# MD&M practice problem of the week



## Slay the P.E.

*SPOILER ALERT*: Attempt to solve it before scrolling down and reading the responses.


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## Kloeb222

A. 

But the problem statement says "a load P=3 kip" and the figure shows 2 load P's which would result in a combined load of 6 kip and yield a thickness of .1 in. Unless i'm missing this completely...


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## Slay the P.E.

You have to have the two forces applied as shown for equilibrium. The magnitude of each one is 3 kip.

Looks like you only checked for compressive stress in the cross section. There’s more going on...Make a FBD of a segment of the link by making a cut at section a-a


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## MikeGlass1969

I come up with (C)  ...  

I am not used to doing these types of problems though...


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## TWeatherford

I get C.

Maximum stress = tensile stress + bending stress = F/A + My/I = 3000/(2*t) + 3000lb(3in)/(1/12*t*2^3) = 30000 when t = 0.5.


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## Slay the P.E.

TWeatherford said:


> I get C.
> 
> Maximum stress = tensile stress + bending stress = F/A + My/I = 3000/(2*t) + 3000lb(3in)/(1/12*t*2^3) = 30000 when t = 0.5.


Absolutely correct!


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## Kloeb222

Ah i got it now. I though the problem statement was trying to be tricky so i neglected bending stress. Good problem, thanks.


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## SacMe24

Excellent problem... keep them coming. For those of you who have MERM Rev. 13, this is solved by equation 51.43 for Axial &amp; Eccentric Bending... I got 0.55 in


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## TWeatherford

How is it known from the information given that _t_ is the depth into the page, rather than the width (which would make _I_ = (2*_t^3)/12)?_


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## Slay the P.E.

TWeatherford said:


> How is it known from the information given that _t_ is the depth into the page, rather than the width (which would make _I_ = (2*_t^3)/12)?_


Maybe the figure can use an additional view of the link. Like this:


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## TWeatherford

Slay the P.E. said:


> Maybe the figure can use an additional view of the link. Like this:
> 
> View attachment 10927


Yes, that makes sense.  I wanted to make sure there isn't some way to know the orientation without the additional view.


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## Slay the P.E.

Happy Friday all. Enjoy this one:

The post is fixed to the floor at its base and has a diameter _d_. External loads _P_1 and _P_2 are acting in the _x_ and [SIZE=11.0pt]_y_ [/SIZE]directions, respectively. Point _A_ is located on the surface of the post within cross section _a – a_, which is at a distance _h_ below the line of action of _P_1. Distance _b_ is from the axis of the vertical part of the post to the end of the horizontal part of the post. Of the following statements, select the one that is true:

(A) The bending moment due to _P_1 causes a normal stress distribution at the cross section, and at point _A_, the magnitude of the normal stress due to _P_1 is independent of _h_ .

(B) The shear force due to _P_1 causes a shear stress distribution at the cross section that reaches a maximum at point _A._

(C) The load _P_2 causes a torsional stress distribution at the cross section that reaches a minimum at point _A._

(D) The bending moment due to _P_2 causes a normal stress distribution at the cross section and at point _A_, the magnitude of the normal stress due to _P_2 is independent of _h_.


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## MA_PE

(C)


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## Kloeb222

D.

The bending moment due to P2 causes a normal stress distribution at cross section a-a. At point A the magnitude of normal stress is zero and therefore independent of h.


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## Slay the P.E.

Kloeb222 said:


> D.
> 
> The bending moment due to P2 causes a normal stress distribution at cross section a-a. At point A the magnitude of normal stress is zero and therefore independent of h.


This is correct.


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## emmajuwa

I believe option C is also correct. P2 is creating a torsional stress/twisting effect on the section a-a, and torsional stress is minimum on the external fibers and maximum at the center.


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## Kloeb222

I think you have it backwards. Shear stress due to torsion is zero at the center and increases to a maximum at the external fibers.


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## Slay the P.E.

Kloeb222 said:


> I think you have it backwards. Shear stress due to torsion is zero at the center and increases to a maximum at the external fibers.


This is correct.


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## Slay the P.E.

That’s transverse shear. Option C is about torsional shear. These are totally different.


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## emmajuwa

You are right. I had it wrong


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## Engineer_562

TWeatherford said:


> I get C.
> 
> Maximum stress = tensile stress + bending stress = F/A + My/I = 3000/(2*t) + 3000lb(3in)/(1/12*t*2^3) = 30000 when t = 0.5.


Hello,

Thanks for posting this problem. I have two questions about the solution tho. 

1) why is the area normal to force P=t*2? The way I see it is t is the thickness which should not be consider for the normal area to P

2) why is x=3in in the moment for bending stress?


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## Engineer_562

A=t*2


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## Slay the P.E.

The problem statement indicates that the section of interest (where the stress should not exceed 30 ksi) is section a-a. 

So, split the link into two pieces by the cross section. Draw the free body diagram of any of the two pieces. What are the reactions acting on the cross section?

1. A force of magnitude P normal to the cross section. The area of the cross section is A=(2in)xt, so the normal stress is P/A

2. A moment of magnitude Px(3in) because the external load P is applied 3 inches away from the centroid of the cross-section.


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## SacMe24

With respect to the 2nd problem (post attached to the floor)... I also believe the answer is C....


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## Slay the P.E.

jvanoye said:


> With respect to the 2nd problem (post attached to the floor)... I also believe the answer is C....


Option C is about the torsional stress due to P2, not the transverse shear. Torsional stress is highest at the outermost fibers and zero at the center.


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## SacMe24

Slay the P.E. said:


> Option C is about the torsional stress due to P2, not the transverse shear. Torsional stress is highest at the outermost fibers and zero at the center.


I see that now...thanks.  If Option C had said:

The load _P_2 causes a torsional stress distribution at the cross section that reaches a MAXIMUM  at point _A._

Then it would've been true... correct? Thinking of Shear=(Txr)/J


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## Slay the P.E.

jvanoye said:


> I see that now...thanks.  If Option C had said:
> 
> The load _P_2 causes a torsional stress distribution at the cross section that reaches a MAXIMUM  at point _A._
> 
> Then it would've been true... correct? Thinking of Shear=(Txr)/J


Correct. Torsional shear stress from P2 is at its highest at point A.


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## Engineer_562

Slay the PE,

thank you for your response. This was an excellent problem to solve. If you have more, please post more. Thanks again


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## SacMe24

Slay the P.E. said:


> Correct. Torsional shear stress from P2 is at its highest at point A.


Thank you !


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## Slay the P.E.

Happy Thursday. Here's the problem of the week;

*SPOILER ALERT: *Try to solve it before scrolling down and reading the discussion.


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## birdboatboy

(A) for the buckling problem.


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## Slay the P.E.

birdboatboy said:


> (A) for the buckling problem.


Correct!


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## emmajuwa

Correct. Thanks Slay the PE


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## TWeatherford

Good problem.  Thanks for posting.


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## SacMe24

I also got A for the answer.... great problem and thanks for helping us keep our skills sharp for the exam...


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## Slay the P.E.

jvanoye said:


> I also got A for the answer.... great problem and thanks for helping us keep our skills sharp for the exam...


Thanks all for serving as test subjects as we beta test problems for a Slay the PE MDM practice exam!


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## Slay the P.E.

Happy Thursday. Here's one we're cooking for a future SlaythePE MDM practice exam. Would love your feedback.

The experimental setup shown in the figure has been devised to determine the coefficient of friction between a cord and a steel cylinder of 0.2 m in diameter. The steel cylinder is fixed and does not rotate. A small force is applied at end of the cord. This force is slowly increased until the block is set in motion (sliding horizontally). Any stretching of the cord is negligible. The experiment was performed maintaining a constant angle of contact, ϕ=90°, between the cord and the cylinder. The force _F_ required to initiate movement of the block was measured to be 65 N. The mass of the block is 20 kg, and the static coefficient of friction between the block and the horizontal surface on which it slides is 0.25.

The coefficient of friction between the rope and the cylinder is nearest:
(A) 0.003
(B) 0.180
(C) 0.750
(D) Cannot be calculated with the information provided.


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## Kloeb222

(B) 0.180.

Ftight/Fslack = e^(mu*phi)

Ftight=65N

Fslack=m*g*mu_s

solve for mu

Another good problem. Thanks!


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## Slay the P.E.

Kloeb222 said:


> (B) 0.180.
> 
> Ftight/Fslack = e^(mu*phi)
> 
> Ftight=65N
> 
> Fslack=m*g*mu_s
> 
> solve for mu


Correct!

Thanks.


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## SacMe24

My answer was also B but the calculated value I got was 0.23...answered in 6:23 min....


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## Slay the P.E.

jvanoye said:


> My answer was also B but the calculated value I got was 0.23...answered in 6:23 min....


But you should get 0.179 — there has to be an error somewhere.


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## emmajuwa

Slay the P.E. said:


> But you should get 0.179 — there has to be an error somewhere.


Maybe the calculator 

Sent from my SM-N950U1 using Tapatalk


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## SacMe24

Slay the P.E. said:


> But you should get 0.179 — there has to be an error somewhere.


Yes.. instead of 49.05N for the friction force I wrote 45.05 when I took the natural log...that did it... I do get 0.179...whew....gotta watch out for these simple mistakes on the exam...thanks !


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## Slay the P.E.

Last one before the big day next Friday. Good luck to all!!!!

An external point load of 3 kip is applied at point C on the steel bar shown. The bar is supported by two springs at its ends A and B. Each spring has the same stiffness and is originally un-stretched. You may assume the weight of the bar is negligible, and that Young's modulus is E=29,000 ksi. For the bar, I=12 in^4. The vertical displacement at the point of application of the load shall not exceed 1.51 inches. Under these conditions, the stiffness (kip/ft) of each spring is most nearly:

(A) 15

(B) 20

(C) 25

(D) 30


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## emmajuwa

Is this question set up right? Deflections at the ends will be zero. Maybe I'm missing something?


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## Slay the P.E.

emmajuwa said:


> Is this question set up right? Deflections at the ends will be zero. Maybe I'm missing something?


Deflections at the ends are not zero, because the supports are compression springs which will be shortened upon the application of the load.

But, even if they were zero at the ends... the deflection at the point of application of the load is non-zero and is given as 1.51 inches.

What the problem is asking for is the spring stiffness, k.


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## MikeGlass1969

I come up with (A)..  15kip/ft

Found the deflection of the beam at the point load.  Then subtracted the deflection from the total deflection.  Created a slope intercept form equation for the slope of the beam due to the uneven deflections of the springs.


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## Kloeb222

I get A as well. Took me a bit too long though. about 12 min because I didn't have any beam tables readily accessible (at work) and had to search through my bookmarked resources online

Same way as Mikeglass describes above.


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## TWeatherford

I get A as well.  Maybe it's just me, but I found it to be a challenging problem, but a good one.


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## Slay the P.E.

A is the correct answer for the beam deflection problem. Use the principle of superposition to add the displacement of the load application point assuming rigid supports with an elastic beam plus the displacement of the load application point assuming a rigid beam but elastic supports.


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## emmajuwa

Slay the P.E. said:


> A is the correct answer for the beam deflection problem. Use the principle of superposition to add the displacement of the load application point assuming rigid supports with an elastic beam plus the displacement of the load application point assuming a rigid beam but elastic supports.


Please do you guys mind sharing your calculation? Thanks.


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## Kloeb222

Here ya go. Hopefully this is clear.


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## emmajuwa

Kloeb222 said:


> Here ya go. Hopefully this is clear.
> 
> View attachment 11035


Thank you so much. i got halfway and got stuck.


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## Slay the P.E.

Kloeb222 said:


> Here ya go. Hopefully this is clear.
> 
> View attachment 11035


You beat me to it. Good work!


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## Kloeb222

Is anyone else excited for practice problems of the week to start again? lol


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## monty01

Bring it!


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## squaretaper LIT AF PE

Kloeb222 said:


> Here ya go. Hopefully this is clear.
> 
> View attachment 11035


As a thermal-fluids guy, this makes my stomach hurt. :rotflmao:


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## monty01

squaretaper said:


> As a thermal-fluids guy, this makes my stomach hurt. :rotflmao:


Deflection?  Easy!  Now, enthalpy and entropy?  WTH is that all about?!?!?!?  Adiabati-wha????


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## SacMe24

Kloeb222 said:


> Is anyone else excited for practice problems of the week to start again? lol


I'll probably still look at these just for kicks... :bananalama:


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## Kloeb222

monty01 said:


> Deflection?  Easy!  Now, enthalpy and entropy?  WTH is that all about?!?!?!?  Adiabati-wha????


My thoughts exactly. lol. To each their own..


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## Vel2018

Kloeb222 said:


> My thoughts exactly. lol. To each their own..


Me: Entropy, Enthalpy, Polytropic, Adiabatic, Carnot Cycle, Re-heat Regen Cycle, Gas Turnine - Steam cogen.. Easy!     Deflection, Kinematics, Vibration Tri-Axialllalalal Stressss what??? lol


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## monty01

Vel2018 said:


> Me: Entropy, Enthalpy, Polytropic, Adiabatic, Carnot Cycle, Re-heat Regen Cycle, Gas Turnine - Steam cogen.. Easy!     Deflection, Kinematics, Vibration Tri-Axialllalalal Stressss what??? lol


LOLOLOL!!!!


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## monty01

I really enjoyed all of the Thermo (Heat Transfer, Thermal Fluid, etc.) classes in college.  But i just don't delve into those subjects much anymore for work, so i'd be wayyyy out of practice.  I wouldn't mind refreshing on that stuff again, but studying for the MDM (a bazillion topics) is already enough for this year.


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## Kloeb222

Lol I actually did better in thermo and heat transfer than statics and dynamics in school. Using machine design stuff at work over the last 6 years helped keep the basics relatively fresh in my head and made studying easier.

On another note, finally got my license number today  great start to a vacation week.


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## Slay the P.E.

Brand new practice problem of the week:

Platform _P _ is supported by four springs, each having a stiffness of 55 pounds-force per foot and has a load rigidly attached to it.  The load and platform have a combined mass of 44 pounds-mass. The floor is subjected to a vertical displacement with amplitude 0.4 inches and a frequency of 1.27 Hertz. Under these conditions, the amplitude (inches) of the steady state vibration is most nearly:

(A) 0.40

(B) 0.66

(C) 0.86

(D) 1.04

​


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## MikeGlass1969

OOF!!!   I can not remember how to approach these problems...  Partial Credit Given?


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## monty01

Haha...support &amp; isolation is actually next weeks lesson.  I'll have to take a stab at this later today.


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## Slay the P.E.

MikeGlass1969 said:


> OOF!!!   I can not remember how to approach these problems...  Partial Credit Given?


Sure!

go for it.


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## monty01

I get (B) for this one, with a ratio |X/Y| = 1.65.  Zeta is zero (no damping).


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## Slay the P.E.

monty01 said:


> I get (B) for this one, with a ratio |X/Y| = 1.65.  Zeta is zero (no damping).


(B) is correct, but it would be nice if you showed more of your work for the benefit of other readers 

Maybe upload a picture or scanned PDF?


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## monty01

Wow, now we have to show our work!?!?!  Hard chargers around here!  Give me a couple of minutes and i'll see if i can get the solution uploaded.

View attachment Support Vibration.pdf


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## Slay the P.E.

monty01 said:


> Wow, now we have to show our work!?!?!  Hard chargers around here!  Give me a couple of minutes and i'll see if i can get the solution uploaded.
> 
> View attachment 11929


Beautiful. 

One observation: The spring constant is given in pounds-force per foot. The mass of the system was given in pounds-mass. Therefore, the calculation of the natural frequency is sqrt[ (18.33lbf/in) / (44lbm) ] 

From your work it appears you thought the 44 were pounds-force (a weight) because you replaced "m" with "W/g" and used W=44lb.  It turns out to numerically be the same. I'm just worried that you're using the symbol "lb" for "lbf" and also for "lbm".


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## MikeGlass1969

monty01 said:


> Wow, now we have to show our work!?!?!  Hard chargers around here!  Give me a couple of minutes and i'll see if i can get the solution uploaded.
> 
> View attachment 11929


Nice!

I am envious of your knowledge and penmanship.


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## Slay the P.E.

MikeGlass1969 said:


> Nice!
> 
> I am envious of your knowledge and penmanship.


This is true. @monty01 has beautiful penmanship. A skill that seems to be in decline...


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## monty01

I'm with Dr. Tom when it comes to lbm and lbf.  I use slugs instead, which is where the W/g came from.


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## monty01

I think i have OCD when it comes to writing.  If my radical lines aren't straight, it throws me off!  My straight edges are gonna be burning up on test day!


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## monty01

Now that I'm thinking about it, this might be helpful to people to avoid unit issues with lbm and lbf:


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## Workx

Slay the P.E. said:


> Happy Thursday. Here's the problem of the week;
> 
> *SPOILER ALERT: *Try to solve it before scrolling down and reading the discussion.
> 
> View attachment 10959


I am getting 2.5...can you someone help to how to do? Calculate the Euler stress and compare with Sy/2?


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## Slay the P.E.

Workx said:


> I am getting 2.5...can you someone help to how to do? Calculate the Euler stress and compare with Sy/2?


Can you show your work?


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## MikeGlass1969

Workx said:


> I am getting 2.5...can you someone help to how to do? Calculate the Euler stress and compare with Sy/2?


I tried this too,  and I am not getting it, either.


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## monty01

I'm getting 2.0 for the SF on the buckling problem assuming it's within the Euler range.  The safety factor I used was Pcrit/P.


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## Slay the P.E.

monty01 said:


> I'm getting 2.0 for the SF on the buckling problem assuming it's within the Euler range.  The safety factor I used was Pcrit/P.


Correct. For buckling, the Factor of Safety is Pcrit/P.


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## Slay the P.E.

Workx said:


> I am getting 2.5...can you someone help to how to do? Calculate the Euler stress and compare with Sy/2?


The critical load formula is derived by assuming that buckling will occur at that load, that is when P = Pcrit. Therefore, the factor of safety for buckling is defined as  Pcrit/P. Try it this way.


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## Slay the P.E.

In the engine system shown, crank AB is 3 inches long and has a constant clockwise angular velocity of 2000 rpm, and connecting rod BC is 8 inches long. When angle _α_ is 40 degrees, the velocity (ft/s) of piston P is most nearly:

(A) 3.7

(B) 21

(C) 41

(D) 44


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## finbean5

Most nearly D, 44 ft/sec. I had Vb=628in/sec and Vc/b=497 in/sec. Some trig happens and I'm left with 43.5 ft/sec.


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## Slay the P.E.

finbean5 said:


> Most nearly D, 44 ft/sec. I had Vb=628in/sec and Vc/b=497 in/sec. Some trig happens and I'm left with 43.5 ft/sec.


That is correct. The actual answer is 43.5 ft/s


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## monty01

Here is a copy of the crank/slider solution if anyone is interested.  I definitely wouldn't consider this to be a 6 minute problem.  I had to do a fair amount of angle finding and then break up the vector equation into x/y components, being careful to keep things straight.  Hopefully somebody else has a quicker solution.

I'd have to say this one is a "C" and I'd only go back to it if I had some time toward the end.


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## Slay the P.E.

monty01 said:


> Here is a copy of the crank/slider solution if anyone is interested.  I definitely wouldn't consider this to be a 6 minute problem.  I had to do a fair amount of angle finding and then break up the vector equation into x/y components, being careful to keep things straight.  Hopefully somebody else has a quicker solution.
> 
> I'd have to say this one is a "C" and I'd only go back to it if I had some time toward the end.
> 
> View attachment 11954


It might be a little quicker if you approach it by finding the instantaneous center of rotation for rod BC.


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## monty01

Not a bad idea.  It's been a while since I've done any IC problems.  Does anyone have a solution using the IC method?


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## nosi9wrx

See equation 57.59 in MERM


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## monty01

There it is!  Nice work!


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## monty01

Wow, a one-liner for a crank/slider problem.  Nice!  What reference did you find that in?


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## tx2011

The one liner equation for a crank slider problem can also be found in the reference book by Timothy Kennedy.  Just got his book last week and its great. The equation is much simpler than the one above but it uses only x and y values instead of theta so you do have to do basic trig to figure out the different lengths.


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## monty01

Ahhh, that's a time saver for sure.  I tried to get familiar with the IC method in the MERM, but it didn't really make sense to me so I moved on pretty quickly.  Maybe if i have some extra time, i'll give it another shot.


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## Workx

I got this from one of my friends' notes from his class. I also used to Timothy's book formula but I feel this is much easier. 

I also like the IC method. In case the question is asking other parameters, we need to use MERM.


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## Slay the P.E.

We won't be adding any more problems to the "practice problem of the week" threads. However, the threads are still here in the boards and are relevant, not just because of the problems themselves but also because of the lively discussion.


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## 23and1

Slay the P.E. said:


> Happy Thursday. Here's the problem of the week;
> 
> *SPOILER ALERT: *Try to solve it before scrolling down and reading the discussion.
> 
> View attachment 10959


Hey errrbody,

So I've been crunching on this one over and over. What is it that I'm missing with the buckling check? I keep getting a FS of 2.66. Is there an eccentric component that I'm missing to the bending (MERM equation 53.14)? I've attached my work for reference (sorry if it seems a little disorganized...) Any help would be appreciated!

View attachment Buckling FS.pdf


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## Castaway81

I solved for the reaction at B by taking the moment about A and got 7800 lbf.   Then your FS is 1.99


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## 23and1

Ah, I see...
So the reaction at the column isn't R=0.375*w*L (from propped cantilever equation in MERM Appendix 51.A)? Is that because the end on the wall is pinned instead of built in?


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## Castaway81

That's what it looks like to me.  This would be the simple supported beam in the appendix where R=wl/2. 

I should have used this rather than the moment equation which would have been a bit quicker.


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## YW55




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## Slay the P.E.

YW55 said:


> View attachment 12735


LOL.

No. I'm afraid not. There are like 6 or 7 problems in this thread. Did you work them all?


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## 23and1

Castaway81 said:


> That's what it looks like to me.  This would be the simple supported beam in the appendix where R=wl/2.
> 
> I should have used this rather than the moment equation which would have been a bit quicker.


Whoops...

It only took me 10x looking at that sketch to realize that pinned connection. Another silly mistake to avoid during the real thing this Friday. Thanks for your help @Castaway81


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## Castaway81

I am glad you brought it up as I wasn't even paying close attention to the end connections either.   I think my biggest focus going into Friday is making sure I don't make any small mistakes.   Especially since you know that answer will be one of the options.  Best of luck with the rest of your preparation.


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## Slay the P.E.

Slay the P.E. said:


> *SPOILER ALERT*: Attempt to solve it before scrolling down and reading the responses.
> 
> View attachment 10923


Friendly reminder that this thread is still relevant.

Cheers, and good luck


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## Sam N

Its asking for _normal _stress


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## Slay the P.E.

Sam N said:


> Its asking for _normal _stress


Correct. The normal stress is due to two sources: the axial force normal to the section, and the bending moment due to the external load.


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## Sam N

I think its been answered before.

Someone said (C) - 0.5.

Im getting the same.


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## Slay the P.E.

Correct, (C)

yes, all problems in the thread have been answered and thoroughly discussed.


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## Abogos

Slay the P.E. said:


> Happy Thursday. Here's the problem of the week;
> 
> *SPOILER ALERT: *Try to solve it before scrolling down and reading the discussion.
> 
> View attachment 10959


Could you please post the solution to this problem? I am not getting 2 for the factor of safety. 

Edit: I got it, thanks.


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## Slay the P.E.

Abogos said:


> Could you please post the solution to this problem? I am not getting 2 for the factor of safety.
> 
> Edit: I got it, thanks.


Excellent.


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