Finding a maximum force in a beam type system before failure

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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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  • #2
haruspex
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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?
 
  • #3
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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.
 
  • #4
haruspex
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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.
 
  • #5
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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!
 
  • #6
haruspex
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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?
 
  • #7
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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.. Im 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
 
  • #8
haruspex
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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.
 
  • #9
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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
haruspex
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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
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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
haruspex
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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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