Determinacy And Indeterminacy – GATE CE Structural AnalysisPractice Questions & PYQs

From inextensibility of BF and BA, $\Delta \mathrm{B}_{\mathrm{x}}=0 \quad \Delta \mathrm{B}_{\mathrm{y}}=0$ From inextensibility of $\mathrm{BE}$ and $\mathrm{EF}$ , $\Delta \mathrm{E}_{\mathrm{x}}=0 \quad \Delta \mathrm{E}_{\mathrm{y}}=0$ From inextensibility of CE and $\mathrm{BC}$ $\Delta \mathrm{C}_{\mathrm{x}}=0 \quad \Delta \mathrm{C}_{\mathrm{y}}=0$ From inextensibility of $\mathrm{BD}$ and $\mathrm{CD}$ , $\Delta \mathrm{D}_{\mathrm{x}}=0 \quad \Delta \mathrm{D}_{\mathrm{y}}=0$ From inextensibility of DG, $\Delta \mathrm{G}_{\mathrm{X}}=0$ From inextensibility of $\mathrm{EH}$ and $\mathrm{HF}$ , $\Delta \mathrm{H}_{\mathrm{x}}=0 \quad \Delta \mathrm{H}_{\mathrm{y}}=0$ From inextensibility $\mathrm{FI}$ , $\Delta \mathrm{l}_{\mathrm{x}}=0$ $\Rightarrow$ If at joint $(F), F B, F E, F H, F I$ are rigidly connected then possible displacements at $\mathrm{F}=\theta_{\mathrm{F}}$ . If at joint I, FI and IH are rigidly connected, possible displacements at $\mathrm{I}=\theta_{\mathrm{I}}$ . Hence unknown joint displacements are $\theta_{\mathrm{B}}, \theta_{\mathrm{C}}, \theta_{\mathrm{D}}, \theta_{\mathrm{E}}, \theta_{\mathrm{F}}, \theta_{\mathrm{G}}, \Delta_{\mathrm{Gy}}, \theta_{\mathrm{H}}, \theta_{\mathrm{I}}$ $\Rightarrow \mathrm{D}_{\mathrm{k}}=9$ Note: However, if somebody assumes all members at joint $(F)$ to be connected with pin then at $(F)$ we have unknown joint displacements as $\theta_{\mathrm{FB}}, \theta_{\mathrm{FE}}, \theta_{\mathrm{FH}}, \theta_{\mathrm{Fl}}$ . Similarly, if at joint I if somebody assumes $\mathrm{FI}$ and $\mathrm{HI}$ to be pin connected then unknown joint displacements at I are $\theta_{\mathrm{IF}}, \theta_{\mathrm{IH}}$ . Hence, $D_k$ will increase by 4 . $\Rightarrow \mathrm{D}_{\mathrm{k}}=9+4=13$

Unstability in beam can be checked (i) If support reactions are not enough. (ii) If reactions are concurrent. (iii) If reactions are parallel. (iv) If there is mechanism. where as for truss also along with the above given points the triangular panels are generally stable. But, a virtual inspection must be conducted to understand the stability (I) It is unstable as it shows mechanisms. Also to understand if we apply a vertical force at slider side, the deflected shape will be as follows. (II) The frame is internally stable but all the reactions are concurrent and meeting at hinge A, and the frame can rotate about A. Hence, it is unstable. (III) The frame is unstable because it cannot resist shear in DE and AB since DE and AB are slender members.

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.

Kinematic indeterminacy,

Currently no explanation available

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

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,

Degree of kinematic indeterminacy or degree of freedom,

Static indeterminacy, $D=D_{s \theta}+D_{s i}$

Alternatively,

Kinematic indeterminacy, $k=\left(3 j-r_{e}\right)-m^{\prime}$ where $m^{\prime}=$ total number of axially rigid members. $\Rightarrow \quad k=(3 \times 10-6)-11=13$

Degree of static indeterminacy,

Degree of kinematic indeterminacy,