Transcript Document

2.0 Bending of Beams
☻2.1
Revision – Bending Moments
2.2 Stresses in Beams
(Refer: B,C & A –Sec’s 6.3-6.6)
sx
sx
P
x
Mxz
2.3 Combined Bending and Axial Loading
P1
Mxz
(Refer: B,C & A –
Sec’s 6.11, 6.12)
P2
2.4 Deflections in Beams
2.5 Buckling
(Refer: B,C & A –Sec’s 7.1-7.4)
(Refer: B,C & A –Sec’s 10.1, 10.2)
MECHENG242 Mechanics of Materials
Bending of Beams
2.2 Stresses in Beams
2.2.1 The Engineering Beam Theory
Compression
Mxz
C
y
(Refer: B, C & A–Sec 6.3,
6.4, 6.5, 6.6)
Mxz
D
y’
x
NA
Neutral Axis
y’
z
No Stress
A
y
B
dx
Tension
dq
R
Mxz
C’
A’
MECHENG242 Mechanics of Materials
Mxz
D’ y’
B’
sx=0 on the
Neutral Axis.
In general we
must find the
position of the
Neutral Axis.
Bending of Beams
Mxz
C
Mxz
D
y’
A
B
dq
R
Mxz
Mxz
C’
A’
D’
y’
B’
Assumptions
AB  A' B'  dx  R  dq
Plane surfaces remain plane
Beam material is elastic
sy  sz  0
MECHENG242 Mechanics of Materials
and only
sx  0
Bending of Beams
Geometry of Deformation:
C' D'CD
L

x 
CD
L0
CD  AB  A' B'  dx  R  dq
C' D'  R  y'  dq

R  y' dq
x 
Rdq
Hookes Law:
 Rdq
y'

R

1
  x  s x  s y  s z
E

sy  sz  0
 x 
MECHENG242 Mechanics of Materials
sx
E
and
E
 s x    y'
R 
1
Bending of Beams
E
 s x    y'
R 
1
dx
y
Linear Distribution of sx
x
(Eqn 1 )
y’
NA
0
Neutral Axis
-ve
+ve
sx
Note:
E is a Material Property
 1
 
R 
y
x
is Curvature
MECHENG242 Mechanics of Materials
Mxz
dx
Mxz
Bending of Beams
Equilibrium:
s x  dA  dFxx
Let
y
F
But
dA

Mxz
sx
x
If

Area, A

A
s
A
x
0
 dA  0
E
   y'dA  0
R  A
y’
z
x

A
y'dA  First Moment of Area
y'dA  0, Then y’ is measured from the centroidal axis
of the beam cross-section.
y
 “Neutral Axis”
coincides with the
XZ plane through
the centroid.
y’
NA
x
z
y’
Centroid
Neutral Axis
MECHENG242 Mechanics of Materials
Bending of Beams
y
Equilibrium:
M
z
dA
 dFxx  y'dMxz  0
0
y’
   s x  y'dA   dMxz  Mxz
A
Let IZ 

A
y'2 dA
x
A
E
   y'2 dA  Mxz
R  A
as
sx
Mxz
z
Area, A
E
 s x    y'
R 
1
=The 2nd Moment of Area about Z-axis
E
  Iz  Mxz
R 

Mxz E

Iz
R
2
THE SIMPLE BEAM THEORY:
1 & 2
MECHENG242 Mechanics of Materials
Mxz
s
E
 x 
Iz
y' R
Bending of Beams
Mxz
sx E


Iz
y' R
Mxz - Applied Bending Moment
- N.m
Iz
- Property of Cross-Sectional Area
- m4
sx
- Stress due to Mxz
- N/m2 or Pa
y'
E
R
- Distance from the Neutral Axis
-m
- Young’s Modulus of Beam Material - N/m2 or Pa
- Radius of Curvature due to Mxz
-m
y
Mxz
sx  
 y'
Iz
y’
NA
x
z
y’
o
Neutral Axis
MECHENG242 Mechanics of Materials
Bending of Beams
2.2.2 Properties of Area (Refer: B, C & A–Appendix A, p598-601)
y
Mxz
sx E


Iz
y' R
dA
sx
y’
o
z
y’ is measured from the
Centroidal or Neutral Axis, z.
x
Iz is the 2nd Moment of Area about
the Centroidal or Neutral Axis, z.
Mxz
Position of Centroidal or Neutral Axis:
dA
z
y
Centroidal
Axis
y’
o
y
dA
z
y’
o
Area, A
y
A y   y'dA
A
n
i.e.

A
y'dA  0, (Definition)
MECHENG242 Mechanics of Materials
y
1
y'dA

A A
Bending of Beams
Example:
(Dimensions in mm)
y
200
10
z
120
y  89.6 m m
60
125
o
Centroidal
Axis
1
y   y'dA
A A
n
20
1
200 10125  120 2060
y
200  10  120  20
1
394,000
250,000  144,000 
 89.55 m m
y
4,400
4,400
 89.6  103 m
MECHENG242 Mechanics of Materials
Bending of Beams
2nd Moment of Area:
y
dA
z’
Also, I
o
Example:
IZ   y'2 dA
A
y’
z
Definition:
z
dy
2
y’
o
d
2
d
3
2


bd
y

Iz   y'2 b  dy  b 
12
3  d

d
2
2
2
3

db 3 
 Also , Iy 

12 

2
b
A

y
d
d
y
  z 2  dA
b
2
MECHENG242 Mechanics of Materials
Bending of Beams
The Parallel Axis Theorem:
y
Definition:
In  Iz  Ay
2
z
o
y
n
Example:
b
d
z
2
2
y
b
2
dy
y’
o
d
n
2
d
3


y
bd
In   y'2 b  dy   b   
3
 3 0
0
d
y
MECHENG242 Mechanics of Materials
3
bd
 d
 bd 
 Iz  In  Ay 
3
2
2
3
2
bd3
 Iz 
12
Bending of Beams
y
Example:
(Dimensions in mm)
200
• What is Iz?
• What is maximum sx?
10
z
120
30.4
o
In  Iz  Ay
89.6
2
200
20
2
10
3
20
30.4
35.4
bd3 2089.6
89.6
1
Iz ,1 

 4.79  106 mm4
3
3
3
bd3 2030.4 
I z ,2 

 0.19  106 mm4
20
3
3
3
3
2
bd



200
10
2
6
4

 Ay 

3
.
28

10
mm



 200  10 35.4
12
12
3
I z ,3
MECHENG242 Mechanics of Materials
Bending of Beams
y
Example:
(Dimensions in mm)
200
• What is Iz?
• What is maximum sx?
10
z
120
30.4
o
In  Iz  Ay
89.6
2
200
20
2
1
10
3
20
30.4
35.4
89.6
Iz  Iz,1  Iz,2  Iz,3
20
 Iz  8.26  106 m m4  8.26  106 m4
MECHENG242 Mechanics of Materials
Bending of Beams
Maximum Stress:
y
40.4
Mxz
x
NA
89.6
sx  
s x ,Max
 s x ,Max
Mxz
 y'
Iz
Mxz

 yMax
Iz


Mxz
3



89
.
6

10
8.26  106

MECHENG242 Mechanics of Materials

(N/m2 or Pa)
Bending of Beams
The Perpendicular Axis Theorem:
y
2
2
2
R

y
'

z
'
dA
z’
2
2
2

R

dA

y
'

dA

z
'
dA
R
z y’
A
A
A

o


Jx  Iz  Iy
The Polar 2nd Moment of Area (About the X-axis)
d
Example:
y
dR
z
2
Jx   R2  dA   R 2 2R  dR 
A
0
From Symmetry,
R
o
d 4
32
Iz  Iy
Jx  Iz  Iy  2Iy
J x d 4
 Iy 

2
64
MECHENG242 Mechanics of Materials
Bending of Beams
2.2.3 Summary
The Engineering Beam Theory determines the axial stress distribution
generated across the section of a beam. It is applicable to long, slender
load carrying devices.
Mxz
s
E
 x 
Iz
y' R
Calculating properties of beam cross sections is a necessary part of
the analysis.
• Neutral Axis Position, y
• 2nd Moments of Area, Iy, Iz, Jx
Properties of Areas are discussed in Appendix A of B, C & A.
MECHENG242 Mechanics of Materials
Bending of Beams