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 ?
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.
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 %
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
M
Hypothesis of plane section
aa
(+y)d -d y
= d
=
(a)
M
d
M
2. Physical relation
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.
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.
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
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 ?
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
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
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
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
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.
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:
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
(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:
,
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.
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
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.