Free PE Civil — Structural practice problems

Ten real questions from our PE Civil: Structural bank, drawn from 10 of its 14 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

9 with figures10 topicsFull worked solutions

1
hard
Component Design — reinforced and unreinforced masonry (TMS ASD)
A fully grouted 8-in. CMU wall (actual thickness 7.625 in.) spans vertically and is reinforced with No. 5 bars at 12 in. on center, centered in the wall, giving $A_s = 0.31$ in² per foot and $d = 3.81$ in. The masonry has $f'_m = 2{,}000$ psi and $E_m = 1{,}800$ ksi; the steel has $E_s = 29{,}000$ ksi. The allowable stresses are $F_b = 900$ psi in the masonry and $F_s = 32{,}000$ psi in the steel. By allowable stress design, the allowable out-of-plane moment per foot of wall is most nearly:
Figure — drag or pinch to move around
1. 1970 lb-ft
2. 2120 lb-ft
3. 2420 lb-ft
4. 2760 lb-ft
Show the answer & worked solution
Answer: B.2120 lb-ft
Cracked transformed section, using the service-flexure relations of the PE Civil Reference Handbook §4.3.2.3 with the masonry modulus $E_m$ in place of $E_c$ : $n = E_s/E_m = 29000/1800 = 16.11$ ; $\rho = A_s/(bd) = 0.31/(12 \times 3.81) = 0.00678$ ; $k = \sqrt{(\rho n)^2 + 2\rho n} - \rho n = 0.371$ ; $j = 1 - k/3 = 0.876$ . Steel at $F_s$ : $M_s = A_s F_s\,jd = 0.31(32000)(0.876)(3.81) = 33120$ in-lb, or 2760 lb-ft. Masonry at $F_b$ : from $f_b = 2M/(kjbd^2)$ , $M_m = F_b\,kjbd^2/2 = 900(0.371)(0.876)(12)(3.81)^2/2 = 25470$ in-lb, or 2120 lb-ft. The masonry reaches its allowable stress first, so the allowable moment is 2120 lb-ft per foot of wall. Distractors: put the compression resultant at $kd/2$ , a rectangular-block habit, so $j = 1 - k/2$ instead of $1 - k/3$ (1970 lb-ft); used the full depth $d$ as the lever arm ( $j = 1$ ) (2420 lb-ft); and took the steel-controlled moment (2760 lb-ft) without checking that the masonry limit is lower.
2
medium
Forces & Load Effects — diagrams, reactions, determinacy
The beam shown is pinned at A and supported on a roller at B, $L=16$ ft apart, and overhangs B by $c=6$ ft. It carries a uniform load $w=3$ k/ft over its entire 22-ft length and a concentrated load $P=3$ k at the free end. Neglect self-weight. The maximum shear magnitude in the beam is most nearly:
Figure — drag or pinch to move around
1. 27.4 kip
2. 28.5 kip
3. 31.9 kip
4. 49.5 kip
Show the answer & worked solution
Answer: B.28.5 kip
Superpose the overhanging-beam cases of handbook §4.1.7 (UDL over the whole length, and a load at the end of the overhang), or simply take $\Sigma M_A=0$ : $R_B=\dfrac{w(L+c)^2/2+P(L+c)}{L}=\dfrac{3(22)^2/2+3(22)}{16}=49.5$ k, and $R_A=w(L+c)+P-R_B=69-49.5=19.5$ k. Shear diagram: $V=19.5$ k just right of A, falling at $w$ to $19.5-3(16)=-28.5$ k just left of B; the reaction lifts it to $wc+P=21$ k just right of B, which then falls to $P=3$ k at the tip. The largest magnitude is 28.5 kip, just left of B. Distractors: took the larger reaction ( $R_B=49.5$ k) as the maximum shear, the simple-span habit, but at an overhang support $R_B$ is shared between the shear just left of B and the shear just right of it; left the end load $P$ out, so the shear just left of B is the uniform load's alone (27.4 k) and misses $P$ 's share there, $Pc/L=1.13$ k; and placed the UDL resultant $w(L+c)$ at $L/2$ from A instead of at the middle of the loaded length, $(L+c)/2$ .
3
medium
Component Design — RC flexure and shear
A rectangular reinforced concrete beam is $b=12$ in wide with an effective depth $d=20$ in (overall depth $h=22.5$ in). It is reinforced in tension with four No. 8 Grade 60 bars ( $f_y=60$ ksi), and $f'_c=4{,}000$ psi. The design flexural strength $\phi M_n$ is most nearly:
Figure — drag or pinch to move around
1. 218 kip-ft
2. 251 kip-ft
3. 279 kip-ft
4. 307 kip-ft
Show the answer & worked solution
Answer: B.251 kip-ft
Handbook §4.3.1: $A_s=4(0.79)=3.16$ in². Stress block $a=\dfrac{A_s f_y}{0.85f'_c b}=\dfrac{3.16(60)}{0.85(4)(12)}=4.65$ in. Strain check with $\beta_1=0.85$ (handbook §4.3.2.1): $c=a/\beta_1=5.47$ in and $\varepsilon_t=0.003\,\dfrac{d-c}{c}=0.003\,\dfrac{20-5.47}{5.47}=0.0080$ , above the 0.005 tension-controlled limit, so ACI 318 gives $\phi=0.90$ . $M_n=A_s f_y\,(d-a/2)=3.16(60)(20-2.32)/12=279.3$ kip-ft, so $\phi M_n=0.90(279.3)=251$ kip-ft. Distractors: left out $\phi$ and reported $M_n$ ; used $(d-a)$ for the lever arm instead of $(d-a/2)$ ; and read the bar diameter (1.000 in) from the table instead of its area (0.79 in²), so $A_s=4.00$ in².
4
medium
Component Design — bolted, welded, and anchored connections (AISC)
Two plates are lapped and joined by four 3/4-in. Group A bolts in standard holes, designed as a slip-critical connection with Class A faying surfaces and no fillers. AISC 360 data: $\mu=0.30$ , $D_u=1.13$ , minimum bolt pretension $T_b=28$ kips. The LRFD design slip resistance of the connection is most nearly:
Figure — drag or pinch to move around
1. 28.5 kips
2. 33.6 kips
3. 38.0 kips
4. 75.9 kips
Show the answer & worked solution
Answer: C.38.0 kips
AISC 360 §J3.8: $\phi R_n=\phi\,\mu D_u h_f T_b n_s$ per bolt, with $h_f=1.0$ (no fillers), $n_s=1$ (a lap joint has one faying surface) and $\phi=1.00$ (standard holes). Per bolt: $0.30(1.13)(1.0)(28)(1)=9.49$ kips. $\phi R_n=4(9.49)=38.0$ kips. Distractors: left out $D_u$ (took it as 1.0); applied the bolt-shear factor $\phi=0.75$ instead of $\phi=1.00$ for slip in standard holes; and counted two slip planes because two plates are joined, but a lap has one faying surface ( $n_s=1$ ).
5
medium
Deflection — methods, formulas, serviceability
A steel cantilever beam ( $E=29{,}000$ ksi, $I=300$ in $^4$ ) projects $L=10$ ft from a fixed support. It carries a uniform service load $w=1.5$ k/ft over its full length and a concentrated service load $P=3$ k at its free end. Neglecting shear deformation, the deflection at the free end is most nearly:
Figure — drag or pinch to move around
1. 0.372 in
2. 0.509 in
3. 0.571 in
4. 1.07 in
Show the answer & worked solution
Answer: C.0.571 in
Superpose handbook §4.1.7 case 19 (cantilever, uniform load: $\Delta=wl^4/8EI$ ) and case 22 (load at the free end: $\Delta=Pl^3/3EI$ ), with $l=120$ in and $w=1.5/12=0.125$ k/in: $\Delta=\dfrac{0.125(120)^4}{8(29{,}000)(300)}+\dfrac{3(120)^3}{3(29{,}000)(300)}=0.372+0.199=0.571$ in. Distractors: left out the end load ( $wL^4/8EI$ only, 0.372 in); replaced the uniform load by its resultant $wL$ at mid-length (giving $5wL^4/48EI$ ), which is valid for reactions but not for deflection; and swapped the coefficients ( $wL^4/3EI+PL^3/8EI$ ).
6
medium
Component Design — shallow foundations
A continuous (strip) footing $B=6$ ft wide is founded at depth $D_f=2$ ft in a soil with cohesion $c=150$ psf, friction angle $\phi=24^\circ$ , and unit weight $\gamma=120$ pcf. The same soil lies above and below the base, and groundwater is far below the footing. Using the bearing capacity factors in the PE Civil Reference Handbook, the ultimate bearing capacity $q_{ult}$ is most nearly:
Figure — drag or pinch to move around
1. 6.28 ksf
2. 8.58 ksf
3. 9.70 ksf
4. 12.0 ksf
Show the answer & worked solution
Answer: B.8.58 ksf
Handbook §3.4.2 (strip footing): $q_{ult}=cN_c+qN_q+0.5\gamma B_fN_\gamma$ with $q=\gamma_aD_f$ . For $\phi=24^\circ$ the handbook's Bearing Capacity Factors table gives $N_c=19.3$ , $N_q=9.6$ , $N_\gamma=9.4$ . Here $q=0.12(2)=0.24$ ksf, so $q_{ult}=0.15(19.3)+0.24(9.6)+0.5(0.12)(6)(9.4)=2.90+2.30+3.38=8.58$ ksf. Distractors: left out the surcharge term $qN_q$ ; dropped the 0.5 on the self-weight term ( $\gamma B N_\gamma$ ); and applied the handbook's square-footing shape factors ( $B_f/L_f=1$ ), but a strip has $B_f/L_f\to0$ and every shape factor is 1.
7
medium
Loads & Load Applications — gravity loads, tributary areas, load paths
Interior floor girders, spaced 12 ft on center, span 36 ft and carry a one-way slab; each girder supports a single floor. The floor dead load (slab, framing, and superimposed) is 90 psf and the unreduced live load is 50 psf. Live-load reduction per ASCE 7 is permitted, with $K_{LL}=2$ . Using the LRFD combination $1.2D+1.6L$ , the factored uniform load on one girder is most nearly:
Figure — drag or pinch to move around
1. 1.54 k/ft
2. 1.72 k/ft
3. 2.03 k/ft
4. 2.26 k/ft
Show the answer & worked solution
Answer: C.2.03 k/ft
Tributary width = girder spacing = 12 ft, so $A_T=12(36)=432$ ft² and $K_{LL}A_T=2(432)=864$ ft² $\ge 400$ ft²: the reduction applies. $L=L_0\left(0.25+\dfrac{15}{\sqrt{K_{LL}A_T}}\right)=50\left(0.25+\dfrac{15}{\sqrt{864}}\right)=50(0.760)=38$ psf, above the $0.50L_0=25$ psf limit for a member supporting one floor. $w_u=[1.2(90)+1.6(38)](12)/1000=2.03$ k/ft. Distractors: skipped the live-load reduction (used $L_0=50$ psf); applied the reduction factor 0.760 to the dead load as well; and left the load unfactored ( $D+L$ ).
8
medium
Materials & Material Properties — soil classification and properties
A Shelby-tube sample of silty clay has a total (moist) unit weight of 118 pcf and a water content of 22%. The specific gravity of the solids is $G_s = 2.70$ . The in-situ void ratio is most nearly:
Figure — drag or pinch to move around
1. 0.428
2. 0.594
3. 0.742
4. 0.908
Show the answer & worked solution
Answer: C.0.742
Weight–volume relationships (PE Civil Reference Handbook §3.8.3): $\gamma_d = \dfrac{\gamma_t}{1+w} = \dfrac{118}{1+0.22} = 96.7$ pcf, and $\gamma_d = \dfrac{G_s\gamma_w}{1+e}$ gives $e = \dfrac{G_s\gamma_w}{\gamma_d} - 1 = \dfrac{2.70(62.4)}{96.7} - 1 = 0.742$ . Check: $S = wG_s/e = 0.80$ , so the sample is not saturated. Distractors: used the total unit weight as if it were the dry unit weight (0.428); assumed the sample saturated and took $e = wG_s$ (0.594); and solved the saturated-unit-weight relation $\gamma_{sat} = (G_s + e)\gamma_w/(1+e)$ with the total unit weight (0.908).
9
medium
Component Design — sawn lumber and glulam (NDS ASD)
Sawn-lumber 2×10 floor joists (actual size 1.5 in. × 9.25 in.) at 16 in. on center span 14 ft, simply supported. The floor dead load, including the joists, is 12 psf and the floor live load is 40 psf. The adjusted allowable bending stress is $F_b' = 1{,}100$ psi. Under the full dead plus live load, the maximum bending stress in a joist is most nearly:
Figure — drag or pinch to move around
1. 206 psi
2. 612 psi
3. 715 psi
4. 953 psi
Show the answer & worked solution
Answer: D.953 psi
Tributary width $= 16/12 = 1.333$ ft, so $w = (12 + 40)(1.333) = 69.3$ lb/ft. $M = wL^2/8 = 69.3(14)^2/8 = 1699$ lb-ft. $S = bd^2/6 = 1.5(9.25)^2/6 = 21.39$ in³. $f_b = M/S = 1699(12)/21.39 = 953$ psi, within $F_b' = 1{,}100$ psi (OK). Distractors: took a 1-ft tributary width, using the psf loads as lb/ft (715 psi); used the nominal 2 in. × 10 in. instead of the actual dressed size, $S = 33.3$ in³ (612 psi); and divided the moment by $I = bd^3/12$ instead of $S$ (206 psi).
10
medium
Materials & Material Properties — properties and test methods
A precast concrete girder acts compositely with a cast-in-place slab. The girder is normal-weight concrete with a unit weight of 145 pcf and $f'_c = 6{,}000$ psi; the slab is lightweight concrete with a unit weight of 115 pcf and $f'_c = 4{,}000$ psi. For a transformed-section analysis by ACI 318, the modular ratio of the girder concrete to the slab concrete is most nearly:
1. 1.22
2. 1.54
3. 1.73
4. 2.12
Show the answer & worked solution
Answer: C.1.73
By the PE Civil Reference Handbook §4.3.2.1, $E_c = 33\,w_c^{1.5}\sqrt{f'_c}$ (psi, with $w_c$ in pcf). Girder: $E_c = 33(145)^{1.5}\sqrt{6000} = 4463$ ksi. Slab: $E_c = 33(115)^{1.5}\sqrt{4000} = 2574$ ksi. The girder is the stiffer material, so $n = E_1/E_2 = 4463/2574 = 1.73$ (§1.6.7.3). For the slab, $57000\sqrt{f'_c}$ would overstate the modulus of a lightweight concrete; it is only valid for normal-weight concrete. Distractors: ignored the unit weights (as with $57000\sqrt{f'_c}$ for both concretes), leaving only $\sqrt{f'_{c,g}/f'_{c,s}}$ (1.22); took the unit-weight ratio to the first power instead of the 1.5 power (1.54); and used the strength ratio without its square root (2.12).

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