SME
Reinforced Concrete Design · Problem type
Factored column load
Use this for “A column carries 120 kips dead load and 80 kips live load. What service load sizes its footing, and what factored load Pᵤ checks it?”
Chart
Dead load D = 120 kip
Live load L = 80 kip
Factored load Pᵤ = 272 kip
This is an example. Type your own values to start a new problem.
U₁ (kip)
=
1.4
×
D (kip)
U₂ (kip)
=
1.2
×
D (kip)
+
1.6
×
L (kip)
Pᵤ (kip)
=
max(
U₁ (kip)
,
U₂ (kip)
)
P
Service load
calculated · kip
Formulas
P
=
D
+
L
200 = 120 + 80
U₁
=
1.4
×
D
168 = 1.4 × 120
U₂
=
1.2
×
D
+
1.6
×
L
272 = 1.2 × 120 + 1.6 × 80
Pᵤ
=
max(
U₁
,
U₂
)
272 = max(168, 272)
Assumptions
•
ACI 318-19 strength combinations (from ASCE 7) with dead and live load only: 1.4D and 1.2D + 1.6L.
•
D and L are service (unfactored) loads; their sum P sizes the footing on the allowable soil pressure.
•
The larger combination is Pᵤ; 1.4D governs only when L is less than D ÷ 8.
Step-by-step
WE KNOW
D = 120 kip, L = 80 kip
FIND
service load (P), combination 1 (1.4D) (U₁), combination 2 (1.2D + 1.6L) (U₂), factored load (Pᵤ)
Step 1 · Find service load (P)
Use P = D + L
The load the soil carries day to day: dead and live, unfactored.
P
=
120
+
80
P
=
200
kip
Step 2 · Find combination 1 (1.4D): U₁
Use U₁ = 1.4 × D
Dead load alone, raised by 1.4: it governs when there is little live load.
U₁
=
1.4
×
120
U₁
=
168
kip
Step 3 · Find combination 2 (1.2D + 1.6L): U₂
Use U₂ = 1.2 × D + 1.6 × L
Live load varies more than dead load, so it takes the larger factor, 1.6.
U₂
=
1.2
×
120
+
1.6
×
80
U₂
=
144
+
128
U₂
=
272
kip
Step 4 · Find factored load (Pᵤ)
Use Pᵤ = max(U₁, U₂)
The footing must carry the worse of the two, so the larger combination is Pᵤ.
Pᵤ = max(168, 272)
Pᵤ = 272 kip
Check
Put the numbers back in. Both sides should match.
200 = 120 + 80 ✓
168 = 1.4 × 120 ✓
272 = 1.2 × 120 + 1.6 × 80 ✓
272 = max(168, 272) ✓
Topic check
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