SME
Mechanics of Materials (Solid Mechanics I) · Problem type
An eccentric load on a short post
Use this for “A 150 × 300 mm post carries 120 kN pressing down 40 mm off center along the 300 mm side. Find the stress at each edge.”
Chart
σA = σavg − σb
−2.667 − 2.133 = −4.8
The flows in add and the flows out take away: σA = −4.8.
This is an example. Type your own values to start a new problem.
σA (MPa)
=
−
P (kN)
b (mm)
×
h (mm)
×
1
+
6
×
e (mm)
h (mm)
σB (MPa)
=
−
P (kN)
b (mm)
×
h (mm)
×
1
−
6
×
e (mm)
h (mm)
σavg
Average stress
worked out (not typed)
σb
Bending stress
worked out (not typed)
eₖ
Kern limit
worked out (not typed)
Formulas
σavg
=
−
P
b
×
h
−2.6667
=
−
120,000
150 × 300
σb
=
6
×
P
×
e
b
×
h
2
2.1333
=
6
×
120,000
×
40
150 ×
300
2
σA
=
−(
P
b
×
h
)
×
(1
+
6
×
e
h
)
−4.8
=
−(
120,000
150 × 300
)
×
(1
+
6
×
40
300
)
σB
=
−(
P
b
×
h
)
×
(1
−
6
×
e
h
)
−0.5333
=
−(
120,000
150 × 300
)
×
(1
−
6
×
40
300
)
eₖ
=
h
6
50
=
300
6
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
A short post, so it does not buckle; e lies on the section’s center line along h.
•
Plane sections stay plane: σ = −P ÷ A ∓ Mc ÷ I with M = Pe. Compression is −.
•
Past e = h ÷ 6 (the kern) the far edge goes into tension.
Step-by-step
WE KNOW
P = 120 kN, e = 40 mm, b = 150 mm, h = 300 mm
FIND
average stress (σavg), bending stress (σb), near-edge stress (σA), far-edge stress (σB), kern limit (eₖ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
P = 120 kN = 120,000 N (1 kN = 1000 N)
Step 2 · Find average stress (σavg)
Use σavg = −P ÷ (b × h)
The load spread evenly over the area; − because it squeezes.
σavg
=
−
120,000
150 × 300
σavg
=
−
120,000
45,000
σavg = −2.6667 MPa
Step 3 · Find bending stress (σb)
Use σb = 6 × P × e ÷ (b × h²)
The
moment
Pe
bends
the
post:
Mc
I
with
c
=
h
2
and
I
=
bh
³
12
.
σb
=
6
×
120,000
×
40
150 ×
300
2
σb
=
6
×
120,000
×
40
150 × 90,000
σb
=
6
×
120,000
×
40
13,500,000
σb
=
720,000
×
40
13,500,000
σb
=
28,800,000
13,500,000
σb = 2.1333 MPa
Step 4 · Find near-edge stress (σA)
Use σA = −(P ÷ (b × h)) × (1 + 6 × e ÷ h)
Mc
I
=
Pe
(
h
÷ 2)
bh
³ ÷ 12
=
(
P
bh
)(
6
e
h
);
on
the
load’s
side
it
adds
to
the
squeeze.
σA
=
−(
120,000
150 × 300
)
×
(1
+
6
×
40
300
)
σA
=
−(
120,000
45,000
)
×
(1
+
240
300
)
σA
=
−(
120,000
45,000
)
×
(1
+
0.8)
σA
=
−(
120,000
45,000
)
×
1.8
σA = −2.667 × 1.8
σA = −4.8 MPa
Step 5 · Find far-edge stress (σB)
Use σB = −(P ÷ (b × h)) × (1 − 6 × e ÷ h)
On the far side the moment pulls, so the bending part takes away from the squeeze.
σB
=
−(
120,000
150 × 300
)
×
(1
−
6
×
40
300
)
σB
=
−(
120,000
45,000
)
×
(1
−
240
300
)
σB
=
−(
120,000
45,000
)
×
(1
−
0.8)
σB
=
−(
120,000
45,000
)
×
0.2
σB = −2.667 × 0.2
σB = −0.5333 MPa
Step 6 · Find kern limit (eₖ)
Use eₖ = h ÷ 6
The
kern’s
edge:
at
e
=
h
6
the
far-edge
stress
is
just
0.
eₖ
=
300
6
eₖ = 50 mm
Check
Put the numbers back in. Both sides should match.
−2.6667
=
−
120,000
150 × 300
✓
2.1333
=
6
×
120,000
×
40
150 ×
300
2
✓
−4.8
=
−(
120,000
150 × 300
)
×
(1
+
6
×
40
300
)
✓
−0.5333
=
−(
120,000
150 × 300
)
×
(1
−
6
×
40
300
)
✓
50
=
300
6
✓
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