PASCO ME-9891 Flexible I-beam - Manual (Page 5)

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M o d e l N o . M E - 9 8 9 1 T h e o r y
5
Deflection Functions
The deflection of a beam under load can be found by combining Equation 1 with the
moment function (or functions) and integrating twice to solve for y. Below are two
examples of this method used to derive the deflection functions of the beams in Fig-
ure 6.
To solve the equation for the cantilevered
beam in Figure 6b, combine M = F(x - L)
with Equation 1:
Integrate once:
Because the fixed support restricts rotation,
we know that the slope at the support is zero:
dy/dx = 0 at x = 0; therefore C
1
= 0.
Integrate again:
Because the fixed support restricts displace-
ment, we know that y = 0 at x = 0; therefore
C
2
= 0, and
The maximum displacement occurs at x = L:
EI
F
------
d
2
y
dx
2
-------- x L – =
EI
F
------
dy
dx
------
x 2
2
----- Lx – C
1
+ =
EI
F
------ y
x 3
6
-----
Lx 2
2
--------- – C
2
+ =
y
F
6EI
--------- x 3 3Lx 2 –
( )
=
y
max
FL 3
3EI
---------- – =
For the left half of the beam in Figure 6a,
M
1
= Fx/2. Combine that function with
Equation 1:
Integrate once:
From symmetry, we know that the slope of
the beam at its center is zero: dy/dx = 0 at
x = L/2; therefore C
1
= -L
2
/16.
Integrate again:
Because the roller support restricts displace-
ment, y = 0 at x = 0; therefore C
2
= 0, and
The maximum displacement occurs at
x = L/2:
The right side of the beam can be solved for
using Equation 1 and M
2
= -F(x - L)/2; how-
ever this is not necessary if we are interested
only in the maximum displacement at the
center of the beam.
EI
F
------
d
2
y
dx
2
--------
x
2
--- = 0 x L 2
⁄ ≤ ≤
EI
F
------
dy
dx
------
x 2
4
----- C
1
+ =
EI
F
------ y
x 3
12
------
L 2 x
16
--------- C
2
+ – =
y
F
48EI
------------ 4x 3 3L 2 x –
( )
= 0 x L 2
⁄ ≤ ≤
y
max
FL 3
48EI
------------ – =
Example 2: Cantilevered Beam Example 1: Three-point Bending