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Tutorial 2 Solution: Kinematics and Dynamics of Machines (Me 316)

This document provides solutions to tutorial questions on kinematics and dynamics of machines. Question 1 calculates the degree of freedom for five mechanisms based on the number of links, lower pairs, and higher pairs. Question 2 demonstrates the graphical construction of a slider crank mechanism, including its velocity diagram with scales and acceleration diagram with given values.

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Pratik Shirsath
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136 views5 pages

Tutorial 2 Solution: Kinematics and Dynamics of Machines (Me 316)

This document provides solutions to tutorial questions on kinematics and dynamics of machines. Question 1 calculates the degree of freedom for five mechanisms based on the number of links, lower pairs, and higher pairs. Question 2 demonstrates the graphical construction of a slider crank mechanism, including its velocity diagram with scales and acceleration diagram with given values.

Uploaded by

Pratik Shirsath
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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KINEMATICS AND DYNAMICS OF MACHINES (ME 316)

Tutorial 2 Solution

Q1.

No. of Links (N )= 7
Number of Lower pairs (L) = 8
Number of higher pairs (H) = 1
Degree of freedom (DOF) = 3(N − 1) − 2L − H = 3(7-1)-2(8)-1 = 1

No. of Links (N )= 9
Number of Lower pairs (L) = 11
Number of higher pairs(H) = 1 (Between wheel and runway)
Degree of freedom (DOF) = 3(N − 1) − 2L − H = 3(9-1)-2(11)-1 = 1

i
C

No. of Links (N )= 7
Number of Lower pairs (L) = 7 (Six turning pairs and one sliding pair)
Number of higher pairs(H) = 1 (Fork joint)
Degree of freedom (DOF) = 3(N − 1) − 2L − H = 3(7-1)-2(7)-1 = 3

No. of Links (N )= 6
Number of Lower pairs (L) = 7
Number of higher pairs(H) = 0
Degree of freedom (DOF) = 3(N − 1) − 2L − H = 3(6-1)-2(7)-0 = 1

ii
No. of Links (N )= 9
Number of Lower pairs (L) = 12
Number of higher pairs(H) = 0
Degree of freedom (DOF) = 3(N − 1) − 2L − H = 3(9-1)-2(12)-0 = 0

Q2.
Construction of Slider crank mechanism

Scale = 10 cm= 1 cm
Steps: From the figure

• Lay out an xy-axis system with its origin at pivot, A.

• Draw link 2 from point A with same scale at its given angle (60◦ ). Angle is
measured from line AC.

• The AB line shows the length of crank (48 cm) in scaled form.

• Mark the arc 160 cm length from point B to x axis.

• Connect the point B to C. This is the link 3.

Construction of velocity diagram


Given:

ωB = 20rad/s
Radius of crank (AB) = 48cm/s
VB = ωB × AB
VB = 960cm/s

Scale = 100 cm/s= 1 cm


The direction of VB is perpendicular to link 2. Graphically, we can obtain VBC and VC
Steps: From the figure

• Choose a convenient velocity scale and layout the known vector VA (line A’B’).

iii
• From the tip of VB (point B’), draw a construction line with the perpendicular
direction of BC, magnitude unknown.
• From the tail of VB (point A’), draw a construction line with the parallel of AC,
magnitude unknown.
• The intersection of these two lines gives the point C’.
• where A’B’ is the VB , B’C’ is the VBC and C’A’ is the VC (slider velocity).
measuring from figure
VBC = length of B’C’ × scale = 5.25 × 100 = 525 cm/s
VC = 9.7 × 100 = 970cm/s
ωAB = VBC /BC = 525/160 = 3.28 rad/s

Construction of acceleration diagram

Given:
ωB = 20rad/s
αB = 0
ωAB = 3.28rad/s
Radius of crank (AB) = 48cm/s
AnB = (ωB )2 × AB = 19200cm/s2
AnBC = (ωAB )2 × BC = 1721.34cm/s2

Scale = 5000 cm/s2 = 1 cm


Steps: From the figure

iv
• Choose a convenient acceleration scale and draw the vectors with known magnitude
and direction.

• From the origin A”, Draw the magnitude of AnB at an angle (180+60)◦ . Extreme
point of the AnB is B”.

• Draw B”B1 ” parallel to the BC with the magnitude of AnBC .

• From point B1 ”, mark a arc to x axis and draw a line B1 ”C.

AtBC = 3.34 × 5000 = 16700 cm/s2


αBC = AtBC /BC = 16700/160 = 104.37 rad/s2
AC = 1.4 ×5000 = 7200 cm/s2

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