Static and Kinematic Indeterminacy of Structure.

Static and Kinematic
Indeterminacy of
Structure
PRITESH PARMAR
D.D.UNIVERSITY,NADIAD
priteshparmar5.6.1998@gmail.com

Structure
• A structure refers to a system of connected parts used to support a load.
• A structure defined as an assembly of different members connected to each other which
transfers load from space to ground.
• Mainly of two types :
1. Load Bearing Structure
2. Framed Structure

Support System
• Supports are used in structures to provide it stability and strength.
• Main types of support :
1. Fixed Support
2. Hinged or Pinned Support
3. Roller Support
4.Vertical Guided Roller Support
5. Horizontal Guided Roller Support

2-D Support
Fixed Support :
No. of Reaction - 3 (RCX ,RCY ,MCZ)
• RCX -Reaction at joint ‘ C ’ in x-direction
• RCY -Reaction at joint ‘ C ’ in y-direction
• MCZ -Moment at joint ‘ C ’ about z-direction
Displacement in x-direction at joint ‘ C ‘ is zero ( i.e yCX = O )
Displacement in y-direction at joint ‘ C ‘ is zero ( i.e yCY = O )
Rotation about z-direction at joint ‘ C ‘ is zero ( i.e θCZ = O )

2-D Support
Hinged or Pinned Support :
No. of Reaction - 2 (RAX ,RAY )
• RAX -Reaction at joint ‘A ’ in x-direction
• RAY -Reaction at joint ‘ A ’ in y-direction
Displacement in x-direction at joint ‘ A ‘ is zero ( i.e yAX = O )
Displacement in y-direction at joint ‘ A ‘ is zero ( i.e yAY = O )
Rotation about z-direction at joint ‘ A ‘ is not zero ( i.e θAZ ≠ O )

2-D Support
Roller Support :
No. of Reaction - 1 (RBY )
• RBY -Reaction at joint ‘ B ’ in y-direction
Displacement in x-direction at joint ‘ B ‘ is not zero ( i.e yBX≠O )
Displacement in y-direction at joint ‘ B ‘ is zero ( i.e yBY = O )
Rotation about z-direction at joint ‘ B ‘ is not zero ( i.e θBZ ≠ O )

2-D Support
Vertical Guided Roller Support :
No. of Reaction - 2 (RAX , MAZ)
• RAX -Reaction at joint ‘ A ’ in x-direction
• MAZ -Moment at joint ‘ A ’ about z-direction
Displacement in y-direction at joint ‘ A ‘ is not zero ( i.e yAY≠ O )
Rotation about z-direction at joint ‘ A ‘ is zero ( i.e θAZ =O )

2-D Support
Horizontal Guided Roller Support :
No. of Reaction - 2 (RAY , MAZ)
• MAZ -Moment at joint ‘ A ’ about z-direction
Displacement in x-direction at joint ‘ A ‘ is not zero ( i.e yAX ≠ O )

2-D Support
Spring Support :
No. of Reaction - 1 (RAY )

2-D INTERNAL JOINTS
Internal Hinge or Pin :
Characteristics :
• Moment at ‘ C ‘ is zero (i.e. M@C = O)
Displacement in y-direction at joint ‘ C ‘ is not zero ( i.e yCY ≠ O )
Displacement in x-direction at joint ‘ C ‘ is not zero ( i.e yCX ≠ O )
Rotation about z-direction at joint ‘ C ‘ are may be different at either side ( i.e θC1 ≠ θC2 )

2-D INTERNAL JOINTS
Internal Roller :
• Can’t transfer horizontal reaction (axial thrust)(i.e. FBX = O)
Displacement in y-direction at joint ‘ B ‘ may not be zero ( i.e yBY ≠ O )
Displacement in x-direction at joint ‘ B ‘ is not zero ( i.e yBX ≠ O )

2-D INTERNAL JOINTS
Internal Link :
• Portion ‘ BC ‘ is known as internal link.
• Two internal pins at B & C
• Portion BC contains only axial load because moment at B and C is zero.(i.e. M@B &
M@C = O)
Displacement in y-direction at joint ‘ B & C ‘ is not zero ( i.e YBY , yCY ≠ O )
Displacement in x-direction at joint ‘ B & C ‘ is not zero ( i.e YBx ,YCX ≠ O )

2-D INTERNAL JOINTS
Torsional Spring Support :
• θZ -Rotational resistance at joint in z-direction

3-D Support
Fixed Support :
No. of Reaction - 6 (RCX ,RCY ,RCZ ,MCX ,MCY ,MCZ )
• RCX ,RCY ,RCZ - Reaction at joint ‘ C ’ in x,y,z-direction
• MCX ,MCY ,MCZ -Moment at joint ‘ C ’ about x,y,z-direction
Displacement in x,y,z-direction at joint ‘ C ‘ is zero ( i.e yCX , yCY ,yCZ = O )
Rotation about x,y,z-direction at joint ‘ C ‘ is zero ( i.e θCX , θCY , θCZ = O )

3-D Support
No. of Reaction - 3 (RCX ,RCY ,RCZ)
• RCX ,RCY ,RCZ - Reaction at joint ‘ C ’ in x,y,z-direction
Rotation about x,y,z-direction at joint ‘ C ‘ is not zero ( i.e θCX , θCY , θCZ ≠ O )

3-D Support
Roller Support :
No. of Reaction - 1 (RCY)
• RCY - Reaction at joint ‘ C ’ in y-direction
Displacement in x,z-direction at joint ‘ C ‘ is not zero ( i.e yCX , yCZ ≠ O )

Equilibrium Equation
• When a body is in static equilibrium, no translation or rotation occurs in any
direction.
• Since there is no translation, the sum of the forces acting on the body must
be zero.
• Since there is no rotation, the sum of the moments about any point must be
zero.

PIN JOINT PLANE FRAME (2-DTruss)
No. of Equilibrium Equation : 2
• ∑ Fx = O
• ∑ Fy = O

PIN JOINT SPACE FRAME (3-DTruss)
• ∑ Fx = O
• ∑ Fy = O
• ∑ Fz = O

RIGID JOINT PLANE FRAME (2-D Frame)
• ∑ Fx = O
• ∑ Fy = O
• ∑ Mz = O

RIGID JOINT SPACE FRAME (3-D Frame)
• ∑ Fx = O
• ∑ Fy = O
• ∑ Fz = O
• ∑ Mx = O
• ∑ My = O
• ∑ Mz = O
NOTE :Above equilibrium equations are used to find members forces and moments , To find out support reaction
equilibrium equation for any type of structure always remains 3(i.e. ∑ Fx = O ∑ Fy = O ∑ Mz = O )for 2-D and 6 for 3-D
structure.
•

Static Indeterminacy
Statical Determinant Structure :
• If condition of static equilibrium are sufficient to analyse the structure , it is called
Statical Determinant Structure.
• Bending moment and Shear force are independent of material properties and cross
section.
• Stresses are not induced due to temp. changes and support settlement.

Statical Indeterminant Structure :
• If condition of static equilibrium are not sufficient to analyse the structure , it is called
Statical Indeterminant Structure.
• Bending moment and Shear force are dependent on material properties and cross
section.
• Stresses are induced due to temp. changes and support settlement.

Static Indeterminacy = External Indeterminacy + Internal Indeterminacy
Ds = Dse + Dsi
External Indeterminacy : If no. of reactions are more than equilibrium equation is
known as Externally Indeterminant Structure.
No of Reactions = 4 Equilibrium Equations=3 for 2-D and 6 for 3-D structure.
Beams is externally indeterminate to the first degree

Internal Indeterminacy : If no. of Internal forces or stresses can’t evaluated
based on equilibrium equation is known as Internally
Indeterminant Structure.
• Member forces ofTruss can not be determined based on statics alone, forces in the
members can be calculated based on equations of equilibrium.Thus, structures is
internally indeterminate to first degree.

(A) Rigid Jointed Plane Frame :
• External Indeterminacy,Dse : R-E
• Internal Indeterminacy,Dsi : 3C-r’
OR R = No. of external unknown reaction
Ds = 3m+R-3j-r’ E = No. of Equilibrium Equation = 3
m = No. of members , j = joints
C = No. of close loop
r’ = Total no. of internal released or
= No. of members
connected -1
with internal hinge
=(m’-1)

• Some of the example for the r’ :
• r’ = 2 ( Moment and Horizontal Reaction Released
• at joint ‘ B ‘ )
• r’ = 1 (Only Vertical Reaction Released at joint
• ‘ B ‘)
r’ = 2-1 =1 (i.e. member connected to hinges = 2)
•
• r’ = 3-1 =2 (i.e. member connected to hinges = 3)
B
B

(B) Rigid Jointed Space Frame :
• Internal Indeterminacy,Dsi : 6C-r’
Ds = 6m+R-6j-r’ E = No. of Equilibrium Equation = 6
m = No. of members , j = joints
C = No. of close loop
= No. of members
3 * connected -1
with internal hinge
= 3(m’-1)

(C) Pinned Jointed Plane Frame :
• Internal Indeterminacy,Dsi : m+E-2j
Ds = m+R-2j E = No. of Equilibrium Equation = 3
m = No. of members
j = joints

(D) Pinned Jointed Space Frame :
• Internal Indeterminacy,Dsi : m+E-3j
Ds = m+R-3j E = No. of Equilibrium Equation = 6
m = No. of members
j = joints

• Ds < 0 : Unstable & statically determinant structure
Deficient Frame or Structure
• Ds = 0 : Stable & statically determinant structure
Perfect Frame or Structure
• Ds > 0 : Stable & statically indeterminant structure
Redundant Frame or Structure

Kinematic Indeterminacy
• Kinematic Indeterminacy = Degree of Freedom
• If the displacement component of joint can’t be determined by
Compatibility Equation , it is called Kinematic Indeterminant Structure.
Degree of Kinematic Indeterminacy(Dk) :
• It is defined as total number of unrestrained displacement (translation
and rotation) component at joint.

2-D Support
Fixed Support :
Degree of Freedom - O
Displacement in x-direction at joint ‘ C ‘ is zero ( i.e yCX = O )
Rotation about z-direction at joint ‘ C ‘ is zero ( i.e θCZ = O )

2-D Support
Degree of Freedom - 1 (θAZ )
• θAZ -Rotation about z-direction at joint ‘ A ‘
Rotation about z-direction at joint ‘ A ‘ is not zero ( i.e θAZ ≠ O )

2-D Support
Roller Support :
Degree of Freedom - 2(θBZ ,yBX )
• yBX -Displacement in x-direction at joint ‘ B ‘
• θBZ -Rotation about z-direction at joint ‘ B ‘
Displacement in x-direction at joint ‘ B ‘ is not zero ( i.e yBX≠O )
Displacement in y-direction at joint ‘ B ‘ is zero ( i.e yBY = O )

2-D Support
Vertical Guided Roller Support :
Degree of Freedom - 1(yAY)
• yAY -Displacement in y-direction at joint ‘ A ‘
Displacement in x-direction at joint ‘A ‘ is zero ( i.e yAX = O )
Displacement in y-direction at joint ‘ A ‘ is not zero ( i.e yAY≠ O )

2-D Support
Horizontal Guided Roller Support :
Degree of Freedom - 1(yAX)
• yAX -Displacement in x-direction at joint ‘ A ‘
Displacement in y-direction at joint ‘A ‘ is zero ( i.e yAY = O )
Displacement in x-direction at joint ‘ A ‘ is not zero ( i.e yAX ≠ O )

3-D Support
Fixed Support :
Degree of Freedom - O
Rotation about x,y,z-direction at joint ‘ C ‘ is zero ( i.e θCX , θCY , θCZ = O )

3-D Support
Degree of Freedom - 3 (θCX , θCY , θCZ )
• θCX , θCY , θCZ -Rotation about x,y,z-direction at joint ‘ C ‘

3-D Support
Roller Support :
Degree of Freedom - 5 (yCX , yCZ , θ CX , θCY , θCZ )
• yCX , yCZ -Displacement in x,z-direction at joint ‘ C ‘
• θCX , θCY , θCZ - Rotation about x,y,z-direction at joint ‘ C ‘
Displacement in x,z-direction at joint ‘ C ‘ is not zero ( i.e yCX , yCZ ≠ O )

2-D INTERNAL JOINTS
Internal Hinge or Pin :
Degree of Freedom – 4(yCX ,yCY ,θC1 ,θC2 )
Rotation about z-direction at joint ‘ C ‘ are may be different at either side and not zero (
i.e θC1 ≠ θC2 )

2-D INTERNAL JOINTS
Free End :
Degree of Freedom – 3(yBX , yBY , θ BZ )
Displacement in x-direction at joint ‘ B ‘ is not zero ( i.e yBX ≠ O )
Displacement in y-direction at joint ‘ B ‘ is not zero ( i.e yBY ≠ O )
B

2-D INTERNAL JOINTS
AxialThrust Release:
Degree of Freedom – 4(yCX1 , yCX2 ,yCY , θ CZ )
Displacement in x-direction at joint ‘ C ‘ is not zero ( i.e yCX1 , yCX2 ≠ O )
Rotation about z-direction at joint ‘ C ‘ is not zero ( i.e θCZ ≠ O )

2-D INTERNAL JOINTS
Shear Release:
Degree of Freedom – 4(yCY1 , yCY2 ,yCX , θ CZ )
Displacement in y-direction at joint ‘ C ‘ is not zero ( i.e yCY1 , yCY2≠ O )
Rotation about z-direction at joint ‘ C ‘ is not zero ( i.e θCZ ≠ O )

2-D INTERNAL JOINTS
Frame Joint:
Degree of Freedom – 5(yOY ,yOX , θ OAZ , θ OBZ , θ OCZ )
Displacement in x-direction at joint ‘ O ‘ is not zero ( i.e yOX ≠ O )
Displacement in y-direction at joint ‘ O ‘ is not zero ( i.e yOY ≠ O )
Rotation about z-direction at joint ‘ O ‘ is not zero ( i.e θ OAZ , θ OBZ , θ OCZ ≠ O )

2-D INTERNAL JOINTS
Internal Roller:
Degree of Freedom – 5(yCX1 , yCX2 , θ CZ1 , θ CZ2 , yCY)
Displacement in x-direction at joint ‘ C ‘ is not zero ( i.e yCX1 , yCX2≠ O )
Rotation about z-direction at joint ‘ C ‘ is not zero ( i.e θ CZ1 , θ CZ2 ≠ O )

(A) Rigid Jointed Plane Frame :
• Dk : 3j-R+r’
• Dk(NAD) : Dk-m’
R = No. of external unknown reaction
NAD=Neglecting Axial Deformations m’ = No. of axially rigid members
(Beams are azially rigid or stiffness is r’ = Total no. of internal released or
infinite) = No. of members
connected -1
with internal hinge
=(m-1)

(B) Rigid Jointed Space Frame :
• Dk : 6j-R+r’
• Dk(NAD) : Dk-m’
NAD=Neglecting Axial Deformations m’ = No. of axially rigid members
= No. of members
3* connected -1
with internal hinge
=3(m-1)

(C) Pinned Jointed Plane Frame :
• Dk : 2j-R
• Dk(NAD) : 0
NAD=Neglecting Axial Deformations j = No. of joints

(D) Pinned Jointed Space Frame :
• Dk : 3j-R
• Dk(NAD) : 0
NAD=Neglecting Axial Deformations j = No. of joints

Stability of Structure
External stability :
• For any 2-D structure 3 no. of reactions and foe 3-D structure 6 no. of
reactions are required to keep structure in stable condition.
• All reactions should not be Parallel.
• All reactions should not be Concurrent (line of action meets at one point).
Unstable Structure because all
reactions are parallel.
Unstable Structure because all
reactions are concurrent.

Internal stability :
• No part of structure can move rigidly releative to other part.
• For geometric stability there should not be any condition of mechanism.
• Static Indeterminacy should not be less than zero.(i.e. Ds >=0)(But it is not
mandatory, sometimes structure is not stable though this conditions satisfied)
• For internal stability following conditions should be satisfied :
(1) Pinned Jointed Plane Truss : m>=2j-3 m = No. of members
(2) Pinned Jointed Space Truss : m>=3j-6 j =No. of joints
(3) Rigid Jointed Plane Frame : 3m>=3j-3
(4) Rigid Jointed Space Frame : 6m>=6j-6

Example of unstable structure :

Example of unstable structure :
Unstable because of local member failure.
Geometric unstable because of no
diagonal member.

Examples
* Calculate static indeterminacy and comment on stability of structure :
1) Dse =R-E =3-3=0 Ds=0
Dsi =3C-r’= 0 (no close loop) Stable and Determinate Structure
2) Dse =R-E =7-3=4 Ds=3 (internal hinge is part of internal
indeterminacy)
Dsi =3C-r’=3*0-(2-1)=-1 (no close loop)
Stable and Indeterminate Structure
3) Dse=R-E=6-3=3 Ds=1
Dsi=3C-r’=3*0-(2*(2-1))=-2 (no close loop)

Examples
4) Dse =R-E =7-3=4 Ds=2
Dsi =3C-r’=3*0-2*(2-1)=-2 (no close loop,
two hinges & not link,link is only vertical)
5) Dse =R-E =6-3=3 Ds=2 (Guided roller)
Dsi =3C-r’=3*0-(2-1)=-1 (no close loop and one releases)
6) Dse=R-E=6-3=3 Ds=1
Dsi=3C-r’=3*0-2=-2 (no close loop and two releases)

Examples
7) Dse =R-E =5-2=3(no axial load) Ds=3
Dsi =3C-r’=0 (no close loop)
8) Dse =R-E =10-3=7 Ds=6
Dsi =3C-r’=3*0-(2-1)=-1 (no close loop)
9) Dse=R-E=4-3=1 Ds=12
Dsi=3C-r’=3*4-(2-1)=11 (4 close loop and one hinge)

Examples
10) Dse =R-E =3-3=3 Ds=1
Dsi =3C-r’=-2 (no close loop and two releases)
11) Dse =R-E =5-3=2 Ds=2
Dsi =3C-r’=3*0-0=0 (no close loop)
12) Dse=R-E=8-3=5 Ds=3
Dsi=3C-r’=3*0-2=-2 (0 close loop and two releases)

Examples
13) Dse =R-E =8-3=5 Ds=14
Dsi =3C-r’=3*3-0=9 ( 3close loop and no releases because guided roller
is support not internal joint)
14) Dse =R-E =11-3=8 Ds=7
Dsi =3C-r’=3*2-7=-1 (two close loop and 7 releases)
15) Dse=R-E=8-3=5 Ds=16
Dsi=3C-r’=3*6-7=11 (6 close loop and 7 releases)

Examples
16) Dse =R-E =12-6=6 Ds=12
Dsi =6C-r’=6*1-0=6 ( one close loop and zero releases)
17) Dse =R-E =16-6=10 Ds=13
Dsi =6C-r’=6*1-3(2-1)=3 (one close loop and 3 releases)
18) Dse=R-E=3-3=0 Ds=2
Dsi=m+E-2j=13+3-2*7=2 (13 members and 7 joints)

Examples
19) Dse =R-E =4-3=1 Ds=3
Dsi =m+E-2j=17+3-2*9=2 (17 members and 9 joints)
20) Dse =R-E =6-3=3 Ds=0
Dsi =m+E-2j=12+3-2*9=-3 (12 members and 9 joints)
Stable and Determinate Structure
21) Dse=R-E=3-3=0 Ds=2
Dsi=m+E-2j=19+3-2*10=2 (19 members and 10 joints)

Examples
* Calculate kinematic indeterminacy structure :
1) Dk =3j-R+r’ =3*4-4=8
Dk(NAD)=Dk-m=8-3=5
2) Dk =3j-R+r’ =3*5-5+2*(2-1)=12
Dk(NAD)=Dk-m=12-4=8
3) Dk =3j-R+r’ =3*4-7=5
Dk(NAD)=Dk-m=5-3=2

Examples
4) Dk =3j-R+r’ =3*8-6+2=20
Dk(NAD)=Dk-m=20-7=13
5) Dk =3j-R+r’ =3*4-3=9
Dk(NAD)=Dk-m=9-3=6
6) Dk =3j-R+r’ =3*9-7=20
Dk(NAD)=Dk-m=20-10=10

Examples
7) Dk =2j-R =2*6-4=8
Dk(NAD)=0
8) Dk =2j-R =2*7-3=9
Dk(NAD)=0
9) Dk =3j-R+r’ =3*9-6=21
Dk(NAD)=Dk-m=21-10=11
10) Dk =3j-R+r’ =3*3-5=4
Dk(NAD)=Dk-m=4-2=2

Static and Kinematic Indeterminacy of Structure.

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