Statically determined group (of links)

Statically determined group (of links)

kinematic system in equilibrium

every link in equilibrium

any group of links in equlibrium

forces may be fully determined using the six (three) equations of 3D (2D) statics without

using deflection and stiffness criteria.

j

i

F_ijx F_ijy

i

j Fijx

Fijy

i

j Fijx

Fijy

1 unknown 2 unknowns 2 unknowns

Number of equations

= Number of unknowns

3k = 2p₁ + p₂

In kinetostatics unknowns are joint forces

( ¹ ) ²

¹

3 n p p

DOF = − − − 0 1

2

3 k − p

− p

=

Mobility

Condition of static determination

k 1 2

...

4

p₁ 1 3

...

6

p₂ 1 0

...

0

(k-p₁–p₂) (1-1-1) (2-3-0) (460)

Examples of static. determined groups

group (1-1-1) group (2-3-0)

II class - R or T II class - K or J or Z

I kl - R lub T

1-1-1 stat. determined group (one link)

3 1

2 F₁₂

F₂

F₃₂ F₃₂ + F₁₂ + F₂ = 0

2-link groups (k-p

-p

= 230)

RRR

RTR RTT

RRT

TRT

EXAMPLE 1

Group RRR

Group RRR – forces

1

2

3

4 F₂

F3

F^t12

F^t₄₃

Fⁿ43

Fⁿ₁₂ A

B

C h₂

h₃

F₂

F^t₁₂

F₃

F^t₄₃

Fⁿ₄₃ Fⁿ₁₂

Group RRR – forces

1

2

3

4 F₂

F3

F^t12

F^t₄₃

Fⁿ43

Fⁿ₁₂ A

B

C h₂

h₃

F₂

F^t₁₂

F₃

F^t₄₃

Fⁿ₄₃ Fⁿ₁₂

1 2 ^t = ?

F ^F _{4 3} ^{t = ?}

= 0 +

+ +

+

+

t

12

F F F F F

F

Equilibrium equation (forces) of group (links 2 and 3):

Group RRR – forces

1

2

3

4 F₂

F3

F^t12

F^t₄₃

Fⁿ43

Fⁿ₁₂ A

B

C h₂

h₃

F₂

F^t₁₂

F₃

F^t₄₃

Fⁿ₄₃ Fⁿ₁₂

AB t

AB t B

l h F F

l F h

F M

2 2 12

12 2

2

2

0 0

=

→

= +

−

→

 =

BC t

BC t B

l h F F

l F h

F M

3 3 43

43 3

3

3

0 0

=

→

= +

−

→

 =

Equilibrium equation (torques) of link 3 relative to point B:

Equilibrium equation (torques) of link 2 relative to point B:

1

2

3

4 F2

F3

F^t12

F^t₄₃

Fⁿ₄₃ Fⁿ12

A

B

C h2

h3

F₂

F^t₁₂

F₃

F^t₄₃

Fⁿ₄₃ Fⁿ₁₂

Group RRR – forces

= 0 +

+ +

+

+

t

12

F F F F F

F

= 0 +

+

12

F F

F

= 0 +

+

23

F F

F

Equilibrium equation (forces) of link 2:

Equilibrium equation (forces) of link 3:

Equilibrium equation (forces) of group (links 2 and 3):

EXAMPLE 2

Group RRT

M₁ =?

3

0

2

1 Fc c

Fb

F03 c

0

= 0 +

+ _{b 12}

32 F F

F

= 0 +

+ _{03 23}

C F F

F

Equilibrium equation (forces) of link 3:

Equilibrium equation (forces) of link 2:

3

2 F_c C c

B

Fc c

Fb

c F03

c

F_b

c

F12 t

c

F12t c

F12n c

F12n c

h_F

c

12

0

12

03

+ + + =

+

c

F F F F

F

 ^{= → − = → =}

BC F t b

BC t F

b C

l h F F

l F h

F

M

0

0

Links 2-3 group RRT

Equilibrium equation (forces) of group (links 2 and 3):

Equilibrium equation (torques) of link 2 relative to point C:

2

+

= 0

3

F

F

because

3

2 Fc c

C

B

F_c

c

Fb

c F₀₃

c

F_b

c

F₁₂^t

c

F12 t

c

F12 n

c

F12 n

c

hF c

= 0 +

+ +

+

c

F F F F

F

Equilibrium equation (forces) of group (links 2 and 3):

F

M₁

0

1

F₁₀

F₂₁ F₁₂

F₀₁

h₁ A

 ^M

^{= 0 → M}

^{− F}

^h

^{= 0}

Equilibrium equation (forces) of link 1:

Equilibrium equation (torques) of link 1 relative to point A:

= 0 +

21

F

F

EXAMPLE 3

Group RTR

3

0 1

2

3

0 1

2

M₃

M₂=?

F_b2

Data: M

, F

Unknown: M

dynamics, kinetostatics 20

3

0 1

2

M₃

M₂=?

F_b2

stat determined group (2-3)

dynamics, kinetostatics 21

3

1

2

M₃ F_b2

0 C

F

F

B

dynamics, kinetostatics 22

3

1

2

M₃ F_b2

0 C

F

F

B

3

0

2 2

12

h − F h − M =

F

2

h

Equilibrium equation (torques)

of group (links 2 and 3) relative to point C:

CB b t b

h

M h

F

F

+

=

3

1

2

M₃ F_b2

0 C

F

F

B

32

0

2 12

12

+ F + F + F =

F

2

h

Equilibrium equation (forces) of link 2:

3

1

2

M₃ F_b2

0 C

F

F

B

2

h

32

0

2 12

12

+ F + F + F =

F

F

F

d .

b2

F

.F

d

3

1

2

M₃ F_b2

0 C

F

F

B

2

h

F

F

F

F

32

0

2 12

12

+ F + F + F =

F

3

1

2

M₃ F_b2

0 C

F

F

B

2

h

F

F

F

F

b2

F

32

0

2 12

12

+ F + F + F =

F

3

1

2

M₃ F_b2

0 C

F

F

B

2

h

F

F

F_b2

F

F

F

The Three-Force Theorem

3

1

2

M₃ F_b2

0 C

F

F

B

2

h

F

0 3

= ?

F

EXAMPLE 4

Group RRR

M

B

A

D C

2

1

4 3

F

F

M

Data: F

, F

, M

Unknown: M

, F

, F

, F

, F

Group 2.3.0

B

D C

2 4

3

F

F

M

Equilibrium equation (torques) of link 3 relative to point C:

M₁ + F₂ h₂ - F₂₃^t BC = 0

Equilibrium equation (torques) of link 4 relative to point C:

d.F

d.F

d.F

d.F

h

F

Equilibrium equation (forces) of group (links 3 and 4):

F₁ + F₂ + F₂₃^t + F₂₃ⁿ + F₁₄^t + F₁₄ⁿ = 0 F₂₃ = F₂₃^t + F₂₃ⁿ

F₁₄ = F₁₄^t + F₁₄ⁿ

-F₁ h₁+ F₁₄^t DC = 0

F

h

F

F

F

F

d.F

d.F

F

F

F

F

F

F

F

F

F₂₃^t = (M₂ + F₂ h₂)/BC

F₁₄^t = F₁ h₁/DC

A M

2

1 B

F

3

F

= - F

Equilibrium equation (torques) of link 2 relative to point A:

F

F

h

- M

= 0 h

2

F

M

= F

h

Equilibrium equation (forces) of link 2:

F

+ F

= 0

F

= -F

B

A

D C

2

1

4 3

F

F

M

M

F

F

F

EXAMPLE 5

Group 1.1.1 + RRT

B

A C

M 2

1 4

3

F

M

5

Data: F

, M

Unknown : M

, F

, F

, F

, F

, F

, F

1 Group 1.1.1

2 Group 2.3.0 (RRT) 3

D

3

F

5

Equilibrium equation (forces) of link 5:

F

+ F

+ F

= 0

The Three-Force Theorem 1

d.F

d.F

F

d.F

d.F

F

F

F

F

D

B

C 2

4 3

5 F

1

D Equilibrium equation (torques)

of link 3 relative to point C:

F

d.F

d.F

M

M

+ F

h

h

F

BC = 0

F

-

Equilibrium equation (forces) of group (links 3 and 4):

F

+ F

+ F

+F

= 0

Equilibrium equation (torques) of link 4 relative to point C:

-F

h

=0 , h

= 0 F

= F

+ F

d.F

F

d.F

F

d.F

F

F

F

F

h

F

F

F

F

= - F

F

= (M

+ F

h

)/BC

A M

2

1 B

F

3

F

= - F

Equilibrium equation (torques) of link 2 relative to point A:

F

F

h

- M

= 0 h

Equilibrium equation (forces) of link 2:

F

+ F

= 0 F

= -F

F

M

= F

h

B

A C

2

1 4

3

F

M

D 5

F

F

F

M

F

F

Statically determined groups - examples

of solutions