Question 10.20: Determine the size of the bolt for the joint as shown in the......

\bar{x}=20 \mathrm{~mm} \text { and } \bar{y}=20 \mathrm{~mm} (due to symmetry of the bolts arrangement) The distance of load from CG of the bolt area is

e=100+20=120

T=F \times e=8000 \times 120=960 \times 10^3 \mathrm{~N} \cdot \mathrm{mm}=960 \mathrm{~N} \cdot \mathrm{m}

The eccentric load will produce primary and secondary forces and hence secondary shear stresses on each bolt. It is assumed that all bolts are equally loaded.

The magnitude of the direct shear force on each bolt is obtained as

f_d=\frac{F}{n}

where n is the number of bolts.

In this problem, n = 4, F = 8 kN

f_{d 1}=f_{d 2}=f_{d 3}=f_{d 4}=f_d=\frac{8000}{4}=2000 \mathrm{~N}

The secondary shear force due to twisting moment T is given as

f_s=\frac{T r}{\sum r^2}=\frac{960 \times 1000 \times 28.284}{4 \times 28.284^2}=8485.3 \mathrm{~N}

where r = 28.284 mm, the radial distance from CG to the bolt concerned. Its calculation is shown below.

r=\sqrt{20^2+20^2}=28.284 \mathrm{~mm}

The forces are shown in the figure below.

Considering bolt no. 2, the resultant force is obtained as the vector addition of primary and secondary forces and obtained as follows.

R=\sqrt{f_d^2+f_s^2+2 f_d f_s \cos \theta}

=\sqrt{2000^2+8485.3^3+2 \times 2000 \times 8485.3 \times 0.707}=9999.82 \mathrm{~N}

\cos \theta=\frac{20}{28.284}=0.707

It is seen that bolts 2 and 3 are heavily loaded and magnitude is the same. If A is the bolt cross sectional area, the shear stress is

\tau=\frac{R}{A}=\frac{9999.82}{A}              (10.77)

Allowable yield strength is \tau_y=0.577 \sigma_y=0.577 \times 420=242.34 MPa for class 5.8 bolt material, assuming a factor of safety 2.

\tau_{\text {all }}=\frac{242.34}{2}=121.17 \mathrm{~MPa}                     (10.78)

Equating (10.77) and (10.78)

\frac{9999.82}{A}=121.17

A=82.53 \mathrm{~mm}^2

M 12 × 1.5 with tensile stress area 88.1 mm² is selected.