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, 1 bar diameter)
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 = [(πd²) / 4] (no. of pcs of bars) = [(π(8/8)²) / 4] (4) A_s = 3.16 in²
Solve T:
T = A_s · f_y = 3.16 in² (60 ksi) T = 189.9 kips
2. C = T
C = volume of compression zone = 0.85 · f'_c · b · a
Solve a:
0.85 (4 ksi)(12 in)(a)(4 bars) = 189.9 kips a = 4.66 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
- c. For f'_c greater than 8000 psi, β₁ = 0.65
a = β₁ · C 4.66 in = 0.85 (C) C = 5.48 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 − 5.48) = (0.003 / 5.48) ε_s = 0.0067
Therefore, ε_s ≥ ε_y, ASSUMPTION OK
4. Compute Nominal Moment (Mₙ)
Mₙ = T · [d − (a/2)] = 189.6 kips [17.5 in − (4.66 in / 2)] Mₙ = 2876.23 kips-in divide by 12 Mₙ = 239.69 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 A.
A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c) = (200 / 60000 psi)(12 in)(17.5 in) A_s,min = 0.7 Therefore, A_s,min = 0.7
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 (standard if not stated, 1 bar diameter)
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 = [(πd²) / 4] (no. of pcs of bars) = [(π(9/8)²) / 4] (4) A_s = 3.98 in²
Solve T:
T = A_s · f_y = 3.98 in² (60 ksi) T = 238.56 kips
2. C = T
C = volume of compression zone = 0.85 · f'_c · b · a
Solve a:
0.85 (4 ksi)(12 in)(a)(4 bars) = 189.9 kips a = 5.85 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
- c. For f'_c greater than 8000 psi, β₁ = 0.65
a = β₁ · C 5.85 in = 0.85 (C) C = 6.88 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 − 6.88) = (0.003 / 6.88) ε_s = 0.0046
Therefore, ε_s ≥ ε_y, ASSUMPTION OK
4. Compute Nominal Moment (Mₙ)
Mₙ = T · [d − (a/2)] = 238.56 kips [17.5 in − (5.85 in / 2)] Mₙ = 3477.017 kips-in divide by 12 Mₙ = 289.751 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 A.
A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c) = (200 / 60000 psi)(12 in)(17.5 in) A_s,min = 0.7 Therefore, A_s,min = 0.7
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 = 0.65 + (ε_t − 0.002)(250/3) = 0.65 + (0.0046 − 0.002)(250/3) φ = 0.87
7. Solve φ(Mₙ)
φ(Mₙ) = 0.87 (289.751 kip-ft) φ(Mₙ) = 252.08 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 (standard if not stated, 1 bar diameter)
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
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. ≈ 16.8 in.
1. Assume tension bars are yielded
f_s = f_y, ε_s ≥ ε_y
Solve A_s:
A_s = [(πd²) / 4] (no. of pcs of bars) = [(π(9/8)²) / 4] (6 bars) A_s = 5.96 in²
Solve T:
T = A_s · f_y = 5.96 in² (60 ksi)
T = 357.85 kips
2. C = T
C = volume of compression zone = 0.85 · f'_c · b · a
Solve a:
0.85 (4 ksi)(12 in)(a)(4 bars) = 357.85 kips a = 8.77 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
- c. For f'_c greater than 8000 psi, β₁ = 0.65
a = β₁ · C 8.77 in = 0.85 (C) C = 10.32 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 − 8.77) = (0.003 / 8.77) ε_s = 0.00188
Therefore, ε_s < ε_y, ASSUMPTION NOT CONFIRMED, OVER REINFORCED STEEL
4. Compute Nominal Moment (Mₙ) by enforcing strain compatibility and section equilibrium
Solve new value C
T = A_s · f_s T = A_s · E_s · ε_s T = A_s · E_s · [(d − C) / C] · ε_cu
357.85 kips = 5.96 in² (29000 ksi) [(16.8 − C) / C] (0.003)
C = 9.94
cc = 0.85 · f'_c · (b)(β₁ · C) T = cc
section equilibrium. Referring to Fig. 4-18, we must now assume that the steel stress, f_s, is an unknown but is equal to the steel strain, ε_s, multiplied by the steel modulus, E_s. Strain compatibility as expressed in Eq. (4-18) still applies, so the steel stress and thus the tension force can be expressed as a function of the unknown neutral axis depth, c.
T = A_s · f_s = A_s · E_s · ε_s = A_s · E_s · ((d − c) / c) · ε_cu
Similarly, the concrete compression force can be expressed as a function of the neutral axis depth, c.
C_c = 0.85 · f'_c · b · β₁ · c
Enforcing equilibrium by setting T = C_c, we can solve a second degree equation for the unknown value of c. The solution normally results in one positive and one negative value for c; the positive value will be selected. Using all of the given section and material properties and recalling that E_s = 29,000 ksi and ε_cu = 0.003, the resulting value for c is 10.1 in. Using this value, the authors obtained:
T = 346 kips ≅ C_c = 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 = 4350 k-in. = 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)
6. Compute ε_t Again using new value of C
7. Solve Strength Reduction Factor φ
(Because ε_t is between 0.002 and 0.005, this is a transition zone section)
Thus, φ = 0.65 = 0.65 + (ε_t − 0.002)(250/3)
8. Solve φ(Mₙ)
3. Confirm that tension steel area exceeds A_s,min. For this beam section, the concrete compressive strength is 4000 psi, as was the case for Beam 1. However, the effective flexural depth d has been reduced to 16.8 in. Using this new value of d, the value for A_s,min is 0.67 in., which is well below the provided tension steel area A_s.
4. Compute the strength reduction factor, φ, and the resulting value of φMₙ. For this section, the value of ε_t will be slightly larger than ε_s and should be used to determine the value of φ. The strain compatibility of Eq. (4-18) can be modified to calculate ε_t by using d_t in place of d. Then,
ε_t = ((d_t − c) / c) · ε_cu = ((17.5 − 10.1) / 10.1) · 0.003 = 0.00220
This is an interesting result, because we have previously considered this to be an over-reinforced section based on the strain, ε_s, calculated at the centroid of the tension reinforcement. However, 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.
φ = 0.65 + (0.00220 − 0.002) · (250 / 3) = 0.67
Then,
φMₙ = 0.67 × 363 = 243 k-ft