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Bài giảng Strength of materials - Chương 8: Beam deflection - Nguyễn Sỹ Lâm

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CHAPTER 8: BEAM DEFLECTION  
8.1 Deflection of a beam under transverse loading  
8.2 Equation of the elastic curve  
8.3 Direct determination of the elastic curve from the loading  
8.4 Statically indeterminate beams  
8.5 Examples  
8.6 Statically indeterminate beams  
8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE  
LOADING  
• Relationship between bending moment and  
curvature for pure bending remains valid for  
general transverse loadings.  
1 M (x)  
EI  
• Cantilever beam subjected to concentrated  
load at the free end,  
1
Px  
EI  
   
• Curvature varies linearly with x  
1
0, ρ    
• At the free end A,  
• At the support B,  
A
ρ
A
B
1
EI  
0,   
B
PL  
8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE  
LOADING  
• Overhanging beam  
• Reactions at A and C  
• Bending moment diagram  
• Curvature is zero at points where the bending  
moment is zero, i.e., at each end and at E.  
1 M (x)  
EI  
• Beam is concave upwards where the bending  
moment is positive and concave downwards  
where it is negative.  
• Maximum curvature occurs where the moment  
magnitude is a maximum.  
• An equation for the beam shape or elastic curve  
is required to determine maximum deflection  
and slope.  
8.2 EQUATION OF THE ELASTIC CURVE  
• From elementary calculus, simplified for beam  
parameters,  
2
d y  
2
2
1
d y  
dx  
3 2  
2
2
dx  
dy  
   
1  
   
dx  
   
• Substituting and integrating,  
2
1
d y  
EI EI  
M  
x
   
2
dx  
x
dy  
x dx C  
EIEI M  
1
dx  
0
x
x
x dx C x C  
EI y dxM  
1 2  
0
0
8.2 EQUATION OF THE ELASTIC CURVE  
• Constants are determined from boundary  
conditions  
x
x
x dx C x C  
EI y dxM  
1 2  
0 0  
• Three cases for statically determinant beams,  
– Simply supported beam  
y 0, y 0  
A
B
– Overhanging beam  
y 0, y 0  
A
B
– Cantilever beam  
y 0, 0  
A
A
• More complicated loadings require multiple  
integrals and application of requirement for  
continuity of displacement and slope.  
8.3 DIRECT DETERMINATION OF THE ELASTIC CURVE  
FROM THE LOAD DISTRIBUTION  
• For a beam subjected to a distributed load,  
2
dM  
dx  
d M dV  
V  
x
   
 w x  
2
dx  
dx  
• Equation for beam displacement becomes  
2
4
d M  
d y  
EI  
   
 w x  
2
4
dx  
dx  
• Integrating four times yields  
EI y  
x
x dx  
 dxdxdxw  
3
2
1
1
C x C x C x C  
1
2
3
4
6
2
• Constants are determined from boundary  
conditions.  
8.4 STATICALLY INDETERMINATE BEAMS  
• Consider beam with fixed support at A and roller  
support at B.  
• From free-body diagram, note that there are four  
unknown reaction components.  
• Conditions for static equilibrium yield  
F 0  
F 0  
M 0  
A
x
y
The beam is statically indeterminate.  
• Also have the beam deflection equation,  
x
x
x dx C x C  
EI y dxM  
1 2  
0 0  
which introduces two unknowns but provides  
three additional equations from the boundary  
conditions:  
At x 0, 0 y 0  
At x L, y 0  
8.5 EXAMPLES  
EXAMPLE 9.01  
SOLUTION:  
•Develop an expression for M(x) and  
derive differential equation for elastic  
curve.  
4
6
•Integrate differential equation twice  
and apply boundary conditions to  
obtain elastic curve.  
W14 68  
I 723in E 29 10 psi  
a 4 ft  
P 50 kips L 15ft  
For portion AB of the overhanging beam,  
(a) derive the equation for the elastic curve,  
(b) determine the maximum deflection,  
(c) evaluate ymax.  
•Locate point of zero slope or point of  
maximum deflection.  
•Evaluate corresponding maximum  
deflection.  
8.5 EXAMPLES  
EXAMPLE 9.01  
SOLUTION:  
•Develop an expression for M(x) and derive  
differential equation for elastic curve.  
-Reactions:  
Pa  
L
a
R   
R P 1   
   
B
A
L
-From the free-body diagram for section AD,  
a
M  P x  
0 x L  
L
-The differential equation for the elastic  
curve,  
2
d y  
a
EI  
 P x  
2
L
dx  
8.5 EXAMPLES  
EXAMPLE 9.01  
•Locate point of zero slope or point of  
maximum deflection.  
2
dy  
dx  
PaL  
x
L
   
m
0   
13  
x   
0.577L  
   
m
6EI  
L
   
3
•Evaluate corresponding maximum  
deflection.  
3
2
PaL x x  
   
y   
   
6EI L L  
   
2
PaL  
3
y
0.577   
0.577  
max  
6EI  
2
PaL  
y
0.0642  
max  
6EI  
2
6
50 kips48in 180in  
y
0.0642  
max  
6
4
29 10 psi
723in  
y
0.238in  
max  
8.5 EXAMPLES  
EXAMPLE 9.02  
SOLUTION:  
•Develop the differential equation for the  
elastic curve (will be functionally  
dependent on the reaction at A).  
•Integrate twice and apply boundary  
conditions to solve for reaction at A and  
to obtain the elastic curve.  
For the uniform beam, determine the  
reaction at A, derive the equation for  
the elastic curve, and determine the  
slope at A. (Note that the beam is  
statically indeterminate to the first  
degree)  
•Evaluate the slope at A.  
8.5 EXAMPLES  
EXAMPLE 9.02  
•Consider moment acting at section D,  
M 0  
D
2
1 w x x  
0
R x   
M 0  
A
2 L 3  
3
w x  
0
M R x   
A
6L  
•The differential equation for the elastic curve,  
2
3
d y  
w x  
0
EI  
M R x   
A
2
6L  
dx  
8.5 EXAMPLES  
EXAMPLE 9.02  
•Integrate twice  
4
dy  
1
w x  
0
2
EI EIR x   
C  
A
1
dx  
2
24L  
C x C  
2
5
1
w x  
0
3
EI y R x   
A
1
6
120L  
•Apply boundary conditions:  
2
3
d y  
w x  
0
EI  
M R x   
A
at x 0, y 0 : C 0  
2
2
6L  
dx  
3
1
2
w L  
2
0
at x L,0 :  
at x L, y 0 :  
R L   
C 0  
1
A
24  
4
1
6
w L  
3
0
R L   
C L C 0  
A
1
2
120  
•Solve for reaction at A  
1
3
1
1
3
4
R L w L 0  
R w L   
A
0
A
0
30  
10  
8.5 EXAMPLES  
EXAMPLE 9.02  
•Substitute for C1, C2, and RA in the elastic  
curve equation,  
5
1 1  
w x  
1
3
3
0
EI y   
w L x   
w L x  
0
0
6 10  
120L 120  
w
5
2 3  
4
0
y   
x 2L x L x  
120EIL  
•Differentiate once to find the slope,  
dy  
   
dx 120EIL  
w
4
2 2  
4
0
5x 6L x L  
3
w L  
0
at x = 0,  
  
A
120EI  
8.6 METHOD OF SUPERPOSITION  
Principle of Superposition:  
•Deformations of beams subjected to  
combinations of loadings may be  
obtained as the linear combination of  
the deformations from the individual  
loadings  
•Procedure is facilitated by tables of  
solutions for common types of  
loadings and supports.  
8.6 METHOD OF SUPERPOSITION  
EXAMPLE 9.03  
For the beam and loading shown,  
determine the slope and deflection at  
point B.  
SOLUTION:  
Superpose the deformations due to Loading I and Loading II as shown.  
8.6 METHOD OF SUPERPOSITION  
EXAMPLE 9.03  
Loading I  
3
4
wL  
wL  
   
y
   
B
B
I
I
6EI  
8EI  
Loading II  
3
4
wL  
wL  
y
C
C
II  
II  
48EI  
128EI  
In beam segment CB, the bending moment is  
zero and the elastic curve is a straight line.  
3
wL  
B
C
II  
II  
48EI  
4
3
4
wL  
wL L  
7wL  
   
   
y
   
B
II  
128EI 48EI 2 384EI  
8.6 METHOD OF SUPERPOSITION  
EXAMPLE 9.03  
Combine the two solutions,  
3
3
3
wL  
wL  
7wL  
  
   
  
B
B
B
B
I
II  
6EI 48EI  
48EI  
4
4
4
wL  
7wL  
41wL  
y   
y
y
   
y   
B
B
B
B
I
II  
8EI 384EI  
384EI  
8.6 METHOD OF SUPERPOSITION  
APPLICATION OF SUPERPOSITION TO STATICALLY  
INDETERMINATE BEAMS  
•Method of superposition may be applied •Determine the beam deformation  
to determine the reactions at the supports without the redundant support.  
of statically indeterminate beams.  
•Treat the redundant reaction as an  
unknown load which, together with the  
other loads, must produce deformations  
compatible with the original supports.  
•Designate one of the reactions as  
redundant and eliminate or modify the  
support.  
8.6 METHOD OF SUPERPOSITION  
EXAMPLE 9.04  
For the uniform beam and loading shown,  
determine the reaction at each support and  
the slope at end A.  
SOLUTION:  
•Release the “redundant” support at B, and find deformation.  
•Apply reaction at B as an unknown load to force zero displacement at B.  
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