Question

A bracket plate is connected to the flange of a column using 6 - 24 mm diameter bolts of grade 4.6 as shown below. The maximum magnitude of load $P_u$ that the joint can transmit without failure.

• A. 66.82 kN
• B. 129.8 kN
• C. 113.7 kN
• D. 181.6 kN

Answer 1

The correct option is C 113.7 kN

Bolts 1 and 2 are highly stressed and critical

$\text{eccentricity}=150+\frac{125}{2}=212.5mm$

$r_1=\sqrt{{75}^2+{62.5}^2}=97.63mm$

$\sum r^2=4\times {97.63}^2+2\times {62.5}^2=45938.97m{m}^2$

$\text{Direct shear force per bolt}=\frac{P}{6}=0.167P$

And shear force due to torque:

$\frac{Pe{r}_1}{\sum r^2}=\frac{P\times 212.5\times 97.63}{(4\times {97.63}^2+2\times {62.5}^2)}=0.4516P$

$\cos \theta =\frac{62.5}{97.63}=0.640$

Result force in critical bolt:

$=\sqrt{(0.167P{)}^2+(0.4516P{)}^2+2\times 0.167P\times 0.4516P\times 0.640}$

$=0.573P....\left(1\right)$

Strength of bolt in shear

$V_{dsb}=\frac{n_n{A}_{nb}{f}_{ub}}{\sqrt{3}{\gamma }_{mb}}=\frac{1\times 0.78\times \frac{\pi }{4}\times {24}^2\times 400}{\sqrt{3}\times 1.25}\times {10}^{-3}kN$

$\Rightarrow V_{dsb}=65.16kN$

$k_b=min\{\frac{60}{3\times 26},\frac{75}{3\times 26}-0.25,\frac{400}{410},1\}$

$k_b=0.712$

$V_{dpb}=\frac{2.5k_{b}dt{f}_u}{{\gamma }_{mb}}=\frac{2.5\times 0.712\times 24\times 12\times 410}{1.25}=168.14kN$

$\text{Bolt value}{V}_{db}=65.16kN....\left(2\right)$

From equation (1) & (2) , we get

Maximum factored load allowed,

$0.573P=65.16$

$\Rightarrow P=113.72kN$