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Question 1

Singly-Reinforced Beams

Formula Summary

$$U=1.4D$$ $$U=1.2D+1.6L+0.5(L_r\text{ or }S\text{ or }R)$$ $$U=1.2D+1.6(L_r\text{ or }S\text{ or }R)+(L\text{ or }0.5W)$$ $$U=1.2D+1.0W+L+0.5(L_r\text{ or }S\text{ or }R)$$ $$U=1.2D+1.0E+L+0.2S$$ $$U=0.9D+1.0W$$ $$U=0.9D+1.0E$$

$$a=\frac{A_s f_y}{0.85f'_c b}$$ $$a=\beta_1c$$ $$\beta_1= \begin{cases} 0.85, & f'_c\leq 28\text{ MPa}\\ 0.85-\frac{0.05(f'_c-28)}{7}, & 28\text{ MPa}\lt f'_c\lt 55\text{ MPa}\\ 0.65, & f'_c\geq 55\text{ MPa} \end{cases}$$ $$M_n=A_s f_y\left(d-\frac{a}{2}\right)$$ $$M_u=\phi M_n$$

$$\omega=\rho\frac{f_y}{f'_c}$$ $$M_u=\phi f'_cbd^2\omega(1-0.59\omega)$$ $$\rho=\omega\frac{f'_c}{f_y}$$ $$A_s=\rho bd$$

$$A_{s,min}=\frac{0.25\sqrt{f'_c}}{f_y}b_wd$$ $$A_{s,min}=\frac{1.4}{f_y}b_wd$$ $$A_{s,min}=\text{larger value of the two equations above}$$ $$\rho_b=0.85\beta_1\frac{f'_c}{f_y}\left(\frac{600}{600+f_y}\right)$$ $$\rho_{max}=0.75\rho_b$$

$$f_s=\frac{600(d-c)}{c}$$ $$\epsilon_t=0.003\frac{d-c}{c}$$ $$\phi=0.75+0.15\left(\frac{f_s-f_y}{1000-f_y}\right) \quad \text{for spiral members}$$ $$\phi=0.65+0.25\left(\frac{f_s-f_y}{1000-f_y}\right) \quad \text{for other tension-controlled members}$$ $$M_n=0.85f'_cab\left(d-\frac{a}{2}\right) \quad \text{when steel does not yield}$$

Answer

See images:

Singly Reinforced Beam | Principles of Reinforced Concrete – Problem 1: Design of a Reinforced Concrete Beam – Diagram

Answer

See images:

Singly Reinforced Beam | Principles of Reinforced Concrete – Problem 2: Triangular Beam | Investigation Problem – Diagram

Answer

Beam Section and Ratio and Proportion based on strain converted to stresses:

Singly Reinforced Beam | Principles of Reinforced Concrete – Problem 3: Rectangular Beam | Investigation Problem – Diagram

Zoomed-in View of the computation of the centroidal cover:

Compute c and check fs if steel yields. If fs > fy, steel yields and we can proceed computing the nominal moment capacity. However, if fs < fy, we must recompute c and express C=T in terms of c.

Since steel yields, we keep the value of a and compute the nominal moment capacity by taking the couple moment: $Force\times Distance$. Note that C and T are equal in magnitude and opposite in direction, so they induce a couple moment equivalent to any one of the forces multiplied by the distance between them.

To compute the additional service live load at the midspan that the beam can carry in addition to a service dead load of 20kN/m, we first compute M_u since we must analyze beams at ultimate condition. To obtain M_u, simply multiply the nominal moment M_n by the reduction factor, Φ. Since fs exceeds 1000MPa, Φ=0.9 (tension-controlled). At ultimate condition, we also use the factored load combination: $1.2DL + 1.6LL$. Since the loadings are common, we can use the formulas: $\frac{PL}{4}$ and $\frac{w\cdot L^2}{8}$ for the maximum moments.

Answer

See images:

Singly Reinforced Beam | Principles of Reinforced Concrete – Problem 4: Rectangular Beam | Investigation Problem – Diagram

Answer

See images:

Singly Reinforced Beam | Principles of Reinforced Concrete – Problem 5: – Diagram

Answer

#1 — Effective Depth

$$d = h - c_c - d_s - \frac{d_b}{2} = 500 - 40 - 10 - \frac{22}{2} = \boxed{439 \text{ mm}}$$

#2 — Stress Block and Neutral Axis

$$A_s = 4\times\frac{\pi}{4}(22)^2 = 4\times 380.13 = 1520.5 \text{ mm}^2$$ $$a = \frac{A_s f_y}{0.85\,f'_c\,b} = \frac{1520.5\times 414}{0.85\times 21\times 280} = \frac{629{,}487}{4998} = 125.9 \text{ mm}$$

β₁ = 0.85 (f'_c = 21 MPa ≤ 28 MPa)

$$c = \frac{a}{\beta_1} = \frac{125.9}{0.85} = \boxed{148.1 \text{ mm}}$$

#3 — Strain Check (Tension-Controlled?)

$$\varepsilon_t = 0.003\,\frac{d - c}{c} = 0.003\times\frac{439 - 148.1}{148.1} = 0.003\times 1.963 = 0.00589$$

ε_t = 0.00589 > 0.005 → Tension-controlled, φ = 0.90 ✓

#4 — Design Moment Capacity

$$\phi M_n = \phi\,A_s f_y\!\left(d - \frac{a}{2}\right) = 0.90\times 1520.5\times 414\times\left(439 - \frac{125.9}{2}\right)$$ $$= 0.90\times 629{,}487\times 376.1 = \boxed{213.2 \text{ kN·m}}$$

Answer

#1 — ρ_max and Design ρ

β₁ = 0.85 (f'_c = 28 MPa)

$$\rho_{max} = \frac{0.85\,f'_c\,\beta_1}{f_y}\left(\frac{3}{8}\right) = \frac{0.85(28)(0.85)}{414}\left(\frac{3}{8}\right) = \frac{20.23}{414}(0.375) = 0.01833$$ $$\rho = 0.5\,\rho_{max} = 0.5(0.01833) = 0.009163$$

At ρ = 0.5ρ_max, c/d = 0.5 × (3/8) = 0.1875 → ε_t ≈ 0.013 ≫ 0.005 → tension-controlled, φ = 0.90 ✓

#2 — Required Effective Depth (ω Method)

$$\omega = \rho\,\frac{f_y}{f'_c} = 0.009163\times\frac{414}{28} = 0.13549$$ $$M_u = \phi\,f'_c\,b\,d^2\,\omega(1 - 0.59\omega)$$ $$240\times10^6 = 0.9(28)(400)\,d^2(0.13549)(1 - 0.59\times0.13549)$$ $$240\times10^6 = 0.9(28)(400)(0.13549)(0.9201)\,d^2 = 1256.7\,d^2$$ $$d = \sqrt{\frac{240\times10^6}{1256.7}} = \boxed{437 \text{ mm} \rightarrow \text{use } d = 440 \text{ mm}}$$

#3 — Steel Area and Bar Count

$$A_s = \rho\,bd = 0.009163(400)(440) = 1612.7 \text{ mm}^2$$ $$n = \frac{1612.7}{\pi/4\,(25)^2} = \frac{1612.7}{490.87} = 3.29 \rightarrow \boxed{4\text{ bars},\; A_{s,\text{prov}} = 1963.5 \text{ mm}^2}$$

Spacing check (4 – 25mm bars, b = 400 mm, cc = 40, ds = 12): clear width used = 2(40)+2(12)+4(25)+3(26.67) = 284 mm < 400 mm ✓

Check A_s,min = larger of $\frac{0.25\sqrt{28}}{414}(400)(440)=566$ mm² and $\frac{1.4}{414}(400)(440)=598$ mm² → 598 mm² < 1613 mm² ✓

Answer

Step 1 — Moment Capacity φM_n

A_s = 3 × π/4(25)² = 1472.6 mm² | β₁ = 0.85

$$a = \frac{A_s f_y}{0.85\,f'_c\,b} = \frac{1472.6(414)}{0.85(27.5)(300)} = \frac{609{,}856}{7012.5} = 86.97 \text{ mm}$$ $$c = \frac{a}{\beta_1} = \frac{86.97}{0.85} = 102.3 \text{ mm}$$ $$\varepsilon_t = \frac{0.003(460-102.3)}{102.3} = 0.01050 > 0.005 \implies \phi = 0.90$$ $$\phi M_n = \frac{0.90(609{,}856)(460-43.49)}{10^6} = \frac{0.90(609{,}856)(416.51)}{10^6} = \boxed{228.6 \text{ kN·m}}$$

Step 2 — Maximum Service P

For a simply supported beam with uniform DL and concentrated LL at midspan:

$$M_u = \frac{1.2\,w_{DL}\,L^2}{8} + \frac{1.6\,P\,L}{4} \le \phi M_n$$ $$\frac{1.2(12)(6)^2}{8} + \frac{1.6\,P(6)}{4} = 228.6$$ $$64.8 + 2.4\,P = 228.6$$ $$P = \frac{228.6 - 64.8}{2.4} = \boxed{68.25 \text{ kN}}$$

Answer

#1 — Factored Moment

$$w_u = 1.2(16) + 1.6(20) = 19.2 + 32.0 = 51.2 \text{ kN/m}$$ $$M_u = \frac{w_u L^2}{8} = \frac{51.2(6)^2}{8} = 230.4 \text{ kN·m}$$

#2 — Required Effective Depth

β₁ = 0.85 (f'_c = 27.5 ≤ 28 MPa)

$$\rho_{max} = \frac{0.85\,f'_c\,\beta_1}{f_y}\left(\frac{3}{8}\right) = \frac{0.85(27.5)(0.85)}{414}(0.375) = 0.01800$$ $$\rho = 0.5\,\rho_{max} = 0.009000 \qquad \omega = \rho\frac{f_y}{f'_c} = 0.009\times\frac{414}{27.5} = 0.13549$$

At ρ = 0.5ρ_max, c/d = 0.5 × (3/8) = 0.1875 → ε_t ≈ 0.013 ≫ 0.005 → tension-controlled, φ = 0.90 ✓

$$M_u = \phi f'_c b d^2\,\omega(1 - 0.59\omega)$$ $$230.4\times10^6 = 0.9(27.5)(300)\,d^2(0.13549)(1-0.59\times0.13549)$$ $$230.4\times10^6 = 0.9(27.5)(300)(0.13549)(0.9201)\,d^2 = 925.6\,d^2$$ $$d = \sqrt{\frac{230.4\times10^6}{925.6}} = \boxed{499 \text{ mm} \rightarrow \text{use } d = 500 \text{ mm}}$$

#3 — Steel Area and Bar Count

$$A_s = \rho\,bd = 0.009(300)(500) = 1350 \text{ mm}^2$$ $$n = \frac{A_s}{\pi/4\,(20)^2} = \frac{1350}{314.16} = 4.30 \rightarrow \boxed{5\text{ bars},\; A_{s,\text{prov}} = 1571 \text{ mm}^2}$$

Check A_s,min = larger of $\frac{0.25\sqrt{27.5}}{414}(300)(500) = 475$ mm² and $\frac{1.4}{414}(300)(500) = 507$ mm² → 507 mm² < 1350 mm² ✓

Answer

#1 — Compression Block and Neutral Axis

Effective depth: d = 500 − 40 − 10 − 11 = 439 mm (cc + stirrup + half bar dia)

A_s = 4 × π/4(22)² = 1520.5 mm²

$$a = \frac{A_s f_y}{0.85 f'_c b} = \frac{1520.5(414)}{0.85(28)(300)} = \frac{629{,}487}{7140} = 88.16 \text{ mm}$$ $$c = \frac{a}{\beta_1} = \frac{88.16}{0.85} = 103.7 \text{ mm}$$

#2 — Design Moment Capacity

$$\varepsilon_t = \frac{0.003(d-c)}{c} = \frac{0.003(439-103.7)}{103.7} = 0.00970 > 0.005 \implies \phi = 0.9$$ $$\phi M_n = \frac{0.9(1520.5)(414)(439-44.08)}{10^6} = \frac{0.9(629{,}487)(394.92)}{10^6} = \boxed{223.6 \text{ kN·m}}$$

#3 — Adequacy Check

$$w_u = 1.2(14) + 1.6(18) = 16.8 + 28.8 = 45.6 \text{ kN/m}$$ $$M_{u,\text{demand}} = \frac{45.6(6)^2}{8} = 205.2 \text{ kN·m} < 223.6 \text{ kN·m} = \phi M_n$$

The beam IS adequate. The capacity exceeds the demand by 18.4 kN·m.

Answer

#1 — Moment Capacity

d = 500 − 40 − 10 − 14 = 436 mm | A_s = 3 × π/4(28)² = 1847.3 mm² | β₁ = 0.85

$$a = \frac{1847.3(414)}{0.85(21)(350)} = \frac{764{,}782}{6247.5} = 122.4 \text{ mm} \qquad c = \frac{122.4}{0.85} = 144.0 \text{ mm}$$ $$\varepsilon_t = \frac{0.003(436-144.0)}{144.0} = 0.00608 > 0.005 \implies \phi = 0.9$$ $$\phi M_n = \frac{0.9(764{,}782)(436-61.2)}{10^6} = \frac{0.9(764{,}782)(374.8)}{10^6} = \boxed{257.9 \text{ kN·m}}$$

#2 — Maximum Service Live Load

$$\phi M_n = \frac{w_u L^2}{8} \implies w_u = \frac{8(257.9)}{8^2} = 32.24 \text{ kN/m}$$ $$w_u = 1.2(DL) + 1.6(LL) \implies 32.24 = 1.2(10) + 1.6(LL)$$ $$LL = \frac{32.24 - 12}{1.6} = \boxed{12.65 \text{ kN/m}}$$

Answer

#1 — Steel Ratio vs ρ_max

A_s = 5 × π/4(25)² = 2454.4 mm² | β₁ = 0.85 − 0.05(30−28)/7 = 0.8357

$$\rho = \frac{2454.4}{280(430)} = 0.02039$$ $$\rho_{max} = \frac{0.85\,f'_c\,\beta_1}{f_y}\left(\frac{3}{8}\right) = \frac{0.85(30)(0.8357)}{415}(0.375) = 0.01926$$ $$\rho = 0.02039 > \rho_{max} = 0.01926 \implies \textbf{Section is in the transition zone (steel still yields, but } \phi < 0.90\textbf{)}$$

#2 — Compression Block and Strain

$$a = \frac{2454.4(415)}{0.85(30)(280)} = \frac{1{,}018{,}576}{7140} = 142.65 \text{ mm} \qquad c = \frac{142.65}{0.8357} = 170.7 \text{ mm}$$ $$\varepsilon_t = \frac{0.003(430-170.7)}{170.7} = 0.00456$$

Since ε_y = 415/200,000 = 0.002075 < ε_t = 0.00456 < 0.005 → Transition zone

#3 — Reduced φ and φM_n

$$\phi = 0.65 + 0.25\cdot\frac{\varepsilon_t - \varepsilon_y}{0.005 - \varepsilon_y} = 0.65 + 0.25\cdot\frac{0.00456 - 0.002075}{0.005 - 0.002075} = 0.65 + 0.212 = 0.862$$ $$\phi M_n = \frac{0.862(1{,}018{,}576)(430-71.33)}{10^6} = \frac{0.862(1{,}018{,}576)(358.7)}{10^6} = \boxed{315.1 \text{ kN·m}}$$

Note: Using φ = 0.9 would give 328.2 kN·m — a 4.1% overestimate. Always check ε_t against 0.005.

Answer

#1 — ρ_max and Design ρ

β₁ = 0.85 (f'_c = 21 < 28 MPa)

$$\rho_{max} = \frac{0.85\,f'_c\,\beta_1}{f_y}\left(\frac{3}{8}\right) = \frac{0.85(21)(0.85)}{415}(0.375) = 0.013710$$ $$\rho = 0.6\,\rho_{max} = 0.6(0.013710) = 0.008226$$

At ρ = 0.6ρ_max, c/d = 0.6 × (3/8) = 0.225 → ε_t ≈ 0.010 ≫ 0.005 → tension-controlled, φ = 0.90 ✓

#2 — Required Depth

$$\omega = 0.008226\times\frac{415}{21} = 0.16258 \qquad 1-0.59\omega = 1 - 0.09592 = 0.90408$$ $$300\times10^6 = 0.9(21)(300)\,d^2(0.16258)(0.90408) = 833.6\,d^2$$ $$d = \sqrt{\frac{300\times10^6}{833.6}} = \boxed{600 \text{ mm}}$$

#3 — Steel Area

$$A_s = \rho\,bd = 0.008226(300)(600) = 1480.7 \text{ mm}^2$$ $$n = \frac{1480.7}{\pi/4\,(25)^2} = \frac{1480.7}{490.87} = 3.02 \rightarrow \boxed{4\text{ bars},\; A_{s,\text{prov}} = 1963.5 \text{ mm}^2}$$

Check A_s,min = larger of $\frac{0.25\sqrt{21}}{415}(300)(600) = 496$ mm² and $\frac{1.4}{415}(300)(600) = 607$ mm² → 607 mm² < 1481 mm² ✓

Answer

#1 — Check if Steel Yields

β₁ = 0.85 (f'_c = 20.7 < 28 MPa)

$$\rho = \frac{3600}{200(350)} = 0.05143$$ $$\rho_{max} = \frac{0.85\,f'_c\,\beta_1}{f_y}\left(\frac{3}{8}\right) = \frac{0.85(20.7)(0.85)}{275}(0.375) = 0.02039$$ $$\rho = 0.05143 \gg \rho_{max} = 0.02039 \implies \textbf{Overreinforced — steel does NOT yield}$$

#2 — Find Neutral Axis (C = T, with f_s unknown)

Since f_s < f_y, use $f_s = 600(d-c)/c$ and set C = T:

$$0.85f'_c\beta_1 c\,b = A_s\cdot\frac{600(d-c)}{c}$$ $$0.85(20.7)(0.85)(200)\,c = 3600\cdot\frac{600(350-c)}{c}$$ $$2993.6\,c^2 = 2{,}160{,}000(350-c)$$ $$2993.6\,c^2 + 2{,}160{,}000\,c - 756{,}000{,}000 = 0$$

Solving via quadratic formula:

$$c = \frac{-2{,}160{,}000 + \sqrt{(2{,}160{,}000)^2 + 4(2993.6)(756{,}000{,}000)}}{2(2993.6)} = \frac{-2{,}160{,}000 + 3{,}704{,}709}{5987.2} = \boxed{258 \text{ mm}}$$ $$a = \beta_1 c = 0.85(258) = 219.3 \text{ mm}$$ $$f_s = \frac{600(350-258)}{258} = 213.9 \text{ MPa} < f_y = 275 \text{ MPa} \checkmark$$

#3 — Design Moment Capacity

ε_t = 0.003(350−258)/258 = 0.00107 < ε_y = 0.001375 → compression-controlled → φ = 0.65

$$M_n = 0.85f'_c\,a\,b\left(d-\frac{a}{2}\right) = \frac{0.85(20.7)(219.3)(200)(350-109.65)}{10^6} = 185.6 \text{ kN·m}$$ $$\phi M_n = 0.65(185.6) = \boxed{120.6 \text{ kN·m}}$$

An overreinforced beam is not permitted by NSCP for new designs (ρ must not exceed ρ_max). This problem illustrates the analysis procedure when such a section is encountered in investigation.

Question 2

Problem 1: Design of a Reinforced Concrete Beam

Design a reinforced concrete beam to carry a service dead load of 20kN/m (including its own weight) and service live load of 25kN/m plus a concentrated service live load of 30kN shown. Use f'c=21MPa, fy=276MPa, b=d/2, and ⌀=28mm main bars. The beam is to be reinforced in tension only.

Question 3

Problem 2: Triangular Beam | Investigation Problem

For the beam shown below, f'c=27MPa, fy=270MPa, steel cover to centroid of bars = 100mm. A_s=4-32mm diameter bars.
a. Determine the depth of the compression block.
b. What is the compressive strength of concrete?
c. What is the nominal moment capacity of the beam?

Question 4

Problem 3: Rectangular Beam | Investigation Problem

The beam shown is reinforced with 3-28mm⌀ and its width is 300mm. The total height of the beam is 600mm. Assume that the clear cover is 40mm and 10mm⌀ stirrups are used. Use f'c=27.5MPa and fy=414MPa.
a. Compute the depth of the compression block in mm.
b. Determine the ultimate moment capacity of the beam section.
c. What additional service live load at the midspan can the beam carry in addition to a service dead load of 20kN/m (including its own weight)?

Question 5

Problem 4: Rectangular Beam | Investigation Problem

A reinforced concrete beam with width of 300mm and d=410mm is reinforced for tension with a reinforcement area of 3700mm². Use f'c=27MPa and fy=415MPa.
a. Determine the distance (mm) of the steel reinforcements to the neutral axis.
A. 248
B. 162
C. 262
D. 148
b. Compute the tensile strain in the reinforcements.
A. 0.00196
B. 0.00169
C. 0.00122
D. 0.00244
c. If the beam is 8m long and is simply supported, find the concentrated service live load acting at the midspan that can be supported by the beam if it already carries a total service uniform dead load of 24kN/m.
A. 32.33
B. 52.32
C. 17.78
D. 66.13

Question 6

Problem 5: Singly Reinforced Beam Investigation | Two Layers (3 along first row and 2 along second row)

The beam section shown has the following properties:
bar diameter, d_b = 25mm
total height, h = 600mm
center to center distance between first and second row bars, a = 27mm
f'c = 32MPa
f_y = 414MPa
beam width, b_w = 300mm
diameter of stirrups, d_s = 12mm
clear cover, c_c = 40mm
Determine the following:
a. The ultimate moment, M_u, that the beam can carry
b. The ultimate uniform load that the beam can carry if it has a simply supported span of 7m
c. The ultimate uniform dead load (w_uDL) carried by the beam if the service live load is 10kN/m
d. What additional concentrated service live load at the midspan (P_LL can the beam carry if it has a service dead load of 20kN/m (including its own weight) and a service live load of 12kN/m?

Question 7

Problem 6: Investigation — Design Moment Capacity φM_n

A singly reinforced rectangular beam has the following properties:
b = 280 mm, total height h = 500 mm, 4 – 22 mm ∅ bars (single row)
f'_c = 21 MPa, f_y = 414 MPa, clear cover = 40 mm, stirrups = 10 mm ∅

Determine the effective depth d.
Find the depth of the stress block a and the neutral axis depth c.
Check whether the section is tension-controlled and determine φ.
Compute the design moment capacity φM_n.

Question 8

Problem 7: Design — Using ρ_max (ω Method)

Design a simply supported singly reinforced rectangular beam to carry a factored moment M_u = 240 kN·m. Use b = 400 mm, f'_c = 28 MPa, f_y = 414 MPa, and design for ρ = 0.5ρ_max. Use 25 mm ∅ bars.

Compute ρ_max and the design ρ.
Find the required effective depth d.
Find the required A_s and number of bars.

Question 9

Problem 8: Investigation — Maximum Concentrated Live Load at Midspan

A simply supported beam has a span of 6 m and the following section properties: b = 300 mm, d = 460 mm, 3 – 25 mm ∅ bars (single row). f'_c = 27.5 MPa, f_y = 414 MPa. The service dead load (including self-weight) is 12 kN/m. Determine the maximum service concentrated live load P that can be applied at midspan.

Question 10

Problem 9: Design — ω Method (Uniform Dead + Live Load)

A simply supported beam has a span of 6 m and carries a service dead load of 16 kN/m (including self-weight) and a service live load of 20 kN/m. Use f'_c = 27.5 MPa, f_y = 414 MPa, b = 300 mm, and design for ρ = 0.5ρ_max.

Determine the factored design moment M_u.
Find the required effective depth d.
Find the required steel area A_s and the number of 20 mm ∅ bars needed.

Question 11

Problem 10: Investigation — Adequacy Check

A rectangular beam has b = 300 mm, total height h = 500 mm, and is reinforced with 4 – 22 mm ∅ bars. Clear cover = 40 mm, stirrups = 10 mm ∅. Use f'_c = 28 MPa, f_y = 414 MPa.

Compute the depth of the compression block a and neutral axis depth c.
Compute the design moment capacity φM_n.
Is the beam adequate for a 6 m simply supported span carrying service DL = 14 kN/m (including self-weight) and service LL = 18 kN/m?

Question 12

Problem 11: Investigation — Maximum Allowable Live Load

A simply supported beam spanning 8 m has b = 350 mm, h = 500 mm, and is reinforced with 3 – 28 mm ∅ bars at the bottom. Clear cover = 40 mm, stirrups = 10 mm ∅. Use f'_c = 21 MPa, f_y = 414 MPa. The service dead load (including self-weight) is 10 kN/m.

Compute the design moment capacity φM_n of the beam.
Determine the maximum allowable service live load (kN/m) the beam can carry.

Question 13

Problem 12: Investigation — Transition Zone φ

A singly reinforced rectangular beam has b = 280 mm, d = 430 mm, and is reinforced with 5 – 25 mm ∅ bars. Use f'_c = 30 MPa, f_y = 415 MPa.

Compute ρ and compare with ρ_max; determine whether the section is tension-controlled.
Compute a, c, and the net tensile strain ε_t.
Determine the appropriate φ factor and compute φM_n.

Question 14

Problem 13: Design — Given M_u Directly

Design a singly reinforced rectangular beam to resist a design moment of M_u = 300 kN·m. Use f'_c = 21 MPa, f_y = 415 MPa, b = 300 mm. Design for ρ = 0.6ρ_max.

Compute ρ_max and the design ρ.
Find the required effective depth d.
Find the required A_s and select 25 mm ∅ bars.

Question 15

Problem 14: Investigation — Steel Does NOT Yield (Overreinforced Beam)

A rectangular beam has b = 200 mm, d = 350 mm, and is reinforced with A_s = 3600 mm². Use f'_c = 20.7 MPa, f_y = 275 MPa.

Compare ρ with ρ_max and determine whether the steel yields at ultimate.
If steel does not yield, find the neutral axis depth c by solving the equilibrium equation.
Compute the design moment capacity φM_n.

Singly-Reinforced Beams

Problem 1: Design of a Reinforced Concrete Beam

Problem 2: Triangular Beam | Investigation Problem

Problem 3: Rectangular Beam | Investigation Problem

Problem 4: Rectangular Beam | Investigation Problem

Problem 5: Singly Reinforced Beam Investigation | Two Layers (3 along first row and 2 along second row)

Problem 6: Investigation — Design Moment Capacity φMn

#1 — Effective Depth

#2 — Stress Block and Neutral Axis

#3 — Strain Check (Tension-Controlled?)

#4 — Design Moment Capacity

Problem 7: Design — Using ρmax (ω Method)

#1 — ρmax and Design ρ

#2 — Required Effective Depth (ω Method)

#3 — Steel Area and Bar Count

Problem 8: Investigation — Maximum Concentrated Live Load at Midspan

Step 1 — Moment Capacity φMn

Step 2 — Maximum Service P

Problem 9: Design — ω Method (Uniform Dead + Live Load)

#1 — Factored Moment

#2 — Required Effective Depth

#3 — Steel Area and Bar Count

Problem 10: Investigation — Adequacy Check

#1 — Compression Block and Neutral Axis

#2 — Design Moment Capacity

#3 — Adequacy Check

Problem 11: Investigation — Maximum Allowable Live Load

#1 — Moment Capacity

#2 — Maximum Service Live Load

Problem 12: Investigation — Transition Zone φ

#1 — Steel Ratio vs ρmax

#2 — Compression Block and Strain

#3 — Reduced φ and φMn

Problem 13: Design — Given Mu Directly

#1 — ρmax and Design ρ

#2 — Required Depth

#3 — Steel Area

Problem 14: Investigation — Steel Does NOT Yield (Overreinforced Beam)

#1 — Check if Steel Yields

#2 — Find Neutral Axis (C = T, with fs unknown)

#3 — Design Moment Capacity

Problem 6: Investigation — Design Moment Capacity φM_n

Problem 7: Design — Using ρ_max (ω Method)

#1 — ρ_max and Design ρ

Step 1 — Moment Capacity φM_n

#1 — Steel Ratio vs ρ_max

#3 — Reduced φ and φM_n

Problem 13: Design — Given M_u Directly

#1 — ρ_max and Design ρ

#2 — Find Neutral Axis (C = T, with f_s unknown)