# Help with understanding Shear Force and Bending Moment

1. Nov 22, 2013

### NoobeAtPhysics

1. The problem statement, all variables and given/known data

What is:

i) In diagram 1, the magnitude of the normal force at A

ii) In diagram 2, the magnitude of the shear force at B

iii) In diagram 3, the magnitude of the bending moment at C

iv) In diagram 4, the magnitude of the bending moment at the support

v) In diagram 1, the magnitude of the shear force at A

2. Relevant equations

3. The attempt at a solution

i) 2Fcos(30)

ii) 0

iii)-L/2 * F*cos(30)

iv) 2LF

v) 2Fsin(30)

Any help or advice would be great!

Thanks,
Pete

2. Nov 22, 2013

### NoobeAtPhysics

Hmm any ideas?

3. Nov 22, 2013

### haruspex

Do they mean normal to the wall or normal to the beam? (I'm really not sure.)
Shear force is not the same as net force.
A shear force is a couple acting along parallel surfaces of a volume. So it consists of a pair of equal and opposite forces acting along parallel lines. These create a torque if about an axis, but in this case they are being applied along parallel surfaces of a rigid volume.
But AFAIK, one doesn't normally speak of the magnitude of a shear force, unless you simply mean the magnitude of each force of the couple. More usually one speaks of shear stress. See e.g. http://en.wikipedia.org/wiki/Shear_stress.
(But I don't think the explanation there is very clear. It defines:
but a force vector only defines a line, so being parallel to it does not serve to define the area. A more precise way would be as the torque divided by the volume.) ​
None of that is going to help here since we don't know all the dimensions of the beam.
I've a nasty suspicion they're looking for the answer F sin(30), but one could equally justify F. It all depends what shape you care to make the sheared volume. For a beam, I suppose it's natural to take a rectangular volume with one axis along the beam.
In statics, if you want the bending moment at a point of a beam you only have to consider the part of the beam from that point to one end. Just add up the moments of all the forces on that part of the beam about that point. It doesn't matter which end you choose because, things being static, the two sums of moments must be equal and opposite. In this case, it's obviously simpler to pick the free end.
Yes.
That's probably what's wanted, but you could also defend 2F.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}