T ORSION OF T HIN -W ALLED B EAMS
Chapter 16 Bending of open and closed,
20.3 Effect of idealization on the analysis of open and closed section beams
20.3.2 Shear of open section beams
The derivation of Eq. (17.14) for the shear flow distribution in the cross-section of an open section beam is based on the equilibrium equation (17.2). The thickness t in this
C
Fig. 20.6(a) Elemental length of shear loaded open section beam with booms; (b) equilibrium of boom element.
equation refers to the direct stress carrying thickness tDof the skin. Equation (17.14) may therefore be rewritten
qs= −
SxIxx−SyIxy IxxIyy−Ixy2
s 0
tDx ds−
SyIyy−SxIxy IxxIyy−Ixy2
s 0
tDy ds (20.3)
in which tD=t if the skin is fully effective in carrying direct stress or tD=0 if the skin is assumed to carry only shear stresses. Again the section properties in Eq. (20.3) refer to the direct stress carrying area of the section since they are those which feature in Eqs (16.18) and (16.19).
Equation (20.3) makes no provision for the effects of booms which cause discon- tinuities in the skin and therefore interrupt the shear flow. Consider the equilibrium of the rth boom in the elemental length of beam shown in Fig. 20.6(a) which carries shear loads Sxand Syacting through its shear centre S. These shear loads produce direct stresses due to bending in the booms and skin and shear stresses in the skin. Suppose that the shear flows in the skin adjacent to the rth boom of cross-sectional area Brare q1and q2. Then, from Fig. 20.6(b)
σz+ ∂σz
∂z δz
Br−σzBr+q2δz−q1δz=0
which simplifies to
q2−q1= −∂σz
∂zBr (20.4)
20.3 Effect of idealization on the analysis of open and closed section beams 565
Substituting forσzin Eq. (20.4) from (16.18) we have q2−q1= −
(∂My/∂z)Ixx−(∂Mx/∂z)Ixy IxxIyy−Ixy2
Brxr
−
(∂Mx/∂z)Iyy−(∂My/∂z)Ixy IxxIyy−Ixy2
Bryr or, using the relationships of Eqs (16.23) and (16.24)
q2−q1= −
SxIxx−SyIxy IxxIyy−Ixy2
Brxr−
SyIyy−SxIxy IxxIyy−Ixy2
Bryr (20.5) Equation (20.5) gives the change in shear flow induced by a boom which itself is subjected to a direct load (σzBr). Each time a boom is encountered the shear flow is incremented by this amount so that if, at any distance s around the profile of the section, n booms have been passed, the shear flow at the point is given by
qs= −
SxIxx−SyIxy IxxIyy−Ixy2
s 0
tDx ds+ n r=1
Brxr
−
SyIyy−SxIxy IxxIyy−Ixy2
s 0
tDy ds+ n r=1
Bryr
(20.6)
Example 20.3
Calculate the shear flow distribution in the channel section shown in Fig. 20.7 produced by a vertical shear load of 4.8 kN acting through its shear centre. Assume that the walls of the section are only effective in resisting shear stresses while the booms, each of area 300 mm2, carry all the direct stresses.
The effective direct stress carrying thickness tDof the walls of the section is zero so that the centroid of area and the section properties refer to the boom areas only. Since Cx (and Cy as far as the boom areas are concerned) is an axis of symmetry Ixy=0; also Sx=0 and Eq. (20.6) thereby reduces to
qs= −Sy Ixx
n r=1
Bryr (i)
in which Ixx=4×300×2002=48×106mm4. Substituting the values of Syand Ixx in Eq. (i) gives
qs= −4.8×103 48×106
n r=1
Bryr= −10−4 n r=1
Bryr (ii)
At the outside of boom 1, qs=0. As boom 1 is crossed the shear flow changes by an amount given by
q1= −10−4×300×200= −6 N/mm
Fig. 20.7Idealized channel section of Example 20.3.
Hence q12= −6 N/mm since, from Eq. (i), it can be seen that no further changes in shear flow occur until the next boom (2) is crossed. Hence
q23= −6−10−4×300×200= −12 N/mm Similarly
q34= −12−10−4×300×(−200)= −6 N/mm while, finally, at the outside of boom 4 the shear flow is
−6−10−4×300×(−200)=0
as expected. The complete shear flow distribution is shown in Fig. 20.8.
It can be seen from Eq. (i) in Example 20.3 that the analysis of a beam section which has been idealized into a combination of direct stress carrying booms and shear
Fig. 20.8Shear flow in channel section of Example 20.3.
20.3 Effect of idealization on the analysis of open and closed section beams 567
O
Fig. 20.9Curved web with constant shear flow.
stress only carrying skin gives constant values of the shear flow in the skin between the booms; the actual distribution of shear flows is therefore lost. What remains is in fact the average of the shear flow, as can be seen by referring to Example 20.3. Analysis of the unidealized channel section would result in a parabolic distribution of shear flow in the web 23 whose resultant is statically equivalent to the externally applied shear load of 4.8 kN. In Fig. 20.8 the resultant of the constant shear flow in the web 23 is 12×400=4800 N=4.8 kN. It follows that this constant value of shear flow is the average of the parabolically distributed shear flows in the unidealized section.
The result, from the idealization of a beam section, of a constant shear flow between booms may be used to advantage in parts of the analysis. Suppose that the curved web 12 in Fig. 20.9 has booms at its extremities and that the shear flow q12in the web is constant. The shear force on an elementδs of the web is q12δs, whose components horizontally and vertically are q12δs cosφand q12δs sinφ. The resultant, parallel to the x axis, Sx, of q12is therefore given by
Sx = 2
1
q12cosφds or
Sx =q12 2
1
cosφds which, from Fig. 20.9, may be written
Sx=q12 2
1
dx=q12(x2−x1) (20.7) Similarly the resultant of q12parallel to the y axis is
Sy=q12(y2−y1) (20.8)
Thus the resultant, in a given direction, of a constant shear flow acting on a web is the value of the shear flow multiplied by the projection on that direction of the web.
The resultant shear force S on the web of Fig. 20.9 is
S= S2x+Sy2=q12 (x2−x1)2+(y2−y1)2 i.e.
S=q12L12 (20.9)
Therefore, the resultant shear force acting on the web is the product of the shear flow and the length of the straight line joining the ends of the web; clearly the direction of the resultant is parallel to this line.
The moment Mqproduced by the shear flow q12about any point O in the plane of the web is, from Fig. 20.10
Mq= 2
1
q12p ds=q12 2
1
2 dA or
Mq=2Aq12 (20.10)
in which A is the area enclosed by the web and the lines joining the ends of the web to the point O. This result may be used to determine the distance of the line of action of the resultant shear force from any point. From Fig. 20.10
Se=2Aq12 from which
e= 2A S q12
Fig. 20.10Moment produced by a constant shear flow.
20.3 Effect of idealization on the analysis of open and closed section beams 569
Substituting for q12from Eq. (20.9) gives e= 2A
L12