김경종
26 KiB
On considering the third, fourth, and sixth equations of Eqs. (4.2.3) corresponding to the rows with unknown degrees of freedom and using Eqs. (4.2.4), we obtain
\left\{ \begin{array}{c} - 1 0 0 0 \\ 1 0 0 0 \\ 0 \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c} 2 4 & 0 & 6 L \\ 0 & 8 L ^ {2} & 2 L ^ {2} \\ 6 L & 2 L ^ {2} & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} d _ {2 y} \\ \phi_ {2} \\ \phi_ {3} \end{array} \right\} \tag {4.2.5}
where F _ { 2 y } = - 1 0 0 0 lb, M _ { 2 } = 1 0 0 0 1 \mathrm { b } { \cdot } \mathrm { f t } , and M _ { 3 } = 0 have been substituted into the reduced set of equations. We could now solve Eq. (4.2.5) simultaneously for the unknown nodal displacement d _ { 2 y } and the unknown nodal rotations \phi _ { 2 } and \phi _ { 3 } . We leave the final solution for you to obtain. Section 4.3 provides complete solutions to beam problems.
4.3 Examples of Beam Analysis Using the Direct Stiffness Method
We will now perform complete solutions for beams with various boundary supports and loads to illustrate further the use of the equations developed in Section 4.1.
Example 4.1
Using the direct stiffness method, solve the problem of the propped cantilever beam subjected to end load P in Figure 4–8. The beam is assumed to have constant EI and length 2L. It is supported by a roller at midlength and is built in at the right end.
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P L L 1 2 3
Figure 4–8 Propped cantilever beam
We have discretized the beam and established global coordinate axes as shown in Figure 4–8. We will determine the nodal displacements and rotations, the reactions, and the complete shear force and bending moment diagrams.
Using Eq. (4.1.14) for each element, along with superposition, we obtain the structure total stiffness matrix by the same method as described in Section 4.2 for obtaining the stiffness matrix in Eq. (4.2.3). The \underline { { K } } is
\underline {{{K}}} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} d _ {1 y} & \phi_ {1} & d _ {2 y} & \phi_ {2} & d _ {3 y} & \phi_ {3} \\ 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 \\ & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 \\ & & 1 2 + 1 2 & - 6 L + 6 L & - 1 2 & 6 L \\ & & & 4 L ^ {2} + 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ & & & & 1 2 & - 6 L \\ \text { Symmetry } & & & & & 4 L ^ {2} \end{array} \right] \tag {4.3.1}
The governing equations for the beam are then given by
\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 y} \\ M _ {3} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 \\ - 1 2 & - 6 L & 2 4 & 0 & - 1 2 & 6 L \\ 6 L & 2 L ^ {2} & 0 & 8 L ^ {2} & - 6 L & 2 L ^ {2} \\ 0 & 0 & - 1 2 & - 6 L & 1 2 & - 6 L \\ 0 & 0 & 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 y} \\ \phi_ {1} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 y} \\ \phi_ {3} \end{array} \right\} \tag {4.3.2}
On applying the boundary conditions
d _ {2 y} = 0 \quad d _ {3 y} = 0 \quad \phi_ {3} = 0 \tag {4.3.3}
and partitioning the equations associated with unknown displacements [the first, second, and fourth equations of Eqs. (4.3.2)] from those equations associated with known displacements in the usual manner, we obtain the final set of equations for a longhand solution as
\left\{ \begin{array}{c} - P \\ 0 \\ 0 \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c} 1 2 & 6 L & 6 L \\ 6 L & 4 L ^ {2} & 2 L ^ {2} \\ 6 L & 2 L ^ {2} & 8 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} d _ {1 y} \\ \phi_ {1} \\ \phi_ {2} \end{array} \right\} \tag {4.3.4}
where F _ { 1 \nu } = - P , M _ { 1 } = 0 , and M _ { 2 } = 0 have been used in Eq. (4.3.4). We will now solve Eq. (4.3.4) for the nodal displacement and nodal slopes. We obtain the transverse displacement at node 1 as
d _ {1 y} = - \frac {7 P L ^ {3}}{1 2 E I} \tag {4.3.5}
where the minus sign indicates that the displacement of node 1 is downward.
The slopes are
\phi_ {1} = \frac {3 P L ^ {2}}{4 E I} \quad \phi_ {2} = \frac {P L ^ {2}}{4 E I} \tag {4.3.6}
where the positive signs indicate counterclockwise rotations at nodes 1 and 2.
We will now determine the global nodal forces. To do this, we substitute the known global nodal displacements and rotations, Eqs. (4.3.5) and (4.3.6), into Eq. (4.3.2). The resulting equations are
\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 y} \\ M _ {3} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 \\ - 1 2 & - 6 L & 2 4 & 0 & - 1 2 & 6 L \\ 6 L & 2 L ^ {2} & 0 & 8 L ^ {2} & - 6 L & 2 L ^ {2} \\ 0 & 0 & - 1 2 & - 6 L & 1 2 & - 6 L \\ 0 & 0 & 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} - \frac {7 P L ^ {3}}{1 2 E I} \\ \frac {3 P L ^ {2}}{4 E I} \\ 0 \\ \frac {P L ^ {2}}{4 E I} \\ 0 \\ 0 \end{array} \right\} \tag {4.3.7}
Multiplying the matrices on the right-hand side of Eq. (4.3.7), we obtain the global nodal forces and moments as
F _ {1 y} = - P \quad M _ {1} = 0 \quad F _ {2 y} = \frac {5}{2} P \tag {4.3.8}
M _ {2} = 0 \quad F _ {3 y} = - \frac {3}{2} P \quad M _ {3} = \frac {1}{2} P L
The results of Eqs. (4.3.8) can be interpreted as follows: The value of F _ { 1 \nu } = - P is the applied force at node 1, as it must be. The values of F _ { 2 y } , F _ { 3 y } , , and M _ { 3 } are the reactions from the supports as felt by the beam. The moments M _ { 1 } and M _ { 2 } are zero because no applied or reactive moments are present on the beam at node 1 or node 2.
It is generally necessary to determine the local nodal forces associated with each element of a large structure to perform a stress analysis of the entire structure. We will thus consider the forces in element 1 of this example to illustrate this concept (element 2 can be treated similarly). Using Eqs. (4.3.5) and (4.3.6) in the \underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { \hat { d } } } equation for element 1 [also see Eq. (4.1.13)], we have
\left\{ \begin{array}{l} \hat {f} _ {1 y} \\ \hat {m} _ {1} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \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} - \frac {7 P L ^ {3}}{1 2 E I} \\ \frac {3 P L ^ {2}}{4 E I} \\ 0 \\ \frac {P L ^ {2}}{4 E I} \end{array} \right\} \tag {4.3.9}
where, again, because the local coordinate axes of the element coincide with the global axes of the whole beam, we have used the relationships \underline { { d } } = \hat { \underline { { d } } } and \underline { { k } } = \underline { { \hat { k } } } (that is, the local nodal displacements are also the global nodal displacements, and so forth). Equation (4.3.9) yields
\hat {f} _ {1 y} = - P \quad \hat {m} _ {1} = 0 \quad \hat {f} _ {2 y} = P \quad \hat {m} _ {2} = - P L \tag {4.3.10}
A free-body diagram of element 1, shown in Figure 4 { \mathrm { - } } 9 ( { \mathrm { a } } ) , should help you to understand the results of Eqs. (4.3.10). The figure shows a nodal transverse force of negative P at node 1 and of positive P and negative moment PL at node 2. These values are consistent with the results given by Eqs. (4.3.10). For completeness, the free-body diagram of element 2 is shown in Figure 4–9(b). We can easily verify the element nodal forces by writing an equation similar to Eq. (4.3.9).
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P ① L P PL
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3P/2 ② 3P/2 PL L PL/2
Figure 4–9 Free-body diagrams showing forces and moments on (a) element 1 and (b) element 2
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P 1 2 3 3/2 P PL/2 L 5/2 P L
Figure 4–10 Nodal forces and moment on the beam
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V 1
- P 2 3/2 P L L
Figure 4–11 Shear force diagram for the beam of Figure 4–10
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M 1 L L 2 -PL 3 PL/2
Figure 4–12 Bending moment diagram for the beam of Figure 4–10
From the results of Eqs. (4.3.8), the nodal forces and moments for the whole beam are shown on the beam in Figure 4–10. Using the beam sign conventions established in Section 4.1, we obtain the shear force V and bending moment M diagrams shown in Figures 4–11 and 4–12.
In general, for complex beam structures, we will use the element local forces to determine the shear force and bending moment diagrams for each element. We can then use these values for design purposes. Chapter 5 will further discuss this concept as used in computer codes.
Example 4.2
Determine the nodal displacements and rotations, global nodal forces, and element forces for the beam shown in Figure 4–13. We have discretized the beam as indicated by the node numbering. The beam is fixed at nodes 1 and 5 and has a roller support at node 3. Vertical loads of 10,000 lb each are applied at nodes 2 and 4. Let E ¼ 3 0 \times 1 0 ^ { 6 } psi and I = 5 0 0 \mathrm { i n } ^ { 4 } throughout the beam.
We must have consistent units; therefore, the 10-ft lengths in Figure 4–13 will be converted to 120 in. during the solution. Using Eq. (4.1.10), along with superposition of the four beam element stiffness matrices, we obtain the global stiffness matrix
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Elastic curve 1 2 10,000 lb 3 4 10,000 lb 5 10 ft 10 ft 10 ft 10 ft
Figure 4–13 Beam example
and the global equations as given in Eq. (4.3.11). Here the lengths of each element are the same. Thus, we can factor an L out of the superimposed stiffness matrix.
\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 y} \\ M _ {3} \\ F _ {4 y} \\ M _ {4} \\ F _ {5 y} \\ M _ {5} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c c c c c} d _ {1 y} & \phi_ {1} & d _ {2 y} & \phi_ {2} & d _ {3 y} & \phi_ {3} & d _ {4 y} & \phi_ {4} & d _ {5 y} & \phi_ {5} \\ 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 & 0 & 0 & 0 & 0 \\ - 1 2 & - 6 L & 1 2 + 1 2 & - 6 L + 6 L & - 1 2 & 6 L & 0 & 0 & 0 & 0 \\ 6 L & 2 L ^ {2} & - 6 L + 6 L & 4 L ^ {2} + 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 2 & - 6 L & 1 2 + 1 2 & - 6 L + 6 L & - 1 2 & 6 L & 0 & 0 \\ 0 & 0 & 6 L & 2 L ^ {2} & - 6 L + 6 L & 4 L ^ {2} + 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 \\ 0 & 0 & 0 & 0 & - 1 2 & - 6 L & 1 2 + 1 2 & - 6 L + 6 L & - 1 2 & 6 L \\ 0 & 0 & 0 & 0 & 6 L & 2 L ^ {2} & - 6 L + 6 L & 4 L ^ {2} + 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ 0 & 0 & 0 & 0 & 0 & 0 & - 1 2 & - 6 L & 1 2 & - 6 L \\ 0 & 0 & 0 & 0 & 0 & 0 & 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 y} \\ \phi_ {1} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 y} \\ \phi_ {3} \\ d _ {4 y} \\ \phi_ {4} \\ d _ {5 y} \\ \phi_ {5} \end{array} \right\} \tag {4.3.11}
For a longhand solution, we reduce Eq. (4.3.11) in the usual manner by application of the boundary conditions
d _ {1 y} = \phi_ {1} = d _ {3 y} = d _ {5 y} = \phi_ {5} = 0
The resulting equation is
\left\{ \begin{array}{c} - 1 0, 0 0 0 \\ 0 \\ 0 \\ - 1 0, 0 0 0 \\ 0 \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c} 2 4 & 0 & 6 L & 0 & 0 \\ 0 & 8 L ^ {2} & 2 L ^ {2} & 0 & 0 \\ 6 L & 2 L ^ {2} & 8 L ^ {2} & - 6 L & 2 L ^ {2} \\ 0 & 0 & - 6 L ^ {2} & 2 4 & 0 \\ 0 & 0 & 2 L ^ {2} & 0 & 8 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \\ \phi_ {3} \\ d _ {4 y} \\ \phi_ {4} \end{array} \right\} \tag {4.3.12}
The rotations (slopes) at nodes 2–4 are equal to zero because of symmetry in loading, geometry, and material properties about a plane perpendicular to the beam length and passing through node 3. Therefore, \phi _ { 2 } = \phi _ { 3 } = \phi _ { 4 } = 0 , and we can further reduce Eq. (4.3.12) to
\left\{ \begin{array}{l} - 1 0, 0 0 0 \\ - 1 0, 0 0 0 \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c} 2 4 & 0 \\ 0 & 2 4 \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ d _ {4 y} \end{array} \right\} \tag {4.3.13}
Solving for the displacements using L ¼ 120 in., E = 3 0 \times 1 0 ^ { 6 } psi, and I = 5 0 0 \mathrm { i n . } ^ { 4 } in Eq. (4.3.13), we obtain
d _ {2 y} = d _ {4 y} = - 0. 0 4 8 \text { in. } \tag {4.3.14}
as expected because of symmetry.
As observed from the solution of this problem, the greater the static redundancy (degrees of static indeterminacy or number of unknown forces and moments that cannot be determined by equations of statics), the smaller the kinematic redundancy (unknown nodal degrees of freedom, such as displacements or slopes)—hence, the fewer the number of unknown degrees of freedom to be solved for. Moreover, the use of symmetry, when applicable, reduces the number of unknown degrees of freedom even further. We can now back-substitute the results from Eq. (4.3.14), along with the numerical values for E; I , and L, into Eq. (4.3.12) to determine the global nodal forces as
F _ {1 y} = 5 0 0 0 \mathrm{lb} \quad M _ {1} = 2 5, 0 0 0 \mathrm{lb-ft}
F _ {2 y} = 1 0, 0 0 0 \mathrm{lb} \quad M _ {2} = 0
F _ {3 y} = 1 0, 0 0 0 \mathrm{lb} \quad M _ {3} = 0 \tag {4.3.15}
F _ {4 y} = 1 0, 0 0 0 \mathrm{lb} \quad M _ {4} = 0
F _ {5 y} = 5 0 0 0 \mathrm{lb} \quad M _ {5} = - 2 5, 0 0 0 \mathrm{lb-ft}
Once again, the global nodal forces (and moments) at the support nodes (nodes 1, 3, and 5) can be interpreted as the reaction forces, and the global nodal forces at nodes 2 and 4 are the applied nodal forces.
However, for large structures we must obtain the local element shear force and bending moment at each node end of the element because these values are used in the design/analysis process. We will again illustrate this concept for the element connecting nodes 1 and 2 in Figure 4–13. Using the local equations for this element, for which all nodal displacements have now been determined, we obtain
\left\{ \begin{array}{l} \hat {f} _ {1 y} \\ \hat {m} _ {1} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \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}{l} \hat {d} _ {1 y} = 0 \\ \hat {\phi} _ {1} = 0 \\ \hat {d} _ {2 y} = - 0. 0 4 8 \\ \hat {\phi} _ {2} = 0 \end{array} \right\} \tag {4.3.16}
Simplifying Eq. (4.3.16), we have
\left\{ \begin{array}{l} \hat {f} _ {1 y} \\ \hat {m} _ {1} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \end{array} \right\} = \left\{ \begin{array}{l} 5 0 0 0 \mathrm{lb} \\ 2 5, 0 0 0 \mathrm{lb-ft} \\ - 5 0 0 0 \mathrm{lb} \\ 2 5, 0 0 0 \mathrm{lb-ft} \end{array} \right\} \tag {4.3.17}
If you wish, you can draw a free-body diagram to confirm the equilibrium of the element.
Finally, you should note that because of reflective symmetry about a vertical plane passing through node 3, we could have initially considered one-half of this beam and used the following model. The fixed support at node 3 is due to the
slope being zero at node 3 because of the symmetry in the loading and support conditions.
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10,000 lb 1 2 3
Example 4.3
Determine the nodal displacements and rotations and the global and element forces for the beam shown in Figure 4–14. We have discretized the beam as shown by the node numbering. The beam is fixed at node 1, has a roller support at node 2, and has an elastic spring support at node 3. A downward vertical force of P = 5 0 kN is applied at node 3. Let E ¼ 210 GPa and I = 2 \times 1 0 ^ { - 4 } \ \mathrm { m } ^ { 4 } throughout the beam, and let k = 2 0 0 \mathrm { k N } / \mathrm { m } .
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P = 50 kN 1 ① 2 ② 3 3 m 3 m ③ k = 200 kN/m 4
Figure 4–14 Beam example
Using Eq. (4.1.14) for each beam element and Eq. (2.2.18) for the spring element as well as the direct stiffness method, we obtain the structure stiffness matrix as
\underline {{{K}}} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c c} d _ {1 y} & \phi_ {1} & d _ {2 y} & \phi_ {2} & d _ {3 y} & \phi_ {3} & d _ {4 y} \\ 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 & 0 \\ & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 & 0 \\ & & 2 4 & 0 & - 1 2 & 6 L & 0 \\ & & & 8 L ^ {2} & - 6 L & 2 L ^ {2} & 0 \\ & & & & 1 2 + \frac {k L ^ {3}}{E I} & - 6 L & - \frac {k L ^ {3}}{E I} \\ & & & & & 4 L ^ {2} & 0 \\ & & & & & & \frac {k L ^ {3}}{E I} \\ \text {Symmetry} \end{array} \right] \tag {4.3.18a}
where the spring stiffness matrix \underline { { k } } _ { s } given below by Eq. (4.3.18b) has been directly added into the global stiffness matrix corresponding to its degrees of freedom at nodes 3 and 4.
\underline {{{{k}}}} _ {s} = \left[ \begin{array}{c c} d _ {3 y} & d _ {4 y} \\ k & - k \\ - k & k \end{array} \right] \tag {4.3.18b}
It is easier to solve the problem using the general variables, later making numerical substitutions into the final displacement expressions. The governing equations for the beam are then given by
\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 y} \\ M _ {3} \\ F _ {4 y} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c c} 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 & 0 \\ & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 & 0 \\ & & 2 4 & 0 & - 1 2 & 6 L & 0 \\ & & & 8 L ^ {2} & - 6 L & 2 L ^ {2} & 0 \\ & & & & 1 2 + k ^ {\prime} & - 6 L & - k ^ {\prime} \\ & & & & & 4 L ^ {2} & 0 \\ \text {Symmetry} & & & & & & k ^ {\prime} \end{array} \right] \left\{ \begin{array}{l} d _ {1 y} \\ \phi_ {1} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 y} \\ \phi_ {3} \\ d _ {4 y} \end{array} \right\} \tag {4.3.19}
where k ^ { \prime } = k L ^ { 3 } / ( E I ) is used to simplify the notation. We now apply the boundary conditions
d _ {1 y} = 0 \quad \phi_ {1} = 0 \quad d _ {2 y} = 0 \quad d _ {4 y} = 0 \tag {4.3.20}
We delete the first three equations and the seventh equation (corresponding to the boundary conditions given by Eq. (4.3.20)) of Eqs. (4.3.19). The remaining three equations are
\left\{ \begin{array}{c} 0 \\ - P \\ 0 \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c} 8 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 6 L & 1 2 + k ^ {\prime} & - 6 L \\ 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} \phi_ {2} \\ d _ {3 y} \\ \phi_ {3} \end{array} \right\} \tag {4.3.21}
Solving Eqs. (4.3.21) simultaneously for the displacement at node 3 and the rotations at nodes 2 and 3, we obtain
d _ {3 y} = - \frac {7 P L ^ {3}}{E I} \left(\frac {1}{1 2 + 7 k ^ {\prime}}\right) \quad \phi_ {2} = - \frac {3 P L ^ {2}}{E I} \left(\frac {1}{1 2 + 7 k ^ {\prime}}\right) \tag {4.3.22}
\phi_ {3} = - \frac {9 P L ^ {2}}{E I} \left(\frac {1}{1 2 + 7 k ^ {\prime}}\right)
The influence of the spring stiffness on the displacements is easily seen in Eq. (4.3.22). Solving for the numerical displacements using P = 5 0 ~ \mathrm { k N } , L = 3 ~ \mathrm { m } , E = 2 1 0 ~ \mathrm { G P a } ( = 2 1 0 ^ { - } \times 1 0 ^ { 6 } \mathrm { k N } / \mathrm { m } ^ { 2 } ) , I = 2 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 4 } , and k ^ { \prime } = 0 . 1 2 9 in Eq. (4.3.22), we obtain
d _ {3 y} = \frac {- 7 (5 0 \mathrm{kN}) (3 \mathrm{m}) ^ {3}}{(2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2}) (2 \times 1 0 ^ {- 4} \mathrm{m} ^ {4})} \left(\frac {1}{1 2 + 7 (0 . 1 2 9)}\right) = - 0. 0 1 7 4 \mathrm{m} \tag {4.3.23}
Similar substitutions into Eq. (4.3.26) yield
\phi_ {2} = - 0. 0 0 2 4 9 \mathrm{rad} \quad \phi_ {3} = - 0. 0 0 7 4 7 \mathrm{rad} \tag {4.3.24}
We now back-substitute the results from Eqs. (4.3.23) and (4.3.24), along with numerical values for P , E , I , L , and k ^ { \prime } , into Eq. (4.3.19) to obtain the global nodal forces as
F _ {1 y} = - 6 9. 9 \mathrm{kN} \quad M _ {1} = - 6 9. 7 \mathrm{kN} \cdot \mathrm{m}
F _ {2 y} = 1 1 6. 4 \mathrm{kN} \quad M _ {2} = 0. 0 \mathrm{kN} \cdot \mathrm{m} \tag {4.3.25}
F _ {3 y} = - 5 0. 0 \mathrm{kN} \quad M _ {3} = 0. 0 \mathrm{kN} \cdot \mathrm{m}
For the beam-spring structure, an additional global force F _ { 4 y } is determined at the base of the spring as follows:
F _ {4 y} = - d _ {3 y} k = (0. 0 1 7 4) 2 0 0 = 3. 5 \mathrm{kN} \tag {4.3.26}
This force provides the additional global y force for equilibrium of the structure.
other
| Position | Load (kN) | Moment (kN) | |---|---|---| | 1 | 69.9 | 69.7 | | 2 | 50 | 116.4 | | 3 | 3.5 | 3.5 |Figure 4–15 Free-body diagram of beam of Figure 4–14
A free-body diagram, including the forces and moments from Eqs. (4.3.25) and (4.3.26) acting on the beam, is shown in Figure 4–15.
Example 4.4
Determine the displacement and rotation under the force and moment located at the center of the beam shown in Figure 4–16. The beam has been discretized into the two elements shown in Figure 4–16. The beam is fixed at each end. A downward force of 10 kN and an applied moment of 20 kN-m act at the center of the beam. Let E ¼ 210 GPa and I = 4 \times 1 0 ^ { - 4 } \ m ^ { 4 } throughout the beam length.
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10 kN 3 m 2 3 m ① 20 kN-m ② 3
Figure 4–16 Fixed-fixed beam subjected to applied force and moment
Using Eq. (4.1.14) for each beam element with L ¼ 3 m, we obtain the element stiffness matrices as follows:
\underline {{{{k}}}} ^ {(1)} = \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 L ^ {2} \\ & & 1 2 & - 6 L \\ & \text {Symmetry} & & 4 L ^ {2} \end{array} \right] \quad \underline {{{{k}}}} ^ {(2)} = \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 L ^ {2} \\ & & 1 2 & - 6 L \\ & \text {Symmetry} & & 4 L ^ {2} \end{array}\right] \tag {4.3.27}
The boundary conditions are given by
d _ {1 y} = \phi_ {1} = d _ {3 y} = \phi_ {3} = 0 \tag {4.3.28}
The global forces are F _ { 2 \nu } = - 1 0 { , } 0 0 0 \mathrm { \ N } and M _ { 2 } = 2 0 \mathrm { , 0 0 0 \ : N – m } .
Applying the global forces and boundary conditions, Eq. (4.3.28), and assembling the global stiffness matrix using the direct stiffness method and Eqs. (4.3.27), we obtain the global equations as:
\left\{ \begin{array}{c} - 1 0, 0 0 0 \\ 2 0, 0 0 0 \end{array} \right\} = \frac {(2 1 0 \times 1 0 ^ {9}) (4 \times 1 0 ^ {- 4})}{3 ^ {3}} \left[ \begin{array}{c c} 2 4 & 0 \\ 0 & 8 (3 ^ {2}) \end{array} \right] \left\{ \begin{array}{c} d _ {2 y} \\ \phi_ {2} \end{array} \right\} \tag {4.3.29}
Solving Eq. (4.3.29) for the displacement and rotation, we obtain
d _ {2 y} = - 1. 3 3 9 \times 1 0 ^ {- 4} \mathrm{m} \text { and } \phi_ {2} = 8. 9 2 8 \times 1 0 ^ {- 5} \mathrm{rad} \tag {4.3.30}
Using the local equations for each element, we obtain the local nodal forces and moments for element one as follows:
\left\{ \begin{array}{l} f _ {1 y} ^ {(1)} \\ m _ {1} ^ {(1)} \\ f _ {2 y} ^ {(1)} \\ m _ {2} ^ {(1)} \end{array} \right\} = \frac {(2 1 0 \times 1 0 ^ {9}) (4 \times 1 0 ^ {- 4})}{3 ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 (3) & - 1 2 & 6 (3) \\ 6 (3) & 4 (3 ^ {2}) & - 6 (3) & 2 (3 ^ {2}) \\ - 1 2 & - 6 (3) & 1 2 & - 6 (3) \\ 6 (3) & 2 (3 ^ {2}) & - 6 (3) & 4 (3 ^ {2}) \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ - 1. 3 3 3 9 \times 1 0 ^ {- 4} \\ 8. 9 2 8 \times 1 0 ^ {- 5} \end{array} \right\} \tag {4.3.31}
Simplifying Eq. (4.3.31), we have
f _ {1 y} ^ {(1)} = 1 0, 0 0 0 \mathrm{N}, \quad m _ {1} ^ {(1)} = 1 2, 5 0 0 \mathrm{N} - \mathrm{m}, \quad f _ {2 y} ^ {(1)} = - 1 0, 0 0 0 \mathrm{N}, \quad m _ {2} ^ {(1)} = 1 7, 5 0 0 \mathrm{N} - \mathrm{m} \tag {4.3.32}
Similarly, for element two the local nodal forces and moments are
f _ {2 y} ^ {(2)} = 0, \quad m _ {2} ^ {(2)} = 2 5 0 0 \mathrm{N} \cdot \mathrm{m}, \quad f _ {3 y} ^ {(2)} = 0, \quad m _ {3} ^ {(2)} = - 2 5 0 0 \mathrm{N} \cdot \mathrm{m} \tag {4.3.33}
Using the results from Eqs. (4.3.32) and (4.3.33), we show the local forces and moments acting on each element in Figure 4–16 as follows:
Using the results from Eqs. (4.3.32) and (4.3.33), or Figure 4–17, we obtain the shear force and bending moment diagrams for each element as shown in Figure 4–18.
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12,500 N-m 10,000 N 17,500 N-m 10,000 N 2500 N-m 0 2500 N-m 0 0
Figure 4–17 Nodal forces and moments acting on each element of Figure 4–15