김경종
23 KiB
other
| Node | Value (kip) | Direction (k-in.) |
|---|---|---|
| 1 | 5.83 | 15 kip |
| 2 | 2.44 | 275 k-in. |
| 3 | 0.687 | 50 ft |
| 4 | 0.877 | 312 k-in. |
| 5 | 1.68 | 589 k-in. |
| 6 | 7.59 | 1058 k-in. |
| 7 | 5.03 | 15 kip |
| 8 | 2.44 | 158 k-in. |
| 9 | 0.877 | 137 k-in. |
Figure 5–9 Free-body diagrams of all elements of the frame in Figure 5–8(a)
in element 1 as
\underline {{{\hat {f}}}} ^ {(1)} = \left\{ \begin{array}{r} 1. 6 7 \\ - 0. 8 8 \\ - 1 5 8 \\ - 1. 6 7 \\ 0. 8 8 \\ - 3 1 1 \end{array} \right\} - \left\{ \begin{array}{r} - 3. 3 6 \\ 6. 7 1 \\ 9 0 0 \\ - 3. 3 6 \\ 6. 7 1 \\ - 9 0 0 \end{array} \right\} = \left\{ \begin{array}{r} 5. 0 3 \text { kip } \\ - 7. 5 9 \text { kip } \\ - 1 0 5 8 \text { k - in. } \\ 1. 6 8 \text { kip } \\ - 5. 8 3 \text { kip } \\ 5 8 9 \text { k - in. } \end{array} \right\} \tag {5.2.38}
Similarly, we can use Eqs. (5.2.35) and (5.1.8) for elements 3 and 2 to obtain the local nodal forces in these elements. Since these elements do not have any applied loads on them, the final nodal forces in local coordinates associated with each element are given by \underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { T } } \underline { { d } } . . These forces have been determined as follows:
Element 3
\hat {f} _ {4 x} = - 4. 1 2 \text {kip} \quad \hat {f} _ {4 y} = - 0. 6 8 7 \text {kip} \quad \hat {m} _ {4} = - 2 7 5 \text {k - in}. \tag {5.2.39}
\hat {f} _ {3 x} = 4. 1 2 \text { kip } \quad \hat {f} _ {3 y} = 0. 6 8 7 \text { kip } \quad \hat {m} _ {3} = - 1 3 7 \text { k - in. }
Element 2
\hat {f} _ {2 x} = - 2. 4 4 \text {kip} \quad \hat {f} _ {2 y} = - 0. 8 7 7 \text {kip} \quad \hat {m} _ {2} = - 1 5 8 \text {k - in}. \tag {5.2.40}
\hat {f} _ {4 x} = 2. 4 4 \mathrm{kip} \quad \hat {f} _ {4 y} = 0. 8 7 7 \mathrm{kip} \quad \hat {m} _ {4} = - 3 1 2 \mathrm{k-in}.
Free-body diagrams of all elements are shown in Figure 5–9. Each element has been determined to be in equilibrium, as often occurs even if errors are made in the longhand calculations. However, equilibrium at node 4 and equilibrium of the whole frame are also satisfied. For instance, using the results of Eqs. (5.2.38)–(5.2.40) to check equilibrium at node 4, which is implicit in the formulation of the global
equations, we have
\begin{array}{l} \sum M _ {4} = 5 8 9 - 2 7 5 - 3 1 2 = 2 \mathrm{k} \text {-in.} \quad (\text { close to zero }) \\ \sum F _ {x} = 1. 6 8 (0. 4 4 7) + 5. 8 3 (0. 8 9 5) - 2. 4 4 (0. 4 4 7) \\ - 0. 8 7 7 (0. 8 9 5) - 4. 1 2 = - 0. 0 2 7 \text { kip } \quad \text {(close to zero)} \\ \sum F _ {y} = 1. 6 8 (0. 8 9 5) - 5. 8 3 (0. 4 4 7) + 2. 4 4 (0. 8 9 5) \\ - 0. 8 7 7 (0. 4 4 7) - 0. 6 8 7 = 0. 0 0 4 \text { kip } \quad \text {(close to zero)} \\ \end{array}
\begin{array}{l} \sum F _ {x} = 1. 6 8 (0. 4 4 7) + 5. 8 3 (0. 8 9 5) - 2. 4 4 (0. 4 4 7) \\ - 0. 8 7 7 (0. 8 9 5) - 4. 1 2 = - 0. 0 2 7 \text { kip } \quad \text {(close to zero)} \\ \end{array}
Thus, the solution has been verified to be correct within the accuracy associated with a longhand solution.
To illustrate the solution of a problem involving both bar and frame elements, we will solve the following example.
Example 5.4
The bar element 2 is used to stiffen the cantilever beam element 1, as shown in Figure 5–10. Determine the displacements at node 1 and the element forces. For the bar, let A = 1 . 0 \times 1 0 ^ { - 3 } \mathrm { m } ^ { 2 } . For the beam, let A = 2 \times 1 0 ^ { - 3 } ~ \mathrm { m } ^ { 2 } , ~ I = 5 \times 1 0 ^ { - 5 } ~ \mathrm { m } ^ { 4 } , and L ¼ 3 m. For both the bar and the beam elements, let E = 2 1 0 \mathrm { G P a } . Let the angle between the beam and the bar be 4 5 ^ { \circ } . A downward force of 500 kN is applied at node 1.
For brevity’s sake, since nodes 2 and 3 are fixed, we keep only the parts of \underline { { k } } for each element that are needed to obtain the global \underline { { K } } matrix necessary for solution of the nodal degrees of freedom. Using Eq. (3.4.23), we obtain \underline { { k } } for the bar as
\underline {{k}} ^ {(2)} = \frac {(1 \times 1 0 ^ {- 3}) (2 1 0 \times 1 0 ^ {6})}{(3 / \cos 4 5 ^ {\circ})} \left[ \begin{array}{c c} 0. 5 & 0. 5 \\ 0. 5 & 0. 5 \end{array} \right]
or, simplifying this equation, we obtain
\underline {{k}} ^ {(2)} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{l l} 0. 3 5 4 & 0. 3 5 4 \\ 0. 3 5 4 & 0. 3 5 4 \end{array} \right] \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.2.41}
text_image
y ② 3 45° ① 1 x 2 3 m 500 kN
Figure 5–10 Cantilever beam with a bar element support
Using Eq. (5.1.11), we obtain k for the beam (including axial effects) as
\begin{array}{c c c c} d _ {1 x} & d _ {1 y} & \phi_ {1} \end{array}
\underline {{k}} ^ {(1)} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{l l l} 2 & 0 & 0 \\ 0 & 0. 0 6 7 & 0. 1 0 \\ 0 & 0. 1 0 & 0. 2 0 \end{array} \right] \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.2.42}
where ( E / L ) \times 1 0 ^ { - 3 } has been factored out in evaluating Eq. (5.2.42).
We assemble Eqs. (5.2.41) and (5.2.42) in the usual manner to obtain the global stiffness matrix as
\underline {{K}} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{l l l} 2. 3 5 4 & 0. 3 5 4 & 0 \\ 0. 3 5 4 & 0. 4 2 1 & 0. 1 0 \\ 0 & 0. 1 0 & 0. 2 0 \end{array} \right] \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.2.43}
The global equations are then written for node 1 as
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {1 y} \\ M _ {1} \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ - 5 0 0 \\ 0 \end{array} \right\} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{l l l} 2. 3 5 4 & 0. 3 5 4 & 0 \\ 0. 3 5 4 & 0. 4 2 1 & 0. 1 0 \\ 0 & 0. 1 0 & 0. 2 0 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ \phi_ {1} \end{array} \right\} \tag {5.2.44}
Solving Eq. (5.2.44), we obtain
d _ {1 x} = 0. 0 0 3 3 8 \mathrm{m} \quad d _ {1 y} = - 0. 0 2 2 5 \mathrm{m} \quad \phi_ {1} = 0. 0 1 1 3 \mathrm{rad} \tag {5.2.45}
In general, the local element forces are obtained using \underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { T } } \underline { { d } } . . For the bar element, we then have
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {3 x} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c c} C & S & 0 & 0 \\ 0 & 0 & C & S \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} \tag {5.2.46}
The matrix triple product of Eq. (5.2.46) yields (as one equation)
\hat {f} _ {1 x} = \frac {A E}{L} (C d _ {1 x} + S d _ {1 y}) \tag {5.2.47}
Substituting the numerical values into Eq. (5.2.47), we obtain
\hat {f} _ {1 x} = \frac {(1 \times 1 0 ^ {- 3} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2})}{4 . 2 4 \mathrm{m}} \left[ \frac {\sqrt {2}}{2} (0. 0 0 3 3 8 - 0. 0 2 2 5) \right] \tag {5.2.48}
Simplifying Eq. (5.2.48), we obtain the axial force in the bar (element 2) as
\hat {f} _ {1 x} = - 6 7 0 \mathrm{kN} \tag {5.2.49}
where the negative sign means \hat { f } _ { 1 x } is in the direction opposite x^ for element 2. Similarly, we obtain
\hat {f} _ {3 x} = 6 7 0 \mathrm{kN} \tag {5.2.50}
text_image
670 kN 670 kN x̂ 3 ② 473 kN 0.0 kN · m x̂ 3 m ① 78.3 kN · m 473 kN 26.5 kN 26.5 kN
Figure 5–11 Free-body diagrams of the bar (element 2) and beam (element 1) elements of Figure 5–10
which means the bar is in tension as shown in Figure 5–11. Since the local and global axes are coincident for the beam element, we have \underline { { \hat { f } } } = \underline { { f } } and \hat { \underline { d } } = \underline { d } . . Therefore, from Eq. (5.1.6), we have at node 1
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \\ \hat {m} _ {1} \end{array} \right\} = \left[ \begin{array}{c c c} C _ {1} & 0 & 0 \\ 0 & 1 2 C _ {2} & 6 C _ {2} L \\ 0 & 6 C _ {2} L & 4 C _ {2} L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ \phi_ {1} \end{array} \right\} \tag {5.2.51}
where only the upper part of the stiffness matrix is needed because the displacements at node 2 are equal to zero. Substituting numerical values into Eq. (5.2.51), we obtain
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \\ \hat {m} _ {1} \end{array} \right\} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{c c c} 2 & 0 & 0 \\ 0 & 0. 0 6 7 & 0. 1 0 \\ 0 & 0. 1 0 & 0. 2 0 \end{array} \right] \left\{ \begin{array}{c} 0. 0 0 3 3 8 \\ - 0. 0 2 2 5 \\ 0. 0 1 1 3 \end{array} \right\}
The matrix product then yields
\hat {f} _ {1 x} = 4 7 3 \mathrm{kN} \quad \hat {f} _ {1 y} = - 2 6. 5 \mathrm{kN} \quad \hat {m} _ {1} = 0. 0 \mathrm{kN} \cdot \mathrm{m} \tag {5.2.52}
Similarly, using Eq. (5.1.6), we have at node 2,
\left\{ \begin{array}{c} \hat {f} _ {2 x} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \end{array} \right\} = 7 0 \times 1 0 ^ {3} \left[ \begin{array}{c c c} - 2 & 0 & 0 \\ 0 & - 0. 0 6 7 & - 0. 1 0 \\ 0 & 0. 1 0 & 0. 1 0 \end{array} \right] \left\{ \begin{array}{c} 0. 0 0 3 3 8 \\ - 0. 0 2 2 5 \\ 0. 0 1 1 3 \end{array} \right\}
The matrix product then yields
\hat {f} _ {2 x} = - 4 7 3 \mathrm{kN} \quad \hat {f} _ {2 y} = 2 6. 5 \mathrm{kN} \quad \hat {m} _ {2} = - 7 8. 3 \mathrm{kN} \cdot \mathrm{m} \tag {5.2.53}
To help interpret the results of Eqs. (5.2.49), (5.2.50), (5.2.52), and (5.2.53), freebody diagrams of the bar and beam elements are shown in Figure 5–11. To further verify the results, we can show a check on equilibrium of node 1 to be satisfied. You should also verify that moment equilibrium is satisfied in the beam. 9
5.3 Inclined or Skewed Supports—Frame Element
For the frame element with inclined support at node 3 in Figure 5–12, the transformation matrix T used to transform global to local nodal displacements is given by Eq. (5.1.10).
In the example shown in Figure 5–12, we use T applied to node 3 as follows:
\left\{ \begin{array}{c} d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \\ \phi_ {3} ^ {\prime} \end{array} \right\} = \left[ \begin{array}{c c c} \cos \alpha & \sin \alpha & 0 \\ - \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{c} d _ {3 x} \\ d _ {3 y} \\ \phi_ {3} \end{array} \right\}
The same steps as given in Section 3.9 then follow for the plane frame. The resulting equations for the plane frame in Figure 5–12 are (see also Eq. (3.9.13))
[ T _ {i} ] \{f \} = [ T _ {i} ] [ K ] [ T _ {i} ] ^ {T} \{d \}
or
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {1 y} \\ M _ {1} \\ F _ {2 x} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 x} ^ {\prime} \\ F _ {3 y} ^ {\prime} \\ M _ {3} \end{array} \right\} = [ T _ {i} ] [ K ] [ T _ {i} ] ^ {T} \left\{ \begin{array}{c} d _ {1 x} = 0 \\ d _ {1 y} = 0 \\ \phi_ {1} = 0 \\ d _ {2 x} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} = 0 \\ \phi_ {3} ^ {\prime} = \phi_ {3} \end{array} \right\}
where
[ T _ {i} ] = \left[ \begin{array}{c c c} [ I ] & [ 0 ] & [ 0 ] \\ [ 0 ] & [ I ] & [ 0 ] \\ [ 0 ] & [ 0 ] & [ t _ {3} ] \end{array} \right]
and
[ t _ {3} ] = \left[ \begin{array}{c c c} \cos \alpha & \sin \alpha & 0 \\ - \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array} \right]
text_image
y 2 1 x y' x' 3 α
Figure 5–12 Frame with inclined support
5.4 Grid Equations
A grid is a structure on which loads are applied perpendicular to the plane of the structure, as opposed to a plane frame, where loads are applied in the plane of the structure. We will now develop the grid element stiffness matrix. The elements of a grid are assumed to be rigidly connected, so that the original angles between elements connected together at a node remain unchanged. Both torsional and bending moment continuity then exist at the node point of a grid. Examples of grids include floor and bridge deck systems. A typical grid structure subjected to loads F _ { 1 } , F _ { 2 } , F _ { 3 } , and F _ { 4 } is shown in Figure 5–13.
We will now consider the development of the grid element stiffness matrix and element equations. A representative grid element with the nodal degrees of freedom and nodal forces is shown in Figure 5–14. The degrees of freedom at each node for a grid are a vertical deflection \hat { d } _ { i y } (normal to the grid), a torsional rotation \hat { \phi } _ { i x } about the x^ axis, and a bending rotation \hat { \phi } _ { i z } about the z^ axis. Any effect of axial displacement is ignored; that is, \hat { d } _ { i x } = 0 . The nodal forces consist of a transverse force \hat { f } _ { i y } , a torsional moment \hat { m } _ { i x } about the x^ axis, and a bending moment \hat { m } _ { i z } about the z^ axis. Grid elements do not resist axial loading; that is \hat { f } _ { i x } = \bar { 0 } .
To develop the local stiffness matrix for a grid element, we need to include the torsional effects in the basic beam element stiffness matrix Eq. (4.1.14). Recall that Eq. (4.1.14) already accounts for the bending and shear effects.
We can derive the torsional bar element stiffness matrix in a manner analogous to that used for the axial bar element stiffness matrix in Chapter 3. In the derivation, we simply replace \hat { f } _ { i x } with \hat { m } _ { i x } , \hat { d } _ { i x } with \hat { \phi } _ { i x } , E with G (the shear modulus), A with J (the torsional constant, or stiffness factor), s with t (shear stress), and e with \gamma (shear strain).
text_image
y x F₁ F₂ z F₃ F₄
Figure 5–13 Typical grid structure
text_image
ŷ m̂₁z, φ̂₁z 1 m̂₁x, φ̂₁x x̂ L 2 m̂₂x, φ̂₂x m̂₂z, φ̂₂z ẑ f̂₁y, d̂₁y f̂₂y, d̂₂y
Figure 5–14 Grid element with nodal degrees of freedom and nodal forces
text_image
1 \hat{m}{1x}, \hat{\phi}{1x} \hat{m}{2x}, \hat{\phi}{2x} 2 \hat{x} L
text_image
1 \hat{m}_x, \hat{\phi}_x \hat{m}_x, \hat{\phi}_x L 2 \hat{x}
Figure 5–15 Nodal and element torque sign conventions
The actual derivation is briefly presented as follows. We assume a circular cross section with radius R for simplicity but without loss of generalization.
Step 1
Figure 5–15 shows the sign conventions for nodal torque and angle of twist and for element torque.
Step 2
We assume a linear angle-of-twist variation along the x^ axis of the bar such that
\hat {\phi} = a _ {1} + a _ {2} \hat {x} \tag {5.4.1}
Using the usual procedure of expressing a _ { 1 } and a _ { 2 } in terms of unknown nodal angles of twist \hat { \phi } _ { 1 x } and \hat { \phi } _ { 2 x } . , we obtain
\hat {\phi} = \left(\frac {\hat {\phi} _ {2 x} - \hat {\phi} _ {1 x}}{L}\right) \hat {x} + \hat {\phi} _ {1 x} \tag {5.4.2}
or, in matrix form, Eq. (5.4.2) becomes
\hat {\phi} = [ N _ {1} \quad N _ {2} ] \left\{ \begin{array}{l} \hat {\phi} _ {1 x} \\ \hat {\phi} _ {2 x} \end{array} \right\} \tag {5.4.3}
with the shape functions given by
N _ {1} = 1 - \frac {\hat {x}}{L} \quad N _ {2} = \frac {\hat {x}}{L} \tag {5.4.4}
Step 3
We obtain the shear strain g/angle of twist \hat { \phi } relationship by considering the torsional deformation of the bar segment shown in Figure 5–16. Assuming that all radial lines, such as OA, remain straight during twisting or torsional deformation, we observe that the arc length \widehat { A B } is given by
\widehat {A B} = \gamma_ {\mathrm{max}} d \hat {x} = R d \hat {\phi}
Solving for the maximum shear strain \gamma _ { \mathrm { { m a x } } } . , we obtain
\gamma_ {\mathrm{max}} = \frac {R d \hat {\phi}}{d \hat {x}}
text_image
ŷ r A C O' D γmax B dφ̂ x̂ ẑ dˆx
Figure 5–16 Torsional deformation of a bar segment
Similarly, at any radial position r, we then have, from similar triangles OAB and OCD,
\gamma = r \frac {d \hat {\phi}}{d \hat {x}} = \frac {r}{L} (\hat {\phi} _ {2 x} - \hat {\phi} _ {1 x}) \tag {5.4.5}
where we have used Eq. (5.4.2) to derive the final expression in Eq. (5.4.5).
The shear stress t/shear strain g relationship for linear-elastic isotropic materials is given by
\tau = G \gamma \tag {5.4.6}
where G is the shear modulus of the material.
Step 4
We derive the element stiffness matrix in the following manner. From elementary mechanics, we have the shear stress related to the applied torque by
\hat {m} _ {x} = \frac {\tau J}{R} \tag {5.4.7}
where J is called the polar moment of inertia for the circular cross section or, generally, the torsional constant for noncircular cross sections. Using Eqs. (5.4.5) and (5.4.6) in Eq. (5.4.7), we obtain
\hat {m} _ {x} = \frac {G J}{L} (\hat {\phi} _ {2 x} - \hat {\phi} _ {1 x}) \tag {5.4.8}
By the nodal torque sign convention of Figure 5–15,
\hat {m} _ {1 x} = - \hat {m} _ {x} \tag {5.4.9}
or, by using Eq. (5.4.8) in Eq. (5.4.9), we obtain
\hat {m} _ {1 x} = \frac {G J}{L} (\hat {\phi} _ {1 x} - \hat {\phi} _ {2 x}) \tag {5.4.10}
Similarly, \hat { m } _ { 2 x } = \hat { m } _ { x } ð5:4:11Þ
or \hat { m } _ { 2 x } = \frac { G J } { L } ( \hat { \phi } _ { 2 x } - \hat { \phi } _ { 1 x } ) m^2x ¼ ð5:4:12Þ
Expressing Eqs. (5.4.10) and (5.4.12) together in matrix form, we have the resulting torsion bar stiffness matrix equation:
\left\{ \begin{array}{l} \hat {m} _ {1 x} \\ \hat {m} _ {2 x} \end{array} \right\} = \frac {G J}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {\phi} _ {1 x} \\ \hat {\phi} _ {2 x} \end{array} \right\} \tag {5.4.13}
Hence, the stiffness matrix for the torsion bar is
\underline {{\hat {k}}} = \frac {G J}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {5.4.14}
The cross sections of various structures, such as bridge decks, are often not circular. However, Eqs. (5.4.13) and (5.4.14) are still general; to apply them to other cross sections, we simply evaluate the torsional constant J for the particular cross section. For instance, for cross sections made up of thin rectangular shapes such as channels, angles, or I shapes, we approximate J by
J = \sum \frac {1}{3} b _ {i} t _ {i} ^ {3} \tag {5.4.15}
where b _ { i } is the length of any element of the cross section and t _ { i } is the thickness of any element of the cross section. In Table 5–1, we list values of J for various common cross sections. The first four cross sections are called open sections. Equation (5.4.15) applies only to these open cross sections. (For more information on the J concept, consult References [2] and [3], and for an extensive table of torsional constants for various cross-sectional shapes, consult Reference [4].) We assume the loading to go through the shear center of these open cross sections in order to prevent twisting of the cross section. For more on the shear center consult References [2] and [5].
On combining the torsional effects of Eq. (5.4.13) with the shear and bending effects of Eq. (4.1.13), we obtain the local stiffness matrix equation for a grid element as
\left\{ \begin{array}{l} \hat {f} _ {1 y} \\ \hat {m} _ {1 x} \\ \hat {m} _ {1 z} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2 x} \\ \hat {m} _ {2 z} \end{array} \right\} = \left[ \begin{array}{c c c c c c} \frac {1 2 E I}{L ^ {3}} & 0 & \frac {6 E I}{L ^ {2}} & \frac {- 1 2 E I}{L ^ {3}} & 0 & \frac {6 E I}{L ^ {2}} \\ & \frac {G J}{L} & 0 & 0 & \frac {- G J}{L} & 0 \\ & & \frac {4 E I}{L} & \frac {- 6 E I}{L ^ {2}} & 0 & \frac {2 E I}{L} \\ & & & \frac {1 2 E I}{L ^ {3}} & 0 & \frac {- 6 E I}{L ^ {2}} \\ & & & & \frac {G J}{L} & 0 \\ \text {Symmetry} & & & & & \frac {4 E I}{L} \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 y} \\ \hat {\phi} _ {1 x} \\ \hat {\phi} _ {1 z} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2 x} \\ \hat {\phi} _ {2 z} \end{array} \right\} \tag {5.4.16}
Table 5–1 Torsional constants J and shear centers SC for various cross sections
| Cross Section | Torsional Constant |
| 1. Channel | |
| $J = \frac{t^{3}}{3}(h + 2b)$ | |
 | $e = \frac{h^{2}b^{2}t}{4I}$ |
| 2. Angle | |
| $J = \frac{1}{3}(b_{1}t_{1}^{3} + b_{2}t_{2}^{3})$ | |
 | |
| 3. Z section | |
 | $J = \frac{t^{3}}{3}(2b + h)$ |
| 4. Wide-flanged beam with unequal flanges | |
| $J = \frac{1}{3}(b_{1}t_{1}^{3} + b_{2}t_{2}^{3} + ht_{w}^{3})$ | |
 | |
5. Solid circular | $J = \frac{\pi}{2}r^{4}$ |
| 6. Closed hollow rectangular | |
 | $J = \frac{2tt_{1}(a - t)^{2}(b - t_{1})^{2}}{at + bt_{1} - t^{2} - t_{1}^{2}}$ |