Give step-by-step solution with explanation and final answer:
Design the tension reinforcement for the following doubly reinforced beam.Compressive strength of concrete is
5
0
0
0
psi
.
Yield

Question

Give step-by-step solution with explanation and final answer:
 ﻿Design the tension reinforcement for the following doubly reinforced beam.Compressive strength of concrete is 
5
0
0
0
 ﻿psi 
.
 ﻿Yield strength of steel is 
\
(
 
4
0
,
0
0
0
 
\
mathrm
{
psi
}
 
\
)
 ﻿for both tension and compression reinforcement. Width 
=
 
1
6
 ﻿in
.
 ﻿Thickness 
=
 
3
0
 ﻿in
.
 ﻿Effective depth for tension steel 
=
 
2
7
 ﻿in
.
 ﻿Effective depth for compression steel 
=
 
3
 ﻿in
.
 ﻿Compression steel: 
2
 
-
 
\
# 
9
 ﻿bars placed in single layer. Ultimate moment 
=
 
4
,
9
5
0
,
0
0
0
 ﻿in
-
lb
.
 ﻿Use 
\
# 
9
 ﻿bars for tension reinforcement.

VIPSolutions · Accepted Answer

# Step-by-Step Solution: Doubly Reinforced Beam Design

Given:

- **Concrete compressive strength:** $ f'_c = 500 $ psi  
- **Steel yield strength:** $ f_y = 40,000 $ psi (both tension and compression)
- **Beam width:** $ b = 16 $ in
- **Overall thickness:** $ h = 30 $ in
- **Effective depth to tension steel:** $ d = 27 $ in
- **Effective depth to compression steel:** $ d' = 3 $ in
- **Compression steel:** 2 #9 bars in single layer
- **Ultimate moment:** $ M_u = 4,950,000 $ in-lb
- **Use #9 bars for tension steel**

Let's proceed step by step.

---

## 1. **Convert Units**

$$
M_u = 4,950,000~\text{in-lb} = \frac{4,950,000}{12} = 412,500~\text{ft-lb} = 4,950~\text{kip-in}
$$

But we'll keep it in in-lb since other dimensions are in inches.

---

## 2. **Area of #9 Bar**

Standard #9 bar area ($ A_b $):  
$$
A_b = 1.00~\text{in}^2 \quad \text{(from ACI Table)}
$$

---

## 3. **Area of Compression Reinforcement**

2 #9 bars:
$$
A_s' = 2 \times 1.00 = 2.00~\text{in}^2
$$

---

## 4. **Reduction Factor ($ \phi $)**

ACI 318 for flexure (tension-controlled) beams:
$$
\phi = .9
$$

---

## 5. **Calculate Factored Moment $ M_n $**

$$
\phi M_n \ge M_u \implies M_n = \frac{M_u}{\phi} = \frac{4,950,000}{.9} = 5,500,000~\text{in-lb}
$$

---

## 6. **Check Limiting Moment for Singly Reinforced Beam**

Maximum singly reinforced moment:
$$
M_{n,\text{max}} = .90 \times A_{s,\text{max}} \times f_y \times (d - a/2)
$$
But let's use the classic limit for singly reinforced:

$$
M_{n,\text{max}} = .85 f'_c b a_{\text{max}} (d - a_{\text{max}}/2)
$$

Where $ a_{\text{max}} = \beta_1 c_{\text{max}} $, and $ c_{\text{max}} $ is the distance to the neutral axis at yield strain limit.

For $ f'_c = 500 $ psi:
$$
\beta_1 = .85 - .05 \left(\frac{f'_c - 400}{100}\right) = .85 - .05(1) = .80
$$
(Limit min .65, max .85)

For tension-controlled section ($ \varepsilon_t \ge .005 $),  
$ c_{\text{max}} = \frac{\varepsilon_{cu}}{\varepsilon_{cu} + \varepsilon_y}d $

But for quick estimation, let’s assume singly reinforced capacity is not enough and proceed to doubly reinforced calculation.

---

## 7. **Force in Compression Steel ($ A_s' $)**

Assume compression steel yields:

$$
F_s' = A_s' f_y = 2.00 \times 40,000 = 80,000~\text{lb}
$$

---

## 8. **Assume Neutral Axis Depth, $ c $, and Compute $ a $**

Let the total compressive force be:

$$
C = .85 f'_c b a + F_s'
$$
$$
T = A_s f_y
$$
$$
C = T
$$

Let’s denote $ a $ as the depth of equivalent rectangular stress block.

$$
T = .85 f'_c b a + A_s' f_y
$$
$$
A_s = \frac{.85 f'_c b a}{f_y} + A_s'
$$

---

## 9. **Relate $ a $ to $ M_n $**

Total moment capacity:

$$
M_n = .85 f'_c b a (d - \frac{a}{2}) + A_s' f_y (d - d')
$$

Set $ M_n = 5,500,000 $ in-lb. This is a nonlinear equation in $ a $. We'll solve by iteration.

Let’s denote:

- $ b = 16 $ in
- $ d = 27 $ in
- $ d' = 3 $ in
- $ f'_c = 500 $ psi
- $ f_y = 40,000 $ psi
- $ A_s' = 2.00 $ in$^2$

Let’s try $ a = 4 $ in (initial guess).

$$
C_c = .85 \times 500 \times 16 \times 4 = .85 \times 500 \times 64 = .85 \times 320,000 = 272,000~\text{lb}
$$
$$
M_{c} = C_c \times (d - a/2) = 272,000 \times (27 - 2) = 272,000 \times 25 = 6,800,000~\text{in-lb}
$$
$$
M_{s'} = 80,000 \times (27 - 3) = 80,000 \times 24 = 1,920,000~\text{in-lb}
$$
But for moment, **compression steel acts in opposite direction**. The total moment is:

$$
M_n = C_c(d - a/2) + F_s'(d - d')
$$

But in the ACI approach, the *net* moment capacity is:

$$
M_n = .85 f'_c b a (d - a/2) + A_s' f_y (d - d')
$$

Plugging in $ a = 4 $ in:

$$
M_n = 272,000 \times 25 + 80,000 \times 24 = 6,800,000 + 1,920,000 = 8,720,000~\text{in-lb}
$$

But this is much higher than needed ($ 5,500,000 $). Let's try a smaller $ a $.

Try $ a = 2.5 $ in:

$$
C_c = .85 \times 500 \times 16 \times 2.5 = .85 \times 500 \times 40 = .85 \times 200,000 = 170,000~\text{lb}
$$
$$
M_n = 170,000 \times (27 - 1.25) + 80,000 \times 24 = 170,000 \times 25.75 + 1,920,000 = 4,377,500 + 1,920,000 = 6,297,500~\text{in-lb}
$$

Still high. Try $ a = 2 $ in:

$$
C_c = .85 \times 500 \times 16 \times 2 = .85 \times 500 \times 32 = .85 \times 160,000 = 136,000~\text{lb}
$$
$$
M_n = 136,000 \times (27 - 1) + 1,920,000 = 136,000 \times 26 + 1,920,000 = 3,536,000 + 1,920,000 = 5,456,000~\text{in-lb}
$$

Very close to required $ 5,500,000 $. Try $ a = 2.1 $:

$$
C_c = .85 \times 500 \times 16 \times 2.1 = .85 \times 500 \times 33.6 = .85 \times 168,000 = 142,800~\text{lb}
$$
$$
M_n = 142,800 \times (27 - 1.05) + 1,920,000 = 142,800 \times 25.95 + 1,920,000 = 3,704,460 + 1,920,000 = 5,624,460~\text{in-lb}
$$

Too high. Try $ a = 2.05 $:

$$
C_c = .85 \times 500 \times 16 \times 2.05 = .85 \times 500 \times 32.8 = .85 \times 164,000 = 139,400~\text{lb}
$$
$$
M_n = 139,400 \times (27 - 1.025) + 1,920,000 = 139,400 \times 25.975 + 1,920,000 = 3,621,215 + 1,920,000 = 5,541,215~\text{in-lb}
$$

That's very close to our target ($ 5,500,000 $).

---

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