CE Board Exam Randomizer

Question 1

Wide Beam Shear: One-Way Action and Diagonal Tension

Wide beam shear is the one-way shear check of the footing slab. The critical section is a vertical plane located a distance $d$ from the face of the column, pedestal, or wall. The shear crack is treated as diagonal tension, similar to beam action across the full footing width.

$$V_u=q_u A_{outside}$$ $$\phi V_c=\phi(0.17\lambda\sqrt{f'_c})b_wd$$ $$\text{Requirement: } V_u\leq \phi V_c$$

For a rectangular isolated footing, check both principal directions when the column or footing is rectangular:

$$V_{u,x}=q_uB\left(\frac{L-c_2}{2}-d\right)$$ $$V_{u,y}=q_uL\left(\frac{B-c_1}{2}-d\right)$$

Here $b_w$ is the width of the resisting section, usually the full footing width perpendicular to the shear span. Use consistent units: if $f'_c$ is in MPa and dimensions are in mm, $V_c$ is in N.

Answer

1. Factored soil pressure

$$q_u=\frac{1800}{2.4(2.4)}=312.5\text{ kPa}$$

2. One-way shear at $d$ from face

$$V_u=q_uB\left(\frac{L-c}{2}-d\right)$$ $$V_u=312.5(2.4)(1.0-0.46)=405\text{ kN}$$ $$\phi V_c=0.75(0.17\sqrt{28})(2400)(460)=746\text{ kN}$$

Since $405\text{ kN}<746\text{ kN}$, one-way shear is adequate.

3. Punching shear at $d/2$ from face

$$b_o=2(400+400+2(460))=3440\text{ mm}$$ $$A_{inside}=(0.4+0.46)(0.4+0.46)=0.7396\text{ m}^2$$ $$V_u=1800-312.5(0.7396)=1569\text{ kN}$$ $$\phi V_c=0.75(0.33\sqrt{28})(3440)(460)=2072\text{ kN}$$

Since $1569\text{ kN}<2072\text{ kN}$, punching shear is adequate.

4. Critical moment at column face

$$l=\frac{2.4-0.4}{2}=1.0\text{ m}$$ $$m_u=\frac{312.5(1.0)^2}{2}=156.25\text{ kN-m per m}$$

5. Singly reinforced design for a 1-m strip

$$R_n=\frac{156.25(10^6)}{0.90(1000)(460^2)}=0.821\text{ MPa}$$ $$\rho=\frac{0.85(28)}{420}\left(1-\sqrt{1-\frac{2(0.821)}{0.85(28)}}\right)=0.00199$$ $$A_s=0.00199(1000)(460)=915\text{ mm}^2/\text{m}$$ $$A_{s,min}=0.0018(1000)(550)=990\text{ mm}^2/\text{m}$$

Minimum steel controls. Provide at least $990\text{ mm}^2/\text{m}$ each way at the bottom. For 16 mm bars, $A_b=201\text{ mm}^2$ and $s=201(1000)/990=203$ mm, so use 16 mm bars at 200 mm on center each way, subject to development length and detailing checks.

Answer

$$P=880+520+140=1540\text{ kN}$$ $$M=48+28+180=256\text{ kN-m}$$ $$e=\frac{M}{P}=\frac{256}{1540}=0.1662\text{ m}<\frac{L}{6}=0.6667\text{ m}$$

Since the eccentricity lies inside the kern, the soil pressure is trapezoidal and there is no uplift.

$$q_{max}=\frac{P}{BL}+\frac{6M}{BL^2}+\gamma_sT+\gamma_cH$$ $$q_{max}=\frac{1540}{3(4)}+\frac{6(256)}{3(4^2)}+17(1.5)+24(0.75)=203.83\text{ kPa}=0.2038\text{ MPa}$$ $$q_{min}=\frac{P}{BL}-\frac{6M}{BL^2}+\gamma_sT+\gamma_cH$$ $$q_{min}=139.83\text{ kPa}=0.1398\text{ MPa}$$

Using the net trapezoid, $R_1=q_{min}L=559.33\text{ kN/m}$ and $R_2=\frac{1}{2}L(q_{max}-q_{min})=128\text{ kN/m}$. Thus:

$$R=687.33\text{ kN/m}$$ $$x=\frac{R_1(L/2)+R_2(L/3)}{R}=1.876\text{ m}$$

Answer

$$M=144(2.5)=360\text{ kN-m}$$ $$e=\frac{M}{P}=\frac{360}{1200}=0.30\text{ m}<\frac{L}{6}=0.50\text{ m}$$ $$q_{max,net}=\frac{P}{bd}+\frac{6M}{bd^2}=256\text{ kPa}$$ $$q_{min,net}=\frac{P}{bd}-\frac{6M}{bd^2}=64\text{ kPa}$$

Add the footing and soil surcharge to convert the maximum net pressure to required gross bearing pressure.

$$q_{allow}=q_{max,net}+24(0.70)+17(1.5)=298.3\text{ kPa}$$

Answer

$$P_u=580+440=1020\text{ kN}$$ $$q_u=\frac{P_u}{2^2}=255\text{ kPa}$$

Punching shear:

$$V_u=255\left[2^2-(0.4+d)^2\right]$$ $$b_o=4(0.4+d)$$ $$1.76=\frac{V_u(1000)}{0.75b_od(1000^2)}$$ $$d=0.2606\text{ m}\quad\Rightarrow\quad t=100+260.6=360.6\text{ mm}$$

Wide beam shear:

$$V_u=255\left(\frac{2-0.4}{2}-d\right)(2)$$ $$0.88=\frac{V_u(1000)}{0.75(2)d(1000^2)}$$ $$d=0.223\text{ m}\quad\Rightarrow\quad t=323\text{ mm}$$

Moment:

$$M_u=\frac{1}{2}(255)(2)\left(\frac{2-0.4}{2}\right)^2=163.2\text{ kN-m}$$ $$w=\rho\frac{f_y}{f'_c}=0.021\frac{415}{28}=0.31125$$ $$163.2(10^6)=0.9(28)(2000)d^2(0.31125)(1-0.59(0.31125))$$ $$d=112.9\text{ mm}\quad\Rightarrow\quad t=212.9\text{ mm}$$

Punching shear controls; use a footing thickness of about $361\text{ mm}$, rounded upward in design.

Answer

Two-way punching shear:

$$V_u=\frac{P_u}{2^2}\left[2^2-(0.4+0.35)^2\right]$$ $$b_o=4(0.4+0.35)=3.0\text{ m}$$ $$1.76=\frac{V_u(1000)}{0.75b_od(1000^2)}$$ $$P_u=1612.8\text{ kN}$$

Wide beam shear:

$$V_u=\frac{P_u}{2^2}\left(\frac{2-0.4}{2}-0.35\right)(2)$$ $$0.88=\frac{V_u(1000)}{0.75(2)(0.35)(1000^2)}$$ $$P_u=2053.33\text{ kN}$$

Steel for moment:

$$300(10^6)=0.9(28)(2000)(350^2)w(1-0.59w)$$ $$w=0.05007$$ $$\rho=w\frac{f'_c}{f_y}=0.003378$$ $$A_s=\rho bd=0.003378(2000)(350)=2364.7\text{ mm}^2$$ $$n=\frac{A_s}{\pi(20^2)/4}=7.53$$

Use 8-20 mm bars.

Answer

$$P_u=1.2(175)+1.6(89)=352.4\text{ kN/m}$$ $$q_u=\frac{P_u}{1(1.75)}=201.37\text{ kPa}$$ $$a=\frac{1.75-0.25}{2}=0.75\text{ m}$$ $$M_u=\frac{q_u}{2}(1)(a^2)=56.64\text{ kN-m}$$

Solving the singly reinforced section gives $w=0.08016$, $\rho=0.003979$, and:

$$A_s=\rho bd=0.003979(1000)(200)=795.8\text{ mm}^2/\text{m}$$ $$s=\frac{1000(\pi16^2/4)}{795.8}=252.65\text{ mm}$$

Use 16 mm bars at about 250 mm on center.

$$V_u=q_u(0.55)(1)=110.75\text{ kN}$$ $$v_w=\frac{110.75(1000)}{0.75(1)(0.20)(1000^2)}=0.738\text{ MPa}$$ $$96=23.6(0.25)+\frac{175+89}{L}\quad\Rightarrow\quad L=2.93\text{ m}$$

Use a footing width of about $3.0\text{ m}$ for the service bearing check.

Answer

$$P_u=1.2(900)+1.6(450)=1800\text{ kN}$$ $$R=\frac{P_u}{9}=200\text{ kN per pile}$$

Beam shear:

$$0.76=\frac{3R(1000)}{0.75(3)d(1000^2)}$$ $$d=0.3509\text{ m}\quad\Rightarrow\quad t=d+0.25=0.601\text{ m}$$

Punching shear:

$$V_u=P_u-R=1600\text{ kN}$$ $$1.52=\frac{V_u(1000)}{0.75[4(0.4+d)]d(1000^2)}$$ $$d=0.4252\text{ m}\quad\Rightarrow\quad t=0.675\text{ m}$$

Moment steel:

$$M_u=3R(1.2-0.2)=600\text{ kN-m}$$ $$600(10^6)=0.9(21)(3000)(425^2)w(1-0.59w)$$ $$w=0.06076,\quad \rho=0.003075$$ $$\rho_{min}=\frac{1.4}{415}=0.003373$$ $$A_s=0.003373(3000)(425)=4301\text{ mm}^2$$ $$n=\frac{4301}{\pi(20^2)/4}=13.69$$

Use 14-20 mm bars. Punching shear controls the footing thickness.

Question 2

Punching Shear: Two-Way Action

Punching shear checks whether the column tends to punch through the footing. The critical perimeter is located at $d/2$ from the face of the loaded area. For an interior rectangular column with dimensions $c_1$ and $c_2$:

$$b_o=2(c_1+c_2+2d)$$ $$A_{inside}=(c_1+d)(c_2+d)$$ $$V_u=P_u-q_uA_{inside}$$

The design concrete punching shear strength is commonly taken as the least of the applicable limits:

$$\phi V_c=\phi(0.33\lambda\sqrt{f'_c})b_od$$ $$\phi V_c=\phi\left[0.17\left(1+\frac{2}{\beta_c}\right)\lambda\sqrt{f'_c}\right]b_od$$ $$\phi V_c=\phi\left[0.083\left(\frac{\alpha_s d}{b_o}+2\right)\lambda\sqrt{f'_c}\right]b_od$$

Use $\beta_c$ as the ratio of long side to short side of the column. For $\alpha_s$, common values are 40 for interior columns, 30 for edge columns, and 20 for corner columns. In many board problems, the simple $0.33\lambda\sqrt{f'_c}b_od$ limit controls or is the only limit requested.

Question 3

Critical Moment at the Column Face

The critical flexural section for an isolated footing is at the face of the column, pedestal, or wall. Treat each side of the footing projection as a cantilever loaded by upward soil pressure.

$$l_x=\frac{L-c_2}{2}$$ $$l_y=\frac{B-c_1}{2}$$ $$M_{u,x}=\frac{q_uB l_x^2}{2}=\frac{q_uB(L-c_2)^2}{8}$$ $$M_{u,y}=\frac{q_uL l_y^2}{2}=\frac{q_uL(B-c_1)^2}{8}$$

For design by one-meter strips, use:

$$m_u=\frac{q_ul^2}{2} \quad \text{per unit strip width}$$

The larger moment usually controls the required effective depth and reinforcement. Check the direction of the bars: bars resisting moment from projection $l_x$ are placed perpendicular to that projection.

Question 4

Design as a Singly Reinforced Section

A footing slab is normally designed as a singly reinforced rectangular section because tension steel is placed near the bottom and compression is carried by concrete above the neutral axis.

$$\phi M_n\geq M_u$$ $$a=\frac{A_s f_y}{0.85f'_c b}$$ $$M_n=A_s f_y\left(d-\frac{a}{2}\right)$$

For direct solution of required steel area, use the design equation with $M_u$ in N-mm:

$$a=d-\sqrt{d^2-\frac{2M_u}{\phi(0.85f'_c)b}}$$ $$A_s=\frac{0.85f'_cba}{f_y}$$

Equivalent reinforcement-ratio form:

$$R_n=\frac{M_u}{\phi bd^2}$$ $$\rho=\frac{0.85f'_c}{f_y}\left(1-\sqrt{1-\frac{2R_n}{0.85f'_c}}\right)$$ $$A_s=\rho bd$$

Minimum footing steel is commonly checked using the slab shrinkage and temperature requirement:

$$A_{s,min}=0.0018bh \quad \text{for Grade 420 deformed bars}$$

After selecting bars, compute spacing by $s=A_b(1000)/A_s$ for a 1-m strip. Also check clear cover, bar spacing, development length past the critical section, and whether the required bars fit within the footing width.

Question 5

Pile Footing Reactions

For pile-supported footings, distribute the factored axial load and moment to the piles before checking beam shear, punching shear, and flexural moment. Let $n$ be the number of piles. If all piles have the same size and stiffness, the pile area is uniform and cancels out of the reaction formula, so the pile-group inertia can be computed using only pile distances and number of piles.

$$R_{avg}=\frac{P_u}{n}$$ $$I_x=\sum y_i^2 \quad \text{for moment about the } x\text{-axis}$$ $$I_y=\sum x_i^2 \quad \text{for moment about the } y\text{-axis}$$

Here $x_i$ and $y_i$ are pile distances from the centroidal axes of the pile group. If pile area $A_p$ is shown in a derivation, $I=A_p\sum d_i^2$ and the force increment is $M(A_pc)/I$, which simplifies to $Mc/\sum d_i^2$ for uniform piles.

$$R_i=\frac{P_u}{n}\pm\frac{M_x y_i}{I_x} \quad \text{for moment about the } x\text{-axis}$$ $$R_i=\frac{P_u}{n}\pm\frac{M_y x_i}{I_y} \quad \text{for moment about the } y\text{-axis}$$ $$R_i=\frac{P_u}{n}\pm\frac{M_x y_i}{I_x}\pm\frac{M_y x_i}{I_y} \quad \text{for biaxial moment}$$

$R_{max}$ is the reaction of the pile or pile row on the compressive side of the moment. $R_{min}$ corresponds to the farthest pile or pile row on the tension side, where the moment reduces compression and may cause uplift if the result becomes negative.

$$R_{max}=\frac{P_u}{n}+\frac{M_uc}{I}$$ $$R_{min}=\frac{P_u}{n}-\frac{M_uc}{I}$$ $$R_{mid}=\frac{P_u}{n} \quad \text{for piles on the centroidal axis of bending}$$

Question 6

Board-Exam-Style Reminders

Use service loads for soil bearing size unless the problem gives a factored bearing pressure or asks directly for strength checks.
Use factored loads and factored soil pressure for shear and moment design.
One-way shear is checked at $d$ from the column face; punching shear is checked at $d/2$ from the column face.
Critical moment is at the column face, not at $d$ or $d/2$.
Increasing footing thickness increases $d$, which helps both one-way and punching shear.
Footing flexural steel is bottom steel because the upward soil pressure causes tension at the bottom near the column face.
Do not mix meters and millimeters in the same formula. Convert moments to N-mm when using MPa and mm.

Question 7

Sample Design of an Isolated Footing

A square tied column carries a factored axial load $P_u=1800$ kN. The column is $400 \text{ mm} \times 400 \text{ mm}$ and is supported by a $2.4 \text{ m} \times 2.4 \text{ m}$ square footing. Use $f'_c=28$ MPa, $f_y=420$ MPa, normal-weight concrete, $h=550$ mm, effective depth $d=460$ mm, and $\phi=0.75$ for shear, $\phi=0.90$ for flexure. Check shear and find the required bottom steel per meter strip.

Question 8

Eccentric Footing Soil Pressure and Resultant

A footing has dimensions $B \times L \times H=3\text{ m}\times4\text{ m}\times0.75\text{ m}$. The column is $0.60\text{ m}\times0.75\text{ m}$. Earth fill above the footing is $1.5\text{ m}$, soil unit weight is $17\text{ kN/m}^3$, and concrete unit weight is $24\text{ kN/m}^3$. The service column loads are: dead load $880\text{ kN}$ with $48\text{ kN-m}$, live load $520\text{ kN}$ with $28\text{ kN-m}$, and earthquake load $140\text{ kN}$ with $180\text{ kN-m}$ about the $y$-axis. Determine the maximum soil pressure, minimum soil pressure, and location of the resultant from the end of the footing.

Question 9

Net and Gross Bearing Pressure with Moment

A rectangular footing is $2.5\text{ m}$ wide along the $y$-axis and $3\text{ m}$ long along the $x$-axis. It supports a circular pedestal $0.45\text{ m}$ in diameter. A horizontal force of $144\text{ kN}$ acts at the top of the pedestal. The total axial service load is $1200\text{ kN}$, footing thickness is $0.70\text{ m}$, backfill height above the footing is $1.5\text{ m}$, and the pedestal height from the footing base is $2.5\text{ m}$. Use $\gamma_c=24\text{ kN/m}^3$ and $\gamma_s=17\text{ kN/m}^3$. Find the maximum net soil pressure, minimum net soil pressure, and required gross bearing capacity.

Question 10

Minimum Footing Thickness by Shear and Moment

A $2\text{ m}\times2\text{ m}$ footing carries a concentric $0.40\text{ m}\times0.40\text{ m}$ column. Factored column loads are $D=580\text{ kN}$ and $L=440\text{ kN}$. Cover to the centroid of footing reinforcement is $100\text{ mm}$, $f'_c=28\text{ MPa}$, and $f_y=415\text{ MPa}$. Allowable ultimate stresses are $1.76\text{ MPa}$ for diagonal tension or two-way action shear, $0.88\text{ MPa}$ for wide beam shear, and maximum steel ratio for moment is $0.021$. Determine the minimum footing thickness required for punching shear, wide beam shear, and moment.

Question 11

Allowable Axial Load and Required Bars

A $2\text{ m}\times2\text{ m}$ footing, $0.45\text{ m}$ thick, carries a centered $0.40\text{ m}\times0.40\text{ m}$ square column. Effective depth is $0.35\text{ m}$, $f'_c=28\text{ MPa}$, and $f_y=415\text{ MPa}$. Allowable stresses at ultimate loads are $1.76\text{ MPa}$ for two-way action shear and $0.88\text{ MPa}$ for wide beam shear. Determine the allowable axial load based on punching shear, the allowable axial load based on wide beam shear, and the number of 20 mm bars required if $M_u=300\text{ kN-m}$.

Question 12

Wall Footing Bar Spacing and Shear

A footing supports a $250\text{ mm}$ thick concrete wall. Wall loads are $D=175\text{ kN/m}$ and $L=89\text{ kN/m}$. Use $f'_c=20.6\text{ MPa}$, $f_y=415\text{ MPa}$, concrete unit weight $23.6\text{ kN/m}^3$, and $U=1.2D+1.6L$. The footing width is $1.75\text{ m}$, footing thickness is $300\text{ mm}$, and cover to the centroid of reinforcement is $100\text{ mm}$. Find the required spacing of 16 mm bars due to moment, the wide beam shear stress, and the footing width if allowable soil pressure is $96\text{ kPa}$ with a $250\text{ mm}$ footing thickness.

Question 13

Pile-Supported Footing Thickness and Bars

A $400\text{ mm}\times400\text{ mm}$ column is supported by a square footing on 9 piles. Dimensions are $a=0.6\text{ m}$, $b=1.8\text{ m}$, and $d=0.6\text{ m}$. The column carries $D=900\text{ kN}$ and $L=450\text{ kN}$. Use $f'_c=21\text{ MPa}$, $f_y=415\text{ MPa}$, allowable beam shear stress $0.76\text{ MPa}$, allowable punching shear stress $1.52\text{ MPa}$, and minimum cover to centroid of reinforcement $0.25\text{ m}$. Determine required footing thickness for beam shear, required thickness for punching shear, and number of 20 mm bars required for maximum moment about the $x$-axis.

Reinforced Concrete Footings