# Patient Lifter Boom design for my ME Senior Project

• charliebrown
Similarly, since the cylinder force is always in the direction of the cylinder, a pinned joint is not needed. This means that there is only one pinned joint needed in the FBD. In summary, the conversation discusses a communication problem regarding the use of the term "fixed" in the design of a crane boom. The solution is to create a free body diagram and show the correct forces and supports. The conversation also touches on the importance of correctly indicating units and lengths in the FBD. The main focus is on correctly identifying the section modulus and considering the piston force in the analysis.

#### charliebrown

Hello, I'm a student of mechanical engineering and I'm doing a project for my university in order to graduate.
Right now we are doing the design for the boom of the crane but I have some doubts according to the analysis we could say is the FBD part my team says that the beam is considered as fix in 3 parts. the part where it bends, where the piston it's located, and where it's pinned. I don´t think that is correct even our adviser says so but I really don't think so I did the analysis no real measures and no real results it's just an example but I just want to know if the analysis is correct i think it's an easy problem but now I'm doubtful and stressed because we don't have much time to finish the project just one month. could you please see if my analysis is correct. thank you very much in advance!

#### Attachments

• patient lifter boom design.pdf
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It sounds like there is a communication problem here. I'm guessing, but you seem to understand that the beam is loaded in three point bending, while the others are using the term "fixed" to refer to the three points of support. The way to solve this is to make a free body diagram of the beam. Make the diagram exactly like you were taught in your statics class - a single line for the beam, and the proper symbols for fixed, pinned, or simple supports and forces. Your marked up photo is the correct way to start, next make the free body diagram.

jrmichler said:
It sounds like there is a communication problem here. I'm guessing, but you seem to understand that the beam is loaded in three point bending, while the others are using the term "fixed" to refer to the three points of support. The way to solve this is to make a free body diagram of the beam. Make the diagram exactly like you were taught in your statics class - a single line for the beam, and the proper symbols for fixed, pinned, or simple supports and forces. Your marked up photo is the correct way to start, next make the free body diagram.

Yes exactly, for me there are two pinned connections as in the photo, so according to my statics class we only have reactions in the x and y directions for those pinned connections, the part where there is a bending of the bar I'm not sure if I'm correct but i got it's component in the x direction to make a straight bar and to make the analysis easier that's why I wrote Lcos45°, what is most important for me really is to specify the section modulus and also to know if i should consider the piston force or not to determine the section modulus.

the FBD it's below the photo i don't know if you saw it? thank you very much for you reply.

You are right, I missed the FBD. Getting the FBD right is key to solving these problems. With that in mind, Fp is not quite vertical, therefore Rb is not vertical. The difference is small, but should be included so that you get in the habit of doing it correctly.

The moment arm from the cylinder to the 150 kg load is shown correctly in the photo, while the FBD is confusing. You are correctly showing the load and the cylinder as forces, not as pinned or fixed joints. You should always show units. The FBD should include the bend in the arm, but you only need to indicate the horizontal distance from the cylinder pin to the load pin. Note that the horizontal distance will change as the arm is raised or lowered. Show actual lengths on the FBD.

If somebody wants to argue about fixed vs simple vs pinned supports, show them the FBD. Point out that the load is a vertical force (gravity is vertical), the force of the cylinder is aligned with the cylinder centerline, and the force on the pivot pin is the resultant of the first two forces. If the arm is lifted or lowered, the load force will stay the same, and will continue to be vertical. Since it is a force, a pinned joint is not necessary. Similarly, since the cylinder force is always in the direction of the cylinder, a pinned joint is not needed. The direction of the cylinder force will change, but the force is always applied at the same point. That leaves one pinned joint.

jrmichler said:
You are right, I missed the FBD. Getting the FBD right is key to solving these problems. With that in mind, Fp is not quite vertical, therefore Rb is not vertical. The difference is small, but should be included so that you get in the habit of doing it correctly.

The moment arm from the cylinder to the 150 kg load is shown correctly in the photo, while the FBD is confusing. You are correctly showing the load and the cylinder as forces, not as pinned or fixed joints. You should always show units. The FBD should include the bend in the arm, but you only need to indicate the horizontal distance from the cylinder pin to the load pin. Note that the horizontal distance will change as the arm is raised or lowered. Show actual lengths on the FBD.

If somebody wants to argue about fixed vs simple vs pinned supports, show them the FBD. Point out that the load is a vertical force (gravity is vertical), the force of the cylinder is aligned with the cylinder centerline, and the force on the pivot pin is the resultant of the first two forces. If the arm is lifted or lowered, the load force will stay the same, and will continue to be vertical. Since it is a force, a pinned joint is not necessary. Similarly, since the cylinder force is always in the direction of the cylinder, a pinned joint is not needed. The direction of the cylinder force will change, but the force is always applied at the same point. That leaves one pinned joint.

Hi friend thank you for your answer! Actually I'm going to consider the piston as vertical, the column yes you're right I missed that It's going to be Rbsin75 to make it vertical, I'm sorry for the FBD i redo it could you please see if i got it right. I left just one pinned joint that is the one of the column and since you ask me to use units, I'm going to do it using the imperial system which is the one I'm using for the project. i just transform the values of the international system to the imperial one and the force 140 kg = 308.647 lbs not 150 kg I'm sorry :( everything else remains the same.

Yes I told them that they were forces, but their argument is that as you can see in the photo of the patient lifter the part where the piston is pinned it's welded to the boom but I don't think because the sides of the boom are welded that should be considered as a fixed beam,

#### Attachments

• boom.pdf
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charliebrown said:
Yes I told them that they were forces, but their argument is

Your FBD is better, but still not correct. Hint: Show vertical and horizontal force components, then make sure they both sum to zero. Also, it's a good idea to round off forces to the nearest pound (or maybe tenth) on the FBD. It makes it more readable.

When you find yourself in a situation where you need convince somebody else who does not see it your way, and I promise it will happen many times in your career, you need to go back a step. Review your statics book, and do the FBD of the arm exactly the way the book does it. The time to do this is not wasted because you need their approval to move ahead. Also, you really need to review FBD's.

charliebrown
jrmichler said:
Your FBD is better, but still not correct. Hint: Show vertical and horizontal force components, then make sure they both sum to zero. Also, it's a good idea to round off forces to the nearest pound (or maybe tenth) on the FBD. It makes it more readable.

When you find yourself in a situation where you need convince somebody else who does not see it your way, and I promise it will happen many times in your career, you need to go back a step. Review your statics book, and do the FBD of the arm exactly the way the book does it. The time to do this is not wasted because you need their approval to move ahead. Also, you really need to review FBD's.

Oh ok I'm going to correct it, and upload it again
jrmichler said:
Your FBD is better, but still not correct. Hint: Show vertical and horizontal force components, then make sure they both sum to zero. Also, it's a good idea to round off forces to the nearest pound (or maybe tenth) on the FBD. It makes it more readable.

When you find yourself in a situation where you need convince somebody else who does not see it your way, and I promise it will happen many times in your career, you need to go back a step. Review your statics book, and do the FBD of the arm exactly the way the book does it. The time to do this is not wasted because you need their approval to move ahead. Also, you really need to review FBD's.

Hello friend thanks again for your answer I redid the fbd considering the force components, but I have a problem with the reactions at x they don't sum to zero what do you think would be the problem thanks!

#### Attachments

• fbd.pdf
137.8 KB · Views: 237
charliebrown said:
Oh ok I'm going to correct it, and upload it again

Hello friend thanks again for your answer I redid the fbd considering the force components, but I have a problem with the reactions at x they don't sum to zero what do you think would be the problem thanks!

Excuse I made a mistake I considered the forces in the x axis, but since its base it's movable, then there are only forces in the y-axis i reuploaded it.

#### Attachments

• fbd1.pdf
111 KB · Views: 317
The FBD is looking much better. You are almost there. When the reactions do not sum to zero in both directions, then you did something wrong. Things to check:
1) Gravity is always vertical.
2) The force of a member pinned at both ends (the cylinder) is always in line with the member.
3) It is helpful to show the actual force components on the FBD. In lbs in your case.
4) You do not know the direction of the force vector through a pinned joint until you find the components, then do a vector sum.
5) Look only at the forces on the arm until you have it fully solved. The forces and moments on the column and the base are a separate problem that cannot be solved until you have fully solved the arm.

charliebrown said:
Hi sir Michler! sorry for the delay but i got really busy, I think i got it, mind you check if i got it right, also i have another problem, I now need to get the maximum moment, but i don't know how can i do the shear and moment diagram for the geometry of the arm since it is not horizontal I got really confused how to do it:(do you have any idea? thank you in advance.

I made my bar horizontal and
jrmichler said:
The FBD is looking much better. You are almost there. When the reactions do not sum to zero in both directions, then you did something wrong. Things to check:
1) Gravity is always vertical.
2) The force of a member pinned at both ends (the cylinder) is always in line with the member.
3) It is helpful to show the actual force components on the FBD. In lbs in your case.
4) You do not know the direction of the force vector through a pinned joint until you find the components, then do a vector sum.
5) Look only at the forces on the arm until you have it fully solved. The forces and moments on the column and the base are a separate problem that cannot be solved until you have fully solved the arm.

Hi sir michler I think I got it could you please see if everything's ok? thank you :)

#### Attachments

• fbd3.pdf
184.5 KB · Views: 250
charliebrown said:
I made my bar horizontal and

Hi sir michler I think I got it could you please see if everything's ok? thank you :)
charliebrown said:
I made my bar horizontal and

Hi sir michler I think I got it could you please see if everything's ok? thank you :)

this is the analysis considering the bar horizontal, I just need to do the analysis considering max and min angles and tension and compresión forces on the piston.

#### Attachments

• fbd3REV.pdf
141.8 KB · Views: 225
I did not check your calculations, but the FBD's (both of them) look good. So do the shear and moment diagrams.

It is not necessary to show Cx in the correct direction. If you get it wrong, as in fbd3.pdf, then the force will calculate to a negative number.

jrmichler said:
I did not check your calculations, but the FBD's (both of them) look good. So do the shear and moment diagrams.

It is not necessary to show Cx in the correct direction. If you get it wrong, as in fbd3.pdf, then the force will calculate to a negative number.

Oh Ok thank you very much sir you’ve been so helpful thanks a lot! If I have any other doubts can I contact you?

charliebrown said:
If I have any other doubts can I contact you?
In general, it's best to post your questions and follow-ups in the open forums, rather than trying to start Private Conversations. Thanks.

jrmichler said:
I did not check your calculations, but the FBD's (both of them) look good. So do the shear and moment diagrams.

It is not necessary to show Cx in the correct direction. If you get it wrong, as in fbd3.pdf, then the force will calculate to a negative number.

Hi Sir Michler sorry for bothering you again, but I went with my advisers and they're telling me that my previous analysis was wrong, that the arm it's considered as a member subjected to combined loadings do you agree with that? I really couldn't understand why it was considered like that, they're telling that I need to do the arm in two parts but if that is the case then i don't know how to tackle the second arm that is at a 25° angle. I hope you can help me thank you in advance.

#### Attachments

• arm analysis.pdf
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Please, I'm not a Sir, just an old fart retired guy who occasionally wanders over to PF. Your advisers are technically correct, and they want you to do an analysis that is completely correct. If you refer back to your statics book, it tells you that you can cut any system or member at any location to find the forces at the cut location. You did this in the bottom diagram on the first page of your PDF. That diagram is missing only the tension force of 198 lbs at the fixed point, but you did include it in your calculations. It's important to get the FBD correct, any mistake jumps out at the person grading it, and it's real easy to mark the whole analysis wrong if the FBD has any mistakes at all.

I see two different sets of calculations for My/I, with the same value of M and two different values for y and I. If the arm is a rectangular tube, and the I and y values are for the two different directions (the short dimension and long dimension of the tube cross section), then one set of calculations is wrong because the tube has a bending moment in one direction. If you look vertically down on the arm from above, the 309 lb force does not show up, so is zero in a FBD in that direction.

The Von Mises criterion applies to stresses at a point. On the top surface of your beam, you have a tensile stress from bending and a tensile stress from P/A. Those two stress components add, as you did. Assume a Cartesian coordinate system parallel to that part of the beam. Then the X direction stress is My/I + P/A, the Y direction (across the beam) is zero, and the Z direction (down into the beam) is also zero. There is a shear stress in the beam, but that is zero at the top and bottom surfaces, and maximum in the center. All of this is in your strength of materials book.

Keep plugging away, you are getting closer to the finish.

jrmichler said:
Please, I'm not a Sir, just an old fart retired guy who occasionally wanders over to PF. Your advisers are technically correct, and they want you to do an analysis that is completely correct. If you refer back to your statics book, it tells you that you can cut any system or member at any location to find the forces at the cut location. You did this in the bottom diagram on the first page of your PDF. That diagram is missing only the tension force of 198 lbs at the fixed point, but you did include it in your calculations. It's important to get the FBD correct, any mistake jumps out at the person grading it, and it's real easy to mark the whole analysis wrong if the FBD has any mistakes at all.

I see two different sets of calculations for My/I, with the same value of M and two different values for y and I. If the arm is a rectangular tube, and the I and y values are for the two different directions (the short dimension and long dimension of the tube cross section), then one set of calculations is wrong because the tube has a bending moment in one direction. If you look vertically down on the arm from above, the 309 lb force does not show up, so is zero in a FBD in that direction.

The Von Mises criterion applies to stresses at a point. On the top surface of your beam, you have a tensile stress from bending and a tensile stress from P/A. Those two stress components add, as you did. Assume a Cartesian coordinate system parallel to that part of the beam. Then the X direction stress is My/I + P/A, the Y direction (across the beam) is zero, and the Z direction (down into the beam) is also zero. There is a shear stress in the beam, but that is zero at the top and bottom surfaces, and maximum in the center. All of this is in your strength of materials book.

Keep plugging away, you are getting closer to the finish.

Oh I'm sorry I'll call you michler, I said sir in a sign of respect and appreaciation for your help.
Yes I think so too but at first I doubted it the thing is that the arm is going to be welded I'll send you a photo, I already corrected the diagram.

Yes there are two different calculations there first the profile is not rectangular it's squared but first we calculated the section modulus s=.1004 I went to a catalog of squared profiles and the nearest value was .162 and a square profile of 1 1/4 in we then calculated the Factor os security and it gave 1.40≈1 since we considered this FS to be very low, we recalculated the FS with a bigger profiles so we went to a profile of 1 5/8 and then it gave a FS= 3

I did the beam profile diagram so that the combined loadings analysis gets easier and added the shear stress. Could you please review it thank you friend.

#### Attachments

• armanalysis.pdf
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• welded patient lifter.png
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Your use of Von Mises is wrong. At the top of the beam there is a tensile stress of 8220 PSI. The stress in the two directions perpendicular to the tensile stress are zero, and there is zero shear stress at the top of the beam.

The shear stress is maximum at the neutral axis (where the bending stress is zero), decreasing to zero at the extreme fibers (where the bending stress is maximum). Review the derivation of the shear stress in your strength of materials book. They have a diagram showing this.

The Von Mises criteria applies to stress at a single point. For example, if you were to bend and twist your beam, then you would have a shear stress at the points of maximum bending stress. The Von Mises equation would then be used to find the maximum stress. Your beam is a simpler case; the maximum bending stress is at one location, the maximum shear stress at a different location.

Check the units of your shear stress.

Hi my friend! thank you very much for your help I think I made the necessary corrections, do you mind checking if everythings it's ok? thank you very much I consideres the most crítical point to be at b, because it has shear and bending stress.

#### Attachments

• armanalysis1.pdf
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