SME
Structural Analysis
Deflections
Use this for “A wall bracket juts 2 m out: a horizontal tie on top and an inclined strut meeting the wall 1.5 m lower. Both bars are 800 mm² steel (E = 200 GPa). How far does the tip drop under 12 kN?”
Diagram
ΣFᵧ = 0: 12 + 0 = 12 kN, the load.
Every joint balances.
T pulls on its joints, C pushes on them.
δ
=
ΣFfL
AE
=
126
kN·
m
AE
=
0.7875
mm,
f
from
a
load
of
1
at
C.
The dashed shape is drawn far bigger than life.
Vertical deflection of C δ = 0.7875 mm
This is an example. Type your own values to start a new problem.
FAC (kN)
=
−
P (kN)
×
LAC (m)
÷
h (m)
FBC (kN)
=
P (kN)
×
b (m)
÷
h (m)
δ (mm)
=
(
(
FAC (kN)
)
2
×
LAC (m)
+
(
FBC (kN)
)
2
×
b (m)
)
÷
(
P (kN)
×
A (mm²)
×
E (GPa)
)
Formulas
LAC
=
√
b
2
+
h
2
2.5
=
√
2
2
+
1.5
2
FAC
=
−
P
×
LAC
h
−20
=
−12
×
2.5
1.5
FBC
=
P
×
b
h
16
=
12
×
2
1.5
δ
=
FAC
2
× LAC +
FBC
2
×
b
P
×
A
×
E
0.0007875
=
(
(−20)
2
×
2.5
+
16
2
×
2)
÷
(12
×
800
×
200)
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Pin-jointed and linear elastic; both members share one area A and one modulus E.
•
Virtual work (unit load): δ = ΣFfL ÷ (AE), with f each member’s force under 1 kN at C.
•
Tension +, compression −: the horizontal tie pulls (T), the inclined strut pushes (C).
•
mm² × GPa is kN, so kN²·m over P × AE gives δ in m: the steps turn it into mm last.
Step-by-step
WE KNOW
b = 2 m, h = 1.5 m, P = 12 kN, A = 800 mm², E = 200 GPa
FIND
strut length (LAC), strut force (FAC), tie force (FBC), vertical deflection of C (δ)
Step 1 · Find strut length (LAC)
Use LAC = √(b² + h²)
The strut is the hypotenuse of the reach and the height.
LAC
=
√
2
2
+
1.5
2
LAC
=
√
4 + 2.25
LAC
=
√
6.25
LAC = 2.5 m
Step 2 · Find strut force (FAC)
Use FAC = −P × LAC ÷ h
At
C
only
the
strut
can
hold
up
P
:
its
upward
push,
−FAC
×
h
LAC
,
equals
P
(C).
FAC
=
−12
×
2.5
1.5
FAC
=
−
30
1.5
FAC = −20 kN
Step 3 · Find tie force (FBC)
Use FBC = P × b ÷ h
The
strut
also
pushes
C
away
from
the
wall
by
−FAC
×
b
LAC
=
Pb
h
;
the
tie
pulls
it
back
(T).
FBC
=
12
×
2
1.5
FBC
=
24
1.5
FBC = 16 kN
Step 4 · Find vertical deflection of C (δ)
Use δ = (FAC² × LAC + FBC² × b) ÷ (P × A × E)
Put
1
kN
at
C
along
δ
;
each
member’s
force
is
then
f
=
F
P
,
and
δ
=
ΣFfL
AE
.
δ
=
(
(−20)
2
×
2.5
+
16
2
×
2)
÷
(12
×
800
×
200)
δ
=
400 × 2.5 + 256 × 2
9,600 × 200
δ
=
1,000 + 512
1,920,000
δ
=
1,512
1,920,000
δ
=
0.0007875
m
Step 5 · Convert the answers
Change each answer back to the unit you chose.
δ = 0.0007875 m = 0.7875 mm (1 m = 1000 mm)
Check
Put the numbers back in. Both sides should match.
2.5
=
√
2
2
+
1.5
2
✓
−20
=
−12
×
2.5
1.5
✓
16
=
12
×
2
1.5
✓
0.0007875
=
(
(−20)
2
×
2.5
+
16
2
×
2)
÷
(12
×
800
×
200)
✓
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Spring system by the direct stiffness method
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Deflection under an off-centre point load
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