SME
Structural Analysis
Determinate structures
Use this for “An 8 m simply supported beam carries 40 kN 3 m from A. Find the reactions and the largest moment.”
Diagram
ΣF = 0: RA + RB = 25 + 15 = 40 kN, the total load.
M peaks where V crosses 0: M = 75 kN·m at x = 3 m.
The area under V between two points is the change in M.
This is an example. Type your own values to start a new problem.
RA (kN)
=
P (kN)
×
b (m)
÷
L (m)
RB (kN)
=
P (kN)
×
a (m)
÷
L (m)
Mₘₐₓ (kN·m)
=
P (kN)
×
a (m)
×
b (m)
÷
L (m)
Formulas
b
=
L
−
a
5 = 8 − 3
RA
=
P
×
b
L
25
=
40
×
5
8
RB
=
P
×
a
L
15
=
40
×
3
8
Mₘₐₓ
=
P
×
a
×
b
L
75
=
40
×
3
×
5
8
Assumptions
•
Simply supported: a pin at A and a roller at B; the beam’s own weight is ignored.
•
Reactions are + up; the moment is largest under the load, where the shear changes sign.
Step-by-step
WE KNOW
L = 8 m, P = 40 kN, a = 3 m
FIND
load from B (b), reaction at A (RA), reaction at B (RB), largest moment (Mₘₐₓ)
Step 1 · Find load from B (b)
Use b = L − a
The load’s distance from B is what is left of the span.
b
=
8
−
3
b
=
5
m
Step 2 · Find reaction at A (RA)
Use RA = P × b ÷ L
Take
moments
about
B:
RA
×
L
=
P
×
b
,
so
divide
P
×
b
by
L
.
RA
=
40
×
5
8
RA
=
200
8
RA = 25 kN
Step 3 · Find reaction at B (RB)
Use RB = P × a ÷ L
Take
moments
about
A:
RB
×
L
=
P
×
a
,
so
divide
P
×
a
by
L
.
RB
=
40
×
3
8
RB
=
120
8
RB = 15 kN
Step 4 · Find largest moment (Mₘₐₓ)
Use Mₘₐₓ = P × a × b ÷ L
Under
the
load,
M
=
RA
×
a
=
(
Pb
L
)
×
a
.
Mₘₐₓ
=
40
×
3
×
5
8
Mₘₐₓ
=
120
×
5
8
Mₘₐₓ
=
600
8
Mₘₐₓ = 75 kN·m
Check
Put the numbers back in. Both sides should match.
5 = 8 − 3 ✓
25
=
40
×
5
8
✓
15
=
40
×
3
8
✓
75
=
40
×
3
×
5
8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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