MD&M practice problem of the week

[Engineer_562]

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? 

 

[Engineer_562]

A=t*2

 

[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.

 

[SacMe24]

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

 

[Slay the P.E.]

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.

 

[SacMe24]

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 P2 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

 

[Slay the P.E.]

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

The load P2 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.

 

[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 

 

[SacMe24]

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

Thank you !

 

[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.

 

[birdboatboy]

(A) for the buckling problem.

 

[Slay the P.E.]

(A) for the buckling problem.

Correct!

 

[emmajuwa]

Correct. Thanks Slay the PE

 

[TWeatherford]

Good problem.  Thanks for posting.

 

[SacMe24]

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

 

[Slay the P.E.]

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! 

 

[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.
 

 

[Kloeb222]

(B) 0.180.

Ftight/Fslack = e^(mu*phi)

Ftight=65N

Fslack=m*g*mu_s

solve for mu

Another good problem. Thanks!

 

[Slay the P.E.]

(B) 0.180.

Ftight/Fslack = e^(mu*phi)

Ftight=65N

Fslack=m*g*mu_s

solve for mu

Correct!

Thanks.

 

[SacMe24]

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