SME
Structural Analysis · Problem type
Shear and moment of a uniform load
Use this for “A 10 m simply supported beam carries 12 kN/m along its length. Find the reactions, the largest shear and the largest moment.”
Diagram
ΣF = 0: RA + RB = 60 + 60 = 120 kN, the total load.
|M| is largest at x = 5 m: 150 kN·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.
R (kN)
=
w (kN/m)
×
L (m)
÷
2
Mₘₐₓ (kN·m)
=
w (kN/m)
×
(
L (m)
)
2
÷
8
Vₘₐₓ
Largest shear
calculated
Formulas
R
=
w
×
L
2
60
=
12
×
10
2
Vₘₐₓ
=
R
60 = 60
Mₘₐₓ
=
w
×
L
2
8
150
=
12 ×
10
2
8
Assumptions
•
Simply supported: a pin at A and a roller at B; w covers the whole span and includes the beam’s weight.
•
The shear falls in a straight line from R to −R; the moment is a parabola peaking at midspan.
Step-by-step
WE KNOW
w = 12 kN/m, L = 10 m
FIND
each reaction (R), largest shear (Vₘₐₓ), largest moment (Mₘₐₓ)
Step 1 · Find each reaction (R)
Use R = w × L ÷ 2
The whole load wL is shared equally by the two supports.
R
=
12
×
10
2
R
=
120
2
R
=
60
kN
Step 2 · Find largest shear (Vₘₐₓ)
Use Vₘₐₓ = R
Just
inside
each
support
the
shear
is
the
reaction;
it
falls
by
w
each
meter.
Vₘₐₓ = 60 kN
Step 3 · Find largest moment (Mₘₐₓ)
Use Mₘₐₓ = w × L² ÷ 8
At
midspan
V
=
0,
and
M
=
R
×
L
2
−
w
(
L
2
)
2
÷
2
=
wL
²
8
.
Mₘₐₓ
=
12 ×
10
2
8
Mₘₐₓ
=
12
×
100
8
Mₘₐₓ
=
1,200
8
Mₘₐₓ = 150 kN·m
Check
Put the numbers back in. Both sides should match.
60
=
12
×
10
2
✓
60 = 60 ✓
150
=
12 ×
10
2
8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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