김경종
25 KiB
Figure 4–18 Shear force (a) and bending moment (b) diagrams for each element
Example 4.5
To illustrate the effects of shear deformation along with the usual bending deformation, we now solve the simple beam shown in Figure 4–19. We will use the beam stiffness matrix given by Eq. (4.1.15o) that includes both the bending and shear deformation contributions for deformation in the xy plane. The beam is simply supported with a concentrated load of 10,000 N applied at mid-span. We let material properties be E ¼ 207 GPa and G = 8 0 \mathrm { G P a } . The beam width and height are b ¼ 25 mm and h ¼ 50 mm, respectively.
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P = 10,000 N 200 mm 400 mm h b
Figure 4–19 Simple beam subjected to concentrated load at center of span
We will use symmetry to simplify the solution. Therefore, only one half of the beam will be considered with the slope at the center forced to be zero. Also, one half of the concentrated load is then used. The model with symmetry enforced is shown in Figure 4–20.
The finite element model will consist of only one beam element. Using Eq. (4.1.15o) for the Timoshenko beam element stiffness matrix, we obtain the global
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P/2 1 ① 200 mm
Figure 4–20 Beam with symmetry enforced
equations as
\frac {E I}{L ^ {3} (1 + \varphi)} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & (4 + \varphi) L ^ {2} & - 6 L & (2 - \varphi) L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & (2 - \varphi) L ^ {2} & - 6 L & (4 + \varphi) L ^ {2} \end{array} \right] \left\{ \begin{array}{c} d _ {1 y} = 0 \\ \phi_ {1} \\ d _ {2 y} \\ \phi_ {2} = 0 \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 y} \\ 0 \\ - P / 2 \\ 0 \end{array} \right\} \tag {4.3.34}
Note that the boundary conditions given by d _ { \mathrm { l } y } = 0 and \phi _ { 2 } = 0 have been included in Eq. (4.3.34).
Using the second and third equations of Eq. (4.3.34) whose rows are associated with the two unknowns, \phi _ { 1 } and d _ { 2 y } , we obtain
d _ {2 y} = \frac {- P L ^ {3} (4 + \varphi)}{2 4 E I} \quad \text { and } \quad \phi_ {1} = \frac {- P L ^ {2}}{4 E I} \tag {4.3.35}
As the beam is rectangular in cross section, the moment of inertia is
I = b h ^ {3} / 1 2
Substituting the numerical values for b and h, we obtain I as
I = 0. 2 6 \times 1 0 ^ {- 6} \mathrm{m} ^ {4}
The shear correction factor is given by
\varphi = \frac {1 2 E I}{k _ {s} A G L ^ {2}}
and k _ { s } for a rectangular cross section is given by k _ { s } = 5 / 6
Substituting numerical values for E , I , G , L . ; and k _ { s } , , we obtain
\varphi = \frac {1 2 \times 2 0 7 \times 1 0 ^ {9} \times 0 . 2 6 \times 1 0 ^ {- 6}}{5 / 6 \times 0 . 0 2 5 \times 0 . 0 5 \times 8 0 \times 1 0 ^ {9} \times 0 . 2 ^ {2}} = 0. 1 9 3 8
Substituting for P = 1 0 { , } 0 0 0 \ \mathrm { N } , L = 0 . 2 { \mathrm { ~ m ~ } } , and \varphi = 0 . 1 9 3 8 into Eq. (4.3.35), we obtain the displacement at the mid-span as
d _ {2 y} = - 2. 5 9 7 \times 1 0 ^ {- 4} \mathrm{m} \tag {4.3.36}
If we let l ¼ the whole length of the beam, then l = 2 L and we can substitute L = l / 2 into Eq. (4.3.35) to obtain the displacement in terms of the whole length of the beam as
d _ {2 y} = \frac {- P l ^ {3} (4 + \varphi)}{1 9 2 E I} \tag {4.3.37}
For long slender beams with l about 10 or more times the beam depth, h, the transverse shear correction term \varphi is small and can be neglected. Therefore, Eq. (4.3.37) becomes
d _ {2 y} = \frac {- P l ^ {3}}{4 8 E I} \tag {4.3.38}
Equation (4.3.38) is the classical beam deflection formula for a simply supported beam subjected to a concentrated load at mid-span.
Using Eq. (4.3.38), the deflection is obtained as
d _ {2 y} = 2. 4 7 4 \times 1 0 ^ {- 4} \mathrm{m} \tag {4.3.39}
Comparing the deflections obtained using the shear-correction factor with the deflection predicted using the beam-bending contribution only, we obtain
\% \text { change } = \frac {2 . 5 9 7 - 2 . 4 7 4}{2 . 4 7 4} \times 1 0 0 = 4. 9 7 \% \text { difference }
4.4 Distributed Loading
Beam members can support distributed loading as well as concentrated nodal loading. Therefore, we must be able to account for distributed loading. Consider the fixed-fixed beam subjected to a uniformly distributed loading w shown in Figure 4–21. The reactions, determined from structural analysis theory [2], are shown in Figure 4–22. These reactions are called fixed-end reactions. In general, fixed-end reactions are those reactions at the ends of an element if the ends of the element are assumed to be fixed—that is, if displacements and rotations are prevented. (Those of you who are unfamiliar with the analysis of indeterminate structures should assume these reactions as given and proceed with the rest of the discussion; we will develop these results in a subsequent presentation of the work-equivalence method.) Therefore, guided by the results from structural analysis for the case of a uniformly distributed load, we replace the load by concentrated nodal forces and moments tending to have the same
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w(lb/ft) L
Figure 4–21 Fixed-fixed beam subjected to a uniformly distributed load
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M = \frac{wL^2}{12} \quad M = \frac{wL^2}{12} \n \frac{wL}{2} \quad L \quad \frac{wL}{2}
Figure 4–22 Fixed-end reactions for the beam of Figure 4–20
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w 1 2 L 3 4
(a)
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wL²/12 wL/2 L wL/2 wL²/12
(b)
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wℓ/2 + wℓ/2 wℓ²/12 wℓ²/12 wℓ²/12 wℓ²/12 wℓ/2 1 2 ℓ 5 ℓ 3 4
Figure 4–23 (a) Beam with a distributed load, (b) the equivalent nodal force system, and (c) the enlarged beam (for clarity’s sake) with equivalent nodal force system when node 5 is added to the midspan
effect on the beam as the actual distributed load. Figure 4–23 illustrates this idea for a beam. We have replaced the uniformly distributed load by a statically equivalent force system consisting of a concentrated nodal force and moment at each end of the member carrying the distributed load. That is, both the statically equivalent concentrated nodal forces and moments and the original distributed load have the same resultant force and same moment about an arbitrarily chosen point. These statically equivalent forces are always of opposite sign from the fixed-end reactions. If we want to analyze the behavior of loaded member 2–3 in better detail, we can place a node at midspan and use the same procedure just described for each of the two elements representing the horizontal member. That is, to determine the maximum deflection and maximum moment in the beam span, a node 5 is needed at midspan of beam segment 2–3, and work-equivalent forces and moments are applied to each element (from node 2 to node 5 and from node 5 to node 3) shown in Figure 4–23 (c).
Work-Equivalence Method
We can use the work-equivalence method to replace a distributed load by a set of discrete loads. This method is based on the concept that the work of the distributed load wðx^Þ in going through the displacement field v^ðx^Þ is equal to the work done by nodal loads \dot { f } _ { i y } and \hat { m } _ { i } in going through nodal displacements \hat { d } _ { i y } and \hat { \phi } _ { i } for arbitrary nodal displacements. To illustrate the method, we consider the example shown in Figure 4–24. The work due to the distributed load is given by
W _ {\text { distributed }} = \int_ {0} ^ {L} w (\hat {x}) \hat {v} (\hat {x}) d \hat {x} \tag {4.4.1}
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ŷ, v̂ w(x̂) 1 2 x̂ L
(a)
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f̂₁y, d̂₁y m̂₁, φ̂₁ 1 L 2 f̂₂y, d̂₂y m̂₂, φ̂₂
(b)
Figure 4–24 (a) Beam element subjected to a general load and (b) the statically equivalent nodal force system
where \hat { v } ( \hat { x } ) is the transverse displacement given by Eq. (4.1.4). The work due to the discrete nodal forces is given by
W _ {\text { discrete }} = \hat {m} _ {1} \hat {\phi} _ {1} + \hat {m} _ {2} \hat {\phi} _ {2} + \hat {f} _ {1 y} \hat {d} _ {1 y} + \hat {f} _ {2 y} \hat {d} _ {2 y} \tag {4.4.2}
We can then determine the nodal moments and forces \hat { m } _ { 1 } , \hat { m } _ { 2 } , \hat { f } _ { 1 y } , , and \hat { f } _ { 2 y } used to replace the distributed load by using the concept of work equivalence—that is, by setting W _ { \mathrm { d i s t r i b u t e d } } = W _ { \mathrm { d i s c r e t e } } for arbitrary displacements \hat { \phi } _ { 1 } , \hat { \phi } _ { 2 } , \hat { d } _ { 1 y } , and \hat { d } _ { 2 y } .
Example of Load Replacement
To illustrate more clearly the concept of work equivalence, we will now consider a beam subjected to a specified distributed load. Consider the uniformly loaded beam shown in Figure 4–25(a). The support conditions are not shown because they are not relevant to the replacement scheme. By letting W _ { \mathrm { d i s c r e t e } } = W _ { \mathrm { d i s t r i b u t e d } } and by assuming arbitrary \hat { \phi } _ { 1 } , \hat { \phi } _ { 2 } , \hat { d } _ { 1 y } , and \hat { d } _ { 2 y } , we will find equivalent nodal forces \hat { m } _ { 1 } , \hat { m } _ { 2 } , \hat { f } _ { 1 y } , and \hat { f } _ { 2 y } . Figure 4–25(b) shows the nodal forces and moments directions as positive based on Figure 4–1.
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w 1 2 L
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f̂₁ᵧ, d̂₁ᵧ φ̂₁, m̂₁ 1 2 L f̂₂ᵧ, d̂₂ᵧ m̂₂, φ̂₂
(b)
Figure 4–25 (a) Beam subjected to a uniformly distributed loading and (b) the equivalent nodal forces to be determined
Using Eqs. (4.4.1) and (4.4.2) for W _ { \mathrm { d i s t r i b u t e d } } = W _ { \mathrm { d i s c r e t e } } . , we have
\int_ {0} ^ {L} w (\hat {x}) \hat {v} (\hat {x}) d \hat {x} = \hat {m} _ {1} \hat {\phi} _ {1} + \hat {m} _ {2} \hat {\phi} _ {2} + \hat {f} _ {1 y} \hat {d} _ {1 y} + \hat {f} _ {2 y} \hat {d} _ {2 y} \tag {4.4.3}
where \hat { m } _ { 1 } \hat { \phi } _ { 1 } and \hat { m } _ { 2 } \hat { \phi } _ { 2 } are the work due to concentrated nodal moments moving through their respective nodal rotations and \hat { f } _ { 1 y } \hat { d } _ { 1 y } and \hat { f } _ { 2 y } \hat { d } _ { 2 y } are the work due to the nodal forces moving through nodal displacements. Evaluating the left-hand side of
Eq. (4.4.3) by substituting w ( \hat { x } ) = - w and v^ðx^Þ from Eq. (4.1.4), we obtain the work due to the distributed load as
\begin{array}{l} \int_ {0} ^ {L} w (\hat {x}) \hat {v} (\hat {x}) d \hat {x} = - \frac {L w}{2} \left(\hat {d} _ {1 y} - \hat {d} _ {2 y}\right) - \frac {L ^ {2} w}{4} \left(\hat {\phi} _ {1} + \hat {\phi} _ {2}\right) - L w \left(\hat {d} _ {2 y} - \hat {d} _ {1 y}\right) \\ + \frac {L ^ {2} w}{3} (2 \hat {\phi} _ {1} + \hat {\phi} _ {2}) - \hat {\phi} _ {1} \left(\frac {L ^ {2} w}{2}\right) - \hat {d} _ {1 y} (w L) \tag {4.4.4} \\ \end{array}
Now using Eqs. (4.4.3) and (4.4.4) for arbitrary nodal displacements, we let \hat { \phi } _ { 1 } = 1 ; \hat { \phi } _ { 2 } = 0 , \hat { d } _ { 1 y } = 0 , and \hat { d } _ { 2 y } = 0 and then obtain
\hat {m} _ {1} (1) = - \left(\frac {L ^ {2} w}{4} - \frac {2}{3} L ^ {2} w + \frac {L ^ {2}}{2} w\right) = - \frac {w L ^ {2}}{1 2} \tag {4.4.5}
Similarly, letting \hat { \phi } _ { 1 } = 0 , \hat { \phi } _ { 2 } = 1 , \hat { d } _ { 1 y } = 0 , and \hat { d } _ { 2 y } = 0 yields
\hat {m} _ {2} (1) = - \left(\frac {L ^ {2} w}{4} - \frac {L ^ {2} w}{3}\right) = \frac {w L ^ {2}}{1 2} \tag {4.4.6}
Finally, letting all nodal displacements equal zero except first \hat { d } _ { 1 y } and then \hat { d } _ { 2 y } , we obtain
\hat {f} _ {1 y} (1) = - \frac {L w}{2} + L w - L w = - \frac {L w}{2} \tag {4.4.7}
\hat {f} _ {2 y} (1) = \frac {L w}{2} - L w = - \frac {L w}{2}
We can conclude that, in general, for any given load function w ( \hat { x } ) , we can multiply by v^ðx^Þ and then integrate according to Eq. (4.4.3) to obtain the concentrated nodal forces (and/or moments) used to replace the distributed load. Moreover, we can obtain the load replacement by using the concept of fixed-end reactions from structural analysis theory. Tables of fixed-end reactions have been generated for numerous load cases and can be found in texts on structural analysis such as Reference [2]. A table of equivalent nodal forces has been generated in Appendix D of this text, guided by the fact that fixed-end reaction forces are of opposite sign from those obtained by the work equivalence method.
Hence, if a concentrated load is applied other than at the natural intersection of two elements, we can use the concept of equivalent nodal forces to replace the concentrated load by nodal concentrated values acting at the beam ends, instead of creating a node on the beam at the location where the load is applied. We provide examples of this procedure for handling concentrated loads on elements in beam Example 4.7 and in plane frame Example 5.3.
General Formulation
In general, we can account for distributed loads or concentrated loads acting on beam elements by starting with the following formulation application for a general structure:
\underline {{F}} = \underline {{K}} \underline {{d}} - \underline {{F}} _ {o} \tag {4.4.8}
where \underline { { F } } are the concentrated nodal forces and \underline { { F } } _ { o } are called the equivalent nodal forces, now expressed in terms of global-coordinate components, which are of such magnitude that they yield the same displacements at the nodes as would the distributed load. Using the table in Appendix D of equivalent nodal forces \underline { { \hat { f } } } _ { o } expressed in terms of localcoordinate components, we can express \underline { { F } } _ { o } in terms of global-coordinate components.
Recall from Section 3.10 the derivation of the element equations by the principle of minimum potential energy. Starting with Eqs. (3.10.19) and (3.10.20), the minimization of the total potential energy resulted in the same form of equation as Eq. (4.4.8) where \underline { { F } } _ { o } now represents the same work-equivalent force replacement system as given by Eq. (3.10.20a) for surface traction replacement. Also, \underline { { F } } = \underline { { P } } \left[ \underline { { P } } \right. from Eq. (3.10.20)] represents the global nodal concentrated forces. Because we now assume that concentrated nodal forces are not present ( \underline { { F } } = 0 ) , as we are solving beam problems with distributed loading only in this section, we can write Eq. (4.4.8) as
\underline {{F}} _ {o} = \underline {{K}} \underline {{d}} \tag {4.4.9}
On solving for \underline { d } in Eq. (4.4.9) and then substituting the global displacements \underline { d } and equivalent nodal forces \underline { { F } } _ { o } into Eq. (4.4.8), we obtain the actual global nodal forces \underline { { F } } . For example, using the definition of \underline { { \hat { f } } } _ { o } and Eqs. (4.4.5)–(4.4.7) (or using load case 4 in Appendix D) for a uniformly distributed load w acting over a one-element beam, we have
\underline {{F}} _ {o} = \left\{ \begin{array}{l} \frac {- w L}{2} \\ \frac {- w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.10}
This concept can be applied on a local basis to obtain the local nodal forces \underline { { \hat { f } } } in individual elements of structures by applying Eq. (4.4.8) locally as
\underline {{\hat {f}}} = \underline {{\hat {k}}} \underline {{\hat {d}}} - \underline {{\hat {f}}} _ {o} \tag {4.4.11}
where \underline { { \hat { f } } } _ { o } are the equivalent local nodal forces.
Examples 4.6–4.8 illustrate the method of equivalent nodal forces for solving beams subjected to distributed and concentrated loadings. We will use globalcoordinate notation in Examples 4.6–4.8—treating the beam as a general structure rather than as an element.
Example 4.6
For the cantilever beam subjected to the uniform load w in Figure 4–26, solve for the right-end vertical displacement and rotation and then for the nodal forces. Assume the beam to have constant EI throughout its length.
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y w L x
(a)
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-wL/2 1 ① L wL²/12 wL²/12 wL/2
(b)
Figure 4–26 (a) Cantilever beam subjected to a uniformly distributed load and (b) the work equivalent nodal force system
We begin by discretizing the beam. Here only one element will be used to represent the whole beam. Next, the distributed load is replaced by its work-equivalent nodal forces as shown in Figure 4–26(b). The work-equivalent nodal forces are those that result from the uniformly distributed load acting over the whole beam given by Eq. (4.4.10). (Or see appropriate load case 4 in Appendix D.) Using Eq. (4.4.9) and the beam element stiffness matrix, and realizing \underline { { \hat { k } } } = \underline { { k } } as the local x^ axis is coincident with the global x axis, we obtain
\frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ & 4 L ^ {2} & - 6 L ^ {2} & 2 L ^ {2} \\ & & 1 2 & - 6 L \\ & & & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 y} \\ \phi_ {1} \\ d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 y} - \frac {w L}{2} \\ M _ {1} - \frac {w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.12}
where we have applied the work equivalent nodal forces and moments from Figure 4–26(b).
Applying the boundary conditions d _ { \mathrm { l } y } = 0 and \phi _ { 1 } = 0 to Eqs (4.4.12) and then partitioning off the third and fourth equations of Eq. (4.4.12), we obtain
\frac {E I}{L ^ {3}} \left[ \begin{array}{c c} 1 2 & - 6 L \\ - 6 L ^ {2} & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{l} - \frac {w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.13}
Solving Eq. (4.4.13) for the displacements, we obtain
\left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \frac {L}{6 E I} \left[ \begin{array}{l l} 2 L ^ {2} & 3 L \\ 3 L & 6 \end{array} \right] \left\{ \begin{array}{l} \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.14a}
Simplifying Eq. (4.4.14a), we obtain the displacement and rotation as
\left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {- w L ^ {4}}{8 E I} \\ \frac {- w L ^ {3}}{6 E I} \end{array} \right\} \tag {4.4.14b}
The negative signs in the answers indicate that d _ { 2 y } is downward and \phi _ { 2 } is clockwise. In this case, the method of replacing the distributed load by discrete concentrated loads gives exact solutions for the displacement and rotation as could be obtained by classical methods, such as double integration [1]. This is expected, as the workequivalence method ensures that the nodal displacement and rotation from the finite element method match those from an exact solution.
We will now illustrate the procedure for obtaining the global nodal forces. For convenience, we first define the product { \underline { { K d } } } to be \underline { { F } } ^ { ( e ) } , where \underline { { F } } ^ { ( e ) } are called the effective global nodal forces. On using Eq. (4.4.14) for \underline { { d } } , we then have
\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ \frac {- w L ^ {4}}{8 E I} \\ \frac {- w L ^ {3}}{6 E I} \end{array} \right\} \tag {4.4.15}
Simplifying Eq. (4.4.15), we obtain
\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \left\{ \begin{array}{l} \frac {w L}{2} \\ \frac {5 w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.16}
We then use Eqs. (4.4.10) and (4.4.16) in Eq. (4.4.8) ( \underline { { F } } = \underline { { k } } \ : \underline { { d } } - \underline { { F } } _ { o } ) to obtain the correct global nodal forces as
\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {w L}{2} \\ \frac {5 w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} - \left\{ \begin{array}{l} \frac {- w L}{2} \\ \frac {- w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} = \left\{ \begin{array}{l} w L \\ \frac {w L ^ {2}}{2} \\ 0 \\ 0 \end{array} \right\} \tag {4.4.17}
In Eq. (4.4.17), F _ { 1 y } is the vertical force reaction and M _ { 1 } is the moment reaction as applied by the clamped support at node 1. The results for displacement given by Eq. (4.4.14b) and the global nodal forces given by \operatorname { E q . } (4.4.17) are sufficient to complete the solution of the cantilever beam problem.
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M₁ = \frac{wL^2}{2} 1 w L F₁y = wL 2
\sum F _ {y} \colon F _ {1 y} - w L = 0
\sum M _ {1} \colon \frac {w L ^ {2}}{2} - (w L) \frac {L}{2} = 0
Figure 4–26 (c) Free-body diagram and equations of equilibrium for beam of Figure 4–(26)a.
A free-body diagram of the beam using the reactions from Eq. (4.4.17) verifies both force and moment equilibrium as shown in Figure 4–26(c).
The nodal force and moment reactions obtained by Eq. (4.4.17) illustrate the importance of using Eq. (4.4.8) to obtain the correct global nodal forces and moments. By subtracting the work-equivalent force matrix, \underline { { F } } _ { o } from the product of K times d, we obtain the correct reactions at node 1 as can be verified by simple static equilibrium equations. This verification validates the general method as follows:
- Replace the distributed load by its work-equivalent as shown in Figure
4 - 2 6 ( \mathsf b )to identify the nodal force and moment used in the solution. - Assemble the global force and stiffness matrices and global equations illustrated by Eq. (4.4.12).
- Apply the boundary conditions to reduce the set of equations as done in previous problems and illustrated by Eq. (4.4.13) where the original four equations have been reduced to two equations to be solved for the unknown displacement and rotation.
- Solve for the unknown displacement and rotation given by Eq. (4.4.14a) and Eq. (4.4.14b).
- Use Eq. (4.4.8) as illustrated by Eq. (4.4.17) to obtain the final correct global nodal forces and moments. Those forces and moments at supports, such as the left end of the cantilever in Figure 4–26(a), will be the reactions.
We will solve the following example to illustrate the procedure for handling concentrated loads acting on beam elements at locations other than nodes.
Example 4.7
For the cantilever beam subjected to the concentrated load P in Figure 4–27, solve for the right-end vertical displacement and rotation and the nodal forces, including reactions, by replacing the concentrated load with equivalent nodal forces acting at each end of the beam. Assume EI constant throughout the beam.
We begin by discretizing the beam. Here only one element is used with nodes at each end of the beam. We then replace the concentrated load as shown in