12.8 USE OF THE MOMENT-AREA METHOD Procedures:
• Construct separately the M/EI diagrams for each
applied force or moment, and each redundant as well.
• Then use the method of superposition and apply the two moment area theorems to obtain the proper
relationship between the tangents on the elastic curve in order to meet the conditions of displacement and/or slope at the supports of the beam or shaft.
USE OF THE MOMENT-AREA METHOD (cont) Procedures:
EXAMPLE 12.19
The beam is subjected to the concentrated force shown in Fig.
12–39a. Determine the reactions at the supports. EI is constant.
EXAMPLE 12.19 (cont)
• The free-body diagram is shown in Fig. 12–39b.
• Using the method of superposition, the separate M/EI diagrams for the redundant reaction By and the load P are shown in Fig. 12–39c.
• The elastic curve for the beam is shown in Fig. 12–39d.
Solutions
EXAMPLE 12.19 (cont)
• Applying Theorem 2, we have
• Using this result, the reactions at A on the free-body diagram, Fig. 12–
39b, are
Solutions
Ans) (
5
. 2
2 0 1 3
2 2
2 1 3
2
/
P B
EI L L PL
EI L PL L L
EI L L B
t
y
y A
B
(Ans)
0
; 0
F A
EXAMPLE 13.1
EXAMPLE 13.1
12.9 USE OF THE METHOD OF SUPERPOSITION Procedures:
Elastic Curve
• Specify the unknown redundant forces or moments that must be removed from the beam in order to make it statistically determinate and stable.
• Using the principle of superposition, draw the
statistically indeterminate beam and show it equal to a sequence of corresponding statistically determinate
beams.
USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Elastic Curve (cont)
• The first of these beams, the primary beam, supports the same external loads as the statistically
indeterminate beam, and each of the other beams
“added” to the primary beam shows the beam loaded with a separate redundant force or moment.
• Sketch the deflection curve for each beam and indicate
USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Compatibility Equations
• Write a compatibility equation for the displacement or slope at each point where there is a redundant force or moment.
• Determine all the displacements or slopes using an appropriate method as explained in Secs. 12.2 through 12.5.
USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Compatibility Equations (cont)
• Substitute the results into the compatibility equations and solve for the unknown redundant.
• If the numerical value for a redundant is positive, it has the same sense of direction as originally assumed.
Similarly, a negative numerical value indicates the redundant acts opposite to its assumed sense of direction.
USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Equilibrium Equations
• Once the redundant forces and/or moments have been determined, the remaining unknown reactions can be found from the equations of equilibrium applied to the loadings shown on the beam’s free body diagram.
EXAMPLE 12.21
Determine the reactions at the roller support B of the beam shown in Fig. 12–44a, then draw the shear and moment diagrams. EI is constant.
EXAMPLE 12.21 (cont)
• By inspection, the beam is statically indeterminate to the first degree.
• Taking positive displacement as downward, the compatibility equation at B is
• Displacements can be obtained from Appendix C.
Solutions
0 vB v'B (1)
EI B EI
v PL
EI PL EI
v wL
y B
B
3 3
3 3
4
m 9 ' 3
EI m kN 25 . 83 48
5 8
EXAMPLE 12.21 (cont)
• Substituting into Eq. 1 and solving yields
Solutions
kN 25 . 9
25 9 . 0 83
y
y
B
EI B EI
EXAMPLE 13.1
EXAMPLE 13.1
EXAMPLE 13.1
EXAMPLE 13.1
EXAMPLE 12.23
Determine the moment at B for the beam shown in Fig. 12–46a.
EI is constant. Neglect the effects of axial load.
EXAMPLE 12.23 (cont)
• Since the axial load on the beam is neglected, there will be a vertical force and moment at A and B.
• Referring to the displacement and slope at B, we require
Solutions
0 ' '' (1) 0 ' '' (2)
B B B
B B B
v v v
EXAMPLE 12.23 (cont)
• Use Appx C to calculate slopes and displacements,
Solutions
EI M EI
ML
EI B EI
v PL
EI B EI
PL
EI EI
v wL
EI wL
B B
y B
y B
B B
'' 4
33 . 21 ' 3
8 ' 2
m kN 42 384
7
EI m kN 2 1 48
3 2
3 4
3 3
EXAMPLE 12.23 (cont)
• Substituting these values into Eqs. 1 and 2 and cancelling out the common factor EI, we get
• Solving these equations simultaneously gives
Solutions
y BB y
M B
M B
8 33
. 21 42
0
4 8
12 0
3.375 kN
3.75 kN m (Ans)
y B
B M
Chapter Objectives
Understand the behavior of columns and concept of critical load and buckling
Determine the axial load needed to buckle a so-called
‘ideal’ column
APPLICATIONS
APPLICATIONS (cont)
13.1 CRITICAL LOAD
• Long slender members subjected to an axial
compressive force are called columns, and the lateral deflection that occurs is called buckling.
• The maximum axial load that a column can support when it is on the verge of buckling is called the critical load.
CRITICAL LOAD (cont)
• From the free-body diagram:
• For small θ, tan θ ≈ θ,
• Note:
This loading (Pcr = kL/4) represents a case of the
mechanism being in neutral equilibrium. Since Pcr is independent of θ, any slight disturbance given to the
/2
tan
2P F k k L
) of nt (independe
4 /
2 / 2
kL P
L k P
13.2 IDEAL COLUMN
• Ideal column
– It is perfectly straight before loading.
– Both ends are pin-supported.
– Loads are applied throughout the centroid of the cross section.
• Behavior
– When P < Pcr, the column remains straight.
– When P = Pcr,
P v
d
Pv dx M
v EI d
2 2 2
IDEAL COLUMN (cont)
• Since v = 0 at x = 0, then C2 = 0
• Since v = 0 at x = L, then
• Therefore,
• Which is satisfied if
• Or
0
1sin
L
EI C P
0
sin
L
EI P
n EI L
P
2 2
2 where 1, 2, 3,...
n EI
P n
L
IDEAL COLUMN (cont)
• Smallest value at P is when n = 1, thus
• Corresponding stress is
• Where r = √ (I/A) is called ‘radius of gyration’
• (L/r) is called the ‘slenderness ratio’.
• The critical-stress curves are hyperbolic, valid only for σcr is below yield stress
2 2
L Pcr EI
22
/ r KL
E
cr
EXAMPLE 13.1
The A-36 steel W200×46 member shown in Fig. 13–8 is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields.
EXAMPLE 13.1 (cont)
• From Appendix B,
• By inspection, buckling will occur about the y–y axis.
• When fully loaded, the average compressive stress in the column is
Solutions
1887.6 1000 2
320.5 N/mm (MPa) 5890
cr cr
P
A
4 6
4 6
2, 45.5 10 mm , 15.3 10 mm
mm
5890
Ix Iy
A
2 6
2
2
2 2
200 15.3 10
1887.6 kN 4 1/ 1000
cr
P EI
L
13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS
• one fixed end and one pined end
Vx Pv
M
Vx EI Pv
EI M dx
v
d22 1 ( ) 0 ,
0 :
'
.C s at x v B
0 ,
L v x
0
,
dx
L dv
xP
V y
• one fixed end and one pined end
' 'x B A
vp
sin cos
h
P P
v A x B x
EI EI
h p
v v v
EI x B V
x EI A
P ( ' ')
0 ' ,
'
B
P A V
P
V y
• one fixed end and one pined end
0
; 0 ,
0 :
'
.C s at x v B B
0 sin
; 0
,
L
P L V EI A P
v L x at
, ' 0; P cos P V 0
at x L v A L
EI EI P
L V A
P L L V A
cos sin
L L
tan 1
P
V y
• one fixed end and one pined end
---
• The smallest critical load occurs when n = 1, thus
2 2
2 or 2 with 2
cr cr
EI EI
P P K
error
&
by trial Solving
EI L L 49344. P
2 2 2
19 . 20
e
cr L
EI L
P EI
Then, Le 0.7L
• K for various end conditions:
