happens What when a beam bends?
g
Originally, the line segments ab, ab and cd are the same length, but after bending, the
top surface (ab) gets shorter and the bottom (ab) longer, and cd stays the same length.
If you draw a short segment, de that is parallel to aa, the the length of the line ae is the
same as the length of the line cd, and the beam has stretched (in that location) a length
1
ef. But by how much?
O
Triangle OCD
is similar to
triangle DEF.
l ef
df
y
= =
=
=
l
cd Od
If we assume Hooke' s law :
= E = E
y
g
2
= E = E
y
=M
I
EI
M=
y
Bending Stress
M
=
EI
1
Curvature
Material
Property
Applied Moment
Geometric
Property
3
From calculus, curvature =
1
=
d2y
dx 2
(1 + ( dx ) ) 2
(Note, here y is deflection, not distance from neutral axis).
dy
2
3
For small deflections, (dy dx ) << 1, so denom ~ 1.
2
d2y
=2
dx
1
d2y M
=
2
dx
EI
or
d2y
M = EI 2 Euler - Bernoulli Equation for Bending
dx
4
What if we have a cantilevered beam with an
applied point force (P) on the end
From our equations of static equilibrium we
can find that there is a reaction force P and a
reaction moment PL at x=0.
Therefore the moment along the length of the
beam is
M ( x) = PL + Px
5
d2y
= M = PL + Px
dx 2
Integrating once
EI
PL3
y ( L) = max deflection =
3EI
PL2
slope at the end y ' ( L) =
2 EI
dy
Px 2
EI
= PLx +
+ C1
2
dx
The beam must have no slope at x = 0 so
dy
dx
= 0 C1 = 0
x =0
Integrating again
PLx 2 Px 3
EIy =
+
+ C2
2
6
again applying the boundary condition that at x = 0 there is no beam deflection ( y = 0) C2 = 0
Therefore
P x 3 Lx 2
y=
2
EI 6
6
What about a beam with a distributed load, w
d2y
w
2
EI 2 = (L x )
dx
2
dy w
3
= (L x ) + C1
EI
dx 6
dy
w
= 0 C1 = L3
dx x =0
6
w
dy w
3
= (L x ) L3
dx 6
6
w
w
4
EIy = (L x ) L3 x + C2
24
6
wL4
y (0 ) = 0 C2 =
24
w
w3
wL4
4
EIy = (L x ) L x +
24
6
24
w
4
(L x ) 4 L3 x + L4
y ( x) =
24 EI
wL4
y ( L) = max deflection =
8 EI
EI
w
2
M = (L x )
2
(
)
7
So our spring constant is just the Force/deflection
For a constant end force
3EI
F= 3 y
L
Effective spring
constant k
For a distributed load
8 EI
F = qL = 3 y
L
Effective spring
constant k
8
How can biomechanics be applied?
Cellular Mechanics
9
What about a beam in 3 point bending with symmetry?
The bending moment for the first
half of the beam is
M ( x) = R1 x =
C2=0 because there is no
deflection at x=0
P
x
2
0< x<
d2y P
EI 2 = x
2
dx
dy P x 2
=
+ C1
EI
dx 2 2
P x3
+ C1 x + C2
EIy =
26
Px 3
+ C1 x
EIy =
12
L
2
10
Because the beam is symmetric, the slope at x=L/2
must be zero
dy
EI
dx
P (L / 2 )
=
+ C1 = 0
22
2
L/2
PL2
C1 =
16
Px 3 PL2
EIy =
x
12
16
11
Shear Stress
Shearing Stresses
For any beam subjected to a shearing force, V, both
horizontal and vertical shear stresses, , are produced
12
The shear stress at a distance yo from the
neutral axis is given by
c
V
= yda
Ib yo
So for a rectangular beam of height h
h
2
V
V
bydy = = y 2
Ib yo
2I
h
2
yo
V h2
2
= y0
2I 4
From this we see the maximum shear
stress occurs at the neutral axis (yo=0)
and has magnitude
Vh 2
bh 3
max =
8I
I=
12
max
12V 3V
=
=
8bh 2bh
13
Beams/Stress Summary
Tension or Compression:
stress=F/A
stress is same throughout
Bending:
stress=My/I
maximum=Mc/I
Shear:
stress=
max = 3/2 V/bh
So, given some complex loading
at the same time!
Vc
ydA
situation,yall of
Ib 0
F
Fy
Fx
these stresses may be acting
Produces:
1) Compression (Fx)
2) Shear (Fy)
3) Bending (Fy)
These stresses combine!
14
Example
F
Ry
Rx
Fx = 0 : Fx Rx = 0; Rx = Fx
Fy = 0 : Fy + R y = 0; R y = Fy
M = 0 : Fy L + M = 0; M = Fy L
h
M
w
1
wh 3
12
A = wh
I=
What stresses do each of these cause?
15
Fx = 0 : Fx Rx = 0; Rx = Fx
Fx
Rx
h
w
Fx and Rx cause uniaxial compression.how much?
1
wh 3
12
A = wh
I=
Fx
Fx
=
=
A
wh
Is it uniform across the cross-section? YES
Is it uniform along the length of the beam? YES
16
Fy
Fy = 0 : Fy + R y = 0; R y = Fy
M = 0 : Fy L + M = 0; M = Fy L
M
h
Fx = 0 : Fx R x = 0; R x = Fx
Fy and Ry cause shear.how much?
Fy and M cause bending.how much?
Ry
1
wh 3
12
A = wh
I=
w
Fy
VM
Fy = 0; Fy V = 0;V = Fy
M = 0; Fy x + M = 0; M = Fy x
17
Fy = 0 : Fy + R y = 0; R y = Fy
Fy
Fy = 0; Fy V = 0;V = Fy
M = 0; Fy x + M = 0; M = Fy x
V
-Fy
Note: Results consistent with
FBD of whole beam/end conditions
= CLOSURE
M
-(Fy)(L)
18
What stress does V produce?
Shear stress,
How much?
VQ
Ib
Is it constant across the cross-section?
NOMax at neutral axis, zero at top and bottom
3V
3 Fy
max =
=
2 bh
2 wh
=
Is it distributed evenly along the length of the beam?
YES
Why?
Because from our shear diagram we know the shear force, V is a constant
V=-Fy
19
What stress does M produce?
Normal stress,
How much?
My
=
I
Is it constant across the cross-section?
NOMax at top and bottom, zero at neutral axis
Also, one side of the beam will be in tension (top) and one side in
compression (bottom)
1 Mh
2I
Is it distributed evenly along the length of the beam?
NO!
Why?
Because from our moment diagram we know the internal bending moment, M
is a function of the distance, x
M=-Fy x
Where is it max?
max/ min =
max/ min
1 Fy Lh 6 Fy L
=
=
2I
bh 2
1
wh 3
12
A = wh
I=
20
So..combined we have (max values shown):
From Fx :
Fx
x =
bh
Uniform compression across xsection and along length
From Fy :
3 Fy
=
2 bh
At neutral axis (y = 0) and uniform
along length
x =
6 Fy L
bh
x =
At x = L and y = h/2
2
6 Fy L
At x = L and y = -h/2
bh 2
21
So..combined we have (max values shown):
From Fx :
F
x = x
bh
x=L
Uniform compression
From Fy :
3 Fy
=
2 bh
x =
+
6 Fy L
bh
x =
Tension on top
Compression on bottom
Zero at y=0
2
6 Fy L
=
bh 2
6 Fy L Fx
Tension on top : x =
2
bh
bh
Compression on bottom : x =
6 Fy L
bh 2
Fx
bh
COMBINED:
Less Tension on top
More Compression on
bottom
NOT Zero at y=0
22