SME
Structural Analysis · Problem type
Truss member force by the method of sections
Use this for “A six-panel Pratt truss, 3 m panels and 3 m deep, carries 10 kN at each inner joint. Find the chord force at midspan.”
Diagram
ΣM about L3 = 0: F × h = M, so 45 × 3 = 135 kN·m.
ΣFᵧ = 0 on the free body: Fd sin θ = V, so 7.07 × sin 45° = 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.
R (kN)
=
(
n
−
1)
×
P (kN)
÷
2
F (kN)
=
M (kN·m)
÷
h (m)
d
Panel length
example
Formulas
R
=
(
n
−
1)
×
P
2
25
=
(6
−
1)
×
10
2
M
=
R
×
n
×
d
2
−
P
×
d
×
n
×
n
− 2
8
135
=
25
×
6
×
3
2
−
10
×
3
×
6
×
6 − 2
8
F
=
M
h
45
=
135
3
Assumptions
•
A Pratt truss of n equal panels on a pin and a roller, with P at every inner bottom joint.
•
Cut left of midspan and take moments about the lower midspan joint: only the top chord turns about it.
•
Pin joints and joint loads only, so members pull or push along their lines; the top chord is squeezed (C).
Step-by-step
WE KNOW
n = 6, d = 3 m, h = 3 m, P = 10 kN
FIND
each reaction (R), moment at midspan (M), top chord force (F)
Step 1 · Find each reaction (R)
Use R = (n − 1) × P ÷ 2
n
−
1
inner
joints
carry
P
;
the
two
supports
share
it
equally.
R
=
(6
−
1)
×
10
2
R
=
5
×
10
2
R
=
50
2
R
=
25
kN
Step 2 · Find moment at midspan (M)
Use M = R × n × d ÷ 2 − P × d × n × (n − 2) ÷ 8
Moments
about
the
lower
midspan
joint
of
everything
left
of
the
cut:
R
at
arm
nd
2,
less
each
joint
load
P
at
its
own
arm.
M
=
R
×
9
−
P
×
6
−
P
×
3
M
=
25
×
9
−
10
×
6
−
10
×
3
M
=
225
−
60
−
30
M
=
135
kN·m
Step 3 · Find top chord force (F)
Use F = M ÷ h
Only
the
top
chord
turns
about
the
lower
midspan
joint,
with
arm
h
.
F
=
135
3
F
=
45
kN
Check
Put the numbers back in. Both sides should match.
25
=
(6
−
1)
×
10
2
✓
135
=
25
×
6
×
3
2
−
10
×
3
×
6
×
6 − 2
8
✓
45
=
135
3
✓
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