Question 10.11: Determine (a) the suitable bolt size, (b) the preload, (c) t......

Assumptions:

(a) Assume both the clamped parts are steel having E = 207GPa

(b) Preload as 90% of proof strength (a trial value)

(c) A bolt M20 coarse (a trial value), type 5.8 class

From properties we have, the minimum proof strength = 380 MPa, the minimum yield strength = 420 MPa, and the minimum tensile strength = 520 MPa

The bolt stiffness can be written as

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

The major diameter area for M20 bolt can be written as

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

Therefore, the major diameter area is given as

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

From Table 10.2 the tensile stress area is given as

A_t=245 \mathrm{~mm}^2 \text { and } d_r=16.933 \mathrm{~mm}

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}

Length of the bolt is calculated as follows.

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

L_G=50 \mathrm{~mm}

Length of the bolt is

L=50+17.5=67.5 \mathrm{~mm}

Take standard length L = 10 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=70-46=24 \mathrm{~mm}

l_t=L_G-l_d

Therefore, the threaded length in the grip is obtained as

l_t=50-24=26 \mathrm{~mm}

The modulus of elasticity for the steel bolt is given as E = 207 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 24+314.16 \times 26}=1134.14 \mathrm{~N} / \mathrm{m}

The stiffness of the clamped material is obtained as follows.

k_m=\frac{0.5774 \pi E d}{2 \ln \left(5 \frac{0.5774 l+0.5 d}{0.5774 l+2.5 d}\right)}

k_m=\frac{0.5774 \pi \times 207 \times 10^3 \times 20}{2 \ln \left(5 \times \frac{0.5774 \times 50+0.5 \times 20}{0.5774 \times 50+2.5 \times 20}\right)}=4163.5 \mathrm{~N} / \mathrm{m}

The stiffness constant of the joint can be written as

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

Therefore, the stiffness constant of the joint is given as

C=\frac{4163.5}{1134.14+4163.5}=0.786

This implies that the bolt takes 78.6% of the total applied load.

The proof load can be written as

F_p=A_t \sigma_p

Therefore, the proof load is given as

F_p=245 \times 380=93.1 \mathrm{~kN}

The preload force is assumed as 90% of the proof load.

Therefore, the preload force is given as

F_i=0.9 F_p

Therefore, the preload force is given as

F_i=0.9 \times 93.1=83.79 \mathrm{~kN}

The portion of the applied load F taken by bolt is

F_b=C F=0.786 \times 10=7.86 \mathrm{~kN}

F_m=(1-C) F=(1-0.786) \times 10=2.14 \mathrm{~kN}

The resulting loads in bolt and the material after the load F is applied

F_b=F_i+C F=83.79+7.86=91.65 \mathrm{~kN}

F_m=F_i-(1-C) F=83.79-2.14=81.65 \mathrm{~kN}

Maximum tensile stress in the bolt is given as

\sigma_b=\frac{F_b}{A_t}=\frac{91.65 \times 1000}{245}=374.08 \mathrm{~N} / \mathrm{mm}^2

Safety factor against yielding is

N_y=\frac{\sigma_y}{\sigma_b}=\frac{420}{374.08}=1.123

The load to cause joint separation can be written as

F_0=\frac{F_i}{1-C}=\frac{83.79}{1-0.786}=391.54 \mathrm{~kN}

The factor of safety against separation

N_{\text {separation }}=\frac{F_0}{F}=\frac{391.54}{10}=39.1

With other percentage of proof loads, the calculations are made and both factor of safety against yielding and separation is presented in Table 10.8.

The results shown in the above tables and figures indicate that factor of safety against separation increases linearly, whereas the factor of safety against yielding shows a non-linear relation with percentage of proof load. Feasible solutions are shown within the dotted lines. At high preload, the factor of safety against yielding is low and factor of safety against separation is high. Two lines crosses at a preload of 65% of proof strength. If the goal is to prevent the failure against overload, the larger preload is desirable. For 85% of the preload, the factor of safety against separation is 2.16 which is acceptable and at this overload 16% of the load is still reserved for yielding failure (FOS 1.16 at yielding). Hence it is suggested that a preload of 85% is suitable for this application.

The Matlab program used for calculation of all parameters at this preload is given below.