Question 10.9: For the bolted joint as shown in Figure 10.30, (a) determine......

From Table 10.7, we have the minimum proof strength = 380 MPa, the minimum yield strength = 420 MPa, and the minimum tensile strength = 520 MPa

The modulus of elasticity for the steel member, E = 207 GPa

(a) The member stiffness can be found using Equation 10.59(a) (conical frusta model). The member stiffness can be written as

k_m \approx E d\left(\frac{0.707+0.654(d / l)}{1-0.12(d / l)}\right)

The length of the grip is specified as

l=L_G=75 \mathrm{~mm}

d = 10 mm

k_m \approx 207 \times 10^3 \times 10\left(\frac{0.707+0.654(10 / 75)}{1-0.12(10 / 75)}\right)=1.67 \times 10^6 \mathrm{~N} / \mathrm{mm}

Using the model given by Shigley [Equation 10.59(b)]

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)} (10.59b)

k_m=\frac{0.5774 \pi \times 207 \times 10^3 \times 10}{2 \ln \left(5 \times \frac{0.5774 \times 75+0.5 \times 10}{0.5774 \times 75+2.5 \times 10}\right)}=1.49 \times 10^6 \mathrm{~N} / \mathrm{mm}

(b) 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 M10 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 10^2=78.54 \mathrm{~mm}^2

From Table 10.2, the tensile stress area is given as

A_t=58 \mathrm{~mm}^2 \text { and } d_r=8.16 \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 10+6=26 \mathrm{~mm}

The length of the bolt is calculated as follows.

Taking two additional threads (i.e. assuming two threads projected from the nut after tightening)

Length of two additional threads = 2 × 1.5 = 3 mm

H=\frac{7}{8} d=\frac{7}{8} \times 10=8.75 \mathrm{~mm}

Length of the bolt is

L=75+3+8.75=86.75 \mathrm{~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=86.75-26=60.75 \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=75-60.75=14.25 \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{78.54 \times 58 \times 207 \times 10^3}{78.54 \times 14.25+58 \times 60.75}=203.1 \mathrm{~N} / \mathrm{m}

(c) 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{203.1 \times 10^3}{203.1 \times 10^3+1.67 \times 10^6}=0.108

This implies that the bolt takes 10.8% of the total applied load and 0.892, i.e 89 .2% by members.

(d) The proof load can be written as

F_p=A_t \sigma_p

Therefore, the proof load is given as

F_p=58 \times 380=22.04 \mathrm{~kN}

The preload force is specified to be 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 22.04=19.84 \mathrm{~kN}

We have

F_b=F_i+C F and for bolt separation, F_b=F

Hence, F=F_i+C F

F=F_0=\frac{F_i}{1-C}

The load to cause joint separation can be written as

F_0=\frac{F_i}{1-C}=\frac{19.84}{1-0.108}=22.24 \mathrm{~kN}

(e) The preload force is specified to be 75% of the proof load.

Therefore, the preload force is given as

F_i=0.75 \sigma_p

Therefore, the preload force is given as

F_i=0.75 \times 22.04=16.53 \mathrm{~kN}

The load to cause joint separation can be written as

F_0=\frac{F_i}{1-C}=\frac{16.53}{1-0.108}=18.53 \mathrm{~kN}