Question

A steel flat of rectangular section of size 70 x 6 mm is connected to a gusset plate by three bolts each having a shear capacity of 15 kN in holes having diameter 11.5 mm. If the allowable tensile stress in the flat is 150 MPa, the maximum tension that can be applied to the flat is ________.

Answer 1

t = 6 mm
B = 70 mm
d = 11.5 mm
$S_t=150MPa$
p = 35 mm
g = 20 mm

Shear capacity of each bolt,
F = 15 kN
= 15000 N

Maximum force applied on flat = T

Case I : By failure of all the bolts in shear
T = 3 x 15
= 45 kN

Only option below 45 kN is option a (42.3 kN)
So answer is (a)
Diameter of hole,
d' = 11.5mm

Case II : By tearing of Flat at section (1)
$T=(B-d^{\prime }).t.{\sigma }_t$
$T=(70-11.5)\times 6\times 150$
$=52650N$
$=52.65kN$

Case III : By tearing of flat at section (2)
$T=(B-d^{\prime }).t.{\sigma }_t+F$
$=\left[\right(70-2\times 11.5)\times 6\times 150+15000]N$
$=57.3kN$

Case IV : By tearing of flat along section (3)

$Netarea=(B-3d^{\prime })\times t+(\frac{p^2}{4g}+\frac{p^2}{4g})\times t$

$=(70-3\times 11.5)\times 6+\frac{2\times {35}^2}{4\times 20}\times 6$

$=396.75m{m}^2$

$StrengthofFlat=NetArea\times {\sigma }_t$
$=396.75\times 150$
$\Rightarrow$ T = 59.512 kN

Out of various mechanism of failure, minimum value of T is 45 kN, but option 45 kN is only a (42.3 kN) so we will adopt this option. Here failure is taking place by some other mechanism and for that mechanism sufficient data is not given. So T = 42.3 kN