Shear in Bending
Internal stresses and forces due to shear within a beam bending situation.
Lecture Notes: shear in bending  worked exercises.pdf empty.one
Lecture Video: FEA Shear in bending of I beam (4 mins) or jump to embedded video
Transverse Shear Force in Bending
In the bending moment chapter we looked at the Shear Force Diagram. This is actually the traverse shear force that is being determined  for a horizontal beam with vertical loads, this means it is the vertical shear force.
Transverse Shear Force
We can cut the beam to find the Bending Moment (by doing moment equation of ONE side).But what if we don't know where the maximum is? Are we going to pick lots of places to "cut" until we eventually find it?
There is another way. We can use shear force to find bending moment, using a diagram method.
But first  a definition of shear force in a beam.
Imagine ONLY the sliding aspect of the beam, not the bending. This is a bit wierd since beams don't slide apart, they bend apart. But anyway, we still need to get this shear force thing happening.
Imagine the beam as a stack of magnets. They can slide OK but not bend.
Now if one hand pushes up and the other down, it slides. (shears)
Of course, it doesn't matter which magnet slides, they all want to slide. This one just happened to have the least friction, but in fact every slice has the same shear force.
Positive Shear Force
In a diagram we would show it like this;
When the left hand side (LHS) goes up then this is called positive shear force.
A Shear Force Diagram (SFD) is a graph of the shear force all the way along a beam.
We can construct a shear force diagram for a loaded beam, but the shear force at a particular point along the beam is actually the average for the crosssection. For more information about Shear Force Diagrams, see Bending Moment
Unfortunately, this is shear force cannot be used to determine the shear stress with the simple "stress equals force on area" equation. Why not? Because, as you might have guessed, shear stress it is not uniform throughout the crosssection.
Drat.
Longitudinal Shear in Bending
In addition to the transverse shear force, a longitudinal shear force also exists in the beam. Evidence of this longitudinal shear force (i.e. a force parallel to the beam axis) is shown below.
(a) Three planks not glued together. There is slippage between the planks.  (b) Planks are glued together. No slippage occurs. Glue carries a shear load. 
The glue in the beam in Figure (b) carries a shear load parallel to the axis of the beam. This load produces a shear stress called the longitudinal (or horizontal) shear stress.
Relationship Between Vertical And Horizontal Shear Stress
The transverse and longitudinal shear forces produce stresses called the vertical and horizontal shear
stresses. These stresses are shown acting on a small part of the beam in the figure above.
At any particular point in the beam the horizontal shear stress is equal to the vertical shear stress.
At any point in the beam;
τ_{v} = τ_{h}
Where;
τ_{v} = vertical shear stress
τ_{h} = horizontal shear stress
Proof To prove τ_{v} = τ_{h} τ_{h} (ab)c = τ_{v} (cb) a So τ_{v} = τ_{h} (Note: This works even if the element is a rectangular rather than a square prism) 
Calculating The Shear Stress
Below shows a typical beam (a simply supported beam)
However, the shear stress is not uniform throughout this cross section. The shear stress is largest at C, then a bit lower at B; and lowest at A for this particular beam.
So when we investigate the section itself, the shear force varies with height.
Now we focus on calculating the longitudinal shear stress at longitudinal plane BB; (shown below in the cross sectional view)
The shear stress at BB on the cross section can be calculated using:
Where τ = shear stress (MPa), which is the same as τ_{v} or τ_{h} for a small element at section BB
Q = first moment of area (mm^{3}) of the upper crosssection of the beam (above the BB plane). Where Q = A.y
V = vertical shear force (N)
I = 2nd moment of area of beam (mm^{4}), which is bh^{3}/12 for a rectangular beam
b = width of beam at BB (mm)
In most parts of a beam the shear stress is low, it is highest at the centre of a beam.
For I beams, the web thickness is selected so that on the neutral axis (which is the maximum longitudinal shear plane) has a longitudinal (or transverse) shear stress (at that point in the crosssection), that is not excessive.
Composite Cross sections
A beam with a builtup cross section is shown. To calculate the shear stress at BB, you need the value of Q for the beam above BB. To calculate Q, break the beam down into separate areas, in this case Area 1 and Area 2. Q = A_{1}y_{1} + A_{2}y_{2} A_{1}= Area 1 (mm^{2}) y_{1}= distance from neutral axis to centroid of Area 1 (mm) A_{2}= Area 2(mm^{2}) y_{2}= distance from neutral axis to centroid of Area 2 (mm) 
Shear Stress using FEA:
The stress labelling convention is;
 XX, YY, ZZ are the normal/planar stresses
 XY, XZ, YZ are Shear stresses in each of those 3 planes
Note that the FEA analysis may give a positive of negative shear stress, depending on the orientation of the X,Y & Z axis with respect to the job. We can ignore the negative sign for now, it is only needed for heavyduty 3D stress analysis which is beyond the scope of our study here.
Solve Example 3 Using FEA1. Draw up beam in Inventor (give it 500mm long Z axis typical) 2. In Stress Analysis, use steel, but do not turn on gravity. 3. Constrain "fixed" at the wall, apply 25kN to the free end. 4. Simulate it, then probe at (X,Y,Z) = (0,30,100) using Stress YZ. The result is a shear stress in the YZ plane of 18.75MPa. (using an average mesh size of 1% of its length). 
FEA analysis of the beam in Ex3. Note that the probe point is 100mm away from the fixed constraint. This avoids the weirdness of a surface of immovable elements  
This becomes very important when analysing the shear stress between individual members in a composite crosssection. e.g. the shear stress in a welded joint between web and flange of an I beam.
This is very interesting. It says that just above the junction BB the long shear stress in the FLANGE is only 1.5 MPa, but immediately below that plane, in the WEB, the stress suddenly increases to 15 MPa. Obviously this is because the area is suddenly reduced from 100mm wide to 10mm wide  hence the stress increases 10 times. Makes sense.
Here, FEA matching the results will be more difficult because the shear stress is being constrained by proximity to the flange.
Solve Example 4 Using FEAUsing the same mesh refinement, the stress near the flange is sensitive to location. As we measure stress in the web as we approach the flange, the stress starts to deviate in the last few mm before it meeets the flange;

BEAMS WITH A CIRCULAR CROSS SECTION.
For a beam with a circular cross section the maximum shear stress occurs on the neutral axis. The maximum stress is calculated below.
Shear Stress Variation In Beams
From the previous examples you can see that the shear stress is not constant, but varies from the top to the bottom of the beam. If you were to calculate the shear stress at all points from top to bottom of a beam cross section and graph the magnitude of the stress you would get shear. stress diagrams as shownbelow.
Glues are often used today for making fabricated beams.
The shear strength of the glue should be greater than the
stress at the location of the glued joint. Alternatively
select a glue with a shear stress greater than the maximum
shear stress in the beam.
Calculating The Longitudinal Shear Force
The transverse shear force can be calculated using the procedure for shear force diagrams. The longitudinal (or horizontal) shear force cannot be calculated as directly. However, having calculated the shear stresses, the horizontal shear force can then be calculated. The horizontal shear force is used to find weld sizes and nail sizes in builtup beams. (Amoung other things)
Above shows a typical beam. You will note that the bending moment varies along the beam. If you look at part of a beam closely (below), you can calculate the bending stress acting on the left hand side and the right hand side of the segment.
The stresses acting on the right hand side of the segment are larger than those acting on the left hand side. This occurs because the bending moment is larger at the position of the right hand side.
For equilibrium of the segment in Figure 13,
Σ Horizontal forces = 0
R1 + S R2 = 0
Hence the horizontal shear force balances out the difference in the bending forces.
But...the horizontal shear force can be calculated from the shear stress equation! Tadaa!
Now stress is force over area. From Figure at right (yet another view of that element of the beam), the shear force is S1, and it acts over an area δx.b. Hence the shear stress becomes, 
Example:
FABRICATED BEAMS
Beams are often made up, or in some cases strengthened, by adding top and bottom plates.
In the nailed beam the nail carries the shear force on A.A. The weld carries the shear force between the top plate and the flange.
Following those numbers (above) in more detail:
A = 150 * 50 = 7500 mm2
y = 50mm
Q = Ay = 375E3 mm3
V = 1000N
I = 53E6 mm4
S/dx = QV/I = 7.08 so if we invert this,
dx/S = 0.141
But S=450N, so multiply by S;
dx = 63.6mm
Where does 965N/mm come from?
It is the load per mm of a 10mm weld, where, from the welded joints formula (fillet welds);
F = 0.707*stress*length*size
F = 0.707*stress*1*10
F = 7.07*stress
So rearranging to give;
Stress = F / 7.07 = 965/7.07 = 136 MPa (Which is the standard weld stress used in Ivanoff),
With S.F = 3, gives electrode rating of 136*3 = 410 MPa.
In other words;
F = 0.707*stress*length*size
F = 0.707 * (410/3) * 1mm * 10mm
= 965 N/mm