SME
Structural Analysis · Problem type
Deflection under an off-centre point load
Use this for “A 5 m simply supported steel beam (I = 30 × 10⁶ mm⁴) carries 15 kN 1.5 m from one end. How far does it sag under the load and at midspan?”
Diagram
ΣF = 0: RA + RB = 10,500 + 4,500 = 15,000 N, the total load.
The bent shape is drawn bigger than life; it meets every support.
δₐ = 4.594 mm at a = 1.5 m: the bent shape’s drop there.
This is an example. Type your own values to start a new problem.
b (m)
=
L (m)
−
a (m)
δₐ (mm)
=
P (kN)
×
(
a (m)
)
2
×
(
b (m)
)
2
÷
(3
×
E (GPa)
×
I (mm⁴)
×
L (m)
)
δmid (mm)
=
P (kN)
×
a (m)
×
(3
×
(
L (m)
)
2
−
4
×
(
a (m)
)
2
)
÷
(48
×
E (GPa)
×
I (mm⁴)
)
Formulas
b
=
L
−
a
3,500 = 5,000 − 1,500
δₐ
=
P
×
a
2
×
b
2
3 ×
E
×
I
×
L
4.5938
=
15,000
×
1,
500
2
×
3,
500
2
3 × 200,000 × 30,000,000 × 5,000
δmid
=
P
×
a
×
3 ×
L
2
− 4 ×
a
2
48 ×
E
×
I
5.1563
=
15,000
×
1,500
×
3 × 5,
000
2
− 4 × 1,
500
2
48 × 200,000 × 30,000,000
Formulas are worked in N, mm, MPa. The step-by-step shows the conversions.
Assumptions
•
Simply supported (pin and roller), constant EI, small deflections; the steps work in N, mm and MPa.
•
a is measured from the nearer support (a ≤ L ÷ 2), so the midspan formula holds.
•
The largest sag lies between the load and midspan, slightly more than δmid.
Step-by-step
WE KNOW
P = 15 kN, L = 5 m, a = 1.5 m, E = 200 GPa, I = 30,000,000 mm⁴
FIND
distance to the far support (b), deflection under the load (δₐ), midspan deflection (δmid)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N, mm, MPa).
P = 15 kN = 15,000 N (1 kN = 1000 N)
L = 5 m = 5,000 mm (1 m = 1000 mm)
a = 1.5 m = 1,500 mm (1 m = 1000 mm)
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find distance to the far support (b)
Use b = L − a
The rest of the span, from the load to the far support.
b
=
5,000
−
1,500
b
=
3,500
mm
Step 3 · Find deflection under the load (δₐ)
Use δₐ = P × a² × b² ÷ (3 × E × I × L)
Virtual
work
with
the
unit
load
at
the
load
point
(∫Mm
dx
EI
over
both
parts)
gives
Pa²
b
2
3
EIL
.
δₐ
=
15,000
×
1,
500
2
×
3,
500
2
3 × 200,000 × 30,000,000 × 5,000
δₐ
=
15,000
×
1,
500
2
×
3,
500
2
600,000 × 30,000,000 × 5,000
δₐ
=
15,000
×
1,
500
2
×
3,
500
2
18,000,000,000,000 × 5,000
δₐ
=
15,000
×
1,
500
2
×
3,
500
2
9 ×
10
16
δₐ
=
15,000
×
2,250,000
×
12,250,000
9 ×
10
16
δₐ
=
33,750,000,000
×
12,250,000
9 ×
10
16
δₐ
=
4.134 ×
10
17
9 ×
10
16
δₐ = 4.5938 mm
Step 4 · Find midspan deflection (δmid)
Use δmid = P × a × (3 × L² − 4 × a²) ÷ (48 × E × I)
Virtual
work
again,
with
the
unit
load
at
midspan:
∫Mm
dx
EI
over
the
three
stretches
gives
Pa
(3
L
² − 4
a
²)
48
EI
.
δmid
=
15,000
×
1,500
×
3 × 5,
000
2
− 4 × 1,
500
2
48 × 200,000 × 30,000,000
δmid = 15,000 × 1,500 × (3 × 25,000,000 − 4 × 2,250,000) ÷ (9,600,000 × 30,000,000)
δmid
=
15,000
×
1,500
×
75,000,000 − 9,000,000
288,000,000,000,000
δmid
=
15,000
×
1,500
×
66,000,000
288,000,000,000,000
δmid
=
22,500,000
×
66,000,000
288,000,000,000,000
δmid
=
1,485,000,000,000,000
288,000,000,000,000
δmid = 5.1563 mm
Step 5 · Convert the answers
Change each answer back to the unit you chose.
b = 3,500 mm = 3.5 m (1 m = 1000 mm)
Check
Put the numbers back in. Both sides should match.
3,500 = 5,000 − 1,500 ✓
4.5938
=
15,000
×
1,
500
2
×
3,
500
2
3 × 200,000 × 30,000,000 × 5,000
✓
5.1563
=
15,000
×
1,500
×
3 × 5,
000
2
− 4 × 1,
500
2
48 × 200,000 × 30,000,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.