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
☻2.4
P2
Deflections in Beams
2.5 Buckling
(Refer: B,C & A –Sec’s 10.1, 10.2)
MECHENG242 Mechanics of Materials
Bending of Beams
2.5 Bars Under Axial Compression - Buckling
(Refer: B, C & A–Sec 10.1, 10.2)
Euler, Leonard (1741)
For bars or columns under compressive loading, we may witness a
sudden failure, known as buckling.
This is a form of instability.
P
P
Consider load, P, plotted against
horizontal displacement, v.
Point of Bifurcation
P
Short, Thick
Long, Slender
MECHENG242 Mechanics of Materials
v
Bending of Beams
Consider a long slender bar under axial compression:
y
v
P
y
P
L
x
z
d
b
P
x
P
Mxz=-P.v
From the Engineering Beam Theory:
d2 v
EI z  2  Mxz  P  v
dx
d2 v
 EI z  2  P  v  0
dx
Let
P
 
EI z
2
MECHENG242 Mechanics of Materials
d2 v
2


v0
2
dx
Euler Equation
Bending of Beams
General Solution: v  A Sinx  B Cosx
d2 v
2


v0
2
dx
Boundary Conditions:
From (i):
(i) @ x=0
v0
(ii) @ x=L
v0
(Also, (iii) @x=L/2,
dv
0)
dx
0  A Sin 0  B Cos 0
 B0
 v  A Sinx
From (ii):
0  A SinL
• If A=0, v=0 for all values of x
(i.e. The bar remains straight – NO BUCKLING)
• If Sin L=0, then (possible buckling solutions)
 L  0,  , 2 , 3 , ...
MECHENG242 Mechanics of Materials
Bending of Beams
The Lowest Real Value:
Since
L  
2
P
2
 
EI z
PC
 2 
L
EI z
Where PC is the Critical or Euler Buckling Load.
For buckling (i.e. at point of bifurcation):
P
PC 
L  2
 2EI z
L2
(Iz is the smallest 2nd
Moment of Area)
P
v
L  
P
MECHENG242 Mechanics of Materials
Points of
Bifurcation
4 2EI z
PC 
L2
PC 
 2EI z
L2
v
Bending of Beams
Let
PC
sC 
A
 sC 
Let Kz2 (Radius of Gyration) 
AL2
L
Iz
, and S (Slenderness Ratio) 
Kz
A
2
L
 S2  2
Kz
sC
 sC 
 2 E 
S2
Unstable
sYield
Short,
Thick
 2 EI z 
Empirical
Euler Stress
Material
Yielded
Long,
Slender
0
MECHENG242 Mechanics of Materials
Stable
S
Bending of Beams
Free or Pinned Ends
End Constraints:
PC 
P
P
 2 EI z 
L2
L
Fixed Ends
L/2
P
P
L
4 2 EI z 
PC 
L2
One End Free
& One End Fixed
2L
P
P
L
P
L
MECHENG242 Mechanics of Materials
4L2
One End Pinned
& One End Fixed
 2L
P
PC 
 2 EI z 
22 EI z 
PC 
L2
Bending of Beams
Example: Find the shortest length L of a Pin-Ended strut, having a
cross-section of 60mm x 100mm for which the Euler
Equation applies. Assume E=200 Gpa and sYield=250 MPa
sC 
Euler Eqn.,
Iz
Min. value,
sC
i.e. Iz
 L
sYd
3

10060

12
AL2
 1.8  106 m m4
 2 200  109 1.8  10 6 
250  10 0.1 0.06
6
Euler
Theory
=250 MPa
 2 EI z 
Iz
L
2
& Kz 
Note: S 
A
Kz
 S
0
90
MECHENG242 Mechanics of Materials
 1.54 m
S
1.54
 1.8  10 
 0.1 0.06 


6
 90
Bending of Beams