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1. THE RELATIONSHIP BETWEEN BENDING MOMENT AND SHEAR FORCE
Consider a beam subject to bending and transverse shear. At some distance along the x direction further
consider a short length δx. Over this length the bending moment increases by dM and the shear force
increases by dF.
Figure 1
If we take this section out of the beam we must add the forces and moments that were previously exerted
by the material in order to maintain equilibrium. Equilibrium of forces and moments exists at all points soit is convenient to look at the equilibrium of moments at the bottom right corner.
The shear force should be equal to the integration of the shear stress over the section.
dF = τ dA = τ b dy The product τ b is called the shear flow and denoted q.∫=top
bottom
dy bτF ∫=top
bottom
dyqF
For standard sections such as ‘I’, ‘T’, ‘U’ and ‘L’, there is vertical shear
flow in the flange and horizontal shear flow in the web in the horizontaldirection. For this reason ‘t’ is often used for the thickness of the flange or
web. and the same applies to any direction such asdyqF yy ∫= dzqF zz ∫=
The importance of this work will be clear later in the tutorial. The units of q
are N/m. If the cross section is not a uniform shape, e.g. a Tee section, t or b
is not constant and the maximum shear stress occurs where τy/b is a
maximum.
Figure 5
WORKED EXAMPLE No.1
Derive an equation for the shear stress and shear flow distribution on a rectangular cross section.
Compare the average and maximum shear stress. Show that the shear force is obtained by integrating
the shear flow over the section.
SOLUTION
Let D/2 = C. In this case the shaded area is B (C – y) and y = (C + y)/2 by = B
Iz = BD3/12 A = BD
( ) ( ) ( ) ( )( )
( ) ( )⎟⎟
⎠
⎞⎜⎜
⎝
⎛ −=
−=
−=
+−=
+−=
2
2
2
22
2
22
3z
yy
C
y1
2A
3F
4AC
yCF6
AD
yCF6
DB2
yCyCF12
2BI
yCyCBFτ
Plotting τy against y we find the shear stress varies as a parabola from zero at the bottom to a
maximum at the centroid and zero at the top.
Figure 6
The maximum shear stress occurs at y = 0 3F/2Aτmax =
The transverse shear force on the section is Fy and the mean shear stress is F/A so 3/2xττ meanmax =
The shear flow is easily obtained since the width is constant.⎟⎟
A circular section can be solved with some difficulty in the same way. For a circular section of radius R:
4
22
yR π3
yR F4τ
−= mean2max τ
3
4
R π3
F4τ ==
5.2 NON UNIFORM SECTIONS
If the shape is hollow or does not have a constant
width b, the problem is more complex. For example
consider a triangular section.
bI
yAFτ
yz
yy =
Because b is a function of y, the maximum shear stress
does not occur at the centroid but at the point shown.
Figure 7
6 SUDDEN DISCONTINUITIES
With ‘T’, ‘I’, ‘U’ and ‘L’ sections, the width ‘b’ or ‘t’ suddenly changes at the junctions of the web and
flange so the shear stress suddenly changes as the ratio of the widths. This is best illustrated with a
worked example.
WORKED EXAMPLE No.2
Determine the shear stress at the junction of the top flange and web for the section shown when ashear force of 40 kN acts vertically down on the section.
Figure 8
SOLUTION
First calculate the second moment of area. The tabular method is used here. Divide the shape into
three sections A, B and C. First determine the position of the centroid from the bottom edge.
Consider a cantilever beam with a point load that acts vertically but not through the centroid. The beam
bends but in addition it twists as shown.
Figure 22
This produces additional torsional stress in the beam. In the case of a symmetrical beam like that shown,
the solution is simply to apply the load so that it acts through the centroid. In other cases we must apply
the load so that it acts through some other point that results in no twisting and this point is called thecentre of shear or shear centre.
The shear centre is that point through which the loads must act if there is to be no twisting, or torsion.
The shear centre is always located on the axis of symmetry; therefore, if a member has two axes of
symmetry, the shear centre will be the intersection of the two axes. If there is only one axis of
symmetry, the shear centre is somewhere on that axis.
Here are some examples of sections that are symmetrical in two axis.
Figure 23
A ‘U’ section is a good example of one where the shear centre is difficult to find. As the following
example shows, it occurs off the section altogether. Note that when the sections are made from thin sheetsof material, the problems are easier to resolve and most of this work concerns the shear stress in thin
Determine the position of the shear centre in terms of dimension ‘a’ for the ‘U’ section shown madefrom thin metal sheet of thickness ‘t’. The force is applied vertically.
Figure 24
The centre of shear for the ‘U’ channel is to the left of the section as drawn. First calculate the position of the centroid. This must be on the horizontal centre line so we need to calculate the
position from the vertical edge.Area z A z
A at a/2 a2t/2
B at a/t a2t/2
C 2at t/2 at2
Total 4at a2t + at
2
For the section z = ( a2t + at2)/4at if t is small the t2 term may be ignored so z =a2t /4at = a/4
Next we calculate the shear stress in the section due to transverse shear force F
bI
yFAτ
z= Iz is the second moment of area of the section about the z axis.
For the vertical section I = t(2a)3/12 = (2/3)a
3t
For the flanges we may approximate with I = A x a2 where A = a t so I = 2 x a
3t
Adding we get Iz = (2/3)a3t + 2 x a3t
ta3
8I 3
z = tb8a
y3FA
bI
yFAτ
3z
==
SHEAR DISTRIBUTION IN THE FLANGE
Figure 25
Area A = (a-z)t y = atb8a
y3FA τ
3= and in this case b = t the thickness of the flange.