SME
Reinforced Concrete Design · Problem type
Axial strength of a spiral column
Use this for “A 16 in square column has 8 #8 bars in a spiral, f′c = 5 ksi, fᵧ = 60 ksi. Find φPn,max.”
Diagram
Gross area Ag = 256 in²
8 bars of 0.79 in², Aₛₜ = 6.32 in².
Concrete strength f′c = 5 ksi
Steel yield stress fᵧ = 60 ksi
Squash load P₀ = 1,440.34 kip
Design axial strength φPn,max = 918.2 kip
Steel ratio ρg = 2.4688%
This is an example. Type your own values to start a new problem.
P₀ (kip)
=
0.85
×
f′c (ksi)
×
(
Ag (in²)
−
Aₛₜ (in²)
)
+
fᵧ (ksi)
×
Aₛₜ (in²)
φPn,max (kip)
=
0.85
×
0.75
×
P₀ (kip)
P₀ is the squash load: the concrete at 0.85f′c on Ag − Aₛₜ plus the bars at their yield stress fᵧ. Ag = h² is the whole section’s area.
h
Column width
example · in
n
Number of bars
example
Ab
Area of one bar
example · in²
ρg
Steel ratio
calculated · %
Formulas
Ag
=
h
2
256
=
16
2
Aₛₜ
=
n
×
Ab
6.32 = 8 × 0.79
P₀
=
0.85
×
f′c
×
(Ag
−
Aₛₜ)
+
fᵧ
×
Aₛₜ
1,440.34 = 0.85 × 5 × (256 − 6.32) + 60 × 6.32
φPn,max
=
0.85
×
0.75
×
P₀
918.2 = 0.85 × 0.75 × 1,440.34
ρg
=
100
×
Aₛₜ
Ag
2.4688
=
100
×
6.32
256
Assumptions
•
ACI 318-19: a short spiral column, loaded nearly through its center; every bar has yielded.
•
φ = 0.75 for a spiral column; the 0.85 cap covers a small accidental eccentricity.
•
A spiral holds at least 6 bars; ρg, bar area over gross area, is from 1% to 8%.
Step-by-step
WE KNOW
h = 16 in, n = 8, Ab = 0.79 in², f′c = 5 ksi, fᵧ = 60 ksi
FIND
gross area (Ag), steel area (Aₛₜ), squash load (P₀), design axial strength (φPn,max), steel ratio (ρg)
Step 1 · Find gross area (Ag)
Use Ag = h²
A square column: side times side.
Ag
=
16
2
Ag = 256 in²
Step 2 · Find steel area (Aₛₜ)
Use Aₛₜ = n × Ab
n
bars,
each
of
area
Ab.
Aₛₜ = 8 × 0.79
Aₛₜ = 6.32 in²
Step 3 · Find squash load (P₀)
Use P₀ = 0.85 × f′c × (Ag − Aₛₜ) + fᵧ × Aₛₜ
The concrete at 0.85f′c on the area the bars leave, plus the steel at fᵧ.
P₀
=
0.85
×
5
×
(256
−
6.32)
+
60
×
6.32
P₀
=
0.85
×
5
×
249.68
+
60
×
6.32
P₀
=
4.25
×
249.68
+
379.2
P₀
=
1,061.14
+
379.2
P₀
=
1,440.34
kip
Step 4 · Find design axial strength (φPn,max)
Use φPn,max = 0.85 × 0.75 × P₀
A spiral column: the cap times φ, applied to the squash load.
φPn,max = 0.85 × 0.75 × 1,440.34
φPn,max = 0.6375 × 1,440.34
φPn,max = 918.2 kip
Step 5 · Find steel ratio (ρg)
Use ρg = 100 × Aₛₜ ÷ Ag
The steel’s share of the gross area, as a percent.
ρg
=
100
×
6.32
256
ρg
=
632
256
ρg = 2.4688%
Check
Put the numbers back in. Both sides should match.
256
=
16
2
✓
6.32 = 8 × 0.79 ✓
1,440.34 = 0.85 × 5 × (256 − 6.32) + 60 × 6.32 ✓
918.2 = 0.85 × 0.75 × 1,440.34 ✓
2.4688
=
100
×
6.32
256
✓
Topic check
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