材料力学第五章A

From the Table 3, we get: Ix : Sx = 17.2 cm, d = 7 mm
FSSz()max
50103
max =
Iz
=
= 41.5 MPa
17.210-2710-3
Average shearing stress of web is:
m = FS / h = 50103 / [(200-211.4)7] = 40.3 MPa e = 2.87 %
c,max= Fl y1/Iz = 150001.2(0.14-0.045)/(8.8410-6) = 193.4 MPa t,max= Fl yC /Iz= 150001.2 0.045/(8.8410-6) = 91.6 MPa
3. Maximum shearing stresses in bending:
(3) For the tubular section:
Iy = Iz = D4(1- 4)/64 where = d / D .
O
z
y
z dA y
b
h
Cz y
dy y
D dd
zz C
yy
5. Moment of inertia for the composite area
(1) Parallel axis theorem:
z
A
b
C• a zC
yC dA
zC
yC
1•
• • 2
y
z1
z z2
5.2 Normal stresses in bending
dx
1. Geometrical relation
M来自百度文库
M
Hypothesis of plane section
aa
(+y)d -d y
= d
=
(a)
M
d
M
2. Physical relation
Example 5.1
F
Know: F = 15 kN, l = 1.2 m
Find: t,max , c,max , max
Solution:
l
1. Centroid and moment of inertia:
120 z1
20
yC
120
20 y1z
y
yC = [(0.120.02)0.01+(0.020.12)(0.02+0.06)]/(20.120.02) = 0.045m
uniformly across the width).
z
max
Limitation about the theory :
h / b 1.5
error 5 %
yy
2. For - shaped section:
b
For the web of I beam, (5.10) can be used: why ?
Iz2 = a (3a)3/12 + 3a2 (5a/2 - 3a/2)2 = 21a4/4
Iz = Iz1 + Iz2 = 17a4 / 2
3a
1
yC
z 2
a y
Problem 5.2
Plot the distribution of bending normal stresses in following cross sections. If the positive moments act on the beams.
Problem 5.5
Establish the relation about the maximum bending normal stresses in following three beams loaded by the same bending moment M.
2a
max
=
max
a a
<
,
For example:
yC =
Sy = Ai zi
zC =
Ai zi Ai
A1 y1 + A2 y2 A1 + A2
O
z
1• zi
y
2 y• i • • Ci n
z
y2 y1
1•
•
2•
y
z1 yC zC
z2
4. Moment of inertia of the area
Iz = A y2 dA , Iy = A z2 dA the moment of inertia of area to z and y.
(5.10) F1
dA
y
z
F2
y*
where Sz() = b(h/2-y)(h/2+y)/2 = b(h2/4-y2)/2
b dx
(y) =
FSSz() Iz b
Where: Sz() = b (h2/4 - y2) / 2
From above equations we get :
(y) =
Ip = A 2dA = A(y2 + z2)dA = Iz + Iy
(1) For the rectangular section: Iz = A y2 dA = -hh//22 y2bdy = bh3/12 Similarly, Iy = hb3/12
(2) For the circular section: From Iy = Iz , we get: Iy = Iz = Ip / 2 = d 4/64
3 FS 2bh
(1
-
4y2 h2 )
max =
3FS 2bh
=
3 FS 2A
(5.11) (5.12)
Analysis of error :
when h / b 2 , error 1 %
b
when h / b = 1 , error 10 %
Why ? Assumption about (distributed h FS
O
Iy = Iyc + a2 A Iz = Izc + b2 A
( 5.9a) ( 5.9b)
(2) For the composite area:
y
Iz(1) = Iz1 + y12 A1
Iz(2) = Iz2 + y22 A2
y1
Iz = Iz(1) + Iz(2)
y2
= Iz1 + y12 A1 + Iz2 + y22A2
Problem 5.3
Know: F, l, a, E Which bending rigidity is bigger ? Which bending strength is bigger ?
Fa a
l
(a)
aa
(b)
Ia = Ib = a4/12 E Ia= E Ib The rigidity of (a) is equal to that of (b).
5.1 Geometrical properties of plane areas
1. Static moment of an area
Sz = A y dA , Sy = A z dA 2. Centroid
yC = A y dA /A = Sz /A , Sz = yC A zC = A z dA /A = Sy /A , Sy = zC A
Sz,max = (0.02+0.12-0.045)20.02/2 = 9.0310-5 m3
max=FSz,max/bIz = (150009.0310-5) / [(8.8410-6)0.02] = 7.66 MPa
We get: c,max : max = 25.2 , t,max : max = 12.0
a,max = 6Fl /a3 b,max = 62Fl /a3 The strength of (a) is bigger.
Problem 5.4
Know: h, b, M, E
Find: max , EI . And compare
with that of the whole beam.
M
h h/2
From (b),(d),
E
A
y2dA
=
E
Iz
=
M
z
Mz Oy
x
y max
y
EIz: flexural rigidity
dA
From (5.2),(b), we get:
1M
= EIz
(5.2)
= M y / Iz (5.3)
= M y / Iz
(5.3)
The maximum normal stress in bending:
Wz = d4/64/(d/2) = d3/32 (3) For the tubular section:
Wz = D3 (1- 4) / 32
Problems of Chapter 5 :
5 . 4 (b) 5.7
Problem 5.1
Know: a and section T shown in the Fig.
Chapter 5 Stresses in Bending
Contents:
1. Geometrical properties of plane areas 2. Normal stresses in bending 3. Shearing stresses in bending 4. Strength of beams 5. Principal moment of inertia 6. Unsymmetrical bending 7. Shear center
FSSz()max
max =
Iz
z
h0 h
max
y
y
From the Table 3 in the page 351, we
can look up Ix : Sx and calculate max .
For example: calculate max of No. 20a steel I beam, if FS = 50kN.
max
5.3 Shearing stresses in bending
b
1. For the rectangular section:
Assumption: all shearing stresses are parallel to shearing force and uniformly distributed across the width of the section.
M
Solution:
max = (M/2)/[b(h/2)2/6] = 12M/(bh2) EI = 2E[b(h/2)3/12] = Ebh3/48
For whole beam: w,max = 6M/ bh2 EIw = Ebh3/12
max /w,max = 2 stress
EI / EIw= 0.25 rigidity
a
Find: yC and Iz
4a
Solution:
1. Find yC
A1 = 3a2 , y1 = a / 2 ;
A2 = 3a2 , y2 = 5a / 2
yC = (A1 y1 + A2 y2) / (A1+A2) = 3a / 2 2. Find Iz
Iz1 = 3a a3/12 + 3a2 (3a/2 - a/2)2 = 13a4/4
A
O
zC
yC •C
y
z
z dA
y
the static moment of the area with respect to its centroid axis is zero.
3. Calculations for the composite area
Sz = Ai yi ,
yC =
Ai yi Ai
FS
C
h (y)
z
y
y
F1 = dA =
M Iz
y* dA
=
M Sz() Iz
M M+dM y
F2 =
(M+dM) Sz() Iz
=
(M+FSdx) Iz
Sz()
FS FS
y
dx
Fx= 0, F2 - F1 - bdx = 0
(y) = =
F2 - F1 bdx
=
FSSz() Iz b
(y)
Iz = (0.120.023)/12+(0.120.02)(0.045-0.01)2 + (0.020.123)/12+(0.120.02)(0.08-0.045)2 = 8.8410-6 m4
2. Maximum normal stresses in bending: Sections ? Points ?
y
= E =E
(b)
3. Statical relation
FN = A dA = 0
(c)
y a
a
Neutral layer
dx
Mz = A ( dA) y = M
(d)
From (b), (c), A y dA = Sz = 0
Neutral axis z must be centroidal axis.
max =
M ymax Iz
=
M Wz
(5.4)
where
Wz
=
Iz ymax
is called the section modulus to z in bending .
(1) For the rectangular section:
Wz = bh3/12/(h/2) = bh2/6 (2) For the circular section: