FE Civil Statics practice problems

Sum the components by axis. Along x: $120 - 30 = 90$ N. Along y: 120 N. The resultant is $R = \sqrt{90^2 + 120^2} = 150$ N (a 3-4-5 triple scaled by 30), and the angle from the horizontal is $\arctan(120/90) = 53.1\degree$. The 270 N option adds all three magnitudes ($120 + 120 + 30 = 270$), which is invalid for non-collinear forces. The 150 N at $36.9\degree$ option has the correct magnitude but uses $\arctan(90/120) = 36.9\degree$, the angle measured from the vertical instead of the horizontal. The 90 N at $53.1\degree$ option has the correct angle but mistakes the x-component $R_x = 90$ N for the full resultant, forgetting to combine it with $R_y = 120$ N.

1. Sum the x-components (the negative-x force subtracts):
   $R_x = 120 - 30 = 90\,\text{N}$
2. Sum the y-components:
   $R_y = 120\,\text{N}$
3. Resultant magnitude:
   $R = \sqrt{90^2 + 120^2} = \sqrt{8{,}100 + 14{,}400} = \sqrt{22{,}500} = 150\,\text{N}$
4. Direction from the horizontal:
   $\theta = \arctan\!\left(\frac{120}{90}\right) = 53.1\degree$

FE Reference Handbook 10.6, p. 95

The fastest route is the scalar cross product $M_z = x F_y - y F_x$. Resolve the force: $F_x = 200\cos 30\degree = 173.2$ N, $F_y = 200\sin 30\degree = 100$ N. Then $M_z = (4)(100) - (3)(173.2) = 400 - 519.6 = -119.6$ N$\cdot$m, where the negative sign means clockwise. The 400 N$\cdot$m counterclockwise choice keeps only the $x F_y$ term (400) and drops $y F_x$. The 519.6 N$\cdot$m clockwise choice keeps only the $y F_x$ term (519.6). The 392.8 N$\cdot$m counterclockwise choice swaps the trig functions, using $F_x = 100$ and $F_y = 173.2$ to get $(4)(173.2) - (3)(100) = 392.8$.

1. Resolve the force into components:
   $F_x = 200\cos 30\degree = 173.2\,\text{N}, \quad F_y = 200\sin 30\degree = 100\,\text{N}$
2. Apply the 2D cross product about O:
   $M_z = x F_y - y F_x = (4)(100) - (3)(173.2)$
3. Evaluate:
   $M_z = 400 - 519.6 = -119.6\,\text{N}\cdot\text{m}$
4. The negative result indicates a clockwise moment of 119.6 N$\cdot$m.

FE Reference Handbook 10.6, p. 95

Sum moments about B to eliminate $B_y$ and find $A_y$ directly, or sum moments about A to find $B_y$ first. Taking moments about A: $B_y(8) - 10(10) = 0$, so $B_y = 12.5$ kN. Then $\sum F_y = 0$: $A_y + 12.5 - 10 = 0$, giving $A_y = -2.5$ kN, i.e. 2.5 kN downward. The overhang load actually pulls the pin down (it acts as a tie-down). The 12.5 kN upward choice reports $B_y$ instead of $A_y$. The 5.0 kN choice splits the load evenly, valid only for a midspan load between the supports. The 2.5 kN upward choice has the right magnitude but the wrong direction, and the sign matters here because an overhang can reverse the reaction.

1. Sum moments about A (CCW positive), load is 10 m from A, $B_y$ is 8 m from A:
   $B_y(8) - 10(10) = 0 \quad \Rightarrow \quad B_y = 12.5\,\text{kN}$
2. Apply vertical equilibrium:
   $A_y + B_y - 10 = 0 \quad \Rightarrow \quad A_y = 10 - 12.5 = -2.5\,\text{kN}$
3. The negative sign means $A_y$ acts downward, so the pin holds the beam down against the overhang load:
   $A_y = 2.5\,\text{kN (downward)}$

FE Reference Handbook 10.6, p. 95, Equilibrium Requirements

Sum moments about the hinge A so the pin reactions drop out. The weight acts at the 2 m tip (clockwise), and the cable tension provides a counterclockwise moment through its vertical component $T\sin 30\degree$ acting at the 2 m tip. The horizontal component of the cable passes along the bar through... it acts at the tip too, but its line of action is horizontal through A-level, so its moment arm about A is zero (the bar is horizontal). So $T\sin 30\degree (2) = 600(2)$, giving $T = 600/\sin 30\degree = 1{,}200$ N. The 692.8 N choice uses $\cos 30\degree$ instead of $\sin 30\degree$ ($600/\cos 30\degree = 692.8$). The 300 N choice applies $\sin 30\degree$ to the weight ($600 \times 0.5$) instead of the tension. The 519.6 N choice multiplies the weight by $\cos 30\degree$ ($600\cos 30\degree = 519.6$) instead of dividing by $\sin 30\degree$.

1. Sum moments about A (the weight and the cable both act at the 2 m tip):
   $\sum M_A = 0: \quad T\sin 30\degree\,(2) - 600(2) = 0$
2. The 2 m arm cancels:
   $T\sin 30\degree = 600$
3. Solve for the tension:
   $T = \frac{600}{\sin 30\degree} = \frac{600}{0.5} = 1{,}200\,\text{N}$

FE Reference Handbook 10.6, p. 95, Equilibrium Requirements

The load is off-center, so the reactions are unequal. Find them first. Taking moments about A: $30(3) - C_y(9) = 0$, so $C_y = 10$ kN and $A_y = 30 - 10 = 20$ kN. At joint A the only members are diagonal AE and bottom chord AB. Diagonal AE runs from A(0,0) to E(3,4), length 5 (a 3-4-5 triangle), direction cosines (0.6, 0.8). Vertical equilibrium at A: $20 + F_{AE}(0.8) = 0$, so $F_{AE} = -25$ kN, and the negative sign means compression. The 20.0 kN choice is the reaction $A_y$, not the member force. The 15.0 kN tension choice is the bottom-chord force AB ($-F_{AE}(0.6) = 15$ kN tension), the other member at the joint. The 33.3 kN choice swaps the direction cosines, dividing the reaction by 0.6 instead of 0.8.

1. Reactions (load is off midspan). Sum moments about A:
   $30(3) - C_y(9) = 0 \quad \Rightarrow \quad C_y = 10\,\text{kN}, \quad A_y = 20\,\text{kN}$
2. Diagonal AE: A(0,0) to E(3,4), length $= 5$, direction cosines (0.6, 0.8). At joint A, $\sum F_y = 0$ (tension assumed):
   $20 + F_{AE}(0.8) = 0 \quad \Rightarrow \quad F_{AE} = -25\,\text{kN (compression)}$
3. Check via horizontal equilibrium (gives the bottom chord, a different member):
   $F_{AB} + F_{AE}(0.6) = 0 \quad \Rightarrow \quad F_{AB} = 15.0\,\text{kN (tension)}$

FE Reference Handbook 10.6, p. 98, Plane Truss: Method of Joints

The zero-force rule (Rule 2) says that when three members meet at an unloaded joint and two are collinear, the third (non-collinear) member is zero-force. The critical word is unloaded. Here a load $P$ is applied at the joint, so the rule does not apply. Summing vertical forces at K: the two horizontal top-chord members have no vertical component, so the vertical member alone must balance the load: $F_{vert} = P$. The "zero-force member" choice is the trap: applying the zero-force rule without checking for the applied load. The "P/2 split" choice wrongly splits the load into the collinear members, which carry only horizontal force. The "2P" choice has no physical basis.

1. Rule 2 (zero-force) applies only when the joint is unloaded. Here a load $P$ acts at the joint, so the rule is void.
2. The two collinear top-chord members are horizontal and contribute no vertical force component.
3. Vertical equilibrium at K:
   $\sum F_y = 0: \quad F_{vert} - P = 0 \quad \Rightarrow \quad F_{vert} = P$

FE Reference Handbook 10.6, p. 98, Plane Truss: Method of Joints

First check whether the block slides. The driving force down the incline is $W\sin\theta = 400\sin 20\degree = 136.8$ N. The maximum available friction is $\mu_s N = 0.40(400\cos 20\degree) = 0.40(375.9) = 150.4$ N. Since $136.8 < 150.4$, the block does not slide, so friction only develops as much as it needs to: it equals the down-slope component, 136.8 N. The 150.4 N option is the maximum possible friction, but friction does not reach its cap unless motion impends. The 160.0 N option uses $\mu_s W$ ($0.40 \times 400$) forgetting the $\cos\theta$ in the normal force. The 375.9 N option is the normal force itself.

1. Down-slope gravity component (the friction demand):
   $W\sin\theta = 400\sin 20\degree = 136.8\,\text{N}$
2. Maximum available friction:
   $F_{\max} = \mu_s N = 0.40(400\cos 20\degree) = 0.40(375.9) = 150.4\,\text{N}$
3. Since $136.8 < 150.4$, the block is static and friction equals the demand:
   $F = 136.8\,\text{N}$

FE Reference Handbook 10.6, p. 97, Friction

The block tends to slide down, so friction acts up the incline at its maximum. A horizontal $P$ increases the normal force because part of $P$ presses into the surface: $N = W\cos\theta + P\sin\theta$. At impending downward motion, the up-slope resistances (the up-slope component of $P$ plus friction) balance the down-slope weight component: $W\sin\theta = P\cos\theta + \mu_s N$. Substituting $N$ and solving gives $P = W(\sin\theta - \mu_s\cos\theta)/(\cos\theta + \mu_s\sin\theta) = 228.9$ N. The 461.9 N choice ignores friction entirely ($P = W\tan\theta = 461.9$). The 261.9 N choice uses $N = W\cos\theta$ only, forgetting that $P$ also adds to the normal force. The 773.5 N choice points friction the wrong way (down-slope): using $W\sin\theta = P\cos\theta - \mu_s N$ gives $P = W(\sin\theta + \mu_s\cos\theta)/(\cos\theta - \mu_s\sin\theta) = 773.5$ N, but with the block on the verge of sliding down, friction must resist that motion by acting up-slope.

1. Normal force includes the inward component of $P$:
   $N = W\cos\theta + P\sin\theta$
2. Impending sliding down, so balance along the incline ($P$ up-slope component plus friction resist the weight):
   $W\sin\theta = P\cos\theta + \mu_s(W\cos\theta + P\sin\theta)$
3. Solve for $P$:
   $P = \frac{W(\sin\theta - \mu_s\cos\theta)}{\cos\theta + \mu_s\sin\theta} = \frac{800(0.5 - 0.25 \times 0.866)}{0.866 + 0.25 \times 0.5}$
4. Evaluate:
   $P = \frac{800(0.2835)}{0.991} = \frac{226.8}{0.991} = 228.9\,\text{N}$

FE Reference Handbook 10.6, p. 97, Friction

Treat the notch as negative area. The full rectangle has $A_1 = 100 \times 120 = 12{,}000$ mm$^2$ with centroid at $y_1 = 60$ mm. The removed notch has $A_2 = 40 \times 40 = 1{,}600$ mm$^2$ with its centroid at $y_2 = 120 - 20 = 100$ mm. Using subtraction: $\bar{y} = (12{,}000(60) - 1{,}600(100))/(12{,}000 - 1{,}600) = 560{,}000/10{,}400 = 53.8$ mm. Removing material from the top pulls the centroid down below mid-height. The 60 mm option ignores the notch. The 50.0 mm option is the geometric mid-height. The 46.2 mm option over-corrects, as if material were added at the top instead of removed.

1. Full rectangle:
   $A_1 = 100 \times 120 = 12{,}000\,\text{mm}^2, \quad y_1 = 60\,\text{mm}$
2. Removed notch (centroid 20 mm below the top):
   $A_2 = 40 \times 40 = 1{,}600\,\text{mm}^2, \quad y_2 = 100\,\text{mm}$
3. Composite centroid (subtract the hole):
   $\bar{y} = \frac{12{,}000(60) - 1{,}600(100)}{12{,}000 - 1{,}600} = \frac{720{,}000 - 160{,}000}{10{,}400} = \frac{560{,}000}{10{,}400} = 53.8\,\text{mm}$

FE Reference Handbook 10.6, p. 96, Centroid of Area

Split into a rectangle and a triangle. Rectangle: $A_1 = 60 \times 40 = 2{,}400$ mm$^2$, centroid at $y_1 = 20$ mm. Triangle: $A_2 = \frac{1}{2}(60)(30) = 900$ mm$^2$; its centroid sits one-third of its height above its base, so $y_2 = 40 + 30/3 = 50$ mm. Then $\bar{y} = (2{,}400(20) + 900(50))/(2{,}400 + 900) = 93{,}000/3{,}300 = 28.2$ mm. The 29.5 mm choice uses the triangle midpoint ($40 + 15 = 55$) instead of the one-third point. The 35.0 mm choice is the simple average of 20 and 50, ignoring the area weighting. The 25.0 mm choice underweights the triangle.

1. Rectangle:
   $A_1 = 60 \times 40 = 2{,}400\,\text{mm}^2, \quad y_1 = 20\,\text{mm}$
2. Triangle (centroid one-third of height above its base at 40 mm):
   $A_2 = \tfrac{1}{2}(60)(30) = 900\,\text{mm}^2, \quad y_2 = 40 + \frac{30}{3} = 50\,\text{mm}$
3. Composite centroid:
   $\bar{y} = \frac{2{,}400(20) + 900(50)}{2{,}400 + 900} = \frac{48{,}000 + 45{,}000}{3{,}300} = \frac{93{,}000}{3{,}300} = 28.2\,\text{mm}$

FE Reference Handbook 10.6, p. 96, Centroid of Area

Because the void is concentric, both rectangles share the same centroidal axis, so you simply subtract: $I = \frac{b_o h_o^3 - b_i h_i^3}{12}$. Outer: $100(150)^3 = 337.5 \times 10^6$. Inner: $60(110)^3 = 79.86 \times 10^6$. Difference $= 257.64 \times 10^6$, divided by $12 = 21.5 \times 10^6$ mm$^4$. The $28.1 \times 10^6$ choice is the solid outer rectangle alone ($337.5/12 = 28.1$). The $6.7 \times 10^6$ choice is the inner void contribution by itself (subtracting areas rather than the cubed-height terms). The $34.8 \times 10^6$ choice mistakenly mixes the dimensions or adds instead of subtracting.

1. Outer rectangle contribution:
   $\frac{b_o h_o^3}{12} = \frac{100 \times 150^3}{12} = \frac{337.5 \times 10^6}{12} = 28.125 \times 10^6\,\text{mm}^4$
2. Inner void contribution (same centroidal axis):
   $\frac{b_i h_i^3}{12} = \frac{60 \times 110^3}{12} = \frac{79.86 \times 10^6}{12} = 6.655 \times 10^6\,\text{mm}^4$
3. Subtract:
   $I = 28.125 \times 10^6 - 6.655 \times 10^6 = 21.5 \times 10^6\,\text{mm}^4$

FE Reference Handbook 10.6, p. 99

A tangent axis is offset from the centroid by one radius, $d = r = 50$ mm. The centroidal moment is $\bar{I} = \pi r^4/4 = \pi(50)^4/4 = 4.91 \times 10^6$ mm$^4$. The transfer term is $A d^2 = \pi(50)^2(50)^2 = \pi(2{,}500)(2{,}500) = 19.63 \times 10^6$. Adding: $I = 4.91 + 19.63 = 24.5 \times 10^6$ mm$^4$. The $4.9 \times 10^6$ choice forgets the transfer term and reports $\bar{I}$ only. The $19.6 \times 10^6$ choice reports the transfer term only. The $83.4 \times 10^6$ choice uses $d =$ diameter $= 100$ mm instead of the radius in $Ad^2$.

1. Centroidal moment of inertia for a circle:
   $\bar{I} = \frac{\pi r^4}{4} = \frac{\pi (50)^4}{4} = 4.91 \times 10^6\,\text{mm}^4$
2. Transfer distance equals the radius ($d = 50$ mm):
   $A d^2 = \pi (50)^2 (50)^2 = \pi (2{,}500)(2{,}500) = 19.63 \times 10^6\,\text{mm}^4$
3. Parallel axis theorem:
   $I = \bar{I} + A d^2 = 4.91 \times 10^6 + 19.63 \times 10^6 = 24.5 \times 10^6\,\text{mm}^4$

FE Reference Handbook 10.6, p. 96

Loaded at only two points → two-force member, so the forces are equal, opposite, and directed along the member axis. That immediately fixes the force direction and simplifies the frame analysis. The "bending with perpendicular pin forces" choice describes a multi-force member. The "carries no force at all" choice is false (it can carry axial force). The "force direction cannot be determined" choice is wrong because the direction IS known.

1. Two load points → two-force member.
2. Equilibrium makes the forces collinear with the two pins.
3. So the force acts along the member axis.