Structural Mechanics
2.080 Recitation 6
Semester Yr
Recitation 6: Energy Methods
Example 1
A circular plate simply supported at its periphery is loaded in the center with point force,
P.
Find the center deflection, w0 .

1 Assume a deflection shape: w(r) = w0 1 Q
⇣ r ⌘2 ]
R
R
e dA
2 Calculate U : U = S U
Q
eb = D [(11 + 22 )2 - 2(1 - ⌫)(11 22 - 2 )] (eqn. 4.112)
For a plate, U
12
2
In our case,
@2w
@r2
1 @w
= ✓ = r @r
= r✓ = 0
11 = r = -
(6.1)
22
(6.2)
12
@2w
2w0
=- 2
2
@r
R
9
@2w
2w0 >
r = - 2 = 2 =
@r
R
 = ✓
1 @w
2w0 > r
✓ = = 2
;
r @r
R
@w
r
= -2w0 2 ,
@r
R
So
eb = D [(r + ✓ )2 - 2(1 - ⌫)r ✓ ]
U
2
D 2
w2
= [4r - 2(1 - ⌫)2r ] = D(1 + ⌫)2r = 40(1 + ⌫) o4
2
R
6-1
(6.3)
(6.4)
(6.5)
(6.6)
Structural Mechanics
2.080 Recitation 6
U=
Z
S
Z
2⇡
Z
Semester Yr
R
w02
rdrd✓
R4
0
0
w2
R2
w2
= 4D(1 + ⌫) 04 · 2⇡
= 4⇡D(1 + ⌫) 02
2
R
R
eb dA =
U
4D(1 + ⌫)
(6.7)
3 Calculate work of external forces, W
Q
(6.8)
W = P w0
4 Calculate ⇧= U - W : ⇧ = 4⇡D(1 + ⌫)
Q
5 Find w0 (P ) by setting
Q
w02
- P w0
R2
@⇧
= 0:
@w0
8⇡D(1 + ⌫)
w0
-P =0
R2
(6.9)
P R2
8⇡D(1 + ⌫)
! w0 =
(6.10)
Example 2
Same plate as Example 1, but with uniform pressure load, p. Find center deflection, and
compare to the exact solution in Lecture 7.

1 Deflected shape should be the same: w(r) = w0 1 Q
2 U = 4⇡D(1 + ⌫)
Q
3 W =
Q
=
R
pwdA
ZS
2⇡ Z R
0
0
= 2⇡pw0
✓
R
w02
R2

pw0 1 r2
r4
2
4R2
⇣ r ⌘2 ]
R
◆R
0
4 ⇧ = U - W = 4⇡D(1 + ⌫)
Q
5
Q
⇣ r ⌘2 ]
rdrd✓
= 2⇡pw0
✓
R2 R2
2
4
◆
=
⇡
pw0 R2
2
w02 ⇡
- pw0 R2
2
R2
w0
@⇧
pR4
⇡
= 0 ! 8⇡D(1 + ⌫) 2 - pR2 = 0 ! w0 =
R
2
16D(1 + ⌫)
@w0
6-2
Structural Mechanics
2.080 Recitation 6
Semester Yr
Exact soln:
]

⇣ r ⌘2 3 + ⌫
5+⌫
pR2 ⇣ r ⌘4
-2
(eqn. 7.28)
w(r) =
+
64D
R
R 1+⌫
1+⌫
✓
◆
pR4 5 + ⌫
w0 = w(r = 0) =
64D 1 + ⌫
◆
✓
4
pR
w0,approx.
64
4
16D(1 + ⌫)
◆=
=✓ 4
=
' 0.75
16(5 + ⌫)
5.3
w0,exact
pR (5 + ⌫)
|{z}
⌫=0.3
64D(1 + ⌫)
(6.11)
(6.12)
(6.13)
Example 3
Find the out-of-plane force required to open a keyring.
Compare to the required in-plane force for the same opening.
This is a combined bending-torsion problem.
Introduction to torsion
— Torque, T , causes rotation d<
def
↵=
d<
dz
(like  =
6-3
d✓
)
dx
(6.14)
Structural Mechanics
2.080 Recitation 6
Shear modulus,
G=
E
2(1 + ⌫)
Polar moment of inertia,
def
J = Ip =
For a solid circle, J =
⇡r4
2
T = GJ↵
Strain energy,
U=
Z
L
Z
(6.15)
r2 dA
(6.16)
A
(like bending, M = EI)
1
T ↵dz
2
T
↵=
GJ
Semester Yr
(like U =
)
U=
Z
Z
L
L
(6.17)
1
M dx)
2
(6.18)
T2
dz
2GJ
(6.19)
In our keyring problem:
a = R sin ✓
b = R(1 - cos ✓)
M✓ = T = P b = P R(1 - cos ✓)
Mr = Mb = P a = P R sin ✓
Z
Z
Mb2
T2
U=
ds +
ds
(ds = Rd✓)
L 2GJ
L 2EI
◆
Z 2⇡ ✓ 2
Mb2
T
=
+
Rd✓
2GJ
2EI
0
]

Z
P 2 R3 2⇡
(1 - cos ✓)2 sin2 ✓
d✓
=
+
2
GJ
EI
0
◆
✓
Z
P 2 R3 2⇡ 1 - 2 cos ✓ + cos2 ✓ sin2 ✓
d✓
=
+
2
GJ
EI
0
✓
◆
1 ⌘
P 2 R3 2⇡
⇡
⇡
⇡P 2 R3 ⇣ 3
=
+
+
=
+
GJ
EI
2
GJ
GJ
EI
2
|{z}
|{z}
Torsion
6-4
Bending
(6.20)
(6.21)
Structural Mechanics
2.080 Recitation 6
Semester Yr
Which contribution is larger?
G=
J=
So
⇡r4
,
2
E
E
=
2(1 + ⌫)
2.6
4
⇡r
I=
! J = 2I
4
(6.22)
(6.23)
4
3
3(2.6)
=
'
) torsion is ⇠4x bending
GJ
E(2I)
EI
CAstigliano:
]

1
@U
3 3(2.6)
w0 =
= ⇡P R
+
@P
2EI
EI
! P =
EIw0
4.9⇡R3
(6.24)
(6.25)
In-plane force:
Bending only:
Z
2⇡
Mb = P R(1 - cos ✓)
Mb2
Rd✓
0
2EI
Z
P 2 R3 2⇡
P 2 R3
3⇡P 2 R3
=
(1 - 2 cos ✓ + cos2 ✓)d✓ =
(2⇡ + ⇡) = 2EI 0
2EI
2EI
U=
@U
3⇡P R3
=
@P
EI
EIw0
-! P =
3⇡R3
w0 =
Compare:
Pout-of-plane
Pin-plane
◆
EIwo
3
4.9⇡R3
◆ =
= ✓
' 0.6
EIwo
4.9
3⇡R3
✓
6-5
(6.26)
(6.27)
(6.28)
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2.080J / 1.573J Structural Mechanics
Fall 2013
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