Free PE Civil — Structural practice problems

Ten real questions from our PE Civil: Structural bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

9 with figures10 knowledge areasFull worked solutions

1
medium
Forces & Load Effects — diagrams, reactions, determinacy
A simply supported beam of span $L=24$ ft carries a uniform load $w=2$ k/ft over its full length plus a concentrated load $P=12$ k located $a=6$ ft from support A. The maximum shear magnitude in the beam is most nearly:
1. 27 kip
2. 33 kip
3. 9 kip
4. 24 kip
Show the answer & worked solution
Answer: B.33 kip
Reactions: $R_A=wL/2+Pb/L=33$ k, $R_B=wL/2+Pa/L=27$ k. Max shear $=\max(R_A,R_B)=33$ kip, just inside that support. Distractors: took the smaller reaction; used the UDL-only shear $wL/2$ ; and used only the point-load reaction.
2
medium
Component Design — RC flexure and shear
A rectangular RC beam ( $b=14$ in, $d=21.5$ in, $A_s=4$ in², $f_y=60$ ksi, $f'_c=4$ ksi) is tension-controlled, so $\phi=0.90$ (use the 0.85 stress-block factor). The factored demand is $M_u=320$ kip-ft. The design flexural strength $\phi M_n$ is most nearly:
1. 341.6 kip-ft
2. 387 kip-ft
3. 284.7 kip-ft
4. 379.6 kip-ft
Show the answer & worked solution
Answer: A.341.6 kip-ft
$a=5.04$ in; $M_n=A_s f_y(d-a/2)/12=379.6 kip-ft$ . $\phi M_n=0.90(379.6)=341.6 kip-ft$ , which ≥ $M_u=320$ kip-ft (adequate). Distractors: omit $\phi$ ; use the shear $\phi=0.75$ ; and apply $\phi$ but use lever arm $d$ (no $a/2$ ).
3
medium
Component Design — bolted, welded, and anchored connections (AISC)
A slip-critical connection has 4 bolts on a Class A faying surface ( $\mu = 0.3$ , $D_u = 1.13$ , $h_f = 1$ , pretension $T_b = 28$ kip, slip planes $n_s = 1$ ). Using $R_n = \mu D_u h_f T_b n_s$ per bolt and LRFD $\phi = 1.00$ , the design slip resistance of the connection is most nearly:
1. 28.5 kip
2. 38 kip
3. 9.5 kip
4. 33.6 kip
Show the answer & worked solution
Answer: B.38 kip
Per bolt $R_n = \mu D_u h_f T_b n_s = 0.3(1.13)(1)(28)(1) = 9.49$ kip; with $\phi = 1.00$ the connection design slip resistance $= 4(9.49) = 38$ kip. Distractors give one bolt only, omit $D_u$ (treat it as 1.0), and apply the bearing-bolt $\phi = 0.75$ to a slip-critical check.
4
medium
Deflection — methods, formulas, serviceability
A cantilever beam of length $L=10$ ft carries a uniform service load $w=1.5$ k/ft over its full length. With $E=29000$ ksi and $I=250$ in $^4$ , the tip deflection $\delta=\dfrac{wL^4}{8EI}$ is most nearly:
1. 0.447 in
2. 0.0466 in
3. 0.000259 in
4. 1.19 in
Show the answer & worked solution
Answer: A.0.447 in
$\delta=\dfrac{w L^4}{8EI}=\dfrac{0.125(120)^4}{8(29000)(250)}=0.447 in$ with $L=120$ in, $w=0.125$ k/in. Distractors: span left in feet; used the simply-supported $5wL^4/384EI$ ; and used the tip-point-load $PL^3/3EI$ with $P=wL$ .
5
medium
Component Design — shallow foundations
A continuous (strip) footing $B = 4$ ft wide bears at depth $D_f = 3$ ft in a soil with cohesion $c = 0.4$ ksf and total unit weight $\gamma = 0.12 \text{ kcf} = 120$ pcf. The Terzaghi bearing-capacity factors are $N_c = 17.7$ , $N_q = 7.4$ , $N_\gamma = 5$ . The ultimate bearing capacity $q_{ult} = cN_c + qN_q + 0.5\,\gamma B N_\gamma$ is most nearly:
1. 8.28 ksf
2. 30.5 ksf
3. 12.1 ksf
4. 10.9 ksf
Show the answer & worked solution
Answer: D.10.9 ksf
$q = \gamma D_f = 0.12(3) = 0.36$ ksf. $q_{ult} = 0.4(17.7) + 0.36(7.4) + 0.5(0.12)(4)(5) = 10.9 ksf$ . Distractors drop the surcharge $qN_q$ term, forget the 0.5 on the self-weight term, and use the depth $D_f$ in place of the surcharge stress $\gamma D_f$ .
6
medium
Loads & Load Applications — gravity loads, tributary areas, load paths
A girder carries a tributary strip 12 ft wide of a floor with dead pressure $D=90$ psf and live pressure $L=50$ psf. Using the LRFD combination $1.2D+1.6L$ , the factored uniform line load on the girder is most nearly:
1. 2.26 k/ft
2. 1.68 k/ft
3. 2.35 k/ft
4. 2.45 k/ft
Show the answer & worked solution
Answer: A.2.26 k/ft
Factored pressure $=1.2(90)+1.6(50)=188$ psf; times the 12 ft width gives $w_u=2.26$ k/ft. Distractors use the unfactored service load, swap the 1.2/1.6 factors, and apply a single 1.4 factor to (D+L).
7
medium
Component Design — reinforced and unreinforced masonry (TMS ASD)
A reinforced CMU wall strip has vertical steel $A_s=0.31$ in² with an allowable steel tensile stress $F_s=32000$ psi. The flexural lever-arm factor is $j=0.9$ and the effective depth is $d=3.81$ in. By steel-controlled TMS ASD, the allowable moment $M=A_s F_s\,j\,d$ is most nearly:
1. 34016 lb-ft
2. 3150 lb-ft
3. 2835 lb-ft
4. 1417 lb-ft
Show the answer & worked solution
Answer: C.2835 lb-ft
$M=A_s F_s\,j\,d=0.31(32000)(0.9)(3.81)=34016$ in-lb $=2835 lb-ft$ . Distractors omit $j$ (full $d$ lever arm), forget the in→ft conversion (report the in-lb magnitude, 12× too large, tagged lb-ft), and use a $d/2$ lever arm by analogy to the concrete stress block.
8
medium
Materials & Material Properties — soil classification and properties
A laboratory test on a silty clay sample gives a specific gravity of solids $G_s = 2.7$ , a gravimetric water content $w = 18\%$ , and a degree of saturation $S = 90\%$ . The void ratio is most nearly:
1. 0.2
2. 0.486
3. 0.54
4. 0.437
Show the answer & worked solution
Answer: C.0.54
From $S\,e = w\,G_s$ : $e = \dfrac{w\,G_s}{S} = \dfrac{0.18\times2.7}{0.9} = 0.54$ . Distractors omit the $\div S$ (treats the soil as saturated), MULTIPLY by $S$ instead of dividing, and drop $G_s$ entirely (using $w/S$ ).
9
medium
Component Design — sawn lumber and glulam (NDS ASD)
A simply supported sawn beam of actual cross-section $b=1.5$ in by $d=9.25$ in spans $L=14$ ft and carries a uniform load $w=70$ lb/ft (total). The maximum moment is $M=wL^2/8$ . The adjusted allowable bending stress is $F_b'=1100$ psi. The actual maximum bending stress $f_b=M/S$ is most nearly:
1. 5933 psi
2. 962 psi
3. 481 psi
4. 80 psi
Show the answer & worked solution
Answer: B.962 psi
$M=wL^2/8=70(14)^2/8=1715$ lb-ft; $S=bd^2/6=1.5(9.25)^2/6=21.4 in³$ . $f_b=M/S=1715(12)/21.4=962 psi$ (vs $F_b'=1100$ psi, OK). Distractors: forget the lb-ft→lb-in conversion; use $wL^2/16$ ; and swap $b$ and $d$ in $S=bd^2/6$ (compute about the weak axis).
10
medium
Materials & Material Properties — properties and test methods
A reinforced-concrete section is transformed using a modular ratio. The steel modulus is $E_s = 29000$ ksi and the concrete strength is $f'_c = 4000$ psi (use $E_c = 57000\sqrt{f'_c}$ , psi). The modular ratio $n = E_s/E_c$ is most nearly:
1. 13.89
2. 8.04
3. 9.67
4. 4.02
Show the answer & worked solution
Answer: B.8.04
$E_c = 57000\sqrt{4000} = 3600000$ psi $= 3605$ ksi, so $n = E_s/E_c = 29000/3605 = 8.04$ . Distractors use the wrong coefficient ( $E_c = 33000\sqrt{f'_c}$ , the $w_c^{1.5}$ base value), use a rounded flat $E_c = 3000$ ksi, and halve the steel modulus ( $E_s/2$ ). Every option is a single/low-double-digit ratio.

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