Calculating Maximum Weight Capacity of a Tripod with Given Dimensions and Angle

[woodywheel]

Homework Statement

well guys i am in 10 th standard...jus out of nowhere i challenged a person that i can find out the maximum weight a trpod can take...now i have no clue how to solve this problem
pleasez help me out
its 3.5 m hollow shaft with 2 inch outer dia and 5 mm thickness and angle of tropd to the ground is 60 degree ..any help will b appreciated
thnk you

Homework Equations

The Attempt at a Solution

 

[woodywheel]

no replies ...pleasez help guys :)

 

[nvn]

woodywheel: Text messaging shorthand is not allowed on this forum. See the PF rules link at the top of each page. Only grammatically correct sentences are allowed, with correct punctuation and spelling. E.g., b for be is not allowed.

You must list relevant equations yourself. We are not allowed to tell you how to approach or solve your problem. You must post a valid attempt. We only check math here. Give it a try.

 

[woodywheel]

i am really sorry nvn...since this was my first post...i have been doing some research lately through your wonderful site and found out the stress can be found out by stress=My/I
and I for hollow shaft is pi*(Do^4-Di^4)/32
is my approach correct ? doesn't it depend on any material property of aluminum ... and is this the maximum load a shaft of a tripod stand...and if i am short of information can you suggest me some online link to refer
thank you

 

[rock.freak667]

i am really sorry nvn...since this was my first post...i have been doing some research lately through your wonderful site and found out the stress can be found out by stress=My/I
and I for hollow shaft is pi*(Do^4-Di^4)/32
is my approach correct ? doesn't it depend on any material property of aluminum ... and is this the maximum load a shaft of a tripod stand...and if i am short of information can you suggest me some online link to refer
thank you

Ok, well you have the formula for bending stress, if the force is P and is at the extreme end of the tripod (the other end is on the ground), then what is the bending moment M that the end touching the ground experiences?

M=force*perpendicular distance

Your force is simply P, but what is your perpendicular distance? (Hint: reduce the leg of the tripod to being the hypotenuse of a triangle).

 

[Inferior89]

i am really sorry nvn...since this was my first post...i have been doing some research lately through your wonderful site and found out the stress can be found out by stress=My/I
and I for hollow shaft is pi*(Do^4-Di^4)/32
is my approach correct ? doesn't it depend on any material property of aluminum ... and is this the maximum load a shaft of a tripod stand...and if i am short of information can you suggest me some online link to refer
thank you

Tip: Sentences end with one (1) dot and start with a capitalized letter. It makes everything easier to read and you are much more likely to get an answer to your question.

 

[pongo38]

Can you draw a detail showing how the load gets into the tripod at the top?

 

[woodywheel]

Thanks a lot rockfreak.I will calculate and post again, but here i am confused about one thing that how does aluminum properties effect the maximum load taken by the tripod and how ... what will be the maximum P here so that tripod does not break ?
@pongo- the wait is suspended at the top of tripod by means of a chain block which is hooked at the top(meeting point of legs)
Thank You

 

[pongo38]

Are the three feet tied together, or can they slide?

 

[rock.freak667]

Thanks a lot rockfreak.I will calculate and post again, but here i am confused about one thing that how does aluminum properties effect the maximum load taken by the tripod and how ... what will be the maximum P here so that tripod does not break ?
@pongo- the wait is suspended at the top of tripod by means of a chain block which is hooked at the top(meeting point of legs)
Thank You

Right well, your design stress will depend on your factor of safety,FS, and your yield point (this is where the material properties come into play).

 

[woodywheel]

@pongo- they are tied together with all legs making an angle of 60 degrees with respect to ground

 

[pongo38]

Woodywheel: What do think are the magnitudes and directions of the reactions at the feet?

 

[nvn]

woodywheel: Where are the legs/feet tied together?

Your formula in post 4 is close but is slightly incorrect, and should instead be I = pi*(Do^4 - Di^4)/64. You have Do = 50.8 mm, and Di = 40.8 mm.

You could use yield factor of safety, FSy = 1.50, and ultimate factor of safety, FSu = 2.0. Aluminum 6061-T6 has tensile yield strength, Sty = 255 MPa, and tensile ultimate strength, Stu = 290 MPa. And E = 69 000 MPa.

 

[woodywheel]

@pongo- the direction will be opposite to that of the force applied and magnitude will be calculated by stress formula(i guess )

@nvn-the legs are tied at the extreme bottom by means of chain only... and what about neutral axis shall i take the distance as radius of the shaft?
and how to use these values of fsy etc.?
please let me know
thank you again

 

[rock.freak667]

@pongo- the direction will be opposite to that of the force applied and magnitude will be calculated by stress formula(i guess )

@nvn-the legs are tied at the extreme bottom by means of chain only... and what about neutral axis shall i take the distance as radius of the shaft?
and how to use these values of fsy etc.?
please let me know
thank you again

Working/Design stress = yield stress/FS

 

[pongo38]

Before you even get to stresses, you need to know the axial force N, and the bending moment M. Do you know the formulas for normal stress due to N and to M, and how to combine them? Actually this is getting a bit out of hand because you haven't demonstrated how you will get the M. That is why I asked you for the detail at the top. That will determine whether there is an eccentric load there, or not. I suppose you could ignore the self weight, especially as it is aluminium. But the material does not affect the distribution of loads in the tripod, as you implied, because it is statically determinate. Do you know what that means?

 

[nvn]

The legs are tied at the extreme bottom by means of chain only.

woodywheel: Great. Therefore, lucky for you, you have no primary bending moment on these legs (other than self weight). Therefore, to solve the problem, compute the Euler critical load[/color] of one leg tube, using a column effective length factor of k = 1.10. Then divide the Euler critical load by FSu = 2.0, to obtain the axial load capacity, P, of one leg. After that, use a trigonometric function to obtain the vertical component of P, which you can call P_z. After that, multiply P_z by 3, because you have three legs.

 

[woodywheel]

@pongo- well the load is eccentric because all three legs are hinged to a flange(which has welded hooks, its made such that all legs make equal angles) and as nvn made it clear that its a case of buckling so i think there's no need of bending moment here.
@NVN-I calculated the final load by eulers critical load and its coming out to be 2.7tons for for all three legs(mterial used here was 6063-t6)
angle of legs with ground = 76 degrees
Do=63.5
Di=53.5
I=0.396x10^-6 m^4
critical load=26940.7 N
Pz=Pcos14
finally 2.7 tons

Pz=

 

[nvn]

woodywheel: You got the correct final answer for the tripod capacity, 2.70 metric tonnes. However, one of your lines before your final answer currently looks incorrect. It should instead be as follows.

P = critical load of one leg = 9096.1 N
Pz = P*sin(76 deg) = 8825.9 N
3*Pz = 26 478 N
3*Pz/(9.8067*1000) = 2.700 metric tonnes