SME
Reinforced Concrete Design · Problem type
Short or slender column
Use this for “A 16 in square column has 12 ft between floors, k = 1, in a sway frame. Is it slender?”
Diagram
k = 1: pinned at the bottom, pinned on top.
The whole column bends in one half-wave: kℓᵤ = ℓᵤ.
Column depth h = 16 in
Radius of gyration r = 4.8 in
Column type S = slender
This is an example. Type your own values to start a new problem.
kℓᵤ ÷ r
=
×
ℓᵤ (ft)
r (in)
h
Column depth
example · in
S
Column type
worked out (not typed)
Formulas
r
=
0.3
×
h
4.8 = 0.3 × 16
kℓᵤ
r
=
k
×
12
×
ℓᵤ
r
30
=
1
×
12
×
12
4.8
S
=
short
if
kℓᵤ
r
≤
22,
else
slender
30 > 22, so slender
Assumptions
•
ACI 318-19: a rectangular section, so r ≈ 0.3h, with h the depth in the direction it bends.
•
In a sway frame the column is short while kℓᵤ ÷ r ≤ 22; above that, slenderness must be counted.
•
k comes from the frame (0.5 fixed–fixed, 1 pinned–pinned, 2 fixed–free, more when it sways); ℓᵤ is in feet.
•
A braced frame allows more, 34 − 12M₁ ÷ M₂ (at most 40), which this page does not use.
Step-by-step
WE KNOW
k = 1, ℓᵤ = 12 ft, h = 16 in
FIND
radius of gyration (r), slenderness ratio (kℓᵤ ÷ r), column type (S)
Step 1 · Find radius of gyration (r)
Use r = 0.3 × h
For
a
rectangle,
r
is
about
0.3
times
the
depth
in
the
direction
it
bends.
r
=
0.3
×
16
r
=
4.8
in
Step 2 · Find slenderness ratio (kℓᵤ ÷ r)
Use kℓᵤ ÷ r = k × 12 × ℓᵤ ÷ r
Turn
ℓᵤ
into
inches
(×
12),
multiply
by
k
,
then
divide
by
r
.
kℓᵤ
r
=
1
×
12
×
12
4.8
kℓᵤ
r
=
144
4.8
kℓᵤ
r
=
30
Step 3 · Find column type (S)
Use S = short if kℓᵤ ÷ r ≤ 22, else slender
Above 22 the column bends enough to add moment, so its design must count slenderness.
30 > 22: slender
S
=
short
if
30
≤
22,
else
slender
S
=
slender
Check
Put the numbers back in. Both sides should match.
4.8 = 0.3 × 16 ✓
30
=
1
×
12
×
12
4.8
✓
30 > 22, so slender ✓
Topic check
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