P
a
a
2a
Rigid
g
L
Al
St
1
2
Al
3
Two aluminum bars and a steel bar are loaded as shown. Assuming the
load P closes the gap, g, determine the deformations and forces in the
bars. The bars all have the same areas and Est = 3E Al
strain energy for a single bar
U=
1 EA 2 1 2
∆ = k∆
2 L
2
∆ = elongation
1
1
1
k Al ∆12 + k St ∆ 22 + k Al ∆ 32
2
2
2
1
3
1
= k Al ∆12 + k Al ∆ 22 + k Al ∆ 32
2
2
2
U=
total strain energy
2a
2a
x
∆1
∆
g
∆3
∆2
∆ − ∆1 ∆ 3 − ∆1
=
x
4a
∆=
At the steel bar
x ( ∆ 3 − ∆1 )
+ ∆1
4
x = 2a
∆=
∆1 + ∆ 3
= ∆2 + g
2
U ( ∆1 , ∆ 3 ) =
At the load
1
3
⎡ ∆ + ∆3
⎤ 1
k Al ∆12 + k Al ⎢ 1
− g ⎥ + k Al ∆ 32
2
2
⎣ 2
⎦ 2
2
x=a
∆ = ∆P =
3∆1 ∆ 3
+
4
4
Principle of virtual work
δU = δW
∂∆ P
∂∆ P
∂U
∂U
δ∆1 +
δ∆ 3 = Pδ∆ P = P
δ∆1 + P
δ∆ 3
∂∆1
∂∆ 3
∂∆1
∂∆ 3
for all
δ∆1 , δ∆ 3
∂∆
∂U
=P P
∂∆1
∂∆1
∂∆
∂U
=P P
∂∆ 3
∂∆ 3
U ( ∆1 , ∆ 3 ) =
1
3
⎡ ∆ + ∆3
⎤ 1
k Al ∆12 + k Al ⎢ 1
− g ⎥ + k Al ∆ 32
2
2
⎣ 2
⎦ 2
2
∂∆
∂U
=P P
∂∆1
∂∆1
∆ = ∆P =
3∆1 ∆ 3
+
4
4
∂∆
∂U
=P P
∂∆ 3
∂∆ 3
⎛ ∆1 + ∆ 3
⎞ ⎛ 1 ⎞ 3P
− g ⎟⎜ ⎟ =
k Al ∆1 + 3k Al ⎜
⎝ 2
⎠⎝ 2 ⎠ 4
⎛ ∆1 + ∆ 3
⎞⎛ 1 ⎞ P
− g ⎟⎜ ⎟ =
k Al ∆ 3 + 3k Al ⎜
⎝ 2
⎠⎝ 2 ⎠ 4
or, equivalently
7 ∆1 3∆ 3 3P 3 g
+
=
+
4
4
4k Al 2
3∆1 7 ∆ 3
3g
P
+
=
+
4
4
4k Al 2
Solving, we find
0.45 P
∆1 =
+ 0.6 g
k Al
∆3 =
−0.05 P
+ 0.6 g
k Al
(Compressional) forces in the bars are
P1 = k Al ∆1 = 0.45 P + 0.6k Al g
⎡ ∆1 + ∆ 3
⎤
P2 = 3k Al ⎢
− g ⎥ = 0.6 P − 1.2k Al g
⎣ 2
⎦
P3 = k Al ∆ 3 = −0.05 P + 0.6k Al g
Note: for
P2 > 0
we need
P > 2k Al g
Otherwise the gap will not be closed