Design of bolted joint :-

CASE I: Load acts perpendicular to bolt axis:

(1) Due to the loading as shown in the fig., bolts are subjected to direct shear (primary effect).

(2) Bolts are also subjected to tensile stress due to turning moment about tilting edge (secondary effect).

• Design Procedure:

Let, n = no. of bolts.

(I) Primary effect:

(a) Shear force per bolt, $F_{s}$

$F_{s}$ = $\frac{P}{n}$

$\therefore$ $\tau$ = $\frac{F_{s}}{A}$; (A = core area)

(II) Secondary effect:

$\;\;\;$ (a) Tensile force per unit length from tilting edge --

$F = \frac{P . e}{l_{1}^2 + l_{2}^2 + l_{3}^2 + l_{4}^2}$

$\;\;\;$ (b) Maximum tensile force --

$F_{tmax} = F_{3} = F_{4} = F . l_{3} = F . l_{4}$

$\;\;\;$ (c) $\sigma_{t}$ = $\frac{F_{tmax}}{A}$

$\;\;\;$ (d) Maximum tensile stress theory --

$\sigma_{tmax}$ = $\frac{1}{2}$[$\sigma_{t}$ + $\sqrt{\sigma_{t}^2 + 4\;\tau^2}$]

$\;\;\;$ (e) Maximum shear stress theory --

$\tau_{tmax}$ = $\frac{1}{2}$[$\sqrt{\sigma_{t}^2 + 4\;\tau^2}$]

Q.1 A bracket is fixed to a steel column as shown in fig. A vertical load of $12\ kN$ is acting on the bracket. The bracket is fixed by 5 bolts (3 in top row and 2 in bottom row).

Determine 1) size of bolt. 2) size of c/s area of bracket which is rectangular.

Assume allowable stresses in tension and shear for bolt and bracket material to be $85\ N/mm^2$ and $52\ N/mm^2$ respectively.

Soln. Given: $n = 5$

$P = 12\ kN$

$\sigma_{tmax} = 85\ N/mm^2$

$\tau_{tmax} = 52\ N/mm^2$

Soln: (I) Primary effect

(a) $F_{s} = \frac{P}{h} = \frac{12\times 10^{3}}{5} = 2.4 \times 10^3 N$

$\therefore$ $\tau$ = $\frac{F_{s}}{A}$ = $\frac{2.4 \times 10^3}{A}$

(II) Secondary effect

(a) $F = \frac{P . e}{l_{1}^2 + l_{2}^2 + l_{3}^2 + l_{4}^2 + l_{5}^2 }$

$\therefore F = \frac{12 \times 10^3 \times 400}{50^2 + 50^2 + 400^2 + 400^2 + 400^2 }$

$\therefore F = 9.8969\ N/mm$

(b) $F_{tmax} = F_{3} = F_{4} = F_{5} = F . l_{3} = F . l_{4} = F . l_{5}$

$= 9.8969 \times 400$

$= 3.9588 \times 10^3\ N$

(c) $\sigma_{t}$ = $\frac{F_{tmax}}{A}$ = $\frac{3.9588 \times 10^3}{A}$

(d) Max tensile stress theory --

$\sigma_{tmax}$ = $\frac{1}{2}$[$\sigma_{t}$ + $\sqrt{\sigma_{t}^2 + 4\;\tau^2}$]

$\therefore 85 = \frac{1}{2}[\frac{3.9588 \times 10^3}{A} + \sqrt{(\frac{3.9588 \times 10^3}{A}) + 4(\frac{2.4 \times 10^3}{A})^2}]$

$\therefore A = 59.8865\ mm^2$

(e) Max. shear stress theory

$\tau_{tmax}$ = $\frac{1}{2}$[$\sqrt{\sigma_{t}^2 + 4\;\tau^2}$]

$52 = \frac{1}{2}$[$\sqrt{(\frac{3.9588 \times 10^3}{A})^2 + 4\;(\frac{2.4 \times 10^3}{A})^2}$]

$\therefore A = 59.826\ mm^2$

Select, $A = 59.8865\ mm^2$

$\therefore \frac{\pi}{4}d_{i}^2 = 59.8865$

$\therefore d_{i} = 8.7321\ mm$

From PSG 5.42, from coarse series,

select M12.

Design of bracket:-

$\sigma_{t}$ = $\frac{M}{I}$.Y = $\frac{M}{(\frac{I}{Y})}$ = $\frac{M}{Z}$

$Z = \frac{I}{Y}$ = [$\frac{\frac{t(300)^3}{12}}{\frac{300}{2}}$]

$\therefore z = 15000t$

$\therefore 85 = \frac{12 \times 10^3 \times 400}{15000t}$

$\therefore t = 3.7647\ mm \approx 4 mm.$

Q.2 Fig. shows pulley bracket supported to a vertical wall by four bolts to each at location A and B. The pull 'P' on each side of wire rope over the pulley is $22\ kN$. Determine the diameter of bolt if $\tau_{max}$ is $20\ MPa$. Select the bolt from the following table.

Soln: Given: $n = 4$

$P = 22\ kN$

$\tau_{tmax} = 30\ Mpa = 30\ N/mm^2$

Soln: (I) Primary effect

(a) Shear force per bolt,

$F_{s} = \frac{P}{h} = \frac{2 \times 22 \times 10^3}{4} = 11 \times 10^3$

$\therefore$ $\tau$ = $\frac{F_{s}}{A}$ = $\frac{11 \times 10^3}{A}$

(II) Secondary effect

(a) Tensile force per unit length from tilting edge,

$F = \frac{P . e}{l_{1}^2 + l_{2}^2 + l_{3}^2 + l_{4}^2}$ = $\frac{44 \times 10^3 \times 450}{75^2 + 75^2 + 525^2 + 525^2}$

$\therefore F = 35.2\ N/mm$

(b) Maximum tensile force,

$F_{tmax} = F_{3} = F_{4} = F . l_{3} = F . l_{4}$

$= 35.2 \times 525$

$= 18.48 \times 10^3\ N$

(c) $\sigma_{t}$ = $\frac{F_{tmax}}{A}$ = $\frac{18.48 \times 10^3}{A}$

(d) Max shear stress theory --

$\tau_{tmax}$ = $\frac{1}{2}$[$\sqrt{\sigma_{t}^2 + 4\;\tau^2}$]

$30 = \frac{1}{2}[\sqrt{(\frac{18.48 \times 10^3}{A})^2 + 4\;(\frac{11 \times 10^3}{A})^2}]$

$\therefore A = 478.8616\ mm^2$

From table, we select M30

$\therefore A = 561$

$\therefore \frac{\pi}{4}d_{c}^2 = 561$

$\therefore d_{c} = 25.7261\ mm$

CASE II: Load is in the plane of bolt system but away from C.G of bolt system, i.e, when load acts in the plane containing bolt

Due to this kind of loading

(1) Direct shear load or force will be produced in each bolt (primary).

(2) Shear load or force due to turning moment of external load will be produced (secondary shear)

(a) Primary shear force :

$F_{P} = \frac{P}{n}$

(II) Secondary effect :

(a) Shear force per unit length --

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2 + r_{4}^2}$

(b) Secondary shear force = F$_{s}$ :

F$_{S_{1}}$ = F . r$_{1}$

F$_{S_{2}}$ = F . r$_{2}$

F$_{S_{3}}$ = F . r$_{3}$

F$_{S_{4}}$ = F . r$_{4}$

(c) F$_{R_{1}}$ = $\sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

F$_{R_{2}}$ = $\sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

F$_{R_{3}}$ = $\sqrt{F_{P}^2 + F_{S_{3}}^2 + 2 F_{P} . F_{S_{3}} . cos\theta_{3}}$

F$_{R_{4}}$ = $\sqrt{F_{P}^2 + F_{S_{4}}^2 + 2 F_{P} . F_{S_{4}} . cos\theta_{4}}$

From above eqns, we get F$_{R_{max}}$.

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

Q3] Find diameter of bolt for the configuration as shown in fig. Assume material for bolt -- Sy = $\sigma_{y}$ = 350MPa, FOS = 3.5

Soln. (I) Primary effect :

(a) Primary shear force --

$F_{P} = \frac{P}{n} = \frac{40 \times 10^3}{4} = 10 \times 10^3N$

(II) Secondary effect :

(a) Shear force per unit length --

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2 + r_{4}^2}$

$\therefore F = \frac{40 \times 10^3 \times 300}{{4}{(75)^2}}$

$\therefore F = 533.3333\ N/mm$

(b) Secondary shear force :

$F_{S_{1}} = F . r_{1} = 533.3333 \times 75 = 40 \times 10^3\ N$

$F_{S_{2}} = F . r_{2} = 40 \times 10^3N$

$F_{S_{3}} = F . r_{3} = 40 \times 10^3N$

$F_{S_{4}} = F . r_{4} = 40 \times 10^3N$

(c) $F_{R_{1}} = \sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

$= \sqrt{(10^4)^2 + (40 \times 10^3)^2 + 2(10^4)(40 \times 10^3) . cos0}$

$= 50 \times 10^3\ N.$

$F_{R_{2}} = \sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

$= \sqrt{(10 \times 10^3)^2 + (40 \times 10^3)^2 + 2(10 \times 10^3)(40 \times 10^3) . cos90}$

$= 41.2311 \times 10^3\ N$.

$F_{R_{3}} = \sqrt{F_{P}^2 + F_{S_{3}}^2 + 2 F_{P} . F_{S_{3}} . cos\theta_{3}}$

$= \sqrt{(10 \times 10^3)^2 + (40 \times 10^3)^2 + 2(10 \times 10^3)(40 \times 10^3) . cos180}$

$= 30 \times 10^3\ N$

$F_{R_{4}} = \sqrt{F_{P}^2 + F_{S_{4}}^2 + 2 F_{P} . F_{S_{4}} . cos\theta_{4}}$

= $\sqrt{(10 \times 10^3)^2 + (40 \times 10^3)^2 + 2(10 \times 10^3)(40 \times 10^3) . cos90}$

$= 41.2311 \times 10^3N$.

$\therefore F_{R_{max}} = F_{R_{1}} = 50 \times 10^3\ N$.

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

$\tau_{max} = \frac{0.5\sigma_{y}}{FOS} = \frac{0.5 \times 350}{3.5} = 50$

$\therefore 50 = \frac{50 \times 10^3}{A}$

$\therefore A = 10^3\ mm^2$

$\therefore \frac{\pi}{4}d_{c}^2 = 10^3$

$\therefore d_{c} = 35.6825\ mm$

From PSG 5.42,

$d_{c} = 35.6825\ mm$ for bolt is not available, therefore using imperical relation,

$d_{c} = 0.84\ d$

$\therefore 35.6825 = 0.84\ d$

$\therefore d = 42.4792\ mm$

Q.4 A steel plate subjected to a force of $5\ kN$ and fixed to a channel by means of 3 identical bolts are made of carbon steel whose $\sigma_{y} = 380\ N/mm^2, FOS = 3$. Specify size of bolt.

Given $P = 5 \times 10^3\ N$

$\tau_{max} = \frac{0.5\sigma_{y}}{FOS} = \frac{0.5 \times 380}{3} = 63.3333\ N/mm^2$

$n = 3$

Soln. (I) Primary effect --

(a) Primary shear force --

$F_{P} = \frac{P}{n} = \frac{5 \times 10^3}{3} = 1.6667 \times 10^3N$

(II) Secondary effect --

(a) Shear force per unit length --

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2}$

$\therefore F = \frac{5 \times 10^3 \times 305}{75^2 + 0^2 + 75^2}$

$\therefore F = 135.5556\ N/mm$

(b) Secondary shear force --

$F_{S_{1}} = F . r_{1} = 135.5556 \times 75 = 10.1667 \times 10^3N$

$F_{S_{2}} = F . r_{2} = 0\ N$

$F_{S_{3}} = F . r_{3} = 10.1667 \times 10^3\ N$

(c) $F_{R_{1}} = \sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

$= \sqrt{(1.6667 \times 10^3)^2 + (10.1667 \times 10^3)^2 + 2(1.6667 \times 10^3)(10.1667 \times 10^3) . cos\theta_{1}}$

$= 8.6 \times 10^3\ N$.

$F_{R_{2}} = \sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

$= \sqrt{(1.6667 \times 10^3)^2 + 0^2 + 2F_{P}(0)cos\theta_{2}}$

$= 1.6667 \times 10^3\ N$

$F_{R_{3}} = \sqrt{F_{P}^2 + F_{S_{3}}^2 + 2 F_{P} . F_{S_{3}} . cos\theta_{3}}$

= $\sqrt{(1.6667 \times 10^3)^2 + (10.1667 \times 10^3)^2 + 2(1.6667 \times 10^3)(10.1667 \times 10^3) . cos\theta_{3}}$

$= 11.8334 \times 10^3N$.

$\therefore F_{R_{max}} = 11.8334 \times 10^3\ N$.

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

$\therefore 63.3333 = \frac{11.8334 \times 10^3}{A}$

$\therefore A = 186.8433\ mm^2$

$\therefore \frac{\pi}{4}d_{c}^2 = 186.8433$

$\therefore d_{c} = 15.4239\ mm$.

From PSG 5.42, from coarse series,

select M20.

Q.5 A steel plate is subjected to a force of $5\ kN$ and fixed to a channel by means of 3 identical bolts. The bolts are made up of C20 steel. Assume FOS = 3. Determine size of bolt.

Soln. Given $P = 5 \times 10^3\ N$

$\tau_{max} = \frac{0.5\sigma_{y}}{FOS} = \frac{0.5 \times 260}{3} = 43.3333\ N/mm^2$

$e = 250\ mm$

$n = 3$

(I) Primary effect

(a) Primary shear force --

$F_{P} = \frac{P}{n} = \frac{5 \times 10^3}{3} = 1.6667 \times 10^3\ N$

(II) Secondary effect --

(a) Shear force per unit length --

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2}$

$\therefore F = \frac{5 \times 10^3 \times 250}{75^2 + 0^2 + 75^2}$

$\therefore F = 0.1111 \times 10^3\ N/mm$

(b) Secondary shear force --

$F_{S_{1}} = F . r_{1} = 0.111 \times 10^3 \times 75 = 8.3325 \times 10^3N$

$F_{S_{2}} = F . r_{2} = 0\ N$

$F_{S_{3}} = F . r_{3} = 8.3325 \times 10^3\ N$

(c)

$F_{R_{1}} = \sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

$= \sqrt{(1.6667 \times 10^3)^2 + (8.3325 \times 10^3)^2 + 2(1.6667 \times 10^3)(8.3325 \times 10^3) . cos90}$

$= 8.4976 \times 10^3N$

$F_{R_{2}} = \sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

$= \sqrt{(1.6667 \times 10^3)^2 + 0 + 2F_{P}(0)cos\theta_{2}}$

$= 1.6667 \times 10^3\ N$.

$F_{R_{3}} = \sqrt{F_{P}^2 + F_{S_{3}}^2 + 2 F_{P} . F_{S_{3}} . cos\theta_{3}}$

$= \sqrt{(1.6667 \times 10^3)^2 + (8.3325 \times 10^3)^2 + 2(1.6667 \times 10^3)(8.3325 \times 10^3) . cos90}$

$= 8.4976 \times 10^3\ N$.

$\therefore F_{R_{max}} = 8.4976 \times 10^3\ N$

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

$\therefore 43.3333 = \frac{8.4976 \times 10^3}{A}$

$\therefore A = 196.0986\ mm^2$

$\therefore \frac{\pi}{4}d_{i}^2 = 196.0986$

$\therefore d_{i} = 15.8013\ mm$

From PSG 5.42, from coarse series,

select M20.

$d_{c} = 16.933$

$d = 20$.

Q.6 A steel plate is subjected to a force of $3\ kN$ and fixed to a vertical channel by 4 identical bolts. The bolts are made up of carbon steel whose $S_{yt} = \sigma_{y} = 380\ N/mm^2$ and FOS = 4. Determine the diameter of bolt.

(I) Primary effect

(a) Primary shear force --

$F_{P} = \frac{P}{n} = \frac{3 \times 10^3}{4} = 750\ N$

(II) Secondary effect --

(a) Shear force per unit length --

By pythagoras theorem,

$r_{1} = \sqrt{50^2 + 50 ^2}$

$\therefore r_{1} = 70.7107\ mm$.

$r_{1} = r_{2} = r_{3} = r_{4} = 70.7107\ mm$

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2 + r_{4}^2}$

$\therefore F = \frac{3 \times 10^3 \times 250}{4(70.7107)^2}$

$\therefore F = 37.5\ N/mm$

(b) Secondary shear force --

$F_{S_{1}} = F . r_{1} = 37.5(70.7107) = 2.6517 \times 10^3\ N$

$F_{S_{2}} = F . r_{2} = 2.6517 \times 10^3\ N$

$F_{S_{3}} = F . r_{3} = 2.6517 \times 10^3\ N$

$F_{S_{4}} = F . r_{4} = 2.6517 \times 10^3\ N$

(c)

$F_{R_{1}} = \sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

$= \sqrt{(750)^2 + (2.6517 \times 10^3)^2 + 2(750)(2.6517 \times 10^3) . cos135}$

$= 2.1867 \times 10^3\ N$.

$F_{R_{2}} = \sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

$= \sqrt{(750^2 + (2.6517 \times 10^3)^2 + 2(750)(2.6517 \times 10^3).cos45}$

$= 3.2259 \times 10^3\ N$.

$F_{R_{3}} = 3.2259 \times 10^3\ N$.

$F_{R_{4}} = 2.1867 \times 10^3\ N$.

$\therefore F_{R_{max}} = 3.2259 \times 10^3\ N$

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

$\tau_{max}$ = $\frac{0.5\sigma_{y}}{FOS}$ = $\frac{0.5 \times 380}{4}$

$\therefore \tau_{max} = 47.5$

$\therefore 47.5 = \frac{3.2259 \times 10^3}{A}$

$\therefore A = 67.9137\ mm^2$

$\therefore \frac{\pi}{4}d_{c}^2 = 67.9137$

$\therefore d_{c} = 9.2989\ mm$.

From PSG 5.42, from coarse series,

select M12.

$d_{c} = 9.853\ mm$

$d = 12\ mm$

Q.7 Find diameter of rivet in shown arrangement. The max. shear stress in rivet is limited to $65\ N/mm^2$. The max. load carred by the joint is $109\ kN$. Assume that all rivets are of same size.

(I) Primary effect

(a) Primary shear force --

$F_{P} = \frac{P}{n} = \frac{109 \times 10^3}{4} = 27.25 \times 10^3N$

(II) Secondary effect --

(a) Shear force per unit length --

$F = \frac{P . e}{r_{1}^2 + r_{2}^2 + r_{3}^2 + r_{4}^2}$

$sin 30 = \frac{e}{112.5}$

$\therefore e = 56.25\ mm$.

$F = \frac{109 \times 10^3 \times 56.25}{(112.5)^2 + (37.5)^2 + (37.5)^2 + (112.5)^2}$

$\therefore F = 218\ N/mm$

(b) Secondary shear force -- $F_{s}$ --

$F_{S_{1}} = F . r_{1} = 218 \times 112.5 = 24.525 \times 10^3\ N$

$F_{S_{2}} = F . r_{2} = 218 \times 37.5 = 8.175 \times 10^3\ N$

$F_{S_{3}} = F . r_{3} = 218 \times 37.5 = 8.175 \times 10^3\ N$

$F_{S_{4}} = F . r_{4} = 218 \times 112.5 = 24.525 \times 10^3\ N$

(c)

$F_{R_{1}} = \sqrt{F_{P}^2 + F_{S_{1}}^2 + 2 F_{P} . F_{S_{1}} . cos\theta_{1}}$

$= \sqrt{(27.25 \times 10^3)^2 + (24.525 \times 10^3)^2 + 2(27.25 \times 10^3)(24.525 \times 10^3) . cos60}$

$= 44.8592 \times 10^3\ N$.

$F_{R_{2}} = \sqrt{F_{P}^2 + F_{S_{2}}^2 + 2 F_{P} . F_{S_{2}} . cos\theta_{2}}$

$= \sqrt{(27.25 \times 10^3)^2 + (8.175 \times 10^3)^2 + 2(27.25 \times 10^3)(8.175 \times 10^3)cos60}$

$= 32.1273 \times 10^3\ N$.

$F_{R_{3}} = \sqrt{F_{P}^2 + F_{S_{3}}^2 + 2 F_{P} . F_{S_{3}} . cos\theta_{3}}$

$= \sqrt{(27.25 \times 10^3)^2 + (8.175 \times 10^3)^2 + 2(27.25 \times 10^3)(8.175 \times 10^3) . cos120}$

$= 24.2203 \times 10^3\ N$.

$F_{R_{4}} = \sqrt{F_{P}^2 + F_{S_{4}}^2 + 2 F_{P} . F_{S_{4}} . cos\theta_{4}}$

$= \sqrt{(27.25 \times 10^3)^2 + (24.525 \times 10^3)^2 + 2(27.25 \times 10^3)(24.525 \times 10^3) . cos120}$

$= 25.9948 \times 10^3\ N$.

$\therefore F_{R_{max}} = 44.8592 \times 10^3\ N$.

(d) $\tau_{max}$ = $\frac{F_{R_{max}}}{A}$

$\therefore 65 = \frac{44.8592 \times 10^3}{A}$

$\therefore A = 690.1415\ mm^2$

$\therefore \frac{\pi}{4}d_{i}^2 = 690.1415$

$\therefore d_{i} = 29.6431\ mm$.