Degree of Static Indeterminacy - INTRODUCTION TO STRUCTURAL ANALYSIS

Chapter 1 INTRODUCTION TO STRUCTURAL ANALYSIS

1.9 Degree of Static Indeterminacy

The degree of static indeterminacy of a structure, denoted by DI, is defined as a number of independent static quantities (i.e. support reactions and the internal force) that must be prescribed in addition to available static equilibrium equations in order to completely describe a static state of the entire structure (a state where all support reactions and internal forces at any locations within the structure are known) or, equivalently, to render the structure statically determinate.

From this definition, the degree of static indeterminacy is equal to the number of independent static unknowns subtracted by the number of independent static equilibrium equations. Thus, the degree of static indeterminacy of a statically determinate structure is equal to zero while the degree of static indeterminacy of a statically indeterminate structure is always greater than zero. The degree of static indeterminacy is also known as the degree of static redundancy and the corresponding extra, static unknowns exceeding the number of static equilibrium equations are termed as the redundants.

1.9.1 General formula for computing DI

The degree of static indeterminacy of a statically stable structure can be computed from the general formula: c j m a n n n r DI    (1.18) where ra is the number of all components of the support reactions, nm is the number of components of the internal member force, nj is the number of independent equilibrium equations at all nodes or joints, and nc is the number of static conditions associated with all internal releases present within the structure. It is evident that the term ra + nm represents the number of all static unknowns while the term nj + nc represents the number of all available equilibrium equations (including static conditions at the internal releases).

1.9.1.1 Number of support reactions

The number of all components of support reaction at a given structure can be obtained using the following steps: 1) identify all supports within the structures, 2) identify the type and a number of

(a) (b) u1=0 1=0 v1=0 u3=0 3=0 v3=0 u2=0 2=0 v2=0 _{Node 3} Node 2 Node 1 u2 2 v2 u1=0 1=0 v1=0 u3=0 3=0 v3=0 Node 3 Node 2 Node 1

components of the support reaction at each support (see section 1.1.5), and 3) sum the number of components of support reactions over all supports.

It is emphasized here that for a beam structure, the component of the support reaction in the direction of the beam axis must not be counted in the calculation of ra since the beam is subjected only to transverse loads and there is no internal axial force at any cross section. For instance, the number of support reactions of the structure shown in Figure 1.42(a), Figure 1.42(b) and Figure 1.42(c) is 3, 4, and 8, respectively.

Figure 1.42: Schematics indicating all components of support reaction

1.9.1.2 Number of internal member forces

As clearly demonstrated in subsection 1.5.2, the number of independent components of the internal force for an axial member, a flexural member, and a two-dimensional frame member are equal to 1, 2 and 3, respectively. Thus, the number of components of the internal forces for the entire structure (nm) can simply be obtained by summing the number of components of the internal forces for all individual members. It is worth noting that nm depends primarily on both the number and the type of constituting members of the structure. For instance, nm for the structure shown in Figure 1.42(a) is equal to 14(1) = 14 since it consists of 14 axial members; nm for the structure shown in Figure 1.42(b) is equal to 2(2) = 4 since it consists of 2 flexural members (by considering all supports as joints or nodes); and nm for the structure shown in Figure 1.42(c) is equal to 20(3) = 60 since it consists of 20 frame members (by considering supports and connections between columns and beams as joints or nodes).

1.9.1.3 Number of joint equilibrium equations

To compute nj, it is required to know both the number and the type of joints present in the structure. The number of independent equilibrium equations at each joint depends primarily on the type of the joint. Here, we summarize standard joints found in the idealized structures.

1.9.1.3.1 Truss joints

A truss joint is an idealized joint used for modeling connections of a truss structure. The truss joint behaves as a hinge joint so it cannot resist any moment and allows all members joining the joint to rotate freely relative to each other. Since the truss member possesses only the internal axial force, when the truss joint is separated from the structure to sketch the FBD, all forces acting to the joint are concurrent forces as shown in Figure 1.43; in particular, P1 and P2 are external loads and P1, P2 and P3 are internal axial forces from the truss members. As a consequence, the number of

(a) (c)

(b)

independent equilibrium equations per one truss joint is equal to 2 (i.e. FX = 0 and FY = 0; see also subsection 1.7.2).

Figure 1.43: FBD of the truss joint

1.9.1.3.2 Beam joints

A beam joint is an idealized joint used for modeling connections of a flexural or beam structure. The beam joint behaves as a rigid joint so it can resist the external applied moment and can also transfer the moment among ends of members joining that joint. Since the flexural or beam member possesses only two components of the internal force, i.e. the shear force and the bending moment, when the beam joint is separated from the structure to sketch the FBD, all forces and moments acting to the joint form a set of transverse loads as shown in Figure 1.44; in particular, P and Mo are external loads and V1, M1, V2 and M2 are internal forces from the beam members. As a consequence, the number of independent equilibrium equations per one beam joint is equal to 2 (i.e. FY = 0 and MZ = 0; see also subsection 1.7.3).

Figure 1.44: FBD of the beam joint

1.9.1.3.3 Frame joints

A frame joint is an idealized joint used for modeling connections of a frame structure. The frame joint behaves as a rigid joint so it can resist the external applied moment and can also transfer the moment among ends of the members joining the joint. Since the frame member possesses three components of the internal force, i.e. the axial force, the shear force and the bending moment, when the frame joint is separated from the structure to sketch the FBD, all forces and moments acting to the joint form a set of general 2D loads as shown in Figure 1.45; in particular, P1, P2 and Mo are external loads and F1, V1, M1, F2, V2, M2, F3, V3 and M3 are internal forces from the frame members. As a result, the number of independent equilibrium equations per one frame joint is equal to 3 (i.e. FX = 0, FY = 0 and MZ = 0).

1.9.1.3.4 Compound joints

A compound joint is an idealized joint used for modeling connections where more than one types of members are connected. When the compound joint is separated from the structure to sketch the

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FBD, all forces and moments acting to the joint can form a set of general 2D loads as shown in Figure 1.46. As a result, the number of independent equilibrium equations per one compound joint is generally equal to 3 (i.e. FX = 0, FY = 0 and MZ = 0).

Figure 1.45: FBD of the frame joint

Figure 1.46: FBD of the compound joint

The number of independent joint equilibrium equations of the structure (nj) can simply be obtained by summing the number of independent equilibrium equations available at each joint.

1.9.1.4 Internal releases

An internal release is a point within the structure where certain components of the internal force such as axial force, shear force and bending moment are prescribed. Presence of the internal releases within the structure provides extra equations in addition to those obtained from static equilibrium. Here, we summarize various types of internal releases that can be found in the idealized structure.

1.9.1.4.1 Moment release or hinge

A moment release or hinge is an internal release where the bending moment is prescribed equal to zero or, in the other word, the bending moment cannot be transferred across this point (see Figure 1.47). At the moment release, the displacement is continuous while the rotation or slope is not. For this particular type of internal releases, it provides 1 additional equation per one hinge, i.e. M = 0 at the hinge point.

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Truss member

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Figure 1.47: Schematics of moment releases or hinges

1.9.1.4.2 Axial release

An axial release is an internal release where the axial force is prescribed equal to zero or, in the other word, the axial force cannot be transferred across this point (see Figure 1.48). At the axial release, the longitudinal component of the displacement is discontinuous while the transverse component and the rotation are still continuous. For this particular type of internal releases, it provides 1 additional equation per one release, i.e. F = 0 at the axial release.

Figure 1.48: Schematic of axial release

1.9.1.4.3 Shear release

A shear release is an internal release where the shear force is prescribed equal to zero or, in the other word, the shear force cannot be transferred across this point (see Figure 1.49). At the shear release, the transverse component of the displacement is discontinuous while the longitudinal component of the displacement and the rotation are still continuous. For this particular type of internal releases, it provides 1 additional equation per one release, i.e. V = 0 at the shear release.

Figure 1.49: Schematic of shear release

1.9.1.4.4 Combined release

A combined release is an internal release where two or more components of the internal force are prescribed equal to zero (see Figure 1.50). Behavior of the combined release is the combination of behavior of the moment release, axial release, or shear release. For this particular type of internal releases, it provides two or more additional equations per one release depending on the number of prescribed components of the internal force.

1.9.1.4.5 Full moment release joint

A joint or node where the bending moment at the end of all members jointing that joint is prescribed equal to zero is termed as a full moment release joint (see Figure 1.51). This joint has the same behavior and characteristic as the hinge joint. For truss structures, while all joints are full moment release joints, they provide no additional equation since presence of such joints has been considered in the reduction of the number of internal forces per member from three to one (i.e. only axial force is present). For beam or frame structures, presence of a full moment release joint provides n – 1 additional equations where n is the number of member joining the joint; for instance, a full moment release joint shown in Figure 1.51 provides 4 – 1 = 3 additional equations.

Figure 1.51: Schematic of full moment release joint

1.9.1.4.6 Partial moment release joint

A joint or node where the bending moment at the end of certain but not all members jointing that joint is prescribed equal to zero is termed as a partial moment release joint (see Figure 1.52). This type of releases can be found in beam and frame structures. A partial moment release joint provides n additional equations if the bending moment at the end of n members are prescribed equal to zero; for instance, a partial moment release joint shown in Figure 1.52 provides 2 additional equations.

Figure 1.52: Schematic of partial moment release joint

The number of static conditions associated with all internal releases present in the structure (nc) can simply be obtained by summing the number of additional equations provided by each internal release.

Example 1.1 Determine the degree of static indeterminacy (DI) of the following structures

 ra = 2 + 1 = 3

 25 truss members  nm = 25(1) = 25  14 truss joints  nj = 14(2) = 28  No internal release  nc = 0  DI = 3 + 25 – 28 – 0 = 0  Statically determinate structure

 Statically indeterminate structure

 ra = 3(3) + 2 + 1 = 12

 28 frame members  nm = 28(3) = 84

 21 frame joints  nj = 21(3) = 63

 No internal release  nc = 0

 DI = 12 + 84 – 63 – 0 = 33  Statically indeterminate structure

 ra = 4(3) = 12

 28 frame members and 12 truss members  nm = 28(3) + 12(1) = 96

 6 frame joints and 14 compound joints  nj = 6(3) + 14(3) = 60

 No internal release  nc = 0

 DI = 12 + 96 – 60 – 0 = 48  Statically indeterminate structure

1.9.2 Check of external static indeterminacy

For a given statically stable structure, let ra be the number of all 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 and only if

a et cr

r n n (1.19) and the structure is externally, statically indeterminate if and only if

a et cr

r n n (1.20) This check of external static indeterminacy is essential when the support reactions of the structure are to be determined.

In general, for plane structures, the number of independent equilibrium equations that can be set up for the entire structure (net) is equal to 3, except for beam structures where the number of

independent equilibrium equations reduces to 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 “moment release or hinge”, “shear release”, and “axial release”. It is important to note that not all the static conditions can be incorporated in the counting of ncr but ones that introduce no additional unknowns other than the support reactions can be

counted. 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.

To clearly demonstrate the check of external static indeterminacy, let consider a frame structure as shown in Figure 1.53. For this structure, we obtain ra = 2(2) = 4, nm = 6(3) = 18, nj =

6(3) = 18, nc = 2(1) = 2  DI = 4 +18 – 18 – 2 = 2; thus the structure is statically determinate. In

addition, net = 3 for frame structure and ncr = 1 since one additional equation (without introducing

additional unknowns other than support reactions) can be set up by sectioning at the hinge A and then enforcing moment equilibrium about the point A of one part of the structure. The static condition associated with the hinge B cannot be included in ncr since no new equation can be set up

without introducing additional unknown internal forces along the cut. It is evident that ra = 4 = net +

ncr  the structure is externally, statically determinate and, therefore, all support reactions can be

determined from static equilibrium. Since the structure is also statically indeterminate, from the definition provided above, this implies that the structure is internally, statically indeterminate.

Figure 1.53: Schematic of externally statically determinate structure

1.9.3 DI of truss structures

Consider a statically stable truss structure that consists of m members and n joints. For this particular structure, we obtain nm = m(1) = m, nj = n(2) = 2n and nc = 0. Upon using the general

formula (1.18), the degree of static indeterminacy of a truss is given by

2n m r

DI a   (1.21)

It is important to emphasize that there cannot be an internal release at interior points of all truss members since each member possesses only one component of internal forces; presence of the (axial) internal release will render the structure statically unstable.

B

Example 1.2 Determine the degree of static indeterminacy (DI) of the following statically stable

truss structures  ra = 2 + 2 = 4  m = 35  n = 18  DI = 4 + 35 – 2(18) = 3  Statically indeterminate  net = 3  ncr = 0  ra = 4 > net + ncr = 3

 Externally statically indeterminate

 ra = 2 + 1 = 3  m = 14  n = 8  DI = 3 + 14 – 2(8) = 1  Statically indeterminate  net = 3  ncr = 0  ra = 3 = net + ncr

 Externally statically determinate

 ra = 2 + 2 = 4  m = 10  n = 7  DI = 4 + 10 – 2(7) = 0  Statically indeterminate  net = 3  ncr = 1  ra = 4 = net + ncr

 Externally statically determinate (or implies from DI as well)

1.9.4 DI of beam structures

Consider a statically stable beam structure that consists of m members and n joint. For this particular structure, we obtain nm = m(2) = 2m and nj = n(2) = 2n. Upon using the general formula

(1.18), the degree of static indeterminacy of a beam is given by

c a 2(m n) n

DI    (1.22) It is important to emphasize that in the determination of ra, components of the support reactions in

the direction parallel to the beam axis must be ignored since there is no internal axial force in any beam members. In addition, the number of members of a given beam is not unique but it depends

primarily on the choice of joints or nodes considered; in general, joints are located at the supports and free ends. However, the choice of joints and members does not affect the final value of DI.

Example 1.3 Determine the degree of static indeterminacy (DI) of the following statically stable

beam structures  ra = 2(2) + 3(1) = 7  m = 4  n = 5  nc = 2  DI = 7 + 2(4 – 5) – 2 = 3  Statically indeterminate  net = 2  ncr = 2  ra = 7 > net + ncr = 4

 Externally statically indeterminate  ra = 2 + 2(1) = 4  m = 3  n = 4  nc = 2  DI = 4 + 2(3 – 4) – 2 = 0  Statically determinate  net = 2  ncr = 2  ra = 4 = net + ncr

 Externally statically determinate  ra = 2 + 3(1) = 5  m = 4  n = 5  nc = 0  DI = 5 + 2(4 – 5) – 0 = 3  Statically indeterminate  net = 2  ncr = 0  ra = 5 > net + ncr= 2

 Externally statically indeterminate

1.9.5 DI of frame structures

Consider a statically stable frame structure that consists of m members and n joints. For this particular structure, we obtain nm = m(3) = 3m and nj = n(3) = 3n. Upon using the general formula

(1.18), the degree of static indeterminacy of a frame is given by

c a 3(m n) n

DI    (1.23) Note that the free end must be treated as a joint or node.

Example 1.4 Determine the degree of static indeterminacy (DI) of the following statically stable

frame structures  ra = 2(3) + 2 = 8  m = 7  n = 8  nc = 3 – 1 = 2  DI = 8 + 3(7 – 8) – 2 = 3  Statically indeterminate  net = 3  ncr = 2  ra = 8 > net + ncr = 5

 Externally statically indeterminate

 ra = 2 + 1 = 3  m = 7  n = 6  nc = 0  DI = 3 + 3(7 – 6) – 0 = 6  Statically indeterminate  net = 3  ncr = 0  ra = 3 = net + ncr

 Externally statically determinate

 ra = 3 + 1 = 4  m = 3  n = 4  nc = 1  DI = 4 + 3(3 – 4) – 1 = 0  Statically determinate  net = 3  ncr = 1  ra = 4 = net + ncr = 4

 Externally statically determinate

In document Fundamental Structural Analysis (2) (Page 47-57)