In this article, you going to learn about beam design, **how to calculate the area of tension reinforcement in a beam** and **how to determine a beam-unknown cross-section** by using ACI 318-19 design code. We also give a beam design example for better understanding and also gave beam design formulas with reference.

## Table of Contents

**Step 1:** Minimum depth of nonprestressed beams: **(ACI Table 9.3.1.1)**

**Step 2:** Ultimate Load: W_{u}

**Step 3:** Ultimate Moment: M_{u}

\tag{4.26d} {{\rho }_{\max }}=0.85{{\beta }_{1}}\frac{f_{c}^{'}}{{{f}_{y}}}\frac{{{\varepsilon }_{u}}}{{{\varepsilon }_{u}}+{{\varepsilon }_{t,\min }}}

**Step 5: **Calculate bd^{2} from Nominal Moment: \phi {{M}_{n}}=\phi \rho {{f}_{y}}b{{d}^{2}}\left( 1-0.59\frac{\rho {{f}_{y}}}{f_{c}^{'}} \right) (4.34)

**Step 6:** Find the trial size of the beam

**Step 7:** Calculate the area of steel

**Step 8:** Check for 𝜙 = 0.9

**Step 9:** Rebar calculation & Detailing

Find the concrete cross-section and the steel area required for a simply supported rectangular beam with a** span of 20 ft** that is to carry a calculated **dead load of 1.5 kips∕ft** and a service **live load of 2.6 kips∕ft**, as shown in Figure below. Material strengths are **fc′ = 4000 psi** and **fy = 60,000 psi.**

## Solution

Given, **L = 20 ftDead Load = 1.5 kips/ftLive Load = 2.6 kips/ftUnit weight of concrete, γ**

_{c}= 150 pcf (lb/ft

^{3})f’c = 4000 psify = 6000 psiβ

_{1 }= 0.85

**Step 1: Minimum depth of nonprestressed beams: (ACI Table 9.3.1.1)**

Assume, Beam depth-width ratio = 1.5 (h/b=1.5)

h_{min}= L/16 = (20*12)/16= 15 inch

b_{min}= 15/1.5 = 10 inch

**Step 2: Ultimate Load: W _{u}**

**Self-weight W _{o} =** {{\gamma }_{c}}*{{h}_{\min }}{{b}_{\min }}=150\left( \frac{15}{12} \right)\left( \frac{10}{12} \right)=156.25lb/ft=0.156kips/ft

**Ultimate load** **W _{u}** = 1.2DL + 1.6LL = 1.2(w

_{o}+DL)+1.6LL

= 1.2(0.156+1.5)+1.6(2.6)

**W**

_{u}= 6.147 kips/ft**Step 3: Ultimate Moment: M _{u}**

For simply supported beam

Maximum moment M_{u}:

\begin{align} & {{M}_{u}}=\frac{{{W}_{u}}{{L}^{2}}}{8}=\frac{(6.147){{(20)}^{2}}}{8} \\ & {{M}_{u}}=307.35kips-ft \\ \end{align} \tag{}

**Step 4: Reinforcement ratio:**

\begin{align} & {{\rho }_{\max }}=0.85{{\beta }_{1}}\frac{f_{c}^{'}}{{{f}_{y}}}\frac{{{\varepsilon }_{u}}}{{{\varepsilon }_{u}}+{{\varepsilon }_{t,\min }}} \\ & \to {{\rho }_{\max }}=0.85*0.85*\frac{4}{60}*\frac{0.003}{0.003+0.005} \\ & \to {{\rho }_{\max }}=0.0180 \\ \end{align} \tag{}

Note: Mobile users must use desktop mode to view equations correctly.

For Economy **Use 60%-80% of the maximum reinforcement ratio**. This is required for Underenforcement beam design.

{{\rho }_{design}}=0.8*{{\rho }_{\max }}=0.0145

**Step 5: Calculate bd ^{2} from Nominal Moment:**

\phi {{M}_{n}}=\phi \rho {{f}_{y}}b{{d}^{2}}\left( 1-0.59\frac{\rho {{f}_{y}}}{f_{c}^{'}} \right)

\tag{}\begin{align} & {{M}_{u}}=\phi {{M}_{n}} \\ & 307.35*12=0.9*0.0145*60*b{{d}^{2}}\left( 1-0.59\frac{0.0145*60}{4} \right) \\ & \therefore b{{d}^{2}}=5403.79i{{n}^{3}} \\ \end{align}

**Step 6: Find the trial size of the beam**

Used,

d = 1.5b

d^{2 }= 2.25b^{2 }(Square both side)

bd^{2 }= 2.25b^{3} (both sides multiply by b)

Now,

2.25b^{3 }= 5403.79 in^{3}

b = 13.4 inch

So, d= 1.5b= 1.5*13.4 = 20.1 inch

So, trial size of the beam

**b = 13.4 inch****d = 20.1 inch**

**Note: You can assume b=14 in & d=20.5 in for more practical usages. For this example, we are using it as it is.**

**Step 7: Calculate the area of steel**

{{A}_{s}}={{\rho }_{design}}bd=0.0145*13.4*20.1=3.90\text{ }i{{n}^{2}}

**Minimum flexural reinforcement (ACI 9.6.1.2):**

{{A}_{s,\min }}\text{ larger of }\left\{ \begin{matrix} \frac{3\sqrt{f_{c}^{'}}}{{{f}_{y}}}{{b}_{w}}d=0.85\text{ i}{{\text{n}}^{\text{2}}} \\\\ \frac{200}{{{f}_{y}}}{{b}_{w}}d=0.90\text{ i}{{\text{n}}^{\text{2}}} \\ \end{matrix} \right. \tag{4.37a}

\therefore {{A}_{s,\min }}=0.90\text{ i}{{\text{n}}^{2}}\text{ OK}\text{.}

Note: f_{c}^{'}\text{ and }{{f}_{y}} must be in psi.

**Step 8: Check for 𝛟 = 0.9 (tension control)**

a=\frac{{{A}_{s}}{{f}_{y}}}{0.85f_{c}^{'}b}=\frac{3.90*60}{0.85*4*13.4}=5.13\text{ in} \tag{4.28}

c=\frac{a}{{{\beta }_{1}}}=\frac{5.13}{0.85}=6.03\text{ in}

{{\varepsilon }_{t}}={{\varepsilon }_{u}}\frac{d-c}{c}=0.003*\frac{20.1-6.03}{6.03}=0.007>0.005\text{ OK}

If you do not reduce the value of {{\rho }_{\max }} then It becomes exactly 0.005 or maybe less. This means the reduction factor 𝛟 = 0.9 it doesn’t applicable. So you must reduce the {{\rho }_{\max }} value as {{\rho }_{design}} .

**Step 9: Rebar calculation & Detailing**

**Used 4-#9 (4.00 in ^{2}) bar**

{{A}_{s,provide}}(4.00\text{ }i{{n}^{2}})>{{A}_{s,required}}(3.90\text{ }i{{n}^{2}})\text{ OK}

Assuming 2.5 in. concrete cover from the centroid of the bars,

h = d+2.5” = 20.1+2.5=22.6 inch**Beam Size:b=13.4 inchh=22.6 inch**

## Reference

- Design of Concrete Structures Sixteenth Edition by David Drawin & Charles W. Dolan.
- Building Code Requirements for Structural Concrete (ACI 318-19)
- ASCE 7_22 Minimum Design Loads and Associated Criteria for Buildings

Check this **Beam Design Calculator ACI 318-19**

See more Article:

## Discussion about this post