Structural Mechanics
2.080 Recitation 3
Semester Yr
Recitation 3
3.1
Summary of Beam Equations
Equilibrium:
Hooke’s Law:
dN
=0
dx9
dV
>
+ g = 0 = d2 M
dx
+g =0
2
dM
; dx
=V >
dx
M = EI
(3.1)
(3.2)
(3.3)
Geometry:
d✓
d2 w
=- 2
dx
dx
d2 w
M = -EI 2
dx
d3 w
V = -EI 3
dx
=
3-1
(3.4)
(3.5)
(3.6)
Structural Mechanics
2.080 Recitation 3
Semester Yr
d2 M
d4 w
+
q
=
0
!
EI
=q
dx4
dx2
3.2
(3.7)
Methods of Solution
1. Direct Integration: EI
d4 w
=q
dx4
2. Uncoupled solution: find M (x), then integrate M (x) = -EI
d2 w
dx2
3. Rayleigh-Ritz:
- Assume shape function
- Apply BC’s
- Calclate ⇧= U - V , find w(x) to minimize ⇧ 4. Castigliano’s Theorem: w =
@U
@P
Example
Find the max. deflection for the following cantilever beam:
Direct integration: EI
d4 w
=q
dx4
d3 w
1
(qx + C1 )
=
3
EI
dx
1 x2
d2 w
2nd integ:
(q + C1 x + C2 )
=
EI 2
dx2
dw
1 x3
x2
3rd integ:
(q + C1 + C2 x + C3 )
=
dx
EI 6
2
1 x4
x3
x2
4th integ: w =
(q + C1 + C2 + C3 x + C4 )
EI 24
6
2
1st integ:
3-2
(3.8)
(3.9)
(3.10)
(3.11)
Structural Mechanics
2.080 Recitation 3
Semester Yr
Need 4 BC’s to evaluate C1 , C2 , C3 , and C4
1
@w(0)
= 0 ! C4 = 0
(3.12)
0
2
(0) = 0 ! C3 = 0
@w
3 (L) = -EI
@M
(3.13)
d2 w
L2
L2
(L) = 0 ! q
+ C1 L + C2 = 0 ! C2 = -q
- C1 L
2
2
dx2
d3 w
4 (L) = -EI 3 (L) = 0 ! qL + C1 = 0 ! C1 = -qL
@V
dx
qL2
qL2
C2 = + qL2 =
2
2
So
1
w(x) =
EI
Max. at
x = L ! w(L) =
q
EI
✓
✓
x4 qLx3 qL2 x2
q +
24
6
4
L 4 L4 L4
+
24
6
4
◆
=
◆
qL4
3qL4
=
24EI
8EI
(3.14)
(3.15)
(3.16)
(3.17)
(3.18)
Example
Find w(x) for the following beam:
Find M (x), then use M (x) = -EI
d2 w
:
dx2
First find reaction forces:
M0
L
M0
⌃Fy = 0: A + B = 0 ! A = -B =
L
+⌃MA = 0: - M0 + BL = 0 ! B =
— M (x) will be discontinuous at L/2 ! need to evaluate both sections.
3-3
(3.19)
(3.20)
Structural Mechanics
2.080 Recitation 3
For x < L/2:
M0
L
M0
M =x
L
For x > L/2:
M0
L
M0
M0
M=
(L - x) = M0 x
L
L
V =-
Semester Yr
SIGN CONVENTION
V =-
8
M
>
< - 0x
(x < L/2)
l
M (x) =
>
: M0 - M0 x (x > L/2)
l
— M (x) = -EI
d2 w
! Ingegrate twice to get w(x):
dx2
M0
d2 w
x = -EI 2
dx
l
2
dw
M0 x
1st int:
=
+ C1
dx
EIL 2
M0 x3
2nd int: w =
+ C1 x + C 2
EIL 6
For x < L/2: -
M0
d2 w
x = -EI 2
l
dx
dw
M0 x
M0 x2
1st int:
=+
+ C3
dx
EI
EIL 2
M0 x 3
M0 x 2
2nd int: w = +
+ C3 x + C4
EI 2
EIL 6
For x > L/2: M0 -
— We need 4 BC’s to evaluate C1 , C2 , C3 , and C4 .
1 w(0) = 0 ! C2 = 0 @
M L2
M0 L 3
M0 L2
2 w(L) = 0 ! - 0
+
+ C 3 L + C4 = 0 ! C 4 =
@
- C3 L
EI 2
EIL 6
3EI
3-4
(3.21)
Structural Mechanics
3
@
2.080 Recitation 3
Semester Yr
dw
L
at
must be continuous
dx
2
M0 (L/2)
M0 (L/2)2
M0 L
M0 (L/2)2
+ C1 = +
+ C 3 ! C 1 = C3 EIL 2
EI
EIL 2
2EI
4 w at L/2 must be continuous
@
M0 (L/2)3
L
M0 (L/2)2
M0 (L/2)3
L
M0 L2
+ C1 ( ) = +
+ C3 ( ) +
- C3 L
EIL 6
2
EI
2
EIL 6
2
|3EI {z
}
C4
— Substitute for C1 , solve for C3 :
(C3 -
M0 L L
L
5 M0 L 2
) = -C3 ( ) +
2EI 2
2
24 EI
11 M0 L
C3 =
24 EI
C1 =
C4 =
=)
M0 L
11 M0 L M0 L
= 24 EI
2EI
24EI
M0 L 2
M0 L2 11 Mo L2
= 3EI
24 EI
8EI
8 M
0
>
x3 <
6EIL
w(x) =
>
: M0 x3 6EIL
M0 L
x
(x 6 L/2)
24EI
M0 2 11 M0 L
M0 L2
x +
x2EI
24 EI
8EI
Example
Find w(x) for the following beam:
— Statically indeterminant ! must use direct integration
3-5
(3.22)
(x > L/2)
Structural Mechanics
1st int:
2nd int:
3rd int:
4th int:
2.080 Recitation 3
x < L/2
d4 w
=0
dx4
d3 w
= C1
dx3
d2 w
= C1 x + C 2
dx2
dw
C1 2
=
x + C 2 x + C3
dx
2
C1 3 C2 2
x +
x + C 3 x + C4
w=
6
2
Semester Yr
x > L/2
d4 w
=0
dx4
d3 w
dx3 = C5
d2 w
= C5 x + C 6
dx2
dw
C5 2
=
x + C 6 x + C7
dx
2
C 5 3 C6 2
x +
x + C 7 x + C8
w=
6
2
— We need 8 BS’s to evaluate C1 –C8 :
1 w(0) = 0 ! C4 = 0
@
2 w0 (0) = 0 ! C3 = 0
@
d2 w
a
(L) = 0 ! C5 L + C6 = 0 @
dx2
C5 3 C6
2
b
L +
w(L) = 0 !
L + C7 L + C 8 = 0 @
6
2
C 1 L 3 C2 L 2 C 5 L 3 C6 L 2
L
c
w(L/2) is continuous:
( ) +
( ) =
( ) +
( ) + C 7 ( ) + C8 @
6 2
2 2
6 2
2 2
2
C1 L 2
L
C5 L 2
L
d
w0 (L/2) is continuous:
( ) + C2 ( ) =
( ) + C 6 ( ) + C7 @
2 2
2
2 2
2
L
L
e
M (L/2) is continuous: C1 ( ) + C2 = C5 ( ) + C6 @
2
2
LL+
f
)=V(
) + P ! -EIC1 - EIC5 + P @
V(
2
2
3 M (L) = -EI
@
4
@
5
@
6
@
7
@
8
@
(Step jump due to point load)
a -@)
f to solve for C1 , C2 , C5 -C8
— Now we have 6 eons (@
(Omit algebra)
C1 = -
11 P
3 PL
5 P
5 PL
P L2
P L3
, C2 =
, C5 =
, C6 = , C7 =
, C8 = 8EI
48EI
16 EI
16 EI
16 EI
16 EI
3-6
Structural Mechanics
So
2.080 Recitation 3
✓
◆
8
> P - 11 x3 + 3 x2 L
>
(x 6 L/2)
< EI
96
32
◆
✓
w(x) =
>
1 2
1 3
P
5 3
5 2
>
:
x - x L + xL - L
(x 6 L/2)
EI 96
32
8
48
3-7
Semester Yr
(3.23)
MIT OpenCourseWare
http://ocw.mit.edu
2.080J / 1.573J Structural Mechanics
Fall 2013
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