Question 10.25: A pipe of nominal diameter 125 mm is joined by eight M20 × 2......

Inner diameter of pipe = 125 mm

Outer diameter of pipe = 150 mm

Outer diameter of gasket = 150 mm

Thickness of gasket = 4 mm

Thickness of flange = 30 mm

Proof stress of 5.8 class material = 380 MPa

Ultimate strength of 5.8 class = 520 MPa

Tensile area of bolt = 245 mm² (coarse threads)

Width of the gasket =\frac{150-125}{2}=12.5 \mathrm{~mm}

Effective width of gasket =2.52 \sqrt{\left(\frac{\text { width }}{2}\right)}=2.52 \sqrt{\left(\frac{12.5}{2}\right)}=6.3 \mathrm{~mm}

Hence the effective inner diameter of gasket = 150 – 2 × 6.3 = 137.4 mm

Effective area of gasket =\frac{\pi}{4}\left(150^2-137.4^2\right)=2844.12 \mathrm{~mm}^2

Proof load of the bolt =F_p=380 \times 245=93100 \mathrm{~N}

Assuming tightening load is 75% of the proof load, bolt tightening load is

F_i=0.75 \times 93100=69.825 \mathrm{~kN} \text { per bolt }

The stiffness of the bolt can be written as

k_b=\frac{A_d A_t E}{A_d l_t+A_t l_d}

The area of the unthreaded portion of the bolt is

A_d=\frac{\pi d^2}{4}

Therefore, the area of the unthreaded portion of the bolt (M20) is given as

A_d=\frac{\pi}{4} \times 20^2=314.16 \mathrm{~mm}^2

From table, the tensile stress area is given as

A_t=245 \mathrm{~mm}^2

The thread length, for a bolt length less than 125 mm, can be written as

L_t=2 d+6 \mathrm{~mm}=2 \times 20+6=46 \mathrm{~mm}

H=\frac{7}{8} d=\frac{7}{8} \times 20=17.5 \mathrm{~mm}

Length of the grip, L_G=2 \times 30+4=64 \mathrm{~mm}

Length of the bolt is

L=64+17.5=81.5 \mathrm{~mm}

Length of standard bolt is L = 85 mm

The length of the unthreaded portion in the grip, can be written as

l_d=L-L_T

Therefore, the length of the unthreaded portion in the grip is given as

l_d=85-46=39 \mathrm{~mm}

The threaded length in the grip, l_t=L_G-L_d

Therefore, the threaded length in the grip is obtained as

l_t=64-39=25 \mathrm{~mm}

The modulus of elasticity for the steel bolt is given as E = 201 GPa

Therefore, the stiffness of the bolt is given as

k_b=\frac{A_t A_d E}{A_t l_d+A_d l_t}=\frac{245 \times 314.16 \times 207 \times 10^3}{245 \times 39+314.16 \times 25}=915.195 \mathrm{~N} / \mathrm{m}

The portion of each flange in direct compression will be modelled as a 30 mm long conical frustum and using the following relation, we have

k_{\text {flange }}=E d\left(\frac{0.707+0.654(d / l)}{1-0.12(d / l)}\right)

k_{\text {flange }}=207 \times 10^3 \times 20\left(\frac{0.707+0.654(20 / 30)}{1-0.12(20 / 30)}\right)=5143.5 \mathrm{~N} / \mathrm{m}

The limiting stiffness value of gasket material is in the range of 270 to 1800 MPa/mm. The unit of force can be obtained by multiplying the gasket area.

A_g=\frac{1}{8} \times 2844.12=356 \mathrm{~mm}^2

96 \leq k_{\text {gasket }} \leq 641 \mathrm{~kN} / \mathrm{mm}

Taking the higher value, k_{\text {gasket }}=600 \mathrm{~N} / \mathrm{m}

The member stiffness is

\frac{1}{k_m}=\frac{2}{k_{\text {flange }}}+\frac{1}{k_{\text {gasket }}}

\frac{1}{k_m}=\frac{2}{5143.5}+\frac{1}{600}=2.056 \times 10^{-3}

k_m=486.5 \mathrm{~N} / \mathrm{m}

The stiffness constant of the joint,

C=\frac{k_b}{k_b+k_m}

Therefore, the stiffness constant of the joint is given as

C=\frac{915.195}{915.1952+486.5}=0.653

Bolt fatigue

S_e=0.5 \times 520 \times \frac{1}{2.2}=118.18 MPa based on rolled threads

F=\frac{\pi}{4} \times 125^2 \times 7=85.9 \mathrm{~kN}

\sigma_a=\frac{C F}{2 A_t}

=\frac{0.653 \times \frac{\pi}{4} \times 125^2 \times 7}{2 \times 8 \times 245}=14.3 \mathrm{~MPa}

\sigma_m=\frac{F_i}{A_t}+\sigma_a=\frac{69825 \times 8}{245 \times 8}+14.3=299.3 \mathrm{~MPa}

As per Goodman criterion

\frac{\sigma_a}{S_e}+\frac{\sigma_m}{\sigma_u}=\frac{1}{N}

\frac{1}{N}=\frac{14.3}{118.18}+\frac{299.3}{520}=0.697

N = 1.436