Transcript Document

2.0 Bending of Beams
☻2.1
☻2.2
Revision – Bending Moments
Stresses in Beams
sx
sx
P
x
☻2.3
Mxz
Mxz
Combined Bending and Axial Loading
P1
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
2.4 Beam Deflection
(Refer: B, C & A–Sec 7.1, 7.2, 7.3, 7.4)
Recall: THE ENGINEERING BEAM THEORY

sx
y'

Mxz E

Iz
R
2.4.1 Moment-Curvature Equation
v (Deflection)
y
x
A
B
NA
x
A’
s
B’
If deformation is small (i.e. slope is “flat”):
s  x
MECHENG242 Mechanics of Materials
Bending of Beams
I d
 
R dx
R.  S  x

R
B’
S
A’
v

v
and  
x
(slope is “flat”)
I d2 v
 
R dx2
Alternatively: from Newton’s Curvature Equation
v
R
v  f (x)
x
MECHENG242 Mechanics of Materials
I

R
2
 d2 v 
 2 
 dx 
  dv 
1  
  dx 

2




3
2
 dv 
  1
if 
 dx 
I d2 v
 
R dx2
Bending of Beams
From the Engineering Beam Theory:
Mxz E

Iz
R
1 Mxz d2 v


R EI z dx2
d2 v
 EI z  2  Mxz
dx
Flexural
Stiffness
Curvature
Mxz
Bending
Moment
Recall, for Bars under
axial loading:
K  u  Load
Flexural
Stiffness
MECHENG242 Mechanics of Materials
1 d2 v

R dx2
Axial
Stiffness
Extension
Bending of Beams
d2 v  1 
M xz
 
Since,
2
dx
 EIz 

Curvature
dv  1 
  Mxz  dx  C1
 
dx  EI z 
 1 
  Mxz  dx  dx 
 v  
 EI z 
Slope
C
1
 dx  C2
Deflection
Where C1 and C2 are found using the boundary conditions.
Curvature
R
MECHENG242 Mechanics of Materials
Slope
dv
dx
Deflection
v
Bending of Beams
v = Deflection
Example:
y
P.L
P
L
A
B
v
x
vMax
P
Deflected
Shape
x
P.L
Mxz
P
Qxy
d2 v
EI z  2  Mxz  Px  PL
dx
dv
x2
 EIz 
P
 PLx  C1
dx
2
x 3 PLx2
 C1x  C2
 EI z v  P 
6
2
MECHENG242 Mechanics of Materials
 Mxz  Px  PL
Bending of Beams
P
x 3 PLx2
 EI z v  P

 C1x  C2
6
2
To find C1 and C2:
Boundary conditions:
(i) @ x=0
(ii) @ x=0
dv
0
dx
v0
 C1  0 & C2  0
Equation of the deflected shape is:
x 3 PLx2
EI z v  P 
6
2
vMax occurs at x=L
vMax
MECHENG242 Mechanics of Materials
1 PL3

3 EI z
Bending of Beams
2.4.2 Macaulay’s Notation
P
a
b
Example:
y
L
x
Pb
L
Pa
L
x
P
Pb
L
Mxz
Qxy
 Mxz  Pb
d2 v
 EI z  2  Mxz  Pb x   P x  a 
L
dx
dv Pb  x 2  P
2
  

 EI z 

x  a   C1
L 2
2
dx
 

EI z v  Pb 6L x3   P 6 x  a3
MECHENG242 Mechanics of Materials

x   P x  a 
L
 C1x   C2
Bending of Beams

EI z v  Pb 6L x3   P 6 x  a3
Boundary conditions:
From (i):
C2  0
From (ii):
0  Pb
 C1x   C2
(i) @ x=0
v0
(ii) @ x=L
v0

L   P L  a 
6L
6
3
3
 C1  Pb
b
6L
2
 L2
 C1 L

Since (L-a)=b
Equation of the deflected shape is:

 



1 Pb
3
3
P

 v
x 
x  a   Pb
b2  L2 x 
6L
6
6L
EI z
MECHENG242 Mechanics of Materials
Bending of Beams
To find vMax:
vMax occurs where
dv
 0 (i.e. slope=0)
dx
2

 P
x
2
2
2
Pb
Pb




i.e. EI z 0 

x

a

b

L
L 2 
2
6L
 

Assuming vMax will be at x<a,

dv
0
dx
when
x  a2
i.e.



0

x 2   1 b2  L2  1 L2  b 2
3
3

This value of x is then substituted into the above equation of the
deflected shape in order to obtain vMax.
Note:
L
if a  b 
2
 vMax
PL3

48EI z
MECHENG242 Mechanics of Materials
L
2
P
L
2
vMax
Bending of Beams
2.4.3 Summary
After considering stress caused by bending, we have now looked at the
deflections generated. Keep in mind the relationships between
Curvature, Slope, and Deflection, and understand what they are:
d2 v
1
I

Mxz 
2
dx
EI z
R
• Curvature
• Slope
dv
dx
• Deflection
v
Apart from my examples and problems:
• B, C & A
Worked Examples, pg 185-201
Problems, 7.1 to 7.15, pg 207
MECHENG242 Mechanics of Materials
Bending of Beams