Please do problem $5$ using problem $4$ as shown. The answer is already given. Just use a different method to get to it$.$ Thanks!

SAMPLE PROBLEM $10\mathrm{.}5$

Select Bolts for Bracket Attachment, Neglecting

Friction, and Assuming Shear Forces Are Carried

by the Bolts

Repeat Sample Problem $10\mathrm{.}4,$ except neglect the frictional forces.

SOLUTION

Known: Three SAE class $9\mathrm{.}8$ steel bolts having a specified safety factor are used to attach a bracket of

known geometry that supports a known vertical load.

Find: Select an appropriate bolt size.

Schematic and Given Data: See Sample Problem $10\mathrm{.}4$ and Figure $10\mathrm{.}32.$

Assumptions:

The shear forces caused by the eccentric vertical load are carried completely by the bolts.

The vertical shear load is distributed equally among the three bolts.

The tangential shear force carried by each bolt is proportional to its distance from the center of gravity

of the group of bolts.Analysis:

Neglecting friction has no effect on bolt stresses in the threaded region, where attention was focused

on Sample Problem $10\mathrm{.}4.$ For this problem, attention is shifted to the bolt shear plane $($at the interface

between bracket and fixed plate$).$ This plane experiences the tensile force of $87\mathrm{.}27$ kN calculated in

Sample Problem $10\mathrm{.}4$ in addition to the shear force calculated in the following step $2.$

The applied eccentric shear force of $24kN(6)=144kN$ tends to displace the bracket downward and

alsorotate it clockwise about the center of gravity of the bolt group cross section. For three bolts of equal

size, the center of gravity corresponds to the centroid of the triangular pattern, as shown in Figure $10\mathrm{.}33.$

This figure shows the original applied load $($dotted vector$)$ replaced by an equal load applied at the

centroid $($solid vector$)$ plus a torque that is equal to the product of the force and the distance it was

moved. As assumed, each bolt carries one$-$third of the vertical shear load, plus a tangential force $($with

respect to rotation about the center of gravity$)$ that is proportional to its distance from the center of

gravity. Calculations on the figure show this tangential force to be $37\mathrm{.}1$ kN for each of the top bolts.

The vector sum of the two shear forces is obviously greatest for the upper right bolt. Routine calculation

shows $V=81\mathrm{.}5kN.$

The critical upper right bolt is thus subjected to a tensile stress, $=87,\frac{270}{A},$ and a shear stress,

$=81,\frac{500}{?}$ A$.$ Substitution in the distrition energy equation gives an equivalent tensile stress of

${}_e=\sqrt[\phantom{2}]{{}^2+3{}^2}=\frac{1}{A}\sqrt[\phantom{2}]{(87,270{)}^2+3(81,500{)}^2}=\frac{166,000}{A}$

Equating this to the proof stress gives

$\frac{166,000}{A}=S_p=650$MPa

Therefore,

$A=255m{m}^2$

Finally,

$A=\frac{d^2}{4},or,d=\sqrt[\phantom{2}]{\frac{4A}{}}=\sqrt[\phantom{2}]{\frac{4(255)}{}}=18\mathrm{.}03mm$

Thus, a shank diameter of $18$ mm is required.