Determinacy and Indeterminacy

Static indeterminacy, SI=r-3=(3+2)-3=2
Kinematic indeterminacy=0+1=1

Static indeterminacy D_s = D_{se} + D_{si}-\text{Force releases}
External indeterminacy, D_{se} = r - s
No. of support reactions, r = 7
Number of equilibrium equations, s = 3
D_{se} = 7-3=4
Internal indeterminacy D_{si}= 3\times \text{No of closed boxes}
= 3\times 4 = 12
Force releases = 1 [At the internal hinge]
D_S= 4 + 12 - 1 = 15

D_{k} (when extensible) =14
D_{k}( when inextensible )=D_{k}( when extensible) - No. of axially rigid member
=14-6=8
Shear deformation are already neglected. They have no significance on D_{k}. So, reduction in D_{k} is 6.

\begin{aligned} \text{External Indeterminacy}&=r-s=4-3=1 \text { degree } \\ \text{Internal indeterminacy}&=m-(2 j-3)=15-(2 \times 8-3)\\ & =2 \text{ degrees} \end{aligned}

Kinematic indeterminacy,
\begin{aligned} D_{k}&=2 j-r_{e} \\ &=2 \times 7-3=11 \end{aligned}

Number of member,
m=4
Number of external reaction,
r_{e}=4
Number of joint,
j=5
Number of reaction released,
r_{r}=1
Degree of static -indeterminacy
\begin{aligned} D_{s} &=3 m+r_{e}-3 j-r_{r} \\ &=3 \times 4+4-3 \times 5-1 \\ &=0 \end{aligned}

\begin{aligned} D_{s} &=D_{s_{e}}+D_{s_{i}} \\ &=\left(r_{e}-3\right)+3 C-r_{r} \\ &=(4-3)+3 \times 0-1 \\ &=0 \end{aligned}
Since cable is provided, it implies that there is only vertical loading.
Hence the structure is stable.

In general, total number of equilibrium equations for a 3-D rigidly jointed frame are 6. i.e.
\Sigma F_{x}=0, \Sigma F_{y}=0, \Sigma F_{2}=0, \Sigma M_{x}=0, \Sigma M_{y}=0
\Sigma M_{z}=0
But for vertical loading, these equilibrium equations reduces to 3.
I.e. \Sigma F_{y}=0, \Sigma M_{x}=0\text{ and }\Sigma M_{z}=0
Also, at fixed support in a 3 -D rigidly jointed frame
total number of external reactions are 6 i.e..
R_{x^{\prime}}, R_{y^{\prime}} R_{z^{\prime}} M_{x^{\prime}} M_{y^{\prime}} M_{z}
But for vertical loading, these external reactions reduce to 3 i.e.,
R_{y^{\prime}} M_{x^{\prime}} M_{z}
Thus for two fixed supports total external reactions will be 6
\therefore Degree of static indeterminacy
= Total external reactions - Total number of available equilibrium equations
=6-3=3

The degree of static indeterminacy for a rigid hybrid frame is given by,
\begin{aligned} D_{s} &=3 m+r_{e}-r_{r}-3(j+7) \\ \text { Where, } m &=\text { total number of members }=9 \\ r_{e} &=\text { total number of external reactions } \\ &=2+1+1=4\\ r_{r}&= \text{total number of released reactions at }\\ &\text{hybrid joint}\\ &=\Sigma\left(m_{j}-1\right)=(2-1)+(2-1)=2 \\ j &=\text { total number of rigid joints }=6 \\ j &=\text { total number of hybrid joints }=2 \\ \therefore \quad D_{S} &=(3 \times 9)+4-2-3(6+2) \\ &=27+4-2-24=31-26=5 \end{aligned}