Chapter 23 Wings 607 Chapter 24 Fuselage frames and wing ribs 638
21.2 Open and closed section beams
Fig. 21.3Shear flow (N/mm) distribution at Section AA in Example 21.1.
21.2 Open and closed section beams
We shall now consider the more general case of a beam tapered in two directions along its length and comprising an arrangement of booms and skin. Practical examples of such a beam are complete wings and fuselages. The beam may be of open or closed section; the effects of taper are determined in an identical manner in either case.
Figure 21.4(a) shows a short lengthδz of a beam carrying shear loads Sx and Sy at the section z; Sx and Syare positive when acting in the directions shown. Note that if the beam were of open cross-section the shear loads would be applied through its shear centre so that no twisting of the beam occurred. In addition to shear loads the beam is subjected to bending moments Mx and My which produce direct stressesσz in the booms and skin. Suppose that in the rth boom the direct stress in a direction parallel to the z axis isσz,r, which may be found using either Eq. (16.18) or Eq. (16.19). The component Pz,rof the axial load Prin the rth boom is then given by
Pz,r=σz,rBr (21.8)
where Bris the cross-sectional area of the rth boom.
From Fig. 21.4(b)
Py,r =Pz,rδyr
δz (21.9)
Further, from Fig. 21.4(c)
Px,r =Py,rδxr δyr or, substituting for Py,r from Eq. (21.9)
Px,r =Pz,rδxr
δz (21.10)
The axial load Pris then given by
Pr=(P2x,r+Py,r2 +P2z,r)1/2 (21.11)
Fig. 21.4Effect of taper on the analysis of open and closed section beams.
or, alternatively
Pr =Pz,r(δx2r+δy2r+δz2)1/2
δz (21.12)
The applied shear loads Sxand Syare reacted by the resultants of the shear flows in the skin panels and webs, together with the components Px,r and Py,rof the axial loads in the booms. Therefore, if Sx,wand Sy,ware the resultants of the skin and web shear flows and there is a total of m booms in the section
Sx=Sx,w+ m r=1
Px,r Sy =Sy,w+ m r=1
Py,r (21.13)
21.2 Open and closed section beams 589
Fig. 21.5Modification of moment equation in shear of closed section beams due to boom load.
Substituting in Eq. (21.13) for Px,r and Py,rfrom Eqs (21.10) and (21.9) we have Sx =Sx,w+
m r=1
Pz,rδxr
δz Sy =Sy,w+ m r=1
Pz,rδyr
δz (21.14)
Hence
Sx,w=Sx− m r=1
Pz,rδxr
δz Sy,w =Sy− m r=1
Pz,rδyr
δz (21.15)
The shear flow distribution in an open section beam is now obtained using Eq. (20.6) in which Sx is replaced by Sx,wand Syby Sy,wfrom Eq. (21.15). Similarly for a closed section beam, Sxand Syin Eq. (20.11) are replaced by Sx,wand Sy,w. In the latter case the moment equation (Eq. (17.17)) requires modification due to the presence of the boom load components Px,r and Py,r. Thus from Fig. 21.5 we see that Eq. (17.17) becomes
Sxη0−Syξ0=
qbp ds+2Aqs,0− m r=1
Px,rηr+ m r=1
Py,rξr (21.16) Equation (21.16) is directly applicable to a tapered beam subjected to forces positioned in relation to the moment centre as shown. Care must be taken in a particular problem to ensure that the moments of the forces are given the correct sign.
Example 21.2
The cantilever beam shown in Fig. 21.6 is uniformly tapered along its length in both x and y directions and carries a load of 100 kN at its free end. Calculate the forces in the booms and the shear flow distribution in the walls at a section 2 m from the built-in end if the booms resist all the direct stresses while the walls are effective only in shear.
Each corner boom has a cross-sectional area of 900 mm2 while both central booms have cross-sectional areas of 1200 mm2.
The internal force system at a section 2 m from the built-in end of the beam is Sy=100 kN Sx =0 Mx = −100×2= −200 kN m My =0
Fig. 21.6(a) Beam of Example 21.2; (b) section 2 m from built-in end.
The beam has a doubly symmetrical cross-section so that Ixy=0 and Eq. (16.18) reduces to
σz= Mxy
Ixx (i)
in which, for the beam section shown in Fig. 21.6(b)
Ixx=4×900×3002+2×1200×3002=5.4×108mm4 Then
σz,r= −200×106 5.4×108 yr or
σz,r = −0.37yr (ii)
Hence
Pz,r= −0.37yrBr (iii)
The value of Pz,r is calculated from Eq. (iii) in column in Table 21.1; Px,r and Py,rfollow from Eqs (21.10) and (21.9), respectively in columnsand. The axial load Pr, column, is given by [2+2+2]1/2and has the same sign as Pz,r(see Eq. (21.12)). The moments of Px,r and Py,r are calculated for a moment centre at the centre of symmetry with anticlockwise moments taken as positive. Note that in Table 21.1, Px,r and Py,r are positive when they act in the positive directions of the section x and y axes, respectively; the distancesηr andξr of the lines of action of Px,r and
21.2 Open and closed section beams 591 Table 21.1
Pz,r δxr/δz δyr/δz Px,r Py,r Pr ξr ηr Px,rηr Py,rξr
Boom (kN) (kN) (kN) (kN) (m) (m) (kN m) (kN m)
1 −100 0.1 −0.05 −10 5 −101.3 0.6 0.3 3 −3
2 −133 0 −0.05 0 6.7 −177.3 0 0.3 0 0
3 −100 −0.1 −0.05 10 5 −101.3 0.6 0.3 −3 3
4 100 −0.1 0.05 −10 5 101.3 0.6 0.3 −3 3
5 133 0 0.05 0 6.7 177.3 0 0.3 0 0
6 100 0.1 0.05 10 5 101.3 0.6 0.3 3 −3
Py,r from the moment centre are not given signs since it is simpler to determine the sign of each moment, Px,rηr and Py,rξr, by referring to the directions of Px,r and Py,r individually.
From column
6 r=1
Py,r =33.4 kN From column
6 r=1
Px,rηr =0 From column
6 r=1
Py,rξr=0 From Eq. (21.15)
Sx,w=0 Sy,w =100−33.4=66.6 kN
The shear flow distribution in the walls of the beam is now found using the method described in Section 20.3. Since, for this beam, Ixy=0 and Sx=Sx,w=0, Eq. (20.11) reduces to
qs= −Sy,w Ixx
n r=1
Bryr+qs,0 (iv)
We now ‘cut’ one of the walls, say 16. The resulting ‘open section’ shear flow is given by
qb= −66.6×103 5.4×108
n r=1
Bryr
or
qb= −1.23×10−4 n r=1
Bryr (v)
Thus
qb,16=0
qb,12=0−1.23×10−4×900×300= −33.2 N/mm qb,23= −33.2−1.23×10−4×1200×300= −77.5 N/mm qb,34= −77.5−1.23×10−4×900×300= −110.7 N/mm qb,45= −77.5 N/mm (from symmetry)
qb,56= −33.2 N/mm (from symmetry)
giving the distribution shown in Fig. 21.7. Taking moments about the centre of symmetry we have, from Eq. (21.16)
−100×103×600=2×33.2×600×300+2×77.5×600×300 +110.7×600×600+2×1200×600qs,0
from which qs,0= −97.0 N/mm (i.e. clockwise). The complete shear flow distribution is found by adding the value of qs,0to the qbshear flow distribution of Fig. 21.7 and is shown in Fig. 21.8.
Fig. 21.7‘Open section’ shear flow (N/mm) distribution in beam section of Example 21.2.
Fig. 21.8Shear flow (N/mm) distribution in beam section of Example 21.2.