SME
Statics
Trusses and frames
Use this for “A triangular roof truss spans 8 m and rises 3 m with 10 kN at the apex. Find the force in each member.”
Diagram
ΣFᵧ = 0: 5 + 5 = 10 kN, the load.
Every joint balances.
T pulls on its joints, C pushes on them.
This is an example. Type your own values to start a new problem.
θ
°
=
tan⁻¹
2
×
h (m)
L (m)
FAC (kN)
=
−
P (kN)
2
×
sin(
°)
FAB (kN)
=
P (kN)
2
×
tan(
°)
The roof rises h over half the span, so tan θ = 2h ÷ L. At the apex the two sloping members hold up P: 2 × FAC × sin θ = −P. At A the bottom chord balances their level pull: FAB = −FAC × cos θ.
R
Each reaction
calculated
Formulas
θ
=
tan⁻¹(2
×
h
L
)
36.8699
=
tan⁻¹(2
×
3
8
)
R
=
P
2
5
=
10
2
FAC
=
−
P
2 × sin(
θ
)
−8.3333
=
−
10
2 × sin(36.8699°)
FAB
=
P
2 × tan(
θ
)
6.6667
=
10
2 × tan(36.8699°)
Assumptions
•
Pin joints and loads only at the joints, so each member is a two-force member pulling or pushing along its line.
•
A pin at A and a roller at B; the truss is symmetric, so both inclined members carry the same force.
•
Sign convention: + is tension (T), − is compression (C).
Step-by-step
WE KNOW
L = 8 m, h = 3 m, P = 10 kN
FIND
slope of the inclined members (θ), each reaction (R), inclined member force (FAC), bottom chord force (FAB)
Step 1 · Find slope of the inclined members (θ)
Use θ = tan⁻¹(2 × h ÷ L)
The
apex
is
h
above
the
middle
of
the
span,
so
tan
θ
=
h
L
÷ 2
.
θ
=
tan⁻¹(2
×
3
8
)
θ
=
36.8699°
Step 2 · Find each reaction (R)
Use R = P ÷ 2
ΣFᵧ = 0 for the whole truss, and by symmetry each support takes half the load.
R
=
10
2
R
=
5
kN
Step 3 · Find inclined member force (FAC)
Use FAC = −P ÷ (2 × sin(θ))
ΣFᵧ
=
0
at
the
apex
C:
the
two
inclined
members
hold
up
P
,
so
2FAC
sin
θ
=
−
P
.
FAC
=
−
10
2 × sin(36.8699°)
FAC
=
−
10
2 × 0.6
FAC
=
−
10
1.2
FAC = −8.3333 kN
Step 4 · Find bottom chord force (FAB)
Use FAB = P ÷ (2 × tan(θ))
ΣFₓ
=
0
at
joint
A:
the
chord
balances
the
inclined
member’s
sideways
part,
R
tan
θ
.
FAB
=
10
2 × tan(36.8699°)
FAB
=
10
2 × 0.75
FAB
=
10
1.5
FAB = 6.6667 kN
Check
Put the numbers back in. Both sides should match.
36.8699
=
tan⁻¹(2
×
3
8
)
✓
5
=
10
2
✓
−8.3333
=
−
10
2 × sin(36.8699°)
✓
6.6667
=
10
2 × tan(36.8699°)
✓
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Reactions of a simply supported beam
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Chord and diagonal forces by the method of sections
Topic check
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