The tool TFS-02 determines the static response of a frame, including nodal displacements and reactions at supports. The tool also plots the undeformed and deformed configurations of the frame, and the distributions of displacements and internal forces of any selected frame member.

In addition, a tutorial on how to use the tool in computation is provided and a subject review on fundamental theories and useful formulas is presented.

**1. Data Preparation for Tool TFS-02**

The online tool TFS-02 is useful and efficient in static analysis of plane frames. Before using TFS-02, one needs to know how to prepare input data by the format shown in the main page of the tool. We shall explain this by considering one example in Fig. 1T, where a frame is subject to three external loads.

**Fig. 1T** Nodal displacements and forces in the local and global coordinates

The first thing in data preparation is to label members and nodes with identification numbers. Assigned in Fig. 1T are three member numbers in circles and four node members. With the node and member numbers, the frame and external loads are described as follows.

**Coordinates of Nodes**

The global coordinates of nodes of the undeformed frame can be arranged in the following form

where the ith row gives the coordinates of the ith node. Note that coordinates of different nodes are separated by semicolons (;). To save space, coordinates of two or more nodes can be put in a row as long as they are separated by semicolons. Thus, we can have

The parameters of a node are separated by space or comma. For the frame in Fig. 1T, we have the node input data as follows

**Parameters of Members**

The parameters of the frame members are arranged as follows

where E_{i}, A_{i} and I_{i} are the Young’s modulus, cross-section area and area of moment of inertia of the ith member, respectively; node_{Li} and node_{Ri} are the node numbers of the left and
right ends of the ith member. Note that the parameters of different members are separated by semicolons (;). Again, to save space,
parameters of two or more members can put in one row as long as they are separated by semicolons. The parameters of a member are separated
by space or comma. For the example in Fig. 1T, assume that all the members have the same Young’s modulus and cross-section area:
E=70GPa=70x10^{9} N/m^{2}, A = 1.8x10^{-4} m^{2}, I = 10^{-4} m^{4}.
The member data are given as follows

In the above data, 70E9 = 70x10^{9}, 1.8E-4 = 1.8x10^{-4} and 1E-4 = 10^{-4}.

**Support Specification**

Assume that the frame in consideration has r supports, which means that the frame is supported at r nodes. The supports of the frame are described by

where the first parameter of each row is the node number of a support; the second parameter is an integer identifying the type of the support as follows

Type 1: Fixed, support type = 1

Type 2: Pin or hinge, support type = 2

Type 3: Roller on a horizontal surface, support type = 3

Type 4: Roller on a vertical surface, support type = 4

Type 5: Roller on an incline with an angle α, support type = 5

Type 6: Slider on a horizontal surface, support type = 6

Type 7: Slider on a vertical surface, support type = 7

Type 8: Slider on an incline with an angle α, support type = 8

and the third parameter is zero for supports of Types 1 to 4, 6 and 7, and is the incline angle α in degrees for supports of Types 5 and 8. The parameters of different supports are separated by semicolons. As before, parameters of two or more supports can be placed in one row as long as they are separated by semicolons. For the frame in Fig. 1T, the support data are given as follows

**Load Specification**

**(i)** Assume that **external forces are applied at q nodes** of the frame. These loads are specified by

where Fx and Fy are the horizontal and vertical components of a load __in the global coordinate system__; τ is an external torque. As before, loads at different nodes can
be described in a row through use of semicolons. For the example in Fig. 1T, the load applied at Node 3 is described by

**(ii)** Assume that **external forces are applied at the interior points of members** of the frame. These loads are specified by

where the first parameter of each row is the member number; the second parameter is an integer between -1 and 4 identifying the type of loads;
p1 to p4 are parameters describing the location and form of the load __in the local coordinates (x, y) of the member__. The format of loads is as follows

As before, loads at different nodes can be described in a row through use of semicolons. For the example in Fig. 1T, the loads applied at the interior of members are described by

**2. Results**

The deformed configuration of the frame is shown as follows.

The analysis results are also shown as follows

(2a) Nodal Displacements U, V and θ

Node 1: U = 0, V = 0, θ = 0 degrees

Node 2: U = -5.0105024E-2, V = 0, θ = 0.86543061 degrees

Node 3: U = -1.0766634E-1, V = -4.1871943E-3, θ = 0.70172838 degrees

Node 4: U = 0, V = 0, θ = -0.10272351 degrees

(2b) Maximum Response of the Frame

Axial displacement: u_{max} = -1.0766634E-1, on member = 3, at x = 0

Axial force: S_{max} = 6.7829791E5, on member = 3, at x = 0

Transverse displacement: w_{max} = 1.0766634E-1, on member = 2, at x = 4

Rotation: (dw/dx)_{max} = 2.9012761E-2, on member = 1, at x = 2.48451997

Bending moment: M_{max} = 1.5995632E5, on member = 1, at x = 0

Shear force: Q_{max} = 1.1580784E5, on member = 3, at x = 0

(2c) Reactions

At node 1 (fixed): R_{x} = 1.2170209E5, R_{y} = 1.0261817E5, M_{c} = -1.5995632E5

At node 2 (roller on horizontal surface): R_{y} = 1.3189662E4

At node 4 (hinge): R_{x} = 6.7829791E5, R_{y} = -1.5807836E4

Frames are flexible structures commonly seen in buildings, bridges, highway interchanges, automobiles, machines and mechanical devices. A frame is an assembly of prismatic members that are rigidly connected at their ends. The joints at which members are interconnected and the points at which the frame is supported are called nodes. External loads are applied to the frame either at the nodes or at the interior points of the members, and as such a frame member may undergo both longitudinal and transverse deflections. Figure 1 shows a plane frame subject to a concentrated load at a node and a distributed load within a member.

**Fig. 1** A plane frame structure

A fundamental problem of a plane frame is to determine its static response (displacements at nodes, internal forces and elastic deformation of members, and reactions at supports) subject to external loads. The tool TFS-02 provides solutions to such a problem. In sequel, a mathematical model for static analysis of plane frames is presented.

A frame member is modeled as an elastic prismatic element that can sustain both transverse and longitudinal deformations; see Fig. 2 where u(x) and w(x) are the longitudinal transverse displacements of a typical member. The member displacements are governed by the differential equations

where

**Fig. 2** A frame member undergoing longitudinal transverse deflections

The internal forces of the member with sign convention are shown in Fig. 3 and which are described by

**Fig. 3** Internal forces of a member with sign convention

**2. Nodal Forces and Nodal Displacements**

A frame member is rigidly connected to two nodes (joints), say nodes i and j as shown in Fig. 2. The nodes impose displacements and rotations at the two ends of the member; see Fig. 4, where the rotations are given by

These displacements are called nodal displacements. Corresponding to these nodal displacements, nodal forces are developed at the ends of the member; see Fig. 5, where the sign convention for the nodal forces is different from that for the internal forces in Fig. 3. In other words,

**Fig. 4** Nodal displacements of a frame member

**Fig. 5** Nodal forces of a frame member

**3. Supports**

In this mathematical model of plane frames, eight types of supports for frames are considered in Fig. 6,
where R_{x}, R_{y}, R_{n} and M_{c} are the reaction forces and moments. Types T3 and T4 are special cases of T5 with α=0 and 90^{o}, respectively; Types T6 and T7 are
special cases of T8 with α=0 and 90^{o}, respectively.

**Fig. 6** Types of supports of plane frames: T1 - fixed end; T2 - pin or hinge; T3 - roller on a horizontal surface;
T4 - roller on a vertical surface; T5 - roller on an incline of angle α; T6 - slider on a horizontal surface; T7 - slider on a
vertical surface; T8 - slider on an incline of angle α;

**4. Solution Method**

Several methods are available for static solutions of plane frames, including the stiffness method (also called the displacement method), the finite element method (FEM) and the Distributed Transfer Function Method (DTFM). Tool TFS-01 adopts the DTFM as it delivers exact analytical solutions without approximation (see Reference 5 for more detail). Static analysis of a frame structure by the DTFM takes the following four steps.

**Step 1.** Derivation of state space formulation for member equations

In the DTFM, the governing equations (1) and (2) for a frame member are cast in spatial state form

where the state vectors are defined as

the state matrices are given by

and {e_{2}} = (0 1)^{T}, {e_{4}} = (0 0 0 1)^{T}. The boundary conditions for a member connected to nodes i and j are given by

with

where the nodal displacements are shown in Fig. 4.

**Step 2.** Determination of member stiffness matrices and equivalent nodal forces

The solutions of Eqs. (10) and (11) take the form (Reference 5)

where

for j = 1, 2. The exponential matrices in Eq. (13) are

The G_{j}(x, ξ) and H_{j}(x) are the Green’s functions and boundary influence functions of the member, respectively.

By Eqs. (3) the internal forces of the member can be expressed by

With Eqs. (12), (13) and (15), it can be shown that the nodal forces and nodal displacements are related by

where the nodal force and nodal displacement vectors are given by

the member stiffness matrix is a six-by-six stiffness matrix of the member of the form

and the transmitted force vector due to external forces applied at the interior points of the member is

with

The integrals in Eq. (20) can be obtained by exact quadrature. It can be shown that the member stiffness matrix in Eq. (18) is the same as that obtained by the stiffness method

**Step 3.** Formation of global equilibrium equation and determination of static response

The formation of a global equilibrium equation is the same as that in FEM. In other word, by coordinate transformation of Eq. (16) into the global coordinate system and through force balance at all the nodes of the frame, a global equilibrium equation is obtained as follows

where

{U} = vector of all nodal displacements

{F} = vector consisting of external loads and transmitted forces at nodes

[K] = global stiffness matrix assembled from member stiffness matrices

**Step 4.** Determination of frame response

Before determination of the frame response, the conditions of the frame supports are introduced, to eliminate all known displacements at the supports from {U}, the corresponding entries from {F}, and the corresponding rows and columns from [K]. This yields a modified equilibrium equation

where {U} only contains independent unknown nodal displacements. It follows that

Substitution of {U} into Eqs. (12) and (15) gives the static response of every member and the reactions of the frame.

1. Hibbeler, R. C., 2001, Structural Analysis, 5th edition, Prentice Hall, Upper Saddle River, New Jersey.

2. Kassimali, A., 1999, Matrix Analysis of Structures, Brooks Cole.

3. Mikhelson, I., 2004, Structural Engineering Formulas, McGraw-Hill, New York.

4. West, H. H., and Geschwindner, L. F., 2002, Fundamentals of Structural Analysis, 2nd edition, Wiley Text Books.

5. Yang, B., 2005, Stress, Strain, and Structural Dynamics: An Interactive Handbook of Formulas, Solutions, and MATLAB Toolboxes, Elsevier Science, Boston.