Indeterminate Trusses & Industrial Frames - 2

Example 35.2
Determine bar forces in the 3-panel truss of the previous example (shown in Fig. 35.2a) assuming that the diagonals can carry only tensile forces.

Solution:
When the diagonals can resist only tension, any diagonal that would take compression in the exact solution must be assumed to carry zero force. Therefore the panel shear must be entirely resisted by the tensile diagonal(s). The support reactions of the truss remain the same as in the previous example and are given by,

Consider the equilibrium of the portion of the truss left of section A-A. The free-body diagram of this part is shown in Fig. 35.3a.

Apply the vertical equilibrium ∑F_y = 0 to the free body. The algebraic expression for vertical equilibrium is shown in the figure:

(2)

From the geometry of the truss it is easily seen that

Consider the vertical equilibrium of joint L₀. From the joint equilibrium we obtain

(3)

Because the diagonals are inclined at 45° to the horizontal, the vertical and horizontal components of any diagonal force are equal. This relation simplifies component equations and is used repeatedly in the following steps.

Now take equilibrium of the truss left of section C-C (refer to Fig. 35.3b). In this panel the shear to be carried is 3.33 kN. Considering vertical equilibrium of the free body yields

(4)

Take moments of all forces about U₁.

(6)

Consider the joint equilibrium at L₁ (see Fig. 35.3c). From the two component equilibrium equations at that joint we obtain

(7)

Now consider the equilibrium of the right side of section B-B (refer to Fig. 35.3d). Using equilibrium relations for that part, the forces in the members of the 3rd panel are evaluated and expressed by the relations below.

(11)

(12)

The complete set of member forces obtained by the approximate tensile-only diagonal assumption is shown schematically in Fig. 35.3f. For reference, bar forces from the exact (full) analysis are shown in brackets on the same diagram so that comparison is possible.

Industrial frames and portals

Common types of industrial frames are shown in Figs. 35.4a and 35.4b. These structures commonly consist of two columns and a roof truss spanning between the columns. They are normally subjected to vertical loads (dead and live) and horizontal loads (wind). For analysis it is convenient to adopt different support and connection assumptions for vertical and horizontal load cases.

When analysing gravity (vertical) loads it is usual to assume that the roof truss is simply supported on the column heads; each end of the truss rests on the column but is free to rotate. When analysing horizontal (wind) loads the truss is assumed to be rigidly connected to the columns so that moment transfer between truss and columns is possible. The column bases may be assumed hinged or fixed depending on the foundation stiffness: small concrete footings justify hinged bases while massive foundations justify fixed bases.

Before analysing structures for wind loads, consider portal frames that are used both as industrial end frames and as end portals of bridges (see Fig. 35.5). Portals behave similarly to industrial trusses and again may be assumed fixed or hinged at their bases according to foundation stiffness.

Consider a portal hinged at the bases, as in Fig. 35.5a. This portal is statically indeterminate to degree one (one redundant reaction). To solve it by static equilibrium alone an assumption is needed. If the two columns have nearly equal stiffness, it is standard to assume that the horizontal shear at the base of each column is equal. If the column stiffnesses differ, assume the shear at each base is proportional to that column's stiffness. This proportional-stiffness assumption is an approximation based on elastic behaviour and usually gives good engineering estimates.

Reactions and bending moments (hinged bases)

Under the equal-shear assumption (for equal column stiffness) the shear at the base of each column is given by the relation shown in Fig. 35.6.

Taking moments about hinge D gives the equations depicted below.

And

The bending moment diagram corresponding to this assumed distribution is shown in Fig. 35.7.

From the moment diagram it is clear that an imaginary hinge forms at the midpoint of the girder. Therefore the assumption that shear at the column bases is equal is equivalent to assuming a hinge at the mid-span of the girder; both yield the same internal moment pattern for this simplified analysis.

Fixed bases and multiple indeterminacy

Now consider a portal frame fixed at the bases as shown in Fig. 35.5b. This frame is statically indeterminate to third degree and three independent assumptions are required to solve by equilibrium alone. Again, if column stiffnesses are equal, a practical assumption is that the shear at the base of each column is equal.

The deformed shape of a hinged-base portal is shown in Fig. 35.8a and the deformed shape of a typical industrial frame is shown in Fig. 35.8b. In the fixed-base case, the bending moment at the column base produces tension at the outside fibres of the column cross-section, whereas the bending moment at the top of the column produces tension on the inside fibres. Hence the bending moment changes sign somewhere along the column height and there is an inflection point where moment is zero. For approximate analysis this inflection point is usually taken near mid-height of the column. Using this and similar assumptions yields three independent relations that permit estimation of reactions and internal moments.

In many practical industrial frames the inflection points are assumed to occur at mid-height between A and B on the columns.

Take moments of all forces to the left of hinge 1 about hinge 1 (see Fig. 35.9a). The resulting relation is shown below.

Similarly, taking moments of all forces to the left of hinge 2 about hinge 2 yields

Taking moments of all forces to the right of hinge 1 about hinge 1 gives

Similarly, taking moments of forces to the right of hinge 2 about hinge 2 gives

The bending moment diagram for this assumed hinge and inflection-point arrangement is shown in Fig. 35.9b.

If the column base is partially fixed the inflection point is commonly assumed at one-third the column height measured from the base. Note that when the base is hinged the inflection point occurs at the support, while for a fully fixed base it is assumed to occur at mid-height. These assumptions provide a practical set of constraints to reduce the degrees of indeterminacy.

Example 35.3

Example 35.3
Determine approximately forces in the member of a truss portal shown in Fig. 35.10a.

Taking moments of all forces to the right of hinge 2 about hinge 2 gives the relation shown below.

(2)

Similarly, the moment M_A at the support A is

M_A = kN·m 20

(3)

Taking moments of all forces to the right of hinge 1 about hinge 1 gives

Similarly, another moment relation follows:

(4)

Member forces in the truss may be calculated either by the method of sections or the method of joints. For instance, consider equilibrium of the truss portion left of section A-A as shown in Fig. 35.10d.

(5)

Taking moments about U₀ yields the relation shown below.

(6)

Taking moments about L₁ gives

(7)

Summary

Before performing exact indeterminate analysis (stiffness method, flexibility/force method, etc.), one requires information about relative stiffnesses and material properties of members. Such detailed data are typically not available during preliminary design. Therefore approximate methods, based on reasonable assumptions about stiffness distribution and likely deformation shapes, are used to reduce indeterminate structures to statically determinate ones. These assumptions are chosen to reflect the expected structural behaviour under the applied loads. The determinate reduced structure is then solved by equations of statics. When the assumptions are valid, the approximate results compare favourably with exact methods, as demonstrated by the numerical examples in this chapter.