Transcript Slide 1
d
dx
1
d
2
v dx
2
M EI
For small deflections
EI d
4
v dx
4
w
d
3
v EI dx
3
V EI d
2
v dx
2
M
Load Equation Shear Equation Moment Equation Integrate 4 times to find v(x) Integrate 3 times to find v(x) Integrate 2 times to find v(x)
For each integration there will be a constant of integration. These constants can often be found by applying boundary, continuity or symmetry conditions.
Boundary conditions (see left) Continuity conditions
At the connection of two regions of integration the deflection curve is physically continuous.
Symmetry conditions
The cantilevered beam is subjected to a vertical load
P
at its end. Determine the equation of the elastic curve. EI is constant.
M
Px
EI d
2
v
Px dx
2
EI dv dx
Px
2 2
C
1
EIv
Px
3 6
C
1
x
C
2
The cantilevered beam is subjected to a vertical load
P
at its end. Determine the equation of the elastic curve. EI is constant.
0
PL
2 2
C
1
C
1
PL
2 2 0
PL
3 6
C
1
L
C
2
C
2
PL
3 3 dv/dx is 0 at x=L v is 0 at x=L
P
2
EI
L
2
x
2 ;
v
P
6
EI
x
3 3
L
2
x
2
L
3
Determine the maximum deflection of the beam. EI is constant Sketch the elastic curve
M
1
P x
1 3
M
2
P x
2 3
P
x
2 2
a
2
P
3
a
x
2 3
M 1
EI d
2
v
1
dx
1 2
P x
1 3
dv EI dx
1 1
P x
1 2 6
C
1
EIv
1
Px
1 3
C
1
x
1
C
2 18
Boundary Conditions
v 1 =0 at x 1 =0; v 2 =0 at x 2 =3a
M 2
EI d
2
v
2
dx
2 2
EI dv dx
2 2 2
P
3 3
a
x
2 2
P
3 3
ax
2
x
2 2 2
C
3
EIv
2 2
P
3 3 2
ax
2 2
x
2 3 6
C
3
x
2
C
4
Continuity Conditions
dv 1 /dx 1 = dv 2 /dx 2 at x 1 =2a and v 1 =v 2 at x=2a
dv
1
dx
1
P
6
EI x
1 2 4
Pa
2 9
EI dv dx
2 2 2
Pa x
2
EI
P
3
EI x
2 2 22
Pa
2 9
EI v
1
P
18
EI x
1 3 4
Pa
2 9
EI x
1
v
2
P EI x
2 2
P
9
EI x
2 3 22
Pa
2 9
EI x
2 4
Pa
3 3
EI
Maximum occurs between A&B when slope is 0
dv
1
dx
1 0
P x
1 2 6 4
Pa
2
x
1 9 1 .
633
a v
max 0 .
484
Pa
3
EI
Determine the equations of the elastic curve using the x 1 and x 2 coordinates. EI is constant.