Bending moment

Computing the moment of force

An important part of determining bending moments in practical problems is the computation of moments of force. Let ${\displaystyle \mathbf{F}}$ be a force vector acting at a point A in a body. The moment of this force about a reference point (O) is defined as

${\displaystyle \mathbf{M}=\mathbf{r}\times \mathbf{F}}$

where ${\displaystyle \mathbf{M}}$ is the moment vector and ${\displaystyle \mathbf{r}}$ is the position vector from the reference point (O) to the point of application of the force (A). The ${\displaystyle \times }$ symbol indicates the vector cross product. For many problems, it is more convenient to compute the moment of force about an axis that passes through the reference point O. If the unit vector along the axis is ${\displaystyle \mathbf{e}}$, the moment of force about the axis is defined as

${\displaystyle M=\mathbf{e}\cdot \mathbf{M}=\mathbf{e}\cdot (\mathbf{r}\times \mathbf{F})}$

where ${\displaystyle \cdot }$ indicates the vector dot product.

Example

The adjacent figure shows a beam that is acted upon by a force ${\displaystyle F}$. If the coordinate system is defined by the three unit vectors ${\displaystyle {\mathbf{e}}_x,{\mathbf{e}}_y,{\mathbf{e}}_z}$, we have the following

${\displaystyle \mathbf{F}=0{\mathbf{e}}_x-F{\mathbf{e}}_y+0{\mathbf{e}}_z\text{and}\mathbf{r}=x{\mathbf{e}}_x+0{\mathbf{e}}_y+0{\mathbf{e}}_z.}$

Therefore,

${\displaystyle \mathbf{M}=\mathbf{r}\times \mathbf{F}=\left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ x& 0& 0\\ 0& -F& 0\end{array}\right|=-Fx{\mathbf{e}}_z.}$

The moment about the axis ${\displaystyle {\mathbf{e}}_z}$ is then

${\displaystyle M_z={\mathbf{e}}_z\cdot \mathbf{M}=-Fx.}$

Sign conventions

The negative value shows that a moment that tends to rotate a body clockwise around an axis pointed towards the viewer should have a negative sign. This is further illustrated by the following example with the axis pointed away from the viewer. We choose another right handed coordinate system with ${\displaystyle {\mathbf{E}}_x={\mathbf{e}}_x,{\mathbf{E}}_y=-{\mathbf{e}}_z,{\mathbf{E}}_z={\mathbf{e}}_y}$, we have

${\displaystyle \mathbf{F}=0{\mathbf{E}}_x+0{\mathbf{E}}_y-F{\mathbf{E}}_z\text{and}\mathbf{r}=x{\mathbf{E}}_x+0{\mathbf{E}}_y+0{\mathbf{E}}_z.}$

Then,

${\displaystyle \mathbf{M}=\mathbf{r}\times \mathbf{F}=\left|\begin{array}{ccc}{\mathbf{E}}_x& {\mathbf{E}}_y& {\mathbf{E}}_z\\ x& 0& 0\\ 0& 0& -F\end{array}\right|=Fx{\mathbf{E}}_y\text{and}{M}_y={\mathbf{E}}_y\cdot \mathbf{M}=Fx.}$

For this new choice of axes, a positive moment tends to rotate body clockwise around an axis, since the axis points away from the viewer. Alternatively, if we were to view the same system with the axis pointed towards us, from this perspective the load tends to rotate the body anticlockwise.

Computing the bending moment

In a rigid body or in an unconstrained deformable body, the application of a moment of force causes a pure rotation. But if a deformable body is constrained, it develops internal forces in response to the external force so that equilibrium is maintained. An example is shown in the figure below. These internal forces will cause local deformations in the body.

For equilibrium, the sum of the internal force vectors is equal to the negative of the sum of the applied external forces, and the sum of the moment vectors created by the internal forces is equal to the negative of the moment of the external force. The internal force and moment vectors are oriented in such a way that the total force (internal + external) and moment (external + internal) of the system is zero. The internal moment vector is called the bending moment.

Though bending moments have been used to determine the stress states in arbitrary shaped structures, the physical interpretation of the computed stresses is problematic. However, physical interpretations of bending moments in beams and plates have a straightforward interpretation as the stress resultants in a cross-section of the structural element. For example, in a beam in the figure, the bending moment vector due to stresses in the cross-section A perpendicular to the x-axis is given by

${\displaystyle {\mathbf{M}}_x={\int }_A\mathbf{r}\times ({\sigma }_{xx}{\mathbf{e}}_x+{\sigma }_{xy}{\mathbf{e}}_y+{\sigma }_{xz}{\mathbf{e}}_z)dA\text{where}\mathbf{r}=y{\mathbf{e}}_y+z{\mathbf{e}}_z.}$

Expanding this expression we have,

${\displaystyle {\mathbf{M}}_x={\int }_A\left(-y{\sigma }_{xx}{\mathbf{e}}_z+y{\sigma }_{xz}{\mathbf{e}}_x+z{\sigma }_{xx}{\mathbf{e}}_y-z{\sigma }_{xy}{\mathbf{e}}_x\right)dA=:M_{xx}{\mathbf{e}}_x+M_{xy}{\mathbf{e}}_y+M_{xz}{\mathbf{e}}_z.}$

We define the bending moment components as

${\displaystyle \left[\begin{array}{c}{M}_{xx}\\ M_{xy}\\ M_{xz}\end{array}\right]:={\int }_A\left[\begin{array}{c}y{\sigma }_{xz}-z{\sigma }_{xy}\\ z{\sigma }_{xx}\\ -y{\sigma }_{xx}\end{array}\right]dA.}$

The internal moments are computed about an origin that is at the neutral axis of the beam or plate and the integration is through the thickness (${\displaystyle h}$)

Example

In the beam shown in the adjacent figure, the external forces are the applied force at point A (${\displaystyle -F{\mathbf{e}}_y}$) and the reactions at the two support points O and B (${\displaystyle {\mathbf{R}}_O=R_O{\mathbf{e}}_y}$ and ${\displaystyle {\mathbf{R}}_B=R_B{\mathbf{e}}_y}$). For this situation, the only non-zero component of the bending moment is

${\displaystyle {\mathbf{M}}_{xz}=-\left[{\int }_z\left[{\int }_0^{h}y{\sigma }_{xx}dy\right]dz\right]{\mathbf{e}}_z.}$

where ${\displaystyle h}$ is the height in the ${\displaystyle y}$ direction of the beam. The minus sign is included to satisfy the sign convention.

In order to calculate ${\displaystyle {\mathbf{M}}_{xz}}$, we begin by balancing the forces, which gives one equation with the two unknown reactions,

${\displaystyle R_O+R_B-F=0.}$

To obtain each reaction a second equation is required. Balancing the moments about any arbitrary point X would give us a second equation we can use to solve for ${\displaystyle R_0}$ and ${\displaystyle R_B}$ in terms of ${\displaystyle F}$. Balancing about the point O is simplest but let's balance about point A just to illustrate the point, i.e.

${\displaystyle -{\mathbf{r}}_A\times {\mathbf{R}}_O+({\mathbf{r}}_B-{\mathbf{r}}_A)\times {\mathbf{R}}_B=\mathbf{0}.}$

If ${\displaystyle L}$ is the length of the beam, we have

${\displaystyle {\mathbf{r}}_A=x_A{\mathbf{e}}_x\text{and}{\mathbf{r}}_B=L{\mathbf{e}}_x.}$

Evaluating the cross-products:

${\displaystyle \left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ -x_A& 0& 0\\ 0& R_0& 0\end{array}\right|+\left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ L-x_A& 0& 0\\ 0& R_B& 0\end{array}\right|=-x_A{R}_0{\mathbf{e}}_z+(L-x_A)R_B{\mathbf{e}}_z=0.}$

If we solve for the reactions we have

${\displaystyle R_O=\left(1-\frac{x_A}{L}\right)F\text{and}{R}_B=\frac{x_A}{L}F.}$

Now to obtain the internal bending moment at X we sum all the moments about the point X due to all the external forces to the right of X (on the positive ${\displaystyle x}$ side), and there is only one contribution in this case,

${\displaystyle {\mathbf{M}}_{xz}=({\mathbf{r}}_B-{\mathbf{r}}_X)\times {\mathbf{R}}_B=\left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ L-x& 0& 0\\ 0& R_B& 0\end{array}\right|=\frac{F{x}_A}{L}(L-x){\mathbf{e}}_z.}$

We can check this answer by looking at the free body diagram and the part of the beam to the left of point X, and the total moment due to these external forces is

${\displaystyle \mathbf{M}=({\mathbf{r}}_A-{\mathbf{r}}_X)\times \mathbf{F}+(-{\mathbf{r}}_X)\times {\mathbf{R}}_O=\left[(x_A-x){\mathbf{e}}_x\right]\times \left(-F{\mathbf{e}}_y\right)+\left(-x{\mathbf{e}}_x\right)\times \left(R_O{\mathbf{e}}_y\right).}$

If we compute the cross products, we have

${\displaystyle \mathbf{M}=\left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ x_A-x& 0& 0\\ 0& -F& 0\end{array}\right|+\left|\begin{array}{ccc}{\mathbf{e}}_x& {\mathbf{e}}_y& {\mathbf{e}}_z\\ -x& 0& 0\\ 0& R_0& 0\end{array}\right|=F(x-x_A){\mathbf{e}}_z-R_{0}x{\mathbf{e}}_z=-\frac{F{x}_A}{L}(L-x){\mathbf{e}}_z.}$

Thanks to the equilibrium, the internal bending moment due to external forces to the left of X must be exactly balanced by the internal turning force obtained by considering the part of the beam to the right of X

${\displaystyle \mathbf{M}+{\mathbf{M}}_{xz}=\mathbf{0}.}$

which is clearly the case.

Sign convention

In the above discussion, it is implicitly assumed that the bending moment is positive when the top of the beam is compressed. That can be seen if we consider a linear distribution of stress in the beam and find the resulting bending moment. Let the top of the beam be in compression with a stress ${\displaystyle -{\sigma }_0}$ and let the bottom of the beam have a stress ${\displaystyle {\sigma }_0}$. Then the stress distribution in the beam is ${\displaystyle {\sigma }_{xx}(y)=-y{\sigma }_0}$. The bending moment due to these stresses is

${\displaystyle M_{xz}=-\left[{\int }_z{\int }_{-h/2}^{h/2}y(-y{\sigma }_0)dydz\right]={\sigma }_{0}I}$

where ${\displaystyle I}$ is the area moment of inertia of the cross-section of the beam. Therefore, the bending moment is positive when the top of the beam is in compression.

Many authors follow a different convention in which the stress resultant ${\displaystyle M_{xz}}$ is defined as

${\displaystyle {\mathbf{M}}_{xz}=\left[{\int }_z{\int }_{-h/2}^{h/2}y{\sigma }_{xx}dydz\right]{\mathbf{e}}_z.}$

In that case, positive bending moments imply that the top of the beam is in tension. Of course, the definition of top depends on the coordinate system being used. In the examples above, the top is the location with the largest ${\displaystyle y}$-coordinate.