MIT - 16.20
Fall, 2002
Unit 17
The Beam-Column
Readings:
Theory of Elastic Stability, Timoshenko (and Gere),
McGraw-Hill, 1961 (2nd edition), Ch. 1
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics
and Engineering Systems
Paul A. Lagace © 2001
MIT - 16.20
Fall, 2002
Thus far have considered separately:
•
•
beam -- takes bending loads
column -- takes axial loads
Now combine the two and look at the “beam-column”
(Note: same geometrical restrictions as on others:
l >> cross- sectional dimensions)
Consider a beam with an axial load (general case):
Figure 17.1
Representation of beam-column
(could also have py for
bending in y direction)
Consider 2-D case:
Paul A. Lagace © 2001
Unit 17 - 2
MIT - 16.20
Fall, 2002
Cut out a deformed element dx:
Figure 17.2
Loads and moment acting on deformed infinitesimal
element of beam-column
Assume small angles such that:
dw
dw
sin
≈
dx
dx
dw
cos
≈1
dx
Paul A. Lagace © 2001
Unit 17 - 3
MIT - 16.20
Fall, 2002
Sum forces and moments:
• ∑ Fx = 0
+
:
dF
dx + px dx
dx
dw 
dS dw
d 2w 
−S
+ S +
+
dx 
dx = 0
2
dx 
dx   dx
dx

−F+F+
This leaves:
(dx)2
 dS dw
dF
d 2w 
dx + px dx + 
+ S 2  dx + H.O.T. = 0
dx 
dx
 dx dx
⇒
dF
d  d w
= − px −
S

dx
dx  dx 
(17-1)
new term
Paul A. Lagace © 2001
Unit 17 - 4
MIT - 16.20
Fall, 2002
•
∑F
= 0 +:
z

dw
dF   d w
d 2w 
−F
+ F +
+
dx 
dx
2
dx
dx   dx
dx



dS 
+ S − S +
dx + pz dx = 0
dx 

This results in:
dS
d  dw 
= pz +
F

dx 
dx
dx
(17-2)
new term
•
∑M
y
= 0
+:
dM
dx
dx + pz dx
dx
2

dw d x
dS 
− px dx
− S +
dx dx = 0
dx 2
dx 

−M + M +
(using the previous equations) this results in:
Paul A. Lagace © 2001
Unit 17 - 5
MIT - 16.20
Fall, 2002
dM
= S
dx
(17-3)
Note: same as before (for Simple Beam Theory)
Recall from beam bending theory:
d 2
w
(17-4)
M = EI 2
dx
Do some manipulating - place (17-4) into (17-3):
d
S =
dx

d 2
w 
 EI 2 

dx 
(17-5)
and place this into (17-2) to get:
d2
dx 2

d 2
w 
d  d w
 EI 2  −
F
 = pz

dx 
dx  d x 
(17-6)
Basic differential equation for Beam-Column -(Bending equation -- fourth order differential equation)
Paul A. Lagace © 2001
Unit 17 - 6
MIT - 16.20
Fall, 2002
--> To find the axial force F(x), place (17-5) into (17-1):
dF
d d w d 
d 2w  
= − px −
 EI 2  

dx
dx  d x d x 
dx  
For w small, this latter part is a second order term in w and
is therefore negligible
Thus:
dF
= − px
dx
(17-7)
Note: Solve this equation first to find F(x)
distribution and use that in equation (17-6)
Examples of solution to Equation (17-7)
•
End compression Po
Figure 17.3
Simply-supported column under end compression
px = 0
Paul A. Lagace © 2001
Unit 17 - 7
MIT - 16.20
Fall, 2002
dF
= 0 ⇒ F = C1
dx
find C1 via boundary condition @x = 0, F = -Po = C1
⇒ F = -Po
•
Beam under its own weight
Figure 17.4
Representation of end-fixed column under its own weight
px = -mg
dF
= + mg ⇒ F = mgx + C1
dx
boundary condition: @ x = l, F = 0
So: mgl + C1 = 0
Paul A. Lagace © 2001
⇒ C1 = -mgl
Unit 17 - 8
MIT - 16.20
Fall, 2002
⇒ F = -mg (l - x)
Helicopter blade
Figure 17.5
Representation of helicopter blade
•
(radial force due to rotation)
similar to previous case
Once have F(x), proceed to solve equation (17-6). Since it is fourth
order, need four boundary conditions (two at each end of the beamcolumn)
--> same possible boundary conditions as previously enumerated
Notes:
•
When EI --> 0, equation (17-6) reduces to:
d  d w
F
 = pz
dx  d x 
this is a string (second order ⇒ only need two boundary conditions
-- one at each end)
−
Paul A. Lagace © 2001
Unit 17 - 9
MIT - 16.20
Fall, 2002
(also note that a string cannot be clamped
since it cannot carry a moment)
• If F = 0, get:
d 2  d 2
w 
 EI 2  = pz
dx 2 
dx 
and for EI constant:
d 4
w
EI
= pz
4
dx
•
(basic bending equation)
For pz = 0, EI constant, and F constant (= -P), get:
d 4w
d 2
w
EI 4 + P 2
= 0
dx
dx
(basic buckling equation)
Buckling of Beam-Column
Consider the overall geometry (assume beam-column initially straight)
Paul A. Lagace © 2001
Unit 17 - 10
MIT - 16.20
Fall, 2002
Figure 17.6
Representation of general configuration of beam-column
Cut the beam-column:
Figure 17.7
Representation of beam-column with cut to determine
stress resultants
∑M
= 0 :
Μ − Μ primary + P w = 0
due to transverse loading
Paul A. Lagace © 2001
secondary moment (due to
deflection)
Unit 17 - 11
MIT - 16.20
Fall, 2002
gives:
d 2w
Μ = EI
2 = Μ primary − P w
dx
for transverse loading:
d2 
d 2w 
d  dw 
 EI 2  −
F
 = pz
dx 2 
dx 
dx  dx 
integrate twice with F =
P = C1
d 2w
EI 2 + P w = M primary
dx
same equation as by doing equilibrium
Solve this by:
• getting homogenous solution for w
• getting particular solution for Mprimary
• applying boundary condition
Paul A. Lagace © 2001
Unit 17 - 12
MIT - 16.20
Fall, 2002
Figure 17.8
Representation of moment(s) versus applied load for
beam-column
large moment!
Examples
• “Old” airplanes w/struts
Paul A. Lagace © 2001
Unit 17 - 13
MIT - 16.20
Fall, 2002
•
Space structure undergoing rotation
inertial
loading
Final note: The beam-column is an important concept and the moments
in a beam-column can be much worse/higher than beam
theory or a perfect column alone
Paul A. Lagace © 2001
Unit 17 - 14