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Slay the P.E.

MD&M practice problem of the week

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

Screen Shot 2018-03-13 at 8.47.23 PM.png

Edited by Slay the P.E.
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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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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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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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3 hours ago, 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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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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Excellent problem... keep them coming. For those of you who have MERM Rev. 13, this is solved by equation 51.43 for Axial & Eccentric Bending... I got 0.55 in

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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)?

Edited by TWeatherford

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3 hours ago, 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:

MDM link.PNG

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20 hours ago, Slay the P.E. said:

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

MDM link.PNG

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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Happy Friday all. Enjoy this one:

The post is fixed to the floor at its base and has a diameter d. External loads P1 and P2 are acting in the x and y 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 P1. 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 P1 causes a normal stress distribution at the cross section, and at point A, the magnitude of the normal stress due to P1 is independent of h .

(B) The shear force due to P1 causes a shear stress distribution at the cross section that reaches a maximum at point A.

(C) The load P2 causes a torsional stress distribution at the cross section that reaches a minimum at point A.

(D) The bending moment due to P2 causes a normal stress distribution at the cross section and at point A, the magnitude of the normal stress due to P2 is independent of h.

8131590.jpg

Edited by Slay the P.E.

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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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3 minutes ago, 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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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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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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10 minutes ago, 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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That’s transverse shear. Option C is about torsional shear. These are totally different.

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On 3/14/2018 at 5:37 AM, 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? 

Edited by Engineer_562

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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.

Edited by Slay the P.E.

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

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4 minutes ago, 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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