Free FE Practice Test

## Free FE Civil Example Practice Problems

We've selected 10 diverse practice problems from our question bank that you can use to review for the Civil engineering FE exam and give you an idea about some of the content we provide.

### 1) A table has been set up to aid in the drawing of an influence line. What is most nearly the internal force in member AF when the influence line point load is placed at joint B? Assume the moving point load is equal to 1kN.$$\begin{array}{c|c}\text{Point Load Location} & F_{AF} \\\hline\text{Joint A} & 0\si{kN} \\\text{Joint B} & \\\text{Joint C} & 0\si{kN} \\\end{array}$$

◯A. 

1 kN

◯B. 

0 kN

◯C. 

0.5 kN

◯D. 

2.5 kN

### 2) What is most nearly the bending moment in the beam when shear equals zero?

◯A. 

4.21 kN-m

◯B. 

1.69 kN-m

◯C. 

3.88 kN-m

◯D. 

11.22 kN-m

### 3) What is the angle of rotation needed to get the stress element to the maximum principal stress state?

◯A. 

$36^\circ$ clockwise

◯B. 

$24^\circ$ clockwise

◯C. 

$24^\circ$ counterclockwise

◯D. 

$48^\circ$ counterclockwise

### 4) A level circuit was run from Station 1+00 to Station 2+00 and back. Partial leveling notes were taken as shown on the table below. What is most nearly the elevation at Station 2+00?$$\begin{array}{|c|c|c|}\text{Station} & \text{BS}& \text{HI}& \text{FS} & \text{Elev.}\\\hline\ 1+00 & 8.12' & \text{XX} &\text{XX}&450.21' \\\ 2+00 & 7.11' & &5.92'& \\\end{array}$$

◯A. 

456.13 ft

◯B. 

458.33 ft

◯C. 

457.32 ft

◯D. 

452.41 ft

### 5) A geotechnical investigation at a proposed road construction site spotted a 7 ft thick weak clay layer, 10 feet below the ground surface. On top of the weak clay layer, there is 10 feet of fill material. A sample was obtained from the drilling investigation and was subsequently sent to a lab for testing. The lab results for the weak clay layer and the fill material layer can be seen below. Assume building a road on top of the soil profile will increase vertical stress by 100 psf at the midpoint of the weak clay layer and the clay layer is a normally consolidated clay. What is most nearly the primary consolidation of the weak clay layer? __Fill layer__

$H=10\si{ft}$

$\gamma=121\si{lb/ft^3}$__Weak Clay layer__

$\gamma=119\si{lb/ft^3}$

$C_c=0.252$

$e_o=0.85$

$H=7\si{ft}$

◯A. 

0.3 in

◯B. 

0.1 in

◯C. 

0.9 in

◯D. 

1.2 in

### 6) A 20-acre suburban commercial lot has stores and parking lots. The corresponding NRCS curve number is 97. What is most nearly the amount of runoff created after a 2.75" rain event over the course of 24 hours?

◯A. 

2.4 in

◯B. 

1.9 in

◯C. 

1.7 in

◯D. 

3.2 in

### 7) A vertical curve connects a 3% downslope with a 2.5% upslope. What is most nearly the minimum curve length required to meet AASHTO's standard headlight sight distance requirements? Assume the sight distance is 750 ft.

◯A. 

93 ft

◯B. 

4,550 ft

◯C. 

1,023 ft

◯D. 

950 ft

### 8) Evaluate the following expression: $\int_1^2 5x^2dx$

◯A. 

-8.1

◯B. 

8.1

◯C. 

-11.7

◯D. 

11.7

### 9) What is most nearly the area moment of inertia about the x-axis of the composite shape shown below?

◯A. 

$10,249\si{m^4}$

◯B. 

$7,232\si{m^4}$

◯C. 

$4,521\si{m^4}$

◯D. 

$9,256\si{m^4}$

### 10) Select all that apply. Which of the following is part of an activated sludge system when treating wastewater?

◯A. 

Aeration basin

◯B. 

Clarifier

◯C. 

Hardness treatment

◯D. 

Disinfection

## Solutions

### 1) A table has been set up to aid in the drawing of an influence line. What is most nearly the internal force in member AF when the influence line point load is placed at joint B? Assume the moving point load is equal to 1kN.$$\begin{array}{c|c}\text{Point Load Location} & F_{AF} \\\hline\text{Joint A} & 0\si{kN} \\\text{Joint B} & \\\text{Joint C} & 0\si{kN} \\\end{array}$$

A.1 kN

B.0 kN

C.0.5 kN

D.2.5 kN

The correct answer is C.

### Explanation:

Refer to the Influence Lines for Beams and Trusses - Structural Analysis section in the Civil Engineering chapter of the FE Reference Handbook.

An influence line shows the effect a point load has on a truss member as the point load moves along the truss. In other words, what's the internal force in a truss member when the point load is at joint A, joint B, joint C, etc.

We can tabulate the internal force of member AF as the point load goes from joint A to joint C.$$\begin{array}{c|c}\text{Point Load Location} & F_{AF} \\\hline\text{Joint A} & 0\si{kN} \\\text{Joint B} & 0.5\si{kN} \\\text{Joint C} & 0\si{kN} \\\end{array}$$

__When point load is at joint A__

$$\curvearrowleft +\sum M_C=-A_y(5+5\si{m})+1\si{kN}(5+5\si{m})=0\\\therefore A_y=1\si{kN}$$Method of joints at joint A:

$$\sum F_y=F_{AF}-P+A_y=0\\\sum F_y=F_{AF}-1\si{kN}+1\si{kN}=0\\\therefore F_{AF}=0$$

__When point load is at joint B__

$$\curvearrowleft +\sum M_C=-A_y(5+5\si{m})+1\si{kN}(5\si{m})=0\\\therefore A_y=0.5\si{kN}$$Method of joints at joint A:

$$\sum F_y=F_{AF}+A_y=0\\\sum F_y=F_{AF}+0.5\si{kN}=0\\\therefore F_{AF}=-0.5\si{kN}$$

__When point load is at joint C__

$$\curvearrowleft +\sum M_C=-A_y(5+5\si{m})=0\\\therefore A_y=0\si{kN}$$Method of joints at joint A:

$$\sum F_y=F_{AF}+A_y=0\\\sum F_y=F_{AF}+0\si{kN}=0\\\therefore F_{AF}=0\si{kN}$$

### 2) What is most nearly the bending moment in the beam when shear equals zero?

A.4.21 kN-m

B.1.69 kN-m

C.3.88 kN-m

D.11.22 kN-m

The correct answer is B.

### Explanation:

The FE Reference Handbook provides equations that you can use to quickly calculate the maximum moment or shear for *certain* beam loading-support conditions. Search for *simply supported beam* on the FE Reference Handbook to find a table that has max shear and max moment equations for different types of beams. In this particular example, this beam loading-support condition is not part of the table.

Therefore, we have no alternative but to use shear and moment diagrams to determine the maximum bending moment in the beam.

Steps to draw shear and moment diagrams:

1. Draw a free body diagram of the beam.

2. Determine all the reactions and moments by using the equilibrium equations.

3. Draw the shear and moment diagrams by slicing the beam into sections.

Step 1: Draw a FBD.

Roller supports will have a vertical reaction force. Pin supports will have a vertical and a horizontal reaction force.

Step 2: Use equilibrium equations to solve for all the forces on the FBD. $$\sum M=\text{Force}\cdot\text{Distance}=0\\\sum F_x=0, \sum F_y=0$$

$$\sum M_A=\left[(-6\si{kN/m})(2\si{m}+1\si{m})\right](\frac{2\si{m}+1\si{m}}{2})+B_y(2\si{m})=0\\B_y=13.5\si{kN}$$Side note: we divide the distance arm (2m+1m) by 2 to get the resultant for the moment generated by the 6kN/m load.$$\sum F_y=\left[(-6\si{kN/m})(2\si{m}+1\si{m})\right]+13.5\si{kN}+A_y=0\\A_y=4.5\si{kN}$$$$\sum F_x=A_x=0\\A_x=0$$

Step 3: Develop the shear and moment diagrams by slicing the beam into sections.

Looking at slice #1. Develop a shear equation when $0\leq x \leq2\si{m}$$$\sum F_y = 4.5\si{kN}-(6\si{kN/m}\cdot x)-v=0\\v=4.5\si{kN}-(6\si{kN/m}\cdot x)$$Looking at slice #1. Develop a moment equation when $0\lt x \leq2\si{m}$$$\sum M_{\text{right side}} = \left[(+6\si{kN/m})(x)\right](\frac{x}{2})-4.5\si{kN}(x)+M=0\\M=(+4.5\si{kN})(x)-(6\si{kN/m}\cdot x)(\frac{x}{2})$$Looking at slice #2. Develop a shear equation when $2\lt x \leq 3\si{m}$$$\sum F_y = 4.5\si{kN}-(6\si{kN/m}\cdot x)-v+13.5\si{kN}=0\\v=18\si{kN}-(6\si{kN/m}\cdot x)$$Looking at slice #2. Develop a moment equation when $2\lt x \leq 3\si{m}$$$\sum M_{A} = \left[(-6\si{kN/m})(x)\right](\frac{x}{2})+13.5\si{kN}(2\si{m})\\+M-\left[18\si{kN}-(6\si{kN/m}\cdot x)\right](x)=0$$

$$M=3x^2-27+18x-6x^2$$Now that we have the shear and moment equations for different sections of the beam, we can use these equations to plot the shear and moment diagrams.

$$\begin{array}{c|c|c}\text{x} & \text{Shear (v)}& \text{Moment (M)} \\\hline0\si{m} & 4.5\si{kN} & 0\si{kNm} \\1\si{m} & -1.5\si{kN} & 1.5\si{kNm} \\1.99\si{m} & -7.44\si{kN} & -2.92\si{kNm} \\2.10\si{m} & 5.40\si{kN} & -2.43\si{kNm} \\3\si{m} & 0\si{kN} & 0\si{kNm} \\\end{array}$$We know the maximum moment occurs when shear is equal to zero. And based on the table above, we know zero shear occurs somewhere where $0\leq x \leq 2\si{m}$. Therefore, if we set shear equal to zero and use the shear equation when $0\leq x \leq 2\si{m}$ then we can find where along the beam shear=0 occurs. And then if we know the x value, then we use the moment equation to determine the moment for the given x.

Looking at slice #1 $0\leq x \leq2\si{m}$$$\sum F_y = 4.5\si{kN}-(6\si{kN/m}\cdot x)-v=0\\\sum F_y = 4.5\si{kN}-(6\si{kN/m}\cdot x)-(0)=0\\x=0.75\si{m} \text{ when shear = 0}$$Max moment occurs when shear is zero, therefore$$M=(+4.5\si{kN})(x)-(6\si{kN/m}\cdot x)(\frac{x}{2})\\M=(+4.5\si{kN})(0.75)-(6\si{kN/m}\cdot 0.75)(\frac{0.75}{2})\\M_{max}=1.69\si{kNm}$$

### 3) What is the angle of rotation needed to get the stress element to the maximum principal stress state?

A.$36^\circ$ clockwise

B.$24^\circ$ clockwise

C.$24^\circ$ counterclockwise

D.$48^\circ$ counterclockwise

The correct answer is B.

### Explanation:

Refer to the Mohr's Circle section in the Mechanics of Materials chapter of the FE Reference Handbook.

Ultimately, we must determine the angle rotation needed to get the stress element to a principal stress state.

__Steps to determine the angle of rotation needed:__

1. Draw a quick Mohr's circle based on what's shown in the stress element.

2. Draw a right triangle from the center of the Mohr's circle to a known point along the circle. Determine the Mohr's circle's angle between the known point and the sigma axis (x-axis), which is where the principal stress occurs. That will be the Mohr's circle rotation angle needed to get to a principal stress state.

3. Convert the Mohr's circle's rotation angle needed to get to a principal stress state to the stress element's angle needed to get to a principal stress state.

Step 1. Draw a quick Mohr's circle based on what's shown in the stress element - Identify all points on the Mohr's circle.

Refer to the stress element shown in the FE Reference Handbook to understand its sign notation as well as this problem's explanation image. Play close attention to the sign notation as it's often the number one mistake when drawing Mohr's circles. Normal stresses ($\sigma_x,\sigma_y)$ in tension (going away from the stress element) are positive and normal stresses in compression (going into the stress element) are negative.

For shear stresses ($\tau_{xy}$), look at the shear arrow for the X face of the stress element and for the Y face of the stress element. If a shear arrow causes clockwise rotation to the stress element, then the shear for that stress element face (x or y face) is positive. If a shear arrow causes counterclockwise rotation to the stress element, then the shear for that stress element face (x or y face) is negative. Therefore, two points along the circle can be plotted as:$$\text{X face}= (\sigma_x,\tau_{xy})=(+400\si{MPa},+530\si{MPa})\\\text{Y face}= (\sigma_y,\tau_{xy})=(-550\si{MPa},-530\si{MPa})$$ Determine the center of the Mohr's circle.$$C=\frac{\sigma_x+\sigma_y}{2}=\frac{400+(-550)}{2}=-75\si{MPa}$$Refer to the explanation image to see the final result. Plot the center of the Mohr's circle as well as the X face and Y face of the stress element.

Step 2. Plot a right triangle from the center of the Mohr's circle to the known point along the circumference of the circle.

Use Pythagoras Theorem to solve for the hypotenuse of the triangle, which is also the radius of the circle. $$a^2+b^2=hypotenuse^2\\(+530)^2+(400-(-75))^2=hypotenuse^2\\hypotenuse=radius=711.7\si{MPa}$$The angle, 2$\theta$, is the **Mohr's circle's angle** to get to a principal stress state. The rotation angle needed to get the **stress element** to a principal stress state is $\theta$. This problem asks for the angle to get the **stress element** to a principal stress state.

Calculate the Mohr's circle's angle to a principal stress state. Use soh-cah-toa. $$\sin{\left[\frac{opposite}{hypotenuse}\right]}^{-1}=angle\\\sin{\left[\frac{530}{711.6}\right]}^{-1}=48^\circ$$Step 3. Convert the Mohr's circle's rotation angle needed to get to a principal stress state to the stress element's angle needed to get to a principal stress state.

$$\text{Mohr's Circle's angle}=2\theta=48^\circ$$Stress element's rotation angle to a principal stress state $\theta=24^\circ$ clockwise. Refer to blue arrow in the explanation image to understand why the rotation is clockwise to get to a max principal stress point.

### 4) A level circuit was run from Station 1+00 to Station 2+00 and back. Partial leveling notes were taken as shown on the table below. What is most nearly the elevation at Station 2+00?$$\begin{array}{|c|c|c|}\text{Station} & \text{BS}& \text{HI}& \text{FS} & \text{Elev.}\\\hline\ 1+00 & 8.12' & \text{XX} &\text{XX}&450.21' \\\ 2+00 & 7.11' & &5.92'& \\\end{array}$$

A.456.13 ft

B.458.33 ft

C.457.32 ft

D.452.41 ft

The correct answer is D.

### Explanation:

The easiest way to solve leveling problems during the FE exam is by sketching them out. The sketch always looks like this, a survey instrument in the middle with a measuring rod to the left and to the right of the survey instrument. Refer to the explanation image in this problem.

HI = Elevation of the surveying instrument from datum

BS = Backsight reading

FS = Foresight reading

Elev. = Station elevation from datum

Ultimately, we must solve for the elevation at station 2+00. By looking at the figure, we can infer that $$\text{Elevation}_\text{2+00}=\text{HI}_\text{1+00 to 2+00}-\text{FS}_\text{2+00}$$First, find $\text{HI}_\text{1+00 to 2+00}$. By looking at the figure, we can infer that $$\text{HI}_\text{1+00 to 2+00}=\text{Elev}_\text{1+00}+\text{BS}_\text{1+00}$$$$\text{HI}_\text{1+00 to 2+00}=450.21'+8.12'=458.33'$$Now, go back and solve for the elevation at station 2+00$$\text{Elevation}_\text{2+00}=458.33'-5.92'=452.41'$$$$\begin{array}{|c|c|c|}\text{Station} & \text{BS}& \text{HI}& \text{FS} & \text{Elev.}\\\hline\ 1+00 & 8.12' & \text{XX} &\text{XX}&450.21' \\\ 2+00 & 7.11' &458.33' &5.92'&452.41' \\\end{array}$$During the actual FE, you will have to quickly sketch out the figure above in order to come up with a solution. Just remember what HI, BS, and FS mean and you'll be prepared for this question.

### 5) A geotechnical investigation at a proposed road construction site spotted a 7 ft thick weak clay layer, 10 feet below the ground surface. On top of the weak clay layer, there is 10 feet of fill material. A sample was obtained from the drilling investigation and was subsequently sent to a lab for testing. The lab results for the weak clay layer and the fill material layer can be seen below. Assume building a road on top of the soil profile will increase vertical stress by 100 psf at the midpoint of the weak clay layer and the clay layer is a normally consolidated clay. What is most nearly the primary consolidation of the weak clay layer? __Fill layer__

$H=10\si{ft}$

$\gamma=121\si{lb/ft^3}$__Weak Clay layer__

$\gamma=119\si{lb/ft^3}$

$C_c=0.252$

$e_o=0.85$

$H=7\si{ft}$

A.0.3 in

B.0.1 in

C.0.9 in

D.1.2 in

The correct answer is A.

### Explanation:

Refer to the Geotechnical section in the Civil Engineering chapter of the FE Reference Handbook.

Determine $p_0$, $\Delta p$, and $p_c$ to determine which consolidation/settlement equation to use. There are three possible equations for consolidation/settlement. We must choose the right one.

$p_0 $ is the initial effective consolidation stress, $\sigma_0'$, i.e. the total stress at the **midpoint** of the layer in question minus the pore water pressure.$$p_o=\sigma_0'=\sigma_0-u$$where $\sigma_0$ is the total stress at the mid point of the layer in question. In this case, the stress at the mid point of the clay layer will be the sum of the stresses above the mid point. $$\sigma_0=\sum(H \cdot \gamma)$$$$=\sigma_{\text{fill layer}}+\sigma_{\text{half of clay layer}}$$$$=\left[ (10')(121\si{pcf})+(\frac{7}{2}')(119\si{pcf}) \right]$$$$\sigma_0=1,626.5\si{psf}$$Since we were not told about a ground water table, we will assume the ground is dry and the water table is somewhere below our layers in question. Therefore, porewater pressure, $u$, is zero.

Go back and solve for $p_0$.$$p_0=1,626.5-0=1,626.5\si{psf}$$We were given $\Delta p$ to be equal to 100 psf.

We were told in the problem statement this was a normally consolidated clay layer, therefore $p_c=0$.

Determine the appropriate settlement equation to use. $$p_0\geq p_c \text{ and }p_0+\Delta p\geq p_c$$$$1,626.5\si{psf}\geq 0\si{psf} \text{ and }1,626.5\si{psf}+100\si{psf}\geq 0\si{psf}\checkmark$$Thus$$\Delta H=\frac{H_0}{1+e_0}\left[C_c\log(\frac{p_0+\Delta p}{p_0})\right]$$$$\Delta H=\frac{7'}{1+0.85}\left[0.252\log(\frac{1,626.5+100}{1,626.5})\right]$$$$\Delta H =0.0247\si{ft}=0.3\si{in}$$

### 6) A 20-acre suburban commercial lot has stores and parking lots. The corresponding NRCS curve number is 97. What is most nearly the amount of runoff created after a 2.75" rain event over the course of 24 hours?

A.2.4 in

B.1.9 in

C.1.7 in

D.3.2 in

The correct answer is A.

### Explanation:

Refer to the Hydrology/Water Resources section in the Civil Engineering chapter of the FE Reference Handbook.

Ultimately, we must solve for runoff. $$\text{Runoff}=Q=\frac{(P-0.2S)^2}{P+0.8S}$$Calculate the unknown $S$ and $P$.$$S=\frac{1000}{CN}-10$$$$S=\frac{1000}{97}-10=0.3093\si{in}$$$$\text{Precipitation}=P=2.75\si{in}$$Calculate the amount of runoff, $Q$.$$Q=\text{Runoff}=\frac{\left[2.75\text{"}-0.2(0.3093\text{"})\right]^2}{2.75\text{"}+0.8(0.3093\text{"})}$$$$Q=\text{Runoff}=2.4\si{in}$$

### 7) A vertical curve connects a 3% downslope with a 2.5% upslope. What is most nearly the minimum curve length required to meet AASHTO's standard headlight sight distance requirements? Assume the sight distance is 750 ft.

A.93 ft

B.4,550 ft

C.1,023 ft

D.950 ft

The correct answer is C.

### Explanation:

Refer to Vertical Curves: Sight Distance Related to Curve Length table - Transportation section in the Civil Engineering chapter of the FE Reference Handbook.

We must determine, $L$, when $S\leq L$ and when $S\gt L$.

When $S\gt L$, sag vertical curves - headlight criteria.$$L=2(S) - \left[ \frac{400+3.5(S)}{A} \right]$$$$A=\lvert \text{grade difference}\rvert= \lvert 2.5 -(-3)\rvert= 5.5\%$$Side Note: $A$ and $g$ in **Vertical Curves** equations use % units. Do not divide $A$ or $g$ by 100 in **Vertical Curves**.$$L=2(750\si{ft}) - \left[ \frac{400+3.5(750\si{ft})}{5.5\%} \right]=950'$$Check if $S\gt L \Rightarrow 750' \lt 950'$. Condition not true. Reject.

When $S\leq L$, sag vertical curves - headlight criteria$$L=\frac{AS^2}{400+3.5(S)}$$$$L=\frac{(5.5\%)(750\si{ft})^2}{400+3.5(750\si{ft})}=1,022.73'$$Check if $S\leq L \Rightarrow 750' \leq 1,022.73'$ Condition is true. Accept.

Therefore the curve length, $L=1,023\si{ft}$

### 8) Evaluate the following expression: $\int_1^2 5x^2dx$

A.-8.1

B.8.1

C.-11.7

D.11.7

The correct answer is D.

### Explanation:

Refer to the Indefinite Integrals section in the Calculus chapter of the FE Reference Handbook. There, you will find a long list of typical integrals.

According to this list, the integral of $x^mdx$ is $\frac{x^{m+1}}{m+1}$

So if we take the integral rule stated above and apply it to $\int5x^2$, then we get:$$\int5x^2dx\\=(5)\int\ x^2dx\\=(5)\frac{x^{2+1}}{2+1}\\=\frac{5x^3}{3}$$Now, evaluate the integral [1,2] via substitution.$$\frac{5(2)^3}{3}-\frac{5(1)^3}{3}\\=11.7$$

Alternatively, you could solve this entire problem using the integration feature in your TI-36X Pro calculator .

### 9) What is most nearly the area moment of inertia about the x-axis of the composite shape shown below?

A.$10,249\si{m^4}$

B.$7,232\si{m^4}$

C.$4,521\si{m^4}$

D.$9,256\si{m^4}$

The correct answer is D.

### Explanation:

Refer to the Moment of Inertia Parallel Axis Theorem section in the Statics chapter of the FE Reference Handbook.

Use the parallel axis theorem to find the moment of inertia about the x axis for this **composite** shape. Ultimately, we must solve for:$$I_{x}=I_{xc}+d^2_yA$$To determine the moment of inertia about the x axis of the composite shape, we must break up the shape into common smaller shapes. In this case, the composite shape shown in the problem can be broken down into a triangle (1) and a rectangle (2). Refer to the Area & Centroid table in the Statics chapter of the FE Reference Handbook. There, you will find $I_{xc},d_y,A$ equations for commons shapes.

1st. Calculate the $I_{xc},d_y,A$ values for the triangle

$A=bh/2=(12)(6)/2=36\si{m^2}\\I_{xc}=bh^3/36=(12)(6)^3/36=72\si{m^4}$

$d$= distance b/w desired axis (x-axis) and shape's centroid

$d=10\si{m}+6\si{m}/3=12\si{m}\\d^2A=(12^2)(36)=5,184\si{m^4}$

2nd. Calculate the $I_{xc},d_y,A$ values for the rectangle.

$A=bh=(12)(10)=120\si{m^2}\\I_{xc}=bh^3/12=(12)(10)^3/12=1,000\si{m^4}\\d=\text{distance b/w desired axis (x-axis) and shape's centroid}\\d=10\si{m}/2=5\si{m}\\d^2A=(5^2)(120)=3,000\si{m^4}$

Plug in all values into the parallel axis theorem:$$I_{x}=\sum I_{xc}+\sum d^2_yA$$$$I_{x}=\left[72\si{m^4}+1000\si{m^4}\right]\\+\left[5,184\si{m^4}+3,000\si{m^4}\right]\\I_{x}=9,256\si{m^4}$$Notes: The FE Reference Handbook has moments of inertia equations for several different shapes e.g. circles, rectangles, triangles. This particular problem is a composite shape (a rectangle plus a triangle). Because the Reference Handbook doesn'thave a moment of inertia equation for this composite shape, we have no choice but to use the Parallel Axis Theorem to calculate the moment of inertia. In other words: if the overall shape is found in the FE Reference Handbook tables (squares, triangles, etc), use the moment of inertia equation the Reference Handbook has for that shape. If the Handbook doesn't have the shape you're looking for, then you'll have to use the Parallel Axis Theorem to solve for the momentof inertia.

Another situation where you would have to use the Parallel Axis Theorem over the Reference Handbook's standard moment of inertia equations is if you're asked to calculate the moment of inertia about an axis that is not the centroidal, x, or y axis! Let's say you're asked to calculate the moment of inertia of a square about an axis that is 1 cm off the x axis. The Reference Handbook doesn't have a moment of inertia equation for this custom axis, so you'll have to use the Parallel Axis Theorem to calculate the moment of inertia.

### 10) Select all that apply. Which of the following is part of an activated sludge system when treating wastewater?

A.Aeration basin

B.Clarifier

C.Hardness treatment

D.Disinfection

The correct answers are A and B.

### Explanation:

An activated sludge system is the most popular secondary treatment step of wastewater at a wastewater treatment plant. It is made of an aeration basin and a clarifier.

When wastewater enters a wastewater treatment plant, it often goes through these steps: Preliminary treatment $\rightarrow$ primary clarifier $\rightarrow$ secondary treatment (activated sludge) $\rightarrow$ disinfection or discharge to lakes, rivers, ocean, etc

An activated sludge system consists of first treating the wastewater to an aeration basin and then sending the water to a clarifier. The goal of the aeration basin is to create enough dissolved oxygen (DO) in the water via air diffusers to where bacteria in the water can remain alive. In turn, this bacteria eats up the organic matter that is present in the wastewater. After enough dissolved oxygen has been created in the aeration basin via air bubbles, the wastewater is sent to another clarifier. In the clarifier, the wastewater is held for some time so the organic matter that is now infused with bacteria can settle to the bottom of the tank. We call this precipitate "sludge". A portion of the sludge is actually sent again back to the aeration basin to help out break down new influent organic matter. And the water that remains in the clarifier is of low BOD levels and can be discharged off to a disinfection step or back to the environment if clean enough.

Although invisible to our eyes, water in waterways e.g. rivers, lakes, etc has small amounts of dissolved oxygen (DO) mixed with the water. This dissolved oxygen mixed with the water is what keeps aquatic life alive. Without enough DO, aquatic life will quickly die. When wastewater plants are done treating their wastewater, they will release the treated wastewater back to the environment e.g. rivers, lakes, etc. This treated wastewater comes with bacteria that themselves feed off DO. The quantity of DO bacteria present in the treated wastewater consume is called biological oxygen demand (BOD). One of the most important goals of a wastewater treatment plant is to treat wastewater in a way that will lower the BOD from the bacteria present in the wastewater. That way, the effluent (discharge) treated wastewater's bacteria don't consume all the DO that aquatic life needs to remain alive. If a wastewater treatment plant releases treated wastewater with a high BOD from the bacteria, fish from the receiving streams will be left out of DO and will then die.

## FAQs

### Is the FE practice exam enough to pass? ›

Although the questions on the NCEES Practice Exam are great, **there just aren't enough of them for a good FE exam review**. We recommend you add on with other sources that have more practice problems like PrepFE.

**How many questions do I need to get right on the FE civil exam? ›**

NCEES generally does not provide a specific number of questions required to answer correctly for one to pass. However after doing the research and referring to different articles listed above it has been said that out of 110 problems, to pass, you will need to answer **approximately 50 to 60 percent of them correctly**.

**What is the hardest FE exam? ›**

A **54% pass rate** is a hard exam, the hardest among the FE exams. The FE Mechanical and FE Environmental exams now have the highest pass rate at 66% (see our chart in the next section below), and a 34% failure rate is something we consider very demanding.

**What percent do you need to pass the FE civil exam? ›**

FE Civil: **73%** FE Electrical and Computer: 73% FE Environmental: 79%

**Is the FE practice exam harder than the actual exam? ›**

NCEES Practice Exam

This is one that I recommend everyone use. It comes in all disciplines, so it is suitable for all engineering students. **Some students tell me that this exam is easier than the actual exam**, but they still recommend it.

**What is the easiest FE exam to pass? ›**

Among all the NCEES® FE exams, the **FE Other (General) exam** is the easiest FE exam to pass, especially for those who are confused about choosing the right engineering domain.

**Is the FE civil exam curved? ›**

In short, **no, the FE Exam is not curved**. But it is definitely important to understand how it is scored – let's lay it out as it is defined for us by the NCEES. When you take the FE Exam, you will complete 110 questions. Of those 110 questions, you are going to get a certain number correct.

**How many people fail the FE? ›**

To put it in perspective, the FE Civil Exam only has a **62% pass rate** and continues to go down year after year. The FE is a comprehensive exam that tests your knowledge of all the fundamentals of engineering. You have to put in the hard work and dedication to pass.

**How many pass the FE exam first try? ›**

The passing score of the FE exam on average is **70%**.

It is imperative to remember that these numbers can change very quickly.

**How long should you study for the FE civil exam? ›**

We recommend planning to study for at least **2 to 3 months** before your FE exam date. Once you get started practicing problems from all of the different topics covered, you'll have a better idea if you need to spend more or less time preparing.

### How long did you study for the FE civil exam? ›

In general, we recommend **12 to 14 weeks** of serious study for a minimum of three hours per day. However, FE exam prep time varies depending on which FE exam you plan to take. Each exam has different questions and a different number of test-takers.

**How can I pass FE exam civil? ›**

**The Top 10 Tips to Ace the Civil FE Exam**

- Understand Your Own Learning Style. ...
- Know What's on the Test. ...
- Manage Your Time Wisely. ...
- Use the FE Reference Handbook. ...
- Practice, Practice, Practice... ...
- Get a Good Night's Sleep Before Taking an Exam. ...
- Don't Try to 'Wing It' at the Last Minute. ...
- Stay Organized.

**How much time do you get per question on the FE exam? ›**

FE Exam Duration

The total time you'll have to actually answer the exam questions is 5 hours and 20 minutes. The problem-solving pace works out to **slightly less than 3 minutes per question**, and you may work through the questions (in that session) in any sequence.

**Can you skip questions on the FE exam? ›**

The great thing about the FE exam being computer-based is that **you can flag or skip a question you don't know** and you will be able to review all of your answers before submitting each half of the exam.

**Why do I keep failing the FE exam? ›**

**Failing to set aside enough time to prepare**

The best way to pass the FE exam is proper preparation. Start studying for the test at least one month in advance. Buy reference materials and practice tests early to find enough time to prepare. Time yourself with practice tests so you can enter that test with confidence.

**Is the FE exam harder the second time? ›**

A question we often get coming in from those who are getting back on the horse after failing is whether or not the FE Exam gets harder to pass the more you take it. The short answer is No. But, that isn't the whole story. **The FE Exam actually can get harder to pass the more you attempt it.**

**Do you lose points for wrong answers on the FE exam? ›**

**There are no deductions for wrong answers**. The score is then converted to a scaled score, which adjusts for any minor differences in difficulty across the different exam forms.

**Is Fe or PE harder? ›**

**The FE exam is actually harder than the Professional Engineer (PE) exam**, simply because it is much more broad. It's best to take it as close to graduation as you can, with all that schooling still fresh in your mind.

**How many times can you fail the FE exam? ›**

After all, you can retake the FE exam. How many times per year, exactly? Well, NCEES® policy states that you can attempt the test once per every testing window, up to three times in 12 months.

**How many people take the FE exam every year? ›**

One available program-assessment tool is the NCEES Fundamentals of Engineering (FE) exam. **Approximately 55,000** people take this exam each year. Most of them are college seniors within one year of graduating or are recent graduates.

### What percentage of engineers pass the FE? ›

According to the January/February and April/May 2015 FE-CBT Civil exam results, 70% of 4874 test takers passed. The aforementioned examinees took the exam for the first time, after having attended EAC/ABET-accredited engineering programs, within 12 months of graduation.

**Can employers see if you failed the FE? ›**

**Unless you give your NCEES username and password to your employer, then they will see the failure notices**.

**How do you pass the FE for the first time? ›**

**It is to work hard and smart.**

- Review the list of FE Exam topics and classify the exam topics into buckets. Familiarize yourself with the published list of topics that could be asked on the Fundamentals of Engineering (FE) exam. ...
- Develop a study schedule. Develop a study schedule that takes into account: ...
- Practice. Practice.

**Does FE exam discipline matter? ›**

Does it matter which FE exam I take? **No.** **You may take any FE exam discipline to get EIT certification**, and it does not matter for your PE license.

**How many hours a week should I study for the FE? ›**

You should dedicate at least **14 to 21 hours a week** to studying for the FE exam, and that is roughly two-three hours per day. Reading the reference handbook, going through an exam preparation course, and practicing multiple problems should be the focus of your study hours.

**How many hours did you study for FE exam? ›**

How many total hours will you study for the FE Exam? Now regardless of the stage you are in, we have seen that most students will spend somewhere around **150-350 hours** preparing for the FE Exam.

**What is the best way to pass the FE exam? ›**

**It is to work hard and smart.**

- Review the list of FE Exam topics and classify the exam topics into buckets. Familiarize yourself with the published list of topics that could be asked on the Fundamentals of Engineering (FE) exam. ...
- Develop a study schedule. Develop a study schedule that takes into account: ...
- Practice. Practice.

**How long should you study before taking the FE exam? ›**

How long you should study for the FE exam depends on you as an individual. In general, we recommend **12 to 14 weeks of serious study for a minimum of three hours per day**.

**What percentage of people pass the FE exam on the first try? ›**

Everyone has different study habits and test-taking skills, and what works for someone else may not work for you. To put it in perspective, the FE Civil Exam only has a **62%** pass rate and continues to go down year after year.

**How long should I give myself to study for the FE exam? ›**

We recommend planning to study for at least **2 to 3 months** before your FE exam date. Once you get started practicing problems from all of the different topics covered, you'll have a better idea if you need to spend more or less time preparing.

### What is the average FE exam passing score? ›

6. The FE Exam Passing Score. The passing score of the FE exam on average is **70%**.

**How many hours should I study for the FE civil exam? ›**

You should dedicate at least **14 to 21 hours a week** to studying for the FE exam, and that is roughly two-three hours per day. Reading the reference handbook, going through an exam preparation course, and practicing multiple problems should be the focus of your study hours.

**Can you bring a cheat sheet to FE exam? ›**

Because the exam is administered via computer, **you are not permitted to bring any books to the testing center**. However, you will be provided with the NCEES FE Reference Handbook to search through during the exam.