Problem 3.15. Cantilever subjected to end torque

A cantilever beam (L = 0.5 m) is subjected to an end torque, T. The shear modulus is G = 80 GPa. (Neglect restrained warping.)
a) Determine the maximum allowable end torque for the given cross sections (Figures a)-d)). The shear strength is 120 MPa.
b) Determine the rotation of the beam end when the torque is T = 8 kNm for all the given cross sections (Figures a)-d))

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Cross section a)

a) Allowed torque, T [kNm]=

b) Rotation of beam end ψ [1/m]=

Cross section b)

a) Allowed torque, T [kNm]=

b) Rotation of beam end ψ [1/m]=

Cross section c)

a) Allowed torque, T [kNm]=

b) Rotation of beam end ψ [1/m]=

Cross section d)

a) Allowed torque, T [kNm]=

b) Rotation of beam end ψ [1/m]=

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Steps

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Step 1. Calculate torsional stiffness of the cross sections

Step 2. Express allowed torque from the maximum shear stress

Cross section a)

Check torsional stiffnessHide torsional stiffness

$G I_{t} = G I_{0} = G \frac{π R^{4}}{2} = 80 \times 10^{3} \frac{π \times 25^{4}}{2} = 4.909 \times 10^{10} {Nmm}^{2}$

Eq.(3-79)

Check allowed torqueHide allowed torque

\begin{array}{r} T = G I_{t} ϑ \\ τ_{\max} = ϑ G R = \frac{T}{G I_{t}} G R = f \end{array}\} \to T_{allowed} T_{allowed} = \frac{f}{G R} G I_{t} = \frac{120}{80 \times 10^{3} \times 25} 4.909 \times 10^{10} = 2.945 \times 10^{6} Nmm = 2.945 kNm

Figure 3.51. and Eq.(3-79)

Cross section b)

Check torsional stiffnessHide torsional stiffness

Assuming thin walled cross section: $G I_{t} = G 2 t {(R - \frac{t}{2})}^{3} π = 80 \times 10^{3} \times 2 \times 2 {(25 - \frac{2}{2})}^{3} π = 1.39 \times 10^{10} {Nmm}^{2}$ Assuming thick walls: $G I_{t} = G (\frac{R^{4} π}{2} - \frac{{(R - t)}^{4} π}{2}) = 80 \times 10^{3} (\frac{25^{4} π}{2} - \frac{{(25 - 2)}^{4} π}{2}) = 1.39 \times 10^{10} {Nmm}^{2}$

The results are close to each other, thus both approximations of can be used.

Eqs.(3-79) and (3-112)

Check allowed torqueHide allowed torque

\begin{array}{r} T = G I_{t} ϑ \\ τ_{\max} = ϑ G R = \frac{T}{G I_{t}} G R = f \end{array}\} \to T_{allowed} T_{allowed} = \frac{f}{G R} G I_{t} = \frac{120}{80 \times 10^{3} \times 25} 1.39 \times 10^{10} = 8.35 \times 10^{5} Nmm = 0.835 kNm

Cross section c)

Check torsional stiffnessHide torsional stiffness

$G I_{t} = G c_{2} b a^{3} = 80 \times 10^{3} \times 0.196 \times 75 \times 50^{3} = 1.47 \times 10^{11} {Nmm}^{2}$

where factor, c₂ = 0.196 belongs to the ratio b/a = 1.5.

See Table 3.9 and Eq.(3-95)

Check allowed torqueHide allowed torque

τ_{\max} = \frac{T}{c_{1} b a} \to T_{allowed} T_{allowed} = f c_{1} b a^{2} = 120 \times 0.231 \times 75 \times 50^{2} = 5.198 \times 10^{6} Nmm = 5.198 kNm

where factor, c ₁= 0.231 belongs to the ratio b/a = 1.5.

See Table 3.9 and Eq.(3-96)

Cross section d)

Check torsional stiffnessHide torsional stiffness

G I_{t} = G \sum_{k - 1}^{5} \frac{b_{k} t_{k}^{3}}{3} = 80 \times 10^{3} \times (2 \frac{13.5 \times 3^{3}}{3} + 2 \frac{57 \times 3^{3}}{3} + \frac{87 \times 3^{3}}{3}) = 1.641 \times 10^{8} {Nmm}^{2}

See Eq.(3-100)

Check allowed torqueHide allowed torque

τ_{\max} = ϑ G t = \frac{T}{G I_{t}} G t \to T_{allowed} T_{allowed} = \frac{f}{G t} G I_{t} = \frac{120}{80 \times 10^{3} \times 3} 1.641 \times 10^{8} = 8.208 \times 10^{4} Nmm = 0.08208 kNm

Figure 3.51 and Eq.(3-79)

Step 3. Calculate the rotation of the beam end for the given torque.

Cross section a)

Check rotationHide rotation

The torque and rate of twist functions are constant along the beams length. Thus the rotation of the beam end results in the rate of twist multiplied by the beam’s length:

See geometrical equation in Table 3.7.

$ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{4.909 \times 10^{10}} 500 = 0.0815$

Cross section b)

Check rotationHide rotation

ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.39 \times 10^{10}} 500 = 0.287

See geometrical equation in Table 3.7.

Cross section c)

Check rotationHide rotation

ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.47 \times 10^{11}} 500 = 0.0272

See geometrical equation in Table 3.7.

Cross section d)

Check rotationHide rotation

ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.641 \times 10^{8}} 500 = 24.37

See geometrical equation in Table 3.7.

Results

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Problem a)

First the torsional stiffnesses of each cross sections are calculated. Then the allowed torque is expressed from the maximum shear stress.

Cross section a)

$G I_{t} = G I_{0} = G \frac{π R^{4}}{2} = 80 \times 10^{3} \frac{π \times 25^{4}}{2} = 4.909 \times 10^{10} {Nmm}^{2}$

Eq.(3-79)

\begin{array}{r} T = G I_{t} ϑ \\ τ_{\max} = ϑ G R = \frac{T}{G I_{t}} G R = f \end{array}\} \to T_{allowed} T_{allowed} = \frac{f}{G R} G I_{t} = \frac{120}{80 \times 10^{3} \times 25} 4.909 \times 10^{10} = 2.945 \times 10^{6} Nmm = 2.945 kNm

Figure 3.51.

Cross section b)

The results are close to each other, thus both approximations of can be used.

Eqs.(3-79) and (3-112)

\begin{array}{r} T = G I_{t} ϑ \\ τ_{\max} = ϑ G R = \frac{T}{G I_{t}} G R = f \end{array}\} \to T_{allowed} T_{allowed} = \frac{f}{G R} G I_{t} = \frac{120}{80 \times 10^{3} \times 25} 1.39 \times 10^{10} = 8.35 \times 10^{5} Nmm = 0.835 kNm

Cross section c)

$G I_{t} = G c_{2} b a^{3} = 80 \times 10^{3} \times 0.196 \times 75 \times 50^{3} = 1.47 \times 10^{11} {Nmm}^{2}$

where factor, c₂ = 0.196 belongs to the ratio b/a = 1.5.

See Table 3.9 and Eq.(3-95)

τ_{\max} = \frac{T}{c_{1} b a} \to T_{allowed} T_{allowed} = f c_{1} b a^{2} = 120 \times 0.231 \times 75 \times 50^{2} = 5.198 \times 10^{6} Nmm = 5.198 kNm

where factor, c ₁= 0.231 belongs to the ratio b/a=1.5.

See Table 3.9 and Eq.(3-96)

Cross section d)

$G I_{t} = G \sum_{k - 1}^{5} \frac{b_{k} t_{k}^{3}}{3} = 80 \times 10^{3} \times (2 \frac{13.5 \times 3^{3}}{3} + 2 \frac{57 \times 3^{3}}{3} + \frac{87 \times 3^{3}}{3}) = 1.641 \times 10^{8} {Nmm}^{2}$

See Eq.(3-100)

τ_{\max} = ϑ G t = \frac{T}{G I_{t}} G t \to T_{allowed} T_{allowed} = \frac{f}{G t} G I_{t} = \frac{120}{80 \times 10^{3} \times 3} 1.641 \times 10^{8} = 8.208 \times 10^{4} Nmm = 0.08208 kNm

Figure 3.51 and Eq.(3-79)

Problem b)

The torque and rate of twist functions are constant along the beams length. Thus the rotation of the beam end results in the rate of twist multiplied by the beam’s length. The rotation of the beam end for the given torque is given below for the different cross sections.

See geometrical equation in Table 3.7.

Cross section a)

$ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{4.909 \times 10^{10}} 500 = 0.0815$

Cross section b)

$ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.39 \times 10^{10}} 500 = 0.287$

Cross section c)

$ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.47 \times 10^{11}} 500 = 0.0272$

Cross section d)

$ϑ = \frac{d ψ}{d x} = \frac{T}{G I_{t}} \to ψ = ϑ L = \frac{T}{G I_{t}} L = \frac{8 \times 10^{6}}{1.641 \times 10^{8}} 500 = 24.37$