T ORSION OF T HIN -W ALLED B EAMS
Chapter 16 Bending of open and closed,
17.2 Shear of open section beams
∂vt
∂z =pdθ
dz −xRsinψdθ
dz +yRcosψdθ
dz (17.9)
Also from Eq. (17.7)
∂vt
∂z =pdθ dz +du
dz cosψ+ dv
dzsinψ (17.10)
Comparing the coefficients of Eqs (17.9) and (17.10) we see that xR= −dv/dz
dθ/dz yR= du/dz
dθ/dz (17.11)
17.2 Shear of open section beams
The open section beam of arbitrary section shown in Fig. 17.5 supports shear loads Sx and Sy such that there is no twisting of the beam cross-section. For this condition to be valid the shear loads must both pass through a particular point in the cross-section known as the shear centre.
Since there are no hoop stresses in the beam the shear flows and direct stresses acting on an element of the beam wall are related by Eq. (17.2), i.e.
∂q
∂s +t∂σz
∂z =0
We assume that the direct stresses are obtained with sufficient accuracy from basic bending theory so that from Eq. (16.18)
∂σz
∂z = [(∂My/∂z)Ixx−(∂Mx/∂z)Ixy]
IxxIyy−Ixy2 x+ [(∂Mx/∂z)Iyy−(∂My/∂z)Ixy] IxxIyy−Ixy2 y
Fig. 17.5Shear loading of open section beam.
Using the relationships of Eqs (16.23) and (16.24), i.e. ∂My/∂z=Sx, etc., this expression becomes
∂σz
∂z = (SxIxx−SyIxy)
IxxIyy−Ixy2 x+ (SyIyy−SxIxy) IxxIyy−Ixy2 y Substituting for∂σz/∂z in Eq. (17.2) gives
∂q
∂s = −(SxIxx−SyIxy)
IxxIyy−Ixy2 tx− (SyIyy−SxIxy)
IxxIyy−Ixy2 ty (17.12) Integrating Eq. (17.12) with respect to s from some origin for s to any point around the cross-section, we obtain
s
0
∂q
∂sds= −
SxIxx−SyIxy IxxIyy−Ixy2
s 0
tx ds−
SyIyy−SxIxy IxxIyy−Ixy2
s 0
ty ds (17.13)
If the origin for s is taken at the open edge of the cross-section, then q=0 when s=0 and Eq. (17.13) becomes
qs= −
SxIxx−SyIxy IxxIyy−Ixy2
s 0
tx ds−
SyIyy−SxIxy IxxIyy−Ixy2
s 0
ty ds (17.14) For a section having either Cx or Cy as an axis of symmetry Ixy=0 and Eq. (17.14) reduces to
qs= −Sx Iyy
s
0
tx ds− Sy Ixx
s
0
ty ds
Example 17.1
Determine the shear flow distribution in the thin-walled Z-section shown in Fig. 17.6 due to a shear load Syapplied through the shear centre of the section.
The origin for our system of reference axes coincides with the centroid of the section at the mid-point of the web. From antisymmetry we also deduce by inspection that the shear centre occupies the same position. Since Sy is applied through the shear centre then no torsion exists and the shear flow distribution is given by Eq. (17.14) in which Sx=0, i.e.
qs= SyIxy IxxIyy−Ixy2
s
0
tx ds− SyIyy IxxIyy−Ixy2
s
0
ty ds or
qs= Sy IxxIyy−Ixy2
Ixy
s
0
tx ds−Iyy s
0
ty ds
(i)
17.2 Shear of open section beams 509
Fig. 17.6Shear loaded Z-section of Example 17.1.
The second moments of area of the section have previously been determined in Example 16.14 and are
Ixx= h3t
3 , Iyy= h3t
12, Ixy= h3t 8 Substituting these values in Eq. (i) we obtain
qs= Sy h3
s
0
(10.32x−6.84y)ds (ii)
On the bottom flange 12, y= −h/2 and x= −h/2+s1, where 0≤s1≤h/2. Therefore q12= Sy
h3 s1
0
(10.32s1−1.74h)ds1 giving
q12= Sy
h3(5.16s21−1.74hs1) (iii) Hence at 1 (s1=0), q1=0 and at 2 (s1=h/2), q2=0.42Sy/h. Further examination of Eq. (iii) shows that the shear flow distribution on the bottom flange is parabolic with a change of sign (i.e. direction) at s1=0.336h. For values of s1<0.336h, q12is negative and therefore in the opposite direction to s1.
In the web 23, y= −h/2+s2, where 0≤s2≤h and x=0. Then q23= Sy
h3 s2
0
(3.42h−6.84s2)ds2+q2 (iv)
Fig. 17.7Shear flow distribution in Z-section of Example 17.1.
We note in Eq. (iv) that the shear flow is not zero when s2= 0 but equal to the value obtained by inserting s1 = h/2 in Eq. (iii), i.e. q2= 0.42Sy/h. Integration of Eq. (iv) yields
q23= Sy
h3(0.42h2+3.42hs2−3.42s22) (v) This distribution is symmetrical about Cx with a maximum value at s2=h/2( y=0) and the shear flow is positive at all points in the web.
The shear flow distribution in the upper flange may be deduced from antisymmetry so that the complete distribution is of the form shown in Fig. 17.7.
17.2.1 Shear centre
We have defined the position of the shear centre as that point in the cross-section through which shear loads produce no twisting. It may be shown by use of the reciprocal theorem that this point is also the centre of twist of sections subjected to torsion. There are, however, some important exceptions to this general rule as we shall observe in Section 26.1. Clearly, in the majority of practical cases it is impossible to guarantee that a shear load will act through the shear centre of a section. Equally apparent is the fact that any shear load may be represented by the combination of the shear load applied through the shear centre and a torque. The stresses produced by the separate actions of torsion and shear may then be added by superposition. It is therefore necessary to know the location of the shear centre in all types of section or to calculate its position. Where a cross-section has an axis of symmetry the shear centre must, of course, lie on this axis. For cruciform or angle sections of the type shown in Fig. 17.8 the shear centre is located at the intersection of the sides since the resultant internal shear loads all pass through these points.
17.2 Shear of open section beams 511
Fig. 17.8Shear centre position for type of open section beam shown.
Example 17.2
Calculate the position of the shear centre of the thin-walled channel section shown in Fig. 17.9. The thickness t of the walls is constant.
The shear centre S lies on the horizontal axis of symmetry at some distanceξS, say, from the web. If we apply an arbitrary shear load Sy through the shear centre then the shear flow distribution is given by Eq. (17.14) and the moment about any point in the cross-section produced by these shear flows is equivalent to the moment of the applied shear load. Sy appears on both sides of the resulting equation and may therefore be eliminated to leaveξS.
For the channel section, Cx is an axis of symmetry so that Ixy=0. Also Sx=0 and therefore Eq. (17.14) simplifies to
qs= −Sy Ixx
s
0
ty ds (i)
where
Ixx=2bt h
2 2
+th3 12 = h3t
12
1+ 6b h
Fig. 17.9Determination of shear centre position of channel section of Example 17.2.
Substituting for Ixxin Eq. (i) we have qs= −12Sy
h3(1+6b/h) s
0
y ds (ii)
The amount of computation involved may be reduced by giving some thought to the requirements of the problem. In this case we are asked to find the position of the shear centre only, not a complete shear flow distribution. From symmetry it is clear that the moments of the resultant shears on the top and bottom flanges about the mid-point of the web are numerically equal and act in the same rotational sense. Furthermore, the moment of the web shear about the same point is zero. We deduce that it is only necessary to obtain the shear flow distribution on either the top or bottom flange for a solution. Alternatively, choosing a web/flange junction as a moment centre leads to the same conclusion.
On the bottom flange, y= −h/2 so that from Eq. (ii) we have q12= 6Sy
h2(1+6b/h)s1 (iii)
Equating the clockwise moments of the internal shears about the mid-point of the web to the clockwise moment of the applied shear load about the same point gives
Syξs=2 b
0
q12h 2ds1 or, by substitution from Eq. (iii)
Syξs=2 b
0
6Sy h2(1+6b/h)
h 2s1ds1 from which
ξs= 3b2
h(1+6b/h) (iv)
In the case of an unsymmetrical section, the coordinates (ξS,ηS) of the shear centre referred to some convenient point in the cross-section would be obtained by first deter- miningξSin a similar manner to that of Example 17.2 and then findingηSby applying a shear load Sxthrough the shear centre. In both cases the choice of a web/flange junction as a moment centre reduces the amount of computation.