Question 10.23: Bolts are used to hold the cover plate and the cylinder. The......

Assumptions

1. Load is shared equally by all bolts

2. Initial bolt tension is taken as the lowest value, i.e. 0.55 \sigma_p A_t·

3. 8.8 class rolled threads are used.

4. A higher factor of safety 4 is used in the design though it increases the cost, but failure may be costlier.

Data given are

p_s=1.1 MPa, Factor of safety N = 4.0, D = 1200 mm, d_p=1400 \mathrm{~mm}

If n is the number of bolts used in the cylinder joint, the maximum value of force F is

F=\frac{\pi}{4} D^2 \times p_s \times N=\frac{\pi}{4} \times 1200^2 \times 1.1 \times 4=4976.3 \mathrm{~kN}

This force will be shared by number of bolts.

The modulus of elasticity for the bolts assuming they are made of steel, is given as

E=207 \mathrm{~GPa}

k_b=\frac{A_b E_b}{l_b}=\frac{A_b \times 207}{40}=5.175 A_b

The net stiffness of the member is

k_m=\frac{k_1 k_2}{k_1+k_2}

where

k_1=\frac{A_1 E_1}{l_1}=\frac{5 A_b \times 100}{20}=25 A_b \text { given } A_m=5 A_b \text {, i.e } A_1=A_2=5 A_b

k_2=\frac{A_2 E_2}{l_2}=\frac{5 A_b \times 70}{20}=17.5 A_b

Therefore, the net stiffness of the member is given as

k_m=\frac{25 \times 17.5 \times A_b^2}{25 A_b+17.5 A_b}=10.29 A_b

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{5.175}{5.175+10.29}=0.335

Increased bolt force is

\Delta F_b=C F=0.335 \times 4976.3=1667.06 \mathrm{~kN}

Alternating force is

F_a=\frac{\Delta F_b}{2}=\frac{1667.06}{2}=833.5 \mathrm{~kN}

Equating alternating stress to endurance strength, we get

S_e=\frac{F_a}{A_t}

The endurance strength for 8.8 class is 129 MPa

A_t=\frac{833.5 \times 1000}{129}=6461.47 \mathrm{~mm}^2

where A_t is the total area of all bolts used.

Hence n A_t=6461.47 \mathrm{~mm}^2

For different n values the standard bolt (fine series) is calculated as shown.

From the bolt spacing condition, the ratio of bolt spacing is

3 \leq \frac{\pi d_b}{N_b \times d} \leq 6

3 \leq \frac{\pi \times 1400}{n d} \leq 6

the calculated bolt spacing ratio is

Bold numbers are desirable. The optimum size can be obtained from the other consideration such as cost optimization, figure of merit or factor of safety.

The factor of safety N guarding against fatigue failure for a bolt can be written as

N=\frac{S_a}{\sigma_a}

The alternating component of the bolt stress can be written as

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

Note that stress concentration is NOT included in this stress calculation, because the threads are assumed to be formed by a rolling process. A fully corrected (including k_f) endurance limit will be used. The maximum external load acting on each bolt is given as

P=\frac{\text { effective area } \times \text { pressure }}{\text { number of bolts }}

Therefore, the external load acting on each bolt (taking Ml2, 80 number of bolts) is given as

F=4976.3 / 80 \mathrm{~kN}=62.3 \mathrm{~kN}

Therefore, the alternating component of the bolt stress is given as

\sigma_a=\frac{0.335 \times 62.2 \times 1000}{2 \times 88.1}=118.26 \mathrm{~MPa}

The mean and alternating components using the Gerber criterion can be written as

S_m=\frac{\sigma_u^2}{2 S_e}\left\{-1+\sqrt{1+\frac{4 S_e}{\sigma_u^2}\left(S_e+\sigma_i\right)}\right\}

S_a=S_m-\sigma_i

The tensile strength for ISO class 8.8, M12 bolt is given as

\sigma_u=830 \mathrm{~MPa}

The endurance strength for ISO class 8.8, M12 bolt is

S_e=129 \mathrm{~MPa}

This endurance limit has been fully corrected for bolt geometry and surface treatment.

\sigma_i=\frac{F_i}{A_t}

The preload force is specified to be 55% of the proof load.

Therefore, the preload force can be written as

F_i=0.55 F_p

The proof load is

F_p=A_t \sigma_p

The proof strength for ISO class 8.8, M12 bolt, \sigma_p=600 \mathrm{~MPa}

Therefore, the proof load is given as F_p=86 \times 600=52.86 \mathrm{~kN}

Therefore, the preload force is given as F_i=0.55 \times 88.1 \times 600=29.073 \mathrm{~kN}

Therefore, the minimum stress in the bolts is given as

\sigma_i=\frac{29.073 \times 1000}{88.1}=330 \mathrm{~MPa}

Therefore, the mean component is

S_m=\frac{830^2}{2 \times 129}\left\{-1+\sqrt{1+\frac{4 \times 129}{830^2}(129+330)}\right\}=425.15 \mathrm{~MPa}

Alternating component is

S_a=S_m-\sigma_i=425.15-330=95.15 \mathrm{~MPa}

Therefore, the fatigue factor of safety using the Gerber criterion is given as

N=\frac{95.15}{118.26}<1

This is not a feasible solution as it will fail under fatigue loading.

Other combinations are to be tried to obtain a feasible solution. The rest of the solution is left for the reader to write a computer program to make a decision on the optimum solution of the joint. The procedure for design of bolted joint with a bolt circle is illustrated below.

Algorithm

Step 1 Select a bolt sized as a trial diameter. The size may be close to the half of the clamp length.

Step 2 Calculate the minimum and maximum value of N_b from the relation 3 \leq \frac{\pi \times 1400}{N_b d} \leq 6.

Step 3 Calculate the joint coefficient C.

Step 4 Calculate the torque, F_i \text {, and } \sigma_i,

Step 5 Select thread type and determine the endurance strength as per procedure explained.

Step 6 Calculate the mean and variable stresses.

Step 7 Find S_m \text { and } S_a using the fatigue diagram.

Step 8 Find the factor of safety N \text { and } N_0\left(\text { for } C=1 \text { and } F_i=0\right).

Step 9 Compare the factor of safety.

Step 10 If the factor of safety is less than the desirable factor of safety, go to Step 1.

Step 11 Perform the cost analysis.

Step 12 Prepare table for the other bolt size.

Step 13 Obtain a feasible domain of solution in which condition of bolt spacing and the factor of safety satisfy the desirable conditions.

Step 14 From the table, recommend the optimum design.