Deflections of Beam
Deflections of Beam
Deflections of Beam
MATERIALES II
SEMESTRE 2017 - II
DEFLECTION OF BEAMS
1 EI
At the support B, 0, B
B PL
9
-
5
x
dy
M x dx C1
dx
EI EI
0
x x
EI y dx M x dx C1x C2
0 0
9
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7
Overhanging beam
y A 0, yB 0
Cantilever beam
y A 0, A 0
16 C1x3 12 C2 x 2 C3 x C4
SOLUTION:
Develop an expression for M(x)
and derive differential equation for
elastic curve.
- Reactions:
Pa a
RA RB P1
L L
2 at x 0, y 0 : C2 0
d y a
EI P x 1 a 1
dx 2 L at x L, y 0 : 0 P L3 C1L C1 PaL
6 L 6
Substituting,
dy PaL x
2
dy 1 a 1
EI P x 2 PaL 1 3
dx 2 L 6 dx 6 EI L
1 a 1
EI y P x3 PaLx
PaL2 x x
3
6 L 6
y
6 EI L L
9
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1
3
SAMPLE PROBLEM 9.1
Locate point of zero slope or point
of maximum deflection.
PaL xm
2
dy L
0 1 3 xm 0.577 L
dx 6 EI L 3
50 kips 48 in 180 in 2
ymax 0.0642
6 29 106 psi 723 in 4
ymax 0.238in
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1
4
SAMPLE PROBLEM 9.3
SOLUTION:
Develop the differential equation for
the elastic curve (will be functionally
dependent on the reaction at A).
w0 x3
M RA x
6L
y
w0
120 EIL
x5 2 L2 x3 L4 x
dy
w0
dx 120 EIL
5 x 4 6 L2 x 2 L4
w0 L3
at x = 0, A
120EI
9
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1
8
METHOD OF SUPERPOSITION
Principle of Superposition:
Deformations of beams subjected to Procedure is facilitated by tables of
combinations of loadings may be solutions for common types of
obtained as the linear combination of loadings and supports.
the deformations from the individual
loadings
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1
9
SAMPLE PROBLEM 9.7
SOLUTION:
Superpose the deformations due to Loading I and Loading II as shown.
9
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2
0
SAMPLE PROBLEM 9.7
Loading I
wL3 wL4
B I yB I
6 EI 8EI
Loading II
wL3 wL4
C II yC II
48EI 128EI