SME
Statics · Problem type
Chord and diagonal forces by the method of sections
Use this for “A Pratt truss has four 3 m panels, 4 m deep, with 10 kN at each lower inner joint. Cut the second panel: find the top chord and diagonal forces.”
Diagram
ΣM about L2 = 0: F × h = M, so 15 × 4 = 60 kN·m.
ΣFᵧ = 0 on the free body: Fd sin θ = V, so 6.25 × sin 53.1° = 5 kN.
The dashed 0 member meets an unloaded joint where nothing else can balance it.
This is an example. Type your own values to start a new problem.
M (kN·m)
=
R (kN)
×
x (m)
−
P (kN)
×
(
x (m)
−
a (m)
)
F (kN)
=
M (kN·m)
h (m)
Fd (kN)
=
V (kN)
sin(
°)
x is the distance from A to the cut joint L2, two panels along, so x = 2a.
Formulas
R
=
3
×
P
2
15
=
3
×
10
2
x
=
2
×
a
6 = 2 × 3
θ
=
tan⁻¹(
h
a
)
53.1301
=
tan⁻¹(
4
3
)
M
=
R
×
x
−
P
×
(
x
−
a
)
60 = 15 × 6 − 10 × (6 − 3)
V
=
R
−
P
5 = 15 − 10
F
=
M
h
15
=
60
4
Fd
=
V
sin(
θ
)
6.25
=
5
sin(53.1301°)
Assumptions
•
A Pratt truss of four equal panels a, P at each of the three lower inner joints; pin and roller.
•
The cut passes through the second panel; take moments about the lower joint where the other two cut members meet.
•
The top chord is in compression (C) and the diagonal in tension (T); sizes are shown.
Step-by-step
WE KNOW
a = 3 m, h = 4 m, P = 10 kN
FIND
left reaction (R), cut joint from A (x), slope of the diagonal (θ), moment at the cut joint (M), shear in the cut panel (V), top chord force (F), diagonal force (Fd)
Step 1 · Find left reaction (R)
Use R = 3 × P ÷ 2
ΣFᵧ
=
0
for
the
whole
truss:
three
loads
of
P
,
shared
by
two
supports
by
symmetry.
R
=
3
×
10
2
R
=
30
2
R
=
15
kN
Step 2 · Find cut joint from A (x)
Use x = 2 × a
The cut joint is the second lower joint, two panels in.
x
=
2
×
3
x
=
6
m
Step 3 · Find slope of the diagonal (θ)
Use θ = tan⁻¹(h ÷ a)
The
diagonal
rises
h
over
one
panel.
θ
=
tan⁻¹(
4
3
)
θ
=
53.1301°
Step 4 · Find moment at the cut joint (M)
Use M = R × x − P × (x − a)
ΣM
=
0
about
the
cut
joint
for
the
part
left
of
the
cut:
R
turns
one
way,
P
the
other.
M
=
15
×
6
−
10
×
(6
−
3)
M
=
15
×
6
−
10
×
3
M
=
90
−
30
M
=
60
kN·m
Step 5 · Find shear in the cut panel (V)
Use V = R − P
Up minus down, left of the cut.
V
=
15
−
10
V
=
5
kN
Step 6 · Find top chord force (F)
Use F = M ÷ h
Of
the
three
cut
members
only
the
top
chord
turns
about
the
cut
joint,
with
arm
h
.
F
=
60
4
F
=
15
kN
Step 7 · Find diagonal force (Fd)
Use Fd = V ÷ sin(θ)
ΣFᵧ = 0 left of the cut: only the diagonal has an up-and-down part to carry the shear.
Fd
=
5
sin(53.1301°)
Fd
=
5
0.8
Fd = 6.25 kN
Check
Put the numbers back in. Both sides should match.
15
=
3
×
10
2
✓
6 = 2 × 3 ✓
53.1301
=
tan⁻¹(
4
3
)
✓
60 = 15 × 6 − 10 × (6 − 3) ✓
5 = 15 − 10 ✓
15
=
60
4
✓
6.25
=
5
sin(53.1301°)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.