How does this statics design FBD look?

[jklops686]

I have a FBD I'm not quite sure about. I attached what i have along with the original picture. 2 pin supports plus the other one on the hydraulic jack. How does it look? Here's my problem:

The chute of a concrete truck for delivering wet concrete to a construction site is shown. The length of the chute may be changed by adding or removing segments BC and CD. Chute segments AB, BC and CD each weigh 50 lb, and the maximum length of the chute is 144 inches. The chute has semicircular shape with 8 inch inside radius, and the hydraulic cylinder GF is used to raise and lower the chute such that. I have to specify the force capacity of the hydraulic cylinder GH.

I have to use the max weight on the chute to figure critical weight on the jack. How do I do this without using distributed weight for the chute? I'm not sure of my force arrows at D and H.

 

[SteamKing]

I have to use the max weight on the chute to figure critical weight on the jack. How do I do this without using distributed weight for the chute? I'm not sure of my force arrows at D and H.

Why don't you want to use the distributed weight of the chute? BTW, doesn't the cylinder have to support both the weight of the chute and any cement carried by the chute?

 

[jklops686]

Why don't you want to use the distributed weight of the chute? BTW, doesn't the cylinder have to support both the weight of the chute and any cement carried by the chute?

Right, so first i have to figure the weight of the chute full of wet cement. And I haven't learned how to use distributed weight so I figured there must be a way to solve it a different way.

 

[SteamKing]

Well, the weight of the chute and the cement contained within won't be located at the end.

Using distributed weight is not that hard. The trick is to replace the weight distribution with the total weight acting at the center of gravity of the distribution. For a constant weight distribution, like you have here, the center of gravity will be in the center of the distribution. For example, each section of chute weighs 50 lbs. and is 48 in. long. The distributed weight for one section of the chute alone is 50/48 = 1.042 lb/in, and the center of gravity is 25 in. from one end.

The weight distribution of the cement can be figured in a similar manner by calculating the volume of the chute and using the density of cement.

 

[jklops686]

Well, the weight of the chute and the cement contained within won't be located at the end.

Using distributed weight is not that hard. The trick is to replace the weight distribution with the total weight acting at the center of gravity of the distribution. For a constant weight distribution, like you have here, the center of gravity will be in the center of the distribution. For example, each section of chute weighs 50 lbs. and is 48 in. long. The distributed weight for one section of the chute alone is 50/48 = 1.042 lb/in, and the center of gravity is 25 in. from one end.

The weight distribution of the cement can be figured in a similar manner by calculating the volume of the chute and using the density of cement.

Okay great, thank you, so I just find the weight at the center of gravity of the chute with cement and then use that to take a moment about H?

 

[jklops686]

is the capacity of the jack just the weight of the chute with cement in the HG direction?