Ground Sliding full joint Note There are no rollslide half j

Ground Sliding full joint Note There are no roll-slide (half) joints in this linkage Ground L=8, J=10 DOF = 1 Multiple joint 4 2 Ground (link 1) Ground (a) Linkage with full and multiple joints Ground Multiple joint 4 L=6, J=7.5 DOF = 0 Half joint Ground Ground (link 1) (b) Linkage with full, half, and multiple joints

Solution

Part:- 1

Gruebler’s equation   :   DOF = 3(L - 1) - 2P₁ - P₂

Number of Fixed LInk = 1 (Ground), Slider is a Link = 1

Number of binary link = 5,   Number of ternary link = 1

So, Total Number of Links, L = 8  

P₁ = N + L - 1    (Here N = Number of LInk , L = Number of Loops)

( If we join link number 7, 3, 4 and 6 to the ground three loops can be made )

P₁ = 8 + 3 - 1 = 10

Number of Joints Having one degree of freedom, P₁ = 10

Number of Joints Having two or half degree of freedom, P₂ = 0

So, DOF = 3(8 - 1) - 2 x 10 - 0 = 1

Part:- 2

Gruebler’s equation   :   DOF = 3(L - 1) - 2P₁ - P₂

Number of Fixed LInk = 1 (Ground),

Number of binary link = 3,   Number of ternary link = 2

So, Total Number of Links, L = 6

P₁ = N + L - 1    (Here N = Number of LInk , L = Number of Loops)

(If we join link number 2, 3, 4 and 5 one loop will be created, and joining link number 2, 6 at ground one more loops will be created. So, total loops is two)

P₁ = 6 + 2 - 1 = 7

Number of Joints Having one degree of freedom, P₁ = 7

Apart from this there is one half joint

So, total number of joints = 7 + 0.5 = 7.5

Number of Joints Having two or half degree of freedom, P₂ = 1

So, DOF = 3(6 - 1) - 2 x 7 - 1 = 1