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A linkage is shown below in the figure in which links ABC and DEF are ternary Jinks whereas AF, BE and CD are binary links.
The degrees of freedom of the linkage when link ABC is fixed are
[IES-2002]
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
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A linkage is shown below in the figure in which links ABC and DEF are ...
EB is not redundant , if u remove EB then its dof will be 1... its dof is 0 , u can also visualise this.
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A linkage is shown below in the figure in which links ABC and DEF are ...
To calculate the degrees of freedom (DOF) of the given linkage, we can use Gruebler's equation for planar mechanisms:
F=3(n−1)−2j−h
Where:
- F is the degrees of freedom.
- n is the number of links in the mechanism.
- j is the number of lower pairs (revolute or prismatic joints).
- h is the number of higher pairs (if any, typically ignored in simple linkages).
Step 1: Count the number of links (nnn)
In the figure:
- Links ABC and DEF are ternary links (each connects three joints), so we count each as one link.
- Links AF, BE, and CD are binary links (each connects two joints).
Thus, the total number of links nnn is:
n=2(ternary links)+3(binary links)=5
Step 2: Count the number of joints (j)
In the figure:
- There are six revolute joints, one at each intersection of the links: A, F, B, E, C, and D.
Thus, the total number of lower pairs (joints) j is:
j=6
Step 3: Apply Gruebler’s equation
Substitute the values into Gruebler’s equation:
F=3(n−1)−2j
F=3(5−1)−2(6)
F=3(4)−12
F=12−12=0
Conclusion:
The degrees of freedom of the linkage is F=0, meaning the linkage is a structure and does not have any mobility when the link ABC is fixed.
Thus, the correct answer is:
Option 1: 0.
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A linkage is shown below in the figure in which links ABC and DEF are ternary Jinks whereas AF, BE and CD are binary links.The degrees of freedom of the linkage when link ABC is fixed are[IES-2002]a)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer?
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