For the bolted bracket connection shown in figure, the force (in kN round off to two decimals) experienced by bolt 3 will be ____.
95

Open in App

Solution

The correct option is A 95
$c o s θ = \frac{60}{100} = 0.6$
$s i n θ = 0.8$
Direct force in bolt 3, $F_{D} = \frac{200}{5} = 40 k N$
Force due to torsion, $F_{T} = \frac{T r}{\sum r^{2}}$
$r = \sqrt{(60)^{2} + (80)^{2}} = 100 m m$
$F_{T} = \frac{200 \times 250 \times 100 \times 10^{3}}{4 \times (100)^{2}} = 125 k N$
Resultant force on bolt 3 $(F_{r}) = \sqrt{(F_{D})^{2} + (F_{T})^{2} + 2 F_{D} F_{T} c o s (90 + θ)}$
$= \sqrt{(40)^{2} + (125)^{2} - 2 \times 40 \times 125 \times 0.8}$
$[∵ c o s (90 + θ) = - c o s θ]$
$= 96.05 k N$

Suggest Corrections

Similar questions

A bracket plate connected to a column flange transmits a load of 120 kN as shown in the figure. The maximum force for which the bolts should be designed is kN

Q. A steel plate, connnected to a fixed channel using three identical bolts

A, B

and

C

, carries a load of

6 k N

as shown in the figure. Considering the effect of direct load and moment,the magnitude of resultant shear force (in kN) on bolt

C

A bolted joint has four bolts arranged as shown in figure. the cross sectional area of each bolt is 25 m m^{2} . A torque T=200 Nm is acting on the joint. Neglecting friction due to clamping force, maximum shear stress in a bolt is

Q. A steel bar of

10 \times 50 m m

is cantilevered with two

M 12

bolts (P and Q) to support a static load of

4 k N

as shown in the figure.

The primary and secondary shear loads on bolt

P

, respectively, are

A cantilever bracket is bolted to a column using three M 12 \times 1.75 bolts, P, Q and R. The value of maximum shear stress developed in the bolt P (in MPa) is

Join BYJU'S Learning Program

For the bolted bracket connection shown in figure, the force (in kN round off to two decimals) experienced by bolt 3 will be ____.95

For the bolted bracket connection shown in figure, the force (in kN round off to two decimals) experienced by bolt 3 will be ____.
95