Hello guys,
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
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 = 45209.336 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
Then add R (45209.336) to the concrete weight (after converting it into KN) and multiplying the answer by (0.5775). Then dividing by 0.55 to get the W figure?
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
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 = 45209.336 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
Then add R (45209.336) to the concrete weight (after converting it into KN) and multiplying the answer by (0.5775). Then dividing by 0.55 to get the W figure?