sm10 10 11

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PROBLEMS (10.10 and 10.11) A cantilever beam is loaded as depicted in Fig. P10.10 and
P10.11. Using the double-integration approach, determine:
(a) The equation of the deflection curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.10)
(a)
y
PL
P
A
L/2
C
L/2
B
x
M 1 = PL
(0 ≤ x ≤ L 2)
3
L
M 2 = PL − Px
( ≤ x ≤ L)
2
2
Segment AC
EIv1 '' = PL
EIv1 ' = PLx + C1
1
EIv1 = PLx 2 + C1 x + C2
2
Segment BC
3
EIv2 '' = PL − Px
2
3
1
EIv2 ' = PLx − Px 2 + C3
2
2
3
1
EIv2 = PLx 2 − Px3 + C3 x + C4
4
6
Boundary conditions:
3 2 1 2
v2 '( L) = 0 :
PL − PL + C3 = 0, C3 = − PL2
2
2
3 3 1 3
5
v2 ( L) = 0 :
PL − PL − PL3 + C4 = 0, C4 = PL3
4
6
12
Thus,
P
(3Lx − x 2 − 2 L2 )
v2 ' =
2 EI
P
(9 Lx 2 − 2 x 3 − 12 L2 x + 5 L3 )
v2 =
12 EI
(1)
(2)
Continued on next slide
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Continuity conditions:
1 2
3
L
L
v1 '( ) = v2 '( ) :
PL + C1 = − PL2 ,
2
2
2
8
5
L
L
PL3
,
v1 ( ) = v2 ( ) : − PL3 + C2 = −
2
2
16
12
7
C1 = − PL2
8
19
C2 = − PL2
48
Therefore,
P
(−8 Lx − 7 L2 )
8 EI
P
(24 Lx 2 − 42 L2 x + 19 L3 )
v1 =
48 EI
v1 ' =
(3)
(4)
(b) x=0 in Eq. (3):
7 PL2 7 PL2
θA = −
=
8 EI 8 EI
(c) x=0 in Eq. (4):
19 PL3
vA = −
↑
48 EI
SOLUTION (10.11)
y
(a)
w
MA=2wa2
A
2a
C
a
B
x
RA=2wa
M1 =
1
w(−4a 2 + 4ax − x 2 )
2
M2 = 0
(0 ≤ x ≤ 2a )
L
( ≤ x ≤ L)
2
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
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
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Segment AC
w
EIv1 '' = (−4a 2 + 4ax − x 2 )
2
w
x3
EIv1 ' = (−4a 2 x + 2ax 2 − ) + C1
2
3
2 3 x4
w
2 2
EIv1 = (−2a x + ax − ) + C1 x + C2
2
3
12
Boundary conditions:
v1 '(0) = 0 : C1 = 0,
v1 (0) = 0 : C2 = 0
Thus,
w
(−12a 2 x + 6ax 2 − x 3 )
v1 ' =
6 EI
w
(−24a 2 x 2 + 8ax 3 − x 4 )
v1 =
24 EI
Segment BC
EIv2 '' = 0,
EIv2 ' = C3 ,
(1)
(2)
EIv2 = C3 x + C4
Continuity conditions:
4
v1 '(2a) = v2 '(2a) : − wa 3 = C3
3
8
v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C4 ,
3
2
C4 = − wa 4
3
Therefore,
4 wa 3
3 EI
wa 3
(−4 x + 2a)
v2 =
3EI
v2 ' = −
(3)
(4)
(b) x=3a in Eq. (3):
4 wa 3 4 wa 3
θB = −
=
3 EI
3 EI
(c) x=3a in Eq. (4):
10 wa 4 10 wa 4
vB = −
=
↓
3 EI
3 EI
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
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