Question 1 |

Consider a beam PQ fixed at P, hinged at Q, and subjected to a load F as shown in figure (not drawn to scale). The static and kinematic degrees of
indeterminacy, respectively, are

2 and 1 | |

2 and 0 | |

1 and 2 | |

2 and 2 |

Question 1 Explanation:

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

Kinematic indeterminacy=0+1=1

Question 2 |

The degree of static indeterminacy of the plane frame as shown in the figure is ___

10 | |

15 | |

20 | |

25 |

Question 2 Explanation:

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

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

Question 3 |

Consider the frame shown in the figure:

If the axial and Shear deformation in different members of the frame are assumed to be negligible. The reduction in the degree of Kinematical indeterminacy would be equal to

If the axial and Shear deformation in different members of the frame are assumed to be negligible. The reduction in the degree of Kinematical indeterminacy would be equal to

5 | |

6 | |

7 | |

8 |

Question 3 Explanation:

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.

Question 4 |

A Planar truss tower structure is shown in the figure.

Consider the following statements about the external and internal determinacies of the truss.

(P) Externally Determinate

(Q) External Static Indeterminacy = 1

(R) External Static Indeterminacy = 2

(S) Internally Determinate

(T) Internal Static Indeterminacy = 1

(U) Internal Static Indeterminacy = 2

Which one of the following options is correct?

Consider the following statements about the external and internal determinacies of the truss.

(P) Externally Determinate

(Q) External Static Indeterminacy = 1

(R) External Static Indeterminacy = 2

(S) Internally Determinate

(T) Internal Static Indeterminacy = 1

(U) Internal Static Indeterminacy = 2

Which one of the following options is correct?

P-False; Q-True; R-False; S-False; T-False; U-True | |

P-False; Q-True; R-False; S-False; T-True; U-False | |

P-False; Q-False; R-True; S-False; T-False; U-True | |

P-True; Q-True; R-False; S-True; T-False; U-True |

Question 4 Explanation:

\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}

Question 5 |

The kinematic indeterminacy of the plane truss shown in the figure is

11 | |

8 | |

3 | |

0 |

Question 5 Explanation:

Kinematic indeterminacy,

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

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

Question 6 |

A guided support as shown in the figure below is represented by three springs (horizontal, vertical and rotational) with stiffness k_{x},k_{y}\: and \: k_{\theta } respectively. The limiting values of k_{x},k_{y}\: and \: k_{\theta } are:

\infty ,0,\infty | |

\infty ,\infty ,\infty | |

0 ,\infty ,\infty | |

\infty ,\infty ,0 |

Question 7 |

The static indeterminacy of the two-span continuous beam with an internal hinge,shown below,is_________

0 | |

1 | |

2 | |

-1 |

Question 7 Explanation:

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}

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}

Question 8 |

The degree of static indeterminacy of a rigid jointed frame PQR supported as shown in the figure is

zero | |

one | |

two | |

unstable |

Question 8 Explanation:

\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.

Since cable is provided, it implies that there is only vertical loading.

Hence the structure is stable.

Question 9 |

The degree of static indeterminacy of rigidity jointed frame in a horizontal plane
and subjected to vertical load only, as shown in figure below, is

6 | |

4 | |

3 | |

1 |

Question 9 Explanation:

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

\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

Question 10 |

The degree of static indeterminacy of the rigid frame having two internal hinges as shown in the figure below, is

8 | |

7 | |

6 | |

5 |

Question 10 Explanation:

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}

\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}

There are 10 questions to complete.