P - Department of Mechanical and Aerospace Engineering
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Transcript P - Department of Mechanical and Aerospace Engineering
Skew Loads and Non-Symmetric
Cross Sections (Notes + 3.10)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
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Skew Loads & Non-Symmetric XSections
Introduction
Will perform advanced stress and deflection analysis of
beams with skew loads and non-symmetric cross
P
α
P
sections.
Skew load
x
y
y
Challenge: Need to calculate
Non-Symmetric
moments of inertia – Iyy, Izz,
and Iyz – and principal moments of inertia.
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P
z
Skew Loads & Non-Symmetric XSections
z
y
Moments of Inertia
For any cross-section shape
I yy z 2dA, I zz y 2dA, I yz yz dA
A
A
dA
C
A
z
y
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Skew Loads & Non-Symmetric XSections
Moments of Inertia
The moments of inertia can be transformed to y1-z1
coordinates by
I y1 y1
I z1z1
I y1z1
I yy I zz
2
I yy I zz
2
I yy I zz
2
I yy I zz
2
I yy I zz
2
cos 2 I yz sin 2
cos 2 I yz sin 2
sin 2 I yz cos 2
dA
C
Does this look familiar??
Skew Loads & Non-Symmetric XSections
z
θ
y
4
z1
y1
Moments of Inertia
Similar to transformation of stress, principal angle (angle
to the principal axes of inertia) can be found from
tan2 P
2I yz
I yy I zz
Where θP is the angle at which Iyz is zero.
dA
z1
z
C
θ
y1
y
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Skew Loads & Non-Symmetric XSections
Example
Find Iyy and Izz for a rectangle
b
h
C
z
dA = dydz
y
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Skew Loads & Non-Symmetric XSections
Example
Find Iyy and Izz for a Z-section (non-symmetric about y-z)
Let b = 7 in, t = 1 in, and h = 16 in.
b
t (all)
h/2
C
z
h/2
y
b
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Skew Loads & Non-Symmetric XSections
Example
Find Iyy and Izz for an L-section (non-symmetric about yz) Let b = 4 in, t = 0.5 in, and h = 6 in.
t
h
C
z
y
t
b
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Skew Loads & Non-Symmetric XSections
Skew Loads (3.10)
Skew loads for doubly symmetric cross sections
Beam will bend in two directions
Py = P cos α
Pz = P sin α
Pz C
Py
α
P
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z
y
x
(origin of x-axis at fixed end)
Skew Loads & Non-Symmetric XSections
Skew Loads (3.10)
Find bending moments
Side view
x
L-x
C
Pz
z (in)
Mz Mz
Py
y
From statics: Mz = Py(L-x) = P cos α (L-x)
Why is Mz positive?
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beam is curving in direction of positive y
Skew Loads & Non-Symmetric XSections
Py
z
α
x
P
y
Skew Loads (3.10)
Find bending moments
Top view
z
x
L-x
C
Pz
Py
z
α
x
y (in)
My My
Pz
From statics: My = Pz(L-x) = P sin α (L-x)
Why is My positive or negative?
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Skew Loads & Non-Symmetric XSections
P
y
Skew Loads (3.10)
Bending stress
From side view
xx
From top view
z
α
x
Combine to get
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C
Pz
Py
xx
Mz y
I zz
P
Myz
I yy
xx
Myz
I yy
Mz y
I zz
Skew Loads & Non-Symmetric XSections
y
Skew Loads
What about the neutral axis?
When there is only vertical bending, σxx=0 because y=0
at the neutral axis.
xx
My
0 at y 0
I
no stress
on this line
C
z
N.A. (y=0)
y
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Skew Loads & Non-Symmetric XSections
Skew Loads
But with a skew load:
xx
M yz
I yy
Mzy
0
I zz
y M y I zz
t an
z M z I yy
C
z
β
no stress
on this line
y
It turns out deflection will be perpendicular to this line.
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Skew Loads & Non-Symmetric XSections
Skew Loads
Curvature due to moment
From side view
M z ( x) EI zz
d 2v y
dx2
Py
From top view
C
Pz
α
x
P
d 2vz
M y ( x) EI yy
dx2
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z
Where vy and vz are deflections in the positive y and z
directions, respectively.
Skew Loads & Non-Symmetric XSections
y
Skew Loads
Find deflection at free end.
EI zz
d 2v y
dx2
M z ( x) P cos ( L x)
x2
EI zz
P cos Lx c1
dx
2
Lx 2 x 3
EI zz v y P cos
c1 x c2
2
6
dvy
Py
x
P
Apply B.C.’s: vy(0)=0 & vy’(0)=0
(0) 2
c1 0 c1 0
(0) P cos L(0)
dx
2
dvy
The tip deflection in the y-direction is
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z
α
L(0) 2 (0)3
c1 (0) c2 0 c2 0
v y (0) P cos
2
6
C
Pz
PL3 cos
v y ( L)
3EIzz
Skew Loads & Non-Symmetric XSections
y
Skew Loads
Continued…
d 2vz
EI yy
M y ( x) P sin ( L x)
dx2
dvz
x2
EI yy
P sin Lx c1
dx
2
Lx 2 x 3
EI yyvz P sin
c1 x c2
2
6
Py
x
P
Apply B.C.’s: vz(0)=0 & vz’(0)=0
dvz
(0) 2
c1 0 c1 0
(0) P sin L(0)
dx
2
The tip deflection in the z-direction is
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z
α
L(0) 2 (0)3
c1 (0) c2 0 c2 0
vz (0) P sin
2
6
C
Pz
PL3 sin
v z ( L)
3EI yy
Skew Loads & Non-Symmetric XSections
y
Skew Loads
The resultant tip deflection is
v y 2 vz 2
cos sin
2
2
I zz
I yy
3
2
PL
3E
Py
C
z
β
vy
δ
N.A.
y
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Skew Loads & Non-Symmetric XSections
z
α
x
P
vz
C
Pz
2
y
Example
Consider a cantilever beam with the cross-section and load shown below.
Find the stress at A and the tip deflection when α = 0o and α=1o. Let L =
12 ft, P = 10 kips, E = 30x106 psi and assume an S24x80 rolled steel beam is
used.
A (z=3.5 in, y=-12 in)
C
z
y
P
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α
Skew Loads & Non-Symmetric XSections
Non-Symmetric Cross-Sections
Bending of non-symmetric
cross-sections
C
Iyz ≠ 0
Iyy & Izz are not principal axes
y
My
Use generalized flexure formula
x
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z
( M y I zz M z I yz ) z ( M z I yy M y I yz ) y
I yy I zz I yz
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Skew Loads & Non-Symmetric XSections
Mz
Non-Symmetric Cross-Sections
Generalized moment-curvature
formulas
C
d 2v y
2
dx
M z I yy M y I yz
E ( I yy I zz I yz )
y
2
M y I zz M z I yz
d 2vz
2
2
dx
E ( I yy I zz I yz )
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z
Skew Loads & Non-Symmetric XSections
My
Mz
Non-Symmetric Cross-Sections
A special case – which we discussed previously – is when
Iyz = 0 and y & z are the principal axes.
xx
M yz
I yy
Mzy
I zz
d 2v y
Mz
2
dx
EI zz
My
d 2vz
2
dx
EI yy
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Skew Loads & Non-Symmetric XSections
Example
Analysis choices
Work in principal coordinates – simple formulas
Work in arbitrary coordinates – more complex formulas
Calculate the stress at A and the tip deflection for the beam
shown below.
A (z=-0.99 in, y=-4.01 in)
L = 10 ft
Mz = 10,000 in-lbs
(pure bending)
Cross-section dimensions:
6 x 4 x 0.5 in
C
z
y
Iyy = 6.27 in4
Izz = 17.4 in4
Iyz = 6.07 in4
E = 30 x 106 psi
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Skew Loads & Non-Symmetric XSections