Problem 5.7
Please determine the probable locations of the shear centers of the following sections.
•
•
•
• •
Probons of shear flows and the probable distribution of shearing stresses of following sections.
If the origin of coordinates is at the centroid, y1 and z1 are called the centroidal principal axes.
If the origin of coordinates is at the centroid, Imax and Imin are called the centroidal principal moments of inertia.
z0 z
y0 y
Mz (Iy y0–Iyz z0)+My (Iz z0–Iyz y0) = 0
(Mz Iy-My Iyz)y0+(My Iz-Mz Iyz)z0= 0 —— the function of neutral axis.
z0
tan = y0 = -
Mz Iy-My Iyz My Iz-Mz Iyz
Read the Example 5.10 in page 129 and answer questions.
Problems of Chapter 12 :
5 . 25 5 . 42
Problem 5.6
An unsymmetrical thin-walled beam is subjected to the transverse loads, for only producing plane bending and not torsion the acting conditions of the loads are ( C ).
z1
O
z
y1 y
B
E
Cz
D
z1
dA
A
y y1
We get
Iy1z1 =
Iy – 2
Iz
sin2
+
Iyz cos2
(5.23)
Similarly
Iy1 = Iz1 =
Iy + 2
Iz
+
Iy + Iz 2
–
Iy
– 2
Iz
cos2
–
Iyz
sin2
Iy
– 2
Izcos2
+ Iyz sin2
(5.21) (5.22)
11 70
80 11
80 C
1z
z1
11 y y1
Iy = 2Iy1+Iy2= 2 (11593/12+5911352) +160 113/12 = 1.9810-6 m4 Iyz = 2Iyz1+Iyz2= 2 [ 59 11 (-74.5) 35 ] + 0 = – 3.3810-6 m4
2. Determine Imax and Imin
Example 5.4
Find: the centroidal principal moments of
inertia of the area as shown in the figure.
Solution:
1. Determine Iy , Iz and Iyz
Iz = 2Iz1+Iz2= 2 (59113/12+591174.52) + 111603/12 = 10.97106 mm4= 10.9710-6 m4
C. Through the shear center and parallel to the centroidal principal plane of inertia.
C is correct.
A is not complete; B is generally impossible to realize.
5.5 Principal moment of inertia
1. Product of inertia of the section
O
Iyz = A yz dA
A
Iyz may be positive, negative or zero.
z y
If y or z is symmetrical axis, Iyz must be zero.
(2)
Similarly, (only My )
=
My ( Iz z - Iyz y ) Iy Iz - Iyz2
(3)
Mz and My
=
Mz (Iyy Iy Iz -
-Iyzz) Iyz2
+
My (Izz Iy Iz -
-Iyzy) Iyz2
tan = Iy / Iyz (1) From = 0 and (4)
Imax Imin
=
Iy + Iz 2
+–
Iy
– 2
Iz
2
+
Iyz2
=
1.98+10.97
2
+–
1.98-10.97 2
2
+(-3.38)2
10-6
=
12.110-6 m4 0.8510-6 m4
3. Determine 1
tan21 = –2Iyz / (Iy–Iz)= 23.38/(1.98 –10.97)= – 0.752 1 = –18.5
3. Which method was used in order to enduce the formula of ?
The equilibrium method of infinitesimal element. 4. How can the shearing stresses of thin-walled beams be calculated ? According to (5.27), the key is to calculate Sz . 5. Whats the definition of the shear center ? The intersection of lines on which FSy and FSz act respectively.
z y dA
2. Parallel axis theorem
O
z
y
Iyz = A yz dA = A (a + yC) (b + zC) dA
-z z
= A yC zCdA + aA zCdA + bA yCdA + AabdA
y
= Iyczc+ a Syc+ b Szc+ abA = Iyczc+ abA
Chapter 12 Bending of Special Beams
Contents:
1. Normal stresses in unsymmetrical bending 2. Shearing stresses in open thin-walled members ;
Shear center
A. The plane the loads act on is parallel to or superposes on the centroidal principal plane of inertia;
B. Through the shear center and act in the principal plane of inertia;
z
2. Physical relation
= E =E
3. Statical relation
From (b) :
axaixsis
(b)
=
E
(
ysin
-
zcos
)
(f )
From (d),(f ) :
zz
O yy
z z ddAA
FN = A dA = 0 My = A z dA = 0 Mz = A y dA From (b), (c):A dA=
We can find: Iy1 + Iz1 = Iy + Iz = constant There must be the coordinate system, the moment of inertia about
its one axis is maximum and that about its another axis is minimum.
(c) (d) (e)
My = =
E
E
E
A
dA
=
0
yy
(sin AzydA-cos Az2dA)
( Iyz sin - Iy cos ) = 0 tan = Iy / Iyz
From (e),(f ) :
(1)
A dA = 0
Mz =
E
(sin
A y2dA-cos
A yzdA)
Neutral axis must be across centroid.
4. Principal moments of inertia
z1
From
d Iy1
d
=
0，
d Iz1
d
=
0，we
get：
Iy1z1=
0
And we obtained :