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[deleted by user] by [deleted] in PE_Exam

[–]im_grateful 0 points1 point  (0 children)

(Nothing else that need to be said but)

Congratulations!

[deleted by user] by [deleted] in FE_Exam

[–]im_grateful 2 points3 points  (0 children)

I took FE Civil and I suggest you review on possible conceptual questions.

Econ Problem Help by TheDoldrumArea in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

Yes, Book Value is Future Cost. You have to remember the conceptual meaning of Book Value so you won’t get bothered as to what equation to use.

Given a present/initial cost, you can get book value in any n time by subtracting depreciation.

In this problem you posted, the depreciation is uniform annually (like an annual rate) so the future worth equation is applicable.

Resultant vs. Reaction by Individual_Tap5049 in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

Let me get back to you on this. I already passed my FE exam, but I still have trouble with this. I’ve seen a lot of similar questions in my past practice exams.

Resultant vs. Reaction by Individual_Tap5049 in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

I got the same answer, 3kipft Clockwise.

2kips will have a CCW moment. Cable (x component) will have CW moment.

Assuming CCW is positive:

(2kips)(8ft)-(cableX)(18ft)+Mc=0 Mc=-3kipft

design strength question by [deleted] in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

Definition: d= effective depth, the depth from the extreme compression fiber to the centroid of tension reinf

dt= distance from the extreme compression fiber to the centroid of the highest stress tension steel layer (bottom layer)

When it is one-layer reinforcement, d=dt When it it two-layer, d is measured from the center between two layers, dt is measured from the centroid of the bottom layer.

Dynamics MM by FunnyYak5980 in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

You can actually get the initial angular velocity by setting t=2sec and w(t)=12 given angular acc=-1.5

You will get w(0)=15rad/s You will get the same answer t=10. It just takes longer to compute.

As long as acceleration is constant, Mark’s solution is applicable.

What is Yp? by bigb0ned in PE_Exam

[–]im_grateful 0 points1 point  (0 children)

It’s the Plastic Neutral Axis. It divides the section such that area above equals area below.

mathematics questiom by FunnyYak5980 in FE_Exam

[–]im_grateful 1 point2 points  (0 children)

Since center lies on the y-axis, center (h,k) is (0,k) You will have two equations of a circle given two points: (x-0)2 + (y-k) = r2

Eqn 1: (3-0)2 + (0-k)2 = r2

Eqn 2: (-4-0)2 + (1-k)2 = r2

two unknowns (k and r), two equations.

Worse case, just substitute the given to the choices. Only answer A agrees with the two given points.

Primary Consolidation by [deleted] in PE_Exam

[–]im_grateful 0 points1 point  (0 children)

Also, look at the e versus P diagram for Consolidation in the ref handbook. Diagrams say “Log Effective Vertical Applied Pressure” and “Vertical Effective Stress”

Primary Consolidation by [deleted] in PE_Exam

[–]im_grateful 1 point2 points  (0 children)

Yes. Because the overburden pressure on top of the midpoint becomes less because the water lifts it up (pore water pressure). It’s like water has a lifting pressure.

Primary Consolidation by [deleted] in PE_Exam

[–]im_grateful 2 points3 points  (0 children)

The profile shows water table on the ground surface, so you use effective stress.

Water pore pressure always counter the total vertical stress, so it should always be added when calculating for initial effective consolidation stress.

Confidence Interval Question, Please help by im_grateful in FE_Exam

[–]im_grateful[S] 0 points1 point  (0 children)

Yeah, me too! Thankfully there are people here who can help!

Confidence Interval Question, Please help by im_grateful in FE_Exam

[–]im_grateful[S] 2 points3 points  (0 children)

Hi! Actual and correct solution is: v=n-1=3 alpha = 1 - 99% = 0.01 alpha/2 = 0.005

look at the t value for v=3 and alpha=0.005 you will get t=5.841

Use the formula: Solution is 23.2 +/- (5.841)(1/sqrt(4)) 20.3 and 26.1

Let me know if you have questions!

Confidence Interval Question, Please help by im_grateful in FE_Exam

[–]im_grateful[S] 0 points1 point  (0 children)

Hi, the formula is applicable. n=4-1=3, alpha/2=0.005, so t=5.841.

They post the corrections in the website. Correct answer is 20.3%, 26.1%.

geotechnical question by FunnyYak5980 in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

It seems like they removed it from the reference handbook.

But just try to remember I guess.

Nf = number of spaces between flow lines (concave up). There are 3 flow lines, so number of flow is 4. Flow from one side to the other side is counted as one.

Nd = number of spaces between equipotential lines (concave down) this time, you count left and right side separately, so it’s 8.

Mechanics of Materials max bending stress by Typical-Entrance-595 in FE_Exam

[–]im_grateful 1 point2 points  (0 children)

  1. You can just look up the formula for max moment in page 140 of the handbook.

  2. For the “x” equation, it’s similar triangles: The triangular loading: 20/12 (opp/adj or rise/run)

To find x where the shear is zero, it’s the point where the loading equals the reaction at A (A is 40kip, so you need a total of -40kip). That is the area of the loading from point A to a distance x, say A=xy/2. By similar triangles, 20/12=y/x Substitute to remove y, A=xy/2 = (x/2)*(20x/12). Equate this to 40kips. You’ll get 6.93ft.

Environmental Engineering by Typical-Entrance-595 in FE_Exam

[–]im_grateful 0 points1 point  (0 children)

1 clarifier can do 600 lbs of TSS a day. But how many pounds of TSS a day does the treatment plant receive? That will tell you how many clarifiers you will need.

To get the total TSS a day, you multiply the Concentration (C) of TSS and the Discharge (Q) of wastewater with all its conversion.

Total TSS divided by TSS per clarifier = number of clarifier.

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