5.2 The governing equation for a beam-column

Consider the deformed beam-column element in Fig. 5.1. We must differentiate the forces in the deformed and undeformed axis of the beam. The axial force $S$ versus normal force $N$ and the transverse shear force $H$ versus shear force $Q$ are depicted in Fig. 5.1.

Figure 5.1: Normal, shear, axial and transverse shear forces

The beam-column theory uses the Bernoulli^5.1-Euler^5.2 theory kinematic assumption that the curvature $\psi_{y}$ can be considered equal to the second derivative of the deflected longitudinal axis $w^{\prime \prime}$.

$${\psi_{y}} = \frac{-w^{\prime\prime}}{\left(1 + w^{\prime 2}\right)^{3/2}} \approx {\,} {-w^{\prime\prime}}$$ (5.1)

The relationship of Eq. (A.26) between the forces referred to the deformed and undeformed axis is

$$\left[ \begin{array}{c} S \\ H \end{array}\right] = \left[ \begi... ...s\varphi} \end{array}\right] \left[ \begin{array}{c} N \\ Q \end{array}\right]$$ (5.2)

The ``small slope'' assumption, $\varphi \ll 1$, allows us to conclude that

$$\cos\varphi \approx {\,} 1, \quad \sin\varphi \approx {\,} tan\varphi \approx {\,} \varphi = - w^{\prime}$$ (5.3)

and the relationship between the forces referred to the deformed and undeformed axis can be rewritten according to [RH95] and [PW94] in the form

$$\left[ \begin{array}{c} S \\ H \end{array}\right] = \left[ \begi... ...phi & {1} \end{array}\right] \left[ \begin{array}{c} N \\ Q \end{array}\right]$$ (5.4)

Figure 5.2: Deformed beam elements with the axial and normal forces

Now we consider Fig. 5.2 c, and project the forces onto the direction of z-axis:

$$\Sigma Z = 0:{\qquad} N\underbrace{\sin{\varphi}}_{\approx{\,} \v... ...i}}_{\approx{\,} 1} + q_{z}dx {\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}$$

$$- \left(N + dN\right)\underbrace{\sin \left({\varphi} + d{\varphi... ...os \left({\varphi} + d{\varphi}\right)}_{\approx{\,} 1} = 0 \quad \qquad \qquad$$ (5.5)

Taking into account the assumption of Eq. (5.3), we rewrite Eq. (5.5):

$$\Sigma Z = 0:{\qquad\qquad\qquad} N{\varphi} - Q + q_{z}dx {\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}$$

$$- N{\varphi} - Nd{\varphi} - dN{\varphi} - \underbrace{dNd\varphi}_{\approx {\,} 0} + Q + dQ = 0 {\qquad\qquad\qquad}$$ (5.6)

After reducing Eq. (5.6),

$$- N\frac{d{\varphi}}{dx} - {\varphi}\frac{dN}{dx} + \frac{dQ}{dx} + q_{z}\left(x\right) = 0$$ (5.7)

or

$$\frac{d}{dx}\left(- N{\varphi} + Q\right) + q_{z}\left(x\right) = 0$$ (5.8)

From Eq. (5.4) we have

$$- N{\varphi} + Q = H$$ (5.9)

Using the previous equation, we can rewrite the equilibrium equation (5.8) in the form

$$\frac{d}{dx}H + q_{z}\left(x\right) = 0$$ (5.10)

Now we project the forces onto the direction of x-axis:

$$\Sigma X = 0:{\quad\qquad} - N\underbrace{\cos{\varphi}}_{\approx... ...pprox{\,} \varphi} + q_{x}dx {\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$$

$$+ \left(N + dN\right)\underbrace{\cos \left({\varphi} + d{\varphi... ... d{\varphi}\right)}_{\approx{\,} \varphi + d{\varphi} } = 0 \quad \qquad \qquad$$ (5.11)

With the assumption of Eq. (5.3),

$$\Sigma X = 0:{\qquad\qquad\qquad} - N - Q{\varphi} + q_{x}dx {\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}$$

$$+ N + dN + Q{\varphi} + dQ{\varphi} + {\varphi}dQ + \underbrace{dQd\varphi}_{\approx {\,} 0} = 0 {\qquad\qquad\quad}$$ (5.12)

Reducing this equation gives

$$\frac{d}{dx}N + Q\frac{d}{dx}{\varphi} + {\varphi}\frac{d}{dx}Q + q_{x}\left(x\right) = 0$$ (5.13)

or

$$\frac{d}{dx}\left(N + Q{\varphi}\right) + q_{x}\left(x\right) = 0$$ (5.14)

From Eq. (5.4)

$$N + \underbrace{Q\varphi}_{{\ll} N} = S_{x} {\cong} N$$ (5.15)

Using Eqs. (5.15) and (5.14), we get

$$\frac{d}{dx}S_{x} + q_{x}\left(x\right) = 0$$ (5.16)

Next we write the equation of the sum of the moments and the moments of the forces acting about the end point of the element shown in Fig. 5.2 c:

$$\Sigma M_{y} = 0:{\qquad\qquad\qquad} M + dM - M - Qds = 0$$ (5.17)

Here, $ds$ is the length of the element. Hence $ds \approx {\,} dx$, and we rewrite Eq. (5.17):

$$\frac{d}{dx}M - Q = 0$$ (5.18)

Due to the orthogonality of Eq. (A.27), we can write an inverse relationship to Eq. (5.4)

$$\left[ \begin{array}{c} N \\ Q \end{array}\right] = \left[ \begi... ...phi & {1} \end{array}\right] \left[ \begin{array}{c} S \\ H \end{array}\right]$$ (5.19)

where the expression for the shear force $Q$ is

$$Q = S{\varphi} + H$$ (5.20)

Figure 5.3: Components of the normal force

Replacing from Eq. (5.18) by the expression (5.20), we obtain

$$\frac{d}{dx}M - S{\varphi} - H = 0$$ (5.21)

Hence $S_{x} {\cong} N$, and from Eq. (5.16)

$$N^{\prime} + q_{x}\left(x\right) = 0$$ (5.22)

Differentiating Eq. (5.21) we have

$$M^{\prime\prime} -\left(S_{x}{\varphi}\right)^{\prime} - H^{\prime} = 0$$ (5.23)

Replacing $H^{\prime}$ in the above equation by the expression from Eq. (5.10), and ${\varphi}$ by the expression ${\varphi} = {\beta} = - w^{\prime}$, we obtain

$$M^{\prime\prime} +\left(S_{x}{w}^{\prime}\right)^{\prime} + q_{z}\left(x\right) = 0$$ (5.24)

or

$$M^{\prime\prime} + S^{\prime}_{x}{w}^{\prime} + \left(S_{x}{w}^{\prime}\right)^{\prime} + S_{x}{w}^{\prime} + q_{z}\left(x\right) = 0$$ (5.25)

For a constant axial force $S_{x}$ and a constant $EI_{y}$ we obtain

$$M^{\prime\prime} + S_{x}{w}^{\prime\prime} + q_{z}\left(x\right) = 0$$ (5.26)

$$M = - EI_{y}{w}^{\prime\prime}$$ (5.27)

The governing differential equations take the form

$$M^{\prime\prime}_{y} {\mp} {\vert}\frac{S_{x}}{EI_{y}}{\vert}M_{y} + q_{z}\left(x\right) = 0 \quad$$ (5.28)

where the minus sign denotes a tensile axial force ($S \geq 0$) and the plus sign a compressive axial force ($S \leq 0$).

Figure 5.4: Bending moments of a beam-column

The governing equations (5.28) for the beam-column transverse displacement can be written as

$$\left[EI_{y}{w}^{\prime\prime}\right]^{\prime\prime} {\mp} {\vert} S_{x}{\vert}{w}^{\prime\prime} - q_{z}\left(x\right) = 0$$ (5.29)

and for a constant $EI_{y}$, we obtain

$$\left[{w}^{\prime\prime}\right]^{\prime\prime} {\mp} {\vert} \fra... ...EI_{y}}{\vert}{w}^{\prime\prime} - \frac{q_{z}\left(x\right)}{EI_{y}} = 0 \quad$$ (5.30)

where the minus sign marks a tensile axial force ($S \geq 0$) and the plus sign a compressive axial force ($S \leq 0$).

Equation (5.30) is the differential equation governing the deflection w of a beam-column member with a constant $EI_{y}$, subjected to a constant axial force at any end restraint.

In a second-order analysis, the total deflection w is calculated (see Fig. 5.4). The total bending moment $M^{II}$ is the sum of the bending moment $M^{I}$ of undeformed geometry of the member and the moment Sw (S - axial load, w - displacement) due to the deformed geometry of the member shown in Fig. 5.4 b. The moment is said to be amplified.

$$M^{II} = M^{I} + S_{x}{w}$$ (5.31)

The axial force $S$ can be presented with the displacements $u$ and $w$:

$$S {\approx} {\,} N = - EA\left[u^{\prime} + \frac{1}{2}\left(w^{\prime}\right)^{2}\right]$$ (5.32)

where $\left[u^{\prime} + \frac{1}{2}\left(w^{\prime}\right)^{2}\right]$ is a large Kármán^5.3-type deflection and $EA$ is the axial stiffness.