FE Civil Structural Engineering practice problems

For a plane frame, compare 3m + r to 3j + c. Two fixed supports give $r = 3 + 3 = 6$ reaction components. Then $3m + r = 3(6) + 6 = 24$, and $3j + c = 3(6) + 0 = 18$. The degree of indeterminacy is $24 - 18 = 6$. The value 3 results from counting only one fixed support ($r = 3$): $21 - 18 = 3$. The value 4 results from treating each base as a pin ($r = 4$): $22 - 18 = 4$. The value 0 wrongly applies the truss formula $m + r$ vs 2j.

1. Count reactions (two fixed supports):
   $r = 3 + 3 = 6$
2. Frame left side:
   $3m + r = 3(6) + 6 = 24$
3. Frame right side:
   $3j + c = 3(6) + 0 = 18$
4. Degree of indeterminacy:
   $24 - 18 = 6$

FE Reference Handbook 10.6, p. 277

Each pin provides 2 reaction components, so $r = 2 + 2 = 4$. A planar beam has 3 independent equilibrium equations ($\Sigma F_x = 0$, $\Sigma F_y = 0$, $\Sigma M = 0$). The external degree of indeterminacy is $r - 3 = 4 - 3 = 1$. Two pins introduce one redundant horizontal restraint that equilibrium alone cannot resolve. The "stable and statically determinate" choice would require exactly $r = 3$, which is the familiar pin-plus-roller case. The "unstable mechanism" choice would require $r < 3$. The "indeterminate to the 3rd degree" choice overcounts the redundancy, since only one extra reaction component exists beyond the three equilibrium equations.

1. Reaction components (two pins):
   $r = 2 + 2 = 4$
2. Equilibrium equations for a planar beam:
   3
3. External indeterminacy:
   $r - 3 = 4 - 3 = 1$

FE Reference Handbook 10.6, p. 277

ASD uses unfactored loads with the 0.75 companion factor: $D + 0.75L + 0.75S = 40 + 0.75(30) + 0.75(24) = 40 + 22.5 + 18 = 80.5$ kips. The 94 kips option is the plain sum $D + L + S = 40 + 30 + 24$, forgetting the 0.75 factors. The 84 kips option is the LRFD combination $1.2D + 1.6L = 48 + 48$, which is wrong because the problem specifies ASD. The 62.5 kips option drops the snow term and uses only $D + 0.75L = 40 + 22.5$.

1. ASD combination:
   D + 0.75L + 0.75S
2. Substitute:
   $40 + 0.75(30) + 0.75(24) = 40 + 22.5 + 18$
3. Service load:
   $= 80.5 \text{ kips}$

FE Reference Handbook 10.6, p. 278

Combination 1 is $1.4D = 1.4(100) = 140$ kips. Combination 2 is $1.2D + 1.6L = 1.2(100) + 1.6(10) = 120 + 16 = 136$ kips. Because dead load dominates and live load is small, Combination 1 (140) produces the larger factored load and controls. This is the case the 1.4D combination exists for. The 136 kips option is Combination 2, which is smaller here. The 120 kips option keeps only the 1.2D term and drops live load. The 110 kips option is the unfactored service load $D + L$.

1. Combination 1:
   $1.4D = 1.4(100) = 140 \text{ kips}$
2. Combination 2:
   $1.2D + 1.6L = 1.2(100) + 1.6(10) = 136 \text{ kips}$
3. Controlling: Combination 1 at 140 kips (dead load dominates).

FE Reference Handbook 10.6, p. 278

The IL for $R_A$ is a straight line from 1.0 at A to 0 at B, so the ordinate at distance $x$ is $(L - x)/L$. Over the loaded region (0 to 10 ft) the ordinate runs from 1.0 down to 0.5. The area under the IL there is the trapezoid $\tfrac{1.0 + 0.5}{2}(10) = 7.5$. For a distributed load, $R_A = w \times \text{area} = 3 \times 7.5 = 22.5$ kips. Check by statics: the half-span load is $3(10) = 30$ kips acting at 5 ft from A, so $R_A = 30(20 - 5)/20 = 22.5$ kips. The 30 kips option is the total applied load, not the reaction. The 15 kips option is the right reaction $R_B$. The 7.5 option is the IL area without multiplying by $w$.

1. IL ordinate at A and at midspan:
   $\eta_A = 1.0, \quad \eta_{mid} = \frac{20 - 10}{20} = 0.5$
2. Area under IL over the loaded half:
   $A_{IL} = \frac{1.0 + 0.5}{2}(10) = 7.5$
3. Left reaction:
   $R_A = w \times A_{IL} = 3 \times 7.5 = 22.5 \text{ kips}$

FE Reference Handbook 10.6, p. 274

The IL for shear at midspan is antisymmetric: a positive triangle on the right half and an equal negative triangle on the left half. For a uniform load over the whole span, the shear equals $w$ times the net area under the IL. Because the positive and negative areas are equal in magnitude, they cancel and the net shear at midspan is zero. This matches the familiar result that shear is zero at the point of maximum moment in a symmetric beam. The "maximum positive" and "maximum negative" choices assume the load sits on only one half of the span. The choice equating shear to the midspan moment divided by L confuses the shear-moment relationship.

1. Midspan shear IL: positive triangle (right half) and negative triangle (left half) of equal area.
2. Net area for a full-span uniform load:
   $A^{+} + A^{-} = 0$
3. Shear:
   $V_{mid} = w \times 0 = 0$

FE Reference Handbook 10.6, p. 274

ACI requires $A_{s,\min}$ to be the larger of $\frac{3\sqrt{f_c'}}{f_y} b_w d$ and $\frac{200}{f_y} b_w d$, with stresses in psi. First term: $\frac{3\sqrt{4000}}{60000}(12)(20) = \frac{189.7}{60000}(240) = 0.759$ in$^2$. Second term: $\frac{200}{60000}(12)(20) = \frac{200}{60000}(240) = 0.80$ in$^2$. The larger value governs, so $A_{s,\min} = 0.80$ in$^2$. The 0.76 in$^2$ choice is the first term alone, but the $200/f_y$ term controls for $f_c' \leq 4{,}444$ psi. The 0.40 in$^2$ choice drops $b_w$. The 1.60 in$^2$ choice doubles the area by a unit slip.

1. First expression:
   $\frac{3\sqrt{4000}}{60000}(12)(20) = \frac{189.7}{60000}(240) = 0.759 \text{ in}^2$
2. Second expression:
   $\frac{200}{60000}(12)(20) = 0.80 \text{ in}^2$
3. Governing minimum (larger value):
   $A_{s,\min} = 0.80 \text{ in}^2$

FE Reference Handbook 10.6, p. 281

$V_s = \frac{A_v f_y d}{s} = \frac{0.40 \times 60 \times 22}{8} = \frac{528}{8} = 66$ kips. The 33 kips option uses only one stirrup leg ($A_v = 0.20$) instead of the two-leg area. The 132 kips option halves the spacing to $s = 4$ in. by mistake. The 30 kips option drops $d$ and uses an incorrect lever term.

1. Stirrup shear capacity:
   $V_s = \frac{A_v f_y d}{s} = \frac{0.40 \times 60 \times 22}{8}$
2. Compute:
   $V_s = \frac{528}{8} = 66 \text{ kips}$

FE Reference Handbook 10.6, p. 282

$A_g = 12 \times 18 = 216$ in$^2$. Inside the brackets: $0.85(4)(216 - 4.0) + 4.0(60) = 3.4(212) + 240 = 720.8 + 240 = 960.8$ kips. For a tied column apply both factors: $\phi P_n = 0.80(0.65)(960.8) = 0.52 \times 960.8 = 500$ kips. The 961 kips choice omits both the 0.80 and $\phi$ factors. The 625 kips choice applies only $\phi = 0.65$. The 769 kips choice applies only the 0.80 factor.

1. Gross area:
   $A_g = 12 \times 18 = 216 \text{ in}^2$
2. Inside brackets:
   $0.85(4)(212) + 4.0(60) = 720.8 + 240 = 960.8 \text{ kips}$
3. Apply factors:
   $\phi P_n = 0.80(0.65)(960.8) = 0.52 \times 960.8 = 500 \text{ kips}$

FE Reference Handbook 10.6, p. 284

ACI limits the longitudinal steel ratio to $0.01 \leq \rho_g \leq 0.08$ (1 to 8 percent). A 10 percent ratio exceeds the 8 percent maximum. The upper limit prevents reinforcement congestion that makes it impossible to place and consolidate concrete around the bars and at lap splices. The "acceptable, no upper limit" choice is wrong because an upper limit exists. The "below the 1 percent minimum" choice has the direction backwards: 10 percent is far above, not below, the minimum. The "only a problem for spiral columns" choice is wrong because the limits apply to both tied and spiral columns.

1. ACI limits:
   $0.01 \leq \rho_g \leq 0.08$
2. Proposed ratio:
   $\rho_g = 0.10 > 0.08$
3. Exceeds the 8 percent maximum, which causes bar congestion and placement problems.

FE Reference Handbook 10.6, p. 284

With full bracing, $M_n = M_p = F_y Z_x = 50 \times 100 = 5{,}000$ kip-in $= 416.7$ kip-ft. For ASD, divide by $\Omega_b = 1.67$: $M_n/\Omega_b = 416.7/1.67 = 249.5 \approx 250$ kip-ft. The 417 kip-ft choice is the nominal $M_n$ without applying $\Omega_b$. The 375 kip-ft choice is the LRFD design strength $\phi_b M_n = 0.90 \times 416.7$, but the problem asks for ASD. The 278 kip-ft choice uses an incorrect $\Omega_b = 1.50$.

1. Plastic moment:
   $M_p = F_y Z_x = 50 \times 100 = 5{,}000 \text{ kip-in} = 416.7 \text{ kip-ft}$
2. ASD allowable strength:
   $\frac{M_n}{\Omega_b} = \frac{416.7}{1.67} = 250 \text{ kip-ft}$

FE Reference Handbook 10.6, p. 287

Web area $A_w = d \times t_w = 18 \times 0.30 = 5.40$ in$^2$. Nominal shear $V_n = 0.6 F_y A_w C_{v1} = 0.6(36)(5.40)(1.0) = 116.6$ kips. With $\phi_v = 1.00$: $\phi_v V_n = 117$ kips. The 194 kips choice omits the 0.6 coefficient ($36 \times 5.40 = 194$). The 162 kips choice uses $0.5 F_y$ instead of $0.6 F_y$ on a larger web area. The 105 kips choice applies $\phi_v = 0.90$ instead of 1.00 ($0.9 \times 116.6 = 105$).

1. Web area:
   $A_w = d \times t_w = 18 \times 0.30 = 5.40 \text{ in}^2$
2. Nominal shear:
   $V_n = 0.6(36)(5.40)(1.0) = 116.6 \text{ kips}$
3. Design shear ($\phi_v = 1.00$):
   $\phi_v V_n = 117 \text{ kips}$

FE Reference Handbook 10.6, p. 287

Euler stress: $F_e = \frac{\pi^2 E}{(KL/r)^2} = \frac{\pi^2 (29{,}000)}{140^2} = \frac{286{,}218}{19{,}600} = 14.6$ ksi. For elastic buckling, $F_{cr} = 0.877 F_e = 0.877(14.6) = 12.8$ ksi. The 14.6 ksi choice is $F_e$ itself, which forgets the 0.877 elastic-buckling reduction. The 25.9 ksi choice misapplies the inelastic formula $0.658^{F_y/F_e} F_y$. The 9.6 ksi choice applies the 0.658 base directly.

1. Euler buckling stress:
   $F_e = \frac{\pi^2 (29{,}000)}{140^2} = \frac{286{,}218}{19{,}600} = 14.6 \text{ ksi}$
2. Elastic critical stress:
   $F_{cr} = 0.877 F_e = 0.877(14.6) = 12.8 \text{ ksi}$

FE Reference Handbook 10.6, p. 287

The dividing line between inelastic and elastic column buckling is $4.71\sqrt{E/F_y} = 4.71\sqrt{29{,}000/50} = 4.71\sqrt{580} = 4.71(24.08) = 113.4$. Since the column slenderness $KL/r = 90$ is less than 113.4, the column buckles inelastically and uses $F_{cr} = 0.658^{F_y/F_e} F_y$. The "elastic buckling, because $KL/r = 90 > 113$" choice has the inequality backwards. The "$KL/r > 50$ always means elastic" choice uses an arbitrary cutoff of 50, which is not the AISC threshold. The "cannot be determined without the cross-sectional area" choice is wrong because the regime depends only on $KL/r$, $E$, and $F_y$, not on area.

1. Threshold slenderness:
   $4.71\sqrt{\frac{E}{F_y}} = 4.71\sqrt{\frac{29{,}000}{50}} = 4.71(24.08) = 113.4$
2. Compare:
   $\frac{KL}{r} = 90 < 113.4$
3. Since $KL/r$ is below the threshold, inelastic buckling governs.

FE Reference Handbook 10.6, p. 287

For a staggered (zig-zag) failure path, the net width is $b_g - \Sigma d_h + \Sigma \frac{s^2}{4g}$. Each hole diameter is $d_h = d_b + 1/8 = 7/8 + 1/8 = 1.0$ in. The three-hole path crosses 3 holes and has 2 diagonal segments, so add $\frac{s^2}{4g}$ twice: net width $= 10 - 3(1.0) + 2\left(\frac{1.5^2}{4(2.5)}\right) = 10 - 3 + 2(0.225) = 10 - 3 + 0.45 = 7.45$ in. Net area $= 7.45 \times 0.375 = 2.79$ in$^2$. The 3.00 in^2 choice takes the straight transverse cut through only the two in-line holes, (10 - 2.0)(0.375), ignoring the third hole and the stagger. The 2.93 in^2 choice adds the two $s^2/4g$ terms but forgets the 1/8 in. hole allowance, using $d_h = 0.875$: (10 - 3(0.875) + 0.45)(0.375). The 2.63 in^2 choice deducts all three holes but omits the $s^2/4g$ correction: (10 - 3.0)(0.375).

1. Effective hole diameter:
   $d_h = \frac{7}{8} + \frac{1}{8} = 1.0 \text{ in.}$
2. Net width along the staggered path (3 holes, 2 diagonals):
   $b_n = 10 - 3(1.0) + 2\left(\frac{1.5^2}{4(2.5)}\right) = 10 - 3 + 0.45 = 7.45 \text{ in.}$
3. Net area:
   $A_n = 7.45 \times 0.375 = 2.79 \text{ in}^2$

FE Reference Handbook 10.6, p. 288

The shear lag factor is $U = 1 - \bar{x}/L = 1 - 1.20/9.0 = 1 - 0.1333 = 0.867$. The effective net area is $A_e = U A_n = 0.867 \times 4.00 = 3.47$ in$^2$. The 4.00 in$^2$ choice assumes $U = 1.0$, which only applies when all cross-sectional elements are connected. The 0.87 in$^2$ choice reports $U$ itself, not $A_e$. The 3.20 in$^2$ choice subtracts $\bar{x} \times A_n / L$ incorrectly or uses $U = 0.80$.

1. Shear lag factor:
   $U = 1 - \frac{\bar{x}}{L} = 1 - \frac{1.20}{9.0} = 0.867$
2. Effective net area:
   $A_e = U A_n = 0.867 \times 4.00 = 3.47 \text{ in}^2$

FE Reference Handbook 10.6, p. 289

Cantilever under UDL: $\delta = wL^4/8EI = 6(4^4)/(8 \times 30{,}000) = 1536/240{,}000 = 0.0064\text{ m} = 6.4\text{ mm}$. The 25.6 mm choice uses the wrong denominator (2EI instead of 8EI); the 1.6 mm choice divides the correct answer by 4 (e.g. using L² instead of L⁴ scaling).

1. Standard case: cantilever, UDL.
2. Apply:
   $\delta = \frac{wL^4}{8EI} = \frac{6(4)^4}{8(30{,}000)}$
3. Compute:
   $\delta = \frac{1536}{240{,}000} = 0.0064\text{ m} = 6.4\text{ mm}$

Fixed-fixed beam, central point load: the fixed-end moment is $PL/8 = 24(6)/8 = 18\text{ kN·m}$. The 36 kN·m choice uses PL/4 (the simple-span midspan moment). The 144 kN·m choice forgets to divide entirely.

1. Standard result for a fixed-fixed beam under a central point load.
2. Apply:
   $M_{FE} = \frac{PL}{8} = \frac{24(6)}{8}$
3. Compute:
   $M_{FE} = \frac{144}{8} = 18\text{ kN·m}$