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- Homework Statement
- Calculate the concrete volume that is required to keep the sluice gate AB closed, assuming that the only resisting force is the friction between the concrete block and soil. The sluice door is square (900 x 900 mm) and hinged at point A (refer to Figure 1).

Use a factor of safety of 1.5.

Take density of water 1000 kg/m3 and concrete 2500 kg/m3. Assume the friction between the concrete block and soil is 0.55 x concrete mass

Height/H = 5.4m

Degree = 140

- Relevant Equations
- My working out is:

Sine 140 = AC/0.9 AC = 0.9*sin40 = 0.579m

y = 5.4 + (0.579/2) = 5.6895

A = 0.9*0.9 = 0.81m^2

R = pgyA = 1000*9.81*0.81*5.6895 = 45.2093 kN

D/Magnitude = sin^2 (IGG + A (y/sin)^2 / A *y)

sin140^2 = 0.413

IGG = bd^3 / 12 = 0.9*0.9^3 / 12 = 0.0547m^4

D/Magnitude = 0.413 (0.0547 + 0.081(5.6895/0.643)^2 / 0.81*5.6895) = 5.688m

I am now perplexed an confused on how to find the volume of concrete to keep the gate shut.

Would it be:

R*(FE + EA) = W*0.55

sin140 = D - y / EF EF = D - y / sin140 = 5.688 - 5.1105 / sin140 = 0.5775m

EA (Middle of gate) = 0.9/2 = 0.45

45.2093*(0.5775 + 0.45) = W*0.55

W = 84.459 KN x (1000/9.81) = 8609.5 kg

I do not know how to get the concrete volume and how will I use the factor of safety

Please can you help with a question I am struggling with. I have done as much working out as I could until I was completely stumped:

Calculate the concrete volume that is required to keep the sluice gate AB closed, assuming that the only resisting force is the friction between the concrete block and soil. The sluice door is square (900 x 900 mm) and hinged at point A (refer to Figure 1).

Use a factor of safety of 1.5.

Take density of water 1000 kg/m3 and concrete 2500 kg/m3. Assume the friction between the concrete block and soil is 0.55 x concrete mass

Height/H = 5.4m

Degree = 140

I have tried again and got this

My working out is:

Sine 140 = AC/0.9 AC = 0.9*sin40 = 0.579m

y = 5.4 + (0.579/2) = 5.6895

A = 0.9*0.9 = 0.81m^2

R = pgyA = 1000*9.81*0.81*5.6895 = 45.2093 kN

D/Magnitude = sin^2 (IGG + A (y/sin)^2 / A *y)

sin140^2 = 0.413

IGG = bd^3 / 12 = 0.9*0.9^3 / 12 = 0.0547m^4

D/Magnitude = 0.413 (0.0547 + 0.081(5.6895/0.643)^2 / 0.81*5.6895) = 5.688m

I am now perplexed an confused on how to find the volume of concrete to keep the gate shut.

Would it be:

R*(FE + EA) = W*0.55

sin140 = D - y / EF EF = D - y / sin140 = 5.688 - 5.1105 / sin140 = 0.5775m

EA (Middle of gate) = 0.9/2 = 0.45

45.2093*(0.5775 + 0.45) = W*0.55

W = 84.459 KN x (1000/9.81) = 8609.5 kg

I do not know how to get the concrete volume and how will I use the factor of safety

Calculate the concrete volume that is required to keep the sluice gate AB closed, assuming that the only resisting force is the friction between the concrete block and soil. The sluice door is square (900 x 900 mm) and hinged at point A (refer to Figure 1).

Use a factor of safety of 1.5.

Take density of water 1000 kg/m3 and concrete 2500 kg/m3. Assume the friction between the concrete block and soil is 0.55 x concrete mass

Height/H = 5.4m

Degree = 140

I have tried again and got this

My working out is:

Sine 140 = AC/0.9 AC = 0.9*sin40 = 0.579m

y = 5.4 + (0.579/2) = 5.6895

A = 0.9*0.9 = 0.81m^2

R = pgyA = 1000*9.81*0.81*5.6895 = 45.2093 kN

D/Magnitude = sin^2 (IGG + A (y/sin)^2 / A *y)

sin140^2 = 0.413

IGG = bd^3 / 12 = 0.9*0.9^3 / 12 = 0.0547m^4

D/Magnitude = 0.413 (0.0547 + 0.081(5.6895/0.643)^2 / 0.81*5.6895) = 5.688m

I am now perplexed an confused on how to find the volume of concrete to keep the gate shut.

Would it be:

R*(FE + EA) = W*0.55

sin140 = D - y / EF EF = D - y / sin140 = 5.688 - 5.1105 / sin140 = 0.5775m

EA (Middle of gate) = 0.9/2 = 0.45

45.2093*(0.5775 + 0.45) = W*0.55

W = 84.459 KN x (1000/9.81) = 8609.5 kg

I do not know how to get the concrete volume and how will I use the factor of safety