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Doubly Reinforced Formula Summary

Use these as the quick board-exam checklist before solving the worked problems.

$$d=h-c_c,\qquad d'=c'_c,\qquad A_s=n\left(\frac{\pi}{4}d_b^2\right)$$ $$c=\frac{a}{\beta_1}$$ $$f_s=600\left(\frac{d-c}{c}\right),\qquad f'_s=600\left(\frac{c-d'}{c}\right)$$ $$\text{Use }f_s\leq f_y\text{ and }f'_s\leq f_y\text{ in force equilibrium.}$$
$$T=C_c+C_s$$ $$A_sf_s=0.85f'_cab+A'_sf'_s$$ $$a=\frac{A_sf_s-A'_sf'_s}{0.85f'_c b}$$ $$M_n=0.85f'_cab\left(d-\frac{a}{2}\right)+A'_sf'_s(d-d')$$ $$M_u=\phi M_n$$
$$\rho_{max}=\frac{3}{8}\left(\frac{0.85f'_c\beta_1}{f_y}\right)$$ $$A_{s1}=\rho_{max}bd$$ $$M_{n1}=0.85f'_cab\left(d-\frac{a}{2}\right)$$ $$M_{n2}=\frac{M_u}{\phi}-M_{n1}$$ $$A_{s2}=\frac{M_{n2}}{f_y(d-d')},\qquad A_s=A_{s1}+A_{s2},\qquad A'_s=A_{s2}\frac{f_y}{f'_s}$$

Problem 1:

Design a rectangular beam to carry a uniform live load of 20kN/m and a dead load of 25kN/m (excluding its own weight) on a simple span of 6m. The beam is limited in cross-section to 250x450mm. Using f'c=21MPa, fy=345MPa, determine the reinforcement. Use a steel covering of 60mm top and bottom.

Diagram A Diagram B

See images:

Solution Diagram 1 Solution Diagram 2 Solution Diagram 3

Problem 2:

A simply supported beam is reinforced with 4-28mm bars at the bottom and 2-28mm bars at the top. Steel covering to the centroid of reinforcement is 70mm at the top and bottom of the beam. The beam has a total depth of 400mm and a width of 300mm. Use f'c=30MPa and fy=420MPa.

  1. Determine the depth of compression block, mm
    A. 96.930    B. 67.61    C. 80.90    D. 115.102
  2. Determine the design strength of the beam in kN-m
    A. 256.436    B. 262.714    C. 248.146    D. 260.611
  3. Determine the concentrated service live load (kN) at the midspan in addition to a total service dead load of 20kN/m if it has a span of 6 meters. Use U = 1.2 + 1.6L
    A. 61.848    B. 64.041    C. 62.333    D. 69.471
Diagram A Diagram B

See images:

Solution Diagram 1 Solution Diagram 2 Solution Diagram 3

CE Board 2013: Doubly Reinforced Beam Strength

A concrete beam, 350 mm wide by 400 mm deep, is simply supported on a span of 5 m. The beam is reinforced with 4-28 mm diameter bars in tension and 2-28 mm diameter bars in compression. Use $f'_c=20.7\text{ MPa}$, $f_y=415\text{ MPa}$, concrete cover to the centroid of reinforcements $=70\text{ mm}$, balanced steel ratio $\rho_b=0.021$, and $U=1.2D+1.6L$.

  1. Calculate the depth of the rectangular stress block based on a uniform compressive stress of $0.85f'_c$ and maximum concrete strain $\epsilon_c=0.003$. For simplification, assume both tension and compression steel yield.
  2. Determine the design bending strength $M_u$ in kN-m using $\phi=0.90$.
  3. If the nominal bending strength is $400\text{ kN-m}$, determine the safe concentrated service live load at midspan in addition to a total dead load of $20\text{ kN/m}$.

Compute the steel areas and effective depth:

$$A_s=4\left(\frac{\pi}{4}28^2\right)=2463.01\text{ mm}^2$$ $$A'_s=2\left(\frac{\pi}{4}28^2\right)=1231.50\text{ mm}^2$$ $$d=h-c_c=400-70=330\text{ mm}$$ $$d'=70\text{ mm}$$

With both tension and compression steel assumed yielding:

$$T=C_1+C_2$$ $$A_s f_y=0.85f'_cab+A'_s f_y$$ $$a=\frac{(A_s-A'_s)f_y}{0.85f'_c b}=82.99\text{ mm}$$

Design bending strength:

$$M_n=0.85f'_cab\left(d-\frac{a}{2}\right)+A'_s f_y(d-d')$$ $$M_u=0.90M_n=252.29\text{ kN-m}$$

For the concentrated live load, use $M_u=0.90(400)=360\text{ kN-m}$:

$$360=1.2\left(\frac{20(5^2)}{8}\right)+1.6\left(\frac{P(5)}{4}\right)$$ $$P=142.5\text{ kN}$$

CE Board 2010: Propped Beam Moment Capacity

A 12 m simply supported beam is provided with an additional support at midspan. The beam has width $b=300\text{ mm}$ and total depth $h=450\text{ mm}$. It is reinforced with 4-25 mm diameter bars at the tension side and 2-25 mm diameter bars at the compression side, with 70 mm cover to centroid of reinforcements. Use $f'_c=30\text{ MPa}$, $f_y=415\text{ MPa}$, and $0.75\rho_b=0.023$.

  1. Determine the depth of the rectangular stress block.
  2. Determine the nominal bending moment $M_n$.
  3. Determine the total factored uniform load including beam weight, considering moment capacity reduction of $0.90$.

Compute steel area and section data:

$$A_s=4\left(\frac{\pi}{4}25^2\right)=1963.50\text{ mm}^2$$ $$A'_s=2\left(\frac{\pi}{4}25^2\right)=981.75\text{ mm}^2$$ $$d=450-70=380\text{ mm}$$ $$\rho=\frac{A_s}{bd}=\frac{1963.50}{300(380)}=0.0172$$

Since $\rho\lt0.023$, the section can be analyzed as singly reinforced for the compression block:

$$a=\frac{A_s f_y}{0.85f'_c b}=106.52\text{ mm}$$ $$\beta_1=0.85-\frac{0.05(f'_c-28)}{7}=0.8357$$ $$c=\frac{a}{\beta_1}=127.46\text{ mm}$$ $$f_s=\frac{600(d-c)}{c}=1188.86\text{ MPa}>f_y$$

Nominal moment capacity:

$$M_n=0.85f'_cab\left(d-\frac{a}{2}\right)=266.25\text{ kN-m}$$

For a beam with an intermediate support at midspan, use $M_u=WL^2/32$:

$$0.90M_n=\frac{WL^2}{32}$$ $$W=\frac{0.90(266.25)(32)}{12^2}=53.25\text{ kN/m}$$

Doubly Reinforced Design for Factored Moment

A beam section is limited to $b=250\text{ mm}$ and total depth $h=500\text{ mm}$. It is subjected to a factored moment $M_u=448\text{ kN-m}$. Use $f'_c=27.6\text{ MPa}$, $f_y=415\text{ MPa}$, 80 mm concrete cover to the center of reinforcing bars, and $\rho_b=0.028$.

  1. Determine the ultimate moment capacity allowed for the section as singly reinforced.
  2. Determine the total required area of tension reinforcement if compression bars are needed.
  3. Determine the required area of compression reinforcement.

Use the maximum tension-controlled steel ratio for the singly reinforced part:

$$d=500-80=420\text{ mm},\quad d'=80\text{ mm},\quad \beta_1=0.85$$ $$\rho_{max}=\frac{3}{8}\left(\frac{0.85f'_c\beta_1}{f_y}\right)=0.018$$ $$\omega=\rho_{max}\frac{f_y}{f'_c}=0.2709$$ $$M_{n1}=f'_cbd^2\omega(1-0.59\omega)=277.06\text{ kN-m}$$

The required nominal strength is:

$$M_n=\frac{M_u}{0.90}=\frac{448}{0.90}=497.78\text{ kN-m}$$ $$M_{n2}=M_n-M_{n1}=220.72\text{ kN-m}$$

Find the steel for the concrete compression block portion:

$$A_{s1}=\rho_{max}bd=0.018(250)(420)=1891.99\text{ mm}^2$$ $$c=\frac{A_{s1}f_y}{0.85f'_c\beta_1b}=157.5\text{ mm}$$ $$f'_s=\frac{600(c-d')}{c}=295.24\text{ MPa}$$

For the additional couple:

$$M_{n2}=A_{s2}f_y(d-d')$$ $$A_{s2}=\frac{220.72(10^6)}{415(420-80)}=1564.27\text{ mm}^2$$ $$A_s=A_{s1}+A_{s2}=3456.27\text{ mm}^2$$ $$A'_s=A_{s2}\frac{f_y}{f'_s}=1564.27\left(\frac{415}{295.24}\right)=2198.81\text{ mm}^2$$
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