Confusion about sign of beam's moment force

Question

A positive bending moment bends a beam concave upward(or towards the positive y direction), whereas a negative bending moment bends a beam concave downward(or towards the negative y direction). (source)

Excuse me if the question is so simple, but in the photo below it is not clear why there is a negative sign for the moment in both $M_x$ and $M_z$. Also, I understand that moment in general is equal to the cross product of force and distance from axis of rotation which are both perpendicular to the moment. But why is $M_x$ written in terms of two stresses while the other moments are written in terms of one stress, knowing that each moment also have two stresses that are respectively perpendicular to it.

Edit: (I have corrected the reference source, I have previously entered the wrong one)

Answer 1

If the moment is about a single axis only, the "smiley face" is a good way to indicate a positive moment, however, in the 3D universe, I don't think it is useful. Rather, you shall use a convenient/simple sign convention as indicated below and follow it through.

So, let's set up the sign convention as - the stress is positive when acting in the positive direction of its axis; also, a moment is positive when the rotation produces a vector in direction of the positive axis (follows the right-hand rule).

$M_x = \int_A (-\sigma_{xy}z - \sigma_{xz}y)dA$

$M_y = \int_A (\sigma_{xx}z - \sigma_{xz}x)dA$

$M_z = \int_A (\sigma_{xy}x + \sigma_{xx}y)dA$

Correction: The equations above represent 3D body forces. For plane forces, in this case (a plane bounded by y & z axes), $x = 0$, so the equations reduce to:

$M_x = \int_A (-\sigma_{xy}z - \sigma_{xz}y)dA$

$M_y = \int_A (\sigma_{xx}z)dA$

$M_z = \int_A (\sigma_{xx}y)dA$

Note: The difference in sign is mainly due to sign convention. Example of a popular sign convention - a moment is positive when produces tension on the bottom-most fiber, compression on the upper-most fiber, and the element bend to a shape that resembles a smiley face:

Answer 2

The infintismal small disk in the diagram has only three stresses. $\sigma_{xx}, \ \sigma_{yz},\ \sigma_{zy}$ which we use as force for calculating moments.

• Again $\sigma_{xx}$ cross multiplied with $i$ will be equal to zero.

• The only two stresses that are left are $\sigma_{yz}$ and $\sigma_{zy}$ which are multiplied by their distance from x and summed up (algebraic).

• As for the y and z-axis moments generated by element $dA$ the equations seem to be correct, just look at the stresses.

• as for the moment sign, I have this mental image: a positive moment about the x-axis while the y-axis lies in the page and directs up produces a smiley beam.

I highlight your disk so you see the stresses as shown on your question.

Answer 3

If you notice the image

there are only three stresses that are used i.e.:

• the normal to the surface $\sigma_{xx}$ and
• the shear stresses $\sigma_{xy}$ and $\sigma_{xz}$.

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Reminder on the notation.

A stress denoted as $\sigma_{ik}$ is applied on a surface that is normal to the axis $i$, and the direction of the stress is parallel to axis $k$

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I assume (the image is placed out of context), that this derivation is to determine either the neutral axis or something relating to the deflection stresses/displacement equation. As such this calculation pertains only to the cross-section at hand. So the internal reactions (namely the moments in this case) are calculated only on the cross-section surface. Therefore, the only stresses that are relevant are the stresses which are applied on the cross-section (see above).

sign

Finally, regarding the sign, I recently wrote the following answer which I am replicating in part here. Note that in the place of a Force you can put a stress times the area that it is applied.

When a force is multiplied by a distace to generate the moment the sign is depended on both the signs of the distance and the force. In the 2D case of the xy plane (forces are only on that plane), the moments of a force are given by the following equation (notice the minus) :

$$M_z = -F_x\cdot y + F_y\cdot x$$

So for horizontal forces:

	Positive y	Negative y
Positive $F_x$	- M	[![enter image description here][2]][2] +M
Negative $F_x$	[![enter image description here][3]][3] +M	[![enter image description here][4]][4] -M

Similarly, vertical forces:

	Positive x	Negative x
Positive $F_y$	[![enter image description here][5]][5] + M	[![enter image description here][6]][6] -M
Negative $F_y$	[![enter image description here][7]][7] -M	[![enter image description here][8]][8] + M

In the generic case 3D case the equation is more easily derived by the determinant form

$$\vec{M} = \left|\begin{matrix}\vec{i} &\vec{j} &\vec{k}\\ x & y & z\\ F_x & F_y&F_z \end{matrix}\right|= \vec{i}(y\cdot F_z - z\cdot F_y) + \vec{j}(z\cdot F_x - x\cdot F_z) + \vec{k}(x\cdot F_y - y\cdot F_x)$$

So when you are calculating on the yz plane, then the moment would be: $$\vec{M}_x = \vec{i}(y\cdot F_z - z\cdot F_y) $$

additionally for the $M_z$

$$\vec{M}_z = \vec{k}(x\cdot F_y - y\cdot F_x)$$

however because $x=0$ (on the plane of the cross-section this devolves to:

$$\vec{M}_z = - y\cdot F_x\cdot \vec{k}$$