 0) + cos2(0 0) sin2(0 + 0)] = 0 L L y
B
2
The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). When i = 0, i = 1, 2, 3, 4, 5, 6, these equations become: n n Mx = mBR 2 cos ai cos(i i)sin i sin ai sin(i i )sin i i=1 i=1 n n + R cos2 ai cos2(i i )sin i R sin2 ai sin2(i i)sin i L L
i=1 i=1
and the y component is given by n n My = mBR 2 cos ai cos(i i )cos i sin ai sin(i i)cos i i=1 i=1 n n + R cos2 ai cos2(i i )cos i R sin2 ai sin2(i i)cos i L L
i=1 i=1
For 1 = 0˚; 2 = 90˚; 1 = 0˚; 2 = 60˚ and a1 = 0; a2 = a
and
Mx = mBR 2 cos(0 0) sin(0 + 0) + R cos2 (0 0) R sin2 (0 + 0) = 0 L L
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My = mBR 2 cos (0 a) sin (0 + 0) + R cos2 (0 a) R sin2 (0 + 0) L L R = mBR 2 cos (a) + cos2 (a) L
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Therefore, the primary and secondary shaking moments are unbalanced, but the primary and secondary shaking forces are balanced directly.
Problem 15.25 The six-cylinder V engine shown below has identical cranks, connecting rods, and pistons. The rotary masses are perfectly balanced. Derive an expression for the shaking forces and shaking moments for the angles and offset values indicated. The V angle is = 60˚, and the phase angles 1, 2, and 3 are 0˚, 120˚, and 240˚, respectively. Are the primary or secondary shaking forces balanced? What about the primary and secondary shaking moments? X
B5
B6
L
B3
ψ φ2
A 2 ,A 5
Cylinders 1 and 4
B2
Cylinders 2 and 5
B4
Cylinders 3 and 6
B1
θ A 1, A 4
φ3
R a 2a
A 3, A 6
Solution The x and y components of the shaking force for a general engine with n cylinders are given by Eqs. (15.37) and (15.38). n n fx = mBR 2 cos cos(i i )cos i sin sin(i i)cos i i=1 i=1 n n + R cos2 cos2(i i )cos i R sin2 sin2(i i)cos i L L
i=1 i=1
and
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n n fy = mB R 2 cos cos(i i )sin i sin sin(i i)sin i i=1 i=1 n n + R cos2 cos2(i i )sin i R sin2 sin2(i i)sin i L L
i=1 i=1
For this problem,
1 = 0˚; 2 = 120˚; 3 = 240˚; 4 = 0˚; 5 = 120˚; 6 = 240˚. and
1 = 0˚; 2 = 0˚; 3 = 0˚; 4 = 60˚; 5 = 60˚; 6 = 60˚ Therefore, the shaking force components can be written as cos (1 0.5 0.5+ 0.25 + 0.25 0.5) sin (0 + 0.866 0.866 0.433+ 0.433+ 0) =0 fx = mBR 2 + R cos2(1 0.5 0.5 0.25 0.25 + 0.5) L R L sin2(0 0.866 + 0.866 0.433 + 0.433 + 0) and cos (0 + 0 + 0 + 0.433+ 0.433 0.866) sin (0 + 0 + 0 0.750 + 0.750 + 0) R fy = mBR 2 + cos2 (0 + 0 + 0 0.433 0.433 + 0.866) = 0 L R L sin2 (0 + 0 + 0 0.75 + 0.75 + 0) The x and y components of the shaking moments are given by Eqs. (15. 47) and (15.48). n n Mx = mBR 2 cos ai cos(i i)sin i sin ai sin(i i )sin i i=1 i=1 n n + R cos2 ai cos2(i i )sin i R sin2 ai sin2(i i)sin i L L
i=1 i=1
and the y component is given by n n My = mBR 2 cos ai cos(i i )cos i sin ai sin(i i)cos i i=1 i=1 n n + R cos2 ai cos2(i i )cos i R sin2 ai sin2(i i)cos i L L
i=1 i=1
For 1 = 0˚; 2 = 120˚; 3 = 240˚; 4 = 0˚; 5 = 120˚; 6 = 240˚, and a1 = 0; a2 = a; a3 = 2a; a4 = 0; a5 = a; a6 = 2a
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cos(0 + 0 + 0 + 0 + 0.433a 1.732a) sin (0 + 0 + 0 + 0 + 0.75a + 0) Mx = mB R 2 + R cos2 (0 + 0 + 0 + 0 0.433a +1.732a) L R L sin2 (0 + 0 + 0 + 0 + 0.75a + 0) = mBR 2 cos (1.299a) sin(0.75a) + R cos2(1.299a) R sin2 (0.75a) L L
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and cos (0 0.5a a + 0 + 0.25a a) sin (0 + 0.866a + 1.732a + 0 + 0.433a + 0) My = mBR 2 + R cos2(0 + 0.5a a + 0 0.25a + a) L R L sin2(0 0.866a +1.732a + 0 + 0.433a + 0) = mBR 2 cos (2.25a) sin (0.433a)+ R cos2 (0.75a) R sin2 (1.299a) L L
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Therefore, the primary shaking moment is unbalanced, but the primary and secondary shaking forces and secondary shaking momenet are balanced directly.
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