Problem 3.16. Welded tube

Maximum allowed torque, T_allowed [kNm] =

Step 1. Calculate the torsional stiffness of the tube cross section.

Check torsional stiffness assuming thin wallsHide torsional stiffness

$G{I}_t=G2t{\left(R–\frac{t}{2}\right)}^3\pi =G2\times 6{\left(50–\frac{6}{2}\right)}^3\pi =3.914\times {10}^{6 }G$

Eq.(3-112)

Check torsional stiffness assuming thick wallsHide torsional stiffness

Step 2.  Write the maximum shear stress in the function of the unknown torque.

Check shear stressHide shear stress

${\tau }_{\max }=\frac{T}{G{I}_t}GR$

Figure 3.51 and Eq.(3-79)

Step 3.  Perform stress transformation into a coordinate system attached to the weld’s direction. 

Check stress transformationHide stress transformation

The rotation angle is 45°. Shear results in tension and compression in 45° degree direction.

Eq.(2-9)

$\left\{\begin{array}{c}\sigma {‘}_x\\ \sigma {‘}_y\\ \tau {‘}_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}0\\ 0\\ {\tau }_{xy}\end{array}\right\}=\left\{\begin{array}{c}{\tau }_{xy}\\ –{\tau }_{xy}\\ 0\end{array}\right\}$

Step 4.  Express allowed torque from Rankine criterion. 

Check allowed torqueHide allowed torque

No shear arises in the rotated coordinate system, the axis are in the principal directions. Rankine criterion yields

Eq.(2-22)

$\left.\begin{array}{r}\left|{\sigma }_1\right|<f\\ \left|{\sigma }_2\right|<f\end{array}\right\} \to \left|\pm {\tau }_{xy}\right|<f \to {\tau }_{xy}=\frac{T}{G{I}_t}GR<f \to T_{\mathrm{allowed}}=\frac{f}{GR}G{I}_t$

Assuming thin walled cross section the maximum allowed torque is:
${\mathit{T}}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{o}\mathbf{w}\mathbf{e}\mathbf{d}}=\frac{f}{GR}G{I}_t=\frac{80}{50}3.93\times {10}^6=6.26\times {10}^6\mathrm{Nmm}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{ }\mathbf{kNm}$

Assuming thick walled cross section the maximum allowed torque is:${\mathit{T}}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{o}\mathbf{w}\mathbf{e}\mathbf{d}}=\frac{f}{GR}G{I}_t=\frac{80}{50}3.914\times {10}^6=6.29\times {10}^6\mathrm{Nmm}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{29}\mathbf{ }\mathbf{kNm}$

Results are close two each other, thus both approximations are appropriate.

The torsional stiffness of the tube cross section can be calculated assuming thin walled or thick walled cross section. Both calculations will be presented. Assuming thin walled beam:

$G{I}_t=G2t{\left(R–\frac{t}{2}\right)}^3\pi =G2\times 6{\left(50–\frac{6}{2}\right)}^3\pi =3.914\times {10}^{6 }G$

Eq.(3-112)

Assuming thick walls:$G{I}_t=G\left(\frac{R^4\mathrm{\pi }}{2}–\frac{{\left(R–t\right)}^4\mathrm{\pi }}{2}\right)=G\left(\frac{{50}^4\mathrm{\pi }}{2}–\frac{{\left(50–6\right)}^4\mathrm{\pi }}{2}\right)=3.93\times {10}^{6 }G$

The maximum shear stress can be given in the function of the unknown torque:

${\tau }_{\max }=\frac{T}{G{I}_t}GR$

Figure 3.51 and Eq.(3-79)

Now stress transformation is performed into a coordinate system attached to the weld’s direction. The rotation angle is 45°. Shear results in tension and compression in 45° degree direction.

Eq.(2-9)

No shear arises in the rotated coordinate system, the axis are in the principal directions. Rankine criterion yields

Eq.(2-22)

$\left.\begin{array}{r}\left|{\sigma }_1\right|<f\\ \left|{\sigma }_2\right|<f\end{array}\right\} \to \left|\pm {\tau }_{xy}\right|<f \to {\tau }_{xy}=\frac{T}{G{I}_t}GR<f \to T_{\mathrm{allowed}}=\frac{f}{GR}G{I}_t$

Assuming thin walled cross section the maximum allowed torque is:
${\mathit{T}}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{o}\mathbf{w}\mathbf{e}\mathbf{d}}=\frac{f}{GR}G{I}_t=\frac{80}{50}3.93\times {10}^6=6.26\times {10}^6\mathrm{Nmm}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{ }\mathbf{kNm}$

Assuming thick walled cross section the maximum allowed torque is:
${\mathit{T}}_{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{o}\mathbf{w}\mathbf{e}\mathbf{d}}=\frac{f}{GR}G{I}_t=\frac{80}{50}3.914\times {10}^6=6.29\times {10}^6\mathrm{Nmm}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{29}\mathbf{ }\mathbf{kNm}$

Results are close two each other, thus both approximations are appropriate.