1.050 – Content overview
I. Dimensional analysis
1.
2.
1.050 Engineering Mechanics I
On monsters, mice and mushrooms
Similarity relations: Important engineering tools
Lectures 1-3
Sept.
II. Stresses and strength
3.
4.
Lecture 26
Beam elasticity – how to sketch the solution
Another example
Transversal shear in beams
Stresses and equilibrium
Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain
5.
6.
How strain gages work?
How to measure deformation in a 3D
structure/material?
Lectures 16-19
Oct.
IV. Elasticity
7.
8.
Handout
Lectures 4-15
Sept./Oct.
Elasticity model – link stresses and deformation
Variational methods in elasticity
Lectures 20-31
Oct./Nov.
V. How things fail – and how to avoid it
9.
10.
11.
1
Elastic instabilities
Plasticity (permanent deformation)
Fracture mechanics
1.050 – Content overview
Lectures 32-37
Dec.
2
Drawing approach
I. Dimensional analysis
• Start from
f z = EIξ z'''' , then work your way up…
II. Stresses and strength
• Note sign changes:
III. Deformation and strain
ξ z''' ~ −Qz
IV. Elasticity
Lecture 20:
Lecture 21:
Lecture 22:
Lecture 23:
Lecture 24:
Lecture 25:
Lecture 26:
…
ξ z'''' ~ f z
+→−
ξ z'' ~ −M y
Introduction to elasticity (thermodynamics)
Generalization to 3D continuum elasticity
Special case: isotropic elasticity
Applications and examples
Beam elasticity Applications and examples (beam elasticity)
… cont’d and closure
ξ z' ~ −ω y
ξ z ~ ξ z
−→+
• At each level of derivative, first plot extreme cases at ends of beam
• Then consider zeros of higher derivatives; determine points of local min/max
V. How things fail – and how to avoid it
•
3
ξz
represents physical shape of the beam (“beam line”)
4
1
Review: Finding min/max of functions
Example solved in lecture 25:
Example
z
p = force/length
f (x) = x 2
f (x) function of x
x
l length
f ' (x) = 0
necessary condition for
min/max
f ' ' (x) < 0
local maximum
f ' ' (x) > 0
local minimum
EI
f ' (x) = 2x
f ' ' (x) = 0 inflection point
f ' ' (x) = 2
5 ⎞
⎛
Qz (x) = p ⎜ x − l ⎟
8 ⎠
⎝
ω y (x) =
⎛1
x2 5 ⎞
M y (x) = p ⎜⎜ l 2 +
− lx ⎟
2 8 ⎟⎠
⎝8
ξ z (x) = −
p
EI
⎛1 2
⎞
x3 5
⎜⎜ l x + − lx 2 ⎟⎟
6 16 ⎠
⎝8
p ⎛ 1 2 2 x4 5 3 ⎞
⎜ l x +
− lx ⎟
EI ⎜⎝ 16
24 48 ⎟⎠
5
6
−p
f z (x) = − p ~ ξ z''''
−
5
pl
8
3
pl
8
pl 2
8
5 ⎞
⎛
Qz (x) = p⎜ x − l ⎟ ~ −ξ z'''
8 ⎠
⎝
⎛1
x2 5 ⎞
M y (x) = p ⎜⎜ l 2 +
− lx ⎟ ~ −ξ z''
8
2 8 ⎟⎠
⎝
min
max
ω y (x) =
−
pl 3 1
EI 48
p
EI
ξ z (x) = −
⎛1 2
x3 5 2 ⎞
'
⎜⎜ 8 l x + 6 − 16 lx ⎟⎟ ~ −ξ z
⎝
⎠
p
EI
⎛ 1 2 2 x 4 5 3 ⎞
⎜⎜ l x +
− lx ⎟
24 48 ⎟⎠
⎝ 16
7
ξ z ( x) = −
p
EI
⎛ 1 2 2 x4 5 3 ⎞
⎜⎜ l x +
− lx ⎟
24 48 ⎟⎠
⎝ 16
8
min
2
Illustration of various BCs
Example with point load
z
Free end
r
F =0
r
M =0
ξz = 0
P
l
x
My = 0
Concentrated force
ξx = 0
ωy = 0
Qz = −P
Step 1: BCs
x=0
ξ z (0) = 0
ω y (0) = 0
x=l
Qz (l) = −P
M y (l) = 0
P
Hinge (bending)
r
ξ =0
ωy = 0
My = 0
Step 2: Governing equation
d 4ξ z
=0
dx 4
9
10
Example with point load (cont’d)
Example with point load (cont’d)
Step 3: Integrate
Qz
EI
M
ξ z'' = C1 x + C2 = − y
EI
x2
ξ z' = C1 + C2 x + C3 = −ω y
2
x3
x2
ξ z = C1 + C2 + C3 x + C4
6
2
ξ z'''' = 0, ξ z''' = C1 = −
f z = 0
Qz = −P
M y = P (l − x )
⎛ P x2
Step 4: Determine integration constants by applying BCs
ξ z (0) = 0 → C4 = 0
ω y = −ξ z' (0) = 0 → C3 = 0
Pl
P
M y (l) = EI ( l + C2 ) = 0 → C2 = −
EI
EI
P
Qz (l) = −C1EI = −P → C1 =
EI
11
Pl ⎞
ω y = −⎜⎜
−
x ⎟⎟
⎝ EI 2 EI ⎠
ξz =
P x 3 Pl x 2
−
EI 6 EI 2
12
3
f z (x) = 0 ~ ξ z''''
−P
Qz (x) = −P ~ −ξ z'''
M y (x) = P (l − x ) ~ −ξ z''
Pl
max
1 Pl 2
2 EI
max
⎛ P x 2 Pl ⎞
−
x ⎟⎟ ~ −ξ z'
EI
2
EI
⎝
⎠
P x 3 Pl x 2
ξ z (x) =
−
~ ξz
EI 6 EI 2
ω y (x) = −⎜⎜
1 Pl 3
−
3 EI
P
ξ z (x) =
13
Plotting stress distribution in beam’s
cross-section
14
Example: Plotting stress distribution in
beam’s cross-section
z
Given: Section quantities known as a function of position x
Want: Calculate stress distribution in the section
P x 3 Pl x 2
−
EI 6 EI 2
Fixed x:
σ xx = E (ε xx0 + ϑ y z )
with:
N = ESε xx0
M y = EIϑ y
⎛ N (x) M y (x) ⎞ N (x) M y (x)
+
+
z ⎟⎟ =
z
EI
S
I
⎝ ES
⎠
15
σ xx (z; x) = E ⎜⎜
σ xx (z)
N > 0, M y > 0
σ xx (z) =
N My
+
z
S
I
16
4