SME
Statics · Problem type
Resultant of a triangular load
Use this for “A load falls from 6 kN/m to 0 over 4.5 m. Find its resultant and where it acts.”
Diagram
ΣF = 0: RA + RB = 9 + 4.5 = 13.5 kN, the total load.
This is an example. Type your own values to start a new problem.
FR (kN)
=
w₀ (kN/m)
×
L (m)
2
x̄ (m)
=
L (m)
3
Formulas
FR
=
w₀
×
L
2
13.5
=
6
×
4.5
2
x̄
=
L
3
1.5
=
4.5
3
Assumptions
•
The load falls in a straight line from w₀ at the high end A to 0 at B.
•
The resultant is the area under the load, and it acts at that area’s centroid.
•
The resultant gives the same reactions; inside the span, shear and moment still need the real load.
Step-by-step
WE KNOW
w₀ = 6 kN/m, L = 4.5 m
FIND
resultant (FR), resultant from the high end (x̄)
Step 1 · Find resultant (FR)
Use FR = w₀ × L ÷ 2
The resultant is the triangle’s area: half the base times the height.
FR
=
6
×
4.5
2
FR
=
27
2
FR = 13.5 kN
Step 2 · Find resultant from the high end (x̄)
Use x̄ = L ÷ 3
A triangle’s centroid is a third of its base from the tall side.
x̄
=
4.5
3
x̄
=
1.5
m
Check
Put the numbers back in. Both sides should match.
13.5
=
6
×
4.5
2
✓
1.5
=
4.5
3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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