Question 3.5: Two sections (AB, BC) of steel drill pipe, joined by bolted ...

(a) Internal torques T(x). First, we must find the reactive torque at A using statics (see Section 1.2, Example 1-3). Summing torsional moments about the x axis of the structure, we find

\Sigma M_{x}=0 \quad R_{A}+2 T_{0}-t_{0}\left(\frac{2 L}{5}\right)=0

so    R_{A}=-2 T_{0}+\left(\frac{3 T_{0}}{L}\right)\left(\frac{2 L}{5}\right)=\frac{-4 T_{0}}{5}                 (a)

Reaction R_{A} is negative, which means that the reactive torsional moment vector is in the (-x) direction based on a statics sign convention. We now can draw free-body diagrams (FBD) of segments of the pipe to find inter-nal torsional moments T(x) over the length of the pipe.

From the FBD of segment 1 (Fig. 3-20a), we see that the internal torsional moment is constant and is equal to reactive torque R_{A}. Torque T_{1}(x) is positive, because the torsional moment vector points away from the cut section of the pipe; we refer to this as a deformation sign convention:

T_{1}(x)=\frac{4}{5} T_{0} \quad \quad 0 \leq x \leq \frac{2}{5} L                    (b)

Next, a FBD of segment 2 of the pipe structure (Fig. 3-20b) gives

T_{2}(x)=\frac{4}{5} T_{0}-2 T_{0}=\frac{-6}{5} T_{0} \quad \quad \frac{2}{5} L \leq x \leq \frac{3}{5} L                         (c)

where T_{2}(x) is also constant and the minus sign means that T_{2}(x) actually points in the negative x direction.
Finally, the FBD of segment 3 of the pipe structure (Fig. 3-20c) provides the following expression for internal torsional moment T_{3}(x):

T_{3}(x)=\frac{4}{5} T_{0}-2 T_{0}+t_{0}\left(x-\frac{3}{5} L\right)=3 T_{0}\left(\frac{x}{L}-1\right) \quad \quad \frac{3}{5} L \leq x \leq L                     (d)

Evaluating Eq. (d) at B and C, we see that at B we have

T_{3}\left(\frac{3}{5} L\right)=3 T_{0}\left(\frac{3}{5}-1\right)=\frac{-6}{5} T_{0}

and at C, we have

T_{3}(L)=3 T_{0}(1-1)=0

We can now plot Eqs. (b), (c), and (d) to get a torsional moment diagram (Fig. 3-21) (TMD) which displays the variation of internal torsional moment over the length of the pipe structure (x = 0 to x = L).

(b) Maximum shear stress in pipe \tau_{\max }. We will use the torsion formula [Eq. (3-13) \tau_{\max }=\frac{T r}{I_{P}}] to compute the shear stress in the pipe. The maximum shear stress is on the surface of the pipe. The polar moment of inertia of each pipe is computed as

\begin{aligned}I_{p A B} &=\frac{\pi}{32}\left[\left(d+2 t_{A B}\right)^{4}-(d)^{4}\right] \\&=\frac{\pi}{32}\left[[250\mathrm{~mm}+2(19 \mathrm{~mm})]^{4}-(250 \mathrm{~mm})^{4}\right]=2.919 \times 10^{-4} \mathrm{~m}^{4}\end{aligned}

and

\begin{aligned}I_{p B C} &=\frac{\pi}{32}\left[\left(d+2 t_{B C}\right)^{4}-(d)^{4}\right] \\&=\frac{\pi}{32}\left[[250\mathrm{~mm}+2(16 \mathrm{~mm})]^{4}-(250 \mathrm{~mm})^{4}\right]=2.374 \times 10^{-4} \mathrm{~m}^{4}\end{aligned}

The shear modulus G is constant, so the torsional rigidity of AB is 1.23 times that of BC. From the TMD (Fig. 3-21), we see that the maximum torsional moments in both AB and BC (each equal to 6T_{0}/5) are near joint B. Applying the torsion formula to pipes AB and BC near B gives

\begin{aligned}\tau_{\max A B} &=\frac{\left(\frac{6}{5} T_{0}\right)\left(\frac{d+2 t_{A B}}{2}\right)}{I_{p A B}} \\&=\frac{\left(\frac{6}{5} 226 \ \mathrm{kN} \cdot \mathrm{m}\right)\left[\frac{250 \mathrm{~mm}+2(19 \mathrm{~mm})}{2}\right]}{2.919 \times 10^{-4} \mathrm{~m}^{4}}=133.8 \ \mathrm{MPa} \\\tau_{\max B C} &=\frac{\left(\frac{6}{5} T_{0}\right)\left(\frac{d+2 t_{B C}}{2}\right)}{I_{p B C}} \\&=\frac{\left(\frac{6}{5} 226 \ \mathrm{kN} \cdot \mathrm{m}\right)\left[\frac{250 \mathrm{~mm}+2(16 \mathrm{~mm})}{2}\right]}{2.374 \times 10^{-4} \mathrm{~m}^{4}} =161.1 \ \mathrm{MPa}  \end{aligned}

So the maximum shear stress in the pipe is just to the right of the flange plate connection at joint B. “Just to the right of” means that we must move an appropriate distance away from the connection to avoid any stress concentration effects at the point of attachment of the two pipes in accordance with St. Venant’s principle (see Section 3.12).

(c) Twist rotations \phi (x) . Next, we use the torque-displacement relation, Eqs. (3-24) through (3-27) , to find the variation of twist rotation \phi overthe  length of the pipe structure. Support A is fixed, so \phi_{A} = \phi (0) = 0. The internal torque from x = 0 to x = 2L/5 (segment 1) is constant, so we use Eq. (3-24) to find twist rotation \phi_{1}(x) which varies linearly from x = 0 to x = 2L/5:

\phi=\sum \limits_{i=1}^{n} \phi_{i}=\sum \limits_{i=1}^{n} \frac{T_{i} L_{i}}{G_{i}\left(I_{P}\right)_{i}}                 (3-24)

d\phi = \frac{Tdx}{GI_P(x)}                      (3-25)

\phi = \int_{0}^{L}{d\phi } = \int_{0}^{L}{\frac{Tdx}{GI_P(x)}}                   (3-26)

\phi=\int_{0}^{L} d \phi=\int_{0}^{L} \frac{T(x) d x}{G I_{P}(x)}                        (3-27)

\phi_{1}(x)=\frac{T_{1}(x)(x)}{G I_{p A B}}=\frac{\left(\frac{4 T_{0}}{5}\right)(x)}{G I_{p A B}}=\frac{4 T_{0} x}{5 G I_{p A B}} \quad \quad 0 \leq x \leq \frac{2 L}{5}                               (e)

Evaluating Eq. (e) at x = 2L/5, we find the twist rotation at the point of application of torque 2T_{0} to be

\phi_{1}\left(\frac{2 L}{5}\right)=\frac{T_{1}\left(\frac{2 L}{5}\right)\left(\frac{2 L}{5}\right)}{G I_{p AB}}=\frac{\left(\frac{4 T_{0}}{5}\right)\left(\frac{2 L}{5}\right)}{G I_{p A B}}=\frac{8 T_{0} L}{25 G I_{p AB}}=\frac{0.32 T_{0} L}{G I_{p A B}}                  (f)

Next, we find an expression for the variation of twist angle \phi_{2}(x) from x = 2L/5 to x = 3L/5 (point B). As with \phi_{1}(x), twist \phi_{2}(x) varies linearly over segment 2, because torque T_{2} (x) is constant (Fig. 3-21). Using Eq. (3-24) , we get

\begin{aligned}\phi_{2}(x)=\phi_{1}\left(\frac{2 L}{5}\right)+\frac{T_{2}(x)\left(x-\frac{2 L}{5}\right)}{G I_{p A B}} &=\frac{8 T_{0} L}{25 G I_{p A B}}+\frac{\left(\frac{-6}{5} T_{0}\right)\left(x-\frac{2 L}{5}\right)}{G I_{p A B}} \\&=\frac{2 T_{0}(2 L-3 x)}{5 G I_{p A B}} \quad \frac{2 L}{5} \leq x \leq \frac{3 L}{5}\end{aligned}       (g)

Finally, we develop an expression for twist over segment 3 (or pipe BC). We see that the internal torsional moment now has a linear variation (Fig. 3-21), so an integral form of the torque-displacement relation [Eq. (3-27) ] is required. We insert the expression for T_{3}(x) from Eq. (d) and add the torsional displacement at B to get a formula for the varia-tion of twist in BC

\begin{aligned}\phi_{3}(x) &=\phi_{2}\left(\frac{3 L}{5}\right)+\int_{\frac{3 L}{5}}^{x} \frac{\left[3 T_{0}\left(\frac{\zeta}{L}-1\right)\right]}{G I_{p B C}} d \zeta \\&=\frac{2 T_{0}\left[2 L-3\left(\frac{3 L}{5}\right)\right]}{5 G I_{p A B}}+\int_{\frac{3 L}{5}}^x \frac{\left[3 T_{0}\left(\frac{\zeta}{L}-1\right)\right]}{G I_{p B C}} d \zeta\end{aligned}

Torque T_{3}(x) has a linear variation, so evaluating the integral yields a quadratic expression for variation of twist in BC:

\phi_{3}(x)=\frac{2 L T_{0}}{25 G I_{p A B}}+\frac{3 T_{0}\left(21 L^{2}-50 L x+25 x^{2}\right)}{50 G I_{p B C} L} \quad \frac{3 L}{5} \leq x \leq L               (h)

Substituting x = 3L/5, we obtain the twist at B:

\phi_{3}\left(\frac{3 L}{5}\right)=\frac{2 L T_{0}}{25 G I_{P A B}}

At x = L, we get the twist at C :

\phi_{3}(L)=\frac{2 L T_{0}}{25 G I_{p A B}}-\frac{6 L T_{0}}{25 G I_{p B C}}=-0.215 \frac{T_{0} L}{G I_{p A B}}

If we assume that I_{p A B} = 1.23I_{p A B} (based on the numerical properties here), we can plot the variation of twist over the length of the pipe structure (Fig. 3-22), noting that \phi_{\max } occurs at x = 2L/5 [see Eq. (f)].

Finally, if we restrict \phi_{\max } to the allowable value of 0.5°, we can solve for the maximum permissible value of load variable T_{0} (kN · m) using the numerical properties given previously

\begin{aligned}T_{0 \max }=\frac{G I_{p A B}}{0.32 L}\left(\phi_{\text {allow }}\right) &=\frac{(81 \ \mathrm{GPa})\left(2.919 \times 10^{-4} \mathrm{~m}^{4}\right)}{0.32(3 \mathrm{~m})}\left(0.5^{\circ}\right) \\&=215 \ \mathrm{kN} \cdot \mathrm{m}\end{aligned}              (i)

(d) Number of bolts required in flange plate. We now use T_{0, \max } from Eq. (i) to find the required number of d_{b} = 22 mm diameter bolts at radius r = 380 mm in the flange plate connection at B. The allowable shear stress in the bolts is \tau_{a } = 190 MPa. We assume that each bolt carries an equal share of the torque at B, so each of n bolts carries shear force F_{b} at distance r from the centroid of the cross section (Fig. 3-23)

The maximum shear force F_{b} per bolt is \tau_{a} times the bolt cross-sectional area A_{b}, and the total torque at B is 6 T_{0, \max }/5 (see TMD in Fig. 3-21), so we find

\begin{aligned}&n F_{b} r=\frac{6}{5} T_{0 \max } \text { or } n=\frac{\frac{6}{5} T_{0 \max }}{\tau_{\mathrm{a}} A_{b} r}=\frac{\frac{6}{5}(215 \ \mathrm{kN} \cdot \mathrm{m})}{(190 \ \mathrm{MPa})\left[\frac{\pi}{4}(22 \mathrm{~mm})^{2}\right](380 \mathrm{~mm})}\\&=9.4\end{aligned}

Use ten 22-mm diameter bolts at a radius of 380 mm in the flange plate connection at B.