Calculating Tension in a Rigid Beam Supported by Three Vertical Rods

[aznboywunder]

Homework Statement

A rigid weightless horizontal 5ft long beam is supported by three vertical tension rods. Two of the rods are at the ends of the beam, the third one is in the middle. The beam has a 6000lb load which is applied halfway between the center and the right end. Calculate the tension in each bar. All bars are 6ft long and .375" diameter and are made of steel.

The Attempt at a Solution

I've been on this for an hour now and can't seem to figure out an answer. Is it true that bar 2 and 3 are carrying 3000lbs?

 

[PhanthomJay]

Homework Statement

A rigid weightless horizontal 5ft long beam is supported by three vertical tension rods. Two of the rods are at the ends of the beam, the third one is in the middle. The beam has a 6000lb load which is applied halfway between the center and the right end. Calculate the tension in each bar. All bars are 6ft long and .375" diameter and are made of steel.

The Attempt at a Solution

I've been on this for an hour now and can't seem to figure out an answer. Is it true that bar 2 and 3 are carrying 3000lbs?

No, that is not correct. This is an example of a rigid beam on elastic supports; the beam can rotate but not deform, and the tension rod supports may be considered as elastic springs. If the middle and right supports each carried 3000 lbs, then the left support would carry no load, which would violate the geometry of the rigid beam rotation.
This is a statically indeterminate problem. You must solve it accordingly, and note the geometry of the beam rotation and corresponding relationship of the rod deflections. The right support carries the larger share of the 6000 pound load, while the left support carries the least share of the load. The middle support carries one half the sum of the left and right support reactions.

 

[pongo38]

I agree with Jay, but there is an alternative approach in this case, which is to 'move' the load to the centre, but also add a moment of 6000 x 1.25. The central load is then shared equally between the three bars, (2000 each) and the outer ones take the moment (6000x1.25/5 = 1500). Then add up the effects separately. Left side is 2000-1500= 500 middle is 2000 right end is 2000 + 1500 = 3500. This appoach is possible because the three rods are the same cross-section.

 

[PhanthomJay]

I agree with Jay, but there is an alternative approach in this case, which is to 'move' the load to the centre, but also add a moment of 6000 x 1.25. The central load is then shared equally between the three bars, (2000 each) and the outer ones take the moment (6000x1.25/5 = 1500). Then add up the effects separately. Left side is 2000-1500= 500 middle is 2000 right end is 2000 + 1500 = 3500. This appoach is possible because the three rods are the same cross-section.

Yes, that does make the calculation a lot easier by using the equivalent force -couple at mid-span approach.

 

[berkeman]

I agree with Jay, but there is an alternative approach in this case, which is to 'move' the load to the centre, but also add a moment of 6000 x 1.25. The central load is then shared equally between the three bars, (2000 each) and the outer ones take the moment (6000x1.25/5 = 1500). Then add up the effects separately. Left side is 2000-1500= 500 middle is 2000 right end is 2000 + 1500 = 3500. This appoach is possible because the three rods are the same cross-section.

Please be careful not to do the original poster's (OP's) work for them. We are here to provide tutorial hints, and to find mistakes in their posted work. Please do not work out the equations for them.

 

[hjsun329]

Hi. For your alternative solution, you wrote that "the outer ones take the moment (6000x1.25/5 = 1500). Then add up the effects separately." I don't get the calculation part. Can you explain how you came up with this calculation?(6000x1.25/5)