Chapter 26 Closed section beams 679 Chapter 27 Open section beams 718
26.3 Thin-walled rectangular section beam subjected
In Example 18.2 we determined the warping distribution in a thin-walled rectangular section beam which was not subjected to structural constraint. This free warping distri- bution (w0) was found to be linear around a cross-section and uniform along the length of the beam having values at the corners of
w0= ± T 8abG
b tb − a
ta
The effect of structural constraint, such as building one end of the beam in, is to reduce this free warping to zero at the built-in section so that direct stresses are induced which subsequently modify the shear stresses predicted by elementary torsion theory. These direct stresses must be self-equilibrating since the applied load is a pure torque.
The analysis of a rectangular section beam built-in at one end and subjected to a pure torque at the other is simplified if the section is idealized into one comprising four corner booms which are assumed to carry all the direct stresses together with shear–stress-only carrying walls. The assumption on which the idealization is based is that the direct stress distribution at any cross-section is directly proportional to the warping which has been suppressed. Therefore, the distribution of direct stress is linear around any cross-section and has values equal in magnitude but opposite in sign at opposite corners of a wall. This applies at all cross-sections since the free warping will be suppressed to some extent along the complete length of the beam. In Fig. 26.5(b) all the booms will have the same cross-sectional area from anti-symmetry and, from Eq.
(20.1) or (20.2)
B= ata
6 (2−1)+ btb
6 (2−1)= 1
6(ata+btb)
To the boom area B will be added existing concentrations of area such as connecting angle sections at the corners. The contributions of stringers may be included by allowing for their direct stress carrying capacity by increasing the actual wall thickness by an amount equal to the total stringer area on one wall before idealizing the section.
We have seen in Chapter 20 that the effect of structural idealization is to reduce the shear flow in the walls of a beam to a constant value between adjacent booms.
Fig. 26.5Idealization of a rectangular section beam subjected to torsion: (a) actual; (b) idealized.
Fig. 26.6ldealized rectangular section beam built-in at one end and subjected to a torque at the other.
In Fig. 26.6 suppose that the shear flows in the covers and webs at any section are qa and qb, respectively; from antisymmetry the shear flows in both covers will be qaand in both webs qb. The resultant of these shear flows is equivalent to the applied torque so that
T =
qp ds=2qaab
2 +2qbba 2 or
T =ab(qa+qb) (26.4)
We now use Eq. (17.19), i.e.
q=Gt ∂w
∂s +∂v
∂z
to determine qaand qb. Since the beam cross-section is doubly symmetrical the axis of twist passes through the centre of symmetry at any section so that, from Eq. (17.8)
∂vt
∂z =pRdθ
dz (26.5)
Therefore for the covers of the beam
∂vt
∂z = b 2
dθ
dz (26.6)
and for the webs
∂vt
∂z = a 2
dθ
dz (26.7)
26.3 Thin-walled rectangular section beam 689
Fig. 26.7Shear distortion of (a) an element of the top cover; (b) an element of the right hand web.
The elements of lengthδz of the covers and webs of the beam will warp into the shapes shown in Fig. 26.6 if T is positive (anticlockwise) and b/tb>a/ta. Clearly there must be compatibility of displacement at adjacent edges of the elements. From Fig. 26.7(a)
∂w
∂s = −w
a/2 (26.8)
and from Fig. 26.7(b)
∂w
∂s = w
b/2 (26.9)
Substituting for∂w/∂s and∂vt/∂z in Eq. (17.19) separately for the covers and webs, we obtain
qa=Gta −2w
a + b 2
dθ dz
qb=Gtb 2w
b + a 2
dθ dz
(26.10) Now substituting for qaand qbin Eq. (26.4) we have
T =abG ta −2w
a + b 2
dθ dz
+tb
2w b + a
2 dθ dz
Rearranging
dθ
dz = 4w(bta−atb)
ab(bta+atb) + 2T
abG(bta+atb) (26.11) If we now substitute for dθ/dz from Eq. (26.11) into Eqs (26.10) we have
qa= −4wGtbta
bta+atb + Tta
a(bta+atb) qb= 4wGtbta
bta+atb + Ttb
b(bta+atb) (26.12) Equations (26.11) and (26.12) give the rate of twist and the shear flows (and hence shear stresses) in the beam in terms of the warping w and the applied torque T . Their derivation is based on the compatibility of displacement which exists at the cover/boom/web
Fig. 26.8Equilibrium of boom element.
junctions. We shall now use the further condition of equilibrium between the shears in the covers and webs and the direct load in the booms to obtain expressions for the warping displacement and the distributions of boom stress and load. Thus, for the equilibrium of an element of the top right-hand boom shown in Fig. 26.8
σz+∂σz
∂zδz
B−σzB+qaδz−qbδz=0 i.e.
B∂σz
∂z +qa−qb=0 (26.13)
Now
σz=E∂w
∂z (see Chapter 1) Substituting forσzin Eq. (26.13) we obtain
BE∂2w
∂z2 +qa−qb=0 (26.14)
Replacing qaand qbfrom Eqs (26.12) gives BE∂2w
∂z2 − 8Gtbta
bta+atbw= −T ab
(bta−atb) (bta+atb) or
∂2w
∂z2 −à2w= − T abBE
(bta−atb)
(bta+atb) (26.15)
26.3 Thin-walled rectangular section beam 691 where
à2= 8Gtbta BE(bta+atb)
The differential equation (26.15) is of standard form and its solution is w=C coshàz+D sinhàz+ T
8abG b
tb − a ta
(26.16) in which the last term is seen to be the free warping displacement w0of the top right- hand corner boom. The constants C and D in Eq. (26.16) are found from the boundary conditions of the beam. In this particular case the warping w = 0 at the built-in end and the direct strain∂w/∂z=0 at the free end where there is no direct load. From the first of these
C = − T 8abG
b tb − a
ta
= −w0 and from the second
D=w0tanhàL Then
w=w0(1−coshàz+tanhàL sinhàz) (26.17) or rearranging
w=w0 1− coshà(L−z) coshàL
(26.18) The variation of direct stress in the boom is obtained fromσz=E∂w/∂z and Eq. (26.18), i.e.
σz=àEw0sinhà(L−z)
coshàL (26.19)
and the variation of boom load P is then
P=Bσz=BàEwosinhà(L−z)
coshàL (26.20)
Substituting for w in Eqs (26.12) and rearranging, we obtain the shear stress distribution in the covers and webs. Thus
τa= qa ta = T
2abta 1+ (bta−atb) (bta+atb)
coshà(L−z) coshàL
(26.21) τb= qb
tb = T
2abtb 1− (bta−atb) (bta+atb)
coshà(L−z) coshàL
(26.22) Inspection of Eqs (26.21) and (26.22) shows that the shear stress distributions each comprise two parts. The first terms, T/2abtaand T/2abtb, are the shear stresses pre- dicted by elementary theory (see Section 18.1), while the hyperbolic second terms
Fig. 26.9Shear stress distributions along the beam of Fig. 11.5.
represent the effects of the warping restraint. Clearly, for an anticlockwise torque and bta>atb, the effect of this constraint is to increase the shear stress in the covers over that predicted by elementary theory and decrease the shear stress in the webs. It may also be noted that for bta to be greater than atb for the beam of Fig. 26.6, in which a>b, then ta must be appreciably greater than tb so that T/2abta<T/2abtb. Also at the built-in end (z=0), Eqs (26.21) and (26.22) reduce toτa=T/a(bta+atb) and τb=T/b(bta+atb) so that even thoughτb is reduced by the axial constraint andτa increased,τbis still greater thanτa. It should also be noted that these values ofτaand τbat the built-in end may be obtained using the method of Section 26.2 and that these are the values of shear stress irrespective of whether the section has been idealized or not. In other words, the presence of intermediate stringers and/or direct stress carrying walls does not affect the shear flows at the built-in end since the direct stress gradient at this section is zero (see Section 26.2 and Eq. (17.2)) except in the corner booms.
Finally, when both z and L become large, i.e. at the free end of a long, slender beam τa→ T
2abta and τb→ T 2abtb The above situation is shown in Fig. 26.9.
In the particular case when bta=atbwe see that the second terms on the right-hand side of Eqs (26.21) and (26.22) disappear and no constraint effects are present; the direct stress of Eqs (26.19) is also zero since w0=0 (see Example 18.2).
The rate of twist is obtained by substituting for w from Eq. (26.18) in Eq. (26.11).
Thus dθ
dz = T
2a2b2G b
tb + a ta 1−
bta−atb bta+atb
2
coshà(L−z) coshàL
(26.23) in which we see that again the expression on the right-hand side comprises the rate of twist given by elementary theory, T (b/tb+a/ta)/2a2b2G (see Section 18.1), together with a correction due to the warping restraint. Clearly the rate of twist is always reduced by the constraint since (bta−atb)2is always positive. Integration of Eq. (26.23) gives the distribution of angle of twist along the length of the beam, the boundary condition in this case beingθ=0 at z=0.
26.3 Thin-walled rectangular section beam 693
Example 26.3
A uniform four boom box of span 5 m is 500 mm wide by 20 mm deep and has four corner booms each of cross-sectional area 800 mm2, its wall thickness is 1.0 mm. If the box is subjected to a uniformly distributed torque loading of 20 Nm/mm along its length and it is supported at each end such that complete freedom of warping exists at the end cross-sections calculate the angle of twist at the mid-span section. Take G=20 000 N/mm2and G/E=0.36.
The reactive torques at each support are=20×5000/2=50 000 Nm
Taking the origin for z at the mid-span of the beam the torque at any section is given by
T (z)=20(2500−z)−50 000= −20z Nm Substituting in Eq. (26.16) we obtain
w=C coshàz+D sinhàz− 20zì103(b−a) 8abGt The boundary conditions are:
w=0 when z=0 from symmetry and∂w/∂z=0 when z=L (L=2500 mm) From the first of these C=0 while from the second
D= 20×103(b−a) 8àabGt coshàL Therefore
w= 20(b−a)×103 8abGt
sinhàz àcoshàL −z
(i) Further
à2= 8Gt
AE(b+a) = 8×0.36×1.0
800(200+500) =5.14×10−6 so that Eq. (i) becomes
w= −3.75ì10−4(3.04 sinhàz−z) (ii) Substituting for w, etc. in Eq. (26.11)
dθ
dz =10−8(1.95 sinhàz−3.49z) Hence
θ=10−8 1.95
à coshàz−1.75z2+F
(iii) When z=L (2500 mm) θ=0. Then, from Eq. (iii)
F=10.8×106
so that
θ=10−8(859 coshàz−1.75z2+10.8ì106) (iv) At mid-span where z=0, from Eq. (iv)
θ=0.108 rad or θ=6.2◦