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Question 4.9: Design of a Torsion Bar Problem: Determine the best cross-se...

Design of a Torsion Bar

Problem:    Determine the best cross-sectional shape for a hollow torsion bar to be made from a sheet of steel of known dimensions in order to withstand a pure torsional load with minimum angular deflection. Also find the maximum shear stress.

Given:    The applied torque is 10 N-m. The sheet of steel has length ll = 1 m, width ww = 100 mm, and thickness tt = 1 mm. GG = 80.8 GPa.

Assumptions:    Try four different cross-sectional shapes: unformed flat plate, open-circular section, closed-circular section, and closed-square section. The open-circular shape is rolled but is not welded at the seam. The closed shapes’ seams are welded to create a continuous cross section. Assume a mean diameter or mean perimeter consistent with the sheet width.

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See Figure 4-29 and Table 4-3.

1    Equations 4.26 will apply for all sections, provided that we substitute JJ for KK and J/rJ/r for QQ in the case of the closed-circular section.

τmax⁡=TQ \tau_{\max }=\frac{T}{Q}       (4.26a)

θ=TlKG \theta=\frac{T l}{K G}          (4.26b)

2    The unformed flat plate behaves as a solid rectangular section as shown in Figure 4-29a and Table 4-3. It has dimensions aa = w / 2 = 0.05 m and bb = t / 2 = 0.000 5 m:

K=ab3[163−3.36ba⟮1−b412a4⟯]=(0.05)(0.0005)3[5.333−3.360.00050.05⟮1−(0.0005)412(0.05)4⟯]K=3.312E−11 m4=33.123 mm4θ=TlGK=(10)(1)(8.08E10)(3.312E−11)=3.736 rad=214.1∘ \begin{aligned} K &=a b^3\left[\frac{16}{3}-3.36 \frac{b}{a}\left\lgroup1-\frac{b^4}{12 a^4}\right\rgroup \right] \\ &=(0.05)(0.0005)^3\left[5.333-3.36 \frac{0.0005}{0.05}\left\lgroup1-\frac{(0.0005)^4}{12(0.05)^4}\right\rgroup \right] \\ K &=3.312 E-11  m ^4=33.123  mm ^4 \\ \theta &=\frac{T l}{G K}=\frac{(10)(1)}{(8.08 E 10)(3.312 E-11)}=3.736  rad =214.1^{\circ} \end{aligned}                       (a)

This is obviously a rather large angular deflection, indicating that the flat plate has been wound into a corkscrew by this torsional load.

Q=8a2b23a+1.8b=8(0.05)2(0.0005)33(0.05)+1.8(0.0005)Q=3.313E−8 m3=33.13 mm3τ=TQ=103.313E−8=301.8 MPa=43772 psi \begin{aligned} Q &=\frac{8 a^2 b^2}{3 a+1.8 b} \\ &=\frac{8(0.05)^2(0.0005)^3}{3(0.05)+1.8(0.0005)} \\ Q &=3.313 E-8  m ^3=33.13  mm ^3 \\ \tau &=\frac{T}{Q}=\frac{10}{3.313 E-8}=301.8  MPa =43772  psi \end{aligned}         (b)

The maximum shear stress is 300 MPa, which would require a material with a tensile yield strength of over 520 MPa (75 000 psi) in order not to yield and take a set. This requires a high-strength steel. (See Section 5.1 for discussion of this relationship between tensile and shear strength as defined in equation 5.9b. on p. 251.)

Sys=0.577Sy S_{y s}=0.577 S_y     (5.9b)

3    The open-circular shape is now formed into a 3.18-cm diameter tube, but its longitudinal seam is left unwelded and open as shown in Figure 4-29b. The expressions for KK and QQ from Table 4-3 are

K=23πrt3=23π(wπ−t)2t3=13(w−πt)t3=13(0.1−0.001π)0.0013K=3.2286E−11 m4=32.286 mm4θ=TlGK=(10)(1)(8.08E10)(3.2286E−11)=3.833 rad=219.6∘ \begin{aligned} K &=\frac{2}{3} \pi r t^3 \\ &=\frac{2}{3} \pi \frac{\left(\frac{w}{\pi}-t\right)}{2} t^3=\frac{1}{3}(w-\pi t) t^3=\frac{1}{3}(0.1-0.001 \pi) 0.001^3 \\ K &=3.2286 E-11  m ^4=32.286  mm ^4 \\ \theta &=\frac{T l}{G K}=\frac{(10)(1)}{(8.08 E 10)(3.2286 E-11)}=3.833  rad =219.6^{\circ} \end{aligned}        (c)

This is as large an angular deflection as that of the flat plate.

Q=4π2r2t26πr+1.8t=4π2(0.01542)2(0.001)36π(0.01542)+1.8(0.001)Q=3.209E−8 m3=32.09 mm3τ=TQ=103.209E−8=311.6 MPa=45201 psi \begin{aligned} Q &=\frac{4 \pi^2 r^2 t^2}{6 \pi r+1.8 t} \\ &=\frac{4 \pi^2(0.01542)^2(0.001)^3}{6 \pi(0.01542)+1.8(0.001)} \\ Q &=3.209 E-8  m ^3=32.09  mm ^3 \\ \tau &=\frac{T}{Q}=\frac{10}{3.209 E-8}=311.6  MPa =45201  psi \end{aligned}     (d)

The stress and deflection are unacceptable. It is just as bad a design as the flat plate.

4    The closed-square tube is formed by folding the sheet into a square section with side dimension s=a=w/4s = a = w / 4. The seam is welded as shown in Figure 4-29c. From Table 4-3, KK and QQ, and the stress and deflection are now

K=2t2(a−t)42at−2t2=2t2⟮w4−t⟯42w4t−2t2=2(0.001)2⟮0.14−0.001⟯42⟮0.14⟯(0.001)−2(0.001)2K=1.382E−8 m4=13824 mm4θ=TlGK=(10)(1)(8.08E10)(1.382E−8)=0.00895 rad=0.51∘ \begin{array}{l} K=\frac{2 t^2(a-t)^4}{2 a t-2 t^2}=\frac{2 t^2\left\lgroup\frac{w}{4}-t\right\rgroup^4}{2 \frac{w}{4} t-2 t^2}=\frac{2(0.001)^2\left\lgroup\frac{0.1}{4}-0.001\right\rgroup^4}{2\left\lgroup\frac{0.1}{4}\right\rgroup(0.001)-2(0.001)^2} \\ K=1.382 E-8  m ^4=13824  mm ^4 \\ \theta=\frac{T l}{G K}=\frac{(10)(1)}{(8.08 E 10)(1.382 E-8)}=0.00895  rad =0.51^{\circ} \end{array}       (e)

Q=2t(a−t)2=2t⟮w4−t⟯2=2(0.001)⟮0.14−0.001⟯2Q=1.152E−6 m3τ=TQ=101.152E−6=8.7 MPa=1259 psi \begin{array}{l} Q=2 t(a-t)^2=2 t\left\lgroup\frac{w}{4}-t\right\rgroup ^2=2(0.001)\left\lgroup\frac{0.1}{4}-0.001\right\rgroup ^2 \\ Q=1.152 E-6  m ^3 \\ \tau=\frac{T}{Q}=\frac{10}{1.152 E-6}=8.7  MPa =1259  psi \end{array}      (f)

This angular deflection of the square tube is much less than that of either open section, and the maximum shear stress is now much more reasonable.

5    The closed-circular shape is formed into a 3.18-cm outside diameter tube and its longitudinal seam is welded shut as shown in Figure 4-29d. We can either use equations 4.24 and 4.25 (p. 178) or use the general equations 4.26 (p. 178) involving KK and QQ, which are now a function of JJ for this circular shape. For the deflection,

θ=TlJG \theta=\frac{T l}{J G}       (4.24)

J=πd432 J=\frac{\pi d^4}{32}      (4.25a)

J=π(do4−di4)32 J=\frac{\pi\left(d_o^4-d_i^4\right)}{32}      (4.25b)

K=J=π⟮do4−di4⟯32;do=wπ,di=do−2t=π[⟮wπ⟯4−⟮wπ−2t⟯4]32=π[⟮0.1π⟯4−⟮0.1π−2{0.001}⟯4]32K=J=2.304E−8 m4=23041 mm4θ=TlGK=(10)(1)(8.08E10)(2.304E−8)=0.0054 rad=0.31∘ \begin{aligned} K &=J=\frac{\pi\left\lgroup d_o^4-d_i^4\right\rgroup }{32} ; \quad d_o=\frac{w}{\pi}, \quad d_i=d_o-2 t \\ &=\frac{\pi\left[\left\lgroup \frac{w}{\pi} \right\rgroup ^4-\left\lgroup \frac{w}{\pi}-2 t\right\rgroup ^4\right]}{32}=\frac{\pi\left[\left\lgroup \frac{0.1}{\pi}\right\rgroup ^4-\left\lgroup\frac{0.1}{\pi}-2\{0.001\}\right\rgroup ^4\right]}{32} \\ K &=J=2.304 E-8  m ^4=23041  mm ^4 \\ \theta &=\frac{T l}{G K}=\frac{(10)(1)}{(8.08 E 10)(2.304 E-8)}=0.0054  rad =0.31^{\circ} \end{aligned}          (g)

and for the maximum shear stress at the outer surface,

Q=Jr=π⟮do4−di4⟯32ro=π[⟮wπ⟯4−⟮wπ−2t⟯4]32⟮w2π⟯=π[⟮0.1π⟯4−⟮0.1π−2{0.001}⟯4]32⟮0.12π⟯Q=1.448E−6 m3τ=TQ=101.448E−6=6.91 MPa=1002 psi \begin{aligned} Q=\frac{J}{r}=\frac{\pi\left\lgroup d_o^4-d_i^4\right\rgroup }{32 r_o} &=\frac{\pi\left[\left\lgroup\frac{w}{\pi}\right\rgroup ^4-\left\lgroup\frac{w}{\pi}-2 t\right\rgroup ^4\right]}{32\left\lgroup\frac{w}{2 \pi}\right\rgroup }=\frac{\pi\left[\left\lgroup\frac{0.1}{\pi}\right\rgroup ^4-\left\lgroup\frac{0.1}{\pi}-2\{0.001\}\right\rgroup ^4\right]}{32\left\lgroup\frac{0.1}{2 \pi}\right\rgroup } \\ Q &=1.448 E-6  m ^3 \\ \tau &=\frac{T}{Q}=\frac{10}{1.448 E-6}=6.91  MPa =1002  psi \end{aligned}      (h)

6    This closed circular design has the smallest stress and deflection and is clearly the best choice of the four designs presented. The wall thickness could be increased to further reduce stress and deflection if desired. The design needs to be checked for possible torsional buckling as well. The files EX04-09 can be opened in the program of your choice if desired.

Table 4-3 Expressions for K and Q for Some Cross-Section Shapes in Torsion
The Black Dots Indicate Points of Maximum Shear Stress (Source: Ref. 4 with Permission)
Shape K Q
solid square

 K=2.25a4 K=2.25 a^4 Q=a30.6 Q=\frac{a^3}{0.6}
hollow square

 K=2t2(a−t)42at−2t2 K=\frac{2 t^2(a-t)^4}{2 a t-2 t^2} Q=2t(a−t)2 Q=2 t(a-t)^2
inside corners may have higher stress if corner radius is small
solid rectangle

 K=ab3[163−3.36ba⟮1−b412a4⟯] K=a b^3\left[\frac{16}{3}-3.36 \frac{b}{a}\left\lgroup1-\frac{b^4}{12 a^4}\right\rgroup \right] Q=8a2b23a+1.8b Q=\frac{8 a^2 b^2}{3 a+1.8 b}
hollow rectangle

 K=2t2(a−t)2(b−t)2at+bt−2t2 K=\frac{2 t^2(a-t)^2(b-t)^2}{a t+b t-2 t^2} Q=2t(a−t)(b−t) Q=2 t(a-t)(b-t)
inside corners may have higher stress if corner radius is small
solid ellipse

 K=πa3b3a2+b2 K=\frac{\pi a^3 b^3}{a^2+b^2} Q=πab22 Q=\frac{\pi a b^2}{2}
hollow ellipse

 K=πa3b3a2+b2[1−⟮1−ta⟯4] K=\frac{\pi a^3 b^3}{a^2+b^2}\left[1-\left\lgroup1-\frac{t}{a}\right\rgroup ^4\right] Q=πab22[1−⟮1−ta⟯4] Q=\frac{\pi a b^2}{2}\left[1-\left\lgroup1-\frac{t}{a}\right\rgroup ^4\right]
open circular tube

 K=23πrt3;t≪r K=\frac{2}{3} \pi r t^3 ; \quad t \ll r Q=4π2r2t26πr+1.8t;t≪r Q=\frac{4 \pi^2 r^2 t^2}{6 \pi r+1.8 t} ; \quad t \ll r
open arbitrary shape

U = length of median line		 K=13Ut3;t≪U K=\frac{1}{3} U t^3 ; \quad t \ll U Q=U2t23U+1.8t;t≪U Q=\frac{U^2 t^2}{3 U+1.8 t} ; \quad t \ll U
t must be much smaller than minimum radius of curvature
F4-29

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