Static Indeterminacy

2101-310 Structural Analysis I 1 - 15

Static Indeterminacy

Idealized structures of interest

Consider planar (2D) structures consisting of 1D structural components or members and subjected to in-plane loadings

Truss structures (Fig1)

Consisting of axial members connected by pinned joints Subjected to forces only at joints Internal member forces: axial force 2D geometry

Beam structures (Fig2)

Consisting of flexural members Subjected only to transverse loading Internal member forces: shear force and bending moment 1D geometry

Frame structures (Fig3)

Consisting of flexural-axial members Subjected to transverse and longitudinal loading Internal member forces: axial force, shear force, and bending moment 2D geometry or 1D geometry with additional longitudinal loading

Compound structures (Fig4)

Combination of 1 and/or 2 and/or 3 Fig1. Truss structure Fig2. Beam structure


Fig3. Frame structure Fig4. Compound structure

Classification of structures using static stability criteria

The structure is called a statically stable structure if and only if it can resist any actions (e.g. applied loads and kinematics conditions) without the development of mechanisms or rigid body displacement (e.g. rigid translation and rigid rotation). All desirable structures fall into this category, e.g. Fig1-Fig4.

The structure is called a statically unstable structure if and only if it is not a

statically stable structure or, in the other word, it is a structure that exhibits mechanisms or rigid body displacement (e.g. rigid translation and rigid rotation) for the entire structure or within any parts of the structure when subjected to a particular action, e.g. Fig5-Fig8.

The structure is called an externally, statically unstable structure

if it is a statically unstable structure with the development of the mechanism on the entire structure, e.g. Fig5-Fig6.

The structure is called an internally, statically unstable structure

if it is a statically unstable structure with the development of the mechanism only on certain parts of components of the structure, e.g. Fig7-Fig8.

Fig5. Fig6.


Fig7. Fig8.

Classification of structures using static indeterminacy criteria

The statically stable structure is called an externally, statically determinate structure if and only if all support reactions can be determined by using only static equilibrium equations, e.g. Fig9-Fig11.

The statically stable structure is called a statically determinate structure if and only if all support reactions and internal forces at any locations within the structure can be completely determined by using only static equilibrium equations, e.g. Fig9-Fig10. Note that all statically determinate structures are always externally, statically determinate structures.

Fig9. Fig10. Fig11.

The statically stable structure is called a statically indeterminate structure if and only if it is not a statically determinate structure or, in the other word, static equilibrium equations are insufficient for determining support reactions and/or internal member forces at any locations within the structure, e.g. Fig11-Fig13.

The statically stable structure is called an internally, statically indeterminate structure if and only if it is statically indeterminate but externally, statically determinate, e.g. Fig11-Fig12. For this particular type of structures, all support reactions can be determined from static equilibrium equations while the internal forces can not be completely obtained.


Fig12. Fig13. Degree of static indeterminacy (DI)

Terms and Definitions

The degree of static indeterminacy is a number of static unknowns (support reactions and/or internal forces) that exceeds the number of independent equilibrium equations and static conditions available. The static conditions are additional equations associated with the prescribed static unknowns at certain locations within the structure; e.g. presence of the hinge at a particular point indicates that the bending moment at that point vanishes. The degree of static indeterminacy is also known as the degree of static redundancy. If the degree of static indeterminacy is equal to zero, the structure is statically determinate; if the degree of static indeterminacy is greater than zero, the structure is statically indeterminate.

General Formula

For a given statically stable structure, the degree of static indeterminacy, denoted by DI, can be determined from the general formula

DI = ra + nm – nj - nc

where ra is the total number of components of the support reactions

nm is the total number of internal member forces

nj is the total number of independent equilibrium equations

nc is the total number of internal static conditions

Support Reactions

Unknown reactions exerted to a structure by the supports

Total number of independent components of support reactions depending on

Number of supports Number of components of support reactions at each support.

Types of supports


Roller support 1 direction of translational constraint 1 component of support reactions in the direction of constraint

Pinned support 2 directions of translational constraint 2 components of support reactions in the directions of constraint

Fixed support 2 directions of translational constraint 1 direction of rotational constraint 3 components of support reactions in the directions of constraint

Guide support 1 direction of translational constraint 1 direction of rotational constraint 2 components of support reactions in the directions of constraint

Nonrotational support 1 direction of rotational constraint 1 component of support reactions in the directions of constraint


Example 1: Find the total number of components of support reactions (Ns) of the structures shown in Fig1-Fig4.

Truss structure shown in Fig1. 1 roller support 2 pinned supports

ra = 1(1) + 2(2) = 5

Beam structure shown in Fig2. 3 roller supports 1 fixed support

ra = 3(1) + 1(2) = 5

Frame structure shown in Fig3. 1 roller support 1 pinned support 3 fixed supports

ra = 1(1) + 1(2) + 3(3) = 12


Compound structure shown in Fig4. 1 roller support 1 guide support 1 fixed support

ra = 1(1) + 1(2) + 1(3) = 6

Remark: For beam structures, there is no longitudinal loading and therefore the number of support reactions for pinned and fixed supports reduce by one.

Internal Member Forces

Stress resultants at any cross sections of the member, e.g.

Axial force (F) Bending moment (M) Shear force (V) Torque (T)

Internal member forces induced depend on type of members

Type of members

Axial or truss member

1D straight member; members of truss structures Internal force at any cross section is axial force F(x) {F1,F2} are internal forces at both ends of the member Member equilibrium (1 equation) F1 = F2 and F(x) = F1 Only 1 independent internal force per member

F2 F(x) x F1F1

Flexural or beam member

1D straight member; members of beam structures Subjected only to transverse loading Internal forces at any cross sections are shear force V(x) and

bending moment M(x) {V1, M1 ,V2, M2} are internal forces at both ends of the member Member equilibrium (2 equations) V2, M2, V(x), M(x) in terms

of V1, M1


Only 2 independent internal forces per member

Frame member

1D straight member; members of frame structures Subjected to both longitudinal & transverse loading Internal forces at any cross sections are axial force F(x), shear

force V(x), and bending moment M(x) {F1, V1, M1, F2, V2, M2} are internal forces at both ends of the

member Member equilibrium (3 equations) V2, M2, V(x), M(x) in terms

of V1, M1 Only 3 independent internal forces per member

Static Equilibrium Equations

Ensure the conservation of

Linear momentum Angular momentum

Equilibrium equations

Force equilibrium, i.e. ΣF = 0 Moment equilibrium, i.e. ΣM = 0

Member equilibrium

Already taken into account to obtain the number of independent internal forces

Joint equilibrium

Pinned or hinge joints

V2

x

V1

M1 M2

V1

M1 M(x)

V(x)

M2

F2

V2

V1

M1

F1

M(x) F(x) x

V1

M1

V(x)

F1


Connecting adjacent truss members in truss structures No moment restraint at the joint No preservation of angles between each member joining at the

joint Only forces can be applied at the joint Members transferring internal forces to the joint only in terms of

axial force Only 2 independent equilibrium equations per joint, i.e. ΣFx = 0

and ΣFy = 0

x

y

Beam rigid joint

Connecting two adjacent beam members in beam structures Full moment restraint at the joint Preservation of angle between two members joining at the joint Moment and transverse force can be applied at the joint Members transferring internal forces to the joint in terms of shear

force and bending moment Only 2 independent equilibrium equations per joint, i.e. ΣFy = 0

and ΣΜ = 0

x

y

Frame rigid joint

Connecting adjacent frame members in frame structures Full moment restraint at the joint Preservation of angles between each member joining at the joint Moment and forces can be applied at the joint Members transferring internal forces to the joint in terms of axial

force, shear force and bending moment 3 equilibrium equations per joint, i.e. ΣFx = 0, ΣFy = 0 and ΣΜ = 0

x

y


Compound joints

Connecting more than one type of members in compound structures

Moment and forces can be applied at the joint All joining members may transfer all type of internal forces to the

joint, i.e. axial force, shear force and bending moment 3 equilibrium equations per joint, i.e. ΣFx = 0, ΣFy = 0 and ΣΜ = 0

x

y

Frame member

Truss member

Frame member

Internal static conditions or Internal Releases

Static quantities (e.g. internal forces) are prescribed at interior points of the structure

Additional equations to static equilibrium equations are introduced

Such points are called internal releases

Types of internal static conditions

Bending moment release

Bending moment is prescribed at a point, i.e. M = Mo For special case, if M = 0, it is known as a “hinge” Rotation at hinge point is generally discontinuous 1 additional equation is provided

1 additional equation

Shear force release

Shear force is prescribed at a point, i.e. V = Vo For special case, if V = 0, it is known as a “shear release” Transverse displacement at shear release is generally

discontinuous 1 additional equation is provided



Axial force release

Axial force is prescribed at a point, i.e. F = Fo For special case, if F = 0, it is known as an “axial release” Longitudinal displacement at axial release is generally

discontinuous 1 additional equation is provided


Combined internal release

Internal releases shown above are combined Additional equations are provided depending on the

number of released components 2 additional equations

Full moment release at joint

“Hinge” is present at the joint and it is called “hinge joint” No transferring of bending moment at this joint “Hinge joint” cannot be subjected to applied moment For truss structures, all joints are always hinge joint and no

additional equation is provided for each joint since presence of such joints has been incorporated into the reduction of the number of internal forces per member to one (i.e. only axial force is present)

For beam, frame and compound structures, n-1 additional equations are provided per hinge joint where n is the number of members joining at that joint.

4-1 = 3 additional equations

Partial moment release at joint

Bending moment is released at the end of certain members joining the joint

Transferring of bending moment at this joint is still possible by the joining members with no end moment release.

This joint can be subjected to applied moment This type of release can be found in frame and compound

structures n additional equations are provided per joint where n is the

number of members whose end moment is released excluding the truss members.


2 additional equations

Example 2: Find the degree of static indeterminacy of the structures shown in Fig1-Fig4.

Truss structure shown in Fig1

ra = 1(1) + 2(2) = 5 33 truss members nm = 33(1) = 33 18 hinge joints nj = 18(2) = 36 No internal release nc = 0 DI = ra +nm –nj –nc = 5 +33 –36 –0 = 2 Structure is statically indeterminate

Beam structure shown in Fig2

ra = 3(1) + 1(2) = 5 3 beam members nm = 3(2) = 6 4 beam joints nj = 4(2) = 8 1 internal hinge nc = 1 DI = ra +nm –nj –nc = 5 +6 –8 –1 = 2 Structure is statically indeterminate

Frame structure shown in Fig3

ra = 1(1) + 1(2) + 3(3) = 12 28 frame members nm = 28(3) = 84 21 frame joints nj = 21(3) = 63 No internal release nc = 0 DI = ra +nm –nj –nc = 12 +84 –63 –0 = 33 Structure is statically indeterminate

Compound structure shown in Fig4

ra = 2(1) + 1(1) + 3(1) = 6


17 truss members + 3 frame members nm = 17(1) + 3(3) = 26 8 hinge joints + 3 frame joints + 2 compound joints nj = 8(2) + 3(3) + 3(2) = 31 No internal release nc = 0 DI = ra +nm –nj –nc = 6 +26 –31 –0 = 1 Structure is statically indeterminate Note that although bending moment is fully released at the two compound joints,

there is only one frame member joining those joints; therefore, number of additional equations at each compound joint = 1 – 1 = 0.

Check of External Static Indeterminacy

Compound joints

For a given statically stable structure, let ra be the number of all independent components of the support reactions, net be the number of independent equilibrium equations available for the entire structure and ncr be the number of additional static conditions that can be set up without introducing new static unknowns. The structure is externally, statically determinate if an only if

ra = net + ncr

and the structure is externally, statically indeterminate if an only if

ra > net + ncr

This check of external static indeterminacy is essential when the support reactions of the structure are to be determined. Remarks:

The number of independent components of support reactions ra can be

readily obtained in the same manner as described above.

In general, for planar structures, the number of independent equilibrium equations that can be set up for the entire structure net is equal to three, i.e. net = 3, except for beam structures (that there is no axial force present in all members) where the number of independent equilibrium equations reduces to 2, i.e. net = 2 (the equilibrium of forces in the direction along the beam axis is automatically satisfied).


Additional static conditions are typically the conditions associated with internal releases present within the structure; for instance, points where components of internal forces are prescribed such as “hinge”, “shear release”, and “axial release”. It is important to note that not all the static conditions can be included in the counting of ncr but ones that introduce no additional unknowns other than the support reactions can be included. These additional equations can be set up in terms of equilibrium equations of certain parts of the structure resulting from proper sectioning the structure at the internal releases.

- For this structure, we obtain - ra = 2(2) = 4 - 6 frame members nm = 6(3) = 18 - 6 frame joints nj = 6(3) = 18 - 2 hinges nc = 2 - DI = 4 + 18 – 18 – 2 = 2 - Structure is statically indeterminate - net = 3 (3 equilibrium equations) - ncr = 1 - ra = net + ncr = 4 - Structure is externally statically determinate - All support reactions can be determined

Not included in ncr

Truss Structure

A truss structure is a structure that consists of only axial members being connected by pinned or hinge joints. All loads are assumed to be applied directly to the joints. For a given statically stable truss structure that consists of j hinge joints and m axial members, we obtain

nm = m

nj = 2j

nc = 0

and, therefore, the degree of static indeterminacy of the truss is given by

DI = ra + m – 2j

Remark: There is no internal release at interior points of all axial members since each member possesses only one component of internal forces; presence of the (axial) internal release will render the structure statically unstable.


Example 3: Determine the degree of static indeterminacy (DI) of the following statically stable truss structures.

Structure I Structure II Structure III Structure IV

Structure I

ra = 2(2) = 4 14 truss members m = 14 7 joints j = 7 DI = ra + m – 2j = 4 + 14 – 2(7) = 0 Structure is statically determinate Structure is externally, statically determinate

Structure II

ra = 1(2) + 1(1) = 3 9 truss members m = 9 6 joints j = 6 DI = ra + m – 2j = 3 + 9 – 2(6) = 0 Structure is statically determinate Structure is externally, statically determinate


Structure III

ra = 1(2) + 1(1) = 3 14 truss members m = 14 8 joints j = 8 DI = ra + m – 2j = 3 + 14 – 2(8) = 1 Structure is statically indeterminate ra = 3, net = 3, ncr = 0 ra = net + ncr = 3 Structure is externally, statically determinate Structure is internally, statically indeterminate

Structure III

ra = 2(2) + 1(1) = 5 39 truss members m = 39 20 joints j = 20 DI = ra + m – 2j = 5 + 39 – 2(20) = 4 Structure is statically indeterminate

Static Indeterminacy

Documents

Transcript of Static Indeterminacy