# 2.001 - MECHANICS AND MATERIALS I Lecture 11/21/2006 Prof. Carol Livermore

```2.001 - MECHANICS AND MATERIALS I
Lecture #21
11/21/2006
Prof. Carol Livermore
Recall from last time:
Beam Bending
y = 0 on neutral axis
xx = −ρy (Note: purely geometric, no material properties)
σxx = xx E (All other σ are equal to 0)
So:
σxx =
−Ey
ρ
Force Equilibrium:
A
Fx = 0
σxx dA = 0
1
If E is constant in y then
A
A
Ey
dA = 0
ρ
ydA − 0.
Moment Equilibrium
Mz = 0
M =−
σxx ydA
A
M=
A
Ey 2
dA
ρ
Special case: E constant:
M=
1
EI
ρ
y 2 dA
I=
A
New this time:
Recall:
σxx =
−Ey
ρ
For constant E (special case):
M=
EI
ρ
So:
E
M
−σxx
=
=
ρ
I
y
−M y
I
EXAMPLE: Find location of neutral axis for rectangular beam
σxx =
2
E is constant across cross-section. Recall force equilibrium.
Ey
dA = 0
A ρ
E
ydA = 0
ρ A
b E 2 h−a
ydydz = 0
ρ −b
a
2
2b 2
y
biggl]ha−a dz = 0
−b
2
2
b
1 2
[(h − a)2 − a2 ]dz = 0
2 −b
2
2b
1
(h2 − 2ha)dz = 0
2 −b
2
2b
1
2
(h − 2ha)z
=0
2
−b
2
1 2
b
b
(h − 2ha)( + ) = 0
2
2 2
b 2
(h − 2ha) = 0
2
h2 = 2ha
h
a=
2
So the neutral axis is in the center of the beam.
What if E2 > E1 in:
a is the distance to neutral axis.
3
1
ρ
b
2
−b
2
−a+ h
2
−a
A
Ey
dA = 0
ρ
E(y)ydA = 0
A
E1 ydy +
Note:
Example: Moment of Inertia
One material rectangular beam
4
h−a
−a+ h
2
E2 ydy dz = 0
y 2 dA
I=
I=
I=
b
2
−b
2
A
b
2
−b
2
dz
−h
2
y 2 dy
h2
dzbiggl[f racy 3
−h
2
b
2
−b
2
I=
Example:
h
2
3
I=
Beam Design
dz
bh3
12
h3
12
Find σxx .
FBD:
Fy = 0
Vy − P = 0
Vy = P
M∗ = 0
−Mz − P (a − x) = 0
Mz = −P (a − x) = 0
Mz (x)
1
=
ρ
EI
Distortion of planar sections of beam. The can be ignored for slender (long
and skinny beams)
6
σxx =
−M y
−Mz (x)y
P (a − x)y
=
=
I
I
I
σxxm ax =
P a h2
6P a
=
1
3
bh2
bh
12
Solve for I
a. Do areas of integration
b h3
b h3
b. I = 0120 − i12i
of integral during derivation.
This does not work because E was take out
7
Example: Skis
Get good stiﬀness (bending) but give up axial stiﬀness and lower weight.
Other examples:
Plants
Bird bones
Airplanes
Recall, x-axis is all for ”pure bending”
8
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