SME
Reinforced Concrete Design
Flexure
Use this for “A beam 12 in wide with d = 20 in has 3 #8 bars, f′c = 4000 psi and fᵧ = 60 ksi. Find φMₙ and check it is tension-controlled.”
Diagram
3 bars of 0.79 in², Aₛ = 2.37 in².
h drawn as d plus half a bar, the stirrup and the cover.
c = 4.1 in, a = 3.485 in
a = β₁c, and the strain line crosses zero at c.
This is an example. Type your own values to start a new problem.
b
Beam width
example · in
d
Depth to the bars
example · in
n
Number of bars
example
Ab
Area of one bar
example · in²
Aₛ
Steel area
calculated · in²
f′c
Concrete strength
example
fᵧ
Steel yield stress
example · ksi
β₁
Stress-block factor
worked out (not typed)
a
Depth of the stress block
calculated · in
c
Depth to the neutral axis
worked out (not typed) · in
εₜ
Strain in the steel
calculated
φMₙ
Design moment strength
calculated
φ
Strength reduction factor
worked out (not typed)
T
Section
worked out (not typed)
Formulas
Aₛ
=
n
×
Ab
2.37 = 3 × 0.79
a
=
Aₛ
×
fᵧ
0.85 × f′c ×
b
3.4853
=
2.37
×
60
0.85 × 4 × 12
β₁
=
0.85
0.85 = 0.85
c
=
a
β₁
4.1003
=
3.4853
0.85
εₜ
=
0.003
×
d
−
c
c
0.01163
=
0.003
×
20 − 4.1003
4.1003
φ
=
0.9
0.9 = 0.9
φMₙ
=
φ
×
Aₛ
×
fᵧ
×
d
−
a
÷ 2
12
194.7
=
0.9
×
2.37
×
60
×
20 − 3.4853 ÷ 2
12
φMₙ
=
0.9
×
Aₛ
×
fᵧ
×
(
d
−
Aₛ
×
fᵧ
÷
(1.7
×
f′c
×
b
))
÷
12
194.7 = 0.9 × 2.37 × 60 × (20 − 2.37 × 60 ÷ (1.7 × 4 × 12)) ÷ 12
T
=
tension-controlled
if
εₜ
≥
0.005,
transition
zone
if
0.002
<
εₜ
<
0.005,
else
compression-controlled
0.01163 ≥ 0.005, so tension-controlled
Assumptions
•
ACI 318-19: plane sections stay plane, the concrete crushes at a strain of 0.003, and the steel has yielded.
•
The Whitney block: 0.85f′c over a depth a = β₁c; β₁ = 0.85 up to f′c = 4 ksi, 0.05 less for each ksi above, and 0.65 from 8 ksi.
•
φ = 0.9 while εₜ ≥ 0.005, 0.65 at εₜ ≤ 0.002 (εₜᵧ for Grade 60), a straight line between (tied).
•
Bar areas are the standard sizes #3 to #11; kip·in ÷ 12 gives kip·ft.
Step-by-step
WE KNOW
n = 3, Ab = 0.79 in², b = 12 in, d = 20 in, f′c = 4 ksi, fᵧ = 60 ksi
FIND
steel area (Aₛ), depth of the stress block (a), stress-block factor (β₁), depth to the neutral axis (c), strain in the steel (εₜ), strength reduction factor (φ), design moment strength (φMₙ), section (T)
Step 1 · Find steel area (Aₛ)
Use Aₛ = n × Ab
n
bars
of
area
Ab
each.
Aₛ = 3 × 0.79
Aₛ = 2.37 in²
Step 2 · Find depth of the stress block (a)
Use a = Aₛ × fᵧ ÷ (0.85 × f′c × b)
The steel’s pull Aₛfᵧ equals the block’s push 0.85f′c × ab.
a
=
2.37
×
60
0.85 × 4 × 12
a
=
2.37
×
60
3.4 × 12
a
=
2.37
×
60
40.8
a
=
142.2
40.8
a
=
3.48529
in
Step 3 · Find stress-block factor (β₁)
Use β₁ = 0.85
Up to 4 ksi the stress block reaches 0.85 of the way to the neutral axis.
4 ≤ 4: up to 4 ksi
β₁
=
0.85
Step 4 · Find depth to the neutral axis (c)
Use c = a ÷ β₁
The
block
is
β₁
of
the
depth
to
the
neutral
axis.
c
=
3.48529
0.85
c
=
4.1003
in
Step 5 · Find strain in the steel (εₜ)
Use εₜ = 0.003 × (d − c) ÷ c
The
strain
line
runs
from
0.003
at
the
top
through
zero
at
c
,
so
similar
triangles
give
it
at
the
bars.
εₜ
=
0.003
×
20 − 4.1003
4.1003
εₜ
=
0.003
×
15.8997
4.1003
εₜ
=
0.047699
4.1003
εₜ = 0.01163
Step 6 · Find strength reduction factor (φ)
Use φ = 0.9
The
bars
stretch
well
past
yield
before
the
concrete
crushes,
so
the
full
φ
=
0.9
applies.
0.01163 ≥ 0.005: tension-controlled
φ
=
0.9
Step 7 · Find design moment strength (φMₙ)
Use φMₙ = φ × Aₛ × fᵧ × (d − a ÷ 2) ÷ 12
The
pull
Aₛfᵧ
times
its
lever
arm
d
−
a
2
to
the
block’s
centre,
then
times
φ
and
12
for
kip·ft.
φMₙ
=
0.9
×
2.37
×
60
×
20 − 3.48529 ÷ 2
12
φMₙ
=
0.9
×
2.37
×
60
×
20 − 1.743
12
φMₙ
=
0.9
×
2.37
×
60
×
18.257
12
φMₙ
=
2.133
×
60
×
18.257
12
φMₙ
=
127.98
×
18.257
12
φMₙ
=
2,337
12
φMₙ = 194.7 kip·ft
Step 8 · Find section (T)
Use T = tension-controlled if εₜ ≥ 0.005, transition zone if 0.002 < εₜ < 0.005, else compression-controlled
The bars stretch well past yield before the concrete crushes, so the beam warns before it fails.
T
=
tension-controlled
if
εₜ
≥
0.005
T
=
tension-controlled
if
0.01163
≥
0.005
T
=
tension-controlled
Check
Put the numbers back in. Both sides should match.
2.37 = 3 × 0.79 ✓
3.48529
=
2.37
×
60
0.85 × 4 × 12
✓
0.85 = 0.85 ✓
4.1003
=
3.48529
0.85
✓
0.01163
=
0.003
×
20 − 4.1003
4.1003
✓
0.9 = 0.9 ✓
194.7
=
0.9
×
2.37
×
60
×
20 − 3.48529 ÷ 2
12
✓
194.7 = 0.9 × 2.37 × 60 × (20 − 2.37 × 60 ÷ (1.7 × 4 × 12)) ÷ 12 ✓
0.01163 ≥ 0.005, so tension-controlled ✓
Topic check
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