김경종
26 KiB
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ŷ, v̂ w(x̂) a b x̂ c d
ˆ(a) Undeformed beam under load w(x)
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a' b' c' d'
(b) Deformed beam due to applied loading
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w(x̂) M 1 2 M + dM V d̂x V + dV
(c) Differential beam element
Figure 4–3 Beam under distributed load
Beam Stiffness Matrix Based on Euler-Bernouli Beam Theory (Considering Bending Deformations Only)
The differential equation governing elementary linear-elastic beam behavior [1] (called the Euler-Bernoulli beam as derived by Euler and Bernoulli) is based on plane cross sections perpendicular to the longitudinal centroidal axis of the beam before bending occurs remaining plane and perpendicular to the longitudinal axis after bending occurs. This is illustrated in Figure 4–3, where a plane through vertical line ac (Figure 4–3(a)) is perpendicular to the longitudinal x^ axis before bending, and this same plane through a ^ { \prime } { - } c ^ { \prime } (rotating through angle \hat { \phi } in Figure 4–3(b)) remains perpendicular to the bent x^ axis after bending. This occurs in practice only when a pure couple or constant moment exists in the beam. However it is a reasonable assumption that yields equations that quite accurately predict beam behavior for most practical beams.
The differential equation is derived as follows. Consider the beam shown in Figure 4–3 subjected to a distributed loading wðx^Þ (force/length). From force and moment equilibrium of a differential element of the beam, shown in Figure 4–3(c), we have
\Sigma F _ {y} = 0: V - (V + d V) - w (\hat {x}) d x = 0 \tag {4.1.1a}
Or, simplifying Eq. (4.1.1a), we obtain
- w d \hat {x} - d V = 0 \quad \text { or } \quad w = - \frac {d V}{d \hat {x}} \tag {4.1.1b}
\Sigma M _ {2} = 0: - V d x + d M + w (\hat {x}) d \hat {x} \left(\frac {d \hat {x}}{2}\right) = 0 \quad \text { or } \quad V = \frac {d M}{d \hat {x}} \tag {4.1.1c}
The final form of Eq. (4.1.1c), relating the shear force to the bending moment, is obtained by dividing the left equation by dx^ and then taking the limit of the equation as dx^ approaches 0. The wðx^Þ term then disappears.
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w(x̂) v̂(x̂) x̂
(a) Portion of deflected curve of beam
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ρ φ̂ v̂(x̂)
(b)Radius of deflected curve at (x)
Figure 4–4 Deflected curve of beam
Also, the curvature k of the beam is related to the moment by
\kappa = \frac {1}{\rho} = \frac {M}{E I} \tag {4.1.1d}
where r is the radius of the deflected curve shown in Figure 4–4b, v^ is the transverse displacement function in the y^ direction (see Figure 4–4a), E is the modulus of elasticity, and I is the principal moment of inertia about the z^ axis (where the z^ axis is perpendicular to the x^ and y^ axes).
The curvature for small slopes \hat { \phi } = d \hat { v } / d \hat { x } is given by
\kappa = \frac {d ^ {2} \hat {v}}{d \hat {x} ^ {2}} \tag {4.1.1e}
Using Eq. (4.1.1e) in (4.1.1d), we obtain
\frac {d ^ {2} \hat {v}}{d \hat {x} ^ {2}} = \frac {M}{E I} \tag {4.1.1f}
Solving Eq. (4.1.1f ) for M and substituting this result into (4.1.1c) and (4.1.1b), we obtain
\frac {d ^ {2}}{d \hat {x} ^ {2}} \left(E I \frac {d ^ {2} \hat {v}}{d \hat {x} ^ {2}}\right) = - w (\hat {x}) \tag {4.1.1g}
For constant EI and only nodal forces and moments, Eq. (4.1.1g) becomes
E I \frac {d ^ {4} \hat {v}}{d \hat {x} ^ {4}} = 0 \tag {4.1.1h}
We will now follow the steps outlined in Chapter 1 to develop the stiffness matrix and equations for a beam element and then to illustrate complete solutions for beams.
Step 1 Select the Element Type
Represent the beam by labeling nodes at each end and in general by labeling the element number (Figure 4–1).
Step 2 Select a Displacement Function
Assume the transverse displacement variation through the element length to be
\hat {v} (\hat {x}) = a _ {1} \hat {x} ^ {3} + a _ {2} \hat {x} ^ {2} + a _ {3} \hat {x} + a _ {4} \tag {4.1.2}
The complete cubic displacement function Eq. (4.1.2) is appropriate because there are four total degrees of freedom (a transverse displacement \hat { d } _ { i y } and a small rotation \hat { \phi } _ { i } at each node). The cubic function also satisfies the basic beam differential equation— further justifying its selection. In addition, the cubic function also satisfies the conditions of displacement and slope continuity at nodes shared by two elements.
Using the same procedure as described in Section 2.2, we express v^ as a function of the nodal degrees of freedom \hat { d } _ { 1 y } , \hat { d } _ { 2 y } , \hat { \phi } _ { 1 } , and \hat { \phi } _ { 2 } as follows:
\hat {v} (0) = \hat {d} _ {1 y} = a _ {4}
\frac {d \hat {v} (0)}{d \hat {x}} = \hat {\phi} _ {1} = a _ {3} \tag {4.1.3}
\hat {v} (L) = \hat {d} _ {2 y} = a _ {1} L ^ {3} + a _ {2} L ^ {2} + a _ {3} L + a _ {4}
\frac {d \hat {v} (L)}{d \hat {x}} = \hat {\phi} _ {2} = 3 a _ {1} L ^ {2} + 2 a _ {2} L + a _ {3}
where \hat { \phi } = d \hat { v } / d \hat { x } for the assumed small rotation \hat { \phi } . Solving Eqs. (4.1.3) for a _ { 1 } through a _ { 4 } in terms of the nodal degrees of freedom and substituting into Eq. (4.1.2), we have
\hat {v} = \left[ \frac {2}{L ^ {3}} \left(\hat {d} _ {1 y} - \hat {d} _ {2 y}\right) + \frac {1}{L ^ {2}} \left(\hat {\phi} _ {1} + \hat {\phi} _ {2}\right) \right] \hat {x} ^ {3}
+ \left[ - \frac {3}{L ^ {2}} (\hat {d} _ {1 y} - \hat {d} _ {2 y}) - \frac {1}{L} (2 \hat {\phi} _ {1} + \hat {\phi} _ {2}) \right] \hat {x} ^ {2} + \hat {\phi} _ {1} \hat {x} + \hat {d} _ {1 y} \tag {4.1.4}
In matrix form, we express Eq. (4.1.4) as
\hat {v} = [ N ] \{\hat {d} \} \tag {4.1.5}
where
\{\hat {d} \} = \left\{ \begin{array}{l} \hat {d} _ {1 y} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} \tag {4.1.6a}
and where [ N ] = [ N _ { 1 } N _ { 2 } N _ { 3 } N _ { 4 } ] (4.1.6b)
and N _ { 1 } = \frac { 1 } { L ^ { 3 } } ( 2 \hat { x } ^ { 3 } - 3 \hat { x } ^ { 2 } L + L ^ { 3 } ) \qquad N _ { 2 } = \frac { 1 } { L ^ { 3 } } ( \hat { x } ^ { 3 } L - 2 \hat { x } ^ { 2 } L ^ { 2 } + \hat { x } L ^ { 3 } ) ð4:1:7Þ
N _ {3} = \frac {1}{L ^ {3}} (- 2 \hat {x} ^ {3} + 3 \hat {x} ^ {2} L) \quad N _ {4} = \frac {1}{L ^ {3}} (\hat {x} ^ {3} L - \hat {x} ^ {2} L ^ {2})
N _ { 1 } , N _ { 2 } , N _ { 3 } , and N _ { 4 } are called the shape functions for a beam element. These cubic shape (or interpolation) functions are known as Hermite cubic interpolation (or cubic
spline) functions. For the beam element, N _ { 1 } = 1 when evaluated at node 1 and N _ { 1 } = 0 when evaluated at node 2. Because N _ { 2 } is associated with \hat { \phi } _ { 1 } , we have, from the second of Eqs. (4.1.7), ( d N _ { 2 } / d \hat { x } ) = 1 when evaluated at node 1. Shape functions N _ { 3 } and N _ { 4 } have analogous results for node 2.
Step 3 Define the Strain=Displacement and Stress=Strain Relationships
Assume the following axial strain/displacement relationship to be valid:
\varepsilon_ {x} (\hat {x}, \hat {y}) = \frac {d \hat {u}}{d \hat {x}} \tag {4.1.8}
where u^ is the axial displacement function. From the deformed configuration of the beam shown in Figure 4–5, we relate the axial displacement to the transverse displacement by
\hat {u} = - \hat {y} \frac {d \hat {v}}{d \hat {x}} \tag {4.1.9}
where we should recall from elementary beam theory [1] the basic assumption that cross sections of the beam (such as cross section ABCD) that are planar before bending deformation remain planar after deformation and, in general, rotate through a small angle ( d \hat { v } / d \hat { x } ) . Using Eq. (4.1.9) in Eq. (4.1.8), we obtain
\varepsilon_ {x} (\hat {x}, \hat {y}) = - \hat {y} \frac {d ^ {2} \hat {v}}{d \hat {x} ^ {2}} \tag {4.1.10}
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ŷ, ŷ D A C x̂, ŷ B d̂x ẑ
（a）
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-ŷ \frac{d\hat{v}}{d\hat{x}} = \hat{\phi}
(c)
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D A C B dØ dX̂
（b)
Figure 4–5 Beam segment (a) before deformation and (b) after deformation; (c) angle of rotation of cross section ABCD
From elementary beam theory, the bending moment and shear force are related to the transverse displacement function. Because we will use these relationships in the derivation of the beam element stiffness matrix, we now present them as
\hat {m} (\hat {x}) = E I \frac {d ^ {2} \hat {v}}{d \hat {x} ^ {2}} \quad \hat {V} = E I \frac {d ^ {3} \hat {v}}{d \hat {x} ^ {3}} \tag {4.1.11}
Step 4 Derive the Element Stiffness Matrix and Equations
First, derive the element stiffness matrix and equations using a direct equilibrium approach. We now relate the nodal and beam theory sign conventions for shear forces and bending moments (Figures 4–1 and 4–2), along with Eqs. (4.1.4) and (4.1.11), to obtain
\begin{array}{l} \hat {f} _ {1 y} = \hat {V} = E I \frac {d ^ {3} \hat {v} (0)}{d \hat {x} ^ {3}} = \frac {E I}{L ^ {3}} \left(1 2 \hat {d} _ {1 y} + 6 L \hat {\phi} _ {1} - 1 2 \hat {d} _ {2 y} + 6 L \hat {\phi} _ {2}\right) \\ \hat {m} _ {1} = - \hat {m} = - E I \frac {d ^ {2} \hat {v} (0)}{d \hat {x} ^ {2}} = \frac {E I}{L ^ {3}} \left(6 L \hat {d} _ {1 y} + 4 L ^ {2} \hat {\phi} _ {1} - 6 L \hat {d} _ {2 y} + 2 L ^ {2} \hat {\phi} _ {2}\right) \tag {4.1.12} \\ \end{array}
\hat {f} _ {2 y} = - \hat {V} = - E I \frac {d ^ {3} \hat {v} (L)}{d \hat {x} ^ {3}} = \frac {E I}{L ^ {3}} \left(- 1 2 \hat {d} _ {1 y} - 6 L \hat {\phi} _ {1} + 1 2 \hat {d} _ {2 y} - 6 L \hat {\phi} _ {2}\right)
\hat {m} _ {2} = \hat {m} = E I \frac {d ^ {2} \hat {v} (L)}{d \hat {x} ^ {2}} = \frac {E I}{L ^ {3}} \left(6 L \hat {d} _ {1 y} + 2 L ^ {2} \hat {\phi} _ {1} - 6 L \hat {d} _ {2 y} + 4 L ^ {2} \hat {\phi} _ {2}\right)
where the minus signs in the second and third of Eqs. (4.1.12) are the result of opposite nodal and beam theory positive bending moment conventions at node 1 and opposite nodal and beam theory positive shear force conventions at node 2 as seen by comparing Figures 4–1 and 4–2. Equations (4.1.12) relate the nodal forces to the nodal displacements. In matrix form, Eqs. (4.1.12) become
\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} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} \tag {4.1.13}
where the stiffness matrix is then
\underline {{\hat {k}}} = \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] \tag {4.1.14}
Equation (4.1.13) indicates that \underline { { \hat { k } } } relates transverse forces and bending moments to transverse displacements and rotations, whereas axial effects have been neglected.
In the beam element stiffness matrix (Eq. (4.1.14) derived in this section, it is assumed that the beam is long and slender; that is, the length, L , to depth, h, dimension ratio of the beam is large. In this case, the deflection due to bending that is predicted by using the stiffness matrix from Eq. (4.1.14) is quite adequate. However, for short, deep beams the transverse shear deformation can be significant and can
have the same order of magnitude contribution to the total deformation of the beam. This is seen by the expressions for the bending and shear contributions to the deflection of a beam, where the bending contribution is of order \left( L / h \right) ^ { 3 } , whereas the shear contribution is only of order ( L / h ) . A general rule for rectangular cross-section beams, is that for a length at least eight times the depth, the transverse shear deflection is less than five percent of the bending deflection [4]. Castigliano’s method for finding beam and frame deflections is a convenient way to include the effects of the transverse shear term as shown in [4]. The derivation of the stiffness matrix for a beam including the transverse shear deformation contribution is given in a number of references [5–8]. The inclusion of the shear deformation in beam theory with application to vibration problems was developed by Timoshenko and is known as the Timoshenko beam [9–10].
Beam Stiffness Matrix Based on Timoshenko Beam Theory (Including Transverse Shear Deformation)
The shear deformation beam theory is derived as follows. Instead of plane sections remaining plane after bending occurs as shown previously in Figure 4–5, the shear deformation (deformation due to the shear force V ) is now included. Referring to Figure 4–6, we observe a section of a beam of differential length dx^ with the cross section assumed to remain plane but no longer perpendicular to the neutral axis
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d\hat{x} V \hat{v}(\hat{x}) \beta(\hat{x}) \hat{\phi}(\hat{x}) M + \frac{\partial M}{\partial \hat{x}} \hat{x}
(a)
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d\hat{v}_2^{(1)} / d\hat{x} \hat{\phi}_2^{(1)} \hat{\phi}_2^{(2)} d\hat{v}_2^{(2)} / d\hat{x} Element 2 \hat{\phi}_2^{(1)} = \hat{\phi}_2^{(2)} Element 1 \frac{d\hat{v}_2^{(1)}}{d\hat{x}} \neq \frac{d\hat{v}_2^{(2)}}{d\hat{x}}
(b)
Figure 4–6 (a) Element of Timoshenko beam showing shear deformation. Cross sections are no longer perpendicular to the neutral axis line. (b) Two beam elements meeting at node 2
(x^ axis) due to the inclusion of the shear force resulting in a rotation term indicated by b. The total deflection of the beam at a point x^ now consists of two parts, one caused by bending and one by shear force, so that the slope of the deflected curve at point x^ is now given by
\frac {d \hat {v}}{d \hat {x}} = \hat {\phi} (\hat {x}) + \beta (\hat {x}) \tag {4.1.15a}
where rotation due to bending moment and due to transverse shear force are given, respectively, by \hat { \phi } \left( \hat { x } \right) and \beta ( \hat { x } ) .
We assume as usual that the linear deflection and angular deflection (slope) are small.
The relation between bending moment and bending deformation (curvature) is now
M (\hat {x}) = E I \frac {d \hat {\phi} (\hat {x})}{d \hat {x}} \tag {4.1.15b}
and the relation between the shear force and shear deformation (rotation due to shear) (shear strain) is given by
V (\hat {x}) = k _ {s} A G \beta (\hat {x}) \tag {4.1.15c}
The difference in d \hat { v } / d \hat { x } and \hat { \phi } represents the shear strain \gamma _ { y z } ( = \beta ) of the beam as
\gamma_ {y z} = \frac {d \hat {v}}{d \hat {x}} - \hat {\phi} \tag {4.1.15d}
Now consider the differential element in Figure 4–3c and Eqs. (4.1.1b) and (4.1.1c) obtained from summing transverse forces and then summing bending moments. We now substitute Eq. (4.1.15c) for V and Eq. (4.1.15b) for M into Eqs. (4.1.1b) and (4.1.1c) along with \beta from Eq. (4.1.15a) to obtain the two governing differential equations as
\frac {d}{d \hat {x}} \left[ k _ {s} A G \left(\frac {d \hat {v}}{d \hat {x}} - \hat {\phi}\right) \right] = - w \tag {4.1.15e}
\frac {d}{d \hat {x}} \left(E I \frac {d \hat {\phi}}{d \hat {x}}\right) + k _ {s} A G \left(\frac {d \hat {v}}{d \hat {x}} - \hat {\phi}\right) = 0 \tag {4.1.15f}
To derive the stiffness matrix for the beam element including transverse shear deformation, we assume the transverse displacement to be given by the cubic function in Eq. (4.1.2). In a manner similar to [8], we choose transverse shear strain \gamma consistent with the cubic polynomial for ^vðx^Þ, such that g is a constant given by
\gamma = c \tag {4.1.15g}
Using the cubic displacement function for ^v, the slope relation given by Eq. (4.1.15a), and the shear strain Eq. (4.1.15g), along with the bending moment-curvature relation, Eq. (4.1.15b) and the shear force-shear strain relation Eq. (4.1.15c), in the bending moment–shear force relation Eq. (4.1.1c), we obtain
c = 6 a _ {1} g \tag {4.1.15h}
where g = E I / k _ { s } A G and k _ { s } A is the shear area. Shear areas, A _ { s } vary with crosssection shapes. For instance, for a rectangular shape A _ { s } is taken as 5 / 6 times the
cross section A , , for a solid circular cross section it is taken as 0.9 times the cross section, for a wide-flange cross section it is taken as the thickness of the web times the full depth of the wide-flange, and for thin-walled cross sections it is taken as two times the product of the thickness of the wall times the depth of the cross section.
Using Eqs. (4.1.2), (4.1.15a), (4.1.15g), and (4.1.15h) allow \phi to be expressed as a polynomial in x^ as follows:
\hat {\phi} = a _ {3} + 2 a _ {2} \hat {x} + (3 \hat {x} ^ {2} + 6 g) a _ {1} \tag {4.1.15i}
Using Eqs. (4.1.2) and (4.1.15i), we can now express the coefficients a _ { 1 } through a _ { 4 } in terms of the nodal displacements \hat { d } _ { 1 y } and \hat { d } _ { 2 y } and rotations \hat { \phi } _ { 1 } and \hat { \phi } _ { 2 } of the beam at the ends \hat { x } = 0 and \hat { x } = L as previously done to obtain Eq. (4.1.4) when shear deformation was neglected. The expressions for a _ { 1 } through a _ { 4 } are then given as follows:
\begin{array}{l} a _ {1} = \frac {2 \hat {d} _ {1 y} + L \hat {\phi} _ {1} - 2 \hat {d} _ {2 y} + L \hat {\phi} _ {2}}{L (L ^ {2} + 1 2 g)} \\ a _ {2} = \frac {- 3 L \hat {d} _ {1 y} - (2 L ^ {2} + 6 g) \hat {\phi} _ {1} + 3 L \hat {d} _ {2 y} + (- L ^ {2} + 6 g) \hat {\phi} _ {2}}{L \left(L ^ {2} + 1 2 g\right)} \tag {4.1.15j} \\ \end{array}
a _ {3} = \frac {- 1 2 g \hat {d} _ {1 y} + (L ^ {3} + 6 g L) \hat {\phi} _ {1} + 1 2 g \hat {d} _ {2 y} - 6 g L \hat {\phi} _ {2}}{L (L ^ {2} + 1 2 g)}
a _ {4} = \hat {d} _ {1 y}
Substituting these \boldsymbol { a ^ { \prime } \mathrm { s } } into Eq. (4.1.2), we obtain
\begin{array}{l} \hat {v} = \frac {2 \hat {d} _ {1 y} + L \hat {\phi} _ {1} - 2 \hat {d} _ {2 y} + L \hat {\phi} _ {2}}{L (L ^ {2} + 1 2 g)} \hat {x} ^ {3} \\ \frac {- 3 L \hat {d} _ {1 y} - (2 L ^ {2} + 6 g) \hat {\phi} _ {1} + 3 L \hat {d} _ {2 y} + (- L ^ {2} + 6 g) \hat {\phi} ^ {2}}{L (L ^ {2} + 1 2 g)} \hat {x} ^ {2} \\ \frac {- 1 2 g \hat {d} _ {1 y} + (L ^ {3} + 6 g L) \hat {\phi} _ {1} + 1 2 g \hat {d} _ {2 y} - 6 g L \hat {\phi} _ {2}}{L (L ^ {2} + 1 2 g)} \hat {x} + \hat {d} _ {1 y} \tag {4.1.15k} \\ \end{array}
In a manner similar to step 4 used to derive the stiffness matrix for the beam element without shear deformation included, we have
\begin{array}{l} \hat {f} _ {1 y} = \hat {V} (0) = 6 E I a _ {1} = \frac {E I \left(1 2 \hat {d} _ {1 y} + 6 L \hat {\phi} _ {1} - 1 2 \hat {d} _ {2 y} + 6 L \hat {\phi} _ {2}\right)}{L \left(L ^ {2} + 1 2 g\right)} \\ \hat {m} _ {1} = - \hat {m} (0) = - 2 E I a _ {2} = \frac {E I \left[ 6 L \hat {d} _ {1 y} + \left(4 L ^ {2} + 1 2 g\right) \hat {\phi} _ {1} - 6 L \hat {d} _ {2 y} + \left(2 L ^ {2} - 1 2 g\right) \hat {\phi} _ {2} \right]}{L \left(L ^ {2} + 1 2 g\right)} \tag {4.1.151} \\ \end{array}
\hat {f} _ {2 y} = - \hat {V} (L) = \frac {E I (- 1 2 \hat {d} _ {1 y} - 6 L \hat {\phi} _ {1} + 1 2 \hat {d} _ {2 y} - 6 L \hat {\phi} _ {2})}{L (L ^ {2} + 1 2 g)} \tag {4.1.151}
\hat {m} _ {2} = \hat {m} (L) = \frac {E I \left[ 6 L \hat {d} _ {1 y} + \left(2 L ^ {2} - 1 2 g\right) \hat {\phi} _ {1} - 6 L \hat {d} _ {2 y} + \left(4 L ^ {2} + 1 2 g\right) \hat {\phi} _ {2} \right]}{L \left(L ^ {2} + 1 2 g\right)}
where again the minus signs in the second and third of Eqs. ð4:1:15 lÞ are the result of opposite nodal and beam theory positive moment conventions at node l and opposite
nodal and beam theory positive shear force conventions at node 2, as seen by comparing Figures 4–2 and 4–7. In matrix form Eqs ð4:1:15 lÞ become
\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 \left(L ^ {2} + 1 2 g\right)} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & \left(4 L ^ {2} + 1 2 g\right) & - 6 L & \left(2 L ^ {2} - 1 2 g\right) \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & \left(2 L ^ {2} - 1 2 g\right) & - 6 L & \left(4 L ^ {2} + 1 2 g\right) \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 y} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} \tag {4.1.15m}
where the stiffness matrix including both bending and shear deformation is then given by
\hat {\underline {{k}}} = \frac {E I}{L (L ^ {2} + 1 2 g)} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & (4 L ^ {2} + 1 2 g) & - 6 L & (2 L ^ {2} - 1 2 g) \\ 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & (2 L ^ {2} - 1 2 g) & - 6 L & (4 L ^ {2} + 1 2 g) \end{array} \right] \tag {4.1.15n}
In Eq. (4.1.15n) remember that g represents the transverse shear term, and if we set g = 0 , we obtain Eq. (4.1.14) for the beam stiffness matrix, neglecting transverse shear deformation. To more easily see the effect of the shear correction factor, we define the nondimensional shear correction term as \varphi = 1 2 E I / ( k _ { s } A G L ^ { 2 } ) = 1 2 g / L ^ { 2 } and rewrite the stiffness matrix as
\underline {{\hat {k}}} = \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] \tag {4.1.15o}
Most commercial computer programs, such as [11], will include the shear deformation by having you input the shear area, A _ { s } = k _ { s } A .
4.2 Example of Assemblage of Beam Stiffness Matrices
Step 5 Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions
Consider the beam in Figure 4–7 as an example to illustrate the procedure for assemblage of beam element stiffness matrices. Assume EI to be constant throughout the beam. A force of 1000 lb and a moment of 1000 lb-ft are applied to the beam at midlength. The left end is a fixed support and the right end is a pin support.
First, we discretize the beam into two elements with nodes 1–3 as shown. We include a node at midlength because applied force and moment exist at midlength and, at this time, loads are assumed to be applied only at nodes. (Another procedure for handling loads applied on elements will be discussed in Section 4.4.)
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y 1 x ① 2 1000 lb-ft ② 3 L L 1000 lb
Figure 4–7 Fixed hinged beam subjected to a force and a moment
Using Eq. (4.1.14), we find that the global stiffness matrices for the two elements are now given by
d _ {1 y} \quad \phi_ {1} \quad d _ {2 y} \quad \phi_ {2}
\underline {{k}} ^ {(1)} = \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] \tag {4.2.1}
d _ {2 y} \quad \phi_ {2} \quad d _ {3 y} \quad \phi_ {3}
and kð2Þ \underline { { { k } } } ^ { ( 2 ) } = \underline { { { E } } } \underline { { { I } } } \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 } \rfloor } } \end{array} ð4:2:2Þ
where the degrees of freedom associated with each beam element are indicated by the usual labels above the columns in each element stiffness matrix. Here the local coordinate axes for each element coincide with the global x and y axes of the whole beam. Consequently, the local and global stiffness matrices are identical, so hats ð^Þ are not needed in Eqs. (4.2.1) and (4.2.2).
The total stiffness matrix can now be assembled for the beam by using the direct stiffness method. When the total (global) stiffness matrix has been assembled, the external global nodal forces are related to the global nodal displacements. Through direct superposition and Eqs. (4.2.1) and (4.2.2), the governing equations for the beam are thus 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 & 1 2 + 1 2 & - 6 L + 6 L & - 1 2 & 6 L \\ 6 L & 2 L ^ {2} & - 6 L + 6 L & 4 L ^ {2} + 4 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 {4.2.3}
Now considering the boundary conditions, or constraints, of the fixed support at node 1 and the hinge (pinned) support at node 3, we have
\phi_ {1} = 0 \quad d _ {1 y} = 0 \quad d _ {3 y} = 0 \tag {4.2.4}