Short Notes: Indeterminacy & Stability of a Structure | Short Notes for Civil Engineering - Civil Engineering (CE) PDF Download

 Page 1


 
 
 
 INDETERMINACY & STABILITY OF A STRUCTURE 
1. External Indeterminacy 
Mathematically, external indeterminacy can be expressed as follows. 
Sc
D r s =- 
Where, 
r = total number of unknown support reactions. 
S = total number of equilibrium equations available. 
 S=3 (For 2D structure) and 6 (For 3D structure) 
2. Internal Indeterminacy 
Case 1: Beam 
There is no internal indeterminacy for beams because if we know the support reactions, we can 
find the axial force, shear force and bending moment at any section in the beam. 
Case 2: Trusses 
The internal indeterminacy for the trusses can be determined by following expression. 
DSi = m – (2j – 3); for plane truss 
DSi = m – (3j – 6); for space truss 
Where, 
m = number of members 
j = number of joints 
3. KINEMATIC INDETERMINACY (DK):  
        It is defined as the number of independent displacements at all joints in a structure. 
Displacements are counted always only at the joints. Displacement includes slopes and 
deflection. Wherever the cross-section area, changes or material changes then it is treated as 
a joint in any structure. 
The kinematic indeterminacy can be determined for various cases as follows. 
Case 1: Beams 
Example: 
 
? Displacement at A and B in x-direction is zero 
? Displacement at A and B y-direction is zero 
? Rotation at A and B is zero 
? Degree of freedom = Dk = 0 
Dk(inextensible) = Dk (extensible) – Number of independent displacements prevented. 
 
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 INDETERMINACY & STABILITY OF A STRUCTURE 
1. External Indeterminacy 
Mathematically, external indeterminacy can be expressed as follows. 
Sc
D r s =- 
Where, 
r = total number of unknown support reactions. 
S = total number of equilibrium equations available. 
 S=3 (For 2D structure) and 6 (For 3D structure) 
2. Internal Indeterminacy 
Case 1: Beam 
There is no internal indeterminacy for beams because if we know the support reactions, we can 
find the axial force, shear force and bending moment at any section in the beam. 
Case 2: Trusses 
The internal indeterminacy for the trusses can be determined by following expression. 
DSi = m – (2j – 3); for plane truss 
DSi = m – (3j – 6); for space truss 
Where, 
m = number of members 
j = number of joints 
3. KINEMATIC INDETERMINACY (DK):  
        It is defined as the number of independent displacements at all joints in a structure. 
Displacements are counted always only at the joints. Displacement includes slopes and 
deflection. Wherever the cross-section area, changes or material changes then it is treated as 
a joint in any structure. 
The kinematic indeterminacy can be determined for various cases as follows. 
Case 1: Beams 
Example: 
 
? Displacement at A and B in x-direction is zero 
? Displacement at A and B y-direction is zero 
? Rotation at A and B is zero 
? Degree of freedom = Dk = 0 
Dk(inextensible) = Dk (extensible) – Number of independent displacements prevented. 
 
 
 
Note: It not given in the question, then assume that members are extensible. 
Example:  
 
Sol. 
Degree of freedom DK = 2×3-5+4 (Due to internal Hinge) = 5  
Ignoring axial deformation, DK = 5-2 =3 
Case 2: Truss 
At each joint in a truss number of independent displacements are only two (horizontal and 
vertical displacement). Rotation of a member in a truss is not considered because it implies 
that the member buckled. Rigid body rotation is not counted because it is not unknown. 
 
Dk at A = 0 
Dk at B = 1 
Dk at C = 1 
Dk at D = 2 
Dk at E = 2 
Dk at F = 2 
So, degree of freedom = 0 + 1 + 1 + 2 + 2 + 2 = 8 
Case 3: Frames 
(i) Count only one rotation for all members meeting at a rigid joint. 
(ii) Count rotation of all members meeting at a pin joint. 
 
Dk at A = 0 
Dk at B = 2 
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 INDETERMINACY & STABILITY OF A STRUCTURE 
1. External Indeterminacy 
Mathematically, external indeterminacy can be expressed as follows. 
Sc
D r s =- 
Where, 
r = total number of unknown support reactions. 
S = total number of equilibrium equations available. 
 S=3 (For 2D structure) and 6 (For 3D structure) 
2. Internal Indeterminacy 
Case 1: Beam 
There is no internal indeterminacy for beams because if we know the support reactions, we can 
find the axial force, shear force and bending moment at any section in the beam. 
Case 2: Trusses 
The internal indeterminacy for the trusses can be determined by following expression. 
DSi = m – (2j – 3); for plane truss 
DSi = m – (3j – 6); for space truss 
Where, 
m = number of members 
j = number of joints 
3. KINEMATIC INDETERMINACY (DK):  
        It is defined as the number of independent displacements at all joints in a structure. 
Displacements are counted always only at the joints. Displacement includes slopes and 
deflection. Wherever the cross-section area, changes or material changes then it is treated as 
a joint in any structure. 
The kinematic indeterminacy can be determined for various cases as follows. 
Case 1: Beams 
Example: 
 
? Displacement at A and B in x-direction is zero 
? Displacement at A and B y-direction is zero 
? Rotation at A and B is zero 
? Degree of freedom = Dk = 0 
Dk(inextensible) = Dk (extensible) – Number of independent displacements prevented. 
 
 
 
Note: It not given in the question, then assume that members are extensible. 
Example:  
 
Sol. 
Degree of freedom DK = 2×3-5+4 (Due to internal Hinge) = 5  
Ignoring axial deformation, DK = 5-2 =3 
Case 2: Truss 
At each joint in a truss number of independent displacements are only two (horizontal and 
vertical displacement). Rotation of a member in a truss is not considered because it implies 
that the member buckled. Rigid body rotation is not counted because it is not unknown. 
 
Dk at A = 0 
Dk at B = 1 
Dk at C = 1 
Dk at D = 2 
Dk at E = 2 
Dk at F = 2 
So, degree of freedom = 0 + 1 + 1 + 2 + 2 + 2 = 8 
Case 3: Frames 
(i) Count only one rotation for all members meeting at a rigid joint. 
(ii) Count rotation of all members meeting at a pin joint. 
 
Dk at A = 0 
Dk at B = 2 
 
 
Dk at C = 1 
Dk at D = 1 
Dk at E = 5 
Dk at F = 6 
Dk at G = 3 
Dk at H = 3 
Dk at I = 3 
Dk at J = 3 
Dk at K = 3 
Dk at L = 3 
? Dk when extensible = 0 + 2 + 1 + 1 + 5 + 6 + 3 + 3 + 3 + 3 + 3 + 3 = 33 degree 
Dk when inextensible 
= Dk(extensible) – Number of independent displacements prevented. 
= 33 – 14 = 29 degrees. 
Example:  
 
Sol. 
Total degree of freedom Dk= 3×5-3 +4 (Due to internal Hinge) = 16 
If members are considered inextensible then, Dk = 16-5=8 
4. STABILITY OF STRUCTURE 
The stability of structure includes external stability and internal stability. The external stability 
deals with support reaction and internal stability deals within the structure. 
4.1. External Stability 
Minimum number of reactions required for a structure to be stable externally is 3. These 
three reactions must be non-concurrent and non-parallel. 
If three reactions are parallel then rigid body translation take place. If they are 
concurrent, then rigid body rotation takes place. 
If the structure becomes unstable due to the improper arrangement of three reactions, 
then it is known as geometrically unstable structure. 
If structure becomes unstable due to less than 3 support reactions, then it is called 
statically unstable structure. 
4.2. Internal Stability  
Internal stability of various cases is explained through the following examples: 
Case 1: Beams 
? Internal floating hinge 
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 INDETERMINACY & STABILITY OF A STRUCTURE 
1. External Indeterminacy 
Mathematically, external indeterminacy can be expressed as follows. 
Sc
D r s =- 
Where, 
r = total number of unknown support reactions. 
S = total number of equilibrium equations available. 
 S=3 (For 2D structure) and 6 (For 3D structure) 
2. Internal Indeterminacy 
Case 1: Beam 
There is no internal indeterminacy for beams because if we know the support reactions, we can 
find the axial force, shear force and bending moment at any section in the beam. 
Case 2: Trusses 
The internal indeterminacy for the trusses can be determined by following expression. 
DSi = m – (2j – 3); for plane truss 
DSi = m – (3j – 6); for space truss 
Where, 
m = number of members 
j = number of joints 
3. KINEMATIC INDETERMINACY (DK):  
        It is defined as the number of independent displacements at all joints in a structure. 
Displacements are counted always only at the joints. Displacement includes slopes and 
deflection. Wherever the cross-section area, changes or material changes then it is treated as 
a joint in any structure. 
The kinematic indeterminacy can be determined for various cases as follows. 
Case 1: Beams 
Example: 
 
? Displacement at A and B in x-direction is zero 
? Displacement at A and B y-direction is zero 
? Rotation at A and B is zero 
? Degree of freedom = Dk = 0 
Dk(inextensible) = Dk (extensible) – Number of independent displacements prevented. 
 
 
 
Note: It not given in the question, then assume that members are extensible. 
Example:  
 
Sol. 
Degree of freedom DK = 2×3-5+4 (Due to internal Hinge) = 5  
Ignoring axial deformation, DK = 5-2 =3 
Case 2: Truss 
At each joint in a truss number of independent displacements are only two (horizontal and 
vertical displacement). Rotation of a member in a truss is not considered because it implies 
that the member buckled. Rigid body rotation is not counted because it is not unknown. 
 
Dk at A = 0 
Dk at B = 1 
Dk at C = 1 
Dk at D = 2 
Dk at E = 2 
Dk at F = 2 
So, degree of freedom = 0 + 1 + 1 + 2 + 2 + 2 = 8 
Case 3: Frames 
(i) Count only one rotation for all members meeting at a rigid joint. 
(ii) Count rotation of all members meeting at a pin joint. 
 
Dk at A = 0 
Dk at B = 2 
 
 
Dk at C = 1 
Dk at D = 1 
Dk at E = 5 
Dk at F = 6 
Dk at G = 3 
Dk at H = 3 
Dk at I = 3 
Dk at J = 3 
Dk at K = 3 
Dk at L = 3 
? Dk when extensible = 0 + 2 + 1 + 1 + 5 + 6 + 3 + 3 + 3 + 3 + 3 + 3 = 33 degree 
Dk when inextensible 
= Dk(extensible) – Number of independent displacements prevented. 
= 33 – 14 = 29 degrees. 
Example:  
 
Sol. 
Total degree of freedom Dk= 3×5-3 +4 (Due to internal Hinge) = 16 
If members are considered inextensible then, Dk = 16-5=8 
4. STABILITY OF STRUCTURE 
The stability of structure includes external stability and internal stability. The external stability 
deals with support reaction and internal stability deals within the structure. 
4.1. External Stability 
Minimum number of reactions required for a structure to be stable externally is 3. These 
three reactions must be non-concurrent and non-parallel. 
If three reactions are parallel then rigid body translation take place. If they are 
concurrent, then rigid body rotation takes place. 
If the structure becomes unstable due to the improper arrangement of three reactions, 
then it is known as geometrically unstable structure. 
If structure becomes unstable due to less than 3 support reactions, then it is called 
statically unstable structure. 
4.2. Internal Stability  
Internal stability of various cases is explained through the following examples: 
Case 1: Beams 
? Internal floating hinge 
 
 
 
The above structure is internally unstable. 
Case 2: Trusses 
In case of trusses if following condition exist then it is classified as unstable truss. 
   m (2j – 3) ? 
Where,  
m = number of members in truss structure. 
j = number of joints in truss structure. 
Case 3: Frames 
If reactions are parallel to each other, then the frame structure will be termed as 
unstable. 
 
The above shown structure is unstable due to presence of reactions which are parallel.