材料力学第五章B

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 Example 5.8 in page 124
Problems of Chapter 12 :
5 . 23 5 . 24
5.7 Shear center
Read §5.8 from page 126 to 129 in 8 minutes.
Answer the following questions:
1. How can the direction and distribution of be figured in F.5.33b?
Theorem of complementary shearing stresses and the thin wall. 2. The hypothesis of §5.8 compares with that of §5.3 , which is more accurate and more reasonable ? Why ? Because of the thin wall the hypothesis of §5.8 is more reasonable.
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.
5.6 Normal stresses in unsymmetrical bending z z
1. Geometrical relation Hypothesis of plane section Mz
dx
a
Mz
y
(+)d -d
= d
=
a a
y
(a)
= y sin - z cos NNeueutrtarlal
4. Principal moments of inertia
z1
From
d Iy1
d
=
0，
d Iz1
d
=
0，we
get：
Iy1z1=
0
And we obtained :
tan21 = –
2 Iyz Iy – Iz
(5.24)
CO •
z
1
When Iy1z1= 0, y1 and z1 are called the principal axes. y
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
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.
FS
max
max
FS
max
FS
FS
max
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
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.
(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)
y1
The extreme values of the moments of inertia are:
Imax Imin
=
Iy + Iz 2
+–
Iy
– 2
Iz
2
+
Iyz2
Where Imax and Imin are called the principal moments of inertia.
(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.
=
E
(
Izsin
- Iyzcos )
(g)
Mz =
E
(
Izsin
- Iyzcos )
(g)
E
=
Mz
Izsin - Iyzcos
=
E
(
ysin
-
zcos
)
(f )
=
Mz (ysin -zcos Izsin - Iyzcos
)
From (1)
=
Mz ( Iy y - Iyz z ) Iy Iz - Iy2z
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=
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
Iyz = Iyczc + abA
O
z
C•
a yC
b zC dA
zC
y yC
3. Rotational axis theorem
y1 = OC + CD = y cos + z sin z1 = AE – CB = z cos – y sin
Iy1z1 = A ( ycos + zsin ) ( zcos – ysin ) dA = cos2 A yzdA + sin2 /2 ( Az2dA – A y2dA )
Chapter 12 Bending of Special Beams
Contents:
1. Normal stresses in unsymmetrical bending 2. Shearing stresses in open thin-walled members ;
Shear center
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 ).
5.5 Principal moment of inertia
1. Product of inertia of the section
O
Iyz = A yz dA
ABiblioteka Baidu
Iyz may be positive, negative or zero.
z y
If y or z is symmetrical axis, Iyz must be zero.
(5)
1. If Mz 0, My = 0 and Iyz= 0
From (4) , = Mz y / Iz , = /2
It is called plane bending 2. If Mz 0, My 0, and Iyz= 0
= Mz y + My z
(5.25)
Iz
Iy
(4)
tan = - (Mz Iy) / (My Iz ) (5.26)
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
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;
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.
Problem 5.7
Please determine the probable locations of the shear centers of the following sections.
•
•
•
• •
Problem 5.8
Draw the directions of shear flows and the probable distribution of shearing stresses of following sections.