Question 10.19: Find the size of the bolt required for the bolted joint as s......

Free body diagram showing forces on the three bolts is shown below.

The bracket is presumed to be rigid, and equilibrium is taken to be achieved by shear of the bolt shanks. The centroid of the three equal shear areas A_t is located at G as shown. Since G does not lie on the line of action of the external load, a rotational tendency (torsion) exists.

The analysis is similar to the traditional analysis of fillet welds in which force equilibrium is achieved by uniform direct shear, while torsional equilibrium arises from the secondary shear whose sense is perpendicular to the radius from the centroid to the location of the secondary shear, and which is proportional to that radius, given by the torsion equation.

f_s=A \tau=A\left(\frac{T r}{J}\right)=\frac{A T r}{\sum A r^2}

Since all A’s are the same here

f_s=\frac{T r}{\sum r^2}

\bar{y}=\frac{A \times 180+A \times 180+A \times 0}{3 A}=120 \mathrm{~mm}

\bar{x}=\frac{160 A+80 A}{3 A}=80 \mathrm{~mm}

So the torsional shear forces are

T=F e=12 \times 75=900 \mathrm{~N} \cdot \mathrm{m}

r_1=r_2=\sqrt{60^2+80^2}=100 \mathrm{~mm}

r_3=120 \mathrm{~mm}

f_{s 1}=f_{s 2}=\frac{12 \times 75 \times 100}{\sum\left\{120^2+2\left(60^2+80^2\right)\right\}}=2.62 \mathrm{~kN}

f_{s 3}=\frac{12 \times 75 \times 120}{\sum\left\{120^2+2\left(60^2+80^2\right)\right\}}=3.139 \mathrm{~kN}

and direct shear is

f_{d 1}=f_{d 2}=f_{d 3}=\frac{12}{3}=4 \mathrm{~kN}

The resultant load is

R_1=\sqrt{\left(f_{s 1}\right)^2+\left(f_{d 1}\right)^2+f_{s 1} \times f_{d 1} \times \cos \theta_1}

R_1=\sqrt{(2.62)^2+(4)^2+2.62 \times 4 \times 4 / 5}=6.295 \mathrm{~kN}

\cos \theta_1=\frac{80}{\sqrt{60^2+80^2}}=4 / 5

Equating R_1 with allowable load

\tau_{\mathrm{all}}=\frac{650}{\sqrt{3}}=\frac{6.295 \times 10^3}{A_t}

Solving, we get

A_t=16.77 \mathrm{~mm}^2

d=\sqrt{\frac{16.77 \times 4}{\pi}}=4.6 \mathrm{~mm}=5 \mathrm{~mm}