Chapter 1, Problem 1.8.13P

A steel riser pipe hangs from a drill rig located offshore in deep water (see figure). Separate segments are joined using bolted flange plages (see figure part b and photo). Assume that there are six bolts at each pipe segment connection. Assume that the total length of the riser pipe is L = 5000 ft: outer and inner diameters are d₂= l6in.and d₁= 15 in.; flange plate thickness t₁= 1.75 in.; and bolt and washer diameters are d_b= 1.125 in..and d_W. = 1.875 in., respectively.

(a) If the entire length of the riser pipe is suspended in air. find the average normal stress a in each bolt, the average bearing stress a_bbeneath each washer, and the average shear stress t through the flange plate at each bolt location for the topmost bolted connection.

(b) If the same riser pipe hangs from a drill rig at sea. what are the normal, bearing, and shear stresses in the connection? Obtain the weight densities of steel and sea water from Table I-1. Appendix I. Neglect the effect of buoyant foam casings on the riser pipe

To determine

The stresses on a steel riser pipe suspended in air.

Answer to Problem 1.8.13P

The average normal stress in each bolt is $\underset{\_}{69500\text{psi}}$ .

The bearing stress for each washer is $\underset{\_}{39100\text{psi}}$ .

The average shear stress at each bolt location in flange is $\underset{\_}{21000\text{psi}}$ .

Explanation of Solution

Given information:

Steel riser pipe length is $5000\text{ft}$ , inside pipe diameter is $15\text{in}$ , outside pipe diameter is $16\text{in}$ , flange thickness is $1.75\text{in}$ diameter of the bolt is $1.125\text{in,}$ and diameter of washer is $1.875\text{in}$ .

Write the expression for average normal stress in each bolt for steel riser pipe suspended in air.

  ${\sigma }_b=\frac{A_p\times L_P\times {\rho }_P}{A_{{}_b}\times n}$  .........(I)

Here, the area of the riser pipe is $A_p$ , the length of pipe is $L_P$ , weight density of pipe is ${\rho }_P$ , the area of a bolt is $A_{{}_b},$ and number of bolts are $n$ .

Write the expression for the area of pipe.

  $A_p=\frac{\pi }{4}\left(d_2{}^2-d_1{}^2\right)$  .........(II)

Here, the area of pipe is $A_p$ , the inner diameter of pipe is $d_1$ and the outer diameter of pipe is $d_2$ .

Write the expression for the area of a bolt.

  $A_b=\frac{\pi }{4}\left(d_b{}^2\right)$   .........(III)

Here, the diameter of bolt is $d_b$ .

Write the expression for bearing stress for each washer.

  ${\sigma }_w=\frac{A_p\times L_P\times {\rho }_P}{A_{{}_w}\times n_w}$  .........(IV)

Here, the area of a washer is $A_w$ and number of washers are $n_w$ .

Write the expression for the washer area.

  $A_w=\frac{\pi }{4}\left(d_b{}^2-d_w{}^2\right)$  .........(V)

Here, the diameter of bolt is $d_b$ and the diameter of washer is $d_w$ .

Write the expression for average shear stress at each bolt location in flange.

  $\tau =\frac{A_p\times L_P\times {\rho }_P}{t_f\times d_w\times n_w}$   .........(VI)

Here, the thickness of flange is $t_f$ .

Calculation:

Substitute $16\text{in}$ for $d_2$ and $15\text{in}$ for $d_1$ in Equation (II).

  $\begin{array}{c}{A}_p=\frac{\pi }{4}\left({\left(16\text{in}\right)}^2-{\left(15\text{in}\right)}^2\right)\\ =\frac{\pi }{4}\left(256{\text{in}}^2-225{\text{in}}^2\right)\\ =24.347{\text{in}}^2\end{array}$

Substitute $1.125\text{in}$ for $d_b$ in Equation (III).

  $\begin{array}{c}{A}_b=\frac{\pi }{4}{\left(1.125\text{in}\right)}^2\\ =\left(0.7857\times 1.265625\right)\text{in}{}^2\\ =0.994{\text{in}}^2\end{array}$

Substitute $5000\text{ft}$ for $L_p$ , $0.2836{\text{lb/ft}}^{\text{3}}$ for ${\rho }_P$ , $0.994{\text{in}}^2$ for $A_b$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n$ in Equation (I).

  $\begin{array}{c}{\sigma }_b=\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.2836{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\times 0.2836{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =\frac{414288.552\text{lb}}{0.994{\text{in}}^2\times 6}\\ =69500{\text{lb/in}}^{\text{2}}\left(\frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\end{array}$

  $=69500\text{psi}$

Substitute $1.875\text{in}$ for $d_b$ and $1.125\text{in}$ for $d_w$ in Equation (V).

  $\begin{array}{c}{A}_w=\frac{\pi }{4}\left({\left(1.875\text{in}\right)}^2-{\left(1.125\text{in}\right)}^2\right)\\ A_w=0.785\left(3.515625{\text{in}}^2-1.265625{\text{in}}^2\right)\\ A_w=1.7671{\text{in}}^2\end{array}$

Substitute $5000\text{ft}$ for $L_p$ , $0.2836{\text{lb/in}}^{\text{3}}$ for ${\rho }_P$ , $1.7671{\text{in}}^2$ for $A_w$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n_w$ in Equation (IV).

  $\begin{array}{c}{\sigma }_w=\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.2836{\text{lb/in}}^{\text{3}}}{1.7671{\text{in}}^2\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\text{ft}\left(\frac{\text{12}\text{in}}{\text{1}\text{ft}}\right)\times 0.2836{\text{lb/in}}^{\text{3}}}{1.7671{\text{in}}^2\times 6}\\ =39100\left({\text{lb/in}}^{\text{2}}\times \frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\\ =39100\text{psi}\end{array}$

Substitute $5000\text{ft}$ for $L_p$ , $0.2836{\text{lb/in}}^{\text{3}}$ for ${\rho }_P$ , $1.875\text{in}$ for $d_w$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n_w$ and $1.75\text{in}$ for $t_w$ in Equation (IV).

  $\begin{array}{c}\tau =\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.2836{\text{lb/in}}^{\text{3}}}{1.75\text{in}\times 1.875\text{in}\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\times 0.2836{\text{lb/in}}^{\text{3}}}{1.75\text{in}\times 1.875\text{in}\times 6}\\ =\frac{414288.552\text{lb}}{1.75\text{in}\times 1.875\text{in}\times 6}\\ =21000{\text{lb/in}}^{\text{2}}\left(\frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\end{array}$

  $=21000\text{psi}$

Conclusion:

The average normal stress in each bolt is $\underset{\_}{69500\text{psi}}$ .

The bearing stress for each washer is $\underset{\_}{39100\text{psi}}$ .

The average shear stress at each bolt location in flange is $\underset{\_}{21000\text{psi}}$ .

To determine

The stresses on a steel riser pipe hanging from a drill rig at sea water.

Answer to Problem 1.8.13P

The average normal stress in each bolt is $\underset{\_}{21000\text{psi}}$ .

The average bearing stress beneath each washer is $\underset{\_}{34000\text{psi}}$ .

The average shear stress at each bolt location in flange is $\underset{\_}{21000\text{psi}}$ .

Explanation of Solution

Write the expression for average normal stress in each bolt for a steel riser pipe hanging from a drill rig in sea water.

  ${\sigma }_b=\frac{A_p\times L_P\times {\rho }_{NP}}{A_{{}_b}\times n}$  .........(VII)

Here, the net weight density is ${\rho }_{NP}$ .

Write the expression for average bearing stress beneath each washer.

  ${\sigma }_w=\frac{A_p\times L_P\times {\rho }_{NP}}{A_{{}_w}\times n_w}$  .........(VIII)

Here, the area of a washer is $A_w$ and number of washers are $n_w$ .

Write the expression for average shear stress at each bolt location in flange.

  $\tau =\frac{A_p\times L_P\times {\rho }_{NP}}{t_f\times d_w\times n_w}$   .........(IX)Calculation:

Substitute $5000\text{ft}$ for $L_p$ , $0.28{\text{lb/in}}^{\text{3}}$ for ${\rho }_P$ , $0.994{\text{in}}^2$ for $A_b$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n$ in Equation (VII).

  $\begin{array}{c}{\sigma }_b=\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.28{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\times 0.28{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =60400{\text{lb/in}}^{\text{2}}\left(\frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\\ =60400\text{psi}\end{array}$

Substitute $5000\text{ft}$ for $L_p$ , $0.28{\text{lb/in}}^{\text{3}}$ for ${\rho }_{NP}$ , $1.7671{\text{in}}^2$ for $A_w$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n_w$ in Equation (VIII).

  $\begin{array}{c}{\sigma }_b=\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.28{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\left(\frac{\text{12}\text{in}}{\text{1}\text{ft}}\right)\times 0.28{\text{lb/in}}^{\text{3}}}{0.994{\text{in}}^2\times 6}\\ =34000{\text{lb/in}}^{\text{2}}\left(\frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\\ =34000\text{psi}\end{array}$

Substitute $5000\text{ft}$ for $L_p$ , $0.28{\text{lb/in}}^{\text{3}}$ for ${\rho }_{NP}$ , $1.875\text{in}$ for $d_w$ , $24.347{\text{in}}^2$ for $A_p$ and $6$ for $n_w$ and $1.75\text{in}$ for $t_w$ in Equation (IX).

  $\begin{array}{c}\tau =\frac{24.347{\text{in}}^2\times 5000\text{ft}\times 0.28{\text{lb/in}}^{\text{3}}}{1.75\text{in}\times 1.875\text{in}\times 6}\\ =\frac{24.347{\text{in}}^2\times 5000\text{ft}\left(\frac{\text{12}\text{in}}{\text{1}\text{ft}}\right)\times 0.28{\text{lb/in}}^{\text{3}}}{1.75\text{in}\times 1.875\text{in}\times 6}\\ =18300\left({\text{lb/in}}^{\text{2}}\times \frac{1\text{psi}}{\text{1}{\text{lb/in}}^{\text{2}}}\right)\\ =21000\text{psi}\end{array}$

Conclusion:

The average normal stress in each bolt $\underset{\_}{21000\text{psi}}$ .

The average bearing stress beneath each washer $\underset{\_}{34000\text{psi}}$

The average shear stress at each bolt location in flange $\underset{\_}{21000\text{psi}}$ .