Question

A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)

Answer 1

Answer:

Force on the bolt = 24.525 kN

Force on the 1st hinge = 8.35 kN

Force on the 2nd hinge = 16.17 kN

Explanation:

Given:

height = 2 m

width =1 m

depth of the door from the water surface = 1.5 m

Therefore,

$\bar{y}$ =1.5+1 = 2.5 m

Hydrostatic force acting on the door is

$F= \rho \times g\times \bar{y}\times A$

$F= 1000 \times 9.81\times 2.5\times 2\times 1$

         = 49050 N

         = 49.05 kN

Now finding the Moment of inertia of the door about x axis

$I_{xx}=\frac{1}{12}\times b\times h^{3}$

$I_{xx}=\frac{1}{12}\times1\times 2^{3}$

               = 0.67

Now location of force, $y^{*}$

$y^{*}=\bar{y}+\frac{I_{xx}}{A\times \bar{y}}$

$y^{*}=2.5+\frac{0.67}{2\times 1\times 2.5}$

             = 2.634

Therefore, calculating the unknown forces

$F=F_{A}+R_{B}+R_{C} = 49.05$  ------------------(1)

Now since $\sum M_{R_{A}}=0$

∴ $R_{B}\times L+R_{C}\times L-F\times \frac{1}{2}=0$

  $R_{B}+R_{C}-F\times \frac{1}{2}=0$

  $R_{B}+R_{C}=\frac{F}{2}$

  $R_{B}+R_{C}=24.525$        -----------------------(2)

From (1) and (2), we get

$R_{A} = 49.05-24.525$

                = 24.525 kN

This is the force on the Sliding bolt

Taking $\sum M_{R_{C}}=0$

$F\times 0.706-R_{A}\times 0.84-R_{B}\times 1.68 = 0$

$49.05\times 0.706-24.525\times 0.84-R_{B}\times 1.68 = 0$

$R_{B}$ =8.35 kN

This is the reaction force on the 1st hinge.

Now from (1), we get

$R_{C}$ =16.17 kN

This is the force on the 2nd hinge.