Help with Statics and Strengths of Materials

[nvn]

grandnat_6: I have almost always used Ssu = shear ultimate strength = 0.60*Stu, and Ssy = shear yield strength = 0.577*Sty, where Stu = tensile ultimate strength, and Sty = tensile yield strength. Most textbooks claim the above. I currently do not know why Machinery's Handbook instead says Stu = 0.75*Stu.

The above values are shear strength, and tensile strength, the point where the material shears or ruptures. These material strength values do not include a safety factor. They are material strength values, not allowable stress.

I am still currently leaning toward a yield factor of safety of FSy = 3 for your arm beams, and perhaps FSy = 5 for the pins. The factor of safety is the same in tension and compression. The allowable tensile (or compressive) stress is Sta = Sty/FSy. The allowable shear stress is Ssa = Ssu/FSu = 0.60*Stu/FSu, where FSu = ultimate factor of safety. Because your current FSy values are so high, you can just use FSu = FSy, for now.

We do not yet know the tensile yield strength (Sty) of your A513 steel tubes, because you did not state an SAE steel grade designation yet. A513 covers a lot of SAE steel grade designations (SAE 1008, 1010, 1020, 4130, 4140, just to name a few). You (your supplier) must state the SAE steel grade designation, before we can look up the strength of your A513. If no SAE grade designation for A513 tubes is stated, then we would be forced to assume SAE 1008.

And, your supplier must state whether the A513 steel tube thermal condition is as-welded (not annealed), normalized, DOM, or DOM stress-relieved, before we can look up the strength. If no thermal condition for A513 tubes is stated, then we would be required to assume normalized, or perhaps as-welded, depending on the SAE grade designation. However, A500, grade B, on the other hand, specifically defines a strength.

 

[grandnat_6]

nvn,

I was unable to get any info from my supplier on the tubing, he seems like he does not want to share this with me. Since it was though email, I'll give him a few days, perhaps he is working on it.

I think I understand what you are saying in your last post. I have done an example, please see attached.

Beam 1 in position 1 has the most stress, therefore I only sized up the beam to fit position 1 as it will be more than enough for position 2 and 3. I have also found the yield shear strength, yield tension strength, and yield compressive strength. Since the material is isotropy, in this situation only the tension would need to be calculated. The allowable shear in yield is the most burdening factor, I would have to say the shear stress would be the only thing needed to be calculated. Once the beam meets the shear criteria, it will be well within the tension and compressive stress allowable stress.

I did not use FSu=FSy for now, because I don't understand what you mean by "for now".

Since the 500# and 850# force is double the load per arm, since each arm ideally carries the full load most times; I have divided the axial force and bending moment by 2 and have sized up another beam for that was well. Because there are instances where one arm may carry the majority of the load, I am unsure to use the safety factor of 3 for the double load or the single load.

Please let me know if my math and assumptions are correct.



Thank you,

 

[nvn]

grandnat_6: Forget about shear yield strength; you virtually never need to use it. Secondly, you do not need to compute shear stress for sizing the global cross section of beam 1 or 2; therefore, delete shear stress and shear strength from your current calculations.

All you need to check for globally sizing beam 1 or 2 is the axial stress plus bending stress, and compare it to tensile yield strength, Sty. You can let the full load, in position 1, be applied to only one arm. And I think you can use FSy = 3. Let's say you have a 4 x 2.5 x 0.3125 rectangular tube for beam 1. Even though the axial force reduces immediately to the right-hand side of point D, let's use the axial force immediately to the left-hand side of point D, regardless, which is P = 2.871 kip. If Sty = 50 ksi, the calculations would be as follows.

(1) sigma = (2.871/3.23) + [(45.350 kip*in)/3.06] = 15.709 ksi.

(2) Ny = Sty/(FSy*sigma) = (50 ksi)/(3.0*15.709) = 1.0610 > 1.0; therefore, not overstressed.

 

[grandnat_6]

nvn,

Thank you for the help, It didn't really make sense to use shear in yield anyway. I'd assume you would want to use Shear ultimate strength when you want to design something to shear like a key or a shear pin? You wouldn't want a safety factor for this either, correct?

I was unable to get pricing from my fabricator yet, but I decided to go ahead and make some beams out of 100XF. Attached is my work. on the left side of the shear/moment diagram I have calculated the beam size. The number I use for moment of inertia and area, was obtained directly from the CAD program. I subtracted the OD area and moment of inertia from the ID to obtain my numbers. I then made a beam to these numbers. I hope it is all correct.

To the left of the shear/moment diagram I have tried to make a tapper-ed beam. I started by using the beam selected from the left hand side and used that size for that particular moment. Moving to the left 10" from that moment I made a smaller beam to the same width and thickness. Calculating out a size for that moment, and plotting it on the shear/moment diagram it gave me the angle for the beam. Now on the end of the far left of the shear/moment diagram where zero is at, there I'd assume there is only the axial force of 2871# acting on that area. But it is so small the radii of the beam would not fit. According to a steel website that offers 100XF states to use a bend radii of 1.75 x T. At the point of zero there is no way the beam can be made that small keeping the minimum radii, so I was not able to check the 2871#F with the cross sectional area. So I would have to widen my tapper to a minimal length to satisfy the bending operation and to make sure the cross sectional area will fall within the safe allowable stress. The only other way to make this better is to manipulate the P/A +/- MC/I=St equation to solve for A,C, and I, while using St =32614psi from the previous equation. Almost impossible.



Does this all sound correct?

Thank you!

 

[nvn]

grandnat_6: Yes, you would use shear ultimate strength (Ssu) for pins and keys. And yes, use an ultimate factor of safety (FSu) for this.

At the tapered end of beam 1 (point E), you have axial force, P = 2871, and shear force, V = 1063. I did not know you would make a tapered beam. Therefore, you need to check axial stress and shear stress at end E. Therefore, to check the general beam size at end E (not necessarily connection details), you can use the following.

(1) Axial stress, sigma = P/A. Ensure Ny = Sty/(FSy*sigma) ≥ 1.0.

(2) Shear stress, tau = 1.50*V/(2*h*t), where h = beam cross-sectional depth, and t = wall thickness. Ensure Ny = 0.60*Sty/(FSy*tau) ≥ 1.0.

 

[grandnat_6]

nvn,

For the tapered beam, I'm keeping my options open, and never done one in school. So I would like to know how to make one.

I understand where you are getting the axial stress from. It makes total sense.

I have not seen the equation in #2 before.

(2) Shear stress, tau = 1.50*V/(2*h*t), where h = beam cross-sectional depth, and t = wall thickness. Ensure Ny = 0.60*Sty/(FSy*tau) ≥ 1.0.

I have done some google research. On wikipedia I found a formula for that is different, but I can tell it will not work in this situation because there is no moment. tau = VQ/It.

Where does the 1.50 come from?
What is the 2 from in (2*h*t)?
For H, is that cross sectional depth from E to D? So, 32.1404? Or is it the distance from the centroid to the extreme fiber?

Thank you.

 

[nvn]

grandnat_6: 1.50 comes from V*Q/(I*t) for a rectangular cross section. The 2 is because there are two webs (two sides) in a rectangular tube cross section. The h is two times the distance from the centroid to the extreme fiber.

 

[grandnat_6]

nvn,

Attached I have done the math for Point E. I am kinda limited on this because of the bend radii. I think my fabricator will be able to make the bends, but I'll find out from him. It's not going to matter in this situation because I'm well below the safe shear value, but should I be adding P/A to 1.50*v/(2*h*t)?

A question that now arises, When I get the two halves bent; where should the seams be ideally? I want to say the seams should go on the YY section. They say the welding bead is stronger than the base metal, but I feel since XX is doing the main work, it might be possible to change the metalurgy of the 100XF metal and weaken it. What do you think?

In post 64 it was mentioned the axial force for beam 3 is P=2774#, could you please show that math for this? I can not figure out how you obtained that.

Thank you.

 

[nvn]

grandnat_6: Do not add P/A to 1.50*V/(2*h*t). You misunderstood dimension h; it is cross-sectional depth, not extreme fiber distance (c). Try rereading post 77. I am not sure which faces are best for a seam. I currently guess the side faces would be best, instead of the top and bottom faces. However, check a few existing electric-resistance-welded rectangular tubes, to see where the seam is located; I would currently guess it is on one side, not the top nor bottom face.

The axial force in member BG is the axial component of forces FBx and FBy. Therefore, for position 3, P = -2504.66*cos(37.6009 deg) - 1294.11*sin(37.6009 deg) = -2774.00.

 

[grandnat_6]

nvn,

If I read post 77 correctly now. h should equal 1.625". If this is correct, tau will equal 2610 psi.

Beam 2 position 1 carries the most stress from the other two positions, the stress in beam 2 is very close to the stress in beam 1, and meets the minimium safety factor. Therefore I might as well make beam 1 and beam 2 with the same ends. Beam 2 will just have a slightly larger taper (2.19 degrees vs beam 1 1.67 degree) due to the difference in length betweeen beam 1 and beam 2.

Thank you for showing me the math for position 3 beam 2, I was using the compliment angle of the 1294.11# force.

Thank you for your help!

 

[grandnat_6]

nvn,

Attached I have done the vector analysis of the upright in position 1. Vector hj is a link connection on the backside of the upright at point h and transfers force to a pin at the front of the tractor at pin j.

g is a connection point to the sub frame, and is colinear vertically to pin E.

Please ignore in my calcuations the missing # and * signs as for some reason either the lap top or the program would not allow me to enter those signs, even after a reboot.

Is this look correct?

Thanks.

 

[nvn]

grandnat_6: No, that is wrong. You did not reverse the direction of forces at point E. And, it seems you did not even bother to check whether or not your forces and moments (your answers) are in equilibrium. Check your signs. Also, try to not address general questions specifically to me (unless it is something specific to me), so you could possibly get help from anyone reading this forum.

 

[grandnat_6]

yep, thought I forgot something.

It balances out now. This correct?

Thank you.

 

[nvn]

grandnat_6: Your answers in post 83 look correct.

 

[grandnat_6]

Attached I have the shear/moment diagram, and have also sized up a beam. Is this correct?

Also, I have made an alternative beam. It's a U-shape. In this one, Pin E is offset from Pin g by 1" this makes a very minute change to the forces. I assume because of this it should be ok to use the shear moment force diagram from the rectangular tube to size up the U-shape beam. I do show a wall inside the beam that will close the U into a tube. One upright will have this to hold hydraulic oil to run the system the other will not.

Since the shape is not fully closed, I understand my tension and compressive will be different. Will this effect my moment of ineria for calcuation purposes?

Thanks.

 

[nvn]

grandnat_6: alternative_upright.pdf in post 85 looks correct. In your second file in post 85, you omitted M_11.5700+ in your moment diagram, and moment diagram calculations. Your axial force P is wrong, and is not the maximum axial force in the vicinity of your maximum moment.

Regarding your second question, here you still use the moment of inertia of the cross section, in a similar manner, along with the correct extreme fiber distances. Check textbooks, and example problems in textbooks.

 

[grandnat_6]

SolidElast,

Thanks for the link, I'll have to play around with it some day.
_____________________________________________________________________________

I fixed my moment diagram and calculations for the upright. I have also done the shear moment diagram for the alterative upright.

Is P going to be zero, or is it the product of Fgy and FEy? If it is neither. How is its force calculated? Would this be the same for the alterative beam since point g and E are not in colinear?

Thanks.

 

[nvn]

grandnat_6: P = Fgy + FFy, in the vicinity of a point F section cut. It is roughly the same even if points g and F are noncollinear.

 

[grandnat_6]

The attached I have sized up a beam for upright1. I also sized up a beam for the alternative upright beam. This was done using sections where the pins are located.

A few notes, I didn't size the beam though pin g because I am planning on welding a formed C-channel to the bottom of it. I know this beam could be a lot more narrow, but I wanted to make room to use it as a hydraulic tank.

I didn't check for shear (tau)yet because I did not know if the value in the equation would be different due to the U-shape compared to the 1.50 used for the rectangular tube. If it is different what should the value be?

What factor of safety is recommended for the bearing, tension, shear loads on the 3/16" thick walls for the pins?

Thanks.

 

[nvn]

grandnat_6: You cannot pretend the beam cross section shear force is a bending moment, and use it for the bending stress formula. You must use bending moment in the bending stress formula.

In the bending stress formula, use maximum extreme fiber distance, not minimum extreme fiber distance. You currently used minimum extreme fiber distance, which is incorrect.

You can use the 1.50 factor for shear stress for the U-shaped cross section.

For the pin material, you can use a yield factor of safety of 3, and an ultimate factor of safety of 5. For the beam material, you can use a yield factor of safety of 3.

File shear_momentforcesuprightp1.pdf in post 89 currently looks correct.

 

[grandnat_6]

Attached is the reworked alternative upright beam. How does this look?

In my mind I had it since the distance was shorter it would be the worst case senario, Now I see the actual extreme fiber is the worst case senario.

nvn,

Just so we are on the same page, in the lower right corner of the drawing in a purple box; pin connection F shows how the hole can be elongated in the upright. In post 90 the factors of safety will apply to the elongation of the holes?

Thank you.

 

[nvn]

grandnat_6: Yes, hole elongation involves the beam material. For the beam material, you can currently use a yield factor of safety of 3.

Your file in post 91 now looks correct, except 3.49 is wrong. Parameter h = cross-sectional depth = 2.69, not 3.49.

 

[grandnat_6]

moving on the the upright position 2 I encounterd a problem, and noticed I made the mistake in position 1.

My dimensions show pin g is offset by 1" by the rest of the pins.

I was able to balance out the forces. Next I rotated the beam to make the shear and moment diagrams, but in my final M29 shows I have a positive moment of 556.799. My work is attached.

Can someone tell me where I went wrong? I've been working with it for the past 6 hours and can't get it to come out.

nvn, re-reading in the past I see in post 77 it was mentioned h= 2*distance from centroid to extreme fiber, but in post 78 it was mentioned it is the cross sectional depth. So I should just be using the total distance from one end of the cross section to the other end parallel to the shearing plane?

Thank you,

 

[nvn]

grandnat_6: Yes, h is the total height (depth) of a cross section.

You moved holes F and h closer to a vertical line passing through point g, but it currently appears you did not update your perpendicular distances from point g to force vectors Ff and Fh, when you moved the holes.

 

[grandnat_6]

Doh! That was the problem, I think I have it now. Thank you nvn.

I have updated the fix.

I also atempted to work on upright Position 2, I made sure the distances did not change. I trasfered over the forces on pin E, and Pin F, along with the angle of the cylinder for position 2. I balanced out the forces and then checked for shear and it works out. I went ahead and went straight to checking M29 for balance, but it did not. I went back and checked the perpendicular distance and everything comes out correctly. Again I'm at a lost and don't know where I messed up. Where did I go wrong this time?

Thank you,

 

[nvn]

grandnat_6: In your first attached file in post 95, M23.2592+ is now wrong.

In your second attached file, you failed to copy force Ff correctly from the top of the page to the bottom of the page. Remember to always proofread what you write or type.

After you make this correction, then to check the moment balance, you can compute M_29.0000+. Moment M_29.0000 should be nonzero, whereas M_29.0000+ should be approximately zero.

 

[grandnat_6]

AH! I did not realize I needed a M29.0000+. It took me a little bit, but it makes sense. The horizonal force is 1" above the line of action. Thank you nvn for pointing this out to me.

Sorry, I do proof read my work, I think because I work on it so long it all looks good, I try to be carefull with my work but sometimes i jump around and should not.

Attached I have the math and shear/moment forces for P1, P2, P3.

I feel confident P1, and P3 are correct. P2 I am unsure because my value at M29.0000+ is about 11lbs. This might be high, It appears I have everything right.

How do these look?

Thank you.

 

[nvn]

grandnat_6: Your p1 file in post 97 looks correct. In your p3 file in post 97, M_23.2592 and M_23.2592+ are erroneously labeled as negative on the moment diagram.

In your p2 file in post 97, you failed to copy force Ff correctly from the top of the page to the bottom of the page, which I mentioned in post 96. Remember to always proofread what you write or type.

 

[grandnat_6]

I see my mistakes now.

The attached are corrected.

In P3 the moment diagram crosses the zero line. Where it crosses, is it proper to check the shear in those area's by using the -411.636#'s? Or does it not matter where the moment diagram crosses the zero line, or is it just the shear diagram crossing the zero line that matters and should be checked?


Going backwards a bit. The arm.pdf. We made the cross section for beam 1 by using the maximium moment at point G and then using shear to find point E. I mentioned since beam 2 is very close to the maximium bending moment of beam 1, that I'll just make beam 2 exactly like beam 1. I'm pretty sure I was wrong on this, since B has a different shear force. I should use my shear diagrams for all three positions and find the maximium shear and calculate the size beam at point B.

As I have it drawn, I would like to cut out some solid steel with a radi on them and weld them to the ends of the beams. For beam 1 and beam 2, would I just dimension my moment diagrams in all three positions and find the maximium moment and size the end of the beam to the measured moment where my weld will be to join the radius and the beam together?

Thanks.

 

[grandnat_6]

I've been reviewing the plate in post 66. besides the holes, there must be something else that needs to be done to size up the plate. Do I add the force vectors tip to tail to get a resultant and use that to find the yield and shear strength of the material?

Thanks.

 

[nvn]

grandnat_6: Generally, check a beam where the moment diagram is maximum, which is often where the shear diagram is zero. And/or check at any cross section(s) of interest. If you have textbooks that explain this, follow the guidance therein.

In general, you can check the axial stress, bending stress, and shear stress at each cross section you want to design.

 

[grandnat_6]

Did some calcuations on beam 1 and beam 2. Found out the shear stress is more than the maximum tension and compression stress.

At this point with beam 1, beam 2, and the uprights. Should anything else be checked like defection, buckling, ect. or is it ok to make production drawings of these?

nvn, for the textbook I have a fourth edition Statics and Strength of Materials by Bassin, Brodsky, and Wolkoff, copy right 1988.

My instructor back then said it's an old book but is the best he has found. I know myself and a lot of other students found it rather hard to follow without instructor guidence. If you have a suggestion for a textbook please let me know the title, author, or ISBN # and I'll will gladly pick it up somewhere.

 

[nvn]

grandnat_6: Yes, there are other things to check. At the moment, I have exceeded the amount of time I can currently afford for one thread. Perhaps others may choose to answer.

 

[grandnat_6]

alright nvn, Thank you for all the help this far. I've learned a lot!