Bending of Beams by Terminal Couples

Consider deformations of a straight prismatic bar made of a homogeneous linear elastic isotropic material due to a pair of couples of magnitude M applied to the ends of the beam. Because of symmetry, the problem is equivalent to that a cantilever beam loaded by a couple at one end. The couple M is caused by distributed surface tractions acting on the end faces. The resultant force of these tractions is zero, and their moment equals M about the X₂-axis (see Fig. 12.1). The line passing through the centroids of the cross-sections of the beam is called the central line. Assume that plane sections of the beam normal to the central line before deformation remain plane and normal to the deformed central line. For the co-ordinate axes shown in Fig. 12.1 with X₃-axis coincident with the central line, assume that the stresses in the beam are given by

$$\sigma_{33} = -\frac{MX_1}{I},\ \sigma_{22} = \sigma_{11} = \sigma_{12} = \sigma_{23} = \sigma_{31} = 0\tag{12.1}$$ (12.1)

where I is the moment of inertia of the cross-section about the X₂-axis.

The assumed stress field satisfies the equilibrium equations (3.11). To see if the boundary conditions are also satisfied, we note that on the lateral surface of the bar, n = (n₁,n₂,0),

$$\sigma_{ji}n_j = \sigma_{1i}n_1 + \sigma_{2i}n_2,\ i = 1,2,3. \tag{12.2}$$ (12.2)

For the stress field given by (12.1),

$$\sigma_{ji} n_j = 0,\ i = 1,2,3, \tag{12.3}$$ (12.3)

on the lateral surface of the bar. On the end face X₃ = L,
$$\begin{gather}\begin{split} &\vec n = (0,0,1),\\ &\sigma_{ji}n_j = \sigma_{3i},\ i = 1,2,3, \end{split}\tag{12.4} \end{gather}$$
which is non-zero only for i = 3. For i= 3, $\sigma_{33} = - \displaystyle\frac{MX_1}{I}$ and the boundary condition of zero resultant axial force requires that

$$0 = \int_{\cal A}\frac{MX_1}{I}dA = \frac{M}{I}\bar X_1{\cal A}\tag{12.5}$$ (12.5)

where ${\cal A}$ is the area of cross-section of the bar and $\bar X_1$ is the X₁ coordinate of the centroid of the cross-section. Since X₃-axis coincides with the central line, $\bar X_1 = 0$. The moment about the X₂-axis of surface tractions on the end-face X₃=L should equal the applied bending moment M. Thus

$$M = \int_{\cal A}\left(\frac{MX_1}{I}dA\right) X_1 = \frac{M}{I}\int_{\cal A}X^2_1dA = M.\tag{12.5}$$ (12.5)

Since point-wise traction type boundary conditions are not satisfied on the end face X₃ = L, the assumed stress state (12.1) and hence the displacements computed below from it are not valid in the immediate vicinity of this surface. According to St. Venant, however, the solution (12.1) is good at points far away from the end face X₃ = L.

We now compute the displacement field, find the deformed shape of the bar, and satisfy displacement type boundary conditions at X₃ = 0. Using Hooke's law (6.3) with C given by (6.4), we obtain
$$\begin{gather}\begin{split}&e_{33} = - \frac{MX_1}{EI},\ e_{22} = \frac{\nu MX_1... ...u MX_1}{EI},\\ &e_{12} = e_{23} = e_{31} = 0.\end{split}\tag{12.7} \end{gather}$$
Substitution from (12.7) into the strain-displacement relations (5.13) gives
$$\begin{gather}\begin{split} &\frac{\partial u_3}{\partial X_3} = - \frac{MX_1}{E... ... X_1} + \frac{\partial u_1}{\partial X_3} = 0. \end{split}\tag{12.8}\end{gather}$$
An integration of (12.8)₁ results in

$$u_3 = - \frac{M}{EI}X_1X_3 + f(X_1,X_2). \tag{12.9}$$ (12.9)

Substituting from (12.9) into (12.8)₅ and (12.8)₆ we obtain

$$\frac{\partial u_2}{\partial X_3} = - \frac{\partial f}{\part... ... \frac{M}{EI}X_3 - \frac{\partial f}{\partial X_1}.\tag{12.10}$$ (12.10)

Hence

$$u_2 = - \frac{\partial f}{\partial X_2}X_3 + g(X_1,X_2),\ u_1... ...3 - \frac{\partial f}{\partial X_1}X_3 + h(X_1,X_2)\tag{12.11}$$ (12.11)

where g and h are unknown functions of X₁ and X₂. We now substitute from (12.11) into (12.8)₂ and (12.8)₃ to obtain

$$-\frac{\partial^2f}{\partial X^2_2}X_3 + \frac{\partial g}{\p... ...ac{\partial h}{\partial X_1} = \frac{\nu MX_1}{EI}.\tag{12.12}$$ (12.12)

Since these equations hold for all values of X₃, therefore

$$-\frac{\partial^2f}{\partial X^2_2} = 0,\ \frac{\partial g}{\... ...rac{\partial h}{\partial X_1} = \frac{\nu MX_1}{EI}\tag{12.13}$$ (12.13)

An integration of these equations gives
$$\begin{gather}\begin{split} &f = \beta X_1 + \gamma X_2 + c + X_1X_2d,\\ &g = \... ...,\\ &h = \frac{\nu M}{2EI}X^2_1 + h_0(X_2). \end{split}\tag{12.14} \end{gather}$$
Substituting from (12.13) into (12.10) and the result into (12.7)₄ we arrive at

$$-2X_3d + \frac{dh_0}{dX_2} + \frac{\nu M}{EI}X_2 + \frac{dg_0}{dX_1} = 0. \tag{12.15}$$ (12.15)

This equation holds at every point in the bar if only if

$$d = 0,\ h_0 = - \frac{\nu M}{2EI}X^2_2 + \alpha X_2 + a,\ g_0 = -\alpha X_1 + b.\tag{12.16}$$ (12.16)

Thus
$$\begin{gather}\begin{split}&u_1 = \frac{M}{2EI}X^2_3 - \beta X_3 + \frac{\nu M}{... ...{M}{EI} X_1X_3 + \beta X_1 + \gamma X_2 + c. \end{split}\tag{12.17} \end{gather}$$
Constants $a,\ b,\ c,\ \alpha,\ \beta$ and $\gamma$ represent the rigid motion of the bar. In order to determine these, we fix the beam at the origin, fix an element of the X₃-axis, and an element of the X₁X₃-plane at the origin. Thus

$$u_1 = u_2 = u_3 = \frac{\partial u_1}{\partial X_3} = \frac{\... ...\partial u_2}{\partial X_2} = 0\ {\rm at}\ (0,0,0).\tag{12.18}$$ (12.18)

Conditions (12.17) require that

$$a = b = c = \beta = \gamma = \alpha = 0,\tag{12.19}$$ (12.19)

and the displacement field in the beam is given by
$$\begin{gather}\begin{split} &u_1 = \frac{M}{2EI} (X^2_3 + \nu X^2_1 - \nu X^2_2)... ...M}{EI}X_1X_2,\\ &u_3 = - \frac{M}{EI}X_1X_3.\end{split}\tag{12.20} \end{gather}$$
Points on the central line (X₁ = X₂ = 0) of the beam are deformed into the curve

$$x_1 = \frac{M}{2EI}X^2_3,\ x_2 = 0,\ x_3 = 0,\tag{12.21}$$ (12.21)

or into the parabola

$$x_1 = \frac{M}{2EI}x^2_3.\tag{12.22}$$ (12.22)

We now analyze the problem by the finite element method. The goal is to find the deformed shape of the central line. We denote the vertical displacement of a point by w instead of u₁. The first step is to find the potential energy of the system.

Equation (8.9) gives the strain energy stored in the body. For the present problem
$$\begin{align}W =\ & \frac{1}{2}\sigma_{33}e_{33} = \frac{1}{2}\frac{MX_1}{I}\cdo... ...EI}dX_3 = \int^L_0 \frac{EI}{2}(w^{\prime\prime})^2 dX_3,\tag{12.23} \end{align}$$
where we have assumed that the cross-section of the beam is uniform, and $M = EI\displaystyle\frac{\partial^2u_1}{\partial X^2_3} = EI\frac{\partial^2w}{\partial X^2_3} = EIw^{\prime\prime}$ which follows from (12.19)₁. The potential energy of the beam loaded by a couple or moment at the end X₃ = L is given by

$$V = \frac{1}{2}\int^L_0 EI(w^{\prime\prime})^2 dX_3 - M(w^\prime )\bigg\vert _{X_3 = L}\tag{12.24}$$ (12.24)

where $w^\prime = \displaystyle\frac{\partial w}{\partial X_3}$.

For the potential energy given by (12.24) to have a finite value, the second-order derivatives of w must be square integrable. Thus $w^\prime$must be continuous. Recall that the first-order derivatives of the finite element basis functions used in Section 11 are discontinuous at the node points. It implies that a different set of finite element basis functions is needed.

In order to simplify the notation, we set x₃ = X₃ = x, and $w^\prime = \partial w/\partial x_3 = \partial w/\partial x$.

Divide the domain [0,L] into n finite elements with nodes at the end-points of each element. Thus the coordinates of (n + 1) nodes can be denoted by $x_1 = 0,\ x_2,\ x_3,\ldots,\ x_{n+1} = L$. We use two sets of basis functions - one to make the deflections wcontinuous and the other to make first-order derivatives of w or the slope $\theta = w^\prime$continuous. Recalling that the basis function corresponding to a node i can be be obtained by patching together the shape functions defined on adjoining elements meeting at node i, we generate the shape functions below.

On the element $\Omega_e$ with left node x_e and right node x_e+1, we write

$$w^e(x) = w_eN^0_e(x) + w_{e+1}N^0_{e+1}(x) + \theta_eN^1_e(x) + \theta_{e+1}N^1_{e+1}(x).\tag{12.25}$$ (12.25)

Here w^e(x) is the unknown function w defined on the element $\Omega_e$, shape functions N⁰_e and N⁰_e+1 ensure the continuity of w^e across inter-element boundaries, and shape functions N¹_e and N¹_e+1 are meant to make $\theta = w^\prime$ continuous across inter-element boundaries. In order to meet these continuity requirements, we set

$$w^e(x_e) = w_e,\ w^e(x_{e+1}) = w_{e+1},\ w^{e\prime}(x_e) = \theta_e,\ w^{e\prime}(x_{e+1}) = \theta_{e+1}.\tag{12.26}$$ (12.26)

These are fulfilled if
$$\begin{align}&N^0_e(x_e) = 1,\ N^0_{e+1}(x_e) = 0,\ N^1_e(x_e) = 0,\ N^1_{e+1}(x... ...ime}_{e+1}(x_{e+1}) = 0,\ N^{1^\prime}_{e+1}(x_{e+1}) = 1.\tag{12.30}\end{align}$$
Since each shape function satisfies four conditions, we assume that
$$\begin{gather}\begin{split} &N^0_e(x) = c_1 + c_2x + c_3x^2 + c_4x^3,\ x_e\le x\... ... d_2x + d_3x^2 + d_4x^3,\ x_e\le x\le x_{e+1},\end{split}\tag{12.31}\end{gather}$$
and similar expressions for N⁰_e+1(x) and N¹_e+1(x). We use (12.27)₁, (12.28)₁, (12.29)₁ and (12.30)₁ to obtain four equations for the determination of $c_1,\ c_2,\ c_3$ and c₄. Expressions for the shape functions so obtained are given below.
$$\begin{gather}\begin{split}&N^0_e(x) = 1 - 3\left(\frac{x - x_e}{h_e}\right)^2 +... ...{h_e}\right)^2 - \frac{x - x_e}{h_e}\right]. \end{split}\tag{12.32} \end{gather}$$
Note that N⁰_e and N⁰_e+1 are dimensionless but N¹_eand N¹_e+1 have dimensions of length. Since N¹_e and N¹_e+1multiply slopes and N⁰_e and N⁰_e+1 multiply deflections, every term in (3.4.1) has the same dimension. Shape functions (12.32) are sketched in Fig. 12.2, and are called Hermitian.

Writing (12.24) as

$$V = \frac{1}{2} \sum^n_{i=1}\int^{x_{i+1}}_{x_i} EI(w^{\prime\prime})^2 dx - M(w^\prime)\bigg\vert _{x=L} \tag{12.33}$$ (12.33)

substituting for w from (12.25), and after carrying out the integration, we will have V as a function of $w_1,w^\prime_1,w_2,w^\prime_2,\ldots w_{n+1},w^\prime_{n+1}$. For V to be stationary,

$$\frac{\partial V}{\partial w_e} = 0,\ \frac{\partial V}{\part... ...l V}{\partial\theta_e} = 0,\ e = 1,2,\ldots , n+1.\tag{12.3.4}$$ (12.3.4)

Equations (12.34) are equivalent to

KW = F (12.35)

where
$$\begin{align}&\mathbf{K} = \sum_e\mathbf{K}^e,\nonumber\\ &\mathbf{K}^e = \int^... ...ht] dx. \nonumber\\ &\mathbf{F} = (0,0,0,0,\ldots , 0,1)M.\nonumber \end{align}$$
For the case of constant EI (e.g. a homogeneous beam of uniform cross-section), the element stiffness matrix K^e is given below.

$$K^e = \frac{2EI}{h^3_e}\left[\begin{array}{cccc} 6 & -3h_e & ... ...],\ EI = {\rm const},\ h_e = \mbox{element length} \tag{12.36}$$ (12.36)

We now apply essential boundary conditions $W_1 = w^\prime_1 = 0$ in (12.35) by one of the two methods discussed earlier in Section 11 (e.g. see page 27), and solve (12.35) for W. Knowing deflections and slopes at the node points, we use eqn. (12.25) to find deflections, slopes and curvatures at any point within an element. The value of the bending stress can then be computed from $EIw^{\prime\prime}$.