SME
Advanced Solid Mechanics (Solid Mechanics II) · Problem type
Truss joint deflection by the unit-load method
Use this for “A triangular truss spans 8 m with a 3 m rise and carries 10 kN at the apex. With A = 1000 mm² and E = 200 GPa, how far does the apex drop?”
Diagram
ΣFᵧ = 0: 5 + 5 = 10 kN, the load.
Every joint balances.
T pulls on its joints, C pushes on them.
δ
=
ΣFfL
AE
=
105
kN·
m
AE
=
0.525
mm,
f
from
a
load
of
1
at
C.
The dashed shape is drawn far bigger than life.
This is an example. Type your own values to start a new problem.
P
Apex load
example
A
Member area
example
E
Young’s modulus
example · GPa
Fincl
Inclined member force
calculated
Fbot
Bottom chord force
calculated
δ
Apex deflection
calculated
Formulas
Fincl
=
−5
×
P
6
−8.3333
=
−5
×
10
6
Fbot
=
2
×
P
3
6.6667
=
2
×
10
3
δ
=
1,000
×
(2
×
Fincl
×
(−5
÷
6)
×
5
+
Fbot
×
(2
÷
3)
×
8)
÷
(
A
×
E
)
0.525 = 1,000 × (2 × (−8.3333) × (−5 ÷ 6) × 5 + 6.6667 × (2 ÷ 3) × 8) ÷ (1,000 × 200)
Assumptions
•
Pin-jointed, linear elastic; all three members share one area A and one modulus E.
•
Span 8 m, rise 3 m: each inclined member is 5 m long, so sin θ = 0.6.
•
Tension +, compression −; f is each member’s force under a load of 1 at the apex.
•
kN, m, mm² and GPa give δ in mm with a factor of 1000.
Step-by-step
WE KNOW
P = 10 kN, A = 1,000 mm², E = 200 GPa
FIND
inclined member force (Fincl), bottom chord force (Fbot), apex deflection (δ)
Step 1 · Find inclined member force (Fincl)
Use Fincl = −5 × P ÷ 6
At
the
apex
the
two
inclined
members
hold
up
P
:
2F
sin
θ
=
−
P
with
sin
θ
=
0.6.
Fincl
=
−5
×
10
6
Fincl
=
−
50
6
Fincl = −8.3333 kN
Step 2 · Find bottom chord force (Fbot)
Use Fbot = 2 × P ÷ 3
At
a
support
the
reaction
is
P
2,
and
the
chord
pulls
R
tan θ
with
tan
θ
=
0.75.
Fbot
=
2
×
10
3
Fbot
=
20
3
Fbot = 6.6667 kN
Step 3 · Find apex deflection (δ)
Use δ = 1,000 × (2 × Fincl × (−5 ÷ 6) × 5 + Fbot × (2 ÷ 3) × 8) ÷ (A × E)
Each member adds F × f × L: two inclined members of 5 m and the 8 m chord.
δ
=
1,000
×
(2
×
(−8.3333)
×
(−5
÷
6)
×
5
+
6.6667
×
(2
÷
3)
×
8)
÷
(1,000
×
200)
δ
=
1,000
×
(2
×
(−8.3333)
×
(−0.8333)
×
5
+
6.6667
×
0.6667
×
8)
÷
(1,000
×
200)
δ
=
1,000
×
((−16.6666)
×
(−0.8333)
×
5
+
4.445
×
8)
÷
200,000
δ
=
1,000
×
13.89 × 5 + 35.56
200,000
δ
=
1,000
×
69.45 + 35.56
200,000
δ
=
1,000
×
105.01
200,000
δ
=
105,010
200,000
δ
=
0.525
mm
Check
Put the numbers back in. Both sides should match.
−8.3333
=
−5
×
10
6
✓
6.6667
=
2
×
10
3
✓
0.525 = 1,000 × (2 × (−8.3333) × (−5 ÷ 6) × 5 + 6.6667 × (2 ÷ 3) × 8) ÷ (1,000 × 200) ✓
Topic check
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