displacement method

6-1

Chapter 6 DISPLACEMENT OR STIFFNESS METHOD OF STRUCTURAL ANALYSIS 6.1 GENERAL

We are now in a position to formulate a systematic method of analysis of structures. In this chapter we will discuss the Stiffness or Displacement method of analysis. This is called the stiffness method since it involves writing down a system of equations in the form [P] = [K][U] for the structure – which have been discussed in the previous chapter. It is also known as the displacement method since the primary unknowns are the displacements [U] required to satisfy the equilibrium and compatibility conditions.

6.2 SUPPORT CONDITIONS AND BASIC FORMULATION The stiffness matrix with respect to the absolute degrees of freedom is singular since no support restraints to prevent rigid body motion have been provided. Therefore we need to introduce supports before we can determine displacements (invert the stiffness equations). Mathematically, the supports are imposed by recognizing that the displacement vector [U] consist of both relative and support degrees of freedom. The structure force vector contains the corresponding forces along the structure degrees of freedom and can be partitioned correspondingly. Thus

[ ] [ ]

=

=

s

r

s

r

P

PP

U

UU and

The above implies that the relative degrees of freedom are numbered prior to support degrees of freedom. Partitioning the structure stiffness matrix correspondingly, the force displacement relationship becomes

=

s

r

ssrs

rsrr

s

r

U

U

KK

KK

P

P (6-1)

Multiplying gives us two equations:

Pr = Krr Ur + Krs Us (6-2)

Ps = Ksr Ur + Kss Us (6-3)

6-2 Notes on Matrix Structural Analysis

ABJ Jr - Jun 2011

Recall that the stiffness matrices are independent of loading. It is assumed that the geometry of the structure, member connectivity, material and section properties of all members to be known at this point such that [K] can be determined. [Us] by definition of support degrees of freedom contains known displacement components. [Pr] are nodal forces applied along the relative degrees of freedom and represents the actions for which the response is to be determined. The only unknown in equation (6-2) is the displacements along the relative degrees of freedom [Ur]. Solving for [Ur]

Ur = [Krr]-1 { Pr - Krs Us } (6-4) The quantity (-Krs Us) represents the equivalent nodal forces along the relative degrees of freedom due to support movement. Recognize that [Krr]-1 is actually the structure flexibility matrix. To determine nodal displacements, this has to be multiplied by nodal forces. The concept of equivalent nodal forces will be discussed further in this chapter. The second equation, (6-3), is used to determine the vector [Ps] after [Ur] is determined from equation (6-2). [Ps] should be recognized as the support reactions due to the imposed actions. Note that loads applied directly along support degrees of freedom are not included in the structural analysis since no structural response is generated aside from a change in the values of the support reactions. In actual implementation of the method, equation (6-3) is rarely used since it is more efficient to apply equilibrium at the support joints after determining all member end forces: the support reactions are simply the sum of all member end forces connected to the support. The determination of [Ur] completes the determination of the primary unknown, [U]T = [Ur Us]. The analysis responses of interest for design purposes are the internal stress functions for each member. This is accomplished by determining the member nodal displacements from the compatibility requirements. In matrix form:

[Um] = [C] [U] = [E]T [U] (6-5)

In actual implementation, recall that we do not actually generate the compatibility or equilibrium matrices. Therefore, rather than multiplying out the matrices the member nodal displacements are extracted from the structure nodal displacement vector using the member incidence or destination table. Knowing the member nodal displacements, the member force end forces are determined using the member stiffness relationship.

[Pm] = [Km] [Um] (6-6)

Recall that the force displacement relationship for the members is in the global coordinates. [Pm] above is also in the global coordinates and must be transformed into the member coordinates before the stress function can be determined from equilibrium. Once the stress resultant functions are known, displacements between member ends can be determined, if required. Equations (6-5) and (6-6) are the matrix equation representations for determining final member end forces. Actual implementation of the above is normally done member by member using the member incidence or destination table.

Displacement Method 6-3

Institute of Civil Engineering, UP Diliman

Example 6.1 Plane Truss with Joint Loads. Determine the member forces for the planar truss for the loading shown. The top and bottom chords have as AE = 300,000 kN, and other members have an AE = 210,000 kN. Solution: The truss is statically indeterminate to the first degree so that the method of joints cannot be used to determine the member forces. This shall be solved using the direct stiffness method. For primary unknowns:

Ur = [Krr]-1 { Pr - Krs Us } with Us = Φ

Note that the stiffness method can also solve determinate problems. If member forces are all that is required for a determinate truss, the method of joints will be computationally be more efficient since no force-displacement equations are required. If displacements are also to be determined for a determinate truss, then use of the stiffness method is advantageous.

The solution shown below is basically oriented towards a procedure which involves the use of some mathematical software to do some repetitive tasks, rather than one intended for manual calculations. Since the routines used are not shown, the reader is encourage to verify the intermediate results using whatever tool is available to them. First, we need to identify the members and their connectivity. This is accomplished by numbering the members and joints in a figure such as shown below, indicating its incidence (direction from A- to B-end by an arrow head). The information shown in the figure is translated into input data tables: Joint Coordinates and Member Incidences,

Table of Joint Coordinates (m) Table of Member Incidences

Joint x y 1 0. 0. 2 8. 0. 3 16. 0. 4 24. 0. 5 8. 5. 6 16. 5.

Member JA JB 1 1 2 2 2 3 3 3 4 4 5 6 5 1 5 6 4 6 7 2 5 8 3 6 9 2 6

10 3 5

50 kN 50 kN 8 m 8 m 8 m

5 m

J2 J3 J4 J1

J5 J6

1 2 3

5

4

6 7 8 10 9


ABJ Jr - Jun 2011

These two tables allow the determination of the length and direction cosines of the members. The direction cosines define the orientation of the member with respect to the global axes. For each member for the planar case: ∆x = xB - xA, ∆y = yB - yA, L = (∆x2 + ∆y2)1/2, cos(θ) = ∆x /L and sin(θ) = ∆y /L.

From the above information, the member stiffness matrix in global coordinates for an axial force member is calculated as:

−−−−

−−−−

=

)(sin)cos()sin()(sin)cos()sin()cos()sin()(cos)cos()sin()(cos

)(sin)cos()sin()(sin)cos()sin()cos()sin()(cos)cos()sin()(cos

][

θθθθθθθθθθθθ

θθθθθθθθθθθθ

22

22

22

22

LAEKG

Only the member stiffness matrices for diagonal members 5 and 10 are shown below. Although the full 4x4 matrix is shown, we need to calculate only say [KBB] and recognize that [KAA] = -[KBA] = -[KAB] = [KBB]

35 10

93.829.1493.829.1429.1487.2229.1487.2293.829.1493.829.1429.1487.2229.1487.22

][

−−−−

−−−−

=mK kN/m

310 10

25.600.1025.600.1000.1001.1600.1001.1625.600.1025.600.1000.1001.1600.1001.16

][

−−−−−−

−−

=mK

Note that only the matrix [Krr] is required. The member incidence table cannot be used in the direct assembly because some of the joints have support degrees of freedom. Therefore, the relative degrees of freedom are identified in a figure and the destination table is generated. Member end DOF connected to support DOF are given ‘0’ values.

Members L (m) cos(θ) sin(θ) AE (kN) AE/L (kN/m) 1, 2, 3, 4 8 1.0 0. 300,000 37,000

5 9.434 0.848 0.530 300,000 31,800 6 9.434 -0.848 0.530 300,000 31,800

7, 8 5 0. 1.0 210,000 42,000 9 9.434 0.848 0.530 210,000 22,260 10 9.434 -0.848 0.530 210,000 22,260

θ 1

2 3

4

A

B

∆x

∆y



Structure Relative Degrees of Freedom

Destination Table - RDOF The structure stiffness matrix, [Krr], is generated by direct assembly resulting in the following:

310

19.5729.4.0.029.1400.42.025.600.1029.437.76.050.3787.22.0.000.1001.16

.0.019.5729.4.025.60.1000.42.0

.050.3729.437.76.000.1001.16.0.029.1487.2200.0.037.60.05.37.0.000.42.025.600.10.025.480.10.0.0.0.000.1001.1650.3700.1001.91.050.3725.600.1000.42.0.0.0.025.4800.1000.1001.16.0.0.0.050.3700.1001.91

][

−−−−−−−−−

−−−−−−

−−−−−−−

−−−−−−

=rrK

The structure stiffness matrix is independent of the action applied, and the above can be used for other actions. For this example, the action involves only joint loads and all that is required is to assemble [Pr].

[Pr]T = [ 0. -50. 0. -50. 0. 0. 0. 0. 0. ] kN

Solving for [Ur] = [Krr]-1 [Pr]

[Ur]T = [ 2.13 -13.50 4.06 -13.50 6.20 4.27 -12.42 1.93 -12.42 ] 10-3 m With the structure nodal displacements known, the member end nodal displacements are extracted using the destination table – supports have zero displacements.

Member 1 2 3 4 1 0 0 1 2 2 1 2 3 4 3 3 4 5 0 4 6 7 8 9 5 0 0 6 7 6 5 0 8 9 7 1 2 6 7 8 3 4 8 9 9 1 2 8 9 10 3 4 6 7

5 1 3

8 9 7 6

2 4


ABJ Jr - Jun 2011

Member nodal displacements, [Um], in global coordinates DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10

1 0 2.13 4.06 4.27 0 6.2 2.13 4.06 2.13 4.06 2 0 -13.5 -13.5 -12.4 0 0 -13.5 -13.5 -13.5 -13.5 x 10-3 3 2.13 4.06 6.20 1.93 4.27 1.93 4.27 1.93 1.93 4.27 4 -13.5 -13.5 0 -12.4 -12.4 -12.4 -12.4 -12.4 -12.4 -12.4

The member forces can now be determined from [Pm] = [Km] [Um]. Note that these are still in global coordinates. Only B-end forces are shown.

DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 3 80.0 72.44 80.0 -87.56 -80.0 80.0 0. 0. 7.56 -7.56 4 0. 0. 0. 0. -50.0 -50. 45.28 45.28 4.72 4.72

A rotational transformation is required to transform the above forces into the local coordinates so that we get the axial force induced on the member. Recognizing that the end forces are equal and opposite, only the B-end forces are transformed. The B-end is used since a positive value will indicate a tensile force on the bar which is the usual convention. [PB(local)] = [RLG] [PB(global)] where for this planar case [RLG] = [cosθ sinθ]. Note that the above rotational transformation only finds the local x-component of the forces from the global x- and y-components. The local y-component may also be determined as a check, since this must be equal to zero, by using the 2x2 rotational transformation matrix.

[𝑅𝑅𝐿𝐿𝐿𝐿 ] = � cos 𝜃𝜃 sin𝜃𝜃−sin𝜃𝜃 cos 𝜃𝜃�

The final member B-end forces in local coordinates are:

B-end member total forces in local coordinates: DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10

2 80.0 72.4 80.0 -87.6 -94.3 -94.3 45.3 45.3 8.9 8.9 kN If support reactions are required, rather than using Equation (6-3) we can simply take a free body diagram of the support showing the negative of the member end forces (in global coordinates) of members connected to the joint. Applying equilibrium, it should be clear that the support reactions are simply the sum of the member end forces of the members connected to the support. For our example, at the hinge the reactions are sum of the A-end forces of members 1 and 5; and at the roller support the reactions are the sum of the B-end forces of member 3 and the A-end forces of member 6. Thus:

Ps1 = P1Ax + P5Ax = -80 + 80 = 0. kN

Ps2 = P1Ay + P5Ay = 0 + 50 = 50. kN

Ps3 = P3By + P6Ay = 0 + 50 = 50. kN

Ps2 Ps1

Ps3

-P1Ax = 80 kN

-P5Ay = -50 kN

-P5Ax = -80 kN -P3Bx = -80 kN -P6Ax = 80 kN

-P6Ay = -50 kN



Example 6.2 Gable Frame with Joint Loads. Determine the final member forces for the frame due to the joint loads. Columns have an AE = 2,100.0 x 103 kN and EI = 75.0 x 103 kN-m2. and the beams have an AE = 1,700.0 x 103 kN and EI = 70.0 x 103 kN-m2. Solution: The solution follows essentially that of the previous example. Define joints, members, member incidence and relative degrees of freedom in figures.

Joints, Members, Member Incidences Relative Degrees of Freedom

Table of Joint Coordinates (m) Table of Member Incidences Member data required to determine member stiffness matrices

Member Stiffness matrices (only [KBB] is shown) in local coordinates. Note, [K1BB] = [K4BB].

−

−=

LEI

LEI

LEI

LEI

LAE

K

zz

zzBB

460

6120

00

][

2

23 31 100.505.1205.122.40

00350][ xK BB

−−=

Joint x y 1 0. 0. 2 0. 6. 3 12. 8.5 4 28. 6. 5 28. 0.

Member JA JB 1 1 2 2 2 3 3 3 4 4 5 4

Members L (m) cos(θ) sin(θ) AE (kN) EI (kN-m2) 1, 4 6 0. 1.0 2,100. x 103 75. x 103 2 12.26 0.979 0.204 1,700. x 103 70. x 103 3 16.19 0.988 -0.154 1,700. x 103 70. x 103

J210

J1

J3

J5

M1

M2 M3

M4

J4

1

2

3

4 5

6

7 8

9

12 m 16 m

2.5 m kN

5 kN

6 m

15 kN


ABJ Jr - Jun 2011

32 108.228.208.25.00

007.138][ xK BB

−−= 33 10

3.176.106.12.00

00105][ xK BB

−−=

Member Stiffness matrices [KBB] in global coordinates.

−

−

+

−

−

+

=

LEI

LEI

LEI

LEI

LEI

LAE

LEI

LAE

LEI

LEI

LAE

LEI

LAE

KBB

4cos6sin6

cos6cos12sinsincos12

sin6sincos12sin12cos

23

23

32

32

θθ

θθθθθ

θθθθθ

Note that an alternative way of getting the other submatrices of the member stiffness matrix given [KBB] is by using the equilibrium requirement for each column and symmetry of [K]: i.e. [KAB] = -[HAB] [KBB], [KBA] = [KAB]T , and [KAA] = -[HBA] [KBA]. Also, for the columns only the B-end stiffness matrix is required

][100.5005.12

035005.1202.4

][ 431BBBB KxK =

=

32 10

8.227.26.04.117.26.07.22.66.277.22.66.276.06.279.1326.06.279.1324.117.26.08.227.26.07.22.66.277.22.66.276.06.279.1326.06.279.132

][ xK m

−−−−−−

−−−−−−

−−−−

=

33 10

3.176.12.06.86.12.06.17.20.166.17.20.162.00.165.1022.00.165.1026.86.12.03.176.12.06.17.20.166.17.20.162.00.165.1022.00.165.102

][ xK m

−−−−−−−−−−

−−−−

−−

=

Generate the destination table and assemble structure stiffness matrix and load vector.

Destination Table - RDOF

Member 1 2 3 4 5 6 1 0 0 0 1 2 3 2 1 2 3 4 5 6 3 4 5 6 7 8 9 4 0 0 0 7 8 9



310

3.676.12.126.86.12.0.0.0.06.17.3520.166.17.20.16.0.0.02.120.166.1062.00.165.102.0.0.06.86.12.01.402.18.04.117.26.06.17.20.162.19.86.117.22.66.272.00.165.1028.06.114.2356.06.279.132.0.0.04.117.26.08.727.29.11.0.0.07.22.66.277.22.3566.27.0.0.06.06.279.1329.116.271.137

][

−−−−−

−−−−−−−−−−−−

−−−−−−

−−−

=rrK

[Pr]T = [ 15. 0. 0. | 5. 0. 0. | 10. 0. 0. ]


[Ur]T = [ 8.49 0.006 -1.60 | 8.56 -0.29 0.81 | 8.56 -0.006 -1.69 ] x10-3

Member nodal displacements in global coordinates

DOF M1 M2 M3 M4 1 0 8.49 8.56 0 x10-3 m

2 0 0.006 -1.29 0 x10-3 m

3 0 -1.6 0.81 0 x10-3 rad

4 8.49 8.56 8.56 8.56 x10-3 m

5 0.006 -0.29 -0.006 -0.006 x10-3 m

6 -1.60 0.81 -1.69 -1.69 x10-3 rad

Member nodal forces in global coordinates: [Pm] = [Km] [Um]

DOF M1 M2 M3 M4 1 -15.43 -0.43 4.57 -14.57 kN

2 -2.20 -2.20 -2.20 2.2 kN

3 66.23 -26.35 -1.08 64.79 kN-m

4 15.43 0.43 -4.57 14.57 kN

5 2.20 2.20 2.20 -2.20 kN

6 26.35 1.08 -22.63 22.63 kN-m

Member nodal forces in local coordinates: [Pm(local)] = [RLG] [Pm(global)]

[𝑅𝑅𝐿𝐿𝐿𝐿 ] = �𝐿𝐿 ΦΦ 𝐿𝐿� 𝑤𝑤𝑤𝑤𝑤𝑤ℎ [𝐿𝐿] = �

cos 𝜃𝜃 sin𝜃𝜃 0−sin𝜃𝜃 cos 𝜃𝜃 0

0 0 1�

DOF M1 M2 M3 M4 1 -2.20 -0.87 4.85 2.20 kN

2 15.43 -2.06 -1.46 14.57 kN

3 66.23 -26.35 -1.08 64.79 kN-m

4 2.20 0.87 -4.85 -2.20 kN

5 -15.43 2.06 1.46 -14.57 kN

6 26.35 1.08 -22.63 22.63 kN-m


ABJ Jr - Jun 2011

The stress resultant functions for member 2 are:

N(x) = 0.87 kN

V(x) = 2.06 kN

M(x) = 26.36 – 2.06 x kN-m It should be recognized that the support reactions are equal to the member end forces of the columns connected to the supports – so that Equation (6-3) is not required. Example 6-3 Gable Frame subject to support movement. For the frame in Example 6-2, determine the response if the right support is subject to a support movement equal to a horizontal translation of 0.050 m to the right, a downward translation of 0.025 m and a clockwise rotation of 0.001 rad. Solution: The structure stiffness is the same as that of the previous example. The difference is the applied action, and therefore the only the load vector need to be determined. The equivalent joint load shall be determined from [PrE] = -[Krs][Us]. Only non-zero support movement will produce equivalent joint loads and therefore only the degrees of freedom with nonzero support movement will be defined (numbered)as degrees of freedom 10 to 12. Support displacement vector is thus defined as [Us]T = [ 0.050 m -0.025 m -0.001 rad ]. To generate [Krs] the entries to the destination table for member 4 has to be modified to include the support degrees of freedom.

DT_M4 = [ 10 11 12 7 8 9 ]

Assembling [Krs] and determine [PrE] = -[Krs][Us]

12 m 12

16 m 16

2.5 m 3 6.0 m 6

[Us]

1

2

3

4

5 6

7

8 9

10 1

11 1 12 1



310

0.2505.120.3500

5.12017.4000000000000000000

][ xKrs

−−

−

=

[PrE]T = [ 0. 0. 0. | 0. 0. 0. | 220.8 -8750. 650. ]

Solving for [Ur] = [Krr]-1 [PrE]

[Ur]T = [ 8.49 0.006 -1.60 | 8.56 -0.29 0.81 | 8.56 -0.006 -1.69 ] x10-3 Member nodal displacements in global coordinates

DOF M1 M2 M3 M4 1 0 21.09 32.37 50.0 x10-3 m

2 0 0.0 -53.43 -25.00 x10-3 m

3 0 -5.41 -1.91 -1.00 x10-3 rad

4 21.09 32.37 37.00 37.00 x10-3 m

5 0.0 -53.43 -25.00 -25.00 x10-3 m

6 -5.41 -1.91 3.72 3.72 x10-3 rad

Member nodal forces in global coordinates: [Pm] = [Km] [Um]

DOF M1 M2 M3 M4 1 -20.19 -20.19 -20.19 20.19 kN

2 0.31 0.31 0.31 -0.31 kN

3 128.25 7.11 -47.09 -119.54 kN-m

4 20.19 20.19 20.19 -20.19 kN

5 -0.31 -0.31 -0.31 0.31 kN

6 -7.11 47.09 1.60 -1.60 kN-m

Member nodal forces in local coordinates: [Pm(local)] = [RLG] [Pm(global)]

DOF M1 M2 M3 M4 1 0.31 -19.70 -20.00 -0.31 kN

2 20.19 4.42 -2.81 -20.19 kN

3 128.25 7.11 -47.09 -119.54 kN-m

4 -0.31 19.7 20.00 0.31 kN

5 -20.19 -4.42 2.81 20.19 kN

6 -7.11 47.09 1.60 -1.60 kN-m


ABJ Jr - Jun 2011

6.3 STRUCTURES WITH MEMBER LOADING AND OTHER ACTIONS - GENERAL The analysis procedure discussed in the previous section is applicable only for forced applied along the structure degrees of freedom, also known as joint loads, and for support movement. Truss structures can thus be handled by the above procedure because these typically involve only joint loads. For building structures, wind and earthquake loads are also applied as lateral joint loads. Gravity loads, self weight and occupancy loads, however, are applied as concentrated or distributed loads along the members. Further, there may also be a need to determine the response of the structure to other actions such as volumetric changes due to changes in temperature, creep or shrinkage, and imposed displacements. Since these actions do not involve joint loads, the analysis procedure must be modified to handle these actions. For structures with linearly elastic material, the principle of superposition applies and will be used.

6.4 STRUCTURES WITH MEMBER LOADS Consider the plane frame with a member load in Figure 6-1. Applying superposition, the solution may be divided into two parts: the Fixed-End part and the Equivalent Joint load part. The fixed-end part consist of the original structure and loading to which is added a set of fictitious nodal forces, [PF]. The [PF] forces are of such magnitudes that prevents any nodal displacements in the structure, i.e. [U] = Φ for this case. The equivalent joint load part consist of the original structure subject to nodal forces [PE] to remove the effect of the fictitious [PF] forces in the fixed end part. Necessarily [PE] is just the negative of [PF], i.e. [PE] = -[PF]. Actual Problem = Fixed-End Part + Equivalent Joint Load Part Figure 6-1 Superposition Solution for Structures with Member Loads The nodal forces and displacements for the problem can therefore be written as the sum of the two parts of the solution.

[P] = [PF] + [PE] (6-7)

[U] = [UF] + [UE] (6-8)

Note that by definition of [PF], [UF] = Φ and therefore

[U] = [UE] (6-9)

This is the reason [PE] is referred to as an equivalent nodal forces, i.e. the nodal forces, [PE], produces the same nodal displacements as the original

= +

PE PF



actions. Note, only the nodal displacements are the same, the displacements between nodes is the sum of the response of the two parts of the solution. It should be recognized that the equivalent joint load part consist of the structure subject to nodal forces only. Therefore, the basic force-displacement equation applies, i.e.

[PE] = [K][UE] (6-10)

or from (6-9) [PE] = [K] [U] (6-11)

Thus, equation (6-7) can be written as [P] = [PF] + [K] [U] (6-12)

This is the modified force displacement relationship for the stiffness method considering member loads. Partitioning this with respect to relative and support degrees of freedom we get

+

=

s

r

ssrs

rsrr

Fs

Fr

s

r

U

U

KK

KK

P

P

P

P (6-13)

Multiplying gives: Pr = PFr + Krr Ur + Krs Us (6-14)

Ps = PFs + Ksr Ur + Kss Us (6-15)

Equation (6-14) is used to determine the displacements along the relative degrees of freedom.

Ur = [Krr]-1 [ Pr - PFr - Krs Us ] (6-16)

This may be written as:

Ur = [Krr]-1 [ PEr ] (6-17)

where [ PEr ] = [ Pr - PFr - Krs Us ] Equation (6-15) is used to determine the support reactions. Again in practice the equation is rarely used as a more efficient procedure is available for determining support reactions. To complete the analysis member end displacements are again determined from compatibility requirements. These nodal member displacements should be recognized as the response to the equivalent joint loads. The final member end forces are determined as the sum of the fixed-end and equivalent joint cases.

[Um] = [C] [U] = [E]T [U] (6-5)

[Pm] = [PmF] + [Km] [Um] (6-18)

As before, these forces are still with respect to the global coordinates. A rotational transformation to the member coordinates will be required before determining the member stress functions. Determination of [PF]. What remains to be discussed is the determination of the fixed-end force vector, [PF]. By definition, the [PF] forces are of such magnitude that prevents any nodal displacements in the structure. The response (internal forces and displacements) of each member should be


ABJ Jr - Jun 2011

recognized as identical to the response of the member removed from the structure and provided with fixed-end supports. This greatly simplifies the analysis as each member can be considered independently, rather than analyzing the whole structure. To determine [PF], recognize that to prevent nodal displacements, the member fixed-end forces and the nodal forces acting at a node must be in static equilibrium.

Figure 6-2 Member and Joint Forces for Fixed-End Case

Referring to Figure 6-2, equilibrium at joints 1 and 2 requires that:

P1F = P1

mfB + P2mfA

and P2F = P2

mfB In general the nodal forces at a node are the sum of the fixed-end forces of the member ends connected to that node. As indicated in the figure, the fixed-end forces are already with respect to the global coordinates so that the equilibrium of the joints can be written as a direct sum. In matrix form the [PF] vector may be determined by the relationship

[PF] = [ E ] [PMF]

Where [PMF] is the vector containing the fixed-end forces (in global coordinates) of all the members due to the given action, and [E] is the same equilibrium matrix used in the derivation of the structure stiffness matrix. In actual implementation the [PF] vector can be directly assembled using the member incidence or destination table in the same way as the structure stiffness. Note that the figure only shows the forces required at the free joints (not containing support restraints). Equilibrium of the support joints will determine the elements of [PFs]. Example 6-4 Continuous Beam with Member Loads. Determine the internal stress functions for the middle beam of the three-span continuous beam for the gravity loading shown. Members are prismatic with EI = 200,000 kN-m2

M3 M1 M2

P2mfA P2

mfB

-P2mfA -P2

mfB

J2

P1F PFJ2

-P1mfB

P1mfB

P1mfA

P2F P1F

M1 M3

M2



Solution: Define members and relative degrees of freedom. Note that for the relative degrees of freedom, it is recognized that the loading is parallel planar such that no axial forces are induced in the members. With no axial forces the members will not change length and therefore there are no horizontal displacements at the joints (essentially becomes support degrees of freedom since their final values are known to be zero). Therefore only the rotational relative degrees of freedom exist at the joints. The stiffness matrix for the simply supported member is all that is required to generate the structure stiffness, [Krr].

[𝐾𝐾𝑚𝑚 ] = 4 𝐸𝐸𝐸𝐸𝐿𝐿

� 1 0.50.5 1 �

[𝐾𝐾𝑚𝑚1 ] = [𝐾𝐾𝑚𝑚3 ] = �133.33 66.6766.67 133.33� 𝑥𝑥 103 [𝐾𝐾𝑚𝑚2 ] = �100.0 50.0

50.0 100.0� 𝑥𝑥 103

Destination Table Assemble Structure stiffness

31033.13367.66.067.6633.23300.50.00.5033.233

][ xKrr

=

For the load vector, we will require the fixed-end forces for the members.

[𝑃𝑃𝑚𝑚𝑚𝑚 ] = 𝑤𝑤 𝐿𝐿2

12 � 1 −1 � [𝑃𝑃𝑚𝑚𝑚𝑚1 ] = [𝑃𝑃𝑚𝑚𝑚𝑚3 ] = � 75.0

−75.0 � [𝑃𝑃𝑚𝑚𝑚𝑚2 ] = � 213.3 −213.3� kN-m

Assemble load vector, [PEr] = -[PFr]

[𝑃𝑃𝑚𝑚𝐹𝐹 ] = �−75.0 + 213.3−213.3 + 75−75

� = �138.33

−138.33−75.00

� [𝑃𝑃𝐸𝐸𝐹𝐹 ] = �−138.33

138.3375.00

�

Nodal displacements : [Ur] = [Krr]-1 [Per]

[𝑈𝑈𝐹𝐹 ] = �−0.741

0.689 0.218

� 𝑥𝑥 10−3 rads

Member A B 1 0 1 2 1 2 3 2 3

1

2

M1 M2 M3

3

8 m 6 m 6 m

25 kN/m 25 kN/m 40 kN/m


ABJ Jr - Jun 2011

Member forces are calculated as [Pm] = [PmF] + [Km] [Um]. Since shears are also required in this example, the degrees of freedom shall include those along the end shears. Thus for member 2:

332 10689.0741.0

10

00.10000.5075.1875.1800.5000.10075.1875.18

33.21300.16033.21300.160

][ −

−

−−+

−

= xxPm

mkN

kN

mkN

kN

mP

−

−

−

=

−

−

+

−

=

4.1810.1617.1730.159

9.310.16.390.1

33.21300.16033.21300.160

][ 2

Summary of member final nodal forces

DOF M1 M2 M3 1 50.3 159.0 105.2 kN

2 25.6 173.7 181.4 kN-m

3 99.7 161.0 44.8 kN

4 -173.7 -181.4 0.0 kN-m

Stress resultant functions for member 2: V(x) = -159.0 + 40 x (note: equations are with respect to the right hand coordinates, while diagrams are with respect to the strength of materials sign convention.) M(x) = -173.7 + 159.0 x - 20 x2 Example 6-5 Gable Frame with Member Load. Determine the stress resultant function for member 2 of the gable frame in Example 6-2 if subject a gravity uniform distributed load of 15 kN/m acting on members 2 and 3. Solution: Again, since the structure geometry, members and materials are the same as in Example 6-2, the structure stiffness is the same. What is required is the determination of the structure nodal forces [ PEr ] = [ Pr - PFr - Krs Us ]. With no nodal forces and support movement, [ PEr ] = [-PFr], with [PF] = [ E ] [PMF].

12 m 12

16 m 16

2.5 m 2.5 6.0 m 6

15 kN/m 12

15 kN/m 12

3.975

159 kN

161 kN

173.7 kN-m

181.4 kN-m

142.3 kN-m



The member fixed-end forces are conveniently determined with respect to the member local axis. The member loads in local coordinates are determined by a rotational transformation: [wix wiy]T = [Ri] [wy(Global)] with [Ri]T = [sinθi cosθi]

Load wx wy L cosθ sinθ M1 -3.059 -14.685 12.258 0.979 0.204 M2 2.316 -14.82 16.194 0.988 -0.154

The fixed-end forces for a prismatic member subject to uniform distributed load

−

−

−

=

1222

1222

][

2

2

LwLwLw

LwLwLw

P

y

y

x

y

y

x

mF

Member Fixed-End Forces in Local coordinates:

DOF M1 M2 M3 M4 1 0 18.75 -18.75 0 kN

2 0 90.00 120.00 0 kN

3 0 183.86 323.86 0 kN-m

4 0 18.75 -18.75 0 kN

5 0 90.00 120.00 0 kN

6 0 -183.86 -323.88 0 kN-m

Member Fixed-End Forces Global coordinates: [PmG] = [RGL] [PmL]

[𝑅𝑅𝐿𝐿𝐿𝐿] = �𝐿𝐿 ΦΦ 𝐿𝐿� 𝑤𝑤𝑤𝑤𝑤𝑤ℎ [𝐿𝐿] = �

cos 𝜃𝜃 −sin𝜃𝜃 0sin𝜃𝜃 cos𝜃𝜃 0

0 0 1�

DOF M1 M2 M3 M4 1 0 0 0 0 kN

2 0 91.93 121.46 0 kN

3 0 183.86 323.88 0 kN-m

4 0 0 0 0 kN

5 0 91.93 121.46 0 kN

6 0 -183.86 -323.88 0 kN-m

[PFr] is assembled using the destination table elements.

[PFr]T = [ 0. 91.93 183.86 | 0. 213.39 140.02 | 0. 121.46 -323.88 ]

and [PEr]T = -[PFr]T

[PEr]T = [ 0. -91.93 -183.86 | 0. -213.39 -140.02 | 0. -121.46 323.88 ]

From Example 6-2


ABJ Jr - Jun 2011

310

3.676.12.126.86.12.0.0.0.06.17.3520.166.17.20.16.0.0.02.120.166.1062.00.165.102.0.0.06.86.12.01.402.18.04.117.26.06.17.20.162.19.86.117.22.66.272.00.165.1028.06.114.2356.06.279.132.0.0.04.117.26.08.727.29.11.0.0.07.22.66.277.22.3566.27.0.0.06.06.279.1329.116.271.137

][

−−−−−

−−−−−−−−−−−−

−−−−−−

−−−

=rrK


[Ur]T = [ -45.85 -0.61 -2.25 | 2.22 -239.82 -11.29 | 37.37 -0.61 5.08] x10-3

Final member forces are the sum of the fixed-end case and the equivalent joint load case: [Pm] = [PmF] + [Km] [Um] Member nodal displacements, [Um], in global coordinates

DOF M1 M2 M3 M4 1 0 -45.85 -2.25 0 x10-3 m

2 0 -0.61 2.22 0 x10-3 m

3 0 -2.25 -239.82 0 x10-3 rad

4 -45.85 -2.25 37.37 37.37 x10-3 m

5 -0.61 2.22 -0.61 -0.61 x10-3 m

6 -2.25 -239.82 5.08 5.08 x10-3 rad

Member nodal forces in global coordinates: [PmE] = [Km] [Um]

DOF M1 M2 M3 M4 1 219.16 219.16 219.16 -219.16 kN

2 212.61 120.67 -92.71 214.17 kN

3 -629.36 501.72 -538.50 594.04 kN-m

4 -219.16 -219.16 -219.16 219.16 kN

5 -212.61 -120.67 92.71 -214.17 kN

6 -685.59 398.48 -397.03 720.91 kN-m

Member total nodal forces in global coordinates: [Pm] = [PmF] + [PmE]

DOF M1 M2 M3 M4 1 219.16 219.16 219.16 -219.16 kN

2 212.61 212.61 28.74 214.17 kN

3 -629.36 685.59 -214.61 594.04 kN-m

4 -219.16 -219.16 -219.16 219.16 kN

5 -212.61 -28.74 214.17 -214.17 kN

6 -685.59 214.61 -720.91 720.91 kN-m

Note, the A-end forces for the columns in global coordinates are equal to the support reactions for the structure. Member total nodal forces in local coordinates: [Pm(local)] = [RLG] [Pm(global)]



DOF M1 M2 M3 M4 1 212.61 257.91 212.09 214.17 kN

2 -219.16 163.44 62.23 219.16 kN

3 -629.36 685.59 -214.61 594.04 kN-m

4 -212.61 -220.41 -249.59 -214.17 kN

5 219.16 16.56 177.77 -219.16 kN

6 -685.59 214.61 -720.91 720.91 kN-m

The stress resultant functions for member 2 are: N(x) = -257.91 + 3.059 x kN V(x) = -163.44 + 14.685 x kN M(x) = -685.59 + 163.44 x - 7.3425 x2 kN-m

6.5 STRUCTURES WITH OTHER ACTIONS The same superposition procedure used for member loads can be applied to the problem of the structure subject to other types of action. The fixed-end case will consist of the structure subject to the given action, with the addition of nodal forces that prevent any nodal displacements. The equivalent load case involves the structure with the negative of the fixing nodal forces. Support Movement. Support movement problems can also be handled using the superposition approach rather than determining the equivalent nodal forces by -[Krs] [Us]. In general, use of superposition is simpler and more computationally efficient than the later procedure. Thus for practical purposes the nodal displacement may be determined using [Ur] = [Krr]-1 [Pr - PFr] or [Ur] = [Krr]-1 [Pr + PEr] for any actions. The form involving [PEr] = -[PFr], stresses the point that for actions other than nodal forces, the actions have to be converted into an equivalent set of nodal forces. Consider the example of the frame shown in Figure 6-3 subject to a vertical support settlement at the right support. The fixed-end part consist of the structure with the given support movement and fixing forces to prevent any nodal displacement. It should be clear that the fixing force consist of a vertical upward force as shown to prevent that joint from moving downward (only none zero nodal forces along the relative degrees of freedom are shown in the figure).

-257.9 kN

-220.4 kN

163.4 kN

16.6 kN 11.13 m

685.6 kN-m

214.6 kN-m

223.7 kN-m


ABJ Jr - Jun 2011

Actual Problem = Fixed-End Part + Equivalent Joint Load Part Figure 6-3 Superposition Solution for Structures with Support Movement

Imposed Discrete Displacement. Discrete displacements include problems involving members being too long or too short (compared to that required), members with a kink or lateral displacement of the centroid (when member is intended to be straight) as shown in Figure 6-4. Some authors refer to these types of problems as fabrication errors. However, imposed displacements may also be used in determining the influence lines for structures using the Meuller-Breslau Principle. a) Member too Short b) Member Kink c) Transverse Displacement

Figure 6-4 Imposed Discrete Displacement Distributed Displacements. These actions include volumetric changes due to changes in temperature, creep and shrinkage. Creep and shrinkage is particular to reinforced concrete construction. Creep refers to the deformation due to sustained loading. As concrete has a very low tensile strength, creep strains occurs for concrete in compression. Shrinkage of concrete is primarily due to loss of moisture due to drying. The solution procedure for these types of action is identical to member loads. The difference will be in the determination of the fixed-end forces for these actions.

6.6 MEMBER FIXED-END FORCES

The analysis procedure for all the above actions requires the determination of the member fixed-end forces. For a prismatic member several references are available where the fixed end forces are given for several types of loading. Most, if not all the equations found in the references assume that shear deformations are negligible. Chapter 8 of these notes presents a discussion of the determination of fixed-end forces, particularly for member loads, using the Flexibility or Force Method. While at the member level, it is always convenient to use the member or local coordinates. Prior to generating the [PF] vector of the structure, a rotational transformation of the fixed-end forces will be required. The following are special cases which can be presented without a formal discussion of the Flexibility Method.

required length ∆ θ

∆y

= +

∆

PF

∆

PE



Support Movement. Assume that the B-end of the member shown in Figure 6-9 is given the support displacement [Us].

Figure 6-9 Fixed-End Problem – Support Movement

It should be recognized that this problem is identical to imposing a specific set of nodal displacements on the member. Thus the fixed-end forces for this problem can be determined directly by applying the force-displacement relationships.

[PF] = [Km] [U] (6-29) where [U] for the member in Figure 6-9 is given by

Φ=

=

sB

A

UU

UU ][

and [ ] [ ]sBB

AB

sBBBA

ABAAF U

K

K

UKK

KKP

=

Φ

=

or again we can determine [PFB] = [KBB] [Us] and using equilibrium [PFA] = -[HAB] [PFB] Imposed Discrete Displacements. These types of problems can be solved using the method of consistent deformation or superposition. The method involves replacing the problem by an equivalent one in which the fixed-end forces at the B-end is taken as forces (still unknown) then dividing the problem into two cases: Case 0 for the member with the imposed displacement and Case 1 for the member with the unknown forces at ‘B’. Without the forces at ‘B’, point ‘B’ will displace by an amount [UB0] in Case 0. The problem becomes one of determining the set of end forces that produces a displacement at ‘B’ of [UB1] = -[U0]. The problem becomes the same at that of support movement above, such that

[PF] = -[Km] [U0] (6-30) Figures 6-10 to 12 illustrates the procedure for a fixed-free primary structure. For a straight line element, a change in length produces only an axial fixed end force.

a) fixed-end case = b) case ‘0’ + c) case ‘1’ Figure 6-10 Fixed-End Forces – Member too Short

Verify that the fixed-end forces (axial) for a prismatic member which is fabricated ∆L (positive for an increase in length) are:

−

∆=1

1][ L

LAEPF

+ A B

= UB0 = -∆

PB

UB

A

B UB = Us


ABJ Jr - Jun 2011

The kink and transverse displacement will not produce a change in length assuming small displacements, and therefore the axial terms are not shown in the figures. a) fixed-end case = b) case ‘0’ + c) case ‘1’

Figure 6-11 Fixed-End Forces – Member with Kink For a prismatic member with a kink the fixed-end forces are:

θEI

LaLLaL

LaLLaL

PF

−−−

−−

=

2

3

2

3

/)62(/)2(6

/)64(/)2(6

][

a) fixed-end case = b) case ‘0’ + c) case ‘1’ Figure 6-12 Fixed-End Forces – Member with Transverse Displacement

For a prismatic member with a discrete transverse displacement the fixed-end forces are:

∆

−=

2

3

2

3

/6/12

/6/12

][

LEILEI

LEILEI

PF

Change in Temperature. This type of problem requires the function of the variation of temperature throughout the member. The simplest case is when the change is temperature is the same throughout the depth, width and length of the member. For members that can be modeled as line elements, the change in the cross-sectional dimension is normally not considered as these will be much smaller than the change in length of the member. The change is length is calculated as ∆LT = αT ∆T L. Where αT is the coefficient of thermal expansion (which is strain per degree temperature change), ∆T is the change in temperature (positive for an increase in temperature – equivalent to applying a tensile force) and L is the length of the member. Once the change in length is determined, the solution is identical to members being too long or short.

A B a

∆

= + a

∆ [ ]

∆=

00BU PB

U1B

=

[ ]

−=

θ

θ)( aLU B0

A B a θ

θ

a +

PB

UB



7.4 Fixed-End Forces of Members with Releases by Static Condensation. The following examples determine the fixed-end forces for members with a release, given the fixed-end forces for the member without a release. Example 7-3 Fixed-end forces for member with hinge release by static condensation. Determine the fixed-end forces for the prismatic member with a hinge release at the right end from the fixed-end forces of a prismatic member without a hinge. (this is applicable for any actions producing fixed-end forces in general although a uniform distributed load is shown)

a) Member with hinge release b) Member without a release Solution: P*

u = [Pu - Kuc Kcc-1 Pc ]

[ ] [ ]

−

= 4

34

24

14

443

2

11

mf

mf

mf

mf

u P

k

k

k

kP

P

P

P *

or using COFAB = k24/k44

[ ]( )

( )[ ]

+−

+

−

= 4

3

2

1

1

1

mf

AB

AB

AB

mf

mf

mf

u P

LCOF

COF

LCOF

P

P

P

P

/

/*

If the hinge release is introduced at the A-end rather than the B-end.

[ ] [ ]

−

= 2

42

32

12

224

3

11

mf

mf

mf

mf

u P

k

k

k

kP

P

P

P *

or using COFBA = k42 /k22

[ ]( )( ) [ ]

+−

+

−

= 2

4

3

1

1

1

mf

BA

BA

BA

mf

mf

mf

u P

COF

LCOF

LCOF

P

P

P

P /

/*

For a prismatic member with a hinge at the B-end.

[ ] [ ] ( )4

51

50

511

mfcccuc P

L

L

PKK

−

=−

/.

.

/.

][

1 2

3

Pmf2 Pmf4

Pmf1 Pmf3

1 2

3 Pmf2 Pmf4

Pmf1 Pmf3


ABJ Jr - Jun 2011

For a uniform distributed load:

[ ] [ ]

−

−

−

= 12

51

50

51

2

12

222 /

/.

.

/.

/

/

/* wL

L

L

wL

wL

wL

P u

or

=

8/38/8/5

][ 2*

wLwLwL

P u

Equation (7-6), P*

u = [Pu - Kuc Kcc-1 Pc ], represents the redistribution in the

nodal forces when the nodal force along the condensed degrees of freedom is released (force resultant is made zero as required). Similar to the stiffness matrix, the physical interpretation of the equation P*

u = [Pu - Kuc Kcc

-1 Pc ] can be visualized as the sum of two cases as shown in Figure 7-14 for a prismatic member. The first case is the represents condition when both ends are fixed, while the second case represents removal of the moment at the hinge by applying the negative of the fixed-end moment at that location.

Figure 7-14 Interpretation of in P*u = [Pu - Kuc Kcc

-1 Pc ] Example 7-3 Example 7-4 Fixed-end forces for member with guide release by static condensation. Determine the fixed-end forces for the line element with a guide release at the right end from the fixed-end forces of a prismatic member without a release. Solution: P*

u = [Pu - Kuc Kcc-1 Pc ]

[ ] [ ]

−

= 3

43

23

13

334

2

11

mf

mf

mf

mf

u P

k

k

k

kP

P

P

P *

or

wL2/12 - wL2/12

wL/2 wL/2

w wL2/12 wL2/24

wL/8 wL/8 = +

w wL2/8

5wL/8 3wL/8

Pmf2 Pmf4

Pmf1 Pmf3

2 3

1

w



[ ] [ ]

−

−

= 3

3343

3323

4

2

1 1

mf

mf

mf

mf

u P

kk

kk

P

P

P

P

/

/*

For a prismatic member the fixed-end forces for the member with a guide release at the B-end is as follows.

[ ] [ ]

+

= 3

4

2

1

50

50

1

mf

mf

mf

mf

u P

L

L

P

P

P

P

.

.*

if the load is a uniformly distributed load:

[ ] [ ]

+

−

= 2

50

50

1

12

12

2

2

2 /

.

.

/

/

/* wL

L

L

wL

wL

wL

P u

=

6

32

2

/

/][ *

wL

wL

wL

P u

Example 6-6 Gable Frame subject to support movement. This is the same problem as Example 6-3 where the right support is subject to a support movement equal to a horizontal translation of 0.050 m to the right, a downward translation of 0.025 m and a clockwise rotation of 0.001 rad. However, rather than determining the load vector as [PEr] = -[Krs][Us], superposition will be used, i.e. [PEr] = -[PFr] Solution: The fixed-end case is shown in the figure below. This involves the structure subject to the given action (support movement, [Us]T = [ 0.050 m -0.025 m -0.001 rad ]) and nodal forces, [PF], which prevents any nodal displacements. Only members connected to supports with non-zero support movement will generate end forces. Considering member 4 for the example, the fixed-end forces are the forces required to produce a displacement at its A-end, [UmA] = [Us]

Us

PF

PmfB

PmfA

UmA = Us

12 m 16 m

2.5 m

6 m

Us

Pmf2 = wL2/3 Pmf3 = wL2/6

Pmf1 = wL

w


ABJ Jr - Jun 2011

Thus the fixed-end forces for member 4 is equal to [Km][Um]. The member stiffness matrix and [Us] are in global coordinates so that the calculated end forces are also already in global coordinates. Recognizing that UmB = Φ,

[𝑃𝑃𝑚𝑚𝑚𝑚 ] = �𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚� [𝑈𝑈𝑚𝑚𝑚𝑚 ]

−

−−

−

=

−−

−−

−−

−

=

0.6500.750,883.2200.6750.750,883.220

001.0025.0050.0

10

0.25.05.12.00.350.05.12.017.40.50.05.12.00.350.05.12.017.4

][ 34 xP mF

[PFr] is assembled using the destination table elements for member 4.

[PFr]T = [ 0. 0. 0. | 0. 0. 0. | -220.8 8,750. -650. ]


[PEr]T = [ 0. 0. 0. | 0. 0. 0. | 220.8 -8,750. 650. ]

Example 6-7 Planar truss subject to a change in temperature. Determine the member forces for the plane truss of Example 6-1 if members 4, 5 and 6 are subjected to an increase in temperature of 20OC. The coefficient of thermal expansion, αT = 12.0 x 10-6 /OC. Solution: Since the structure stiffness has been determined in Example 6-1, the solution here starts with determination of the load vector [PEr] due to the change in temperature. The fixed-end forces for the affected members are due to the change in lengths resulting from the change in temperature. In local coordinates:

[PBF] = -[KBB] [∆LT] = -(AE/L) (αT ∆T L) = -αT AE ∆T = -72 kN

Members L (m) cos(θ) sin(θ) AE (kN) AE/L (kN/m) 4 8 1.0 0. 300,000 37,000 5 9.434 0.848 0.530 300,000 31,800 6 9.434 -0.848 0.530 300,000 31,800

m1 m2 m3

m5

m4

m6 m7 m8 m10 m9



Member fixed –end forces in global coordinates:

DOF m4 m5 m6 1 72.0 61.06 -61.06 kN

2 0 38.16 38.16

3 -72.0 -61.06 61.06

4 0 -38.16 -38.16

Assemble [PFR]

[PFr]T = [ 0. 0. 0. | 0. -61.06 10.94 | -38.16 -10.94 -38.16 ]


[PEr]T = [ 0. 0. 0. | 0. 61.06 -10.94 | 38.16 10.94 38.16 ]


[Ur]T = [ 0. 5.95 -0.26 | 5.95 -0.26 -0.96 | 5.81 0.70 5.81 ] x10-3 m

Final member forces are the sum of the fixed-end case and the equivalent joint load case: [Pm] = [PmF] + [Km] [Um] Member nodal displacements, [Um], in global coordinates

DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 1 0 0 -0.26 -0.96 0 -0.26 0 -0.26 0 -0.26 2 0 5.95 5.95 5.81 0 0 5.95 5.95 5.95 5.95 x 10-3 3 0 -0.26 -0.26 0.70 -0.96 0.70 -0.96 0.70 0.70 -0.96 4 5.95 5.95 0 5.81 5.81 5.81 38.16 5.81 5.81 5.81

B-end member nodal forces in global coordinates: [PmE] = [Km] [Um]

DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 3 0 -9.75 0 62.25 61.06 -61.06 0 0 9.75 -9.75 4 0 0 0 0 38.16 38.16 -6.1 -6.1 6.1 6.1

B-end member total nodal forces in global coordinates: [Pm] = [PmF] + [PmE]

DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 3 0 -9.75 0 -9.75 0 0 0 0 9.75 -9.75 4 0 0 0 0 0 0 -6.1 -6.1 6.1 6.1

B-end member total forces in local coordinates:

DOF m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 2 0 -9.75 0 -9.75 0 0 -6.1 -6.1 11.5 11.5 kN

5 1 3

8 9 7 6

2 4

displacement method

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