Free FE Practice Test

# PrepFE™

## 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.

## Solutions

A.1 kN
B.0 kN
C.0.5 kN
D.2.5 kN

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

A.4.21 kN-m
B.1.69 kN-m
C.3.88 kN-m
D.11.22 kN-m

### 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.

$$\begin{array}{c|c|c} \text{x} & \text{Shear (v)}& \text{Moment (M)} \\ \hline 0\si{m} & 4.5\si{kN} & 0\si{kNm} \\ 1\si{m} & -1.5\si{kN} & 1.5\si{kNm} \\ 1.99\si{m} & -7.44\si{kN} & -2.92\si{kNm} \\ 2.10\si{m} & 5.40\si{kN} & -2.43\si{kNm} \\ 3\si{m} & 0\si{kN} & 0\si{kNm} \\ \end{array}$$ 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

### 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.

A.456.13 ft
B.458.33 ft
C.457.32 ft
D.452.41 ft

### 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
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.

A.0.3 in
B.0.1 in
C.0.9 in
D.1.2 in

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

A.2.4 in
B.1.9 in
C.1.7 in
D.3.2 in

### 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, $Q$. $$Q=\text{Runoff}=\frac{\left[2.75\text{"}-0.2(0.3093\text{"})\right]^2}{2.75\text{"}+0.8(0.3093\text{"})}$$ $$Q=\text{Runoff}=2.4\si{in}$$

A.93 ft
B.4,550 ft
C.1,023 ft
D.950 ft

### 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, $L=1,023\si{ft}$

A.-8.1
B.8.1
C.-11.7
D.11.7

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

### 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't have 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 moment of 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?

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.