Determinacy and Indeterminacy MCQ Quiz - Objective Question with Answer for Determinacy and Indeterminacy - Download Free PDF

For the given beam:

The free body diagram (FBD) is as follows:

No. of reactions = 7 = n

No. of equilibrium equation = 3 (∑M = 0, ∑ F_x = 0, ∑ F_y = 0)

Static indeterminacy = n – 3 = 7 – 3 = 4

Important Points:

In a beam, if loading is given, take care of the given loading.

If only vertical load (lateral loading) is there, there will be no horizontal reaction. So H_A = H_C = 0 & “∑ F_x = 0” condition can’t be applied.

e.g:

Here H_A = H_C = 0

So, here equation conditions are ∑ M = 0, ∑ F_y = 0

Reactions = V_A, V_C,V_B M_A & M_C

D_si = 5 – 2 = 3

Explanation:

The degree of static indeterminacy is given by

Ds = 3 x number of cuts required to make tree like structure fixed - number of reaction required to make joint

Ds = 3 x 4 - 3 = 9

Static Indeterminacy:

Total number of support reaction = 3 (for fixed support) + 2 (for hinge support) + 1 (for roller support)

Total number of support reaction = 6

Total number of equilibrium equations = 3

Concept:

A two-dimensional structure in general is classified as a statically indeterminate structure if it cannot be analyzed by conditions of static equilibrium.

The conditions of equilibrium for 2D structures are:

1. The Sum of vertical forces is zero (∑Fy = 0).
2. The Sum of horizontal forces is zero (∑Fx = 0).

• The Sum of moments of all the forces about any point in the plane is zero (∑Mz = 0).

​
Simply supported beam:

Number of unknowns = 3

Degree of static indeterminacy = 3 - 3 = 0. Hence it is statically determinate.

Cantilever beam:

Number of unknowns = 3

Degree of static indeterminacy = 3 - 3 = 0. Hence it is statically determinate.

Three hinged arches:

Number of unknown = 4

Degree of static indeterminacy = 4 - 3 -1 = 0. (Additional equation due to internal hinge ∵ B.M = 0)

Hence it is statically determinate.

Two hinged arches:

Number of unknown = 4

Degree of static indeterminacy = 4 - 3 = 1.

Hence it is statically indeterminate.

Concept:

Kinematic Indeterminacy:

It is the total number of possible degrees of freedom of all the joints.

Dk = 3J - r + h (For beam & portal frame)

Dk = 2J - r + h (For truss structure)

Where,

Dk = Kinematic Indeterminacy,

r = No. of unknown reactions

h = No. of plastic hinges

J = No. of joints

Calculation:

Given;

J = 2

r = 1 + 3 = 4 (1 vertical reaction at roller support, and 1 vertical, 1 horizontal and 1 moment reaction at fixed support)

h = 0

D_k = 3 × 2 - 4 = 2

D_k = 2

For the external stability of structures following conditions should be satisfied:

1) All reactions should not be parallel

2) All reactions should not be concurrent

3) The reaction should be nontrivial

4) There should be a minimum number of externally independent support reactions

5) For stability in 3D structures, all reactions should be non-coplanar, non-concurrent and non-parallel

∴ If all the reactions acting on a planar system are concurrent in nature, then the system is unstable.

Explanation

Given, 3 joints and 4 members so it signifies it a frame

For a given frame:

We know in a frame the relation between members and joints is given by 

m = 2j - 3 

Where m = members , j = joints

Given, m = 4, j = 3 

Let's check the relation

m = 2 × 3 - 3 = 3, so we get m = 3

But we have 4 members i.e 1 in excess

∴ the answer is redundant.

Kinematic Indeterminacy (D_k):

Dk = 3J - R - n

Where,

J = Number of joints = 4,

R = Number of reactions = 6

n = Number of inextensible members = 3

Dk = (3 × 4) - 6 - 3 = 3

Dk = 3

D_s = D_se + D_si - R

D_se = (2 + 1) - 3 = 0

D_si = m - (2j - 3) = 10 - (2 × 5 - 3) = 3

Where m = no of members

J = no of pin joints

R = 0

D_s = 3

Concept:

Total Indeterminacy is given by,

Total Indeterminacy = External indeterminacy (D_se) + Internal indeterminacy (D_si)

External Indeterminacy (D_se) = R - 3

Where, R = Number of external Reaction

Internal indeterminacy (D_si) = 3C

Where, C = Number of closed loop

Calculation:

Given,

R = 6, C = 3

External Indeterminacy (D_se) = R - 3 = 6 - 3 = 3

Internal Indeterminacy (D_si) = 3C = 3 × 3 = 9

Total Indeterminacy = D_se + D_si = 3 + 9 = 12

Concept:

The degree of static determinacy is given by –

\({D_s} = 3m - 3j + {R_e}\)

Here,

m – number of members

j – number of joints

R_e – number of external reactions

Calculation:

Given,

m = 5

j = 6 (including internal hinge)

R_e  = 10 ( i.e. 3 at each fixed support and one at roller support)

∴  \({D_s} = 3\left( 5\right) - 3\left( 6 \right) + 10 = 7\)

Since,

There is one internal hinge, which will provide one compatibility equation.  

∴ we have to reduce indeterminacy by 1.

Hence,

Total degree of indeterminacy

D_s = 7 - 1 = 6

Concept:

For a general system of loading, for a fixed beam,

For each end, the no. unknowns are 3 which are horizontal reaction, vertical reaction, and moment at the fixed end, so a total of 6 unknowns.

We have 3 available equation of equilibrium which are

\({\rm{\;}}\sum {{\rm{F}}_{\rm{x}}} = 0,\sum {{\rm{F}}_{\rm{y}}} = 0,{\rm{and\;}}\sum {{\rm{M}}_{\rm{z}}} = 0{\rm{\;}}\)

Static Indeterminacy, E = No. of unknowns (R) – No. of equilibrium equations (r)

R = 6 and r = 3

Static Indeterminacy = 6 – 3 = 3

Note:

But  for fixed beam loaded transversely or only vertical loading,

R = 4 ( 2 for each support)

r = 2 (\({\rm{\;}}\sum {{\rm{F}}_{\rm{y}}} = 0,{\rm{and\;}}\sum {{\rm{M}}_{\rm{z}}} = 0{\rm{\;}}\))

Static Indeterminacy = 4 – 2 = 2

Where,

m = Number of members, R = Number of support reactions, J = Number of joints, m' = Number of axially rigid members, r = Number of internal support reactions released

Static indeterminacy:

Number of additional reactions required to analyse a structure is called static indeterminacy.

D_s = D_se + D_si