T ORSION OF T HIN -W ALLED B EAMS
Chapter 16 Bending of open and closed,
18.2 Torsion of open section beams
18.2.1 Warping of the cross-section
We saw in Section 3.4 that a thin rectangular strip suffers warping across its thickness when subjected to torsion. In the same way a thin-walled open section beam will warp across its thickness. This warping, wt, may be deduced by comparing Fig. 18.10(b) with Fig. 3.10 and using Eq. (3.32), thus
wt =nsdθ
dz (18.14)
18.2 Torsion of open section beams 539 In addition to warping across the thickness, the cross-section of the beam will warp in a similar manner to that of a closed section beam. From Fig. 17.3
γzs= ∂w
∂s +∂vt
∂z (18.15)
Referring the tangential displacementvtto the centre of twist R of the cross-section we have, from Eq. (17.8)
∂vt
∂z =pRdθ
dz (18.16)
Substituting for∂vt/∂z in Eq. (18.15) gives γzs = ∂w
∂s +pRdθ dz from which
τzs=G ∂w
∂s +pRdθ dz
(18.17) On the mid-line of the section wallτzs = 0 (see Eq. (18.9)) so that, from Eq. (18.17)
∂w
∂s = −pR
dθ dz
Integrating this expression with respect to s and taking the lower limit of integration to coincide with the point of zero warping, we obtain
ws= −dθ dz
s
0
pRds (18.18)
From Eqs (18.14) and (18.18) it can be seen that two types of warping exist in an open section beam. Equation (18.18) gives the warping of the mid-line of the beam; this is known as primary warping and is assumed to be constant across the wall thickness.
Equation (18.14) gives the warping of the beam across its wall thickness. This is called secondary warping, is very much less than primary warping and is usually ignored in the thin-walled sections common to aircraft structures.
Equation (18.18) may be rewritten in the form ws= −2ARdθ
dz (18.19)
or, in terms of the applied torque ws= −2AR T
GJ (see Eq. (18.12)) (18.20)
in which AR=12s
0pRds is the area swept out by a generator, rotating about the centre of twist, from the point of zero warping, as shown in Fig. 18.11. The sign of ws, for a given direction of torque, depends upon the sign of ARwhich in turn depends upon the sign of
Fig. 18.11Warping of an open section beam.
pR, the perpendicular distance from the centre of twist to the tangent at any point. Again, as for closed section beams, the sign of pR depends upon the assumed direction of a positive torque, in this case anticlockwise. Therefore, pR(and therefore AR) is positive if movement of the foot of pRalong the tangent in the assumed direction of s leads to an anticlockwise rotation of pR about the centre of twist. Note that for open section beams the positive direction of s may be chosen arbitrarily since, for a given torque, the sign of the warping displacement depends only on the sign of the swept area AR. Example 18.3
Determine the maximum shear stress and the warping distribution in the channel sec- tion shown in Fig. 18.12 when it is subjected to an anticlockwise torque of 10 N m.
G=25 000 N/mm2.
From the second of Eqs (18.13) it can be seen that the maximum shear stress occurs in the web of the section where the thickness is greatest. Also, from the first of Eqs (18.11)
J = 13(2×25×1.53+50×2.53)=316.7 mm4 so that
τmax= ±2.5×10×103
316.7 = ±78.9 N/mm2
The warping distribution is obtained using Eq. (18.20) in which the origin for s (and hence AR) is taken at the intersection of the web and the axis of symmetry where the warping is zero. Further, the centre of twist R of the section coincides with its shear centre S whose position is found using the method described in Section 17.2.1, this givesξS=8.04 mm. In the wall O2
AR= 12×8.04s1 (pRis positive) so that
wO2= −2×12 ×8.04s1× 10×103
25 000×316.7 = −0.01s1 (i)
18.2 Torsion of open section beams 541
Fig. 18.12Channel section of Example 18.3.
i.e. the warping distribution is linear in O2 and
w2= −0.01×25= −0.25 mm In the wall 21
AR= 12×8.04×25−21×25s2
in which the area swept out by the generator in the wall 21 provides a negative contribution to the total swept area AR. Thus
w21= −25(8.04−s2) 10×103 25 000×316.7 or
w21= −0.03(8.04−s2) (ii)
Again the warping distribution is linear and varies from−0.25 mm at 2 to+0.54 mm at 1. Examination of Eq. (ii) shows that w21changes sign at s2=8.04 mm. The remain- ing warping distribution follows from symmetry and the complete distribution is shown in Fig. 18.13. In unsymmetrical section beams the position of the point of zero warping is not known but may be found using the method described in Section 27.2 for the restrained warping of an open section beam. From the derivation of Eq. (27.3) we see that
2AR =
sect2AR,Ot ds
sectt ds (18.21)
in which AR,O is the area swept out by a generator rotating about the centre of twist from some convenient origin and ARis the value of AR,Oat the point of zero warping.
As an illustration we shall apply the method to the beam section of Example 18.3.
Fig. 18.13Warping distribution in channel section of Example 18.3.
Suppose that the position of the centre of twist (i.e. the shear centre) has already been calculated and suppose also that we choose the origin for s to be at the point 1.
Then, in Fig. 18.14
sect
t ds=2×1.5×25+2.5×50=200 mm2 In the wall 12
A12= 12×25s1 (AR,Ofor the wall 12) (i) from which
A2= 12 ×25×25=312.5 mm2 Also
A23=312.5−12×8.04s2 (ii)
and
A3=312.5− 12×8.04×50=111.5 mm2 Finally
A34=111.5+12 ×25s3 (iii)
18.2 Torsion of open section beams 543
Fig. 18.14Determination of points of zero warping.
Substituting for A12, A23and A34from Eqs (i)–(iii) in Eq. (18.21) we have 2AR= 1
200 25
0
25×1.15s1ds1+ 50
0
2(312.5−4.02s2)2.5 ds2 +
25
0
2(111.5+12.5s3)1.5 ds3
(iv) Evaluation of Eq. (iv) gives
2AR=424 mm2
We now examine each wall of the section in turn to determine points of zero warping.
Suppose that in the wall 12 a point of zero warping occurs at a value of s1equal to s1,0. Then
2×12×25s1,0=424 from which
s1,0=16.96 mm
so that a point of zero warping occurs in the wall 12 at a distance of 8.04 mm from the point 2 as before. In the web 23 let the point of zero warping occur at s2=s2,0. Then
2× 12×25×25−2× 12×8.04s2,0=424
which gives s2,0=25 mm (i.e. on the axis of symmetry). Clearly, from symmetry, a further point of zero warping occurs in the flange 34 at a distance of 8.04 mm from the
point 3. The warping distribution is then obtained directly using Eq. (18.20) in which AR=AR,O−AR
Problems
P.18.1 A uniform, thin-walled, cantilever beam of closed rectangular cross-section has the dimensions shown in Fig. P.18.1. The shear modulus G of the top and bottom covers of the beam is 18 000 N/mm2while that of the vertical webs is 26 000 N/mm2.
Fig. P.18.1
The beam is subjected to a uniformly distributed torque of 20 N m/mm along its length. Calculate the maximum shear stress according to the Bred–Batho theory of torsion. Calculate also, and sketch, the distribution of twist along the length of the cantilever assuming that axial constraint effects are negligible.
Ans. τmax=83.3 N/mm2, θ=8.14×10−9
2500z−z2 2
rad.
P.18.2 A single cell, thin-walled beam with the double trapezoidal cross-section shown in Fig. P.18.2, is subjected to a constant torque T=90 500 N m and is constrained to twist about an axis through the point R.Assuming that the shear stresses are distributed according to the Bredt–Batho theory of torsion, calculate the distribution of warping around the cross-section.
Problems 545 Illustrate your answer clearly by means of a sketch and insert the principal values of the warping displacements.
The shear modulus G=27 500 N/mm2and is constant throughout.
Ans. w1= −w6= −0.53 mm, w2= −w5=0.05 mm, w3= −w4=0.38 mm.
Linear distribution.
Fig. P.18.2
P.18.3 A uniform thin-walled beam is circular in cross-section and has a constant thickness of 2.5 mm. The beam is 2000 mm long, carrying end torques of 450 N m and, in the same sense, a distributed torque loading of 1.0 N m/mm. The loads are reacted by equal couples R at sections 500 mm distant from each end (Fig. P.18.3).
Calculate the maximum shear stress in the beam and sketch the distribution of twist along its length. Take G=30 000 N/mm2and neglect axial constraint effects.
Ans. τmax=24.2 N/mm2, θ= −0.85×10−8z2rad, 0≤z≤500 mm,
θ=1.7×10−8(1450z−z2/2)−12.33×10−3rad, 500≤z≤1000 mm.
Fig. P.18.3
P.18.4 The thin-walled box section beam ABCD shown in Fig. P.18.4 is attached at each end to supports which allow rotation of the ends of the beam in the longitudinal vertical plane of symmetry but prevent rotation of the ends in vertical planes perpen- dicular to the longitudinal axis of the beam. The beam is subjected to a uniform torque
loading of 20 N m/mm over the portion BC of its span. Calculate the maximum shear stress in the cross-section of the beam and the distribution of angle of twist along its length, G=70 000 N/mm2.
Ans. 71.4 N/mm2, θB=θC=0.36◦, θat mid-span=0.72◦.
4 mm
4 mm
6 mm 6 mm
350 mm
200 mm 20Nm/mm
4 m 1 m
1 m A B
C D
Fig. P.18.4
P.18.5 Figure P.18.5 shows a thin-walled cantilever box beam having a constant width of 50 mm and a depth which decreases linearly from 200 mm at the built-in end to 150 mm at the free end. If the beam is subjected to a torque of 1 kN m at its free end, plot the angle of twist of the beam at 500 mm intervals along its length and determine the maximum shear stress in the beam section. Take G=25 000 N/mm2.
Ans. τmax=33.3 N/mm2.
50 mm
200 mm 2.0 mm
2500 mm 150
1 kN m mm
Fig. P.18.5
P.18.6 A uniform closed section beam, of the thin-walled section shown in Fig. P.18.6, is subjected to a twisting couple of 4500 N m. The beam is constrained to twist about a longitudinal axis through the centre C of the semicircular arc 12. For the curved wall 12 the thickness is 2 mm and the shear modulus is 22 000 N/mm2. For the plane walls 23, 34 and 41, the corresponding figures are 1.6 mm and 27 500 N/mm2. (Note: Gt=constant.)
Calculate the rate of twist in rad/mm. Give a sketch illustrating the distribution of warping displacement in the cross-section and quote values at points 1 and 4.
Problems 547 Ans. dθ/dz=29.3×10−6rad/mm, w3= −w4= −0.19 mm,
w2= −w1= −0.056 mm.
Fig. P.18.6
P.18.7 A uniform beam with the doubly symmetrical cross-section shown in Fig.
P.18.7, has horizontal and vertical walls made of different materials which have shear moduli Gaand Gb, respectively. If for any material the ratio mass density/shear modulus is constant find the ratio of the wall thicknesses taand tb, so that for a given torsional stiffness and given dimensions a, b the beam has minimum weight per unit span. Assume the Bredt–Batho theory of torsion is valid.
If this thickness requirement is satisfied find the a/b ratio (previously regarded as fixed), which gives minimum weight for given torsional stiffness.
Ans. tb/ta=Ga/Gb, b/a=1.
Fig. P.18.7
P.18.8 The cold-formed section shown in Fig. P.18.8 is subjected to a torque of 50 N m. Calculate the maximum shear stress in the section and its rate of twist.
G=25 000 N/mm2.
Ans. τmax=220.6 N/mm2, dθ/dz=0.0044 rad/mm.
2 mm
20 mm 25 mm
25 mm 25 mm
15 mm 15 mm
Fig. P.18.8
P.18.9 Determine the rate of twist per unit torque of the beam section shown in Fig.
P.17.11 if the shear modulus G is 25 000 N/mm2. (Note that the shear centre position has been calculated in P.17.11.)
Ans. 6.42×10−8rad/mm.
P.18.10 Figure P.18.10 shows the cross-section of a thin-walled beam in the form of a channel with lipped flanges. The lips are of constant thickness 1.27 mm while the flanges increase linearly in thickness from 1.27 mm where they meet the lips to
Fig. P.18.10
Problems 549 2.54 mm at their junctions with the web. The web has a constant thickness of 2.54 mm.
The shear modulus G is 26 700 N/mm2throughout.
The beam has an enforced axis of twist RR and is supported in such a way that warping occurs freely but is zero at the mid-point of the web. If the beam carries a torque of 100 N m, calculate the maximum shear stress according to the St. Venant theory of torsion for thin-walled sections. Ignore any effects of stress concentration at the corners. Find also the distribution of warping along the middle line of the section, illustrating your results by means of a sketch.
Ans. τmax= ±297.4 N/mm2, w1= −5.48 mm= −w6. w2=5.48 mm= −w5, w3=17.98 mm= −w4.
P.18.11 The thin-walled section shown in Fig. P.18.11 is symmetrical about the x axis. The thickness t0of the centre web 34 is constant, while the thickness of the other walls varies linearly from t0at points 3 and 4 to zero at the open ends 1, 6, 7 and 8.
Determine the St. Venant torsion constant J for the section and also the maximum value of the shear stress due to a torque T . If the section is constrained to twist about an axis through the origin O, plot the relative warping displacements of the section per unit rate of twist.
Ans. J=4at03/3, τmax= ±3T/4at20, w1= +a2(1+2√ 2).
w2= +√
2a2, w7= −a2, w3=0.
Fig. P.18.11
P.18.12 The thin walled section shown in Fig. P.18.12 is constrained to twist about an axis through R, the centre of the semicircular wall 34. Calculate the maximum shear
stress in the section per unit torque and the warping distribution per unit rate of twist.
Also compare the value of warping displacement at the point 1 with that corresponding to the section being constrained to twist about an axis through the point O and state what effect this movement has on the maximum shear stress and the torsional stiffness of the section.
Ans. Maximum shear stress is±0.42/rt2per unit torque.
w03= +r2θ, w32= +r
2(πr+2s1), w21= −r
2(2s2−5.142r).
With centre of twist at O1w1= −0.43r2. Maximum shear stress is unchanged, torsional stiffness increased since warping reduced.
1
2
5
6 4
O R 3
r
r
r
r r
t
r
Fig. P.18.12
P.18.13 Determine the maximum shear stress in the beam section shown in Fig.
P.18.13 stating clearly the point at which it occurs. Determine also the rate of twist of the beam section if the shear modulus G is 25 000 N/mm2.
Ans. 70.2 N/mm2on underside of 24 at 2 or on upper surface of 32 at 2.
9.0×10−4rad/mm.
100 mm
3 mm
25 mm 2 4 3
1
80 mm 1 kN
2 mm
Fig. P.18.13
19
Combined open and closed section beams
So far, in Chapters 16–18, we have analysed thin-walled beams which consist of either completely closed cross-sections or completely open cross-sections. Frequently aircraft components comprise combinations of open and closed section beams. For example the section of a wing in the region of an undercarriage bay could take the form shown in Fig. 19.1. Clearly part of the section is an open channel section while the nose portion is a single cell closed section. We shall now examine the methods of analysis of such sections when subjected to bending, shear and torsional loads.