Degree of static indeterminacy of a rigid-jointed plane frame having 15 members, 3 reaction components and 14 joints is

M=15
R=3
J=14
we know that,
Degree of static indeterminacy for plane frames (Ds)=(3m+R)-(3j+A)
but here is no internal hinge so A=0
Ds=(3x15+3)-(3x14)
Ds =6

static indeterminacy =static internal indeterminacy+ static external indeterminacy
i.e. Ds=DSi+ DSe
Dsi=Re-3 and Dsi=3m-(3j-3)
use the above prescribed formulae you will get answer as 6

3m+r-3j u will get

3(15)+3-3(14)=6

For a rigid jointed 2D plane, the formula is 3m+re-3j
Where, m= no of members
Re= reaction
J= no of joints

Re=3
M=15
J=14
Ds=(Re+3m)-3j
=(3+3×15)-3×14
=6

=3m+r-3j
=(3*15)+3-(3*14)
=6

Degree of static indeterminacy of rigid joint plane frame having 20 members, 7 reaction components and 15 joint is

3m+r-3j-no. of releases[due to inernat pin joint]

3m+r-3j-ec(ec=no.of internal hybrid joint or internal hinge)

3m+r-3j-(no of releases)
=3*15+3-3*14-0=6
No hybrid joints present hence the release is 0

3m+r-3j=15*3+3-3*14=48- 42=6

Ds=3m+r-3j {3×15+3-3×14}=6
m=member
r= reaction
J=joints

3 b +r _3j
3×15+3 _3×14 = 6

Any one could explain me