sm10 34 35

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PROBLEM (*10.34 and *10.35) An overhanging beam is loaded as shown in Figs. P10. 34 and
P10.35. Using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.34)
(a)
y
w
A
L
3wL/8
B
L/2
C
x
9wL/8
Segment AB
3
1
EIv1 '' = M 1 = wLx − wx 2
8
2
3
1
EIv1 ' = wLx 2 − wx3 + C1
16
6
1
1
EIv = wLx3 − wx 4 + C1 x + C2
16
24
(a)
(b)
Segment BC
3
wx 2 9
EIv2 '' = M 2 = wLx −
+ wL( x − L)
8
2
8
3
3
9
wx
EIv2 ' = wLx 2 −
+ wL( x − L) 2 + C3
16
6 16
3
4
3
wLx wx
EIv2 =
−
+ wL( x − L)3 + C3 x + C4
16
24 16
Boundary & continuity conditions:
wL3
v1 (0) = 0 : C2 = 0,
v1 ( L) = 0 : C1 = −
48
v1 '( L) = v2 '( L) : C1 = C3 , v2 ( L) = 0 : C4 = 0
(c)
(d)
Continued on next slide
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Then, Eqs.(a) through (d) become
w
(9 Lx 2 − 8 x 3 − L3 )
48 EI
wx
(3Lx 2 − 2 x 3 − L3 )
v1 =
48 EI
w
[9 Lx 2 − 8 x 3 − 27( x − L) 2 − L3 ]
v2 ' =
48 EI
w
[3Lx 3 − 2 x 4 − L3 x + 9 L( x − L)3 ]
v2 =
48 EI
v1 ' =
(1)
(2)
(3)
(4)
(b) Let x=L/2 in Eq.(2):
wL4
vM =
↓
192 EI
(c) Let x=3L/2 in Eq.(4):
wL4
wL4
(v2 ) max = vC = −
= 0.0078
↓
128EI
EI
To find (v2 ) max , make v1 ' = 0 in Eq.(1): 9 Lx 2 − 8 x 3 − L3 = 0
Solving, by trial and error: xm = 0.422 L
Then Eq.(2) yields
wL4
(v1 ) max = 0.0054
↓
EI
(d) Let x=3L/2 in Eq.(3):
1 wL3
θC =
12 EI
SOLUTION (*10.35)
(a)
y
C
P
P
P
L/2
A
3P/2
E
L
B
L/2 D
x
3P/2
Due to symmetry only one-half the beam need be considered.
Continued on next slide
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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Segment CA
EIv1 '' = M 1 = − Px
1
EIv1 ' = − Px 2 + C1
2
1
EIv1 = − Px 3 + C1 x + C2
6
(a)
(b)
Segment AE
EIv1 '' = M 2 = − Px +
3P
1
3
L
( x − ) = Px − PL
2
2
2
4
1 2 3
Px − PLx + C3
4
4
1
3
EIv2 = Px3 − PLx 2 + C3 x + C4
12
8
EIv2 ' =
(c)
(d)
Boundary & continuity conditions
1 2 3 2
1
v2 '( L) = 0 :
PL − PL + C3 = 0,
C3 = PL2
4
4
2
2
2
2
5
L
L
PL
PL 3PL PL2
,
v1 '( ) = v2 '( ) : −
+ C1 =
−
+
C1 = PL2
2
2
8
16
8
2
16
3
3
3
13PL
L
PL 5 PL
v1 '( ) = 0 :
+
+ C2 = 0,
C2 =
2
48
32
96
3
3
3
L
PL 3PL PL
PL3
v2 ( ) = 0 :
−
+
C4 = 0,
C4 = −
2
96
32
4
6
Equations (a), (b), and (d):
P
(−8 x 2 + 5 L2 )
v1 ' =
16 EI
P
(−16 x 3 + 30 L2 x − 13L3 )
v1 =
96 EI
P
(8 x 3 − 36 Lx 2 + 48 L2 x − 16 L3 )
v2 =
96 EI
(1)
(2)
(3)
(b) Making x=L in Eq.(3):
PL3
vE =
↑
24 EI
Continued on next slide
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( c)
Making x=0 in Eq.(2):
13PL3
vC =
↓
96 EI
Thus, vC = vmax
(d) Making x=0 in Eq.(1):
5 PL2
θC = −θ D =
16 EI
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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