Introduction to Stiffness Analysis displacements. The number of unknowns in the stiffness method of The stiffness method of analysis is analysis is known as the degree of the basis of all commercial kinematic indeterminacy, which structural analysis programs. refers to the number of node/joint Focus of this chapter will be displacements that are unknown development of stiffness equations and are needed to describe the that only take into account bending displaced shape of the structure. deformations, i.e., ignore axial One major advantage of the member, a.k.a. slope-deflection stiffness method of analysis is that method. the kinematic degrees of freedom In the stiffness method of analysis, are well-defined. we write equilibrium equations in 1 2 terms of unknown joint (node)
Definitions and Terminology Stiffness Analysis Procedure Positive Sign Convention: Counterclockwise moments and The steps to be followed in rotations along with transverse forces and displacements in the performing a stiffness analysis can positive y-axis direction. be summarized as: 1. Determine the needed displace- Fixed-End Forces: Forces at the “fixed” supports of the kinema- ment unknowns at the nodes/ tically determinate structure. joints and label them d1, d2, …, d in sequence where n = the Member-End Forces: Calculated n forces at the end of each element/ number of displacement member resulting from the unknowns or degrees of applied loading and deformation freedom. of the structure.
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1 2. Modify the structure such that it fixed-end forces are vectorially is kinematically determinate or added at the nodes/joints to restrained, i.e., the identified produce the equivalent fixed-end displacements in step 1 all structure forces, which are
equal zero. labeled Pfi for i = 1, 2, …, n later 3. Calculate the member fixed-end in the notes. forces in this kinematically 4. Introduce a unit displacement at restrained state at the nodes/ each displacement degree of joints of the restrained structure freedom identified in step 1 one due to the member applied at a time with all others equal to loads. Tables of member fixed- zero and without any loading on
end forces due to member loads the structure, i.e., di = 1 with d1,
for the kinematically restrained …, di-1, di+1, …, dn = 0 for i = 1, members are available later in 2, …, n. Sketch the displaced these notes. The member 5 6
structure for each of these cases. 5. Eliminate the error introduced in Determine the member-end forces step 3 to permit the displace- introduced as result of each unit ment at the nodes/joints. This is displacement for the kinematically accomplished by applying the restrained structure. These negative of the forces calculated member-end forces define the in step 3 and defines the kine- member-end stiffness coeffi- matically released structure. cients, i.e., forces per unit 6. Calculate the unknown node/ displacement. joint displacements. The member-end stiffness coefficients are vectorially added 7. Calculate the member-end at the nodes/joints to produce the forces. structure stiffness coefficients, which are labeled S for i = 1, 2, ij 7 8 …, n and j = 1, 2, …, n.
2 Illustration 1: Determine the degree of kinematic indeterminacy. The only To illustrate the stiffness method of unknown node/joint displacement analysis, we will first consider occurs at node B and it is a continuous beam structures. Start rotational displacement. Thus, the off by considering the two-span rotation at node B is labeled d1. beam shown in Figure 1. 2: Kinematically restrain the structure such that the displace- ments identified in step 1 equal zero. See Figure 2.
Fig. 1 – Two-Span Continuous Beam
Figure 2 – Kinematically Restrained Two-Span Beam of Figure 1
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The heavy vertical line drawn through the horizontal roller support at B signifies that node B is “fixed” against displacement. Thus, the Figure 3 – “Fixed-End” Forces for the rotational displacement d = 0 for Kinematically Restrained Two-Span Beam of 1 Figure 1 the kinematically restrained Since span element (member) A-B structure of Figure 2. is not loaded, it will not produce any 3: Calculate the element/member fixed-end forces. However, fixed-end forces for the kinema- element (member) B-C is loaded tically restrained structure and and the “fixed-end” forces are vectorially add to obtain the fixed- labeled in Figure 3. They are end forces for the structure. simply the support reactions for the “fixed-fixed” beam.
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3 Calculate the fixed-end forces for 4: Impose a unit displacement at the structure by vectorially adding each kinematic degree of freedom the member-end fixed-end forces. (DOF) to establish the structure Figure 4 – Joint Equilibrium at stiffness equations. the Kinematic Degree of Freedom for the Restrained Two-Span Beam of Figure 1
Figure 4 shows that qL2 M0 Pf1 B 12
Pf1 is drawn counterclockwise in Figure 4 since our sign convention is counterclockwise Figure 5 – Kinematically Restrained Two-Span moments are positive. Beam of Figure 1 Subjected to a Unit Displacement d = 1 13 1 14
Figure 5 shows the displaced shape The structure stiffness equations of the two-span beam for d1 = 1 as are expressed as well as the displaced shapes and [S] {d} = {P} – {Pf} member-end stiffness coefficients where [S] is the structure stiffness for the two elements comprising the matrix; {d} is the structure two-span beam of Figure 1. displacement vector; {P} is the Member-end stiffness coefficients applied structure concentrated are defined as the member-end force vector; and {Pf} is the forces resulting from the imposition structure fixed-end force vector of the single unit displacement for calculated in step 3. The applied the structure as shown in Figure 5. structure concentrated force vector Derivation of the member-end {P} lists the point forces for each stiffness coefficients (forces) shown structure displacement DOF. It in Figure 5 and others will be contains nonzero entries only at covered later in the notes. 15 16
4 the displacement DOF where a point force or moment is applied at Sij force at displacement DOF i the corresponding displacement due to a unit displacement at
DOF. DOF j (i.e., dj = 1) with all other displacement DOF equal to
The structure stiffness matrix zero (i.e., di = 0 for i = 1, …, j-1, coefficients are obtained by j+1, …, n). Stiffness coeffi- performing equilibrium at the nodes cients have units of force/ displacement (or moment/ for each structure DOF using the rotation). member-end stiffness coefficients. These structure stiffness matrix The structure stiffness coefficients are obtained by performing coefficients are designated as Sij and i = 1, 2, …, n and j = 1, 2, …, n. equilibrium calculations at the structure displacement degrees of freedom. 17 18
5: Eliminate the error introduced in For example structure: the kinematically restrained
{d} = {d1} = d1 unknown structure: {P} = {0} = 0 [S] {d} = {P} – {Pf} {P } = {P } = qL2/12 f f1 6: Calculate the unknown structure Figure 6 – Equilibrium at Kinematic DOF 1 for displacements the Two-Span Beam of -1 Figure 1 {d} = [S] ({P} – {Pf}) For the example structure: Performing node equilibrium at 2 displacement DOF 1 gives (see d1 = L/8EI (-qL /12) Figure 6) gives = -qL3/96EI
AB BC S11 = (4EI/L) + (4EI/L) 7: Calculate the member-end
= 8EI/L 19 forces. 20
5 Vfb Mfb = Member Fixed-End {Qfb } Vfe Force Vector Mfe
The member fixed-end forces are The beam member stiffness defined as equations can be written as Qfi = Qi in the kinematically determinate {Qbbbbb }4x1 [k ] 4x4 {u } 4x1 {Q f } 4x1 state due to member loading.
Vb vb Mb = Member-End Force b {Qb } {ub } = Member-End Ve Vector ve Displacement Vector M e e
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12 6L 12 6L Simplified Member End 22 Force Calculations EI 6L 4L 6L 2L [k ] bb 3 12 6L 12 6L 12EI 6EI L V(vv)()Vb be be fb LL32 6L 2L22 6L 4L (internal shear at beginning = Member Bending Stiffness node b) Matrix 6EI 2EI M(vv)(2)Mb be be fb L2 L The ij member stiffness coefficient (internal moment at beginning can be expressed mathematically node b) as 12EI 6EI V(vv)()Vebebefe32 kQij i LL u1j (internal shear at end node e)
6EI 2EI M(vv)(2)Mebebefe all other uk 0 (k j) L2 L (internal moment at end node e)
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6 qL 6EI In order to apply equation (1) to the 2 16 AB L 2 calculation of element end forces, Vb 2EI qL 3 AB Mb L qL 48 compatibility between element {Q} V 6EI 96EI qL displacements u (i=1,2,3,4) and the e i 2 Me L 16 structure displacements dj (j=1,2, 4EI qL2 …, n) must be established. For the L 24
example beam: 6EI AB BC L2 e1b1d; d;allothers 0 BC Vb 4EI 1 AB 3 {Qfb } {0} BC Mb L qL qL L/6 {Q} BC T Ve 6EI96EI 2 1 {Qfb } qL/2 1L/61 L/6 M L2 L/6 e 2EI Since only one displacement is L T nonzero for each member, the 21qL qL22 27qL 5qL member end forces are 48 24 48 48 25 26
The shear force and bending Mathematical Expression of moment diagrams are given below Stiffness Superposition for the example structure. As stated in step 5, the structure stiffness equations are expressed as
[S] {d} + {Pf} = {P} [S] structure stiffness matrix {d} structure displacement vector
{Pf} structure fixed-end force vector {P} structure node/joint force vector
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7 Referencing the previous figure on stiffness superposition, the first column of the figure represents the kinematically determinate state, i.e. {d} 0 {Pf }{P} The second column of the figure represents the kinematically released state, i.e.
[S] {d} = {P} - {Pf} from which we can calculate the displacements
-1 {d} = [S] ({P} - {Pf})
29 The last column of the figure is the30 addition of the first two columns.
Example Continuous Beam: 2 DOF (EI = constant)
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8 Modified Stiffness Calculations – Zero End Moment Simplification of the stiffness analysis is possible if we take into account the fact that the bending moment at an end simple support is zero (node C in the previous example). This leads to a reduc- tion of one rotational degree of freedom for each zero moment location. Inclusion of this modify- cation results in a reduction of the number of member displacement and force degrees of freedom from 33 4 to 3. 34
Commensurate with this reduction is a change in the member stiffness coefficients as well as the member fixed-end forces.
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9 Modified Bending Modified Member End Stiffness Matrices Force Calculations For the propped cantilever beam MMM of a – Simple-Fixed (S-F) Beam: {Qbbbb }3x1 [k ] 3x3 {u } 3x1 M 11L {Qf3x1b } S-F 3EI [kbb ] 1 1 L L3 M modified member- 2 {Qb } LLL end force vector M modified member For the propped cantilever beam {Qfb } of b – Fixed-Simple (F-S) Beam: fixed-end force vector 1L 1 Superscript M S-F for the F-S3EI 2 [kbb ] L L L “simple-fixed” beam L3 1L1 Superscript M F-S for the 37 “fixed-simple” beam 38
Simple-Fixed Beam Example Continuous Beam – Modified Equations EI = constant SF3EI 3EI SF V(vv)V q bbeefb32 A LL C B SF3EI 3EI SF LL V(vv)Vebeefe32 LL d1 = 1 SF3EI 3EI SF M(vv)Mebeefe L2 L AB BC Fixed-Simple Beam S11 = (4EI/L) + (3EI/L) = 7EI/L d = (S )-1 (-P ) = L/(7EI) (-qL2/8) FS3EI 3EI FS 1 11 f1 V(vv)Vbbebfb 3 LL32 = -qL /56EI FS3EI 3EI FS Compatibility: M(vv)Mb be b fb L2 L AB BC FS3EI 3EI FS e1bd V(vv)Vebebfe LL32
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10 Stiffness Coefficient Summary Member AB End Forces 6EI 3qL Vdb1 In stiffness analysis of structures L2 28 2 that ignore axial deformation, Md2EI qL b1L28 there are only four sets of 6EI 3qL member – displacement stiffness Vde1 L2 28 results that need to be applied for 2 Md4EI qL prismatic members (elements). e1L14 Member BC End Forces 3EI FS 4qL VdVb1fb L2 7 2 MdM3EI FS qL b1fbL14 (a) “Fixed-Fixed” Beam Subjected to a Unit Transverse Displacement VdV3EI FS 3qL e1fe2 7 12EI 6EI L FF ;FF FS5qL FS qL2 FS 3qL 1332 24 V;M;V LL42 fb888 fb fe
(b) “Fixed-Fixed” Beam Subjected to a (c) “Fixed-Simple” Beam Subjected to Unit Rotational Displacement a Unit Transverse Displacement 6EI 4EI FF ;F2F 3EI 3EI 132 24 FF13 ;F 2 L L LL32 Calculations for displacements at What would the forces (stiffness end e are similar to those shown coefficients) be a unit for end b for the “fixed-fixed” displacement at end e? beam.
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11 Braced Frame Example
(d) “Fixed-Simple” Beam Subjected to a Unit Rotational Displacement 3EI 3EI FF13 ;F 2 L2 L Calculations for the “simple-fixed” displacements at end e are similar to the “fixed-simple” beam.
Results from figures (a) – (d) can be applied directly to frames provided axial deformation is ignored. 45 46
Unbraced Frame Example Unbraced Frame Example - Model
40 kN
12 kN 12 kN 3 kN/m
24 m-kN 24 m-kN
LAB =LDC = 12 m LBC = 20 m IAB = IDC = I IBC = 2I
Equivalent Loaded Structure
d3 d1
d2
Kinematic DOF
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