discussion/main_corrected.md

# Note:

- **"ε"** is epsilon
- **β₁** is beta₁

- **A_s** is the area of reinforcement near the tension face of the beam, tension reinforcement, in.².



- **A'_s** is the area of reinforcement on the compression side of the beam, compression reinforcement, in.².

- **b** is a general symbol for the width of the compression zone in a beam, in. This is illustrated in Fig. 4-2 for positive and negative moment regions. For flanged sections this symbol will normally be replace with b_e or b_w.

- **b_e** is the effective width of a compression zone for a flanged section with compression in the flange, in.



- **b_w** is the width of the web of the beam (and may or may not be the same as b), in.

- **d** is the distance from the extreme fiber in compression to the centroid of the longitudinal reinforcement on the tension side of the member, in. In the positive-moment region (Fig. 4-2a), the tension steel is near the bottom of the beam, while in the negative-moment region (Fig. 4-2b) it is near the top.

- **d'** is the distance from the extreme compression fiber to the centroid of the longitudinal compression steel, in.

- **d_t** is the distance from the extreme compression fiber to the farthest layer of tension steel, in. For a single layer of tension reinforcement, d_t = d, as shown in Fig. 4-2b.



- **f'_c** is the specified compressive strength of the concrete, psi.



- **f_c** is the stress in the concrete, psi.



- **f_s** is the stress in the tension reinforcement, psi.



- **f_y** is the specified yield strength of the reinforcement, psi.



- **h** is the overall height of a beam cross section.



- **jd** is the *lever arm*, the distance between the resultant compressive force and the resultant tensile force, in.



- **j** is a dimensionless ratio used to define the lever arm, *jd*. It varies depending on the moment acting on the beam section.



- **ε_cu** is the assumed maximum useable compression strain in the concrete.



- **ε_s** is the strain in the tension reinforcement.



- **ε_t** is the strain in the extreme layer of tension reinforcement.



- **ρ** is the longitudinal tension reinforcement ratio, ρ = A_s / bd.

---

# Calculation of Mₙ for a Singly Reinforced Rectangular Section

For the beam shown in Fig. 4-19a, calculate Mₙ and confirm that the area of tension steel
exceeds the required minimum steel area given by Eq. (4-11). The beam section is made of
concrete with a compressive strength, f'_c = 4000 psi, and has four No. 8 bars with a yield

strength of f_y = 60 ksi.

For this beam with a single layer of tension reinforcement, it is reasonable to assume that
the effective flexural depth, d, is approximately equal to the total beam depth minus 2.5 in.
This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup
(typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforcement.
Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the
steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design
work unless adjustments in reinforcement location are required to avoid rebar interference at
connections with other members. Small bars are often used in the compression zone to hold the
stirrups in position, but these bars normally are ignored unless they were specifically designed
to serve as compression-zone reinforcement.

## Given:

- f'_c = 4000 psi

- f_y = 60000 psi
- b = 12 in
- h = 20 in
- 4 No. 8 bars
- concrete cover: 2.5 in (standard if not stated, distance to centroid)

## Goal is to solve for:

a. d
b. A_s

c. T

d. a

e. c

f. ε_y
g. ε_s

h. Mₙ

i. A_s,min

> solve d

d = h − concrete cover
= 20 − 2.5
**d = 17.5 in**

## STEPS

### 1. Assume tension bars are yielded

f_s = f_y, ε_s ≥ or equal to ε_y
= 60000 psi

> Solve A_s:



A_s = (4 bars) × (0.79 in²/bar from standard nominal area)
**A_s = 3.16 in²**



> Solve T:



T = A_s · f_y

= 3.16 in² (60 ksi)

**T = 189.6 kips**



### 2. C_c = T



C_c = volume of compression zone = 0.85 · f'_c · b · a



> Solve a:



0.85 (4 ksi)(12 in)(a) = 189.6 kips

40.8 (a) = 189.6

**a = 4.65 in**

> Solve c

(Note: Whitney Stress Block)
- a. f'_c up to and including 4000 psi, β₁ = 0.85

- b. For 4000 psi < f'_c ≤ 8000 psi
- c. For f'_c greater than 8000 psi, β₁ = 0.65



a = β₁ · c

4.65 in = 0.85 (c)

**c = 5.47 in**





### 3. Check Assumption



> Solve ε_y

ε_y = f_y / E_s

ε_y = 60 ksi / 29000 ksi
**ε_y = 0.00207**



> Solve ε_s



(ε_s / (d − c)) = (ε_cu / c)

(ε_s / (17.5 − 5.47)) = (0.003 / 5.47)

**ε_s = 0.0066**

Therefore, ε_s ≥ ε_y,
**ASSUMPTION OK**

### 4. Compute Nominal Moment (Mₙ)

Mₙ = T · [d − (a/2)]
= 189.6 kips [17.5 in − (4.65 in / 2)]
Mₙ = 2877 kips-in
divide by 12
**Mₙ = 239.75 kip-ft**

### 5. Check if it exceeds the required minimum of the code (A_s,min)



> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)



> check 3√(f'_c)
= 3√4000 psi
= 189.74 < 200 , therefore A. governs, use 200.

> A. A_s,min = (200 / f_y)(b)(d)
= (200 / 60000 psi)(12 in)(17.5 in)
A_s,min = 0.70 in²

Therefore, **A_s,min = 0.70 in²**



Since A_s (3.16 in²) > A_s,min (0.70 in²), **OK**.



---



# SINGLE LAYER REINFORCEMENT



Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three

singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.

As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small

longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.

These bars typically are ignored in the calculation of the section nominal moment strength.

Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the

distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.



**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²

## Given:

- f'_c = 4000 psi

- f_y = 60 ksi
- b = 12 in
- h = 20 in
- 4 No. 9 bars

- concrete cover: 2.5 in (distance to centroid)

## Goal is to solve for:

a. d
b. A_s

c. T

d. a

e. c

f. ε_y
g. ε_s

h. Mₙ

i. A_s,min

> solve d

d = h − concrete cover
= 20 − 2.5
**d = 17.5 in**

## STEPS

### 1. Assume tension bars are yielded

f_s = f_y, ε_s ≥ ε_y

> Solve A_s:



A_s = (4 bars) × (1.00 in²/bar from nominal area)
**A_s = 4.00 in²**



> Solve T:



T = A_s · f_y

= 4.00 in² (60 ksi)

**T = 240 kips**



### 2. C_c = T



C_c = volume of compression zone = 0.85 · f'_c · b · a



> Solve a:



0.85 (4 ksi)(12 in)(a) = 240 kips

40.8 (a) = 240

**a = 5.88 in**

> Solve c

(Note: Whitney Stress Block)
- a. f'_c up to and including 4000 psi, β₁ = 0.85



a = β₁ · c

5.88 in = 0.85 (c)

**c = 6.92 in**



### 3. Check Assumption



> Solve ε_y

ε_y = f_y / E_s

ε_y = 60 ksi / 29000 ksi
**ε_y = 0.00207**



> Solve ε_s



(ε_s / (d − c)) = (ε_cu / c)

(ε_s / (17.5 − 6.92)) = (0.003 / 6.92)

**ε_s = 0.00459**

Therefore, ε_s ≥ ε_y,
**ASSUMPTION OK**

### 4. Compute Nominal Moment (Mₙ)

Mₙ = T · [d − (a/2)]
= 240 kips [17.5 in − (5.88 in / 2)]
Mₙ = 3494 kips-in
divide by 12
**Mₙ = 291.17 kip-ft**

### 5. Check if it exceeds the required minimum of the code (A_s,min)



> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)



> check 3√(f'_c)
= 3√4000 psi
= 189.74 < 200 , therefore A. governs, use 200.

> A. A_s,min = (200 / f_y)(b)(d)
= (200 / 60000 psi)(12 in)(17.5 in)
**A_s,min = 0.70 in²**



Since A_s (4.00 in²) > A_s,min (0.70 in²), **OK**.



### 6. Solve Strength Reduction Factor φ



(Note: for a single layer of tension reinforcement, ε_t = ε_s. Because ε_t is between

0.002 and 0.005, this is a transition zone section)



Thus, φ = 0.65 + (ε_t − 0.002)(250/3)

= 0.65 + (0.00459 − 0.002)(250/3)

**φ = 0.87**



### 7. Solve φ(Mₙ)



φ(Mₙ) = 0.87 **(291.17 kip-ft)**

**φ(Mₙ) = 253.3 k-ft**



---



# DOUBLE / MULTIPLE LAYER Reinforcement



Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three

singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.

As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small

longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.

These bars typically are ignored in the calculation of the section nominal moment strength.

Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the

distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.



**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²



**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers

## Given:

- f'_c = 4000 psi

- f_y = 60 ksi
- b = 12 in
- h = 20 in
- 6 No. 9 bars
- concrete cover: 2.5 in (distance to bottom layer centroid)

## Goal is to solve for:

a. d
b. g
c. A_s

d. T

e. a

f. c

g. ε_y
h. ε_s

i. Mₙ

j. A_s,min
k. ε_t

l. φ

m. φ(Mₙ)



> solve d



d = h − g



**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers

(Fig. 4-27). For this section, ε_t will be larger than ε_s and will be calculated using the distance

to the extreme layer of tension reinforcement, d_t, with the same cover and size of stirrup
(d_t = 17.5 in.) as used for d in Beams 1 and 2. The value of d for this section involves a

centroid calculation for the six No. 9 bars. ACI Code Section 7.6.2 requires a clear spacing between

layers of reinforcement greater than or equal to 1 in. Thus, we can assume that the second

layer of steel (two bars) is one bar diameter plus 1 in. above the lowest layer — or a total of

2.5 in. + 1.128 in. + 1 in. ≈ 4.63 in. from the extreme tension fiber. A simple calculation is

used to find the distance from the bottom of the beam to the centroid of the tension reinforcement,

g, and then find the value of d = h − g.



g = (4.0 in.² × 2.5 in. + 2.0 in.² × 4.63 in.) / 6.0 in.² = 3.21 in.



d = h − g = 20 in. − 3.21 in.

**d = 16.8 in.**



## STEPS



### 1. Assume tension bars are yielded



f_s = f_y, ε_s ≥ ε_y



> Solve A_s:

A_s = (6 bars) × (1.00 in²/bar from nominal area)

**A_s = 6.00 in²**



> Solve T:



T = A_s · f_y

= 6.00 in² (60 ksi)

**T = 360 kips**



### 2. C_c = T

C_c = volume of compression zone = 0.85 · f'_c · b · a

> Solve a:

0.85 (4 ksi)(12 in)(a) = 360 kips
40.8 (a) = 360
**a = 8.82 in**

> Solve c

(Note: Whitney Stress Block)
- a. f'_c up to and including 4000 psi, β₁ = 0.85



a = β₁ · c

8.82 in = 0.85 (c)

**c = 10.38 in (Use 10.4 in)**



### 3. Check Assumption



> Solve ε_y

ε_y = f_y / E_s

ε_y = 60 ksi / 29000 ksi
**ε_y = 0.00207**



> Solve ε_s



(ε_s / (d − c)) = (ε_cu / c)

(ε_s / (16.8 − 10.4)) = (0.003 / 10.4)

**ε_s = 0.00185**

Therefore, ε_s < ε_y,
**ASSUMPTION NOT CONFIRMED, OVER REINFORCED SECTION**

### 4. Compute Nominal Moment (Mₙ) by enforcing strain compatibility and section equilibrium

> Solve new value for c

T = A_s · f_s
T = A_s · E_s · ε_s

T = A_s · E_s · [(d − c) / c] · ε_cu

Similarly, the concrete compression force is:
C_c = 0.85 · f'_c · b · β₁ · c

Enforcing section equilibrium, T = C_c:

6.00 in² (29000 ksi) [(16.8 − c) / c] (0.003) = 0.85 (4 ksi)(12 in)(0.85 · c)

522 [(16.8 − c) / c] = 34.68 c



Enforcing equilibrium by setting T = C_c, we can solve a second-degree equation for the unknown
value of c. The result is the following quadratic equation:
34.68 c² + 522 c - 8769.6 = 0

Solving for the positive root yields:
**c = 10.1 in**

Using this value:
T = 6.00 in² (29000 ksi) [(16.8 - 10.1) / 10.1] (0.003) = 346 kips
C_c = 34.68 (10.1) = 350 kips



An average value of T = C_c = 348 kips will be used to calculate Mₙ. Then, using
a = β₁ · c = 0.85 × 10.1 in. = 8.59 in., calculate Mₙ using the more general expression in Eq. (4-20).

Mₙ = T · (d − a/2) = 348 kips (16.8 in. − 8.59 in. / 2)
Mₙ = 4351 k-in.
divide by 12
**Mₙ = 363 k-ft**

### 5. Check if it exceeds the required minimum of the code (A_s,min)



> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)



> check 3√(f'_c)
= 3√4000 psi
= 189.74 < 200 , therefore A. governs, use 200.

> A. A_s,min = (200 / f_y)(b)(d)
= (200 / 60000 psi)(12 in)(16.8 in)
**A_s,min = 0.67 in²**



Since A_s (6.00 in²) > A_s,min (0.67 in²). **OK.**



### 6. Compute ε_t using extreme tension layer

The value of ε_t is calculated using the distance to the extreme layer of tension reinforcement,

d_t = 17.5 in.



ε_t = [(d_t − c) / c] · ε_cu = [(17.5 − 10.1) / 10.1] · 0.003

**ε_t = 0.00220**

### 7. Solve Strength Reduction Factor φ

Because of the difference between d and d_t, we now have found the value of ε_t to be between
0.002 and 0.005. Thus, this is a transition-zone section, and we must use Eq. (4-28) to calculate φ.

Thus, φ = 0.65 + (ε_t − 0.002)(250/3)

φ = 0.65 + (0.00220 − 0.002)(250/3)

**φ = 0.67**



### 8. Solve φ(Mₙ)



φ(Mₙ) = 0.67 × 363 k-ft

**φ(Mₙ) = 243.2 k-ft**