The plane frame shown in the figure has fixed support at joint A, hinge support at joint $\mathrm{F}$ , and roller support at joint 1 . In the figure, $A$ to I indicate joints of the frame. If the axial deformations are neglected, the degree of kinematic indeterminacy is (integer).

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📖 Explanation

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 . $\Rightarrow \mathrm{D}_{\mathrm{k}}=9+4=13$

Q14GATE 2004MCQ

For the plane frame with an overhang as shown below, assuming negligible axial deformation, the degree of static indeterminacy, d, and the degree of kinematic indeterminacy, k, are

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📖 Explanation

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

\begin{aligned} \text{where, }D_{s \theta}&=\text{ external static indeterminacy }\\ D_{s i} &=\text { internal static indeterminacy } \\ \text { alsp. } D_{s e} &=r_{\theta}-3 \\ &\quad \left[\because r_{\theta}=\right.\text { total external reactions }] \\ &=3+2+1-3=3 \\ \text { and } & D_{s i}=3 C \\ &\quad[\because C=\text { total number of closed loops }] \\ &=3 \times 2=6 \\ \therefore \quad D_{s} &=3+6=9 \end{aligned}

Alternatively,

\begin{aligned} D_{s} &=3 m+r_{e}-3 \mathrm{j} \\ m &=\text { total members }=11 \\ r_{e} &=\text { total external reactions } \\ &=3+2+1=6 \\ j &=\text { total number of joints }=10 \\ \therefore \quad D_{s} &=(3 \times 11)+6-(3 \times 10)=9 \end{aligned}

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

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