A beam engine utilizes 2 main mechanisms: Four-bar and slider-crank. A schematic showing the four-bar linkage in a typical beam engine is shown in the following figure. Given: • Link lengths: Link O₂A = 5 cm, O4B = 15 cm, 0204 = 25 cm, AB = 20cm Positions: 0 145°, 8=35°. a) Draw the vector loop and write down the vector loop equation b) Calculate the local angular position of the crank O2A. c) Calculate the local angular position of the rocker O4B. d) Calculate the global angular position of the coupler AB. e) Deduce the configuration (opened or crossed). Y 02 B 04 Position, Velocity and Accleration Analysis: Pin-jointed Fourbar linkage 002 041,2 = 2 arctan • A • • VB VBA VBA VB VA (b) R3 b R4 A 03 x C 0 a R2 02 R1 Position Analysis Ол 004 - X -B±√B2 4AC -E±√E² 4DF 031.2 =2 arctan 2A 2D • D= cos 2-K₁ + K4cos02 + K5 E = -2sin02 FK₁(K41) cos02 + K5 cose₂-K₁- K₂cosе₂ + K3 • B = -2sin02 CK₁(K₂+1)cos0₂ + K3 • K₁₁ = • K₂ = a²-b²+c²+d² . K3 = 2ac K₁ = • K₁₁ = b c²-d²-a²-b2 K5 = 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos02 + ccos04 + d bsin03 = -asin02 + csin04 aw2 sin(02-03) 004 = c sin(04-03) Velocity Analysis aw2 sin(04-02) 003 = b sin(03-04)
A beam engine utilizes 2 main mechanisms: Four-bar and slider-crank. A schematic showing the four-bar linkage in a typical beam engine is shown in the following figure. Given: • Link lengths: Link O₂A = 5 cm, O4B = 15 cm, 0204 = 25 cm, AB = 20cm Positions: 0 145°, 8=35°. a) Draw the vector loop and write down the vector loop equation b) Calculate the local angular position of the crank O2A. c) Calculate the local angular position of the rocker O4B. d) Calculate the global angular position of the coupler AB. e) Deduce the configuration (opened or crossed). Y 02 B 04 Position, Velocity and Accleration Analysis: Pin-jointed Fourbar linkage 002 041,2 = 2 arctan • A • • VB VBA VBA VB VA (b) R3 b R4 A 03 x C 0 a R2 02 R1 Position Analysis Ол 004 - X -B±√B2 4AC -E±√E² 4DF 031.2 =2 arctan 2A 2D • D= cos 2-K₁ + K4cos02 + K5 E = -2sin02 FK₁(K41) cos02 + K5 cose₂-K₁- K₂cosе₂ + K3 • B = -2sin02 CK₁(K₂+1)cos0₂ + K3 • K₁₁ = • K₂ = a²-b²+c²+d² . K3 = 2ac K₁ = • K₁₁ = b c²-d²-a²-b2 K5 = 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos02 + ccos04 + d bsin03 = -asin02 + csin04 aw2 sin(02-03) 004 = c sin(04-03) Velocity Analysis aw2 sin(04-02) 003 = b sin(03-04)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
very urgent i need her solving all parts with using the formulas attached below quick please
Transcribed Image Text:A beam engine utilizes 2 main mechanisms: Four-bar and slider-crank. A schematic showing
the four-bar linkage in a typical beam engine is shown in the following figure.
Given:
•
Link lengths: Link O₂A = 5 cm, O4B = 15 cm, 0204 = 25 cm, AB = 20cm
Positions: 0 145°, 8=35°.
a) Draw the vector loop and write down the vector loop equation
b) Calculate the local angular position of the crank O2A.
c) Calculate the local angular position of the rocker O4B.
d) Calculate the global angular position of the coupler AB.
e) Deduce the configuration (opened or crossed).
Y
02
B
04
Transcribed Image Text:Position, Velocity and Accleration Analysis: Pin-jointed Fourbar
linkage
002
041,2
= 2 arctan
•
A
•
•
VB
VBA
VBA
VB
VA
(b)
R3
b
R4
A
03
x
C
0
a
R2 02
R1
Position Analysis
Ол
004
- X
-B±√B2
4AC
-E±√E²
4DF
031.2
=2 arctan
2A
2D
• D= cos 2-K₁ + K4cos02 + K5
E = -2sin02
FK₁(K41) cos02 + K5
cose₂-K₁- K₂cosе₂ + K3
•
B = -2sin02
CK₁(K₂+1)cos0₂ + K3
•
K₁₁ =
•
K₂ =
a²-b²+c²+d²
.
K3 =
2ac
K₁ =
•
K₁₁ =
b
c²-d²-a²-b2
K5 =
2ab
PS: if is calculated before you can use one of the equations below to solve for 03
bcos03=-acos02 + ccos04 + d
bsin03 = -asin02 + csin04
aw2 sin(02-03)
004
=
c sin(04-03)
Velocity Analysis
aw2 sin(04-02)
003 =
b sin(03-04)
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