Finding a maximum force in a beam type system before failure

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SUMMARY

The discussion centers on calculating the maximum force (P) that a beam-type system can withstand before failure, specifically focusing on a steel rod and a brass block. The maximum force for the steel rod is determined to be 201.06 kN, which serves as a critical design parameter. Through the application of static equilibrium equations and stress calculations, participants derive that the total force P can reach up to 308 kN, factoring in the reactions at the supports and the failure modes of the materials involved. The conversation emphasizes the importance of understanding load distribution and failure points in structural analysis.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of stress and strain calculations
  • Familiarity with material yield strengths, specifically for steel and brass
  • Ability to perform moment calculations in beam systems
NEXT STEPS
  • Learn about advanced beam theory and load distribution methods
  • Study failure modes in structural materials, focusing on yield and ultimate strength
  • Explore the application of finite element analysis (FEA) for complex load scenarios
  • Investigate the design principles for safety factors in structural engineering
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Engineering students, structural engineers, and anyone involved in the design and analysis of beam-type systems in mechanical or civil engineering contexts.

ridiculoid
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Homework Statement


See attached photo

Homework Equations


Stress = F/A
Area = pi d^3 / 4
Sum F h = 0, Sum F v = 0, Sum M = 0

The Attempt at a Solution


Hey everyone, after reading the question a number of times and not knowing for certain where to start, I figured because the stress and area values were given I could find the maximum force each component could take before failure. I found that the steel rod would fail first after 201.06kN so I assume the force P would have to be designed around this.

After drawing a free body diagram with this value, I'm just a bit confused as to what the next step should be to find force P and which of the other forces would be reacting in the rigid bar. I assume it is a case of finding the 2 reactive forces from force P to achieve equilibrium, I'm just not 100% sure I'm on the right track.

Any help would be greatly appreciated!
 

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ridiculoid said:
I found that the steel rod would fail first after 201.06kN so I assume the force P would have to be designed around this.
No necessarily. It depends how the load P gets shared. How will it be shared between the brass block and the steel rod?
 
haruspex said:
No necessarily. It depends how the load P gets shared. How will it be shared between the brass block and the steel rod?

Thanks for the response, I understand what you're saying, will the force P be distributed between the pins / rods and block up to 201kN? I'm just trying to figure out a solid value to work from. Am I to multiply the force at the pin by 0.8 for the moment to the left of P? I'm still confused as to how this would lead me to the reactive force at the block.
 
ridiculoid said:
how this would lead me to the reactive force at the block.
The first part is a very basic statics problem. You have a force P down at a known distance from the end supports, and two unknown upward forces from those supports. Take moments about one support to find the force at the other.
 
haruspex said:
The first part is a very basic statics problem. You have a force P down at a known distance from the end supports, and two unknown upward forces from those supports. Take moments about one support to find the force at the other.

Right, so I take moments about the 201.06kN force and find the reactive force at the brass block is 107kN which sums to P being 308kN. Does this seem correct? Am I right in taking the maximum force of the steel rod (201.06kN) as my left reactive force as this will fail before the pins? (814.3kN)

Again thanks for your help!
 
ridiculoid said:
the 201.06kN
What 201kN force? You are presupposing the answer.
You have an unknown load P. In terms of that, what are the reaction forces at the end supports?
 
haruspex said:
What 201kN force? You are presupposing the answer.
You have an unknown load P. In terms of that, what are the reaction forces at the end supports?

Seeing as the metal rod has a yield stress of 640MPa, I used stress = force / area to find the the amount of force before yield which is 201kN? I used the formula on the other components as well as since the steel rod yeilded at the lowest force I thought I should start there.. I am not sure which defined value to start with.

The reaction forces in terms of P, when I take moments about the left support P = 2.875R and when I take moments about the right support P = 1.533L.

Sorry for the hassle, my lecturer mentioned that this was a bit over our current curriculum
 
ridiculoid said:
I used stress = force / area to find the the amount of force before yield which is 201kN?
Ok. Seems a bit backwards to me, but let's go with that. And yes, that corresponds to P= 308kN.
So now you need to calculate the P values corresponding to the other two failure modes.
 
haruspex said:
Ok. Seems a bit backwards to me, but let's go with that. And yes, that corresponds to P= 308kN.
So now you need to calculate the P values corresponding to the other two failure modes.

Right, I should say force = stress * area.

When I calculated P around the maximum force that can be applied to the brass (4.8MN) and steel pins (814.3kN) I got P as 13.8MN and 1.25MN respectively, which by this point the steel rod will have failed.

So to clarify, am I going down the right track with P being 308kN?
 
  • #10
ridiculoid said:
Right, I should say force = stress * area.

When I calculated P around the maximum force that can be applied to the brass (4.8MN) and steel pins (814.3kN) I got P as 13.8MN and 1.25MN respectively, which by this point the steel rod will have failed.

So to clarify, am I going down the right track with P being 308kN?
Looks right, except that I get a somewhat smaller numbers for the steel pins. I'm no engineering expert so maybe I am not calculating that correctly. Aren't they (18/20)2(800/640)*2 * the numbers for the steel rod? You seem to be getting about double that.
 
  • #11
haruspex said:
Looks right, except that I get a somewhat smaller numbers for the steel pins. I'm no engineering expert so maybe I am not calculating that correctly. Aren't they (18/20)2(800/640)*2 * the numbers for the steel rod? You seem to be getting about double that.

Great, thanks so much for the help. Got there in the end hahah. The doubled value probably comes from the fact the pins are under double shear, 2 pins with 2 shear points each puts the shear value for the pins at force = stress * 4 * Area. That's how I interpreted it anyway, It still comes in at over the failure for the steel rod so should be ok!
 
  • #12
ridiculoid said:
2 pins with 2 shear points each
No, that doesn't work. Each pin has two shear points, but each pin is independently under the same load. It only takes one pin to fail.
 

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