T ORSION OF T HIN -W ALLED B EAMS
Chapter 16 Bending of open and closed,
18.1 Torsion of closed section beams
18.1.1 Displacements associated with the Bredt–Batho
The relationship between q and shear strainγestablished in Eq. (17.19), namely q=Gt
∂w
∂s + ∂vt
∂z
18.1 Torsion of closed section beams 529
Fig. 18.3Sign convention for swept areas.
is valid for the pure torsion case where q is constant. Differentiating this expression with respect to z we have
∂q
∂z =Gt ∂2w
∂z∂s+ ∂2vt
∂z2
=0 or
∂
∂s ∂w
∂z
+∂2vt
∂z2 =0 (18.2)
In the absence of direct stresses the longitudinal strain∂w/∂z(=εz) is zero so that
∂2vt
∂z2 =0 Hence from Eq. (17.7)
pd2θ dz2 + d2u
dz2 cosψ+ d2v
dz2 sinψ=0 (18.3)
For Eq. (18.3) to hold for all points around the section wall, in other words for all values ofψ
d2θ
dz2 =0, d2u
dz2 =0, d2v dz2 =0
It follows that θ=Az+B, u=Cz+D, v=Ez+F, where A, B, C, D, E and F are unknown constants. Thusθ, u andvare all linear functions of z.
Equation (17.22), relating the rate of twist to the variable shear flow qsdeveloped in a shear loaded closed section beam, is also valid for the case qs=q=constant. Hence
dθ dz = q
2A ds
Gt
which becomes, on substituting for q from Eq. (18.1) dθ
dz = T 4A2
ds
Gt (18.4)
The warping distribution produced by a varying shear flow, as defined by Eq. (17.25) for axes having their origin at the centre of twist, is also applicable to the case of a constant shear flow. Thus
ws−w0=q s
0
ds Gt −AOs
A q ds
Gt Replacing q from Eq. (18.1) we have
ws−w0= Tδ 2A
δOs δ − AOs
A
(18.5) where
δ= ds
Gt and δOs = s
0
ds Gt
The sign of the warping displacement in Eq. (18.5) is governed by the sign of the applied torque T and the signs of the parametersδOsand AOs. Having specified initially that a positive torque is anticlockwise, the signs ofδOsand AOs are fixed in thatδOsis positive when s is positive, i.e. s is taken as positive in an anticlockwise sense, and AOs is positive when, as before, p (see Fig. 18.3) is positive.
We have noted that the longitudinal strainεzis zero in a closed section beam subjected to a pure torque. This means that all sections of the beam must possess identical warp- ing distributions. In other words longitudinal generators of the beam surface remain unchanged in length although subjected to axial displacement.
Example 18.1
A thin-walled circular section beam has a diameter of 200 mm and is 2 m long; it is firmly restrained against rotation at each end. A concentrated torque of 30 kN m is applied to the beam at its mid-span point. If the maximum shear stress in the beam is limited to 200 N/mm2and the maximum angle of twist to 2◦, calculate the minimum thickness of the beam walls. Take G=25 000 N/mm2.
The minimum thickness of the beam corresponding to the maximum allowable shear stress of 200 N/mm2is obtained directly using Eq. (18.1) in which Tmax=15 kN m.
Then
tmin= 15×106×4
2×π×2002×200 =1.2 mm The rate of twist along the beam is given by Eq. (18.4) in which
ds
t = π×200 tmin
18.1 Torsion of closed section beams 531 Hence
dθ dz = T
4A2G ×π×200
tmin (i)
Taking the origin for z at one of the fixed ends and integrating Eq. (i) for half the length of the beam we obtain
θ= T
4A2G ×200π tmin z+C1
where C1is a constant of integration. At the fixed end where z=0,θ=0 so that C1=0.
Hence
θ= T
4A2G × 200π tmin z
The maximum angle of twist occurs at the mid-span of the beam where z=1 m. Hence tmin= 15×106×200×π×1×103×180
4×(π×2002/4)2×25 000×2×π =2.7 mm
The minimum allowable thickness that satisfies both conditions is therefore 2.7 mm.
Example 18.2
Determine the warping distribution in the doubly symmetrical rectangular, closed section beam, shown in Fig. 18.4, when subjected to an anticlockwise torque T .
From symmetry the centre of twist R will coincide with the mid-point of the cross- section and points of zero warping will lie on the axes of symmetry at the mid-points of the sides. We shall therefore take the origin for s at the mid-point of side 14 and measure s in the positive, anticlockwise, sense around the section. Assuming the shear modulus G to be constant we rewrite Eq. (18.5) in the form
ws−w0= Tδ 2AG
δOs δ −AOs
A
(i)
Fig. 18.4Torsion of a rectangular section beam.
where
δ= ds
t and δOs = s
0
ds t In Eq. (i)
w0=0, δ=2 b
tb + a ta
and A=ab From 0 to 1, 0 ≤s1≤b/2 and
δOs = s1
0
ds1 tb = s1
tb AOs = as1
4 (ii)
Note thatδOsand AOsare both positive.
Substitution forδOsand AOsfrom Eq. (ii) in (i) shows that the warping distribution in the wall 01, w01, is linear. Also
w1= T 2abG2
b tb + a
ta
b/2tb
2(b/tb+a/ta)− ab/8 ab
which gives
w1= T 8abG
b tb − a
ta
(iii) The remainder of the warping distribution may be deduced from symmetry and the fact that the warping must be zero at points where the axes of symmetry and the walls of the cross-section intersect. It follows that
w2 = −w1= −w3=w4
giving the distribution shown in Fig. 18.5. Note that the warping distribution will take the form shown in Fig. 18.5 as long as T is positive and b/tb>a/ta. If either of these conditions is reversed w1and w3will become negative and w2and w4positive. In the case when b/tb=a/tathe warping is zero at all points in the cross-section.
Fig. 18.5Warping distribution in the rectangular section beam of Example 18.2.
18.1 Torsion of closed section beams 533
Fig. 18.6Arbitrary origin fors.
Suppose now that the origin for s is chosen arbitrarily at, say, point 1. Then, from Fig. 18.6,δOsin the wall 12=s1/taand AOs=12s1b/2=s1b/4 and both are positive.
Substituting in Eq. (i) and setting w0=0 w12 = Tδ
2abG s1
δta − s1 4a
(iv) so that w12varies linearly from zero at 1 to
w2 = T 2abG2
b tb + a
ta
a
2(b/tb+a/ta)ta −1 4
at 2. Thus
w2= T 4abG
a ta − b
tb
or
w2= − T 4abG
b tb − a
ta
(v) Similarly
w23= Tδ 2abG
1 δ
a ta + s2
tb
− 1
4b(b+s2)
(vi) The warping distribution therefore varies linearly from a value−T (b/tb−a/ta)/4abG at 2 to zero at 3. The remaining distribution follows from symmetry so that the complete distribution takes the form shown in Fig. 18.7.
Comparing Figs 18.5 and 18.7 it can be seen that the form of the warping distribution is the same but that in the latter case the complete distribution has been displaced axially.
The actual value of the warping at the origin for s is found using Eq. (17.26).
Fig. 18.7Warping distribution produced by selecting an arbitrary origin fors.
Thus
w0 = 2
2(ata+btb) a
0
w12tads1+ b
0
w23tbds2 (vii) Substituting in Eq. (vii) for w12 and w23 from Eqs (iv) and (vi), respectively, and evaluating gives
w0= − T 8abG
b tb − a
ta
(viii) Subtracting this value from the values of w1(=0) and w2(= −T (b/tb−a/ta)/4abG) we have
w1= T 8abG
b tb − a
ta
, w2= − T 8abG
b tb − a
ta
as before. Note that setting w0=0 in Eq. (i) implies that w0, the actual value of warping at the origin for s, has been added to all warping displacements. This value must therefore be subtracted from the calculated warping displacements (i.e. those based on an arbitrary choice of origin) to obtain true values.
It is instructive at this stage to examine the mechanics of warping to see how it arises.
Suppose that each end of the rectangular section beam of Example 18.2 rotates through opposite anglesθ giving a total angle of twist 2θ along its length L. The corner 1 at one end of the beam is displaced by amounts aθ/2 vertically and bθ/2 horizontally as shown in Fig. 18.8. Consider now the displacements of the web and cover of the beam due to rotation. From Figs 18.8 and 18.9 (a) and (b) it can be seen that the angles of rotation of the web and the cover are, respectively
φb=(aθ/2)/(L/2)=aθ/L and
φa =(bθ/2)/(L/2)=bθ/L
18.1 Torsion of closed section beams 535
Fig. 18.8Twisting of a rectangular section beam.
Fig. 18.9Displacements due to twist and shear strain.
The axial displacements of the corner 1 in the web and cover are then b
2 aθ
L, a 2
bθ L
respectively, as shown in Fig. 18.9(a) and (b). In addition to displacements produced by twisting, the webs and covers are subjected to shear strainsγbandγacorresponding to the shear stress system given by Eq. (18.1). Due toγbthe axial displacement of corner 1 in the web isγbb/2 in the positive z direction while in the cover the displacement is γaa/2 in the negative z direction. Note that the shear strainsγbandγa correspond to the shear stress system produced by a positive anticlockwise torque. Clearly, the total axial displacement of the point 1 in the web and cover must be the same so that
−b 2
aθ L +γbb
2 = a 2
bθ L −γaa
2 from which
θ= L
2ab(γaa+γbb)
The shear strains are obtained from Eq. (18.1) and are γa= T
2abGta, γb= T 2abGtb whence
θ= TL 4a2b2G
a ta + b
tb
The total angle of twist from end to end of the beam is 2θ, therefore 2θ
L = TL 4a2b2G
2a ta + 2b
tb
or
dθ dz = T
4A2G ds
t as in Eq. (18.4).
Substituting forθin either of the expressions for the axial displacement of the corner 1 gives the warping w1at 1. Thus
w1= a 2 b L
TL 4a2b2G
a ta + b
tb
− T 2abGta
a 2 i.e.
w1= T 8abG
b tb − a
ta
as before. It can be seen that the warping of the cross-section is produced by a combination of the displacements caused by twisting and the displacements due to the shear strains; these shear strains correspond to the shear stresses whose values are fixed by statics. The angle of twist must therefore be such as to ensure compatibility of displacement between the webs and covers.