Static Indeterminacy 2c51d

2101-310 Structural Analysis I

1 - 15

Static Indeterminacy Idealized structures of interest Consider planar (2D) structures consisting of 1D structural components or and subjected to in-plane loadings ¾ Truss structures (Fig1) ƒ ƒ ƒ ƒ

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

¾ Beam structures (Fig2) ƒ ƒ ƒ ƒ

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

¾ Frame structures (Fig3) ƒ ƒ ƒ ƒ

Consisting of flexural-axial 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

¾ Compound structures (Fig4) ƒ

Combination of 1 and/or 2 and/or 3

Fig1. Truss structure

Fig2. Beam structure

2101-310 Structural Analysis I

1 - 16

Fig3. Frame structure

Fig4. Compound structure

Classification of structures using static stability criteria ¾ 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. ¾ 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. ƒ

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.

ƒ

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.

2101-310 Structural Analysis I

1 - 17

Fig7.

Fig8.

Classification of structures using static indeterminacy criteria ¾ The statically stable structure is called an externally, statically determinate structure if and only if all reactions can be determined by using only static equilibrium equations, e.g. Fig9-Fig11. ¾

The statically stable structure is called a statically determinate structure if and only if all 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.

¾ 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 reactions and/or internal member forces at any locations within the structure, e.g. Fig11-Fig13. ¾ 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 reactions can be determined from static equilibrium equations while the internal forces can not be completely obtained.

2101-310 Structural Analysis I

1 - 18

Fig12.

Fig13.

Degree of static indeterminacy (DI) and Definitions The degree of static indeterminacy is a number of static unknowns ( 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 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

Reactions ¾ Unknown reactions exerted to a structure by the s ¾ Total number of independent components of reactions depending on ƒ ƒ

Number of s Number of components of reactions at each .

¾ Types of s

2101-310 Structural Analysis I

1 - 19

ƒ

Roller 9 1 direction of translational constraint 9 1 component of reactions in the direction of constraint

ƒ

Pinned 9 2 directions of translational constraint 9 2 components of reactions in the directions of constraint

ƒ

Fixed 9 2 directions of translational constraint 9 1 direction of rotational constraint 9 3 components of reactions in the directions of constraint

ƒ

Guide 9 1 direction of translational constraint 9 1 direction of rotational constraint 9 2 components of reactions in the directions of constraint

ƒ

Nonrotational 9 1 direction of rotational constraint 9 1 component of reactions in the directions of constraint

2101-310 Structural Analysis I

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

Truss structure shown in Fig1. 1 roller 2 pinned s Æ ra = 1(1) + 2(2) = 5

Beam structure shown in Fig2. 3 roller s 1 fixed Æ ra = 3(1) + 1(2) = 5

Frame structure shown in Fig3. 1 roller 1 pinned 3 fixed s Æ ra = 1(1) + 1(2) + 3(3) = 12

1 - 20

2101-310 Structural Analysis I

1 - 21

Compound structure shown in Fig4. 1 roller 1 guide 1 fixed Æ ra = 1(1) + 1(2) + 1(3) = 6

Remark: For beam structures, there is no longitudinal loading and therefore the number of reactions for pinned and fixed s reduce by one. Internal Member Forces ¾ Stress resultants at any cross sections of the member, e.g. ƒ ƒ ƒ ƒ

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

¾ Internal member forces induced depend on type of ¾ Type of ƒ

Axial or truss member 9 9 9 9 9

1D straight member; 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) Æ F1 = F2 and F(x) = F1 Only 1 independent internal force per member

F2 x F1

ƒ

F1

Flexural or beam member 9 1D straight member; of beam structures 9 Subjected only to transverse loading 9 Internal forces at any cross sections are shear force V(x) and bending moment M(x) 9 {V1, M1 ,V2, M2} are internal forces at both ends of the member 9 Member equilibrium (2 equations) Æ V2, M2, V(x), M(x) in of V1, M1

F(x)

2101-310 Structural Analysis I

1 - 22

9 Only 2 independent internal forces per member

M1 V1

ƒ

M1

M2

x

V1

V2

M(x)

V(x)

Frame member 9 1D straight member; of frame structures 9 Subjected to both longitudinal & transverse loading 9 Internal forces at any cross sections are axial force F(x), shear force V(x), and bending moment M(x) 9 {F1, V1, M1, F2, V2, M2} are internal forces at both ends of the member 9 Member equilibrium (3 equations) Æ V2, M2, V(x), M(x) in of V1, M1 9 Only 3 independent internal forces per member M2 F2

M(x)

V2

M1

F(x)

x

M1

V(x) F1

V1

F1

V1

Static Equilibrium Equations ¾ Ensure the conservation of ƒ ƒ

Linear momentum Angular momentum

¾ Equilibrium equations ƒ ƒ

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

¾ Member equilibrium ƒ

Already taken into to obtain the number of independent internal forces

¾ t equilibrium ƒ

Pinned or hinge ts

2101-310 Structural Analysis I

1 - 23

9 Connecting adjacent truss in truss structures 9 No moment restraint at the t 9 No preservation of angles between each member ing at the t 9 Only forces can be applied at the t 9 transferring internal forces to the t only in of axial force 9 Only 2 independent equilibrium equations per t, i.e. ΣFx = 0 and ΣFy = 0 y

x ƒ

Beam rigid t 9 9 9 9 9

Connecting two adjacent beam in beam structures Full moment restraint at the t Preservation of angle between two ing at the t Moment and transverse force can be applied at the t transferring internal forces to the t in of shear force and bending moment 9 Only 2 independent equilibrium equations per t, i.e. ΣFy = 0 and ΣΜ = 0 y

x

ƒ

Frame rigid t 9 9 9 9 9

Connecting adjacent frame in frame structures Full moment restraint at the t Preservation of angles between each member ing at the t Moment and forces can be applied at the t transferring internal forces to the t in of axial force, shear force and bending moment 9 3 equilibrium equations per t, i.e. ΣFx = 0, ΣFy = 0 and ΣΜ = 0 y

x

2101-310 Structural Analysis I

ƒ

1 - 24

Compound ts 9 Connecting more than one type of in compound structures 9 Moment and forces can be applied at the t 9 All ing may transfer all type of internal forces to the t, i.e. axial force, shear force and bending moment 9 3 equilibrium equations per t, i.e. ΣFx = 0, ΣFy = 0 and ΣΜ = 0 y Frame member Truss member

x Frame member

Internal static conditions or Internal Releases ¾ Static quantities (e.g. internal forces) are prescribed at interior points of the structure ¾ Additional equations to static equilibrium equations are introduced ¾ Such points are called internal releases ¾ Types of internal static conditions ƒ

Bending moment release 9 Bending moment is prescribed at a point, i.e. M = Mo 9 For special case, if M = 0, it is known as a “hinge” 9 Rotation at hinge point is generally discontinuous 9 1 additional equation is provided

1 additional equation

ƒ

Shear force release 9 Shear force is prescribed at a point, i.e. V = Vo 9 For special case, if V = 0, it is known as a “shear release” 9 Transverse displacement at shear release is generally discontinuous 9 1 additional equation is provided 1 additional equation

2101-310 Structural Analysis I

ƒ

1 - 25

Axial force release 9 Axial force is prescribed at a point, i.e. F = Fo 9 For special case, if F = 0, it is known as an “axial release” 9 Longitudinal displacement at axial release is generally discontinuous 9 1 additional equation is provided

1 additional equation

ƒ

Combined internal release 9 Internal releases shown above are combined 9 Additional equations are provided depending on the number of released components

2 additional equations

ƒ

Full moment release at t 9 “Hinge” is present at the t and it is called “hinge t” 9 No transferring of bending moment at this t 9 “Hinge t” cannot be subjected to applied moment 9 For truss structures, all ts are always hinge t and no additional equation is provided for each t since presence of such ts has been incorporated into the reduction of the number of internal forces per member to one (i.e. only axial force is present) 9 For beam, frame and compound structures, n-1 additional equations are provided per hinge t where n is the number of ing at that t.

4-1 = 3 additional equations

ƒ

Partial moment release at t 9 Bending moment is released at the end of certain ing the t 9 Transferring of bending moment at this t is still possible by the ing with no end moment release. 9 This t can be subjected to applied moment 9 This type of release can be found in frame and compound structures 9 n additional equations are provided per t where n is the number of whose end moment is released excluding the truss .

2101-310 Structural Analysis I

1 - 26

2 additional equations

Example 2: Find the degree of static indeterminacy of the structures shown in Fig1Fig4.

Truss structure shown in Fig1 ƒ ƒ ƒ ƒ ƒ ƒ

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

Beam structure shown in Fig2 ƒ ƒ ƒ ƒ ƒ ƒ

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

Frame structure shown in Fig3 ƒ ƒ ƒ ƒ ƒ ƒ

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

Compound structure shown in Fig4 ƒ

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

2101-310 Structural Analysis I

ƒ ƒ ƒ ƒ ƒ ƒ

1 - 27

17 truss + 3 frame Æ nm = 17(1) + 3(3) = 26 8 hinge ts + 3 frame ts + 2 compound ts Æ nj = 8(2) + 3(3) + 3(2) = 31 No internal release Æ 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 ts, there is only one frame member ing those ts; therefore, number of additional equations at each compound t = 1 – 1 = 0.

Compound ts

Check of External Static Indeterminacy For a given statically stable structure, let ra be the number of all independent components of the 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 reactions of the structure are to be determined. Remarks: ¾ The number of independent components of reactions ra can be readily obtained in the same manner as described above. ¾ 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 ) 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).

2101-310 Structural Analysis I

1 - 28

¾ 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 reactions can be included. These additional equations can be set up in 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 Æ nm = 6(3) = 18 - 6 frame ts Æ nj = 6(3) = 18 - 2 hinges Æ nc = 2 - DI = 4 + 18 – 18 – 2 = 2 - Structure is statically indeterminate - net = 3 (3 equilibrium equations) Not included in ncr - ncr = 1 - ra = net + ncr = 4 - Structure is externally statically determinate - All reactions can be determined Truss Structure A truss structure is a structure that consists of only axial being connected by pinned or hinge ts. All loads are assumed to be applied directly to the ts. For a given statically stable truss structure that consists of j hinge ts and m axial , 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 since each member possesses only one component of internal forces; presence of the (axial) internal release will render the structure statically unstable.

2101-310 Structural Analysis I

1 - 29

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 ƒ ƒ ƒ ƒ ƒ ƒ

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

Structure II ƒ ƒ ƒ ƒ ƒ ƒ

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

2101-310 Structural Analysis I

Structure III ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ

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

Structure III ƒ ƒ ƒ ƒ ƒ

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

1 - 30