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
1
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.

2
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
3
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
5
Skew Loads & Non-Symmetric XSections
Example
Find Iyy and Izz for a rectangle

b
h
C
z
dA = dydz
y
6
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
7
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
8
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
9
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?

10
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?
11
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 
12
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
13
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.
14
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

15
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
16
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
17
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
18
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
19
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 
20
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
2
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 )
21
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
22
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
23
Skew Loads & Non-Symmetric XSections